Important Questions – Page 8 – Maharashtra Board Solutions

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

July 30, 2024July 29, 2024 by Bhagya

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

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

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.

Animal	Respiratory organ
(1) Fishes	(a) Trachea
(2) Birds/Reptiles	(b) Moist cuticle
(3) Insects	(c) Gills
(4) Earthworm	(d) Lungs

Answer:

Animal	Respiratory organ
(1) Fishes	(c) Gills
(2) Birds/Reptiles	(d) Lungs
(3) Insects	(a) Trachea
(4) Earthworm	(b) Moist cuticle

Question 2.

Respiratory organs	Alternative name
(1) Larynx	(a) Lid of larynx
(2) Trachea	(b) Air sacs
(3) Alveoli	(c) Sound box
(4) Epiglottis	(d) Windpipe

Answer:

Respiratory organs	Alternative name
(1) Larynx	(c) Sound box
(2) Trachea	(d) Windpipe
(3) Alveoli	(b) Air sacs
(4) Epiglottis	(a) Lid of larynx

Question 3.

Respiratory capacities	Respiratory 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 capacities	Respiratory 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.

Disease	Symptoms
(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:

Disease	Symptoms
(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 heart	Location
(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 heart	Location
(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 vessel	Functions
(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 vessel	Functions
(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.

Cells	Functions
(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:

Cells	Functions
(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 ECG	Heart 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 ECG	Heart 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 cycle	Time 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 cycle	Time 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 A	Column B
(1) Plasma proteins	————
(2) Nitrogenous waste	————
(3) Inorganic salts	————

Answer:

Column A	Column B
(1) Plasma proteins	Serum albumin Fibrinogen
(2) Nitrogenous waste	Urea, Uric acid
(3) Inorganic salts	Bicarbonates, 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 A	Column B
(1) External gills	————
(2) Book lungs	————
(3) Lungs	————

Answer:

Column A	Column B
(1) External gills	Amphibian tadpoles of frog, Salamanders
(2) Book lungs	Scorpion, Spiders
(3) Lungs	Reptiles, Birds

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 A	Column B
(1) Occupational disorders	————
(2) Disorders due to smoking and air pollution	————
(3) Disorders due to viruses and bacteria	————

Answer:

Column A	Column B
(1) Occupational disorders	Asbestosis, Silicosis
(2) Disorders due to smoking and air pollution	Emphysema, Chronic bronchitis
(3) Disorders due to viruses and bacteria	Pneumonia, 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 A	Column B
(1) Phagocytic cells	————
(2) Cells involved in giving immune response	————
(3) Cells that increase during allergic and anti-allergic responses	————

Answer:

Column A	Column B
(1) Phagocytic cells	Neutrophils Monocytes
(2) Cells involved in giving immune response	T-lymphocytes-B-lymphocytes
(3) Cells that increase during allergic and anti-allergic responses	Eosinophils, 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 HCO₃^–and 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.

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.

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

Pushing oxygen into the lungs
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:

Transport of oxygen and carbon dioxide
Transport of food
Transport of waste product
Transport of hormones
Maintenance of pH
Water balance
Transport of heat
Defence against infection
Temperature regulation
Blood clotting/coagulation
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.

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:

Pharynx	Laryix
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:

Inspiration	Expiration
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 respiration	Internal 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 O₂	Transport 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 Lung	Total 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 artery	T.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:

Erythrocytes	Leucocytes
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:

Eosinophils	Basophils
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:

Neutrophils	Eosinophils
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:

Lymphocytes	Monocytes
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:

Granulocytes	Agranulocytes
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 circulation	Double 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 circulation	Diastolic 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:

Atria	Ventricles
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. Node	A.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 circulation	Systemic 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 valves	Semilunar 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:

Hypertension	Hypotension
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:

During cellular respiration, the oxidation of food (glucose) takes place.
This happens in the mitochondria using the oxygen present in the blood.
ATP molecules are formed during this oxidation.
ATP is used for various vital body processes and also for maintaining the body temperature to constancy.
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:

Vestibule is the anterior most part of the nasal chamber.
The hairs present in this region trap the dust particles and prevent them from entering into the interiors of the respiratory passage.
Therefore, the vestibule of nasal chambers has fine hair.

Question 3.
Glottis is guarded by a flap called epiglottis.
Answer:

The oesophagus and trachea lie side by side.
There is possibility that food particles may enter respiratory passage at the time of gulping.
However, the epiglottis prevents the entry of food into the respiratory passage by closing it temporarily.
Thus, for preventing the entry of food particles into respiratory passage, the glottis is guarded by a flap called epiglottis.

Question 4.
Alveoli are very flexible.
Answer:

Alveoli are made up of collagen and elastin fibres.
They are very thin (0.0001 mm) and lined by non-ciliated squamous epithelium.
All the above structural components make the alveoli very flexible.

Question 5.
Expiration is called a passive process.
Answer:

During expiration, intercostal muscles relax. This results in pulling the ribs inwards.
Diaphragm also relaxes and returns to its normal dome shape.
The collective contraction of ribs and diaphragm results in the reduction of
thoracic cavity and hence automatically air is pushed out of the lungs.
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:

Pericardium protects the heart. It is double layered peritoneum, having outer fibrous and inner serous pericardium layers.
Fibrous pericardium being tough gives protection to the heart.
Serous pericardium has two layers, parietal and visceral layer or epicardium.
In between these two layers, there is pericardial fluid, which helps to absorb shocks and provide nourishment.
In this way pericardium acts as a defence wall.

Question 7.
Valves are present in veins.
Answer:

Veins carry blood to the heart.
At that time the backward flow of the blood should be prevented.
Therefore, valves are present in veins.

Question 8.
Atria are thin walled than ventricles.
Answer:

Atria are receiving chambers, while ventricles are distributing chambers.
The blood is driven out from ventricles.
Ventricles are therefore, strong and with thicker walls.
Atria are thin walled as compared to ventricles.

Question 9.
Heart is called a pump.
Answer:

The heart acts as a pumping organ. It shows continuous pumping action.
The rhythmic contraction or systole and relaxation or diastole of heart forms one heartbeat.
Such heartbeats occur about 72 times per minute.
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:

In normal human heart, there is completely formed atrioventricular septum.
This septum keeps the deoxygenated and oxygenated blood separate.
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:

When the blood gushes through the blood vessels, the walls of blood vessels -can expand a little due to their elasticity.
But as the age advances, the elasticity is reduced and then the blood vessels do not expand.
Hence the flowing blood gets more resistance and the blood pressure can rise.
Lesser the elasticity more will be the blood pressure, whereas more the elasticity of the vessel wall, then the pressure will not rise.
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:

About 70% of CO₂ is transported in the form of sodium bicarbonates/potassium bicarbonates from tissue cells to lungs.
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.
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.
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.
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.
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:

Carbon monoxide poisoning is caused when carbon monoxide is combined with haemoglobin.
Haemoglobin is said to have 250 times more affinity for carbon monoxide than that for the oxygen.
Therefore, haemoglobin with carbon monoxide forms a stable compound, the carboxyhemoglobin.
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.
Carbon monoxide poisoning occurs in closed rooms with incompletely burning substances such as stove burners or furnaces and garages having running automobile engines.
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:

Erythrocytes or red blood corpuscles. They are circular, biconcave, enucleated cells.
The RBC size : 7 pm in diameter and 2.5 pm in thickness.
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.
The average life span of RBC : 120 days.
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.
The old and worn out RBCs are destroyed in liver and spleen.
Polycythemia is an increase in number of RBCs while erythrocytopenia is decrease in their (RBCs) number.

Functions of RBCs:

Transport of oxygen from lungs to tissues and carbon dioxide from tissues to lungs with the help of haemoglobin.
Maintenance of blood pH as haemoglobin acts as a buffer.
Maintenance of the viscosity of blood.

Question 6.
Heartbeat.
Answer:

The rhythmic contraction and relaxation of the heart is called heartbeat.
Each heartbeat includes one systole and one diastole. During systole the heart contracts and during diastole it relaxes.
The rate with which the heart beats is called heart rate. The normal heart rate is 72 beats per minute.
Tachycardia means faster heart rate of about more than 100 beats per minute.
Bradycardia means slower heart rate that is below 60 beats per minute.

Question 7.
Pulse.
Answer:

A pressure wave that travels through the arteries after each ventricular systole is called a pulse.
The pulse can be felt in any artery that lies near the surface of the body.
The radial artery at the wrist is most commonly used to feel the pulse.
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.

Question 8.
Peacemaker.
Answer:

Pacemaker is the region in tile heart which initiates the beating.
The natural pacemaker of the heart is sinoatrial node (SA node).
The pacemaker is autorhythmic, it is able to repeatedly and rhythmically generate impulses.
SA node is responsible for initiation of cardiac excitation. Therefore, it is called a pacemaker.

Question 9.
Blood pressure.
Answer:

Blood pressure is the pressure exerted by the flowing blood on the walls of arteries.
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.
Diastolic blood pressure is the minimum pressure of blood when heart undergoes diastole. Normal diastolic pressure is 80 mm Hg.
Blood pressure is represented as 120/80 mm Hg for a normal human being.

Question 10.
Hypertension.
Answer:

In a normal healthy person the blood pressure values are 120 mm Hg (systolic)/ 80 mm Hg (diastolic).
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.
Excessively high blood pressure is very dangerous as high blood pressure of about 220/120 mm Hg may cause rupturing of blood vessels.
Rupture of eye blood vessels can lead to blindness.
If blood vessels of kidney are affected then nephritis is caused.
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.
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:

Coronary artery disease is a condition caused due to problems like atherosclerosis.
In this disease, coronary arteries are narrowed due to deposition of fatty substances.
Due to this the blood flow to the heart is reduced.
In coronary heart disease, the heart muscle is damaged because of an inadequate amount of blood due to obstruction of its blood supply.
The symptoms of CAD depend upon the degree of obstruction.
Symptoms are mild chest pain or angina pectoris.
In severe cases it results in heart attack known as myocardial infarction.

Question 12.
Angina pectoris.
Answer:

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.
Atherosclerosis and arteriosclerosis can cause this problem. Basically, the coronary arteries are affected during angina pectoris.
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.
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:

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.
Hypertension increase the after load on the heart leading to significant enlargement of the heart.
This finally results in heart failure.
Factors responsible for heart failure are advanced age, malnutrition, chronic infections, toxins, severe anaemia or hyperthyroidism, etc.
Any problem leading to degeneration of heart muscle, may result in heart failure.

Question 14.
Atherosclerosis.
Answer:

Atherosclerosis is the deposition of fatty substances and cholesterol on the inner lining of eateries.
This deposition results in the formation of atherosclerotic plaque.
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.
Atherosclerosis of the coronary arteries results in decrease in the blood flow to the heart muscles.
Due to such condition, coronary heart disease is caused.

Question 15.
ECG.
Answer:

Electrocardiogram or ECG is the graphic v record of electrical variations produced by the heart during one heartbeat or cardiac cycle.
ECG is taken with the help of an instrument called electrocardiograph or ECG machine. Electrocardiograph records the action potentials generated by the heart muscles.
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.
A normal ECG is a graph having series of ridges and furrows. There are waves such as P-wave, QRS complex and T-wave.
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.
QRS-complex wave begins as a downward deflection, continues as a large upright triangular wave and ends’ as a downward wave.
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.
T-wave is a broad upward wave which represents ventricular repolarization resulting in ventricular relaxation.
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:

Angiography is an X-ray imaging of the cardiac blood vessels to locate the position of blockages.
Depending upon the degree of blockage, remedial procedures like angioplasty or by¬pass surgery are performed.
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:

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.
Symptoms of silent heart attack are so mild that a person often confuses it for regular « discomfort and thereby ignores it.
Men are more affected by silent heart attack than women.

Question 18.
Heart Transplant.
Answer:

Heart transplant is the replacement of severely damaged heart by normal heart from brain-dead or recently dead donor,
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:

Respiration is the biochemical process in which organic compound such as glucose are oxidized to liberate chemical energy.
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
ATP is the biologically useful energy. ATP drives most of the life process.
When cell requires the energy, ATP is hydrolyzed and is converted into ADP with subsequent release of energy.
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:

Haemoglobin is a respiratory pigment present in cytoplasm of RBCs. About 97% of oxygen is transported by these haemoglobin molecules from lungs to tissues.
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)
Oxyhaemoglobin is transported from lungs to the tissues where it readily dissociates to release O₂.
Hb (4O₂) → Hb + 4O₂
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:

Blood is the fluid connective tissue that circulates in the body.
Blood is derived from mesoderm.
It is bright red, slightly alkaline fluid having pH about 7.4. It is salty, viscous fluid heavier than water.
The average sized adult has about 5 litres of blood in his/her body which constitutes about 8% of the total body weight.

Question 5.
Describe the structure and the function of thrombocytes.
Answer:

Thrombocytes or platelets are non- nucleated, round and biconvex blood corpuscles.
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.
Their life span is about 5 to 10 days.
Thrombocytes are formed from megakaryocytes of bone marrow. They break from these cells as fragments during the process of thrombopoiesis.
Thrombocytosis is the increase in platelet count while thrombocytopenia is decrease in platelet count.
Thrombocytes possess thromboplastin which helps in clotting of blood.
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:

The heart wall is composed of three layers, viz. outer epicardium, middle myocardium and inner endocardium.
Epicardium is composed of single layer of mesothelium having flat epithelial cells.
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.
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:

In one normal heartbeat the heart sounds like lubb and dup are produced once each.
The rhythmic contraction (Systole) and relaxation (diastole) forms are heartbeat. The heart sounds are due to closure of valves.
Lubb sound is produced during ventricular systole when the cuspid valves close both the atrioventricular apertures preventing blood flow into atria.
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:

In human beings, there is double circulation because blood passes twice through the heart during one cardiac cycle.
The blood follows two routes viz. pulmonary and systemic.
Pulmonary circulation is circulation between heart and lungs. Systemic circulation is the circulation between the heart and body organs (except lungs).
During pulmonary circulation, the blood passes from the right ventricle to the left atrium of the heart through lungs.
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.
During systemic circulation, the blood from the left ventricle passes to the right atrium of heart through body organs.
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:

Normal activities of the heart are auto regulated. The specialized muscles help in this regulation.
The heart is said to be myogenic due to this ability.
In the medulla oblongata of brain, there is cardiovascular centre,
From this centre, sympathetic and parasympathetic nerves innervate the sinoatrial node.
Sympathetic nerves secrete adrenaline and it stimulates and increases the heartbeat.
Parasympathetic nerves secrete acetylcholine and it decreases the heart rate.
Rate of heartbeat is controlled in response to inputs from various receptors like proprio-receptors.
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:

Hypoxia, acidosis, alkalosis cause decrease in cardiac activity.
Hormones like epinephrine and nor epinephrine enhance the cardiac activity.
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:

It should have a large surface area.
It should be thin, highly vascular and permeable to allow exchange of gases.
It should be moist.

Question 12.
What is the co-relationship between activeness of organism and complexity of transport system?
Answer:

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.
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.
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:

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.
Eosinophils : Granules of eosinophil contains a unique toxic basic protein receptors that bind to IgE used to help in killing parasites.
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:

The normal haemoglobin in adult male is 13-18 mg/100 ml of blood.
In a normal adult female, it is about 11.5-16.5 mg/100 ml of blood.
In anaemic individuals there is lesser amount of haemoglobin.
Functions of haemoglobin is to transport oxygen from lungs to tissues and carbon dioxide from tissues of lungs.
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:

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.
When the breath is held beyond a limit, the urge to breath becomes irresistible.
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:

Diapedesis is the movement of leucocytes through the wall of blood capillaries into the tissue space.
Leucocytes perform diapedesis as an important part of their reaction to tissue injury or infection.
This process forms the part of the innate immune response, involving recruitment of non-specific leucocytes.
Monocytes also use this process during their development into macrophages.
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:

Physical compression of the kidneys by fat.
Activation of the renin-angiotensin – aldosterone system
Increased sympathetic nervous system activity.
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:

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.
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:

Large animals have various organ systems which always work in a coordinated manner.
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.
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.
Therefore, large animals cannot carry out respiration without the help of circulatory system.

Question 21.
What is immunity? Name its types.
Answer:

Immunity is the general ability of a body to recognize, neutralize or destroy and eliminate foreign substances or resist a particular infection or disease.
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:

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.
The dengue virus also links with platelets in the blood when there is a virus-specific antibody present in the human body.
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.
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:

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.
This parasite attacks host tissues including both peripheral and central lymphoid tissues.
This causes systemic acute response in host body which the parasite tries to overcome. The parasite in fact weakens both innate and acquired immunity.
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:

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.
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.
In people, the antigens or markers that identity their immune system are referred to as the human leukocyte antigen (HLA).
Antigens that are recognized as unfriendly invaders stimulate an immune response to destroy them.
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:

Organism	Habitat	Respiratory surface/organ
1. Insects	Terrestrial	—————
2. Amphibian tadpoles of frog, salamanders	—————-	—————-
3. Fish	Aquatic	————–
4. Reptiles, Birds and Mammals	—————-	—————-

Answer:

Organism	Habitat	Respiratory surface/organ
1. Insects	Terrestrial	Tracheal tubes and spiracles
2. Amphibian tadpoles of frog, salamanders	Aquatic	External gills
3. Fish	Aquatic	Internal gills
4. Reptiles, Birds and Mammals	Terrestrial	Lungs

Question 2.

Partial pressure of gases	Alveolar air	Pulmonary, capillaries
PPO₂	—————	—————
PPCO₂	—————-	—————-

Answer:

Partial pressure of gases	Alveolar air	Pulmonary, capillaries
PPO₂	104 mm Hg	40 mm Hg
PPCO₂	40 mm Hg	45 mm Hg

Question 3.
Maharashtra Board Class 12 Biology Important Questions Chapter 8 Respiration and Circulation 3
Answer:

Question 4.
Complete the following flow chart
Maharashtra Board Class 12 Biology Important Questions Chapter 8 Respiration and Circulation 5
Answer:

Question 5.
Complete the following Table

Waves on ECG	Heart Activity	Caused due to
P wave	Atrial contraction	—————-
QRS wave	—————–	Ventricular depolarization
T wave	—————–	Ventricular repolarization

Answer:

Waves on ECG	Heart Activity	Caused due to
P wave	Atrial contraction	Atrial depolarization
QRS wave	Ventricular contraction	Ventricular depolarization
T wave	Ventricular contraction	Ventricular repolarization

Question 6.

Cardiovascular disorders	Symptom
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 Attack	Myocardial infarction without ———————-

Answer:

Cardiovascular disorders	Symptom
Coronary Artery Diseases (CAD)	Deposition of calcium, fat, cholesterol and fibrous tissues in blood vessels.
Angina pectoris	Pain in chest resulting from reduction in the blood supply to the cardiac muscles.
Silent Heart Attack	Myocardial infarction without showing symptoms of classical heart attack.

Question 7.

Instrument / Technique	Purpose of use
Sphygmomanometer	———————
—————-	X-ray imaging of the cardiac blood vessels to locate the position of blockages.
—————	To measure ECG.

Answer:

Instrument / Technique	Purpose of use
Sphygmomanometer	To measure blood pressure.
Angiography	X-ray imaging of the cardiac blood vessels to locate the position of blockages.
Electrocardiograph	To 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:

Name	Function
(A) Epiglottis	Epiglottis prevents the entry of food into trachea.
(B) Tracheal cartilage	Tracheal 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

Why the circulation shown in the above diagram is called double circulation?
What are the two main routes of double circulation?
Which blood vessels carry oxygented blood to heart and deoxygenated blood to lungs?
Which blood vessels carry deoxygented blood to heart and oxygenated blood to body organs?

Answer:

During circulation, blood passes twice through the heart, therefore it is called double circulation.
(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.
Oxygenated blood is carried to the heart by pulmonary veins. Dexoygenated blood is carried to the lungs by pulmonary artery.
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.

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

Which phases of cardiac cycle are shown in the above diagrammatic representation ?
How much time is taken for entire heart to be in diastole?
How much longer is ventricular systole as compared to atrial systole?

Answer:

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.
Diastole of entire heart is called joint diastole, which is for about 0.4 second.
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

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.
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.
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.
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.
Respiratory part is the second region which is richly supplied with the capillaries. Air is made warm and moist in this region.
Sensory part is lined by sensory epithelium. It is concerned with the detection of smell.

2. Pharynx:

Pharynx is a short and vertical tube measuring about 12 cm in length. In pharynx the respiratory and food passages cross each other.
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.
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:

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.
From pharynx air enters the larynx. The opening through which it enters is called glottis. Glottis is guarded by a flap called epiglottis.
Epiglottis prevents the entry of food particles into the trachea.
TWo folds of elastic tissue called vocal cords are seen along the side of glottis. When they vibrate the sound is produced.

4. Trachea:

The trachea or wind pipe is about 12 cm long and 2.5 cm wide.
It is situated in front of the oesophagus and runs downwards in the thorax through the neck.
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.
Internally the tracheal wall bears ciliated epithelium and mucous glands.
When any foreign particle enters the trachea inadvertently. It is thrown out by coughing action.
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:

At the distal end, the trachea divides into two bronchi (Singular – bronchus). Bronchi lie below the sternum or breast bone.
Each bronchus has a complete ring of cartilage for support. The two bronchi enter into the lungs on either side.
After entering into the lungs each bronchus divides into secondary and tertiary bronchi. The tertiary bronchi divide and re-divide to form minute bronchioles.
Bronchioles do not have cartilages in their walls. Each bronchiole ends into a balloon like alveolus.
Owing to the presence of alveoli the lungs become spongy and elastic.

6. Lungs:

Lungs are principal respiratory organs located in the thoracic cavity.
They are pinkish, soft, hollow, paired, elastic and distensible organs.
Each lung is enclosed in a pleural sac which consists of two membranes, viz. an outer parietal and inner visceral.
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.
Lungs are highly vascular as they are richly supplied with blood capillaries.
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.
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.
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.
The alveolar wall is 0.0001 mm thick and made up of simple, non-ciliated, squamous epithelium. It has collagen and elastin fibres.
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:

Inspiration is the process in which the air containing oxygen is taken inside the lung.
Inspiration is the active process which is possible due to intercostal muscles, sternum and diaphragm.
During inspiration, intercostal muscles contract, ribs are pulled outward as a result of which the space in the thoracic cavity is increased.
At the same time the lower part of the breast bone is raised and diaphragm flattens by contraction.
The volume of thoracic cavity is thus increased.
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:

Expiration is the process in which air containing carbon dioxide and water vapour is expelled out of the lungs.
Expiration is a passive process.
During expiration, intercostal muscles relax and ribs are pulled inwards.
The diaphragm relaxes and becomes dome shaped. Intercostal muscles contract simultaneously and due to these events, the volume of the thoracic cavity is reduced.
The pressure on the lungs is thus increased as a result of which they are compressed.
Due to this, air rushes out of the lungs and is expelled out through the nose.

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:

Following are some respiratory disorders : Emphysema is caused due to alveolar abnormalities. Chronic bronchitis results into coughing and shortness of breath.
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.
Allergens like pollen or pet dander can cause asthma. In asthma constriction of bronchioles takes place causing periodic wheezing and difficulty in breathing.
Occupational hazards cause respiratory diseases like silicosis or asbestosis. In these disorders there is inflammation fibrosis leading to lung damage.

Treatments of respiratory diseases:

Bacterial diseases can be completely cured by specific antibiotics.
Viral diseases need to be taken care of by using vaporizers and decongestants.
Asthma needs treatment by inhalers and nebulizers.
For occupational disorders proper mask and other protective gear is a must.
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:

Only 3% of the total oxygen is transported in a dissolved state by the plasma.
The remaining 97% is transported in the form of oxyhaemoglobin in the RBCs.
Hemoglobin present is RBCs combines with oxygen to form oxyhaemoglobin.
Hb + 4O₂ → Hb (4O₂)
Oxyhaemoglobin is transported from lungs to the tissues where it readily dissociates to release O₂.
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.

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.
By bicarbonate ions (70%) : Nearly 70% of carbon dioxide is transported in the form of sodium bicarbonate/potassium bicarbonate in the plasma.
RBCs contains an enzyme, carbonic anhydrase. In the presence of this enzyme CO₂ combines with water to form carbonic acid.
Carbonic anhydrase also brings about dissociation of carbonic acid immediately tending to large accumulation of HCO^3- ions inside the RBCs.
CO₂ + H₂O Carbonic anhydrese H₂CO₂ Carbonic anhydrase H₊ + HCO^3-
The bicarbonate ions moves out of RBCs and this would bring about imbalance of the charge inside the RBCs.
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.
HCO^3- ions from the plasma then joins to Na⁺/K⁺ forming NaHCO₃/KHCO₃ (to maintain PH of blood).
HCO^3- + Na⁺ → NaHCO₃ Sodium bicarbonate
H⁺ is taken up by haemoglobin to form Reduced Hb (HHb).
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⁺ + HCO_3- Carbonic anhydrase H₂CO₃ Carbonic anhydrase CO₂ + H₂O.

3. By red blood cells (23%):

Carbon dioxide binds with the amino group of the haemoglobin and form a loosely bound compound carbaminohaemoglobin Hb + CO₂ = HbCO₂
Due to low partial pressure of CO₂ at alveolus carbaminohaemoglobin decomposes releasing the carbon dioxide.

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

Leucocytes or White Blood Corpuscles (WBCs) are colourless, nucleated, amoeboid and phagocytic cells.
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.
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.
Leucocytes are mainly divided into two types, viz., granulocytes and agranulocytes.
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:

In neutrophils, the cytoplasmic granules take up neutral stains.
Their nucleus is three to five lobed.
It may undergo changes in structure hence they are called polymorphonuclear leucocytes or polymorphs.
Neutrophils are about 70% of total WBCs.
They are phagocytic in function and engulf microorganisms.

