Maharashtra Board Class 12 Biology Important Questions Chapter 10 Human Health and Diseases

Balbharti Maharashtra State Board 12th Biology Important Questions Chapter 10 Human Health and Diseases Important Questions and Answers.

Maharashtra State Board 12th Biology Important Questions Chapter 10 Human Health and Diseases

Multiple Choice Questions

Question 1.
The infectious stage of Plasmodium is …………………
(a) trophozoite
(b) sporozoite
(c) cryptozoite
(d) metacercaria
Answer:
(b) sporozoite

Question 2.
After birth, antibodies are transferred from mother to infant through …………………
(a) colostrum
(b) placenta
(c) blood
(d) tissue fluid
Answer:
(a) colostrum

Maharashtra Board Class 12 Biology Important Questions Chapter 10 Human Health and Diseases

Question 3.
Which cells give rise to T-lymphocytes?
(a) Thymocytes
(b) Bone marrow cells
(c) Erythrocytes
(d) Leucocytes
Answer:
(a) Thymocytes

Question 4.
Where is antigen D is present?
(a) On Rhesus factor
(b) On the surface of RBCs
(c) On A-antigen
(d) On AB-antigen
Answer:
(b) On the surface of RBCs

Question 5.
Erythroblastosis foetalis is caused when mother is …………………
(a) Rh +ve
(b) with antibody ‘a’
(c) Rh -ve
(d) with antibody ‘b’
Answer:
(c) Rh -ve

Question 6.
Which of the following is NOT a parasitic vector insect?
(a) Mosquito
(b) Housefly
(c) Honey bee
(d) Head louse
Answer:
(c) Honey bee

Question 7.
Which is the proper sequence in the developmental stages of Plasmodium?
(a) Merozoites → Sporozoite → Trophozoites → Schizonts
(b) Trophozoites → Merozoites → Sporozoite → Schizonts
(c) Sporozoite → Merozoites → Trophozoites → Schizonts
(d) Schizonts → Merozoites → Sporozoite → Trophozoites.
Answer:
(c) Sporozoite → Merozoites → Trophozoites → Schizonts

Question 8.
There is no vaccination on this disease till today.
(a) Typhoid
(b) Tuberculosis
(c) Polio
(d) AIDS
Answer:
(d) AIDS

Question 9.
Charas, hashish, ganja are obtained from …………………
(a) Papaver somnijerum
(b) Erythroxylum coca
(c) Atropa belladorta
(d) Cannabis sativa
Answer:
(d) Cannabis sativa

Question 10.
………………. Plant is used to obtain cocaine alkaloid.
(a) Marijuana
(b) Papaver somntferum
(c) Cannabis sativa
(d) Coca
Answer:
(d) Coca

Question 11.
………………… fish is released in the waterbody to prevent the spread of Malaria and Filaria.
(a) Pomfret
(b) Tilapia
(c) Gambusia
(d) Gold fish
Answer:
(c) Gambusia

Question 12.
The carcinogen that can cause vaginal cancer is …………………
(a) Vinyl chloride
(b) Diethylstilboestrol
(c) Mustard gas
(d) Cadmium oxide
Answer:
(b) Diethylstilboestrol

Question 13.
Prostate cancer can be caused due to exposure to …………………
(a) cadmium oxide
(b) mustard gas
(c) asbestos
(d) Nickel and chromium compounds
Answer:
(a) cadmium oxide

Question 14.
Choose the correct definition of health …………………
(a) Health is not contracting any disorder or disease.
(b) State of complete physical, mental and social well-being.
(c) Health is complete absence of any disease.
(d) Health is feeling good all the time.
Answer:
(b) State of complete physical, mental and social well-being

Question 15.
The interval between infection and appearance of disease symptoms is called …………………
(a) inoculation
(b) penetration
(c) infection period
(d) incubation period
Answer:
(d) incubation period

Question 16.
What is injected in vaccination ?
(a) Half killed pathogen
(b) Dead pathogens
(c) Live pathogens
(d) Readymade antibodies
Answer:
(a) Half killed pathogen

Question 17.
Who among the following is considered as father of immunology ?
(a) Ferdinand Kohn
(b) Robert Koch
(c) Louis Pasteur
(d) Edward Jenner
Answer:
(d) Edward Jenner

Question 18.
Who coined the term antibiotics ?
(a) Charles Darwin
(b) Louis Pasteur
(c) Alexander Fleming
(d) Selman Waksman
Answer:
(d) Selman Waksman

Question 19.
Who coined the term antibody?
(a) Selman Waksman
(b) Alexander Fleming
(c) Paul Ehrlich
(d) Edward Jenner
Answer:
(c) Paul Ehrlich

Question 20.
Widal test is used for the diagnosis of …………………
(a) Malaria
(b) Typhoid
(c) Diabetes mellitus
(d) HIV/AIDS
Answer:
(b) Typhoid

Question 21.
Which bacterial genus out of the following is the common pathogen causing pneumonia ?
(a) Streptococcus sps
(b) Lactobacillus sps
(c) Pseudomonas sps
(d) Salmonella sps
Answer:
(a) Streptococcus sps

Question 22.
Given below are some statements. Which among them are symptoms of pneumonia ?
(i) Greenish, yellow sputum coughed out.
(ii) Hepatomegaly and hypoglycemia.
(iii) High fever with shaking chills.
(iv) Thickening of skin and underlying tissues.
(v) Stabbing chest pain with shortness of breath.
(vi) Mood swings and joint pains along with nausea and vomiting.
(a) (i), (ii), (iii), (iv)
(b) (i), (iii), (v), (vi)
(c) (i), (ii), (iv), (vi)
(d) (ii), (iii), (iv), (v)
Answer:
(b) (i), (hi), (v), (vi)

Question 23.
Which of the following is not the common way to prevent common cold ?
(a) Using hand sanitizers
(b) Blowing nose in open
(c) Staying away from people suffering from cold
(d) Sipping warm water
Answer:
(b) Blowing nose in open

Question 24.
Common cold is not cured by antibiotics because it is …………………
(a) caused by a virus
(b) caused by a Gram-positive bacterium
(c) caused by a Gram-negative bacterium
(d) not an infectious disease
Answer:
(a) caused by a virus

Question 25.
Motile zygote of Plasmodium occurs in …………………
(a) gut of female Anopheles
(b) salivary glands of Anopheles
(c) Human RBCs
(d) Human liver
Answer:
(a) gut of female Anopheles

Question 26.
Haemozoin is …………………
(a) a precursor of haemoglobin
(b) a toxin from Streptococcus
(c) a toxin from Plasmodium
(d) a toxin from Hemophilus
Answer:
(c) a toxin from Plasmodium

Question 27.
Vaccination against malaria is not possible because …………………
(a) they produce antibodies and antitoxins
(b) they do not produce antibodies and antitoxins
(c) antibodies resistant to vaccines are produced
(d) none of these
Answer:
(b) they do not produce antibodies and antitoxins

Question 28.
The active form of Entamoeba histolytica feeds upon …………………
(a) blood only
(b) erythrocytes, mucosa and submucosa of colon
(c) mucosa and submucosa of colon only
(d) food in intestine
Answer:
(b) erythrocytes, mucosa and submucosa of colon

Question 29.
Eating unwashed and raw green leafy vegetables grown along the railway tracks in Mumbai may cause …………………
(a) malaria
(b) influenza
(c) amoebic colitis
(d) ringworm
Answer:
(c) amoebic colitis

Question 30.
Which method of water purification can terminate amoebae ?
(a) Chlorination
(b) Sedimentation
(c) Filtration
(d) Boiling
Answer:
(d) Boiling

Question 31.
Which of the following measures should be taken to control amoebic dysentery ?
(A) Using insecticidal sprays to kill flies.
(B) Not allowing stagnant water to be accumulated over a long time.
(C) To avoid eating uncovered food.
(D) Drinking only boiled water.
(E) Eating plenty of fruits.
(a) (A) (C) (D)
(b) (A) (B) (D)
(c) (B) (C) (E)
(d) (A) (B) (E)
Answer:
(a) (A) (C) (D)

Maharashtra Board Class 12 Biology Important Questions Chapter 10 Human Health and Diseases

Question 32.
Which of the following should be avoided for endemic spread of amoebiasis ?
(a) Cleaning bathroom taps and toilet seats with disinfectants.
(b) Washing hands and using hand sanitizers.
(c) Proper sewage disposal and treatment.
(d) Eating uncovered roadside food.
Answer:
(d) Eating uncovered roadside food.

Question 33.
Name the disease in which the genital organs are grossly affected due to infective helminth.
(a) Ascariasis
(b) Ring worm
(c) Scabies
(d) Filariasis
Answer:
(d) Filariasis

Question 34.
Find the odd organism:
(a) Wuchereria bancrofti
(b) Brugia malayi
(c) Brugia timori
(d) Ascaris lumbricoides
Answer:
(d) Ascaris lumhricoides

Question 35.
When is hydrocele formed in a man ?
(a) When testis are not functioning properly.
(b) When scrotum is infected with filarial worms.
(c) When testis are injured due to accident.
(d) When there is water accumulation in testis.
Answer:
(b) When scrotum is infected with filarial worms

Question 36.
Which medicine is used for eradicating microfilariae from endemic areas ?
(a) Diethyle carbamacine
(b) Mebendazole
(c) Albendazole
(d) Rimfampcin
Answer:
(a) Diethyle carbamacine

Question 37.
Which of the following fungi are causative organisms of ringworm ?
(a) Microsporum
(b) Candida
(c) Thrush
(d) Tinea pedis
Answer:
(a) Microsporum

Question 38.
On which material present on the outer skin surfaces of the human body does the fungus causing infections feed on ?
(a) Melanin
(b) Keratin
(c) Lignin
(d) Suberin
Answer:
(b) Keratin

Question 39.
Which of the following statements is correct ?
(a) Fungus grows well on dry skin.
(b) Fungus cannot survive on the outside of the hair shafts.
(c) Fungus thrives well on the warm and moist skin.
(d) Nails can never show fungal infections.
Answer:
(c) Fungus thrives well on the warm and moist skin.

Question 40.
Which of the following pair is viral diseases ?
(a) Common cold, AIDS
(b) Dysentery, Common cold
(c) Typhoid, Tuberculosis
(d) Ringworm, AIDS
Answer:
(a) Common cold, AIDS

Question 41.
Which one of the following glands is often referred in relation with AIDS ?
(a) Thymus
(b) Adrenal
(c) Thyroid
(d) Pancreas
Answer:
(a) Thymus

Question 42.
The first patient of AIDS was detected in India in …………………
(a) 1980
(b) 1986
(c) 1990
(d) 1996
Answer:
(b) 1986

Question 43.
Why is it said that for AIDS prevention is the only cure ?
(a) AIDS does not have any cure, once it is contracted.
(b) By prevention AIDS cannot be cured.
(c) AIDS can be cured by proper medication and vaccination.
(d) Only prevention helps as there is no cure for AIDS.
Answer:
(d) Only prevention helps as there is no cure for AIDS.

Question 44.
After a person is detected to be having AIDS by ELISA test, which is the next confirmatory test ?
(a) Western blot
(b) Southern blot
(c) PCR
(d) Northern blot
Answer:
(a) Western blot

Question 45.
What is full form of ELISA ?
(a) Enzyme Linked Inductive Assay
(b) Enzyme Linked Iron Sorbent Assay
(c) Enzyme Linked Immunosorbent Assay
(d) None of the above
Answer:
(c) Enzyme Linked Immunosorbent Assay

Question 46.
The possible ways of transmission of AIDS are …………………
(A) Intimate sexual contact
(B) Hugging and kissing
(C) Blood transfusion without properly checking it
(D) Eating from the same plate
(E) Sharing bed linen
(F) Transplacental infection from infected mother
(G) Sharing same tattoo gun and syringes
(a) (A), (C), (F), (G)
(b) (B), (D), (E), (G)
(c) (C), (D), (E), (F)
(d) (A), (B), (C), (D)
Answer:
(a) (A), (C), (F), (G)

Question 47.
Which one of the following statements is correct ?
(a) Benign tumours show the property of metastasis.
(b) Heroin accelerates body functions.
(c) Malignant tumours may exhibit metastasis.
(d) Patients who have undergone surgery are given cannabinoids to relieve pain.
Answer:
(c) Malignant tumours may exhibit metastasis.

Question 48.
Heroin or smack is chemically …………………
(a) diclofenac
(b) diacetyl morphine
(c) benzodiazepine
(d) amphetamines
Answer:
(b) diacetyl morphine

Question 49.
Ecstasy is a drug that is used in most of the Rev parties which is chemically a derivative of …………………
(a) Barbiturates
(b) Amphetamines
(c) Catecholamine
(d) Morphine
Answer:
(b) Amphetamines

Question 50.
From which plant is charas obtained?
(a) Cannabis sativa
(b) Erythroxylum coca
(c) Papaver somniferum
(d) Atropa belladonna
Answer:
(a) Cannabis sativa

Question 51.
Opium is obtained from the latex of the unripe fruits of …………………
(a) Cannabis sativa
(b) Thea siensis
(c) Papaver somniferum
(d) Erythroxylon coca
Answer:
(c) Papaver somniferum

Question 52.
Use of Cannabis products results in …………………
(a) depressed brain activity and feeling of calmness
(b) suppressed brain function and relief of pain
(c) stimulation of nervous system, increased alertness and activity
(d) alteration in perception, thoughts and feelings
Answer:
(d) alteration in perception, thoughts and feelings

Question 53.
Marijuana, ganja and LSD are …………………
(a) narcotics
(b) stimulants
(c) hallucinogens
(d) all of these
Answer:
(c) hallucinogens

Question 54.
What is the source of LSD ?
(a) Poppy seeds
(b) Datura plant
(c) Sugar
(d) Claviceps purpurea
Answer:
(d) Claviceps purpurea

Question 55.
Narcotics are …………………
(a) amphetamines and caffeine
(b) morphine and heroine
(c) LSD and cocaine
(d) barbiturates and benzodiazepine
Answer:
(b) morphine and heroine

Question 56.
Which is an incorrectly matched pair ?
(a) LSD – Ergot fungus
(b) Heroin – Opium
(c) Amphetamines – Depressant
(d) Benzodiazepine – Calmpose tablets
Answer:
(c) Amphetamines – Depressant

Question 57.
Choose the incorrect statement
(a) The excessive use of anabolic steroids cause severe acne.
(b) In both the sexes there is increased aggressiveness and mood swings, due to steroids.
(c) In females, anabolic steroids cause breast enlargement.
(d) In males, anabolic steroids cause enlargement of prostate gland.
Answer:
(c) In females, anabolic steroids cause breast enlargement.

Question 58.
Who are the first ones to note the danger signs of drug or alcohol abuse in the adolescents ?
(a) Alert parents and teachers
(b) Neighbours
(c) Relatives
(d) Doctors
Answer:
(a) Alert parents and teachers

Question 59.
Who can give professional help for the deaddiction?
(a) Highly qualified psychiatrist
(b) Parents
(c) Teachers
(d) Friends
Answer:
(a) Highly qualified psychiatrist

Match the columns

Question 1.

Column IColumn II
(a) Metchnikoff(i) ABO Blood group system
(b) Fleming(ii) Concept of immunity
(c) Edward Jenner(iii) Phagocytic cells
(d) Karl Lands teiner(iv) Lysozyme

Answer:

Column IColumn II
(a) Metchnikoff(iii) Phagocytic cells
(b) Fleming(iv) Lysozyme
(c) Edward Jenner(ii) Concept of immunity
(d) Karl Lands teiner(i) ABO Blood group system

Question 2.

DiseaseVector species
(a) Dengue(i) Anopheles
(b) Malaria(ii) Housefly
(c) Filaria(iii) Culex
(d) Typhoid(iv) Aedes

Answer:

DiseaseVector species
(a) Dengue(iv) Aedes
(b) Malaria(i) Anopheles
(c) Filaria(iii) Culex
(d) Typhoid(ii) Housefly

Classify the following to form Column B as per the category given in Column A

Question 1.
Benzyl penicillin, Chloromycetin, Mebendazole, Levamisole, Pyrimethamine Ampicillin, Ty21a vaccine, Sulfadoxine.

Column A (Disease)Column B (Treatment)
(1) Pneumonia————–
(2) Malaria————–
(3) Ascariasis————–
(4) Typhoid————–

Answer:

Column A (Disease)Column B (Treatment)
(1) PneumoniaBenzyl penicillin, Ampicillin
(2) MalariaPyrimethamine, Sulfadoxine
(3) AscariasisMebendazole, Levamisole
(4) TyphoidChloromycetin, Ty21a vaccine

Question 2.
Lung cancer, Pituitary, Spleen, Skin cancer, Cancer of adipose tissue, lymph nodes, Adrenal, Bone tumour.

Column (A Type of cancer)Column B (Organs affected)
(1) Carcinoma————–
(2) Sarcoma————–
(3) Lymphoma————–
(4) Adenocarcinoma————–

Answer:

Column (A Type of cancer)Column B (Organs affected)
(1) CarcinomaLung cancer, Skin cancer
(2) SarcomaCancer of adipose tissue, Bone tumour
(3) LymphomaSpleen, Lymph nodes
(4) AdenocarcinomaPituitary, Adrenal

Very Short Answer Questions

Question 1.
By which process T-cells and B-cells are produced?
Answer:
T-cells and B-cells are produced from the stem cells called haemocytoblasts, in bone marrow of adults and in liver of the foetus, by the process of haematopoiesis and in the bone marrow in adult.

Question 2.
What is a hinge?
Answer:
Hinge is the region of Y-shaped structure, holding arms and stem of antibody where four polypeptide chains of antibody are held together by disulfide bonds (-s-s-) to form a ‘Y’-shaped structure.

Question 3.
What is epitope?
Answer:
Epitope is antigenic determinant, which is present on antigens.

Maharashtra Board Class 12 Biology Important Questions Chapter 10 Human Health and Diseases

Question 4.
What is paratope?
Answer:
Paratope is antigen binding site that is present on the antibodies.

Question 5.
Which antigen is present in Rh +ve person?
Answer:
Antigen D is present in Rh +ve person.

Question 6.
Give the role of flushing action of lachrymal secretions.
Answer:
The conjunctiva is freed from foreign particles by the flushing action of lachrymal secretions.

Question 7.
What happens when lachrymal secretion is absent in eyes?
Answer:
Eyes become susceptible to infection when lachrymal secretion is absent.

Question 8.
In tears which antibacterial substance is present?
Answer:
Lysozyme is the antibacterial substance present in the tears.

Question 9.
Which kind of immunity is provided by vaccination?
Answer:
Artificial acquired active and passive immunity is provided by vaccination.

Question 10.
Who was Edward Jenner?
Answer:
Edward Jenner was the British scientist who developed cowpox vaccine for the protection against small pox virus.

Question 11.
Mrunmayi is called universal blood acceptor. What is her blood group?
Answer:
Blood group of Mrunmayi is AB.

Question 12.
What are antigens?
Answer:
Different foreign substances that invade the body and are capable of stimulating an immune response are called antigens.

Question 13.
In which animal Rh factor was discovered at first?
Answer:
Rh factor was first discovered in Rhesus monkey for the first time.

Question 14.
What is elephantiasis?
Answer:
Elephantiasis is one of the symptoms of lymphatic filariasis, in which there is thickening of skin and underlying tissues due to presence of malarial parasite.

Question 15.
What is dermatophytosis ?
Answer:
Dermatophytosis is a clinical condition in which fungal infection of skin occur in humans, pets and cattle which is commonly called as ringworm.

Question 16.
What is sporozoite?
Answer:
Sporozoite is a developmental stage of Plasmodium produced by rupture of oocyst. Sporozoite can enter the bloodstream in human body and then infect hepatocytes or liver cells, where they multiply into merozoites.

Question 17.
Why does male mosquito not spread Malaria?
Answer:
Male mosquito feed only on plant sap and not blood of human beings; therefore it does not spread Malaria.

Question 18.
Where does Plasmodium reproduce asexually?
Answer:
Plasmodium reproduce asexually in the liver cells and red blood cells of infected human being.

Question 19.
Where does Plasmodium reproduce sexually?
Answer:
Plasmodium undergoes sexual reproduction by the process of fertilization and development in the intestine of mosquito.

Question 20.
Which fish can be used for mosquito control?
Answer:
Gambusia fish can be used for mosquito control.

Question 21.
What is arthralgia?
Answer:
Arthralgia means joint pains.

Question 22.
What do you mean by hepatomegaly?
Answer:
Hepatomegaly means enlargement of liver.

Question 23.
Which organism yields LSD?
Answer:
Ergot fungus, Claviceps purpurea yield LSD.

Question 24.
Enlist various types of barriers which prevent entry of foreign agents into the body.
Answer:

  1. Epithelial surface
  2. Antimicrobial substances in blood and tissues
  3. Cellular factors in innate immunity
  4. Fever
  5. Acute phase proteins (APPs).

Question 25.
Which is the gastro-intestinal disease by which 15% Indian population is affected?
Answer:
Amoebiasis or amoebic dysentery is the gastro-intestinal disease by which 15% Indian population is affected.

Question 26.
What are the anti-helminthic drugs which are used in treatment of Ascariasis?
Answer:
Anti-helminthic drugs like Piperazine, Mebendazole, Levamisole, Pyrantel are used against Ascaris lumbricoidies.

Question 27.
Which are the diseases that can be avoided by eradication of mosquitoes in your area?
Answer:
Malaria, dengue, chikungunya and filariasis or elephantiasis can be avoided by eradication of mosquitoes.

Define the following

Question 1.
Serology
Answer:
A branch of immunology which deals with the study of antigen-antibody interactions is called serology.

Question 2.
Hygiene
Answer:
Hygiene is the science of health, which aims at preserving, maintaining and improving the health of an individual or the community as a whole.

Question 3.
Disease
Answer:
Disease is a condition of disrupted or deranged functioning of one or more organs or systems of the body caused due to infection, detective diet or heredity.

Question 4.
Immune system
Answer:
The system which protects us from various infectious agents is called immune system.

Question 5.
Resistance
Answer:
Resistance is an ability to ward off damage or disease through our defence mechanism.

Question 6.
Immunity
Answer:
The immunity is defined as the general ability of a body to recognize and neutralize or destroy and eliminate foreign substances or resist a particular infection or disease.

Question 7.
Antibody
Answer:
The protective chemicals produced by immune cells in response to antigens is called antibodies.

Question 8.
Opsonisation
Answer:
The process of coating of bacteria to facilitate their subsequent phagocytosis by macrophages is called opsonisation.

Question 9.
Pathogen
Answer:
Pathogen are living agents such as viruses, ricketssia, bacteria, fungi, protozoans, helminth and certain insect larvae which are capable of causing diseases.

Question 10.
Parasite
Answer:
An organism that lives in or on another organism called host and takes its nourishment from it (host) is called parasite.

Question 11.
Pathogenicity
Answer:
The ability of an organism to enter a host and cause a disease is called pathogenicity.

Question 12.
Infectious disease
Answer:
The disease which is transmitted from infected person to another healthy person either directly or indirectly is called infectious disease or communicable disease.

Question 13.
Non-infectious disease
Answer:
The disease that cannot be transmitted from one infected person to another healthy person, either directly or indirectly is called non infectious or non-communicable disease.

Question 14.
Innate immunity
Answer:
Innate immunity is defined as the resistance to infections that an individual possesses due to his or her genetic make-up and thus it is inborn defence mechanism present naturally in the body.

Question 15.
Acquired immunity
Answer:
The resistance developed during lifetime is called acquired immunity.

Question 16.
APP proteins
Answer:
APP proteins or acute phase proteins are certain collection of plasma proteins which are suddenly increased by the infection caused after injury.

Question 17.
Incubation period
Answer:
Incubation period is the time interval from the invasion of a pathogen to the development of clinical manifestations or symptoms.

Name the following

Question 1.
Name the cell that produces lymphokines.
Answer:
Helper T-cells

Question 2.
Blood group systems in human beings.
Answer:
ABO, Rh, Duffy, Kidd, Lewis, P MNS, Bombay blood group.

Maharashtra Board Class 12 Biology Important Questions Chapter 10 Human Health and Diseases

Question 3.
Types of sarcoma.
Answer:

  1. Osteosarcoma (bone)
  2. Myosarcoma (muscle)
  3. Chondrosarcoma (cartilage)
  4. Liposarcoma (adipose tissue)

Question 4.
Therapies used for treatment of cancer.
Answer:

  1. Chemotherapy
  2. Radiotherapy
  3. Surgery
  4. Immunotherapy
  5. Supportive therapy

Question 5.
Name the term for the transmission of HIV from pregnant mother to foetus.
Answer:
Transplacental.

Question 6.
Factors that maintain good health.
Answer:

  1. Balanced diet
  2. Personal hygiene
  3. Regular exercise
  4. Right attitude of mind
  5. Good habits.

Question 7.
Two examples of ascaricides.
Answer:
Mebendazole and Albendazole, etc.

Question 8.
Two vaccines for typhoid.
Answer:

  1. Oral Ty21a
  2. Injectable Typhoid polysaccharide vaccine or Typhium vi/ Typherix.

Question 9.
Parasites causing lymphatic filariasis.
Answer:

  1. Wuchereria bancrojti
  2. Brugiamalayi
  3. Brugia timori.

Question 10.
Parasites causing Subcutaneous Filariasis.
Answer:

  1. Loa loa
  2. Mansonella spp.

Question 11.
Name the scientists who discovered AB blood group?
Answer:
Decastallo and Sturti

Distinguish between the following

Question 1.
Inborn immunity and acquired immunity.
Answer:

Inborn ImmunityAcquired Immunity
1. Inborn immunity or innate immunity is also called natural immunity.1. Acquired immunity is also called adaptive immunity.
2. Innate immunity is present right from the birth.2. Acquired immunity is not present at birth, but is acquired during lifetime of the individual.
3. Inborn immunity does not depend upon the previous exposure to a pathogen or foreign substance.3. Acquired immunity always depends upon the previous exposure to a pathogen or foreign substance.
4. It is non-specific immunity as it can offer resistance to any pathogen.4. It is specific immunity as it can offer resistance only to a particular pathogen.
5. Innate immunity consists of various types of barriers for defence against the pathogens.5. Acquired immunity consists of various types of cells which are able to produce antibodies.
6. Inborn immunity shows immediate effect in the body.6. Acquired immunity requires several days to become activated.
7. Inborn immunity is seen in all animals.7. Acquired immunity is seen only in vertebrates.
8. Inborn immunity is genetic in nature and is heritable.8. Acquired immunity is non-genetic in nature and is non-heritable.

Question 2.
Communicable and non-communicable diseases.
Answer:

Communicable diseasesNon-communicable diseases
1. Diseases transmitted from infected person to healthy person are called communicable or infectious diseases.1. Diseases that are not passed from one person to other are non-communicable or non-infectious diseases.
2. Communicable diseases spread through pathogens.2. Non-communicable diseases do not spread through pathogens.
3. Communicable diseases are not inherited from parental generation to offspring.3. Non-communicable diseases like cancer can be from parental generation to offspring.
4. Vectors play the major role in spreading disease from one person to another.4. Caused due to allergy, illness, malnutrition or abnormalities in cell proliferation, changes in lifestyle, environment play a significant role.
5. Treated by conventional methods using antibiotics and other drugs.5. Treated conservatively for a long time or surgically.
6. Diseases are acute which develop suddenly due to infections.

E.g. Pneumonia, Tuberculosis, AIDS, Typhoid, Cholera, Malaria.

6. Diseases are chronic which develop and persist for a long time.
E.g. Cancer, Rickets, Allergies, Kwashiorkor, Diabetes, Heart disease, etc.

Question 3.
Ascariasis and Filariasis.
Answer:

AscariasisFilariasis
1. Only one species Ascaris lumbricoid.es cause ascariasis.1. There are many species of nematode that can cause filariasis.
2. Ascaris causes the infection of alimentary canal.2. Wuchereria bancrofti causes the infection of lymphatic system.
3. Ascaris does not cause swellings of upper and lower limbs.3. Filariasis cause swellings of extremities.
4. Ascaris is caused due to faeco-oral transmission.4. Filariasis is caused due to vector transmission (Culex mosquito).
5. Medicines for treatment of Ascariasis are Piperazine, Mebendazole, Levamisole, Pyrantel.5. Medicines for treatment of filariasis are diethyl- carbamazine citrate.

Short answer questions

Question 1.
Despite constant exposure to variety of pathogens, why do most of us remain healthy?
Answer:

  1. All human beings are exposed to various foreign bodies, including infectious agents like bacteria, viruses, etc. which are called pathogens.
  2. But human body can resist almost all types of these pathogens.
  3. For this purpose, there is immune system which protects us from various infectious agents.
  4. There is resistance and prevention of the damage or disease, through our defence mechanisms.
  5. Thus, despite constant exposure to variety of pathogens, most of us remain healthy.

Question 2.
What are the unique features of acquired immunity?
Answer:
Following are the unique features of acquired immunity:

  1. Specificity : Production of specific antibody or T-lymphocyte against a particular antigen/ pathogen is called specificity.
  2. Diversity : Ability to recognize vast variety of diverse pathogens or foreign molecules by immunity is called diversity.
  3. Discrimination between self and non¬self : Acquired immunity can differentiate between own body cells (self) and foreign (non-self) molecules.
  4. Memory : The first immune response upon encounter of a specific foreign agent and its elimination is retained as a memory. This results in quicker and stronger immune response when the same pathogen is encountered again.

Question 3.
Describe the polypeptide chains seen in the structure of an antibody.
Answer:

  1. There are four polypeptide chains which make the antibody.
  2. There are two heavy or H-chains and two light or L-chains.
  3. Each chain has two distinct regions, the variable region and the constant region.
  4. Variable regions carry the antigen binding site or paratope.
  5. This part of antibody recognizes and bindsto the specific antigen to form an antigen- antibody complex.

Question 4.
Which are the antimicrobial substances in blood and tissues?
Answer:

  1. There are more than 30 serum proteins, circulating in the blood in an inactive state, which forms the complement system.
  2. ‘Complement cascade’ is activated by presence of pathogens and thus they eliminate pathogens.
  3. Cells which are affected by viruses secrete interferons which are a class of cytokines. These soluble proteins attack the pathogens.
  4. Some leucocytes stimulate other cells to protect themselves from viral infection.

Question 5.
Is developing fever a bad sign or a good action? Explain.
Answer:

  1. Fever is the innate immunity mechanism. When there is fever, the body rises.
  2. This is in response to infection in a natural way.
  3. Developing fever is a natural defence mechanism.
  4. Fever helps to accelerate the physiological processes by which the invading pathogens are destroyed.
  5. Fever stimulates the production of interferon and helps in recovery from viral infections.
  6. Taking all these points into consideration, getting fever is a good action that innate immunity is working properly, however, entry of pathogen in the body causing illness is a bad sign.

Question 6.
With the help of a chart explain the compatibility of human blood groups.
Answer:
Different types of blood groups:
Maharashtra Board Class 12 Biology Important Questions Chapter 10 Human Health and Diseases 1

  1. Person with blood group A can donate blood to persons having blood group A or AB and receive blood from A or O.
  2. Person with blood group B can donate blood to persons having blood group B or AB and receive blood from B or O.
  3. Person with blood group O can donate blood to all the persons having blood group either A, B, AB or O, because blood group O is universal donor. But can receive blood only from person having blood group O.
  4. Person with blood group AB can donate blood to only AB but can receive blood from all the persons having blood group either, A, B, O or AB because AB is universal recipient.

Question 7.
The blood group of Krutika is O Rh +ve. What would be the possible blood groups of her parents?
Answer:

  1. Krutika has O blood group, therefore her genotype is I°I°.
  2. Her parents can be of following combinations.
  3. They may be having blood group A or B with heterozygous genotype respectively, i.e. IAI° and IBI°. If both of them are heterozygous, having either A or B blood group, Krutika can be of O type.
  4. Other possibility is both the parents have to be O.
  5. For being Rh positive, her at least one parent should be Rh positive. Rh negative is a recessive phenotype and hence needs double dose of these genes.

Question 8.
Can a person with blood group O Rh+ve donate blood to a patient with blood group O Rh-ve? Why?
Answer:
No. blood group O may be common to both donor and recipient but their Rh factor is different. A person with RH+ve blood cannot donate to a patient with Rh-ve blood group. In Rh+ve blood there is antigen D. This antigen D when enters the body of recipient, there are production of anti-RH antibodies in his or her body. These anitibodies with cause agglutination of the Rh +ve blood which will be given to the patient. This agglutination will cause clots inside the vital organ of the recipient and the death may follow.

Question 9.
Why do we suffer from common cold repetitively in our life, but other viral diseases like Influenza or Small pox only once?
Answer:
Influenza infection causes production of antibodies in our body, once the virus attacks us. Therefore, second encounter with the virus may not cause effect. But in case of common cold, large number of different virus families are responsible for developing infection of common cold. Many a times different allergens are also inducing agents for common cold. Thus we may suffer from common cold again and again.

Question 10.
What are the symptoms of pneumonia?
Answer:

  1. Main symptoms of infectious pneumonia are cough producing greenish or yellow sputum or phlegm and a high fever with chills.
  2. Shortness of breath, stabbing chest pain, coughing up blood, headaches, sweaty and clammy skin, loss of appetite, fatigue, blueness of the skin, nausea, vomiting, mood swings and joint pains or muscle aches are some other symptoms.

Question 11.
What is the diagnosis and treatment of filariasis? How can we control this disease ?
Answer:
I. Diagnosis and Treatment : For the patient, diethyl-carbamazine citrate is the drug used for twice a day for three weeks. Thereafter for five days every six months the same treatment is repeated. This becomes effective against filarial worms.

II. Prevention and Control:

  1. Mosquito eradication should be done for controlling filariasis.
  2. In the areas with mosquitoes, avoid mosquito bite by using mosquito nets and insect repellents.

Question 12.
What are the signs and symptoms of filariasis?
Answer:
Signs and symptoms of filariasis:

  1. As the lymphatic drainage does not take place, there is oedema with thickening of skin and underlying tissue.
  2. Extremities like legs, arms, breasts, scrotum, etc. are affected by nematode causing lymphatic filariasis, i. e. Wuchereria bancrofti.
  3. Lymph vessels and lymph nodes are enlarged and swollen.
  4. Elephantiasis is seen in which limbs are swollen like legs of elephant.
  5. Lymphoedema, i.e. accumulation of lymph fluid is seen in tissue causing swelling.
  6. Hydrocele condition develops in which testis are enlarged due to accumulation of lymphatic fluid in testis.

Question 13.
What are the various ways in which mosquitoes can be eradicated from any area?
Answer:
Eradication of mosquitoes:

  1. Removal of all stagnant water pools around the houses.
  2. If such water bodies are there, they should be sprayed with insecticides.
  3. But better option which is eco-friendly is releasing mosquito eating fish like Gambusia or Tilapia.
  4. Use of mosquito repellent plants like Citronella. Use of coils and repellent creams.
  5. Fumigation of the area to kill the mosquito.
  6. Aedes sps. breed in man-made containers, especially plastic and cement tanks. Care should therefore be taken to dispose such containers properly. Water should not be accumulated in them.

Question 14.
What precautions will you take if you are travelling in an area which has lot of mosquitoes?
Answer:

  1. Avoiding areas where mosquitoes are in more concentration.
  2. Carrying mosquito repellent creams or coils.
  3. Use of mosquito nets and other fumigation devices.
  4. Wearing full clothing in light colours.
  5. Staying indoors when mosquitoes are swarming, especially in the evening.
  6. Taking anti-malarial pills as a precautionary measure.

Maharashtra Board Class 12 Biology Important Questions Chapter 10 Human Health and Diseases

Question 15.
Deaddiction may be difficult but not impossible. Collect information about NGOs, working in the field of deaddiction.
Answer:
There are many NGOs that work in the field of deaddiction. In different cities, there are different organizations. Even the central Government has started helpline number 1800-11-0031 for those drug and alcohol addicts who need help to come out of these addictions. Muktangan in Pune is one such reputed organization which does the great work in the field of deaddiction.

Question 16.
What are the most common warning signs of drug and alcohol abuse among youth ?
Answer:
The most warning signs of addictions are as follows:

  1. Drop in academic performance, absenteeism from school or college.
  2. No interest in personal hygiene and hobbies.
  3. Withdrawal from the society, increased tendency of isolation and depression.
  4. Aggressive and rebellious behaviour resulting into strained relationships with family and friends.
  5. Fatigue and change in sleeping and eating habits.
  6. Fluctuations in weight, appetite, deteriorating health, etc.

Question 17.
What are the preventive measures for malaria?
Answer:
Preventive measures of malaria :

  1. Transmission of malarial parasite can be reduced by preventing mosquito bites. Therefore, mosquitoes should be controlled or totally eradicated.
  2. This can be done by using of mosquito nets and insect repellents.
  3. Mosquito control measures such as spraying insecticides inside houses and draining stagnant water where mosquitoes lay their eggs.
  4. The mosquito larvae can be eradicated by releasing Gambusia fish which can feed upon these larvae.
  5. Vaccine against malaria is also under preparation.

Question 18.
How does Entamoeba histolytica causes amoebiosis ?
Answer:

  1. Amoebiosis is spread through ingestion of the cyst form of Entamoeba histolytica. This is a commensal organism.
  2. Cyst is a semi-dormant and hardy structure found in faeces of infected person.
  3. Non-encysted amoebae are called trophozoites. The trophozoites die quickly after leaving the body but may also be present in faeces.
  4. Trophozoites are rarely the source of new infections.
  5. The infection may remain asymptomatic for many days as Amoeba can remain latent in the gastrointestinal tract.

Question 19.
Describe the signs and symptoms of amoebiasis.
Answer:
Amoebiasis shows following common symptoms:

  1. Diarrhoea, flatulence, stool with mucus and abdominal pains (cramps) are common.
  2. Stool sticky with mucus and blood.
  3. Amoebae form cysts in the liver, in such case there is hepatomegaly, i.e. enlargement of liver.
  4. Liver shows amoebic liver abscess accompanied with fever and pain in right side of the abdomen.

Question 20.
How can amoebiasis be prevented?
Answer:
Prevention of amoebiasis is to be done at two levels, viz. at home and at endemic level.
1. Prevention of the spread of amoebiasis at the home level:

  • Washing hands with soap and water after using the toilet or changing a baby’s diaper and before handling and eating food.
  • Cleaning bathrooms and toilets properly with germicides.
  • Avoiding raw vegetables when in endemic areas where they are grown in soil fertilized by human faeces.
  • Boiling and purifying the drinking water.

2. Prevention of the spread of amoebiasis at endemic level:

  • Avoiding consumption of street foods especially in public places.
  • Following good sanitary practice, as well as using proper sewage disposal or treatment.
  • E. histolytica cysts are usually resistant to chlorination; therefore sedimentation and filtration of water supplies are necessary to reduce the incidence of infection.
  • Avoiding shared towels or face washers.

Question 21.
Describe the symptoms of ascariasis.
Answer:

  1. After infection by Ascaris lumbricoides, there is appearance of eggs in stools in 60 – 70 days.
  2. In larval ascariasis, symptoms are seen in 4-16 days after infection.
  3. The final symptoms are gastrointestinal discomfort, colic and vomiting, fever and appearance of live worms in faeces.
  4. Some patients may have pulmonary symptoms. Inflammation of alveolar walls is seen. This is known as pneumonitis.
  5. Some may show neurological disorders during migration of the larvae.
  6. Loss of appetite which reflects in weight loss.
  7. A bolus of worms may obstruct the intestine.
  8. Larvae that migrate may also cause eosinophilia, i.e. increase in number of eosinophils.

Question 22.
What are the preventive measures against ascariasis?
Answer:

  1. Prevention of ascariasis can be done by adopting the following measures :
  2. Use of proper toilet facilities.
  3. Safe disposal of excreta.
  4. Protection of food from dirt and soil.
  5. Washing of vegetables before cooking and avoiding eating raw, unwashed vegetables and fruits.
  6. Hand washing and use of safe food. Observing personal hygiene.
  7. Use of pharmaceutical drugs such as Mebendazole and Albendazole can kill Ascaris.

Question 23.
Discuss the clinical manifestation of AIDS.
Answer:
There are four stages of clinical manifestations or symptoms of AIDS.

  1. Stage I : This is initial infection with the virus and formation of antibodies, usually 2-8 weeks after initial infection.
  2. Stage II : In this stage the person is asymptomatic carrier. Incubation takes place with a period ranging for 6 months to 10 years.
  3. Stage III : This is called AIDS related complex (ARC). In this stage, one or more of the following clinical signs are seen. E.g. Recurrent fever for longer than one month, fatigue, unexplained diarrhoea, night sweats, shortness of breath, loss of more than 10 per cent body weight, etc.
  4. Stage IV : This is the end stage in which patient shows full blown AIDS. Thus it is called the end stage of HIV infection. Life threatening opportunistic infections (like pneumonia, tuberculosis, Kaposi sarcoma, etc.) are easily caught during this period.

Question 24.
Through which modes HIV infection does not spread?
Answer:

  1. HIV does not spread through casual contact such as hugging, etc.
  2. Insect bite such as mosquito bites does not transmit HIV.
  3. Participation in sports is not the mode by which HIV transmits.
  4. Contact between articles used by AIDS patient, a hand shake with him or her does not transmit HIV
  5. HIV infections do not occur through swimming pool or by sharing clothes, utensils, etc.

Question 25.
Write about laboratory diagnosis and treatment of AIDS.
Answer:
I. Laboratory diagnosis :

  1. There are two tests for diagnosis of AIDS.
  2. First test is ELISA (Enzyme-Linked Immunosorbent Assay) which is used to detect the HIV antibodies.
  3. The second confirmatory test is Western Blot, which is used to weed out any false positive results. It is a highly specific test.
  4. It is based on detecting specific antibody to viral core protein and envelope glycoprotein.

II. Treatment of AIDS:

  1. AIDS cannot be cured.
  2. Antiretroviral drugs are used to reduce the viral load and prolong the life of HIV patient. E.g. Antiretroviral therapy (ART) uses drugs such as TDF (tenofovir), EFV (Efavirenz), Lamivudine (3TC), etc.

Question 26.
Describe the structure of HIV with a suitable diagram.
Answer:

  1. Human Immunodeficiency Virus or HIV is spherical and 100 to 140 nm in diameter.
  2. It has centrally located two ss RNA molecules along with reverse transcriptase enzymes.
  3. There are coverings of two layers of proteins. The outer layer formed by matrix protein (pi7) while in inner layer is of capsid protein (p24).
  4. An additional layer of lipids is seen over the matrix protein layers. This layer is impregnated with glycoprotein GP120 and GP41.
    Maharashtra Board Class 12 Biology Important Questions Chapter 10 Human Health and Diseases 2

Question 27.
What are the modes of transmission of HIV or AIDS?
Answer:
The transmission of HIV occurs through following routes:

  1. Sexual relations, mainly unsafe sexual contact including oral, vaginal and anal sex.
  2. Through blood and blood products either by blood transfusions or sharing needles and syringes.
  3. Transplacental From pregnant mother to her foetus through placenta. Nursing mother can also transmit HIV to her baby through lactation.
  4. Spreading the virus is very rare in case of accidental needle injury, artificial insemination with infected donated semen and transplantation with infected organs.
  5. HIV is seen in urine, tears, saliva, breast milk and vaginal secretions but unless these body fluids enter the injuries and wounds, transmission is not easy.

Question 28.
What are the measures of prevention and control of AIDS ?
Answer:

  1. Preventive measures : AIDS has no cure, hence prevention is the best choice. The
    following steps help in preventing this dreadful disease.
  2. High risk group people should be educated about HIV transmission. They should never donate blood.
  3. Use of disposable needles and syringes should be done with proper disposal.
  4. Risky sexual habits should be avoided.
  5. Tooth brushes, razors, other articles that can become contaminated with blood should not be shared.
  6. Blood should be screened before receiving it.
  7. Routine screening of blood and semen donors, organ donors (kidney, liver, lung, cornea), and patients undergoing haemodialysis must be done.
  8. Pregnant women or those women who are contemplating pregnancy should be regularly screened.

Question 29.
Explain the ill-effects of opioids and cannabinoids on health.
OR
What are harmful effects of drug abuse?
Answer:

  1. Opioids bind to specific opioid receptors which are present in central nervous system and gastrointestinal tract. Some opioids e.g. heroin act like depressants and slow down all the body functions.
  2. Cannabinoids have the capacity to interact with receptors present in the brain. Inhalation or ingestion of cannabinoids such as marijuana, hashish, charas and ganja have adverse effect on cardiovascular system.
  3. Cannabinoid like LSD causes hallucinations.
  4. All these substances are addictive and hence cause adverse effects on the body and health.

Question 30.
Give the adverse effects of opioids, cannabinoids and morphine on human health.
Answer:
1. Opioids:

  • Opioids bind to specific opioid receptors present in the central nervous system and in gastrointestinal tract.
  • They are depressants and slow down the body functions.

2. Cannabinoids:

  • Cannabinoids interact with receptors in the brain.
  • They affect cardiovascular system of the body.

3. Morphine:

  • Morphine is an effective sedative and pain killer when used for medicinal purpose.
  • When abused it affects physical, physiological and psychological functions.

Give reasons

Question 1.
Vaccines are safe.
Answer:
During their manufacture, vaccines are rigorously tested. Many rounds of study, examination and research are carried out before they are used for general public. Extensive research and evidences are gathered to check their safety. Sometimes, some vaccines produce side effects but these are rare and mild. Hence vaccines are considered to be safe.

Question 2.
Innate immunity is also known as non-specific immunity.
Answer:
Innate immunity is non-specific because it does not depend on previous exposure to foreign substances. It is inborn capacity of the body to resist the pathogen that causes the disease. It is natural immunity and hence it remains non-specific, trying to protect the body in case of any invasion of foreign body.

Question 3.
Vaccination is important for preventing pneumonia.
Answer:
Vaccinations for pneumonia are available against Haemophilus influenzae and Streptococcus pneumoniae. If given earlier in life, they reduce the chances of catching pneumococcal infections. The deaths can be prevented which are common due to lung infections. Since it is a common and chronic j disease for all age groups, for the prevention one must take vaccination.

Maharashtra Board Class 12 Biology Important Questions Chapter 10 Human Health and Diseases

Question 4.
Common cold is the most frequent ! infectious disease in humans.
Answer:
Common cold is caused by virus which is abundantly present in congested city environments. The upper respiratory tract is infected due to these rhinoviruses and corona viruses. Average adult contracts such infections 2 to 4 times in a year while children capture it 6 to 12 times in a year. Therefore, it is said to be the most frequent infectious disease of human beings.

Question 5.
Typhoid is food and water-borne disease.
Answer:
Typhoid is caused due to Salmonella typhi which isa Gram-negative bacterium, transmitted from a patient or carrier to another healthy person through contaminated food or water. Flying insects, mostly houseflies transmit the bacteria from faeces to the food. Poor hygienic habits and improper public sanitation system spreads typhoid. Therefore, it is said to be food and water-borne disease.

Question 6.
Malignant tumour is more dangerous than benign tumour.
Answer:
Malignant tumour cells can show metastasis and hence can spread far and wide in the body, affecting other healthy cells and tissues. They are difficult to cure by any therapy as one does not know exact location of the cancerous cells. They also produce variety of symptoms depending on their i location. On the contrary, benign tumours can be treated surgically. Since they are covered by a cyst like membrane, the cancerous cells do not spread from them. Thus malignant tumour can be lethal as against the benign tumour.

Question 7.
Prevention is better than cure for AIDS.
Answer:
Till this time, there is no preventive vaccination for AIDS. There is also no cure for AIDS. The medicines are also costly and may not give complete cure. The only way to remain away from AIDS is the complete awareness about it. Thus it should be prevented by not allowing HIV to enter our body. Once HIV finds the entrance, the cure is impossible. Therefore, it is said that prevention is better them cure for AIDS.

Write short notes

Question 1.
Cellular factors in innate immunity.
Answer:

  1. Phagocytic cells ingest and destroy the pathogens.
  2. This is natural defence against the invasion of pathogenic microorganisms and other foreign particles in blood and tissues.
  3. Phagocytic cells are of two types, viz. microphages and macrophages. They can remove foreign particles that enter the body.
  4. Natural killer (NK) cells is a class of lymphocytes which carry out important and non-specific defence against viral infections and tumours.

Question 2.
Acute phase proteins (APPs).
Answer:

  1. Acute phase proteins are involved in innate immune mechanism.
  2. When there is an infection or injury, it leads to a sudden increase in concentration of certain plasma proteins, which are called acute phase proteins or APPs.
  3. These include C Reactive Protein (CRP), Mannose binding protein, Alpha-1-acid glycoprotein, Serum Amyloid P etc.
  4. APPs enhance host resistance, prevent tissue injury and promote repair of inflammatory lesions.

Question 3.
Rh factor.
Answer:
Rh factor:

  1. Rh factor is the term adapted from Rhesus monkey.
  2. In rhesus monkey, there is antigen D on the surface of their RBCs.
  3. Landsteiner and Wiener discovered this antigen and termed it as Rh factor.
  4. Persons having Rh factor or D antigen are called Rh positive while those lacking D antigen or Rh factor are called Rh negative.

Question 4.
Erythroblastosis foetalis.
Answer:

  1. Erythroblastosis foetalis is condition in which there is destruction of the erythrocytes of the foetus. It is the haemolytic disease of the newborn (HDN).
  2. This is caused in foetus, if mother is Rh -ve and father is Rh +ve. Rh +ve is the dominant allele, the foetus becomes Rh +ve, when its father is RH +ve.
  3. Rh +ve blood groups have D antigen which induces a strong immunogenic response when introduced into Rh -ve individuals.
  4. During foetal life, there is connection between mother and foetus through placenta, therefore Rh +ve antigen D from the foetus enters maternal circulation.
  5. This triggers formation of anti-Rh antibodies in mother. Subsequently Rh+ve foetus receives anti-Rh antibodies produced by mother.
  6. This causes agglutination reaction resulting into haemolysis in foetus. In order to prevent HDN, Rh -ve mother is injected with the anti-Rh antibody during all her pregnancies if her husband is Rh +ve.

Question 5.
Common cold.
Answer:

  1. The common cold (nasopharyngitis or rhinopharyngitis) is a viral infectious disease of the upper respiratory system. The causative organisms are rhinoviruses and coronaviruses.
  2. Symptoms include cough, sore throat, runny nose and fever.
  3. There is no known treatment, however, symptoms usually resolve spontaneously in 7 to 10 days.
  4. The best prevention for the common cold is to stay away from infected people and places where infected individuals have been.
  5. Hand washing with plain soap and water is recommended. Also alcohol-based hand sanitizers provide very little protection.

Question 6.
Life cycle of Plasmodium.
Answer:

  1. Anopheles Female mosquito which is a carrier carries sporozoites. When it bites the human, these sporozoites enter human circulation.
  2. Sporozoites undergo asexual reproduction through fission or schizogony in the liver cells or erythrocytes of the human.
  3. It forms merozoites. The cells formed within erythrocytes function as gametocytes. They undergo gamogony.
  4. Upon biting such person, the gametocytes enter into female Anopheles, fertilization occurs in its gut.
  5. Diploid zygote transforms into oocyst. Oocyst forms large number of haploid sporozoites through meiosis (sporogony).
  6. Sporozoites migrate to salivary glands and are ready to infect new human host.
  7. Again Sporozoite → Merozoite → Trophozoite → Schizont sequence is carried on for plasmodial stages in human body.
  8. The sexual life cycle of Plasmodium occurs in mosquito body which acts as a vector. While its asexual phase takes place in human body.

Question 7.
Pneumonia.
Answer:

  1. Pneumonia is an inflammatory condition of alveoli in the lungs causing formation of fluid in the lungs. This condition is called consolidation and exudation.
  2. Causes of pneumonia are infection due to bacteria, viruses, fungi or parasites, chemical burns or physical injury to the lungs.
  3. Influenza virus, adenovirus, para influenza and Respiratory Syncytial Virus (RSV) are some viruses that can cause pneumonia. Bacteria like Streptococcus pneumoniae and fungal pathogens e.g. Pneumocystis jirovecii and Pneumocystis carinii can also spread infection of pneumonia. Chemical burns of physical injury to lungs also cause similar infection.
  4. Main symptoms of infectious pneumonia are cough producing greenish or yellow sputum or phlegm and a high fever with chills.
  5. Shortness of breath or dyspnea, stabbing chest pain, coughing up blood, headaches, sweaty and clammy skin, loss of appetite, fatigue, blueness of the skin, nausea, vomiting, mood swings and joint pains or muscle aches are some other symptoms.
  6. Preventive vaccination against pneumonia is available. Medicines such as Benzyl penicillin, Ampicillin and Chloramphenicol are effective to prevent pneumonia.

Question 8.
Ringworm.
Answer:

  1. Ringworm or Dermatophytosis is a clinical condition caused by Trichophyton and Microsporum fungal infection of the skin. This infection is seen in humans and pets.
  2. Dermatophytes are the fungi that feed on keratin. Keratin is the material found in the outer layer of skin, hair and nails.
  3. These fungi attack various parts of the body. Infections on the body forms enlarged raised red rings. These patches have intense itching. Infection on the skin of the feet may cause athlete’s foot and jock itch.
  4. When the nails are infected it causes onychomycosis. During this the nails thicken, discolour and finally crumble and fall off.
  5. Prevention of ringworm infection is to be done by avoiding sharing of clothing, sports equipment, towels or sheets. Clothes should be washed in hot water with fungicidal soap after suspected exposure to ringworm. One should not walk barefoot but use appropriate footwear.
  6. Diagnosis of ringworm is done by physical examination and treatment is done with uses drugs like nystatin, fluconazole, itraconazole, etc.

Question 9.
Dengue.
Answer:

  1. Dengue is a viral disease causing high fever. It is a painful, debilitating vector-borne disease.
  2. There are four closely related dengue viruses that cause infection.
  3. Vector of Dengue virus is female Aedes mosquito. The mosquito takes up the dengue virus when it sucks blood of a person suffering from dengue.
  4. The spread of dengue is not directly from one person to another person.

Question 10.
Performance enhancers.
Answer:

  1. Performance enhancers are certain drugs used by sportspersons to enhance their performance during competitions.
  2. Narcotic analgesics, anabolic steroids, diuretics and certain hormones are misused by such sportspersons to increase muscle strength and bulk. It also promotes aggressiveness and improve overall performance.
  3. Use of anabolic steroids cause side effects.
  4. Females show masculinization, increased aggressiveness, mood swings, depression, abnormal menstrual cycles, excessive hair growth on the face and body, enlargement of clitoris, deepening of voice.
  5. Males show acne, increased aggressiveness, mood swings, depression, and reduction of size of the testicles, decreased sperm production, kidney and liver dysfunction, breast enlargement, premature baldness, enlargement of the prostate gland.
  6. These effects may be permanent with prolonged use. Using such drugs is illegal and punishable.

Chart Based Questions

Question 1.
Complete the chart of ABO blood group system and answer the questions given below:
Maharashtra Board Class 12 Biology Important Questions Chapter 10 Human Health and Diseases 3
Questions:
(i) Which blood group from the above table is called universal acceptor?
(ii) Which blood group from the above table is called universal donor?
Answer:
Maharashtra Board Class 12 Biology Important Questions Chapter 10 Human Health and Diseases 4
(i) Blood group AB is called universal acceptor.
(ii) Blood group O is called universal donor.

Question 2.
Complete the table

Plasmodium speciesIncubation periodPattern of fever
————————–High fever after 48 hours.
————–28 days————–
————–17 days————-
————————-High fever at irregular intervals between 22 to 48 hours.

Answer:

Plasmodium speciesIncubation periodPattern of fever
P. vivax14 daysHigh fever after 48 hours.
P. malariae28 daysHigh fever after 72 hours interval
P. ovale17 daysHigh fever after 48 hours interval
P. falciparum12 daysHigh fever at irregular intervals between 22 to 48 hours.

Question 3.
Complete the following table
Maharashtra Board Class 12 Biology Important Questions Chapter 10 Human Health and Diseases 5
Answer:
Maharashtra Board Class 12 Biology Important Questions Chapter 10 Human Health and Diseases 6

Question 4.
Complete the following table:

CarcinogenOrgan affected
N-nitrosodimethlene————–
Aflatoxin———-
————–Vagina
————–Urinary bladder
————–Prostate
————–Skin and lungs

Answer:

CarcinogenOrgan affected
N-nitrosodimethleneLungs
AflatoxinLiver
DiethylstilboestrolVagina
2-naphthylamine and 4-aminobiphenylUrinary bladder
Cadmium oxideProstate
Soot, coal tar (2-4 benzopyrene)Skin and lungs

Diagram Based Questions

Question 1.
Label the given diagram
Img 7Maharashtra Board Class 12 Biology Important Questions Chapter 10 Human Health and Diseases 7
Answer:

  1. Antigen binding site.
  2. Variable region of heavy chain
  3. Varible region of light chain
  4. Constant region of light chain
  5. Constant region of heavy chain
  6. Disulphide bond
  7. Hinge
  8. Light chain
  9. Heavy chain

Question 2.
Observe the given diagram and answer the following questions:
Maharashtra Board Class 12 Biology Important Questions Chapter 10 Human Health and Diseases 8
(1) What is I and II in the above diagram?
Answer:
I is a virus which is trying to cause infection, II are two antigen molecules which are trying to attack the virus.

(2) What structures are responsible for antigen and antibody complex? Identify them in the above diagram.
Answer:
(a) is epitope which is antigen determinant and
(b) is a paratope which is part of the antibody. Epitope and paratope are specific to each other and hence they form a complex.

(3) What is the study of antigen-antibody interactions called?
Answer:
The study of antigen-antibody interactions is called serology.

Question 3.
Fill in the blanks after observing the diagram.
Maharashtra Board Class 12 Biology Important Questions Chapter 10 Human Health and Diseases 9
On the right side of the diagram, the stages of plasmodium are passed in the body of ……………….. Whereas on the left side of the diagram, those take place in the body of ………………, ………………. is the stage that is dormant in the liver of human host. From this ………………… and then ………………… is the stage in the erythrocytes, which rupture and gives rise to …………….. Microgamete and macrogamete fuse with each other to form …………………… which later gives rise to ookinete which forms ………………… This enters the salivary glands of mosquito …………………. phase of Plasmodium occurs in mosquito body, whereas …………………… phase is in human body.
Answer:
On the right side of the diagram, the stages of plasmodium are passed in the body of human. Whereas on the left side of the diagram, those take place in the body of mosquito. Hypnozoite is the stage that is dormant in the liver of human host. From this schizont and then merozoites Trophozoite is the stage in the erythrocytes, which rupture and gives rise to gamerocyte. Microgamete and macrogamete fuse with each other to form zygote which later gives rise to ookinete which forms sporozoites. This enters the salivary glands of mosquito. Sexual phase of Plasmodium occurs in mosquito body, whereas asexual phase is in human body.

Question 4.
Observe the given diagram and answer the following questions
Maharashtra Board Class 12 Biology Important Questions Chapter 10 Human Health and Diseases 10
(1) Enlist the stages of Entamoeba histolytica you see in the above diagram.
Answer:
Trophozoite, pre-cystic form, cyst, binucleate cyst, quadrinucleate cyst, Metacycstic amoeba, amoebulae are the different stage of Entamoeba histolytica that are seen in the above diagram.

(2) Where are these stages passed?
Answer:
These stages are passed in the lumen of intestine of the host human being.

(3) How does Entamoeba come out of the body of the host?
Answer:
Encysted Entamoeba pass out with the faecal matter of the host.

Maharashtra Board Class 12 Biology Important Questions Chapter 10 Human Health and Diseases

Question 5.
Observe the given diagram and answer the following questions:
Maharashtra Board Class 12 Biology Important Questions Chapter 10 Human Health and Diseases 11

(1) In life cycle of Ascaris at what stage do they enter the human body?
Answer:
When there is development of infective larva inside the egg of Ascaris, it enters the human body.

(2) How do they enter the human body and through which organ do they enter?
Answer:
Ascaris eggs are deposited in the faeces. They mix in the soil, if faeces is exposed in open. From there, it can enter into nearby water body or it may contaminate vegetables or other food stuffs. Such unhygienic food or unclean hands pass these eggs in the body of human through the mouth.

(3) What are the vital organs affected by the Ascaris during its development within the body of host human ?
Answer:
Ascaris can affect trachea, lungs, heart, brain and eyes too.

(4) In which organ do they copulate and produce fertilized eggs?
Answer:
Adult male and female copulate in the intestine of the host human.

Question 6.
Observe the given diagram and answer the following questions:
Maharashtra Board Class 12 Biology Important Questions Chapter 10 Human Health and Diseases 12

(1) In which host stage II is completed?
Answer:
Stage II is completed in human.

(2) What happens in step 1 and step 2?
Answer:
Humans are infected at step 1 when mosquito bites human and larvae enter blood stream. In step 2 adult Wuchereria worms are formed in lymphatics.

(3) Describe the events in step 3.
Answer:
Mosquito carries the blood as it bites the human in step 4 and ingests microfilariae in human blood. Later the microfilariae start growing in the midgut of mosquito.

(4) In which host stage I is completed?
Answer:
Stage I is completed in mosquito.

Question 7.
Sketch and label the diagram of pathogen that causes typhoid.
Answer:
Pathogen of typhoid is Salmonella typhi.
Maharashtra Board Class 12 Biology Important Questions Chapter 10 Human Health and Diseases 13

Question 8.
Sketch and label disease causing agents of pneumonia.
Answer:
Maharashtra Board Class 12 Biology Important Questions Chapter 10 Human Health and Diseases 14

Question 9.
Sketch and label benign and malignant tumours.
Answer:
Maharashtra Board Class 12 Biology Important Questions Chapter 10 Human Health and Diseases 15

Question 10.
Sketch and label structure of HIV.
Answer:
Maharashtra Board Class 12 Biology Important Questions Chapter 10 Human Health and Diseases 16

Long answer questions

Question 1.
Describe different ways in which epithelial surface offers the innate immunity.
Answer:

  1. Skin and mucous covering, when intact protects the body against the invasion by pathogens. The healthy skin has bactericidal activity due to the salts present in drying sweat.
  2. Sebaceous glands in the skin produce secretions and long chain of fatty acids. These are bactericidal and fungicidal.
  3. Respiratory tract is provided with mucosa which prevents entry of microorganisms to a large extent.
  4. The inhaled particles are arrested through hair in the nasal passage. The particles that pass beyond nasal passage are caught by mucus lining the epithelium. They are swept back to pharynx. Then they are either swallowed or coughed out.
  5. The cough reflex is an important defence mechanism of respiratory tract.
  6. There is saliva in the mouth which has inhibitory effect on microorganisms. Gastric secretions has acidity and hence microorganisms are destroyed in stomach.
  7. The flushing action of urine eliminates bacteria from the urethra. Semen too has antibacterial substances, e.g. Spermine and zinc.

Question 2.
Explain ABO blood group system in human being with a suitable chart.
Answer:

  1. In ABO system, the blood groups are determined by the antigen present on the surface of red blood cells.
  2. The blood groups are of four types, viz. A. B, AB and O.
  3. In person with blood group A there is antigen ‘A’ on the surface of their red blood cells (RBCs) and antibodies ‘b’ in their plasma.
  4. In person with blood group B there is antigen ‘B’ on the surface of their red blood cells (RBCs) and antibodies ‘a’ in their plasma.
  5. In person with blood group AB there are both antigens ‘A’ and ‘B’ on the surface of their RBCs and no antibodies in their plasma.
  6. In person with blood group ‘O’ there are no antigens ‘A’ and ‘B’ on the surface of their RBCs but have both ‘a’ and ‘b’ antibodies in their plasma.
  7. During blood transfusion compatibility of blood has to be taken into consideration.
  8. Person with ‘O’ blood group is called universal donor while the person with ‘AB’ blood group is called universal recipient. Individuals with blood group O can donate blood to anyone, while those individuals with blood group AB can receive blood from any person.
    Maharashtra Board Class 12 Biology Important Questions Chapter 10 Human Health and Diseases 17

Question 3.
What are the main causes of cancer?
Answer:
Causes of Cancer : Following carcinogenic factors are responsible for causing cancer.

  1. Chemicals : Many induce development of cancer. E.g. nicotine, caffeine, polycyclic hydrocarbons and products of combustion of coal and oil. Sex hormone and steroids, if given or secreted in excess, can cause cancer. E.g. Breast cancer.
  2. Radiation : Radiations such as X-rays, gamma-rays, cosmic rays, ultra-violet rays are carcinogenic.
  3. Viruses : Virus possessing oncogenes (v-onc genes) are carcinogenic. E.g. EBV (Epstein-barr virus), HPV (Human papiloma virus) are oncogenic viruses.
  4. Oncogenes : Cellular oncogenes (c-onc genes) or proto-oncogenes can cause cancer. They are present in normal cells but if activated they lead to oncogenic transformation of cells.
  5. Addiction : Addictive substances like cigarette smoke, tobacco lead to cancer of mouth, lips and lungs. Alcohol can cause cancer of oesophagus, stomach, intestine and liver. Drugs like marijuana or anaerobic steroids can also cause cancer.

Maharashtra Board Class 12 Biology Important Questions Chapter 10 Human Health and Diseases

Question 4.
What are the different ways of treating cancer?
Answer:
Cancer treatment consists of combination of a number of therapies which are follows:
(1) Chemotherapy : Chemotherapy means giving certain anticancer drugs. These drugs check cell division by inhibiting DNA synthesis. But these are more toxic to cancerous cell than to normal cells. Chemotherapy shows side effects such as hair loss or anaemia.

(2) Radiotherapy : In addition to chemotherapy, radiations are given. The cancer cells are bombarded with the radiations from radioactive materials such as cobalt, iridium and iodine. The X-rays, gamma rays and charge particles are used to destroy the cancerous tissue or cells. They cause minimum damage to the surrounding normal tissue or cells.

(3) Surgery : Entire cancerous tissue or cells are removed surgically. E.g. breast tumour or uterine tumour. After removing the cancerous tissue, additionally other treatments are also given.

(4) Immunotherapy : For tackling with tumour, patients are given biological response modifiers such as a-interferon which activates their immune system to destroy the tumour.

(5) Supportive therapy : With supportive therapy, patient’s quality of life is increased. To treat symptoms of cancer and side effects of cancer treatments, this therapy is used. This therapy varies depending upon condition of individual patient.

Maharashtra Board Class 12 Biology Important Questions Chapter 9 Control and Co-ordination

Balbharti Maharashtra State Board 12th Biology Important Questions Chapter 9 Control and Co-ordination Important Questions and Answers.

Maharashtra State Board 12th Biology Important Questions Chapter 9 Control and Co-ordination

Multiple choice questions

Question 1.
The supporting cells that produce myelin sheath in the CNS are ……………….
(a) Oligodendrocytes
(b) Satellite cells
(c) Astrocytes
(d) Schwann cells
Answer:
(a) Oligodendrocytes

Question 2.
Human brain develops to its full size at an age of year/s.
(a) 1
(b) 6
(c) 12
(d) 18
Answer:
(b) 6

Maharashtra Board Class 12 Biology Important Questions Chapter 9 Control and Co-ordination

Question 3.
Telencephalon is the other name of ……………….
(a) pons varolii
(b) medulla oblongata
(c) cerebrum
(d) cerebellum
Answer:
(c) cerebrum

Question 4.
Olfactory tracts merge in olfactory area of lobe.
(a) Frontal
(b) Parietal
(c) Occipital
(d) Temporal
Answer:
(d) Temporal

Question 5.
………………. is the largest commissure of the human brain.
(a) Corpora striata
(b) Corpora quadrigemina
(c) Habencular commissure
(d) Corpus callosum
Answer:
(d) Corpus callosum

Question 6.
Grey matter of the brain shows large collection of ……………….
(a) dendrons
(b) cytons
(c) axons
(d) synapsis
Answer:
(b) cytons

Question 7.
Masses of grey matter in white matter of the cerebrum are called ……………….
(a) corpora striata
(b) corpus callosum
(c) paracoel
(d) basal ganglia
Answer:
(d) basal ganglia

Question 8.
Parietal and temporal lobes of cerebrum are separated by sulcus.
(a) lateral
(b) parieto occipital
(c) central
(d) median longitudinal
Answer:
(a) lateral

Question 9.
……………… area is motor speech area
(a) Acoustic
(b) Wernike’s
(c) Somato sensory
(d) Broca’s
Answer:
(d) Broca’s

Question 10.
Maxillary nerve is a branch of nerve.
(a) Occulomotor
(b) Trochlear
(c) Trigeminal
(d) Facial
Answer:
(c) Trigeminal

Question 11.
Spinal accessory is the cranial nerve.
(a) IV
(b) VI
(c) IX
(d) XI
Answer:
(d) XI

Question 12.
Rotation of eye ball is controlled by ……………….
(a) Optic nerve
(b) Pathetic nerve
(c) Auditory nerve
(d) Hypoglossal nerve
Answer:
(b) Pathetic nerve

Question 13.
The spinal nerves emerge out of vertebral column through ……………….
(a) intervertebral foramina
(b) neural canal
(c) central canal
(d) foramen magnum
Answer:
(a) intervertebral foramina

Question 14.
The neuro transmitter is removed by an enzyme called ……………….
(a) noradrenaline
(b) acetylcholine
(c) hyaluronidase
(d) cholinesterase
Answer:
(d) cholinesterase

Question 15.
The reflex action originates in ……………….
(a) sensory neuron
(b) motor neuron
(c) receptor organ
(d) effector organ
Answer:
(c) receptor organ

Question 16.
Cytons of neurons are located in dorsal root ganglion.
(a) afferent
(b) efferent
(c) adjustor
(d) association
Answer:
(a) afferent

Question 17.
Wall of carotid arteries contain ……………….
(a) thermoreceptors
(b) mechanoreceptors
(c) baroreceptors
(d) statoacoustic receptors
Answer:
(c) baroreceptors

Question 18.
The electronegativity inside the membrane is due to ……………….
(a) less anions than cations
(b) less cations than anions
(c) bicarbonates
(d) carbonates
Answer:
(b) less cations than anions

Question 19.
The neuro transmitters stimulate ……………….
(a) presynaptic membrane
(b) cyton
(c) axon terminals
(d) postsynaptic membrane
Answer:
(d) postsynaptic membrane

Question 20.
………………. is an extero-receptor.
(a) Thermoreceptor
(b) Baroreceptor
(c) Proprioreceptor
(d) Enteroreceptor
Answer:
(a) Thermoreceptor

Question 21.
The are described as windows for brain.
(a) sensory neurons
(b) motor neurons
(c) effectors
(d) sense organs
Answer:
(d) sense organs

Question 22.
Otolith organ is formed of ……………….
(a) cochlea and vestibule
(b) sacculus and utriculus
(c) semicircular canals
(d) ear ossicles
Answer:
(b) sacculus and utriculus

Question 23.
Olfactory bulbs are extensions of brain’s ……………….
(a) cerebrum
(b) limbic system
(c) RAS
(d) pons varolii
Answer:
(b) limbic system

Question 24.
Gustatory senses are noted by ……………….
(a) retina
(b) skin
(c) nose
(d) tongue
Answer:
(d) tongue

Question 25.
………………. is attached to the eardrum.
(a) Malleus
(b) Incus
(c) Stapes
(d) Cochlea
Answer:
(a) Malleus

Question 26.
Eustachian tube is present in ……………….
(a) external ear
(b) internal ear
(c) heart
(d) middle ear
Answer:
(d) middle ear

Question 27.
The internal ear is a fluid filled structure called ……………….
(a) cochlea
(b) vestibule
(c) labyrinth
(d) otolith
Answer:
(c) labyrinth

Question 28.
The space within cochlea is known as ……………….
(a) scala vestibule
(b) scala tympani
(c) aqueous chamber
(d) scala media
Answer:
(d) scala media

Question 29.
Thermoregulatory centre in the body is ……………….
(a) hypothalamus
(b) cerebellum
(c) spinal cord
(d) pituitary
Answer:
(a) hypothalamus

Maharashtra Board Class 12 Biology Important Questions Chapter 9 Control and Co-ordination

Question 30.
Which of the following is a sensory nerve?
(a) Vagus
(b) Auditory
(c) Facial
(d) Lumbar
Answer:
(b) Auditory

Question 31.
Chemical transmission in synapse occurs due to ……………….
(a) cholesterol
(b) ADH
(c) acetylcholine
(d) cholinesterase
Answer:
(c) acetylcholine

Question 32.
Voluntary muscular coordination is under the control of ……………….
(a) medulla
(b) pons
(c) hypothalamus
(d) cerebrum
Answer:
(d) cerebrum

Question 33.
All involuntary vital activities are under the control of ……………….
(a) medulla oblongata
(b) cerebellum
(c) cerebral hemispheres
(d) pons Varolii
Answer:
(a) medulla oblongata

Question 34.
Cerebellum is controlling centre for ……………….
(a) muscular strength
(b) memory
(c) equilibrium
(d) muscular coordination
Answer:
(c) equilibrium

Question 35.
Which receptors are present in the retina?
(a) Chemoreceptors
(b) Thermoreceptors
(c) Photoreceptors
(d) Baroreceptors
Answer:
(c) Photoreceptors

Question 36.
Breathing is controlled by ……………….
(a) trachea
(b) medulla oblongata
(c) lungs
(d) hypothalamus
Answer:
(b) medulla oblongata

Question 37.
Corpus callosum is a nerve fibre bridge which connects ……………….
(a) two cerebral hemispheres
(b) cerebrum and cerebellum
(c) cerebellum and medulla
(d) midbrain and hindbrain
Answer:
(a) two cerebral hemispheres

Question 38.
Centre for thirst and hunger are located in ……………….
(a) cerebrum
(b) cerebellum
(c) hypothalamus
(d) medulla
Answer:
(c) hypothalamus

Question 39.
Gyri in the brain are present in ……………….
(a) cerebral cortex
(b) olfactory lobes
(c) medulla oblongata
(d) hypothalamus
Answer:
(a) cerebral cortex

Question 40.
Which of the following is a structure of mesencephalon?
(a) Inferior colliculi
(b) Thalamus
(c) Cerebellum
(d) Pons varolii
Answer:
(a) Inferior colliculi

Question 41.
Third ventricle lies in ……………….
(a) midbrain
(b) forebrain
(c) cerebellum
(d) medulla oblongata
Answer:
(b) forebrain

Question 42.
Medulla oblongata encloses ……………….
(a) third ventricle
(b) fourth ventricle
(c) first ventricle
(d) second ventricle
Answer:
(b) fourth ventricle

Question 43.
Loss of memory may result from injury to the ……………….
(a) corpora quadrigemina
(b) pons varolii
(c) cerebellum
(d) cerebrum
Answer:
(d) cerebrum

Question 44.
Terminal non-nervous part of spinal cord is ……………….
(a) funiculus
(b) filum terminale
(c) cauda equina
(d) conus terminalis
Answer:
(b) filum terminale

Question 45.
Which part of the pituitary is neurohaemal organ?
(a) Pars distalis
(b) Infundibulum
(c) Pars nervosa
(d) Pars intermedia
Answer:
(c) Pars nervosa

Question 46.
Development of secondary sexual characteristics in female is under the control of ……………….
(a) growth hormone
(b) TSH
(c) estrogen
(d) progesterone
Answer:
(c) estrogen

Question 47.
Hypersecretion of STH in children causes ……………….
(a) cretinism
(b) gigantism
(c) dwarfism
(d) myxoedema
Answer:
(b) gigantism

Question 48.
Milk secretion in lactating woman is controlled by ……………….
(a) LH
(b) prolactin
(c) relaxin
(d) oestrogen
Answer:
(b) prolactin

Question 49.
ADH is secreted by
(a) adrenal gland
(b) thyroid
(c) hypothalamus
(d) pancreas
Answer:
(c) hypothalamus

Question 50.
BMR is increased by the administration of ……………….
(a) insulin
(b) GH
(c) thyroxine
(d) testosterone
Answer:
(c) thyroxine

Question 51.
The largest endocrine gland in the body is ……………….
(a) pituitary
(b) adrenal
(c) liver
(d) thyroid
Answer:
(d) thyroid

Question 52.
Diabetes insipidus is caused by the deficiency of ……………….
(a) calcitonin
(b) oxytocin
(c) atrial natriuretic factor
(d) vasopressin
Answer:
(d) vasopressin

Question 53.
Simple goitre is caused by the deficiency of ……………….
(a) TSH
(b) thyrocalcitonin
(c) thyroxine
(d) iodine
Answer:
(d) iodine

Question 54.
Exopthalmic goitre is also known as ……………….
(a) Grave’s disease
(b) Gull’s disease
(c) Simple goitre
(d) Cushing’s disease
Answer:
(a) Grave’s disease

Question 55.
Cushing’s syndrome is developed due to ……………….
(a) hyposecretion of ACTH
(b) hypersecretion of corticoids
(c) hyposecretion of thyroxine
(d) hypersecretion of thyroxine
Answer:
(b) hypersecretion of corticoids

Maharashtra Board Class 12 Biology Important Questions Chapter 9 Control and Co-ordination

Question 56.
Pituitary gland is under the control of ……………….
(a) thyroid
(b) adrenal
(c) pineal
(d) hypothalamus
Answer:
(d) hypothalamus

Question 57.
FSH is secreted by ……………….
(a) pituitary gland
(b) thyroid gland
(c) ovary
(d) adrenal gland
Answer:
(a) pituitary gland

Question 58.
ICSH stimulates ……………….
(a) ovary
(b) Leydig cells
(c) seminiferous tubules
(d) kidney
Answer:
(b) Leydig cells

Question 59.
Which of the following secrete LH ?
(a) Pituitary
(b) Thyroid
(c) Ovary
(d) Adrenal
Answer:
(a) pituitary

Question 60.
TSH regulates secretion.
(a) thyroxine
(b) MSH
(c) androgens
(d) insulin
Answer:
(a) thyroxine

Question 61.
Deficiency of thyroxine in adults cause ……………….
(a) cretinism
(b) myxoedema
(c) diabetes
(d) Cushing’s disease
Answer:
(b) myxoedema

Question 62.
Osmotic pressure and blood pressure are maintained by ……………….
(a) glucocorticoids
(b) aldosterone
(c) TRF
(d) MSH
Answer:
(b) aldosterone

Question 63.
Hormone secreted by corpus luteum is ……………….
(a) aldosterone
(b) progesterone
(c) testosterone
(d) cortisol
Answer:
(b) progesterone

Question 64.
………………. is also called hypophyseal stalk.
(a) Infundibulum
(b) Median eminence
(c) Pars intermedia
(d) Sphenoid bone
Answer:
(a) infundibulum

Question 65.
………………. is like a collar around hypophyseal stalk.
(a) Pars distalis
(b) Pars nervosa
(c) Pars intermedia
(d) Pars tuberalis
Answer:
(d) Pars tuberalis

Question 66.
Herring bodies are the parts of ……………….
(a) hypothalamo-hypophyseal tracts
(b) pituicytes
(c) hypothalamo-hypophyseal portal system
(d) pituitary cleft
Answer:
(a) hypothalamo-hypophyseal tracts

Question 67.
Corticotropin is the other name of ……………….
(a) ACTH
(b) STH
(c) Aldosterone
(d) ADH
Answer:
(a) ACTH

Question 68.
Adrenal failure leads to ……………….
(a) Acromegaly
(b) Simmond’s disease
(c) Midget
(d) Addison’s disease
Answer:
(d) Addison’s disease

Question 69.
Prolactin inhibiting factor is secreted by ……………….
(a) Hypophysis
(b) Hypothalamus
(c) Thyroid
(d) Mammary glands
Answer:
(b) Hypothalamus

Question 70.
Which one of the following is not applicable to prolactin ?
(a) Mammotropin
(b) Lactogenic hormone
(c) Somatotropin
(d) Luteotropin
Answer:
(c) Somatotropin

Question 71.
………………. is a gonadotropic hormone.
(a) STH
(b) LTH
(c) ACTH
(d) FSH
Answer:
(d) FSH

Question 72.
Rhythmic integrated contractions of jejunum are controlled by ……………….
(a) coherin
(b) insulin
(c) glucagon
(d) ADH
Answer:
(a) coherin

Question 73.
Thyroid gland is derived from of embryo.
(a) ectoderm
(b) mesoderm
(c) endoderm
(d) ecto-endoderm
Answer:
(c) endoderm

Question 74.
Deficiency of thyroxine in infants causes ……………….
(a) Cretinism
(b) Grave’s disease
(c) Myxoedema
(d) Exophthalmos
Answer:
(a) Cretinism .

Question 75.
………………. is a hypercalcemic hormone.
(a) PTH
(b) TCT
(c) Thyroxine
(d) ACTH
Answer:
(a) PTH

Question 76.
………………. is a middle layer of adrenal cortex.
(a) Zona fasciculata
(b) Zona pellicida
(c) Zona glomerulosa
(d) Zona reticularis
Answer:
(a) Zona fasciculata

Question 77.
Decrease in the blood calcium level is ……………….
(a) hyperglycemia
(b) hypercalcemia
(c) hypoglycemia
(d) hypocalcemia
Answer:
(d) hypocalcemia

Question 78.
………………. stimulates RBC production.
(a) Aldosterone
(b) Cortisol
(c) Epinephrine
(d) Parathormone
Answer:
(b) Cortisol

Question 79.
Chemicals which are released at the synaptic junction are called ……………….
(a) hormones
(b) neurotransmitters
(c) cerebrospinal fluid
(d) lymph
Answer:
(b) neurotransmitters

Question 80.
Potential difference across resting membrane is negatively charged. This is due to differential distribution of the following ions.
(a) Na+ and K+ ions
(b) Ca++ and Cl ions
(c) Ca++ and Mg++ ions
(d) Mg++ and Cl ions
Answer:
(a) Na+ and K+ ions

Question 81.
Which of the following is not involved in Knee-jerk reflex?
(a) Muscle spindle
(b) Motor neuron
(c) Brain
(d) Inter neurons
Answer:
(c) Brain

Question 82.
An area in the brain which is associated with strong emotions is ……………….
(a) Cerebral cortex
(b) Cerebellum
(c) Limbic system
(d) Medulla
Answer:
(c) Limbic system

Question 83.
Which is the vitamin present in Rhodopsin?
(a) Vitamin A
(b) Vitamin B
(c) Vitamin C
(d) Vitamin D
Answer:
(a) Vitamin A

Question 84.
Wax gland present in the ear canal is modified ……………….
(a) Sweat gland
(b) Vestibular gland
(c) Cowper’s gland
(d) Sebaceous gland
Answer:
(d) Sebaceous gland

Question 85.
The part of internal ear responsible for hearing is ……………….
(a) cochlea
(b) semicircular canal
(c) utriculus
(d) sacculus
Answer:
(a) cochlea

Question 86.
The organ of corti is a structure present in ……………….
(a) external ear
(b) middle ear
(c) semi circular canal
(d) cochlea
Answer:
(d) cochlea

Question 87.
Select the right match of endocrine gland and their hormones among the options given below.
A. Pineal i. Epinephrine
B. Thyroid ii. Melatonin
C. Ovary iii. Estrogen
D. Adrenal medulla iv. Tetraiodothyronine
(a) A-iv, B-ii, C-iii, D-i
(b) A-ii, B-iv, C-i, D-iii
(c) A-iv, B-ii, C-i, D-iii
(d) A-ii, B-iv, C-iii, D-i
Answer:
(d) A-ii, B-iv, C-iii, D-i

Question 88.
Listed below are the hormones of anterior pituitary origin. Tick the wrong entry.
(a) Growth hormone
(b) FSH
(c) Oxytocin
(d) ACTH
Answer:
(c) Oxytocin

Question 89.
Mary is about to face an interview. But during the first five minutes before the interview she experiences sweating, increased rate of heartbeat, respiration, etc. Which hormone is responsible for her restlessness?
(a) Estrogen and progesterone
(b) Oxytocin and vasopressin
(c) Adrenaline and noradrenaline
(d) Insulin and glucagon
Answer:
(c) Adrenaline and noradrenaline

Question 90.
The steroid responsible for balance of water and electrolytes in our body is ……………….
(a) Insulin
(b) Melatonin
(c) Testosterone
(d) Aldosterone
Answer:
(d) Aldosterone

Question 91.
Thymosin is responsible for ……………….
(a) raising the blood sugar level
(b) raising the blood calcium level
(c) increased production of T lymphocytes
(d) decrease in blood RBC
Answer:
(c) increased production of T lymphocytes

Question 92.
In the mechanism of action of a protein hormone, one of the second messengers is ……………….
(a) Cyclic AMP
(b) Insulin
(c) T3
(d) Gastrin
Answer:
(a) Cyclic AMP

Maharashtra Board Class 12 Biology Important Questions Chapter 9 Control and Co-ordination

Question 93.
Leydig cells produce a group of hormones called ……………….
(a) androgens
(b) estrogen
(c) aldosterone
(d) gonadotropins
Answer:
(a) androgens

Question 94.
Corpus luteum secretes a hormone called ……………….
(a) prolactin
(b) progesterone
(c) aldosterone
(d) testosterone
Answer:
(b) progesterone

Question 95.
Cortisol is secreted from ……………….
(a) pancreas
(b) thyroid
(c) adrenal
(d) thymus
Answer:
(c) adrenal

Question 96.
A hormone responsible for normal sleep – wake cycle is ……………….
(a) epinephrine
(b) gastrin
(c) melatonin
(d) insulin
Answer:
(c) melatonin

Question 97.
Match the pairs and choose the correct answer among the following options.
A. Epinephrine
i. Increase in muscle growth
B. Testosterone
ii. Decrease in blood pressure
C. Glucagon
iii. Decrease in liver glycogen content
D. Atrial natriuretic factor
iv. Increase in heartbeat
(a) A-ii, B-i, C-iii, D-iv
(b) A-iv, B-i, C-iii, D-ii
(c) A-i, B-ii, C-iii, D-iv
(d) A-i, B-iv, C-ii, D-iii
Answer:
(b) A-iv, B-i, C-iii, D-ii

Question 98.
Blood calcium level is a resultant of how much dietary calcium is absorbed, how much calcium is lost in the urine, how much bone dissolves releasing calcium into the blood and how much calcium from blood enters tissues. Several factors play an important role in these processes. Mark the one which has no role.
(a) Vitamin D
(b) Parathyroid hormone
(c) Thyrocalcitonin
(d) Thymosin
Answer:
(d) Thymosin

Question 99.
All the following tissues in mammals except one consists of a central ‘medullary’ region surrounded by a cortical region. Mark the wrong entry.
(a) Ovary
(b) Adrenal
(c) Liver
(d) Kidney
Answer:
(c) Liver

Match the columns

Question 1.
Match the hormones with their source

Column AColumn B
(1) Glucagon(i) Neurohypophysis
(2) Adrenaline(ii) Islets of Langerhans
(3) Somato tropins(iii) Adenohypophysis
(4) ADH(iv) Medulla

Answer:

Column AColumn B
(1) Glucagon(ii) Islets of Langerhans
(2) Adrenaline(iv) Medulla
(3) Somato tropins(iii) Adenohypophysis
(4) ADH(i) Neurohypophysis

Question 2.
Match the layer of adrenal with its hormone.