(II) Eosinophils or acidophils:

Cytoplasmic granules of eosinophils take up acidic dyes such as eosin. They have bilobed nucleus.
Eosinophils are about 3% of total WBCs.
They are non-phagocytic in nature.
Their number increases (i.e. eosinophilia) during allergic conditions.
They have antihistamine property.

(III) Basophils:

The cytoplasmic granules of basophils take up basic stains such as methylene blue.
They have twisted nucleus.
In size, they are smallest and constitute about 0.5% of total WBCs.
They too are non-phagocytic.
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:

Agranulocytes with a large round nucleus are called lymphocyte.
They are about 30% of total WBCs.
Agranulocytes are responsible for immune response of the body by producing antibodies.

(II) Monocytes:

Largest of all WBCs having large kidney shaped nucleus are monocytes. They are about 5% of total WBCs.
They are phagocytic in function.
They can differentiate into macrophages for engulfing microorganisms and removing cell debris. Hence they are also called scavengers.
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:

There are two atria which are separated from each other by interatrial septum. They are thin walled receiving chambers on the upper side.
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.
Eustachian valve guards the opening of inferior vena cava while opening of coronary sinus is guarded by Thebesian valve.
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.
Right atrium opens into the right ventricle.

2. Left atrium:

The oxygenated blood from the lungs is brought into left atrium through four openings of pulmonary veins.
Left atrium opens into the left ventricle.

II. Ventricles:

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.
Left ventricle has thickest wall as it pumps blood to all parts of the body.
The inner surface of the ventricle is thrown into a series of irregular muscular ridges called columnae carnae or trabeculae carnae.
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.
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.
From the right ventricle arises pulmonary trunk which carries deoxygenated blood to lungs for oxygenation.
From the left ventricle arise systemic aorta which distributes oxygenated blood to all parts of the body.
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.

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:

Blood clotting is coagulation of blood in order to stop the blood flow and resuting blood loss at the time of injury.
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.
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:

Release of thromboplastin from thrombocytes and injured tissue.
Formation of enzyme prothrombinase in the blood due to initiation of thromboplastin.
Conversion of inactive prothrombin into active thrombin by prothrombinase in the presence of Ca ions.
Conversion of soluble fibrinogen into insoluble fibrin by thrombin.
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:

When cardiac cells are at rest, they are polarized, meaning no electrical activity takes place.
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.
Electrical impulses are generated by specialized cardiac cells automatically.
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.
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.
Depolarization with corresponding contraction of myocardial muscle moves as a wave through the heart. Depolarization thus corresponds with contraction of heart.
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.
Depolarization and repolarization are electrical activities which cause muscular activity.
The electrical changes in the myocardial cell during the depolarization – repolarization cycle is detected on ECG.

Question 13.
How are the signals detected and amplified by electrocardiograph?
Answer:

The action potential created by contractions of the heart wall spreads electrical currents from the heart throughout the body.
The spreading electrical currents create different potentials at points in the body, which can be sensed by electrodes placed on the skin.
The electrodes are made of metals and salts and they act as biological transducers.
Ten electrodes are attached to different points on the body while taking ECG.
There Eire three main leads responsible for measuring the electrical potential difference between arms and legs.
Electrical potential difference between electrodes is recorded.
As in all ECG lead measurements, the electrode connected to the right leg is considered the ground node.
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

July 30, 2024July 29, 2024 by Bhagya

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

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

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

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 A	Column 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 A	Column 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 A	Column 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 A	Column 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 A	Column 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 A	Column 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 A	Column B (Organisms)
1. Symbiotic nitrogen fixation	a. Nitrobacter
2. Denitrification	b. Cyanobacteria
3. Free living nitrogen fixers	c. Rhizobium
4. Nitrification	d. Paracoccus

Answer:

Column A	Column B (Organisms)
1. Symbiotic nitrogen fixation	c. Rhizobium
2. Denitrification	d. Paracoccus
3. Free living nitrogen fixers	b. Cyanobacteria
4. Nitrification	a. 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 A	Column B
(1) Shedding of leaves	————-
(2) Induce flowering in LDP	————
(3) Apical dominance	————-
(4) Induce RNA synthesis	————-

Answer:

Column A	Column B
(1) Shedding of leaves	Abscissic acid
(2) Induce flowering in LDP	GA
(3) Apical dominance	IAA
(4) Induce RNA synthesis	Cytokinin

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 A	Column B
(1) Selective synthetic herbicide	————-
(2) Seedless fruits	————
(3) Flowering in pineapple	————-
(4) Synthetic auxin	————-
(5) Agent orange	————

Answer:

Column A	Column B
(1) Selective synthetic herbicide	2, 4-D
(2) Seedless fruits	IAA
(3) Flowering in pineapple	NAA
(4) Synthetic auxin	IBA
(5) Agent orange	2, 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 A	Column B
(1) Bolting of rosette plants	————-
(2) Epinasty	————
(3) Closure of stomata	————-
(4) Proliferation of callus	————-

Answer:

Column A	Column B
(1) Bolting of rosette plants	GA
(2) Epinasty	Ethylene
(3) Closure of stomata	ABA
(4) Proliferation of callus	Kinetin

Question 4.
Classify the given disease to their cause given in Column B.
(Ethylene, GA, Deficiency of BO, Deficiency of Cu)

Column A	Column B
(1) Brown heart disease	————-
(2) Bakane disease of Rice	————
(3) Die back of shoot	————-
(4) Degreening of Banana	————-

Answer:

Column A	Column B
(1) Brown heart disease	Deficiency of BO
(2) Bakane disease of Rice	GA
(3) Die back of shoot	Deficiency of Cu
(4) Degreening of Banana	Ethylene

Question 5.
Classify the given organisms related to Nitrogen cycle in Column B.
[Nitrobacter, Rhizobium, Pseudomonas, Nitrosococcus)

Column A	Column B
(1) Symbiont in root nodule	————-
(2) Conversion of nitrite to nitrate	————
(3) Denitrification process	————-
(4) Conversion of ammonia to nitrite	————-

Answer:

Column A	Column B
(1) Symbiont in root nodule	Rhizobium
(2) Conversion of nitrite to nitrate	Nitrobacter
(3) Denitrification process	Pseudomonas
(4) Conversion of ammonia to nitrite	Nitrosococcus

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.

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

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.

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.

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 division	Phase 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 plants	Short 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 LDP	4. 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 minerals	Active 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:

Meristematic cells divide and form new cells.
Absorption of water is necessary for maintaining turgidity in the newly formed cells.
Turgidity results in enlargement of cells in phase of cell elongation.
Water is essential component of protoplasm of cells.
biochemical reactions. Therefore it is essential for growth in plants.

Question 2.
2, 4-D is used as herbicide.
Answer:

2, 4-D is a synthetic auxin which kills dicot weeds.
Our most of the food crops are cereals, i.e. monocot plants.
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:

Auxins and cytokinins are growth promoting substances which stimulate cell division and cell enlargement.
A high cytokinin promotes shooting in plants.
A low ratio of cytokinin to auxin induces root development.
A high ratio of cytokinin to auxin induces growth of buds and shoot development.
Thus cytokinin and auxin ratio and their interactions control morphogenesis in plants.

Question 4.
ABA is described as an antitranspirant.
Answer:

ABA is a growth inhibiting hormone.
ABA is responsible for causing efflux of K⁺ ions from guard cells of stomata.
As a result of this, osmotic changes occur and guard cells become flaccid resulting in closure of stomata.
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:

When mineral element is present below a certain critical concentration it is said to be deficient.
Symptoms are indicated in the form of certain morphological changes on the mobility of element.
These symptoms depend on the mobility of element inside the plant body.
When the element is relatively immobile like S, Ca then the symptoms appear first in young leaves.
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:

Minerals exist in soil in the form of charged particles.
Certain negatively charged (anions) get accumulated on the inner side of cell membrane after their entry inside cell.
These anions cannot diffuse out through semipermeable cell membrane.
Thus additional mobile cautions are needed inside the cell to balance these fixed anions.
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:

Absorption of mineral ions from soil against concentration gradient is known as active absorption.
It requires energy (ATP) for absorption by absorbing root cell.
The source of energy is respiration of cells for supply of ATE
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:

Nitrogen is a major nutrient for plant growth.
Proper carbon/nitrogen ratio in soil is necessary for plant growth.
It is component of proteins in the form of amino acids.
Proteins are synthesised from photosynthetic products sugars.
Nitrogen exists in atmosphere but it is inert, non-reactive.
Plants need nitrogen in a reactive form usually nitrate in soil.
This supply need to be maintained through biological and physical nitrogen fixation.
Otherwise productivity is affected hence it is limiting nutrient in the agricultural ecosystem.

Question 9.
Cucumber and sunflower are regarded as photoneutral plants.
Answer:

In cucumber and sunflower, the flower is not controlled by light period.
Both these plants flower in all light periods.
Cucumber and sunflower, therefore, are regarded as photoneutral plants.

Write short notes on the following

Question 1.
Meristems
Answer:

In vascular plants, growth is indeterminate and occurs at specific regions where meristems are located.
Meristems are of three types based on their location – apical, intercalary and lateral.
Meristems are thin walled cells with prominent nucleus with granular cytoplasm, non-vacuolated.
Mitotic divisions take place in meristematic cells.

Question 2.
Phase of cell formation
Answer:

Formative phase is the first phase of growth.
During this phase, the meristematic cells undergo mitosis to form new cells.
During formative phase, the rate of growth is slow.
The phase of cell formation is also called lag phase.
This phase is also known as phases of cell division.

Question 3.
Development
Answer:

Development is progressive changes taking place in shape, form and degree of complexity in an organism.
In plants, it includes all the changes taking place in sequence from seed germination to senescence or death of plant.
Development is an orderly process.
It includes growth, morphogenesis, maturation and senescence.

Question 4.
Plasticity
Answer:

Plasticity is the capacity of plant being molded or formed.
It is ability of plant to develop different kinds of structures in response to environmental factors or stimuli.
Different kinds of structures can be developed in plants due to internal stimuli in different phases, i.e. juvenile and adult.
Heterophylly is shown in plant in different phases or in different environmental conditions.
In coriander and cotton plants, two different kinds of leaves are observed in young (juvenile) and mature (adult) plant.
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:

Phytohormones or plant growth regulators are internal factor that influence growth.
They inhibit, promote or modify the plant growth.
Plant hormones are organic substances produced naturally in plants and required in small amount.
Their place of production and site of the activity are different.
Auxins, gibberellins, cytokinins are growth promoters and ethylene, abscissic acid are growth inhibitors.

Question 6.
Phytochrome
Answer:

These are proteinaceous pigments present
Phytochrome exists in two interconvertible forms. P_r and P_fr.
Phytochromes are located in cell membranes of chlorophyllous cells.
When P_fr absorbs red light it is converted to P_r.
P_fr is accumulated in plants during daytime and inhibits flowering in SDP but initiates flowering in LDP

Question 7.
Venralization
Answer:

Effect of temperature on flowering of plants is known as vernalization.
For inducing early flowering pretreatment of seeds or seedlings is done at 1 to 6 °C for about a months duration.
Shoot apical meristem is believed to be site of vernalization stimulus.
Vernalization stimulus is in a form of chemical named vernalin.
Vernalization is effective in ereals (wheat) and crucifers.

Question 8.
Apical dominance
Answer:

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.
It is believed that apical dominance is controlled by an auxin which is synthesized in the apical bud.
From the apical bud, the auxin migrates to the lateral buds and inhibits their growth.
When apical bud is removed, the lateral buds grow and form branches.
For producing more branches therefore, the apical buds are removed.
Cytokinins reverse apical dominance effect by promoting growth of lateral buds by cell division.

Question 9.
Toxicity of micronutrients or mineral toxicity
Answer:

Micronutrients are required in minute quantities by plants.
Their moderate decrease causes deficiency symptoms while their moderate increase causes toxicity.
The reduction in dry weight of a tissue by 10% by any mineral is known as toxicity.
It is not easy to identify toxicity symptoms.
Most of the time, the excess of an element inhibits the uptake of another element resulting in causing the deficiency symptom of that element.
Manganese inhibits calcium translocation towards apex of stem and exhibits symptoms of chlorosis with grey spots appearing on leaves.
This is because manganese competes with iron and magnesium for uptake.
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:

Day neutral plants are those in which flowering is not affected by the duration of the light period.
Day neutral plants flower in all photoperiods.
Day neutral plants are also known as photoneutral or intermediate plants.
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:

The differentiated cells which are formed may again gain the capacity to divide as per need.
Permanent cells (mature cells) undergo de-differentiation and become meristematic.
This acquired feature of living permanent cells is known as de-differentiation.
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:

F.W. Went named the growth promoting substance as Auxin.
Auxin was also isolated from urine of patient of pellagra.
Indole 3 acetic acid was first hormone discovered in plants.
IAA is most common and natural hormone synthesized at growing tips and responsible for cell elongation.
It is synthesized from tryptophan and shows polar transport – Basipetal transport.

Question 3.
Discuss about discovery of Gibberellins.
Answer:

Gibberellins are growth promoting hormones and were isolated form fungus Gibberella Jujikuroi by Kurasawa.
Rice plants when infected with this fungus show stem elongation i.e. Bakane disease.
Yabuta and Sumuki isolated gibberellins in crystalline form, from fungal culture and named it gibberellins.
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:

Cytokinins are growth promoting hormones that stimulate cell division.
Skoog and Miller discovered first cytokinin when they were investigating nutritional requirements of tobacco callus culture.
It was observed that the callus proliferated when there was addition of coconut milk as supplement.
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:

Abscissic acid (ABA) is a natural growth inhibiting hormone.
It was observed by Carns and Addicott that shedding of cotton balls occur due to chemical substance abscission I and II.
From the buds of Acer, Wareing isolated substance that causes bud dormancy and named it as dormin.
These two chemical substances were identical and now known as abscissic acid.
ABA is synthesized in leaves, fruits and seeds from mevalonic acid.

Question 6.
What is day neutral plant (DNP)? Give any two examples.
Answer:

The plants which do not require specific duration of light period or dark period flowering are day neutral plants (DNP).
They flower throughout the year, as they do not need specific photoperiod.
The flowering in these plants is independent of photoperiod.
Examples – cucumber, tomato, cotton, sunflower, maize and balsam.

Question 7.
Discuss about discovery of phytochromes.
Answer:

Phytochromes are proteinaceous pigments present in cell membrane of green cells.
Phytochromes receive photoperiodic stimulus and induce flowering in plants in response to light duration.
Hendrick and Borthwick observed that in SD plants flowering is inhibited if dark period is interrupted by flash of red light (660 nm).
If flash of far red light (730 nm) is given then again flowering is observed in SD plants.
Pigment system of plant receives photoperiodic stimulus and these pigments exist in two interconvertible forms P^fr and P^r

Question 8.
What is mineral nutrition of plants?
Answer:

Plants require inorganic materials for the synthesis of food.
These elements are obtained by plants in the form of minerals mainly form soil.
Chemical analysis of plant ash reveals that about 40 different minerals are needed by plants which are taken from surroundings, (air, soil and water)
These minerals are absorbed in dissolved form, i.e. ionic form through root system mainly root hairs.
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:

Stunting : Retarded growth and thus stem appears short and condensed.
Chlorosis : This is loss or lack of chlorophyll that result in yellowing of leaf.
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:

Donnan equilibrium is a process of passive absorption of minerals in plants which is without any expenditure of energy.
It is assumed that certain anions after their entry by diffusion into the cell get fixed on inner side of cell membrane.
Additional mobile cations are needed to balance this fixed anions as they cannot diffuse outside.
Concentration of cations thus increases due to accumulation.
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:

Conversion of free nitrogen of air into nitrogenous compounds that are made available to plants for uptake is known as : nitrogen fixation.
Physical nitrogen fixation occurs in step- : wise manner and it takes place in atmosphere j and soil.
Under the influence of electric discharge, lightning and thunder, atmospheric nitrogen combines with oxygen to form nitric oxide.
Nitric oxide is then oxidized to nitrogen peroxide in presence of oxygen.
Nitrogen peroxide combines with rainwater to form nitrous and nitric acid which come on ground as acid rains.
On ground, alkali radicals (mainly of Ca, K) react with nitric acid to produce nitrites and nitrates which are absorbable forms for plants.

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.

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:

Question 2.
Complete the flow chart of development.
Maharashtra Board Class 12 Biology Important Questions Chapter 7 Plant Growth and Mineral Nutrition 8
Answer:

Question 3.
Complete the table of growth hormones.
Maharashtra Board Class 12 Biology Important Questions Chapter 7 Plant Growth and Mineral Nutrition 10
Answer:

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:

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

From which cells newly formed cells are added? How?
How is rate of growth in zone of cell elongation?
What is peculiarity of zone of differentiation? How is rate of growth in this region?

Answer:

Meristematlc cells add new cells by mitotic division.
Rate of growth is at accelerated pace.
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

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:

Long answer questions

Question 1.
What is hydroponics? How is it useful in identifying the role of nutrients?
Answer:

Growing plants in aqueous (soilless) medium is known as hydroponics. [Greek word hudor = water and ponos = work]
It is technique of growing plants by supplying all necessary nutrients in the water supply given to plant.
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.
Hydroponics is of great use in studying the deficiency symptoms of different mineral nutrients.
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.
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.
Any visible change noticed from normal structure and function of the plant is the symptom or hunger sign considered.
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:

Plants absorb minerals from the soil with their root system.
Minerals are absorbed from the soil in the form of charged particles, positively charged cations and negatively charged anions.
The absorption of minerals against the concentration gradient which requires expenditure of metabolic energy is called active absorption.
The ATP energy derived from respiration in root cells is utilized for active absorption.
Ions get accumulated in the root hair against the concentration gradient.
These ions pass into cortical cells and finally reach xylem of roots.
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:

Growth is a permanent increase in size, weight, shape, volume and dry weight of a plant.
The change occurring due to growth is permanent and irreversible.
Growth is an intrinsic process caused due to internal activities.
Growth occurs by cell division and cell elongation followed by cell maturation which lead to the formation of different types of tissues.
Growth in plants is mostly localized, i.e. restricted to some regions of plants possessing meristematic tissues or meris terns.
Growth has a qualitative aspect where development takes place in an orderly manner and differentiation leads to higher and more complex state.

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:

For a proper growth of plant various environmental and physiological conditions are necessary.
Carbon/Nitrogen ratio in soil is having effect on growth as both carbon and nitrogen are structural elements in carbohydrates, proteins and other biomolecules.
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.
Nutrients are necessary for proper growth. Macronutrients and micronutrients have their specific role.
Temperature of 25 – 35 °C is optimum for growth.
Light is essential for seed germination and photosynthesis.
Oxygen is necessary for respiration and supply of energy.
Gravitational force decides direction of growth for root system and shoot system.
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:

Plant growth regulators are organic compounds other than nutrients.
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.
They are required in minute quantities.
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:

Auxins
Gibberellins
Cytokinins
Ethylene and
Abscissic acid.

Role of abscissic acid (ABA) in plants:

Abscissic acid influences abscission and dormancy.
ABA accelerates senescence of leaves, flowers and fruits.
It is a stress hormone as it is produced during drought and other unfavourable climatic conditions.
ABA induces dormancy in seeds, buds and tubers.
It acts as growth inhibitor as it retards growth.
ABA plays an important role in closing of stomata to check transpiration.
It inhibits and delays cell division and suppresses cambial activity by inhibiting mitosis in vascular cambium.
ABA inhibits flowering in LDP and stimulates flowering in short day plants (SDP).

Question 8.
Write an account of auxins as growth regulators.
Answer:

Auxins are plant growth regulators produced naturally by plants.
They are weak organic acids capable of promoting cell elongation during the growth of stem and root.
Auxins are synthesized in shoot and root apices besides young leaf primordia.
Auxins may be natural or synthetic.
Naturally occurring auxins are indole-3- acetic acid (IAA) and its derivatives.
NAA, 2, 4-D and 2,4, 5-T are synthetic auxins.
Auxins in higher concentration promote the growth of stem.
Auxins play an important role in initiation and promotion of cell division.
Auxins help in the formation of adventitious roots from cuttings when applied in lower concentration.
Auxins play an important role in apical dominance.
Auxin prevents abscission by preventing the action of hydrolytic enzymes in abscission layer.
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:

Auxins are growth promoting substances synthesized at meristematic regions of plants.
The primary effect of auxin is cell enlargement and it stimulates growth of stem and root.
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.
Owing to activity of inducing multiplication of cells it is used in plant tissue culture to produce callus.
Auxin stimulates formation of lateral and adventitious roots hence used for rooting propagation of cuttings.
2, 4-D is a synthetic herbicide which kills dicot weeds.
Induced parthenocarpy in fruits-oranges, banana, grapes, lemons is by application of auxin.
Foliar spray of NAA and 2, 4-D induces flowering in litchi and pineapple.
Premature fruit drop of apples, pear and oranges is prevented.
Auxins break seed dormancy and promote germination.
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:

Gibberellins are growth promoting hormone and it is present in root tips, stem tips and seeds.
Gibberellins break dormancy of bud, dormancy of seed.
They promote seed germination in cereals by activating or synthesising enzyme amylase to produce sugar.
Gibberellins induce elongation of the cells in stem hence increase in internode length is noticed.
In rosette plants like cabbage it causes ‘bolting’ that is increase in internode length before flowering.
Gibberellins are more effective in inducing parthenocarpy than auxins in plants like tomato, apple and pear.
It is also used to increase fruit size and length of bunches in grapes.
By its application genetically dwarf plants can be converted to phenotypically tall e.g. Maize.
It overcomes requirement of vernalization, delays senescence and prevents abscission.
Application of gibberellins causes production of male flowers on female plants.

Question 11.
Describe about the physiological effects and applications of cytokinin.
Answer:

Cytokinins are growth promoting hormone that promotes cell division. Kinetin, zeatin are examples of cytokinin.
They promote cell division as well as cell enlargement.
High cytokinin promotes shoot development.
Growth of lateral buds is promoted by cytokinins. Thus it controls apical dominance.
Process of ageing and senescence, abscission of plant organs is delayed by their application.
It promotes formation of interfascicular cambium.
It has a role in breaking seed dormancy and promotes seed germination.
Cytokinins induce RNA synthesis.
Cytokinin and auxin ratio and their interactions control morphogenesis and cell differentiation.

Question 12.
Discuss about physiological effects and applications of ethylene.
Answer:

Ethylene is a gaseous growth inhibitor hormone.
It promotes ripening of fruits like bananas, apples and mangoes. The commercial application of ethephon is done.
It initiates growth of lateral roots.
Dormancy of buds and seeds is broken by its application.
It accelerates formation of abscission layer and thus abscission of leaves, flowers and fruits is observed.
Ethylene is responsible for checking growth of lateral buds thus causes apical dominance and retards flowering.
Process of senescence of plant organs is enhanced.
Epinasty, i.e. drooping of leaves and flowers results due to its application in some plants.
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:

Phytochrome pigments receive photoperiodic stimulus and control flowering in plants.
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.
If this interruption is again exposed to flash of far red light of wavelength 730 nm, then these plants flower.
From this they concluded that some pigment system in plant receives the photoperiodic stimulus.
These pigment proteins are called phytochromes and it exists in two interconvertible forms – P_r and P_fr.
These pigments are located in cell membranes of green cells.
P_fr is biologically active form and during daytime it gets accumulated in the plant.

Question 14.
Give schematic representation of nitrogen cycle and enlist important steps of this cycle.
Answer:

Atmospheric nitrogen is a source of nitrogen cycle.
The important steps of the cycle are Nitrogen fixation, Nitrification, Ammonification, Nitrogen assimilation by plants. Amino acid synthesis and amidation, Denitrification and sedimentation.
Amino acid are building blocks of proteins. Amides are amino acid with two amino groups.
Schematic representation of Nitrogen

Question 15.
What is nitrogen cycle? Describe it briefly.
Answer:

The cyclic movement of nitrogen between the atmosphere, biosphere and geosphere in different forms is called nitrogen cycle.
Nitrogen cycle is one of the most important biogeochemical cycles.
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.
Bacteria such as Nitrosomonas, Nitrosococcus and Nitrobacter are the nitrifying bacteria which play an important role in nitrification.
Denitrification is carried out by the bacteria Pseudomonas denitrifficans. From this it is obvious that bacteria play a major role in nitrogen cycle.
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

July 30, 2024July 29, 2024 by Bhagya

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

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

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 A	Column B (New Terminology)
(1) D.P.D.	(a) Osmotic potential
(2) O.E	(b) Pressure potential
(3) T.P	(c) Water potential

Answer:

Column A	Column 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 A	Column B
Macro elements	——–, ———-, ——–
Micro elements	——–, ———-, ——–

Answer:

Column A	Column B
Macro elements	P, N, S
Micro elements	B, Co, Mn

Question 2.
Capillary water, Combined water, Hygroscopic water, Gravitational water.

Column A	Column 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 A	Column B
(1) Water adsorbed on soil particles	Hygroscopic water
(2) Water penetrated deep in soil	Gravitational water
(3) Water present as hydrated oxides	Combined water
(4) Water present between soil particles	Capillary 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.

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:

Epistomatic leaf – Lotus
Hypostomatic – Nerium.
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

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:

Necessary for absorption of water by root hairs
For absorption of minerals
Conduction of water against gravity
Exchange of gases
Transport and distribution of food.

Question 3.
Turgor pressure.
Answer:

Keeps cells and cell organelles stretched
Provides support
During growth essential for cell enlargement
Maintains shape of cell
Facilitates opening and closing of stomata

Question 4.
Osmosis.
Answer:

Absorption of water into root
Maintains turgidity of cell
Facilitates cell to cell movement of absorbed water
Resistance to drought or frost injury
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:

Removal of excess of water
Helps in passive absorption of water and minerals
Helps in ascent of sap – transpiration pull
Maintains turgor of cells
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:

They contain few chloroplasts so capable of photosynthesis.
They can change their size hence opening and closing of stomatal aperture.

Distinguish between the following

Question 1.
Diffusion and Osmosis.
Answer:

Diffusion	Osmosis
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 absorption	Passive 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 transpiration	Stomatal 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:

Water plays an important role in living organisms.
About 90 – 95% water is present in cell which is functional and structural unit of living organisms.
It helps in maintaining turgidity and shape of cells and cell organelles.
Due to its various properties, it is medium of biochemical reactions, transporting medium and thermal buffer also.
Therefore it is absolutely necessary for life i.e. ‘elixir of life’.

Question 2.
Water is a best transporting medium.
Answer:

Water is in liquid state at room temperature.
It is best solvent for most of the solutes. Thus called universal solvent.
It is inert inorganic compound with neutral pH i.e. pH 7 when in pure form.
Hence it is best medium for dissolved minerals.

Question 3.
Water is significant molecule that connects physical world with biological processes.
Answer:

Water is an important constituent of cell. About 90 – 95% of protoplasm is water.
Water in liquid state is best solvent in which various minerals and food molecules are dissolved and transported.
Water acts as the thermal buffer has high specific heat.
Water molecules have high adhesive and cohesive forces of attraction.
It can rise in capillaries due to high surface tension and adhesive forces, e.g. Ascent of sap in plants.
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:

Water is absorbed from rhizosphere with the help of unicellular root hairs.
Root hairs have plasma membrane and thin, double layer cell wall of pectin and cellulose.
During imbibition water molecules get tightly adsorbed to the wall of hydrophilic colloids.
Cell wall is freely permeable membrane hence through diffusion water passes into the cell.
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.
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:

Secondary roots develop from the pericycle which is inside endodermis.
Protoxylem is situated close to pericycle in root.
Endodermis have suberized layer Casparian strip which forces water to move in the symplast so that it can enter the vascular xylem.
Since secondary roots originate form pericycle, they bypass the Casparian strip.
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:

Water is absolutely necessary for life.
It is a major constituent of protoplasm.
If provides aqueous medium for various metabolic reactions that take place in plant.
It is raw material for photosynthesis.
It helps in maintaining the turgidity of cells.
It is excellent solvent for various organic materials.
It is transporting medium for dissolved minerals.
It is thermal buffer.

Question 2.
Properties of water.
Answer:

Water is a compound and it is in liquid state at room temperature.
It is an inert inorganic compound with neutral pH.
It has high specific heat, high heat of vapourization, high heat of fusion.
It has high surface tension.
Water molecule has good adhesive and cohesive forces of attraction.
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

Small, raised structures, which are composed of loose cells with large intercellular spaces situated on bark are lenticels.
They are observed on bark of old stem, root and on woody pericarp of fruits.
Lenticular transpiration is of very small amount about 0.1% to 1% of total transpiration.
It is very slow process but occurs throughout the day.

Question 4.
Structure of stomatal apparatus.
Answer:

Stomata are minute openings mainly located in the epidermal surfaces of young stem and leaves.
It is composed of two guard cells and accessory cells which form the opening stomatal pore for transpiration.
Guard cells are kidney shaped in dicot plants and dumbbell shaped in grasses.
Guard cells have unevenly thickened wall, inner wall thick, inelastic while outer wall is thin and elastic.
Guard cells are nucleated cells with few chloroplasts and hence can perform photosynthesis.
Accessory cells are specialized cells that surround the guard cells and are reservoirs of K⁺ ions.

Question 5.
Apoplast pathway.
Answer:

The movement of water across the root cells occur by two pathways viz. apoplast and symplast.
Apoplast movement of water in plants occurs exclusively through cell walls and intercellular spaces in living cells of cortex.
Apoplast pathway is up to the endodermis as the suberized layer of Casparian strip obstructs the movement.
Additional apoplastic route giving direct contact to xylem and phloem is through secondary roots which originate at pericycle and bypass endodermis.
The cellulosic walls of root hair, cortical cells are hydrophilic and permeable to water.

Question 6.
Water potential.
Answer:

Chemical potential of water is called water potential.
The Greek letter (ψ) psi represents water potential.
The unit of measurement is in bars/pascals (Pa) or atmosphere.
Water potential of pure water is always zero.
Water potential of protoplasm is equal to D.PD. but it has a negative value.
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:

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.
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:

Water is absorbed by unicellular root hairs from soil.
Presence of capillary water in soil is needed as this water from soil is absorbed by root hairs.
Soil temperature of 20-30°C favours water absorption.
Rate of absorption is decreased by high concentration of solutes in soil.
Soil should be properly aerated, poorly aerated soil shows poor absorption rate.
Increased transpiration accelerates rate of absorption of water.

Question 3.
Explain mechanism of sugar transport through phloem.
Answer:

The part of plant where food is synthesized is known as source and where it is utilized is sink.
Food is translocated through phloem tissue in the form of sucrose along the concentration gradient from source to sink.
Munch’s pressure flow theory or mass flow hypothesis is widely accepted theory for sugar transport.
Glucose is synthesized in photosynthetic cells hence endo-osmosis occurs due to increased osmotic concentration.
With this turgor pressure increases and sugar is forced in sieve tube of vein from photosynthetic cell – Vein loading.
At the sink, sugar is utilized and excess amount converted to starch, hence osmotic concentration is lowered and exosmosis takes place.
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:

Cohesion tension theory is widely accepted theory for translocation of water proposed by Dixon and Joly.
It is based on principles cohesion and adhesion of water molecules along with transpiration pull.
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.
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.
Transpiration pull is developed in the xylem vessel in leaves.
This pull is transmitted downwards and due to suction pressure, water is pulled passively against gravity, i.e. ascent of sap.

Limitations of theory-

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.
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:

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.
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.
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.

Question 6.
What are adhesive and cohesive forces?
Answer:

Adhesive and cohesive forces are attractive forces between molecules of same substance.
Cohesive forces exist between molecules of the same type. E.g. between water-water molecule.
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:

Question 2.
Complete the table.

Event	Physical 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:

Event	Physical process
1. Soaking of Seeds	Imbibition
2. Water entering guard cells	Osmosis
3. Exchange of gases	Diffusion
4. Loss of water in liquid form	Guttation
5. Water coming out through earthen pot	Diffusion
6. Loss of water in vapour form	Transpiration
7. Absorption of solutes passively by carrier	Facilitated Diffusion
8. Spreading of fragrance of incense stick	Diffusion

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

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:

Plants absorb mincral nutrients from their surroundings.
For a proper growth of plants about 35 to 40 different elements are required.
Plants absorb these nutrients in ionic or dissolved form from soil with their root system e.g. Phosphorus as PO₄^–. Sulphur as SO₄^2- etc.
Based on their requirement in quantity. they are classified as major nutrients or macronutrients and those needed In small amounts are minor or micronutrients.
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.
C. H, O are non-mineral major elements obtained from air and water e.g. CO₂ is source of carbon, Hydrogen from water.
Microelements arc required In traces as they mainly have catalytic role as co-factors or activators of enzymes.
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.
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

July 30, 2024July 29, 2024 by Bhagya

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

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

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 time	Animal life
(1) Cambrian	(a) Amphibians
(2) Ordovician	(b) First terrestrial animals
(3) Silurian	(c) Jawless fishes
(4) Devonian	(d) Trilobite

Answer:

Geological time	Animal life
(1) Cambrian	(d) Trilobite
(2) Ordovician	(c) Jawless fishes
(3) Silurian	(b) First terrestrial animals
(4) Devonian	(a) Amphibians

Question 2.

Human stage	Cranial 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 stage	Cranial 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 I	Column II
(1) Homologous organs	————-
(2) Analogous organs	————-
(3) Vestigial organs	————-
(4) Connecting links	————-

Answer:

Column I	Column II
(1) Homologous organs	Flipper of whale and wing of bird, Vertebrate heart and brain
(2) Analogous organs	Wing of insect and wing of bird. Eye of octopus, an eye of mammal
(3) Vestigial organs	Nictitating membrane, wisdom tooth
(4) Connecting links	Seymouria, Lung fish

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 period	Major events
Cambrian	————-
Devonian	————-
Permian	————-
Triassic	————-

Answer:

Geological time period	Major events
Cambrian	Abundance of trilobites, All types of marine algae
Devonian	Diversification of fishes, Formation of forests
Permian	Origin of conifers. Rise of modern insects.
Triassic	Rise 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.

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:

Overproduction
Struggle for existence
Organic variations
Natural selection
Origin of next species.

Question 3.
Five key factors of evolution as suggested by Stebbins.
Answer:

Gene mutations
Mutations in the chromosome structure and number
Genetic recombinations
Natural selection
Reproductive isolation.

Question 4.
Name the four types of chromosomal aberrations.
Answer:

Deletion
Translocation
Duplication
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:

Habitat or ecological isolating mechanism
Seasonal or temporal isolating mechanism
Ethological isolating mechanism
Mechanical isolating mechanism

Question 9.
Post-mating isolating mechanisms.
Answer:

Gamete mortality
Zygote mortality
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:

Marsupials
Monotremes
Eutheria

Question 17.
Name the three subfamilies of family Hominoidea.
Answer:

Hyalobatidae
Pongidae
Hominidae

Give the significance of the following

Question 1.
Significance of Natural selection
Answer:

Natural selection is the main driving force behind the evolution.
Natural selection favours those genetic variations which have better fitness value.
Such organisms are at selective advantage and they produce more offspring than the rest. Such organisms have greater survival and reproductive capacity.
In this way natural selection helps in the evolution of new species.
Natural selection favours differential reproduction of gene and brings about changes in the gene frequency.
Natural selection brings about evolutionary changes.
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 flow	Genetic 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.

Question 2.
Directional, Stabilizing and Disruptive selection.
Answer:

Directional selection	Stabilizing selection	Disruptive 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 organs	Analogous 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 speciation	Sympatric 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:

Darwin took into consideration minute fluctuating variation as principal factors. But these are neither heritable nor are part of evolution.
Darwin did not distinguish somatic and germinal variation and considered all variations are heritable.
‘Arrival of the fittest’ was not explained by him.
Darwin was unable to explain the cause, origin and inheritance of variations and of vestigial organs.
He also could not explain extinction of species.
Gradual accumulation of useful variations forms the new species, but their intermediate forms were not recognised.
Darwin could not explain existence of neutral flowers and the sterility of hybrids.

Question 3.
The main features of mutation theory.
Answer:

Mutations are large, sudden and discontinuous variations in a population.
Changes caused due to mutations are inheritable.
The raw material for organic evolution is provided by mutations.
Mutation can be useful or harmful. Useful mutations are at evolutionary advantage as they are selected by nature.
Accumulation of the useful mutations over a period of time leads to the origin and establishment of new species.
Harmful or non-adaptive mutation may persist or get eliminated by nature.

Question 4.
Directional selection.
Answer:

Natural selection bringing about directional change without disrupting the balance is called directional selection.
In a population when more individuals acquire characters which are other than the mean character value, then it is called directional selection.
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.
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:

Stabilizing selection is the type of natural selection which balances the population, hence it is also known as balancing selection.
In such population more individuals acquire a mean character value.
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.
Stabilizing selection reduces variations.
It tends to maintain phenotypic stability within population, and does not bring about drastic evolutionary changes.
A population showing stabilizing selection is well-adapted to its environment.

Question 6.
Homologous organs.
Answer:

The structural similarities between the homologous organs indicate that they have a common ancestry.
Different homologous organs indicate divergent evolution or adaptive radiation.
Homologous organs help in tracing the phylogenetic relationships.
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:

Analogous organs are similar in function but dissimilar in structural details.
They do not help to trace the relationship in the evolution but help to understand the convergent evolution.
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:

Vestigial organs are rudimentary organs which are imperfectly developed and non¬functional, degenerate structures.
These organs in animals become functionless thus their presence in the body is not required.
But they are simply present as they descend down during evolution and continue to exist.
In the process of evolution, they may disappear totally.
They indicate evolutionary line as they were once functional in the ancestors.

Examples of vestigial organs in human beings:

Caecum and vermiform appendix : These are functional in herbivorous animals where they help in cellulose digestion. In humans they are functionless.
Nictitating membrane situated in the eyes of humans. It is a remnant of third eyelid.
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:

Major changes that took place in evolution of man are as follows :
Increase in size and complexity of brain and enhanced intelligence.
Increase in cranial capacity.
Bipedal locomotion.
Opposable thumb.
Erect posture.
Shortening of forelimbs and lengthening of hind limbs.
Development of chin. Orthognathous face.
Broadening of pelvic girdle and development of lumbar curvature.
Social and cultural development such as articulated speech, art, development of tools, etc.

Question 11.
Dryopithecus.
Answer:

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.
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.
Dryopithecus has a close similarity to chimpanzee and also walked like a modern chimpanzee.
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.
It had arms and legs of the same length and had a semi-erect posture.

Question 12.
Ramapithecus.
Answer:

Ramapithecus was on direct line of evolution of man.
It was called an ape-man like primate.
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.
It existed during late Miocene and early Pliocene epoch about 14 to 12 million years ago.
It walked erect on its hind limbs.
It had close similarity with chimpanzee.
Some scientists believe that Dryopithecus evolved into Ramapithecus.

Question 13.
Australopithecus.
Answer:

Australopithecus is considered as connecting link between ape and man.
Its fossils were obtained from Toung valley in South Africa, from Ethiopia and Tanzania.
It was in late Pliocene or early Pleistocene epoch about 4 to 1.8 million years ago.
It was about 4 feet tall. It had prognathus face, with larger jaws. Chin was absent. Lumbar curvature was present.
It walked upright.
The cranial capacity was about 450 to 600 CC. Therefore, it was called man with ape brain.

Question 14.
Homo habilis.
Answer:

Homo habilis is described as Handy man. His fossils were obtained from Olduvai Gorge in Tanzania, Africa.
He existed in late Pliocene or early Pleistocene about 2.5 to 1.4 million years ago.
He was lightly built.
Fossil of lower jaw was obtained which showed that his dentition was more like modern man with small molars.
He walked erect. His cranial capacity was 640 to 800 cc.
He did not eat meat and made stone tools.

Question 15.
Homo erectus.
Answer:

Homo erectus was also known as Java man or Peking Man due to his fossils obtained from these areas.
He was also called ape man.
He lived in the middle Pleistocene epoch about 1.5 million years ago.
He was 5 feet in height with prognathous face, massive jaws, huge teeth and bony eye brow ridges.
Chin was absent.
He walked erect.
The cranial capacity was 900 cc.
He was omnivorous and probably used fire and ate meat.

Question 16.
Neanderthal man.
Answer:

The scientific name of Neanderthal man is Homo neanderthalensis. He is described as advanced prehistoric man.
It was called Neanderthal man because its first fossil was collected from Neanderthal valley in Germany by Fuhlrott (1856).
It was heavily built and short and had outwardly curved thigh bones.
The facial features were as follows : prominent brow ridges, thick skull bones, low and slanting forehead, deep jaw without a chin, etc.
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.
The cranial capacity of Neanderthal man was about 1400 cc, which was roughly equal to that of modern man. He used hide for dressing.
It showed intellectual development in constructing and using flint tools and fire.
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:

Origin of Earth and Primitive atmosphere.
Formation of ammonia, water and methane.
Formation of simple organic molecules.
Formation of complex organic molecules.
Formation of Nucleic acids.
Formation of Protobionts or Procells.
Formation of first cell.

Question 2.
When did Earth originate? Which transformations took place later?
Answer:

Earth originated about 4.6 billion years ago as a part of the solar system.
When it was formed, it was a rotating cloud of hot gases and cosmic dust. It was then appearing like a nebula.
Later the condensation and cooling started which resulted in stratification.
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.
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:

The primitive broth in which simple organic molecules were suspended, was neutral and free from oxygen.
In this broth polymerisation took place and simple organic molecules aggregated to form new complex organic molecules like polysaccharides, fats, proteins, nucleosides and nucleotides.
Protoproteins were formed by polymerisation of amino acids. These protoproteins later formed proteins.
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:

By the reaction between phosphoric acid, sugar and nitrogenous bases (purines and pyrimidines), nucleotides may have been formed.
These nucleotides joined together to form nucleic acids such as RNA and DNA.
Nucleic acids acquired self-replicating ability which is a fundamental property of living form.
They later formed protobionts. They were the first form of life formed by nucleic acids along with inorganic and organic molecules.
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.

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:

A connecting link is an intermediate or transitional state between two systematic groups of organisms.
It bears characters common to both these groups on either side of its position. Thus it represents an evolutionary line.
Connecting links are also called a missing link.
E.g. Archaeopteryx, the extinct bird is a connecting link between Reptiles and Aves.
Seymouria is a connecting link between Amphibia and Reptilia.
Ichthyostega is a connecting link between Pisces and Amphibia.

Question 10.
What is geological time scale? How is it divided ?
Answer:

Geological time scale is the arrangement of major divisions of geological time into eras, periods and epochs on the time scale.
This division is based on the study of fossilized organisms obtained from the different strata of the earth.
The characteristic significant events that occurred in the organization of organisms helped the geologists to understand the geological time scale.
The major divisions of geological time are called eras.
The eras are divided into periods and the periods into epochs.
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:

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.
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.
Fossils are formed in variety of materials such as sedimentary rocks, amber, volcanic gas, ice, peat bogs, soil, etc.
They provide the true, direct and reliable evidences of evolution.

Question 12.
What are the molecular evidences that show the evolution?
Answer:

Different organisms have basic similarities in their molecules and the cellular constituents.
All living organisms have the same basic structural and functional unit, i.e. cell.
Cell organelles such as endoplasmic reticulum, Golgi bodies, mitochondria, etc. are present in different types of organisms.
Proteins and gene performing different functions have the same basic pattern which shows a common ancestry.
Catabolic activities of liberating energy, synthesis of macromolecules such as proteins, carbohydrates, nucleic acids, etc. are similar in different organisms.
ATP is the common energy currency of all the organisms.
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:

Homo habilis (650-800 cc)
Homo erectus (850-1200 cc)
Neanderthal man (1400 cc)
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.

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:

Protobionts or coacervates were colloidal aggregations of hydrophobic proteins and lipids (lipoid bubbles).
They grew in size by taking up material from surrounding aqueous medium.
During their growth they became thermodynamically unstable and split into smaller units. These were called microspheres.
They were proteinoids formed from colloidal hydrophilic complexes surrounded by water molecules.
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.
Coacervates and microspheres were non-living colloidal aggregations of lipids and proteinoids respectively.
But they showed growth and division like living cells.
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.

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:

Deletion : Loss of genes from chromosome.
Duplication : Genes are repeated or doubled in number on chromosome.
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.
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:

In post-zygotic or post-mating isolating mechanisms, the two individuals can mate but the result of mating is not favourable.
Thus the populations remain isolated without the actual genetic exchange.

Post-mating isolating mechanisms are divided into the following categories:

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.
Zygote mortality : In zygote mortality, the zygote is formed but it fails to thrive. Though the egg is fertilized the zygote does not survive.
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.

Question 10.
What is Hardy-Weinberg equilibrium? Explain it in brief.
Answer:

Hardy and Weinberg were two scientists who proposed a concept of genetic equilibrium popularly known as Hardy- Weinberg principle or equilibrium.
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.
For explaining the concept of equilibrium they assumed that there are two alleles located at a single locus (A and a).
Their respective frequencies are p and q.
The frequency of genotype AA is p, for 2Aa is 2pq and for aa is q.
The equilibrium equation is p² + 2pq + q² = 1
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

July 30, 2024July 29, 2024 by Bhagya

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

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

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 A	Column 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 A	Column 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 A	Column B
1. Initiator codon	————–
2. Stop codons	————-
3. Codon that codes for Phenyl alanin	————–
4. Codon that codes for leucine	————-

Answer:

Column A	Column 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 A	Column 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 A	Column 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.

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:

Transcription
Activation of amino acids and formation of charged t-RNAs,
Synthesis of polypeptide chain:
initiation
elongation and
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

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:

DNA regulates and controls all the cellular activities.
It replicates and gets distributed equally to the daughter cells when the cell divides.
It is a carrier of genetic information.
Heterocatalytic function : DNA directs the synthesis of chemical molecules other than itself. E.g. Synthesis of RNA (transcription), synthesis of protein (Translation), etc.
Autocatalytic function : DNA directs the synthesis of DNA itself. E.g. Replication.
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:

Euchromatin	Heterochromatin.
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 prokaryotes	DNA 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:

DNA	RNA
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-RNA	t-RNA	r-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:

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.
On the other hand, DNA was considered as a small, simple molecule whose composition does not vary much among species.
Variations in the DNA molecules are different than the variation in shape, electrical charge and function shown by proteins.
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:

Griffith obtained live S-strain bacteria from the blood of the dead mice.
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.
Transforming principle allowed R-type bacteria to synthesize capsule and thus they became virulent.
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:

In a typical mammalian cell, length of DNA double helix is approximately 2.2 metres.
The size of typical nucleus is approximately 10^-6 m
Such a long DNA molecule has to be fitted in small nuclear space.
Therefore, DNA is highly condensed, coiled and supercoiled so that it can be accommodated in the nucleus.

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:

Proteins play a significant role in the metabolism of living cells.
The actual phenotypic expression of living cells is dependent on the biochemical reactions.
Each biochemical reaction needs a specific enzyme for its initiation and completion. All the enzymes are proteins.
In a cell there are many structural proteins too. Thousands of structural and catalytic proteins are constantly required within the cell at all times.
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:

It was believed that there are total 20 amino acids in the living world. But 21st amino acid called selenocysteine was discovered later.
This amino acid is coded by UGA which is usually a termination codon.
In both prokaryotic and eukaryotic cells polypeptide chains contain 100-300 amino acids and they are formed by specific arrangement of 21 amino acids.
But formation of selenocysteine requires the availability of element selenium in the cells.
Therefore, only 20 amino acids are considered as standard.

Write Short Notes on the following

Question 1.
Friedrich Miescher’s nuclein
Answer:

Nuclein is an acidic substance, having high phosphorus content and it was isolated by Friedrich Miescher in 1869, from the nuclei of pus cells.
As Nuclein had acidic properties and it was isolated from nucleus, it was called as nucleic acid.
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.
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

Size of cell in E. coli size is 2-3µ.
The nucleoid is small, circular, highly folded, naked DNA (1100 µm long in perimeter and contains about 4.6 million base pairs).
When the negatively charged DNA becomes circular, the size reduces to 350 µm in diameter.
Folding/looping (40-50 domains (loops) further reduce it to 30 µm in diameter.
RNA connectors assist in loop formation.
Further coiling and super coiling of each domain reduces the size to 2 µm in diameter.
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

Eukaryotes have split genes.
Primary transcript or hnRNA is non-functional and contains both exons and introns.
Processing of hnRNA results in functional m-RNA.
The fully processed hnRNA is called m-RNA.
m-RNA comes out of the nucleus through nuclear pore for getting translated.

hnRNA undergoes capping, tailing and splicing.

Capping : Methylated guanosine tri¬phosphate is added to 5’ end of hnRNA.
Tailing : Polyadenylation take place at 3′ end.
Splicing : It is removal of introns.
DNA ligase joins exons in a definite sequence (order).

Question 6.
Cracking of genetic code.
Answer:

M. Nirenberg and Matthaei synthesized artificial poly-U m-RNA.
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.
It suggested that UUU codes for phenylalanine.
Dr. Har Gobind Khorana devised a technique for synthesis of artificial m-RNA with repeated sequences of known nucleotides.
He synthesized artificial RNA consisting of known repeated sequences of two or three nucleotides. E.g. CUC, UCU, CUC, UCU, by using synthetic DNA.
This resulted in formation of polypeptide chain consisting of alternate amino acids leucine and serine.
Synthetic RNA consisting of CUA, CUA, CUA, CUA repeats gave polypeptide chain with only one amino acid – leucine.
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).
Thus all the 64 codons in the dictionary of genetic code were deciphered.

Question 7.
Types of mutations.
Answer:

Chromosomal mutations : Loss (deletion) or gain (insertion/duplication) of a segment of DNA results in alteration in the chromosome.
Point mutations : They involve change in a single base pair of DNA. E.g. Mutation that results in Sickle-cell anaemia.
Deletion or insertion of base pairs of DNA : It causes frame-shift mutations or deletion mutation.
Insertion or deletion of one or two bases changes the reading frame from the point of insertion or deletion.
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:

m-RNA has some additional sequences that are not translated. These sequences are referred as untranslated regions (UTR).
The UTRs are present at both 5′-end (before start codon) and at 3′-end (after stop codon).
They are required for efficient translation process.

Short Answer Questions

Question 1.
Why are Okazaki fragments formed on lagging strand only?
Answer:

The lagging template is the template strand with free 5’ end.
The replication always starts at C-3 end of template strand and proceeds towards C-5 end.
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.
Hence, the lagging templates becomes available for replication only discontinuously in small patches.
The new lagging strand develops discontinuously away from the replicating fork in the form of small Okazaki fragments.
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:

In each of the two daughter DNA molecules thus formed, one strand is parental and the other one is newly synthesized.
50% is contributed by mother DNA.
Hence, DNA replication is described as semiconservative replication.