Column A (Layers of adrenal cortex)Column B (Hormones)
(1) Zona glomerulosa(A) Cortisols
(2) Zona fasciculata(B) Androgens
(3) Zona reticularis(C) Aldosterone

Answer:

Column A (Layers of adrenal cortex)Column B (Hormones)
(1) Zona glomerulosa(C) Aldosterone
(2) Zona fasciculata(A) Cortisols
(3) Zona reticularis(B) Androgens

Question 3.
Match the disorder with the gland associated with it.

Column A (Disorder)Column B (Associated Gland)
(1) Addison’s disease(A) Hypothalamus
(2) Grave’s disease(B) Pituitary
(3) Diabetes insipidus(C) Thyroid
(4) Acromegaly(D) Adrenal

Answer:

Column A (Disorder)Column B (Associated Gland)
(1) Addison’s disease(D) Adrenal
(2) Grave’s disease(C) Thyroid
(3) Diabetes insipidus(A) Hypothalamus
(4) Acromegaly(B) Pituitary

Classify the following to form Column B as per the category given in Column A

Question 1.

Column AColumn B
(1) Forebrain————–
(2) Midbrain————–
(3) Hindbrain————–

Answer:

Column AColumn B
(1) ForebrainOlfactory lobes, Corpus callosum
(2) MidbrainSuperior colliculi, Iter
(3) HindbrainPons varolii, Vermis

Question 2.
Types of nerves
Occulomotor, Facial, Optic, Vagus, Abducens, Vestibulocochlear

Column AColumn B
(1) Sensory————–
(2) Motor————–
(3) Mixed————–

Answer:

Column AColumn B
(1) SensoryOptic, Vestibulocochlear
(2) MotorOcculomotor, Abducens
(3) MixedFacial, Vagus

Question 3.
Hormones
Estrogen, Glucagon, Epinephrine, Relaxin, Somatostatin, Nor-Adrenalin

Column AColumn B
(1) Ovary————
(2) Pancreas————
(3) Adrenal Medulla————

Answer:

Column AColumn B
(1) OvaryEstrogen, Relaxin
(2) PancreasGlucagon, Somatostatin
(3) Adrenal MedullaEpinephrine, Nor-Adrenaline

Question 4.
Disorders
Dwarfism, Myxoedema, Addison’s disease, Cushing’s disease, Gigantism, Goitre

Column AColumn B
(1) Pituitary————
(2) Thyroid————
(3) Adrenal Cortex————

Answer:

Column AColumn B
(1) PituitaryDwarfism, Gigantism
(2) ThyroidMyxoedema, Goitre
(3) Adrenal CortexAddison’s disease, Cushing’s disease

Very short answer questions

Question 1.
What is the need for the control and coordination in multicellular animals?
Answer:
Multicellular animals need control and coordination to maintain constancy of internal environment, i.e. homeostasis.

Question 2.
How do plants carry out control and coordination?
Answer:
Plants carry out control and coordination by sending chemical signals and bringing about various types of movements.

Question 3.
What is the type of nervous system of Earthworm?
Answer:
Earthworm is an annelid having ventral, ganglionated nervous system. It consists mainly of nerve ring, nerve cord and peripheral segmentally arranged nerves.

Question 4.
What kind of nervous system is seen in Hydra, earthworm and cockroach?
Answer:
In Hydra, the nervous system is in the form of nerve net, while in earthworm and cockroach the nervous system is ganglionated.

Question 5.
What is a gland?
Answer:
An organized collection of secretory epithelial cells capable of producing of some secretion is called a gland.

Maharashtra Board Class 12 Biology Important Questions Chapter 9 Control and Co-ordination

Question 6.
What do you mean by discrete endocrine glands ?
Answer:
The glands which are exclusively endocrine in function are discrete endocrine glands.

Question 7.
What is genu and splenium?
Answer:
Genu is anterior and splenium is posterior fold of corpus callosum.

Question 8.
Which is the largest basal nucleus of brain ?
Answer:
Corpus striatum is the largest basal nucleus of the brain.

Question 9.
What is EEG?
Answer:
EEG is electro encephalography. It is used to detect the electrical changes taking place in the brain.

Question 10.
Mention the major sulci present in the cerebral hemispheres.
Answer:
The sulci present in the cerebral hemisphere are – central sulcus, parieto-occipital sulcus and lateral sulcus.

Question 11.
What is the difference in ‘tract’ and ‘nerve’?
Answer:
A bundle of axons within CNS is called a ‘tract’ while the one outside the CNS is called ‘nerve’.

Question 12.
What is a choroid plexus? State its locations.
Answer:
Network of blood capillaries associated with the brain is called choroid plexus, which are anterior choroid plexus present on the roof of epithalamus and posterior choroid plexus located on the roof of medulla oblongata.

Question 13.
What is a synapse ?
Answer:
The interconnection between two neurons or neuron with motor organ is called synapse.

Question 14.
What is a polarised membrane?
Answer:
The cell membrane of a neuron at resting stage is called polarised membrane. In such membrane, the outer side of the cell membrane is more electropositive due to more Na+ ions.

Question 15.
What is summation effect?
Answer:
Many weak stimuli given in a quick succession may produce a response due to summation or additive effect of stimuli, which is known as summation effect.

Question 16.
What is synaptic delay?
Answer:
The time required for a nerve impulse to cross a synapse during transmission of a nerve impulse is called synaptic delay, which is about 0.3-0.5 milliseconds.

Question 17.
What is refractory period?
Answer:
The time interval between two consecutive nerve impulses is called refractory period, during which a nerve cannot be stimulated. Nerve is stimulated only after completion of this period.

Question 18.
What is a synaptic cleft?
Answer:
Small intercellular space between two successive neurons which is about 20¬30 nm in width, is called synaptic cleft.

Question 19.
What is synaptic transmission?
Answer:
The process by which the impulse from the pre-synaptic neuron is conducted to the post-synaptic neuron or cell is called synaptic transmission.

Question 20.
What is a pre-synaptic neuron?
Answer:
The neuron carrying an impulse to the synapse is the pre-synaptic neuron.

Question 21.
What is transmission terminal and generator region?
Answer:
The pre-synaptic membrane of a neuron is transmission terminal while the post- synaptic membrane is called generation region.

Question 22.
What is synaptic fatigue?
Answer:
The synaptic fatigue is the time lag or halting of the transmission of nerve impulse, temporarily at synapse due to exhaustion of its neurotransmitter.

Question 23.
What is the width of synaptic cleft?
Answer:
The width of synaptic cleft is about 20-30 nm.

Question 24.
How much is the resting potential of axon?
Answer:
The resting potential of axon is -70 mV.

Question 25.
How does Na+ – K+ pump work?
Answer:
During repolarisation of certain part of a nerve, Na+ – K+ Pump, pumps out 3 Na+ ions for every 2 K+ ions they pump into the cell.

Question 26.
What is blood-brain barrier?
Answer:
The barrier that keeps a check on passage of ions and large molecules from the blood to the brain tissue is called blood-brain barrier. Endothelial cells lining the blood capillaries help in this process along with the astrocytes.

Question 27.
Enlist meninges of human brain.
Answer:
Dura mater, arachnoid mater or membrane and pia mater are the meninges of human brain.

Question 28.
What is the source of CSF?
Answer:
CSF is secreted by the choroid plexuses of brain and ependymal cells lining the ventricles of brain and central canal of spinal cord.

Question 29.
What is the function of tympanic membrane?
Answer:
Tympanic membrane vibrates on receiving the sound waves and then transfers the vibrations to malleus, the first of the three ear ossicles.

Question 30.
Mention the role of semicircular canals in ear.
Answer:
Semicircular canals help in balancing the equilibrium of the body.

Question 31.
What is the cause for diabetes insipidus?
Answer:
Diabetes insipidus is caused by the deficiency of ADH.

Question 32.
Give role of Parathormone.
Answer:
Parathormone regulates the calcium balance in the body. It increases blood calcium level by increasing reabsorption of calcium from bones.

Question 33.
What is meningitis?
Answer:
Infection of meninges is called meningitis.

Question 34.
Where is general motor area located? What is the function of this area?
Answer:
General motor area is located on precentral gyrus of frontal lobe. It is concerned with all motor activities.

Question 35.
What is the role of Wernicke’s area?
Answer:
Wernicke’s area is the sensory speech area responsible for understanding and formulating written and spoken language.

Question 36.
What is the location of Wernicke’s area?
Answer:
Wernicke’s area is located in the area of contact between temporal, parietal and occipital lobes of cerebrum.

Question 37.
What is the role of Broca’s area?
Answer:
The role of Broca’s area which is the motor speech area that translates thoughts into speech and controls movement of tongue, lips and vocal cords.

Question 38.
What is the location of Broca’s area?
Answer:
Broca’s area is present in the frontal lobe of cerebrum.

Question 39.
Which areas are present in the post central gyrus?
Answer:
General sensory areas concerned with sensation of temperature, touch, pressure, pain and speech are present in the post central gyrus.

Question 40.
Which areas are located on temporal lobe?
Answer:
The areas concerned with sense of taste, sense of hearing and sense of smell are located on temporal lobe.

Question 41.
What is the main function of occipital lobe?
Answer:
Occipital lobe carries sensory visual area which is concerned with the sense of sight. Association visual area is concerned with perception, analysis and storing of information obtained by sight.

Question 42.
What is neuroendocrine system?
Answer:
Neuroendocrine system is an integrated and coordinated system consisting of nervous and endocrine system which brings about the coordination of the body.

Question 43.
What is the quantity of cerebrospinal fluid in adult human being?
Answer:
Cerebrospinal fluid is about 120 ml in adult human being.

Question 44.
How many neurons are present in the brain?
Answer:
There are about 30,000 million neurons in the brain.

Question 45.
Which part of the brain forms 80-85% of the brain?
Answer:
Cerebrum forms 80-85% of the brain.

Question 46.
Which sense is poorly developed in human beings and what is the reason for this?
Answer:
Sense of smell is poorly developed in human beings and it is due to less developed olfactory lobes.

Question 47.
Where are lateral ventricles situated and what is its roof called?
Answer:
Lateral ventricles are the cavities filled with cerebrospinal fluid, present inside the cerebrum and its roof is called pallium.

Question 48.
What is foramen of Monro?
Answer:
Foramen of Monro is a narrow opening that connects lateral ventricle present in cerebrum with third ventricle present in diencephalon.

Question 49.
What is the roof and floor of diencephalon called respectively?
Answer:
Roof of diencephalon is epithalamus and floor of diencephalon is hypothalamus.

Question 50.
Which hormones are produced from pineal gland? Name their functions.
Answer:
Serotonin and melatonin are hormones of pineal gland which are concerned with metabolic activities and regulation of biological rhythm respectively.

Question 51.
What is Habenular commissure?
Answer:
Habenular commissure connects the lateral walls of diencephalon or thalami with each other.

Question 52.
What structure is present on the anterior side of the hypothalamus?
Answer:
Optic chiasma or crossing of two optic nerves is seen on the anterior side of the hypothalamus.

Question 53.
Name the parts of corpora quadrigemina and give their functions.
Answer:
The upper two lobes of corpora quadrigemina are called superior colliculi which receive impulse from optic nerves while the lower two lobes are called inferior colliculi which receive auditory stimuli. Both colliculi control and coordinate head movements.

Question 54.
What is arbour vitae?
Answer:
Arbor vitae is white, branching tree-like processes of white matter that are sent into grey cortex of cerebellum.

Question 55.
Which cranial nerve is called a dentist nerve?
Answer:
Trigeminal, the Vth cranial nerve is called a dentist nerve.

Question 56.
What is central canal?
Answer:
Central canal is the narrow central cavity present inside the spinal cord.

Question 57.
Which cranial nerves originate from the midbrain?
Answer:
Cranial nerves (III) Occulomotor, (IV) Pathetic and (VI) Abducens originate from the midbrain.

Question 58.
What is the other name for the auditory cranial nerve?
Answer:
Vestibulocochlear.

Question 59.
What is the difference between unconditioned and conditioned reflexes?
Answer:
Unconditioned reflexes are inborn or hereditary and permanent while the conditioned reflexes are temporary, learnt or acquired during lifetime.

Question 60.
Give examples of unconditional reflexes.
Answer:
Blinking of eyes, withdrawing of hand upon pricking, suckling to breast by infant, swallowing, knee jerk, sneezing and coughing are some of the unconditional reflexes.

Question 61.
What do you mean by accommodation power of the lens?
Answer:
The ability of the lens by which the light ray from far and near objects is focused on the retina is called accommodation power of the lens. The lens makes fine adjustments to bring such sharp focus on retina.

Question 62.
Where is pituitary located?
Answer:
Pituitary is located inside a depression called sella turcica which is present in sphenoid bone of the skull.

Question 63.
What is the difference between Lorain dwarf and Frohlic dwarf?
Answer:
Lorain dwarf is mentally normal while Frohlic dwarf is mentally abnormal.

Question 64.
How is Simmond’s disease caused?
Answer:
Simmond’s disease is caused due to hyposecretion of GH during adulthood.

Question 65.
How is Addison’s disease caused?
Answer:
Addison’s disease is caused by the hyposecretion of ACTH that leads to adrenal failure.

Question 66.
How is Cushing’s disease caused?
Answer:
Hypersecretion of corticoids causes Cushing’s disease.

Question 67.
What is the main difference between diabetes mellitus and diabetes insipidus?
Answer:
Diabetes mellitus is caused due to deficiency of insulin while diabetes insipidus is caused due to deficiency of ADH.

Question 68.
Which cells secrete coherin? What is the function of coherin?
Answer:
Coherin is the hormone secreted by hypothalamic neurons that brings about prolonged, rhythmic integrated contractions of the jejunum.

Question 69.
What is the difference between Gull’s disease (Myxoedema) and Grave’s disease?
Answer:
Gull’s disease is caused due to hyposecretion of thyroxine while Grave’s disease is caused due to hypersecretion of thyroxine.

Question 70.
In which part thyroid gland stores its hormones?
Answer:
The lumen of thyroid follicles store the thyroid hormones in the form of thyroglobulins.

Question 71.
What are thymosins?
Answer:
Thymosins are hormones secreted by thymus gland which promote the production of antibodies.

Question 72.
Which hormone is secreted by the heart?
Answer:
Heart secretes ANF or Atrial Natriuretic Factor.

Name the following

Question 1.
Name the region consisting of nerve fibres that connects cerebrum and medulla oblongata.
Answer:
Pons Varolii.

Question 2.
Give the names of cranial nerve number VIth and VIIth.
Answer:
VTth cranial nerve is Abducens and Vllth cranial nerve is Facial.

Question 3.
Name the three sulci present on the cerebral hemispheres.
Answer:
Central sulcus, lateral sulcus and parieto¬occipital sulcus.

Question 4.
Name the band of nerve fibres that connect cerebrum, cerebellum and spinal cord.
Answer:
Crura cerebri.

Question 5.
Name the second largest part of the brain.
Answer:
Cerebellum.

Question 6.
Name the three branches of trigeminal nerve.
Answer:
Ophthalmic, Maxillary and Mandibular.

Question 7.
Name the nerve which arises from ventral side of medulla and supplies the tongue.
Answer:
Hypoglossal.

Question 8.
Name of emergency hormones secreted by sympathetic nervous system.
Answer:
Adrenaline and nor-adrenaline are the emergency hormones secreted by sympathetic nervous system.

Question 9.
Name the disorders caused by hyposecretion of thyroxine in children and adults.
Answer:

  1. Hyposecretion of thyroxine in children : Cretinism.
  2. Hyposecretion of thyroxine in adults : Myoxedema.

Question 10.
Name the dual exocrine as well as endocrine gland. Name the hormones secreted by it.
Answer:
Pancreas is the dual gland, exocrine as well as endocrine, it secretes hormones like insulin, glucagon and somatostatin.

Question 11.
Name the four peptide hormones secreted by endocrine cells of alimentary canal.
Answer:
Gastrin, secretin, cholecystokinin and GIPT or Gastric inhibitory peptide.

Question 12.
Name the disorder caused by the under secretion of thyroxine in children.
Answer:
Cretinism is the disorder seen in children due to under secretion of thyroxine.

Give functions of the following

Question 1.
Meninges and CSF.
Answer:
Functions of meninges:

  1. Meninges give protection to the brain and the spinal cord.
  2. They are also nutritive in function.
  3. Cerebrospinal fluid acts as a shock absorber. It protects the brain from mechanical injuries and from desiccation.
  4. It also maintains constant pressure inside and outside the CNS and regulates the temperature.

Functions of CSF:

  1. CSF helps in exchange of nutrients and wastes between the blood and the brain tissue.
  2. CSF supplies oxygen to the brain.

Question 2.
The functions of forebrain.
Answer:
(A) Functions of cerebrum:

  1. The cerebrum controls the voluntary activities.
  2. The cerebrum perceives various sensory stimuli received through vision, taste, smell, sound, touch, speech, etc.
  3. It is the centre of memory, will-power, intelligence, reasoning and learning.
  4. The cerebrum is the centre for emotions, thoughts and feelings, pain, pleasure, fear, fatigue, pressure, temperature, etc.
  5. It is also the centre for micturition, defecation, weeping, laughing, etc

(B) Function of olfactory lobes: Sensation of smell.

(C) Functions of diencephalon:

  1. Diencephalon acts as a relay centre for motor and sensory impulses between spinal cord, brainstem and various areas of cerebral cortex.
  2. Diencephalon consists of epithalamus, thalami and hypothalamus. Therefore it acts as a centre for homeostasis and higher centre of autonomous nervous system.
  3. Hypothalamic nuclei secrete neurohormones which influence the pituitary gland.
  4. Diencephalon regulates heartbeats, blood pressure and water balance.
  5. Anterior choroid plexus which is located in the diencephalon secretes cerebrospinal fluid.
  6. Hypothalmic regions control many involuntary functions such as hunger, thirst, thermo-regulation, fear, anger, sleep, sexual desire, etc.

Question 3.
Write various functions of hindbrain.
Answer:
Functions of hindbrain:
(i) Cerebellum:

  1. Cerebellum is a primary centre for the control of equilibrium, posture, balancing and orientation.
  2. Neuromuscular activities are regulated by the cerebellum.
  3. Coordination of walking, running, speaking, etc. is under the control of hindbrain.

(ii) Pons:

  1. Activities of two cerebellar hemispheres are coordinated by pons.
  2. Nerve fibres cross over in this area and thus the right side of the brain controls the left part of the body and vice versa.
  3. Pons controls the consciousness of the brain.
  4. Breathing centre is located in pons along with medulla.

(iii) Medulla oblongata:

  1. Medulla oblongata controls all the involuntary activities such as heartbeats, respiration, vasomotor activities.
  2. Peristalsis and reflex actions such as coughing, sneezing, swallowing, etc. are also under the control of medulla oblongata.
  3. Medulla oblongata is essential for all the vital functions of the body.

Question 4.
Spinal cord.
Answer:
Functions of spinal cord:

  1. Spinal cord is the main pathway for conduction of sensory and motor nerve impulses.
  2. The sensory impulses travel from the body
    to the brain and the motor impulses travel from the brain to the body.
  3. Spinal reflexes are controlled by spinal cord.
  4. The spinal cord reduces the load on the brain by taking appropriate actions in a reflex way.

Question 5.
Pituitary gland.
Answer:
Functions of pituitary:

  1. Pituitary secretes seven main hormones viz. ACTH, TSH. GH or STH, LTH or Prolactin, MSH and gonadotropins such as FSH, LH or ICSH.
  2. These hormones are secreted upon receiving proper releasing factor from hypothalamus.
  3. These hormones bring about many coordinating functions in the body. Almost all endocrine glands are under the control of these hormones.
  4. The neurohypophysis part stores two hormones, viz. vasopressin and oxytocin which are secreted by hypothalamus.
  5. Important functions such as growth and reproductive processes, pregnancy, childbirth, lactation, etc. are under the control of pituitary.
  6. Neurohypophysis acts as a neurohaemal organ and stores the hormones for some time.

Question 6.
Functions of hormones secreted by the thyroid gland.
Answer:

  1. Thyroxine is the main metabolic hormone in the body.
  2. Thyroxine maintains basal metabolic rate (BMR) by increasing glucose oxidation. It brings about calorigenic effect by energy production.
  3. It also controls normal protein synthesis.
  4. The physical growth, development of gonads and development of mental faculties is under the control of thyroxine.
  5. It controls tissue differentiation during metamorphosis particularly in amphibia.
  6. Body weight, respiration rate, heart rate, blood pressure, temperature, digestion, etc. is regulated by thyroxine.
  7. Another hormone of thyroid, i.e. calcitonin regulates calcium metabolism of the body.

Question 7.
Give significance of relaxin and inhibin.
Answer:
1. Relaxin : Relaxin relaxes the cervix of the pregnant female and ligaments of pelvic girdle during parturition.
2. Inhibin : Inhibin inhibits the FSH and GnRH production.

Question 8.
Enlist hormones secreted by GI tract and state their role.
Answer:
Gastrointestinal tract:
In the gastrointestinal mucosa, certain ells are endocrine in function. These cells produce hormones which play vital role in digestive processes and flow of digestive juices.

  1. Gastrin : This hormone stimulates gastric glands to produce gastric juice.
  2. Secretin : This hormone is responsible for secretion of pancreatic juice and bile from pancreas and liver.
  3. Cholecystokinin CCK/Pancreozymin PZ : This hormone stimulates gall bladder to release bile and stimulates the pancreas to release its enzymes.
  4. Entero-gastrone/Gastric inhibitory peptide (GIP) : It slows gastric contractions and inhibits the secretion of gastric juice.

Distinguish between the following

Question 1.
Electrical and chemical synapses:
Answer:

Electrical synapseChemical synapse
1. The gap between the successive neurons in electrical synapse is very less [3.8 nm],1. The gap between two successive neurons in chemical synapse is larger than electrical synapse [10-20 nm].
2. Transmission across the gap is faster in electrical synapse.2. Transmission across the gap in chemical synapse is relatively slower than electrical synapse.
3. Electrical synapse is less common.3. Chemical synapse is more common.
4. Electrical synapse is found in those places of the body requiring instant response.4. Chemical synapse is found almost everywhere and connects neuron to neuron, muscles or glands.

Question 2.
Sympathetic and parasympathetic nervous system.
Answer:

Sympathetic nervous systemParasympathetic nervous system
1. Sympathetic nervous system is formed by 22 pairs of sympathetic ganglia, 2 sympathetic cords which run parallel to vertebral column.1. Parasympathetic nervous system has nerve fibres which run along with cranial and spinal nerves.
2. Sympathetic nervous system works through neurotransmitter, adrenaline.2. Parasympathetic nervous system works through release of acetylcholine.
3. Sympathetic nervous system enhances all the involuntary functions.3. Parasympathetic nervous system retards all the involuntary functions.
4. It brings about fight, fright and flight responses.4. It brings about relaxation, comfort, pleasure, etc.
5. The pre-ganglionic nerve fibres are short and the post-ganglionic nerve fibres are long in sympathetic nervous system.5. The pre-ganglionic nerve fibres are long and the post-ganglionic nerve fibres are short in parasympathetic nervous system.

Question 3.
Unconditional reflexes and Conditional reflexes
Answer:

Unconditional reflexesConditional reflexes
1. Unconditional reflexes are inborn.1. Conditional reflexes are not inborn, they require training.
2. Unconditional reflexes are permanent.2. Conditional reflexes are temporary.
3. They never disappear and need no previous experience.3. They may disappear after sometime and need proper training for developing it.
4. Unconditional reflexes are heritable.4. Conditional reflexes are non-heritable.
5. Sneezing, coughing, blinking of eye, etc. are unconditional reflexes.5. Cycling, driving, playing games, etc. are due to conditional reflexes.

Question 4.
Epithalamus and hypothalamus.
Answer:

Epithalamushypothalamus
1. Epithalamus is the roof of diencephalon.1. Hypothalamus is the floor of diencephalon.
2. Epithalamus shows pineal stalk to which pineal gland is attached.2. Hypothalamus shows infundibulum to which pituitary gland is attached.
3. Epithalamus is non-nervous in nature.3. Hypothalamus is the higher centre of autonomous nervous system.
4. Epithalamus has anterior choroid plexus which secretes cerebrospinal fluid.4. Hypothalamus has neurons which secrete two endocrine hormones.
5. Epithalamus controls biological rhythm.5. Hypothalamus controls homeostasis of the body.

Question 5.
Dura mater and pia mater.
Answer:

Dura materpia mater
1. Dura mater is the outermost meninx.1. Pia mater is the innermost meninx.
2. Dura mater lies on the innermost side of skull or cranium.2. Pia mater lies on outermost side of the brain.
3. Dura mater is tough, thick and fibrous.3. Pia mater is thin and highly vascular.
4. Dura mater is mainly protective in function.4. Pia mater is mainly nourishing in nature.
5. Below dura mater is subdural space.5. Above pia mater is sub-arachnoidal space.

Question 6.
Cerebrum and cerebellum
Answer:

CerebrumCerebellum
1. The cerebrum is the larger part forming 85% of the brain. It has four lobes.1. The cerebellum is the smaller part forming 11%. of the brain. It has three lobes.
2. The cerebrum coordinates the functions of the sensory and motor areas.2. The cerebellum coordinates the equilibrium of muscular movements during walking and running.
3. The cerebrum plays an important role in receiving the sensory impulses such as touch, pain, heat, cold, etc.3. The cerebellum plays an important role in maintaining the posture and balance of the body.
4. The cerebrum is concerned with higher mental faculties such as memory, will and intelligence.4. The cerebellum is concerned with muscular mechanism.

Question 7.
Cranial nerves and Spinal nerves
Answer:

Cranial nervesSpinal nerves
1. Nerves arising from the brain are cranial nerves.1. Nerves arising from the spinal cord are spinal nerves.
2. There are 12 pairs of cranial nerves.2. There are 31 pairs of spinal nerves.
3. Cranial nerves are of three types, viz sensory, mixed and motor.3. All spinal nerves are of mixed type.
4. Cranial nerves are responsible for cerebral reflexes.4. Spinal nerves are responsible for spinal reflexes.

Question 8.
Extero and entero receptors.
Answer:

ExteroceptorsInteroceptors
1. Receptors receiving stimuli from outer environment of the body are called exteroceptors.1. Receptors receiving stimuli from inside the body are called interoceptors.
2. These are somatic receptors.2. These are visceral receptors.
3. Exteroceptors keep the body informed about . changes in the environment like temperature, pressure, touch, etc.3. Interoceptors keep the homeostasis in the body by receiving stimuli from inside the body.
4. Mechanoreceptors, thermoreceptors, chemical receptors, photoreceptors and statoacoustic receptors are the different types of exteroceptors.4. Propioceptors, enteroceptors, baroceptors are the different types of interoceptors.

Question 9.
Adenohypophysis and neurohypophysis.
Answer:

AdenohypophysisNeurohypophysis
1. Adenohypophysis is the anterior lobe of pituitary.1. Neurohypophysis is the posterior lobe of pituitary.
2. There is portal system between adenohypo-physis and hypothalamus which has blood sinusoids.2. There is axonal knobs and blood vessels that connect neurohypophysis and hypothalamus.
3. Adenohypophysis forms 75% of pituitary.3. Neurohypophysis forms 25% of pituitary.
4. Adenohypophysis has three parts, pars tuberalis, pars distalis and pars intermedia.4. Neurohypophysis has three parts, median eminence, infundibulum and pars nervosa
5. Adenohypophysis has chromophil (acidophil and basophil) and chromophobe cells.5. Neurohypophysis has axonic fibres and pituicytes.
6. Adenohypophysis secretes seven different hormones after receiving an appropriate message from hypothalamus through releasing factors.6. Neurohypophysis does not produce hormones on its own. It is a neurohaemal organ as it receives and stores two hormones from hypothalamus.

Question 10.
FSH and LH
Answer:

FSHLH
1. FSH is follicle stimulating hormone essential for the development of ovary.1. LH is luteinizing hormone responsible for ovulation in females.
2. FSH stimulates ovary (follicular cells) to produce estrogen.2. LH stimulates ovary (corpus luteum) to produce progesterone.
3. FSH in males is responsible for the spermatogenesis.3. LH in females is responsible for the development of corpus luteum.
4. Negative feedback mechanism exists between amounts of FSH and estrogen in females.4. Negative feedback mechanism exists between amounts of LH and progesterone in females.
5. FSH is indirectly responsible for the development of secondary sexual characters in females.5. LH is indirectly responsible for maintenance of pregnancy in females.

Question 11.
Glucocorticoids and mineralcorticoids
Answer:

GlucocorticoidsMineralocorticoids
1. Glucocorticoids control carbohydrate metabolism.1. Mineralocorticoids regulate mineral concentration.
2. These are secreted by the cells of zona fasciculata.2. These are secreted by the cells of zona glomerulosa.
3. These also regulate protein and fat metabolism.3. These regulate salt-water balance.
4. Cortisol is the main glucocorticoid.4. Aldosterone is the main mineralocorticoid.

Give scientific reasons

Question 1.
Number of gyri is related to the degree of intelligence.
Answer:

  1. Gyri are the ridges in the folds present on the cerebral cortex.
  2. The number of folds increase the surface area of the cerebral cortex.
  3. Cerebral cortex has sensory and motor areas such as Wernicke’s area, Broca’s area, etc. If the surface area is greater, the neurons in these areas are also more in number.
  4. Greater the number of neurons greater is the intelligence.
  5. The number of gyri therefore is said to be related to the degree of intelligence.

Question 2.
A drunken person cannot maintain balance of the body.
Answer:

  1. Cerebellum is the primary centre for controlling equilibrium and balance of the body.
  2. Alcohol has an adverse effect on the neurons of cerebellum.
  3. Consciousness of brain is also controlled by cerebellum.
  4. When a person is drunk, the alcohol in his or her blood affects the activities of cerebellum and hence the person cannot maintain the balance of the body.

Question 3.
We are able to understand the smell of the first showers of rain or the sudden changes in the climate.
Answer:

  1. We are able to understand any smell because of our olfactory mucosa and olfactory lobes of the brain.
  2. Volatile substances are received by the olfacto-receptors in the nose.
  3. Nerve impulse generated are carried by olfactory nerve and transmitted to brain where the impulse is interpreted.
  4. The characteristic earthy smell is due to a compound ‘geosmin’.
  5. Geosmin is produced by some species of Streptomyces [ gram positive soil bacterium],
  6. Similarly, sudden change in the climate is easily noticed in the form of temperature change in the surrounding.
  7. This change is detected by caloreceptors of the skin. From these receptors the signal is transmitted to CNS where the change is perceived.

Question 4.
We are able to hear the chirping of the birds and recognize the sound of the bird.
Answer:

  1. The phonoreceptors of the body receive the sound waves and transfer the nerve impulses to the auditory areas of the brain.
  2. The interpretation, of the sound is a combined effort of sensory and association areas [auditory areas] of temporal lobes of the brain.
  3. This is how we are able to hear the chirping of the birds and recognize the sound of the bird.

Question 5.
We can see and enjoy the beautiful colours of the nature after the sunrise.
Answer:

  1. The light receptor are present in the retina of eyes.
  2. These are photosensitive rod and cone cells. Cone cells are sensitive to bright light-and colours.
  3. On receiving the light rays, these cells convert them into nerve impulses.
  4. The cones are of three types, which contain their own characteristic photo-pigments that respond to red, green and blue lights.
  5. Various combinations of these cones and their photo pigments produce sensation of different colours.
  6. These impulses are carried to the visual cortex of the occipital lobe where the image is interpreted.
  7. In this manner, we can see and enjoy the beautiful colours of the nature after the sunrise.

Question 6.
Cerebellum is well developed in humans.
Answer:

  1. Our posture is upright and mode of locomotion is bipedal.
  2. While standing, walking and running, our body has to be in a state of balance.
  3. Cerebellum controls balancing, posture, body equilibrium and orientation.
  4. Thus to control static as well as dynamic equilibrium of the body, cerebellum is well developed.

Question 7.
Mr. Sharma suffered from a stroke and the right side of his body was paralysed. However his response was normal for knee jerk reflex with either leg. Explain how and why?
Answer:

  1. Stroke is damage to the brain due to interruption of its blood supply.
  2. Due to this brain functions and cerebral reflexes are severely affected.
  3. However, spinal reflexes remain largely unaffected.
  4. In the above case. Mr. Sharma suffered from a stroke and developed paralysis. But his response to the knee jerk was normal as it is controlled by spinal cord.

Maharashtra Board Class 12 Biology Important Questions Chapter 9 Control and Co-ordination

Question 8.
The word hormone is a misnomer.
Answer:

  1. Hormone is adapted from the Greek word ‘hormein’ means to excite.
  2. But the hormones secreted by the endocrine glands and cells carry out two fold functions, viz. excitatory and inhibitory.
  3. The word hormone is therefore a misnomer.

Question 9.
Pituitary may be considered as the coordinator of endocrine orchestra but not master endocrine gland.
Answer:

  1. The pituitary gland was formerly considered as a master endocrine gland, because all other endocrine glands are under the control of pituitary hormones.
  2. But now it is known that the pituitary gland itself is under the control of hypothalamus through hypo thalamo – hypophysial axis.
  3. Through various releasing factors and release inhibiting factors, the secretions of pituitary are regulated by hypothalamus.
  4. Pituitary in turn controls growth, secretion and maintenance of glands such as adrenal cortex, thyroid and gonads.

Question 10.
Pituitary gland plays an important role in pregnancy and childbirth.
Answer:
Pituitary gland plays an important role in pregnancy and childbirth for the following reasons:

  1. Luteinizing hormone secreted by the pituitary gland stimulates ovulation, formation of corpus luteum and the synthesis of progesterone which are necessary for pregnancy.
  2. Lactogenic hormone secreted by the pituitary gland promotes breast development during pregnancy and stimulates the secretion of milk after childbirth.
  3. Oxytocin secreted by hypothalamus and sent by pituitary gland stimulates the contraction of the uterus during childbirth.

Question 11.
Fall and rise in blood calcium stimulates secretion of parathyroid.
Answer:

  1. Normal calcium level in the blood is called calcemia.
  2. Increase in the blood calcium is called hypercalcemia while decrease in the level is called hypocalcemia.
  3. Decrease in the blood calcium, stimulates the parathyroid glands to secrete PTH.
  4. PTH stimulates osteoclasts of bone to start bone resorption. This will help to cause increase in the blood calcium level.
  5. Increase in the blood calcium above normal will retard the secretion of PTH. Hence further increase in the blood calcium is stopped.

Question 12.
Pancreas is both exocrine as well as endocrine gland.
Answer:

  1. Pancreas is heterocrine gland i.e. both exocrine and endocrine gland.
  2. The exocrine part is pancreatic acini. The cells of acini secrete pancreatic juice containing digestive enzymes like trypsinogen, chymotrypsinogen, etc.
  3. Endocrine part of pancreas is made up of groups of cells called Islets of Langerhans. There are four kinds of cells in islets of Langerhans which secrete hormones.

Alpha (α) cells (20%) secrete glucagon.
Beta (β)) cells (70%) secrete insulin.
Delta (δ) cell (5%) secrete somatostatin. PP cells or F cells (5%) secrete pancreatic polypeptide (PP).
Hence, pancreas is said to be exocrine as well as endocrine gland.

Question 13.
Patient suffering from hypothyroidism shows increased level of TSH. Why?
Answer:

  1. Hypothyroidism means decrease in the secretion of T3 and T4
  2. Decrease in T4, i.e. thyroxine in the blood triggers negative feedback mechanism.
  3. The hormone receptors in hypothalamus detect this change and secrete TRE
  4. TRF stimulates the pituitary to secrete TSH
  5. Thus, Patient suffering from hypothyroidism shows increased level of TSH due to negative feedback control.

Question 14.
We are advised to use iodized salt.
OR
Why do we use iodized salt?
Answer:

  1. Iodine is needed for synthesis of thyroid hormone.
  2. If there is deficiency of iodine in the diet, it causes enlargement of thyroid gland leading to simple goitre.
  3. This disease is common in hilly areas hence it is also called endemic goitre.
  4. Addition of iodine to table salt prevents this disease.
  5. Therefore we must use iodised salt.

Write short notes on the following

Question 1.
Nervous system of Hydra.
Answer:

  1. Hydra, a cnidarian shows the diffused nervous system in the form of nerve net.
  2. It is the most primitive nervous system.
  3. There are two nerve nets in the mesoglea-one connected towards the epidermis and second towards the gastro-dermis.
  4. Hydra lacks sensory organs, but the sensory cells scattered in the body wall.
  5. The nerve impulse shows no polarity or direction. As all neurons are interconnected the response is seen throughout the body.

Question 2.
Nervous system of Planaria.
Answer:

  1. Planaria is a flatworm and belongs to the phylum Platyhelminthes.
  2. It is the most primitive animal with a Central Nervous System (CNS) located on the ventral side of body.
  3. Nervous system consists of a mass of cerebral or cephalic ganglion appearing like an inverted U-shaped brain.
  4. Ventrally from below the ganglia arise a pair of Ventral Nerve Cords (VNC) or long nerve cords. These are interconnected to each other by transfer nerve or commissure in a ladder like manner.
  5. The PNS include sensory cells arranged in lateral cords in the body.

Question 3.
Resting potential of a nerve fibre.
Answer:

  1. The neurons have a property of excitability.
  2. The transmission of the nerve impulse along the long nerve fibre is a result of electrical changes across the neuronal membrane during conduction of an excitation.
  3. The plasma membrane separates the outer [extra cellular fluid] and inner solutions [intra cellular fluid/cytoplasm] of different chemical compounds.
  4. The external tissue fluid has both Na+ and K+
  5. On the inside there is an excess of K+ along with large amount of negatively charged protein molecules and nucleic acid.
  6. This condition of a resting nerve is also called a polarised state and it is established by maintaining an excess of Na+ on the outer side.

Question 4.
Electrical synapse.
Answer:

  1. In this type of synapse gap between the neighbouring neurons is very narrow.
  2. This electrical link is formed between the pre-and post-synaptic neurons.
  3. At the gap junction, the two cells are within almost 3.8 nm distance of each other.
  4. Transmission across the gap is faster.
  5. Electrical synapses are found in those places of the body requiring fastest response as in the defence reflexes.

Question 5.
Velocity of nerve impulse.
Answer:

  1. The rate of transmission of impulse is higher in long and thick nerves.
  2. It is higher in warm blooded animals [homeotherms] than in cold blooded animal [poikilotherms].
  3. The velocity of transmission is higher in voluntary fibres (100- 120 m/second in man) as opposed to autonomic or involuntary nerves (10-20 m/second).
  4. It is faster in medullated nerve fibre [up to 150 m/s], as the impulse has to jump from one node of Ranvier to the next as compared with non-medullated nerve fibre [10-25 m/s]

Question 6.
Meninges of CNS.
Answer:

  1. Meninges are the connective tissue membranes that cover the brain and the spinal cord. There are three meninges, viz. dura mater, arachnoid mater and pia mater that cover the Central nervous system.
  2. The outermost tough, thick and fibrous meninx is dura mater. It is protective in function as it is attached to the inner side of the cranium.
  3. The middle, thin and vascular membrane formed of reticular connective tissue is called arachnoid mater. It carries out nutritive function and also gives protection to the brain.
  4. The innermost highly vascular and thin membrane is pia mater. It lies in contact with CNS. It is nutritive in function.
  5. There is subdural space between the dura mater and arachnoid mater. It is filled with serous fluid.
  6. Besides, there is a sub-arachnoidal space lying between arachnoid mater and pia mater. It is filled with cerebrospinal fluid.

Question 7.
Ventricles in human brain.
Answer:

  1. Ventricles are the cavities present in different parts of the brain.
  2. There are four ventricles in the human brain. All the ventricles are connected with each other.
  3. They are filled with cerebrospinal fluid.
  4. Paracoel or lateral ventricles-I and-II are present inside the cerebral hemispheres.
  5. The diencephalon has ventricle-III.
  6. Ventricle-III is in connection with lateral ventricles by foramen of Monro.

Question 8.
Sympathetic nervous system.
Answer:

  1. Sympathetic Nervous System is formed by 22 pairs of sympathetic ganglia. These ganglia are linearly arranged on two sympathetic cords. Sympathetic nerve cords run on either side of the vertebral column.
  2. Sympathetic nerve cords are connected to CNS by rami communicans of spinal nerve fibres.
  3. This system works during stress, pain, anger, fear or emergency. It is supposed to bring about fight, flight or fright reactions.
  4. Action of sympathetic nervous system is dependent on adrenaline or noradrenaline. This neurotransmitter is secreted by sympathetic nervous system as an emergency hormone.

Question 9.
Parasympathetic nervous system.
Answer:

  1. Parasympathetic nervous system consists of nerve fibres of some cranial nerves, sacral nerves and parasympathetic ganglia.
  2. These parasympathetic ganglia are present on the sides of visceral organs like heart, lungs, stomach, kidney, etc.
  3. Parasympathetic ganglia gives out parasympathetic fibres which innervates these involuntary organs.
  4. Parasympathetic nervous system works through release of acetylcholine which acts as neurotransmitter. It is an inhibiting neurotransmitter which affects visceral organs.
  5. This system works during rest and brings about relaxation, comfort, pleasure, etc.

Question 10.
Parkinson’s disease.
Answer:

  1. Degeneration of dopamine-producing neurons in the CNS causes Parkinson’s disease.
  2. Symptoms develop gradually over the years.
  3. Symptoms are tremors, stiffness, difficulty in walking, balance and coordination.
  4. Seen in old age and is incurable.

Question 11.
Alzheimer’s disease.
Answer:

  1. Alzheimer’s disease is the most common form of dementia.
  2. Its incidence increases with the age.
  3. Symptoms include the loss of cognitive functioning, thinking, remembering, reasoning and behavioural abilities. It interferes with the person’s daily life and activities.
  4. It occurs due to loss of cholinergic and other neurons in the CNS and accumulation of amyloid proteins.
  5. There is no cure for Alzheimer’s, but treatment slows down the progression of the disease and may improve the quality of life.

Question 12.
Types of reflex actions.
Answer:

  1. The reflex actions are of two types, viz. cerebral and spinal.
  2. Cerebral reflex actions are controlled by the brain.
  3. Spinal reflex actions are controlled by the spinal cord.
  4. In man, most of the reflex actions are controlled by the spinal cord.

Question 13.
Pavlov’s experiment about conditional reflex.
Answer:

  1. Conditional reflex was demonstrated by Pavlov while performing experiments with dogs.
  2. Pavlov offered some food to dog and noticed that the dog starts salivating after smelling and seeing the food.
  3. Simultaneously he rang the bell so that the dog associated the food with the sound of bell. This experiment was repeated many a times by him.
  4. Later he only rang the bell and did not give any food to the dog. But still the dog salivated.
  5. This shows that the dog was conditioned to the sound of bell. The dog learnt that there is a relation between food and sound of bell. This is called conditional response.

Question 14.
Mechanism of vision.
Answer:

  1. The light rays of visible wavelength pass through the cornea and the lens and are focused on the retina of the eye.
  2. The sight is possible due to conjugated proteins present in the rods and the cones.
  3. These are photo pigments which are composed of opsin (a protein) and retinal (Vitamin A derivative).
  4. The light induces dissociation of retinal from the opsin, which causes a change in the structure of the opsin.
  5. This causes the change in the permeability of the retinal cells.
  6. It generates action potential which is carried via bipolar cells and ganglion cells and further conducted by the optic nerves to the visual cortex (vision centre) of the brain.
  7. The neural impulses are analyzed and the image formed on the retina is thus recognized.

Question 15.
Mechanism of hearing.
Answer:

  1. The external ear receives the sound waves and sends to the tympanic membrane. The tympanum vibrates transmitting the vibrations to the chain of three ossicles and then to the oval window.
  2. The vibrations are further passed on to the fluid of cochlea.
  3. The waves in the perilymph and endolymph induces movements in the basilar membrane.
  4. The hair cells of organ of Corti bend and are pressed against the tectorial membrane.
  5. Due to this pressure, the nerve impulses are generated and are sent to the afferent nerve fibres.
  6. The impulses are carried by the auditory nerves to the auditory centre of the brain, where the impulses are analyzed and the sound is perceived.

Question 16.
Types of endocrine systems.
Answer:
There are 3 types of endocrine systems.

  1. Discrete endocrine system : The glands exclusively endocrine in function are called discrete endocrine glands.
  2. Mixed endocrine system : The glands endocrine as well as exocrine in function are called mixed endocrine glands.
  3. Diffused endocrine system : Some endocrine cells scattered in a particular region/gland form diffused endocrine system.

Question 17.
Thymus.
OR
Functions of Thymosin.
Answer:

  1. Thymus is located on the dorsal side of the heart and the aorta.
  2. It consists of many lobules.
  3. The thymus plays major role in the development of the immune system.
  4. Thymus secretes thymosin which plays an important role in the differentiation of T-lymphocytes. These cells built cell mediated immunity.
  5. The thymosin also promotes the production of antibodies providing humoral immunity.
  6. The degeneration of thymus gland occur in old individuals leading to decreased production of thymosin thereby weakening of immune response.

Question 18.
The role of heart and kidney in hormone secretion.
Answer:
(I) Kidney:

  1. Kidney produces renin, erythropoietin and calcitriol (calcitriol is the active form of vitamin cholecalciferol (D3).
  2. Renin along with angiotensin helps in maintaining the blood pressure in the renal artery by vasoconstriction.
  3. Erythropoietin stimulates erythropoiesis.
  4. Calcitriol helps in absorbing calcium from the stomach.

(II) Heart:

  1. Heart walls secrete Atrial natriuretic hormone /ANE
  2. ANF increases sodium excretion [natriuresis] along with water.
  3. It acts along with kidneys and reduces blood pressure by lowering blood volume.

Question 19.
Hormones secreted by adrenal gland.
Answer:

  1. The adrenal cortex secretes corticoids. Corticoids is a group of several hormones that control several vital body functions.
  2. Corticoids are of two types, viz. mineralocorticoids and glucocorticoids.
  3. Small amounts of androgenic steroids are also secreted by the adrenal cortex which have the role in the growth of axial and pubic hairs and facial hairs during puberty.
  4. The mineralocorticoids regulate the electrolyte balance while the glucocorticoids are involved in carbohydrate metabolism.
  5. Adrenal medulla secretes adrenaline or epinephrine and noradrenaline or norepinephrine.

Question 20.
Role of mineralocorticoids.
Answer:

  1. The mineralocorticoids regulate ionic and osmotic balance, by regulating the amounts of electrolyte and water.
  2. Aldosterone is the main mineralocorticoid that acts on the renal tubules.
  3. Aldosterone stimulates the re-absorption of Na+ and water and excretion of K+ and phosphate ions.
  4. The aldosterone helps in the maintenance of electrolytes, body fluid volume, osmotic pressure and blood pressure.

Question 21.
Secretions of adrenal medulla and their role.
Answer:

  1. The adrenal medulla secretes two catecholamine hormones, viz. adrenaline (epinephrine) and noradrenaline (nor-epinephrine).
  2. Adrenaline and noradrenaline increase alertness, dilation of pupils, piloerection, sweating, etc.
  3. Both the hormones increase the rate of heartbeat, strength of heart contraction and rate of respiration.
  4. These hormones also stimulate the breakdown of glycogen, lipids and proteins thereby increasing blood glucose level.
  5. All the above reactions are useful for survival during emergency situations and in stress condition. Therefore, they are called emergency hormones or 3 F hormones of fright, fight or flight.

Question 22.
Cortisols and their role.
Answer:
I. Cortisol : Cortisol is the main glucocorticoid hormone.

II. Role of cortisol:

  1. Cortisol stimulates many metabolic reactions such as gluconeogenesis, lipolysis and proteinolysis.
  2. It inhibits cellular uptake and utilization of amino acids.
  3. Cortisol also plays an important role in maintaining the cardiovascular system and kidney functions.
  4. It is also involved in anti-inflammatory reactions and suppresses the immune response.
  5. Cortisol stimulates the RBC production.

Question 23.
Disorders of adrenal cortical hormones.
Answer:

  1. Disorders of adrenal cortical secretions are of two types, viz. hyposecretion and hypersecretion.
  2. Hyposecretion of corticosteroids causes Addison’s disease.
  3. The symptoms of Addison’s disease are general weakness, weight loss, low body temperature, feeble heart action, low BR acidosis, excessive loss of Na+ and Cl in urine, impaired kidney functioning and kidney failure, etc.
  4. Hypersecretion of corticoids causes Cushing’s disease.
  5. The symptoms of Cushing’s disease are alkalosis, enhancement of total quantity of electrolytes in extracellular fluid, polydipsia, increased BR muscle paralysis, etc.

Question 24.
Hormones of adenohypophysis.
Answer:
I. Somatotropic Hormone (STH) or Growth Hormone (GH):

  1. The secretion of GH is under dual control of hypothalamus through GHRF (Growth hormone releasing factor) and GHIF (Growth hormone inhibiting factor).
  2. The GH brings about general growth of the body.
  3. The principal actions of GH Eire promotion of linear growth in the skeleton, increase in the size of the muscles and connective tissue.
  4. GH enhances the protein synthesis. The lipolysis in adipose tissue to release more fatty acids is also stimulated by GH.
  5. The growth of bones by absorption of calcium takes place due to GH.

II. Thyroid Stimulating Hormone (TSH) or Thyrotropin:

  1. TSH is regulated by TRF (Thyrotropin releasing factor) from hypothalamus’
  2. For inhibition there is a negative feedback between thyroxine level in the blood and secretion of TSH.
  3. TSH stimulates thyroid glands to increase uptake of iodine for synthesis of thyroxine. It brings breakdown of colloid to release thyroxine.

III. Adrenocorticotropic Hormone (ACTH) Corticotropin :

  1. ACTH stimulates growth of adrenal cortex Eind stimulates it to secrete glucocorticoids Eind mineralocorticoids.
  2. The regulation of ACTH secretion is under the control of hypothalamic CRF (Corticotropin releasing factor) and the negative feedback mechanism between plasma level of cortisol and ACTH.

IV. Prolactin (PRL) or Luteotropic hormone (LTH):
1. Secretion of prolactin is under duad control by hypothalamus by two factors such as PRF (Prolactin releasing factor) and PIF (Prolactin inhibiting factor) of hypothalamus.

2. Prolactin performs many functions therefore it has many terms as follows :

  • Development of mammary glands (Mammo tropin).
  • Milk secretion by mammary glands (Lactogenic hormone).
  • Maintenance of corpus luteum so that- it keeps on secreting progesterone during pregnancy (Luteotropin).

3. It may be inhibiting the chances of pregnancy during lactation period.

V. Gonadotropic Hormones (GTH) or Gonadotropins:
1. There are two types of gonadotropins, viz. FSH and LH or ICSH.

2. The secretions of gonadotropin are regulated by gonadotropin releasing factor (GHRF) of hypothalamus.

(3) They are regulated by negative feedback by sex hormones such as testosterone and estrogen.
(a) Follicle Stimulating Hormone (FSH):

  • FSH in female stimulates development of Graafian follicles. It helps in the formation of ovum by stimulating oogenesis.
  • It also stimulates ovarian follicular cells for secretion of female sex hormones, estrogen.
  • Under influence of estrogen, development of secondary sexual characters occurs in female.
  • In males, FSH stimulates germinal epithelium of seminiferous tubules for spermatogenesis and helps in the production and maturation of sperms.
  • Deficiency of FSH leads to infertility in both the sexes.

(b) Luteinizing hormone (LH) in females and Interstitial cell stimulating hormone (ICSH) in males:

  • LH brings about ovulation, i.e. rupture and release of ovum from the mature Graafian follicle. The empty Graafian follicle is transformed into corpus luteum.
  • Corpus luteum is a secondary endocrine source which secretes gestational hormone progesterone. Progesterone is a pregnancy stabilizing hormone.
  • In males, ICSH stimulates interstitial cells of Leydig which in turn secretes male sex hormone, the testosterone.
  • Testosterone develops secondary sexual characters in males.
  • High level of progesterone in female signals negative feedback, to pituitary to stop secretion of LH.
  • In males, high level of testosterone in blood gives negative feedback signal to Inhibit the secretion of FSH.

Short answers questions

Question 1.
Give one point of distinction between nervous coordination and hormonal coordination.
Answer:

  1. The activity of nervous coordination is quick, immediate and fast as it sends the electrical signals.
  2. Hormonal coordination is slow and long lasting as it takes place through the action of hormones.

Question 2.
Write about types of nerves.
Answer:
Nerve is a group of neurons enclosed in a connective tissue sheath epineurium. It is classified as:

  1. Sensory nerve : A nerve having all sensory neurons is called sensory nerve. It carries information from sense organs to CNS. It is also called afferent nerve.
  2. Motor nerve : A nerve having all motor neurons is called motor nerve. It carries information from CNS to effector organs. It is also called efferent nerve.
  3. Mixed nerve : A nerve having with both sensory and motor neurons is called a mixed nerve. Sensory neurons in it carry nerve impulses from sense organs to CNS while motor neurons carry nerve impulses from CNS to effector organs.

Maharashtra Board Class 12 Biology Important Questions Chapter 9 Control and Co-ordination

Question 3.
What is the composition of neural tissue?
Answer:
Neural tissue is derived from embryonic ectoderm. It consists of two types of cells.

  1. Neurons or nerve fibres : These are structural and functional units of nervous tissue. They conduct nerve impulses and coordinate various body activities.
  2. Neuroglial cells : These are supportive cells which protect neurons throughout CNS and PNS. They perform other functions like secretion of myelin sheath, phagocytosis, production of CSF, etc. e.g Schwann cells, astrocytes, satellite cells, etc.

Question 4.
Explain the structure of cyton.
Answer:

  1. Cyton is the main body of a neuron or nerve fibre.
  2. The cyton has a distinct central nucleus with a nucleolus and neuroplasm.
  3. Cytoplasm surrounds the nucleus around which there are neurofibrils, Nissl’s granules and other cell organelles.
  4. Nissl’s granules are rich in ribosomes and proteins.
  5. Neurofibrils play an important role in transmission of nerve impulse.
  6. The cytons are generally found inside the brain, spinal cord (CNS) and in the ganglia.
  7. Cytons within CNS form ‘nuclei’ while those present outside CNS in nerves form ‘ganglia’.

Question 5.
Enlist the various connective tissue layers in a nerve along with their location.
Answer:
Connective tissue layers in a nerve are:

  1. Endoneurium : Covers each nerve fibre.
  2. Perineurium : Covers each nerve bundle having a number of neurons.
  3. Epineurium : Covers many nerve bundles to form a peripheral nerve.

Question 6.
What is a synapse?
Answer:

  1. Synapse is a microscopic functional gap between two successive neurons.
  2. In this telodendrites of pre-synaptic neuron are in close proximity with dendrites of post-synaptic neuron.
  3. This gap is also called synaptic cleft.
  4. During transmission of nerve impulse, the synapse get filled with neurotransmitters like acetyl choline.

Question 7.
What are the three divisions of nervous system? What are their chief functions?
Answer:

  1. The three divisions of nervous system are central nervous system, peripheral nervous system and autonomous nervous system.
  2. The central nervous system (CNS) consists of brain and spinal cord. The brain and spinal cord are the coordinators for all nervous functions.
  3. The peripheral nervous system (PNS) is constituted by several nerves given by the central nervous system to all the body parts. All these nerves carry impulses to the CNS and bring back the responses from them. They are divided into cranial nerves and spinal nerves.
  4. The autonomous nervous system controls all the internal organs and is not under voluntary control.

Question 8.
Does this CSF remain enclosed inside the ventricles? What can be the outcome of such a situation?
Answer:

  1. CSF is present within the CNS as well as around it.
  2. This fluid communicates with each other on the roof of medulla oblongata through 3 apertures, viz. Foramina of Luschka and foramen of Magendie.
  3. This communication ensures constant pressure of CSF within as well as outside the CNS.
  4. In the absence of this communication, there would be a pressure difference within as well as outside the CNS which will result in disturbances in the activities of CNS. Moreover, intercranial pressure would rise.

Question 9.
Enlist the different parts of the brain.
Answer:

  1. There are three divisions of the brain, viz. forebrain (prosencephalon), midbrain mesencephalon) and hindbrain (rhombencephalon).
  2. Forebrain is divided into cerebrum (telencephalon) and diencephalon (thalamencephalon). Underdeveloped of factory lobes (rhinencephalon) can also be seen in the anterior region.
  3. Midbrain consists of corpora quadrigemina and crura cerebri.
  4. Hindbrain has cerebellum (metencephalon) and brain stem. It is divided into pons varolii and medulla oblongata (myelencephalon).

Question 10.
Describe functional areas of cerebral cortex.
Answer:
Functional areas of cerebrum:

  1. There are three functional areas in cerebrum viz., sensory, association and motor area.
  2. In sensory area, sensory receptors bring the sensory inputs. These inputs are analysed in sensory area.
  3. The sensory speech area is located in parietal lobe. It is called Wernicke’s area.
  4. Association area forms the major portion of the cerebrum. It processes, analyses and stores the information given by the inputs. Power of reasoning, will, understanding, memory, etc. are the faculties present in the cerebral cortex.
  5. Motor area is present in the frontal lobe lying anterior to the premotor area. In the lower part of the motor area just above the lateral sulcus lies the Broca’s area or motor speech area. The Broca’s area controls the movements necessary for speech.

Question 11.
Explain in detail the regions associated with the diencephalon.
Answer:
Diencephalon is a part present between forebrain and midbrain. It has three regions epithalamus, thalami and hypothalamus.
(1) Epithalamus : Epithalamus is the roof of diencephalon. It is highly vascular and non- nervous. It forms anterior choroid plexus that secretes cerebrospinal fluid. Pineal body is attached to epithalamus with the help of pineal stalk. Pineal body secretes serotonin and melatonin.

(2) Thalami : The lateral parts of diencephalon which are interconnected with the habencular commissure are called thalami. From thalami all sensory impulses (except olfactory impulses) pass on to the cerebrum.

3. Hypothalamus : Hypothalamus is the floor of the brain. Pituitary gland is attached to this floor by an infundibular stalk. Hypothalamus has many hypothalamic nuclei which are scattered in the white matter.

Question 12.
What is EBG? What information can be obtained from the EEG?
Answer:

  1. EEG stands for electroencephalography.
  2. It refers to the recordings of the brain’s spontaneous electrical activities in certain period of time.
  3. These are recorded using multiple electrodes.
  4. EEG is non-invasive method and measures voltage fluctuations resulting from ionic current within the neurons.
  5. The basic concepts involved in this are similar to ECG.
  6. It is used to diagnose conditions like epilepsy, sleep disorders, encephalopathies, coma, etc.

Question 13.
Find out how different functional areas of the brain can be mapped?
Answer:
Functional areas and status of the brain can be mapped by several imaging techniques available such as-

  1. MRI : Magnetic Resonance Imaging
  2. CT : Computed Tomography
  3. PET : Positron Emission Tomography

Question 14.
Which are silent areas of the brain?
Answer:

  1. Silent areas of the brain refer to association areas of the brain.
  2. One such area is in the prefrontal cortex of brain.
  3. These are the areas of the brain in which pathogenic conditions may occur without producing symptoms.
  4. Injury to these areas is not accompanied by symptoms related to sensory and motor functions.

Question 15.
Is nervous tissue without lymphatic vessels?
Answer:

  1. CSF is the lymph of CNS.
  2. CSF is continuously generated by the ependymal cells lining the ventricles and central canal and simultaneously drained out of the brain into the blood stream.
  3. There are no lymphatic vessels in the nervous system.
  4. But the CSF is drained into peripheral blood circulation with the help of lymph vessels associated with meninges mainly the dura mater.

Question 16.
Explain the structure of spinal cord.
Answer:
Structure of spinal cord:

  1. Spinal cord is a 42 to 45 cm long, 2 cm thick and hollow tube, extending from medulla oblongata to lumbar region.
  2. It lies in the neural canal of vertebral column.
  3. At the other end, it tapers down and is called conus medullaris. The posterior most end is called filum terminale which appears as a thread-like structure.
  4. Beyond the second lumbar vertebra, it forms a horse tail-like structure called cauda equina. Cauda equina is a bunch of dorsal and ventral roots of last pair of spinal nerves.
  5. There are two swellings on the spinal cord. The upper is cervical swelling and lower is lumbar swelling. Accordingly there are two plexuses, the cervical plexus supplying nerves to hands and the lumbar plexus supplying nerves to legs.
  6. 31 pairs of spinal nerves arise from spinal cord.

Question 17.
What is the significance of reflex action?
Answer:
Significance of reflex action :

  1. Reflex action helps the animals to adjust quickly to the changing environment.
  2. Reflex action is for quick actions necessary for survival. The life may have been in danger in the absence of reflex action.
  3. Most of the reflexes are spinal reflexes, i.e. reflexes controlled by spinal cord. Thus the brain is not involved in these actions. This prevents overloading of the brain and brain fatigue.
  4. Some reflexes are inborn and hence training or learning is not required for these.

Question 18.
During extraction of a tooth, the dentist gives an injection of Anaesthesia to the patient before extraction. Is the action potential generated? How does the local anaesthesia work? What is the effect of pain killer on the nervous system?
Answer:

  1. Teeth are innervated by branches of trigeminal nerve [Vth cranial nerve]
  2. Extraction of tooth stimulates this nerve which then carries the impulse [action potential) to the pain centre of the brain where the pain is perceived.
  3. To avoid this, dentists give anaesthesia, to numb the nerve.
  4. Action potential is not generated due to anaesthesia.
  5. Hence the pain is not perceived.
  6. Similarly, some common pain killers act on the nerve endings and pain centres of the brain, preventing generation of action potential.

Question 19.
Give a list of psychological disorders.
Answer:

  1. Autism spectrum disorder.
  2. Bipolar disorder.
  3. Depression.
  4. Anxiety disorder.
  5. ADHD (Attention Deficit Hyperactivity Disorder).
  6. Stress related disorders.

Question 20.
What are endocrine glands?
Answer:

  1. Endocrine glands are ductless glands which are capable of secreting hormones.
  2. The hormones are poured directly into the bloodstream as the endocrine glands do not have duct.
  3. Hormones regulate the function of target tissue or organ.
  4. They either have excitatory effect or have an inhibitory effect.

Question 21.
What are the main endocrine glands in human body?
Answer:
The main endocrine glands in human body are as follows:

  1. Pituitary or hypophysis
  2. Hypothalamus
  3. Thyroid
  4. Parathyroid
  5. Adrenal or suprarenal
  6. Islets of Langerhans in pancreas
  7. Endocrine parts of gonads, i.e. testis and ovary.
  8. Pineal gland and thymus are also endocrine glands of less importance.

Question 22.
What are the common properties of hormones?
OR
State properties of hormones.
Answer:

  1. Hormones are specifically produced in response to a certain stimulus.
  2. Depending on nature and intensity of the stimulus, the rate of secretion of a hormone varies from low to very high.
  3. Hormones are produced in one organ and show their effect on distant ‘target’ organ. The source and the target region may be distantly located.
  4. Hormones are directly poured in blood circulation and always carried through blood.
  5. Hormones are always bound to specific carrier proteins while being transported through the blood.
  6. Hormones have a high degree of target specificity.
  7. Every hormone acts basically by modifying some aspect of cellular metabolism.
  8. The excessive secretions or deficiencies- of hormones may lead to serious disorders. Such disorders are called hyper – and hypo- disorders, respectively.

Question 23.
What are the disorders caused due to hyposecretion and hypersecretion of GH or STH?
Answer:
(1) Hypersecretion is excessive secretion. In children, the hypersecretion of GH causes gigantism. In adults, it causes Acromegaly.

(2) Hyposecretion, i.e. lesser secretion of GH in children cause dwarfism. The person is also referred to as midget. There are two types of dwarfs, viz. Frohlic dwarf who are mentally abnormal and Lorain dwarf who are mentally normal.

(3) Hyposecretion of GH in adults cause Simmond’s disease.

Question 24.
What are the disorders caused due to hyposecretion and hypersecretion of ACTH?
Answer:

  1. Hyposecretion of ACTH leads to Addison’s disease, i.e. adrenal failure. This results in affected carbohydrate metabolism leading to weakness and fatigue.
  2. The hypersecretion leads to excessive growth of adrenal cortex. This causes Cushing’s disease.

Question 25.
Write an account of hormones secreted by the thyroid gland.
Answer:

  1. Thyroid secretes triiodothyronine or T3, tetraiodothyronine or thyroxine or T4 and thyr ocalcitonin.
  2. The thyroid gland synthesize, store and discharge these hormones.
  3. T3 and T4 are iodinated derivatives of amino acid tyrosine which are secreted by thyroid follicular cells and stored in follicles. They have similar function. The secretion of T3 and T4 are regulated by Thyroid stimulating hormone (TSH) or thyrotropin of pituitary gland in negative feedback manner. T3 is more active and T4 is more potent hormone.
  4. Thyrocalcitonin is secreted by the parafollicular cells.
  5. Thyrocalcitonin regulates blood calcium level. It stimulates bones to take up Ca++ from the blood and deposit it in the form of calcium phosphates in the bones, thereby decreasing blood Ca++ level. Increased calcium level of blood stimulates ‘C’ cells to secrete thyrocalcitonin and vice versa.

Question 26.
Describe adrenal glands with respect to morphology, histology and secretions.
Answer:

  1. A pair of adrenal or suprarenal glands are located just on the upper border of kidneys.
  2. The adrenal gland shows two distinct regions, viz. thicker outer cortex and thinner inner medulla.
  3. The adrenal cortex consists of three distinct regions. The outer zona glomerulosa, the middle zona fasciculata and inner zona reticularis.
  4. Adrenal cortex produces corticoids. Corticoids is a collective term for many hormones, such as glucocorticoids, mineralocorticoids and steroid sex hormones.
  5. Adrenal medulla secretes adrenaline or epinephrine and noradrenaline or norepinephrine.

Question 27.
Why are reproductive organs called dual in function?
Answer:

  1. A pair of testes in males and a pair of ovaries in female both secrete hormones which are essential for sexual characters and function.
  2. Besides this, they also produce male and female gametes respectively. Therefore they are said to be dual in function.

Question 28.
What are male hormones? What is their source and functions?
OR
Write a short note on the functions of androgens.
Answer:

  1. Androgens are male hormones. The most significant androgen is the testosterone.
  2. Interstitial cells of Leydig present in the testis of mature man produce androgen. Androgens are steroid in chemical nature.
  3. Androgens regulate and stimulate the development, maturation and functions of the male reproductive organs like seminiferous tubules, epididymis, vas deferens, seminal vesicles, prostate glands and urethra.
  4. Androgens are made sex hormones. They produce secondary sexual characteristics. Low pitch of voice is produced due to changes in the vocal cords which take place due to testosterone, etc. They stimulate muscular growth and growth of facial and axillary hairs.
  5. The mental make up of a man like aggressiveness is due to testosterone.
  6. They stimulate seminiferous tubules for the process of spermatogenesis.

Question 29.
What are female sex hormones? What role do they play?
Answer:
(1) The ovaries secrete two steroid hormones viz., estrogen and progesterone.

(2) The estrogen is secreted by the developing ovarian follicles. It has many roles in stimulation of female reproductive functions and growth of ovaries, fallopian tubes, uterus and vagina.

(3) It also controls female secondary sexual characteristics like high pitch of voice, development of mammary glands, broadening of pelvis, growth of pubic hairs and deposition of subcutaneous fats to produce feminine stature.

(4) The estrogen also regulates female sexual behaviour.

(5) The empty Graafian follicle after ovulation is converted into a structure called corpus luteum which secretes a hormone known as progesterone. Progesterone is a gestational hormone which is essential for maintaining the pregnancy. It also acts on the mammary glands and stimulates them for lactation, milk synthesis and ejection.

Question 30.
Why is pancreas called a dual gland?
Answer:

  1. Pancreas is called a dual gland because it is exocrine as well as endocrine in nature.
  2. The exocrine pancreas secretes digestive enzymes through acini.
  3. The endocrine pancreas secretes hormones through its endocrine cells called Islets of Langerhans.

Question 31.
What are the hormones of pancreas? Describe the functions of pancreatic hormones.
OR
Pancreas plays an important role in controlling diabetes mellitus. Explain.
Answer:

  1. Islets of Langerhans consists of three types of cells known as a-cells, β-cells and δ-cells.
  2. α-cells secrete glucagon while β-cells secrete insulin. δ-cells secrete somatostatin.
  3. The glucagon is a hyperglycemic hormone. It is a peptide hormone which acts mainly on the liver cells. Here it stimulates hepatocytes for glycogenolysis (i.e. breakdown of glycogen) leading to increased level of blood glucose (i.e. hyperglycemia).
  4. It also stimulates gluconeogenesis (i.e. formation of glucose from non-carbohydrate sources). This in turn brings rise in blood glucose level or hyperglycemia.
  5. Glucagon reduces the cellular glucose uptake and utilisation.
  6. Insulin is also a peptide hormone, which plays a major role in maintenance of blood glucose level.
  7. Insulin stimulates hepatocytes and adipocytes for cellular glucose uptake and utilization.

Therefore glucose from the blood decreases causing hypoglycemia. This hormone helps in the conversion of glucose to glycogen (i.e. glycogenesis) that occurs in target cells.

Question 32.
How is blood glucose level maintained?
Answer:
The blood glucose level is maintained by the joint but antagonistic action of insulin and glucagon.
Insulin is hypoglycemic hormone while glucagon is hyper/glycemic hormone.

When there is excess sugar in the blood, more insulin is secreted by the pancreatic islets. This causes the conversion of blood glucose into glycogen. This process is known as glycogenesis. This causes decline in the level of glucose in the blood.

When there is less blood glucose level then the glucagon is secreted. It causes stored glycogen to be converted into glucose. This process is called glycogenolysis.

Question 33.
What happens when there is insufficiency or deficiency of insulin in the body?
Answer:

  1. Due to insufficiency, of insulin level there is prolonged hyperglycemia. This leads to diabetes mellitus.
  2. In this diabetic condition cells are unable to utilize glucose. Therefore, in a diabetic person blood glucose levels are high. The glucose is excreted in urine.
  3. The harmful compounds like ketone bodies are formed leading to ketosis.
  4. Diabetes can be treated by taking insulin injections or tablets (insulin therapy) or with hypoglycemic drugs.

Question 34.
Where are parathyroid glands located? What are their functions?
OR
Write a short note on the functions of Parathyroid hormone (PTH).
Answer:

  1. Parathyroid glands are located on the back side of the thyroid gland.
  2. There are two pairs of parathyroid glands. One pair of parathyroid is in each lobe of thyroid.
  3. Parathyroid glands secrete a peptide hormone known as parathromone or parathyroid hormone (PTH).
  4. The level of Ca++ in the blood regulates the secretion of PTH.
  5. PTH is hypercalcemic hormone, it increases blood calcium level. Thus the calcium balance is maintained by TCT and PTH.

Question 35.
What are the gastrointestinal hormones? Explain the function of each.
Answer:

  1. There are scattered endocrine cells in different parts of alimentary canal.
  2. These cells secrete four peptide hormones which are gastrin, secretin, cholecystokinin (CCK) and gastric inhibitory peptide (GIP).
  3. Gastrin stimulates gastric glands for the secretion of hydrochloric acid and pepsinogen.
  4. The secretin acts on exocrine pancreas and stimulates secretion of water and bicarbonate ions to form pancreatic juice.
  5. CCK acts on pancreas and gall bladder and stimulates the secretion of pancreatic enzymes and bile juice respectively.
  6. GIP inhibit gastric secretion and motility.

Maharashtra Board Class 12 Biology Important Questions Chapter 9 Control and Co-ordination

Question 36.
Name the hormone secreted by the heart. What is its function?
Answer:

  1. The atrial wall of the heart secrete a peptide hormone known as atrial natriuretic factor (ANF).
  2. When the blood pressure increases, ANF hormone is secreted.
  3. It causes dilation of the blood vessels.
  4. Blood then can easily flow with lesser resistance and hence BP decreases.

Question 37.
What are the hormones of kidney? What function do they carry out ?
Answer:

  1. The juxtaglomerular cells of the kidney produce a peptide hormone known as erythropoietin.
  2. Erythropoiesis stimulates bone marrow for the production of RBCs. It thus stimulates the process of erythropoiesis.
  3. Hormone calcitriol from kidney helps in the formation of bones.

Question 38.
Give importance of hypothalamus.
Answer:

  1. Hypothalamus is the controlling centre for hypophysis.
  2. Hypothalamus secretes releasing factors and inhibiting factors and hence regulate the secretions of hypophysis. E.g. Adrenocorticotropin hormone releasing factor or CRF; Thyrotropin releasing factor or TSHRF; GHRF and GHRIE i.e. Growth hormone releasing and release inhibiting factor, etc.
  3. Hypothalamus forms the hypothalamo- hypophyseal axis through which transportation of neurohormones take place.
  4. Hormones like vasopressin and oxytocin are secreted by neurosecretory cells of hypothalamus.
  5. Hypothalamus can register the internal changes in the body as it is a part of diencephalon and thus it accordingly brings about coordination in the body through endocrine system.

Question 39.
Write a brief account of releasing factors secreted by hypothalamus.
Answer:
(1) In the hypothalamus, there are several groups of neurosecretory cells which form different nuclei.

(2) These hypothalamic nuclei are supraoptic, paraventricular, dorso-median and ventromedian nuclei. These neurosecretory cells produce releasing and inhibiting factors.

(3) The hypothalamic neurohormones regulating the release of pituitary hormones are called releasing factors. The following are some of J the important releasing factors:

  • CRF or corticotropin releasing factor or ACTH releasing factor releases secretion of Adrenocorticotropin hormone (ACTH).
  • TRF or TSHRF (Thyroid stimulating hormone releasing factor) stimulates release of TSH.
  • FSH RF (Follicle stimulating hormone ; releasing factor) stimulates release of FSH.
  • GHRF (Growth hormone releasing factor) GHRIF (Growth hormone release inhibiting factor) act on release and regulation of growth hormone.
  • PRF (Prolactin releasing factor) and PRIF ; (Prolactin release inhibiting factor) act on release and regulation of prolactin.
  • MSHRF (Melanocyte stimulating hormone releasing factor) and MSH RIF (Melanocyte stimulating hormone release inhibiting factor) act on release and regulation of MSH.

Question 40.
Hormones are called chemical messengers and regulators. Explain.
Answer:

  1. Hormones bring about coordination in the body with the help of nervous system.
  2. Endocrine system and nervous system together form neuro-endocrine system.
  3. This system works in tune with the external and internal environmental changes.
  4. The hormones are either excitatory or inhibitory. They bring about the actions accordingly to keep the body in homeostasis or equilibrium.
  5. Almost all endocrine glands are controlled by negative feedback inhibition. Some glands are auto-regulatory. Therefore, the concentration of hormones cannot be in excess or in deficiency.
  6. Almost all the functions such as metabolism, growth, reproduction, etc. are under the control of hormones. Therefore hormones are called regulators and messengers.

Chart based/Table based questions

Question 1.
Draw a flow chart of – steps in generation and conduction of a nerve impulse.
Answer:
Maharashtra Board Class 12 Biology Important Questions Chapter 9 Control and Co-ordination 1

Question 2.
Enlist the names of following cranial nerves: I, II, VII, XII
Answer:

NumberNAMENATURE
IOlfactorySensory
IIOpticSensory
VIIFacialMixed
XIIhypoglossalMotor

Question 3.
Enlist the names of following cranial nerves : III, W VI, XI
Answer:

NumberNAMENATURE
IIIOcculomotorMotor
IVTrochlear/PatheticMotor
VIAbducensMotor
XISpinal accessoryMotor

Question 4.
Complete the table.

NumberTypeNo. of PairsRegion
————Cervical————-—————
T1 – T12————-12 pairs————-
L1 – L5————-5 pairsLower back
————-Sacral————-Pelvic
————–Coccygeal—————Tall region

Answer:

NumberTypeNo. of PairsRegion
C1 – C8Cervical8 pairsNeck
T1-T12Thoracic12 pairsThorax / Upper back
L1 -L5Lumber5 pairsLower back
S – S5Sacral5 pairsPelvic
Co1Coccygeal1 pairTall region

Question 5.
Write types of neuroglial cells of CNS and PNS in tabular form.
Answer:

CNS – glial cellsPNS – glial cellsFunctions
Oligodendrocytes [cells with few branches]Schwann cellsSecrete myelin sheath
Astrocytes [star-shaped and most abundant glial cells in CNS]Satellite cellsProtect, cushion and supply nutrients to nearby neurons. Help in maintaining blood-brain barrier.
Microglia

[small cells with few branches]

MacrophagesPhagocytosis
Ependymal cells lining the ventricles of brain [mostly columnar]Ependymal cells lining central canal of spinal cordSecrete cerebrospinal fluid

Question 6.
Enlist the various receptors found at various location in the body.
Answer:

ReceptorsTypeslocations
MechanoreceptorsThermoreceptorsSkin
Tango [touch and pressure] receptorsSkin
Tactile [light touch] receptorsSkin
ChemoreceptorsGustato receptorstongue
Olfacto receptorsOlfactory mucosa
PhotoreceptorsRod and cone cellsRetina of eye
PhonoreceptorsOrgan of CortiCochlea of internal ear
Stato receptorsCristae and maculaeSemicircular canals, utricle, saccule of internal ear

Question 7.
Sketch the concept maps for mechanism of vision and mechanism of hearing.
Answer:
Maharashtra Board Class 12 Biology Important Questions Chapter 9 Control and Co-ordination 2
Maharashtra Board Class 12 Biology Important Questions Chapter 9 Control and Co-ordination 3

Question 8.
Complete the table given below by putting a tick (✓) in the boxes where applicable.

ActionReflexVoluntary
1. Touching a hot object———-———–
2. Releasing saliva on smelling food———–————
3. Applying a car’s brakes in an emergency————————–
4. Blinking of eyes when a small insect touches the eye———–————

Answer:

ActionReflexVoluntary
1. Touching a hot object
2. Releasing saliva on smelling food
3. Applying a car’s brakes in an emergencycan be a conditioned feclex too.
4. Blinking of eyes when a small insect touches the eye————

Question 9.
Complete the following table

ActionReflexVoluntary
1. Optic nerve———–———–
2. Facial———–————
3. Hypoglossal————————–
4. Trigeminal———–————
5. Auditory
6. Glosso-pharyngeal

Answer:

ActionReflexVoluntary
1. Optic nerveSensorySense of vision and light
2. FacialMixedFacial expression, movement of neck, tongue, etc. and saliva secretion
3. HypoglossalMotorMovement of tongue
4. TrigeminalMixedSensation of touch, taste and jaw movements
5. AuditorySensoryHearing and equilibrium
6. Glosso-pharyngealMixedTaste, pharyngeal contractions ‘and saliva secretion

Diagram based questions

Question 1.
Sketch and label – nerve net of Hydra.
Answer:
Maharashtra Board Class 12 Biology Important Questions Chapter 9 Control and Co-ordination 4

Question 2.
Sketch and label – nervous system of Planaria
Answer:
Maharashtra Board Class 12 Biology Important Questions Chapter 9 Control and Co-ordination 5

Question 3.
Sketch and label – depolarization and repolarization along nerve.
Answer:
Maharashtra Board Class 12 Biology Important Questions Chapter 9 Control and Co-ordination 6

Question 4.
Sketch and label ultrastructure of synapse.
Answer:
Maharashtra Board Class 12 Biology Important Questions Chapter 9 Control and Co-ordination 7

Question 5.
Sketch and label – lateral view of brain.
Answer:
Maharashtra Board Class 12 Biology Important Questions Chapter 9 Control and Co-ordination 8

Maharashtra Board Class 12 Biology Important Questions Chapter 9 Control and Co-ordination

Question 6.
Sketch and label – functional areas of Brain?
Answer:
Maharashtra Board Class 12 Biology Important Questions Chapter 9 Control and Co-ordination 9

Question 7.
Sketch and label – ventral view of human brain.
Answer:
Maharashtra Board Class 12 Biology Important Questions Chapter 9 Control and Co-ordination 10

Question 8.
Sketch and label – ventricles of brain.
Answer:
Maharashtra Board Class 12 Biology Important Questions Chapter 9 Control and Co-ordination 11

Question 9.
Sketch and label T.S. of spinal cord.
Answer:
Maharashtra Board Class 12 Biology Important Questions Chapter 9 Control and Co-ordination 12

Question 10.
Sketch and label – formation of spinal nerve.
Answer:
Maharashtra Board Class 12 Biology Important Questions Chapter 9 Control and Co-ordination 13

Question 11.
Sketch and label – mechanism of hormone action.
Answer:
Maharashtra Board Class 12 Biology Important Questions Chapter 9 Control and Co-ordination 14
Maharashtra Board Class 12 Biology Important Questions Chapter 9 Control and Co-ordination 15

Question 12.
Sketch and label – V.S. of pituitary.
Answer:
Maharashtra Board Class 12 Biology Important Questions Chapter 9 Control and Co-ordination 16

Question 13.
Sketch and label – morphology of thyroid
Answer:
Maharashtra Board Class 12 Biology Important Questions Chapter 9 Control and Co-ordination 17

Question 14.
Sketch and label – histology of thyroid
Answer:
Maharashtra Board Class 12 Biology Important Questions Chapter 9 Control and Co-ordination 18

Question 15.
Sketch and label – parathyroid glands
Answer:
Maharashtra Board Class 12 Biology Important Questions Chapter 9 Control and Co-ordination 19

Question 16.
Sketch and label thymus.
Answer:
Maharashtra Board Class 12 Biology Important Questions Chapter 9 Control and Co-ordination 20

Question 17.
Sketch and label – adrenal gland.
Answer:
Maharashtra Board Class 12 Biology Important Questions Chapter 9 Control and Co-ordination 21

Long answer questions

Question 1.
Explain transmission of nerve impulse across a synapse.
OR
Explain how is impulse transmitted through a synapse?
Answer:

  1. The nerve impulse travels along the axon of the pre-synaptic neuron to the axon terminal.
  2. Pre-synaptic neurons or axons have several synaptic knobs at their ends or terminals.
  3. These knobs have membranous sacs, called synaptic vesicles having neurotransmitter molecules.
  4. When an impulse reaches a synaptic knob, voltage sensitive Ca++ channels open and calcium ions (Ca++) diffuse inward from the extracellular fluid.
  5. The increased calcium concentration inside the cells, initiates a series of events that help to fuse the synaptic vesicles with the cell membrane of pre-synaptic neuron, where they release their neurotransmitters by exocytosis.
  6. The neurotransmitters bind to the receptors of the post-synaptic cell,
  7. This action is either excitatory (stimulating) or inhibitory (slowing down/stopping).
  8. Once the impulse has been transferred across the synapse, the enzyme like acetyl cholinesterase destroys the
  9. neurotransmitter and the synapse is ready to receive a new impulse.

Question 2.
Explain transmission of nerve impulse along the axon.
OR
Describe the conduction of a nerve impulse in the neuron.
Answer:
1. Before conduction of nerve impulse, the cell membrane is in the polarized state.

2. When a stimulus is applied at a site on the polarised membrane, the membrane at that site becomes freely permeable to Na+.

3. This leads to a rapid influx of Na+ followed by the reversal of the polarity at that site, i.e., the outer surface of the membrane becomes negatively charged and the inner side becomes positively charged.

4. The polarity of the membrane at that site [site A] is thus reversed and hence depolarised. The electrical potential difference across the plasma membrane at the site of stimulation is called the action potential, which is in fact termed as a nerve impulse.

5. At sites immediately ahead [site B], the axon membrane has a positive charge on the outer surface and a negative charge on its inner surface. As a result, a current flows on the inner surface from site A to site B.
Maharashtra Board Class 12 Biology Important Questions Chapter 9 Control and Co-ordination 22

6. On the outer surface current flows from site B to site A to complete the circuit of current flow. Hence, the polarity at the site is reversed, and an action potential is generated at site B. Thus, the impulse (action potential) generated at site A arrives at site B.

7. The sequence is repeated along the length of the axon and consequently the impulse is conducted.

8. The rise in the stimulus-induced permeability to Na+ is extremely short-lived. It is quickly followed by a rise in permeability to K+.

9. Within a fraction of a second, K+ diffuses outside the membrane and restores the resting potential of the membrane at the site of excitation and the fibre becomes once more responsive to further stimulation.

Maharashtra Board Class 12 Biology Important Questions Chapter 9 Control and Co-ordination

Question 3.
Explain the structure of cerebrum. Structure of cerebrum
Answer:
Answer: Structure of cerebrum:

  1. Cerebrum is the largest part of the brain. It forms 80-85% volume of the brain.
  2. A median longitudinal fissure divides the cerebrum into two cerebral hemispheres. These hemispheres are interconnected by a thick band of transverse nerve fibres called corpus callosum.
  3. The outer part of cerebrum is called cerebral cortex while the inner part is called cerebral medulla.
  4. The roof of cerebrum is called pallium. Pallium is highly folded forming sulci and gyri. Sulci are depressions while gyrl are ridges. The gyri increase the surface area of cerebral cortex.
  5. The ventro-lateral walls of cerebrum are thickened and are called corpora striata.
  6. The cerebral cortex has three deep sulci, the central, lateral and parieto-occipital.
  7. These sulci divide the cerebral hemisphere into four lobes. These are frontal, parietal, occipital and temporal lobes. A fifth median lobe called insula or insular cortex is folded deep within the lateral sulcus.
  8. The central sulcus separates frontal and parietal lobes, the lateral sulcus separates parietal and temporal lobes and the parieto¬occipital sulcus separates parietal and occipital lobe.

Question 4.
Describe the structure and function of midbrain.
Answer:
1. Structure of midbrain:

  • Midbrain is the middle part of the brain situated between the forebrain and the hindbrain. It is present between the pons varolii and diencephalon.
  • It has two distinct regions : Corpora quadrigemina and crura cerebri.
  • Corpora quadrigemina consists of two pairs of lobes, viz., superior colliculi and inferior colliculi. These are located in the thick wall of midbrain.
  • Crura cerebri are thick bands of longitudinal nerve fibres, present on the floor of midbrain.

2. Functions of midbrain:

  • Inferior colliculi control and coordinate auditory reflexes.
  • Superior colliculi control head and eye movements.
  • Crura cerebri connect the cerebrum to cerebellum and spinal cord.

Question 5.
Give an account of structure of hindbrain.
Answer:
Structure of hindbrain:

  1. Hindbrain includes cerebellum, pons varolii and medulla oblongata.
  2. Cerebellum is 11% of the total brain and is the second largest part of the brain.
  3. It has three lobes, median vermis and lateral two cerebral hemispheres. It has outer grey and inner white matter.
  4. Cerebral cortex shows sulci and gyri. The inner white matter of cerebellar medulla shows arbor vitae or branching tree-like processes.
  5. Pons is the part that connects the two cerebral hemispheres. It has outer white and inner grey matter. Pons is made up of nerve fibres which form bridges between cerebrum and medulla oblongata.
  6. Medulla oblongata is the last part of the hindbrain which continues further as a spinal cord. It has outer white and inner grey matter.
  7. Its roof shows has posterior choroid plexus.
  8. Eight pairs of cranial nerves arise from medulla oblongata.

Question 6.
Describe T.S. of spinal cord.
Answer:

  1. Externally there are three meninges covering spinal cord Duramater, arachnoid membrane and pia mater.
  2. Dorsoventrally there are two fissures, the shallow dorsal or posterior fissure and the deeper ventral or anterior fissure.
  3. From dorsal fissure a dorsal septum extends inside.
  4. Neurocoel or central canal is situated in the centre of spinal cord.
  5. The central canal is filled with cerebro¬spinal fluid and is lined by cuboidal epithelial cells called layer of ependyma.
  6. There is inner grey and outer white matter in the spinal cord. This grey matter is in the shape of ‘H’ with two dorsolateral horns and two ventro-lateral horns.
  7. Dorsal horns form dorsal roots and ventral horns form ventral roots.
  8. White matter is divided into three columns, viz., the dorsal funiculi, ventral funiculi and lateral funiculi on either side.
  9. Ascending and descending tracts of nerve fibres arise from dorsal and ventral roots of the spinal cord. Ascending tracts are sensory while descending tracts are motor in nature.

Question 7.
What are the different types of reflexes?
Answer:
1. Based on the location of their action : Thereflexes are divided into somatic reflexes and visceral reflexes.

  • When effector is located in body structures such as skeletal muscles, it is called a somatic reflex.
  • When the effector is located in the visceral organs such as glands or smooth muscles then it is called a visceral reflex.

2. Based on the basis of number of neurons : Reflexes are of two types, viz. monosynaptic reflexes and polysynaptic reflexes.

  • Simple or monosynaptic reflexes are those in which one sensory and one motor neuron are involved in the reflex action.
  • Polysynaptic or complex reflexes are those when more than two neurons are involved in the reflex action.

3. Based on inheritance and experience of learning : The reflexes are subdivided into unconditional or inborn and conditional or acquired.

  • Unconditional or inborn reflexes are inborn or hereditary. They are permanent, never disappear and need no previous experience, e.g. blinking of eyes, suckling, swallowing, knee jerk, sneezing, coughing, etc.
  • Conditional or acquired reflexes are acquired during life by experience or learning. They are based on individual learning or experience. These are not heritable, temporary and may disappear or reappear, e.g. driving, cycling, etc.

Question 8.
Explain the mechanism of reflex action.
Answer:

  1. Mechanism of reflex action: There are series of sequential events in which reflex action is completed:
  2. Stimulus is picked up by any receptors, e.g. pricking of a needle in the hand, causes skin to be a receptor.
  3. Sensory impulse is formed in grey matter of spinal cord. It receives sensory impulse, interprets it and generates motor impulse.
  4. The cyton of motor neuron present in the ventral horn of grey matter and axon conducts motor impulse from spinal cord to effector organ. This is further carried by dendrites innervating the skin.
  5. Impulse is carried to the association neuron by axon of sensory neuron, when impulse reaches the end of the axon there is a synapse.
  6. Transmission takes place by releasing acetylcholine from the synaptic buttons at the end of the axon.
  7. It fills the synaptic gap and helps in chemical transmission of the impulse from axon of one neuron to dendron of the other neuron. Once the impulse reaches the dendrites of association neuron; axonic button releases an enzyme, acetylcholine esterase which neutralizes the acetylcholine and again a synaptic gap is formed. This mechanism helps to receive new impulse or avoid the mixing of different impulses.
  8. The association neuron receives sensory impulse, interprets it, analyses it and generates motor impulse. Motor impulse again travels through synapse between association neuron and motor neuron.
  9. Impulse travels through motor neuron and reaches the effector organ like skeletal muscles or the gland. The effector organ gives a proper response like contraction of the muscles or secretion by the gland.

Question 9.
Define receptors. Enlist different types of receptors.
Answer:
1. Receptors : Receptors are specialized cells, tissues or organs present in the body which receive different stimuli.

2. Types of receptors:

  • Receptors are of two types, viz. exteroceptors and interoceptors.-
  • Exteroceptors receive stimuli directly from the external environment. They are somatic in nature.
  • Interoceptors are located inside the body and are visceral in nature. They respond to internal changes in the body.
  • The various types of exteroceptors and interoceptors, their location and functions have been summarized in the table given below:
    Types of ExteroceptorsLocationFunction
    1. MechanoreceptorsTouch corpuscles in skinTangoreceptors Pressure Tactile receptors- Touch
    2. ThermoreceptorsSkinFrigido receptors (cold) Heat receptors (warmth)
    3. ChemoreceptorsTongue, nasal mucosaGustatoreceptors – Taste Olfactoreceptors- Smell
    4. Statoacoustic receptorsInternal earCochlea – Hearing Semicircular canals-Balance and equilibrium
    5. PhotoreceptorsRetina of the eyeRods and cones interpret images Rods-black and white image. Sensitive to dim light. Cones – Coloured image. Sensitive to bright light.

Question 10.
Describe the different parts of human eye.
OR
Describe briefly the structure of eye.
Answer:

  1. Human eyes are a pair of organs located in sockets of the skull called orbits.
  2. Eyeball is spherical and has three layers.
  3. Sclera is the outer layer of dense connective tissue with anterior transparent cornea.
  4. Choroid is the middle layer. It is bluish in colour containing many blood vessels. The anterior region is thick and forms the ciliary body. Posterior 2/3rd region is thinner.
  5. Iris is the forward segment of the ciliary body which is pigmented and opaque. This part is the visible coloured portion of the eye.
  6. Lens is present anteriorly inside the iris and is held in position by the ligaments of ciliary body.
  7. The aperture surrounded by the iris in front of the lens is known as pupil. The movement of the pupil is regulated by the muscle fibres of iris.
  8. The innermost layer of the eye is the retina having three sub-layers formed by ganglion cells, bipolar cells and photoreceptor cells, which are sensitive to light.
  9. There are two types of photoreceptor cells, viz. rods and cones containing light sensitive proteins. They are termed as photo pigments, rhodopsin which is a derivative of vitamin A (in rods) and iodopsin (in cones).
  10. The cones are responsible for daylight or photopic vision and colour vision. The rods function in dim light giving scotopic vision.
  11. The cones are of three types, each containing its own characteristic photopigments that respond to red, green and blue lights.
  12. The optic nerve leaves the eye at a point slightly away from the median posterior pole of the eyeball. In this region, the rods and cones are absent therefore this region is known as blind spot. Macula lutea, a yellowish pigmented spot is present lateral to the blind spot.
  13. Fovea is a central pit present beside it. Fovea is a thinned out portion of the retina where only the cones are densely packed and therefore have greatest visual acuity (resolution).
  14. A space between the cornea and the lens is called aqueous chamber. It contains a thin watery fluid known as aqueous humor.
    Maharashtra Board Class 12 Biology Important Questions Chapter 9 Control and Co-ordination 24

Question 11.
Describe the internal structure of human ear.
OR
Ear is one of the important sense organs known for its role in hearing and balancing. Describe those structures present in the internal ear which helps in these functions.
Answer:

  1. The ears are the auditory sensory organs, also involved in maintaining equilibrium of the body.
  2. The ear is composed of three divisions namely the outer ear, middle ear and internal ear.
  3. The external ear consists of the pinna and external auditory meatus (canal). The pinna is for the collection of sound waves coming from the environment. The external auditory canal is the circular tube leading inside up to the eardrum or tympanic membrane.
  4. The tympanic membrane or eardrum is formed of connective tissues having outer skin cover and inner mucus membrane.
  5. The middle ear consists of chain of three ossicles called malleus, incus and stapes, The malleus is attached to the tympanic membrane and the stapes is connected to the oval window of the internal ear. They help in the transmission of sound waves from external auditory canal to internal ear.
  6. Connecting middle ear with the pharynx is eustachian tube which helps in equalizing the air pressure on either side of the tympanic membrane.
  7. The internal ear is fluid filled structure called labyrinth. It has two parts, bony and the membranous labyrinth.
  8. The outer bony labyrinth is formed by the series of channels in which the membranous abyrinth containing endolymph fluid is present.
  9. The membranes consist of coiled cochlea, the reissner’s membrane and basilar membranes. These membranes divide the surrounding perilymph filled bony labyrinth into an upper scala vestibule and a lower scala tympani.
  10. The space within cochlea which is known as scala media is filled with endolymph. The scala vestibule ends at the oval window at the base of cochlea.
  11. The scala tympani terminates at the round window which opens to the middle ear. The organ of corti is located on the basilar membrane. It contains the hair cells which act as auditory receptors.
  12. The hair cells are columnar cells present in rows. The basal ends of the hair cells are in close contact with the afferent nerve fibres while their apical end contains numerous cilia. A thin elastic membrane projects above the rows of the hair cells called tectorial membrane.
  13. Above the cochlea, the internal ear also contains vestibular apparatus. It consists of three semicircular canals and the otolith organ formed of the sacculus and utriculus. The semicircular canals lie in different plane at right angles to each other and are suspended in the perilymph.
  14. The bases of canals are swollen and are called ampullae, which contain a projecting ridge known as crista ampullaris which contain hair cells.
  15. The sacculus and utriculus also have projecting ridge called macula. The crista and macula are the specific receptors of vestibular apparatus. They are responsible for maintenance of body posture and the balance.
    Maharashtra Board Class 12 Biology Important Questions Chapter 9 Control and Co-ordination 25

Question 12.
Write an account of position and structure of pituitary gland.
Answer:
Pituitary gland (Hypophysis):
I. Position : Pituitary or Hypophysis is located on the ventral side of brain below the hypothalamus. Infundibulum or hypophyseal stalk attaches pituitary to hypothalamus just behind the optic chiasma. It is well protected in sella turcica which is a depression of the sphenoid bone of the skull.

II. Morphological structure of pituitary- gland: The pituitary gland shows two distinct regions : Anterior lobe or adenohypophysis and posteriorlobe or neuro-hypophysis.
(1) Adenohypophysis or Anterior lobe : It is the largest lobe of the gland and forms about 75% of pituitary gland. It develops as an outgrowth called Rathke’s pouch from the roof of embryonic buccal cavity. It has three divisions, viz. pars tuberalis, pars distalis and pars intermedia.

(i) Pars tuberalis : Tubular region present below the hypothalamus is known as pars tuberalis. It is like a collar around the infundibulum. It is non-secretory in nature.
(ii) Pars distalis : The largest anterior region which is secretory in nature is called pars distalis. It is made up of loose cords of epitheloid secretory cells which are separated by reticular connective tissue containing blood sinusoids. It is connected to the hypothalamus by portal system formed by blood sinusoids.
(iii) Pars intermedia : The narrow cleft between the pars distalis and pars nervosa of neuro – hypophysis is called the intermediate part or pars intermedia. It is reduced, less developed and non-functional in human being.

(2) Neuro-hypophysis or Posterior lobe : The posterior lobe of the pituitary which is attached to hypothalamus by infundibular stalk is called neuro-hypophysis. It is smaller and constitute 25% of pituitary. It has the following three parts:

  1. Median eminence : The swollen median part of the hypothalamus where infundibulum gets attached is called median eminence.
  2. Infundibulum : Infundibulum is the hypophyseal stalk that helps in attachment of pituitary gland to the hypothalamus. It contains mainly the axonic fibres of neurosecretory cells present in hypothalamus. It forms the major connection of hypothalamo-hypophysis axis.
  3. Pars nervosa : The lowermost, larger region of neuro-hypophysis that contains axons is called pars nervosa. It acts as a neurohaerhal organ and contains specialized cells called pituicytes.

Maharashtra Board Class 12 Biology Important Questions Chapter 9 Control and Co-ordination

Question 13.
In a person, Pars distalis part of the Pituitary gland is not producing hormones in sufficient quantity. Explain the effects it will produce with respect to the different hormones.
OR
Give names and functions of hormones secreted by adenohypophysis.
Answer:
Pars distalis of the pituitary gland produces following hormones:
GH, ACTH, TSH, FSH, LH/ICSH, LTH and MSH. If these hormones are produced in deficient quantities, following disorders are seen.
1. GH:
(a) Hyposecretion of GH in childhood leads to dwarfism. Frohlich dwarf or Lorain dwarf may be produced based on mental capacity. Hyposecretion in adulthood causes Simmonds’s disease.
(b) Hypersecretion of GH in childhood causes gigantism and in adulthood it causes acromegaly.

2. TSH :
(a) Hyposecretion of TSH leads to thyroid atrophy.
(b) Hypersecretion of TSH causes excessive functioning of thyroid gland.

3. ACTH :
(a) Hyposecretion of ACTH causes Addisons’ disease, in which adrenal gland shows failure of functions.
(b) Hypersecretion of ACTH causes Cushing’s disease, in which the adrenal cortex undergoes excessive growth.

4. FSH : Hyposecretion of FSH leads to infertility in both the sexes. Hypersecretion of FSH in females cause disturbances in menstrual cycle.

5. LH/ICSH : Hyposecretion of LH in females will cause lack of ovulation. Hyposecretion of ICSH in males cause reduction in masculinity. Sperm production may be affected. Hypersecretion of LH/ICSH can cause disturbances in reproductive cycles.

6. LTH : Corpus luteum is not maintained due to lesser amount of LTH. Lactogenesis will also hamper if there is hyposecretion of LTH.

Question 14.
Describe the hormones of neuro¬hypophysis.
Answer:
Hormones of neuro-hypophysis : Neuro-hypophysis does not secrete any hormone itself but stores the hormones which are secreted by hypothalamic neurons. It stores and releases the following hormones, viz. ADH, Oxytocin and coherin.
1. Anti Diuretic Hormone (ADH) or Vasopressin:

  • ADH brings about anti-diuretic action and also increases blood pressure.
  • It is a regulatory hormone which plays a major role in osmoregulation.
  • It increases the permeability of distal convoluted tubule or collecting tubules of uriniferous tubules of kidney.
  • Higher ADH levels decrease the urine output and helps for water conservation. It helps in the absorption of water from the ultrafiltrate thus regulates the water balance of body fluids.
  • ADH also controls constriction of arterioles and increases blood pressure in kidney which facilitates ultra filtration. Therefore it is also called vasopressin.
  • ADH is regulated by increase or decrease of osmotic pressure of blood in a feedback manner.
  • The osmotic pressure is detected by osmoreceptors in the hypothalamus.

2. Oxytocin (Birth hormone):

  • Oxytocin helps in parturition.
  • It is a powerful stimulant of contraction of uterine myometrium at the end of gestation due to which the labour is initiated.
  • It also stimulates myoepithelial cells of mammary glands for milk ejection during lactation.
  • It also helps in fertilization by powerful contractions of the uterine musculature to drive the sperms upward towards fallopian tubes.
  • Oxytocin also excites musculature of gallbladder, ureters, urinary bladder, intestine, etc. for proper functioning of these organs.

3. Coherin : Coherin induces prolonged, rhythmic integrated contractions of the jejunum.

Question 15.
Describe the morphology of thyroid gland.
Or
With the help of a suitable diagram describe the structure of thyroid gland.
Answer:
Morphology of thyroid gland:

  1. Thyroid is the largest endocrine gland in the body.
  2. It weighs about 25 to 30 g and measures about 5 cm in length and 3 cm in width.
  3. It is located in the neck region anteriorly just below the larynx and situated ventrolaterally to the trachea.
  4. The thyroid is derived from the endoderm of the embryo.
  5. The thyroid can vary in size as per age, sex and diet.
  6. It is reddish brown, bilobed and highly vascular
  7. The two lobes are joined by connective tissue called isthmus which is located at 2nd to 4th tracheal cartilage.
  8. Therefore, the right and left lobe of thyroid are seen on both sides of the trachea.
  9. The gland is H-shaped having butterfly-like appearance.
  10. The structural and functional units of thyroid gland are the thyroid follicles.
  11. From the outer surface there lies a connective tissue capsule. A number of septa arise from the connective tissue which are called trabeculae. They divide the gland into lobules. Each lobule has follicles which store hormone. The number of follicles are about three million.

Question 16.
Describe neurohormonal regulation of pituitary and thyroid gland.
Answer:
Pituitary gland is directly under the influence of neurohormones of hypothalamus while thyroid is indirectly influenced.
I. Neurohormonal regulation of pituitary:

  1. Hypothalamus secretes releasing factors and inhibiting factors and hence regulate the secretions of pituitary (hypophysis).
  2. Hypothalamus forms the hypothalamohypophysial axis through which transportation of neuro-hormones take place.
  3. Pituitary secretes a variety of hormones which influence other endocrine glands of the body. E.g. GH, PRL, TSH, ACTH, Gonadotropins

II. Neurohormonal regulation of thyroid :

  1. Hypothalamus secretes TRF [Thyrotropin releasing factor] which influences the anterior pituitary to release TSH.
  2. TSH in turn stimulates thyroid follicles to produce and release two thyroid hormones – T3 and T4. (Thyroxin)
  3. Increase in T3 and T4 triggers negative feedback mechanism that stops the secretion of TRF.

(4) As the pituitary does not get the signal in the form of TRF TSH secretion is stopped.

Question 17.
Name the hormones secreted by the adrenal cortex and state their role.
Answer:
Adrenal cortex secretes 3 types of corticoids – mineralocorticoids, glucocorticoids and sex corticoids.
I. Mineralocorticoids:

  1. The mineralocorticoids regulate ionic and osmotic balance, by regulating the amounts of electrolyte and water.
  2. Aldosterone is the main mineralocorticoid that acts on the renal tubules.
  3. Aldosterone stimulates the re-absorption of Na+ and water and excretion of K+ and phosphate ions.
  4. The aldosterone helps in the maintenance of electrolytes, body fluid volume, osmotic pressure and blood pressure.

II. Glucocorticoids:

  1. Cortisol is the main glucocorticoid. Cortisol stimulates many metabolic reactions such as gluconeogenesis, lipolysis and proteinolysis.
  2. It inhibits cellular uptake and utilization of amino acids.
  3. Cortisol also plays an important role in maintaining the cardiovascular system and kidney functions.
  4. It is also involved in anti-inflammatory reactions and suppresses the immune response.
  5. Cortisol stimulates the RBC production.

III. Sex corticoids (Gonadocorticoids).

  1. Sex corticoids, Androgens and estradiols are produced by the adrenal cortex.
  2. In males, they have a role in development and maintenance of external sex characters.
  3. Excess sex corticoids in female causes adrenal virilism and hirsutism (excess hair on face)
  4. Excess sex corticoids in males causes gynaecomastia i.e. enlarged breast.

Maharashtra Board Class 12 Biology Important Questions Chapter 8 Respiration and Circulation

Balbharti Maharashtra State Board 12th Biology Important Questions Chapter 8 Respiration and Circulation Important Questions and Answers.

Maharashtra State Board 12th Biology Important Questions Chapter 8 Respiration and Circulation

Multiple-choice Questions

Question 1.
The nasal cavity is divided into right and left nasal chambers by a …………………..
(a) sphenoid
(b) palatine
(c) mesethmoid
(d) zygomatic
Answer:
(c) mesethmoid

Question 2.
The right lung is divided into …………………..
(a) 3 lobes
(b) 2 lobes
(c) 4 lobes
(d) 6 lobes
Answer:
(a) 3 lobes

Maharashtra Board Class 12 Biology Important Questions Chapter 8 Respiration and Circulation

Question 3.
Carbon dioxide is carried in the blood mainly as …………………..
(a) sodium carbonate
(b) sodium bicarbonate
(c) carbaminohaemoglobin
(d) carbonic acid
Answer:
(b) sodium bicarbonate

Question 4.
Transport of oxygen is carried out by …………………..
(a) plasma
(b) lungs
(c) RBCs
(d) nostrils
Answer:
(c) RBCs

Question 5.
Respiration taking place at the alveoli of lungs is called …………………..
(a) internal respiration
(b) external respiration
(c) cellular respiration
(d) tissue respiration
Answer:
(b) external respiration

Question 6.
The volume of air inspired or expired during normal breathing is …………………..
(a) ERV
(b) IRV
(c) TV
(d) VC
Answer:
(c) TV

Question 7.
What is the partial pressure of oxygen and carbon dioxide respectively in the atmospheric air?
(a) PPO2 159 mmHg, PPCO22 0.3 mmHg
(b) PPO2 104 mmHg, PPCO2 40 mmHg
(c) PPO2 40 mmHg, PPCO2 45 mmHg
(d) PPO2 95 mmHg, PPCO2 40 mmHg
Answer:
(b) PPO2 104 mmHg, PPCO2 40 mmHg

Question 8.
The vital capacity of human lung is equal to …………………..
(a) 3500 ml
(b) 4600 ml
(c) 500 ml
(d) 1200 ml
Answer:
(b) 4600 ml

Question 9.
The exchange of gases between alveolar air and alveolar capillaries occurs by …………………..
(a) osmosis
(b) active transport
(c) absorption
(d) diffusion
Answer:
(d) diffusion

Question 10.
Oxygen dissociation curve will shift to right on the decrease of …………………..
(a) acidity
(b) carbon dioxide concentration
(c) temperature
(d) pH
Answer:
(d) pH

Question 11.
Respiratory organs in scorpion are …………………..
(a) gills
(b) book lungs
(c) skin
(d) book gills
Answer:
(b) book lungs

Question 12.
Breakdown of alveoli of lungs resulting in reducing surface area for gas exchange is known as …………………..
(a) emphysema
(b) sneezing
(c) pneumonia
(d) tuberculosis
Answer:
(a) emphysema

Question 13.
During inspiration, the diaphragm …………………..
(a) relaxes
(b) contracts
(c) expands
(d) shows no change
Answer:
(b) contracts

Question 14.
Over inflation of the lungs is prevented due to …………………..
(a) Bohr’s effect
(b) Conditioned reflex
(c) Hering-Breuer reflex
(d) Haldane effect
Answer:
(c) Hering-Breuer reflex

Question 15.
Which of the following prevents collapsing of trachea?
(a) Muscles
(b) Diaphragm
(c) Ribs
(d) Cartilaginous rings
Answer:
(d) cartilaginous rings

Question 16.
Which one of the following produces antibodies ?
(a) Monocytes
(b) Erythrocytes
(c) Lymphocytes
(d) Monocytes
Answer:
(c) Lymphocytes

Question 17.
Plasma protein which initiate blood coagulation is …………………..
(a) prothrombin
(b) fibrinogen
(c) thrombin
(d) fibrin
Answer:
(a) prothrombin

Question 18.
The covering of heart is …………………..
(a) perichondrium
(b) pericardium
(c) periosteum
(d) peritoneum
Answer:
(b) pericardium

Question 19.
Left atrioventricular aperture is guarded by …………………..
(a) tricuspid valve
(b) Eustachian valve
(c) bicuspid valve
(d) semilunar valve
Answer:
(c) bicuspid valve

Question 20.
The pulmonary trunk and systemic aorta are joined by …………………..
(a) chordae tendinae
(b) columnae carnae
(c) ligamentum arteriosum
(d) Purkinje fibres
Answer:
(c) ligamentum arteriosum

Question 21.
Atrioventricular node is located in …………………..
(a) left atrium
(b) right atrium
(c) left ventricle
(d) right ventricle
Answer:
(b) right atrium

Question 22.
…………………. is most commonly used to feel pulse.
(a) Radial vein
(b) Brachial artery
(c) Brachial vein
(d) Radial artery
Answer:
(d) Radial artery

Question 23.
QRS is related to …………………..
(a) atrial contraction
(b) ventricular contraction
(c) atrial relaxation
(d) ventricular relaxation
Answer:
(b) ventricular contraction

Question 24.
Blood is a fluid connective tissue derived from …………………..
(a) ectoderm
(b) mesoderm
(c) endoderm
(d) epithelium
Answer:
(b) mesoderm

Question 25.
What is the increase in number of RBCs called?
(a) Erythropoiesis
(b) Polycythaemia
(c) Erythrocytopenia
(d) Erythroblastosis
Answer:
(b) Polycythaemia

Question 26.
What is the increase in the number of WBCs called?
(a) Leucopoiesis
(b) Leukopenia
(c) Leucocytosis
(d) Leukaemia
Answer:
(c) Leucocytosis

Question 27.
In which of the following diseases there is uncontrolled increase in number of WBCs ?
(a) Leucopoiesis
(b) Leukopenia
(c) Leucocytosis
(d) Leukaemia
Answer:
(d) Leukaemia

Question 28.
What is the decrease in the number of WBCs called?
(a) Leucopoiesis
(b) Leukopenia
(c) Leucocytosis
(d) Leukaemia
Answer:
(b) Leukopenia

Question 29.
Which is the correct arrangement of types of WBCs with respect to their number in blood?
(Consider Neutrophil = N, Eosinophil = E, Basophil = B, Monocyte = M and Lymphocyte = L)
(a) NLMEB
(b) BEMLN
(c) NEBLM
(d) MEBLN
Answer:
(a) NLMEB

Question 30.
Which is the correct order in which the proteins participate in clotting of blood?
(a) Prothrombinase → Prothrombin → Thromboplastin → Thrombin
(b) Thromboplastin → Prothrombinase → Prothrombin → Thrombin
(c) Prothrombin → Thromboplastin → Thrombin → Prothrombinase
(d) Thrombin → Prothrombin → Thromboplastin → Prothrombinase
Answer:
(b) Thromboplastin → Prothrombinase → Prothrombin → Thrombin

Question 31.
Decrease in platelet count is called …………………..
(a) thrombocytopenia
(b) thrombocytosis
(c) thrombokinase
(d) thromboplastin
Answer:
(a) thrombocytopenia

Question 32.
Atrioventricular groove is also called a …………………..
(a) foramen ovale
(b) ligamentum arteriosum
(c) coronary sulcus
(d) ductus arteriosus
Answer:
(c) coronary sulcus

Question 33.
The coronary sinus opens into the …………………..
(a) left atrium
(b) right atrium
(c) left ventricle
(d) right ventricle
Answer:
(b) right atrium

Question 34.
Name the valve from the following that guards the opening of inferior vena cava.
(a) Tricuspid valve
(b) Semilunar valve
(c) Eustachian valve
(d) Thebesian valve
Answer:
(c) Eustachian valve

Question 35.
Name the valve from the following guarding the opening of coronary sinus …………………..
(a) Thebesian valve
(b) Eustachian valve
(c) Tricuspid valve
(d) Semilunar valve
Answer:
(a) Thebesian valve

Question 36.
What is an oval aperture in the interatrial septum of the foetus called?
(a) Fossa ovalis
(b) Foramen ovalis
(c) Ligamentum arteriosum
(d) Ductus arteriosus
Answer:
(b) Foramen ovalis

Question 37.
What is the meaning of stroke volume?
(a) Amount of blood in the body
(b) Pressure of contraction of heart
(c) Amount of blood put out of the ventricles in one minute
(d) Amount of blood put out of the ventricles in one beat
Answer:
(d) Amount of blood put out of the ventricles in one beat

Question 38.
How much amount of blood is put out of the heart during one minute?
(a) Equal to cardiac output
(b) Equal to stroke volume
(c) Equal to half of blood volume
(d) Equal to quarter of blood volume
Answer:
(a) Equal to cardiac output

Question 39.
What is the time taken for one cardiac cycle of normal human being?
(a) 0.1 second
(b) 0.3 second
(c) 0.4 second
(d) 0.8 second
Answer:
(d) 0 .8 second

Question 40.
Deposition of fatty substances in the lining of arteries results in …………………..
(a) arteriosclerosis
(b) atherosclerosis
(c) hyperglycemia
(d) hypotension
Answer:
(b) atherosclerosis

Maharashtra Board Class 12 Biology Important Questions Chapter 8 Respiration and Circulation

Question 41.
Largest number of white blood corpuscles are …………………..
(a) eosinophils
(b) basophils
(c) neutrophils
(d) monocytes
Answer:
(c) neutrophils

Question 42.
Which of the following animal have open circulatory system?
(a) Earthworm
(b) Cockroach
(c) Frog
(d) Rabbit
Answer:
(b) Cockroach

Question 43.
Which of the following leucocytes have unlobed nucleus?
(a) lymphocyte
(b) eosinophils
(c) neutrophils
(d) basophils
Answer:
(a) lymphocyte

Question 44.
Carbonic anhydrase is found in …………………..
(a) WBC
(b) RBCs
(c) thrombocytes
(d) blood plasma
Answer:
(b) RBCs

Question 45.
The typical Lubb – Dup sounds heard in the heart of a healthy person are due to …………………..
(a) closing of cuspid valves followed by the closing of the semilunar valves
(b) closing of semilunar valves
(c) closing of tricuspid valves
(d) closing of bicuspid valves
Answer:
(a) closing of cuspid valves followed by the closing of the semilunar valves

Match the columns

Question 1.

AnimalRespiratory organ
(1) Fishes(a) Trachea
(2) Birds/Reptiles(b) Moist cuticle
(3) Insects(c) Gills
(4) Earthworm(d) Lungs

Answer:

AnimalRespiratory organ
(1) Fishes(c) Gills
(2) Birds/Reptiles(d) Lungs
(3) Insects(a) Trachea
(4) Earthworm(b) Moist cuticle

Question 2.

Respiratory organsAlternative name
(1) Larynx(a) Lid of larynx
(2) Trachea(b) Air sacs
(3) Alveoli(c) Sound box
(4) Epiglottis(d) Windpipe

Answer:

Respiratory organsAlternative name
(1) Larynx(c) Sound box
(2) Trachea(d) Windpipe
(3) Alveoli(b) Air sacs
(4) Epiglottis(a) Lid of larynx

Question 3.

Respiratory capacitiesRespiratory volumes
(1) Residual volume (RV)(a) 500 ml
(2) Vital capacity (VC)(b) 2000 – 3000 ml
(3) Tidal volume (TV)(c) 1100 – 1200 ml
(4) Inspiratory reserve volume (IRV)(d) 4100 – 4600 ml

Answer:

Respiratory capacitiesRespiratory volumes
(1) Residual volume (RV)(c) 1100 – 1200 ml
(2) Vital capacity (VC)(d) 4100 – 4600 ml
(3) Tidal volume (TV)(a) 500 ml
(4) Inspiratory reserve volume (IRV)(b) 2000 – 3000 ml

Question 4.

DiseaseSymptoms
(1) Asthma(a) Fully blown out alveoli
(2) Bronchitis(b) Inflammation of lungs with cough and fever
(3) Emphysema(c) Spasm of Bronchial muscles
(4) Pneumonia(d) Inflammation of bronchi

Answer:

DiseaseSymptoms
(1) Asthma(c) Spasm of Bronchial muscles
(2) Bronchitis(d) Inflammation of bronchi
(3) Emphysema(a) Fully blown out alveoli
(4) Pneumonia(b) Inflammation of lungs with cough and fever

Question 5.

Valves in heartLocation
(1) Bicuspid/Mitral valve(a) Opening of inferior vena cava
(2) Tricuspid valve(b) Opening of coronary sinus
(3) Eustachian valve(c) Left atrioventricular aperture
(4) Thebesian valve(d) Right atrioventricular aperture

Answer:

Valves in heartLocation
(1) Bicuspid/Mitral valve(c) Left atrioventricular aperture
(2) Tricuspid valve(d) Right atrioventricular aperture
(3) Eustachian valve(a) Opening of inferior vena cava
(4) Thebesian valve(b) Opening of coronary sinus

Question 6.

Blood vesselFunctions
(1) Pulmonary aorta(a) Carries oxygenated blood to left atrium
(2) Superior vena cava(b) Carries oxygenated blood to all body parts
(3) Pulmonary vein(c) Carries deoxygenated blood from upper parts of body to right atrium
(4) Aorta(d) Carries deoxygenated blood to lungs

Answer:

Blood vesselFunctions
(1) Pulmonary aorta(d) Carries deoxygenated blood to lungs
(2) Superior vena cava(c) Carries deoxygenated blood from upper parts of body to right atrium
(3) Pulmonary vein(a) Carries oxygenated blood to left atrium
(4) Aorta(b) Carries oxygenated blood to all body parts

Question 7.

CellsFunctions
(1) T-lymphocytes(a) Phagocytic in function
(2) Neutrophils(b) Responsible for Humoral immunity
(3) Eosinophils/Acidophils(c) Responsible for cell-medicated immunity
(4) B-lymphocytes(d) Anti-allergic [Antihistamine] in function

Answer:

CellsFunctions
(1) T-lymphocytes(c) Responsible for cell-medicated immunity
(2) Neutrophils(a) Phagocytic in function
(3) Eosinophils/Acidophils(d) Anti-allergic [Antihistamine] in function
(4) B-lymphocytes(b) Responsible for Humoral immunity

Question 8.

Waves recorded in ECGHeart activity
(1) P wave(a) Ventricular repolarization
(2) QRS complex(b) Atrial depolarization
(3) T wave(c) Isoelectric segment
(4) ST segment(d) Ventricular depolarization

Answer:

Waves recorded in ECGHeart activity
(1) P wave(b) Atrial depolarization
(2) QRS complex(d) Ventricular depolarization
(3) T wave(a) Ventricular repolarization
(4) ST segment(c) Isoelectric segment

Question 9.

Events in cardiac cycleTime duration
(1) Atrial systole(a) 0.3 second
(2) Atrial diastole(b) 0.5 second
(3) Ventricular systole(c) 0.1 second
(4) Ventricular diastole(d) 0.7 second

Answer:

Events in cardiac cycleTime duration
(1) Atrial systole(c) 0.1 second
(2) Atrial diastole(d) 0.7 second
(3) Ventricular systole(a) 0.3 second
(4) Ventricular diastole(b) 0.5 second

Classify the following to form Column B as per the category given in Column A

Question 1.
Classify the following composition of blood plasma given below as per Column ‘A’ and complete Column ‘B’. Select from the given options
(i) Serum albumin
(ii) Bicarbonates
(iii) Urea
(iv) Sulphates of sodium
(v) Fibrinogen
(vi) Uric acid

Column AColumn B
(1) Plasma proteins————
(2) Nitrogenous waste————
(3) Inorganic salts————

Answer:

Column AColumn B
(1) Plasma proteinsSerum albumin Fibrinogen
(2) Nitrogenous wasteUrea, Uric acid
(3) Inorganic saltsBicarbonates, Sulphates of sodium

Question 2.
Classify the following animals having different respiratory organs given below as per Column ‘A’ and complete Column ‘B’.
Select from the given options:
(i) Scorpion
(ii) Reptiles
(iii) Amphibian tadpoles of frog
(iv) Spiders
(vi) Salamanders
(v) Birds

Column AColumn B
(1) External gills————
(2) Book lungs————
(3) Lungs————

Answer:

Column AColumn B
(1) External gillsAmphibian tadpoles of frog, Salamanders
(2) Book lungsScorpion, Spiders
(3) LungsReptiles, Birds

Maharashtra Board Class 12 Biology Important Questions Chapter 8 Respiration and Circulation

Question 3.
Classify the following disorders of respiratory system given below as per Column A and complete Column ‘B’. Select from the given options:
(i) Pneumonia
(ii) Asbestosis
(iii) Emphysema
(iv) Laryngitis
(v) Chronic bronchitis
(vi) Silicosis

Column AColumn B
(1) Occupational disorders————
(2) Disorders due to smoking and air pollution————
(3) Disorders due to viruses and bacteria————

Answer:

Column AColumn B
(1) Occupational disordersAsbestosis, Silicosis
(2) Disorders due to smoking and air pollutionEmphysema, Chronic bronchitis
(3) Disorders due to viruses and bacteriaPneumonia, Laryngitis

Question 4.
Classify the following white blood corpuscles given below as per Column A and complete Column ‘B’. Select from the given options:
(i) Eosinophils
(ii) T-lymphocytes
(iii) Neutrophils
(iv) Basophils
(v) B-lymphocytes
(vi) Monocytes

Column AColumn B
(1) Phagocytic cells————
(2) Cells involved in giving immune response————
(3) Cells that increase during allergic and anti-allergic responses————

Answer:

Column AColumn B
(1) Phagocytic cellsNeutrophils Monocytes
(2) Cells involved in giving immune responseT-lymphocytes-B-lymphocytes
(3) Cells that increase during allergic and anti-allergic responsesEosinophils, Basophils

Very Short Answer Questions

Question 1.
How many molecules of ATP are formed when one molecule of glucose is oxidized?
Answer:
38 molecules of ATP are formed when one molecule of glucose is oxidized.

Question 2.
What are the three regions of nasal chamber?
Answer:
Vestibule, respiratory part and sensory part are the three regions of nasal chamber.

Question 3.
What is meant by respiratory cycle?
Answer:
Alternate inspiration and expiration together make one respiratory cycle.

Question 4.
Why is it dangerous to sleep in a garage where automobiles have running engines?
Answer:
It is dangerous to sleep in a garage where automobiles have running engines because it may cause carbon monoxide poisoning.

Question 5.
In which form major part of CO2 is transported in the blood?
Answer:
CO2 is transported in the blood in the form of sodium and potassium bicarbonates.

Question 6.
Which are the parts of plant that help in the process of gaseous exchange?
Answer:
The parts of plants that help in the process of gaseous exchange are stomata, lenticels, etc.

Question 7.
Which respiratory membranes help in gaseous exchange between the alveolar air and the blood?
Answer:
The layer of squamous epithelium lining the alveolus, basement membrane and a layer of squamous epithelium lining the capillary wall help in gaseous exchange between the alveolar air and the blood.

Question 8.
When will the oxygen dissociation curve shift towards the right?
Answer:
The oxygen dissociation curve will shift towards the right due to increase in H+ concentration, increase in PPCO2 rise in temperature and rise in DPG (2, 3 diphosphoglycerate), formed in RBCs during glycolysis.

Question 9.
What is the action of carbonic anhydrase in the RBCs of blood?
Answer:
In the RBCs, CO2 combines with water in the presence of a Zn containing enzyme, carbonic anhydrase to form carbonic acid. In the presence of carbonic anhydrase carbonic acid immediately dissociates into HCO3and H+ ions leading to large accumulation of HCO3 inside the RBCs.

Question 10.
How much energy is required for the formation of single molecule of ATP ?
Answer:
For the formation of a single molecule of ATP about 7.3 Kcal of energy is required.

Question 11.
What is Hamburger’s phenomena?
Answer:
The diffusion of Chloride ions into the RBCs to main the ionic balance between RBCs and the plasma is called Hamburger’s phenomena or chloride shift.

Question 12.
What is the role of Hering-Breuer reflex in respiration?
Answer:
The Hering-Breuer reflex controls the depth and rhythm of respiration. It also prevents the lungs from inflating to the point of bursting.

Question 13.
How much blood is present in the human body and from which embryonic germ layer is it derived?
Answer:
An average adult has about 4 to 6 litres of blood, which is red coloured fluid connective tissue derived from embryonic mesoderm.

Question 14.
What is the percentage of plasma in the blood and how much water does it contain?
Answer:
There is 55% of plasma in the blood and it contains 90 to 92% water.

Question 15.
What is the average life span of RBCs?
Answer:
RBCs have a life span of about 120 days.

Question 16.
What is normal RBC count and total WBC count?
Answer:
Average RBC count in adult human is 5.1 to 5.8 million per cubic mm and average total WBC count in adult human is 5000 to 9000 per cubic mm.

Question 17.
What is erythropoiesis?
Answer:
The process of formation of Red Blood Cells is called erythropoiesis.

Question 18.
What is increase in the RBC number called?
Answer:
The increase in the number of RBCs is called polycythemia.

Question 19.
What is leucopenia and erythrocytopenia ?
Answer:
The decrease in the number of white blood cells is called leucopenia whereas decrease in the number of red blood cells is called erythrocytopenia.

Question 20.
Where are Eustachian valve and Thebesian valve located?
Answer:
Eustachian valve is present at the opening of inferior vena cava while Thebesian valve is present near the opening of coronary sinus.

Question 21.
What is foramen ovale and how is it related to fossa ovalis?
Answer:
Foramen ovale is an oval opening in the interatrial septum of the foetal heart representing the fossa ovalis which lies as a depression on the right side of interatrial septum.

Question 22.
When is a person described as having hypertension?
Answer:
When the blood pressure values Eire more than 140 mm Hg systolic pressure and more than 90 mm Hg diastolic pressure, then the person is described as having hypertension.

Question 23.
What are the effects of excessive hypertension?
Answer:
Excessive hypertension of values about 220/120 mm Hg can cause blindness, nephritis, stroke or paralysis.

Question 24.
What is the difference between anemia and leukemia?
Answer:
Anemia is disorder caused due to the deficiency of heaemoglobin while leukemia is blood cancer in which there is abnormal increase in the number of white blood cells.

Question 25.
What is the difference between tachycardia and bradycardia?
Answer:
The faster heart rate over 100 beats per minute is called tachycardia, while the slower heart rate below 60 beats per minute is called bradycardia.

Question 26.
What is the difference between chordae tendinae and columnae carnae?
Answer:
Chordae tendinae are chords that connect bicuspid and tricuspid valves with the papillary muscles in ventricles while columnae carnae are series of irregular muscular ridges present on the inner surface of the ventricles.

Question 27.
Which valves prevent the backward flow of blood at the time of ventricular systole?
Answer:
Semilunar valves located at the base of pulmonary artery and systemic aorta prevent the backward flow of blood at the time of ventricular systole.

Maharashtra Board Class 12 Biology Important Questions Chapter 8 Respiration and Circulation

Question 28.
What are the time intervals for atrial systole, ventricular systole and joint diastole?
Answer:
Atrial systole is for 0.1 second, ventricular systole is for 0.3 second and joint diastole is for 0.4 second.

Question 29.
In the electrocardiogram shown below, which wave represents ventricular diastole?
Maharashtra Board Class 12 Biology Important Questions Chapter 8 Respiration and Circulation 1
Answer:
‘T’ wave represents ventricular diastole.

Question 30.
Mention the role of pacemaker in human heart.
Answer:
Pacemaker can generate wave of contraction or cardiac impulse for rhythmic contraction of heart.

Question 31.
Which structure in the heart is called pacemaker?
Answer:
Sinuatrial node [S. A. node] in the heart wall is called a pacemaker.

Question 32.
What is electrocardiograph?
Answer:
The instrument which is used to record action potentials generated in the heart muscles is called an electrocardiograph or ECG machine.

Question 33.
What is angina pectoris?
Answer:
Angina pectoris is the pain in the chest caused due to reduction in blood supply to cardiac muscle caused due to narrowed and hardened coronary arteries.

Question 34.
What is pulse pressure?
Answer:
Difference between systolic and diastolic pressure is called pulse pressure which is normally 40 mm Hg.

Question 38.
What would happen if respiration takes place in one single step?
Answer:
If respiration takes place in one single step, then the chemical energy released at once during that step might result in a brief blast of light and heat and may lead to death of the cell. Hence respiration is a step-wise process.

Question 39.
Why do the veins have valves?
Answer:
The veins have valves at regular intervals to prevent backflow of blood as blood flows through veins with low pressure.

Question 40.
What is Bohr effect?
Answer:
Bohr effect is the shift of oxyhaemoglobin dissociation curve due to change in partial pressure of CO in blood.

Question 41.
What is Haldane effect?
Answer:
Decrease of pH of blood, due to increase in the number of H+ ions, HCO3 changes into H2O and CO2 by the presence of oxyhaemoglobin is called Haldane effect.

Give definitions of the following

Question 1.
Respiration
Answer:
It is a biochemical process of oxidation of organic compounds in an orderly manner for the liberation of chemical energy in the form of ATP.

Question 2.
Breathing
Answer:
It is a physical process by which gaseous exchange takes place between the atmosphere and the lungs. It involves inspiration and expiration.

Question 3.
Tidal Volume (TV)
Answer:
It is the volume of un¬ inspired or expired during normal breathing. It is 500 ml.

Question 4.
Inspiratory reserve volume (IRV)
Answer:
The maximum or the extra volume of air that is inspired during forced breathing in addition to TV (2000 to 3000 ml).

Question 5.
Expiratory reserve volume (ERV)
Answer:
The maximum volume of air that is expired during forced breathing after normal expiration. (1000 to 1100 ml).

Question 6.
Dead space (DS)
Answer:
The volume of air that is present in the respiratory tract (from nose to the terminal bronchioles), but not involved in gaseous exchange (150 ml).

Question 7.
Residual volume (RV)
Answer:
The volume of air that remains in the lungs and the dead space even after maximum expiration (1100 to 1200 ml).

Question 8.
Total lung capacity
Answer:
The maximum amount of air that the lungs can hold after a maximum forceful inspiration (5200 to 5900 ml).

Question 9.
Vital capacity (VC)
Answer:
The maximum amount of air that can be breathed out after of maximum inspiration. It is the sum total of TV, IRV and ERV and is 4100 to 4600 ml.

Question 10.
Oxygen dissociation curve
Answer:
The relationship between HbO2 saturation and oxygen tension (PPO2) is called oxygen dissociation curve.

Question 11.
Phosphorylation
Answer:
The process that involves trapping the heat energy in the form of high energy bond of ATP molecule is called phosphorylation.

Question 12.
Artificial ventilation
Answer:
It is the method of inducing breathing in a person when natural respiration has ceased or is faltering.

Question 13.
Ventilator
Answer:
A ventilator is a machine that supports breathing and is used during surgery, treatment for serious lung diseases or other conditions when normal breathing fails.

Question 14.
Cyclosis
Answer:
Cyclosis is the streaming movement of the cytoplasm shown by almost all living organisms. E.g. Paramoecium, Amoeba, etc.

Question 15.
Single circulation
Answer:
The movement of blood once through the heart during each circulation cycle is called single circulation.

Question 16.
Double circulation
Answer:
The movement of blood twice through the heart during one circulation cycle is called double circulation.

Question 17.
Erythropoiesis
Answer:
The process of formation of RBCs is called erythropoiesis.

Question 18.
Polycythemia
Answer:
The increase in the number of RBCs is called polycythemia.

Question 19.
Erythrocytopenia
Answer:
The decrease in the number of RBCs is called Erythrocytopenia.

Question 20.
Hematocrit
Answer:
The hematocrit is ratio of the volume of RBCs to total blood volume of blood.

Question 21.
Diapedesis
Answer:
Leucocytes perform amoeboid movement. Due to this kind of movement they can move out of the capillary walls. This is called diapedesis.

Question 22.
Leucocytosis
Answer:
Increase in the number of leucocytes or WBCs is called leucocytosis.

Question 23.
Leucopenia
Answer:
The decrease in the number of white blood cells is called leucopenia.

Question 24.
Leukaemia
Answer:
Pathological Increase in the number WBCs is called leukaemia or blood cancer.

Question 25.
Thrombocytopenia
Answer:
Decrease in the number of blood platelets is called thrombocytopenia.

Question 26.
Blood Coagulation
Answer:
Conversion of liquid blood into semisolid jelly is called blood coagulation or blood clotting.

Question 27.
Pericardium
Answer:
Double layered peritoneum that covers the heart from outside is called pericardium.

Question 28.
Pacemaker
Answer:
Pacemaker is the region that has power of generation of wave of contraction. In heart, sinoatrial node is called pacemaker.

Question 29.
Heartbeat
Answer:
The rhythmic contraction and relaxation of the heart is called heartbeat.

Question 30.
Pulse
Answer:
A pressure wave that travels through the arteries after each ventricular systole is called pulse.

Question 31.
Heart rate
Answer:
The rate with which the heart beats per minute is called the heart rate.

Question 32.
Stroke volume
Answer:
The amount of blood thrown out of the ventricles during one systole is called the stroke volume.

Question 33.
Cardiac output
Answer:
The amount of blood thrown out of the ventricles during one minute is called cardiac output.

Question 34.
Tachycardia
Answer:
Higher heart rate over 100 beats per minute is called tachycardia.

Question 35.
Bradycardia
Answer:
Lower heart rate which is lesser than 60 per minute is called bradycardia.

Question 36.
Myogenic
Answer:
When the initiation and further regulation of heartbeats take place in the muscles then such a heart is called myogenic.

Question 37.
Cardiac cycle
Answer:
Consecutive systole and diastole constitutes a single heartbeat or cardiac cycle.

Maharashtra Board Class 12 Biology Important Questions Chapter 8 Respiration and Circulation

Question 38.
Arterial blood pressure
Answer:
The pressure exerted by blood on the wall of artery is called arterial blood pressure.

Question 39.
Angiology
Answer:
Study of blood vessels is called angiology.

Question 40.
Angiography
Answer:
X-ray or imaging of the cardiac blood vessels to locate the position of blockages is called angiography.

Question 41.
Heart transplant
Answer:
Replacement of severely damaged heart by normal heart from brain- dead or recently dead donor is called heart transplant.

Question 42.
Silent Heart Attack
Answer:
Silent heart attack, also known as silent myocardial infarction, is a type of heart attack that lacks the general symptoms of classic heart attack like extreme chest pain, hypertension, shortness of breath, sweating and dizziness.

Question 43.
Electrocardiogram
Answer:
Graphical recording of electrical variations detected at the surface of body during their propagation through the wall of heart is electrocardiogram (ECG).

Question 44.
Lymph
Answer:
It is a fluid connective tissue with almost similar composition to the blood except RBCs, platelets and some proteins.

Give functions of the following

Question 1.
Epiglottis.
Answer:
The epiglottis prevents the entry of food into the trachea by closing the glottis temporarily.

Question 2.
Carbonic anhydrase.
Answer:
Carbonic anhydrase enzyme is found inside the RBCs only to accelerate the rate of formation of carbonic acid from CO2 and H2O.

Question 3.
Ventilators.
Answer:
Ventilators used in hospitals are part of life supporting system, which help in breathing by

  1. Pushing oxygen into the lungs
  2. Removing carbon dioxide from the lungs

Question 4.
Erythrocytes.
Answer:
Erythrocytes carry oxygen to all cells of the body from the lungs and bringing carbon dioxide from all the cells back to lungs.

Question 5.
Neutrophils.
Answer:
Neutrophils are responsible for destroying pathogens by the process of phagocytosis.

Question 6.
Thrombocytes/Platelets.
Answer:
Platelets secrete platelet factors which are essential in blood clotting. They also seal v the ruptured blood vessels by formation of platelet plug/thrombus. They secrete serotonin, a local vasoconstrictor.

Question 7.
Pericardial fluid.
Answer:
Pericardial fluid acts as a shock absorber and protects the heart from mechanical injuries. It also keeps the heart moist and acts as lubricant.

Question 8.
Heart walls.
Answer:
The epicardium and endocardium are protective in function whereas myocardium is responsible for contraction and relaxation of heart.

Question 9.
Valves in heart.
Answer:
Valves in the heart prevent the backflow of the blood at the time of systole and help in maintaining a unidirectional flow of blood.

Question 10.
Chordae tendinae.
Answer:
Chordae tendinae attach the bicuspid and tricuspid valves to the ventricular wall (papillary muscles) and regulate their opening and closing.

Question 11.
Semilunar valves.
Answer:
Semilunar valves prevent the backward flow of blood from pulmonary aorta and the aorta into the respective ventricles.

Question 12.
Sinoatrial node [SA] or Pacemaker.
Answer:
SA node acts as pacemaker of heart because it has the power of generating a new wave of contraction and making the pace of contraction.

Question 13.
Electrocardiogram (ECG).
Answer:
ECG helps to diagnose the abnormality in conducting pathway, enlargement of heart chambers, damage to cardiac muscles, reduced blood supply to cardiac muscles and causes of chest pain.

Question 14.
Blood.
Answer:
Functions of blood:

  1. Transport of oxygen and carbon dioxide
  2. Transport of food
  3. Transport of waste product
  4. Transport of hormones
  5. Maintenance of pH
  6. Water balance
  7. Transport of heat
  8. Defence against infection
  9. Temperature regulation
  10. Blood clotting/coagulation
  11. Helps in healing

Name the following

Question 1.
Name two animals in which moist skin acts as a respiratory surface.
Answer:
Earthworm, Frog

Question 2.
Name the respiratory organs in insects and fish.
Answer:
Insects – Tracheal tubes and spiracles
Fish – Internal gills

Question 3.
Name any two disorders of respiratory system.
Answer:
Asthma and pneumonia are the two disorders of respiratory system.

Question 4.
Name the structural and functional unit of lungs.
Answer:
Alveolus is the structural and functional unit of lungs.

Question 5.
Name the energy currency of cell.
Answer:
ATP is the energy currency of cell.

Question 6.
Name the site where actual exchange of O2 and CO2 takes place between air and blood in the body of man.
Answer:
Alveolus of lung.

Question 7.
Name any two respiratory centres required for regulation of breathing.
Answer:
Inspiratory centre, Expiratory centre, Pneumotaxic centre and Apneustic centre.

Question 8.
Name the muscles which move ribs up and down.
Answer:
External intercostal muscles.

Question 9.
Name two phyla where haemocoel is present.
Answer:
Phylum-Arthropoda and Phylum-Mollusca.

Question 10.
Name the animal-group which show single circulation.
Answer:
Fishes

Question 11.
Name the cells which produce thrombocytes.
Answer:
Megakaryocytes produce thrombocytes.

Question 12.
Name the process of formation of red blood corpuscles.
Answer:
Erythropoiesis

Question 13.
Name the space in which human heart is located.
Answer:
Mediastinum is the space in which human heart is located.

Question 14.
Name the types of lymphocytes depending upon functions.
Answer:
B-lymphocytes and T-lymphocytes.

Question 15.
Name the layers of peritoneum that surrounds the heart sequentially from outside to inside.
Answer:
Fibrous pericardium, parietal layer of serous pericardium and visceral layer of serous pericardium.

Question 16.
Name the connection between the pulmonary trunk and systemic aorta.
Answer:
Ligamentum arteriosum that represents remnant of ductus arteriosus of foetus.

Question 17.
Name the valve between left atrium and left ventricle and give its significance.
Answer:
Between left atrium and left ventricle is mitral or bicuspid valve which maintains the unidirectional flow of blood by preventing hs backflow.

Question 18.
Name the walls of an artery.
Answer:
Outer tunica externa, middle tunica media and inner tunica interna.

Question 19.
Name the instrument used to measure blood pressure.
Answer:
Sphygmomanometer is used to measure blood pressure.

Question 20.
Name the plasma proteins involved in the process of blood clotting.
Answer:
Prothrombin and fibrinogen.

Question 21.
Name the various components of conducting system of the heart.
Answer:
Conducting system of the heart consists of SA node, AV node, bundle of His and Purkinje fibers.

Maharashtra Board Class 12 Biology Important Questions Chapter 8 Respiration and Circulation

Question 22.
Name the neurotransmitters that decrease and increase the heart rate in human beings respectively.
Answer:
Acetylcholine decreases heart rate and adrenaline or epinephrine increases the heart rate in human.

Question 23.
Who discovered ECG?
Answer:
Willem Einthoven discovered ECG.

Distinguish between the following

Question 1.
Pharynx and Larynx.
Answer:

PharynxLaryix
1. Pharynx is a short, vertical tube.1. Larynx is a sound producing organ located at the end of pharynx.
2. Mouth leads to the pharynx.2. Larynx leads to the oesophagus.
3. Vocal cords are absent.3. Vocal cords are present.
4. Pharynx does not increase in size at the time of puberty.4. Larynx increases in size at the time of puberty.
5. Pharynx does not show Adam’s apple.5. Larynx shows Adam’s apple in adult males.

Question 2.
Inspiration and Expiration.
Answer:

InspirationExpiration
1. Inspiration is an active process.1. Expiration is a passive process.
2. During inspiration diaphragm contracts and becomes flattened.2. During expiration diaphragm relaxes and becomes dome shaped.
3. During inspiration intercostal muscles contract.3. During expiration intercostal muscles relax.
4. During inspiration ribs are pulled outwards and sternum is raised.4. During expiration ribs are pulled inwards and sternum is lowered.
5. During inspiration the space in the thoracic cavity increases.5. During expiration the space in the thoracic cavity decreases.
6. During inspiration pressure in the lungs decreases.6. During expiration pressure in the lungs increases.
7. During inspiration the volume of the lungs increase.7. During expiration the volume of the lungs decreases.
8. During inspiration air comes inside the body.8. During expiration air goes out of the body.

Question 3.
External respiration and Internal respiration.
Answer:

External respirationInternal respiration
1. The respiratory processes occurring in lungs is called external respiration.1. The respiratory processes that occur in tissues is called internal respiration.
2. During external respiration O2 from the lungs enters into the lung capillaries by diffusion.2. During internal respiration O2 from the blood enters the tissue cells.
3. During external respiration CO2 from the lung capillaries diffuse into the lungs.3. During internal respiration CO2 from the tissues enters into the blood.
4. During external respiration exchange of gases takes place between the air and the lungs.4. During internal respiration exchange of gases take place between the blood and the tissue.
5. Formation of oxyhaemoglobin takes place during external respiration.5. Oxyhaemoglobin dissociates into oxygen and haemoglobin during internal respiration.
6. During external respiration CO2 is released.6. During internal respiration carbamino haemoglobin is formed which is carried to the lungs.

Question 4.
Transport of O2 and Transport of CO2.
Answer:

Transport of O2Transport of CO2
1. Transport of O2 takes place from lungs to the tissues and cells.1. Transport of CO2 takes place from tissues and cells to the lungs.
2. Oxygen is carried as oxyhaemoglobin to the tissues with the help of RBCs.2. Carbon dioxide is carried as carbaminohaemoglobin from the tissues with the help of plasma and RBCs.
3. Oxygen does not form oxides or other products during its transport.3. CO2 forms bicarbonates with sodium and potassium during its transport.
4. O2 does not form acids during its transport.4. CO2 dissolves in water to form carbonic acid.

Question 5.
Vital Capacity of Lung and Total Lung Capacity.
Answer:

Vital Capacity of LungTotal Lung Capacity
1. It is the maximum amount of air a person can expire and inspire to their maximum extent.1. It is the maximum amount of air that the lungs can hold after a maximum forceful inspiration.
2. It is the sum total of inspiratory reserve volume, tidal volume and expiratory reserve volume.2. It is the sum total of vital capacity and residual volume.
3. It ranges from 4100 to 4600 ml.3. It ranges from 5200 to 5800 ml.

Question 6.
Inspiratory Reserve Volume (IRV) and Expiratory Reserve Volume (ERV).
Answer:

Inspiratory Reserve Volume (IRV)Expiratory Reserve Volume (ERV).
1. It is the maximum volume of air, or the extra volume of air, that is inspired during forced breathing.1. It is the maximum volume of air that is expired during forced breathing.
2. Its value is 2000/3000 ml.2. Its value is 1000/1100 ml.

Question 7.
T. S. of artery and T.S. of vein.
Answer:

T. S. of arteryT.S. of vein
1. Histologically in transverse section of artery there are three walls, tunica externa, tunica media and tunica interna or endothelium.1. Histologically in transverse section of vein there are three walls, tunica externa, tunica media and tunica interna or endothelium.
2. Tunica media is thick and muscular.2. Tunica media is thinner as compared to artery.
3. Lumen of artery is narrow.3. Lumen of vein is broad.

Question 8.
Erythrocytes and Leucocytes.
Answer:

ErythrocytesLeucocytes
1. Erythrocytes have a definite shape which is elliptical or oval.1. Leucocytes do not have definite shape as they are amoeboid.
2. They are enucleated.2. They are nucleated.
3. Erythrocytes contain haemoglobin and hence appear red.3. Leucocytes are devoid of any respiratory pigment and hence appear colourless.
4. The normal erythrocyte count is 4.3 to 5.8 million per cubic mm of blood.4. The normal leucocyte count is 4000 to 11000 per cubic mm of blood.
5. The life span of erythrocytes is 100 to 120 days.5. The life span of leucocytes is 3 to 4 days.
6. The diameter of erythrocytes is 7.2 m and thickness is about 2 to 2.2 m.6. The size of leucocytes varies with its subtypes and is of average size of 8 to 15 m.
7. Erythrocytes are formed by the process of erythropoiesis in red bone marrow.7. Leucocytes are formed by the process of leucopoiesis in bone marrow, tonsils, lymph nodes, spleen, thymus, etc.
8. Erythrocytes transport the respiratory gases.8. Leucocytes help in the formation of antibodies besides fighting against foreign antigens by phagocytic activity.

Question 9.
Eosinophils and Basophils.
Answer:

EosinophilsBasophils
1. Cytoplasmic granules present in eosinophils are stained with acidic stains.1. Cytoplasmic granules present in basophils are stained with basic stains.
2. Nucleus is bilobed.2. Nucleus is twisted.
3. Eosinophils constitute 3% of total WBCs.3. Basophils constitute 0.5% of total WBCs.

Question 10.
Neutrophils and Eosionophils.
Answer:

NeutrophilsEosinophils
1. Cytoplasmic granules present in neutrophils are stained with neutral stains.1. Cytoplasmic granules present in eosinophils are stained with acidic stains.
2. Nucleus is three to five lobes showing polymorphic form.2. Nucleus is bilobed.
3. Neutrophils constitute 62% of total WBCs.3. Eosinophils constitute 3% of total WBCs.

Question 11.
Lymphocytes and Monocytes.
Answer:

LymphocytesMonocytes
1. Large round nucleus but size of the cell is smaller.1. Large kidney shaped nucleus and largest size among WBCs.
2. Lymphocytes form 25-33% of WBCs.2. Monocytes form 3-9% of WBCs.

Question 12.
Granulocytes and Agranulocytes.
Answer:

GranulocytesAgranulocytes
1. WBCs with granular cytoplasm are called granulocytes. Thus, cytoplasmic granules are present.1. WBCs with agranular cytoplasm are called agranulocytes. Thus, cytoplasmic granules are absent.
2. Nuclei of granulocytes are variously lobed.2. Nuclei of agranulocytes are not lobed.

Question 13.
Single circulation and Double circulation.
Answer:

Single circulationDouble circulation
1. Blood flows only once through the heart in a complete cycle.1. Blood flows twice through the heart during one complete circulation. Systemic – to and fro ‘ from heart to body and pulmonary – to and fro from heart to lungs.
2. Heart pumps deoxygenated blood only.2. Heart pumps both deoxygenated and oxygenated blood to lungs and body respectively.
3. Blood is oxygenated in gills.3. Blood is oxygenated in lungs.
4. Occurs only in fishes.4. Occurs in amphibians, reptiles, birds and mammals.

Question 14.
Systolic blood circulation and Diastolic blood circulation.
Answer:

Systolic blood circulationDiastolic blood circulation
1. Blood is passed from right ventricle to lungs by pulmonary artery during systolic circulation. Similarly, from left ventricle the oxygenated blood is given to the entire body through systemic aorta during systolic circulation.1. Blood is passed to left atrium from lungs by pulmonary veins during diastolic circulation. Similarly, deoxygenated blood from entire body is brought back to heart through vena cava during diastolic circulation.
2. Systolic blood circulation is under maximum pressure as heart is forcing the blood to come out of heart.2. Diastolic blood circulation is under minimum blood pressure as heart is relaxed during diastole.

Question 15.
Atria and Ventricles.
Answer:

AtriaVentricles
1. Atria are upper chambers of the heart.1. Ventricles are lower chambers of the heart.
2. Atria are thin walled.2. Ventricles are thick walled.
3. Atria are receiving chambers.3. Ventricles are distributing chambers.
4. Interatrial septum divides the two auricles (atria).4. Interventricular septum divides the two ventricles.
5. Right atrium is larger in size than left atrium.5. Left ventricle is larger in size than the right ventricle.

Question 16.
S.A. Node and A.V. Node.
Answer:

S.A. NodeA.V. Node
1. Sinoatrial node is present in the right ventricle near the opening near the opening of the superior vena cava.1. Atrioventricular node is present in the right ventricle near the opening of the coronary sinus.
2. S.A. node is the pacemaker of the heart and it starts atrial systole.2. A.V. node starts ventricular systole through bundles of His and Purkinje’s fibre system.

Question 17.
Pulmonary circulation and Systemic circulation.
Answer:

Pulmonary circulationSystemic circulation
1. The course of blood from the right ventricle to the left atrium of the heart through the lungs is called pulmonary circulation.1. The course of blood from the left ventricle to the right atrium of the heart through the body is called systemic circulation.
2. Pulmonary circulation is mainly for sending the blood for oxygenation in the lungs from the heart and bringing it back to the heart after oxygenation.2. Systemic circulation is for sending the deoxygenated blood from the body to the heart and sending oxygenated blood from the heart to the body.

Question 18.
Atrio ventricular valves and Semilunar valves.
Answer:

Atrio ventricular valvesSemilunar valves
1. Atrio ventricular valves Eire present between the atria and ventricles. On the right side there is tricuspid valve whereas on the left side there is bicuspid valve.1. Semilunar valves are present at the opening of pulmonary artery and systemic aorta.
2. Atrio ventricular valves prevent the back flow of blood from ventricles to atria at the time of systole.2. Semilunar valves prevent the back flow of blood from pulmonary artery and systemic aorta back to the heart.

Question 19.
Hypertension and Hypotension.
Answer:

HypertensionHypotension
1. Blood pressure values more than 140 mm Hg SP and 90 mm HG DP is called hypertension.1. Blood pressure values less them 120 mm Hg SP and 70 mm HG DP is called hypotension.
2. Excessive hypertension can result into lethal complications such as stroke or paralysis.2. Hypotension may not be lethal if immediate measures are taken to raise the blood pressure.

Give reasons

Question 1.
ATP is called energy currency of the cell.
Answer:

  1. During cellular respiration, the oxidation of food (glucose) takes place.
  2. This happens in the mitochondria using the oxygen present in the blood.
  3. ATP molecules are formed during this oxidation.
  4. ATP is used for various vital body processes and also for maintaining the body temperature to constancy.
  5. Since energy is stored in the form of ATP it is called an energy currency of the cell.

Question 2.
The vestibule of nasal chamber has fine hair.
Answer:

  1. Vestibule is the anterior most part of the nasal chamber.
  2. The hairs present in this region trap the dust particles and prevent them from entering into the interiors of the respiratory passage.
  3. Therefore, the vestibule of nasal chambers has fine hair.

Question 3.
Glottis is guarded by a flap called epiglottis.
Answer:

  1. The oesophagus and trachea lie side by side.
  2. There is possibility that food particles may enter respiratory passage at the time of gulping.
  3. However, the epiglottis prevents the entry of food into the respiratory passage by closing it temporarily.
  4. Thus, for preventing the entry of food particles into respiratory passage, the glottis is guarded by a flap called epiglottis.

Maharashtra Board Class 12 Biology Important Questions Chapter 8 Respiration and Circulation

Question 4.
Alveoli are very flexible.
Answer:

  1. Alveoli are made up of collagen and elastin fibres.
  2. They are very thin (0.0001 mm) and lined by non-ciliated squamous epithelium.
  3. All the above structural components make the alveoli very flexible.

Question 5.
Expiration is called a passive process.
Answer:

  1. During expiration, intercostal muscles relax. This results in pulling the ribs inwards.
  2. Diaphragm also relaxes and returns to its normal dome shape.
  3. The collective contraction of ribs and diaphragm results in the reduction of
    thoracic cavity and hence automatically air is pushed out of the lungs.
  4. Since the pressure on the lungs increase rushing the air to outside, expiration is called a passive process.

Question 6.
Pericardium acts as a defence wall for the heart.
Answer:

  1. Pericardium protects the heart. It is double layered peritoneum, having outer fibrous and inner serous pericardium layers.
  2. Fibrous pericardium being tough gives protection to the heart.
  3. Serous pericardium has two layers, parietal and visceral layer or epicardium.
  4. In between these two layers, there is pericardial fluid, which helps to absorb shocks and provide nourishment.
  5. In this way pericardium acts as a defence wall.

Question 7.
Valves are present in veins.
Answer:

  1. Veins carry blood to the heart.
  2. At that time the backward flow of the blood should be prevented.
  3. Therefore, valves are present in veins.

Question 8.
Atria are thin walled than ventricles.
Answer:

  1. Atria are receiving chambers, while ventricles are distributing chambers.
  2. The blood is driven out from ventricles.
  3. Ventricles are therefore, strong and with thicker walls.
  4. Atria are thin walled as compared to ventricles.

Question 9.
Heart is called a pump.
Answer:

  1. The heart acts as a pumping organ. It shows continuous pumping action.
  2. The rhythmic contraction or systole and relaxation or diastole of heart forms one heartbeat.
  3. Such heartbeats occur about 72 times per minute.
  4. The heart efficiently pumps about 5 litres of blood per minute. Therefore, the heart is called a pump.

Question 10.
In normal human heart, there is no mixing of oxygenated and deoxygenated blood.
Answer:

  1. In normal human heart, there is completely formed atrioventricular septum.
  2. This septum keeps the deoxygenated and oxygenated blood separate.
  3. Hence there is no mixing of the two types of blood.

Question 11.
Blood pressure is inversely related to the elasticity of the blood vessels.
Answer:

  1. When the blood gushes through the blood vessels, the walls of blood vessels -can expand a little due to their elasticity.
  2. But as the age advances, the elasticity is reduced and then the blood vessels do not expand.
  3. Hence the flowing blood gets more resistance and the blood pressure can rise.
  4. Lesser the elasticity more will be the blood pressure, whereas more the elasticity of the vessel wall, then the pressure will not rise.
  5. In this way, the blood pressure is inversely related to the elasticity of the blood vessels.

Write short notes

Question 1.
Chloride shift or Hamburger’s phenomenon.
Answer:

  1. About 70% of CO2 is transported in the form of sodium bicarbonates/potassium bicarbonates from tissue cells to lungs.
  2. In the RBCs, CO2 combines with water in the presence of a Zn containing enzyme, carbonic anhydrase to form carbonic acid. This action is rapid in RBCs as compared to that in the plasma.
  3. Carbonic acid being unstable, immediately dissociates into HCO3 and H+in the presence of same enzyme, leading to large accumulation of HCO3 inside the RBCs. It thus moves out of RBCs. This can bring about imbalance of the charge inside the RBCs.
  4. To maintain the ionic balance between the RBCs and the plasma, Cl diffuses into the RBCs. This movement of chloride ions is known as chloride shift or Hamburger’s phenomenon.
  5. HCO3 that comes in the plasma joins to Na+/K+ forming NaHCO3/KHCO3 which can maintain pH of blood. The remaining H+ ions in the RBCs are buffered by haemoglobin by the formation of oxyhaemoglobin.
  6. At the level of lungs, due to the low partial pressure of carbon dioxide of the alveolar air, hydrogen ion and bicarbonate ions combine to form carbonic acid and under the influence of carbonic anhydrase again yields carbon dioxide and water.

Question 2.
Regulation of breathing.
Answer:
(1) Respiration is under dual control, i.e. nervous and chemical. Normal breathing is an involuntary process. Steady state of respiration is controlled by neurons located in the pons and medulla and are known as the respiratory centres. They regulate the rate and depth of breathing.

(2) These centres are divided into three groups : dorsal group of neurons in the medulla (inspiratory centre), ventro-lateral group of neurons in medulla (inspiratory and expiratory centre) and pneumotaxic centre located in the pons and apneustic centre which is antagonistic in action to pneumotaxic centre.

(3) During inspiration, when the lungs expand to a critical point, the stretch receptors are stimulated and impulses are sent along the vagus nerves to the expiratory centre. It then sends out inhibitory impulses to the inspiratory centre.

(4) The inspiratory muscles relax and expiration follows. As the air leaves but, the lungs are deflated and the stretch receptors are no longer stimulated. Thus, the inspiratory centre is no longer inhibited and a new respiration begins. These events are called the Hering – Breuer reflex. The Hering – Breuer reflex controls the depth and rhythm of respiration. It also prevents the lungs from inflating to the point of bursting.

(5) The respiratory centre has connections with the cerebral cortex that means we can voluntarily change our pattern of breathing. Voluntary control is protective because it enables us to prevent water or irritating gases from entering the lungs.

Question 3.
Carbon monoxide poisoning.
Answer:

  1. Carbon monoxide poisoning is caused when carbon monoxide is combined with haemoglobin.
  2. Haemoglobin is said to have 250 times more affinity for carbon monoxide than that for the oxygen.
  3. Therefore, haemoglobin with carbon monoxide forms a stable compound, the carboxyhemoglobin.
  4. Due to the formation of carboxyhaemoglobin, the haemoglobin no longer carries oxygen to the cells and tissues. Tissues then suffer from oxygen starvation. This leads to asphyxiation and in extreme cases it leads to death.
  5. Carbon monoxide poisoning occurs in closed rooms with incompletely burning substances such as stove burners or furnaces and garages having running automobile engines.
  6. Person suffering from carbon monoxide poisoning has to be administered with oxygen-carbon dioxide mixture, so that high levels of CO2 makes carbon monoxide dissociated from haemoglobin.

Question 4.
Artificial ventilation.
Answer:
(1) Artificial ventilation is the artificial respiration. It is the method of inducing breathing in a person when natural respiration has ceased or is faltering. If used properly and quickly, it can prevent death due to drowning, choking, suffocation, electric shock, etc.

(2) The process involves two main steps:
a. Establishing and maintaining an open air passage from the upper respiratory tract to the lungs.
b. Force inspiration and expiration as in mouth to mouth respiration or by mechanical means like ventilator.

(3) A ventilator is a machine that supports breathing and is used during surgery, treatment for serious lung diseases or other conditions when normal breathing fails.

Question 5.
Erythrocytes.
Answer:

  1. Erythrocytes or red blood corpuscles. They are circular, biconcave, enucleated cells.
  2. The RBC size : 7 pm in diameter and 2.5 pm in thickness.
  3. The RBC count : 5.1 to 5.8 million RBCs/ cu mm of blood in an adult male and 4.3 to 5.2 million/cu mm in an adult female.
  4. The average life span of RBC : 120 days.
  5. RBCs are formed by the process of erythropoiesis. In foetus, RBC formation takes place in liver and spleen whereas in adults it occurs in red bone marrow.
  6. The old and worn out RBCs are destroyed in liver and spleen.
  7. Polycythemia is an increase in number of RBCs while erythrocytopenia is decrease in their (RBCs) number.

Functions of RBCs:

  1. Transport of oxygen from lungs to tissues and carbon dioxide from tissues to lungs with the help of haemoglobin.
  2. Maintenance of blood pH as haemoglobin acts as a buffer.
  3. Maintenance of the viscosity of blood.

Question 6.
Heartbeat.
Answer:

  1. The rhythmic contraction and relaxation of the heart is called heartbeat.
  2. Each heartbeat includes one systole and one diastole. During systole the heart contracts and during diastole it relaxes.
  3. The rate with which the heart beats is called heart rate. The normal heart rate is 72 beats per minute.
  4. Tachycardia means faster heart rate of about more than 100 beats per minute.
  5. Bradycardia means slower heart rate that is below 60 beats per minute.

Question 7.
Pulse.
Answer:

  1. A pressure wave that travels through the arteries after each ventricular systole is called a pulse.
  2. The pulse can be felt in any artery that lies near the surface of the body.
  3. The radial artery at the wrist is most commonly used to feel the pulse.
  4. The pulse rate per minute indicates the heart rate. Since each heartbeat generates one pulse in the arteries, the pulse rate is same as that of heart rate, i.e. 72 times per minute.

Maharashtra Board Class 12 Biology Important Questions Chapter 8 Respiration and Circulation

Question 8.
Peacemaker.
Answer:

  1. Pacemaker is the region in tile heart which initiates the beating.
  2. The natural pacemaker of the heart is sinoatrial node (SA node).
  3. The pacemaker is autorhythmic, it is able to repeatedly and rhythmically generate impulses.
  4. SA node is responsible for initiation of cardiac excitation. Therefore, it is called a pacemaker.

Question 9.
Blood pressure.
Answer:

  1. Blood pressure is the pressure exerted by the flowing blood on the walls of arteries.
  2. Blood pressure described in two terms viz. systolic blood pressure and diastolic blood pressure. Systolic blood pressure is the maximum pressure of blood when heart undergoes ventricular systole. It is responsible for flow of blood in the arteries. Normal systolic pressure is 120 mm Hg.
  3. Diastolic blood pressure is the minimum pressure of blood when heart undergoes diastole. Normal diastolic pressure is 80 mm Hg.
  4. Blood pressure is represented as 120/80 mm Hg for a normal human being.

Question 10.
Hypertension.
Answer:

  1. In a normal healthy person the blood pressure values are 120 mm Hg (systolic)/ 80 mm Hg (diastolic).
  2. When the blood pressure is persistently more than 140 mm Hg systolic pressure and 90 mm Hg diastolic arterial blood pressure then it is said to be hypertension or high blood pressure.
  3. Excessively high blood pressure is very dangerous as high blood pressure of about 220/120 mm Hg may cause rupturing of blood vessels.
  4. Rupture of eye blood vessels can lead to blindness.
  5. If blood vessels of kidney are affected then nephritis is caused.
  6. Hemorrhage occurring in the brain can lead to stroke or paralysis. Therefore, hypertension is commonly called silent killer. It may be present for years with no distinct symptoms.
  7. The factors causing hypertension are arteriosclerosis (reduction of elasticity of blood vessels), atherosclerosis (deposition, of cholesterol inside the blood vessels wall), obesity, physical or emotional stress, alcoholism, smoking, cholesterol rich diet, increased secretion of renin, epinephrine or aldosterone, etc.

Question 11.
Coronary artery disease (CAD).
Answer:

  1. Coronary artery disease is a condition caused due to problems like atherosclerosis.
  2. In this disease, coronary arteries are narrowed due to deposition of fatty substances.
  3. Due to this the blood flow to the heart is reduced.
  4. In coronary heart disease, the heart muscle is damaged because of an inadequate amount of blood due to obstruction of its blood supply.
  5. The symptoms of CAD depend upon the degree of obstruction.
  6. Symptoms are mild chest pain or angina pectoris.
  7. In severe cases it results in heart attack known as myocardial infarction.

Question 12.
Angina pectoris.
Answer:

  1. Angina pectoris is the pain in the chest. It results from a reduction in blood supply to cardiac muscle due to narrowed and hardened coronary arteries.
  2. Atherosclerosis and arteriosclerosis can cause this problem. Basically, the coronary arteries are affected during angina pectoris.
  3. It causes heaviness and severe pain in the chest. The pain can also be felt at the neck, lower jaw, left arm and left shoulder.
  4. Angina pectoris often occurs during exertion, when the heart demands more oxygen and narrowed blood vessels cannot supply. It disappears with rest.

Question 13.
Heart failure.
Answer:

  1. Heart failure is caused due to progressive weakening of the heart muscle. This results in the failure of the heart to pump the blood effectively.
  2. Hypertension increase the after load on the heart leading to significant enlargement of the heart.
  3. This finally results in heart failure.
  4. Factors responsible for heart failure are advanced age, malnutrition, chronic infections, toxins, severe anaemia or hyperthyroidism, etc.
  5. Any problem leading to degeneration of heart muscle, may result in heart failure.

Question 14.
Atherosclerosis.
Answer:

  1. Atherosclerosis is the deposition of fatty substances and cholesterol on the inner lining of eateries.
  2. This deposition results in the formation of atherosclerotic plaque.
  3. It results in the decrease of the lumen of the blood vessels causing increasing resistance for the blood to flow which in turn results in the hypertension.
  4. Atherosclerosis of the coronary arteries results in decrease in the blood flow to the heart muscles.
  5. Due to such condition, coronary heart disease is caused.

Question 15.
ECG.
Answer:

  1. Electrocardiogram or ECG is the graphic v record of electrical variations produced by the heart during one heartbeat or cardiac cycle.
  2. ECG is taken with the help of an instrument called electrocardiograph or ECG machine. Electrocardiograph records the action potentials generated by the heart muscles.
  3. The electrical activity of heart is represented in the form of a graph plotted with time on X-axis against voltage displacement on Y-axis.
  4. A normal ECG is a graph having series of ridges and furrows. There are waves such as P-wave, QRS complex and T-wave.
  5. P-wave is a small upwards wave representing impulse generated by SA node. P-wave is caused by atrial depolarization that results in atrial contraction.
  6. QRS-complex wave begins as a downward deflection, continues as a large upright triangular wave and ends’ as a downward wave.
  7. QRS-complex wave represents spreading of impulse from SA node to AV node, then to bundle of His and Purkinje fibres. It causes ventricular depolarization resulting in ventricular contraction.
  8. T-wave is a broad upward wave which represents ventricular repolarization resulting in ventricular relaxation.
  9. Functions of ECG are mainly for diagnosis and also for prognosis. It is useful to detect abnormal functioning of heart as in coronary artery diseases, heart block, angina pectoris, tachycardia, ischemic heart disease, myocardial infarction, cardiac arrest, etc.

Question 16.
Angiography.
Answer:

  1. Angiography is an X-ray imaging of the cardiac blood vessels to locate the position of blockages.
  2. Depending upon the degree of blockage, remedial procedures like angioplasty or by¬pass surgery are performed.
  3. In angioplasty a stent is inserted at the site of blockage to restore the blood supply while in by-pass surgery, the atherosclerotic region is by-passed with part of vein or artery taken from any other suitable part of the body, like hands or legs.

Question 17.
Silent Heart Attack or silent myocardial infarction.
Answer:

  1. Silent heart attack is a type of heart attack that lacks the general symptoms of classic heart attack like extreme chest pain, hypertension, shortness of breath, sweating and dizziness.
  2. Symptoms of silent heart attack are so mild that a person often confuses it for regular « discomfort and thereby ignores it.
  3. Men are more affected by silent heart attack than women.

Question 18.
Heart Transplant.
Answer:

  1. Heart transplant is the replacement of severely damaged heart by normal heart from brain-dead or recently dead donor,
  2. Heart transplant is necessary in case of patients with end-stage heart disease and severe coronary arterial disease.

Short Answer Questions

Question 1.
What is meant by respiration? How is it useful in the production of energy?
Answer:

  1. Respiration is the biochemical process in which organic compound such as glucose are oxidized to liberate chemical energy.
  2. During respiration energy is released in gradual and step wise process. The released energy is in the form of bonds of ATP (Adenosine Tri Phosphate) molecules are shown below:
    C6H12O6 + 6O2 → 6CO2 + 6H2O + 38 ATP
  3. ATP is the biologically useful energy. ATP drives most of the life process.
  4. When cell requires the energy, ATP is hydrolyzed and is converted into ADP with subsequent release of energy.
  5. The respiratory system, blood and the body cells play an important role in the process of respiration.

Question 2.
How does exchange of gases take place at the alveolar level?
Answer:
1. Exchange of gases between the alveolar air and the blood is known as external respiration.

2. Simple squamous epithelial layer of alveolus is intimately associated with a similar layer lining the capillary wall. Both of these layers are thin walled and together they make up the respiratory membrane through which gaseous exchange occurs between the alveolar air and the blood.