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:

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.
There is single DNA dependent-RNA polymerase that catalyses transcription of all 3 types of RNA in bacteria.
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:

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.
If each codon has two nucleotides, then there will be 4² = 16 codons, which can encode for only 16 different types of amino acids.
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:

Each single base is a part of only one codon.
Adjacent codons do not overlap.
If it had been overlapping type, with 6 bases, there would be 4 amino acid molecules in a chain.
Experimental evidence favours non-overlapping nature of genetic code.

Question 7.
How degeneracy of the code is explained by Wobble hypothesis?
Answer:

In 1966, Crick proposed Wobble hypothesis.
According to this hypothesis, in codon- anticodon base pairing, the third base may not be complementary.
The third base of codon is called wobble base and this position is called wobble position.
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.
For example, GUU, GUC, GUA and GUG codons code for amino acid Valine.
Degeneracy of genetic code means many codons can code for same amino acid.
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:

UAA, UAG and UGA are known as termination codons.
They do not code for any amino acid.
They terminate or stop the process of elongation of polypeptide chain.
Hence, they are known as termination codons.

Question 10.
How many amino acids are required for protein synthesis? From where are they obtained?
Answer:

About 20 different types of amino acids are required for protein synthesis.
They are available in the cytoplasm.

Question 11.
How DNA regulates protein synthesis?
Answer:

DNA regulates protein synthesis by coding for the specific sequence of amino acids in a protein.
This control is possible through transcription of m-RNA.
Genetic code is specific for particular amino acid.

Question 12.
What is the role of ribosomes in protein synthesis?
Answer:

Ribosomes serve as site for protein synthesis.
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).
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:

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.
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:

A set of genes will be switched on when a substrate is to be metabolized.
This phenomenon is called induction and small molecule responsible for this, is known as inducer.
It is positive control of gene regulation.
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:

Small subunit of ribosome binds to the m-RNA at 5’ end.
Start codon is positioned properly at P-site.
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.
The large subunit and smaller subunit of ribosome bind in the presence of Mg⁺⁺.
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:

Structural and functional genomics is used in the improvement of crop plant, human health and livestock.
The knowledge and understanding acquired by genomics research can be applied in medicine, biotechnology and social sciences.
It helps in the treatment of genetic disorders through gene therapy.
It helps in the development of transgenic crops having desirable characters.
Genetic markers have applications in forensic analysis.
Genomics can lead to introduction of new gene in microbes to produce enzymes, therapeutic proteins and biofuels.

Question 18.
Why is HGP important?
Answer:

HGP is associated with rapid development of Bioinformatics.
Knowledge gained about the functions of genes and proteins has a major impact in the fields like Medicine. Biotechnology and the Life sciences.
It has helped in identifying the genes that are associated with genetic characteristics.
The genetic basis of many hereditary diseases can be understood.
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:

Organism	Diploid chromosome number
Mouse	————-
Fruitfly	————
Roundworm	————-
Yeast	————–

Answer:

Organism	Diploid chromosome number
Mouse	40
Fruitfly	8
Roundworm	12
Yeast	32

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

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.
b.

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)

Which step of protein synthesis is shown in the following diagrams?
During initiation, which subunit of ribosome binds with m-RNA?
What are the three binding sites for t-RNA on ribosomes?
On which site of ribosome second and subsequent t-RNA arrives?
Which link is binding amino acids in diagram (b)?
Which chain is being released from ribosome in diagram (c) ?

Answer:

Translation
30S or 40S
P site, A site and E site
A site
Peptide link
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:

Isolation of DNA
Restriction digestion
Gel electrophoresis
Southern blotting
Selection of DNA probe
Hybridization
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.

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.

The human genome contains 3164.7 million nucleotide bases.
The average gene consists of 3000 bases.
Largest known human gene is dystrophin at 2.4 million bases.
Total number of genes is estimated to be 3000.
99.9% nucleotide bases are exactly the same in all people.
The function of about 50% of the discovered genes are unknown.
Less than 2% of the genome codes for proteins.
Repeated sequences make up a very large portion of the human genome. They can shed light on chromosome structure, dynamics and evolution.
Chromosome 1 has most genes (2968) and the Y has the fewest (231).
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.

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

July 30, 2024July 29, 2024 by Bhagya

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

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) I^AI^A
(b) I^AI^B
(c) ii
(d) I^Ai
Answer:
(c) ii

Question 20.
The genotype of human blood group B is ……………….
(a) I^AI^A
(b) I^Bi
(c) I^AI^B
(d) ii
Answer:
(b) I^Bi

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

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 A	Column B
(1) Co-dominance	—————–
(2) Incomplete dominance	—————–
(3) Multiple allelism	—————-
(4) Pleiotropy	—————–
(5) Polygenes	—————-
(6) Autosomal dominance	——————

Answer:

Column A	Column B
(1) Co-dominance	Coat colour of cattle
(2) Incomplete dominance	Flower colour of Mirabelis jalapa
(3) Multiple allelism	Human blood groups
(4) Pleiotropy	Sickle-cell anaemia
(5) Polygenes	Height in human beings
(6) Autosomal dominance	Widow’s peak

Question 2.
Types of sex-linked genes:
[Haemophilia, Ichthyosis, Nephritis, Myopia, Hypertrichosis, Retinitis pigmentosa]

Column A	Column B
(1) Completely X-linked genes	—————–
(2) Completely Y-linked genes	—————–
(3) Incompletely sex-linked genes	—————-

Answer:

Column A	Column B
(1) Completely X-linked genes	Haemophilia, Myopia
(2) Completely Y-linked genes	Ichthyosis,Hypertrichosis
(3) Incompletely sex-linked genes	Nephritis, Retinitis pigmentosa

Question 3.
Genetic Disorders
[Turner’s syndrome, Sickle-cell anaemia, Thalassemia, Edward’s syndrome, Klinefelter’s syndrome, Down’s syndrome]

Column A	Column B
(A) Autosomal disorder	—————–
(B) Sex chromosomal disorder	—————–
(C) Mendelian disorder	—————-

Answer:

Column A	Column B
(A) Autosomal disorder	Edward’s syndrome, Down’s syndrome
(B) Sex chromosomal disorder	Turner’s syndrome, Klinefelter’s syndrome
(C) Mendelian disorder	Sickle-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.

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.

Tt × Tt
TT × tt
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.

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:

Homozygous	Heterozygous
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 cross	Dihybrid 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 characters	Recessive 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:

Phenotype	Genotype
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.

Question 5.
Incomplete dominance and Co-dominance.
Answer:

Incomplete dominance	Co-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 syndrome	Klinefelter’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:

According to the law of segregation, the members of the allelic pair remain together without mixing with each other.
They segregate or separate when the gametes are formed.
Thus the gametes that are formed receive only one of the two factors.
Now it is known that the organisms are diploid and the gametes produced by them are haploid.
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:

The pea plants were true breeding varieties.
The pea plants being annual, it was possible to cross and study many generations within a short period.
The pea plants had a number of distinguishable, contrasting characters such as tall habit and dwarf habit, round seed and wrinkled seed.
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:

The tall habit of the pea plant is dominant over the dwarf habit of the pea plant.
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:

A cross between a homozygous tall (TT) and a homozygous dwarf (tt) gives rise to a heterozygous tall (Tt) plant in the F₁ generation.
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.
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:

Haemophilia is caused due to X-linked recessive gene. Females have double X chromosomes.
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.
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:

Females possess double X-chromosomes in her gametes.
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.
But such female can carry the defective gene to her progeny. Thus she is called carrier of colour-blindness.
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:

Linkage is the tendency of genes to be inherited together because they are present in the same chromosome.
All the genes on a chromosome are linked to one another. In the linkage group some of the genes are included.
The number of linkage groups of a particular species corresponds to its haploid number of chromosomes present in the organism.
In human beings, there are 23 linkage groups which correspond to the pairs of chromosomes found in each cell.
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.
Linkages Eire of two types, viz, complete linkage and incomplete linkage.
Morgan discovered linkage in animals while Bateson and Punnett discovered it in plants.

Question 2.
Multiple alleles.
Answer:

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.
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.
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.
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.
Human blood groups A, B, AB and O Eire also due to series of multiple alleles.

Question 3.
Autosomal inheritance.
Answer:

Transmission of body characters occurs due to autosomes. They are not concerned with sex determination or sex linkage.
All the body characters from parents are passed on to their offspring through autosomes. This is called autosomal inheritance.
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.
Phenyl ketonuria (PKU), Cystic fibrosis and Sickle-cell anaemia are autosomal recessive traits.

Question 4.
Widow’s peak.
Answer:

Widow’s peak is a prominent ‘V’ shaped hairline on forehead.
It is due to autosomal dominant gene.
Widow’s peak occurs in homozygous dominant (WW) and also heterozygous (Ww) individuals.
Individuals with homozygous recessive (ww) genotype do not have widow’s peak but have a straight hair line.
Both males and females have equal chance of inheritance.

Question 5.
Environmental sex determination
Answer:

Environmental sex determination is shown by lower organisms such as Bonellia viridis.
In this animal the environmental factors decide the sex of an offspring.
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.
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.
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:

Holandric means entirely of male sex. Y-linked genes are called holandric genes because they are located on non-homologous region of Y chromosome.
The Y-linked genes are inherited directly from male to male.
These genes are never seen in females due to lack of Y chromosome in them.
Hyper Mchosis and ichthyosis are examples of holandric genes.
Hypertrichosis means excessive development of hair on pinna of ear. This character is transmitted directly from father to son.
Ichthyosis person with rough skin.

Question 7.
X-body.
Answer:

German biologist, Henking in 1891, was studying spermatogenesis of the squash bug (Anasa tristis).
He noted that 50% of sperms receive the unpaired chromosomes while other 50% sperms do not receive it.
Henking gave a name to this structure as the X-body. He was unable to explain its role in sex determination.
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:

Farmers use hybrid seeds for agriculture or horticulture. Hybrid seeds are produced by crossing two unrelated parent plants.
Hybrid seed varieties give improved yields and crop vigour to the farmer.
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.
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:

Single trait is shown due to single gene. Every single gene has two contrasting alleles.
Two alleles are always in interaction in which one is completely dominant while other is completely recessive.
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:

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.
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:

Drosophila cam easily be cultured under laboratory conditions.
Life span of Drosophila is short for about two weeks.
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:

Down’s syndrome is caused due to aneuploidy.
Aneuploidy is due to non-disjunction of chromosome at the time of gamete formation during meiosis. Due to non-disjunction, chromosomes fail to separate.
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:

Typical facial features.
An epicanthal skin fold, over the inner corner of eyes causing downward slanting eyes.
Typical flat face, rounded flat nose, mouth always open with protruding tongue.
Mental retardation.
Poor skeletal development.
Short stature, relatively small skull and arched palate.
Flat hand with simian crease that runs across the palm.

Question 8.
Write a brief account of Turner’s syndrome.
Answer:

Turner’s syndrome is a genetic disorder caused due to monosomy of X chromosome.
It was first described by H. H. Turner.
Turner’s syndrome is caused due to non-disjunction of sex chromosomes which takes place during gamete formation.
Chromosomal complement of Turner’s syndrome is 44 + XO, having a total of 45 chromosomes.

Symptoms of Turner’s syndrome are as follows:

Female phenotype.
Short stature with webbing of neck.
Low posterior hair line.
Secondary sexual characters fail to develop.
Mental retardation.

Question 9.
Write a brief account of Klinefelter’s syndrome.
Answer:

Klinefelter’s syndrome is a genetic disorder caused due to trisomy of X chromosome.
It was first described by Harry Klinefelter.
Klinefelter’s syndrome is caused due to non-disjunction of sex chromosomes which takes place during gamete formation.
Chromosomal complement of Klinefelter’s syndrome is 44+XXY, having a total of 47 chromosomes.

Symptoms of Klinefelter’s syndrome are as follows:

The Klinefelter’s syndrome individuals are tall, thin and eunuchoid.
They are sterile with poorly developed sexual characteristics.
Testes are underdeveloped and small. Spermatogenesis does not take place.
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

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

	R	W
R	RR	RW
W	RW	WW

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

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

	R	R
R	Rr	Rr
R	Rr	Rr

II. In such cross, all the flowers will be violet. II. Unknown flower considering as Rr (heterozygous)
Rr × rr Homozygous recessive with heterozygous

	R	R
R	Rr	Rr
R	Rr	Rr

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

Question 5.
Explain the mechanism of sex determination in case of birds.
Answer:

Sex determination in birds is by ZW-ZZ mechanism.
In birds, males are homogametic while females are heterogametic.
Males produce all similar types of sperms, containing 8 autosomes and ‘Z’ sex chromosome.
Females produce two different types of eggs, one containing 8 autosomes and Z chromosome and the other containing 8 autosomes and W chromosome.
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.

[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

July 29, 2024July 28, 2024 by Bhagya

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

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

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 I	Column 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 I	Column 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 I	Column 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 I	Column 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 I	Column 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 I	Column 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 I	Column II
(1) Mechanical means	(a) Saheli
(2) Physiological device	(b) Jellies
(3) Chemical device	(c) Vasectomy
(4) Permanent method	(d) Diaphragm

Answer:

Column I	Column 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 A	Column B
(1) Mechanical means	————–, ————
(2) Chemical means	————-, ————-
(3) Intra-uterine device	————-, ————

Answer:

Column A	Column B
(1) Mechanical means	Cervical caps, Diaphragms
(2) Chemical means	Foams, Jellies
(3) Intra-uterine device	Lippe’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 A	Column B
(1) Seminal fluid	————–, ————
(2) Prostatic fluid	————-, ————-
(3) Fluid from Cowper’s gland	————-, ————

Answer:

Column A	Column B
(1) Seminal fluid	Prostaglandins, Fructose, Fibrinogen
(2) Prostatic fluid	Acid phosphatase, Citric acid
(3) Fluid from Cowper’s gland	Mucous 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.

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:

Uterus receives ovum from fallopian tubes, develops placenta and provides site for implantation of embryo.
It provides protection and nourishment to the developing embryo.
It also provides path for sperms to ascend.
Due to contractions of uterus, baby is expelled out at the time of parturition.

Question 10.
Vagina
Answer:

Vagina acts as a copulatory passage.
It acts as a birth canal during parturition in normal delivery
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

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:

Dartos muscles
Cremaster muscles

Question 15.
Names of erectile tissues in penis.
Answer:

Corpora cavernosa
Corpus spongiosum

Question 16.
Any two copper releasing IUD.
Answer:

Copper-T, Cu 7
Multiload 375

Question 17.
Any two hormone-releasing IUDs.
Answer:

LNG-20
Progestaert

Question 18.
Two methods of birth control which have high chances of failure.
Answer:

Safe period
Lactational amenorrhea

Question 19.
Uterine walls.
Answer:

Perimetrium
Myometrium
Endometrium

Question 20.
Regions of the uterus.
Answer:

Fundus
Body
Cervix

Question 21.
Parts of fallopian tubes.
Answer:

Infundibulum
Ampulla
Isthmus

Question 22.
Layers of Graafian follicle which enclose antrum.
Answer:

Theca externa
Theca interna
Membrana granulosa

Question 23.
Stages of cells in spermatogenesis.
Answer:

Spermatogonia
Primary spermatocytes
Secondary spermatocytes
Spermatids
Sperms

Question 24.
Stages of cells in oogenesis.
Answer:

Oogonia
Primary oocytes
Secondary oocytes
Ootid
Ovum

Give significance of the following

Question 1.
Fertilization.
Answer:
Significance of fertilization:

Fertilization forms the zygote which eventually produces new offspring.
Fertilization restores diploid number of chromosomes in the zygote as two haploid gametes come together in a zygote.
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.
By fertilization the genetic characters of two parents are mixed. This leads to variation and has significance in evolution.
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:

Corpus luteum is the temporary source of female hormone, progesterone.
Corpus luteum is formed from empty Graafian follicle after the process of ovulation.
Due to progesterone secreted from corpus luteum, the endometrial wall of uterus undergoes repair and increase in thickness.
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:

Due to the process of gastrulation, three germinal layers, viz. ectoderm, mesoderm and endoderm are formed.
Cells of embryonal knob become embryonic disc which develop into embryo due to gastrulation.
Gastrulation is necessary for the formation of amniotic cavity which is filled with amniotic fluid.

Question 6.
Trophoblast of blastocyst.
Answer:

Trophoblast cells help in absorbing nutrition for the developing embryo.
Trophoblast cells at the embryonal knob (cells of Rauber) help in implantation of blastocyst.
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:

Colostrum is the first milk which is sticky and yellowish secreted by the mammary glands soon after the parturition.
Being high protein in its content, it nourishes the newly born child.
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 reproduction	Asexual 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 organs	Secondary / 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 efferentia	Vasa 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 follicle	Corpus 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.

Question 5.
Menarche and Menopause.
Answer:

Menarche	Menopause
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 Phase	Secretory 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:

Spermatogenesis	Oogenesis
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 pellucida	Corona 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:

Morula	Blastula
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:

Blastula	Gastrula
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:

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.
For the development of the sperm, lesser temperature than the body temperature is required.
If the testes remain in the abdominal cavity, then the sperm production does not take place.
This may result in impotency. Therefore, testes are located outside the body cavity.

Question 2.
Urethra is also called urinogenital duct in males.
Answer:

Urinogenital duct means common duct for urine and the genital products.
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.
Since the urethra carries both urine and semen, it is called urinogenital duct.

Question 3.
Proliferative phase is also called follicular phase.
Answer:

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.
The follicular phase of ovaries is due to effect of FSH from adenohypophysis.
The ovaries follicles grow due to FSH and start secreting estrogen.
This estrogen from ovaries bring proliferative effect on the uterus.

Question 4.
Missing of menses is the first indication of pregnancy.
Answer:

Menstruation occurs if there is no fertilization of ovum.
The endometrium of uterus along with unfertilized egg is given out in the form of menstrual flow.
The sloughing off uterine endometrium takes place due to degeneration of corpus luteum.
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.
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:

Progesterone is secreted from corpus luteum which is formed from empty ovarian follicle after the ovulation.
Progesterone has the capacity to maintain pregnancy.
It acts on uterine endometrium and causes it to proliferate and develop in thickness.
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:

In human female, the reproductive period is about 30 – 33 years.
There is menarche at the age of about 13 and menopause at the age of 45-50.
During this span of 30 years, ovaries secrete sex hormones like estrogen and progesterone. After menopause this secretion is suspended.
Due to changes in hormonal level, human females cannot produce eggs later. Moreover, eggs in her ovaries are utilized by the age of 45.
Human female, therefore, has restricted reproductive period.

Question 7.
Zona pellucid is retained for sometime after fertilization.
Answer:

Fertilization of the ovum takes place in fallopian tube where it starts cleavages immediately.
Zona pellucida which remains on the surface of the ovum prevents the implantation of the blastocyst at an abnormal site such as fallopian tube.
Zona pellucida keeps the sticky and phagocytic trophoblast cells unexposed till the ovum reaches the uterine lumen.
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:

The enzyme hyaluronidase secreted by acrosome of sperm dissolves the membranous covering of the ovum to facilitate the entry of sperm into the ovum.
It is a lytic enzyme causing lysis of egg membrane.
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:

Mitochondria provide energy required by sperms for their agile movement.
The agile movement of sperms helps them to reach the vicinity of the ovum at the time of fertilization.
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:

The layer zone pellucida is retained around the embryo and thus, there is no change in the overall size from zygote to morula.
This layer is important because it prevents the implantation of the blastocyst at an abnormal site such as fallopian tube.
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:

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.
Oxygen from mother’s blood is also given to the foetus.
The foetal metabolic wastes such as carbon dioxide, urea and water pass from foetus into the maternal blood.
This exchange takes place through the placenta.
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

Graafian follicle is a mature ovarian follicle.
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.
Next to theca interna, there is membrana granulosa which forms discus proligerous and the corona radiata layer.
Graafian follicle contains an eccentric secondary oocyte. The oocyte is surrounded by a vitelline membrane which produces zona pellucida layer.
In the centre there is antrum which is filled with liquor folliculi fluid.

Question 2.
Mammary glands.
Answer:

Mammary glands are accessory organs of female reproductive system. These glands are essential for lactation after parturition.
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.
Each mammary gland consists of fatty connective tissue and many lactiferous ducts.
Each breast has glandular tissue which is divided into 15-20 irregularly shaped mammary lobes. Each lobe has an alveolar glands and lactiferous duct.
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.
All the lactiferous ducts converge towards the nipple.
Nipple is surrounded by a dark brown coloured and circular area of the skin called areola.

Question 3.
Structure of sperm.
Answer:

Sperm is microscopic, elongated haploid motile male gamete produced by spermatogenesis.
It measures to about 0.055 mm or 60y in length.
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:

Head is the main part which is flat and oval and has a large nucleus and an acrosome.
Acrosome is formed from Golgi complex. It secretes enzyme hyaluronidase which helps in penetration of the egg during fertilization.
The acrosome and anterior half of nucleus is covered by a fibrillar sheath.

Neck : Neck is short region having two centrioles.

The proximal centriole plays a role in first cleavage of zygote.
The axial filament of the sperm is formed by the distal centriole.

Middle piece:

Middle piece acts as a power house for sperm.
It bears many spirally coiled mitochondria or Nebenkern around the axial filament.
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.
Posterior half of nucleus, neck and middle piece of sperm are covered by a sheath.

Tail:

The tail is formed of cytoplasm and is long, slender and tapering structure.
The axial filament is a fine thread-like structure that arises from the distal centriole and traverses the middle piece and tail.
Nine accessory fibres are present surrounding the two central longitudinal axial filaments.
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

The unfertilized egg released through ovary at the time of ovulation is a secondary oocyte.
It is rounded, non-motile and haploid, non- cleidoic and microlecithal female gamete.
The size is approximately 0.1 mm (100 microns).
It has abundant cytoplasm called ooplasm which contains a large eccentric and prominent nucleus called germinal vesicle.
Centrioles are absent in secondary oocyte.
Various coverings seen around the oocyte are (i) vitelline membrane (ii) zona pellucida (iii) Corona radiata.
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:

Implantation is the process by which the blastocyst is embedded into the endometrium of uterus in the fundus region.
The implantation starts 7 days after fertilization and completed by the end of 10th day.
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.
Cytotrophoblast is the inner layer whose cells retain their cell boundaries.
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

hair and nails
sweat glands
conjunctiva
cornea
lens
retina
internal and external ear
enamel of teeth
nasal cavity
adrenal medulla
stomodaeum and proctodaeum
neurohypophysis and
entire nervous system.

(ii) Fate of mesoderm : The mesoderm forms the following derivatives:

All types of muscles
connective tissue
dermis of skin
adrenal cortex
kidney
circulatory system
heart
blood vessels
blood
lymphatic vessels
middle ear and
dentine of teeth.

(iii) Fate of endoderm : The following organs develop from the endoderm:

Epithelium of gut from pharynx to colon
glands of stomach and intestine
tongue and tonsils
lungs, trachea, bronchi, larynx, etc.
urinary bladder, vestibule and vagina
liver and pancreas
adenohypophysis
thymus, thyroid and parathyroid
eustachian tube
epithelium of urethra and associated glands.

Question 7.
Placenta.
Answer:

Placenta is a temporary organ derived from the tissues of the foetus as well as mother.
Human placenta is called chorionic placenta as it is made up of chorion which is an extra-embryonic membrane.
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.
There are two parts of placenta, viz. foetal placenta and maternal placenta.
Foetal placenta is formed of chorionic villi.
Maternal placenta is formed of uterine wall which is in intimate contact with the chorionic villi.
Chorionic villi receive the blood from the embryo by umbilical artery. Umbilical vein returns the blood back to the embryo.
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:

IUDs are plastic or metal objects which act as contraceptive devices. They are placed into the uterus by a doctor or trained nurse.
E.g. Lippe’s loop, copper releasing IUDs (Cu-T, Cu-7, multiload 375) and hormone releasing IUDs (LNG-20, Progestaert).
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.
Copper releasing IUDs suppress sperm motility and the fertilizing capacity of sperms.
The hormone releasing IUDs make the : uterus unsuitable for implantation and ; cervix hostile to the sperms.
Their presence in the uterus acts as a minor irritant and thus makes the ovum to move quickly out of the body.
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:

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.
Woman who is using pills does not release ovum at the time of ovulation and therefore conception does not occur.
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.
These pills also alter the quality of cervical J mucus to prevent the entry of sperms.
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:

Trophoblast cells do not form any part of the embryo proper.
They form ectoderm of the extra-embryonic membrane called chorion.
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.
Thus, these cells have an important role in formation of placenta.

Question 11.
Medical Termination of Pregnancy (MTP).
Answer:

MTP or Medical Termination of Pregnancy is voluntary termination of pregnancy under medical supervision. It is an induced abortion.
Only during first trimester, MTP is safe for mother’s health.
Upon amniocentesis examination, if abnormality is detected, usually MTP is performed.
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
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:

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.
The chromosomes from the foetal cells are sujected to karyotyping. This helps to detect abnormalities in the developing foetus.
Amniocentesis is misused to determine the sex of the unborn child. This is illegal in India because it results into female foeticide.
Another risks involved in amniocentesis are miscarriage, needle injury to foetus, leaking amniotic fluid, infection, etc.
As per MTP Act (1971) the misuse of amniocentesis is curtailed.

Question 13.
ZIFT [Zygote Intra Fallopian Transfer].
Answer:

If there is a blockage in the fallopian tubes due to which fertilization is prevented, then ZIFT treatment is used.
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).
Later the zygote is transferred in fallopian tube to achieve pregnancy.