3. Diffusion of gases will take place from an area of higher partial pressure to an area of lower partial pressure until the partial pressure in the two regions reaches equilibrium.

4. The partial pressure of carbon dioxide of blood entering the pulmonary capillaries is 45 mmHg while partial pressure of carbon dioxide in alveolar air is 40 mmHg. Due to this difference, carbon dioxide diffuses from the capillaries into the alveolus.

5. Similarly, partial pressure of oxygen of blood in pulmonary capillaries is 40 mmHg while in alveolar blood it is 104 mmHg. Due to this difference oxygen diffuses from alveoli to the capillaries.
Maharashtra Board Class 12 Biology Important Questions Chapter 8 Respiration and Circulation 2

Question 3.
What is the role of haemoglobin in the transport of oxygen in the blood?
Answer:

  1. Haemoglobin is a respiratory pigment present in cytoplasm of RBCs. About 97% of oxygen is transported by these haemoglobin molecules from lungs to tissues.
  2. Haemoglobin has a high affinity for Oa and combines with it to form oxyhaemoglobin. One molecule of Hb has four FeT, each of which can pick up a molecule of oxygen (O2). Hb + 4O2 → Hb (4O2)
  3. Oxyhaemoglobin is transported from lungs to the tissues where it readily dissociates to release O2.
    Hb (4O2) → Hb + 4O2
  4. In the alveoli where PPOa is high and PPCO2 is low, oxygen binds with haemoglobin, but in tissues, where PPO2 is lower and PPCO2 is high, Oxyhaemoglobin dissociates and releases O2 for diffusion into the tissue cells.

Question 4.
What is blood? What is the normal quantity of blood in an adult human being?
Answer:

  1. Blood is the fluid connective tissue that circulates in the body.
  2. Blood is derived from mesoderm.
  3. It is bright red, slightly alkaline fluid having pH about 7.4. It is salty, viscous fluid heavier than water.
  4. The average sized adult has about 5 litres of blood in his/her body which constitutes about 8% of the total body weight.

Maharashtra Board Class 12 Biology Important Questions Chapter 8 Respiration and Circulation

Question 5.
Describe the structure and the function of thrombocytes.
Answer:

  1. Thrombocytes or platelets are non- nucleated, round and biconvex blood corpuscles.
  2. They are smallest corpuscles measuring about 2.5 to 5 mm in diameter with a count of about 2.5 lakhs/cu mm of blood.
  3. Their life span is about 5 to 10 days.
  4. Thrombocytes are formed from megakaryocytes of bone marrow. They break from these cells as fragments during the process of thrombopoiesis.
  5. Thrombocytosis is the increase in platelet count while thrombocytopenia is decrease in platelet count.
  6. Thrombocytes possess thromboplastin which helps in clotting of blood.
  7. Therefore, at the site of injury platelets aggregate and form a platelet plug. Here they release thromboplastin due to which further blood clotting reactions take place.

Question 6.
Describe the structure of the heart wall.
Answer:

  1. The heart wall is composed of three layers, viz. outer epicardium, middle myocardium and inner endocardium.
  2. Epicardium is composed of single layer of mesothelium having flat epithelial cells.
  3. Myocardium is composed of cardiac muscle fibres. These muscle fibres perform the function of systole and diastole by showing contraction and relaxation of muscle wall of the heart.
  4. Endocardium is composed of single layer of flat epithelial cells called endothelium.

Question 7.
Name the two heart sounds. How and when are they produced?
Answer:

  1. In one normal heartbeat the heart sounds like lubb and dup are produced once each.
  2. The rhythmic contraction (Systole) and relaxation (diastole) forms are heartbeat. The heart sounds are due to closure of valves.
  3. Lubb sound is produced during ventricular systole when the cuspid valves close both the atrioventricular apertures preventing blood flow into atria.
  4. Dub sound is produced during ventricular diastole when semilunar valves are closed, preventing backflow of blood from pulmonary trunk and systemic aorta into ventricles.

Question 8.
What is double circulation? What is its significance?
Answer:
(1) Double circulation : Movement of blood twice through the heart during one circulation cycle is called double circulation. Body → heart → lungs → heart → body is the course of double circulation.

(2) Significance of double circulation:
a. Double circulation is more effective type of circulation in which oxygenated and deoxygenated type of blood do not intermix.
b. The systemic circulation i.e. from body to heart and back to body while the pulmonary circulation i.e. from heart to lungs and back to heart circulate the blood uniformly.

(3) Coronary and hepatic portal circulation is also achieved due to double circulation.

Question 9.
Describe pulmonary and systemic circulation.
Answer:

  1. In human beings, there is double circulation because blood passes twice through the heart during one cardiac cycle.
  2. The blood follows two routes viz. pulmonary and systemic.
  3. Pulmonary circulation is circulation between heart and lungs. Systemic circulation is the circulation between the heart and body organs (except lungs).
  4. During pulmonary circulation, the blood passes from the right ventricle to the left atrium of the heart through lungs.
  5. The right ventricle pumps deoxygenated blood into the pulmonary trunk which carries it to lungs for oxygenation. The oxygenated blood from the lungs is brought to left atrium by two pairs of pulmonary veins.
  6. During systemic circulation, the blood from the left ventricle passes to the right atrium of heart through body organs.
  7. The left ventricle pumps oxygenated blood into the systemic aorta which carries it to all body organs except lungs. The deoxygenated blood from the body organs is brought to right atrium by superior and inferior venae cavae.

Question 10.
How is cardiac activity regulated?
Answer:

  1. Normal activities of the heart are auto regulated. The specialized muscles help in this regulation.
  2. The heart is said to be myogenic due to this ability.
  3. In the medulla oblongata of brain, there is cardiovascular centre,
  4. From this centre, sympathetic and parasympathetic nerves innervate the sinoatrial node.
  5. Sympathetic nerves secrete adrenaline and it stimulates and increases the heartbeat.
  6. Parasympathetic nerves secrete acetylcholine and it decreases the heart rate.
  7. Rate of heartbeat is controlled in response to inputs from various receptors like proprio-receptors.
  8. These receptors monitor the position of limbs and muscles. There are chemoreceptors which monitor chemical change in blood and baroreceptors that monitor the stretching of main arteries and veins.

Chemical control on heart rate:

  1. Hypoxia, acidosis, alkalosis cause decrease in cardiac activity.
  2. Hormones like epinephrine and nor epinephrine enhance the cardiac activity.
  3. Elevated blood level of K+ and Na+ decreases the cardiac activity.

Question 11.
What are the main features of respiratory surface?
Answer:
The respiratory surface, for the efficient gaseous exchange should have the following features:

  1. It should have a large surface area.
  2. It should be thin, highly vascular and permeable to allow exchange of gases.
  3. It should be moist.

Question 12.
What is the co-relationship between activeness of organism and complexity of transport system?
Answer:

  1. As the size of an organism increases, its surface area to volume ratio decreases. This means it has relatively less surface area available for substances to diffuse through.
  2. Large multicellular organisms therefore cannot rely on diffusion alone to supply their cells with Substances such as food and oxygen and to remove waste products. Large multicellular organisms require specialized transport system.
  3. In short for the organisms to become active, they must be having complex transport system to bring about their vital functions rapidly.

Question 13.
What are the granules in granulocytes?
Answer:

  1. Neutrophils : Granules . of neutrophils contain cationic proteins and other proteins that are used to kill bacteria, some enzymes to breakdown bacterial proteins, lysozymes to breakdown bacterial cell wall. etc.
  2. Eosinophils : Granules of eosinophil contains a unique toxic basic protein receptors that bind to IgE used to help in killing parasites.
  3. Basophils : Granules of basophils contain abundant histamine, heparin and platelet activating factor.

Question 14.
What is haemoglobin count in normal human beings? What is the function of haemoglobin?
Answer:

  1. The normal haemoglobin in adult male is 13-18 mg/100 ml of blood.
  2. In a normal adult female, it is about 11.5-16.5 mg/100 ml of blood.
  3. In anaemic individuals there is lesser amount of haemoglobin.
  4. Functions of haemoglobin is to transport oxygen from lungs to tissues and carbon dioxide from tissues of lungs.
  5. Haemoglobin acts as a buffer and maintains the blood pH.

Question 15.
Why has the heart-recipient to rely upon life-time supply of immunosuppressants?
Answer:
Person who has undergone heart transplant needs lifetime supply of immunosuppressants because in these persons organ rejection is a constant threat. Keeping away the immune system from attacking the transplanted organ requires constant supply of immunosuppressant drugs. These drugs help prevent immune system from attacking (“rejecting”) the donor organ. Typically, these drugs are taken for the life-time for maintaining transplanted organ.

Question 16.
Why is it difficult to hold one’s breath beyond a limit?
Answer:

  1. It is difficult to hold one’s breath beyond a limit because the pressure of oxygen and carbon dioxide in blood changes as one holds his breath.
  2. When the breath is held beyond a limit, the urge to breath becomes irresistible.
  3.  When the breath is held forever, body becomes starved of oxygen and person may fall unconscious and the instinct to breath would take over.

Question 17.
Why and when do the leucocytes perform diapedesis?
Answer:

  1. Diapedesis is the movement of leucocytes through the wall of blood capillaries into the tissue space.
  2. Leucocytes perform diapedesis as an important part of their reaction to tissue injury or infection.
  3. This process forms the part of the innate immune response, involving recruitment of non-specific leucocytes.
  4. Monocytes also use this process during their development into macrophages.
  5. Diapedesis helps leucocytes to perform their functions like phagocytosis, production of antibodies, secretion of inflammatory response chemicals, etc.

Question 18.
Why are obese persons prone to hypertension?
Answer:
(1) Being overweight or obese is a major cause of hypertension, accounting for 65% to 75% of the risk for human primary hypertension.

(2) Following factors play an important role in initiating obesity hypertension:

  1. Physical compression of the kidneys by fat.
  2. Activation of the renin-angiotensin – aldosterone system
  3. Increased sympathetic nervous system activity.
  4. Obesity means more body-surface area. In order to supply blood to these parts. Heart and blood vessels work more resulting into hypertensions.

(3) Blood pressure rises as body weight increases and therefore obese persons are prone to hypertension.

Question 19.
Why does the transplanted heart beats at higher rate than normal?
Answer:

  1. The transplanted heart beats at higher rate than normal (about 100 to 110 beats per minute) because the nerves leading to the heart are cut during the operation. These nerves stimulate the pacemaker i.e. Sinoatrial node.
  2. The new heart also responds more slowly to exercise and does not increase its rate as quickly as before.

Question 20.
Why do large animals cannot carry out respiration without the help of circulatory system?
Answer:

  1. Large animals have various organ systems which always work in a coordinated manner.
  2. These animals provide large respiratory surfaces (numerous alveoli) for the exchange of gases. But these respiratory gases must be carried to the cells of tissues which are away from the respiratory surfaces.
  3. To carry these gases to tissues, there is need of transport system. These gasses are transported from respiratory surfaces to the cells of tissues through blood as a transporting medium.
  4. Therefore, large animals cannot carry out respiration without the help of circulatory system.

Question 21.
What is immunity? Name its types.
Answer:

  1. Immunity is the general ability of a body to recognize, neutralize or destroy and eliminate foreign substances or resist a particular infection or disease.
  2. There are two basic types of immunity, viz. innate immunity and acquired immunity, Acquired immunity is further divided into four types, i.e. Natural acquired active immunity, Natural acquired passive immunity, Artificial acquired active passive immunity and Artificial acquired passive immunity.

Question 22.
Why does the platelet count decrease in dengue patient?
Answer:

  1. The causative pathogen of dengue is dengue virus which induces bone marrow suppression. Since in bone marrow blood cells are formed its suppression causes the deficiency of blood cells leading to low platelet count.
  2. The dengue virus also links with platelets in the blood when there is a virus-specific antibody present in the human body.
  3. When vascular endothelial cell which are infected with dengue virus gets combined with platelets, they tend to destroy platelets. This is one of the major causes of low platelet count in dengue fever.
  4. Even the antibodies that are produced after infection of the dengue virus also cause the destruction of platelets, thus lowering the platelet count.

Question 23.
Why does our immune system fail against pathogens like Trypanosoma cruzi and Plasmodium?
Answer:

  1. Microbes have evolved a diverse range of strategies to destroy the host immune system. The protozoan parasite Trypanosoma cruzi and Plasmodium show similar such adaptations to disturb host defence mechanism.
  2. This parasite attacks host tissues including both peripheral and central lymphoid tissues.
  3. This causes systemic acute response in host body which the parasite tries to overcome. The parasite in fact weakens both innate and acquired immunity.
  4. It interferes with the antigen presenting function of dendritic cells via an action on hosts like lectin receptors. These receptors also induce suppression of CD4+ T cells responses. Therefore, our immune system fail against such pathogens.

Question 24.
What is the relation between immunity and organ transplantation?
Answer:

  1. Those who undergo an organ transplant face the possibility that their immune system will reject their new organ and that they will always be at a higher risk for infections.
  2. The immune system is able to recognize the difference between cells that belong to our body and those that do not by learning to identity protein markers (antigens) that are found on cell and infection surfaces.
  3. In people, the antigens or markers that identity their immune system are referred to as the human leukocyte antigen (HLA).
  4. Antigens that are recognized as unfriendly invaders stimulate an immune response to destroy them.
  5. Therefore, when organ transplantation is done, the immune responses are temporarily stalled. This helps in acceptance of the graft in the recipient’s body.

Question 25.
How do monocytes perform amoeboid movement and phagocytosis?
Answer:
Monocytes can perform phagocytosis. They do this by using intermediary or opsonising proteins such as antibodies or complement that coat the pathogen. They also bind to the microbe directly via pattern-recognition receptors that recognize pathogens. In this way they perform amoeboid movement and indulge in phagocytosis.

Question 26.
How do monocytes modify into macrophages?
Answer:
Monocytes upon having inflammation, selectively travel to the sites of inflammation. Here they produce inflammatory cytokines and contribute to local and systemic inflammation. They are highly infiltrative. They differentiate into inflammatory macrophages, which then remove PAMPs or pathogen-associated molecular patterns and cell debris.

Chart based/Table based questions

Question 1.
Complete the following:

OrganismHabitatRespiratory surface/organ
1. InsectsTerrestrial—————
2. Amphibian tadpoles of frog, salamanders—————-—————-
3. FishAquatic————–
4. Reptiles, Birds and Mammals—————-—————-

Answer:

OrganismHabitatRespiratory surface/organ
1. InsectsTerrestrialTracheal tubes and spiracles
2. Amphibian tadpoles of frog, salamandersAquaticExternal gills
3. FishAquaticInternal gills
4. Reptiles, Birds and MammalsTerrestrialLungs

Question 2.

Partial pressure of gasesAlveolar airPulmonary, capillaries
PPO2——————————
PPCO2—————-—————-

Answer:

Partial pressure of gasesAlveolar airPulmonary, capillaries
PPO2104 mm Hg40 mm Hg
PPCO240 mm Hg45 mm Hg

Question 3.
Maharashtra Board Class 12 Biology Important Questions Chapter 8 Respiration and Circulation 3
Answer:
Maharashtra Board Class 12 Biology Important Questions Chapter 8 Respiration and Circulation 4

Question 4.
Complete the following flow chart
Maharashtra Board Class 12 Biology Important Questions Chapter 8 Respiration and Circulation 5
Answer:
Maharashtra Board Class 12 Biology Important Questions Chapter 8 Respiration and Circulation 6

Question 5.
Complete the following Table

Waves on ECGHeart ActivityCaused due to
P waveAtrial contraction—————-
QRS wave—————–Ventricular depolarization
T wave—————–Ventricular repolarization

Answer:

Waves on ECGHeart ActivityCaused due to
P waveAtrial contractionAtrial depolarization
QRS waveVentricular contractionVentricular depolarization
T waveVentricular contractionVentricular repolarization

Question 6.

Cardiovascular disordersSymptom
Coronary Artery Diseases (CAD)Deposition of calcium, fat, cholesterol and ———————
—————-Pain in chest resulting from reduction in the blood supply to the cardiac muscles.
Silent Heart AttackMyocardial infarction without ———————-

Answer:

Cardiovascular disordersSymptom
Coronary Artery Diseases (CAD)Deposition of calcium, fat, cholesterol and fibrous tissues in blood vessels.
Angina pectorisPain in chest resulting from reduction in the blood supply to the cardiac muscles.
Silent Heart AttackMyocardial infarction without showing symptoms of classical heart attack.

Maharashtra Board Class 12 Biology Important Questions Chapter 8 Respiration and Circulation

Question 7.

Instrument / TechniquePurpose of use
Sphygmomanometer———————
—————-X-ray imaging of the cardiac blood vessels to locate the position of blockages.
—————To measure ECG.

Answer:

Instrument / TechniquePurpose of use
SphygmomanometerTo measure blood pressure.
AngiographyX-ray imaging of the cardiac blood vessels to locate the position of blockages.
ElectrocardiographTo measure ECG.

Diagram based questions

Question 1.
Give the name and function of A and ‘B’ from the diagram given below
Maharashtra Board Class 12 Biology Important Questions Chapter 8 Respiration and Circulation 7
Answer:

NameFunction
(A) EpiglottisEpiglottis prevents the entry of food into trachea.
(B) Tracheal cartilageTracheal cartilage prevents collapse of trachea and always keeps it open.

Question 2.
Label parts A’ and ‘B’ from the following diagram and answer the following questions
Maharashtra Board Class 12 Biology Important Questions Chapter 8 Respiration and Circulation 8
(a) What is the partial pressure of O2 in part ‘B’?
(b) What is the partial pressure of CO2 in part A’?
(c) How many alveoli are present in the lungs?
Answer:
Label A → Blood from pulmonary artery Label B → Alveolus
(a) The partial pressure of O2 in alveolar air is 104 mm Hg.
(b) The partial pressure of CO2 in pulmonary capillaries is 45 mmHg.
(c) There are about 700 million alveoli in the lungs.

Question 3.
Label parts A’ and ‘B’ from the given diagram and give their functions.
Maharashtra Board Class 12 Biology Important Questions Chapter 8 Respiration and Circulation 9
Answer: Part A → Sinoatrial [SA] node
Function : SA node acts as pacemaker of heart because it has the power of generating a new wave of contraction and making the pace of contraction.
Part B → Atrioventricular [AV] node
Function : Atrioventricular [AV] node acts as pace-setter of heart.

Question 4.
Sketch and label the dorsal (posterior) view of human heart.
Answer:
Maharashtra Board Class 12 Biology Important Questions Chapter 8 Respiration and Circulation 10

Question 5.
Sketch and label the ventral (anterior) view of human heart.
Answer:
Maharashtra Board Class 12 Biology Important Questions Chapter 8 Respiration and Circulation 11

Question 6.
Sketch and label – Electrocardiogram or ECG.
Answer:
Maharashtra Board Class 12 Biology Important Questions Chapter 8 Respiration and Circulation 12

Question 7.
Sketch and label – T.S. of Artery, Vein and Capillary.
Answer:
Maharashtra Board Class 12 Biology Important Questions Chapter 8 Respiration and Circulation 13

Question 8.
Observe the diagrams of blood cells and answer the following questions
Maharashtra Board Class 12 Biology Important Questions Chapter 8 Respiration and Circulation 14
(a) Which of the above is agranulocyte ?
(b) Describe its origin and structure.
(c) Mention its types.
(d) Explain its function.
Answer:
(a) The figure ‘D’ is agranulocyte.
(b) Structure : Agranulocytes do not show cytoplasmic granules and their nucleus is not lobed.
(c) Agranulocytes are of two types, viz. lymphocytes and monocytes.
(d) Functions of agranulocytes : Agranulocytes are responsible for immune response of body by producing antibodies and monocytes are phagocytic in function.

Question 9.
Observe the diagrammatic representation of double circulation and answer the given questions.
Maharashtra Board Class 12 Biology Important Questions Chapter 8 Respiration and Circulation 15

  1. Why the circulation shown in the above diagram is called double circulation?
  2. What are the two main routes of double circulation?
  3. Which blood vessels carry oxygented blood to heart and deoxygenated blood to lungs?
  4. Which blood vessels carry deoxygented blood to heart and oxygenated blood to body organs?

Answer:

  1. During circulation, blood passes twice through the heart, therefore it is called double circulation.
  2. (1) Pulmonary circulation which is from heart to lungs and back from lungs to heart.
    (2) Systemic circulation which is from heart to body and back from all body organs to the heart.
  3. Oxygenated blood is carried to the heart by pulmonary veins. Dexoygenated blood is carried to the lungs by pulmonary artery.
  4. Deoxygenated blood is carried to heart by superior and inferior vena cavae. Oxygenated blood is carried to the body organs by systemic or dorsal aorta.

Maharashtra Board Class 12 Biology Important Questions Chapter 8 Respiration and Circulation

Question 10.
Observe the cardiac cycle given below and answer the following questions
Maharashtra Board Class 12 Biology Important Questions Chapter 8 Respiration and Circulation 16

  1. Which phases of cardiac cycle are shown in the above diagrammatic representation ?
  2. How much time is taken for entire heart to be in diastole?
  3. How much longer is ventricular systole as compared to atrial systole?

Answer:

  1. There are four main phases of cardiac cycle shown in the above diagram. They are
    (1) AS : Arterial systole. (2) AD : Atrial diastole. (3) VS : Ventricular systole.
    (4) VD : Ventricular diastole which is along with joint diastole.
  2. Diastole of entire heart is called joint diastole, which is for about 0.4 second.
  3. Ventricular systole is almost for the double time than the atrial systole. Atrial systole is for 0.15 second whereas ventricular systole is for 0.3 second.

Long Answer Questions

Question 1.
Describe the respiratory system of human.
Answer:
Respiratory system of human : Human respiratory system consists of nostrils, nasal chambers, pharynx, larynx, trachea, bronchi, bronchioles, lungs, diaphragm and intercostal muscles.
1. Nostrils and nasal chambers:
Maharashtra Board Class 12 Biology Important Questions Chapter 8 Respiration and Circulation 17

  1. Oxygen rich air is taken in the body through the nostrils or external nares. They are external opening of the nose. Carbon dioxide and water vapour are also released out of the body through the same passage i.e. the nostrils.
  2. Internal nares open into the pharynx. The space between external and internal nares is knows as nasal chamber which is lined internally by mucous membrane and ciliated epithelium.
  3. Nasal chamber is divided into two parts by a cartilage called mesethmoid. Each part of these halves is further divided into three regions, viz. vestibule, respiratory part and sensory part.
  4. Vestibule is the anteriormost part of nasal chamber. In the vestibule fine hairs are present. They filter out the dust particles and prevent them from going inside.
  5. Respiratory part is the second region which is richly supplied with the capillaries. Air is made warm and moist in this region.
  6. Sensory part is lined by sensory epithelium. It is concerned with the detection of smell.

2. Pharynx:

  1. Pharynx is a short and vertical tube measuring about 12 cm in length. In pharynx the respiratory and food passages cross each other.
  2. The upper part of pharynx is known as naso-pharynx which conducts the air. The lower part is called laryngo-pharynx or oro¬pharynx which conducts food to the oesophagus.
  3. Tonsils that are made up of lymphatic tissue are present in the pharynx. They kill the bacteria by trapping them in the mucus.

3. Larynx:

  1. Larynx produces sound. In males it increases in size at puberty. This is termed as Adam’s apple. It is clearly seen in the neck region.
  2. From pharynx air enters the larynx. The opening through which it enters is called glottis. Glottis is guarded by a flap called epiglottis.
  3. Epiglottis prevents the entry of food particles into the trachea.
  4. TWo folds of elastic tissue called vocal cords are seen along the side of glottis. When they vibrate the sound is produced.

4. Trachea:

  1. The trachea or wind pipe is about 12 cm long and 2.5 cm wide.
  2. It is situated in front of the oesophagus and runs downwards in the thorax through the neck.
  3. The trachea is made up of fibrous muscular tissue wall which is supported by ‘C’-shaped cartilages. These cartilaginous rings are 16 to 20 in number.
  4. Internally the tracheal wall bears ciliated epithelium and mucous glands.
  5. When any foreign particle enters the trachea inadvertently. It is thrown out by coughing action.
  6. Mucous and ciliary action remove the dust particles and push them upwards to the larynx. These particles are then gulped and taken into the oesophagus.

5. Bronchi and bronchioles:

  1. At the distal end, the trachea divides into two bronchi (Singular – bronchus). Bronchi lie below the sternum or breast bone.
  2. Each bronchus has a complete ring of cartilage for support. The two bronchi enter into the lungs on either side.
  3. After entering into the lungs each bronchus divides into secondary and tertiary bronchi. The tertiary bronchi divide and re-divide to form minute bronchioles.
  4. Bronchioles do not have cartilages in their walls. Each bronchiole ends into a balloon like alveolus.
  5. Owing to the presence of alveoli the lungs become spongy and elastic.

6. Lungs:

  1. Lungs are principal respiratory organs located in the thoracic cavity.
  2. They are pinkish, soft, hollow, paired, elastic and distensible organs.
  3. Each lung is enclosed in a pleural sac which consists of two membranes, viz. an outer parietal and inner visceral.
  4. The parietal and visceral membranes enclose pleural cavity which is filled with pleural fluid. The pleural fluid lubricates and prevents friction when pleural membranes slide on each other.
  5. Lungs are highly vascular as they are richly supplied with blood capillaries.
  6. The left lung has two lobes while the right lung has three lobes. Each lobe has many bronchioles and alveolar sacs. The alveolar sacs are spherical and thin walled.
  7. Each alveolar sac contains about 20 alveoli. The alveoli appear as a bunch of grapes. The lobule in the lung thus consists of alveolar ducts, alveolar sacs and alveoli.
  8. Each alveolus has thin and elastic walls. It is about 0.1 mm in diameter. Alveoli are covered by network of capillaries from pulmonary artery and pulmonary vein. A network of pulmonary capillaries supply the alveolus.
  9. The alveolar wall is 0.0001 mm thick and made up of simple, non-ciliated, squamous epithelium. It has collagen and elastin fibres.
  10. Every lung has about 700 million alveoli. They increase the surface area of the lungs for exchange of gases.

Question 2.
Describe the process of respiration in man.
OR
Describe the mechanism of respiration in human beings.
Answer:
Respiration includes breathing, external respiration, internal respiration and cellular respiration.
A. Breathing : During breathing air comes in and goes out of the lungs. The rate of gaseous exchange is speeded up by breathing. Breathing involves two processes, viz. inspiration and expiration
1. Inspiration:

  1. Inspiration is the process in which the air containing oxygen is taken inside the lung.
  2. Inspiration is the active process which is possible due to intercostal muscles, sternum and diaphragm.
  3. During inspiration, intercostal muscles contract, ribs are pulled outward as a result of which the space in the thoracic cavity is increased.
  4. At the same time the lower part of the breast bone is raised and diaphragm flattens by contraction.
  5. The volume of thoracic cavity is thus increased.
  6. Pressure in the lungs decreases as the lungs expand and their volume is increased. Owing to this the atmospheric air enters inside the body through respiratory passage and reaches the lungs.

2. Expiration:

  1. Expiration is the process in which air containing carbon dioxide and water vapour is expelled out of the lungs.
  2. Expiration is a passive process.
  3. During expiration, intercostal muscles relax and ribs are pulled inwards.
  4. The diaphragm relaxes and becomes dome shaped. Intercostal muscles contract simultaneously and due to these events, the volume of the thoracic cavity is reduced.
  5. The pressure on the lungs is thus increased as a result of which they are compressed.
  6. Due to this, air rushes out of the lungs and is expelled out through the nose.
    Maharashtra Board Class 12 Biology Important Questions Chapter 8 Respiration and Circulation 18

B. External respiration : The process of external respiration takes place in the lungs where oxygen from the lungs diffuses into the blood capillaries present in the lung tissue. Similarly carbon dioxide from the blood capillaries diffuses out and enters in the alveoli in the lungs.

C. Internal respiration : Internal respiration takes place in the cells of the body. Oxygen brought by the blood is given to the cells and the tissues during internal respiration. Similarly carbon dioxide passes into the blood cells from the cells and tissues.

D. Cellular respiration : Cellular respiration takes place in the mitochondria of the cell, where oxygen is utilised to liberate energy in the form of ATP molecules.

Question 3.
Describe the respiratory disorders.
Answer:

  1. Following are some respiratory disorders : Emphysema is caused due to alveolar abnormalities. Chronic bronchitis results into coughing and shortness of breath.
  2. Viral and bacterial respiratory diseases. Acute bronchitis, sinusitis, laryngitis and pneumonia are some of the inflammatory diseases caused either due to virus or due to bacteria.
  3. Allergens like pollen or pet dander can cause asthma. In asthma constriction of bronchioles takes place causing periodic wheezing and difficulty in breathing.
  4. Occupational hazards cause respiratory diseases like silicosis or asbestosis. In these disorders there is inflammation fibrosis leading to lung damage.

Treatments of respiratory diseases:

  1. Bacterial diseases can be completely cured by specific antibiotics.
  2. Viral diseases need to be taken care of by using vaporizers and decongestants.
  3. Asthma needs treatment by inhalers and nebulizers.
  4. For occupational disorders proper mask and other protective gear is a must.
  5. Lethal diseases like pneumonia should be controlled by medication and rest.

Question 4.
How does transport of O2 and CO2 take place in man?
Answer:
1. Transport of O2:

  1. Only 3% of the total oxygen is transported in a dissolved state by the plasma.
  2. The remaining 97% is transported in the form of oxyhaemoglobin in the RBCs.
  3. Hemoglobin present is RBCs combines with oxygen to form oxyhaemoglobin.
    Hb + 4O2 → Hb (4O2)
  4. Oxyhaemoglobin is transported from lungs to the tissues where it readily dissociates to release O2.
  5. Binding of oxygen with haemoglobin in the alveoli and release of oxygen into the tissue cells depends upon the difference in partial pressure of O2 and CO2.

2. Transport of CO2: Carbon dioxide is transported by RBCs and plasma in three different forms.

  1. By plasma in solution form (7%) : About 7% of CO2 is transported in a dissolved form as carbonic acid (which can be broken down into CO2 and H2O).
    CO2 + H2O = H2CO3.
  2. By bicarbonate ions (70%) : Nearly 70% of carbon dioxide is transported in the form of sodium bicarbonate/potassium bicarbonate in the plasma.
  3. RBCs contains an enzyme, carbonic anhydrase. In the presence of this enzyme CO2 combines with water to form carbonic acid.
  4. Carbonic anhydrase also brings about dissociation of carbonic acid immediately tending to large accumulation of HCO3- ions inside the RBCs.
    CO2 + H2O Carbonic anhydrese H2CO2 Carbonic anhydrase H+ + HCO3-
  5. The bicarbonate ions moves out of RBCs and this would bring about imbalance of the charge inside the RBCs.
  6. To maintain the ionic balance, Cl ions diffuse from plasma into the RBCs. This movement of chloride ions is known as chloride shift or Hamburger’s phenomenon.
  7. HCO3- ions from the plasma then joins to Na+/K+ forming NaHCO3/KHCO3 (to maintain PH of blood).
    HCO3- + Na+ → NaHCO3 Sodium bicarbonate
  8. H+ is taken up by haemoglobin to form Reduced Hb (HHb).
  9. At the level of the lungs due to the low partial pressure of the alveolar air, hydrogen ion and bicarbonate ions recombine to form carbonic acid and in presence of carbonic anhydrase it again yields carbon dioxide and water.
    H+ + HCO3- Carbonic anhydrase H2CO3 Carbonic anhydrase CO2 + H2O.

3. By red blood cells (23%):

  1. Carbon dioxide binds with the amino group of the haemoglobin and form a loosely bound compound carbaminohaemoglobin Hb + CO2 = HbCO2
  2. Due to low partial pressure of CO2 at alveolus carbaminohaemoglobin decomposes releasing the carbon dioxide.

Maharashtra Board Class 12 Biology Important Questions Chapter 8 Respiration and Circulation

Question 5.
Describe different types of leucocytes.
OR
Describe five types of leucocytes, with the help of diagrams. Add a note on their functions.
Answer:
Maharashtra Board Class 12 Biology Important Questions Chapter 8 Respiration and Circulation 19

  1. Leucocytes or White Blood Corpuscles (WBCs) are colourless, nucleated, amoeboid and phagocytic cells.
  2. Their size ranges between 8 to 15 pm. Total WBC count is 5000 to 9000 WBCs/cu mm of blood. The average life span of a WBC is about 3 to 4 days.
  3. They are formed by leucopoiesis in red bone marrow, spleen, lymph nodes, tonsils, thymus and Payer’s patches, whereas the dead WBCs are destroyed by phagocytosis in blood, liver and lymph nodes.
  4. Leucocytes are mainly divided into two types, viz., granulocytes and agranulocytes.
  5. Granulocytes : Granulocytes are cells with granular cytoplasm and lobed nucleus. Based on their staining properties and shape of nucleus, they are of three types, viz. neutrophils, eosinophils and basophils.

(I) Neutrophils:

  1. In neutrophils, the cytoplasmic granules take up neutral stains.
  2. Their nucleus is three to five lobed.
  3. It may undergo changes in structure hence they are called polymorphonuclear leucocytes or polymorphs.
  4. Neutrophils are about 70% of total WBCs.
  5. They are phagocytic in function and engulf microorganisms.

(II) Eosinophils or acidophils:

  1. Cytoplasmic granules of eosinophils take up acidic dyes such as eosin. They have bilobed nucleus.
  2. Eosinophils are about 3% of total WBCs.
  3. They are non-phagocytic in nature.
  4. Their number increases (i.e. eosinophilia) during allergic conditions.
  5. They have antihistamine property.

(III) Basophils:

  1. The cytoplasmic granules of basophils take up basic stains such as methylene blue.
  2. They have twisted nucleus.
  3. In size, they are smallest and constitute about 0.5% of total WBCs.
  4. They too are non-phagocytic.
  5. Their function is to release heparin which acts as an anticoagulant and histamine that is involved in inflammatory and allergic reaction.

Agranulocytes : There are two types of agranulocytes, viz. monocytes and lymphocytes. Agranulocytes do not show cytoplasmic granules and their nucleus is not lobed. They are of two types, viz. lymphocytes and monocytes.
(I) Lymphocytes:

  1. Agranulocytes with a large round nucleus are called lymphocyte.
  2. They are about 30% of total WBCs.
  3. Agranulocytes are responsible for immune response of the body by producing antibodies.

(II) Monocytes:

  1. Largest of all WBCs having large kidney shaped nucleus are monocytes. They are about 5% of total WBCs.
  2. They are phagocytic in function.
  3. They can differentiate into macrophages for engulfing microorganisms and removing cell debris. Hence they are also called scavengers.
  4. At the site of infections they are seen in more enlarged form.

Question 6.
Give an account of external features of the human heart.
Answer:
(1) The heart is hollow, muscular, conical organ about the size of one’s fist with broad base and narrow apex tilted towards left measuring about 12 cm in length. 9 cm in breadth and weighing about 250 to 300 grams.

(2) The human heart has four chambers, two atria which are superior, small, thin walled receiving chambers and two ventricles which are inferior, large, thick walled, distributing chambers.
Maharashtra Board Class 12 Biology Important Questions Chapter 8 Respiration and Circulation 20

(3) Externally there is a transverse groove between the atria and the ventricles which is known as atrioventricular groove or coronary sulcus.

(4) Between the right and left ventricles there is interventricular sulcus (pi. sulci). In these sulci the coronary arteries and coronary veins are present.

(5) Oxygenated blood to the heart is supplied by coronary arteries while coronary veins collect deoxygenated blood from the heart. The coronary veins join to form coronary sinus which opens into the right atrium.

(6) Right atrium is larger in size than the left atrium. Deoxygenated blood from all over the body is brought through superior vena cava and inferior vena cava and poured into right atrium. Oxygenated blood from lungs is brought to heart by two pairs of pulmonary veins which carry it to the left atrium.

(7) Pulmonary trunk is seen arising from the right ventricle, which carries deoxygenated blood to lungs. While systemic aorta arises from the left ventricle and carries oxygenated blood to all parts of the body.

(8) The pulmonary trunk and systemic aorta are connected by ligamentum arteriosum that represents remnant of ductus arteriosus of foetus.

Question 7.
With the help of well labelled diagram describe the internal structure of human heart.
OR
Sketch and label internal view of heart.
Answer:
Maharashtra Board Class 12 Biology Important Questions Chapter 8 Respiration and Circulation 21
The heart shows four chambers with two atria and two ventricles.

I. Atria:
1. Right atrium:

  1. There are two atria which are separated from each other by interatrial septum. They are thin walled receiving chambers on the upper side.
  2. The right atrium receives deoxygenated blood from upper part of body through superior vena cava and from the lower part of the body by inferior vena cava. In the right atrium opens the coronary sinus.
  3. Eustachian valve guards the opening of inferior vena cava while opening of coronary sinus is guarded by Thebesian valve.
  4. On the right side of interatrial septum is seen an oval depression called the fossa ovalis. In the interatrial septum of the foetus there is an oval opening called foramen ovale. Fossa ovalis is remnant of this foramen ovale.
  5. Right atrium opens into the right ventricle.

2. Left atrium:

  1. The oxygenated blood from the lungs is brought into left atrium through four openings of pulmonary veins.
  2. Left atrium opens into the left ventricle.

II. Ventricles:

  1. There are two ventricles which are separated from each other by interventricular septum. They are two thick walled distributing chambers situated on the lower side of the heart.
  2. Left ventricle has thickest wall as it pumps blood to all parts of the body.
  3. The inner surface of the ventricle is thrown into a series of irregular muscular ridges called columnae carnae or trabeculae carnae.
  4. Each atrium opens into the ventricle of its side through atrioventricular aperture. These apertures are guarded by valves made up of connective tissue. The right atrioventricular valve has three flaps hence called tricuspid valve. Left atrioventricular valve has two flaps hence called bicuspid valve or mitral valve.
  5. Bicuspid and tricuspid valves are attached to papillary muscles of ventricles by chordae tendinae. The chordae tendinae prevent the valves from turning back into the atria during the contraction of ventricles.
  6. From the right ventricle arises pulmonary trunk which carries deoxygenated blood to lungs for oxygenation.
  7. From the left ventricle arise systemic aorta which distributes oxygenated blood to all parts of the body.
  8. Pulmonary aorta and systemic aorta has three semilunar valves at the base which prevent backward flow of blood during ventricular diastole.

Question 8.
With the help of suitable diagram, describe the conducting system of human heart.
Answer:
(1) The human heart is myogenic.

(2) Conducting system of the heart consists of sinoatrial node, atrioventricular node, bundles of His and Purkinje’s fibre system.

(3) The pacemaker of the heart is sinoatrial node because here the heartbeat originates. Pacemaker has power of generation of wave of contraction. This is modified cardiac tissue, also called a nodal tissue.

(4) SA node is situated in the wall of right atrium near the opening of superior vena cava. The wave of contraction generated by SA node is conducted by cardiac muscle fibres to both the atria. This results in contraction resulting into atrial systole.
Maharashtra Board Class 12 Biology Important Questions Chapter 8 Respiration and Circulation 22

(5) The atrioventricular node (AV node) is located in the wall of right atrium near the opening of coronary sinus. AV node receives the wave of contraction generated by SA node through intermodal pathways.

(6) Bundle of His arises from AV node and divides into right and left bundle branches. These are located in the interventricular septum.

(7) The bundle branches further form Purkinje fibres which penetrate into myocardium of ventricles.

(8) The bundle of His and Purkinje fibres conduct the wave of contraction from AV node to myocardium of ventricles causing ventricular systole.

Maharashtra Board Class 12 Biology Important Questions Chapter 8 Respiration and Circulation

Question 9.
Describe the detail cardiac cycle.
OR
Explain the working of heart.
Answer:
(1) The working of heart or cardiac cycle is formed by atrial systole, ventricular systole and joint diastole. It takes place in 0.8 second.

(2) During atrial systole from right atrium, the deoxygenated blood is poured into right ventricle through atrioventricular aperture. Similarly, from left atrium, the oxygenated blood enters the left ventricle through atrioventricular aperture. This entire atrial systole lasts for 0.1 second.

(3) After auricular systole follows the ventricular systole. During ventricular systole, the deoxygenated blood from the right ventricle enters the pulmonary trunk, which carries blood to lungs for oxygenation. At the same time, the oxygenated blood from the left ventricle enters the aorta which is then supplied to all parts of the body. The ventricular systole lasts for 0.3 second.

(4) Joint diastole or complete cardiac diastole is the phase taking place after the systole, when the entire heart undergoes relaxation, for 0.4 second.

(5) During joint diastole, the right atrium receives deoxygenated blood from all parts of the body through superior vena cava, inferior vena cava and coronary sinus. The left atrium receives oxygenated blood from the lungs through two pairs of pulmonary veins.

Question 10.
What is blood clotting? How and when does it occur?
Answer:

  1. Blood clotting is coagulation of blood in order to stop the blood flow and resuting blood loss at the time of injury.
  2. When the blood vessel is intact, blood does not clot due to the presence of active anticoagulants like heparin and antithrombin. But when there is an injury causing rupture of a blood vessel, bleeding starts.
  3. This bleeding is stopped by the process of blood clotting during which liquid blood is converted into semisolid jelly.

The events occurring during blood clotting are as follows:

  1. Release of thromboplastin from thrombocytes and injured tissue.
  2. Formation of enzyme prothrombinase in the blood due to initiation of thromboplastin.
  3. Conversion of inactive prothrombin into active thrombin by prothrombinase in the presence of Ca ions.
  4. Conversion of soluble fibrinogen into insoluble fibrin by thrombin.
  5. Formation of a clot by enmeshing platelets, other blood cells and plasma in the fibrin fibres enmesh.

These reactions occur in 2 to 8 minutes. Therefore, clotting time is said to be 2 to 8 minutes.

Question 11.
What is repolarization and depolarization ?
Answer:
Repolarization is a stage of an action potential in which the cell experiences a reduction of voltage due to the efflux of potassium (K+) ions along its electrochemical gradient. This phase occurs after the cell reaches its highest voltage from depolarization.

Depolarization occurs in the four chambers of the heart : both atria first and then both ventricles. The SA node sends the depolarization wave to the atrioventricular (AV) node which-with about a 100 minutes delay to let the atria finish contracting-then causes contraction in both ventricles, seen in the QRS wave.

Question 12.
What is the correlation between depolarization and repolarization as well as contraction and relaxation of the heart?
Answer:
Depolarization and Repolarization:

  1. When cardiac cells are at rest, they are polarized, meaning no electrical activity takes place.
  2. The cell membrane of the cardiac muscle cell separates different concentrations of ions, such as sodium, potassium, and calcium. This is called the resting potential.
  3. Electrical impulses are generated by specialized cardiac cells automatically.
  4. Once an electrical cell generates an electrical impulse, this electrical impulse causes the ions to cross the cell membrane and causes the action potential, also called depolarization.
  5. The movement of ions across the cell membrane through sodium, potassium and calcium channels, is the drive that causes contraction of the cardiac cells/muscle.
  6. Depolarization with corresponding contraction of myocardial muscle moves as a wave through the heart. Depolarization thus corresponds with contraction of heart.
  7. Repolarization is the return of the ions to their previous resting state, which corresponds with relaxation of the myocardial muscle. Repolarization thus corresponds with relaxation of heart.
  8. Depolarization and repolarization are electrical activities which cause muscular activity.
  9. The electrical changes in the myocardial cell during the depolarization – repolarization cycle is detected on ECG.

Maharashtra Board Class 12 Biology Important Questions Chapter 8 Respiration and Circulation

Question 13.
How are the signals detected and amplified by electrocardiograph?
Answer:

  1. The action potential created by contractions of the heart wall spreads electrical currents from the heart throughout the body.
  2. The spreading electrical currents create different potentials at points in the body, which can be sensed by electrodes placed on the skin.
  3. The electrodes are made of metals and salts and they act as biological transducers.
  4. Ten electrodes are attached to different points on the body while taking ECG.
  5. There Eire three main leads responsible for measuring the electrical potential difference between arms and legs.
  6. Electrical potential difference between electrodes is recorded.
  7. As in all ECG lead measurements, the electrode connected to the right leg is considered the ground node.
  8. These ECG signals are acquired using a biopotential amplifier and then displayed using instrumentation software. This is recorded on ECG machine or electrocardiograph. The recorded ECG is anailysed by an expert.

Maharashtra Board Class 12 Biology Important Questions Chapter 7 Plant Growth and Mineral Nutrition

Balbharti Maharashtra State Board 12th Biology Important Questions Chapter 7 Plant Growth and Mineral Nutrition Important Questions and Answers.

Maharashtra State Board 12th Biology Important Questions Chapter 7 Plant Growth and Mineral Nutrition

Multiple-choice questions

Question 1.
For the following, which is the aspect of growth? Example – Increase in length, size, and a number of cells.
(a) Quantitative
(b) Qualitative
(c) Both (a) and (b)
(d) Three dimensional
Answer:
(a) Quantitative

Question 2.
In vascular plants, growth takes place due to ………………..
(a) conducting tissues
(b) embryo
(c) meristems
(d) stem cell
Answer:
(c) meristems

Maharashtra Board Class 12 Biology Important Questions Chapter 7 Plant Growth and Mineral Nutrition

Question 3.
What is growth?
(a) Temporary, irreversible increase in an organism.
(b) Permanent, reversible increase in an organism.
(c) Permanent, irreversible decrease in an organism.
(d) Permanent, irreversible increase in an organism.
Answer:
(d) Permanent, irreversible, increase in an organism

Question 4.
Based on location, plants have these three types of meristems.
(a) Basal, Intercalary and Lateral
(b) Apical, Basal and Lateral
(c) Apical, Interfascicular and Lateral
(d) Apical, Intercalary, Lateral
Answer:
(d) Apical, Intercalary, Lateral

Question 5.
Meristem cells are ………………..
(a) thin walled, vacuolated with prominent nucleus
(b) thick walled, non-vacuolated without nucleus
(c) thin walled, non-vacuolated with prominent nucleus
(d) thick walled, vacuolated with prominent nucleus
Answer:
(c) thin walled, non-vacuolated with prominent nucleus

Question 6.
Select correct sequence of phases of growth.
(a) Phase of formation, phase of elongation, phase of maturation
(b) Phase of cell division, phase of enlargement, phase of elongation
(c) Phase of formation, phase of maturation, phase of elongation
(d) Phase of enlargement, phase of cell division, phase of maturation
Answer:
(a) Phase of formation, phase of elongation, phase of maturation

Question 7.
In which phase of growth, growth rate is at accelerated pace?
(a) Lag phase
(b) Log phase
(c) Steady phase
(d) Stationary phase
Answer:
(b) Log phase

Question 8.
Water is essential for growth because it is necessary ………………..
(a) for turgidity
(b) as nutrient
(c) as raw material
(d) for gravity
Answer:
(a) for turgidity

Question 9.
Which equipment is suitable for measuring linear growth of shoot?
(a) Horizontal microscope
(b) Spectroscope
(c) Crescograph
(d) Auxanometer
Answer:
(d) Auxanometer

Question 10.
Which is correct expression of absolute growth rate (AGR)?
(a) AGR = \(\frac { dn }{ dt }\)
(b) AGR = \(\frac { dt }{ dn}\)
(c) AGR = \(\frac { RGR }{ n }\)
(d) AGR = \(\frac { n }{ RGR }\)
Answer:
(a) AGR = \(\frac { dn }{ dt }\)

Question 11.
Arithmetic growth in plants shows ……………….. graph.
(a) Sigmoid
(b) J-shaped
(c) linear
(d) elliptical
Answer:
(c) linear

Question 12.
What is grand period of growth?
(a) The total time required for all phases to occur
(b) The total time required for exponential phase
(c) The total time required for Lag and Log phase together
(d) The toted time required for stationary phase
Answer:
(a) The total time required for all phases to occur

Question 13.
Which tissue is formed by process of de-differentiation?
(a) Intrafascicular cambium
(b) Secondary phloem
(c) Interfascicular cambium
(d) Secondary xylem
Answer:
(c) Interfascicular cambium

Question 14.
The example of environmental plasticity- heterophylly observed is ………………..
(a) Cotton
(b) Coriander
(c) Larkspur
(d) Buttercup
Answer:
(d) Buttercup

Question 15.
Synthesis of IAA takes place from amino acid ………………..
(a) Methionine
(b) Tryptophan
(c) Valine
(d) Aspartic acid
Answer:
(b) Tryptophan

Question 16.
Find the odd one out.
(a) IAA
(b) 2, 4-D
(c) NAA
(d) IBA
Answer:
(a) IAA

Question 17.
The selective herbicide is ………………..
(a) IBA
(b) GA3
(c) 2, 4 D
(d) NAA
Answer:
(c) 2, 4 D

Question 18.
This hormone promotes rooting in artificial method of cutting ………………..
(a) Gibberellin
(b) Auxin
(c) Cytokinin
(d) Dormin
Answer:
(b) Auxin

Question 19.
Chemically the peculiar structure of gibberellins is ……………….. ring.
la) pyrole ring
(b) purine ring
(c) gibbeane ring
(d) pyrimidine
Answer:
(c) gibbeane ring

Question 20.
First natural cytokinin was obtained from ……………….. by Letham.
(a) Maize grains
(b) Coconut milk
(c) Rice seedling
(d) Tomato
Answer:
(a) Maize grains

Question 21.
A low ratio of cytokinin to auxin induces ……………….. in plants.
(a) rooting
(b) shooting
(c) bud formation
(d) flowering
Answer:
(a) rooting

Question 22.
Apical dominance : Auxin : : Fruit ripening : ?
(a) Gibberellin
(b) Cytokinin
(c) Ethylene
(d) Abscissic acid
Answer:
(c) Ethylene

Question 23.
Which hormone is known as stress hormone ?
(a) Auxin
(b) Gibberellin
(c) Ethylene
(d) Abscissic acid
Answer:
(d) Abscissic acid

Question 24.
Abscissic acid is synthesised from ………………..
(a) Methionine
(b) Malic acid
(c) Mevalonic acid
(d) Mucin
Answer:
(c) Mevalonic acid

Question 25.
Photoperiodic response is because of pigment ………………..
(a) Cytochrome
(b) Phytochrome
(c) Anthocyanin
(d) Phycobilin
Answer:
(b) Phytochrome

Question 26.
The favourable temperature for vernalization is ………………..
(a) 1 to 6 °C
(b) 11 to 16 °C
(c) 10 to 16 °C
(d) – 1 to 1 °C
Answer:
(a) 1 to 6 °C

Question 27.
Identify the group of non-mineral elements needed by plants.
(a) PO4, CO3, SO4
(b) C, H, O
(c) N, P K
(d) C, H, N
Answer:
(b) C, H, O

Maharashtra Board Class 12 Biology Important Questions Chapter 7 Plant Growth and Mineral Nutrition

Question 28.
The deficiency symptoms of these minerals are visible in young leaves ………………..
(a) Ca, CO
(b) S, P
(c) Ca, S
(d) Zn, Mg
Answer:
(c) Ca, S

Question 29.
The symptom chlorosis is observed as ………………..
(a) yellowing of leaf
(b) premature leaf fall
(c) malformation of leaf
(d) localized death of tissue
Answer:
(a) yellowing of leaf

Question 30.
…………….. is constituent of chlorophyll.
(a) Mn
(b) Mg
(c) Mo
(d) Fe
Answer:
(b) Mg

Question 31.
This is essential for O2 evolution in photosynthesis and for proper solute concentration
(a) Ca
(b) Cu
(c) Cl
(d) Co
Answer:
(c) Cl

Question 32.
For active absorption of mineral uptake, energy is supplied by ………………..
(a) respiration
(b) chemosynthesis
(c) photosynthesis
(d) transpiration
Answer:
(a) respiration

Question 33.
Cyanobacteria fix nitrogen in specialised cells called ………………..
(a) Velamen
(b) Haustoria
(c) Heteocysts
(d) Hormogonia
Answer:
(c) Heteocysts

Question 34.
In root nodule, symbiotic nitrogen fixer organism is ………………..
(a) Rhizopus
(b) Pseudomonas
(c) Rhizobium
(d) Nitrosomonas
Answer:
(c) Rhizobium

Question 35.
In plant body, amides are transported through ………………..
(a) sieve tubes
(b) xylem vessels
(c) phloem parenchyma
(d) plasmodesmata
Answer:
(b) xylem vessels

Question 36.
Building blocks of proteins are ………………..
(a) amides
(b) amino acids
(c) carboxylic acid
(d) nitrates
Answer:
(b) amino acids

Question 37.
Flowering plants Aster, Dahlia and Chrysanthemum are ………………..
(a) SDP
(b) LDP
(c) DNP
(d) SDP or LDP
Answer:
(a) SDP

Question 38.
Experimental material of Garner and Allard for discovery of photoperiodism was ………………..
(a) Cucumber and Tomato
(b) Dahlia and Aster
(c) Soybean and Tobacco
(d) Cabbage and Spinach
Answer:
(c) Soybean and Tobacco

Question 39.
Which of the following is used for the production of long seedless grapes ?
(a) Auxin
(b) Cytokinin
(c) Ethylene
(d) Gibberellin
Answer:
(d) Gibberellin

Question 40.
In vernalization, the cold stimulus is perceived by ………………..
(a) axillary bud
(b) floral bud
(c) leaves
(d) apical bud (shoot apex)
Answer:
(d) apical bud (shoot apex)

Question 41.
Xanthium is ………………..
(a) SDP
(b) LDP
(c) DNP
(d) not a flowering plant
Answer:
(a) SDP

Question 42.
Growth starts slowly during the ………………..
(a) lag phase
(b) exponential phase
(c) maturation phase
(d) stationary phase
Answer:
(a) lag phase

Question 43.
Cytokinins induce the formation of ………………..
(a) shoot apex
(b) intrafascicular cambium
(c) cork cambium
(d) interfascicular cambium
Answer:
(d) interfascicular cambium

Question 44.
Bakane disease in rice is associated with the discovery of ………………..
(a) cytokinins
(b) gibberellins
(c) auxins
(d) ethylene
Answer:
(b) gibberellins

Question 45.
ABA is also known as ………………..
(a) antitoxin
(b) antivirulent
(c) antioxidant
(d) antigibber ellin
Answer:
(d) antigibberellin

Question 46.
Gibberellins were first discovered from ………………..
(a) bacteria
(b) fungi
(c) algae
(d) gymnosperms
Answer:
(b) fungi

Question 47.
Which of the following is trace element?
(a) Mg
(b) Nitrogen
(c) Sulphur
(d) Mn
Answer:
(d) Mn

Maharashtra Board Class 12 Biology Important Questions Chapter 7 Plant Growth and Mineral Nutrition

Question 48.
Which of the following is a macronutrient?
(a) Ca
(b) Mn
(c) Zn
(d) Mo
Answer:
(a) Ca

Question 49.
Nitrogen is an important constituent of ………………..
(a) carbohydrates
(b) sugars
(c) proteins
(d) polyphosphates
Answer:
(c) proteins

Question 50.
Deficiency of phosphorus causes ………………..
(a) stunted growth
(b) inward rolling of leaf margin
(c) brittle cell walls
(d) necrotic spots
Answer:
(a) stunted growth

Question 51.
Which of the following is not an essential element for plant?
(a) Sulphur
(b) Boron
(c) Iron
(d) Cadmium
Answer:
(d) Cadmium

Question 52.
…………….. is a constituent of middle lamella.
(a) Mg
(b) K
(c) Ca
(d) P
Answer:
(c) Ca

Match the columns

Question 1.

Column A (Phase of growth)Column B (Condition)
(1) Lag phase(a) Growth rate faster
(2) Log phase(b) Growth rate steady state
(3) Stationary phase(c) Growth rate slow

Answer:

Column A (Phase of growth)Column B (Condition)
(1) Lag phase(c) Growth rate slow
(2) Log phase(a) Growth rate faster
(3) Stationary phase(b) Growth rate steady state

Question 2.
Maharashtra Board Class 12 Biology Important Questions Chapter 7 Plant Growth and Mineral Nutrition 1
Answer:
(1) -(c) Arithmetic growth
(2) – (a) Rate of growth against time
(3) -(b) Geometric growth curve

Question 3.

Column AColumn B
(1) Use of measuring scale(a) Record of primary growth
(2) Crescograph(b) Increase in height of plant
(3) Auxanometer(c) Measure growth in field
(4) Horizontal microscope(d) Measurement of linear growth of shoot

Answer:

Column AColumn B
(1) Use of measuring scale(b) Increase in height of plant
(2) Crescograph(a) Record of primary growth
(3) Auxanometer(d) Measurement of linear growth of shoot
(4) Horizontal microscope(c) Measure growth in field

Question 4.

Column AColumn B
(1) Auxin(a) Bolting in rosette plants
(2) Cytokinin(b) Stimulate flowering in SDP
(3) Gibberellins(c) Promotion of growth of lateral buds
(4) Abscissic acid(d) Apical dominance

Answer:

Column AColumn B
(1) Auxin(d) Apical dominance
(2) Cytokinin(c) Promotion of growth of lateral buds
(3) Gibberellins(a) Bolting in rosette plants
(4) Abscissic acid(b) Stimulate flowering in SDP

Question 5.

Column AColumn B (Symptoms observed)
(1) Deficiency of Cu(a) Malformed leaves
(2) Deficiency of Bo(b) Leaves with yellow edges
(3) Deficiency of Zn(c) Brown heart disease
(4) Deficiency of K(d) Die back of shoot

Answer:

Column AColumn B (Symptoms observed)
(1) Deficiency of Cu(d) Die back of shoot
(2) Deficiency of Bo(c) Brown heart disease
(3) Deficiency of Zn(a) Malformed leaves
(4) Deficiency of K(b) Leaves with yellow edges

Question 6.

Column AColumn B (Organisms)
1. Symbiotic nitrogen fixationa. Nitrobacter
2. Denitrificationb. Cyanobacteria
3. Free living nitrogen fixersc. Rhizobium
4. Nitrificationd. Paracoccus

Answer:

Column AColumn B (Organisms)
1. Symbiotic nitrogen fixationc. Rhizobium
2. Denitrificationd. Paracoccus
3. Free living nitrogen fixersb. Cyanobacteria
4. Nitrificationa. Nitrobacter

Classify the following to form Column B as per the category given in Column A.

Question 1.
Classify the given plant growth regulators as per their specific control of event in plant life cycle in Column A and complete Column B.
(IAA, GA, Cytokinin, Abscissic acid)

Column AColumn B
(1) Shedding of leaves————-
(2) Induce flowering in LDP————
(3) Apical dominance————-
(4) Induce RNA synthesis————-

Answer:

Column AColumn B
(1) Shedding of leavesAbscissic acid
(2) Induce flowering in LDPGA
(3) Apical dominanceIAA
(4) Induce RNA synthesisCytokinin

Maharashtra Board Class 12 Biology Important Questions Chapter 7 Plant Growth and Mineral Nutrition

Question 2.
Classify the given specific effects of different types of auxins in Column A and complete Column B with the examples.
(IAA, NAA, 2, 4-D, IBA, 2, 4, 5-T)

Column AColumn B
(1) Selective synthetic herbicide————-
(2) Seedless fruits————
(3) Flowering in pineapple————-
(4) Synthetic auxin————-
(5) Agent orange————

Answer:

Column AColumn B
(1) Selective synthetic herbicide2, 4-D
(2) Seedless fruitsIAA
(3) Flowering in pineappleNAA
(4) Synthetic auxinIBA
(5) Agent orange2, 4, 5 – T

Question 3.
Classify the given plant hormones as their specific effect observed in plants in Column A and complete Column B.
(ABA, GA, Ethylene, Kinetin)

Column AColumn B
(1) Bolting of rosette plants————-
(2) Epinasty————
(3) Closure of stomata————-
(4) Proliferation of callus————-

Answer:

Column AColumn B
(1) Bolting of rosette plantsGA
(2) EpinastyEthylene
(3) Closure of stomataABA
(4) Proliferation of callusKinetin

Question 4.
Classify the given disease to their cause given in Column B.
(Ethylene, GA, Deficiency of BO, Deficiency of Cu)

Column AColumn B
(1) Brown heart disease————-
(2) Bakane disease of Rice————
(3) Die back of shoot————-
(4) Degreening of Banana————-

Answer:

Column AColumn B
(1) Brown heart diseaseDeficiency of BO
(2) Bakane disease of RiceGA
(3) Die back of shootDeficiency of Cu
(4) Degreening of BananaEthylene

Question 5.
Classify the given organisms related to Nitrogen cycle in Column B.
[Nitrobacter, Rhizobium, Pseudomonas, Nitrosococcus)

Column AColumn B
(1) Symbiont in root nodule————-
(2) Conversion of nitrite to nitrate————
(3) Denitrification process————-
(4) Conversion of ammonia to nitrite————-

Answer:

Column AColumn B
(1) Symbiont in root noduleRhizobium
(2) Conversion of nitrite to nitrateNitrobacter
(3) Denitrification processPseudomonas
(4) Conversion of ammonia to nitriteNitrosococcus

Very short answer questions

Question 1.
Enlist the types of meristems that we observe in plants.
Answer:
In plants, there are apical, intercalary and lateral meristems.

Question 2.
Give characteristic features of meristematic cells.
Answer:
Meristematic cells are thin walled, non- vacuolated with prominent nuclei having granular cytoplasm and are capable of cell division.

Question 3.
What is the role of oxygen in growth?
Answer:
Oxygen is required for respiration of cells and release of energy for the process of growth.

Question 4.
What is the role of water for growth?
Answer:
Water maintains turgidity of the cell and is chief component of protoplasm as well as it is a medium for various biochemical reactions.

Question 5.
Mention the mathematical formula for rate of absolute growth.
Answer:
Absolute Growth Rate = AGR = \(\frac { dn }{ dt }\) where dn is cell number and dt is time interval.

Question 6.
What is exponential phase of growth curve?
Answer:
In exponential phase or log phase, growth rate is faster. It accelerates and reaches its maximum.

Question 7.
Is there any relation between phases of growth and regions of growth curve?
Answer:
Yes, there is relation between the two as initially growth is slow, which accelerates and ultimately it slows down and becomes steady which is observed in growth curve.

Question 8.
Which plant organ does show both arithmetic and geometric growth?
Answer:
Embryo that develops from zygote inside the seed shows initially the growth which is geometric and later on arithmetic.

Maharashtra Board Class 12 Biology Important Questions Chapter 7 Plant Growth and Mineral Nutrition

Question 9.
Describe the example of plasticity related to internal stimuli.
Answer:
In plants like cotton, coriander and larkspur heterophylly is observed where leaves in juvenile stage and adult stage show different forms.

Question 10.
What is the role of growth hormones in plants?
Answer:
Growth hormones inhibit, promote or modify the growth in plants.

Question 11.
What is the peculiarity of growth hormones ?
Answer:
Growth hormones are needed in very small amount to evoke the response and they act at a site away from their place of production.

Question 12.
Which is the first hormone to be discovered in plants?
Answer:
Auxin IAA i.e. Indole acetic acid is the first hormone to be discovered in plants.

Question 13.
Give the full form of 2, 4 – D.
Answer:
It is synthetic auxin 2, 4-dichlorophenoxy acetic acid.

Question 14.
What is the effect of NAA and 2, 4-D foliar spray?
Answer:
Foliar spray of synthetic hormones induce flowering in plants litchi and pineapple and prevent premature fruit drop in apples, pear and oranges.

Question 15.
How cytokinins control apical dominance ?
Answer:
Cytokinins promote growth of lateral buds by cell division and thus control apical dominance.

Question 16.
What was the discovery of Richmond and Lang?
Answer:
Richmond and Lang discovered that cytokinins delay the process of ageing and senescence, abscission in plant organs.

Question 17.
What is ‘epinasty’?
Answer:
It is an effect of ethylene where it causes drooping of leaves and flowers.

Question 18.
Why is auxin called a growth regulator?
Answer:
Auxin is synthesized at meristematic region of plants and it controls cell enlargement, cell elongation and stimulates growth of stem and root, apical dominance. Hence it is growth promoting hormone.

Question 19.
What is effect of gibberellin application on apple?
Answer:
Gibberellin causes parthenocarpy in apple.

Question 20.
How can we overcome apical dominance?
Answer:
By application of cytokinin we can overcome apical dominance effect.

Question 21.
Which is standard bioassay method for auxins?
Answer:
Avena curvature test/Avena coleoptile test is a standard bioassay method for auxins.

Question 22.
ABA is called as stress hormone why?
Answer:
Answer: ABA induces dormancy in seeds by inhibiting growth. Thus plants can tide over adverse environmental conditions. Hence it is called as stress hormone.

Question 23.
What was the plant material for study of photoperiodism by Garner and Allard?
Answer:
The flowering response of Soybean and Maryland mammoth variety of tobacco was studied by Garner and Allard.

Question 24.
What is photomorphogenesis?
Answer:
Control of morphogenesis by light and phytochrome pigment is called photomorphogenesis.

Question 25.
What is critical concentration of minerals ?
Answer:
The concentration of the essential elements below which plant growth is retarded is termed as critical concentration.

Question 26.
What is a role of Sulphur in plants?
Answer:
Sulphur is constituent of amino acids, proteins, vitamins (mainly thaimine, biotin CoA) and Ferredoxin.

Question 27.
What is a role of nitrogen in plants?
Answer:
Nitrogen is constituent of amino acids, proteins, nucleic acid, vitamins, hormones, coenzymes, ATP and chlorophyll molecule.

Question 28.
Which is the process by which mainly we get nitrogen in human tissues?
Answer:
Industrial nitrogen fixation by Haber – Bosch Nitrate process is responsible for nitrogen found in human tissues.

Question 29.
Give equation of Haber – Bosch process.
Answer:
Maharashtra Board Class 12 Biology Important Questions Chapter 7 Plant Growth and Mineral Nutrition 2

Maharashtra Board Class 12 Biology Important Questions Chapter 7 Plant Growth and Mineral Nutrition

Question 30.
What is nitrogen assimilation?
Answer:
Nitrogen present in the soil as nitrates, nitrites and ammonia is absorbed by plants and converted into nitrogenous organic compounds is nitrogen assimilation.

Give definitions of the following

Question 1.
Growth
Answer:
Growth can be defined as permanent, irreversible increase in the bulk of an organism with change in form.

Question 2.
Efficiency index
Answer:
The increased growth per unit time is called efficiency index.

Question 3.
Absolute growth rate (AGR)
Answer:
The measurement and comparison of total growth per unit time is called absolute growth rate.

Question 4.
Relative growth rate (RGR)
Answer:
The growth of a particular system per unit time expressed on a common basis or alternately it is the ratio of growth in the given time per initial growth.

Question 5.
Differentiation
Answer:
A permanent change in structure and function of cells that leads to their maturation is called differentiation.

Question 6.
Redifferentiation
Answer:
When cells produced from de-differentiation lose their capacity to divide and mature for specific function it is known as re-differentiation.

Question 7.
Development
Answer:
The progressive changes in shape, form and degree of complexity which includes growth, maturation and morphogenesis is referred as development.

Question 8.
Growth Hormone or Growth regulators
Answer:
The internal factors that influence growth by inhibiting, promoting or modifying it are called growth hormones or regulators.

Question 9.
Apical dominance
Answer:
In higher plants growing apical bud inhibits the growth of lateral buds. This is known as apical dominance.

Question 10.
Critical photoperiod
Answer:
That length of photoperiod above or below which the plant shows flowering is critical photoperiod.

Question 11.
Photomorphogenesis
Answer:
The control of morphogenesis of plants by light and phytochrome is photomorphogenesis.

Question 12.
Symptom or hunger sign
Answer:
Any visible deviation from the normal structure and function of the plant is called symptom or hunger sign.

Question 13.
Active absorption of minerals
Answer:
Uptake of mineral ions against concentration gradient, with expenditure of energy (ATP) is called active absorption.

Question 14.
Nitrogen fixation
Answer:
Free nitrogen of air N2 is converted to nitrogenous salts so that it is made available to plants is called nitrogen fixation.

Question 15.
Nitrification
Answer:
Soil microbes, mainly : chemoautotrophs convert ammonia into j nitrate, the form of nitrogen which can be used by plants, this process is nitrification.

Question 16.
Amidation
Answer:
Ammonia may be absorbed by amino acids to produce amides. This process is called amidation.

Name the following

Question 1.
Instrument developed by Indian physiologist to measure growth.
Answer:
Crescograph developed by Sir J.C. Bose.

Question 2.
Instrument to measure linear growth of shoot.
Answer:
Auxanometer.

Question 3.
Type of growth curve for geometric growth.
Answer:
J-shaped or exponential growth curve.

Question 4.
The process by which cork cambium is formed.
Answer:
Dedifferentiation.

Maharashtra Board Class 12 Biology Important Questions Chapter 7 Plant Growth and Mineral Nutrition

Question 5.
Ability of plant to form different kinds of structures.
Answer:
Plasticity.

Question 6.
Condition where plant exhibits different types of leaves on same plant.
Answer:
Heterophylly.

Question 7.
Growth promoter hormones.
Answer:
Auxins, gibberellins and cytokinins.

Question 8.
Growth inhibitors of plants.
Answer:
Ethylene and abscissic acid.

Question 9.
First hormone that is discovered in plant and its precursor.
Answer:
Amino acid tryptophan is precursor of Indole acetic acid (IAA).

Question 10.
A synthetic hormone that acts as selective herbicide.
Answer:
2, 4 – D (dichlorophenoxy acetic acid).

Question 11.
A plant hormone discovered from fungus.
Answer:
Gibberellic acid (GA).

Question 12.
A plant hormone also present in fish (Herring) sperm DNA.
Answer:
Kinetin (cytokinin).

Question 13.
A plant hormone also present in urine of person suffering from Pellagra.
Answer:
Auxin.

Question 14.
Most widely used source of ethylene for fruit ripening.
Answer:
Ethephon – 2 chloroethyl phosphoric acid.

Question 15.
A precursor from which GA and ABA are synthesized.
Answer:
Mevalonic acid.

Question 16.
Plant antitranspirant.
Answer:
Abscissic acid.

Question 17.
Chemical stimulant of low temperature effect on flowering of plants.
Answer:
Vernalin.

Question 18.
Methods of synthesis of amino acids.
Answer:
Reductive animation and Transamination.

Question 19.
Common amides present in plants.
Answer:
Asparagine and Glutamine.

Question 20.
Nitrogen fixing prokaryotic organisms.
Answer:
Diazotrophs or Nitrogen fixers.

Question 21.
Special structures of cyanobacteria where N2 fixation occurs.
Answer:
Heterocysts.

Question 22.
A synthetic cytokinin hormone.
Answer:
6 – benzyl adenine.

Give Functions/Significance/Importance of the following

Question 1.
Meristems
Answer:
In plants meristems are situated at specific regions where growth takes place by constant and continuous addition of new cells. Meristems have capacity to divide (Mitotic divisions).

Question 2.
Synthetic auxin 2, 4-D
Answer:
It is selective herbicide which kills dicot weeds and foliar spray of 2, 4-D induces flowering in litchi and pineapple, prevents premature fruit drop in apples, pear, oranges.

Question 3.
Coconut milk
Answer:
Coconut milk contains natural cytokinin substance kinetin which is used as nutritional supplement for callus tissue culture where proliferation is noticed due to promotion of cell division.

Question 4.
Ethylene/Ethephon
Answer:
It is used for fruit ripening and as it causes degreening effect by increasing activity of chlorophyllase enzyme for banana and citrus fruits.

Question 5.
Abscissic acid/ABA
Answer:
It is natural growth inhibiting substance in plants and it acts as plant anti Iranspirant causing closure of stomata. It is stress hormone that induces plant to bear the adverse environmental conditions like drought.

Question 6.
Phytochrome
Answer:
It is a proteinaceous pigment present in leaves which perceives stimulus of light for flowering. As it is interconvertible in two forms, it promotes flowering in SDP and inhibits flowering in LDE.

Maharashtra Board Class 12 Biology Important Questions Chapter 7 Plant Growth and Mineral Nutrition

Question 7.
Vernalization
Answer:
It is response of flowering to low temperature treatment which helps in cultivation of crops in regions where they do not occur naturally and also crops can be produced earlier.

Question 8.
Nitrogen cycle
Answer:
Nitrogen is essential macronutrient for plant growth, it is constituent of amino acids, proteins, nucleic acids, vitamins, hormones, ATP coenzymes, chlorophyll molecule. It is limiting nutrient for plant productivity and agricultural ecosystem. Through food chain it moves to consumers, i.e. animals and decomposers.

Distinguish between the following

Question 1.
Phase of cell division and Phase of cell enlargement
Answer:

Phase of cell divisionPhase of cell enlargement
1. In this phase cells of meristems divide by mitotic division.1. In this phase newly formed cells become vacuolated and turgid, osmotically active.
2. Rate of growth at slow pace.2. Rate of growth at accelerated pace.
3. This is described as lag phase.3. This is described as log phase.
4. There is no synthesis of any new material.4. Synthesis of new wall materials and other materials takes place.

Question 2.
Long day plants and Short day plants
Answer:

Long day plantsShort day plants
1. Plants that flower only when they are exposed to light period longer than their critical photoperiod are called long day plants (LDP).1. Plants that flower only when they are exposed to light period shorter than the critical photoperiod are called short day plants (SDP).
2. The plants usually flower in summer.2. The plants usually flower in winter or late summer.
3. These plants require short night period for flowering. Hence, they are also known as short night plants.3. These plants require long night period for flowering. Hence, they are also known as long night plants.
4. Plants such as Pea, Radish, Sugar, Beet, Cabbage, Spinach, Wheat, poppy are LDP4. Plants such as Dahlia, Aster, Tobacco, Chrysanthemum, Soybean (Glycine max) Cocklebur (Xanthium) are SDP

Question 3.
Passive absorption of Minerals and Active absorption of minerals
Answer:

Passive absorption of mineralsActive absorption of minerals
1. The movement of mineral ions into root cells due to diffusion, without expenditure of energy is called passive absorption.1. The uptake of mineral ions by root cells that requires expenditure of ATP energy is known as active absorption.
2. The movement is according to concentration gradient.2. The movement of mineral ions is against concentration gradient.
3. It takes place by direct ion exchange, diffusion, indirect ion exchange – mass flow and Donnan equilibrium.3. It takes place by mineral ion accumulation in root hair and carrier concept.

Given reasons

Question 1.
Water is essential for growth in plants.
Answer:

  1. Meristematic cells divide and form new cells.
  2. Absorption of water is necessary for maintaining turgidity in the newly formed cells.
  3. Turgidity results in enlargement of cells in phase of cell elongation.
  4. Water is essential component of protoplasm of cells.
  5. biochemical reactions. Therefore it is essential for growth in plants.

Question 2.
2, 4-D is used as herbicide.
Answer:

  1. 2, 4-D is a synthetic auxin which kills dicot weeds.
  2. Our most of the food crops are cereals, i.e. monocot plants.
  3. Weeds are unwanted plants which otherwise lower the productivity hence to kill them. Selective herbicide is used.

Question 3.
In morphogenesis of plants cytokinin auxin ratio is important.
Answer:

  1. Auxins and cytokinins are growth promoting substances which stimulate cell division and cell enlargement.
  2. A high cytokinin promotes shooting in plants.
  3. A low ratio of cytokinin to auxin induces root development.
  4. A high ratio of cytokinin to auxin induces growth of buds and shoot development.
  5. Thus cytokinin and auxin ratio and their interactions control morphogenesis in plants.

Question 4.
ABA is described as an antitranspirant.
Answer:

  1. ABA is a growth inhibiting hormone.
  2. ABA is responsible for causing efflux of K+ ions from guard cells of stomata.
  3. As a result of this, osmotic changes occur and guard cells become flaccid resulting in closure of stomata.
  4. Transpiration mainly occurs through open stomata and due to closure the activity is checked. Hence it is described as antitranspirant.

Question 5.
Some deficiency symptoms of mineral are visible in young leaves while some appear in older leaves.
Answer:

  1. When mineral element is present below a certain critical concentration it is said to be deficient.
  2. Symptoms are indicated in the form of certain morphological changes on the mobility of element.
  3. These symptoms depend on the mobility of element inside the plant body.
  4. When the element is relatively immobile like S, Ca then the symptoms appear first in young leaves.
  5. When the elements are actively mobilised inside plant body, they are transported to young tissues then the symptoms are visible in older, i.e. senescent leaves e.g. N, Mg, K.

Question 6.
In Donnan equilibrium of passive absorption of minerals concentration of cations increases inside the cell.
Answer:

  1. Minerals exist in soil in the form of charged particles.
  2. Certain negatively charged (anions) get accumulated on the inner side of cell membrane after their entry inside cell.
  3. These anions cannot diffuse out through semipermeable cell membrane.
  4. Thus additional mobile cautions are needed inside the cell to balance these fixed anions.
  5. Hence, the concentration of cations increases inside the cell.

Question 7.
A sudden drop in active absorption of minerals is noticed if roots are deprived of oxygen supply.
Answer:

  1. Absorption of mineral ions from soil against concentration gradient is known as active absorption.
  2. It requires energy (ATP) for absorption by absorbing root cell.
  3. The source of energy is respiration of cells for supply of ATE
  4. When the roots are deprived of oxygen, their respiration process is affected and thus energy is not supplied in required amount. Hence, a sudden drop in absorption of minerals is noticed.

Question 8.
Nitrogen is a limiting nutrient in the agricultural system.
Answer:

  1. Nitrogen is a major nutrient for plant growth.
  2. Proper carbon/nitrogen ratio in soil is necessary for plant growth.
  3. It is component of proteins in the form of amino acids.
  4. Proteins are synthesised from photosynthetic products sugars.
  5. Nitrogen exists in atmosphere but it is inert, non-reactive.
  6. Plants need nitrogen in a reactive form usually nitrate in soil.
  7. This supply need to be maintained through biological and physical nitrogen fixation.
  8. Otherwise productivity is affected hence it is limiting nutrient in the agricultural ecosystem.

Question 9.
Cucumber and sunflower are regarded as photoneutral plants.
Answer:

  1. In cucumber and sunflower, the flower is not controlled by light period.
  2. Both these plants flower in all light periods.
  3. Cucumber and sunflower, therefore, are regarded as photoneutral plants.

Write short notes on the following

Question 1.
Meristems
Answer:

  1. In vascular plants, growth is indeterminate and occurs at specific regions where meristems are located.
  2. Meristems are of three types based on their location – apical, intercalary and lateral.
  3. Meristems are thin walled cells with prominent nucleus with granular cytoplasm, non-vacuolated.
  4. Mitotic divisions take place in meristematic cells.

Question 2.
Phase of cell formation
Answer:

  1. Formative phase is the first phase of growth.
  2. During this phase, the meristematic cells undergo mitosis to form new cells.
  3. During formative phase, the rate of growth is slow.
  4. The phase of cell formation is also called lag phase.
  5. This phase is also known as phases of cell division.

Question 3.
Development
Answer:

  1. Development is progressive changes taking place in shape, form and degree of complexity in an organism.
  2. In plants, it includes all the changes taking place in sequence from seed germination to senescence or death of plant.
  3. Development is an orderly process.
  4. It includes growth, morphogenesis, maturation and senescence.

Question 4.
Plasticity
Answer:

  1. Plasticity is the capacity of plant being molded or formed.
  2. It is ability of plant to develop different kinds of structures in response to environmental factors or stimuli.
  3. Different kinds of structures can be developed in plants due to internal stimuli in different phases, i.e. juvenile and adult.
  4. Heterophylly is shown in plant in different phases or in different environmental conditions.
  5. In coriander and cotton plants, two different kinds of leaves are observed in young (juvenile) and mature (adult) plant.
  6. In buttercup, two different kinds of leaves are observed in terrestrial (on land) and aquatic (in water) habitat.

Question 5.
Phytohormones/Plant Growth Regulators
Answer:

  1. Phytohormones or plant growth regulators are internal factor that influence growth.
  2. They inhibit, promote or modify the plant growth.
  3. Plant hormones are organic substances produced naturally in plants and required in small amount.
  4. Their place of production and site of the activity are different.
  5. Auxins, gibberellins, cytokinins are growth promoters and ethylene, abscissic acid are growth inhibitors.

Question 6.
Phytochrome
Answer:

  1. These are proteinaceous pigments present
  2. Phytochrome exists in two interconvertible forms. Pr and Pfr.
  3. Phytochromes are located in cell membranes of chlorophyllous cells.
  4. When Pfr absorbs red light it is converted to Pr.
  5. Pfr is accumulated in plants during daytime and inhibits flowering in SDP but initiates flowering in LDP

Question 7.
Venralization
Answer:

  1. Effect of temperature on flowering of plants is known as vernalization.
  2. For inducing early flowering pretreatment of seeds or seedlings is done at 1 to 6 °C for about a months duration.
  3. Shoot apical meristem is believed to be site of vernalization stimulus.
  4. Vernalization stimulus is in a form of chemical named vernalin.
  5. Vernalization is effective in ereals (wheat) and crucifers.

Maharashtra Board Class 12 Biology Important Questions Chapter 7 Plant Growth and Mineral Nutrition

Question 8.
Apical dominance
Answer:

  1. The presence of apical bud inhibits the growth of lateral buds. This phenomenon in which the apical (terminal) bud is active and lateral buds remain inactive is called apical dominance.
  2. It is believed that apical dominance is controlled by an auxin which is synthesized in the apical bud.
  3. From the apical bud, the auxin migrates to the lateral buds and inhibits their growth.
  4. When apical bud is removed, the lateral buds grow and form branches.
  5. For producing more branches therefore, the apical buds are removed.
  6. Cytokinins reverse apical dominance effect by promoting growth of lateral buds by cell division.

Question 9.
Toxicity of micronutrients or mineral toxicity
Answer:

  1. Micronutrients are required in minute quantities by plants.
  2. Their moderate decrease causes deficiency symptoms while their moderate increase causes toxicity.
  3. The reduction in dry weight of a tissue by 10% by any mineral is known as toxicity.
  4. It is not easy to identify toxicity symptoms.
  5. Most of the time, the excess of an element inhibits the uptake of another element resulting in causing the deficiency symptom of that element.
  6. Manganese inhibits calcium translocation towards apex of stem and exhibits symptoms of chlorosis with grey spots appearing on leaves.
  7. This is because manganese competes with iron and magnesium for uptake.
  8. Therefore what we see as symptoms of manganese toxicity, may be the deficiency symptoms of Fe, Mg and Ca.

Question 10.
Day neutral plants (DNP)
Answer:

  1. Day neutral plants are those in which flowering is not affected by the duration of the light period.
  2. Day neutral plants flower in all photoperiods.
  3. Day neutral plants are also known as photoneutral or intermediate plants.
  4. Plants such as cucumber, shoe-flower, sunflower, tomato, maize, balsam, etc. are day neutral plants.

Short Answer Questions

Question 1.
What is de-differentiation?
Answer:

  1. The differentiated cells which are formed may again gain the capacity to divide as per need.
  2. Permanent cells (mature cells) undergo de-differentiation and become meristematic.
  3. This acquired feature of living permanent cells is known as de-differentiation.
  4. e.g. Formation of interfascicular cambium and cork cambium from parenchyma cells of medullary rays and outer cortical cells respectively.

Question 2.
Discuss about natural Auxin.
Answer:

  1. F.W. Went named the growth promoting substance as Auxin.
  2. Auxin was also isolated from urine of patient of pellagra.
  3. Indole 3 acetic acid was first hormone discovered in plants.
  4. IAA is most common and natural hormone synthesized at growing tips and responsible for cell elongation.
  5. It is synthesized from tryptophan and shows polar transport – Basipetal transport.

Question 3.
Discuss about discovery of Gibberellins.
Answer:

  1. Gibberellins are growth promoting hormones and were isolated form fungus Gibberella Jujikuroi by Kurasawa.
  2. Rice plants when infected with this fungus show stem elongation i.e. Bakane disease.
  3. Yabuta and Sumuki isolated gibberellins in crystalline form, from fungal culture and named it gibberellins.
  4. Gibberellins are synthesized from mevalonic acid at young leaves, seeds, root and stem tips and show non-polar transport.

Question 4.
Discuss about discovery of cytokinin.
Answer:

  1. Cytokinins are growth promoting hormones that stimulate cell division.
  2. Skoog and Miller discovered first cytokinin when they were investigating nutritional requirements of tobacco callus culture.
  3. It was observed that the callus proliferated when there was addition of coconut milk as supplement.
  4. The degraded sample of herring (fish) sperm DNA also showed similar growth of tobacco callus. They named the substance as kinetin.

Question 5.
Discuss about discovery of abscissic acid.
Answer:

  1. Abscissic acid (ABA) is a natural growth inhibiting hormone.
  2. It was observed by Carns and Addicott that shedding of cotton balls occur due to chemical substance abscission I and II.
  3. From the buds of Acer, Wareing isolated substance that causes bud dormancy and named it as dormin.
  4. These two chemical substances were identical and now known as abscissic acid.
  5. ABA is synthesized in leaves, fruits and seeds from mevalonic acid.

Question 6.
What is day neutral plant (DNP)? Give any two examples.
Answer:

  1. The plants which do not require specific duration of light period or dark period flowering are day neutral plants (DNP).
  2. They flower throughout the year, as they do not need specific photoperiod.
  3. The flowering in these plants is independent of photoperiod.
  4. Examples – cucumber, tomato, cotton, sunflower, maize and balsam.