Question 14.
GIFT [Gamete Intra Fallopian Transfer].
Answer:

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.
When the entrance or upper segments of the fallopian tubes is blocked, this technique is used.
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.
GIFT is successful in only 30 per cent cases.

Question 15.
Sterilization operations.
Answer:

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.
These surgical interventions block the gamete transport and thus prevents the pregnancy.
Sterilization operation in males is called vasectomy while in females it is called tubectomy.
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:

Presence of beard, Moustache.
Hair on the Chest, Axillary and Pubic Region.
Muscular body.
Enlarged larynx (Adam’s apple).

(ii) Secondary sexual characters in females:

Breast development.
Broadening of pelvis.
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:

Vestibular glands or Bartholin’s glands are homologous to the Cowper’s glands of the male.
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:

The menstruation is painful in some women as the muscles in the uterus contract or tighten.
Women who experience painful periods can have higher levels of prostaglandins, a natural body chemical that causes contractions of the uterus and blood vessels.
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.
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:

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.
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.
Feeding of mother’s milk reduces the risk of overweight and obesity during childhood.
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:

Keeping the pubic area clean.
Changing the sanitary napkin every 4-5 hours.
Reducing risk of infections by maintaining hygiene.
Proper disposal of soiled sanitary napkin.
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:

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.
Audio-visual and the print media should be used by government and non-government organisations for creating awareness about reproductive health.
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.
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.
Statutory ban on amniocentesis for sex determination is practised. This should be known by all.
Details of child immunization programmes should be understood.
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:

Tobacco, marijuana and other drugs, smoking may cause infertility in both men and women.
Nicotine blocks the production of sperm and decreases the size of testicles.
Alcoholism by men interferes with the synthesis of testosterone and has an impact on sperm count.
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:

Tuberculosis disrupts sexual and reproductive function in patients.
Moreover T.B. patients have to take not less than 4 anti-TB drugs simultaneously for a long time.
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.
Pulmonary TB patient shows, deterioration of all parameters of copulatory act, from sexual desire to orgasm and thus the couple is unable to conceive.
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 Organs	Homology to Male Reproductive Organs
1. Labia majora	————–
2. ————-	Bulbourethral glands/ Cowper’s gland
3. Clitoris	—————

Answer:

Female Reproductive Organs	Homology to Male Reproductive Organs
1. Labia majora	Scrotum
2. Bartholin’s gland/ Vestibular gland	Bulbourethral glands/ Cowper’s gland
3. Clitoris	Penis

Question 2.

Hormones	Functions
1. Testosterone	————–
2. ————-	Stimulates contractions uterine during parturition
3. Progesterone	—————

Answer:

Hormones	Functions
1. Testosterone	Stimulates spermatogenesis
2. Oxytocin	Stimulates contractions uterine during parturition
3. Progesterone	Maintain endometrium of uterus during secretory phase and gestation.

Question 3.
Complete the following chart and rewrite

Hormones	Functions
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:

Hormones	Functions
1. Rapid regeneration of endometrium and maturation of Graafian follicle	Proliferative phase / Follicular phase
2. Secretion of endometrial glands and increased secretion of progesterone	Secretory phase / Luteal phase
3. Breakdown of endometrium in absence of fertilization	Menstrual 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

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:

Seminal vesicle
Ejaculatory duct
Cowper’s glands
Urethra
Epididymis
Testis
Urinary bladder
Prostate gland
Vas deferens
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:

Fallopian tube
Fundus of Uterus
Ampulla of fallopian tube
Ovarian ligament
Uterus
Ovary
Infundibulum with fimbriae
Endometrium of uterus
Cervix
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

Give the name of the surgical method represented in the diagram.
Which part is ligated or cut.
Name the corresponding surgical method conducted in females.
Name the part which is ligated in females and why?

Answer:

Vasectomy
Vas deferens
Tubectomy
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.

Name the structure and its type.
Identify ‘A’. In which technique it is used.
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

Give the name of intra-uterine contraceptive device shown in the diagram.
What is its mode of action?

Answer:

The Intra-uterine contraceptive device shown in the diagram is Lippes loop.
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.

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.)

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

July 29, 2024July 28, 2024 by Bhagya

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

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

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

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

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 A	Column 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 A	Column 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:

Endospermic seed : Castor
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.

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:

It ensures seed formation with food storage for embryo developed from fertilized egg.
Diploid zygote develops into embryo which further forms a new plant.
Triploid PEN forms endosperm which is nutritive tissue for embryo.
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.

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 Reproduction	Sexual 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. Pea	4. Offspring are genetically varied due to recombination. E.g. Food and fibre crops – Maize, Rice.

Question 3.
Hypohydrophily – Epihydrophily
Answer:

Hypohydrophily	Epihydrophily
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:

Embryo sac develops inside the nucellus of ovule from megaspore.
Megaspore mother cell is diploid structure which undergoes meiosis.
After meiosis, tetrad of haploid cells are produced.
The upper three megaspores degenerate and the lower one of the tetrad is functional.
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:

Fertilization is fusion of male and female gametes.
Pollination is transfer of pollen grains which carry non-motile male gametes.
Pollen grains are transferred from anther to stigma of flower where they germinate.
Both male and female gametes are non- motile and they are produced at two different sites.
Therefore the pollination process is necessary for act of fertilization in plants.

Question 3.
Dichogamy favours cross pollination.
Answer:

Maturation of anther (stamen) and stigma (carpel) at different times is called dichogamy.
Dichogamy is of two types, viz, protandry and protogyny.
Maturity of anthers before that of gynoecium is protandry and maturity of carpel before maturity of pollen grains is protogyny.
As this forms barrier for self-pollination, dichogamy favours cross pollination.

Question 4.
Fertilization in angiosperms is double fertilization.
Answer:

In angiosperms, pollen tube carries two non-motile male gametes.
Pollen tube enters the embryo sac in synergids and the contents are relased.
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.
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:

Endosperm, that is developed after fertilization is a nutritive tissue for developing embryo.
Endosperm stores food material.
In some seeds this reserved food is partially utilized by embryo for development, E.g; Castor.
The endosperm remains in the seed and it is utilized further during seed germination. Hence the seed is endospermic or albuminous.

Question 6.
Parthenocarpic fruits are without seeds.
Answer:

In parthenocarpy, fruit is developed without fertilization.
When fertilization takes place ovules in the ovary are transformed into seeds.
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:

Polyembryony is a phenomenon where we get many embryos in the seed.
Polyembryony increases chances of survival of plants as there are multiple seedlings formed.
Nucellar embryos are formed from diploid parental tissue.
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:

It is asexual method of reproduction.
Plants reproduce through their vegetative plant parts.
New plants produced are genetically identical to their parents.
It is very useful in agriculture and horticulture.
Artificial methods like cutting and grafting are useful for propagation of desired varieties as per human needs.

Question 2.
Grafting.
Answer:

It is type of artificial vegetative propagation.
In this method two different plants are joined together.
The part of stem containing one or more buds is scion which is joined on a rooted plant stock.
They grow as one plant, e.g. Apple, Pear, Mango.
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:

It is a functional ability of pollen grain to form male gametophyte by its germination.
Viable pollen grains germinate on stigmatic surface,
Environmental factors mainly temperature and humidity influence its germination.
Viability is low up to 30 minutes in plants like rice and wheat.
Duration of viability is up to months in some plants of family Leguminosae, Rosaceae and Solanaceae.

Question 4.
Seed Dormancy.
Answer:

It is a state of metabolic arrest which helps in survival of organism in unfavourable environmental conditions.
Structure or physiological adaptive measures of seed that are helpful in adverse conditions is called dormancy.
Seeds are dispersed during their dormancy.
When dormancy period of seeds is completed then only the viable seed germinate.

Question 5.
Parthenocarpy.
Answer:

It is a condition in which fruit is developed without event of fertilization.
It is a natural process observed on Pineapple and Banana.
A chemical stimulus in the form of auxin (IAA) is given by placental tissues of unfertilized ovary.
Due to the stimulus, enlargement of ovary takes place to form a fruit.
Parthenocarpic fruits are without seeds.

Question 6.
Polyembryony.
Answer:

It is a condition when more than one embryos are developed inside the seed.
It was first noticed in Citrus by Leeuwenhoek.
When embryos develop from diploid cells of nucellus or integuments, it is described as adventive polyembryony.
When zygote divides into small units which develop into embryos then it is called cleavage polyembryony.
It results in multiple seedlings and is of significance in horticulture.

Question 7.
Anemophily.
Answer:

The transfer of pollen grains through wind is called anemophily.
Plants that are pollinated by wind are called anemophilous plants.
Anemophilous plants bear small and inconspicuous flowers without any bright colours, fragrance and nectar.
Flowers are produced in large numbers.
Stamens are long with versatile anthers.
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:

The transfer of pollen grains with the help of water is called hydrophily.
Plants that are pollinated by water are called hydrophilous plants.
Hydrophilous plants possess small, inconspicuous unisexual flowers.
Flowers lack fragrance, nectar and bright colour.
Pollen grains and other floral parts are protected from getting wet.
Stigma is long and sticky, e.g. Zostera, Vallisneria, etc.

Question 9.
Omithophily.
Answer:

The transfer of pollen grains through birds is called ornithophily.
Bird pollinated plants are called ornithophilous plants.
Ornithophilous plants bear large and showy flowers.
Flowers brightly coloured to attract birds for pollination.
Ornithophilous flowers lack fragrance as birds have poor sense of smell.
Pollen grains are sticky and spiny e.g. Callistemon, Bignonia, Bombax, Butea, etc.

Question 10.
Dichogamy.
Answer:

When stamens and carpels mature at different times in a bisexual flower, the condition is known as dichogamy.
Owing to dichogamy self-pollination is avoided and cross pollination is favoured.
Dichogamy is of two types, viz., protandry and protogyny.
Protandry is seen in sunflower in which pollen grains are released much before stigma becomes receptive.
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:

Egg apparatus is a three celled structure lying at the micropylar end of the embryo sac.
The egg apparatus consists of a median egg cell called oosphere and two lateral cells called synergids.
The embryo sac also consists of three antipodal cells or antipodals towards the chalazal end which degenerate after fertilization.
In the centre, the embryo sac consists of a large central cell consisting of two haploid polar nuclei.
The polar nuclei at a later stage fuse with each other forming a diploid secondary nucleus.
The secondary nucleus develops into endosperm.

Question 12.
Entomophily.
Answer:

Pollination with the help of insects is called entomophily.
The insect pollinated flowers are called J entomophilous flowers.
Entomophilous flowers show the following adaptations:
Flowers are large and attractive.
Flowers are brightly coloured with i pleasant smell.
Flowers produce nectar which is food for the insects.
Pollen grains are spiny and sticky for easy adherance to the rough and sticky stigma.
Entomophily is seen in plants like rose, Jasmine, Oestrum, Salvia, etc.

Question 13.
Endosperm.
Answer:

Endosperm is a nutritive tissue. It nourishes the developing embryo.
The endosperm develops from the primary endosperm nucleus (PEN).
The endosperm is a post fertilization tissue.
There are two types of seeds depending upon the presence or absence of endosperm, viz., endospermic and non-endospermic.
Castor, coconut, maize, etc. are endospermic seeds, while bean, pea, gram, etc. are non-endospermic seeds.

Question 14.
Triple fusion.
Answer:

Triple fusion is also called second fertilization.
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.
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:

Production of offspring without involving fusion of two compatible gametes or sex cells is called asexual reproduction.
Fragmentation : It is a type of asexual reproduction observed in lower plants, e.g. algae.
Multicellular organisms break into small pieces called fragments which develop into new plant.
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:

Vegetative propagation is a kind of asexual reproduction which occurs with the help of vegetative plant parts.
Cutting and grafting are two methods used to propagate desired varieties of plants.
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.
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.
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

Internally it shows four chambers called microsporangia or pollen sacs.
The anther consists of two main parts, viz., anther wall and microsporangium or pollen sac.
The wall of the anther can be differentiated into four layers, viz., epidermis, endothecium, middle layers and tapetum.
The epidermis is the outermost layer of the anther wall. It is made up of flattened cells which are protective in function.
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.
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.
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

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.
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.
The point at which the funicle is attached to the body of the ovule is called hilum.
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.
Micropyle : The integuments do not completely cover the nucellus. They leave a small opening called micropyle at the tip.
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:

When flowers open, their sex organs are exposed for further process of fertilization then it is chasmogamous condition.
Pollinating agents can easily transfer pollen grains in such flowers for self as well as cross pollination.
When flowers are closed, they are self pollinated in bud conditions then this condition is cleistogamy.
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:

The pollination that occurs with the help of bats is called chiropterophily.
In chiropterous plants the flowers are large and stout enough in such a way that bats can hold onto the flowers.
Chiropterous flowers are nocturnal, i.e., they open during the night time only.
Flowers emit rotten fruits like fermenting fruity odours which attract bats.
Flowers produce copious nectar.
Flowers possess large number of stamens which produce large amount of edible pollen grains.
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:

Question 2.
Maharashtra Board Class 12 Biology Important Questions Chapter 1 Reproduction in Lower and Higher Plants 5
Answer:

Question 3.
Maharashtra Board Class 12 Biology Important Questions Chapter 1 Reproduction in Lower and Higher Plants 7
Answer:

II. Complete the following tables

Question 1.
Maharashtra Board Class 12 Biology Important Questions Chapter 1 Reproduction in Lower and Higher Plants 9
Answer:

Question 2.
Complete the table-Related to outbreeding devices.

Type	Description	Example
1. ————	Unisexual flowers/Monoecious or dioecious plants	Papaya, Maize
2. ————	Mechanical device to prevent Self-Pollination – Natural physical barrier	—————-
3. Prepotency	——————–	Apple
4. Heteromorphy	——————–	Primula
5. Protandry	Androecium matures earlier than gynoecium	—————-

Answer:

Type	Description	Example
1. Unisexuality	Unisexual flowers/Monoecious or dioecious plants	Papaya, Maize
2. Herkogamy	Mechanical device to prevent Self-Pollination – Natural physical barrier	Calotropis
3. Prepotency	Pollens of other flower germinate rapidly rather than from same	Apple
4. Heteromorphy	Presence of different forms of flowers with respect to Stigma and anthers	Primula
5. Protandry	Androecium matures earlier than gynoecium	Sunflower disc florets

Question 3.

Hypohydrophily	Epihydrophily
1. Porogamy	————–
2. ————–	Entry of pollen tube in Ovule piercing integuments
3. —————	Entry of pollen tube in Ovule through Chalaza

Answer:

Hypohydrophily	Epihydrophily
1. Porogamy	Entry of pollen tube into Ovule through micropyle
2. Mesogamy	Entry of pollen tube in Ovule piercing integuments
3. Chalazogamy	Entry 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

It is a type of asexual reproduction.
It is of very common occurrence in unicellular organism yeast.
It is observed in favourable condition.
Mother cell produces small outgrowth which is known as bud.
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

It is a tpe of asexual reproduction.
It is of very common occurrence in lower plants.
It occurs by production of motile zoospores that are formed in sporangia.
Flagellated zoospores when liberated can grow independently into new individuals.
Biflagellate zoospores are formed in algae Chlamydomonas.

Question 2.
Describe the structure of a mature pollen grain.
OR
Sketch and label pollen grain.
Answer:

A typical angiospermic pollen grain (mature) is a unicellular, uninucleate, spherical or oval haploid structure.
The pollen grain is also called microspore.
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

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.
The exine is made up of a complex substance called sporopollenin. The sporopollenin protects the pollen grain from physical and biological decomposition.
The exine is spiny in insect pollinated plants, with sculptured pattern or smooth in wind pollinated plants.
The exine is not continuous throughout. It is interrupted, very thin at one or more places by small pores called germ pores.
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.

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

Nuclear endosperm is the most common type of endosperm.
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.
A central vacuole pushes the nuclei towards periphery.
Later on wall formation takes place around these nuclei to form a cellular mass.
It is seen in plants like maize, sunflower, wheat, coconut, etc.

(ii) Cellular endosperm:

In this type of endosperm, the triploid primary endosperm nucleus undergoes nuclear divisions followed by cytokinesis.
Owing to this, the development of endosperm occurs in cellular form.
It is less common and seen in dicot plants like Datura, Petunia, Balsam, Adoxa.

(iii) Helobial endosperm:

In helobial type of endosperm, the first division of the primary endosperm nucleus is followed by the formation of cell wall.
Owing to this, the central cell is divided into a large micropylar cell and a small chalazal cell.
In both micropylar and chalazal chamber, the further development of the endosperm is of nuclear type.
Walls develop between nuclei in micropylar chamber.
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.

Recurrent
Non-recurrent and
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.

Maharashtra Board Class 12 Chemistry Important Questions Chapter 15 Introduction to Polymer Chemistry

July 19, 2024July 18, 2024 by Bhagya

Balbharti Maharashtra State Board 12th Chemistry Important Questions Chapter 15 Introduction to Polymer Chemistry Important Questions and Answers.

Maharashtra State Board 12th Chemistry Important Questions Chapter 15 Introduction to Polymer Chemistry

Question 1.
What are polymers?
Answer:
Polymers are high molecular mass macromolecules (10³ – 10⁷ u) and consist of repeating units of monomers.

Question 2.
Write the classification of polymers based on the source. Give examples.
OR
What are natural and synthetic polymers? Give two examples of each type.
Answer:
Based on the source polymers are classified as natural, semisynthetic and synthetic polymers.

Natural polymers : These polymers are obtained either from plants or animals, e.g., cellulose, jute, linene, rubber, silk.
Semisynthetic polymers : The fibres obtained by giving special chemical treatment to natural fibres (cellulose) and further regenerated are called semisynthetic polymers e.g., acetate rayon, viscose rayon, cuprammonium silk.
Synthetic polymers : The man made fibres prepared by polymerization of one monomer or copolymerization of
two or more monomers are called synthetic polymers, e.g., nylon, terylene, polythene, etc.

Question 3.
Write the classification of polymers based on structure. Give examples
OR
Write the reactions involved in the preparation of (1) Polyvinyl chloride (PVC) (2) Polypropylene.
Answer:
Based on structure polymers are classified as linear chain polymers, branched chain polymers and network or cross linked polymers.

(1) Linear chain polymers : When the monomer molecules are joined together in a linear arrangement, the resulting polymer is straight-chain or long-chain polymer, e.g., polythene, PVC.
Maharashtra Board Class 12 Chemistry Solutions Chapter 15 Introduction to Polymer Chemistry 4
They have high melting points; high densities and high tensile strength.

(2) Branched-chain polymers : These polymers consist of long and straight chain with smaller side chains give rise to branched-chain polymers. They have low density. They have lower melting points and tensile strength. Polypropylene having methyl groups as branches.
Maharashtra Board Class 12 Chemistry Solutions Chapter 15 Introduction to Polymer Chemistry 6

(3) Network or cross-linked polymers : These polymers consist of cross-linking of chains by strong covalent bonds leading to a network-like structure. Cross-linking results from polyfunctional monomers, e.g., melamine, bakelite, vulcanization of rubber. These polymers are hard rigid and brittle.
Maharashtra Board Class 12 Chemistry Solutions Chapter 15 Introduction to Polymer Chemistry 7

Question 3.
How are polymers classified on the basis of mode of polymerization?
OR
Write the classification of polymers based on mode of polymerization.
Answer:
There are three modes of polymerization depending upon the types of reactions taking place between the monomers.

Addition polymerization (or chain growth polymerization)
Condensation polymerization (or step growth polymerization)
Ring opening polymerization
Addition polymerization or chain growth polymerization : It is a process of polymers by the repeated addition of a large number of monomers is called addition polymerization (like alkenes) without loss of any small molecules.
Example : Formation of polyethylene from ethylene is well known example of addition polymerization. It is a chain growth polymerization.
Condensation polymerization or step growth polymerization : The process of formation of polymers from polyfunctional monomers with the elimination of some small molecules such as water, hydrochloric acid, methanol, ammonia is called condensation polymerization.

Example : The formation of terylene, a polyester polymer, from ethylene glycol and terephthalic acid. Condensation polymerization is a step growth polymerization.
Ring opening polymerization : The process of formation of polymers from cyclic compounds (like lactams, cyclic ethers, etc.) by ring opening is called ring opening polymerization.
Example : Polymerization of e-caprolactam.
Ring opening polymerization proceeds by addition of a single monomer unit to the growing chain molecules. It is a step growth polymerization.

Question 4.
Classify the polymers given below as addition, condensation and ring opening polymers:
PVC, polythene, cyclic ethers, polyester, polyacrylonftrile. polystyrene.
Answer:

Addition polymers: PVC, polythene. polyacrylonitrile. polystyrene.
Condensation polymers: Polyester.
Ring opening polymers : Cyclic ethers

Question 5.
Write the classification of polymers based on intermolecular forces. Give examples.
OR
In which dasses, are the polymers classified on the basis of Inter molecular forces?
Answer:
Molecular forces bind the polymer chains either by hydrogen bonds or Vander Waal’s forces. These forces are called intermolecular forces. On the basis of magnitude of intcrmolccular forces, polymers are further classified as ebstomers, fibres, thermoplastic polymers. thermosetting polymers.

(1) Elastomers: Weak van der Waals type of intermolecular forces of attraction between the polymer chains are observed in cbstomcrs. When polymer is stretched, the polymer chain stretches and when the strain is relieved the chain returns to its odginal position, Thus, polymer shows elasticity and is called elastomers. Elastomers, the elastic polymers, have weak van der Waals type of intermolecular forces which permit the polymer to be stretched. Lilastorners are soft and stretchy and used in making rubber bands. E.g.. neoprene, vulcanized rubber, buna.S, buna-N.

(2) Fibres : It consists of strong intermolecular forces of’ attraction due to hydrogen bonding and strong dipole-dipole forces. These polymers possess high tensile strength. Due to these strong intermolecular forces the fibres are crystalline in nature. They are used in textile industries, strung tyres. etc.. e.g., nylon, terylene.

(3) Thermoplastic polymers: These polymer possess moderately strong intermolecular forces of attraction between those of elastomers and fibres. These polymers arc called thermoplastic because they become soft on heating and hard on cooling. They are either linear or branched chain polymers. They can be remoulded and recycled. E.g. polyethenc, PVC, polystyrene.

(4) Thermosetting polymers: These polymers are cross linked or branched molecules and are rigid polymers. During their formation they have property of being shaped on heating. but they get hardened while hot. Once hardened these become infusible, cannot be softened by heating and therefore, cannot be remoulded and recycled.
This shows extensive cross linking by covalent bonds formed in the moulds during hardening/setting process while hot. E.g. Bakelite, urea formaldehyde resin.

Question 6.
What is meant by the term homopolymer?
Answer:
A polymer made from only one type of repeating units is called homopolymer. in some cases the repeating unit is formed by condensation of two distinct monomers. Examples. Polythene, PVC. Nylon-6.

Question 7.
What is meant by the term copolymer?
Answer:
A polymer made from two or more types of repeating units is called a copolymer. The different monomer units are randomly sequenced in the copolymer, e.g., Terylene, Nylon-6, 6, Buna-S, Buna-N.

Question 8.
Write the classification of polymers on the basis of biodegradability?
OR
(1) What are biodegradable polymers?
(2) What are nonbiodegradable polymers?
Answer:
Based un biodegradability, polymers are classified as biodegradable polymer and nonbiodegradable polymers.

(1) Biodegradable polymers: Polymers that are affected by microbes or disintegrate by themselves afler a certain period of time due to environmental degradation are called biodegradable polymers.

Examples: PHBV i.e., Polyhydroxy butyrate-CO-β-hydroxy valerate Dextron. Nylon-2-nylon-6.

(2) Non biodegradable polymers: Synthetic polymers do not disintegrate by themselves after a certain period or not affected by microbes, are called nonbiodegradhle polymers.

Examples: Bakelite, Nylon, Terylene.

Question 9.
Explain the following terms :
Answer:

Branched chain polymer : The polymer consists of long and straight chain with smaller side chains give rise to branched chain polymers, e.g. Polypropylene
Addition polymer : The polymer formed by the repeated addition of a large number of monomers (like alkenes) without loss of any small molecules are called addition polymers, e.g. polythene -[-CH₂ – CH₂-]_n. It is a chain growth polymerization.
Condensation polymer : The polymers formed by the repeated condensation reaction between polyfunctional monomers with the elimination of some molecules such as water, hydrochloric acid, methanol, ammonia are called condensation polymers, e.g. Nylon-6, 6.
Elastomers : Polymers in which the intermolecular forces of attraction between the polymer chains are the weakest. When polymer is stretched, it has ability to stretch and when the strain is relieved it returns to its original position. Thus, polymer shows elasticity and is called elastomers, e.g. natural rubber, neoprene, vulcanized rubber.
Homopolymer : A polymer made from only one type of repeating unit of one monomer is called homopolymer e.g. Polythene, PVC, etc.
Biodegradable polymer : Polymers which are affected by microbes or disintegrate by themselves after a certain period of time due to environmental degradation are called biodegradable polymers.
Example : PHSV i.e. Polyhydroxy butyrate -CO-β-hydroxy valerate Dextron.

Question 10.
What is natural rubber?
Answer:
Natural rubber is a high molecular mass linear polymer of isoprene (2-methyl-1, 3-butadiene).
Maharashtra Board Class 12 Chemistry Solutions Chapter 15 Introduction to Polymer Chemistry 12

Question 11.
Write a note on natural rubber.
Answer:
Natural rubber is manufactured from rubber latex obtained from the rubber tree in the form of colloidal suspension. Reaction involved in the formation of natural rubber by the process of addition polymerization is as follows :
Maharashtra Board Class 12 Chemistry Solutions Chapter 15 Introduction to Polymer Chemistry 13

Natural rubber is -1, 4- polyisoprene. It exhibits elastic properties.