Question 7.
Discuss about discovery of phytochromes.
Answer:

  1. Phytochromes are proteinaceous pigments present in cell membrane of green cells.
  2. Phytochromes receive photoperiodic stimulus and induce flowering in plants in response to light duration.
  3. Hendrick and Borthwick observed that in SD plants flowering is inhibited if dark period is interrupted by flash of red light (660 nm).
  4. If flash of far red light (730 nm) is given then again flowering is observed in SD plants.
  5. Pigment system of plant receives photoperiodic stimulus and these pigments exist in two interconvertible forms Pfr and Pr

Question 8.
What is mineral nutrition of plants?
Answer:

  1. Plants require inorganic materials for the synthesis of food.
  2. These elements are obtained by plants in the form of minerals mainly form soil.
  3. Chemical analysis of plant ash reveals that about 40 different minerals are needed by plants which are taken from surroundings, (air, soil and water)
  4. These minerals are absorbed in dissolved form, i.e. ionic form through root system mainly root hairs.
  5. Some minerals are required in large amounts (major) while some are needed only in traces or small amounts (minor).

Question 9.
What are symptoms of mineral deficiency in plants?
Answer:
Any visible deviation from the normal structure and function of plants due to lack or unavailability of particular element below its critical concentration is deficiency symptom of that mineral element.

The common symptoms observed in plants are as follows:

  1. Stunting : Retarded growth and thus stem appears short and condensed.
  2. Chlorosis : This is loss or lack of chlorophyll that result in yellowing of leaf.
  3. Necrosis : It is localized death of tissue. Mottling : This is appearance of green or non-green patches or spots on leaves. Abscission : This is premature fall of leaves, buds, fruits and flowers.

Question 10.
Enlist the role of following minerals and the symptoms caused due to their deficiency : (a) Calcium (b) Boron and (c) Chlorine.
Answer:
(a) Calcium:
Role : Involved in selective permeability of cell membranes, activator of certain enzymes, required as calcium pectate in middle lamella of cell wall at root and stem apex (for cell division).
Deficiency symptom : stunted growth.

(b) Boron:
Role : Required for uptake and utilization of Calcium (Ca2+), pollen germination, cell differentiation, carbohydrate translocation. Deficiency symptom : Brown heart disease

(c) Chlorine:
Role : Na+ and K+ help to determine solute concentration and anion – cation balance in cells, necessary for oxygen evolution in photosynthesis.
Deficiency symptom : Poor growth of plant.

Question 11.
What is Donnan Equilibrium?
Answer:

  1. Donnan equilibrium is a process of passive absorption of minerals in plants which is without any expenditure of energy.
  2. It is assumed that certain anions after their entry by diffusion into the cell get fixed on inner side of cell membrane.
  3. Additional mobile cations are needed to balance this fixed anions as they cannot diffuse outside.
  4. Concentration of cations thus increases due to accumulation.
  5. This passive absorption of anions or cations from exterior against their concentration gradients so as to neutralize the effect of cations or anion is known as Donnan equilibrium.

Question 12.
Explain physical nitrogen fixation.
Answer:

  1. Conversion of free nitrogen of air into nitrogenous compounds that are made available to plants for uptake is known as : nitrogen fixation.
  2. Physical nitrogen fixation occurs in step- : wise manner and it takes place in atmosphere j and soil.
  3. Under the influence of electric discharge, lightning and thunder, atmospheric nitrogen combines with oxygen to form nitric oxide.
  4. Nitric oxide is then oxidized to nitrogen peroxide in presence of oxygen.
  5. Nitrogen peroxide combines with rainwater to form nitrous and nitric acid which come on ground as acid rains.
  6. On ground, alkali radicals (mainly of Ca, K) react with nitric acid to produce nitrites and nitrates which are absorbable forms for plants.

Maharashtra Board Class 12 Biology Important Questions Chapter 7 Plant Growth and Mineral Nutrition

Question 13.
Give equations of physical nitrogen fixation.
Answer:
(1) Physical nitrogen fixation occurs in stepwise manner in atmosphere and on land (soil)
Maharashtra Board Class 12 Biology Important Questions Chapter 7 Plant Growth and Mineral Nutrition 3

Question 14.
Give the equations of amino acid synthesis.
Answer:
1. Macromolecule proteins are made up of building blocks of amino acids.
2. Amino acids are synthesized by two methods – Reductive animation and transamination.
3. Reductive amination – Ammonia reacts with alpha keto glutaric acid to form glutamic acid.
Maharashtra Board Class 12 Biology Important Questions Chapter 7 Plant Growth and Mineral Nutrition 4
4. Transamination – Amino group of one amino acid is transferred to other carboxylic acid at keto position.
Maharashtra Board Class 12 Biology Important Questions Chapter 7 Plant Growth and Mineral Nutrition 5

Question 15.
Explain lag phase, log phase and steady phase of growth.
Answer:
1. In plants, growth curve is always sigmoid, i.e. S-shaped. This is because growth starts slowly during formative phase, becomes rapid during elongation phase and finally slows down to a steady state during the maturation phase.

2. The standard growth curve shows three phases, viz. lag phase, log phase and stationary phase.
(i) Lag phase or initial growth phase : This is the initial phase of growth. During this phase of growth, the rate of growth is slow. It corresponds to formative phase of growth where new cells are formed due to cell division.

(ii) Log phase or exponential phase : This is the second phase of growth. During this phase, the growth is rapid and maximum. It corresponds to the phase of cell elongation.

(iii) Stationary phase or steady phase : The stationary phase is the third and last phase of growth. In this phase, growth slows down and becomes steady. The cells undergo differentiation during stationary phase.

Chart based or Table based questions

Question 1.
Complete the chart of plasticity.
Maharashtra Board Class 12 Biology Important Questions Chapter 7 Plant Growth and Mineral Nutrition 6
Answer:
Maharashtra Board Class 12 Biology Important Questions Chapter 7 Plant Growth and Mineral Nutrition 7

Question 2.
Complete the flow chart of development.
Maharashtra Board Class 12 Biology Important Questions Chapter 7 Plant Growth and Mineral Nutrition 8
Answer:
Maharashtra Board Class 12 Biology Important Questions Chapter 7 Plant Growth and Mineral Nutrition 9

Question 3.
Complete the table of growth hormones.
Maharashtra Board Class 12 Biology Important Questions Chapter 7 Plant Growth and Mineral Nutrition 10
Answer:
Maharashtra Board Class 12 Biology Important Questions Chapter 7 Plant Growth and Mineral Nutrition 11

Question 4.
Complete the table of mineral nutrition of plants.
Maharashtra Board Class 12 Biology Important Questions Chapter 7 Plant Growth and Mineral Nutrition 12
Answer:
Maharashtra Board Class 12 Biology Important Questions Chapter 7 Plant Growth and Mineral Nutrition 13

Diagram based questions

Question 1.
Draw diagram of photoperiodic response of SDP and LDP.
Answer:
Maharashtra Board Class 12 Biology Important Questions Chapter 7 Plant Growth and Mineral Nutrition 14

Question 2.
With the help of diagram show arithmetic growth and geometric growth.
Answer:
Maharashtra Board Class 12 Biology Important Questions Chapter 7 Plant Growth and Mineral Nutrition 15

Question 3.
Draw the sigmoid growth curve.
Answer:
Maharashtra Board Class 12 Biology Important Questions Chapter 7 Plant Growth and Mineral Nutrition 16

Question 4.
Observe the diagram and answer the questions related to it.
Maharashtra Board Class 12 Biology Important Questions Chapter 7 Plant Growth and Mineral Nutrition 17

  1. From which cells newly formed cells are added? How?
  2. How is rate of growth in zone of cell elongation?
  3. What is peculiarity of zone of differentiation? How is rate of growth in this region?

Answer:

  1. Meristematlc cells add new cells by mitotic division.
  2. Rate of growth is at accelerated pace.
  3. The rate of growth is at steady state and cells become specialised to perform specific function become mature.

Question 5.
Observe the adjacent graph indicating growth. What is correct labelling of A, B and C respectively?
Maharashtra Board Class 12 Biology Important Questions Chapter 7 Plant Growth and Mineral Nutrition 18
Answer:
A. Stationary Phase
B. Exponential Phase
C. Lag Phase

Maharashtra Board Class 12 Biology Important Questions Chapter 7 Plant Growth and Mineral Nutrition

Question 6.
Maharashtra Board Class 12 Biology Important Questions Chapter 7 Plant Growth and Mineral Nutrition 19
(1) Observe the above graph indicating increase in height of plants. Which type of growth indicates this pattern of graph?
(2) Give its mathematical expression.
Answer:
(1) Arithmetic growth

(2) Lt = Lo + rt where
Lt = Length at time t,
Lo = Length at time zero r = growth rate,
t = time of growth

Question 7.
Observe the figure given below of two different leaves ‘A’ and ‘B’ Which leaf shows much higher relative growth rate ?
Maharashtra Board Class 12 Biology Important Questions Chapter 7 Plant Growth and Mineral Nutrition 20
Answer:
Relative Growth Rate (RGR) = Growth per unit time as % of intial size.
RGR = \(\frac { Growth per unit time }{ intial size }\) × 100
Leaf A = 10 – 5 = 5, \(\frac { 5 }{ 5 }\) × 100 = 100%
Leaf B = 55 – 50 = 5, \(\frac { 5 }{ 50 }\) × 100 = 10%
Hence fig. A shows more relative growth.

Question 8.
Observe the diagrams A and ‘B’ showing growth in two leaves. Which diagram shows more relative growth?
Maharashtra Board Class 12 Biology Important Questions Chapter 7 Plant Growth and Mineral Nutrition 21
Answer:
Relative Growth Rate (RGR)
= Growth per unit time as % of intial size.
RGR = \(\frac { Growth per unit time }{ intial size }\) × 100
Leaf A = 20 – 10 = 10, \(\frac { 10 }{ 10 }\) × 100 = 100%
Leaf B = 60 – 50 = 10, \(\frac { 10 }{ 50 }\) × 100 = 20%
Hence Diagram A shows more relative growth.

Question 9.
Identify the type of growth curves observed in plants.
Maharashtra Board Class 12 Biology Important Questions Chapter 7 Plant Growth and Mineral Nutrition 22
Answer:
Constant linear growth curve for the arithmetic growth.
Exponential (J shaped) growth curve for the geometric growth.
Sigmoid growth curve related to distinct phases of growth.

Question 10.
Fill in the blanks in the given nitrogen cycle.
Maharashtra Board Class 12 Biology Important Questions Chapter 7 Plant Growth and Mineral Nutrition 23
Answer:
Maharashtra Board Class 12 Biology Important Questions Chapter 7 Plant Growth and Mineral Nutrition 24

Long answer questions

Question 1.
What is hydroponics? How is it useful in identifying the role of nutrients?
Answer:

  1. Growing plants in aqueous (soilless) medium is known as hydroponics. [Greek word hudor = water and ponos = work]
  2. It is technique of growing plants by supplying all necessary nutrients in the water supply given to plant.
  3. A nutrient medium is prepared by dissolving necessary salts of micronutrients and macronutricnts In desired quantity and roots of plants are suspended in this liquid with appropriate support.
  4. Hydroponics is of great use in studying the deficiency symptoms of different mineral nutrients.
  5. The plants uptake mineral nutrients in the form of dissolved ions with the help of root hairs from the surrounding medium or nutrient solution supplied.
  6. While preparing the required nutrient medium particular nutrient can be totally avoided and then the effect of lack of that nntr’cnt can be studied in variation of plant growth.
  7. Any visible change noticed from normal structure and function of the plant is the symptom or hunger sign considered.
  8. For e.g. Yellowing of leaf is observed due to loss of chlorophyll pigments or Chlorosis is noticed if Magnesium is lacking as it is a structural component of chlorophyll pigment.

Question 2.
Explain the active absorption of minerals.
Answer:

  1. Plants absorb minerals from the soil with their root system.
  2. Minerals are absorbed from the soil in the form of charged particles, positively charged cations and negatively charged anions.
  3. The absorption of minerals against the concentration gradient which requires expenditure of metabolic energy is called active absorption.
  4. The ATP energy derived from respiration in root cells is utilized for active absorption.
  5. Ions get accumulated in the root hair against the concentration gradient.
  6. These ions pass into cortical cells and finally reach xylem of roots.
  7. Along with the water these minerals are carried to other parts of plant.

Question 3.
What is growth? What are its characteristics ?
Answer:
Growth : Growth is a “vital process which brings about an irreversible increase in an organism or its part with respect to its size, weight, form and volume.”

Characteristics of growth:

  1. Growth is a permanent increase in size, weight, shape, volume and dry weight of a plant.
  2. The change occurring due to growth is permanent and irreversible.
  3. Growth is an intrinsic process caused due to internal activities.
  4. Growth occurs by cell division and cell elongation followed by cell maturation which lead to the formation of different types of tissues.
  5. Growth in plants is mostly localized, i.e. restricted to some regions of plants possessing meristematic tissues or meris terns.
  6. Growth has a qualitative aspect where development takes place in an orderly manner and differentiation leads to higher and more complex state.

Maharashtra Board Class 12 Biology Important Questions Chapter 7 Plant Growth and Mineral Nutrition

Question 4.
Describe the phases of growth.
Answer:
There are three phases of growth, viz, formation phase, elongation phase and maturation phase.
(1) Formative phase (Phase of cell division) : This is the first phase of growth. In this phase, the meristematic cells undergo mitosis to produce new cells. Owing to the formation of new cells, there occurs a slight increase in the size of the organ.

(2) Elongation phase (Phase of cell enlargement) : This is the second phase of growth. In this phase, the new cells that are formed, undergo enlargement as a result of which the size and volume of the cells increase. Enlargement of cells occur mostly in linear direction as a result of which the elongation of the root and stem takes place. The enlargement of cells causes a considerable increase in size and weight of an organ and a plant as a whole.

(3) Maturation phase (Phase of cell maturation and differentiation) : Maturation phase is the third and last phase of growth. In this phase, the elongated cells undergo maturation and differentiation to form various types of plant tissues like parenchyma, sclerenchyma, xylem and phloem.

Question 5.
Explain the various conditions of growth that are essential.
Answer:

  1. For a proper growth of plant various environmental and physiological conditions are necessary.
  2. Carbon/Nitrogen ratio in soil is having effect on growth as both carbon and nitrogen are structural elements in carbohydrates, proteins and other biomolecules.
  3. Water is essential component of protoplasm and required for turgidity of cells during cell enlargement phase. It is a medium in which various biochemical reactions ocfcur.
  4. Nutrients are necessary for proper growth. Macronutrients and micronutrients have their specific role.
  5. Temperature of 25 – 35 °C is optimum for growth.
  6. Light is essential for seed germination and photosynthesis.
  7. Oxygen is necessary for respiration and supply of energy.
  8. Gravitational force decides direction of growth for root system and shoot system.
  9. Growth hormones are organic compounds that are involved in various physiological aspects and control of growth.

Question 6.
What are plant growth regulators? Give the characteristics of plant growth regulators.
Answer:
Plant growth regulators : Plant growth regulators or phytohormones or plant hormones as they are also called are organic compounds synthesised by the plants which promote, inhibit or control the growth and other physiological processes.

Characteristics of plant growth regulators:

  1. Plant growth regulators are organic compounds other than nutrients.
  2. They are synthesised at the apices of root, stem and leaves from where they are transported to other parts of plants where they produce their effects.
  3. They are required in minute quantities.
  4. A single plant growth regulator can control or regulate the various aspects of growth.

Question 7.
Enlist the five main types of growth regulators and state the role of abscissic acid in plants.
Answer:
Five main types of growth regulators:

  1. Auxins
  2. Gibberellins
  3. Cytokinins
  4. Ethylene and
  5. Abscissic acid.

Role of abscissic acid (ABA) in plants:

  1. Abscissic acid influences abscission and dormancy.
  2. ABA accelerates senescence of leaves, flowers and fruits.
  3. It is a stress hormone as it is produced during drought and other unfavourable climatic conditions.
  4. ABA induces dormancy in seeds, buds and tubers.
  5. It acts as growth inhibitor as it retards growth.
  6. ABA plays an important role in closing of stomata to check transpiration.
  7. It inhibits and delays cell division and suppresses cambial activity by inhibiting mitosis in vascular cambium.
  8. ABA inhibits flowering in LDP and stimulates flowering in short day plants (SDP).

Question 8.
Write an account of auxins as growth regulators.
Answer:

  1. Auxins are plant growth regulators produced naturally by plants.
  2. They are weak organic acids capable of promoting cell elongation during the growth of stem and root.
  3. Auxins are synthesized in shoot and root apices besides young leaf primordia.
  4. Auxins may be natural or synthetic.
  5. Naturally occurring auxins are indole-3- acetic acid (IAA) and its derivatives.
  6. NAA, 2, 4-D and 2,4, 5-T are synthetic auxins.
  7. Auxins in higher concentration promote the growth of stem.
  8. Auxins play an important role in initiation and promotion of cell division.
  9. Auxins help in the formation of adventitious roots from cuttings when applied in lower concentration.
  10. Auxins play an important role in apical dominance.
  11. Auxin prevents abscission by preventing the action of hydrolytic enzymes in abscission layer.
  12. Auxins are used to produce parthenocarpic fruits in plants like orange, apple, tomato and grapes.

Question 9.
Give an account of physiological effects and application of auxin with examples.
Answer:

  1. Auxins are growth promoting substances synthesized at meristematic regions of plants.
  2. The primary effect of auxin is cell enlargement and it stimulates growth of stem and root.
  3. Apical dominance – The phenomenon where growing apical bud inhibits the growth of lateral bud is apical dominance which is controlled by auxin synthesized at apical bud.
  4. Owing to activity of inducing multiplication of cells it is used in plant tissue culture to produce callus.
  5. Auxin stimulates formation of lateral and adventitious roots hence used for rooting propagation of cuttings.
  6. 2, 4-D is a synthetic herbicide which kills dicot weeds.
  7. Induced parthenocarpy in fruits-oranges, banana, grapes, lemons is by application of auxin.
  8. Foliar spray of NAA and 2, 4-D induces flowering in litchi and pineapple.
  9. Premature fruit drop of apples, pear and oranges is prevented.
  10. Auxins break seed dormancy and promote germination.
  11. Auxins promote early differentiation of xylem and phloem, cell elongation, increase rate of respiration, prevent formation of abscission layer.

Question 10.
Explain the application of gibberellins.
Answer:

  1. Gibberellins are growth promoting hormone and it is present in root tips, stem tips and seeds.
  2. Gibberellins break dormancy of bud, dormancy of seed.
  3. They promote seed germination in cereals by activating or synthesising enzyme amylase to produce sugar.
  4. Gibberellins induce elongation of the cells in stem hence increase in internode length is noticed.
  5. In rosette plants like cabbage it causes ‘bolting’ that is increase in internode length before flowering.
  6. Gibberellins are more effective in inducing parthenocarpy than auxins in plants like tomato, apple and pear.
  7. It is also used to increase fruit size and length of bunches in grapes.
  8. By its application genetically dwarf plants can be converted to phenotypically tall e.g. Maize.
  9. It overcomes requirement of vernalization, delays senescence and prevents abscission.
  10. Application of gibberellins causes production of male flowers on female plants.

Question 11.
Describe about the physiological effects and applications of cytokinin.
Answer:

  1. Cytokinins are growth promoting hormone that promotes cell division. Kinetin, zeatin are examples of cytokinin.
  2. They promote cell division as well as cell enlargement.
  3. High cytokinin promotes shoot development.
  4. Growth of lateral buds is promoted by cytokinins. Thus it controls apical dominance.
  5. Process of ageing and senescence, abscission of plant organs is delayed by their application.
  6. It promotes formation of interfascicular cambium.
  7. It has a role in breaking seed dormancy and promotes seed germination.
  8. Cytokinins induce RNA synthesis.
  9. Cytokinin and auxin ratio and their interactions control morphogenesis and cell differentiation.

Question 12.
Discuss about physiological effects and applications of ethylene.
Answer:

  1. Ethylene is a gaseous growth inhibitor hormone.
  2. It promotes ripening of fruits like bananas, apples and mangoes. The commercial application of ethephon is done.
  3. It initiates growth of lateral roots.
  4. Dormancy of buds and seeds is broken by its application.
  5. It accelerates formation of abscission layer and thus abscission of leaves, flowers and fruits is observed.
  6. Ethylene is responsible for checking growth of lateral buds thus causes apical dominance and retards flowering.
  7. Process of senescence of plant organs is enhanced.
  8. Epinasty, i.e. drooping of leaves and flowers results due to its application in some plants.
  9. It increases activity of chlorophyllase enzyme causing degreening effect in banana and Citrus fruits.

Question 13.
Discuss about experiment of Hendricks and Borthwick for discovery of phytochromes.
Answer:

  1. Phytochrome pigments receive photoperiodic stimulus and control flowering in plants.
  2. Hendricks and Borthwick observed that in SDP flowering is inhibited if continuous dark period is interrupted even by a short duration or flash of red light of wavelength 660 nm.
  3. If this interruption is again exposed to flash of far red light of wavelength 730 nm, then these plants flower.
  4. From this they concluded that some pigment system in plant receives the photoperiodic stimulus.
  5. These pigment proteins are called phytochromes and it exists in two interconvertible forms – Pr and Pfr.
  6. These pigments are located in cell membranes of green cells.
  7. Pfr is biologically active form and during daytime it gets accumulated in the plant.

Maharashtra Board Class 12 Biology Important Questions Chapter 7 Plant Growth and Mineral Nutrition

Question 14.
Give schematic representation of nitrogen cycle and enlist important steps of this cycle.
Answer:

  1. Atmospheric nitrogen is a source of nitrogen cycle.
  2. The important steps of the cycle are Nitrogen fixation, Nitrification, Ammonification, Nitrogen assimilation by plants. Amino acid synthesis and amidation, Denitrification and sedimentation.
  3. Amino acid are building blocks of proteins. Amides are amino acid with two amino groups.
  4. Schematic representation of Nitrogen
    Maharashtra Board Class 12 Biology Important Questions Chapter 7 Plant Growth and Mineral Nutrition 25

Question 15.
What is nitrogen cycle? Describe it briefly.
Answer:

  1. The cyclic movement of nitrogen between the atmosphere, biosphere and geosphere in different forms is called nitrogen cycle.
  2. Nitrogen cycle is one of the most important biogeochemical cycles.
  3. The nitrogen cycle involves many processes such as cycling of nitrogen through the biosphere, atmosphere and geosphere, nitrogen fixation, nitrogen uptake, formation of biomass, ammonification, nitrification and denitrification, etc.
  4. Bacteria such as Nitrosomonas, Nitrosococcus and Nitrobacter are the nitrifying bacteria which play an important role in nitrification.
  5. Denitrification is carried out by the bacteria Pseudomonas denitrifficans. From this it is obvious that bacteria play a major role in nitrogen cycle.
  6. Nitrogen fixation occurs by physical, industrial and biological methods. Prokaryotic organism play an important role in biological nitrogen fixation.

Maharashtra Board Class 12 Biology Important Questions Chapter 6 Plant Water Relation

Balbharti Maharashtra State Board 12th Biology Important Questions Chapter 6 Plant Water Relation Important Questions and Answers.

Maharashtra State Board 12th Biology Important Questions Chapter 6 Plant Water Relation

Multiple Choice Questions

Question 1.
What is the reason behind the various properties of water?
(a) Physical state
(b) Hydrogen bonding
(c) Colour
(d) Fluidity
Answer:
(b) Hydrogen bonding

Question 2.
Which special type of tissue is present in epiphytic roots?
(a) Velamen
(b) Lenticel
(c) Aerenchyma
(d) Haustoria
Answer:
(a) Velamen

Maharashtra Board Class 12 Biology Important Questions Chapter 6 Plant Water Relation

Question 3.
In which zone of root we come across root hairs?
(a) meristematic region
(b) zone of elongation
(c) zone of absorption
(d) maturation zone
Answer:
(c) zone of absorption

Question 4.
In double layered cell wall of root hair, outer layer is of ……………..
(a) cellulose
(b) pectin
(c) cutin
(d) suberin
Answer:
(b) pectin

Question 5.
What is rhizosphere?
(a) Root ball formed by growth of roots
(b) Covering of root tip
(c) Microenvironment around root
(d) region of root hairs
Answer:
(c) Microenvironment around root

Question 6.
A root hair is derived from …………….. cell.
(a) epidermal
(b) cortical
(c) endodermal
(d) pericycle
Answer:
(a) epiderinal

Question 7.
When we keep raisins in water they swell up, due to ……………..
(a) exosmosis
(b) plasmolysis
(c) imbibition
(d) diffusion
Answer:
(c) imbibition

Question 8.
This phenomenon is associated with exchange of gases ……………..
(a) osmosis
(b) diffusion
(c) imbibition
(d) plasmolysis
Answer:
(b) diffusion

Question 9.
What is correct expression for diffusion pressure deficit (D.PD.)?
(a) D.PD. = O.R – S.E
(b) D.PD. = T.E – O.E
(c) D.PD. = W.E – T.R
(d) D.PD. = O.R – T.R
Answer:
(d) D.PD. = O.P – T.P

Question 10.
What is true for a turgid cell?
(a) T.R is zero
(b) T.R = S.R
(c) D.P.D. is zero
(d) W.R = S.R
Answer:
(c) D.P.D. is zero

Question 11.
The particles which easily diffuse through cell membrane are ……………..
(a) lipid soluble
(b) water soluble
(c) hydrophilic
(d) lipophobic
Answer:
(a) lipid soluble

Question 12.
Which proteins help in facilitated diffusion process?
(a) cutin
(b) mucin
(c) aquaporins
(d) lipoproteins
Answer:
(c) aquaporins

Question 13.
D.PD. is now known as ……………..
(a) water potential
(b) solute potential
(c) pressure potential
(d) osmotic potential
Answer:
(a) water potential

Question 14.
Water always flows from ……………..
(a) more negative water potential to less negative water potential
(b) high water potential area to low water potential area
(c) low water potential area to high water potential area
(d) negative water potential area to area of zero water potential
Answer:
(b) high water potential area to low water potential area

Question 15.
Plasmolysed cell becoming turgid is process of ……………..
(a) replasmolysis
(b) incipient plasmolysis
(c) deplasmolysis
(d) exosmosis
Answer:
(c) deplasmolysis

Question 16.
In a fully turgid cell ……………..
(a) T.P = O.P
(b) T.P = S.P
(c) O.P = S.P
(d) T.RP = 0 (zero)
Answer:
(a) T.P = O.P

Question 17.
Casparian strip of endodermis has material ……………..
(a) pectin
(b) suberin
(c) cutin
(d) porin
Answer:
(b) suberin

Question 18.
Water from pericycle is forced into xylem due to ……………..
(a) aquaporin
(b) plasmodesmata
(c) root pressure
(d) ion-channels
Answer:
(c) root pressure

Question 19.
Water absorbed from root hair when passes through intercellular spaces and cell wall it is …………….. pathway.
(a) apoplast
(b) symplast
(c) transmembrane
(d) plasmodesmata
Answer:
(a) apoplast

Question 20.
In root system, secondary roots arise from ……………..
(a) cortical cells
(b) endodermis
(c) pericycle
(d) passage cell
Answer:
(c) pericycle

Question 21.
Active absorption of water occurs during ……………..
(a) early morning
(b) daytime
(c) bright sunlight
(d) night time
Answer:
(d) night-time

Question 22.
The value of root pressure is usually about ……………..
(a) +1 to +12 bars
(b) -1 to +1 bars
(c) +1 to +2 bars
(d) +1 to +21 bars
Answer:
(c) +1 to +2 bars

Question 23.
The ascent of sap in plants takes place through ……………..
(a) xylem
(b) phloem
(c) parenchyma
(d) endodermis
Answer:
(a) xylem

Maharashtra Board Class 12 Biology Important Questions Chapter 6 Plant Water Relation

Question 24.
From the following, which water will show greatest water potential (ψ)
(a) pure water
(b) salt water
(c) sugar water
(d) salt + sugar + water
Answer:
(a) pure water

Question 25.
Due to entry of water in a cell, the pressure potential ……………..
(a) increases
(b) decreases
(c) remains unaffected
(d) becomes zero
Answer:
(a) increases

Question 26.
Which mineral element is not remobilized in plants?
(a) P
(b) K
(c) S
(d) Ca
Answer:
(d) Ca

Question 27.
Phloem sap analysis is done using isotope ……………..
(a) 16O
(b) 14C
(c) 15N
(d) 35S
Answer:
(b) 14C

Question 28.
In plants, food is mainly translocated in form of ……………..
(a) starch
(b) glucose
(c) sucrose
(d) amino acids
Answer:
(c) sucrose

Question 29.
Transport of food through phloem is ……………..
(a) unidirectional
(b) bidirectional
(c) three dimensional
(d) absent
Answer:
(b) bidirectional

Question 30.
Guttation occurs through ……………..
(a) stomata
(b) lenticels
(c) hydathodes
(d) velamen
Answer:
(c) hydathodes

Question 31.
Amount of cuticular transpiration is about ……………..
(a) 0.1-1%
(b) 8 – 10%
(c) 90 – 93%
(d) 2 – 8%
Answer:
(b) 8 – 10%

Question 32.
Epistomatic leaf shows ……………..
(a) stomata on upper side
(b) stomata on lower side
(c) stomata on both surfaces
(d) absence of stomata
Answer:
(a) stomata on upper side

Question 33.
Dumbbell shaped guard cells are found in ……………..
(a) most dicots
(b) grasses
(c) gymnosperms
(d) desert plants
Answer:
(b) grasses

Question 34.
Stomatal transpiration occurs during night time in ……………..
(a) most dicots
(b) grasses
(c) gymnosperms
(d) desert plants
Answer:
(d) desert plants

Question 35.
Reservoir of K+ ions is ……………..
(a) guard cells
(b) epidermal cells
(c) subsidiary cells
(d) mesophyll
Answer:
(c) subsidiary cells

Question 36.
Guard cells are surrounded by ……………..
(a) epidermal hairs
(b) mesophyll cells
(c) accessory cells
(d) lenticels
Answer:
(c) accessory cells

Question 37.
When guard cells close stomata at night which acid prevents uptake of K+ and C ions?
(a) Abscissic acid
(b) Pyruvic acid
(c) Malic acid
(d) Acetic acid
Answer:
(a) Abscissic acid

Question 38.
Which type of injury is noticed in plants if there is excessive transpiration?
(a) Burning
(b) Chlorosis
(c) Necrosis
(d) Wilting
Answer:
(d) Wilting

Question 39.
Maximum transpiration occurs through ……………..
(a) cuticle
(b) lenticels
(c) stomata
(d) bark
Answer:
(c) stomata

Question 40.
What will be the condition of guard cells during night-time?
(a) Show increased turgor pressure
(b) Become flaccid
(c) Increase uptake of K+ and Cl ions
(d) Starch converted to sugar
Answer:
(b) Become flaccid

Question 41.
Cell A has water potential – 10 bars and cell B has – 5 bars, the movement of water will occur from ……………..
(a) A to B
(b) B to A
(c) No movement
(d) Either A to B or B to A
Answer:
(b) B to A

Question 42.
What is the correct symbol and unit of water potential?
(a) ψ and ha
(b) ω and ha
(c) ω and Pa
(d) ψ and Pa
Answer:
(d) ψ and Pa

Question 43.
When root system absorbs water, which is the first physical process concerned with this activity?
(a) Osmosis
(b) Imbibition
(c) Facilitated diffusion
(d) Diffusion
Answer:
(b) Imbibition

Match the columns

Question 1.

Column A (Scientist)Column B (Theory)
(1) Munch(a) Proton transport theory
(2) Bohem(b) Pressure flow theory
(3) J. Pristley(c) Capillary theory
(4) Levitt(d) Root Pressure theory

Answer:

Column A (Scientist)Column B (Theory)
(1) Munch(b) Pressure flow theory
(2) Bohem(c) Capillary theory
(3) J. Pristley(d) Root Pressure theory
(4) Levitt(a) Proton transport theory

Question 2.

Column A (Scientist)Column B (Theory)
(1) Dixon and Joly(a) Starch-sugar                inter conversion
(2) Steward(b) Osmotic absorption theory
(3) Atkins and Pristley(c) Non-osmotic absorption theory
(4) Kramer and Thimann(d) Cohesion tension theory

Answer:

Column A (Scientist)Column B (Theory)
(1) Dixon and Joly(d) Cohesion tension theory
(2) Steward(a) Starch-sugar inter conversion
(3) Atkins and Pristley(b) Osmotic absorption theory
(4) Kramer and Thimann(c) Non-osmotic absorption theory

Question 3.

Column AColumn B (New Terminology)
(1) D.P.D.(a) Osmotic potential
(2) O.E(b) Pressure potential
(3) T.P(c) Water potential

Answer:

Column AColumn B (New Terminology)
(1) D.P.D.(c) Water potential
(2) O.E(a) Osmotic potential
(3) T.P(b) Pressure potential

Classify the following to form Column B as per the category given in Column A

Question 1.
B, Co, Mn, P, N, S

Column AColumn B
Macro elements——–, ———-, ——–
Micro elements——–, ———-, ——–

Answer:

Column AColumn B
Macro elementsP, N, S
Micro elementsB, Co, Mn

Maharashtra Board Class 12 Biology Important Questions Chapter 6 Plant Water Relation

Question 2.
Capillary water, Combined water, Hygroscopic water, Gravitational water.

Column AColumn B
(1) Water adsorbed on soil particles—————
(2) Water penetrated deep in soil————–
(3) Water present as hydrated oxides————-
(4) Water present between soil particles————-

Answer:

Column AColumn B
(1) Water adsorbed on soil particlesHygroscopic water
(2) Water penetrated deep in soilGravitational water
(3) Water present as hydrated oxidesCombined water
(4) Water present between soil particlesCapillary water

Very short answer question

Question 1.
Why water acts as a thermal buffer?
Answer:
The water has high specific heat, high heat of vaporization and high heat of fusion hence it acts as a thermal buffer.

Question 2.
Why water can easily rise in capillaries?
Answer:
Water has high surface tension and high adhesive and cohesive force hence it can easily rise in capillaries.

Question 3.
Enlist the kind of water available in soil environment.
Answer:
Soil environment has gravitational water, hygroscopic water, combined water and capillary water.

Question 4.
Mention examples of imbibition process.
Answer:
Soaking of seeds, swelling up of raisins, kneading of flour and warping of wooden doors in rainy season.

Question 5.
What is the mechanism of imbibition?
Answer:
When imbibition takes place water molecules (imbibate) get tightly adsorbed on imbibant without formation of solution.

Question 6.
Why water enters plant cell by process of diffusion?
Answer:
Water present around cell wall has more diffusion pressure than inner cell sap hence water moves in the cell through freely permeable cell wall by diffusion.

Question 7.
What is isotonic condition in osmotic system?
Answer:
A condition where concentration of solution has neither gain nor loss of water in an osmotic system is isotonic.

Question 8.
What is effect on protoplasm when cell is plasmolysed ?
Answer:
When cell is plasmolysed, protoplast of cell shrinks and recedes from the cell wall thus gap is observed between cell wall and protoplast.

Question 9.
What is apoplast pathway?
Answer:
When water absorbed by root hair passes across the roots through cell wall and intercellular spaces of root cortex it is called apoplast pathway.

Question 10.
Which experiment is a proof for existence of root pressure?
Answer:
When a stem of potted plant is cut above the soil, xylem sap is seen oozing through cut end, that proves presence of root pressure.

Question 11.
Which analysis indicates that minerals are absorbed by plants?
Answer:
Analysis of plant ash contents is indication of absorbed minerals.

Question 12.
Which ions are readily remobilized in plants?
Answer:
Ions of phosphorus, sulphur and nitrogen are remobilized from older plants, parts (leaves) to younger parts.

Question 13.
What is radial and tangential translocation?
Answer:
When lateral translocation of food occurs in root or stem, transport from phloem to pith is radial translocation while that from phloem to cortex is tangential translocation.

Question 14.
What is loading of vein?
Answer:
In turgid cell, due to increased turgor pressure of photosynthetic cell, sugar is forced into the sieve tube of the vein, which is known as loading of vein.

Question 15.
What is unloading of vein?
Answer:
At the sink end, turgor pressure is lowered and hence turgor pressure gradient is developed from sieve-tube which translocates food passively along concentration gradient, this is vein unloading.

Question 16.
What is peculiarity of wall of guard cells?
Answer:
The inner wall of guard cell that faces opening is thick and inelastic while its outer wall or lateral wall is thin and elastic.

Question 17.
Give reaction of starch-sugar interconversion theory.
Answer:
Maharashtra Board Class 12 Biology Important Questions Chapter 6 Plant Water Relation 1

Question 18.
How do land plants absorb water and mineral salts from the soil?
Answer:
Land plants absorb water and mineral salts from soil with the help of their roots.

Question 19.
In which form the water is lost from leaves to the atmosphere?
Answer:
The water is lost in the form of vapour from leaves to the atmosphere.

Question 20.
How does plants lift the water from soil up to canopy without any pump?
Answer:
Plants have vascular tissue system of xylem, mainly through vessels and tracheids. Water is conducted upwards due to pull created in this continuous channel of water.

Question 21.
What is a peculiarity of epiphytic plants like orchids?
Answer:
The epiphytic plants like orchids have epiphytic roots with special water vapour absorbing layer of velamen tissue that absorbs water vapour from air.

Maharashtra Board Class 12 Biology Important Questions Chapter 6 Plant Water Relation

Question 22.
What are root hairs?
Answer:
Root hairs are the extensions of epiblema or epidermal cells in the region of absorption.

Question 23.
Why do the wooden doors become very hard to close and open in rainy season?
Answer:
During rainy season wooden doors swell up due to imbibition of water. The humidity or moisture content of air increases and wood which is hydrophilic takes up this moisture.

Question 24
How does the water come out through the surface of porous earthen pot?
Answer:
Earthen pot has tiny pores through which water diffuses out. Due to evaporation of this water from surface water kept inside becomes cool.

Question 25.
Describe external structure of root hair.
Answer:
A root hair is a cytoplasmic extension of epiblema cell which is colourless, unbranched, ephemeral, very delicate tubular structure of about 1 to 10 mm long.

Question 26.
What is gravitational water?
Answer:
Water present in the soil that percolates deep inside due to gravitational force is called gravitational water.

Question 27.
How can we describe osmotic movement of water related to energy?
Answer:
The osmotic movement of water is on the basis of free energy which is used to do work.

Question 28.
Which process is responsible for transport of minerals and to make them available to cells?
Answer:
The minerals absorbed by roots are transported upwards through sap and from veins by process of diffusion cells uptake them.

Question 29.
Give examples of vertical translocation of food in downward direction and upward direction.
Answer:
Food is translocated in downward direction from leaves (source) to root, while during germination of seed, bulbil, corm it is in upward direction.

Question 30.
What is translocation of food?
Answer:
The transport of food from one part of plant to the other part, i.e. source to sink is called translocation of food.

Question 31.
Which plant tissues are involved in transport of water, minerals and food?
Answer:
A complex plant tissue xylem, mainly with its tracheids and vessels is involved in transport of water and minerals while complex tissue phloem with sieve tubes and companion cells is involved in transport of food.

Question 32.
Which is a direct pathway of transport available in plants through root system?
Answer:
Secondary roots that originate from pericycle that is outer layer of vascular cylinder, bypass endodermis with Casparian strip allow direct apoplast pathway to enter xylem and phloem.

Question 33.
For which type of plants root pressure theory of ascent of sap is applicable?
Answer:
Root pressure theory is applicable to plants having height up to 10 to 20 metres.

Question 34.
What is hydrogen bond?
Answer:
Hydrogen bond results from the attractive force between a hydrogen atom covalently bonded to a very electronegative atom like O or N.

Question 35.
When you burn an incense stick in one corner of room, its fragrance spreads all over the room in a short time. How does it happen?
Answer:
The process of diffusion is responsible for spreading the fragrance of an incense stick in one corner of room.

Give definitions of the following

Question 1.
Facilitated diffusion
Answer:
The passive absorption of solutes when mediated by a carrier is called facilitated diffusion.

Question 2.
Osmotic pressure / Osmotic potential
Answer:
The pressure exerted due to osmosis is osmotic pressure now termed as osmotic potential.

Question 3.
Deplasmolysis
Answer:
When plasmolysed cell is placed in hypotonic solution, endo-osmosis occurs making cell turgid again then this is called deplasmolysis.

Question 4.
Translocation of food
Answer:
The movement of food from one part of plant to the other part is called translocation of food.

Name the following

Question 1.
Condition of cell wall and cell membrane : based on permeability.
Answer:
Cell wall freely permeable and Cell membrane semipermeable.

Question 2.
Weak solution having low osmotic concentration.
Answer:
Hypotonic

Question 3.
Strong solution having high osmotic concentration.
Answer:
Hypertonic

Question 4.
A suberised layer on endodermis.
Answer:
Casparian strip

Question 5.
One example of each epistomatic, hypostomatic and amphistomatic leaf.
Answer:

  1. Epistomatic leaf – Lotus
  2. Hypostomatic – Nerium.
  3. Amphistomatic – Grass

Question 6.
Pathway for entry of water into xylem from endodermis.
Answer:
Symplast pathway

Question 7.
A material deposited on endodermis which forms barrier.
Answer:
Suberin

Question 8.
Water imbibed or adsorbed on soil particles.
Answer:
Hygroscopic water

Maharashtra Board Class 12 Biology Important Questions Chapter 6 Plant Water Relation

Question 9.
Water potential was previously known as.
Answer:
D.ED. or Diffusion pressure deficit

Question 10.
Free energy per molecule in chemical system.
Answer:
Chemical potential

Question 11.
The substance synthesized in cell responsible for increasing osmotic concentration as per Munch theory.
Answer:
Glucose

Question 12.
A waxy substance present in layer on outer surface of epidermis.
Answer:
Cutin in layer cuticle

Question 13.
Anatomical structure through which guttation occurs.
Answer:
Hydathode

Question 14.
Type of leaves in hydrophytes based on distribution of stomata.
Answer:
Epistomatic

Question 15.
Type of leaves in xerophytes based on distribution of stomata.
Answer:
Hypostomatic

Question 16.
The form in which food is transported in plants and then stored in plants.
Answer:
Transported in the form of sugar, sucrose and stored as starch.

Question 17.
Chief vascular element concerned with transport of food in plant.
Answer:
Sieve tubes of phloem.

Give functions/significance of the following

Question 1.
Zone of absorption of root.
Answer:
This zone has unicellular root hairs which absorb water available in rhizosphere.

Question 2.
Diffusion.
Answer:

  1. Necessary for absorption of water by root hairs
  2. For absorption of minerals
  3. Conduction of water against gravity
  4. Exchange of gases
  5. Transport and distribution of food.

Question 3.
Turgor pressure.
Answer:

  1. Keeps cells and cell organelles stretched
  2. Provides support
  3. During growth essential for cell enlargement
  4. Maintains shape of cell
  5. Facilitates opening and closing of stomata

Question 4.
Osmosis.
Answer:

  1. Absorption of water into root
  2. Maintains turgidity of cell
  3. Facilitates cell to cell movement of absorbed water
  4. Resistance to drought or frost injury
  5. Helps in drooping movement of leaflet.
    E.g. Touch me not plant

Question 5.
Root pressure.
Answer:
Hydrostatic pressure developed in living cells of root helps in forcing water from pericycle into xylem
Helps in upward conduction of water against gravity

Question 6.
Transpiration.
Answer:

  1. Removal of excess of water
  2. Helps in passive absorption of water and minerals
  3. Helps in ascent of sap – transpiration pull
  4. Maintains turgor of cells
  5. Imparts cooling effect by reducing temperature 90% – 93% is stomatal transpiration and hence when stomata are open gaseous exchange takes place.

Question 7.
Guard cells.
Answer:

  1. They contain few chloroplasts so capable of photosynthesis.
  2. They can change their size hence opening and closing of stomatal aperture.

Distinguish between the following

Question 1.
Diffusion and Osmosis.
Answer:

DiffusionOsmosis
1. Diffusion is the movement of ions, atoms or molecules of solute, liquids or gases.1. Osmosis Is the special type of diffusion of water or solvent.
2. Diffusion involves the flow of diffusing particles in both the directions.2. Osmosis involves the unidirectional flow of solvent molecules.
3. Diffusion does not occur through a semi- permeable membrance.3. Osmosis occurs through a semi-permeable membrance.
4. Diffusion occurs from a place of higher concentration of diffusing particles to a place of lower concentration.4. Osmosis occurs from a solution of lower concentration to a solution of higher concentration.

Question 2.
Active absorption and Passive absorption.
Answer:

Active absorptionPassive absorption
1. Active absorption is a physiological process.1. Passive absorption is a physical process.
2. Active absorption takes place by the process of osmosis.2. Passive absorption takes place by suction force.
3. It involves the expenditure of energy on the part of absorbing cells.3. It does not involve the expenditure of energy on the part of absorbing cells.
4. Active absorption takes place against the concentration gradient.4. Passive absorption takes place along the concentration gradient.

Question 3.
Cuticular transpiration and Stomatal transpiration.
Answer:

Cuticular transpirationStomatal transpiration
1. Cuticular traspiration takes place through the cuticle.1. Stomatal traspiration takes place through the stomata.
2. Cuticular transpiration accounts for 8 to 10% of total loss of water from plants.2. Stomatal transpiration accounts for 80 to 90% of total loss of water from plants.
3. Cuticular transpiration depends upon the thickness of the cuticle.3. Stomatal traspiration depends upon the number and size of stomata.

Give reasons or explain the statements

Question 1.
Water is considered as ‘elixir of life’.
Answer:

  1. Water plays an important role in living organisms.
  2. About 90 – 95% water is present in cell which is functional and structural unit of living organisms.
  3. It helps in maintaining turgidity and shape of cells and cell organelles.
  4. Due to its various properties, it is medium of biochemical reactions, transporting medium and thermal buffer also.
  5. Therefore it is absolutely necessary for life i.e. ‘elixir of life’.

Maharashtra Board Class 12 Biology Important Questions Chapter 6 Plant Water Relation

Question 2.
Water is a best transporting medium.
Answer:

  1. Water is in liquid state at room temperature.
  2. It is best solvent for most of the solutes. Thus called universal solvent.
  3. It is inert inorganic compound with neutral pH i.e. pH 7 when in pure form.
  4. Hence it is best medium for dissolved minerals.

Question 3.
Water is significant molecule that connects physical world with biological processes.
Answer:

  1. Water is an important constituent of cell. About 90 – 95% of protoplasm is water.
  2. Water in liquid state is best solvent in which various minerals and food molecules are dissolved and transported.
  3. Water acts as the thermal buffer has high specific heat.
  4. Water molecules have high adhesive and cohesive forces of attraction.
  5. It can rise in capillaries due to high surface tension and adhesive forces, e.g. Ascent of sap in plants.
  6. Due to all these important factors it is a significant molecule connecting physical world with biological processes.

Question 4.
Absorption of water by roots from soil is with physical processes inhibition, diffusion and osmosis.
Answer:

  1. Water is absorbed from rhizosphere with the help of unicellular root hairs.
  2. Root hairs have plasma membrane and thin, double layer cell wall of pectin and cellulose.
  3. During imbibition water molecules get tightly adsorbed to the wall of hydrophilic colloids.
  4. Cell wall is freely permeable membrane hence through diffusion water passes into the cell.
  5. Osmosis is a special kind of diffusion of solvent through a semipermeable membrane and as plasma membrane is semipermeable, water enters cell by osmotic mechanism.
  6. Hence all these three physical processes occur sequentially when water is absorbed by roots.

Question 5.
Additional apoplastic pathway through secondary roots is beneficial to plants.
Answer:

  1. Secondary roots develop from the pericycle which is inside endodermis.
  2. Protoxylem is situated close to pericycle in root.
  3. Endodermis have suberized layer Casparian strip which forces water to move in the symplast so that it can enter the vascular xylem.
  4. Since secondary roots originate form pericycle, they bypass the Casparian strip.
  5. Therefore, a direct pathway to xylem and phloem is available without moving into symplast.

Write shorts notes

Question 1.
Role of water or Biological importance of water.
Answer:

  1. Water is absolutely necessary for life.
  2. It is a major constituent of protoplasm.
  3. If provides aqueous medium for various metabolic reactions that take place in plant.
  4. It is raw material for photosynthesis.
  5. It helps in maintaining the turgidity of cells.
  6. It is excellent solvent for various organic materials.
  7. It is transporting medium for dissolved minerals.
  8. It is thermal buffer.

Question 2.
Properties of water.
Answer:

  1. Water is a compound and it is in liquid state at room temperature.
  2. It is an inert inorganic compound with neutral pH.
  3. It has high specific heat, high heat of vapourization, high heat of fusion.
  4. It has high surface tension.
  5. Water molecule has good adhesive and cohesive forces of attraction.
  6. The various properties of water are result of weak hydrogen bonding between the water molecules.

Question 3.
Lenticular transpiration.
Answer:
Maharashtra Board Class 12 Biology Important Questions Chapter 6 Plant Water Relation 2

  1. Small, raised structures, which are composed of loose cells with large intercellular spaces situated on bark are lenticels.
  2. They are observed on bark of old stem, root and on woody pericarp of fruits.
  3. Lenticular transpiration is of very small amount about 0.1% to 1% of total transpiration.
  4. It is very slow process but occurs throughout the day.

Question 4.
Structure of stomatal apparatus.
Answer:

  1. Stomata are minute openings mainly located in the epidermal surfaces of young stem and leaves.
  2. It is composed of two guard cells and accessory cells which form the opening stomatal pore for transpiration.
  3. Guard cells are kidney shaped in dicot plants and dumbbell shaped in grasses.
  4. Guard cells have unevenly thickened wall, inner wall thick, inelastic while outer wall is thin and elastic.
  5. Guard cells are nucleated cells with few chloroplasts and hence can perform photosynthesis.
  6. Accessory cells are specialized cells that surround the guard cells and are reservoirs of K+ ions.

Question 5.
Apoplast pathway.
Answer:

  1. The movement of water across the root cells occur by two pathways viz. apoplast and symplast.
  2. Apoplast movement of water in plants occurs exclusively through cell walls and intercellular spaces in living cells of cortex.
  3. Apoplast pathway is up to the endodermis as the suberized layer of Casparian strip obstructs the movement.
  4. Additional apoplastic route giving direct contact to xylem and phloem is through secondary roots which originate at pericycle and bypass endodermis.
  5. The cellulosic walls of root hair, cortical cells are hydrophilic and permeable to water.

Question 6.
Water potential.
Answer:

  1. Chemical potential of water is called water potential.
  2. The Greek letter (ψ) psi represents water potential.
  3. The unit of measurement is in bars/pascals (Pa) or atmosphere.
  4. Water potential of pure water is always zero.
  5. Water potential of protoplasm is equal to D.PD. but it has a negative value.
  6. Water always moves from less negative potential to more negative water potential.

Short answer questions

Question 1.
Enlist macronutrients and micronutrients required for plant growth.
Answer:

  1. Some minerals which are required in large amounts for plant metabolism are macronutrients. E.g. C, H, O, E N, S, Mg, Ca, K. etc.
  2. Some minerals which are required in small amounts for plant metabolism are micronutrients. E.g. Cu, Co, Mn, B, Zn, Cl, etc.

Question 2.
Explain about the factors that affect water absorption.
Answer:

  1. Water is absorbed by unicellular root hairs from soil.
  2. Presence of capillary water in soil is needed as this water from soil is absorbed by root hairs.
  3. Soil temperature of 20-30°C favours water absorption.
  4. Rate of absorption is decreased by high concentration of solutes in soil.
  5. Soil should be properly aerated, poorly aerated soil shows poor absorption rate.
  6. Increased transpiration accelerates rate of absorption of water.

Question 3.
Explain mechanism of sugar transport through phloem.
Answer:

  1. The part of plant where food is synthesized is known as source and where it is utilized is sink.
  2. Food is translocated through phloem tissue in the form of sucrose along the concentration gradient from source to sink.
  3. Munch’s pressure flow theory or mass flow hypothesis is widely accepted theory for sugar transport.
  4. Glucose is synthesized in photosynthetic cells hence endo-osmosis occurs due to increased osmotic concentration.
  5. With this turgor pressure increases and sugar is forced in sieve tube of vein from photosynthetic cell – Vein loading.
  6. At the sink, sugar is utilized and excess amount converted to starch, hence osmotic concentration is lowered and exosmosis takes place.
  7. Turgor pressure is lowered hence turgor pressure gradient is set and food is translocated passively against concentration gradient – Vein unloading.

Question 4.
Explain the principles of cohesion tension theory and its limitations?
Answer:

  1. Cohesion tension theory is widely accepted theory for translocation of water proposed by Dixon and Joly.
  2. It is based on principles cohesion and adhesion of water molecules along with transpiration pull.
  3. Strong attractive force between water molecules is cohesive force and the strong force of attraction between water molecules and wall of lumen of xylem vessels is called adhesive force.
  4. Owing to these cohesive and adhesive forces a continuous column of water is maintained in xylem elements from root to aerial parts, tips of leaves.
  5. Transpiration pull is developed in the xylem vessel in leaves.
  6. This pull is transmitted downwards and due to suction pressure, water is pulled passively against gravity, i.e. ascent of sap.

Limitations of theory-

  1. For activity of transpiration pull, water column should be maintained constantly and continuously. Owing to changes in temperature during day and night, gas bubbles may be formed in the water channel. This will break the continuity.
  2. According to this theory, tracheids are more efficient than vessels because of their tapering end walls which support water column. Vessels are tubular structure with open ends.

Question 5.
What is the meaning of specific heat, heat of vaporization and heat of fusion?
Answer:

  1. Specific heat : The specific heat is the amount of heat per unit mass required to raise the temperature by one degree. The specific heat of water is 1 calorie/gram degree C.
  2. Heat of vapourization : It is the amount of energy that must be added to a liquid substance to transform a quantity of that substance to gas. It is also known as heat of evaporation.
  3. Heat of fusion : It is amount of energy typically heat, provided to a specific quantity of the substance to change its state from solid to liquid at constant pressure.

Maharashtra Board Class 12 Biology Important Questions Chapter 6 Plant Water Relation

Question 6.
What are adhesive and cohesive forces?
Answer:

  1. Adhesive and cohesive forces are attractive forces between molecules of same substance.
  2. Cohesive forces exist between molecules of the same type. E.g. between water-water molecule.
  3. Adhesive forces exist between dissimilar molecules. E.g. water molecule and lignin deposited wall of xylem element.

Chart based/Table based questions

Question 1.
Complete the table based on types of solution.
Maharashtra Board Class 12 Biology Important Questions Chapter 6 Plant Water Relation 3
Answer:
Maharashtra Board Class 12 Biology Important Questions Chapter 6 Plant Water Relation 4

Question 2.
Complete the table.

EventPhysical process
1. Soaking of Seeds—————-
2. Water entering guard cells—————-
3. Exchange of gases—————
4. Loss of water in liquid form—————
5. Water coming out through earthen pot—————
6. Loss of water in vapour form—————
7. Absorption of solutes passively by carrier—————
8. Spreading of fragrance of incense stick—————

Answer:

EventPhysical process
1. Soaking of SeedsImbibition
2. Water entering guard cellsOsmosis
3. Exchange of gasesDiffusion
4. Loss of water in liquid formGuttation
5. Water coming out through earthen potDiffusion
6. Loss of water in vapour formTranspiration
7. Absorption of solutes passively by carrierFacilitated Diffusion
8. Spreading of fragrance of incense stickDiffusion

Diagram based questions

Question 1.
Zones of root
Answer:
Maharashtra Board Class 12 Biology Important Questions Chapter 6 Plant Water Relation 5

Question 2.
Structure of root hair
Answer:
Maharashtra Board Class 12 Biology Important Questions Chapter 6 Plant Water Relation 6

Question 3.
Diffusion of water
Answer:
Maharashtra Board Class 12 Biology Important Questions Chapter 6 Plant Water Relation 7

Question 4.
Pathway for water uptake
Answer:
Maharashtra Board Class 12 Biology Important Questions Chapter 6 Plant Water Relation 8

Question 5.
L. S. of sieve tube (Transport of food)
Answer:
Maharashtra Board Class 12 Biology Important Questions Chapter 6 Plant Water Relation 9

Maharashtra Board Class 12 Biology Important Questions Chapter 6 Plant Water Relation

Question 6.
Structure of stomata and Types of guard cells
Answer:
Maharashtra Board Class 12 Biology Important Questions Chapter 6 Plant Water Relation 10

Long answer questions

Question 1.
Write a note on macronutrients and micronutrients required for plant growth.
Answer:

  1. Plants absorb mincral nutrients from their surroundings.
  2. For a proper growth of plants about 35 to 40 different elements are required.
  3. Plants absorb these nutrients in ionic or dissolved form from soil with their root system e.g. Phosphorus as PO4. Sulphur as SO42- etc.
  4. Based on their requirement in quantity. they are classified as major nutrients or macronutrients and those needed In small amounts are minor or micronutrients.
  5. Macroclements are required in large amounts, as they play nutritive and structural roles e.g. C, H, O. R Mg. N, K. S and Ca. -Ca pectate cell wall component. Mg component of chlorophyll.
  6. C. H, O are non-mineral major elements obtained from air and water e.g. CO2 is source of carbon, Hydrogen from water.
  7. Microelements arc required In traces as they mainly have catalytic role as co-factors or activators of enzymes.
  8. Microelernents may be needed for certain activity in life cycle of plant e.g. B for pollen germination, Si has protective role during stress conditions and fungal attacks, Al enhances availability of phosphorus.
  9. The Important micronutrients for plant growth are Mn. B. Cu, Zn, Cl.

Maharashtra Board Class 12 Biology Important Questions Chapter 5 Origin and Evolution of Life

Balbharti Maharashtra State Board 12th Biology Important Questions Chapter 5 Origin and Evolution of Life Important Questions and Answers.

Maharashtra State Board 12th Biology Important Questions Chapter 5 Origin and Evolution of Life

Multiple Choice Questions

Question 1.
Which statement is true for the theory of spontaneous generation?
(a) Life came from outer space.
(b) Life can arise from dead matter.
(c) Life can arise from non-living things only
(d) Life arises spontaneously by miracle
Answer:
(c) Life can arise from non-living things only

Question 2.
When man makes the selection during animal husbandry and plant breeding programmes, then it is an example of ……………….
(a) reverse evolution
(b) artificial selection
(c) mutation
(d) natural selection
Answer:
(b) artificial selection

Maharashtra Board Class 12 Biology Important Questions Chapter 5 Origin and Evolution of Life

Question 3.
Palaeontological evidences for evolution are ……………….
obtained in the form of
(a) development of embryo
(b) homologous organs
(c) fossils
(d) analogous organs
Answer:
(c) fossils

Question 4.
The homologous organs such as bones of forelimbs of whale, bat, cheetah and man are similar in structure, because ……………….
(a) one organism has given rise to another
(b) they share a common ancestor
(c) they perform the same function
(d) they have biochemical similarities
Answer:
(b) they share a common ancestor

Question 5.
Which type of evolution gives rise to analogous organs?
(a) divergent evolution
(b) parallel evolution
(c) genetic drift
(d) convergent evolution
Answer:
(d) convergent evolution

Question 6.
Hardy Weinberg’s equation of genetic equilibrium i.e. (p + q)² = p² + 2pq + q² = 1 is used in ……………….
(a) population genetics
(b) Mendelian genetics
(c) biometrics
(d) molecular genetics
Answer:
(a) population genetics

Question 7.
Which type of rocks show maximum fossils ?
(a) Sedimentary rocks
(b) Igneous rocks
(c) Metamorphic rocks
(d) Any type of rock
Answer:
(a) Sedimentary rocks

Question 8.
Industrial melanism observed in moth, Biston bitularia shows ………………. type of natural selection.
(a) stabilising
(b) directional
(c) disruptive
(d) artificial
Answer:
(b) directional

Question 9.
Variations during mutations of meiotic recombinations are ……………….
(a) random and directionless
(b) random and directional
(c) small and directional
(d) random, small and directional
Answer:
(a) random and directionless

Question 10.
Theory of special creation is based on ………………. beliefs.
(a) scientific
(b) religious
(c) traditional
(d) mythological
Answer:
(b) religious

Question 11.
Which sentence holds true for theory of biogenesis?
(a) Living organisms arise from non-living things.
(b) Theory of biogenesis can explain origin of life.
(c) Continuity of life can be explained by this theory.
(d) This theory was disproved by Louis Pasteur.
Answer:
(c) Continuity of life can be explained by this theory

Question 12.
Who gave the Big-Bang theory that explained the origin of life ?
(a) Georges Lemaitre
(b) Oparin and Haldane
(c) Charles Darwin
(d) J.B.S. Haldane
Answer:
(a) Georges Lemaitre

Question 13.
Primitive atmosphere of the earth was of type ……………….
la) oxidizing
(b) reducing
(c) aerobic
(d) oxido-reducing
Answer:
(b) reducing

Question 14.
What is the meaning of protobiogenesis?
(a) The origin of life on the earth.
(b) The origin of protozoans on the earth.
(c) The origin of protists on the earth.
(d) The origin of protons of the earth.
Answer:
(a) The origin of life on the earth

Question 15.
What are the first form of life on the earth called ?
(a) Pre-cells or Protobionts
(b) Protoproteins
(c) Coacervates
(d) Chromophores
Answer:
(a) Pre-cells or Protobionts

Question 16.
Which of the following is a landmark in the origin of life ?
(a) Formation of oxygen
(b) Formation of carbohydrates
(c) Formation of proteins
(d) Formation of water
Answer:
(c) Formation of proteins

Question 17.
The first chemicals formed on the earth were ………………. etc.
(a) oxygen, CFC, ozone
(b) DNA. RNA and nucleotides
(c) salt, sugar, proteins
(d) water, ammonia, methane
Answer:
(d) water, ammonia, methane

Question 18.
The unique feature of hot dilute soup is that there was no free ………………. in it.
(a) nitrogen
(b) oxygen
(c) carbon
(d) sulphur
Answer:
(b) oxygen

Question 19.
On which plant did Hugo de Vries work during his experimentations?
(a) Hibiscus rosa sinensis
(b) Oenothera lamarkiana
(c) Mirabilis jalapa
(d) Pisum sativum
Answer:
(b) Oenothera lamarkiana

Question 20.
Which one out of the following is a connecting link between fish and amphibian ?
(a) Archaeopteryx
(b) Seymouria
(c) Ichthyostega
(d) Dinosaurus
Answer:
(c) Ichthyostega

Question 21.
Homologous organs always lead to ………………. evolution.
(a) convergent
(b) divergent
(c) parallel
(d) radiating
Answer:
(b) divergent

Question 22.
Find the odd one out
(a) Caecum
(b) Nictitating membrane
(c) Coccyx
(d) Sacrum
Answer:
(d) Sacrum

Question 23.
The most common types of fossils are ……………….
(a) moulds
(b) casts
(c) actual remains
(d) model
Answer:
(c) actual remains

Question 24.
In geological time scale which period showed dominance of reptiles ?
(a) Triassic
(b) Jurassic
(c) Cretaceous
(d) Eocene
Answer:
(b) Jurassic

Question 25.
Which was the period of beginning of modern birds ?
(a) Triassic
(b) Jurassic
(c) Cretaceous
(d) Eocene
Answer:
(c) Cretaceous

Question 26.
Which epoch showed mammals at height of evolution ?
(a) Eocene
(b) Oligocene
(c) Miocene
(d) Pliocene
Answer:
(c) Mlocene

Question 27.
When did Holocene began ? (mya = million years ago)
(a) 2 mya
(b) 1 mya
(c) 0.5 mya
(d) 0.01 mya
Answer:
(d) 0.01 mya

Question 28.
Cichlid fishes in Lake Victoria are representatives of ………………. type of speciation.
(a) allopatric
(b) sympatric
(c) isolation
(d) random
Answer:
(b) sympatric

Question 29.
Which is the offspring of male horse and female donkey?
(a) Mule
(b) Hinny
(c) Marino
(d) Baroque
Answer:
(b) Hinny

Question 30.
Which one out of the following is a living fossil ?
(a) Coelacanth
(b) Lung fish
(c) Shark
(d) Rays
Answer:
(a) Coelacanth

Question 31.
Giant cephalopods like Nautilus were present in ………………. period.
(a) Silurian
(b) Devonian
(c) Ordovician
(d) Permian
Answer:
(c) Ordovician

Maharashtra Board Class 12 Biology Important Questions Chapter 5 Origin and Evolution of Life

Question 32.
First modern birds were formed during ………………. period.
(a) Triassic
(b) Jurassic
(c) Cretaceous
(d) Eocene
Answer:
(c) Cretaceous

Question 33.
When did man emerge during evolutionary time period ?
(a) Eocene
(b) Oligocene
(c) Miocene
(d) Pliocene
Answer:
(d) Pliocene

Question 34.
Find the odd monkey out:
(a) Baboons
(b) Gibbons
(c) Macaques
(d) Langurs
Answer:
(b) Gibbons

Question 35.
Who was man with ape-brain ?
(a) Pithecanthropus
(b) Dryopithecus
(c) Australopithecus
(d) Neanderthal man
Answer:
(c) Australopithecus

Question 36.
Who was the first true man ?
(a) Pithecanthropus
(b) Cro-magnon
(c) Australopithecus
(d) Neanderthal man
Answer:
(a) Pithecanthropus

Question 37.
Which one of the following is not present in human beings ?
(a) S curves in vertebral column
(b) Orthognathus face
(c) Simian gap
(d) Chin
Answer:
(c) Simian gap

Question 38.
Which is the correct sequence of human evolution ?
(a) Australopithecus → Ramapithecus → Homo sapiens → Homo habilis
(b) Homo erectus → Homo habilis → Homo sapiens
(c) Ramapithecus → Homo habilis → Homo erectus → Homo sapiens
(d) Australopithecus → Ramapithecus → Homo erectus → Homo habilis → Homo sapiens.
Answer:
(c) Ramapithecus – Homo hablUs -‘ Homo erectus Homo sapiens

Question 39.
Fossils of Homo erectus were obtained in ………………. and ……………….
(a) Kenya, Shivalik hills
(b) Java, Peking
(c) Africa, Asia
(d) Neanderthal valley, Taung
Answer:
(b) Java, Peking

Question 40.
Dentition more like that of the modern man was seen for the first time in ……………….
(a) Dryopithecus
(b) Ramapithecus
(c) Australopithecus
(d) Homo habilis
Answer:
(d) Homo habilis

Question 41.
Lemur and Tarsier belongs to ……………….
(a) Prosimi
(b) Hyalobatidae
(c) Pongidae
(d) Homonidae
Answer:
(a) Prosimi

Match the columns

Question 1.

Geological timeAnimal life
(1) Cambrian(a) Amphibians
(2) Ordovician(b) First terrestrial animals
(3) Silurian(c) Jawless fishes
(4) Devonian(d) Trilobite

Answer:

Geological timeAnimal life
(1) Cambrian(d) Trilobite
(2) Ordovician(c) Jawless fishes
(3) Silurian(b) First terrestrial animals
(4) Devonian(a) Amphibians

Question 2.

Human stageCranial capacity in CC
(1) Homo sapiens(a) 650-800
(2) Homo neanderthalensis(b) 900
(3) Homo habilis(c) 1400
(4) Homo erectus(d) 1450

Answer:

Human stageCranial capacity in CC
(1) Homo sapiens(d) 1450
(2) Homo neanderthalensis(c) 1400
(3) Homo habilis(a) 650-800
(4) Homo erectus(b) 900

Classify the following to form Column B as per the category given in Column A

Question 1.
Nictitating membrane, Seymouria, Lung fish, Flipper of whale and wing of bird, Wing of insect and wing of bird, wisdom
tooth, Eye of octopus, an eye of mammal, vertebrate heart and brain.

Column IColumn II
(1) Homologous organs————-
(2) Analogous organs————-
(3) Vestigial organs————-
(4) Connecting links————-

Answer:

Column IColumn II
(1) Homologous organsFlipper of whale and wing of bird, Vertebrate heart and brain
(2) Analogous organsWing of insect and wing of bird. Eye of octopus, an eye of mammal
(3) Vestigial organsNictitating membrane, wisdom tooth
(4) Connecting linksSeymouria, Lung fish

Maharashtra Board Class 12 Biology Important Questions Chapter 5 Origin and Evolution of Life

Question 2.
Origin of conifers, Abundance of trilobites, Diversification of fishes, All types of marine algae, Formation of forests, Rise of dinosaurs, Extinction

Geological time periodMajor events
Cambrian————-
Devonian————-
Permian————-
Triassic————-

Answer:

Geological time periodMajor events
CambrianAbundance of trilobites, All types of marine algae
DevonianDiversification of fishes, Formation of forests
PermianOrigin of conifers. Rise of modern insects.
TriassicRise of dinosaurs, Extinction of seed ferns

Very short answer questions

Question 1.
What is protobiogenesis?
Answer:
The origin of life on the earth is called protobiogenesis.

Question 2.
What are panspermia?
Answer:
Panspermia or cosmozoa are considered to be spores through which life came on the earth from distant planets.

Question 3.
What are protobionts?
Answer:
Protobionts were first form of life which were formed by nucleic acids along with other inorganic and organic molecules. They have some properties of living form.

Question 4.
What are eobionts?
Answer:
Eobionts or protocells are the first primitive living system which are formed by colloidal aggregations of lipids and proteinoids.

Question 5.
When did universe originate? How?
Answer:
Universe originated about 20 billion years ago by huge titanic explosion called big- bang.

Question 6.
What were the different energy sources during primitive times when earth was cooling?
Answer:
The different available energy sources during primitive times were ultra-violet rays, radiations, lightning and volcanic activities.

Question 7.
How was first cell formed on the primitive earth?
Answer:
In the protocell, when RNA or DNA system developed and they started regulating various metabolic activities, then it was called a first cell.

Question 8.
How was first cell formed on the primitive earth?
Answer:
In the protocell, when RNA or DNA system developed and they started regulating various metabolic activities, then it was called a first cell.

Question 9.
How was first cell on the earth in its metabolism ?
Answer:
First cell was anaerobic, heterotrophic and obtained energy by chemoheterotrophic processes.

Question 10.
What are ribozymes?
Answer:
Ribozymes are catalytic RNA which act as biocatalyst.

Maharashtra Board Class 12 Biology Important Questions Chapter 5 Origin and Evolution of Life

Question 11.
Who disproved Lamarck’s theory?
Answer:
August Weismann disproved Lamarck’s theory.

Question 12.
Write theory of germplasm as suggested by Wallace.
Answer:
The theory of germplasm says that variations produced in germ cells or germplasm are inherited to next generations, but the somatic variations present in somatoplasm or somatic cells are not inherited.

Question 13.
Who are the main contributors of modem : synthetic theory of evolution?
Answer:
R. Fischer, J. B. S. Haldane, T. Dobzhansky, Huxley, E. Mayr, Simpson, Stebbins, Fisher, Sewall Wright, Medel, T. H. Morgan, etc. are the main contributors of modern theory of evolution.

Question 14.
What is variation?
Answer:
The variations are the differences that occur in morphology, physiology, nutrition, habit, behavioural patterns, etc.

Question 15.
What is mutation?
Answer:
Sudden, large, inheritable and drastic change occurring in the genetic constitution is called mutation.

Question 16.
What is gene frequency?
Answer:
Gene frequency is the relative frequency of an allele or a gene at a particular locus in a population, as compared to other genes, expressed as a fraction or percentage.

Question 17.
What is the other name for genetic drift?
Answer:
Sewall wright effect is the other name for genetic drift.

Question 18.
Why is genetic drift called founder’s effect?
Answer:
The allelic frequency of new population which undergoes genetic drift becomes different from the original one, thus the original drifted population becomes different and are called ‘founders’. Since founders are formed therefore genetic drift is called founder’s effect.

Question 19.
What are fossils?
Answer:
Fossils are the dead remains of plants and animals from prehistoric times, which are found in different forms such as moulds, casts, actual remains or compression seen in various geological layers.

Question 20.
What is embryology?
Answer:
Embryology is the branch of biology and medicine which deals with study of embryos and their development.

Give definition of the following

Question 1.
Gene flow
Answer:
The transfer of genes between two genetically different populations among themselves is called gene flow.

Question 2.
Genetic drift
Answer:
Any random fluctuation (alteration) in allele frequency, occurring in the natural population by pure chance, is called genetic drift.

Question 3.
Chromosomal aberrations
Answer:
The structural, morphological change in chromosome due to rearrangement, is called chromosomal aberrations.

Question 4.
Mendelian population
Answer:
Small interbreeding group of a population is defined as Mendelian population.

Question 5.
Crossing over
Answer:
Exchange of genetic material between non-sister chromatids of homologous chromosomes in sexually reproducing organisms, during gamete formation is called crossing over.

Question 6.
Saltation
Answer:
Saltation is defined as single step large mutation.

Name the following

Question 1.
The book written by Charles Darwin after returning from voyage.
Answer:
Origin of species by natural selection.

Question 2.
Five main postulates of Darwinism.
Answer:

  1. Overproduction
  2. Struggle for existence
  3. Organic variations
  4. Natural selection
  5. Origin of next species.

Question 3.
Five key factors of evolution as suggested by Stebbins.
Answer:

  1. Gene mutations
  2. Mutations in the chromosome structure and number
  3. Genetic recombinations
  4. Natural selection
  5. Reproductive isolation.

Maharashtra Board Class 12 Biology Important Questions Chapter 5 Origin and Evolution of Life

Question 4.
Name the four types of chromosomal aberrations.
Answer:

  1. Deletion
  2. Translocation
  3. Duplication
  4. Inversion

Question 5.
Name of the insect that displayed industrial melanism and the name of the scientist who studied them.
Answer:
Kettlewell reported that Biston betularia or peppered moth displayed industrial melanism.

Question 6.
Example of homologous organs in plants.
Answer:
Thorns of Bougainvillea and tendrils of Cucurbita.

Question 7.
Example of analogous organs in plants.
Answer:
Sweet potato and potato.

Question 8.
Pre-mating isolating mechanisms.
Answer:

  1. Habitat or ecological isolating mechanism
  2. Seasonal or temporal isolating mechanism
  3. Ethological isolating mechanism
  4. Mechanical isolating mechanism

Question 9.
Post-mating isolating mechanisms.
Answer:

  1. Gamete mortality
  2. Zygote mortality
  3. Hybrid sterility

Question 10.
Name of connecting link between reptiles and birds.
Answer:
Archaeopteryx is the connecting link between reptiles and birds.

Question 11.
Name of connecting link between amphibians and reptiles.
Answer:
Seymouria is the connecting link between amphibians and reptiles.

Question 12.
Name of connecting link between fish and amphibians.
Answer:
Ichthyostegia is a missing link between fish and amphibians.

Question 13.
Name the period that was dominant for Amphibia.
Answer:
Carboniferous is dominant period for Amphibia.

Question 14.
Name the epoch when mammals were at the height of evolution.
Answer:
Miocene epoch.

Question 15.
Name the era when birds began to origin.
Answer:
Mesozoic era.

Question 16.
Name the three subclasses of Class Mammalia.
Answer:

  1. Marsupials
  2. Monotremes
  3. Eutheria

Question 17.
Name the three subfamilies of family Hominoidea.
Answer:

  1. Hyalobatidae
  2. Pongidae
  3. Hominidae

Give the significance of the following

Question 1.
Significance of Natural selection
Answer:

  1. Natural selection is the main driving force behind the evolution.
  2. Natural selection favours those genetic variations which have better fitness value.
  3. Such organisms are at selective advantage and they produce more offspring than the rest. Such organisms have greater survival and reproductive capacity.
  4. In this way natural selection helps in the evolution of new species.
  5. Natural selection favours differential reproduction of gene and brings about changes in the gene frequency.
  6. Natural selection brings about evolutionary changes.
  7. Natural selection also eliminates the genes carrying harmful mutations. This is called mutation balance in which allele frequency of harmful recessives remain constant generation after generation.

Distinguish between the following

Question 1.
Gene flow and Genetic drift.
Answer:

Gene flowGenetic drift
1. Gene flow is the alteration in the gene frequency due to migrations.1. Genetic drift is alteration in the gene frequency by pure chance.
2. Gene flow occurs due to exchange of genes in the adjacent populations through interbreeding.2. Genetic drift occurs due to accidental and sudden elimination of a particular gene.
3. Larger populations tend to show more migrations and hence more gene flow.3. Smaller populations have greater chances of genetic drift.
4. Gene flow occurs due to emigration and immigration.4. Genetic drift occurs only within the specified population.

Maharashtra Board Class 12 Biology Important Questions Chapter 5 Origin and Evolution of Life

Question 2.
Directional, Stabilizing and Disruptive selection.
Answer:

Directional selectionStabilizing selectionDisruptive selection
1. Natural selection operating in a linear direction is called directional selection.1. Natural selection operating to balance or stabilize the population is called stabilizing selection.1. Natural selection which disrupts the mean characteristics of a population is called disruptive selection.
2. In directional selection, more individuals acquire characters other than the mean character value.2. In stabilizing selection, more individuals of a population acquire a mean character value.2. In disruptive selection, more number of individuals acquire extreme or peripheral character value.
3. Directional selection eliminates one of the extremes of the phenotypic range and favour the other.3. Stabilizing selection tends to favour the intermediate forms and eliminate both the phenotypic extremes.3. Disruptive selection favours extreme phenotypes and eliminate intermediate.
4. It streamlines variations.4. It reduces variations.4. It increases variations.
5. This kind of selection is the most common.5. This kind of selection is common.5. This kind of selection is rare.
6. Directional selection operates for many generations, it results in an evolutionary trend within a population and shifting a peak in one direction.

E.g. Industrial melanism, DDT resistance in mosquito, etc.

6. This selection leads to evolutionary change but tend to maintain phenotypic stability within population.

E.g. All the populations which have adapted to their environment.

6. It ensures the effect on the entire gene pool of a population, considering all mating types or systems.

E.g. African seed cracker finches with different sized beaks

Question 3.
Homologous organs and Analogous organs.
Answer:

Homologous organsAnalogous organs
1. Homologous organs are structurally similar to one another.1. Analogous organs are structurally dissimilar to one another.
2. Homologous organs are functionally dissimilar from one another.2. Analogous organs are functionally similar to one another.
3. Homologous organs help in tracing the evolutionary relationships.3. Analogous organs do not help in tracing the evolutionary relationships.
4. Homologous organs lead to divergent evolution.4. Analogous organs lead to convergent evolution.
5. Animals residing in different habitats but having closer evolutionary relationship show homologous organs.

E.g. Forelimbs of frog, lizard, bird, bat, whale, man, etc.

5. Diverse animals residing in the similar habitat show analogous organs.

E.g. Wing of a bird and wing of an insect.

Question 4.
Allopatric and Sympatric speciation.
Answer:

Allopatric speciationSympatric speciation
1. Allopatric speciation is the formation of a new species due to separation of a segment of population from the original population.1. Sympatric speciation is the formation of species within single population.
2. There is geographical barrier cutting across the species range during such speciation.2. There is no geographical isolation during sympatric speciation.
3. Allopatric speciation does not have physiological barrier.3. Sympatric speciation is due to physiological or reproductive isolating barrier.
4. Migration of individual are also helpful in allopatric speciation. E.g. African elephant and Indian elephant.4. Mutations are helpful in sympatric speciation. E.g. Cichlid fishes in Lake Victoria.

Give scientific reasons

Question 1.
Simple organic molecules did not show decomposition in primitive oceans.
Answer:
Simple organic molecules which were formed during chemical evolution, accumulated at the bottom of water bodies. At that time there was no free oxygen and enzymes. Therefore, simple organic molecules did not show decomposition in primitive oceans.

Question 2.
Archaeopteryx is called connecting link between reptiles and birds.
Answer:
(1) Archaeopteryx shows reptilian as well as avian characters.

(2) Reptilian characters are as follows:

  • Jaws with homodont (all similar) teeth. Bones are nonpneumatic i.e. solid.
  • Ribs have a single head. Sternum without keel.
  • Abdominal ribs present which are like the crocodilian ribs.
  • Forearms have three digits ending in distinct claws while the hind limb has four digits ending in clawed digits.

(3) Avian characters shown by it are as follows:

  • Forearms modified into wings.
  • Feathery exoskeleton.
  • Skull bones are completely fused.
  • Cranium is rounded with large orbits and a single condyle.
  • Jaws are modified into beak.
  • Limb bones have first toe in opposable manner. Foot present with clawed digits. Since it showed characters of both the classes, it is considered as the connecting link between the two.

Question 3.
Birds are glorified reptiles.
Answer:
Huxley, the evolutionary biologist gave this statement after studying the characters of birds and reptiles. The fossil bird, Archaeopteryx was discovered which showed characters of both Reptilia and Aves. It showed transformation of reptilian characters into bird characters. Hence, birds are said to be glorified reptiles with feathery exo-skeleton and other glorious characteristics.

Question 4.
Analogous organs do not have significant role in evolution.
Answer:
Analogous organs lead to convergent evolution, i.e. different organisms show same superficial structural similarities due to similar functions or habitat. But anatomically and structurally they are different. These organs do not help to trace the common ancestry. Therefore, they are said to have no significant role in evolution.

Question 5.
Australopithecus is described as a man with ape brain.
Answer:
(1) Australopithecus can be considered as a connecting link between ape and man due to the following ape-like and man like characteristics shown by it.

(2) The ape-like characteristics of Australopithecus:

  • The jaws and teeth were larger than those of modern man.
  • The face was prognathous, i.e. it had a muzzle like slope
  • The chin was absent
  • The eye-brow ridges projected over the eyes
  • Their cranial capacity ranged from 450-600 c.c.

(3) The man-like characteristics of Australopithecus:

  • It walked nearly or completely straight due to erect posture.
  • The vertebral column had a distinct lumbar curve with broad basin-like pelvic girdle.
  • Dentition was man-like with the smoothly rounded parabolic dental arch.
  • A simian gap was absent. Australopithecus is therefore, rightly described as a man with ape brain.

Write short notes on the following

Question 1.
Evidences of Darwinism.
Answer:
(1) Height of neck of Giraffe : Long-necked Giraffe came into existence in the following way. Long-necked Giraffe could pluck and eat more leaves from tall trees and woody climbers. So it was well adapted to the environment. Short-necked one could not get food and thus perished in the struggle. This adaptation was transmitted to their offspring.

(2) Black colour peppered moths : The example of industrial melanism seen in U.K. is an excellent example of natural selection in action. Black coloured moths evolved gradually as new species from the previous white coloured forms.

(3) DDT resistance in mosquitoes : Intensive DDT spraying destroyed all types of mosquitoes. Some mosquitoes developed resistance to DDT and survived in spite of DDT spray. They reproduced more and were thus selected naturally.

Question 2.
Drawbacks and Objections to Darwinism.
Answer:

  1. Darwin took into consideration minute fluctuating variation as principal factors. But these are neither heritable nor are part of evolution.
  2. Darwin did not distinguish somatic and germinal variation and considered all variations are heritable.
    ‘Arrival of the fittest’ was not explained by him.
  3. Darwin was unable to explain the cause, origin and inheritance of variations and of vestigial organs.
  4. He also could not explain extinction of species.
  5. Gradual accumulation of useful variations forms the new species, but their intermediate forms were not recognised.
  6. Darwin could not explain existence of neutral flowers and the sterility of hybrids.

Question 3.
The main features of mutation theory.
Answer:

  1. Mutations are large, sudden and discontinuous variations in a population.
  2. Changes caused due to mutations are inheritable.
  3. The raw material for organic evolution is provided by mutations.
  4. Mutation can be useful or harmful. Useful mutations are at evolutionary advantage as they are selected by nature.
  5. Accumulation of the useful mutations over a period of time leads to the origin and establishment of new species.
  6. Harmful or non-adaptive mutation may persist or get eliminated by nature.

Maharashtra Board Class 12 Biology Important Questions Chapter 5 Origin and Evolution of Life

Question 4.
Directional selection.
Answer:

  1. Natural selection bringing about directional change without disrupting the balance is called directional selection.
  2. In a population when more individuals acquire characters which are other than the mean character value, then it is called directional selection.
  3. Natural selection usually acts to eliminate one of the extremes of the phenotypic range and favour the other. E.g. systematic elimination of homozygous recessives.
  4. Directional selection operates for many generations, it results in an evolutionary trend within a population and shifting a peak in one direction.
    E.g. Industrial melanism, DDT resistant mosquito, etc.

Question 5.
Stabilizing selection/Balancing selection.
Answer:

  1. Stabilizing selection is the type of natural selection which balances the population, hence it is also known as balancing selection.
  2. In such population more individuals acquire a mean character value.
  3. Such selection tends to favour the intermediate forms and eliminate both the phenotypic extremes.
    E.g. More number of infants with intermediate weight survive better as compared to overweight or underweight infants.
  4. Stabilizing selection reduces variations.
  5. It tends to maintain phenotypic stability within population, and does not bring about drastic evolutionary changes.
  6. A population showing stabilizing selection is well-adapted to its environment.

Question 6.
Homologous organs.
Answer:

  1. The structural similarities between the homologous organs indicate that they have a common ancestry.
  2. Different homologous organs indicate divergent evolution or adaptive radiation.
  3. Homologous organs help in tracing the phylogenetic relationships.
  4. Homologous organs are those organs which are structurally similar but functionally dissimilar.

E.g.
(i) Forelimbs of frog, lizard, bird, bat, whale and man are homologous to each other. All the limbs are morphologically similar in construction such as similar limb bones but are dissimilar in function. Frog limbs are meant for hopping, lizard limbs help in crawling, birds and bats fly with the . help of forelimbs while whale uses it for swimming and man for handling the objects.

(ii) Vertebrate heart and brain. & In plants, thorns of Bougainvillea and tendrils of Cucurbita represent homology.

Question 7.
Analogous organs.
Answer:

  1. Analogous organs are similar in function but dissimilar in structural details.
  2. They do not help to trace the relationship in the evolution but help to understand the convergent evolution.
  3. Structural modifications in the organs are due to similar habitat.

E.g.
(i) Wing of an insect and wing of a bird, both are useful in flight so they are functionally similar but are structurally different. Insect wing is formed by exoskeleton expansion while bird wing is the modified forelimb.

(ii) Eye of the Molluscan octopus and of eye of mammals. They differ in their retinal position, structure of lens and origin of different eye parts, but both perform function of vision.

(iii) The flippers of penguins (birds) and dolphins (mammals).

(iv) Sweet potato which is a root modification and potato which is a stem modification, both perform similar function of storing starchy food.

Question 8.
Vestigial organ.
Answer:

  1. Vestigial organs are rudimentary organs which are imperfectly developed and non¬functional, degenerate structures.
  2. These organs in animals become functionless thus their presence in the body is not required.
  3. But they are simply present as they descend down during evolution and continue to exist.
  4. In the process of evolution, they may disappear totally.
  5. They indicate evolutionary line as they were once functional in the ancestors.

Examples of vestigial organs in human beings:

  1. Caecum and vermiform appendix : These are functional in herbivorous animals where they help in cellulose digestion. In humans they are functionless.
  2. Nictitating membrane situated in the eyes of humans. It is a remnant of third eyelid.
  3. Coccyx or tail vertebrae which shows remnant of tail, Wisdom teeth or 3rd molars. These organs indicate that human beings descended from ape like ancestors.

Question 9.
Types of fossils.
Answer:
There are four main types of fossils : actual remains, moulds, casts and compressions.
1. Actual remains : The most common type of fossil is actual remains in which the plants, animals and human bodies are seen embedded in permafrost of arctic or alpine snow. Due to severe cold temperature, the bodies remain preserved in the actual state, E.g., Fossil of Woolly Mammoth in Siberia. Many insects and smaller arthropods remained embedded and thus preserved in amber or hardened resin.

2. Moulds : Hardened encasements formed in the outer parts of organic remains of animals or plants form moulds. The organisms later decays leaving cavities or the impression in permanent form. E.g. Footprints.

3. Casts : Casts are hardened pieces of mineral matter which is deposited in the cavities of moulds.

4. Compressions : A thin carbon film indicates the outline of external features of ancient organism, but other structural details are not seen.

Question 10.
Major changes that occurred in human evolution.
Answer:

  1. Major changes that took place in evolution of man are as follows :
  2. Increase in size and complexity of brain and enhanced intelligence.
  3. Increase in cranial capacity.
  4. Bipedal locomotion.
  5. Opposable thumb.
  6. Erect posture.
  7. Shortening of forelimbs and lengthening of hind limbs.
  8. Development of chin. Orthognathous face.
  9. Broadening of pelvic girdle and development of lumbar curvature.
  10. Social and cultural development such as articulated speech, art, development of tools, etc.

Question 11.
Dryopithecus.
Answer:

  1. Dryopithecus is also called Proconsul. Leakey discovered the fossils of Dryopithecus, on an island in Lake Victoria of Africa. Also the fossil was found in Haritalyanga in Bilaspur district of Himachal Pradesh.
  2. It was a group of apes that lived in Miocene epoch about 20 to 25 million years ago.’
    Several species of Dryopithecus are available, the important among these is African fossil D. africanus.
  3. Dryopithecus has a close similarity to chimpanzee and also walked like a modern chimpanzee.
  4. The structure of its limbs and wrists show that knuckle walking was lesser in it. It used the flat of its hands like a monkey.
  5. It had arms and legs of the same length and had a semi-erect posture.

Maharashtra Board Class 12 Biology Important Questions Chapter 5 Origin and Evolution of Life

Question 12.
Ramapithecus.
Answer:

  1. Ramapithecus was on direct line of evolution of man.
  2. It was called an ape-man like primate.
  3. Its fossils were obtained in the form of teeth and jaw bones in the rocks of Siwalik Hills in India by Lewis and also in Kenya.
  4. It existed during late Miocene and early Pliocene epoch about 14 to 12 million years ago.
  5. It walked erect on its hind limbs.
  6. It had close similarity with chimpanzee.
  7. Some scientists believe that Dryopithecus evolved into Ramapithecus.

Question 13.
Australopithecus.
Answer:

  1. Australopithecus is considered as connecting link between ape and man.
  2. Its fossils were obtained from Toung valley in South Africa, from Ethiopia and Tanzania.
  3. It was in late Pliocene or early Pleistocene epoch about 4 to 1.8 million years ago.
  4. It was about 4 feet tall. It had prognathus face, with larger jaws. Chin was absent. Lumbar curvature was present.
  5. It walked upright.
  6. The cranial capacity was about 450 to 600 CC. Therefore, it was called man with ape brain.

Question 14.
Homo habilis.
Answer:

  1. Homo habilis is described as Handy man. His fossils were obtained from Olduvai Gorge in Tanzania, Africa.
  2. He existed in late Pliocene or early Pleistocene about 2.5 to 1.4 million years ago.
  3. He was lightly built.
  4. Fossil of lower jaw was obtained which showed that his dentition was more like modern man with small molars.
  5. He walked erect. His cranial capacity was 640 to 800 cc.
  6. He did not eat meat and made stone tools.

Question 15.
Homo erectus.
Answer:

  1. Homo erectus was also known as Java man or Peking Man due to his fossils obtained from these areas.
  2. He was also called ape man.
  3. He lived in the middle Pleistocene epoch about 1.5 million years ago.
  4. He was 5 feet in height with prognathous face, massive jaws, huge teeth and bony eye brow ridges.
    Chin was absent.
  5. He walked erect.
  6. The cranial capacity was 900 cc.
  7. He was omnivorous and probably used fire and ate meat.

Question 16.
Neanderthal man.
Answer:

  1. The scientific name of Neanderthal man is Homo neanderthalensis. He is described as advanced prehistoric man.
  2. It was called Neanderthal man because its first fossil was collected from Neanderthal valley in Germany by Fuhlrott (1856).
  3. It was heavily built and short and had outwardly curved thigh bones.
  4. The facial features were as follows : prominent brow ridges, thick skull bones, low and slanting forehead, deep jaw without a chin, etc.
  5. Neanderthal man existed in late Pleistocene epoch about 1,00,000 to 40,000 years ago. It was widely spread in Europe, Asia and North America. It became extinct about 25,000 years ago.
  6. The cranial capacity of Neanderthal man was about 1400 cc, which was roughly equal to that of modern man. He used hide for dressing.
  7. It showed intellectual development in constructing and using flint tools and fire.
  8. The Neanderthal men used to bury their dead bodies along with their tools and perform ceremonies.

Short answer questions

Question 1.
Enlist the steps in the process of chemical evolution.
Answer:

  1. Origin of Earth and Primitive atmosphere.
  2. Formation of ammonia, water and methane.
  3. Formation of simple organic molecules.
  4. Formation of complex organic molecules.
  5. Formation of Nucleic acids.
  6. Formation of Protobionts or Procells.
  7. Formation of first cell.

Question 2.
When did Earth originate? Which transformations took place later?
Answer:

  1. Earth originated about 4.6 billion years ago as a part of the solar system.
  2. When it was formed, it was a rotating cloud of hot gases and cosmic dust. It was then appearing like a nebula.
  3. Later the condensation and cooling started which resulted in stratification.
  4. Heavier elements like nickel and iron settled to the core. Lighter elements like helium, hydrogen, nitrogen, oxygen, carbon, etc. remained on the surface and they formed the primitive atmosphere.
  5. This atmosphere of the earth was of a reducing type, devoid of free oxygen and very hot.

Question 3.
How were simple organic molecules formed on the earth?
Answer:
1. Initially earth’s temperature was very high but as the cooling process started, lighter elements reacted chemically with each other.

2. The early atmosphere was rich in hydrogen, carbon, nitrogen and sulphur. Hydrogen was most active and hence it reacted with other elements to form chemicals on earth like CH4, NH3, H20 and H2S.

3. With decreasing temperature of the earth, steam condensed into water that resulted in heavy rainfall. This constantiy falling rainwater got accumulated on the land to form different water bodies and especially oceans. It also cooled down the earth.

4. The early molecules of hydrocarbons, ammonia, methane and water underwent reactions like condensation, polymerisation, oxidation and reduction due to different energy sources such as ultra-violet rays, radiations, lightning and volcanic activities.

5. These reactions resulted in formation of simple organic molecules like monosaccharides, amino acids, purines, pyrimidines, fatty acids, glycerol, etc.

Question 4.
How were complex organic molecules formed during chemical evolution?
Answer:

  1. The primitive broth in which simple organic molecules were suspended, was neutral and free from oxygen.
  2. In this broth polymerisation took place and simple organic molecules aggregated to form new complex organic molecules like polysaccharides, fats, proteins, nucleosides and nucleotides.
  3. Protoproteins were formed by polymerisation of amino acids. These protoproteins later formed proteins.
  4. Formation of protein molecules is considered as landmark in the origin of life. Later the enzymes were formed which accelerated the rate of other chemical reactions.

Question 5.
How were protobionts formed with the help of nucleic acids during chemical evolution?
Answer:

  1. By the reaction between phosphoric acid, sugar and nitrogenous bases (purines and pyrimidines), nucleotides may have been formed.
  2. These nucleotides joined together to form nucleic acids such as RNA and DNA.
  3. Nucleic acids acquired self-replicating ability which is a fundamental property of living form.
  4. They later formed protobionts. They were the first form of life formed by nucleic acids along with inorganic and organic molecules.
  5. Protobionts were the prebiotic chemical aggregates having some properties of living system. Aggregation of organic molecules due to coacervation formed these protobionts.

Question 6.
Why variations are seen in population?
Answer:
Variations are seen in population due to gene flow, genetic drift, genetic recombinations that occur at the time of gamete formation, crossing over and sudden drastic changes like gene mutations or chromosomal aberrations. All the above factors are constantly operating over every population. Due to these evolutionary processes, variation take place in a population.

Maharashtra Board Class 12 Biology Important Questions Chapter 5 Origin and Evolution of Life

Question 7.
In which conditions the gene frequency of a population will remain constant?
Answer:
In the condition of no migrations of the organisms, no mutations, no sexually reproduction consisting of crossing over, no genetic drift, no recombinations and variation, the gene frequency of a population will remain constant. Such hypothetical conditions will never exist because even one set of sexually reproducing organisms forms an offspring which is slightly different from its parents. This means that there is constant change of gene frequency.

Question 8.
What is carbon dating and how does it work?
Answer:
Carbon dating is the method to find out the age of the fossil or any other organic matter. In carbon dating, the relative proportions of the carbon isotopes, carbon-12 and carbon-14 which are present in the organic matter, is estimated. The ratio between them changes as radioactive carbon-14 decays and is not replaced by exchange with the atmosphere. From these findings the age of that organic matter can be concluded.

Question 9.
What is a connecting link? Give suitable examples of connecting links.
Answer:

  1. A connecting link is an intermediate or transitional state between two systematic groups of organisms.
  2. It bears characters common to both these groups on either side of its position. Thus it represents an evolutionary line.
  3. Connecting links are also called a missing link.
    E.g. Archaeopteryx, the extinct bird is a connecting link between Reptiles and Aves.
  4. Seymouria is a connecting link between Amphibia and Reptilia.
  5. Ichthyostega is a connecting link between Pisces and Amphibia.

Question 10.
What is geological time scale? How is it divided ?
Answer:

  1. Geological time scale is the arrangement of major divisions of geological time into eras, periods and epochs on the time scale.
  2. This division is based on the study of fossilized organisms obtained from the different strata of the earth.
  3. The characteristic significant events that occurred in the organization of organisms helped the geologists to understand the geological time scale.
  4. The major divisions of geological time are called eras.
  5. The eras are divided into periods and the periods into epochs.
  6. By studying fossils in the earth crust, the evolutionary changes in the organisms have been traced out.

Question 11.
What is meant by palaeontological evidences ?
Answer:

  1. Palaeontology means the study of fossils. Palaeontological evidences are the fossilized forms of various organisms which are obtained from different strata of the earth. They represent the dead remains of plants and animals that lived in the past in various geological layers.
  2. The older and more primitive forms of life are excavated from the lower strata of the soil whereas the recent ones are situated on the upper layers of the soil.
  3. Fossils are formed in variety of materials such as sedimentary rocks, amber, volcanic gas, ice, peat bogs, soil, etc.
  4. They provide the true, direct and reliable evidences of evolution.

Question 12.
What are the molecular evidences that show the evolution?
Answer:

  1. Different organisms have basic similarities in their molecules and the cellular constituents.
  2. All living organisms have the same basic structural and functional unit, i.e. cell.
  3. Cell organelles such as endoplasmic reticulum, Golgi bodies, mitochondria, etc. are present in different types of organisms.
  4. Proteins and gene performing different functions have the same basic pattern which shows a common ancestry.
  5. Catabolic activities of liberating energy, synthesis of macromolecules such as proteins, carbohydrates, nucleic acids, etc. are similar in different organisms.
  6. ATP is the common energy currency of all the organisms.
  7. All the above facts are called molecular evidences in favour of evolution.

Question 13.
Arrange the following stages of the human evolution in the order of their increasing cranial capacity, (a) Neanderthal man (b) Cro-Magnon man (c) Homo erectus (d) Homo habilis.
Answer:

  1. Homo habilis (650-800 cc)
  2. Homo erectus (850-1200 cc)
  3. Neanderthal man (1400 cc)
  4. Cro-Magnon man (1450 cc)

Question 14.
Since your earlier school days you have been solving mysteries/puzzles labelled as use your brain power. Did you ever wonder why human brain has such a capacity? Why and how we evolved along these lines? What is the extent of similarity between humans, chimpanzees and monkeys?
Answer:
Human beings have extremely well- developed brain. Especially the cerebral hemispheres are very large constituting 85% of the brain weight. Due to such cerebrum, there are many, neurons. They are responsible for faculties such as speech, memory, emotions, thought process. The mind or psyche is well developed due to over developed cerebral hemispheres.

That’s why human brain has tremendous capacity of Chimpanzees are also comparatively more intelligent than the monkeys. However, the development of speech and language is lacking in them. The cranial capacity of chimpanzee and monkey is 275-500 cc and 45-50 cc respectively, whereas humans have 1450-1500 cc. Larger the brain, more is the intelligence and all other mental faculties which only humans show.

Maharashtra Board Class 12 Biology Important Questions Chapter 5 Origin and Evolution of Life

Question 15.
Even though the cranium of elephant is larger than that of man, humans are considered more intelligent than elephant. Why is it so?
Answer:
Elephant’s brain is large weighing about 5 kg. But elephant’s body weight too is very high. The proportion of body brain weight is highest in human beings. Therefore, humans are considered more intelligent than elephant. Moreover, the cerebral cortex of elephant is not very well developed, instead they have well developed cerebellum which helps in locomotion and balancing their huge bodies. Human brain has very well-developed cerebrum which brings about cognitive behaviour and intelligence.

Chart based/Table based questions

Question 1.
Give the graphical representation of Hardy-Weinberg’s principle in the form of Punnet square.
Answer:
Maharashtra Board Class 12 Biology Important Questions Chapter 5 Origin and Evolution of Life 1
Genotypes = AA + 2 Aa + aa
Gene frequencies = p² + 2pq + q²

Question 2.
Make a chart showing the types of isolating mechanisms.
Answer:
Maharashtra Board Class 12 Biology Important Questions Chapter 5 Origin and Evolution of Life 2

Diagram based questions

Question 1.
Give diagrammatic representation to show RNA world.
Answer:
Maharashtra Board Class 12 Biology Important Questions Chapter 5 Origin and Evolution of Life 3

Question 2.
Sketch and label four types of chromosomal aberrations
Answer:
Maharashtra Board Class 12 Biology Important Questions Chapter 5 Origin and Evolution of Life 4

Question 3.
Sketch the graphs to show directional and stabilizing selection.
Answer:
Maharashtra Board Class 12 Biology Important Questions Chapter 5 Origin and Evolution of Life 5

Long Answer Questions

Question 1.
Write about four old theories which suggested about how did life originate on the earth.
Answer:
1. Theory of special creation : Theory of special creation is the oldest theory which is based on religious beliefs. According to this theory, all the living organisms were created by supernatural power. However, since there are no scientific proofs to this theory, it is not accepted.

2. Cosmozoic theory/Theory of Panspermia : This theory says that life did not originate on the earth but it was exported from the other planets in the form of biological spores or microorganisms which were named as cosmozoa or panspermia. They may have descended to the earth from other planets. Recently, NASA has reported fossils of bacteria-like organisms on a piece of Martian rock recovered from Antarctica. Such facts may throw some light on the cosmozoic theory.

3. Theory of spontaneous generation or Abiogenesis : There was a belief that life originated from non-living material spontaneously. This theory was later disproved by Louis Pasteur.

4. Theory of biogenesis : This theory says that living organisms can originate only from pre-existing living beings. It is same as reproduction. But this theory of biogenesis was unable to explain origin of life on earth. It explains only the continuity of life.

Question 2.
Haldane described ‘Hot dilute soup’ in his theory. Describe how this soup led to formation of some important molecules.
Answer:
(1) The primitive sea containing molecules of organic substances without free oxygen was described as ‘hot dilute soup or primitive broth’ by Haldane. He proposed the theory of chemical evolution.

(2) According to this theory, the chemical evolution took place in the following steps : (a) Origin of earth and its primitive atmosphere, (b) Formation of ammonia, water and methane. These molecules dissolved in rainwater and formed the seas, (c) Then synthesis of simple organic compounds took place, followed by formation of complex organic compounds such as nucleic acids.

(3) The early molecules underwent chemical reactions such as condensation, polymerization, oxidation and reduction.

(4) The biologically important molecules such as monosaccharides, amino acids, purine, pyrimidine, fatty acids and glycerol were formed due to these reactions, utilizing the sources of energy on the primitive earth.

(5) Since oxygen was lacking, there was no degradation. Enzymes were also absent and hence there was formation of complex molecules in the hot dilute soup.

(6) This further led to the formation of pre-cells or protobiont. These aggregates were called coacervates by Oparin or microspheres by Sidney Fox. This further gave rise to first cells on the earth.

Question 3.
Explain the process of formation of eobionts.
Answer:

  1. Protobionts or coacervates were colloidal aggregations of hydrophobic proteins and lipids (lipoid bubbles).
  2. They grew in size by taking up material from surrounding aqueous medium.
  3. During their growth they became thermodynamically unstable and split into smaller units. These were called microspheres.
  4. They were proteinoids formed from colloidal hydrophilic complexes surrounded by water molecules.
  5. These bodies were like primitive cells having outer double-membrane. Across this membrane diffusion and osmosis may have occurred. They were more stable than coacervates.
  6. Coacervates and microspheres were non-living colloidal aggregations of lipids and proteinoids respectively.
  7. But they showed growth and division like living cells.
  8. These colloidal aggregations turned into first primitive living system called eobionts or protocell.

Question 4.
Describe RNA World hypothesis.
Answer:
Maharashtra Board Class 12 Biology Important Questions Chapter 5 Origin and Evolution of Life 6
(1) RNA world hypothesis is based on discovery of catalytic RNA or ribozymes. It was proposed by Carl Woese, Francis Crick and Leslie Orgel in 1960 whereas Ribozymes were discovered by Sidney Altman and Thomas Cech in 1980.

(2) According to this hypothesis, early life must have been based most probably on RNA.

(3) Factors supporting this hypothesis are:

  • RNA is found abundantly in all living cells.
  • It is structurally related to DNA.
  • Chains of RNA can evolve or undergo mutations, replicate and catalyse reactions.
  • Biomolecules like Acetyl-Co-A have a nucleotide in their molecular structure.
  • Ribosome acts as a protein assembly unit in the cell and is seen in many types of cells.
  • In ribosomes, translation process is catalysed by RNA.

(4) The primitive molecules underwent repeated replication and mutation forming varieties of RNA molecules with varying sizes and catalytic properties.

(5) They later developed their own protein coats and machinery to survive the assembly of primitive cell.

(6) From them DNA was developed which was double stranded stable structure.

(7) It further kept on evolving giving rise to rich biodiversity on earth.

Maharashtra Board Class 12 Biology Important Questions Chapter 5 Origin and Evolution of Life

Question 5.
Explain in brief Darwinism and its five main postulates.
Answer:
Darwinism means theories of natural selection and speciation as put forth by Charles Darwin. The five main postulates of his theories are as follows: Overproduction or prodigality, Struggle for existence, Organic variations, Natural selection, Origin of new species (speciation).
1. Overproduction (Prodigality of nature) : There is a natural tendency to produce more number of progeny in geometric ratio for continuing the species. E.g. Salmon fish produces about 28 lakh eggs in a single season. Single pair of elephants would produce 19,000,000 elephants. But the size of given species in a given area remains relatively constant because of fluctuations that occur seasonally.

2. Struggle for existence Due to over¬production there is struggle for existence between the members of population for limited supply of food or to overcome adverse environmental conditions or for a space or to escape from enemies, etc.

3. Organic variations : There are differences in morphology, physiology, nutrition, habit, behavioural patterns, etc., among the members of same species or members of different species. These variations act as raw material for evolution.

4. Natural selection : Some organisms possess better variations to get adapted and survive under existing environmental conditions, while some do not have. Better adapted organisms are selected by the nature while those with unfavourable variations perish. The principle by which useful variations are preserved by nature, is called ‘Natural Selection’. It is also called ‘survival of fittest’ by H. Spencer.

5. Origin of new species (speciation) : Favourable variations are transmitted from generation to generation, resulting into better adapted generations. Gradually these adaptations with few new modifications become fixed in the life cycle, forming a new species.

Question 6.
Explain modern Synthetic Theory of Evolution in brief.
Answer:
(1) Modern synthetic theory of evolution is the result of modification of Darwinism and theory of mutations by taking into consideration studies of genetics, ecology, anatomy, geography and palaeontology.

(2) Five key factors of modern synthetic theory are gene mutations, mutations in the chromosome structure and number, genetic recombinations, natural selection and reproductive isolation. All these finally contribute in the evolution of new species or process of speciation.

(3) Population or Mendelian population is the small group of ‘interbreeding populations’. For every Mendelian population there is a gene pool which is constituted by total number of genotypes in it. The genotype of an organism in a population is constant, but the gene pool constantly undergoes change due to different factors such as mutations, recombination, gene flow, genetic drift, etc.

(4) Every gene has two alleles. The proportion of a particular allele in the gene pool, to the total number of alleles at a given locus, is called gene frequency. Thus any change in the gene frequency in the gene pool affects population.

(5) The five main factors are broadly divided into three main concepts as follows:
(i) Genetic variations caused due to various aspects of mutation, recombination and migration. Such variations cause change in the gene frequency. Gene mutations or point mutation change the phenotype of the organism, leading to variation. Recombination is caused due to crossing over in which new genetic combinations are produced. Sexual reproduction due to fertilization of gametes also cause recombinations. All these lead to variations, Gene flow is movement of genes into or out of the population, either due to migrations or dispersal of gametes.

Gene flow therefore change the gene frequencies of the population. Genetic drift is a random change which occurs by pure chance. It occurs in small populations but change the gene frequency. Chromosomal aberrations are structural or morphological changes in the chromosomes causing rearrangement of the sequence of genes.

(ii) Natural selection is said to be the main driving force in evolution. It brings about evolutionary changes by selecting favourable gene combinations by differential reproduction of genes. This brings about changes in gene frequency from one generation to next generation.

(iii) Isolation means the separation of the population of a particular species into smaller units which prevents interbreeding between them. This over a long time period leads to speciation or formation of new species.

Question 7.
What are different types of chromosomal aberrations?
Answer:
Maharashtra Board Class 12 Biology Important Questions Chapter 5 Origin and Evolution of Life 7
Chromosomal aberrations:
(1) The structural, morphological change, which take place in chromosome due to rearrangement, is called chromosomal aberrations.

(2) The aberrations change the sequence of the genes. This causes variations. Chromosomal aberrations are mainly of following four types:

  1. Deletion : Loss of genes from chromosome.
  2. Duplication : Genes are repeated or doubled in number on chromosome.
  3. Inversion : A particular segment of chromosome is broken and gets reattached to the same chromosome in an inverted position due to 180° twist. There is no loss or gain of gene complement of the chromosome.
  4. Translocation : Transfer or transposition of a part of chromosome or a set of genes to a non-homologous chromosome is called translocation. It is effected naturally by the transposons present in the cell.

Question 8.
What are the different pre-zygotic isolating mechanisms?
Answer:
(1) Pre-zygotic or pre-mating isolating mechanisms do not allow individuals to mate with each other at all.

(2) By various mechanisms the two groups remain isolated. These mechanisms are of following types:
(i) Habitat isolation : Habitat isolation is the phenomenon in which members of a population living in the same region occupy different habitats. Hence the potential mates do not interbreed among themselves.

(ii) Seasonal isolation : In seasonal isolation, members of a population share the same region but attaining sexual maturity at the different times of the year. They thus remain isolated reproductively preventing interbreeding among themselves.

(iii) Ethological isolation : Ethological isolation is seen when members of two populations have different mating behaviours. This prevents interbreeding.

(iv) Mechanical isolation : Mechanical isolation is seen when the members of two populations have differences in the structure of reproductive organs. Due to such differences interbreeding is not possible.

Question 9.
What are the different post-zygotic isolating mechanisms?
Answer:

  1. In post-zygotic or post-mating isolating mechanisms, the two individuals can mate but the result of mating is not favourable.
  2. Thus the populations remain isolated without the actual genetic exchange.

Post-mating isolating mechanisms are divided into the following categories:

  1. Gamete mortality : In gamete mortality, there is death of gametes. Sperm transfer may take place but the egg is not fertilized due to gamete mortality.
  2. Zygote mortality : In zygote mortality, the zygote is formed but it fails to thrive. Though the egg is fertilized the zygote does not survive.
  3. Hybrid sterility : In this isolation, there is the formation of hybrid as the gametes or zygotes do not die but the hybrid formed is sterile. Sterile hybrid cannot contribute genetically to further generations.

Maharashtra Board Class 12 Biology Important Questions Chapter 5 Origin and Evolution of Life

Question 10.
What is Hardy-Weinberg equilibrium? Explain it in brief.
Answer:

  1. Hardy and Weinberg were two scientists who proposed a concept of genetic equilibrium popularly known as Hardy- Weinberg principle or equilibrium.
  2. This principle states that gene, allele or genotype frequencies remain the same from generation to generation unless disturbed by factors like mutation, non-random mating, genetic drift, etc.
  3. For explaining the concept of equilibrium they assumed that there are two alleles located at a single locus (A and a).
  4. Their respective frequencies are p and q.
  5. The frequency of genotype AA is p, for 2Aa is 2pq and for aa is q.
  6. The equilibrium equation is p² + 2pq + q² = 1
  7. It says that if sum total of gene frequencies is 1, then sum total of genotype frequencies is also equal to 1. When the equilibrium is disturbed then only evolution occurs.

Question 11.
Human being is said to be most evolved, intelligent living being. Yet we are not self-sufficient. Think of various aspects for which we depend on other living beings for our survival.
Answer:
Human brain is evolved and super- specialised but yet in many aspects human beings are much dependent on other natural factors. Human body is not with any protective exoskeleton, or organs of offence and defence.

Unless well dressed, he cannot cope up with severe cold temperatures as he lacks natural protective fur. He cannot run fast as the other animals can. Neither he can digest uncooked food. He has overcome all his shortcomings by using his brain power.

He has managed to take fur and feathers from other animals by killing them. He also uses other natural fibres from plants to cover his body. He has also finished fish from the oceans by over-exploitation and polluting the natural habitats of these creatures.

He takes meat from other animals by killing them and for the purpose he domesticates them to satisfy his hunger. Before technical age, animals were used as beasts of burden and as means of transport. Thus human history has shown excessive use of horses, camels, elephants, etc.

Man snatches milk from other animals like cows and buffaloes which is for their young ones. But the entire diary industry and human needs for dairy products have been taken care of by these herbivores.

Apart from all such uses, man also uses other animals for experimentations and pharmaceutical industries. In this way, man has mastered other animal kingdom due to his intelligence.

Maharashtra Board Class 12 Biology Important Questions Chapter 4 Molecular Basis of Inheritance

Balbharti Maharashtra State Board 12th Biology Important Questions Chapter 4 Molecular Basis of Inheritance Important Questions and Answers.

Maharashtra State Board 12th Biology Important Questions Chapter 4 Molecular Basis of Inheritance

Multiple Choice Questions

Question 1.
How many of the following characteristics are shown by the R-strain of Streptococcus pneumonia? Avirulent, Smooth, Pathogenic, Capsulated ………………..
(a) One
(b) TWo
(c) Three
(d) Four
Answer:
(a) One

Question 2.
Griffith obtained …………….. from the blood of the dead mice.
(a) dead S-strain bacteria
(b) live R-strain bacteria
(c) dead R-strain bacteria
(d) live S-strain bacteria
Answer:
(d) live S-strain bacteria

Maharashtra Board Class 12 Biology Important Questions Chapter 4 Molecular Basis of Inheritance

Question 3.
Oswald T. Avery, Colin M. MacLeod and Maclyn McCarty demonstrated that ………………..
(a) transformation of live S-strain bacteria into R-strain type was because of DNA of bacteria of S-strain.
(b) the transforming substance was either a protein or RNA.
(c) only DNA was able to transform harmless R-strain into virulent S-strain.
(d) when DNA isolated from S-strain bacteria, was digested with DNase, the transformation occurred.
Answer:
(c) only DNA was able to transform harmless R-strain into virulent S-strain

Question 4.
Which of the following was NOT observed in Hershey and Chase experiment?
(a) Viruses grown in the presence of radioactive sulphur, had radioactive protein but not radioactive DNA.
(b) Radioactive ‘P’ remained in suspension.
(c) Only radioactive ‘P’ was found inside the bacterial cells in the pellet.
(d) Viruses grown in the presence of radioactive phosphorus contained radioactive DNA but not radioactive proteins.
Answer:
(b) Radioactive ‘P’ remained in suspension.

Question 5.
Enzymes like ……………….. and DNA topoisomerase-I, play important role in maintaining super-coiled state in prokaryotic DNA.
(a) DNA ligase
(b) DNA gyrase
(c) RNA polymerase
(d) None of these
Answer:
(b) DNA gyrase

Question 6.
Histone octamer of nucleosome has two molecules, each of ……………….. proteins.
(a) H2A, H2B, H3 and H4
(b) H2A, H2B, H3 and H1
(c) H2A, H2B. H3A and H3B
(d) H1A, H2B, H3A and H4
Answer:
(a) H2A, H2B, H3 and H4

Question 7.
Select the CORRECT statement.
(a) Euchromatin is mainly located near centromere and telomeres.
(b) Heterochromatin replicates at faster rate than euchromatin.
(c) Heterochromatin has 2 to 3 times more DNA than in the euchromatin.
(d) Heterochromatin is lightly stained region of chromonema.
Answer:
(c) Heterochromatin has 2 to 3 times more DNA than in the euchromatin

Question 8.
A DNA molecule in which both strands have 14N is allowed to replicate in an environment containing 15N. What will be the exact number of DNA molecules that contain the 14N after three replications?
(a) One
(b) Two
(c) Four
(d) Eight
Answer:
(b) Two

Question 9.
As the base sequence present on one strand of DNA decides the base sequence of other 8 strand, this strand is considered as ………………..
(a) descending strand
(b) leading strand
(c) lagging strand
(d) complementary strand
Answer:
(d) complementary strand

Question 10.
In prokaryotes ……………….. recognizes the promoter sequence.
(a) alpha factor
(b) rho factor
(c) theta factor
(d) sigma factor
Answer:
(d) sigma factor

Question 11.
If the base sequence in DNA is 5′ AAAA 3′, then the base sequence in m-RNA is ………………..
(a) 5′ UUUU 3′
(b) 3′ UUUU 5′
(c) 5′ AAAA 3′
(d) 3′ TTTT 5′
Answer:
(c) 5′ AAAA 3′

Question 12.
During capping, methylated guanosine tri¬phosphate is added to 5′ end of ………………..
(a) m-RNA
(b) t-RNA
(c) hnRNA
(d) r-RNA
Answer:
(c) hnRNA

Question 13.
If each codon has two nucleotides, then there will be ……………….. codons, which can encode for only …………….. different types of amino acids.
(a) 16, 16
(b) 16, 20
(c) 20, 16
(d) 64, 64
Answer:
(a) 16, 16

Question 14.
What would happen if in a gene encoding a polypeptide of 50 amino acids, 25th codon (UAU) is mutated to UAA?
(a) A polypeptide of 24 amino acids is formed.
(b) Two polypeptides of 24 and 25 amino acids will be formed.
(c) A polypeptide of 49 amino acids is formed.
(d) A polypeptide of 25 amino acids is formed.
Answer:
(a) A polypeptide of 24 amino acids is formed.

Question 15.
A strand of DNA has following base sequence – 3′ AAAAGTGAATAGTGA 5′. On transcription it produces an m-RNA. Which of the following anticodon of t-RNA recognizes the third codon of this m-RNA?
(a) AAA
(b) CUG
(c) AAG
(d) CTG
Answer:
(c) AAG

Question 16.
Polynucleotide chain consisting of only CUA repeats will give polypeptide chain with only one amino acid ………………..
(a) tryptophan
(b) leucine
(c) serine
(d) methionine
Answer:
(b) leucine

Question 17.
Select the INCORRECT statement.
(a) Dr. Khorana prepared polyribo-nucleotides chains with known repeated sequences of two or three nucleotides by using synthetic DNA.
(b) M. Nirenberg and Matthaei synthesized artificial m-RNA which was a homopolymer of uracil ribonucleotides.
(c) Enzyme polynucleotide phosphorylase polymerizes RNA with defined sequences in a template-dependent manner.
(d) Evidence for triplet nature of geneticcode, was given by Crick (1961) using “frame-shift mutation”.
Answer:
(c) Enzyme polynucleotide phosphorylase polymerizes RNA with defined sequences in a template-dependent manner.

Question 18.
…………… is/are based on complementarity principle.
(a) Replication and translation
(b) Replication and transcription
(c) Translation
(d) Only replication
Answer:
(b) Replication and transcription

Question 19.
Cysteine has codons, while isoleucin has ……………….. codons.
(a) two, three
(b) three, two
(c) two, four
(d) four, two
Answer:
(a) two, three

Question 20.
Out of 64 codons, only 61 code for the 20 different amino acids. This is known as ……………….. of genetic code.
(a) non-ambiguity
(b) overlapping nature
(c) ambiguity
(d) degeneracy
Answer:
(d) degeneracy

Question 21.
Mutation that results in Sickle-cell anaemia is a ………………..
(a) deletion
(b) frame-shiftmutation
(c) point mutation
(d) insertion
Answer:
(c) point mutation

Question 22.
Initiator charged t-RNA occupies the ……………….. of ribosome first.
(a) A-site
(b) P-site
(c) E-site
(d) either A-site or P-site
Answer:
(b) P-site

Question 23.
It takes ……………….. for formation of peptide bond.
(a) 10 seconds
(b) 0.1 second
(c) less than 0.1 second
(d) 60 seconds
Answer:
(c) less than 0.1 second

Question 24.
Anticodon and codon bind by ………………..
(a) glycosidic bond
(b) hydrogen bond
(c) phosphodiester bond
(d) none of these
Answer:
(b) hydrogen bond

Question 25.
The UTRs are present at ………………..
(a) 5′-end, before start codon and at 3′-end, after stop codon of m-RNA
(b) 5′-end, before start codon and at 3′-end, after stop codon of t-RNA
(c) only at 3′-end, after stop codon of m -RNA
(d) only at 5′-end, before start codon of m-RNA
Answer:
(a) 5′-end, before start codon and at 3′-end, after stop codon of m-RNA

Question 26.
The action of structural genes is regulated by …………….. site with the help of a …………….. protein.
(a) operator, inducer
(b) operator, repressor
(c) regulator, repressor
(d) regulator, inducer
Answer:
(b) operator, repressor

Question 27.
Repressor protein is produced by the action of ………………..
(a) gene z
(b) gene y
(c) gene i
(d) gene o
Answer:
(c) gene i

Question 28.
Select the correct pair.
(a) Gene z – Transacetylase
(b) Gene y – Beta-galactocidase
(c) Gene a – Beta-galactoside permease
(d) Gene I – Repressor
Answer:
(d) gene I – repressor

Maharashtra Board Class 12 Biology Important Questions Chapter 4 Molecular Basis of Inheritance

Question 29.
Structural genomics involves ……………….. of genome.
(a) mapping
(b) sequencing
(c) analysis
(d) all of these
Answer:
(d) all of these

Question 30.
The technique of transferring DNA fragments separated on agarose gel to a synthetic nitrocellulose membrane is known as ………………..
(a) Southern blotting
(b) Autoradiography
(c) Southern hybridization
(d) None of these
Answer:
(a) Southern blotting

Question 31.
Sequence of various steps in DNA fingerprinting is ………………..
i. Southern blotting.
ii. Restriction digestion
iii. Agarose gel electrophoresis
iv. DNA isolation
v. Photography.
vi. Selection of DNA probe
vii. Hybridization
(a) iv, iii, ii, i, v, vi, vii
(b) iv. v, iii, i, vi, vii, ii
(c) iv, ii, iii, i, vi, vii, v
(d) ii, iii, iv, i, vi, vii,
Answer:
(c) iv, ii, iii, i, vi. vii, v

Match the Columns

Question 1.

Column AColumn B
(1) Frederick Griffith(a) Test tube assay
(2) Avery, McCarty and MacLeod(b) Streptococcus pneumoniae
(3) Alfred Hershey and Martha Chase(c) E. coli
(4) Meselson and Stahl(d) Bacteriophages

Answer:

Column AColumn B
(1) Frederick Griffith(b) Streptococcus pneumoniae
(2) Avery, McCarty and MacLeod(a) Test tube assay
(3) Alfred Hershey and Martha Chase(d) Bacteriophages
(4) Meselson and Stahl(c) E. coli

Classify the following to form Column B as per the category given in Column A

Question 1.
(i) UUU
(ii) CUA
(iii) UAA
(iv) AUG
(v) UAG
(vi) UGA

Column AColumn B
1. Initiator codon————–
2. Stop codons————-
3. Codon that codes for Phenyl alanin————–
4. Codon that codes for leucine————-

Answer:

Column AColumn B
1. Initiator codon(iv) AUG
2. Stop codons(iii) UAA, (v) UAG, (vi) UGA
3. Codon that codes for Phenyl alanin(i) UUU
4. Codon that codes for leucine(ii) CUA

Question 2.
(i) Severo Ochoa
(ii) F. Jacob and J. Monod
(iii) Temin and Baltimore
(iv) H. Winkler
(v) T. H. Roderick
(vi) R Kornberg

Column AColumn B
(1) Lac Operon————–
(2) Central dogma in retroviruses————-
(3) Coined the term Genome————–
(4) Coined the term Genomics————-
(5) Enzymatic synthesis of RNA————-
(6) DNA is associated with histones and non-histones————–

Answer:

Column AColumn B
(1) Lac Operon(ii) E Jacob and J. Monod
(2) Central dogma in retroviruses(iii) Temin and Baltimore
(3) Coined the term Genome(iv) H. Winkler
(4) Coined the term Genomics(v) T. H. Roderick
(5) Enzymatic synthesis of RNA(i) Severo Ochoa
(6) DNA is associated with histones and non-histones(vi) R. Kornberg

Very Short Answer Questions

Question 1.
What are the two types of bacteria used by F. Griffith and which one out of these is avirulent?
Answer:
S-type and R-type strains of Streptococcus penumoniae were used by F. Griffith and out of these R-type is avirulent.

Question 2.
Enlist the characteristics of S-strain pneumoniae.
Answer:
S-strain pneumoniae are virulent, smooth and encapsulated.

Question 3.
What is the bacteriophage?
Answer:
Bacteriophage is a virus that infects bacterium and injects its genetic material in the bacterium.

Question 4.
What is the length of DNA double helix molecule in a typical mammalian cell?
Answer:
The length of DNA double helix molecule in a typical mammalian cell is approximately 2.2 meters.

Question 5.
What is the approximate size of a typical nucleus ?
Answer:
Approximate size of a typical nucleus is 10-6 m.

Question 6.
What is size of E. coli cell?
Answer:
The size of E. coli cell size is 2-3 µm.

Question 7.
What determines the charge on protein molecules?
Answer:
A protein acquires its charge depending upon the abundance of amino acid residues with charged side chains.

Question 8.
What is nucleosome core?
Answer:
Nucleosome core is a histone octamer.

Question 9.
Where is H1 histone present?
Answer:
H1 histone binds the DNA thread where it enters and leaves the nucleosome.

Question 10.
How are solenoid fibres formed?
Answer:
Six nucleosomes get coiled and then form solenoid that looks like coiled telephone wire of 30 nm diameter (300Å).

Question 11.
How is chromatin fibre formed?
Answer:
Supercoiling of solenoid fibre forms a looped structure called chromatin fibre.

Question 12.
What is NHC?
Answer:
NHC stands for Nonhistone Chromosomal proteins.

Question 13.
List as many different enzyme activities required during DNA synthesis as you can.
Answer:
Phosphorylase, Helicase, DNA polymerase, Primase, DNA ligase, Super helix relaxing enzyme, Topoisomerase (gyrase) are different enzymes required during DNA synthesis.

Question 14.
How many replicons are present in prokaryotes and eukaryotes respectively?
Answer:
Prokaryotes have one replicon. Several replicons in tandem are present in eukaryotes.

Question 15.
What is the function of SSBP?
Answer:
During replication of DNA SSBP proteins remain attached to both the separated strands and prevent them from coiling back.

Question 16.
During which phases of cell cycle, transcription occurs in the nucleus?
Answer:
Transcription occurs in the nucleus during G1 and G2 phases of cell cycle.

Question 17.
Which strand of transcription unit gets transcribed ?
Answer:
DNA strand having 3’ → 5’ polarity acts as template strand and it gets transcribed.

Question 18.
What is a cryptogram?
Answer:
Cryptogram is a genetic code consisting of triplet codons on m-RNA that code for a specific amino acids.

Question 19.
What is meant by the polarity of genetic code?
Answer:
Genetic code is always read in 5′ → 3’ direction. This is called polarity of genetic code.

Question 20.
Which mutation can result in changes in the reading frame?
Answer:
Insertion or deletion of one or two bases changes the reading frame from the point of insertion or deletion.

Question 21.
Which mutation can result in insertion or deletion of amino acids, but reading frame remains unaltered?
Answer:
Insertion or deletion of three or multiples of three bases results in insertion or deletion of amino acids and reading frame remains unaltered from that point onwards.

Question 22.
Which molecule serves as an intermediate molecules between DNA and protein during proteins synthesis?
Answer:
RNA serves as an intermediate molecule between DNA and protein during proteins synthesis.

Maharashtra Board Class 12 Biology Important Questions Chapter 4 Molecular Basis of Inheritance

Question 23.
What is the function of a groove present between two subunits of ribosome in eukaryotes ?
Answer:
The groove present between two subunits of ribosomes in eukaryotes protects the polypeptide chain from the action of cellular enzymes and also protects m-RNA from the action of nucleases.

Question 24.
Enlist different steps of protein synthesis.
Answer:
Steps in protein synthesis are:

  1. Transcription
  2. Activation of amino acids and formation of charged t-RNAs,
  3. Synthesis of polypeptide chain:
  4. initiation
  5. elongation and
  6. termination of polypeptide chain.

Question 25.
What is translocation?
Answer:
During elongation of polypeptide chain, the ribosome moves along the m-RNA in stepwise manner from start codon to stop codon (5′ → 3′), 1 codon ahead each time t, his movement is called translocation and due to this t-RNA carrying a dipeptide at A-site of the ribosome moves to the p-site.

Question 26.
What is meant by inducible enzymes?
Answer:
Bacteria like E.coli adapt to their chemical environment by synthesizing certain enzymes depending upon the substrate present. Such adaptive enzyme is called inducible enzyme.

Question 27.
What is meant by induction and inducer?
Answer:
A set of genes are switched on when a new substrate is to be metabolized. This phenomenon is called induction and small molecule responsible for this is known as inducer.

Question 28.
What is the role of a repressor gene?
Answer:
The role of a repressor gene is to produce repressor protein. Repressor binds with operator gene and this prevents transcription of structural genes in the operon.

Question 29.
Which molecule does act as inducer molecule in lac operon?
Answer:
Allolactose acts as inducer molecule in lac operon.

Question 30.
In which condition, lac operon is switched off?
Answer:
If E.coli bacteria do not have lactose in the surrounding medium as a source of energy, lac operon is switched off.

Question 31.
What lac operon consists of?
Answer:
Lac operon consists of a regulator promoter, operator and three structural genes z, y and a.

Question 32.
Which gene acts as a regulatory gene in lac operon?
Answer:
Repressor protein is produced by the action of gene i (inhibitor). This gene acts as a regulator gene.

Question 33.
When was Human Genome Project started ? When was it completed ?
Answer:
The Human Genome Project was started in 1990 and was completed in 2003.

Question 34.
What is functional genomics?
Answer:
Functional genomics is a branch of genomics that involves the study of functions of all gene sequences and their expressions in organisms.

Question 35.
What is the advantage of sequencing of genomes of non-human organisms?
Answer:
Sequencing of genomes of non-human model organisms allows researchers to study gene functions in these organisms. Since human beings possess many genes which are like those of flies, roundworms and mice, comparative studies will lead to greater understanding of human evolution.

Question 36.
What are VNTRs?
Answer:
Variable Number of Tandem Repeats (VNTRs) are unusual sequences of 20-100 base pairs, which are repeated several times and are arranged tandency.

Question 37.
Do different organisms have the same DNA?
Answer:
Different organisms differ in their DNA sequence.

Question 38.
What is the amino acid sequence encoded by base sequence UCA, UUU, UCC, GGG, AGU of an m-RNA segment?
Answer:
The amino acid sequence: Ser-Phe – Ser – Gly- Ser

Give definitions of the following

Question 1.
Replicon
Answer:
The unit of DNA in which replication occurs is known as replicon.

Question 2.
Transcription
Answer:
Transcription is defined as the process of copying of genetic information from template strand of DNA into a complementary single stranded RNA transcript.

Question 3.
Gene
Answer:
Gene is defined as the DNA sequence coding for m-RNA/ t-RNA or r-RNA.

Question 4.
Cistron
Answer:
Cistron is defined as a segment of DNA coding for a polypeptide.

Question 5.
Monocistronic gene
Answer:
Gene is called monocistronic, when there is a single structural gene in one transcription unit.

Question 6.
Polycistronic gene
Answer:
Gene is called polycistronic, when there is a set of various structural genes in one transcription unit.

Question 7.
Interrupted or Split genes
Answer:
Interrupted or split genes are the structural genes in eukaryotes which have both exons and introns.

Question 8.
Exons
Answer:
Exons are the coding sequences or express sequences in DNA/hnRNA/ m-RNA.

Question 9.
Introns
Answer:
Introns are the non-coding sequences in DNA or hnRNA.

Question 10.
Anticodon
Answer:
Anticodon is a triplet of nucleotides present on the anticodon loop of t-RNA, which is complementary to codon on m-RNA.

Question 11.
Mutation
Answer:
Mutation is a sudden heritable change in the DNA sequence that results in the change of genotype.

Question 12.
Translation
Answer:
Translation is the process in which sequence of codons of m-RNA is decoded and accordingly amino acids are added in specific sequence to form a polypeptide on ribosomes.

Question 13.
Genomics
Genomics is the study of genomes through analysis, sequencing and mapping of genes along with the study of their functions.

Question 14.
Repressors
Answer:
Repressors are proteins which are able to bind the operator region of operon and prevent the RNA polymerase from transcribing the operon.

Name the following

Question 1.
Enzyme that cleaves DNA.
Answer:
DNase

Question 2.
Enzyme that cleaves proteins.
Answer:
Protease

Question 3.
Enzyme involved in activation of nucleotides.
Answer:
Phosphorylase

Question 4.
Enzyme involved in unwinding of DNA.
Answer:
Helicase

Question 5.
Enzyme involved in synthesis of DNA.
Answer:
DNA polymerase

Question 6.
Enzyme involved in joining of Okazaki fragments.
Answer:
DNA ligase

Question 7.
Enzyme involved in synthesis of RNA primer.
Answer:
Primase

Question 8.
Enzyme involved in removal of RNA primer.
Answer:
DNA polymerase

Question 9.
Enzyme involved in replacement of gaps in prokaryotes.
Answer:
DNA polymerase – I

Maharashtra Board Class 12 Biology Important Questions Chapter 4 Molecular Basis of Inheritance

Question 10.
Enzyme involved in replacement of gaps in eukaryotes.
Answer:
DNA polymerase α

Question 11.
Enzyme involved in formation of double helix in daughter DNA molecules.
Answer:
Topoisomerase

Question 12.
Enzyme involved in releasing strain created by unwinding of DNA.
Answer:
Super helix relaxing enzyme

Question 13.
Enzyme involved in synthesis of hnRNA, m-RNA.
Answer:
RNA polymerase – II

Question 14.
Enzyme involved in synthesis of t-RNA, snRNA.
Answer:
RNA polymerase – III

Question 15.
Enzyme involved in synthesis of r-RNA
Answer:
RNA polymerase – I

Question 16.
Enzyme involved in polymerizing RNA in template independent manner.
Answer:
Polynucleotide phosphorylase

Question 17.
Enzyme involved in peptide bond synthesis.
Answer:
Ribozyme

Question 18.
Enzyme involved in Cutting DNA at specific sites.
Answer:
Restriction endonuclease

Question 19.
Name the initiator codon of protein synthesis.
Answer:
AUG is the initiator codon of protein synthesis.

Question 20.
Name three binding sites of ribosome.
Answer:
Three binding sites for t-RNA on ribosomes are P-site (peptidy t-RNA-site), A-site (aminoacyl – t-RNA-site) and E-site (exit site).

Question 21.
Name the different structural genes in sequence of lac operon.
Answer:
There are 3 structural genes in the sequence lac-Z, lac-Y and lac-A.

Question 22.
Name the organisms whose genomes have been sequenced?
Answer:
The genomes of several organisms such as bacteria e.g. E.coli, Caenorhabditis elegans (a free living non-pathogenic nematode), Saccharomyces cerevisiae (yeast), Drosophila (fruit fly), plants (rice and Arabidopsis), Mus musculus (mouse), etc. have been sequenced.

Give Significance/Functions of the following

Question 1.
DNA.
Answer:

  1. DNA regulates and controls all the cellular activities.
  2. It replicates and gets distributed equally to the daughter cells when the cell divides.
  3. It is a carrier of genetic information.
  4. Heterocatalytic function : DNA directs the synthesis of chemical molecules other than itself. E.g. Synthesis of RNA (transcription), synthesis of protein (Translation), etc.
  5. Autocatalytic function : DNA directs the synthesis of DNA itself. E.g. Replication.
  6. DNA is a master molecule of a cell that initiates, guides, regulates and controls the process of protein synthesis.

Question 2.
Proteins.
Answer:
Proteins serve as structural components, enzymes and hormones.

Distinguish between the following.

Question 1.
Euchromatin and Heterochromatin.
Answer:

EuchromatinHeterochromatin.
1. Euchromatin is loosely packed region of the chromatain.1. Heterochromatin is densely packed region of the chromatin.
2. Euchromatin stains lightly.2. Heterochromatin stains darkly.
3. Euchromatin is transcriptionally active region of the chromatin.3. Heterochromatin is transcriptionally inactive region of the chromatin.

Question 2.
DNA in prokaryotic and eukaryotic cells.
Answer:

DNA in prokaryotesDNA in eukaryotic
1. It is present in the cytoplasm.1. It is present in the nucleus.
2. It is not associated with histones.2. It is associated with histones.
3. It is circular.3. It is linear.
4. Genes do not contain introns.4. Genes contain introns along with exons.
5. Genes are polycostronic.5. Genes are monocistronic.

Question 3.
DNA and RNA
Answer:

DNARNA
1. DNA is deoxyribonucleic acid.1. RNA is ribonucleic acid.
2. DNA is double stranded, helical molecule.2. RNA is single stranded molecule.
3. In DNA, there is deoxyribose sugar.3. In RNA, there is ribose sugar.
4. The pyrimidine nitrogen bases are cytosine and thymine.4. The pyrimidine nitrogen bases are cytosine and uracil.
5. DNA is the genetic material in all types of organisms.5. RNA is genetic material in few viruses only.
6. In eukaryotic cells, DNA is present in nucleus.6. In eukaryotic cells, RNA is present in nucleus as well as cytoplasm.
7. The number of purine : pyrimidine ratio is always 1 : 1 in DNA molecule.7. The number of purine : pyrimidine ratio may not be 1 : 1 in RNA molecule.
8. DNA sends the codon for the synthesis of proteins, but otherwise it does not participate in the protein synthesis.8. RNA takes part in the protein synthesis through transcription and translation.

Question 4.
m-RNA, t-RNA and r-RNA.
Answer:

m-RNAt-RNAr-RNA.
1. m-RNA is a simple molecule which shows linear structure without any folds.1. t-RNA is a single stranded molecule. There is regular pattern of folding shown by this molecule.1. r-RNA is a single stranded RNA that is variously folded upon itself. In the folded regions it shows complementary base pairing.
2. It performs the function of transcription during protein synthesis.2. It performs the function of transferring the amino acids during translation.2. This RNA remains associated with the ribosomes permanently. It gives the binding site for m-RNA during the process of protein synthesis. It also orients the m-RNA molecule so as to read the message on codons properly.
3. Of the total cellular RNA, m-RNA forms 3-5%.3. Of the total cellular RNA, t-RNA forms about 10-20%.3. Of the total cellular RNA, r-RNA forms 80%
4. It has molecular weight of about 5,00,000.4. t-RNA is the smallest RNA having only 73-93 nucleotides and has molecular weight of about 23,000 to 30,000 daltons.4. Molecular weight is about 40,000 to 100,000 daltons.

Give Reasons

Question 1.
Nuclein was called as nucleic acid.
Answer:
Nuclein had acidic properties and it was isolated from nucleus. Hence, it was called as nucleic acid.

Question 2.
Initially proteins (and not DNA) were considered as genetic material.
Answer:

  1. Proteins are large, complex molecules and store information required to govern cell metabolism. Hence it was assumed that variations found in species were caused by proteins.
  2. On the other hand, DNA was considered as a small, simple molecule whose composition does not vary much among species.
  3. Variations in the DNA molecules are different than the variation in shape, electrical charge and function shown by proteins.
  4. Hence, initially proteins (and not DNA) were considered as genetic material.

Question 3.
On injecting a mixture of heat-killed S-bacteria and live R bacteria, the mice died.
Answer:

  1. Griffith obtained live S-strain bacteria from the blood of the dead mice.
  2. In a mixture of live R-bacteria and heat killed S-bacteria, live R-strain bacteria picked up something (transforming principle) from the heat-killed S bacterium and got changed into S-type.
  3. Transforming principle allowed R-type bacteria to synthesize capsule and thus they became virulent.
  4. Hence, on injecting a mixture of heat-killed S bacteria and live R bacteria, the mice died.

Question 4.
Viruses obtained by infecting bacteria having radioactive phosphorus contained radioactive DNA (labelled DNA), but not radioactive proteins.
Answer:
Viruses obtained by infecting bacteria having radioactive phosphorus, contained radioactive DNA (labelled DNA). but not radioactive proteins because DNA contains phosphorus (labelled DNA) but proteins do not.

Question 5.
Viruses obtained by infecting bacteria having radioactive sulphur contained radioactive protein but not radioactive DNA.
Answer:
Viruses obtained by infecting bacteria having radioactive sulphur contained radioactive protein but not radioactive DNA because DNA does not contain sulphur and proteins contain sulphur.

Question 6.
In bacteria, m-RNA does not require any processing.
Answer:
In bacteria, m-RNA does not require any processing because it has no introns and it is synthesized in cytoplasm.

Question 7.
Eukaryotic DNA is condensed and supercoiled.
Answer:

  1. In a typical mammalian cell, length of DNA double helix is approximately 2.2 metres.
  2. The size of typical nucleus is approximately 10-6 m
  3. Such a long DNA molecule has to be fitted in small nuclear space.
  4. Therefore, DNA is highly condensed, coiled and supercoiled so that it can be accommodated in the nucleus.

Maharashtra Board Class 12 Biology Important Questions Chapter 4 Molecular Basis of Inheritance

Question 8.
During translation, complementarity principle is not applicable.
Answer:
During translation, complementarity principle is not applicable as, genetic information is transferred from a polymer of nucleotides to a polymer of amino acids.

Question 9.
Protein synthesis is the most important and essential activity in the living cells.
Answer:

  1. Proteins play a significant role in the metabolism of living cells.
  2. The actual phenotypic expression of living cells is dependent on the biochemical reactions.
  3. Each biochemical reaction needs a specific enzyme for its initiation and completion. All the enzymes are proteins.
  4. In a cell there are many structural proteins too. Thousands of structural and catalytic proteins are constantly required within the cell at all times.
  5. Many functional proteins like hormones are also important for metabolism. Thus, for the synthesis of all such proteins, protein synthesis has become the most important and essential activity of the living cell.

Question 10.
Only 20 amino acids are considered as standard.
Answer:

  1. It was believed that there are total 20 amino acids in the living world. But 21st amino acid called selenocysteine was discovered later.
  2. This amino acid is coded by UGA which is usually a termination codon.
  3. In both prokaryotic and eukaryotic cells polypeptide chains contain 100-300 amino acids and they are formed by specific arrangement of 21 amino acids.
  4. But formation of selenocysteine requires the availability of element selenium in the cells.
  5. Therefore, only 20 amino acids are considered as standard.

Write Short Notes on the following

Question 1.
Friedrich Miescher’s nuclein
Answer:

  1. Nuclein is an acidic substance, having high phosphorus content and it was isolated by Friedrich Miescher in 1869, from the nuclei of pus cells.
  2. As Nuclein had acidic properties and it was isolated from nucleus, it was called as nucleic acid.
  3. E Miescher started working with white blood cells (the major component of pus). He used a salt solution to wash the pus off the bandages. He lysed the cells by adding a weak alkaline solution and isolated nucleic acid from nuclei that precipitated out of the solution.
  4. By the early 1900s, it was known that Miescher’s nuclein was a mixture of proteins and nucleic acids (DNA and RNA).

Question 2.
Packaging in Prokaryotes
Answer:
Maharashtra Board Class 12 Biology Important Questions Chapter 4 Molecular Basis of Inheritance 1

  1. Size of cell in E. coli size is 2-3µ.
  2. The nucleoid is small, circular, highly folded, naked DNA (1100 µm long in perimeter and contains about 4.6 million base pairs).
  3. When the negatively charged DNA becomes circular, the size reduces to 350 µm in diameter.
  4. Folding/looping (40-50 domains (loops) further reduce it to 30 µm in diameter.
  5. RNA connectors assist in loop formation.
  6. Further coiling and super coiling of each domain reduces the size to 2 µm in diameter.
  7. This coiling (packaging) is assisted by positively charged HU (Histone like DNA binding proteins) proteins and enzymes like DNA gyrase and DNA topoisomerase-I, which maintain supercoiled state.

Question 3.
Experimental confirmation of semiconservative replication of DNA.
Answer:
Maharashtra Board Class 12 Biology Important Questions Chapter 4 Molecular Basis of Inheritance 2
(1) Matthew Meselson and Franklin Stahl (1958) used equilibrium – density – gradient – centrifugation technique to experimentally prove semiconservative DNA replication.

(2) They cultured bacteria E.coli in the medium containing 14N (light nitrogen). They obtained equilibrium density gradient band by using 6M CsCl2. The position of this band is recorded.

(3) E. coli cells were then transferred to 15N medium (heavy isotopic nitrogen) and allowed to replicate for several generations. At equilibrium point density gradient band was obtained, by using 6M CsCl2. The position of this band is recorded.

(4) The heavy DNA (15N) molecule can be distinguished from normal DNA by centrifugation in a 6M Cesium chloride (CsCl2) density gradient. At the equilibrium point 15N DNA will form a band. In this both the strands of DNA are labelled with 15N.

(5) Such E. coli cells were then transferred to another medium containing 14N i.e. normal (light) nitrogen. After first generation, the density gradient band for 14N 15N was obtained and its position was recorded. After second generation, two density gradient bands were obtained – one at 14N 15N position and other at 14N position.

(6) The position of bands after two generations clearly proved that DNA replication is semiconservative.

Question 4.
Central Dogma of molecular biology.
Answer:
(1) Central dogma of molecular biology was postulated by EH.C. Crick in 1958.
Maharashtra Board Class 12 Biology Important Questions Chapter 4 Molecular Basis of Inheritance 3

(2) DNA gets transcribe to form m-RNA, m-RNA acts as a messenger and gets translated to form a polypeptide chain (protein) having specific amino acid sequence.

(3) This unidirectional flow of information from DNA to RNA and from RNA to proteins is referred as central dogma of molecular biology.

(4) Temin (1970) and Baltimore (1970) : Central dogma in retroviruses.
Maharashtra Board Class 12 Biology Important Questions Chapter 4 Molecular Basis of Inheritance 4

Question 5.
Processing of hnRNA.
Answer:
Maharashtra Board Class 12 Biology Important Questions Chapter 4 Molecular Basis of Inheritance 5

  1. Eukaryotes have split genes.
  2. Primary transcript or hnRNA is non-functional and contains both exons and introns.
  3. Processing of hnRNA results in functional m-RNA.
  4. The fully processed hnRNA is called m-RNA.
  5. m-RNA comes out of the nucleus through nuclear pore for getting translated.

hnRNA undergoes capping, tailing and splicing.

  1. Capping : Methylated guanosine tri¬phosphate is added to 5’ end of hnRNA.
  2. Tailing : Polyadenylation take place at 3′ end.
  3. Splicing : It is removal of introns.
  4. DNA ligase joins exons in a definite sequence (order).

Question 6.
Cracking of genetic code.
Answer:

  1. M. Nirenberg and Matthaei synthesized artificial poly-U m-RNA.
  2. Using this synthetic poly-U m-RNA and cell free system of protein synthesis, a small polypeptide consisting of only amino acid phenylalanine was obtained.
  3. It suggested that UUU codes for phenylalanine.
  4. Dr. Har Gobind Khorana devised a technique for synthesis of artificial m-RNA with repeated sequences of known nucleotides.
  5. He synthesized artificial RNA consisting of known repeated sequences of two or three nucleotides. E.g. CUC, UCU, CUC, UCU, by using synthetic DNA.
  6. This resulted in formation of polypeptide chain consisting of alternate amino acids leucine and serine.
  7. Synthetic RNA consisting of CUA, CUA, CUA, CUA repeats gave polypeptide chain with only one amino acid – leucine.
  8. Severo Ochoa established that the enzyme (polynucleotide phosphorylase) also helps in polymerizing RNA of defined sequences in a template-independent manner (i.e. enzymatic synthesis of RNA).
  9. Thus all the 64 codons in the dictionary of genetic code were deciphered.

Question 7.
Types of mutations.
Answer:

  1. Chromosomal mutations : Loss (deletion) or gain (insertion/duplication) of a segment of DNA results in alteration in the chromosome.
  2. Point mutations : They involve change in a single base pair of DNA. E.g. Mutation that results in Sickle-cell anaemia.
  3. Deletion or insertion of base pairs of DNA : It causes frame-shift mutations or deletion mutation.
  4. Insertion or deletion of one or two bases changes the reading frame from the point of insertion or deletion.
  5. Insertion or deletion of three or multiples of three bases (insert or delete) results in insertion or deletion of amino acids and reading frame remains unaltered from that point onwards.

Question 8.
Transfer-RNA.
Answer:
Maharashtra Board Class 12 Biology Important Questions Chapter 4 Molecular Basis of Inheritance 6
(1) As t-RNA can read the codon and also can bind with the amino acid, t-RNA is considered as an adapter molecule.

(2) Clover leaf structure (2 dimensional) of t-RNA:

  • t-RNA has four arms – DHU arm (has amino acyl binding loop), middle arm (has anticodon loop), Tif/C arm (has ribosome binding loop) and variable arm.
  • It has G nucleotide at 5’ end.
  • Amino acid acceptor end (3’ end) having unpaired CCA bases (i.e. amino acid binding site).

(3) For every amino acid, there is specific t-RNA.
(4) Initiator t-RNA is specific for methionine.
(5) There are no t-RNAs for stop codons.
(6) In the actual structure, the t-RNA molecule looks like inverted L (3 dimensional : structure)

Question 9.
UTRs.
Answer:

  1. m-RNA has some additional sequences that are not translated. These sequences are referred as untranslated regions (UTR).
  2. The UTRs are present at both 5′-end (before start codon) and at 3′-end (after stop codon).
  3. They are required for efficient translation process.

Short Answer Questions

Question 1.
Why are Okazaki fragments formed on lagging strand only?
Answer:

  1. The lagging template is the template strand with free 5’ end.
  2. The replication always starts at C-3 end of template strand and proceeds towards C-5 end.
  3. Both the strands of the parental DNA are antiparallel and new strands are always formed in 5′ → 3′ direction, i.e. DNA polymerase synthesizes new strand in only one direction i.e. 5′ → 3′ direction.
  4. Hence, the lagging templates becomes available for replication only discontinuously in small patches.
  5. The new lagging strand develops discontinuously away from the replicating fork in the form of small Okazaki fragments.
  6. Hence, Okazaki fragments formed on lagging strand only.

Question 2.
Why t-RNA is called as adapter molecule?
Answer:
t-RNA can read the codon on m-RNA. It also can bind with the amino acid at 3’ end and transport it to m-RNA-ribosome complex during translation. It can bind with specific codon which is complementary to its anticodon. So t-RNA is considered as an adapter molecule.

Question 3.
Why DNA replication is called semiconservative replication?
Answer:

  1. In each of the two daughter DNA molecules thus formed, one strand is parental and the other one is newly synthesized.
  2. 50% is contributed by mother DNA.
  3. Hence, DNA replication is described as semiconservative replication.

Maharashtra Board Class 12 Biology Important Questions Chapter 4 Molecular Basis of Inheritance

Question 4.
What are the functions of three types of RNAs in bacteria? Which enzyme is involved in transcription of DNA to form RNAs in bacteria? What is its function?
Answer:

  1. In bacteria, m-RNA provides the encoded message for protein synthesis; t-RNA brings specific amino acid to the site of translation; r-RNA plays role in providing binding site to m-RNA and t-RNA.
  2. There is single DNA dependent-RNA polymerase that catalyses transcription of all 3 types of RNA in bacteria.
  3. RNA polymerase binds to promoter and initiates transcription (initiation) and synthesizes RNA.

Question 5.
Explain why it was suggested that codon is a sequence of three consecutive nucleotides on m-RNA.
Answer:

  1. If each codon has only one nucleotide, then there will be 41 = 4 codons, which can encode for only four different types of amino acids.
  2. If each codon has two nucleotides, then there will be 4² = 16 codons, which can encode for only 16 different types of amino acids.
  3. If each codon has three nucleotides, then there will be 4³ = 64 codons, which are sufficient to specify 20 different types of amino acids.

Question 6.
Explain Genetic code is Non-overlapping.
Answer:

  1. Each single base is a part of only one codon.
  2. Adjacent codons do not overlap.
  3. If it had been overlapping type, with 6 bases, there would be 4 amino acid molecules in a chain.
  4. Experimental evidence favours non-overlapping nature of genetic code.

Question 7.
How degeneracy of the code is explained by Wobble hypothesis?
Answer:

  1. In 1966, Crick proposed Wobble hypothesis.
  2. According to this hypothesis, in codon- anticodon base pairing, the third base may not be complementary.
  3. The third base of codon is called wobble base and this position is called wobble position.
  4. This results in economy of t-RNA as anticodon of a t-RNA may bind with codon even when only first two bases are complementary.
  5. For example, GUU, GUC, GUA and GUG codons code for amino acid Valine.
  6. Degeneracy of genetic code means many codons can code for same amino acid.
  7. Thus, the degeneracy of genetic code gets explained by Wobble hypothesis.

Question 8.
a. What is meant by universal genetic code? Give example.
b. Why genetic code is called Non- ambiguous?
Answer:
a. Universal genetic code means that the specific codon codes for same amino acid in all living organisms, e.g. codon AUG always specifies amino acid methionine.

b. Genetic code is called non-ambiguous because, each codon specifies a particular amino acid.

Question 9.
Give examples of termination codons. Why are they known as termination codons?
Answer:

  1. UAA, UAG and UGA are known as termination codons.
  2. They do not code for any amino acid.
  3. They terminate or stop the process of elongation of polypeptide chain.
  4. Hence, they are known as termination codons.

Question 10.
How many amino acids are required for protein synthesis? From where are they obtained?
Answer:

  1. About 20 different types of amino acids are required for protein synthesis.
  2. They are available in the cytoplasm.

Question 11.
How DNA regulates protein synthesis?
Answer:

  1. DNA regulates protein synthesis by coding for the specific sequence of amino acids in a protein.
  2. This control is possible through transcription of m-RNA.
  3. Genetic code is specific for particular amino acid.

Question 12.
What is the role of ribosomes in protein synthesis?
Answer:

  1. Ribosomes serve as site for protein synthesis.
  2. A ribosome has one binding site for m-RNA and 3 binding sites for t-RNA. They are P-site (peptidylt-RNA-site), A-site (aminoacyl t-RNA-site) and E-site (exit site).
  3. In Eukaryotes, a groove which is present between two subunits of ribosomes, protects the polypeptide chain from the action of cellular enzymes and also protects m-RNA from the action of nucleases.

Question 13.
Give examples of coordinated regulation or expression, of several sets of genes.
Answer:
Examples of coordinated regulation or expression of several sets of gene are:

  1. If E.colt bacteria do not have lactose in the surrounding medium as a source of energy, then structural genes in Lac operon do not get transcribed and enzyme like β -galactosidase is not synthesized.
  2. The development and differentiation of embryo into an adult organism.

Question 14.
Explain with example what is meant by positive control of gene regulation.
Answer:

  1. A set of genes will be switched on when a substrate is to be metabolized.
  2. This phenomenon is called induction and small molecule responsible for this, is known as inducer.
  3. It is positive control of gene regulation.
  4. For example, lactose acts as an inducer in Lac operon.

Question 15.
If operator gene is deleted due to mutation, how will E.coli metabolise lactose?
Answer:
If operator gene is deleted due to mutation, lac operon cannot be regulated. It will get transcribed continuously and enzymes required for lactose metabolism will get synthesized continuously.

Question 16.
Explain in brief the process of initiation during protein synthesis.
Answer:
Initiation of synthesis of polypeptide chain takes place as follows:

  1. Small subunit of ribosome binds to the m-RNA at 5’ end.
  2. Start codon is positioned properly at P-site.
  3. Initiator t-RNA, (carrying methionine in eukaryotes or formyl methionine in prokaryotes) binds with initiation codon (AUG) of m-RNA by it’s anti-codon (UAC). Codon-anti-codon pairing involves formation of hydrogen bonds.
  4. The large subunit and smaller subunit of ribosome bind in the presence of Mg++.
  5. Thus, initiator charged t-RNA occupies the P-site and A-site is vacant for next charged t-RNA.

Question 17.
What are the application of genomics?
Answer:
Applications of genomics are as follows:

  1. Structural and functional genomics is used in the improvement of crop plant, human health and livestock.
  2. The knowledge and understanding acquired by genomics research can be applied in medicine, biotechnology and social sciences.
  3. It helps in the treatment of genetic disorders through gene therapy.
  4. It helps in the development of transgenic crops having desirable characters.
  5. Genetic markers have applications in forensic analysis.
  6. Genomics can lead to introduction of new gene in microbes to produce enzymes, therapeutic proteins and biofuels.

Question 18.
Why is HGP important?
Answer:

  1. HGP is associated with rapid development of Bioinformatics.
  2. Knowledge gained about the functions of genes and proteins has a major impact in the fields like Medicine. Biotechnology and the Life sciences.
  3. It has helped in identifying the genes that are associated with genetic characteristics.
  4. The genetic basis of many hereditary diseases can be understood.
  5. It has increased the understanding of gene structure and function in other species. As human beings have many of the genes same as those of flies, roundworms and mice, such studies will enhance understanding of human evolution.

Chart based / Table based Questions

Question 1.
Complete the following table:

OrganismDiploid chromosome number
Mouse————-
Fruitfly————
Roundworm————-
Yeast————–

Answer:

OrganismDiploid chromosome number
Mouse40
Fruitfly8
Roundworm12
Yeast32

Diagram Based Questions

Question 1.
a. Identify A and B in the following diagram.
b. Name the scientist who conducted this experiment.
Maharashtra Board Class 12 Biology Important Questions Chapter 4 Molecular Basis of Inheritance 7
Answer:
a. A : Smooth strain (III-S)
B : Rough Strain and Heat-killed Smooth Strain

b. F. Griffith conducted the experiment shown in the diagram.

Question 2.
a. Identify A and B in the given diagram.
b. What is the conclusion of the given experiment?
Maharashtra Board Class 12 Biology Important Questions Chapter 4 Molecular Basis of Inheritance 8
Answer:
a. A : Rough, nonvirulent R-strain B : Heat-killed virulent S-strain
b. When DNA of heat-killed S-strain bacteria is treated with DNase, mouse remains alive as transformation does not take place. This proves that DNA is the genetic material.

Question 3.
Identify A, B, C and name the scientists who carried out the experiment given in the diagram.
Maharashtra Board Class 12 Biology Important Questions Chapter 4 Molecular Basis of Inheritance 9
Answer:
Answer: A : Infection B : Blending C : Centrifugation Scientists who carried out experiment are Alfred Hershey and Martha Chase.

Question 4.
(1) Identify A, B and C.
(2) What are their sizes?
Maharashtra Board Class 12 Biology Important Questions Chapter 4 Molecular Basis of Inheritance 10
Answer:
(1) A : Circular, unfolded chromosome
b : Folded chromosome (40 to 50 loops)
c : Supercoiled, folded chromosome

(2) (A) 350 µ (B) 30 µ (C) 2 µ

Question 5.
Draw a labelled diagram of Nucleosome with H1 histone.
Answer:
Maharashtra Board Class 12 Biology Important Questions Chapter 4 Molecular Basis of Inheritance 11

Question 6.
Identify A and B in the given diagram.
Maharashtra Board Class 12 Biology Important Questions Chapter 4 Molecular Basis of Inheritance 12
Answer:
A : Nucleosome
B : Linker DNA

Question 7.
(a) Identify A in the given diagram.
(b) What is the dimension denoted in ‘B’
Maharashtra Board Class 12 Biology Important Questions Chapter 4 Molecular Basis of Inheritance 13
Answer:
A : Solenoid B : 300 Å

Question 8.
Sketch and label process of formation of beads on strings/fibres from chain of nucleosomes.
Answer:
Maharashtra Board Class 12 Biology Important Questions Chapter 4 Molecular Basis of Inheritance 14

Question 9.
a. Identify A. Which enzyme joined them?
b. Identify B. What is its function ?
Maharashtra Board Class 12 Biology Important Questions Chapter 4 Molecular Basis of Inheritance 15
Answer:
a. A is Okazaki fragment. These fragments are joined by DNA ligase enzyme
b. B is RNA primer. It provides 3’ OH to which nucleotide gets attached by ester bond.

Question 10.
Draw a labelled diagram showing semi-conservative replication of DNA.
Answer:
Maharashtra Board Class 12 Biology Important Questions Chapter 4 Molecular Basis of Inheritance 16

Maharashtra Board Class 12 Biology Important Questions Chapter 4 Molecular Basis of Inheritance

Question 11.
Sketch and label, Meselson and Stahls experiment.
Answer:
Maharashtra Board Class 12 Biology Important Questions Chapter 4 Molecular Basis of Inheritance 17

Question 12.
Identify A, B and C in the following diagram.
Maharashtra Board Class 12 Biology Important Questions Chapter 4 Molecular Basis of Inheritance 18
Answer:
A : Transcription
B : Translation and
C : Reverse Transcription

Question 13.
a. Draw a labelled diagram of transcription unit.
b. What is the sequence of m-RNA and coding strand if sequence of template strand of DNA is
Maharashtra Board Class 12 Biology Important Questions Chapter 4 Molecular Basis of Inheritance 19
Answer:
a. Maharashtra Board Class 12 Biology Important Questions Chapter 4 Molecular Basis of Inheritance 20
b. Maharashtra Board Class 12 Biology Important Questions Chapter 4 Molecular Basis of Inheritance 21

Question 14.
a. The following diagram shows processing of ………….
b. What is capping?
c. Identify A, B and C.
Maharashtra Board Class 12 Biology Important Questions Chapter 4 Molecular Basis of Inheritance 22
Answer:
a. hnRNA
b. Capping is addition of methylated guanosine tri-phosphate at 5′ end of hnRNA.
c. A : Exon and B : Intron C : m-RNA

Question 15.
Draw a labelled diagram of t-RNA carrying Glutamic acid.
Answer:
Maharashtra Board Class 12 Biology Important Questions Chapter 4 Molecular Basis of Inheritance 23

Question 16.
Observe the diagrams (a), (b) and (c)

  1. Which step of protein synthesis is shown in the following diagrams?
  2. During initiation, which subunit of ribosome binds with m-RNA?
  3. What are the three binding sites for t-RNA on ribosomes?
  4. On which site of ribosome second and subsequent t-RNA arrives?
  5. Which link is binding amino acids in diagram (b)?
  6. Which chain is being released from ribosome in diagram (c) ?
    Maharashtra Board Class 12 Biology Important Questions Chapter 4 Molecular Basis of Inheritance 24

Answer:

  1. Translation
  2. 30S or 40S
  3. P site, A site and E site
  4. A site
  5. Peptide link
  6. Polypeptide chain

Question 17.
Draw a labelled diagram – Working of lac operon.
Answer:
Maharashtra Board Class 12 Biology Important Questions Chapter 4 Molecular Basis of Inheritance 25

Question 18.
What is the process shown in the following diagram? Mention all the steps given in the diagram in a proper sequence.
Maharashtra Board Class 12 Biology Important Questions Chapter 4 Molecular Basis of Inheritance 26
Answer:
The process shown in the above diagram is DNA fingerprinting. The sequence of steps in the process are:

  1. Isolation of DNA
  2. Restriction digestion
  3. Gel electrophoresis
  4. Southern blotting
  5. Selection of DNA probe
  6. Hybridization
  7. Photography,

Long Answer Questions

Question 1.
Describe Griffith’s transformation experiment.
OR
In the light of Griffith’s experiment, explain the action of two strains of streptococcus pneumoniae and give his conclusion.
Answer:
Maharashtra Board Class 12 Biology Important Questions Chapter 4 Molecular Basis of Inheritance 27
(1) In 1928, Frederick Griffith, carried out experiments with bacterium Streptococcus pneumoniae (which causes pneumonia in humans and other mammals).

(2) Griffith used two strains of Streptococcus pneumonia:

  • S-type (Virulent, smooth, pathogenic and encapsulated).
  • R-type (Non-virulent, rough, non- pathogenic and non-capsulated).

(3) Experiments carried out by E Griffith:

  • Mice were injected with R-strain bacteria and they survived (no pneumonia).
  • Mice injected with S-strain bacteria developed pneumonia and died.
  • When heat-killed S-strain bacteria were injected in mice, the mice survived.
  • On injecting a mixture of heat-killed S-bacteria and live R bacteria, the mice died.

(4) Griffith obtained live S-strain bacteria from the blood of the dead mice.

(5) He concluded that the live R-strain bacteria must have picked up something (which he called transforming principle) from the heat killed S bacterium, and got changed into S-type. Transforming principle allowed R-type to synthesize capsule and it became virulent.

(6) Thus, F. Griffith first demonstrated genetic transformations.

Question 2.
Describe Avery, McCarty and MacLeod’s experiments.
Answer:
Maharashtra Board Class 12 Biology Important Questions Chapter 4 Molecular Basis of Inheritance 28

  • In 1944, Oswald T. Avery, Colin M. MacLeod and Maclyn McCarty proved that the DNA is a genetic material (transforming principle).
  • They purified DNA, RNA, proteins (enzymes) and other materials from heat killed S-strain cells and mixed them with cells of R-strain bacteria separately to confirm which one could transform living R cells into S cells.
  • Only DNA was able to transform harmless R-strain into virulent S-strain.
  • They also demonstrated that proteases, RNases did not affect transformation. Thus it was proved that the transforming substance was neither a protein nor-RNA.
  • When DNA was digested with DNase, there was no transformation.
  • These experiments proved that the transformation of Live R-strain bacteria into S-strain type was because of DNA of bacteria of S-strain.
  • Thus, they proved that the DNA is transforming principle.

Question 3.
Explain how Hershey – Chase experimentally proved that DNA is the genetic material.
Answer:
Maharashtra Board Class 12 Biology Important Questions Chapter 4 Molecular Basis of Inheritance 29
(1) Hershey and Chase worked with bacteriophages (viruses that infect bacteria and which are composed of DNA and protein coat).

(2) They cultured E. coli bacteria in medium containing radioactive phosphorus 32p. By infecting these bacteria with bacteriophages, Hershey and Chase could develop bacteriophages having DNA labelled with 32p. as DNA contains phosphorus and proteins do not.

(3) They also cultured E. coli bacteria in medium containing radioactive sulphur 35s. By infecting these bacteria with bacteriophages, they developed
bacteriophages whose protein coat was labelled with 35s, as proteins contain sulphur and DNA does not.
[Note : Viruses cannot be cultivated in medium.]

(4) Experiment involved three steps.

  • Infection : Both the types of radioactive phages were allowed to infect E.coli bacteria grown on the medium containing normal ‘P‘ and ‘S’.
  • Blending : Then bacterial cultures were agitated in blender to break contact between bacteria and parts of viruses that did not enter bacterial cells.
  • Centrifugation : It was done to separate bacterial cells as a pellet. Parts of viruses which did not enter bacteria remained in the suspension.

(5) Observation:

  • Radioactive ‘S’ remained in suspension.
  • Only radioactive ‘P’ was found inside the bacterial cell in the pellet.

(6) Thus it was proved that DNA is the genetic material which enters bacterial cell and not protein.

Maharashtra Board Class 12 Biology Important Questions Chapter 4 Molecular Basis of Inheritance

Question 4.
Explain the formation of beads on string, solenoid fibre, chromatin fibre and chromosome.
Answer:
Maharashtra Board Class 12 Biology Important Questions Chapter 4 Molecular Basis of Inheritance 30

  • Beads on string (11 nm in diameter) : Under an electron microscope, nucleosomes in thread like chromatin look like ’beads-on- string’.
  • Solenoid fibre (30 nm in diameter) : Six such nucleosomes get coiled and then form solenoid that looks like coiled telephone wire.
  • Chromatin fibre (200 nm in diameter) : Further supercoiling tends to form a looped structure called chromatin fibre.
  • Chromosome (1400 nm in diameter) : Chromatin fibre further coils and condenses at metaphase stage to form the chromosomes. Each chromatid is 700 nm in diameter.
  • Non-Histone Chromosomal Proteins (NHC) are the additional sets of proteins that contribute to the packaging of chromatin at a higher level.

Question 5.
Explain the components of a transcription unit with the help of a diagram?
Answer:
Maharashtra Board Class 12 Biology Important Questions Chapter 4 Molecular Basis of Inheritance 31
Transcription unit (Each transcribed segment of DNA) consists of the promoter, the structural gene and the terminator.
(1) The promoter:

  • The promoter is located towards 5′ end of structural gene, i.e. upstream.
  • It is a DNA sequence that provides binding site for enzyme RNA polymerase.
  • In prokaryotes, sigma factor sub unit of the enzyme recognizes the promoter.

(2) Structural genes:

  • Template strand (Antisense strand) : DNA strand having 3’→ 5’ polarity acts as template strand as DNA dependent- RNA polymerase catalyses polymerization in 5’ → 3’ direction.
  • Sense strand : The other strand of DNA having 5’ → 3’ polarity is complementary to template strand. It is called as sense strand. The sequence of bases in this strand, is same as in RNA (where Thymine is replaced by Uracil). It is the actual coding strand.

(3) The terminator:

  • The terminator is located at 3’ end of coding strand, i.e. downstream.
  • It defines the end of the transcription process.

Question 6.
What have we learnt from the Human Genome Project?
Answer:
We have learnt following salient features of human genome from the Human Genome Project.

  1. The human genome contains 3164.7 million nucleotide bases.
  2. The average gene consists of 3000 bases.
  3. Largest known human gene is dystrophin at 2.4 million bases.
  4. Total number of genes is estimated to be 3000.
  5. 99.9% nucleotide bases are exactly the same in all people.
  6. The function of about 50% of the discovered genes are unknown.
  7. Less than 2% of the genome codes for proteins.
  8. Repeated sequences make up a very large portion of the human genome. They can shed light on chromosome structure, dynamics and evolution.
  9. Chromosome 1 has most genes (2968) and the Y has the fewest (231).
  10. Single nucleotide polymorphism have been identified at about 1.4 million locations. It is useful in finding chromosomal locations for disease-associated sequences and tracing human history.

Question 7.
Describe the steps involved in DNA fingerprinting ?
Answer:
Maharashtra Board Class 12 Biology Important Questions Chapter 4 Molecular Basis of Inheritance 32
Steps involved in DNA fingerprinting are as follows:
1. Isolation of DNA : The DNA can be isolated even from the small amount of tissue like blood, hair roots, skin, etc.