Question 12.
State properties of natural rubber.
Answer:

Polyisoprene molecule has cis configuration of the C = C double bond. It consists of various chains held together by weak van der Waals forces and has coiled structure.
It can be stretched like a spring and exhibits elastic property.
Its molecular mass varies from 130,000 u to 340,000 u.

Question 13.
Write a note on vulcanization of rubber. OR Discuss the main purpose of vulcanization of rubber.
Answer:
The tensile strength, toughness and elasticity of natural rubber can be increased by adding 3 to 5% sulphur and heating at 100-150°C, resulting in cross linking of cis-1, 4-polypropene chains through disulphide bonds, (-S-S-). This process is known as vulcanization of rubber. The physical properties of rubber can be changed by controlling the amount of sulphur in the vulcanization process. Rubber made with 20-30% sulphur is hard, 3 to 10% sulphur is little harder and is used in making tyres and 1 to 3% sulphur is used in making rubber bands.

Question 14.
Write the name and structure of one of the initiators used in free radical polymerisation.
Answer:
The initiator used in free radical polymerisation is acetyl peroxide.
Maharashtra Board Class 12 Chemistry Solutions Chapter 15 Introduction to Polymer Chemistry 19

Question 15.
What is LDP? How is it prepared? Give its properties and uses.
Answer:
LDP means low-density polyethene. LDP is obtained by polymerization of ethylene under high pressure (1000 – 2000 atm) and temperature (350 – 570 K) in presence of traces of O₂ or peroxide as initiator.
Maharashtra Board Class 12 Chemistry Solutions Chapter 15 Introduction to Polymer Chemistry 20

The mechanism of this reaction involves free radical addition and H-atom abstraction. The latter results in branching. As a result the chains are loosely held and the polymer has low density.

Properties of LDP :

LDP films are extremely flexible, but tough chemically inert and moisture resistant.
It is poor conductor of electricity with melting point 110 °C.

Uses of LDP :

LDP is mainly used in preparation of pipes for agriculture, irrigation, domestic water line connections as well as insulation to electric cables.
It is also used in submarine cable insulation.
It is used in producing extruded films, sheets, mainly for packaging and household uses like in preparation of squeeze bottles, attractive containers, etc.

Question 16.
Whatis HDP ? How is it prepared ? Give its properties and uses ?
Answer:
HDP means high density polyethylene. It is a linear polymer with high density due to close packing.
Maharashtra Board Class 12 Chemistry Solutions Chapter 15 Introduction to Polymer Chemistry 21

HDP is obtained by polymerization of ethene in presence of Zieglar-Natta catalyst which is a combination of triethyl aluminium with titanium tetrachloride at a temperature of 333K to 343K and a pressure of 6-7 atm.

Properties of HDP :

HDP is crystalline, melting point in the range of 144 – 150 °C.
It is much stiffer than LDP and has high tensile strength and hardness.
It is more resistant to chemicals than LDP.

Uses of HDP :

HDP is used in manufacture of toys and other household articles like buckets, dustbins, bottles, pipes, etc.
It is used to prepare laboratory wares and other objects where high tensile strength and stiffness is required.

Question 17.
How is polyacrylonitrile (PAN) prepared? Give its uses.
Answer:
Acrylonitrile (monomer) on polymerization (addition polymerization) in the presence of peroxide initiator gives polyacrylonitrile.
Maharashtra Board Class 12 Chemistry Solutions Chapter 15 Introduction to Polymer Chemistry 23

Uses : Polyacrylonitrile resembles wool and is used as wool substitute and for making orlon or acrilan.

Question 18.
How is nylon-6 prepared?
Write the reaction for the preparation of nylon 6.
Answer:
When epsilon (ε)-caprolactam is heated with water at high temperature it undergoes ring opening polymerization to form the polyamide polymer called nylon-6.
Maharashtra Board Class 12 Chemistry Solutions Chapter 15 Introduction to Polymer Chemistry 25

The name nylon-6 is given on the basis of six carbon atoms present in the monomer unit. Nylon-6 has high tensile strength and luster, nylon-6 fibres are used for manufacture of tyre cords, fabrics and ropes.

Question 19.
Draw the structures of polymers formed using the following monomers.
Maharashtra Board Class 12 Chemistry Solutions Chapter 15 Introduction to Polymer Chemistry 28

Maharashtra Board Class 12 Chemistry Solutions Chapter 15 Introduction to Polymer Chemistry 29

Maharashtra Board Class 12 Chemistry Solutions Chapter 15 Introduction to Polymer Chemistry 30
Answer:

Question 20.
How is Novolac prepared?
OR
Write the reaction to prepare Novolac polymer.
Answer:
The monomers phenol and formaldehyde undergo polymerisation in the presence of alkali or acid as catalyst.
Maharashtra Board Class 12 Chemistry Solutions Chapter 15 Introduction to Polymer Chemistry 33
Phenol reacts with formaldehyde to form ortho or p-hydroxy methyl phenols, which further reacts with phenol to form a linear polymer called Novolac. It is used in paints.

Question 21.
How is bakelite prepared?
Answer:
In the third stage, various articles are shaped from novolac by putting it in appropriate moulds and heating at high temperature (138 °C to 176 °C) and at high pressure forms rigid polymeric material called bakelite. Bakelite is insoluble and infusible and has high tensile strength.
Maharashtra Board Class 12 Chemistry Solutions Chapter 15 Introduction to Polymer Chemistry 34
Bakelite is used in making articles like telephone instrument, kitchenware, electric insulators frying pans, etc.

Question 22.
How is a melamine-formaldehyde polymer (melamine) prepared?
Answer:
The monomers formaldehyde and melamine undergo condensation polymerisation to form cross-linked melamine formaldehyde. It is used in making crockeries, decorative table tops like formica and plastic dinner-ware.
Maharashtra Board Class 12 Chemistry Solutions Chapter 15 Introduction to Polymer Chemistry 36

Question 23.
Write the preparation of the following synthetic rubbers and give their uses :
(1) Buna-S or styrene-butadiene rubber (SBR) (2) Neoprene rubber
Answer:
(1) Buna-S rubber : Its trade name is SBR (Styrene-butadiene rubber) Buna-S is a copolymer of styrene and 1, 3-butadiene. When 75 parts of butadiene and 25 parts of styrene subjected to addition polymerization by the action of sodium.
Maharashtra Board Class 12 Chemistry Solutions Chapter 15 Introduction to Polymer Chemistry 37

Uses : Buna-S is superior to natural rubber with regard to mechanical strength and has abrasion resistance. Hence, it is used in tyre industry.

(2) Neoprene : Neoprene, a synthetic rubber, is a condensation polymer of chloroprene (2-chloro-l, 3-butadiene), Chloroprene polymerizes rapidly in presence of oxygen.
Maharashtra Board Class 12 Chemistry Solutions Chapter 15 Introduction to Polymer Chemistry 38
Vulcanization of neoprene takes place in presence of magnesium oxide.

Uses : Neoprene is resistant to petroleum, vegetable oils, light as well as heat. It is used in making hose pipes for transport of gasoline and making gaskets. It is used for manufacturing insulator cable, jackets, belts for power transmission and conveying.

Question 24.
Write structure of natural rubber and neoprene rubber along with the name and structure of their monomers.
Answer:
Maharashtra Board Class 12 Chemistry Solutions Chapter 15 Introduction to Polymer Chemistry 40

Question 25.
Write the preparation of viscose rayon.
Answer:
Viscose rayon is a semisynthetic fibre. It is a regenerated cellulose. The molecular formula of cellulose is (C₆H₁₀O₅)_n. A modified representation of the molecular formula of cellulose Cell-OH is used in the reactions involved in viscose formation.

Maharashtra Board Class 12 Chemistry Solutions Chapter 15 Introduction to Polymer Chemistry 42

Cellulose (wood pulp) is treated with the concentrated NaOH to form alkali cellulose. It is then converted to xanthate by treating with CS₂. Further xanthate is mixed with dilute NaOH to form viscose solution which is extruted through spinnerates of spinning machine into acid bath, when regenerated cellulose fibres precipitate, i.e. viscose rayon.

Question 26.
How is PHBV polymer prepared?
Answer:
It is a copolymer. The monomers β-hydroxy butyric acid (3-hydroxy butanoic acid) and β-hydroxy valeric acid (3-hydroxy pentanoic acid) undergo polymerization to form PHBV polymer. It has an ester linkage. Hydroxyl group of one monomer forms ester link by reacting with carboxyl group of the other. Thus PHBV is an aliphatic polyester i.e. poly β-hydroxybutyrate-co-β-hydroxy valerate (PHBV). It is a biodegradable polymer.
Maharashtra Board Class 12 Chemistry Solutions Chapter 15 Introduction to Polymer Chemistry 45

Question 27.
Write the name/s of monomer/s, polymer structure and one use of each of the following polymers (trade name) :
Answer:
Maharashtra Board Class 12 Chemistry Solutions Chapter 15 Introduction to Polymer Chemistry 47

Question 28.
Write the names of monomers used in preparing following polymers :
(1) Dacron.
Answer:
Monomers : Ethylene glycol and Dmiethyl terephthalate (DMT)
Maharashtra Board Class 12 Chemistry Solutions Chapter 15 Introduction to Polymer Chemistry 49

(2) Bakelite.
Answer:
Monomers : o-hydroxy methyl phenol and formaldehyde.
Maharashtra Board Class 12 Chemistry Solutions Chapter 15 Introduction to Polymer Chemistry 50

(3) Nylon-6, 8.
Answer:
Monomers : Hexamethylene diamine and Hexamethylene dicarboxylic acid.
H₂N-(CH₂)₆-NH₂ HOOC-(CH₂)₆-COOH

(4) Melamine.
Answer:
Monomers : Melamine and Formaldehyde
Maharashtra Board Class 12 Chemistry Solutions Chapter 15 Introduction to Polymer Chemistry 51

(5) Buna-S.
Answer:
Maharashtra Board Class 12 Chemistry Solutions Chapter 15 Introduction to Polymer Chemistry 52

(6) Buna-N.
Answer:
Maharashtra Board Class 12 Chemistry Solutions Chapter 15 Introduction to Polymer Chemistry 53

(7) Butyl rubber.
Answer:
Maharashtra Board Class 12 Chemistry Solutions Chapter 15 Introduction to Polymer Chemistry 54

(8) Teflon.
Answer:
Monomers : F₂C = CF₂Tetrafluoroethene

(9) Natural rubber.
Answer:
Monomers : 1,3-Butadiene
CH₂ = CH – CH = CH₂

(10) Neoprene.
Answer:
Monomers : Chloroprene or 2-Chloro-l,3-butadiene
Maharashtra Board Class 12 Chemistry Solutions Chapter 15 Introduction to Polymer Chemistry 55

Question 29.
Write the structures of monomers used in the preparation of following polymers.
Maharashtra Board Class 12 Chemistry Solutions Chapter 15 Introduction to Polymer Chemistry 56
Answer:
The monomers used in the preparation of above polymer are :

Maharashtra Board Class 12 Chemistry Solutions Chapter 15 Introduction to Polymer Chemistry 58
Answer:
The monomers used in the preparation of above polymer are :

Answer:
The monomer used in the preparation of above polymer are :

Answer:
The monomer used in the preparation of above polymer is
Maharashtra Board Class 12 Chemistry Solutions Chapter 15 Introduction to Polymer Chemistry 61

Question 30.
Following monomers are used to prepare polymers. Predict the structures of polymers:

(1) Ethylene glycol.
Answer:
Ethylene glycol Maharashtra Board Class 12 Chemistry Solutions Chapter 15 Introduction to Polymer Chemistry 62 is used in the preparation of polyester (terylene or dacron).

Terylene is polyester fibre formed by the polymerization of terephthalic acid and ethylene glycol.

Terylene is obtained by condensation polymerization of ethylene glycol and terephthalic acid in presence of catalyst zinc acetate and antimony trioxide at high temperature.
Maharashtra Board Class 12 Chemistry Solutions Chapter 15 Introduction to Polymer Chemistry 26

Properties :

Terylene has relatively high melting point (265 °C)
It is resistant to chemicals and water.

Uses :

It is used for making wrinkle-free fabrics by blending with cotton (terycot) and wool (terywool), and also as glass reinforcing materials in safety helmets.
PET is the most common thermoplastic which is another trade name of the polyester polyethylenetereph- thalate.
It is used for making many articles like bottles, jams, packaging containers.

(2) ε-Caprolactam.
Answer:
ε-caprolactam is used in the preparation of nylon-6.
Maharashtra Board Class 12 Chemistry Solutions Chapter 15 Introduction to Polymer Chemistry 64

When epsilon (ε)-caprolactam is heated with water at high temperature it undergoes ring opening polymerization to form the polyamide polymer called nylon-6.
Maharashtra Board Class 12 Chemistry Solutions Chapter 15 Introduction to Polymer Chemistry 25

(3) Ethene.
Answer:
Ethene is used in the preparation of polythene
Polymer:

Linear chain polymers : When the monomer molecules are joined together in a linear arrangement, the resulting polymer is straight-chain or long-chain polymer, e.g., polythene, PVC.
Maharashtra Board Class 12 Chemistry Solutions Chapter 15 Introduction to Polymer Chemistry 4
They have high melting points; high densities and high tensile strength.

(4) Formaldehyde.
Answer:
Formaldehye is used in the preparation of bakelite.
Maharashtra Board Class 12 Chemistry Solutions Chapter 15 Introduction to Polymer Chemistry 66

The monomers phenol and formaldehyde undergo polymerisation in the presence of alkali or acid as catalyst.
Maharashtra Board Class 12 Chemistry Solutions Chapter 15 Introduction to Polymer Chemistry 33
Phenol reacts with formaldehyde to form ortho or p-hydroxy methyl phenols, which further reacts with phenol to form a linear polymer called Novolac. It is used in paints.

In the third stage, various articles are shaped from novolac by putting it in appropriate moulds and heating at high temperature (138 °C to 176 °C) and at high pressure forms rigid polymeric material called bakelite. Bakelite is insoluble and infusible and has high tensile strength.
Maharashtra Board Class 12 Chemistry Solutions Chapter 15 Introduction to Polymer Chemistry 34
Bakelite is used in making articles like telephone instrument, kitchenware, electric insulators frying pans, etc.

Question 31.
Classify the following polymers as step growth or chain growth polymers :
(1) Nylon-6,6
(2) Terylene
(3) Polyethene (4) PVC.
Answer:
Step growth polymers : Nylon-6,6, terylene
Chain growth polymers : Polythene, PVC.

Question 32.
Classify the following as linear, branched or cross linked polymers :
(1) Bakelite
(2) Starch
(3) Polythene
(4) Nylon.
Answer:
Linear polymers : Polythene, nylon.
Cross-linked polymers : Bakelite, starch.

Question 33.
Classify the following as addition and condensation polymers :
(1) Bakelite
(2) polyvinyl chloride
(3) polythene
(4) terylene.
Answer:
Addition polymers : Polyvinyl chloride, polythene
Condensation polymers : Bakelite, terylene.

Question 34.
Arrange the polymers in increasing order of their intermolecular forces :
Nylon-6,6, Polythene, Buna-S.
Answer:
The increasing order of their intermolecular forces of attraction follows the order :
Buna-S, Polythene, Nylon-6,6.

Question 35.
Classify the following as natural, synthetic and semisynthetic polymers :
Terylene, cuprammonium silk, jute, melamine
Answer:
Natural polymers : Jute
Synthetic polymers : Terylene, melamine
Semisynthetic polymers : Cuprammonium silk

Question 36.
Complete the following statements :
(1) Chemically teflon is …………………………. .
(2) …………………………. is the catalyst used to obtain HDP by polymerisation of ethene.
(3) Viscose rayon is a …………………………. .
Answer:
(1) polytetrafluoroethylene
(2) Zieglar-Natta
(3) semisynthetic fibre (regenerated fibre).

Question 37.
Answer the following in one sentence each.

(1) Name a polymer used for making LCD screen?
Answer:
The polymer used for making LCD screen is Perspex.

(2) Which of the two is a condensation polymer? Bakelite or Polythene?
Answer:
The condensation polymer is bakelite.

(3) Which of the two is a linear polymer? Nylon or Starch.
Answer:
The linear polymer is nylon.

(4) Which of the two is a step growth polymer? Nylon-6,6 or PVC.
Answer:
The step growth polymer is Nylon-6,6.

(5) Write the use of polyacrylamide gel.
Answer:
Polyacrylamide gel is used in electrophoresis.

(6) Write the use of urea formaldehyde resin.
Answer:
Urea formaldehyde resin is used in making unbreakable dinnerware and decorative laminates.

(7) Give an example of semisynthetic polymer.
Answer:
Semisynthetic polymer : Viscose rayon, cuprammonium silk.

(8) Write the monomer unit of teflon.
Answer:
Monomer unit of teflon : Tetrafluoroethene (F₂C = CF₂).

(9) Write the equation for the preparation of polypropylene.
Answer:
Maharashtra Board Class 12 Chemistry Solutions Chapter 15 Introduction to Polymer Chemistry 72

(10) Name a synthetic polymer which contains amide linkage.
Answer:
Polymer that contains amide linkage : Nylon-6; Nylon 6,6.

(11) Name a synthetic polymer which contains ester linkage.
Answer:
Polymer that contains ester linkage : Terylene or Dacron.

(12) Name one thermosetting and one thermoplastic polymer.
Answer:
Thermosetting polymer : Bakelite.
Thermoplastic polymer : Polythene, polystyrene.

(13) State the uses of biodegradable polymers.
Answer:
Biodegradable polymers are used as orthopaedic devices, implants, sutures and drug release matrices.

(14) Name a copolymer which is used for making nonbreakable crockeries.
Answer:
The polymer used in making nonbreakable crockeries : Melamine formaldehyde polymer.

(15) Write the structure of monomer used in the preparation of Maharashtra Board Class 12 Chemistry Solutions Chapter 15 Introduction to Polymer Chemistry 75
Answer:
The structure of monomer :

(16) Write the structure of melamine.
Answer:
The monomers formaldehyde and melamine undergo condensation polymerisation to form cross-linked melamine formaldehyde. It is used in making crockeries, decorative table tops like formica and plastic dinner-ware.
Maharashtra Board Class 12 Chemistry Solutions Chapter 15 Introduction to Polymer Chemistry 36

(17) What does SBR stand for?
Answer:
SBR stand for styrene(S)butadiene (B) rubber (R).

(18) Draw the structure of repeating unit in nylon-6,10.
Answer:
The structure of repeating unit in nylon-6,10 is
Maharashtra Board Class 12 Chemistry Solutions Chapter 15 Introduction to Polymer Chemistry 77

(19) What are the monomers used to prepare nylon given below?
Maharashtra Board Class 12 Chemistry Solutions Chapter 15 Introduction to Polymer Chemistry 78
Answer:
Monomers used in the preparation of nylon are

(20) Write the monomers used to prepare nylon given below :
Maharashtra Board Class 12 Chemistry Solutions Chapter 15 Introduction to Polymer Chemistry 80
Answer:
Monomers used in the preparation of nylon are

(21) Name the catalyst which is formed from titanium chloride and triethyl aluminium.
Answer:
The catalyst Zieglar Natta is formed from titanium chloride and triethyl aluminium.

(22) Define molecular mass of polymer.
Answer:
Molecular mass of a polymer is an average of the molecular masses of the constituent molecules.

(23) Which factor determines the molecular mass of polymer.
Answer:
Molecular mass of polymer depends upon the degree of polymerization (DP). DP is the number of monomer units in a polymer molecule.

Multiple Choice Questions

Question 38.
Select and write the most appropriate answer from the given alternatives for each subquestion :

1. Which one is the natural polyamide polymer?
(a) Cuprammonium silk
(b) Wool
(c) Perlon-L
(d) Jute
Answer:
(b) Wool

2. The synthetic fibres are prepared from
(a) cellulose
(b) starch
(c) chemical compounds
(d) polymers
Answer:
(c) chemical compounds

3. Which of the following is NOT a vegetable fibre?
(a) Hemp
(b) Jute
(c) Cotton
(d) Keratin
Answer:
(d) Keratin

4. Which of the following fibres are made up of polyamides?
(a) Dacron
(b) Rayon
(c) Nylon
(d) Terylene
Answer:
(c) Nylon

5. An example of natural cellulose fibre is
(a) cotton
(b) wool
(c) silk
(d) rayon
Answer:
(a) cotton

6. Cellulose is the main constituent of
(a) nylon-6
(b) cotton
(c) terylene
(d) wool
Answer:
(b) cotton

7. Of the following, which group contains two cellulosic fibres and one protein fibre?
(a) Cotton, keratin, wool
(b) Linen, keratin, wool
(c) Cotton, linen, rayon
(d) Cotton, keratin, linen
Answer:
(d) Cotton, keratin, linen

8. Which of the following is not a vegetable fibres?
(a) Hemp
(b) Jute
(c) Cotton
(d) Keratin
Answer:
(d) Keratin

9. Which of the following is polyamide?
(a) Teflon
(b) Nylon 6, 6
(c) Terylene
(d) Bakelite
Answer:
(b) Nylon 6, 6

10. The monomer of e-caprolactam is
(a) styrene
(b) amino acid
(c) aminocaproic acid
(d) adipic acid O
Answer:
(c) aminocaproic acid

11. Maharashtra Board Class 12 Chemistry Solutions Chapter 15 Introduction to Polymer Chemistry 83 is the formula of _ Jn
(a) Nylon-66 salt
(b) Nylon-66
(c) DMT
(d) Nylon-6
Answer:
(d) Nylon-6

12. Nylon-6 is a
(a) polyester fibre
(b) protein fibre
(c) poly caprolactum fibre
(d) poly amine fibre
Answer:
(c) poly caprolactum fibre

13. The condensation product of e-caprolactum is
(a) teflon
(b) nylon-6
(c) nylon-66
(d) bakelite
Answer:
(b) nylon-6

14. \(\left[\overline{\mathrm{O}} \mathrm{OC}-\left(\mathrm{CH}_{2}\right)_{4}-\mathrm{COO}-\mathrm{N} \mathrm{H}_{3}-\left(\mathrm{CH}_{2}\right)_{6}-\mathrm{N} \mathrm{H}_{3}\right]\) is the formula of
(a) nylon-6
(b) nylon-6 salt
(c) nylon-66 salt
(d) nylon-66
Answer:
(b) nylon-6 salt

15. The starting material required for the preparation of Nylon-66 is
(a) glycol
(b) α-amino acid
(c) adipic acid and hexamethylene diamine
(d) dimethyl terephthalate and ethylene glycol
Answer:
(c) adipic acid and hexamethylene diamine

16. Terylene is also known as
(a) styrene
(b) butadiene
(c) dacron
(d) teflon
Answer:
(c) dacron

17. Terylene is
(a) vegetable fibre
(b) protein fibre
(c) polyester fibre
(d) polyamide fibre
Answer:
(c) polyester fibre

18. Terylene polymer is formed in
(a) the presence of nitrogen
(b) the presence of hydrogen
(c) the presence of oxygen
(d) the vacuum
Answer:
(d) the vacuum

19. The by-product formed during the preparation of terylene fibre is
(a) glycerol
(b) propylene glycol
(c) ethylene glycol
(d) ethyl alcohol
Answer:
(c) ethylene glycol

20. Nylon polymer cannot be used for making
(a) tyre cords
(b) films
(c) dress materials
(d) fishing nets
Answer:
(b) films

21. Glycol is an important constituent of
(a) nylon-6
(b) nylon-66
(c) terylene
(d) hexamethylene diammonium adipate
Answer:
(c) terylene

22. Terylene is prepared by the process of
(a) halogenation
(b) condensation
(c) esterification
(d) hydrogenation
Answer:
(b) condensation

23. What are the steps during polymerisation to form terylene?
(a) Terephthalic acid is condensed with ethylene glycol at 420 K-460 K.
(b) Ethylene glycol displaces methanol to form dihydroxy diethyl terephthalic acid
(c) Zinc acetate – antimony trioxide is used as catalyst
(d) All of these
Answer:
(d) All of these

24. During polymerisation of nylon salt to nylon-66, the conditions are
(a) room temperature and pressure
(b) temperature 503 K
(c) temperature 553 K in presence of Nitrogen
(d) heating in an autoclave at 373 K
Answer:
(c) temperature 553 K in presence of Nitrogen

25. Which one of the following is a condensation polymer?
(a) Nylon
(b) Polythene
(c) PVC
(d) Teflon
Answer:
(a) Nylon

26. Which one of the following is an addition polymer?
(a) Bakelite
(b) Nylon-6,6
(c) Polystyrene
(d) Terylene
Answer:
(c) Polystyrene

27. A polymer of butadiene and Acrylonitrile is called
(a) Buna-S
(b) Buna-N
(c) Buna-B
(d) Buna-A
Answer:
(b) Buna-N

28. Natural rubber is a polymer of
(a) Styrene
(b) Butadiene
(c) Vinyl chloride
(d) Isoprene
Answer:
(d) Isoprene

29. In which of the following pairs both are copolymers?
(a) PHBV, bakelite
(b) Polythene, terylene
(c) Polyacrylonitrile, nylon-6,6
(d) Polystyrene, melamine
Answer:
(a) PHBV, bakelite