2. Restriction digestion:

  • The isolated DNA is treated with restriction enzymes which cut the DNA at specific sites to form small fragments of variable lengths.
  • Variations in the lengths of restriction fragments are known as Restriction Fragment Length Polymorphism (RFLP).

3. Gel electrophoresis:

  • The DNA samples are loaded on agarose gel and electrophoresis is carried out.
  • Negatively charged DNA fragments move to the positive pole.
  • Separation of fragments depends on their length and it results in formation of bands.
  • dsDNA is then denatured into ssDNA by alkali treatment.

4. Southern blotting : The separated DNA fragments are transferred to a nylon membrane or a nitrocellulose membrane.

Maharashtra Board Class 12 Biology Important Questions Chapter 4 Molecular Basis of Inheritance

5. Selection of DNA probe:

  • DNA Probe is a known sequence of single- stranded DNA.
  • It is obtained from organisms or prepared by cDNA preparation method.
  • The DNA probe is labelled with radioactive isotopes.

6. Hybridization:

  • In this process, probe is added to the nitrocellulose membrane containing DNA fragments.
  • The single-stranded DNA probe pairs with the complementary base sequence of the DNA strand.
  • As a result DNA-DNA hybrids are formed on the nitrocellulose membrane. Unbound single-stranded DNA probe fragments are washed off.

7. Photography : The nitrocellulose membrane is then kept in contact with X-ray film. DNA bands, due to radioactive probe, give photographic image on X-ray film. This is autoradiography.

Maharashtra Board Class 12 Biology Important Questions Chapter 3 Inheritance and Variation

Balbharti Maharashtra State Board 12th Biology Important Questions Chapter 3 Inheritance and Variation Important Questions and Answers.

Maharashtra State Board 12th Biology Important Questions Chapter 3 Inheritance and Variation

Multiple Choice Questions

Question 1.
Which one of the following characters is recessive in the case of the pea plants?
(a) Axial flower
(b) Green pod
(c) Green seed
(d) Inflated pod
Answer:
(c) Green seed

Question 2.
Which of the following trait is dominant in Pisum sativum?
(a) White flowers
(b) Green seeds
(c) Yellow pods
(d) Inflated pods
Answer:
(d) Inflated pods

Maharashtra Board Class 12 Biology Important Questions Chapter 3 Inheritance and Variation

Question 3.
When phenotypic and genotypic ratio is the same, then it is an example of ……………….
(a) incomplete dominance
(b) cytoplasmic inheritance
(c) quantitative inheritance
(d) incomplete dominance and co-dominance
Answer:
(a) incomplete dominance

Question 4.
A pea plant with yellow and round seeds is crossed with another pea plant with green and wrinkled seeds produce 51 yellow round seeds and 49 yellow wrinkled seeds, the genotype of plant with yellow round seeds must be ……………….
(a) YYRr
(b) YyRr
(c) YyRR
(d) YYRR
Answer:
(a) YYRr

Question 5.
When a single gene produces two effects and one of it is lethal, then the ratio is ……………….
(a) 2 : 1
(b) 1 : 1
(c) 1 : 2 : 1
(d) 1 : 1 : 1 : 1
Answer:
(a) 2 : 1

Question 6.
When two genes control a single character and have cumulative effect, the ratio is ……………….
(a) 1 : 1 : 1 : 1
(b) 1 : 4 : 6 : 4 : 1
(c) 1 : 2 : 1
(d) 1 : 6 : 15 : 20 : 15 : 6 : 1
Answer:
(d) 1 : 6 : 15 : 20 : 15 : 6 : 1

Question 7.
Genes located on the same locus but show more than two different phenotypes are called ……………….
(a) polygenes
(b) multiple alleles
(c) co-dominants
(d) pleiotropic genes
Answer:
(b) multiple alleles

Question 8.
Genotype refers to the genetic composition of ……………….
(a) an organism
(b) an organ
(c) chromosomes
(d) germ cells
Answer:
(a) an organism

Question 9.
Individuals having identical alleles of a gene are known as ……………….
(a) homozygous
(b) heterozygous
(c) hybrids
(d) dominants
Answer:
(a) homozygous

Question 10.
If a heterozygous tall plant is crossed with a homozygous dwarf plant, the proportion of dwarf progeny will be ……………….
(a) 100 per cent
(b) 75 per cent
(c) 50 per cent
(d) 25 per cent
Answer:
(c) 50 percent

Question 11.
Inheritance of AB blood group is due to ……………….
(a) incomplete dominance
(b) polyploidy
(c) polygeny
(d) co-dominance
Answer:
(d) co-dominance

Question 12.
The recombination of characters in a dihybrid cross is related to ……………….
(a) law of dominance
(b) incomplete dominance
(c) co-dominance
(d) independent assortment
Answer:
(d) independent assortment

Question 13.
Which one of the following is a true pleiotropic gene?
(a) HbA
(b) Hbs
(c) HbD
(d) HbP
Answer:
(b) Hbs

Question 14.
For demonstrating the law of independent assortment, one should carry out ……………….
(a) back cross
(b) test cross
(c) dihybrid cross
(d) monohybrid cross
Answer:
(c) dihybrid cross

Question 15.
Which one of the following is an example of multiple alleles?
(a) Height in pea plant
(b) Hair colour in cattle
(c) Petal colour in four o’clock plant
(d) Wing-size in Drosophila
Answer:
(d) Wing-size in Drosophila

Question 16.
For the formation of 50 seeds, how many minimum meiotic divisions are necessary?
(a) 25
(b) 50
(c) 75
(d) 63
Answer:
(d) 63

Question 17.
A cross used to verify the unknown genotype of F1 hybrid is a ………………. cross.
(a) test
(b) back
(c) dihybrid
(d) monohybrid
Answer:
(a) test

Question 18.
Appearance of new combinations in F2 generation in a dihybrid cross proves the law of ……………….
(a) dominance
(b) segregation
(c) independent assortment
(d) purity of gametes
Answer:
(c) independent assortment

Question 19.
Genotype of human blood group ‘O’ will be ……………….
(a) IAIA
(b) IAIB
(c) ii
(d) IAi
Answer:
(c) ii

Question 20.
The genotype of human blood group B is ……………….
(a) IAIA
(b) IBi
(c) IAIB
(d) ii
Answer:
(b) IBi

Question 21.
……………… chromosome appears ‘V’-shaped during anaphase.
(a) Metacentric
(b) Acrocentric
(c) Telocentric
(d) Sub-Metacentric
Answer:
(a) Metacentric

Question 22.
The sister chromatids are held together by ……………….
(a) centrioles
(b) chromonemata
(c) chromomere
(d) centromere
Answer:
(d) centromere

Question 23.
Which of the following is not X-linked disorder ?
(a) Haemophilia
(b) Night-blindness
(c) Hypertrichosis
(d) Myopia
Answer:
(c) Hypertrichosis

Question 24.
Which of the following is also called bleeder’s disease ?
(a) Anaemia
(b) Thrombocytopenia
(c) Polycythemia
(d) Haemophilia
Answer:
(d) Haemophilia

Question 25.
The person with Turner’s syndrome has ……………….
(a) 45 autosomes and X sex chromosome
(b) 44 autosomes and XYY sex chromosome
(c) 45 autosomes and Y chromosome
(d) 44 autosomes and X chromosome
Answer:
(d) 44 autosomes and X chromosome

Question 26.
Which of the following is sex chromosomal disorder ?
(a) Colour blindness
(b) Turner’s syndrome
(c) Thalassemia
(d) Down’s syndrome
Answer:
(b) Turner’s syndrome

Question 27.
The word chroma means ……………….
(a) a part of nucleus
(b) a part of chromosome
(c) colour
(d) filamentous body
Answer:
(c) colour

Question 28.
Presence of whole sets of chromosomes is called ……………….
(a) aneuploidy
(b) euploidy
(c) ploidy
(d) chromatography
Answer:
(b) euploidy

Question 29.
The synonymous term for centromere is ……………….
(a) primary constriction
(b) secondary constriction
(c) telomere
(d) satellite
Answer:
(a) primary constriction

Question 30.
Small swellings on the surface of the chromosome are called ……………….
(a) centromeres
(b) chromonemata
(c) chromomeres
(d) telomeres
Answer:
(c) chromomeres

Question 31.
On what basis are the chromosomes usually classified?
(a) On the basis of their function
(b) On the basis of their length
(c) On the basis of the position of the centromere
(d) On the basis of their number
Answer:
(c) On the basis of the position of the centromere

Question 32.
Find the mismatched pair :
(a) Metacentric – V-shaped
(b) Sub-Metacentric – L-shaped
(c) Acrocentric – J-shaped
(d) Telocentric – S-shaped
Answer:
(d) Telocentric – S-shaped

Question 33.
Out of the following combinations which individual will have maximum genetically active DNA?
(a) 44 + XX
(b) 44 + XY
(c) 44 + XYY
(d) Down’s syndrome
Answer:
(a) 44 +XX

Maharashtra Board Class 12 Biology Important Questions Chapter 3 Inheritance and Variation

Question 34.
Crossing over occurs at the time of ……………….
(a) diplotene
(b) pachytene
(c) leptotene
(d) zygotene
Answer:
(b) pachytene

Question 35.
A mature woman has ………………. linkage groups.
(a) 44
(b) 22
(c) 46
(d) 23
Answer:
(d) 23

Question 36.
The pairing of homologous chromosomes is called ……………….
(a) crossing over
(b) terminalisation
(c) synapsis
(d) bivalent
Answer:
(c) synapsis

Question 37.
If only one ‘X’ chromosome is found in a female person, which of the following symptoms will she show?
(a) epicanthal skin fold
(b) webbing of neck
(c) small testis and absence of spermatogenesis
(d) presence of simian crease on the palm
Answer:
(b) webbing of neck

Question 38.
If centromere is situated in the middle of the chromosome, it is called ……………….
(a) metacentric
(b) acrocentric
(c) submetacentric
(d) telocentric
Answer:
(a) metacentric

Question 39.
In which of the following disorders the number of chromosomes present is (extra) 47?
(a) Turner’s syndrome
(b) Cushing’s syndrome
(c) Acquired immuno-deficiency syndrome
(d) Down’s syndrome
Answer:
(d) Down’s syndrome

Question 40.
Myopia is an example of ……………….
(a) complete sex linkage
(b) incomplete sex linkage
(c) recombination
(d) crossing over
Answer:
(a) complete sex linkage

Question 41.
Down’s syndrome is represented by ……………….
(a) n + 1
(b) 2n + 1
(c) 3n + 1
(d) n – 1
Answer:
(b) 2n + 1

Classify the following to form Column B as per the category given in Column A

Question 1.
Types of traits:
[Sickle-cell anaemia, Flower colour of Mirabelis jalapa, Coat colour of cattle, Human blood groups, Widow’s peak, Height in human beings.]

Column AColumn B
(1) Co-dominance—————–
(2) Incomplete dominance—————–
(3) Multiple allelism—————-
(4) Pleiotropy—————–
(5) Polygenes—————-
(6) Autosomal dominance——————

Answer:

Column AColumn B
(1) Co-dominanceCoat colour of cattle
(2) Incomplete dominanceFlower colour of Mirabelis jalapa
(3) Multiple allelismHuman blood groups
(4) PleiotropySickle-cell anaemia
(5) PolygenesHeight in human beings
(6) Autosomal dominanceWidow’s peak

Question 2.
Types of sex-linked genes:
[Haemophilia, Ichthyosis, Nephritis, Myopia, Hypertrichosis, Retinitis pigmentosa]

Column AColumn B
(1) Completely X-linked genes—————–
(2) Completely Y-linked genes—————–
(3) Incompletely sex-linked genes—————-

Answer:

Column AColumn B
(1) Completely X-linked genesHaemophilia, Myopia
(2) Completely Y-linked genesIchthyosis,Hypertrichosis
(3) Incompletely sex-linked genesNephritis, Retinitis pigmentosa

Question 3.
Genetic Disorders
[Turner’s syndrome, Sickle-cell anaemia, Thalassemia, Edward’s syndrome, Klinefelter’s syndrome, Down’s syndrome]

Column AColumn B
(A) Autosomal disorder—————–
(B) Sex chromosomal disorder—————–
(C) Mendelian disorder—————-

Answer:

Column AColumn B
(A) Autosomal disorderEdward’s syndrome, Down’s syndrome
(B) Sex chromosomal disorderTurner’s syndrome, Klinefelter’s syndrome
(C) Mendelian disorderSickle-cell anemia, Thalassemia

Very Short Answer Questions

Question 1.
What is hybrid?
Answer:
Hybrid is a heterozygous individual produced from a cross involving two parents differing in one or more contrasting characters.

Question 2.
What are homologues?
Answer:
Homologues are homologous chromosomes which are morphologically similar to each other.

Question 3.
Which law of Mendelian genetics is universally applicable?
Answer:
The law of segregation of Mendelian genetics is universally applicable.

Question 4.
Which law of Mendelian genetics is not universally applicable?
Answer:
The law of independent assortment of Mendelian genetics is not universally applicable.

Maharashtra Board Class 12 Biology Important Questions Chapter 3 Inheritance and Variation

Question 5.
Give the alternative term for checker board.
Answer:
Punnett’s square is the alternative term for the checker board.

Question 6.
Give the genotypic dihybrid ratio.
Answer:
1 : 1 : 2 : 2 : 4 : 2 : 2 : 1 : 1 is the genotypic dihybrid ratio.

Question 7.
What is a lethal gene?
Answer:
The gene which causes the death of the bearer is called lethal gene.

Question 8.
A pea plant pure for yellow seed colour is crossed with a pea plant pure for green seed colour. In F1 generation, all pea plants were with yellow seed. Which law of Mendel is applicable?
Answer:
Mendel’s law of dominance is applicable in this example.

Question 9.
Identify which one of the following is a test cross.

  1. Tt × Tt
  2. TT × tt
  3. Tt × tt

Answer:
3. Tt × tt is a test cross.

Question 10.
What colouration do roans possess? Why?
Answer:
Roans possess the mixture of red and white colour side by side due to codominant alleles for red and white traits.

Question 11.
What are polygenes?
Answer:
When a character is controlled by two or more than two pairs of genes, the genes are called polygenes.

Question 12.
In which region of chromosomes does crossing over take place?
Answer:
Crossing over takes place in the homologous region of the chromosomes.

Question 13.
What are the four sequential steps of crossing over?
Answer:
There are four sequential steps such as synapsis, tetrad formation, crossing over and terminalisation.

Question 14.
Give one example of complete linkage.
Answer:
X chromosome of Drosophila males show complete linkage.

Question 15.
What is the number of linkage groups found in honey bee?
Answer:
The number of linkage group corresponds to the haploid number of chromosomes. Honey bee’s haploid chromosomes number is 16 and thus it has 16 linkage groups.

Question 16.
Name the term for genes located on non-homologous region of Y chromosomes.
Answer:
The genes located on non-homologous region of Y chromosomes are known as holandric genes or Y-linked genes.

Question 17.
What are linkage groups?
Answer:
The genes present on the same chromosome and inherited together are called linkage group.

Question 18.
How are RBCs changed due to sickle-cell anaemia ?
Answer:
RBCs undergo change in their shape and look like a sickle, resulting in reduced capacity to carry haemoglobin.

Question 19.
Which part of a chromosome is called nucleolar organizer?
Answer:
The secondary constriction present on the chromatid arms of a chromosome is called nucleolar organizer.

Question 20.
Why is Y chromosome genetically less active?
Answer:
Since Y-chromosome possesses small amount of euchromatin that contains DNA or genes, therefore it is genetically less active.

Question 21.
Why hypertrichosis is called holandric gene?
Answer:
Hypertrichosis is Y linked gene which can be seen only in males, therefore it is called holandric gene.

Question 22.
What is the genetic difference between total colour blindness and red-green colour blindness ?
Answer:
Total colour blindness is due to incomplete sex-linked genes while red-green colour blindness is due to complete sex linkage.

Question 23.
What happens if the gene for production of factor VIII and IX becomes recessive?
Answer:
The person having recessive gene for haemophilia is deficient in clotting factors (VIII or IX) in blood, such person’s blood does not clot and he thus becomes a patient of haemophilia.

Question 24.
What is the cause of Thalassemia?
Answer:
Thalassemia is caused due to deletion or mutation of gene which codes for alpha (α) and beta (β) globin chains, causing abnormal synthesis of haemoglobin. Thus it is a quantitative abnormality of polypeptide globin chain synthesis.

Question 25.
What is monosomy? Give one example of the same.
Answer:
Monosomy is lack of one chromosome from the usual chromosomal complement. Turner’s syndrome is the example of monosomy.

Give Definitions

Question 1.
Factor
Answer:
The unit of heredity which is responsible for the inheritance and expression of a character and which is responsible for the genetic character is called a factor.

Question 2.
Gene
Answer:
The specific segment of DNA or sequence of nucleotides which is responsible for the inheritance and expression of that character is called a gene.

Maharashtra Board Class 12 Biology Important Questions Chapter 3 Inheritance and Variation

Question 3.
Alleles or Allelomorphs
Answer:
The two or more alternative forms of a given gene which are present on the identical loci on the homologous chromosomes are called alleles of each other.

Question 4.
Phenotype
Answer:
The external appearance of an individual for any trait is called phenotype for that trait.

Question 5.
Genotype
Answer:
Genetic constitution of an organism with respect to a particular trait is called genotype.

Question 6.
Homologous Chromosomes?
Answer:
The morphologically, genetically and structurally essentially identical chromosomes present in a diploid cell are called homologous chromosomes.

Question 7.
Back cross
Answer:
The cross of Fx progeny with any of the parents, irrespective of being dominant or recessive, is called back cross.

Question 8.
Linkage
Answer:
Linkage is defined as the tendency of the genes to be inherited together because they are present in the same chromosome.

Question 9.
Non-disjunction
Answer:
Non-disjunction is the phenomenon in which chromosomes fail to separate at the time of cell division, resulting in abnormal chromosomal combinations.

Question 10.
Syndrome
Answer:
The appearance of different types of symptoms at the same time in an individual is called a syndrome.

Question 11.
Aneuploidy
Answer:
Addition or deletion of one or two chromosomes in a diploid chromosome set is called aneuploidy.

Distinguish Between

Question 1.
Homozygous and Heterozygous.
Answer:

HomozygousHeterozygous
1. Individuals with similar gene pairs are called homozygous.1. Individuals with different gene pairs are called heterozygous.
2. Homozygous individuals form only one type of gametes.2. Heterozygous individuals form more than one type of gametes.
3. Individuals with similar gene pairs TT, tt, RR and rr are homozygous.3. Individuals with dissimilar gene pairs Tt and Rr are heterozygous.
4. Homozygous are also called pure breed.4. Heterozygous are referred to as hybrids.

Question 2.
Monohybrid cross and Dihybrid cross.
Answer:

Monohybrid crossDihybrid cross
1. Crosses involving a single pair of alleles are called monohybrid crosses.1. Crosses involving two pairs of alleles are called dihybrid crosses.
2. Monohybrid crosses yield a phenotypic ratio of 3 : 1 in the F2 generation.2. Dihybrid crosses yield a 9 : 3 : 3 : 1 ratio in F2 generation.
3. Monohybrid crosses yield 1 : 2 : 1 genotypic ratio in F2 generation.3. Dihybrid crosses yield 1 : 1 : 2 : 2 : 4 : 2 : 2 : 1 : 1 genotypic ratio in F2 generation.
4. Application of the law of independent assortment is not applicable in monohybrid crosses.4. Application of the law of independent assortment is applicable in dihybrid crosses.

Question 3.
Dominant characters and Recessive characters.
Answer:

Dominant charactersRecessive characters
1. The characters that are expressed in the F1 generation are called dominant characters.1. The characters that are not expressed in the F1 generation are called recessive characters. They are prevented from expressing themselves, due to presence of dominant allele.
2. Dominant character is expressed either in homozygous or heterozygous combination.2. Recessive characters are expressed only when they are in homozygous combination.
3. Dominant characters cannot be masked by recessive characters.
E.g. Round seed and yellow seed are dominant characters in pea plant.
3. Recessive characters are masked by dominant characters.

E.g. Wrinkled seed and green seed are recessive characters in pea plant.

Question 4.
Phenotype and Genotype.
Answer:

PhenotypeGenotype
1. Phenotype refers to the outward appearance of an individual such as shape, colour, sex, etc.1. Genotype refers to the genetic composition of an individual.
2. Phenotype can be observed directly in an individual.2. Genotype cannot be seen, but can be found out by modern techniques like DNA fingerprinting.
3. Individuals resembling each other may or may not have the same genotype.3. Individuals possessing the same genotype usually have the same phenotypic expression.
4. The phenotypic ratio obtained in the F2 generation of a monohybrid cross is 3 : 1.4. The genotypic ratio obtained in the F2 generation of a monohybrid cross is 1 : 2 : 1.

Maharashtra Board Class 12 Biology Important Questions Chapter 3 Inheritance and Variation

Question 5.
Incomplete dominance and Co-dominance.
Answer:

Incomplete dominanceCo-dominance
1. Incomplete dominance is seen when the phenotypes of the two parents blend together to create a new phenotype for their offspring.1. Co-dominance is seen when the two parent phenotypes are expressed together in the offspring.
2. Both the genes of an allelomorphic pair express themselves partially in F1 hybrids.2. Both the genes of an allelomorphic pair express themselves equally in F1 hybrids.
3. In incomplete dominance, a mixture of the alleles in the genotype is seen in the phenotype.3. In co-dominance, both alleles in the genotype are seen in the phenotype.
4. The phenotypic effect of one allele is more prominent than the other.4. The phenotypic effect of both the alleles is equally prominent.
5. Blending or intermixing of two alleles can be observed. A white flower and a red flower alleles mix and produce pink flowers.

Example : Pink flowers in Mirabilis jalapa.

5. No intermixing or blending effect of two alleles is observed. The colours don’t mix but are seen in patches.

Example : Roan colour in cattle.

Question 6.
Turner’s syndrome and Klinefelter’s syndrome.
Answer:

Turner’s syndromeKlinefelter’s syndrome
1. Individual with Turner’s syndrome has total 45 chromosomes in each of her cell.1. Individual with Klinefelter’s syndrome has total 47 chromosomes in each of his cell.
2. Turner’s syndrome is XO female, caused due to monosomy of X-chromosome.2. Klinefelter’s syndrome is XXY male, caused due to trisomy of X chromosome.
3. The external phenotype is of female.3. The external phenotype is of male.
4. The stature is short.4. The stature is tall and thin.
5. Secondary sexual characteristics are not developed in Turner’s syndrome.5. Secondary sexual characteristics are poorly developed in Klinefelter’s syndrome.

Give Reasons or Explain the Statements

Question 1.
Law of segregation is universally applicable.
Answer:

  1. According to the law of segregation, the members of the allelic pair remain together without mixing with each other.
  2. They segregate or separate when the gametes are formed.
  3. Thus the gametes that are formed receive only one of the two factors.
  4. Now it is known that the organisms are diploid and the gametes produced by them are haploid.
  5. The law of segregation therefore is universally applicable.

Question 2.
Mendel selected garden pea for his breeding experiments.
Answer:
Mendel selected garden pea for his breeding experiments, because:

  1. The pea plants were true breeding varieties.
  2. The pea plants being annual, it was possible to cross and study many generations within a short period.
  3. The pea plants had a number of distinguishable, contrasting characters such as tall habit and dwarf habit, round seed and wrinkled seed.
  4. The pea plants were easy to handle for breeding experiments.

Question 3.
When Mendel crossed a tall pea plant with a dwarf pea plant the offspring obtained from this cross were all tall.
Answer:

  1. The tall habit of the pea plant is dominant over the dwarf habit of the pea plant.
  2. Hence, when Mendel crossed a tall pea plant with a dwarf pea plant, the offspring obtained from this cross were all tall.

Question 4.
A cross between a homozygous tall plant with a homozygous dwarf plant results in two types of tall plants in the F2 generation.
Answer:

  1. A cross between a homozygous tall (TT) and a homozygous dwarf (tt) gives rise to a heterozygous tall (Tt) plant in the F1 generation.
  2. When the F1 plant is selfed (Tt × Tt), it gives rise to three tall plants of which two- are heterozygous (Tt) tall and one is homozygous (TT) tall.
  3. Hence a cross between a homozygous tall plant with a homozygous dwarf plant results in two types of tall plants in the F2 generation.

Question 5.
Possibility of female becoming a haemophilic is extremely rare.
Answer:

  1. Haemophilia is caused due to X-linked recessive gene. Females have double X chromosomes.
  2. Even if she has haemophilic gene on one of her X-chromosome, the dominant gene on other X-chromosome, suppresses its expression. Female therefore, does not become haemophilic.
  3. If she inherits haemophilic gene on both of her X-chromosomes, this combination becomes lethal. Such embryo is aborted. If born, she dies soon. This makes the possibility of female becoming a haemophilic extremely rare.

Question 6.
Human female is referred to as carrier of colour blindness.
Answer:
Human female is referred to as carrier of colour blindness because of the following reasons:

  1. Females possess double X-chromosomes in her gametes.
  2. If one X-chromosome is carrying recessive gene for colour blindness, her other dominant X hides the expression of colour blindness and hence she does not become a patient.
  3. But such female can carry the defective gene to her progeny. Thus she is called carrier of colour-blindness.
  4. A female having one recessive gene on X-chromosome is a carrier female, while a female possessing both recessive genes on both the X-chromosomes will be colour blind which is very rare.

Write Short Notes

Question 1.
Linkage.
Answer:

  1. Linkage is the tendency of genes to be inherited together because they are present in the same chromosome.
  2. All the genes on a chromosome are linked to one another. In the linkage group some of the genes are included.
  3. The number of linkage groups of a particular species corresponds to its haploid number of chromosomes present in the organism.
  4. In human beings, there are 23 linkage groups which correspond to the pairs of chromosomes found in each cell.
  5. Linkage groups can be separated only at the time of crossing over during meiosis. The linkage group can form a new combination of genes after crossing over.
  6. Linkages Eire of two types, viz, complete linkage and incomplete linkage.
  7. Morgan discovered linkage in animals while Bateson and Punnett discovered it in plants.

Question 2.
Multiple alleles.
Answer:

  1. Multiple alleles are more than two alternative alleles of a gene in a population situated on the same locus on a chromosome or its homologue.
  2. Multiple alleles arise by mutations of the wild type of gene. Series of multiple alleles are formed due to several mutations that take place in the wild type of allele. This series show alternative expression.
  3. Different alleles in a series show dominant-recessive relation or may show co-dominance or incomplete dominance among themselves. Among all the wild type is the most dominant one over all other mutant alleles.
  4. In Drosophila, a large number of multiple alleles are known. E.g. The size of wings from normal wings to vestigial wings is due to one allele (vg) in homozygous condition. The normal wing is dominant and wild type while vestigial wing is recessive type.
  5. Human blood groups A, B, AB and O Eire also due to series of multiple alleles.

Question 3.
Autosomal inheritance.
Answer:

  1. Transmission of body characters occurs due to autosomes. They are not concerned with sex determination or sex linkage.
  2. All the body characters from parents are passed on to their offspring through autosomes. This is called autosomal inheritance.
  3. Some autosomal characters are due to dominant genes while some other are due to recessive genes. E.g. Widow’s peak and Huntington’s disease is also autosomal dominant character, etc.
  4. Phenyl ketonuria (PKU), Cystic fibrosis and Sickle-cell anaemia are autosomal recessive traits.

Question 4.
Widow’s peak.
Answer:

  1. Widow’s peak is a prominent ‘V’ shaped hairline on forehead.
  2. It is due to autosomal dominant gene.
  3. Widow’s peak occurs in homozygous dominant (WW) and also heterozygous (Ww) individuals.
  4. Individuals with homozygous recessive (ww) genotype do not have widow’s peak but have a straight hair line.
  5. Both males and females have equal chance of inheritance.

Maharashtra Board Class 12 Biology Important Questions Chapter 3 Inheritance and Variation

Question 5.
Environmental sex determination
Answer:

  1. Environmental sex determination is shown by lower organisms such as Bonellia viridis.
  2. In this animal the environmental factors decide the sex of an offspring.
  3. There is extreme sexual dimorphism in this worm. Female is about 10 cm long while male is tiny and parasitic in the reproductive parts of mature female.
  4. If larva is reared in vicinity of mature female then it becomes a male. By settling on the proboscis of mature female, larva becomes parasitic, enters the female’s mouth and then takes permanent shelter in the female uterus. Such males then produce gametes and fertilize the eggs.
  5. If larvae are drifted away from mature female or if they settle on the sea bottom, they develop into females. Thus determination of sex is due to environmental factors.

Question 6.
Y-linked or Holandric genes.
Answer:

  1. Holandric means entirely of male sex. Y-linked genes are called holandric genes because they are located on non-homologous region of Y chromosome.
  2. The Y-linked genes are inherited directly from male to male.
  3. These genes are never seen in females due to lack of Y chromosome in them.
  4. Hyper Mchosis and ichthyosis are examples of holandric genes.
  5. Hypertrichosis means excessive development of hair on pinna of ear. This character is transmitted directly from father to son.
  6. Ichthyosis person with rough skin.

Question 7.
X-body.
Answer:

  1. German biologist, Henking in 1891, was studying spermatogenesis of the squash bug (Anasa tristis).
  2. He noted that 50% of sperms receive the unpaired chromosomes while other 50% sperms do not receive it.
  3. Henking gave a name to this structure as the X-body. He was unable to explain its role in sex determination.
  4. Further investigations by other scientists led to conclusion that the ‘X-body’ of Henking was a chromosome and gave the name as X-Chromosome to X-body.

Question 8.
Thalassemia
Answer:
(1) Thalassemia is an autosomal-eeessive disorder. The synthesis of alpha ciiains are controlled by two genes, (HBA1 and HBA2) on chromosome 16. Beta chain synthesis is controlled by gene HBB located on chromosome 11. Two alpha chains and two beta chains together form four polypeptide chains that make heterotetrameric haemoglobin molecule. But when there is defective gene on either of chromosome 16 or 11, there is quantitative abnormality of polypeptide globin chain synthesis. This results into thalassemia.

(2) Depending upon which chain is affected, thalassemia is classified as, alpha (α) thalassemia and beta (β) thalassemia.

(3) The clinical symptoms of thalassemia are as follows:

  • Pale yellow skin.
  • Anaemia due to inability to synthesize haemoglobin.
  • Slow growth and development.
  • Variation in the shape and size of RBCs.

(4) Patients need regular blood transfusions to cope with the disorder.

Short Answer Questions

Question 1.
Write the statements of three laws of inheritance given by Mendel.
Answer:
(1) Statement of Law of Dominance : When two homozygous individuals with one or more sets of contrasting characters are crossed, the alleles that appear in F1 are dominant and those which do not appear in F1 are recessive.

(2) Statement of Law of Segregation or Law of purity of gametes : When F1 hybrid forms gametes, the alleles segregate from each other and enter in different gametes. The gametes formed are pure because they carry only one either dominant allele or recessive allele each. Due to this the law is also called “Law of purity of gametes”.

(3) Statement of Law of Independent Assortment : When hybrid possessing two (or more) pairs of contrasting alleles forms gametes, these alleles in each pair segregate independently of the other pair.

Question 2.
Why are farmers and gardeners advised to buy new F1 hybrid seeds every year?
Answer:

  1. Farmers use hybrid seeds for agriculture or horticulture. Hybrid seeds are produced by crossing two unrelated parent plants.
  2. Hybrid seed varieties give improved yields and crop vigour to the farmer.
  3. Hybrids are made by crossing two highly inbred ‘parent’ plants. First generation hybrids, however, do not breed true to type, meaning that the seed they set may not grow into crops that are identical to the ‘parent’ plants.
  4. This can result in variations in yield and quality therefore many farmers prefer to buy new hybrid seed each year to ensure consistency in their final product.

Question 3.
What are the main generalizations given after Mendel’s experiments on the pea plant?
Answer:
After the Mendel’s laws of inheritance and his experiments, following generalizations were made:

  1. Single trait is shown due to single gene. Every single gene has two contrasting alleles.
  2. Two alleles are always in interaction in which one is completely dominant while other is completely recessive.
  3. Factors which were later called genes for different traits are always present on different chromosomes. These traits can assort independently of each other.

Question 4.
Mention the types of deviations from Mendel’s finding.
OR
Describe Neo-Mendelism in short.
Answer:
As the science of genetics progressed, many changes were seen from Mendel’s generalizations. These are called as Neo- Mendelism.
The deviations from Mendel’s findings can be categorised under following heads:

  1. Intragenic interactions : These interactions : are seen between the alleles of same gene. e.g. incomplete dominance and co-dominance. They are also seen in multiple allele series of a gene.
  2. Intergenic interactions : Intergenic interactions are between the alleles of different genes present on the same or different chromosomes, e.g. pleiotropy, polygenes, epistasis, supplementary and complementary genes, etc.

Question 5.
Why Drosophila is most suitable organism for genetics experiments?
Answer:
Drosophila is most suitable organism because of the following reasons:

  1. Drosophila cam easily be cultured under laboratory conditions.
  2. Life span of Drosophila is short for about two weeks.
  3. Drosophila has high rate of reproduction and hence newer organisms can be obtained rapidly.

Question 6.
What are the causes of Down’s syndrome?
Answer:

  1. Down’s syndrome is caused due to aneuploidy.
  2. Aneuploidy is due to non-disjunction of chromosome at the time of gamete formation during meiosis. Due to non-disjunction, chromosomes fail to separate.
  3. In addition to a homologous pair of 21st chromosome there is an extra 21st, therefore it is called trisomy (2n + 1) of 21st chromosome.

Question 7.
What are the characteristic symptoms of Down’s syndrome?
Answer:
Symptoms of Down’s syndrome:

  1. Typical facial features.
  2. An epicanthal skin fold, over the inner corner of eyes causing downward slanting eyes.
  3. Typical flat face, rounded flat nose, mouth always open with protruding tongue.
  4. Mental retardation.
  5. Poor skeletal development.
  6. Short stature, relatively small skull and arched palate.
  7. Flat hand with simian crease that runs across the palm.

Question 8.
Write a brief account of Turner’s syndrome.
Answer:

  1. Turner’s syndrome is a genetic disorder caused due to monosomy of X chromosome.
  2. It was first described by H. H. Turner.
  3. Turner’s syndrome is caused due to non-disjunction of sex chromosomes which takes place during gamete formation.
  4. Chromosomal complement of Turner’s syndrome is 44 + XO, having a total of 45 chromosomes.

Symptoms of Turner’s syndrome are as follows:

  1. Female phenotype.
  2. Short stature with webbing of neck.
  3. Low posterior hair line.
  4. Secondary sexual characters fail to develop.
  5. Mental retardation.

Maharashtra Board Class 12 Biology Important Questions Chapter 3 Inheritance and Variation

Question 9.
Write a brief account of Klinefelter’s syndrome.
Answer:

  1. Klinefelter’s syndrome is a genetic disorder caused due to trisomy of X chromosome.
  2. It was first described by Harry Klinefelter.
  3. Klinefelter’s syndrome is caused due to non-disjunction of sex chromosomes which takes place during gamete formation.
  4. Chromosomal complement of Klinefelter’s syndrome is 44+XXY, having a total of 47 chromosomes.

Symptoms of Klinefelter’s syndrome are as follows:

  1. The Klinefelter’s syndrome individuals are tall, thin and eunuchoid.
  2. They are sterile with poorly developed sexual characteristics.
  3. Testes are underdeveloped and small. Spermatogenesis does not take place.
  4. They have subnormal intelligence and show partial mental retardation.

Diagram Based Questions

Question 1.
Give the graphical representation of test cross and back cross.
Answer:
(1) Test cross
Answer:
Maharashtra Board Class 12 Biology Important Questions Chapter 3 Inheritance and Variation 1
The F2 generation of test cross consists of 50% heterozygous tall plants and 50% homozygous dwarf plants.

(2) Back cross
Maharashtra Board Class 12 Biology Important Questions Chapter 3 Inheritance and Variation 2
F1 crossed back with its dominant parent
Maharashtra Board Class 12 Biology Important Questions Chapter 3 Inheritance and Variation 3

Question 2.
Give a cross for incomplete dominance using a suitable example.
Answer:
Maharashtra Board Class 12 Biology Important Questions Chapter 3 Inheritance and Variation 4
Result:
Genotypic ratio – 1RR : 2 Rr : 11rr
Phenotypic ratio – 1Red : 2 Pink : 1 White

Question 3.
Give a cross of co-dominance using a suitable example.
Answer:
Coat colour in cattle
Red female RR × White male WW
P1 generation : RR × WW
Gametes : R and W
F1 generation all RW Roan coloured
P2 generation RW × RW

RW
RRRRW
WRWWW

Genotypic ratio : 1 RR : 2 RW : 1 WW
Phenotypic ratio : 1 Red : 2 Roan : 1 White

Question 4.
Draw two crosses to show inheritance of colour blindness, (i) A cross between normal female and colour blind male, (ii) A cross of carrier woman with normal man.
Answer:
(i) A cross between normal female and colour-blind male.
Maharashtra Board Class 12 Biology Important Questions Chapter 3 Inheritance and Variation 5

(ii) A cross of carrier female with normal male.
Maharashtra Board Class 12 Biology Important Questions Chapter 3 Inheritance and Variation 6

Question 5.
Draw the following crosses to show inheritance of haemophilia : Normal female with haemophilic male. Show their progeny. If one of their carrier daughters marries a normal male what would be possible genotypes of this generation.
Answer:
(1) Haemophilic male crossed with normal female:
Maharashtra Board Class 12 Biology Important Questions Chapter 3 Inheritance and Variation 7

(2) Carrier female crossed with normal male :
Maharashtra Board Class 12 Biology Important Questions Chapter 3 Inheritance and Variation 8

Maharashtra Board Class 12 Biology Important Questions Chapter 3 Inheritance and Variation

Question 6.
Sketch and label structure of X and Y chromosomes.
Answer:
Maharashtra Board Class 12 Biology Important Questions Chapter 3 Inheritance and Variation 9

Question 7.
Give the graphical representation of pleiotropy to show inheritance of Sickle-cell anaemia.
Answer:
Maharashtra Board Class 12 Biology Important Questions Chapter 3 Inheritance and Variation 10

Long Answer Questions

Question 1.
There are 16 possible individuals in F2 generation. Try to find out the phenotypes as well as the genotypic and phenotypic ratios.
Answer:
In the above dihybrid cross there are 4 phenotypes such as yellow round, yellow wrinkled, green round, green wrinkled.
There are 9 different genotypes as follows:
1 : RRYY / 2 : RRYy / 1 : RRyy / 2 : RrYY / 4 : RrYy / 2 : Rryy / 1 : rrYY /2 : rrYy / 1 : rryy

Question 2.
Explain with suitable diagram how test cross is used to find out genotype of dominant plant.
Answer:
Test cross is used to find out the exact genotype by crossing the F1 individual with homozygous recessive one.
E.g. To find out the genotype of unknown violet flower obtained in F1 generation, one can conduct two crosses as follows:
I. Unknown flower considering as RR (homozygous dominant)
RR × rr Homozygous dominant

RR
RRrRr
RRrRr

II. In such cross, all the flowers will be violet. II. Unknown flower considering as Rr (heterozygous)
Rr × rr Homozygous recessive with heterozygous

RR
RRrRr
RRrRr

In such cross half the flowers will be violet and half will be white.

Question 3.
Describe briefly Morgan’s Experiments showing linkage and crossing over. (Diagram is not needed)
Answer:
(1) Morgan used Drosophila melanogaster for his experiments.

(2) He carried out several dihybrid cross experiments to study sex-linked genes of Drosophila.

(3) Crosses between yellow-bodied, white-eyed female and brown-bodied, red-eyed males were done in P1 generation. Brown-bodied and red-eyed forms were wild.

(4) Morgan intercrossed their F1 progeny and noted that two genes did not segregate independently of each other and F2 ratio deviated very significantly from Mendelian 9 : 3 : 3 : 1 ratio.

(5) When genes are grouped on the same chromosome, some genes are strongly linked. They show very few recombinations (1.3%).

(6) When genes are loosely linked, i.e. located away from each other on chromosome, they show more (higher) recombinations (37.2%).

(7) For example, the genes for yellow body and white eye were strongly linked and showed only 1.3 per cent recombination (in cross-I).

(8) White-bodied and miniature wings showed 37.2 per cent recombination (in cross-II). Cross I shows crossing over between genes y and w.

(9) Cross II shows crossing over between genes white (w) and miniature wing (m). Here dominant wild type alleles are represented with (+) sign.

(10) Parental combinations occur more due to linkage and new combinations less due to crossing over.

Question 4.
Describe the mechanism of sex determination in human beings with a suitable cross.
Answer:
1. Sex determination in human beings:
Maharashtra Board Class 12 Biology Important Questions Chapter 3 Inheritance and Variation 11
(1) In human beings, the sex is determined with the help of X and Y chromosomes. This chromosomal mechanism of sex determination is of XX-XY type.

(2) In male, the nucleus of each cell contains 46 chromosomes or 23 pairs of chromosomes. Of these 22 pairs are autosomes and one pair of sex chromosomes. Males are thus heteromorphic as they have two different types of sex chromosomes.

(3) Autosomes or somatic chromosomes are responsible for determination of other characters of the body, but not the sex.

(4) In female cells, there are 22 pairs of autosomes and one pair of X chromosomes. Females are thus homomorphic as they have similar sex chromosomes.

(5) Thus the genotypes of female and male are as follows:
Female : 46 chromosomes = 44 autosomes + XX sex chromosomes
Male : 46 chromosomes = 44 autosomes + XY sex chromosomes

(6) During gamete formation, the diploid germ cells in the testes and ovaries undergo meiosis to produce haploid gametes (sperms and eggs). The homologous chromosomes separate and enter into different gametes during this process.

(7) The human male produces two different types of sperms, one containing 22 autosomes and one X chromosome and the other containing 22 autosomes and one Y chromosome. Human males are therefore called heterogametic, i.e. they produce different types of gametes.

(8) The human female produces only one type of eggs containing 22 autosomes and one X chromosome and therefore she is homogametic.

(9) During fertilization, if X containing sperm fertilizes the egg having X chromosome, then a female child with XX chromosomes is conceived.

(10) If Y containing sperm fertilizes the egg having X chromosome then a male child with XY chromosomes is conceived.

(11) The sex of the child thus depends upon the type of sperm fertilizing the egg. The heterogametic parent determines the sex of the child and thus the father is responsible for the determination of the sex of the child and not the mother.
Maharashtra Board Class 12 Biology Important Questions Chapter 3 Inheritance and Variation 12

Maharashtra Board Class 12 Biology Important Questions Chapter 3 Inheritance and Variation

Question 5.
Explain the mechanism of sex determination in case of birds.
Answer:

  1. Sex determination in birds is by ZW-ZZ mechanism.
  2. In birds, males are homogametic while females are heterogametic.
  3. Males produce all similar types of sperms, containing 8 autosomes and ‘Z’ sex chromosome.
  4. Females produce two different types of eggs, one containing 8 autosomes and Z chromosome and the other containing 8 autosomes and W chromosome.
  5. When Z bearing egg is fertilized by a sperm a male offspring is produced. If W bearing egg is fertilized then female offspring is produced.
    Maharashtra Board Class 12 Biology Important Questions Chapter 3 Inheritance and Variation 13

[Note : In domestic fowl chromosome number is 18, with 16 autosomes and two sex chromosomes.]

Maharashtra Board Class 12 Biology Important Questions Chapter 2 Reproduction in Lower and Higher Animals

Balbharti Maharashtra State Board 12th Biology Important Questions Chapter 2 Reproduction in Lower and Higher Animals Important Questions and Answers.

Maharashtra State Board 12th Biology Important Questions Chapter 2 Reproduction in Lower and Higher Animals

Multiple-choice questions

Question 1.
Gemmule formation takes place in ……………….
(a) Hydra
(b) Spongilla
(c) Planaria
(d) Human being
Answer:
(b) Spongilla

Question 2.
Which part of ovary in mammals acts as an endocrine gland after ovulation?
(a) stroma
(b) germinal epithelium
(c) vitelline membrane
(d) graafian follicle
Answer:
(d) graafian follicle

Maharashtra Board Class 12 Biology Important Questions Chapter 2 Reproduction in Lower and Higher Animals

Question 3.
Cessation of menstrual cycle in female is called ……………….
(a) lactation
(b) ovulation
(c) menarche
(d) menopause
Answer:
(d) menopause

Question 4.
Capacitation of sperms occurs in ……………….
(a) vas deferens
(b) vas efferens
(c) vagina
(d) ejaculatory duct
Answer:
(c) vagina

Question 5.
How many sperms are formed from a secondary spermatocyte ?
(a) 4
(b) 8
(c) 2
(d) 1
Answer:
(c) 2

Question 6.
The middle piece of the sperm contains ……………….
(a) proximal centriole
(b) nucleus
(c) mitochondria
(d) distal centriole
Answer:
(c) mitochondria

Question 7.
About which day in a normal human menstrual cycle does rapid secretion of LH (popularly called LH surge) normally occurs ?
(a) 14th day
(b) 20th day
(c) 5th day
(d) 11th day
Answer:
(a) 14th day

Question 8.
Morphogenetic movements occur during ……………….
(a) blastulation
(b) gastrulation
(c) fertilization
(d) cleavage
Answer:
(b) gastrulation

Question 9.
The technique used to block the passage of sperm in male is ……………….
(a) tubectomy
(b) vasectomy
(c) coitus interruptus
(d) rhythm method
Answer:
(b) vasectomy

Question 10.
Planaria reproduces asexually through ……………….
(a) budding
(b) gemmule formation
(c) regeneration
(d) binary fission
Answer:
(c) regeneration

Question 11.
The role of Leydig cells is ……………….
(a) nourishment of sperms
(b) to give motility to sperms
(c) synthesis of testosterone
(d) to undergo spermatogenesis
Answer:
(c) synthesis of testosterone

Question 12.
Chancre are the primary lesions caused by ……………….
(a) Neisseria gonorrhoeae
(b) Treponema pallidum
(c) Plasmodium vivax
(d) Salmonella typhi
Answer:
(b) Treponema pallidum

Question 13.
Smooth muscles lining the wall of scrotum are called ……………….
(a) detrusor muscles
(b) dartos muscles
(c) gluteal muscles
(d) latissimus dorsi muscles
Answer:
(b) dartos muscles

Question 14.
The trophoblast cells in contact with embryonal knob are called ……………….
(a) inner mass cells
(b) blastomere
(c) amniogenic cells
(d) cells of Rauber
Answer:
(d) cells of Rauber

Question 15.
The external layer of collagenous connective tissue of human testis is ……………….
(a) tunica vasculosa
(b) tunica vaginalis
(c) tunica granulosa
(d) tunica albuginea
Answer:
(d) tunica albuginea

Question 16.
Which of the following is mesodermal in origin ?
(a) Retina
(b) Enamel of teeth
(c) Heart
(d) Liver
Answer:
(c) Heart

Question 17.
Pregnancy in second trimester is maintained by ……………….
(a) LH (luteinizing hormone)
(b) progesterone
(c) estrogen
(d) hCG (human Chorionic Gonadotropin)
Answer:
(b) progesterone

Question 18.
In human foetus, the heart begins to beat at developmental age of ……………….
(a) 4th week
(b) 3rd week
(c) 6th week
(d) 8th week
Answer:
(c) 6th week

Question 19.
……………… contribute about 60% of the total volume of the semen.
(a) Prostate gland
(b) Cowper’s glands
(c) Seminal vesicles
(d) Bartholin’s glands
Answer:
(c) Seminal vesicles

Question 20.
Which of the following is hormone releasing IUD?
(a) Lippes loop
(b) Cu 7
(c) LNG 20
(d) Multiload 375
Answer:
(c) LNG 20

Question 21.
Which of the following is incorrect regarding vasectomy?
(a) Vasa deferentia is cut and tied
(b) Irreversible sterility
(c) No sperm occurs in seminal fluid
(d) No sperm occurs in epididymis
Answer:
(d) No sperm occurs in epididymis

Question 22.
The test-tube baby programme employs which one of the following techniques?
(a) Gamete intra fallopian transfer (GIFT)
(b) Zygote intra fallopian transfer (ZIFT)
(c) Intra cytoplasmic sperm injection (ICSI)
(d) Intra uterine insemination (IUI)
Answer:
(b) Zygote intra fallopian transfer (ZIFT)

Question 23.
Medical Termination of Pregnancy (MTP) is considered safe up to how many weeks of pregnancy?
(a) 8 weeks
(b) 12 weeks
(c) 24 weeks
(d) 6 weeks
Answer:
(b) 12 weeks

Question 24.
‘Saheli? an oral contraceptive pill is to be taken ……………….
(a) daily
(b) weekly
(c) quarterly
(d) monthly
Answer:
(b) weekly

Question 25.
The role of copper releasing IUDs is to ……………….
(a) inhibit ovulation
(b) prevent fertilization
(c) inhibit implantation of blastocyst
(d) inhibit gametogenesis
Answer:
(b) prevent fertilization

Question 26.
The phenomenon of nuclear fusion of sperm and egg is known as ……………….
(a) karyogamy
(b) parthenogenesis
(c) vitellogenesis
(d) oogenesis
Answer:
(a) karyogamy

Question 27.
Acrosome of spermatozoa is formed from ……………….
(a) lysosomes
(b) Golgi bodies
(c) ribosomes
(d) mitochondria
Answer:
(b) Golgi bodies

Question 28.
Which of the following undergoes spermiogenesis ?
(a) Spermatids
(b) Spermatogonia
(c) Primary spermatocytes
(d) Secondary spermatocytes
Answer:
(a) Spermatids

Question 29.
In mammals, the estrogens are secreted by the graafian follicle from its ……………….
(a) theca externa
(b) theca interna
(c) membrane granulosa
(d) corona radiata
Answer:
(b) theca interna

Question 30.
Which hormone is essential for maintenance of the endometrium of uterus?
(a) FSH
(b) LH
(c) Progesterone
(d) Estrogen
Answer:
(c) Progesterone

Question 31.
Which of the following cells during gametogenesis is normally diploid?
(a) Spermatid
(b) Spermatogonia
(c) Second polar body
(d) Secondary oocyte
Answer:
(b) Spermatogonia

Question 32.
Fertilization takes place at ……………….
(a) cervix
(b) ampulla
(c) isthmus
(d) vagina
Answer:
(b) ampulla

Question 33.
In mammals, failure of testes to descend into scrotum is known as ……………….
(a) paedogenesis
(b) castration
(c) cryptorchidism
(d) impotency
Answer:
(c) cryptorchidism

Maharashtra Board Class 12 Biology Important Questions Chapter 2 Reproduction in Lower and Higher Animals

Question 34.
Polar body is produced during the formation of ……………….
(a) sperm
(b) secondary oocyte
(c) oogonium
(d) spermatocytes
Answer:
(b) secondary oocyte

Question 35.
Menstrual flow occurs due to lack of ……………….
(a) vasopressin
(b) progesterone
(c) FSH
(d) oxytocin
Answer:
(b) progesterone

Question 36.
Approximately how many eggs are produced by a normal healthy human female up to the age of 25 years if the age of menarche is 12 years?
(a) 169
(b) 416
(c) 240
(d) 100
Answer:
(a) 169

Question 37.
In humans, at the end of the first meiotic division, the male germ cells differentiate into the ……………….
(a) spermatids
(b) spermatozoa
(c) primary spermatocytes
(d) secondary spermatocytes
Answer:
(d) Secondary spermotocytes

Question 38.
The part that carries sperms from testis to epididymis is ……………….
(a) rete testis
(b) vasa efferentia
(c) vasa differentia
(d) ejaculatory ducts
Answer:
(c) vasa differentia

Question 39.
Which period of menstrual cycle is called risky period of conception?
(a) 3rd to 7th day
(b) 7th to 13th day
(c) 10th to 17th day
(d) 15th to 25th day
Answer:
(c) 10th to 17th day

Question 40.
Which hormone confirms pregnancy?
(a) Progesterone
(b) Estrogen
(c) hCG
(d) LH
Ans
(c) hCG

Match the columns

Question 1.

Column I [Organs]Column II [Functions]
(1) Epididymis(a) Transport of sperms
(2) Sertoli cells(b) Copulatory organ
(3) Vas deferens(c) Nourishment to developing sperms
(4) Penis(d) Maturation of sperms

Answer:

Column I [Organs]Column II [Functions]
(1) Epididymis(d) Maturation of sperms
(2) Sertoli cells(c) Nourishment to developing sperms
(3) Vas deferens(a) Transport of sperms
(4) Penis(b) Copulatory organ

Question 2.

Column I [Organ/cells]Column II [Hormones]
(1) Corpus luteum(a) hCG
(2) Interstitial cells / Leydig’s cells(b) Estrogen
(3) Placenta(c) Progesterone
(4) Graafian follicle(d) Testosterone

Answer:

Column I [Organ/cells]Column II [Hormones]
(1) Corpus luteum(c) Progesterone
(2) Interstitial cells / Leydig’s cells(d) Testosterone
(3) Placenta(a) hCG
(4) Graafian follicle(b) Estrogen

Question 3.

Column IColumn II
(1) Graafian follicle(a) Site of implantation
(2) Uterus(b) Birth canal
(3) Fallopian tube(c) Site of fertilization
(4) Vagina(d) Release of secondary oocyte

Answer:

Column IColumn II
(1) Graafian follicle(d) Release of secondary oocyte
(2) Uterus(a) Site of implantation
(3) Fallopian tube(c) Site of fertilization
(4) Vagina(b) Birth canal

Question 4.

Column I [Phases]Column II [Hormonal changes]
(1) Menstrual phase(a) Rapid secretion of LH
(2) Proliferative phase(b) Increased level of FSH and estrogen
(3) Ovulatory phase(c) Increased level of progesterone
(4) Secretory phase(d) Decrease in progesterone and estrogen

Answer:

Column I [Phases]Column II [Hormonal changes]
(1) Menstrual phase(d) Decrease in progesterone and estrogen
(2) Proliferative phase(b) Increased level of FSH and estrogen
(3) Ovulatory phase(a) Rapid secretion of LH
(4) Secretory phase(c) Increased level of progesterone

Question 5.

Column IColumn II
(1) Acrosome(a) Completion of IInd meiotic division of secondary oocyte
(2) Penetration of sperm into ovum(b) Dissolution of zona pellucida
(3) Formation of fertilization membrane(c) Secretion of Hyaluronidase
(4) Acrosin / Zona lysine(d) Prevention of polyspermy

Answer:

Column IColumn II
(1) Acrosome(c) Secretion of Hyaluronidase
(2) Penetration of sperm into ovum(a) Completion of IInd meiotic division of secondary oocyte
(3) Formation of fertilization membrane(d) Prevention of polyspermy
(4) Acrosin / Zona lysine(b) Dissolution of zona pellucida

Question 6.

Column IColumn II
(1) Parturition(a) Attachment of embryo to endometrium
(2) Gestation(b) Release of egg from Graafian follicle
(3) Ovulation(c) Delivery of baby from uterus
(4) Implantation(d) Duration between pregnancy and birth

Answer:

Column IColumn II
(1) Parturition(c) Delivery of baby from uterus
(2) Gestation(d) Duration between pregnancy and birth
(3) Ovulation(b) Release of egg from Graafian follicle
(4) Implantation(a) Attachment of embryo to endometrium

Question 7.

Column I [Contraceptive method]Column II [Mode of action]
(1) Pill(a) Prevents sperms reaching cervix
(2) Condom(b) Prevents implantation
(3) Vasectomy(c) Prevents ovulation
(4) Copper T(d) Semen contains no sperms

Answer:

Column I [Contraceptive method]Column II [Mode of action]
(1) Pill(c) Prevents ovulation
(2) Condom(a) Prevents sperms reaching cervix
(3) Vasectomy(d) Semen contains no sperms
(4) Copper T(b) Prevents implantation

Question 8.

Column IColumn II
(1) Mechanical means(a) Saheli
(2) Physiological device(b) Jellies
(3) Chemical device(c) Vasectomy
(4) Permanent method(d) Diaphragm

Answer:

Column IColumn II
(1) Mechanical means(d) Diaphragm
(2) Physiological device(a) Saheli
(3) Chemical device(b) Jellies
(4) Permanent method(c) Vasectomy

Classify the following to form Column B as per the category given in Column A.

Question 1.
Classify the following contraceptives given below as per Column ‘A’ and complete Column ‘B’. Select from the given options:
(i) Foams
(ii) Lippe’s loop
(iii) Cervical caps
(iv) Multiload 375
(v) Diaphragms
(vi) Jellie

Column AColumn B
(1) Mechanical means————–, ————
(2) Chemical means————-, ————-
(3) Intra-uterine device————-, ————

Answer:

Column AColumn B
(1) Mechanical meansCervical caps, Diaphragms
(2) Chemical meansFoams, Jellies
(3) Intra-uterine deviceLippe’s loop, Multiload 375

Question 2.
Classify the following components of semen given below as per Column ‘A’ and complete the Column ‘B’. Select from the given options
(i) Acid phosphatase
(ii) Mucous like fluid
(iii) Prostaglandins
(iv) Citric acid
(v) Fructose
(vi) Fibrinogen

Column AColumn B
(1) Seminal fluid————–, ————
(2) Prostatic fluid————-, ————-
(3) Fluid from Cowper’s gland————-, ————

Answer:

Column AColumn B
(1) Seminal fluidProstaglandins, Fructose, Fibrinogen
(2) Prostatic fluidAcid phosphatase, Citric acid
(3) Fluid from Cowper’s glandMucous like fluid

Very short answer questions

Question 1.
How many sperms are present in single ejaculation?
Answer:
A single ejaculation contains about 400 millions of sperms.

Question 2.
What is gemmule? How is gemmule formed ?
Answer:
Gemmule is an internal bud formed by aggregation of archeocytes in sponges to overcome unfavourable season.

Question 3.
What is cryptorchidism?
Answer:
Failure of testis to descend into scrotum leading to sterility is called cryptorchidism.

Maharashtra Board Class 12 Biology Important Questions Chapter 2 Reproduction in Lower and Higher Animals

Question 4.
What is the beginning of the menstrual cycle and cessation of menstrual cycle respectively called?
Answer:
The beginning of the menstrual cycle is called menarche while cessation of menstrual cycle is called menopause.

Question 5.
Which men have an increased risk of prostate cancer?
Answer:
Men who are over 50 years of age and have a daily high consumption of fat have an increased risk of prostate cancer.

Question 6.
What is capacitation with reference to sperm?
Answer:
Changes in a mammalian sperm which prepare it for fertilization of ovum is called capacitation.

Question 7.
Give any two examples each of seasonal breeders and continuous breeders among sexually reproducing animals.
Answer:
Example of seasonal breeders : Goat, Sheep and Donkey.
Example of continuous breeders : Humans, apes.

Question 8.
What does IUCD indicate?
Answer:
IUCD means Intra Uterine Contraceptive Device.

Question 9.
What is full form of IVF?
Answer:
IVF means In Vitro Fertilization.

Question 10.
From which germinal layers the nervous system is derived?
Answer:
The nervous system is derived from ectoderm.

Question 11.
A mother of a one-year-old child wanted to space her second child. Her doctor suggested ‘Copper-T’. Explain its contraceptive action.
Answer:
Copper ions released from ‘Copper-T’ suppress sperm motility and the fertilizing capacity of sperms.

Question 12.
Which options are available for infertile couples to have child?
Answer:
Infertile couples have many options to have a child such as fertility drugs, modern techniques such as IVE ZIFT, GIFT, ICSI, artificial insemination, IUI, using surrogate mother or taking the sperm from sperm bank.

Question 13.
How many primary spermatocytes and oocytes are required for the formation of 100 spermatozoa and ova?
Answer:
25 Primary spermatocytes and 100 primary oocytes will be required for the formation of 100 spermatozoa and ova respectively.

Question 14.
The entrance of fallopian tube of a lady is blocked. She wants motherhood. Which method will help her?
Answer:
The method of GIFT or Gamete Intra Fallopian Transfer is the method that will help the lady to have a child.

Question 15.
What is the role of birth control pills?
Answer:
Birth control pills are contraceptive pills that check the ovulation by inhibiting the secretion of FSH and LH.

Question 16.
In T.S. of ovary, can all stages of follicles be seen simultaneously?
Answer:
In T.S. of ovary, all the stages of follicles cannot be seen simultaneously. The stage of follicles develop alternately in the ovary as per timing of menstrual cycle under the influence of hormones of pituitary and ovaries.

Question 17.
What will be marriageable age for boy and girl as per the Indian law?
Answer:
As per the Hindu Marriage Act, minimum age for boy must be 21 years and for a girl 18 years, at the time of marriage.

Question 18.
What is MTP Act?
Answer:
MTP Act is for reducing the incidences of illegal abortions and maternal mortalities.

Question 19.
Which is the time period legally allowed by MTP ACT for terminating pregnancy?
Answer:
According to MTP Act, pregnancy may be terminated within first 12 weeks, more than 12 weeks but lesser than 20 weeks.

Give definitions of the following

Question 1.
Amphimixis
Answer:
It is the process which involves the production of offspring by the formation and fusion of gametes.

Question 2.
Gametogenesis
Answer:
The gametogenesis is the process of formation of gametes in sexually reproducing animals.

Question 3.
Spermiogenesis
Answer:
The process of transformation of non-motile and non¬functional spermatid into a functional and motile spermatozoa is called spermiogenesis.

Question 4.
Insemination
Answer:
The process of deposition of semen into the vagina of the female at the time of coitus or sexual intercourse is called insemination.

Question 5.
Cleavage
Answer:
The process of early mitotic division of the zygote to form a multicellular morula stage is called cleavage.

Question 6.
Implantation
Answer:
The process by which the blastocyst after its formation, gets implanted or embedded into the endometrium of the uterus is called implantation.

Question 7.
Gestation
Answer:
The condition of carrying one or more embryos in the uterus is called gestation.

Question 8.
Placenta
Answer:
A flattened, discoidal organ present in the uterus of pregnant mother and which acts as endocrine source and nutrition provider for growing foetus is called placenta.

Question 9.
Lactation
Answer:
The process of secretion of milk in the mammary glands and expelling it through nipples out to provide nourishment to the growing baby is called lactation.

Question 10.
Parturition
Answer:
Parturition is the process of giving birth to a baby.

Question 11.
Amniocentesis
Answer:
Amniocentesis is a process in which amniotic fluid containing foetal cells is collected using a hollow needle inserted into the uterus under ultrasound guidance.

Question 12.
Infertility
Answer:
Infertility is defined as the inability to conceive naturally after (one year of) regular unprotected intercourse.

Question 13.
IVF (In Vitro Fertilization)
Answer:
It is a process of fertilization where an egg is combined with sperm outside the body in a test tube or glass plate to form a zygote under simulated conditions in the laboratory.

Question 14.
Artificial Insemination (AI)
Answer:
It is the technique during which the sperms are collected from the male and artificially introduced into the cervix of female, for the purpose of achieving a pregnancy through in vivo fertilization (inside the body).

Question 15.
Adoption
Answer:
Adoption is a legal process by which a couple or a single parent gets legal rights, privileges and responsibilities that are associated to a biological child for the upbringing of the adopted child.

Give functions of the following

Question 1.
Corpus luteum.
Answer:
Corpus luteum is a secondary endocrine source that produces progesterone for maintaining pregnancy.

Question 2.
Scrotum.
Answer:
Scrotum protects the testis and also acts as thermoregulator.

Question 3.
Acrosome of sperm.
Answer:
Acrosome of the sperm releases hyaluronidase which digests the zona pellucida surrounding the ovum due to which sperm can fertilize the ovum.

Question 4.
Sertoli cells.
Answer:
Sertoli cells provide nourishment and surface to the sperm bundles during their development.

Question 5.
Interstitial cells / Leydig’s cells.
Answer:
Interstitial cells / Leydig’s cells secrete testosterone or androgen which is a male sex hormone.

Question 6.
Prostate gland.
Answer:
Prostate gland secretes prostatic fluid which forms 30% of semen, Citric acid and acid phosphatase present in this fluid protects the sperms from acidic environment of vagina.

Question 7.
Bulbourethral glands.
Answer:
Bulbourethral glands secrete alkaline, viscous mucus like fluid which provides lubrication during copulation.

Question 8.
Bartholin’s glands.
Answer:
Bartholin glands secrete lubricating mucus like fluid which is released in vestibule.

Question 9.
Uterus.
Answer:

  1. Uterus receives ovum from fallopian tubes, develops placenta and provides site for implantation of embryo.
  2. It provides protection and nourishment to the developing embryo.
  3. It also provides path for sperms to ascend.
  4. Due to contractions of uterus, baby is expelled out at the time of parturition.

Question 10.
Vagina
Answer:

  1. Vagina acts as a copulatory passage.
  2. It acts as a birth canal during parturition in normal delivery
  3. It provides the passage for menstrual flow.

Name the following

Question 1.
The canal through which the testes descend into scrotum just before birth in human male child.
Answer:
Inguinal canal

Question 2.
The structure where sperms are matured.
Answer:
Epididymis

Question 3.
The part where the sperms are produced in the testes.
Answer:
Germinal epithelium of seminiferous tubules.

Question 4.
The gland in females homologous to Cowper’s gland.
Answer:
Bartholin’s glands or Vestibular glands.

Question 5.
Type of cleavage in human zygote
Answer:
Holoblastic, radial and indeterminate

Question 6.
The developmental stage of human being which gets implanted in the endometrium of uterus.
Answer:
Blastocyst

Question 7.
Name the primates who show presence of menstrual cycle.
Answer:
Human being and Apes like gorilla, chimpanzee, orangutan, etc.

Question 8.
Structures which help in transport of secondary oocyte through uterine tube.
Answer:
Ciliated epithelium

Question 9.
Hormones produced in women only during pregnancy.
Answer:
hCG, HPL (Human Placental Lactogen) and relaxin.

Question 10.
The oral contraceptive pill which is now. a part of the National Family Programme in India.
Answer:
Saheli

Question 11.
The scientific term for the animals giving birth to live young ones.
Answer:
Viviparous

Maharashtra Board Class 12 Biology Important Questions Chapter 2 Reproduction in Lower and Higher Animals

Question 12.
The site of fertilization in woman.
Answer:
Ampulla of fallopian tube

Question 13.
The trophoblast cells lying over the embryonal knob.
Answer:
Cells of Rauber

Question 14.
The muscles which form the wall of scrotum.
Answer:

  1. Dartos muscles
  2. Cremaster muscles

Question 15.
Names of erectile tissues in penis.
Answer:

  1. Corpora cavernosa
  2. Corpus spongiosum

Question 16.
Any two copper releasing IUD.
Answer:

  1. Copper-T, Cu 7
  2. Multiload 375

Question 17.
Any two hormone-releasing IUDs.
Answer:

  1. LNG-20
  2. Progestaert

Question 18.
Two methods of birth control which have high chances of failure.
Answer:

  1. Safe period
  2. Lactational amenorrhea

Question 19.
Uterine walls.
Answer:

  1. Perimetrium
  2. Myometrium
  3. Endometrium

Question 20.
Regions of the uterus.
Answer:

  1. Fundus
  2. Body
  3. Cervix

Question 21.
Parts of fallopian tubes.
Answer:

  1. Infundibulum
  2. Ampulla
  3. Isthmus

Question 22.
Layers of Graafian follicle which enclose antrum.
Answer:

  1. Theca externa
  2. Theca interna
  3. Membrana granulosa

Question 23.
Stages of cells in spermatogenesis.
Answer:

  1. Spermatogonia
  2. Primary spermatocytes
  3. Secondary spermatocytes
  4. Spermatids
  5. Sperms

Question 24.
Stages of cells in oogenesis.
Answer:

  1. Oogonia
  2. Primary oocytes
  3. Secondary oocytes
  4. Ootid
  5. Ovum

Give significance of the following

Question 1.
Fertilization.
Answer:
Significance of fertilization:

  1. Fertilization forms the zygote which eventually produces new offspring.
  2. Fertilization restores diploid number of chromosomes in the zygote as two haploid gametes come together in a zygote.
  3. During fertilization, centrioles are passed on to the ovum, due to this secondary oocyte can complete meiosis-II. The fertilization thus concludes the process of oogenesis.
  4. By fertilization the genetic characters of two parents are mixed. This leads to variation and has significance in evolution.
  5. Due to fertilization the sex of young one is determined.

Question 2.
Implantation.
Answer:
Gestation becomes possible due to implantation. Implantation protects the embryo and helps it to derive nourishment from the mother’s body through placenta.

Question 3.
Corpus luteum.
Answer:

  1. Corpus luteum is the temporary source of female hormone, progesterone.
  2. Corpus luteum is formed from empty Graafian follicle after the process of ovulation.
  3. Due to progesterone secreted from corpus luteum, the endometrial wall of uterus undergoes repair and increase in thickness.
  4. Progesterone is a gestational hormone and thus pregnancy is maintained if corpus luteum is well functional.

Question 4.
Fertilization membrane.
Answer:
Fertilization membrane prevents any further entry of other sperms into the egg, i.e. polyspermy is avoided.

Question 5.
Gastrulation.
Answer:

  1. Due to the process of gastrulation, three germinal layers, viz. ectoderm, mesoderm and endoderm are formed.
  2. Cells of embryonal knob become embryonic disc which develop into embryo due to gastrulation.
  3. Gastrulation is necessary for the formation of amniotic cavity which is filled with amniotic fluid.

Question 6.
Trophoblast of blastocyst.
Answer:

  1. Trophoblast cells help in absorbing nutrition for the developing embryo.
  2. Trophoblast cells at the embryonal knob (cells of Rauber) help in implantation of blastocyst.
  3. Synctiotrophoblast helps in implantation of fertilized ovum in the uterine endometrium.

Question 7.
hCG [human chorionic gonadotropin].
Answer:
hCG [human chorionic gonadotropin] is secreted in the pregnant female to extend the life of corpus luteum and stimulates its secretory activity. Presence of hCG in maternal blood and urine is an indication of pregnancy.

Question 8.
Colostrum.
Answer:

  1. Colostrum is the first milk which is sticky and yellowish secreted by the mammary glands soon after the parturition.
  2. Being high protein in its content, it nourishes the newly born child.
  3. The antibodies present in it helps in developing resistance for the newborn baby at a time when its own immune response is not fully developed.

Distinguish between the following

Question 1.
Asexual reproduction and Sexual reproduction.
Answer:

Sexual reproductionAsexual reproduction
1. Asexual reproduction requires single parent.1. Sexual reproduction needs two different parents.
2. Meiosis does not take place in asexual reproduction. Only mitosis takes place.2. Sexual reproduction involves meiosis and mitosis.
3. Gamete formation, fertilization and zygote formation does not take place.3. Gamete formation, fertilization and zygote formation are important processes in sexual reproduction.
4. Progeny and parent Eire identical genetically.4. Progeny and parents are genetically dissimilar.
5. Large number of progeny is developed by asexual reproduction. E.g. Spore formation, gemmule formation, budding, regeneration are the types of a sexual reproduction.5. Limited number of progeny is developed by sexual reproduction. E.g. Sexual reproduction is only by a single method.

Question 2.
Primary sex organs and Secondary sex organs.
Answer:

Primary sex organsSecondary / Accessory sex organs
1. Primary sex organs produce gametes.1. Secondary sex organs do not produce gametes.
2. Primary sex organs secrete sex hormones.2. Secondary sex organs do not secrete sex hormones.
3. Development of these organs is under the control of Gonadotropins released from Pituitary.
E.g. Testes in male and Ovaries in females.
3. Development of these organs is under the control of estrogen and progesterone in females and testosterone in males.
Eg. Prostate, seminal vesicles, vas deferens in males. Fallopian tubes, uterus and vagina in females.

Question 3.
Vasa efferentia and Vasa deferentia.
Answer:

Vasa efferentiaVasa deferentia
1. Vasa efferentia arise from the rete testes and enter the epididymis.1. Vasa deferentia arise from the epididymis and form ejaculatory duct after the union with seminal duct.
2. They are present in 15-20 number and are fine convoluted ductules.2. They are thick and coiled ductules present in a single pair.
3. The spermatozoa are carried from rete testis to epididymis by vasa efferentia.3. The spermatozoa are carried from epididymis to ejaculatory ducts by vasa deferentia.

Question 4.
Graafian follicle and Corpus luteum.
Answer:

Graafian follicleCorpus luteum
1. Graafian follicle is produced by the maturation of the primary follicle.1. Corpus luteum is produced by the cells of ruptured Graafian follicle.
2. It is formed in the ovary before ovulation.2. It is formed in the ovary after ovulation.
3. It produces the hormone estrogen.3. It produces the hormone progesterone.
4. It has secondary oocyte surrounded by follicle cells.4. It has only follicle cells.

Maharashtra Board Class 12 Biology Important Questions Chapter 2 Reproduction in Lower and Higher Animals

Question 5.
Menarche and Menopause.
Answer:

MenarcheMenopause
1. Menarche is the beginning of menstrual cycle.1. Menopause is the stoppage of menstrual cycle.
2. Menarche is at the age of 10 to 14.2. Menopause is at the age of 45 to 50.
3. Menarche begins with secretion of FSH and LH.3. Menopause is caused due to decline of FSH and LH secretion.
4. Menarche is the beginning of the reproductive period.4. Menopause is the end of the reproductive period.

Question 6.
Proliferative Phase and Secretory Phase.
Answer:

Proliferative PhaseSecretory Phase
1. Proliferative phase begins with the repair of endometrium.1. Secretory phase begins with ovulation.
2. Time required for proliferative phase is 5th to 13th day of menstrual cycle.2. Time required for secretory phase is 15th to 28th day of menstrual cycle.
3. Proliferative phase always ends with ovulation.3. Secretory phase ends with menstruation if egg is not fertilized. It continues further if egg is fertilized.
4. Proliferative phase is in uterus which coincides . with follicular phase in ovary during which there is formation of Graafian follicle.4. Secretory phase is in uterus which coincides with luteal phase in ovary during which there is formation of corpus luteum.
5. Proliferative phase is controlled by FSH from anterior pituitary.5. Secretory phase is controlled by LH from anterior pituitary.
6. Hormone estrogen is secreted during this phase.6. Hormone progesterone is secreted during this phase.
7. It causes the development of blood vessels and thickening of endometrium of uterus.7. It causes further thickening and secretory activity of the glands of endometrium of uterus.

Question 7.
Spermatogenesis and Oogenesis.
Answer:

SpermatogenesisOogenesis
1. Spermatogenesis takes place in testis in mature and fertile males.1. Oogenesis takes place in ovaries in mature and fertile females.
2. From one spermatogonium four haploid sperms are formed during spermatogenesis.2. From one oogonium one haploid ovum and a polar body is formed during oogenesis.
3. Spermatid developed undergoes metamorphosis in the process of spermiogenesis.3. There is no such process of metamorphosis in oogenesis.
4. Spermatid development takes place which later becomes a functional sperm.4. Ootid development does not take place during oogenesis. It develops only after fertilization.
5. Spermatogonia, primary and secondary spermatocytes and spermatid are the stages of sperms formed during spermatogenesis.5. Oogonia, primary and secondary oocytes are the stages formed during oogenesis. Ootid formation occurs only after fertilization.

Question 8.
Zona pellucida and Corona radiata.
Answer:

Zona pellucidaCorona radiata
1. Zona pellucida is inner, thin and transparent layer surrounding the secondary oocyte.1. Corona radiata is the outer thick layer surrounding the secondary oocyte.
2. Zona pellucida is a non-cellular layer.2. Corona radiata is a cellular layer.
3. Zona pellucida is secreted by the ovum itself.3. Corona radiata is formed by follicular cells which are glued together by hyaluronic acid.
4. Zona pellucida is retained for more time after fertilization till the ovum gets implanted in the uterus.4. Corona radiata is retained till the ovum gets fertilized.
5. Zona pellucida is digested by zona lysine or acrosin at the time of fertilization.5. Corona radiata is digested by hyaluronidase enzyme at the time of fertilization.

Question 9.
Morula and Blastula.
Answer:

MorulaBlastula
1. Morula is the embryonic stage formed after the completion of cleavage.1. Blastula is the embryonic stage formed after the completion of blastulation.
2. Morula is formed 4 to 6 days after the fertilization.2. Blastula is formed 6 to 7 days after the fertilization.
3. Morula consists of 16 cells.3. Blastula consists of more than 64 cells.
4. Morula is solid ball of cells.4. Blastula is a hollow ball of cells.
5. Morula stage is passed in fallopian tube, once it reaches uterus, it starts developing into the next stage.5. Blastula after reaching the uterus is implanted on the wall of uterus.
6. Morula does not have any distinction of its inner cell structure.6. Blastula has a blastocoel, trophoblast and inner cell mass.

Question 10.
Blastula and Gastrula
OR
Give two differences between blastula and gastrula.
Answer:

BlastulaGastrula
1. Blastula is formed from morula on 7th day after fertilization.1. Gastrula is formed from blastula 15 days after fertilization.
2. Blastula has a blastocoel.2. Gastrula has a gastrocoel or archenteron.
3. Blastula is produced by the process of blastulation.3. Gastrula is produced by the process of gastrulation.
4. Blastula undergoes implantation followed by gastrulation.4. Gastrula undergoes morphogenesis and then forms germs layers.
5. During blastula formation there is no movement of cells.5. Gastrula formation results from the morphogenetic movement of cells.

Give reasons

Question 1.
Testes are located outside the body cavity in scrotal sacs.
Answer:

  1. During early foetal life, the testes develop in the lumbar region of the abdominal cavity just below the kidney but during seventh month of development, they descend permanently into the respective scrotal sacs through a passage called inguinal canal.
  2. For the development of the sperm, lesser temperature than the body temperature is required.
  3. If the testes remain in the abdominal cavity, then the sperm production does not take place.
  4. This may result in impotency. Therefore, testes are located outside the body cavity.

Question 2.
Urethra is also called urinogenital duct in males.
Answer:

  1. Urinogenital duct means common duct for urine and the genital products.
  2. In males, the penis lodges urethra throughout its entire length, through which urine as well as semen are given out of the body during urination or copulation.
  3. Since the urethra carries both urine and semen, it is called urinogenital duct.

Question 3.
Proliferative phase is also called follicular phase.
Answer:

  1. Proliferative phase means there is proliferation of endometrial cells in the uterus. Follicular means there is growth of ovarian follicles in the ovaries. Both these phases are simultaneous.
  2. The follicular phase of ovaries is due to effect of FSH from adenohypophysis.
  3. The ovaries follicles grow due to FSH and start secreting estrogen.
  4. This estrogen from ovaries bring proliferative effect on the uterus.

Question 4.
Missing of menses is the first indication of pregnancy.
Answer:

  1. Menstruation occurs if there is no fertilization of ovum.
  2. The endometrium of uterus along with unfertilized egg is given out in the form of menstrual flow.
  3. The sloughing off uterine endometrium takes place due to degeneration of corpus luteum.
  4. In the absence of functional corpus luteum progesterone levels fall down. However, if the ovum is fertilized, the corpus luteum is maintained and it secretes progesterone which maintains the uterine endometrium. In such case, further growth of ovarian follicles and ovulation remains suspended and woman is said to be pregnant.
  5. Endometrial wall of uterus now thickens and helps in the growth of placenta. Thus during pregnancy, menses will not take place.

Question 5.
Progesterone is called pregnancy hormone.
Answer:

  1. Progesterone is secreted from corpus luteum which is formed from empty ovarian follicle after the ovulation.
  2. Progesterone has the capacity to maintain pregnancy.
  3. It acts on uterine endometrium and causes it to proliferate and develop in thickness.
  4. Corpus luteum keeps on secreting progesterone till the placenta takes up the function of secreting the same.

Question 6.
Human female has restricted reproductive life.
Answer:

  1. In human female, the reproductive period is about 30 – 33 years.
  2. There is menarche at the age of about 13 and menopause at the age of 45-50.
  3. During this span of 30 years, ovaries secrete sex hormones like estrogen and progesterone. After menopause this secretion is suspended.
  4. Due to changes in hormonal level, human females cannot produce eggs later. Moreover, eggs in her ovaries are utilized by the age of 45.
  5. Human female, therefore, has restricted reproductive period.

Question 7.
Zona pellucid is retained for sometime after fertilization.
Answer:

  1. Fertilization of the ovum takes place in fallopian tube where it starts cleavages immediately.
  2. Zona pellucida which remains on the surface of the ovum prevents the implantation of the blastocyst at an abnormal site such as fallopian tube.
  3. Zona pellucida keeps the sticky and phagocytic trophoblast cells unexposed till the ovum reaches the uterine lumen.
  4. Zona pellucida also protects the ovum. Therefore zona pellucida is retained for some time after fertilization.

Question 8.
The acrosome of sperm secretes an enzyme called hyaluronidase at the time of fertilization.
Answer:

  1. The enzyme hyaluronidase secreted by acrosome of sperm dissolves the membranous covering of the ovum to facilitate the entry of sperm into the ovum.
  2. It is a lytic enzyme causing lysis of egg membrane.
  3. Owing to this, the acrosome of sperm secretes an enzyme called hyaluronidase at the time of fertilization.

Question 9.
The middle part of the human sperm is characterized by the presence of a number of mitochondria.
Answer:

  1. Mitochondria provide energy required by sperms for their agile movement.
  2. The agile movement of sperms helps them to reach the vicinity of the ovum at the time of fertilization.
  3. Owing to this, the middle part of the human sperm is characterised by the presence of a number of mitochondria.

Question 10.
The size of morula remains almost same as that of ovum.
Answer:

  1. The layer zone pellucida is retained around the embryo and thus, there is no change in the overall size from zygote to morula.
  2. This layer is important because it prevents the implantation of the blastocyst at an abnormal site such as fallopian tube.
  3. Though the number of blastomeres increase, the size of morula remains almost same as that of ovum till it reaches the uterus by the end of the day 4.

Question 11.
Placenta serves as the nutritive, respiratory and excretory organ of the embryo.
Answer:

  1. Between the foetus and mother there is exchange of several materials. Food in the form of glucose, amino acids, simple proteins, lipids, mineral, salts, vitamins and hormones, antibodies, etc. is sent to foetus by maternal circulation.
  2. Oxygen from mother’s blood is also given to the foetus.
  3. The foetal metabolic wastes such as carbon dioxide, urea and water pass from foetus into the maternal blood.
  4. This exchange takes place through the placenta.
  5. In the placenta, foetal blood comes very close to maternal blood to permit these exchanges. Therefore placenta is said to serve as the nutritive, respiratory and excretory organ of the embryo.

Write short notes

Question 1.
Graafian follicle.
Answer:
Maharashtra Board Class 12 Biology Important Questions Chapter 2 Reproduction in Lower and Higher Animals 1

  1. Graafian follicle is a mature ovarian follicle.
  2. There are following protective layers on the Graffian follicle : The outermost protective and fibrous covering, theca externa. Theca interna is the next layer which can secrete hormone estrogen.
  3. Next to theca interna, there is membrana granulosa which forms discus proligerous and the corona radiata layer.
  4. Graafian follicle contains an eccentric secondary oocyte. The oocyte is surrounded by a vitelline membrane which produces zona pellucida layer.
  5. In the centre there is antrum which is filled with liquor folliculi fluid.

Question 2.
Mammary glands.
Answer:

  1. Mammary glands are accessory organs of female reproductive system. These glands are essential for lactation after parturition.
  2. They are modified sweat glands present in the subcutaneous tissue of the anterior thorax. They are in the pectoral region in the location between 2nd to 6th rib.
  3. Each mammary gland consists of fatty connective tissue and many lactiferous ducts.
  4. Each breast has glandular tissue which is divided into 15-20 irregularly shaped mammary lobes. Each lobe has an alveolar glands and lactiferous duct.
  5. Milk is secreted by alveolar glands and it is stored in the lumen of alveoli. The alveoli open into mammary tubules and these in turn forms a mammary duct.
  6. All the lactiferous ducts converge towards the nipple.
  7. Nipple is surrounded by a dark brown coloured and circular area of the skin called areola.

Maharashtra Board Class 12 Biology Important Questions Chapter 2 Reproduction in Lower and Higher Animals

Question 3.
Structure of sperm.
Answer:

  1. Sperm is microscopic, elongated haploid motile male gamete produced by spermatogenesis.
  2. It measures to about 0.055 mm or 60y in length.
  3. The sperm consists of head, neck, middle piece and tail.

Maharashtra Board Class 12 Biology Important Questions Chapter 2 Reproduction in Lower and Higher Animals 2
Head:

  1. Head is the main part which is flat and oval and has a large nucleus and an acrosome.
  2. Acrosome is formed from Golgi complex. It secretes enzyme hyaluronidase which helps in penetration of the egg during fertilization.
  3. The acrosome and anterior half of nucleus is covered by a fibrillar sheath.

Neck : Neck is short region having two centrioles.

  1. The proximal centriole plays a role in first cleavage of zygote.
  2. The axial filament of the sperm is formed by the distal centriole.

Middle piece:

  1. Middle piece acts as a power house for sperm.
  2. It bears many spirally coiled mitochondria or Nebenkern around the axial filament.
  3. The mitochondria supply energy for the sperm to swim in the female genital tract with a speed of about 1.5 to 3 mm per minute.
  4. Posterior half of nucleus, neck and middle piece of sperm are covered by a sheath.

Tail:

  1. The tail is formed of cytoplasm and is long, slender and tapering structure.
  2. The axial filament is a fine thread-like structure that arises from the distal centriole and traverses the middle piece and tail.
  3. Nine accessory fibres are present surrounding the two central longitudinal axial filaments.
  4. Tail lashes and helps the spermatozoa to swim.

Question 4.
Structure of secondary oocyte.
Answer:
Maharashtra Board Class 12 Biology Important Questions Chapter 2 Reproduction in Lower and Higher Animals 3

  1. The unfertilized egg released through ovary at the time of ovulation is a secondary oocyte.
  2. It is rounded, non-motile and haploid, non- cleidoic and microlecithal female gamete.
  3. The size is approximately 0.1 mm (100 microns).
  4. It has abundant cytoplasm called ooplasm which contains a large eccentric and prominent nucleus called germinal vesicle.
  5. Centrioles are absent in secondary oocyte.
  6. Various coverings seen around the oocyte are (i) vitelline membrane (ii) zona pellucida (iii) Corona radiata.
  7. The cells are glued together with hyaluronic acid. Between vitelline membrane and zona pellucida, there is perivitelline space which lodges first polar body. This end is called 5 animal pole and the opposite is called vegetal pole.

Question 5.
Implantation
Answer:

  1. Implantation is the process by which the blastocyst is embedded into the endometrium of uterus in the fundus region.
  2. The implantation starts 7 days after fertilization and completed by the end of 10th day.
  3. The trophoblast cells of blastocyst at the embryonal knob can stick to the uterine endometrium. The trophoblast layer then divides into inner cytotrophoblast and outer syncytiotrophoblast due to contact with endometrial cells.
  4. Cytotrophoblast is the inner layer whose cells retain their cell boundaries.
  5. Syncytiotrophoblast is the outer layer of cells without plasma membrane. The cells of syncytiotrophoblast appear multinucleate. This layer projects invasively into the endometrium and destroys endometrial cells by releasing lytic enzymes. Due to this blastocyst is buried deeply in the endometrium.

Question 6.
Fate of three germinal layers.
Answer:
Fate of germinal layers : The embryo after gastrulation develops the three germ layers, viz., ectoderm, mesoderm and endoderm. Later a process of histogenesis starts which leads to the development of different tissues and organs.
(i) Fate of ectoderm : Following tissues, structures and organs develop from the ectoderm : Epidermis of the skin, epidermal derivatives such as

  1. hair and nails
  2. sweat glands
  3. conjunctiva
  4. cornea
  5. lens
  6. retina
  7. internal and external ear
  8. enamel of teeth
  9. nasal cavity
  10. adrenal medulla
  11. stomodaeum and proctodaeum
  12. neurohypophysis and
  13. entire nervous system.

(ii) Fate of mesoderm : The mesoderm forms the following derivatives:

  1. All types of muscles
  2. connective tissue
  3. dermis of skin
  4. adrenal cortex
  5. kidney
  6. circulatory system
  7. heart
  8. blood vessels
  9. blood
  10. lymphatic vessels
  11. middle ear and
  12. dentine of teeth.

(iii) Fate of endoderm : The following organs develop from the endoderm:

  1. Epithelium of gut from pharynx to colon
  2. glands of stomach and intestine
  3. tongue and tonsils
  4. lungs, trachea, bronchi, larynx, etc.
  5. urinary bladder, vestibule and vagina
  6. liver and pancreas
  7. adenohypophysis
  8. thymus, thyroid and parathyroid
  9. eustachian tube
  10. epithelium of urethra and associated glands.

Question 7.
Placenta.
Answer:

  1. Placenta is a temporary organ derived from the tissues of the foetus as well as mother.
  2. Human placenta is called chorionic placenta as it is made up of chorion which is an extra-embryonic membrane.
  3. Blood vessels from the allantois vascularize the placenta. Branching villi emerge from the chorion and penetrate in the corresponding pits which are located in the uterine wall.
  4. There are two parts of placenta, viz. foetal placenta and maternal placenta.
  5. Foetal placenta is formed of chorionic villi.
  6. Maternal placenta is formed of uterine wall which is in intimate contact with the chorionic villi.
  7. Chorionic villi receive the blood from the embryo by umbilical artery. Umbilical vein returns the blood back to the embryo.
  8. Human placenta is said to be haemochorial because a part of placenta is from foetus which has chorionic villi. The other highly vascularized part is from uterine wall of mother. Thus foetal and maternal placenta together is called haemocorial placenta.

Question 8.
Intratuterine devices (IUDs).
Answer:

  1. IUDs are plastic or metal objects which act as contraceptive devices. They are placed into the uterus by a doctor or trained nurse.
  2. E.g. Lippe’s loop, copper releasing IUDs (Cu-T, Cu-7, multiload 375) and hormone releasing IUDs (LNG-20, Progestaert).
  3. Plastic double ‘S’ loop is called Lippe’s loop which stimulates accumulation of macrophages in the uterine cavity by attracting them. As phagocytosis increases the sperms are destroyed. Thus it acts as a contraceptive.
  4. Copper releasing IUDs suppress sperm motility and the fertilizing capacity of sperms.
  5. The hormone releasing IUDs make the : uterus unsuitable for implantation and ; cervix hostile to the sperms.
  6. Their presence in the uterus acts as a minor irritant and thus makes the ovum to move quickly out of the body.
  7. However, IUD can cause infection and occasional haemorrhage. It can cause discomfort for woman and may get spontaneously expelled out.

Question 9.
Physiological (Oral) Contraceptive « Devices.
Answer:

  1. Physiological devices are in the form of oral contraceptive pills or birth control pills. 5 They are hormonal preparations and check ovulation by inhibiting the secretion of follicle stimulating hormone and luteinizing hormone.
  2. Woman who is using pills does not release ovum at the time of ovulation and therefore conception does not occur.
  3. Birth control pills have side effects such as nausea, breast tenderness, weight gain and ‘break through’ bleeding, i.e. slight bleeding between the menstrual periods. These health hazards are due to synthetic hormones.
  4. These pills also alter the quality of cervical J mucus to prevent the entry of sperms.
  5. The birth control pills contain progesterone and estrogen. Mala-D to be taken daily and Saheli to be taken weekly are two common birth control pills in India. These pills are non-steroidal.

Question 10.
Fate of trophoblast cells of blastocyst.
Answer:

  1. Trophoblast cells do not form any part of the embryo proper.
  2. They form ectoderm of the extra-embryonic membrane called chorion.
  3. Chorion helps in supply of oxygen and nutrients to foetus from mother’s body. CO2 and nitrogenous wastes are collected from foetus and passed in mother’s blood.
  4. Thus, these cells have an important role in formation of placenta.

Question 11.
Medical Termination of Pregnancy (MTP).
Answer:

  1. MTP or Medical Termination of Pregnancy is voluntary termination of pregnancy under medical supervision. It is an induced abortion.
  2. Only during first trimester, MTP is safe for mother’s health.
  3. Upon amniocentesis examination, if abnormality is detected, usually MTP is performed.
  4. Government of India has legalized MTP There was MTP Act in 1971, which was later amended in 2017, to prevent its misuse, especially female foeticide should never be done through MTP
  5. As per MTP Act, the procedure can be done only in first 12 weeks and never after 20 weeks of pregnancy.

Question 12.
Amniocentesis
Answer:

  1. In amniocentesis, amniotic fluid containing foetal cells is collected using a hollow needle. This needle is inserted into the uterus of pregnant mother, under ultrasound guidance.
  2. The chromosomes from the foetal cells are sujected to karyotyping. This helps to detect abnormalities in the developing foetus.
  3. Amniocentesis is misused to determine the sex of the unborn child. This is illegal in India because it results into female foeticide.
  4. Another risks involved in amniocentesis are miscarriage, needle injury to foetus, leaking amniotic fluid, infection, etc.
  5. As per MTP Act (1971) the misuse of amniocentesis is curtailed.

Maharashtra Board Class 12 Biology Important Questions Chapter 2 Reproduction in Lower and Higher Animals

Question 13.
ZIFT [Zygote Intra Fallopian Transfer].
Answer:

  1. If there is a blockage in the fallopian tubes due to which fertilization is prevented, then ZIFT treatment is used.
  2. The oocyte is removed form woman’s ovary. This oocyte is fertilized outside the body under sterile conditions with the known sperms. This forms zygote. This is In Vitro Fertilization (IVF).
  3. Later the zygote is transferred in fallopian tube to achieve pregnancy.

Question 14.
GIFT [Gamete Intra Fallopian Transfer].
Answer:

  1. When the oocyte is collected from donor and transferred into the fallopian tube of another female, the technique is called GIFT. This female provides suitable environment for further development.
  2. When the entrance or upper segments of the fallopian tubes is blocked, this technique is used.
  3. Ooocytes and sperms are directly injected into regions of the fallopian tubes. Here fertilization takes place forming a blastocyst. It later enters the uterus for implantation.
  4. GIFT is successful in only 30 per cent cases.

Question 15.
Sterilization operations.
Answer:

  1. Sterilization operations are the permanent means for the birth control. These can be performed on both the sexes. Usually these are performed after the couple does not desire another child.
  2. These surgical interventions block the gamete transport and thus prevents the pregnancy.
  3. Sterilization operation in males is called vasectomy while in females it is called tubectomy.
  4. In vasectomy the vas deferens are tied and cut. In tubectomy fallopian tubes are ligated or cut.

Question 16.
Gonorrhoea.
Answer:
(1) Gonorrhoea is a sexually transmitted veneral disease caused by Diplococcus bacterium, Neisseria gonorrhoea.

(2) The incubation period is 2 to 14 days in males and 7 to 21 days in females.

(3) Infection sites are mucous membrane of urino-genital tract, rectum, throat and eye.

(4) Males show following symptoms : Partial blockage of urethra and reproductive ducts, pus from penis, pain and burning sensation : during urination, arthritis, etc.

(5) Symptoms in female include, pelvic inflammation of urinary tract, sterility, arthritis. The children born to affected mother suffer from gonococcal ophthalmia, In girl-child, there is occurrence of gonococcal vulvovaginitis before puberty.

(6) Preventive measures for gonorrhoea are as follows:

  • Sexual hygiene
  • Use of condom during coitus.
  • Sex with unknown partner or multiple partners should be avoided.

(7) Gonorrhoea can be treated with Cefixime which is antibiotic.

Short-answer Questions

Question 1.
What are sexual dimorphic characters? Enlist these characters in human male and female.
Answer:
Sexual dimorphism is the phenomena in which the sexes of the individual can be identified externally. In human beings, even in infancy there is sexual dimorphism, by which one can identify the sex of the infant.

But when the male or female reaches puberty, then secondary sexual characters are developed due to sex hormones. These characters are called sexual dimorphic characters.
(i) Secondary sexual characters in males:

  1. Presence of beard, Moustache.
  2. Hair on the Chest, Axillary and Pubic Region.
  3. Muscular body.
  4. Enlarged larynx (Adam’s apple).

(ii) Secondary sexual characters in females:

  1. Breast development.
  2. Broadening of pelvis.
  3. High pitched voice.

Question 2.
Describe the duct system that transports the sperms from seminiferous tubules to the exterior.
Answer:
(1) All the seminiferous tubules present in the testis show posterior network of tubules called rete testis. Vasa efferentia are the fine tubules which are 12-20 in number, are seen arising from rete testis. From testis to epididymis, the sperm transport is done by vasa efferentia.

(2) Epididymis has three parts, caput, corpus and cauda epididymis. In this long and highly coiled tube sperms undergo physiological maturation.

(3) Then from here sperms enter into vas deferens, which is a tube that arise from epididymis enters the abdominal cavity. On its course, later it joins the duct of seminal vesicle. Both together form the ejaculatory duct.

(4) Ejaculatory duct passes through the prostate gland and then opens into the urethra. Urethra is a common passage for urine and semen and hence it is also called urinogenital duct.

(5) Urethra passes through penis and opens to the outside by an opening called the urethral meatus or urethral orifice.

(6) Thus sperms are transported through vas deferens into urethra via ejaculatory duct and then to the outside through urethral orifice.

Question 3.
What is semen? Describe the composition of semen.
Answer:
(1) Semen is the viscous, alkaline and milky fluid having pH 7.2 to 7.7 ejaculated during sexual intercourse by male.

(2) A single ejaculation of semen i.e. 2.5 to 4 ml semen contains about 400 millions of sperms.

(3) Semen consists of sperms suspended in secretions of the epididymis and the accessory glands (seminal vesicles, prostate gland and Cowper’s gland). The semen nourishes the sperms by fructose, neutralizes acidity by Ca++, ions and bicarbonates and also activates them for movement due to prostaglandins.

Question 4.
Describe in detail the external genitalia of human female reproductive system.
Answer:
The external genital organs of female are located external to the vagina. They have collective name, ‘vulva’ or pudendum. Following are the parts of vulva.
(1) Labia majora : Labia majora are homologous to scrotum of males. They are two large folds which form the boundary of the vulva. They are composed of skin, fibrous tissue and fat. These Eire prominent and longitudinal folds on right and left sides of the vestibule.

(2) Labia minora : Smaller and thinner lip-like folds located just medially are labia minora. Posteriorly the labia minora are fused together to form the fourchette.

(3) Mons veneris : Mons veneris is fleshy elevation above the labia majora.

(4) Clitoris : It is present at the anterior end of the labia minora. It shows the presence of erectile tissues.

(5) Vestibule : Vestibule is a median vertical depression of vulva enclosing vagina and urethral opening.

(6) Hymen : Hymen is a thin layer of mucous membrane which partially occludes the opening of the vagina.

(7) Vestibular glands:

  1. Vestibular glands or Bartholin’s glands are homologous to the Cowper’s glands of the male.
  2. These are paired glands situated on either side of the vaginal opening, secreting lubricating fluid.

Question 5.
How is puberty attained in females? Will a female normally remain reproductively capable even after age 50? If not then what makes her incapable?
Answer:
(1) Puberty is achieved due to gonadotropins such as FSH and LH secreted by the anterior pituitary. These hormones stimulate the ovaries. The ovaries in turn produce estrogen and progesterone, which brings about secondary sexual characters in female. Thus she attains the puberty. The beginning of menstrual cycle or menarche takes place due to these hormonal changes at about 10 to 14 years.

(2) But the women do not remain reproductively active after the age of 50 due to hormonal imbalance. This is called menopause or cessation of reproductive cycles. Absence of enough gonadotropins and unresponsive ovarian cells cause menopause at 45 to 50 years of age.

Question 6.
Why is menstruation painful in some women?
Answer:

  1. The menstruation is painful in some women as the muscles in the uterus contract or tighten.
  2. Women who experience painful periods can have higher levels of prostaglandins, a natural body chemical that causes contractions of the uterus and blood vessels.
  3. Some women have a build-up of prostaglandins which means they experience stronger contractions and therefore due to spasmodic pain in some women menstruation is more painful.
  4. Endometrial sloughing that takes at the time of menstruation also causes painful discomfort.

Question 7.
Why is it said that consumption of mother’s milk is safety for the newborn?
Answer:
Consumption of mother’s milk is safety for the newborn because of the following reasons:

  1. Mother’s milk is the perfect food for babies in the first months of their lives. With the exception of vitamin D, it contains all the nutrients an infant needs.
  2. Mother’s milk supplies antibodies [IgA] that protect the baby’s body organs from infections. Mother’s milk provides immunity and also help in maturation of the infant’s immune system which are lacking in ordinary milk. Natural acquired passive immunity is obtained only through mother’s milk.
  3. Feeding of mother’s milk reduces the risk of overweight and obesity during childhood.
  4. It also creates the bond between mother and child.

Question 8.
Which hygienic practices should be followed by the female during menstruation ?
Answer:
The following personal hygienic practices should be followed by the female during menstruation:

  1. Keeping the pubic area clean.
  2. Changing the sanitary napkin every 4-5 hours.
  3. Reducing risk of infections by maintaining hygiene.
  4. Proper disposal of soiled sanitary napkin.
  5. Not to use damp and dirty clothes which can cause infections and bad odour. A sanitary napkin which is not changed in time can act as a perfect environment for rapid growth of infectious bacteria.

Question 9.
How can the goals of RCH be achieved?
Answer:
The goads of RCH can be achieved by the following ways:

  1. Sex education in schools is introduced. Proper and scientific information about sexual organs and safe sexual acts should be given to students. They should be made aware of sexually transmitted diseases (STD, AIDS), and problems related to adolescence.
  2. Audio-visual and the print media should be used by government and non-government organisations for creating awareness about reproductive health.
  3. Younger generation should be educated about family planning measures, pre-natal and post-natal care of women and care of infant with knowledge about importance of breastfeeding.
  4. Awareness should be spread about problems arising due to uncontrolled population growth, sex abuse and sex related crimes. Necessary steps to prevent these to be taken.
  5. Statutory ban on amniocentesis for sex determination is practised. This should be known by all.
  6. Details of child immunization programmes should be understood.
  7. New parents should get the training for new born care so that infant and maternal mortality rate can be reduced.

Question 10.
How do addictions like smoking, alcoholism and drug abuse contribute in causing infertility in men?
Answer:

  1. Tobacco, marijuana and other drugs, smoking may cause infertility in both men and women.
  2. Nicotine blocks the production of sperm and decreases the size of testicles.
  3. Alcoholism by men interferes with the synthesis of testosterone and has an impact on sperm count.
  4. Use of cocaine or marijuana may temporarily reduce the number and quality of sperm.

Question 11.
Jayesh, a young married man of 26 years is suffering from T.B. for the last 2 years. He and his wife are desirous of a child but unable to have one, what could be the possible reason? Explain.
Answer:
Jayesh, though young, is suffering from TB for last 2 years. His wife is unable to conceive the child may be due to following reasons:

  1. Tuberculosis disrupts sexual and reproductive function in patients.
  2. Moreover T.B. patients have to take not less than 4 anti-TB drugs simultaneously for a long time.
  3. These are basically a very high dose antibiotics which may hamper formation of sperms. In this way the anti-tuberculosis drugs may negatively influence on sexual function.
  4. Pulmonary TB patient shows, deterioration of all parameters of copulatory act, from sexual desire to orgasm and thus the couple is unable to conceive.
  5. Infertility is one of the most common symptoms of genital tuberculosis.

Question 12.
Neeta is 45 years old and the doctor advised her not to go for such a late pregnancy. She however wants to be the biological mother of a child without herself getting pregnant. Is this possible and how?
Answer:
(1) Neeta being 45 years old, she is approaching menopause. Therefore, she will be advised by the doctor to take the help of the modern remedial technique called surrogacy.

(2) In this technique the embryo is formed using intended father’s sperm and intended mother’s egg by In Vitro Fertilization (IVF) technique and then that embryo is implanted in a surrogate mother, sometimes called a gestational carrier.

(3) In surrogacy there is legal arrangement where the surrogate mother agrees to bear child for a couple. Remains pregnant with all the care and nourishment. Later she delivers a baby and hands it over to biological mother.

Chart based /Table based questions

Question 1.
Complete the following chart and rewrite

Female Reproductive OrgansHomology to Male Reproductive Organs
1. Labia majora————–
2. ————-Bulbourethral glands/ Cowper’s gland
3. Clitoris—————

Answer:

Female Reproductive OrgansHomology to Male Reproductive Organs
1. Labia majoraScrotum
2. Bartholin’s gland/ Vestibular glandBulbourethral glands/ Cowper’s gland
3. ClitorisPenis

Question 2.

HormonesFunctions
1. Testosterone————–
2. ————-Stimulates contractions uterine during parturition
3. Progesterone—————

Answer:

HormonesFunctions
1. TestosteroneStimulates spermatogenesis
2. OxytocinStimulates contractions uterine during parturition
3. ProgesteroneMaintain endometrium of uterus during secretory phase and gestation.

Question 3.
Complete the following chart and rewrite

HormonesFunctions
1. Rapid regeneration of endometrium and maturation of Graafian follicle————–
2. Secretion of endometrial glands and increased secretion of progesterone————–
3. Breakdown of endometrium in absence of fertilization————-

Answer:

HormonesFunctions
1. Rapid regeneration of endometrium and maturation of Graafian follicleProliferative phase / Follicular phase
2. Secretion of endometrial glands and increased secretion of progesteroneSecretory phase / Luteal phase
3. Breakdown of endometrium in absence of fertilizationMenstrual phase

Diagram based questions

Question 1.
Sketch and label Human male reproductive system.
Answer:
Maharashtra Board Class 12 Biology Important Questions Chapter 2 Reproduction in Lower and Higher Animals 4

Maharashtra Board Class 12 Biology Important Questions Chapter 2 Reproduction in Lower and Higher Animals

Question 2.
Label the given male reproductive system you have studied.
Maharashtra Board Class 12 Biology Important Questions Chapter 2 Reproduction in Lower and Higher Animals 5
Answer:

  1. Seminal vesicle
  2. Ejaculatory duct
  3. Cowper’s glands
  4. Urethra
  5. Epididymis
  6. Testis
  7. Urinary bladder
  8. Prostate gland
  9. Vas deferens
  10. Penis

Question 3.
Sketch and label human female reproductive system.
Answer:
Maharashtra Board Class 12 Biology Important Questions Chapter 2 Reproduction in Lower and Higher Animals 6

Question 4.
Give labels to given diagram of female reproductive system.
Maharashtra Board Class 12 Biology Important Questions Chapter 2 Reproduction in Lower and Higher Animals 7
Answer:

  1. Fallopian tube
  2. Fundus of Uterus
  3. Ampulla of fallopian tube
  4. Ovarian ligament
  5. Uterus
  6. Ovary
  7. Infundibulum with fimbriae
  8. Endometrium of uterus
  9. Cervix
  10. Vagina

Question 5.
Sketch and label Seminiferous tubules as seen in T.S. of testis.
Answer:
Maharashtra Board Class 12 Biology Important Questions Chapter 2 Reproduction in Lower and Higher Animals 8

Question 6.
Identify ‘A’ and ‘B’ in the diagram below and mention their functions.
Maharashtra Board Class 12 Biology Important Questions Chapter 2 Reproduction in Lower and Higher Animals 9
Answer:
(1) A : Seminiferous tubule
Function : Seminiferous tubules produce sperms by spermatogenesis.

(2) B : Vas deferens
Function : Vas deferens carry sperm from epididymis to ejaculatory duct.

Question 7.
Sketch and label – T.S. of ovary.
Answer:
Maharashtra Board Class 12 Biology Important Questions Chapter 2 Reproduction in Lower and Higher Animals 10

Question 8.
Sketch and label sectional view of mammary gland.
Answer:
Maharashtra Board Class 12 Biology Important Questions Chapter 2 Reproduction in Lower and Higher Animals 11

Question 9.
Sketch and label – Graafian follicle.
Answer:
Maharashtra Board Class 12 Biology Important Questions Chapter 2 Reproduction in Lower and Higher Animals 12

Question 10.
Sketch and label – Process spermatogenesis.
Answer:
Maharashtra Board Class 12 Biology Important Questions Chapter 2 Reproduction in Lower and Higher Animals 13

Question 11.
Sketch and label process of oogenesis.
Answer:
Maharashtra Board Class 12 Biology Important Questions Chapter 2 Reproduction in Lower and Higher Animals 14

Question 12.
Give the name and functions of ‘A’ and ‘B’ from the diagram given below
Maharashtra Board Class 12 Biology Important Questions Chapter 2 Reproduction in Lower and Higher Animals 15
Answer:
(1) A is acrosome.
Function of acrosome : Acrosome produces lytic enzyme, hyalourinidase and thus helps in the penetration of the egg during fertilization.

(2) B is tail of the human sperm.
Function of tail : Tail lashes continuously and helps the movement of the sperm in the female genital tract.

Question 13.
The diagram represents a surgical sterilization method in males. Study the same and answer the questions that follow
Maharashtra Board Class 12 Biology Important Questions Chapter 2 Reproduction in Lower and Higher Animals 16

  1. Give the name of the surgical method represented in the diagram.
  2. Which part is ligated or cut.
  3. Name the corresponding surgical method conducted in females.
  4. Name the part which is ligated in females and why?

Answer:

  1. Vasectomy
  2. Vas deferens
  3. Tubectomy
  4. Fallopian tubes are ligated so that the egg may not meet with the sperms.

Question 14.
Given below is the figure of an important structure developed during pregnancy.

  1. Name the structure and its type.
  2. Identify ‘A’. In which technique it is used.
  3. Identify ‘B’ What is its function?

Maharashtra Board Class 12 Biology Important Questions Chapter 2 Reproduction in Lower and Higher Animals 17
Answer:
(1) The given figure is placenta. The type of placenta in humans is haemochorial placenta.

(2) A is amnio tic fluid. Amnio tic fluid is withdrawn in amniocentesis technique. From this fluid foetal cells can be obtained, which are examined for any chromosomal abnormality by karyotyping.

(3) B is umbilical cord. This is the connection between placenta of mother and growing foetus. Through the umbilical cord, foetus gets nutrition and oxygen. Nitrogenous wastes and carbon dioxide is collected from foetus and brought into maternal circulation.

Question 15.
The diagram given below is that of a intra-uterine contraceptive device. Study the same and then answer the questions that follows:
Maharashtra Board Class 12 Biology Important Questions Chapter 2 Reproduction in Lower and Higher Animals 18

  1. Give the name of intra-uterine contraceptive device shown in the diagram.
  2. What is its mode of action?

Answer:

  1. The Intra-uterine contraceptive device shown in the diagram is Lippes loop.
  2. It is a plastic double ‘S’ loop. It attracts the macrophages stimulating them to accumulate in the uterine cavity. Macrophages increase phagocytosis of sperms within the uterus and acts as a contraceptive.

Question 16.
Identify A in the given diagram. Write the function of the same.
Maharashtra Board Class 12 Biology Important Questions Chapter 2 Reproduction in Lower and Higher Animals 19
Answer:
A in the above diagram is intrauterine device or IUD. It is a contraceptive device inserted in the uterus of woman. This is a hormone releasing IUD. It acts as a mechanical means of contraception and avoids pregnancy.

Maharashtra Board Class 12 Biology Important Questions Chapter 2 Reproduction in Lower and Higher Animals

Long answer questions

Question 1.
With the help of diagrammatic representation, explain the process of spermatogenesis.
Answer:
Maharashtra Board Class 12 Biology Important Questions Chapter 2 Reproduction in Lower and Higher Animals 20
(1) The process of spermatogenesis takes place in the male gonads or testis. The cells of germinal epithelium that line the seminiferous tubules undergo spermatogenesis.

(2) Primordial germ cells or germinal cells pass through three phases, viz. phase of multiplication, phase of growth and phase of maturation.

  • Multiplication phase : Primordial germ cells undergo mitotic divisions to produce many diploid (2n) spermatogonia.
  • Growth phase : Spermatogonium accumulates nutrients and grows in size, giving rise to primary spermatocyte (2n).
  • Maturation phase : The primary spermatocyte undergoes first meiotic division or maturation division. Exchange of genetic material occurs between homologous chromosomes in each spermatocyte.

(3) The meiotic division gives rise to secondary spermatocyte which is haploid (n). At the end of first meiotic division two secondary spermatocytes are formed while at the end of second meiotic division four haploid spermatids are formed.

(4) Spermatids are non-motile. They undergo spermiogenesis and form motile spermatozoan (sperm).

(5) The changes taking place during spermiogeneis are as follows:

  • Increase in length.
  • Formation of proximal and distal centriole.
  • Distal centriole forms the axial filament.
  • Mitochondria become spirally coiled.
  • Acrosome is formed from Golgi complex.

Question 2.
What is oogenesis? Describe it briefly.
Answer:
Maharashtra Board Class 12 Biology Important Questions Chapter 2 Reproduction in Lower and Higher Animals 21
1. Oogenesis is the process of formation of the haploid female gamete, i.e. ovum.

2. The process of oogenesis takes place in the follicular cells inside the ovaries. The germinal epithelium cells undergo oogenesis.

3. They pass through three phases, viz. phase of multiplication, phase of growth and phase of maturation at the time of oogenesis.

  • Multiplication phase : Germinal cells undergo mitotic divisions and produce large number of diploid (2n) oogonia. Oogonia are present in the ovaries of female even before she is born.
  • Growth phase : During puberty changes, the FSH from pituitary makes one oogonium to develop at a time. The growth takes place as the follicle matures and larger primary oocyte (2n) is produced inside the Graafian follicle.
  • Maturation phase : The primary oocyte undergoes first meiotic division. There are equal nuclear divisions during meiosis but the cytoplasm is unequally divided.

4. By the end of first meiotic division, larger haploid secondary oocyte and smaller haploid polar body are produced. Since the embryo develops from the egg, there is provision for more food in the secondary oocyte.

(5) The second meiotic division takes place in the secondary oocyte and polar body. But this division is arrested during metaphase.

(6) The secondary oocyte is released from the ovary in the process of ovulation. Remaining division takes place if and only if ovum is fertilized.

(7) The division is unequal and form functional female gamete or ovum at the time of fertilization.

Question 3.
What is gastrulation? What are the changes that are brought about by gastrulation?
Answer:
(1) Gastrulation : The process of formation of three germ layers by morphogenetic movements and rearrangements of the cells in blastula leading to the formation of gastrula is known as gastrulation.

(2) Cells on the free end of inner cell mass called hypoblasts (primitive endoderm) become flat, divide and grow towards the blastocoel to form endoderm.

(3) Endodermal cells grow within the blastocoel to form a Yolk sac.

(4) The remaining cell of the inner cell mass, in contact with cells of Rauber are called epiblasts (primary ectoderm) which further differentiate to form ectoderm.

(5) Cells of ectoderm divide and re-divide and move in such a way that they enclose the amniotic cavity. The floor of this cavity has the embryonal disc while roof is lined by amniogenic cells. Amnion is an extra embryonic membrane that surrounds and protects the embryo.

(6) Actual gastrulation occurs about days after fertilization.
Maharashtra Board Class 12 Biology Important Questions Chapter 2 Reproduction in Lower and Higher Animals 22

(7) Trilaminar embryonic disc begins with the formation of primitive streak and a shallow groove on the surface called primitive groove. From the site of primitive streak, a third layer of cells called mesoderm extends between ectoderm and endoderm. Anterior end of the primitive groove communicates with yolk sac by an aperture called blastopore (future anus).

(8) The embryonal knob thus finally differentiate into three layers – ectoderm, mesoderm and endoderm.

Question 4.
Explain the major changes taking place during the three trimesters of pregnancy in woman.
Answer:
The pregnancy period of approximately nine months (280 days) is divided into three trimesters of three months each.
1. First Trimester : (From fertilization to 12th week)

  • During first trimester there are radical changes in the body of mother as well as in the embryo.
  • The embryo receives nutrients in the first 2-4 weeks directly from the endometrium.
  • It is the main period of organogenesis and the development of body organs.
  • By the end of eight weeks, the major structures found in the adult are formed in the embryo in a rudimentary form. It is now called foetus and is about 3 cm long.
  • Arms, hands, fingers, feet, toes, CNS, excretory and circulatory system including heart are formed and begins to work.
  • Progesterone level becomes high and menstrual cycle is suspended till the end of pregnancy.
  • At the end of first trimester foetus is about 7-10 cm long.
  • The maternal part of placenta grows, the uterus becomes larger. In this period, the mother experiences morning sickness, (nausea, vomiting, mood swings, etc.)

Maharashtra Board Class 12 Biology Important Questions Chapter 2 Reproduction in Lower and Higher Animals

2. Second Trimester: (From 13th to 26th week)

  • The foetus is very active and grows to about 30 cm.
  • The uterus grows enough for the pregnancy to become obvious.
  • Hormone levels stabilize as hCG declines, the corpus luteum deteriorates and the placenta completely takes over the production of progesterone which maintains the pregnancy.
  • Head has hair, eyebrows and eyelashes appear, pinnae are distinct. Baby’s movement can be easily felt by the mother.
  • The baby reaches half the size of a new born.

3. Third Trimester: (From 27th week till the parturition)

  • Foetus grows to about 50 cm in length and about 3-4 kg in weight.
  • As the foetus grows, the uterus expands around it, the mother’s abdominal organs become compressed and displaced, leading to frequent urination, digestive blockages and strain in the back muscles.
  • At the end of third trimester the foetus becomes fully developed and ready for parturition.

Maharashtra Board Class 12 Biology Important Questions Chapter 1 Reproduction in Lower and Higher Plants

Balbharti Maharashtra State Board 12th Biology Important Questions Chapter 1 Reproduction in Lower and Higher Plants Important Questions and Answers.

Maharashtra State Board 12th Biology Important Questions Chapter 1 Reproduction in Lower and Higher Plants

Multiple-choice questions

Question 1.
In grafting, the rooted plant is used as a ………………
(a) scion
(b) stock
(c) stem
(d) root
Answer:
(b) stock

Question 2.
The method of propagation by root cutting is practised in ………………
(a) Rose
(b) Bougainvillea
(c) Sansevieria
(d) Blackberry
Answer:
(d) Blackberry

Maharashtra Board Class 12 Biology Important Questions Chapter 1 Reproduction in Lower and Higher Plants

Question 3.
Monothecous anther has ……………… pollen sacs.
(a) single
(b) two
(c) three
(d) four
Answer:
(b) two

Question 4.
In wall of mature anther, ……………… shows fibrous thickenings.
(a) epidermis
(b) endothecium
(c) middle layer
(d) tapetum
Answer:
(b) endothecium

Question 5.
Intine consists of ………………
(a) cellulose and pectin
(b) cellulose and chitin
(c) cellulose and starch
(d) cellulose and sporopollenin
Answer:
(a) cellulose and pectin

Question 6.
The stalk of the ovule that attaches to placenta is ……………… which is attached at to the body of ovule.
(a) chalaza, hilum
(b) hilum, funiculus
(c) funiculus, hilum
(d) micropyle, hilum
Answer:
(c) funiculus, hilum

Question 7.
……………… is multicellular structure embedded in nucellus.
(a) Micropyle
(b) Chalaza
(c) Embryo sac
(d) Endothecium
Answer:
(c) Embryo sac

Question 8.
The transfer of pollen grains from the anther to the stigma is called ………………
(a) pollination
(b) fertilization
(c) transpiration
(d) viability
Answer:
(a) pollination

Question 9.
This condition is not favourable for autogamy in flowers.
(a) Bisexuality
(b) Homogamy
(c) Cleistogamy
(d) Herkogamy
Answer:
(d) Herkogamy

Question 10.
From the following, mechanism of pollination by abiotic agent is ………………
(a) Ornithophily
(b) Anemophily
(c) Entomophily
(d) Chiropterophily
Answer:
(b) Anemophily

Question 11.
In which type of flowers, the pollen grains are ribbon like without exine?
(a) Anemophilous
(b) Epihydrophilous
(c) Hypohydrophilous
(d) Entomophilous
Answer:
(c) Hypohydrophilous

Question 12.
Which flower exhibits turn pipe mechanism of pollination?
(a) Salvia
(b) Zostera
(c) Oestrum
(d) Callistemon
Answer:
(a) Salvia

Question 13.
The phenomenon of pollen grains of other flowers germinate rapidly on stigma than the pollen grains of same flower is ………………
(a) Protoandry
(b) Protogyny
(c) Prepotency
(d) Pollination
Answer:
(c) Prepotency

Question 14.
Inhibition of germination of pollen on stigma of same flower is ………………
(a) self-sterility
(b) self-pollination
(c) self compatibility
(d) selfing
Answer:
(a) self-sterility

Question 15.
The stigma provides ……………… for germination of pollen on it.
(a) oxygen
(b) water
(c) pectin
(d) malic acid
Answer:
(b) water

Question 16.
For successful artificial hybridization, these processes are essential.
(a) Disbudding and Bagging
(b) Budding and Bagging
(c) Emasculation and Budding
(d) Emasculation and Bagging
Answer:
(d) Emasculation and Bagging

Question 17.
Continued self-pollination results in ………………
(a) Hybrid vigour
(b) Genetic variability at greater extent
(c) Inbreeding depression
(d) Introduction of desired traits
Answer:
(c) Inbreeding depression

Question 18.
Heterostyly : Primula flowers : : Herkogamy: ?
(a) Gloriosa
(b) Calotropis
(c) Thea
(d) Salvia
Answer:
(b) Calotropis

Question 19.
The substance having key role in recognition and compatibility of pollen in pollen – pistil interaction is ………………
(a) special proteins
(b) special lipids
(c) pollenkitt
(d) sucrose
Answer:
(a) special proteins

Question 20.
This is NOT a type but variation in endosperm.
(a) Cellular
(b) Helobial
(c) Nuclear
(d) Mosaic
Answer:
(d) Mosaic

Question 21.
In monocot embryo, the single cotyledon is ……………… shaped and it is called ………………
(a) oval, scutellum
(b) shield, scutellum
(c) angle, coleoptile
(d) angle, coleorhiza
Answer:
(b) shield, scutellum

Question 22.
Endospermic seed : Maize : : Non-endospermic seed : ?
(a) Castor
(b) Coconut
(c) Wheat
(d) Bean
Answer:
(d) Bean

Question 23.
The integuments of fertilized ovule form the ………………
(a) seed
(b) seed coat
(c) hilum
(d) perisperm
Answer:
(b) seed coat

Maharashtra Board Class 12 Biology Important Questions Chapter 1 Reproduction in Lower and Higher Plants

Question 24.
When diploid sporophytic cell forms diploid gametophyte without meiosis, it is phenomenon of ………………
(a) apogamy
(b) apocarpy
(c) apospory
(d) apoptosis
Answer:
(c) apospory

Question 25.
Which chemical substance is responsible for fruit development by Parthenocarpy?
(a) Malic acid
(b) Sucrose
(c) Boric acid
(d) Indole acetic acid
Answer:
(d) Indole acetic acid

Question 26.
Citrus seeds : Polyembryony : : Papaya fruits : ?
(a) Diplospory
(b) Apogamy
(c) Parthenocarpy
(d) Apospory
Answer:
(c) Parthenocarpy

Question 27.
Who discovered the phenomenon of double fertilization?
(a) Noll
(b) Maheshwari
(c) Leeuwenhoek
(d) Nawaschin
Answer:
(d) Nawaschin

Question 28.
When embryo development takes place the first cell of the suspensor which is towards micropylar end functions as ………………
(a) hypophysis
(b) haustorium
(c) scutellum
(d) plumule
Answer:
(b) haustorium

Question 29.
Point out the odd one.
(a) Coleoptile
(b) Coleorhiza
(c) Scutellum
(d) Perisperm
Answer:
(d) perisperm

Question 30.
Select the plant having both chasmogamous and cleistogamous flowers.
(a) Viola
(b) Primula
(c) Thea
(d) Fritillaria
Answer:
(a) Viola

Question 31.
Identify the mismatched pair.
(a) Cellular endosperm – Balsam
(b) Nuclear endosperm – Wheat
(c) Helobial endosperm – Asphodelus
(d) Mosaic endosperm – Coconut
Answer:
(d) Mosaic endosperm – Coconut

Question 32.
Up to which stage embryo development is similar in dicots and monocots?
(a) Proembryo
(b) Quadrant
(c) Octant
(d) Heart-shaped
Answer:
(c) Octant

Question 33.
The cross pollination within the same species is also called ………………
(a) hybridization
(b) xenogamy
(c) allogamy
(d) autogamy
Answer:
(b) xenogamy

Question 34.
In a recently fertilized ovule, the haploid, diploid and triploid conditions are respectively seen in ………………
(a) endosperm, nucellus, egg
(b) egg, nucellus, endosperm
(c) antipodals, oospore, primary endosperm nucleus
(d) polar nuclei, secondary nucleus, endosperm
Answer:
(c) antipodals, oospore, primary endosperm nucleus

Question 35.
In sunflower, self-pollination is avoided by ………………
(a) protogyny
(b) unisexuality
(c) self-sterility
(d) protandry
Answer:
(d) protandry

Question 36.
A versatile anther is an adaptation for ……………… type of pollination.
(a) anemophilous
(b) entomophilous
(c) hydrophilous
(d) ornithophilous
Answer:
(a) anemophilous

Question 37.
The endosperm cells in an angiospermic plant has 18 chromosomes, the number of chromosomes in its root cells will be ………………
(a) 12
(b) 6
(c) 18
(d) 24
Answer:
(a) 12

Question 38.
In porogamy, the pollen tube enters into the ovule through ………………
(a) micropyle
(b) integuments
(c) chalaza
(d) funicle
Answer:
(a) micropyle

Question 39.
Which of the following is not floral adaptation for entomophily?
(a) Large flowers
(b) Bright coloured flowers
(c) Sweet scented flowers
(d) Small inconspicuous flowers
Answer:
(d) small inconspicuous flower

Question 40.
Pollination through water is called ………………
(a) zoophily
(b) hydrophily
(c) anemophily
(d) entomophily
Answer:
(b) hydrophily

Question 41.
The types of pollination exhibited by Vallisneria and Zea mays respectively are ………………
(a) anemophily and hydrophily
(b) entomophily and hydrophily
(c) hydrophily and anemophily
(d) hydrophily and entomophily
Answer:
(c) hydrophily and anemophily

Question 42.
The union of male gamete with the female gamete is called ………………
(a) autogamy
(b) allogamy
(c) fertilization
(d) pollination
Answer:
(c) fertilization

Question 43.
The secondary nucleus is formed by the fusion of ………………
(a) two polar nuclei
(b) three nuclei
(c) two synergids
(d) two antipodal cells
Answer:
(a) two polar nuclei

Question 44.
A group of three cells situated at the base of the embryo sac are called ………………
(a) tube
(b) generative
(c) synergid
(d) antipodal
Answer:
(d) antipodal

Question 45.
The female gametophyte in angiosperms is a ……………… nucleated structure.
(a) 3
(b) 4
(c) 5
(d) 8
Answer:
(d) 8

Question 46.
In artificial hybridization, pollen grains are pollinated by ………………
(a) wind
(b) insect
(c) birds
(d) hand
Answer:
(d) hand

Question 47.
Which of the following does not occur in the embryo sac of angiosperms?
(a) egg apparatus
(b) secondary nucleus
(c) antipodal cells
(d) raphe
Answer:
(d) raphe

Question 48.
To produce 500 pollen grains, how many microspore mother cells are required ?
(a) 500
(b) 125
(c) 250
(d) 1000
Answer:
(b) 125

Maharashtra Board Class 12 Biology Important Questions Chapter 1 Reproduction in Lower and Higher Plants

Question 49.
How many meiotic divisions are required for the formation of 100 seeds ?
(a) 25
(b) 50
(c) 100
(d) 125
Answer:
(b) 50

Question 50.
During fertilization, male gametes are carried by pollen tube. This is called ………………
(a) syngamy
(b) siphonogamy
(c) mesogamy
(d) polygamy
Answer:
(b) siphonogamy

Question 51.
In bisexual flowers, maturation of gynoecium before androecium is known as ………………
(a) protandry
(b) protogyny
(c) gynandry
(d) dicliny
Answer:
(b) protogyny

Question 52.
……………… is formed in angiosperms by triple fusion.
(a) Testa
(b) Integument
(c) Endosperm
(d) Suspensor
Answer:
(c) Endosperm

Question 53.
The minimum number of meiotic divisions required to produce 120 viable seeds in pea plant is ………………
(a) 150
(b) 60
(c) 120
(d) 90
Answer:
(a) 150

Question 54.
Ornithophily is effected by ………………
(a) snails
(b) insects
(c) bats
(d) birds
Answer:
(d) birds

Question 55.
Synergids are ………………
(a) haploid
(b) triploid
(c) diploid
(d) tetraploid
Answer:
(a) haploid

Question 56.
Egg apparatus consists of ………………
(a) egg and antipodals
(b) egg and polar nuclei
(c) egg and synergids
(d) egg and secondary nucleus
Answer:
(c) egg and synergids

Question 57.
Embryo sac is ………………
(a) microgametophyte
(b) microsporangium
(c) megagame tophyte
(d) megasporangium
Answer:
(c) megagametophyte

Question 58.
If the number of chromosomes in an endosperm cell is 27, what will be the chromosome number in the definitive nucleus?
(a) 9
(b) 18
(c) 27
(d) 36
Answer:
(b) 18

Question 59.
How many meiotic divisions will be needed to produce 44 female gametophytes in angiosperms?
(a) 11
(b) 22
(c) 44
(d) 66
Answer:
(c) 44

Question 60.
Endosperm of angiosperm is ………………
(a) haploid
(b) diploid
(c) triploid
(d) tetraploid
Answer:
(c) triploid

Match the columns

Question 1.

(1) Column A (Asexual)Column B (Examples)
(1) Spore formation(a) Spirogyra
(2) Conidia formation(b) Yeast
(3) Fragmentation(c) Chlamydomonas
(4) Budding(d) Penicillium

Answer:

(1) Column A (Asexual)Column B (Examples)
(1) Spore formation(c) Chlamydomonas
(2) Conidia formation(d) Penicillium
(3) Fragmentation(a) Spirogyra
(4) Budding(b) Yeast

Question 2.

Column A (Artificial Vegetative Propagation)Column B (Examples)
(1) Leaf cutting(a) Blackberry
(2) Stem cutting(b) Apple
(3) Grafting(c) Bougainvillea
(4) Root cutting(d) Sansevieria

Answer:

Column A (Artificial Vegetative Propagation)Column B (Examples)
(1) Leaf cutting(d) Sansevieria
(2) Stem cutting(c) Bougainvillea
(3) Grafting(b) Apple
(4) Root cutting(a) Blackberry

Question 3.

Column A (Part of Anatropous ovule)Column B (Terminology)
(1) Opening at the apex(a) Hilum
(2) Stalk of the ovule(b) Integument
(3) Protective covering(c) Micropyle
(4) Place of attachment of body and stalk(d) Funiculus

Answer:

Column A (Part of Anatropous ovule)Column B (Terminology)
(1) Opening at the apex(c) Micropyle
(2) Stalk of the ovule(d) Funiculus
(3) Protective covering(b) Integument
(4) Place of attachment of body and stalk(a) Hilum

Maharashtra Board Class 12 Biology Important Questions Chapter 1 Reproduction in Lower and Higher Plants

Question 4.

Column A (Adaptation)Column B (Type of pollination)
(1) Sticky, spiny pollen grains non-fragrant flowers(a) Anemophily
(2) Feathery stigma and versatile anther(b) Chiropterophily
(3) Presence of nectar glands and sweet smell(c) Ornithophily
(4) Dull coloured flowers with strong fragrance(d) Entomophily

Answer:

Column A (Adaptation)Column B (Type of pollination)
(1) Sticky, spiny pollen grains non-fragrant flowers(c) Ornithophily
(2) Feathery stigma and versatile anther(a) Anemophily
(3) Presence of nectar glands and sweet smell(d) Entomophily
(4) Dull coloured flowers with strong fragrance(b) Chiropterophily

Question 5.

Column A (Mechanism)Column B (Type of pollination)
(1) Geitonogamy(a) Thea
(2) Herkogamy(b) Gloriosa
(3) Self-sterility(c) Cucurbita
(4) Protogyny(d) Calotropis

Answer:

Column A (Mechanism)Column B (Type of pollination)
(1) Geitonogamy(c) Cucurbita
(2) Herkogamy(d) Calotropis
(3) Self-sterility(a) Thea
(4) Protogyny(b) Gloriosa

Question 6.

Column AColumn B
(1) Nutritive tissue of embryo(a) Perisperm
(2) Remnants of nucellus in seed(b) Cotyledon
(3) Nutritive tissue of developing microspores(c) Endosperm
(4) First photosynthetic organ of embryo(d) Tapetum

Answer:

Column AColumn B
(1) Nutritive tissue of embryo(c) Endosperm
(2) Remnants of nucellus in seed(a) Perisperm
(3) Nutritive tissue of developing microspores(d) Tapetum
(4) First photosynthetic organ of embryo(b) Cotyledon

Very short answer questions

Question 1.
What is budding in plants?
Answer:
Budding in plants is an artificial method of propagation in which a single bud is joined or grafted on the stock plant.

Question 2.
What is the function of flower?
Answer:
Flower is a specialized reproductive structure which produces haploid gametes and ensures that act of fertilization will take place.

Question 3.
Enlist the wall layers of mature anther.
Answer:
Epidermis, endothecium, middle layer and tapetum are observed from outside to inside.

Question 4.
What is the peculiarity of angiospermic gametophytes ?
Answer:
The gametophytes are reduced and develop within the flower.

Question 5.
Enlist the layers of sporoderm and their composition.
Answer:
Outer layer exine is composed of sporopollenin and inner layer intine composed of cellulose and pectin.

Question 6.
What is endosporous development of embryo sac?
Answer:
The development of female gametophyte occurs within the megaspore itself.

Question 7.
Enlist the chief agents responsible for pollination process of plants.
Answer:
Abiotic agents – wind, water and Biotic agents – insects, birds, bats.

Question 8.
Describe the characters of pollens of anemophilous flowers.
Answer:
The pollen grains are produced in large number from versatile anthers and are dry, light in weight for their easy dispersal.

Question 9.
What is hay fever?
Answer:
It is the allergic symptoms observed in people who are sensitive to pollen grains mainly of anemophilous plants.

Question 10.
Enlist the different types of pollination observed in aquatic plants.
Answer:
Aquatic plants have hypohydrophilous, epihydrophilous, anemophilous as well as entomophilous type of pollination.

Question 11.
What is the main role of pistil in pollen- pistil interaction?
Answer:
As pollen grain is deposited on stigma, pistil has the ability to recognise and accept the compatible pollen of same species for further germination.

Question 12.
What type is the endosperm of coconut?
Answer:
Coconut has free nuclear vacuolated endosperm in the centre with multicellular endosperm in the outer part.

Question 13.
What is the origin of embryos in adventive polyembryony ?
Answer:
The embryos develop from diploid cells of nucellus and integuments.

Question 14.
What is vegetative propagation ?
Answer:
The reproduction which occurs with the help of vegetative organs like root, stem, leaf or bud is called vegetative reproduction or vegetative propagation.

Question 15.
What is grafting ?
Answer:
Grafting is an artificial method of vegetative propagation in which the parts of two different plants are combined in such a way that they unite with each other and continue their growth as one plant.

Question 16.
What is triple fusion?
Answer:
The process involving the fusion of a male gamete with the diploid secondary nucleus to form a triploid primary endosperm nucleus is called triple fusion.

Question 17.
What is syngamy ?
Answer:
The fusion of male gamete with the egg or oosphere to form diploid zygote or oospore is called syngamy.

Question 18.
What are the two major modes of reproduction in angiosperms ?
Answer:
The two major modes of reproduction in angiosperms are asexual reproduction and sexual reproduction.

Question 19.
What is the main feature of asexual reproduction ?
Answer:
The main feature of asexual reproduction is that it is uniparental and the offspring produced are genetically identical to the parents.

Question 20.
What is sexual reproduction ?
Answer:
The method of reproduction which involves the formation and fusion of gametes is called sexual reproduction.

Question 21.
Name the initial cells of the male and female gametophytes.
Answer:
The haploid microspores (n) and megaspores (n) are the initial cells of the male and female gametophytes.

Question 22.
At which stage, the pollen grains are liberated in the most angiosperms ?
Answer:
The pollen grains are liberated at 2-celled stage in most angiosperms.

Question 23.
What is an anatropous ovule ?
Answer:
The ovule which has a downwardly directed micropyle is called an anatropous ovule.

Question 24.
Give the scientific term used for water pollinated flowers.
Answer:
The scientific term used for water pollinated flowers is hydrophilous.

Question 25.
Give one example each of dicot endospermic seed and non-endospermic seed.
Answer:

  1. Endospermic seed : Castor
  2. Non-endospermic seed : Bean.

Question 26.
What is dichogamy?
Answer:
Maturation of anther and stigma at different times is called dichogamy.

Question 27.
How is diploid condition restored in angiosperms ?
Answer:
In angiosperms, the diploid condition is restored by the fusion of two haploid gametes.

Maharashtra Board Class 12 Biology Important Questions Chapter 1 Reproduction in Lower and Higher Plants

Question 28.
What is egg apparatus ?
Answer:
The egg apparatus is a three-celled structure lying at the micropylar end of the embryo sac.

Question 29.
Why are some seeds of citrus referred to as polyembryonic?
Answer:
When the seeds of citrus germinate, we notice development of multiple seedling. This is due to adventive embryos formed in the seeds in addition to zygotic embryo.

Give definition/meaning of the following terms

Question 1.
Reproduction
Answer:
Reproduction is the process by which offspring is produced which resembles the parents.

Question 2.
Clones
Answer:
Morphologically and genetically identical individuals are called clones.

Question 3.
Scion
Answer:
The part of the stem containing more than one bud which is joined onto a rooted plant.

Question 4.
Stock
Answer:
Stock is a rooted plant on which part of the stem (scion) is joined in grafting.

Question 5.
Microsporogenesis
Answer:
The process in which each microspore mother cell divides meiotically to form tetrad of haploid microspores (pollen grains).

Question 6.
Pollen viability
Answer:
It is the ability of pollen grains to germinate and develop into male gametophyte.

Question 7.
Pollination
Answer:
Pollination is the transfer of pollen grains from anther to the stigma of flower.

Question 8.
Autogamy
Answer:
Autogamy is a type of pollination in which bisexual flower is pollinated by its own pollen grains.

Question 9.
Dioecism
Answer:
Dioecism is condition in which the plant bears either male or female flower and it is also called unisexuality.

Question 10.
Double fertilization
Answer:
The process of fertilization where both the male gametes participate in the complex fertilization mechanism seen in angiosperms is called double fertilization.

Question 11.
Embryogenesis
Answer:
The process of development of zygote into an embryo is called embryogenesis.

Question 12.
Dormancy
Answer:
Structural or physiological adaptive mechanism for survival is called dormancy.

Question 13.
Polyembryony
Answer:
The condition in which there is development of more than one embryo inside the seed is called polyembryony.

Give significance / importance of the following

Question 1.
Reproduction
Answer:
Reproduction is an essential process that leads to continuation of species. It also maintains the continuity of life.

Question 2.
Asexual reproduction
Answer:
The fusion of sex cells is not involved in this process thus it results in the production of genetically identical progeny from a single parent.

Question 3.
Vegetative Reproduction/Propagation
Answer:
The plants reproduce asexually from their vegetative plant parts and thus new plants formed are genetically similar to their parents.

Question 4.
Sexual reproduction
Answer:
It involves fusion of two compatible gametes and thus it results in production of genetically dissimilar offspring. Variations are set in, which are important from point of view of survival and evolution of species.

Question 5.
Exine
Answer:
Outer thick layer of pollen grain which is made up of complex non- biodegradable sporopollenin that is resistant to chemicals.

Question 6.
Germ pores
Answer:
These are thin areas in the exine, through which developing pollen tube emerges out during pollen germination.

Question 7.
Pollen viability
Answer:
The functional viability of pollen grain to form male gametophyte. It can germinate in favourable environmental conditions of suitable temperature and humidity.

Question 8.
Synergid/Filiform apparatus
Answer:
It is present in egg apparatus of embryo sac (female gametophyte) which directs the pollen tube towards the egg cell due to chemicals secreted.

Question 9.
Pollination
Answer:
Non-motile pollen grains are transferred on stigma of flower with some external abiotic or biotic agents.

Question 10.
Autogamy/Self-pollination
Answer:
In a bisexual flower, when it is pollinated by its own pollen grain the offspring formed are genetically identical to their parents.

Question 11.
Xenogamy/Cross pollination/Outbreeding
Answer:
When cross pollination takes place then that generates genetically, varied offspring.

Question 12.
Heterostyly/Heteroanthy (Heteromorphy)
Answer:
When in some flowers; stigmas and anthers are placed at different levels then it prevents self-pollination by preventing pollens to reach the stigma.

Question 13.
Double fertilization
Answer:

  1. It ensures seed formation with food storage for embryo developed from fertilized egg.
  2. Diploid zygote develops into embryo which further forms a new plant.
  3. Triploid PEN forms endosperm which is nutritive tissue for embryo.
  4. Restoration of diploid condition by syngamy.

Question 14.
Endosperm
Answer:
It is nutritive tissue of embryo developed in post-fertilization changes which also triggers the growth of embryo in proper manner.

Question 15.
Seed formation
Answer:
Seeds are important propagating units of plant and their dispersal helps in distribution of species.

Question 16.
Fruit formation
Answer:
Nourishment to the developing seeds and protection of the immature seeds is role of fruit formation.

Question 17.
Apomixis
Answer:
When embryo(s) are formed through asexual method of reproduction without gamete formation, genetically identical plants can be produced rapidly and effectively by apomixis.

Question 18.
Parthenocarpy
Answer:
Development of fruit without the process of fertilization results in formation of seedless fruit.

Question 19.
Polyembryony
Answer:
As there is development of more than one embryo in the seed it increases the chances of survival of new plants.

Name the following

Question 1.
Condition in flower when androecium matures before that of gynoecium.
Answer:
Protandry

Question 2.
Method of asexual reproduction in sponges.
Answer:
Gemmule formation

Question 3.
Method in which small amount of plant tissue is carefully grown.
Answer:
Tissue culture.

Question 4.
Recent or Modern method of vegetative reproduction of plants form plant tissue.
Answer:
Micropropagation.

Maharashtra Board Class 12 Biology Important Questions Chapter 1 Reproduction in Lower and Higher Plants

Question 5.
Most common type of ovule in angiosperms.
Answer:
Anatropous

Question 6.
A diploid nucleus in central cell of embryo sac in plants.
Answer:
Secondary nucleus or definitive nucleus

Question 7.
A condition of flowers where its sex organs are exposed.
Answer:
Chasmogamy.

Question 8.
Components necessary to induce germination of pollen in synthetic medium.
Answer:
Sucrose and boric acid.

Question 9.
The plant material in which double fertilization was discovered.
Answer:
Liliaceae plants like Lilium and Fritillaria.

Question 10.
A condition in which pollen tube enters the ovule through micropyle, through chalaza or through integuments.
Answer:
Porogamy, chalazogamy and mesogamy respectively.

Question 11.
Non-motile male gametes are carried through hollow tube when pollen grain germinate.
Answer:
Siphonogamy

Question 12.
Layers of seed coat.
Answer:
Outer testa and inner tegmen

Question 13.
A state of metabolic arrest that helps in survival of organism In adverse conditions.
Answer:
Dormancy.

Question 14.
Name the nuclei taking part in triple fusion.
Answer:
The nucleus of male gamete and the secondary nucleus formed by fusion of two polar nucleI.

Question 15.
What do you call the kernel that you eat in tender coconut?
Answer:
Coconut meat (kopra)

Distinguish between

Question 1.
Asexual Reproduction – Sexual Reproduction
Answer:

Asexual ReproductionSexual Reproduction
1. Fusion of sex cells or two compatible gametes is not involved.1. Fusion of sex cells or two compatible gametes is involved.
2. It results in production of genetically identical progeny.2. It results in production of genetically dissimilar offspring.
3. Offspring inherit genes of the parent.3. Offspring have combination genes from both the parents through their gametes.
4. Variations are not observed in progeny.4. Variations due to recombination are observed which are useful for survival and evolution of species.

Question 2.
Autogamy (Self-pollination) – Xenogamy (cross pollination)
Answer:

Autogamy (Self-pollination)Xenogamy (Cross pollination)
1. In self-pollination, bisexual flower is pollinated by its own pollen grains.1. In cross pollination the pollen grains from the anther are carried to the stigma of another flower of same species.
2. Self-pollination does not depend upon external agents for pollination.2. Cross pollination does depend upon external agents for pollination.
3. Self-pollination is economical as there is no wastage of pollen grains.3. Cross pollination is not economical as there is wastage of pollen grains during transfer.
4. Offspring are genetically similar to their parents; E.g. Pea4. Offspring are genetically varied due to recombination. E.g. Food and fibre crops – Maize, Rice.

Question 3.
Hypohydrophily – Epihydrophily
Answer:

HypohydrophilyEpihydrophily
1. Pollination takes place below the surface of water.1. Pollination takes place on the surface of water.
2. Pollen grains are heavier and they sink in water.2. Pollen grains float on the water surface.
3. Pollens are long, ribbon like without exine.3. Pollens have specific gravity equal to water.
4. E.g. Zostera (sea grass)4. E.g. Vallisneria

Give scientific reasons

Question 1.
The development of embryo sac is described as monosporic.
Answer:

  1. Embryo sac develops inside the nucellus of ovule from megaspore.
  2. Megaspore mother cell is diploid structure which undergoes meiosis.
  3. After meiosis, tetrad of haploid cells are produced.
  4. The upper three megaspores degenerate and the lower one of the tetrad is functional.
  5. The entire embryo sac is developed by elongation and then three mitotic divisions of this single megaspore take place hence the development is described as monosporic.

Question 2.
Pollination is prerequisite for fertilization in plants.
Answer:

  1. Fertilization is fusion of male and female gametes.
  2. Pollination is transfer of pollen grains which carry non-motile male gametes.
  3. Pollen grains are transferred from anther to stigma of flower where they germinate.
  4. Both male and female gametes are non- motile and they are produced at two different sites.
  5. Therefore the pollination process is necessary for act of fertilization in plants.

Question 3.
Dichogamy favours cross pollination.
Answer:

  1. Maturation of anther (stamen) and stigma (carpel) at different times is called dichogamy.
  2. Dichogamy is of two types, viz, protandry and protogyny.
  3. Maturity of anthers before that of gynoecium is protandry and maturity of carpel before maturity of pollen grains is protogyny.
  4. As this forms barrier for self-pollination, dichogamy favours cross pollination.

Question 4.
Fertilization in angiosperms is double fertilization.
Answer:

  1. In angiosperms, pollen tube carries two non-motile male gametes.
  2. Pollen tube enters the embryo sac in synergids and the contents are relased.
  3. Out of the two male gametes produced by the male gametophyte, one unites with female gamete i.e syngamy and the other with the secondary nucleus i.e. triple fusion.
  4. Since both the male gametes take part in fertilization which takes place twice, it is called double fertilization.

Question 5.
Castor seed is endospermic or albuminous.
Answer:

  1. Endosperm, that is developed after fertilization is a nutritive tissue for developing embryo.
  2. Endosperm stores food material.
  3. In some seeds this reserved food is partially utilized by embryo for development, E.g; Castor.
  4. The endosperm remains in the seed and it is utilized further during seed germination. Hence the seed is endospermic or albuminous.

Maharashtra Board Class 12 Biology Important Questions Chapter 1 Reproduction in Lower and Higher Plants

Question 6.
Parthenocarpic fruits are without seeds.
Answer:

  1. In parthenocarpy, fruit is developed without fertilization.
  2. When fertilization takes place ovules in the ovary are transformed into seeds.
  3. In parthenocarpy, for fruit development chemical stimulus from placental tissue transforms or stimulates ovary into fruit but it is seedless.

Question 7.
Nucellar polyembryony is significant in horticulture.
Answer:

  1. Polyembryony is a phenomenon where we get many embryos in the seed.
  2. Polyembryony increases chances of survival of plants as there are multiple seedlings formed.
  3. Nucellar embryos are formed from diploid parental tissue.
  4. Thus genetically uniform type of seedlings are obtained which are similar to parents.

Write the short notes on the following

Question 1.
Vegetative reproduction.
Answer:

  1. It is asexual method of reproduction.
  2. Plants reproduce through their vegetative plant parts.
  3. New plants produced are genetically identical to their parents.
  4. It is very useful in agriculture and horticulture.
  5. Artificial methods like cutting and grafting are useful for propagation of desired varieties as per human needs.

Question 2.
Grafting.
Answer:

  1. It is type of artificial vegetative propagation.
  2. In this method two different plants are joined together.
  3. The part of stem containing one or more buds is scion which is joined on a rooted plant stock.
  4. They grow as one plant, e.g. Apple, Pear, Mango.
  5. When a single bud is grafted on a stock plant it is known as bud grafting or budding, e.g. Rose.

Question 3.
Pollen Viability.
Answer:

  1. It is a functional ability of pollen grain to form male gametophyte by its germination.
  2. Viable pollen grains germinate on stigmatic surface,
  3. Environmental factors mainly temperature and humidity influence its germination.
  4. Viability is low up to 30 minutes in plants like rice and wheat.
  5. Duration of viability is up to months in some plants of family Leguminosae, Rosaceae and Solanaceae.

Question 4.
Seed Dormancy.
Answer:

  1. It is a state of metabolic arrest which helps in survival of organism in unfavourable environmental conditions.
  2. Structure or physiological adaptive measures of seed that are helpful in adverse conditions is called dormancy.
  3. Seeds are dispersed during their dormancy.
  4. When dormancy period of seeds is completed then only the viable seed germinate.

Question 5.
Parthenocarpy.
Answer:

  1. It is a condition in which fruit is developed without event of fertilization.
  2. It is a natural process observed on Pineapple and Banana.
  3. A chemical stimulus in the form of auxin (IAA) is given by placental tissues of unfertilized ovary.
  4. Due to the stimulus, enlargement of ovary takes place to form a fruit.
  5. Parthenocarpic fruits are without seeds.

Question 6.
Polyembryony.
Answer:

  1. It is a condition when more than one embryos are developed inside the seed.
  2. It was first noticed in Citrus by Leeuwenhoek.
  3. When embryos develop from diploid cells of nucellus or integuments, it is described as adventive polyembryony.
  4. When zygote divides into small units which develop into embryos then it is called cleavage polyembryony.
  5. It results in multiple seedlings and is of significance in horticulture.

Question 7.
Anemophily.
Answer:

  1. The transfer of pollen grains through wind is called anemophily.
  2. Plants that are pollinated by wind are called anemophilous plants.
  3. Anemophilous plants bear small and inconspicuous flowers without any bright colours, fragrance and nectar.
  4. Flowers are produced in large numbers.
  5. Stamens are long with versatile anthers.
  6. Stigma is feathery, exposed to receive the pollen grains coming along with the wind, e.g. Grasses, maize, Jowar and Palms.

Question 8.
Hydrophily.
Answer:

  1. The transfer of pollen grains with the help of water is called hydrophily.
  2. Plants that are pollinated by water are called hydrophilous plants.
  3. Hydrophilous plants possess small, inconspicuous unisexual flowers.
  4. Flowers lack fragrance, nectar and bright colour.
  5. Pollen grains and other floral parts are protected from getting wet.
  6. Stigma is long and sticky, e.g. Zostera, Vallisneria, etc.

Question 9.
Omithophily.
Answer:

  1. The transfer of pollen grains through birds is called ornithophily.
  2. Bird pollinated plants are called ornithophilous plants.
  3. Ornithophilous plants bear large and showy flowers.
  4. Flowers brightly coloured to attract birds for pollination.
  5. Ornithophilous flowers lack fragrance as birds have poor sense of smell.
  6. Pollen grains are sticky and spiny e.g. Callistemon, Bignonia, Bombax, Butea, etc.

Question 10.
Dichogamy.
Answer:

  1. When stamens and carpels mature at different times in a bisexual flower, the condition is known as dichogamy.
  2. Owing to dichogamy self-pollination is avoided and cross pollination is favoured.
  3. Dichogamy is of two types, viz., protandry and protogyny.
  4. Protandry is seen in sunflower in which pollen grains are released much before stigma becomes receptive.
  5. In protogyny, stigma becomes ready to receive the pollen grains before the anthers mature. It is seen in plants like Gloriosa.

Question 11.
Embryo sac.
Answer:

  1. Egg apparatus is a three celled structure lying at the micropylar end of the embryo sac.
  2. The egg apparatus consists of a median egg cell called oosphere and two lateral cells called synergids.
  3. The embryo sac also consists of three antipodal cells or antipodals towards the chalazal end which degenerate after fertilization.
  4. In the centre, the embryo sac consists of a large central cell consisting of two haploid polar nuclei.
  5. The polar nuclei at a later stage fuse with each other forming a diploid secondary nucleus.
  6. The secondary nucleus develops into endosperm.

Question 12.
Entomophily.
Answer:

  1. Pollination with the help of insects is called entomophily.
  2. The insect pollinated flowers are called J entomophilous flowers.
  3. Entomophilous flowers show the following adaptations:
  4. Flowers are large and attractive.
  5. Flowers are brightly coloured with i pleasant smell.
  6. Flowers produce nectar which is food for the insects.
  7. Pollen grains are spiny and sticky for easy adherance to the rough and sticky stigma.
  8. Entomophily is seen in plants like rose, Jasmine, Oestrum, Salvia, etc.

Question 13.
Endosperm.
Answer:

  1. Endosperm is a nutritive tissue. It nourishes the developing embryo.
  2. The endosperm develops from the primary endosperm nucleus (PEN).
  3. The endosperm is a post fertilization tissue.
  4. There are two types of seeds depending upon the presence or absence of endosperm, viz., endospermic and non-endospermic.
  5. Castor, coconut, maize, etc. are endospermic seeds, while bean, pea, gram, etc. are non-endospermic seeds.

Maharashtra Board Class 12 Biology Important Questions Chapter 1 Reproduction in Lower and Higher Plants

Question 14.
Triple fusion.
Answer:

  1. Triple fusion is also called second fertilization.
  2. Out of the two male gametes in angiosperms, the first one fuses with the egg to form the zygote, while the second one fuses with the secondary nucleus to form primary endosperm nucleus. This is called triple fusion. Since each of the polar nuclei is a sister nucleus of the egg, it is called second fertilization.
  3. First fusion involves the fusion of a male gamete with the egg; the second fusion involves the fusion of two polar nuclei to form the secondary nucleus and the third fusion involves the fusion of the other male gamete with the secondary nucleus.

Short Answer Questions

Question 1.
What is asexual reperoduction? Describe fragmentation.
Answer:

  1. Production of offspring without involving fusion of two compatible gametes or sex cells is called asexual reproduction.
  2. Fragmentation : It is a type of asexual reproduction observed in lower plants, e.g. algae.
  3. Multicellular organisms break into small pieces called fragments which develop into new plant.
  4. These fragments are formed due to different reasons like accidental breakdown, death and decay of cells, etc.

Question 2.
Explain about artificial methods of vegetative reproduction.
Answer:

  1. Vegetative propagation is a kind of asexual reproduction which occurs with the help of vegetative plant parts.
  2. Cutting and grafting are two methods used to propagate desired varieties of plants.
  3. Cutting – small pieces of plant parts having one or more buds are selected for propagation, e.g, Stem cutting – Rose, Root Cutting – Blackberry and Leaf cutting – Sansevieria.
  4. Grafting – In this method two plant parts are joined ogether (Stock – rooted plant and Scion-attached plant and they continue their growth as one plant.
  5. When a single bud is grafted on stock plant it is called as bud grafting, e.g. Rose, Apple, Pear.

Question 3.
What do bananas and figs have in common?
Answer:
Banana and fig, both are edible, soft, pulpy sweet fruits which are rich source of potassium. They are grown commercially.
[Note : Banana is a true fruit, simple fleshy berry, developing from single ovary. It may contain tiny seeds in pulp. Banana is parthenocarpic fruit which is developed by parthenocarpy.

Fig is a composite fruit syconus, developing from hypanthodium inflorescence. It is 5 pollinated by insect. We come across tiny seeds inside pulp. It is a false fruit. Receptacle is edible part which encloses tiny female flowers?]

Question 4.
Describe the T.S. of anther.
OR
Sketch and label the T.S. of undehisced anther.
Answer:
Maharashtra Board Class 12 Biology Important Questions Chapter 1 Reproduction in Lower and Higher Plants 1

  1. Internally it shows four chambers called microsporangia or pollen sacs.
  2. The anther consists of two main parts, viz., anther wall and microsporangium or pollen sac.
  3. The wall of the anther can be differentiated into four layers, viz., epidermis, endothecium, middle layers and tapetum.
  4. The epidermis is the outermost layer of the anther wall. It is made up of flattened cells which are protective in function.
  5. The endothecium lies internal to the epidermis. It is made up of a single layer of cells. The cells of endothecium show fibrous thickenings on radial walls.
  6. Internal to the endothecium, lie 1 to 3 layers of parenchymatous cells forming middle layers of the anther wall. The cells of middle layers degenerate at maturity during the formation of microspores.
  7. The tapetum is the innermost nutritive layer of the wall of the anther, consisting of a single layer of cell surrounding the sporogenous tissue.

Question 5.
Describe the structure of a mature anatropous ovule or a typical angiospermic.
Answer:
Maharashtra Board Class 12 Biology Important Questions Chapter 1 Reproduction in Lower and Higher Plants 2

  1. The ovule which has a bent axis and downwardly directed micropyle is called anatropous ovule.
    It is the most common type of ovule in angiosperms.
  2. The matured anatropous ovule consists of two parts, viz., the stalk and the body. The stalk of the ovule is called the funicle or funiculus. The funicle attaches the ovule with the placenta.
  3. The point at which the funicle is attached to the body of the ovule is called hilum.
  4. Nucellus : It is made up of diploid parenchymatous cells.
    The basal part of the nucellus is called chalaza.
    The protective coverings of the nucellus are called integuments.
  5. Micropyle : The integuments do not completely cover the nucellus. They leave a small opening called micropyle at the tip.
  6. Embryo sac : In a mature ovule, the nucellus shows an oval-shaped structure towards its micropylar end called embryo sac or female gametophyte.

Question 6.
Why do some plants have both chasmogamous and cleistogemous plants flowers?
Answer:

  1. When flowers open, their sex organs are exposed for further process of fertilization then it is chasmogamous condition.
  2. Pollinating agents can easily transfer pollen grains in such flowers for self as well as cross pollination.
  3. When flowers are closed, they are self pollinated in bud conditions then this condition is cleistogamy.
  4. When some plants have both of these types of flowers it ensures pollination and fertilization leading to seed setting. When seeds are formed then perpetuation of species is achieved as new plants will germinate from it.

Question 7.
What is pollination ? What are its two types ?
Answer:
1. Pollination : The transfer of pollen grains from the anther to the stigma is called pollination.

2. Types of pollination : Pollination is of two types, viz., self-pollination and cross pollination.
(i) Self-pollination (Autogamy) : The transfer of pollen grains from the anther to the stigma of the same flower or a different flower possessing the sam genetic make-up is called self-pollination.

(ii) Cross pollination (Allogamy) : The transfer of pollen grains from the anther of a flower to the stigma of another flower borne by a different plant possessing dissimilar genetic make-up is called cross pollination.

Question 8.
What are the different types of cross pollination based on the abiotic pollinating agents?
Answer:
Based on the abiotic pollinating agents, pollination can be either anemophily or hydrophily.
(1) Anemophily : Pollination with the help of wind is called anemophily. The wind pollinated plants are called anemophilous plants. Anemophily is seen in plants like grasses, maize, wheat, rice, palms, etc.

(2) Hydrophily : Pollination effected through the agency of water is called hydrophily. Water pollinated plants are called hydrophilous plants. Hydrophily is of two types viz., hypohydrophily and epihydrophily. Plants such a zostera, Vallisneria, etc. are hydrophilous plants.

Question 9.
What are different types of cross pollination based on the biotic pollinating agents?
Answer:
Cross pollination through biotic agents are entomophily, ornithophily and chiropterophily.
(1) Entomophily : Pollination effected through insects is called entomophily. Insect pollinated plants are called entomophilous. Entomophily is seen in plants like Hibiscus, Rose, Salvia, Oestrum, Jasmine, etc.

(2) Ornithophily : Pollination effected through the agency of birds is called ornithophily. Bird pollinated plants are called ornithophilous plants.
Ornithophily is seen in plants like Callistemon, Bombax, Butea, etc.

(3) Chiropterophily : Pollination effected through bats is called chiropterophily. Bat pollinated plants are called chiropterophilous plants. Chiropterophily is seen in plants like Anthocephalous (Kadamb tree), Adansonia (Baobab tree), Kigellia (Sausage tree).

Question 10.
Give the floral adaptations for chiropterophily.
Answer:

  1. The pollination that occurs with the help of bats is called chiropterophily.
  2. In chiropterous plants the flowers are large and stout enough in such a way that bats can hold onto the flowers.
  3. Chiropterous flowers are nocturnal, i.e., they open during the night time only.
  4. Flowers emit rotten fruits like fermenting fruity odours which attract bats.
  5. Flowers produce copious nectar.
  6. Flowers possess large number of stamens which produce large amount of edible pollen grains.
  7. Anthocephalus, Kigellia and Adansonia are chiropterous flowers.

Chart or Table based guestions

I. Complete the following charts

Question 1.
Maharashtra Board Class 12 Biology Important Questions Chapter 1 Reproduction in Lower and Higher Plants 3
Answer:
Maharashtra Board Class 12 Biology Important Questions Chapter 1 Reproduction in Lower and Higher Plants 4

Question 2.
Maharashtra Board Class 12 Biology Important Questions Chapter 1 Reproduction in Lower and Higher Plants 5
Answer:
Maharashtra Board Class 12 Biology Important Questions Chapter 1 Reproduction in Lower and Higher Plants 6

Question 3.
Maharashtra Board Class 12 Biology Important Questions Chapter 1 Reproduction in Lower and Higher Plants 7
Answer:
Maharashtra Board Class 12 Biology Important Questions Chapter 1 Reproduction in Lower and Higher Plants 8

II. Complete the following tables

Question 1.
Maharashtra Board Class 12 Biology Important Questions Chapter 1 Reproduction in Lower and Higher Plants 9
Answer:
Maharashtra Board Class 12 Biology Important Questions Chapter 1 Reproduction in Lower and Higher Plants 10

Question 2.
Complete the table-Related to outbreeding devices.

TypeDescriptionExample
1. ————Unisexual flowers/Monoecious or dioecious plantsPapaya, Maize
2. ————Mechanical device to prevent Self-Pollination – Natural physical barrier—————-
3. Prepotency——————–Apple
4. Heteromorphy——————–Primula
5. ProtandryAndroecium matures earlier than gynoecium—————-

Answer:

TypeDescriptionExample
1. UnisexualityUnisexual flowers/Monoecious or dioecious plantsPapaya, Maize
2. HerkogamyMechanical device to prevent Self-Pollination – Natural physical barrierCalotropis
3. PrepotencyPollens of other flower germinate rapidly rather than from sameApple
4. HeteromorphyPresence of different forms of flowers with respect to Stigma and anthersPrimula
5. ProtandryAndroecium matures earlier than gynoeciumSunflower disc florets

Maharashtra Board Class 12 Biology Important Questions Chapter 1 Reproduction in Lower and Higher Plants

Question 3.

HypohydrophilyEpihydrophily
1. Porogamy————–
2. ————–Entry of pollen tube in Ovule piercing integuments
3. —————Entry of pollen tube in Ovule through Chalaza

Answer:

HypohydrophilyEpihydrophily
1. PorogamyEntry of pollen tube into Ovule through micropyle
2. MesogamyEntry of pollen tube in Ovule piercing integuments
3. ChalazogamyEntry of pollen tube in Ovule through Chalaza

Diagram based questions

Question 1.
T. S. of anther.
Answer:
Maharashtra Board Class 12 Biology Important Questions Chapter 1 Reproduction in Lower and Higher Plants 11

Question 2.
Development of male gametophyte
Answer:
Maharashtra Board Class 12 Biology Important Questions Chapter 1 Reproduction in Lower and Higher Plants 12

Question 3.
Development of female gametophyte
Answer:
Maharashtra Board Class 12 Biology Important Questions Chapter 1 Reproduction in Lower and Higher Plants 13

Question 4.
Double fertilization
Answer:
Maharashtra Board Class 12 Biology Important Questions Chapter 1 Reproduction in Lower and Higher Plants 14

Question 5.
Maize Seed.
Answer:
Maharashtra Board Class 12 Biology Important Questions Chapter 1 Reproduction in Lower and Higher Plants 15

Question 6.
Entry of pollen tube into Ovule – Porogamy
Answer:
Maharashtra Board Class 12 Biology Important Questions Chapter 1 Reproduction in Lower and Higher Plants 16

Long Answer Questions

Question 1.
Describe the types of reproduction in lower plants (i) Budding and (ii) Sporulation. Illustrate your answer with suitable diagrams.
Answer:
(i) Budding:
Maharashtra Board Class 12 Biology Important Questions Chapter 1 Reproduction in Lower and Higher Plants 17

  1. It is a type of asexual reproduction.
  2. It is of very common occurrence in unicellular organism yeast.
  3. It is observed in favourable condition.
  4. Mother cell produces small outgrowth which is known as bud.
  5. Buds maybe one or more and on separation, they grow as new individual.

(ii) Spore formation/Sporulation:
Maharashtra Board Class 12 Biology Important Questions Chapter 1 Reproduction in Lower and Higher Plants 18

  1. It is a tpe of asexual reproduction.
  2. It is of very common occurrence in lower plants.
  3. It occurs by production of motile zoospores that are formed in sporangia.
  4. Flagellated zoospores when liberated can grow independently into new individuals.
  5. Biflagellate zoospores are formed in algae Chlamydomonas.

Question 2.
Describe the structure of a mature pollen grain.
OR
Sketch and label pollen grain.
Answer:

  1. A typical angiospermic pollen grain (mature) is a unicellular, uninucleate, spherical or oval haploid structure.
  2. The pollen grain is also called microspore.
  3. It is covered and protected by a double layered wall called sporoderm.

Maharashtra Board Class 12 Biology Important Questions Chapter 1 Reproduction in Lower and Higher Plants 19

  1. The outer layer of the wall is thick. It is known as exine. The inner layer of the wall is thin. It is known as intine.
  2. The exine is made up of a complex substance called sporopollenin. The sporopollenin protects the pollen grain from physical and biological decomposition.
  3. The exine is spiny in insect pollinated plants, with sculptured pattern or smooth in wind pollinated plants.
  4. The exine is not continuous throughout. It is interrupted, very thin at one or more places by small pores called germ pores.
  5. The intine which is composed of cellulose and pectin encloses the protoplasm with a single haploid nucleus.

Question 3.
Describe the development of female gametophyte in angiosperms.
OR
What is megasporogenesis ? Give an account of development of the female gametophyte.
OR
With a neat diagram explain the 7-celled, 8-nucleate nature of the female gametophyte.
Answer:
1. Megasporogenesis : The process by which . the diploid megaspore mother cell of nucellus undergoes meiosis to form a tetrad of haploid megaspores is known as megasporogenesis.

2. Development of female gametophyte:
(i) The diploid megaspore mother cell undergoes meiosis to form a linear tetrad consisting of four-haploid megaspores. Generally, the chalazal megaspore becomes the functional megaspore. The other three megaspores degenerate.

(ii) The chalazal megaspore (fertile megaspore) is the first cell of the female gametophyte. It undergoes enlargement and develops into the female gametophyte. The haploid nucellus of chalazal megaspore undergoes three successive free nuclear mitotic divisions to produce eight nuclei. Of these, the first mitotic division results in the formation of two nuclei.

(iii) Both these nuclei undergo two successive mitotic divisions resulting in the formation of four nuclei at both the poles. In the meantime, one nucleus from each pole called polar nucleus moves towards the centre of the embryo sac and fuse to form a diploid nucleus called secondary nucleus.

(iv) The three nuclei at the micropylar end are organised to form a three-celled structure called egg apparatus, while the other three nuclei at the chalazal end reorganise to form three antipodal cells. The egg apparatus consists of a central cell called egg cell or female gamete which is flanked by two lateral cells called synergids.

(v) The female gametophyte consists of an egg apparatus, a secondary nucleus and three antipodal cells, A7 celled 8 nucleated structure.

Maharashtra Board Class 12 Biology Important Questions Chapter 1 Reproduction in Lower and Higher Plants

Question 4.
What are the three types of endosperm? Describe them briefly.
Answer:
There are three types of endosperm, viz., nuclear, cellular and helobial.
(i) Nuclear endosperm:
Maharashtra Board Class 12 Biology Important Questions Chapter 1 Reproduction in Lower and Higher Plants 20

  1. Nuclear endosperm is the most common type of endosperm.
  2. During the formation of nuclear endosperm, the primary endosperm nucleus (PEN) undergoes free nuclear division forming a large number of triploid nuclei which remain freely suspended in the common cytoplasm of central cell.
  3. A central vacuole pushes the nuclei towards periphery.
  4. Later on wall formation takes place around these nuclei to form a cellular mass.
  5. It is seen in plants like maize, sunflower, wheat, coconut, etc.

(ii) Cellular endosperm:

  1. In this type of endosperm, the triploid primary endosperm nucleus undergoes nuclear divisions followed by cytokinesis.
  2. Owing to this, the development of endosperm occurs in cellular form.
  3. It is less common and seen in dicot plants like Datura, Petunia, Balsam, Adoxa.

(iii) Helobial endosperm:

  1. In helobial type of endosperm, the first division of the primary endosperm nucleus is followed by the formation of cell wall.
  2. Owing to this, the central cell is divided into a large micropylar cell and a small chalazal cell.
  3. In both micropylar and chalazal chamber, the further development of the endosperm is of nuclear type.
  4. Walls develop between nuclei in micropylar chamber.
  5. This type of embryo development is seen in plants belonging to order Helobiales of Monocots. e.g. Asphodehis.

Question 5.
What is apomixis? Explain the categories of apomixis.
Answer:
(i) Apomixis : The phenomenon of formation of embryo(s) by asexual methods without formation of gametes and fertilization is termed as apomixis.

(ii) There are three main categories of apomixis.

  1. Recurrent
  2. Non-recurrent and
  3. Adventive embryony.

1. Recurrent apomixis : In this diploid sporophytic cell, archesporial cell or nucellus form embryos, When diploid megaspore mother cell forms embryo sac it is known as diplospory. It is also called apospory.

2. Non-recurrent apomixis : Haploid embryo sac is formed but the embryos arise either from egg cell or any other haploid cell. It is also known as apogamy.

3. Adventive Embryony : In this in addition to normal zygotic embryo, additional embryos develop from nucellus or integuments. It results in polyembryony.