30. The polymer used in paints is
(a) Nylon
(b) Glyptal
(c) Neoprene
(d) Terylene
Answer:
(b) Glyptal

31. Which of the following contains biodegradable polymers only?
(a) Cellulose, dextron, PHBV
(b) Starch, PHBV, PVC
(c) Bakelite, nylon-2-nylon-6, nylon-6,6
(d) Cellulose, starch, terylene
Answer:
(a) Cellulose, dextron, PHBV

32. Thermosetting polymer is
(a) Nylon-6
(b) Nylon-6,6
(c) Bakelite
(d) SBR
Answer:
(c) Bakelite

33. Nylon thread contains the polymer
(a) Polyamide
(b) Polyvinyl
(c) Polyester
(d) Polyethylene
Answer:
(a) Polyamide

34. Polythene, PVC, teflon and neoprene are all
(a) Monomers
(b) Homopolymers
(c) Copolymers
(d) Condensation polymers
Answer:
(b) Homopolymers

35. Which one of the following is NOT a biodegrad¬able polymer?
(a) Starch
(b) Cellulose
(c) Dextron
(d) Decron
Answer:
(d) Decron

36. The polymer used in making blankets (artificial wool) is
(a) Polyester
(b) Polyacrylonitrile
(c) Polythene
(d) Polystyrene
Answer:
(b) Polyacrylonitrile

37. Which one of the following is a linear polymer?
(a) Bakelite
(b) LDP
(c) Nylon
(d) Formaldehyde melamine polymer
Answer:
(c) Nylon

38. Which one of the following is a branched polymer?
(a) PVC
(b) Polyester
(c) Nylon
(d) Polypropylene
Answer:
(d) Polypropylene

39. The polymer used to make non-stick cookware is
(a) Polyethene
(b) Polystyrene
(c) Polytetrafluoroethylene
(d) Polyvinyl chloride
Answer:
(c) Polytetrafluoroethylene

40. The monomer used to prepare orlon is
(a) CH₂ = CH-CN
(b) CH₂ = CHCl
(c) CH₂ = CH-F
(d) CH₂ = CF₂Answer:
(a) CH₂ = CH-CN

41. Buna-N rubber is a copolymer of
Maharashtra Board Class 12 Chemistry Solutions Chapter 15 Introduction to Polymer Chemistry 86
Answer:
(b)

42. Bakelite is a polymer of
(a) Formaldehyde and phenol
(b) Benzaldehyde and phenol
(c) Formaldehyde and benzyl alcohol
(d) Acetaldehyde and phenol
Answer:
(a) Formaldehyde and phenol

43. The process involving heating of natural rubber with sulphur is known as
(a) Sulphonation
(b) Vulcanisation
(c) Galvanisation
(d) Calcination
Answer:
(b) Vulcanisation

44. A polymer of bisphenol and phosgene is called
(a) Polyamide
(b) Glyptal
(c) Polycarbonate
(d) Polystyrene
Answer:
(c) Polycarbonate

45. Thermocol is a homopolymer of
(a) terephthalic acid
(b) acrylonitrile
(c) methyl a-cyanoacrylate
(d) styrene
Answer:
(d) styrene

46. The polymer is used to prepare shatter resistant glass is called
(a) Perspex/acrylic glass
(b) Soda glass
(c) Buna N
(d) Polyacrylamide
Answer:
(a) Perspex/acrylic glass

47. A polymer used in making shoe soles is
(a) Glyptal
(b) Buna-N
(c) Buna-S
(d) Poly carbonate
Answer:
(b) Buna-N

48. The Zieglar-Natta catalyst is used in the preparation of
(a) LDPE
(b) PHBV
(c) PAN
(d) HDPE
Answer:
(d) HDPE

49. Which of the following is natural rubber?
(a) cis-1, 4-polyisoprene
(b) neoprene
(c) Trans-1. 4-polyisoprene
(d) Butyl rubber
Answer:
(a) cis-1, 4-polyisoprene

50. Which one from the following is the Terylene polymer?
Maharashtra Board Class 12 Chemistry Solutions Chapter 15 Introduction to Polymer Chemistry 84
Answer:
(b)

51. Equivalent amount of Hexamethylene diamine and adipic acid on complete neutralization produces :
Maharashtra Board Class 12 Chemistry Solutions Chapter 15 Introduction to Polymer Chemistry 85
Answer:
(a)

52. Polyhydroxy butyrate-CO-β-hydroxy valerate represents
(a) Dextron
(b) Nylon-6, 6
(c) Nylon-2-nylon-6
(d) PHBV
Answer:
(d) PHBV

53. Among the following, the biodegradable polymer is
(a) PVC
(b) Nylon-6, 6
(c) Nylon-2-nylon-6
(d) Neoprene
Answer:
(c) Nylon-2-nylon-6

54. The monomers of Nylon-2-nylon-6 are
(a) glycine and ω-amino caproic acid
(b) lactic acid and glycolic acid
(c) glycolic acid and co-amino caproic acid
(d) isobutylene and isoprene
Answer:
(a) glycine and ω-amino caproic acid

Maharashtra Board Class 12 Chemistry Important Questions Chapter 14 Biomolecules

July 17, 2024July 16, 2024 by Bhagya

Balbharti Maharashtra State Board 12th Chemistry Important Questions Chapter 14 Biomolecules Important Questions and Answers.

Maharashtra State Board 12th Chemistry Important Questions Chapter 14 Biomolecules

Question 1.
What are biomolecules? Give examples of biomolecules.
Answer:
Biomolecules: The lifeless, complex organic molecules which combine in a specific manner to produce life or control biological reactions are called biomolecules.

Examples: Carbohydrates, lipids (fats and oils), nucleic acids, enzymes.

Question 2.
What is the importance of biomolecules?
Answer:
Biomolecules are organic molecules which combine in a particular fashion to give complex substances which help to sustain life and produce identical daughter cells and play an important role in the actions of an organism.

Carbohydrates are the major constituents of food and source of energy.
Proteins help in proper functioning of living beings. They are important constituents of skin, hair, muscles. Enzymes which catalyse chemical reactions that take place in cells are proteins.
Lipids (fats and oils) function as the storehouses of energy.
Nucleic acids, the ribonucleic acid (RNA), and deoxyribonucleic acid (DNA) are responsible for genetic characteristics and synthesis of proteins.

Question 3.
What are carbohydrates?
OR
Define the term : Carbohydrates.
Answer:
Carbohydrates : Carbohydrates are optically active polyhydroxy aldehydes or polyhydroxy ketones, or the compounds which on hydrolysis produce polyhydroxy aldehydes or polyhydroxy ketones.

Examples : Glucose, sucrose, fructose.

Question 4.
What is monosaccharide?
Answer:
The basic unit of all carbohydrates which is a simple carbohydrate and cannot be hydrolysed further is known as monosaccharide. The monosaccharide is crystalline and soluble in water. E.g. Glucose, fructose, ribose.

Question 5.
Mention the names of monosaccharides or simple carbohydrates.
Answer:
Monosaccharides are (1) glucose (2) fructose (3) ribose.

Question 6.
State the basic unit of all carbohydrates.
Answer:
The basic unit of all carbohydrates which is a simple carbohydrate and cannot be hydrolysed further is known as monosaccharide.

Question 7.
How are carbohydrates classified?
OR
Classification of carbohydrates with examples.
Answer:
Carbohydrates are classified as monosaccharides oligosaccharides and polysaccharides.
(1) Monosaccharides : These carbohydrates cannot be further hydrolysed into smaller units. They are basic units of all carbohydrates, and are called monosaccharides.

Examples : Glucose, fructose, ribose

(2) Oligosaccharides : An oligosaccharide is a carbohydrate (sugar) which on hydrolysis gives two to ten monosaccharide units.
Depending on the number of monosaccharides produced on hydrolysis, oligosaccharides are further classified as :
Maharashtra Board Class 12 Chemistry Solutions Chapter 14 Biomolecules 1

Oligosaccharide is homogeneous. In this, each molecule of oligosaccharide contains the same number of monosaccharide units joined together in the same order as every other molecule of the same oligosaccharide.

(3) Polysaccharides : These are carbohydrates which on hydrolysis give a large number of monosaccharides.

Polysaccharides are tasteless, amorphous, insoluble in water. They are long chain, naturally αcurring polymers of carbohydrates.

Example : Cellulose, starch, glycogen.

Question 8.
Classify the following carbohydrates into Monosaccharide, Disaccharide, Oligosaccharide, Polysaccharide.
(1) Glucose
(2) Starch
(3) Sucrose
(4) Maltose
(5) Galactose
(6) Lactose
(7) Ribose.
Answer:

Carbohydrates	Class
(1) Glucose	Monosaccharide
(2) Starch	Polysaccharide
(3) Sucrose	Disaccharide
(4) Maltose	Disaccharide
(5) Galactose	Monosaccharide
(6) Lactose	Disaccharide
(7) Ribose	Monosaccharide

Question 9.
Classify the following carbohydrates.
(1) Cellulose,
(2) Maltose,
(3) Raffinose,
(4) Fructose.
Answer:

Carbohydrates

Class

(1) Cellulose

(2) Maltose

(3) Raffinose

(4) Fructose

Polysaccharide

Disaccharide

Trisaccharide

Monosaccharide

Question 10.
Classify the following into monosaccharides, oligosaccharides and polysaccharides.
(1) Starch
(2) Glucose
(3) Stachyose
(4) Maltose
(5) Raffinose
(6) Cellulose
(7) Sucrose
(8) Lactose.
Answer:

Monosaccharides	Glucose
Oligosaccharides	Stachyose, maltose, raffinose, sucrose, lactose
Polysaccharides	Starch, cellulose

Question 11.
Classify the following into monosaccharides and disaccharides.
Ribose, maltose, galactose, fructose and lactose (~2 mark each)
Answer:

Monosaccharides	Ribose, galactose, fructose
Disaccharides	Maltose, lactose

Question 12.
Give the preparation of glucose from sucrose or cane sugar.
OR
Describe the laboratory method of preparation of glucose.
Answer:
Preparation of glucose from sucrose (cane sugar) : Laboratory method.

Glucose is prepared in the laboratory by hydrolysis of sucrose by boiling it with dilute hydrαhloric acid or dilute sulphuric acid for about two hours. On hydrolysis, sucrose gives one molecule of glucose and one molecule of fructose.
Maharashtra Board Class 12 Chemistry Solutions Chapter 14 Biomolecules 5
Alcohol is added during cooling to separate glucose and fructose since, glucose is almost insoluble in alcohol, hence it crystallizes out first. Fructose remains in the solution as it is more soluble than glucose.

Crystals of glucose are separated out by filtration and purified by recrystallization.

Question 13.
Give the preparation of glucose from starch.
OR
How is glucose prepared on commercial scale?
Answer:
Commercially, on a large scale, glucose is prepared by hydrolysis of starch with dilute sulphuric acid. Starchy material is mixed with water and dilute sulphuric acid and heated at 393 K under 2 to 3-atm pressure. Starch is hydrolysed to give glucose.
Maharashtra Board Class 12 Chemistry Solutions Chapter 14 Biomolecules 6

Question 14.
Explain the structure of glucose.
Answer:
Molecular formula of glucose is C₆H₁₂O₆.
Maharashtra Board Class 12 Chemistry Solutions Chapter 14 Biomolecules 7
Glucose has an aldohexose structure. In other words, glucose molecule contains one aldehydic, that is, formyl group and the remaining five carbons carry one hydroxyl group (-OH) each. The six carbons in glucose form one straight chain.

Question 15.
Describe the action of following reagents on glucose :
(1) HI
(2) Hydroxyl amine (NH₂OH)
(3) Hydrogen cyanide
(4) Bromine water
(5) dil. Nitric acid
(6) Acetic anhydride.
Answer:
(1) Action of HI : Glucose on prolonged heating with HI gives n-hexane, indicates that all the six carbon atoms are linked in straight chain.
Maharashtra Board Class 12 Chemistry Solutions Chapter 14 Biomolecules 8

(2) Action of hydroxyl amine : Glucose reacts with hydroxyl amine in an aqueous solution to form glucose oxime. This indicates the presence of CHO group in glucose.
Maharashtra Board Class 12 Chemistry Solutions Chapter 14 Biomolecules 9

(3) Action of hydrogen cyanide : Glucose reacts with hydrogen cyanide to form glucose cyanohydrin.
Maharashtra Board Class 12 Chemistry Solutions Chapter 14 Biomolecules 10

(4) Action of bromine water : Glucose on oxidation with mild oxidising agent like bromine water gives gluconic acid, which shows that the carbonyl group in glucose is aldehyde group.
Maharashtra Board Class 12 Chemistry Solutions Chapter 14 Biomolecules 11

(5) Action of dll. nitric acid : Glucose on oxidation with dilute nitric acid forms dicarboxylic acid, saccharic acid. This indicates the presence of a primary alcoholic group (-CH₂OH) in glucose.
Maharashtra Board Class 12 Chemistry Solutions Chapter 14 Biomolecules 12

(6) Action of acetic anhydride : When glucose is heated with acetic anhydride in the presence of catalyst pyridine, glucose penta acetate is formed. It indicates that glucose is a stable compound and contains five hydroxyl groups.
Maharashtra Board Class 12 Chemistry Solutions Chapter 14 Biomolecules 13

Question 16.
Write Fischer projection formulae for
(1) Glucose
(2) Gluconic acid
(3) Saccharic acid.
Answer:
Fischer projection formulae of glucose, gluconic acid and saccharic acid :
Maharashtra Board Class 12 Chemistry Solutions Chapter 14 Biomolecules 17

Question 17.
Explain D and L configuration in sugars.
Answer:
The simplest carbohydrates glyceraldehyde is chosen as the standard, to assign D and L configuration to monosaccharides. Glyceraldehyde contains one asymmetric carbon atom and exist in two enantiomeric forms
Maharashtra Board Class 12 Chemistry Solutions Chapter 14 Biomolecules 18

The dextro entantiomer is represented as (+) glyceraldehyde and it is referred as D-configuration i.e., D-glyceraldehyde. The laevo enantiomer of glyceraldehyde is represented as ( -) glyceraldehyde and it corelated as L-configuration i.e., L-glyceraldehyde.

In Fischer projection formula, a monosaccharide is assigned D-configuration if the (- OH) hydroxyl group at the last chiral carbon and lies towards right hand side. On the other hand it is assigned L-configuration if the – OH group on the last chiral carbon atom and lies on the left hand side. In monosaccharides, the most oxidised carbon (i.e., -CHO) is at the top.

Examples :
Maharashtra Board Class 12 Chemistry Solutions Chapter 14 Biomolecules 19

Question 18.
Write Fischer projection formulae for (a) L-( + )-erythrose (b) L-( +) ribulose.
Answer:
Maharashtra Board Class 12 Chemistry Solutions Chapter 14 Biomolecules 23

Question 19.
Is the following sugar, D-sugar or L-sugar?
Maharashtra Board Class 12 Chemistry Solutions Chapter 14 Biomolecules 24
Answer:
The compound is L-sugar.
The compound is L-sugar.

Question 20.
Assign D/L configuration to the following monosaccharides.
Maharashtra Board Class 12 Chemistry Solutions Chapter 14 Biomolecules 25

Answer:

Question 21.
Explain ring structure of glucose.
Answer:
Maharashtra Board Class 12 Chemistry Solutions Chapter 14 Biomolecules 28
Glucose has two cyclic structures (II and III) which are in equilibrium with each other through the open chain structure (I) in aqueous solution.

The ring structure of glucose is formed by reaction between the formyl ( – CHO) group and the alcoholic (- OH) group at C – 5. Thus, the ring structure is called a hemiacetal. The two hemiacetal structures (II and III) differ only in the configuration of C – I (Fig.), the additional chiral centre resulting from ring closure. The two ring structures are called α- and β- anomers of glucose and C-l is called the anomeric carbon. The ring of the cyclic structure of glucose contains five carbons and one oxygen. Thus, it is a six membered ring. It is called pyranose structure, in analogy with the six membered heterαyclic compound pyran (IV). Hence glucose is also called glucopyranose.

Question 22.
Write the structures of α-D-( + )-glucopyranose and β-D-( +) glucopyranose.
Answer:
Maharashtra Board Class 12 Chemistry Solutions Chapter 14 Biomolecules 29

Question 23.
Explain Haworth formula of glycopyranose.
Answer:
Maharashtra Board Class 12 Chemistry Solutions Chapter 14 Biomolecules 30
In the Haworth formula the pyranose ring is considered to be in a perpendicular plane with respect to the plane of paper. The carbons and oxygen in the ring are in the places as they appear in figure. The lower side of the ring is called α-side and the upper side is the β-side. The α-anomer has its anomeric hydroxyl (- OH) group (at C-l) on the α-side, whereas the β-anomer has its anomeric hydroxyl (- OH) group (at C-l) on the β-side. The groups which appear on right side in the Fischer projection formula appear on α-side in the Haworth formula, and the groups which appear on left side in the fischer projection formula appear on a β-side in the Haworth formula.

Question 24.
Explain the structure of fructose.
Answer:
Fructose has molecular formula C₆H₁₂O₆. It contains ketonic functional group at carbon number 2 and six carbon atoms in straight chain. It belongs to D-series and is a laevo rotatory compound. It is written as D-( – )-fructose. Being an α-hydroxy keto compound fructose is a reducing sugar.
Maharashtra Board Class 12 Chemistry Solutions Chapter 14 Biomolecules 31

Question 25.
Draw mirror images of glucose and fructose.
Answer:
(1) Glucose
Maharashtra Board Class 12 Chemistry Solutions Chapter 14 Biomolecules 32
(2) Fructose

Question 26.
Write the two cyclic structures of α-D-( – )-fructofuranose and β-D-( – )-fructofuranose exist in equilibrium with open chain structure.
Answer:
Maharashtra Board Class 12 Chemistry Solutions Chapter 14 Biomolecules 34

Question 27.
Write the Haworth projection formulae for α -D-( -) – Fructofuranose and β – D – ( -) – Fructo- furanose.
Answer:
Maharashtra Board Class 12 Chemistry Solutions Chapter 14 Biomolecules 35

Question 28.
Explain the structure of sucrose.
Answer:
Sucrose is a hexasaccharide and has molecular formula C₁₂H₂₂O₁₁. The structure of shcrose contains glycosidic linkage between C – 1 of α-glucose and C – 2 of β-fructose. Since aldehyde and ketone groups of both monosaccharide units are involved in the formation of glycosidic bond, sucrose is a nonreducing sugar.

Sucrose is dextrorotatory, on hydrolysis with dilute acid or an enzyme invertase gives equimolar mixture of dextrorotatory glucose and laevorotatory fructose.
Maharashtra Board Class 12 Chemistry Solutions Chapter 14 Biomolecules 36
The solution is laevorotatory because laevo rotation of fructose (- 92.4°) is more than dextrorotation of glucose ( + 52.50), hence the sign of rotation is changed from (+) to (-) after hydrolysis, the product is called invert sugar.
Maharashtra Board Class 12 Chemistry Solutions Chapter 14 Biomolecules 37

Question 29.
Explain the structure of maltose.
Answer:
Maltose is another disaccharide obtained by partial hydrolysis of starch or made of two units of D-glucose. In maltose, C-l of one α-D-glucose is linked to C-4 of another α-D-glucose molecule by glycosidic linkage. The glucose ring which uses its hydroxyl group at C-1 is α – 1 → 4 glycosidic linkage. It is a reducing sugar because a free aldehyde group can be produced at C₁ of second glucose molecule. Maltose on hydrolysis with dilute acids gives glucose.
Maharashtra Board Class 12 Chemistry Solutions Chapter 14 Biomolecules 38

Question 30.
Draw a neat diagram for Haworth formula of maltose.

Question 31.
Explain the structure of lactose.
Answer:
Maharashtra Board Class 12 Chemistry Solutions Chapter 14 Biomolecules 39
Lactose (C₁₂H₂₂O₁₁) is a disaccharide. It is found in milk, therefore, it is also known as milk sugar. It is formed from two monosaccharide units, namely D – galactose and D – glucose. The glycosidic linkage is formed between C-l of β-D-galactose and C -4 of glucose. Therefore the linkage in lactose is called β – 1,4 – glycosidic linkage. The hemiacetal group at C-l of the glucose unit is not involved in glycosidic linkage but is free. Hence lactose is a reducing sugar. The above figure shows Haworth formula of lactose.

Question 32.
What are the hydrolysis products of (1) lactose (2) sucrose?
Answer:
(1) Lactose on hydrolysis in presence of an acid or enzyme lactase gives one molecule each of glucose and galactose
Maharashtra Board Class 12 Chemistry Solutions Chapter 14 Biomolecules 40
(2) Sucrose on hydrolysis in the presence of dii. acid or the enzyme invertase gives one molecule each of glucose and fructose.

Question 33.
Explain the structure of starch.
Answer:
Starch is found in cereal grains, roots, tubers, potatoes, etc. It is a polymer of α-D-glucose and consists of two components, amylose and amylopectin.

Amylose is water soluble component forms blue coloured complex with iodine. It constitutes about 20 % of starch. Amylose contains 200 to 1000 α-D-glucose units linked together by glycosidic linkage between C-l of one unit and C-4 of another unit. i.e. α-1, 4 glycosidic linkages.
Maharashtra Board Class 12 Chemistry Solutions Chapter 14 Biomolecules 43
Amylopectin is insoluble in water and constitutes about 80 % starch which forms blue-violet coloured complex with iodine. It is a branched chain polymer. In amylopectin, α-D-glucose molecules are linked together by glycosidic linkage between C₁– of one unit and C-4 of another unit to form long chain and branching αcurs by glycosidic linkage between C-l and C6 glycosidic linkage.
Maharashtra Board Class 12 Chemistry Solutions Chapter 14 Biomolecules 44

Question 34.
What are polysaccharides?
Answer:
A large number of same or different monosaccharides are joined together by glycosidic linkages are called polysaccharides. They have general formula (C₆H₁₀O₅)_n.

Question 35.
Explain the structure of cellulose.
Answer:
Maharashtra Board Class 12 Chemistry Solutions Chapter 14 Biomolecules 45
Cellulose mainly αcurs in plants. Cell wall of plant cells is made up of cellulose. It is a long chain polymer. In cellulose, β-D-glucose units are linked by glycosidic linkage between C₁-of one unit of glucose and C₄ of another glucose unit. Thus cellulose contains 1 → 4β glycosidic linkages like those in cellobiose.

Question 36.
Explain the structure of glycogen.
Answer:
The glucose is stored in animal body in the form of glycogen. It is also known as animal starch because its structure is similar to amylopectin. Glycogen is highly branched. Whenever the body is required glucose, enzymes breaks the glycogen to glucose.

Question 37.
How is glycogen different from starch?
Answer:
Starch is the main storage molecules of plants whereas glycogen is the main storage molecule of animals. Starch is found in cereals, roots, tubers, etc. Glycogen is present in liver, muscles and brain.

Question 38.
What do you understand by the term glycosidic linkage?
Answer:
The linkage between two monosaccharide units through oxygen atom is called glycosidic linkage.
Maharashtra Board Class 12 Chemistry Solutions Chapter 14 Biomolecules 46

Question 39.
What is the basic structural difference between starch and cellulose?
Answer:
Starch is a polymer of a-glucose and consists of two components-amylose and amylopectin. In amylose α-D-D-( + )-glucose units held by C,-C4 glycosidic linkage and in amylopectin, α-D-glucose units held by C1-C4 glycosidic linkage whereas branching αcurs by C₁-C₆ glycosidic linkage. [Refer Question 35 (i) (ii) Fig.] Cellulose is a straight chain polysaccharide composed only of β-D-glucose units held by C₁-C₄ glycosidic linkage. (Refer Question 37 Fig.)

Question 40.
Define the term : Protein OR What are proteins?
Answer:
Chemically proteins are polyamides which are high molecular weight polymers of the monomer units i.e. α-amino acids. OR It can also be defined as Proteins are the biopolymers of a large number of a-amino acids and they are naturally occurring polymeric nitrogenous organic compounds containing 16% nitrogen and peptide linkages (-CO-NH-).

Question 41.
Write the common sources of protein.
Answer:
Common sources of proteins are milk, pulses, peanuts, eggs, fishes, cheese, cereals, etc. They are also the principal materials of muscle, nerves, tendons, skin, blood, enzymes, many hormones and antibiotics.

Question 42.
What are the products of hydrolysis of proteins?
Answer:
On hydrolysis, proteins give a mixture of α-anlino acids.
Maharashtra Board Class 12 Chemistry Solutions Chapter 14 Biomolecules 47
The α-carbon in α-amino acids ohtained by hydrolysis of proteins has ‘L’ configuration.

Question 43.
What are the a-amino acids?
Answer:
α-Amino acids are carboxylic acids having an amino (- NH₂) group bonded to the α-carbon, i.e. the carbon next to the carboxyl (- COOH) group.
Maharashtra Board Class 12 Chemistry Solutions Chapter 14 Biomolecules 48
α-amino acids are derivatives of carboxylic acids, obtained by replacing – H atom by amino group. They are bifunctional compounds containing acidic and basic – NH₂ groups.
Example : (where R is an alkyl group or aryl group).

The amino acids are colourless, crystalline, water soluble, high melting solids. These acids in their aqueous solutions behave like salts due to presence of both acidic, and basic. (- NH₂) groups in the same molecule.

Such a doubly charged ion is known as zwitter ion. Example : Maharashtra Board Class 12 Chemistry Solutions Chapter 14 Biomolecules 52

Question 44.
What are the final products of hydrolysis of proteins?
Answer:
Proteins on hydrolysis with dilute solution of acids, alkalies or enzymes give a mixture of large number of a-amino acids as final products.

For example :
Maharashtra Board Class 12 Chemistry Solutions Chapter 14 Biomolecules 54

Question 45.
Write the classification of amino acids, giving examples.
Answer:
The amino acids are of three types : acidic, basic and neutral. The symbol ‘R’ in the structure of a-amino acids represents side chain and may contain additional functional groups.

(1) Acidic amino acids : If ‘R’ contains a carboxyl (- COOH) group the amino acid is acidic amino acid, i.e. If carboxyl groups are more in number than amino groups, then amino acids are acidic in nature.

Examples : Glutamic acid HOOC-CH₂-CH₂-; Aspartic acid HOO-CH₂–

(2) Basic amino acids : If ‘R’ contains an amino (1°, 2°, or 3°) group, it is called basic amino acid i.e. If amino groups are more in number than carboxyl groups then amino acids are basic in nature.

Examples : Arginine Maharashtra Board Class 12 Chemistry Solutions Chapter 14 Biomolecules 55

(3) Neutral amino acids : The other amino acids having neutral or no functional group in ‘R’ are called neutral amino acids, i.e. The amino acids having equal number of amino and carboxyl groups are neutral amino acids.

Examples : Alanine CH₃-; Valine (CH₃)₂-CH

Question 46.
What are essential and non-essential amino acids? Give two examples of each.
Answer:
The amino acids, which cannot be synthesised in the body and are supplied through diet are called essential amino acids. Examples : Lysine H₂N-(CH₂)₄-; Valine (CH₃)₂CH- The amino acids which are synthesized in the body are called non-essential amino acids.

Examples : Glutamic acid HOO-CH₂-CH₂-; Serine HO-CH₂–

Question 47.
What is meant by Zwitter ion?
Answer:
An a-amino acid molecule contains both acidic carboxyl ( – COOH) group as well as basic amino (- NH₂) group. Proton transfer from acidic group to basic group of amino acid forms a salt, which is a dipolar ion called a zwitterion.
Maharashtra Board Class 12 Chemistry Solutions Chapter 14 Biomolecules 61

Question 48.
Draw zwitter ion of alanine and other two forms.
Answer:
Maharashtra Board Class 12 Chemistry Solutions Chapter 14 Biomolecules 62

Question 49.
What is a peptide bond (peptide linkage)?
OR
Define peptide bond.
Answer:
Proteins are the polymers of a-amino acids and they are connected to each other. The bond that connects a-amino acids to each other is called peptide bond (peptide linkage, – CONH -).

Question 50.
How is peptide linkage (dipeptide linkage) formed in proteins? How is tripeptide formed?
Answer:
Peptide linkage is formed by condensation of acidic group of one molecule of a-amino acid and basic -NH₂ group of other molecule of α-amino acid with elimination of water.
Maharashtra Board Class 12 Chemistry Solutions Chapter 14 Biomolecules 64

When one more molecule of amino acid combines with dipeptide, it forms tripeptide.
Maharashtra Board Class 12 Chemistry Solutions Chapter 14 Biomolecules 65
Thus, it forms tetra, penta and finally a polypeptide chain i.e. proteins. Hence, proteins are basically polypeptides.

Question 51.
Write the structures of all possible dipeptides which can be obtained from glycine and alanine.
Answer:
(1) Dipeptide from glycine :
Carboxylic group of glycine reacting with amino group another molecule of glycine to form dipeptide
Maharashtra Board Class 12 Chemistry Solutions Chapter 14 Biomolecules 73
(2) Dipeptide from alanine :
Carboxylic goup of alanine reaction with amino goup of another molecule of alamine to form dipeptide

(3) Dipeptide from glycine and alanine :
Carboxylic group of glycine reacting with amino group another molecule of alanine to form dipeptide
Maharashtra Board Class 12 Chemistry Solutions Chapter 14 Biomolecules 75

Question 52.
How are proteins classified on the basis of molecular shapes?
Answer:
On the basis of their molecular shapes proteins are classified as :
(1) Fibrous proteins : The proteins in which the polypeptide chains lie parallel (side by side) to form fibre-like structure, are called fibrous proteins. The polypeptide chains held together by hydrogen bonds. These proteins are insoluble in water.

The fibrous proteins are tough and insoluble in water, and dilute acids or bases.

Example : myαin (in muscles), keratin (in hair, nails, skin), fibroin (in silk), collagen (in tendons), etc.

(2) Globular proteins : The proteins have spherical shape. This shape results from coiling around of the polypeptide chain of protein, and have intramolecular hydrogen bonding are called globular proteins.

They are soluble in water and dilute acids or bases.

Example : Haemoglobin (in blood), albumin (in eggs), insulin (in pancreas), etc.

Question 53.
Distinguish between globular and fibrous proteins.
Answer:

Globular proteins	Fibrous proteins
(1) The chains of polypeptides of protein coil around to give a spherical shape. (2) Globular proteins are soluble in water. (3) They are sensitive to small changes of temperature and pH. (4) They possess biological activity.	(1) The proteins in which the polypeptide chains lie parallel to form fibre like structure. (2) Fibrous proteins are insoluble in water. (3) They are stable to moderate changes of temperature and pH. (4) They do not possess biological activity.

Question 54.
Draw a neat labelled diagram for the secondary structure of protein.
Answer:
Secondary structure of proteins : The three-dimensional arrangement of lαalized regions of a long polypeptide chain is called the secondary structure of protein. Hydrogen bonding between N-H proton of one amide linkage and C = O oxygen of another gives rise to the secondary structure. There are two different types of secondary structures i.e. α-helix and β-pleated sheet.

α-Helix : In a-helix structure, a polypeptide chain gets coiled by twisting into a right handed or clαkwise spiral known as a-helixn. The characteristic features of α-helical structure of protein are :
Maharashtra Board Class 12 Chemistry Solutions Chapter 14 Biomolecules 78
(1) Each turn of the helix has 3.6 amino acids.
(2) A C = O group of one amino acid is hydrogen bonded to N – H group of the fourth amino acid along the chain.
(3) Hydrogen bonds are parallel to the axis of helix while R groups extend outward from the helix core.
Myosin in muscle and a-keratin in hair are proteins with almost entire a-helical secondary structure.

β-Pleated sheet : In secondary structure, when two or more polypeptide chains (strands) line up side-by-side is called β-pleated sheets. The β-picate sheet structure of protein consists of extended strands of polypeptide chains held together by intermolecular hydrogen bonding. The characteristics of β-pleated sheet structure are :
Maharashtra Board Class 12 Chemistry Solutions Chapter 14 Biomolecules 79

The C = O and N – H bonds lie in the planes of the sheet.
Hydrogen bonding occurs between the N – H and C = O groups of nearby amino acid residues in the neighbouring chains.
The R groups are oriented above and below the plane of the sheet.

The β-pleated sheet arrangement is favoured by amino acids with small R groups.

Question 55.
What is denaturation of proteins? How is denaturation brought about?
OR
What is the effect of denaturation on the structure of proteins?
Answer:
The prαess by which the molecular shape of protein changes without breaking the amide / peptide bonds that form the primary structure is called denaturation. OR Proteins gets easily precipitated. It is an irreversible change and the prαess is called denaturation of proteins.

Denaturation uncoils the protein and destroys the shape and thus loses their characteristic biological activity. Denaturation is brought about by heating the protein with alcohol, concentrated inorganic acids or by salts of heavy metals. During denaturation secondary and tertiary and quternary structures are destroyed but primary structure remains intact.

Example : Boiling of egg to coagulate egg white, conversion of milk to curd.

Question 56.
Define : Enzymes
Answer:
All biological reactions are catalysed by bio-catalyst in living organisms called enzymes.

Question 57.
What are enzymes? Explain with suitable example.
Answer:
All biological or bio-catalysts which catalyse the reactions in living organisms are called enzymes. Chemically all enzymes are proteins. They are required in very small quantities as they are catalyst also they reduce the activation energy for a particular reaction.

Example : Enzyme maltase converts maltose to glucose.
Maharashtra Board Class 12 Chemistry Solutions Chapter 14 Biomolecules 84

Question 58.
Explain the catalytic action of enzymes.
Maharashtra Board Class 12 Chemistry Solutions Chapter 14 Biomolecules 85
Answer:
Mechanism of enzyme catalysis : Action of an enzyme on a substrate is known as lock-and-key mechanism.

Accordingly, the enzyme has active site on its surface. A substrate molecule can attach to this active site only if it has the right size and shape. Once in the active site, the substrate is held in the correct orientation, enzymes provide functional group which will attack the substrate and forms the products of reaction. The products leave the active site and the enzyme is ready to act as catalyst again.

Question 59.
Give examples of industrial application of enzyme catalysis.
Answer:

Glucose Isomerase (enzyme) is used in conversion of glucose to sweet-tasting fructose.
New antibiotics are manufactured using penicillin acylase (enzyme).
Laundry detergentts are manufactured using proteases (enzyme).
Esters used in cosmetics are manufactured using genetically engineered enzyme.

Question 60.
Draw a neat diagram for enzyme catalysis.
Answer:
Maharashtra Board Class 12 Chemistry Solutions Chapter 14 Biomolecules 85

Question 61.
State the main functions of enzymes.
Answer:
Enzymes are biological catalyst and they are highly specific in nature. The two main functions are as follows :
(1) They lower the requirement of activation energy.
(2) They speed up the rate of reaction.
E.g. Enzyme maltase catalyses maltose to glucose.
\(\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}+\mathrm{H}_{2} \mathrm{O} \stackrel{\text { Maltase }}{\longrightarrow} 2 \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\)

Question 62.
What are nucleic acids?
Answer:
Nucleic acids are unbranched polymers of repeating monomers i.e. nucleotides. In other words, nucleic acids have a polynucleotide structure which in turn consists of a base, a pentose sugar and phosphate moiety.

Nucleic acids are biomolecules which are found in the nuclei of all living cells in the form of nucleoproteins or chromosomes.

(Nucleoproteins = Proteins + Nucleic acid)
(prosthetic group)

Question 63.
State the types of nucleic acids.
Answer:
The types of nucleic acids are : Ribonucleic acid (RNA) and deoxy ribonucleic acid (DNA). DNA molecules contain several million nucleotides while RNA molecules contain a few thousand nucleotides.

Question 64.
Explain chemical composition of nucleic acids.
Answer:
Nucleic acids have a polynucleotide structure. Nucleic acids (RNA and DNA) consists of three components :
(1) monosaccharide (sugar)
(2) nitrogen containing base and
(3) phosphate group.

(1) Monosaccharides : Nucleotides of both RNA consist of five membered monosaccharide ring (furanose), called as simply sugar component.
Maharashtra Board Class 12 Chemistry Solutions Chapter 14 Biomolecules 86
In RNA, the sugar component of nucleotide units is D-ribose and in DNA 2-deoxy-D-ribose.
2 – deoxy means no – OH group at C₂ position.

(2) Nitrogen containing base : Total five nitrogen – containing bases are present in nucleic acids. Three bases with one ring (cytosine, uracil and thymine) are derived from the parent compound pyrimidine. Two bases with two rings (adenine and guanine) are derived from the parent compound purine. Each base in designated by a one-letter symbol. Uracil (U) αcurs only in RNA while thymine (T) ocurs only in DNA.
Maharashtra Board Class 12 Chemistry Solutions Chapter 14 Biomolecules 87

(3) Phosphate group : The sugar units are joined to phosphate through C₃ and C₅ hydroxyl groups.

Question 65.
What is meant by nucleosides?
OR
Write the structure of nucleoside. Give examples.
Answer:
Maharashtra Board Class 12 Chemistry Solutions Chapter 14 Biomolecules 89
A nucleoside contains two basic components of nucleic acids i.e. a pentose sugar and a nitrogenous base.

A nucleoside is formed when 1 -position of a pyrimidine (cytosine, thymine or uracil) or 9-position of guanine or adenine base is attached to C- l of sugar by β-linkage.

Examples: Formation of nucleoside:
Maharashtra Board Class 12 Chemistry Solutions Chapter 14 Biomolecules 90

Question 66.
What is meant by nucleotide?
OR
Write the structure of nucleotide. Give example.
Answer:
Maharashtra Board Class 12 Chemistry Solutions Chapter 14 Biomolecules 91
A nucleotide contains all three basic components of nucleic acids i.e., a pentose sugar, a phosphoric acid and a nitrogenous base. These are obtained by esterification of \(\mathrm{C}_{5}^{1}-\mathrm{OH}\) group of the pentose sugar by phosphoric acid. Nucleotides are joined together through phosphate ester linkage. Thus, nucleotides are monophosphates of nucleosides. Abridged names of some nucleotides are AMP, dAMP, UMP, dTMP and so on. Here, the first capital letter is derived from the corresponding base. MP stands for monophosphate. Small letter ‘d’ in the beginning indicates deoxyribose in the nucleotide.

Example :
Maharashtra Board Class 12 Chemistry Solutions Chapter 14 Biomolecules 92

Question 67.
Write the structure of nucleic acids.
Maharashtra Board Class 12 Chemistry Solutions Chapter 14 Biomolecules 95
Answer:
Nucleic acids, both DNA and RNA, are polymers of nucleotides, formed by joining the 3′ – OH group of one nucleotide with 5′ – phosphate of another nucleotide. Two ends of polynucleotide chain are distinct from each other. One end having free phosphate group of 5′ position is called 5′ end. The other end is 3′ end and has free OH – group at 3′ position.

Question 68.
Draw a schematic representation of polynucleotide structure of nucleic acids.
Answer:
Maharashtra Board Class 12 Chemistry Solutions Chapter 14 Biomolecules 96

Question 69.
Explain double helix.
OR
State the salient features of the Watson and Crick mode of DNA.
Answer:
Maharashtra Board Class 12 Chemistry Solutions Chapter 14 Biomolecules 98
The Salient features are :

DNA is made of two polynucleotide strands that wind into a right-handed double helix.
The two strands run in opposite directions: one from the Y end to the 3’ end, while the other from the 3’ end to the Y end.
Pcrpcndicular to the axis of the helix, the sugar – phosphate backbone lies on the outside of the helix and the bases lic on the inside.
The hydrogen bonding between the hases of the two DNA strands stabilizes the double helix. This gives rise to a ladder-like structure of DNA double helix.
Adenine always forms two hydrogen bonds with thymine and guanine forms three hydrogen bonds with cytosinc. Thus A – T arid C – G arc complementary hase pairs and the Two strands of the double helix arc complementary to each other.

Question 70.
Give scientific reasons :
1. In the preparation of glucose from sucrose, ethyl alcohol is added at the time of cooling.
Answer:
Hydrolysis of sucrose with dilute hydrαhloric acid gives glucose along with fructose.

Ethyl alcohol is added at the time of cooling in the preparation of glucose, to separate glucose from fructose. Glucose being insoluble in alcohol, crystallizes out first, while fructose being more soluble in alcohol, remains in the solution.

Question 71.
Answer in one sentence :

(1) How is glucose stored in the animal body?
Answer:
Glucose is stored in the form of glycogen in the animal body.

(2) Write other term used for carbohydrates.
Answer:
Carbohydrates are often termed as saccharides or sugars.

(3) How many moles of acetic anhydride will be required to form glucose penta acetate from 1 mole of glucose?
Answer:
10 moles of acetic anhydride.

(4) What are reducing sugars?
Answer:
Reducing sugars : Carbohydrates which reduce Fehling solution to red ppt of Cu₂0 or Tollen’s reagent to shining metallic silver are called reducing sugars. All monosaccharides and oligosaccharides except sucrose are reducing sugars.

(5) What are non-reducing sugars?
Answer:
Non-reducing sugars : Carbohydrates which do not reduce Fehling solution and Tollen’s reagent are called non-reducing sugars. E.g. sucrose.

(6) Give an example each of reducing and non-reducing sugars.
Answer:
Reducing sugars : Maltose or lactose
Non-reducing sugars : Sucrose.

(7) Name the linkage which joins two monosaccharide units through oxygen atom.
Answer:
The linkage which joins two monosaccharide units through oxygen atom is called glycosidic linkage.

(8) Name the sugar present in DNA.
Answer:
The sugar present in DNA is deoxyribose.

(9) A nucleotide from DNA containing thymine is hydrolysed. What are the products formed?
Answer:
When nucleotide from DNA containing thymine is hydrolysed, 2-deoxy-D-ribose, thymine and phosphoric acid is obtained.

(10) How is zwitterion formed?
Answer:
In aqueous solution, the carboxyl group loses a proton while the amino group accepts it, as a result, a dipolar or zwitter ion is formed.

(11) Name the amino acids which are synthesized in the body.
Answer:
The amino acids which are synthesized in the body are called non-essential amino acids. Examples : Glutamic acid, serine.

(12) Name the four bases present in DNA which of these is not present in RNA.
Answer:
Purines-adenine (A) and guanine (G); Pyrimidines-thymine (T) and cytosine (C), these four bases are present in DNA. Out of these, thymine (T) is not present in RNA.

(13) What are different types of RNA which are found in the cell?
Answer:
There are three different types of RNA found in the cell. (1) The messenger RNA which carries the message to the ribosome (2) Ribosomal RNA where synthesis of protein takes place (3) The transport RNA.

(14) State the functions of RNA and DNA.
Answer:
RNA and DNA are responsible for generic characteristics : DNA preserves the information and uses it by producing duplicate identical DNA molecules. RNA carries messages and transports them.

Multiple Choice Questions

Question 72.
Select and write the most appropriate answer from the given alternatives for each subquestion :

1. Which of the following is not sugar?
(a) Sucrose
(b) Starch
(c) Fructose
(d) Glucose
Answer:
(b) Starch

2. Which of the following is the example of disaccharide?
(a) Glucose
(b) Raffinose
(c) Cellulose
(d) Sucrose
Answer:
(d) Sucrose

3. Fructose is
(a) aldopentose
(b) aldohexose
(c) ketopentose
(d) ketohexose
Answer:
(d) ketohexose

4. Oxidation product of glucose with bromine water is
(a) sorbitol
(b) gluconic acid
(c) glutamic acid
(d) saccharic acid
Answer:
(b) gluconic acid

5. The general formula of carbohydrates is
(a) C(H₂O)
(b) C_x(H₂O)_y
(c) C_x(H₂O)
(d) C_x(H₂O)_x
Answer:
(b) C_x(H₂O)_y

6. Monosaccharides containing aldehyde group are called
(a) aldoses
(b) ketoses
(c) polysaccharides
(d) disaccharides
Answer:
(a) aldoses

7. Which of the following sugars can be used to prepare glucose on a large scale?
(a) Cellulose
(b) Cane sugar
(c) Galactose
(d) Starch
Answer:
(d) Starch

8. Which of the following carbohydrates cannot undergo hydrolysis?
(a) Glucose
(b) Sucrose
(c) Cellulose
(d) Maltose
Answer:
(a) Glucose

9. Glucose differs from fructose in
(a) the functional group
(b) the number of chiral carbon atoms
(c) the number of carbon atoms
(d) both (a) and (b)
Answer:
(d) both (a) and (b)

10. The example of aldopentose is
(a) arabinose
(b) glucose
(c) fructose
(d) sucrose
Answer:
(a) arabinose

11. Dextrose, grape sugar and blood sugar αcurs in
(a) fructose
(b) glucose
(c) sucrose
(d) starch
Answer:
(b) glucose

12. The example of ketopentose is
(a) galactose
(b) ribose
(c) raffinose
(d) maltose
Answer:
(b) ribose

13. Cane sugar on hydrolysis gives
(a) glucose and maltose
(b) glucose and lactose
(c) glucose and fructose
(d) only glucose
Answer:
(c) glucose and fructose

14. On commerical scale, glucose is prepared from
(a) starch
(b) potato pulp
(c) sucrose
(d) both (a) and (b)
Answer:
(d) both (a) and (b)

15. The number of monosaccharide units formed on hydrolysis of glucose are
(a) zero
(b) one
(c) two
(d) three
Answer:
(a) zero

16. Which of the following is NOT TRUE about glucose?
(a) It is monosaccharide
(b) It is a polyhydroxy aldehyde
(c) It is polyhydroxy ketone
(d) It contains six carbon atoms
Answer:
(c) It is polyhydroxy ketone

17. Final hydrolysis product of simple protein is
(a) carboxylic acid
(b) α-amino acid
(c) mineral acid
(d) acetic acid
Answer:
(b) α-amino acid

18. Haemoglobin is the example of-
(a) simple protein
(b) derived protein
(c) fibrous protein
(d) conjugated protein
Answer:
(d) conjugated protein

19. Protein are also called
(a) polysaccharides
(b) polypeptides
(c) polyglycerides
(d) polyster
Answer:
(b) polypeptides

20. The simplest amino acid is
(a) glycine
(b) oxalic acid
(c) adipic acid
(d) caprolactam
Answer:
(a) glycine

21. Amino acids usually exist in the form of zwitter ion which consist of
(a) the basic group-NH₂ and the acidic group -COOH
(b) the acidic group -N⁺H₃ and the basic group COO^–
(c) the acidic group -COO⁺ and the acidic group NH^3-
(d) acidic or basic group
Answer:
(b) the acidic group -N⁺H₃ and the basic group COO-

22. The water insoluble protein is
(a) casein of milk
(b) albumin
(c) serum albumin
(d) keratin of hair
Answer:
(d) keratin of hair

23. The main structural feature of a protein molecule is the presence of
(a) an ester linkage
(b) an ether linkage
(c) a peptide linkage
(d) all of these
Answer:
(c) a peptide linkage

24. Milk sugar is
(a) sucrose
(b) lactose
(c) maltose
(d) glucose
Answer:
(b) lactose

25. The carbohydrates used for silvering of mirror is
(a) fructose
(b) starch
(c) glucose
(d) cellulose
Answer:
(c) glucose

26. Which one of the following is NOT produced by human body?
(a) DNA
(b) Hormones
(c) Enzymes
(d) Vitamins
Answer:
(c) Enzymes

27. A biological catalyst is essentially
(a) an amino acid
(b) an enzyme
(c) a nitrogen molecule
(d) a carbohydrate
Answer:
(d) a carbohydrate

28. Which one of the following is not a constituent of RNA?
(a) Ribose
(b) Uracil
(c) Thymine
(d) Phosphate
Answer:
(b) Uracil

29. DNA is a polymer of units of
(a) sugars
(b) ribose
(c) amino acids
(d) nucleotides
Answer:
(c) amino acids

30. Which one of the following molecules will form zwitter ion?
(a) CH₃COOH
(b) CH₃CH₂NH₂
(c) CCl₃NO₂
(d) NH₂CH₂COOH
Answer:
(d) NH₂CH₂COOH

31. In metabolic prαess the maximum energy is given by
(a) carbohydrates
(b) proteins
(c) vitamins
(d) fats
Answer:
(d) fats

32. DNA has a structure of helix was reported by
(a) Herman Fischer
(b) Fedrick Sauger
(c) Andreas Marggraf
(d) Watson and Crick
Answer:
(d) Watson and Crick

33. The secondary structure of a protein is determined by
(a) co-ordinate bond
(b) covalent bond
(c) ionic bond
(d) hydrogen bond
Answer:
(d) hydrogen bond

34. In maltose, glycosidic linkage is present between the two glucose units at positions
(a) 1, 2
(b) 1, 1
(c) 1, 3
(d) 1, 4
Answer:
(d) 1, 4

35. Which of the following amino acids is basic in nature?
(a) Valine
(b) Tyrosine
(c) Arginine
(d) Luecine
Answer:
(c) Arginine

36. Sucrose molecules consists of
(a) a glucofuranose and a fructopyranose
(b) a glucofuranose and a fructofuranose
(c) a glucopyranose and a fructopyranose
(d) a glucopyranose and a’ fructofuranose
Answer:
(d) a glucopyranose and a’ fructofuranose

37. Which one of the following statements is not correct about DNA molecule?
(a) It has double helix structure
(b) It serves as hereditary material
(c) The two DNA strands are exactly similar
(d) Its replication is called semi-conservative mode of replication
Answer:
(c) The two DNA strands are exactly similar

38. Glycine on heating forms
Maharashtra Board Class 12 Chemistry Solutions Chapter 14 Biomolecules 107

Answer:
(a)

39. Acidic amino acid is
(a) Glutamine
(b) Glutamic acid
(c) Tyrosine
(d) Lysine
Answer:
(b) Glutamic acid

40. Basic amino acid is
(a) Lysine
(b) Glycine
(c) Cystine
(d) Alanine
Answer:
(a) Lysine

41. Precipitation of protein is referred to as
(a) destruction of proteins
(b) separation of proteins
(c) denaturation of proteins
(d) fragmentation of proteins
Answer:
(c) denaturation of proteins

42. An amino acid containing sulphur is
(a) serine
(b) cysteine
(c) valine
(d) asparagine
Answer:
(b) cysteine

43. Rhamnose is a
(a) carbohydrate
(b) protein
(c) lipid
(d) vitamin
Answer:
(a) carbohydrate

44. Lactose on hydrolysis gives
(a) glucose + glucose
(b) glucose + fructose
(c) glucose + galactose
(d) fructose + galactose
Answer:
(c) glucose + galactose

45. Raffinose on hydrolysis gives
(a) glucose + glucose + galactose
(b) glucose + fructose + galactose
(c) glucose + galactose + galactose
(d) fructose + galactose + galactose
Answer:
(b) glucose + fructose + galactose

46. Naturally αcurring glucose is
(a) dextro rotatory
(b) laevo rotatory
(c) racemic mixture
(d) all of these
Answer:
(a) dextro rotatory

47. Amylopectin is
(a) soluble in water and constitutes about 80% of starch
(b) insoluble in water and constitutes about 80% of starch
(c) Soluble in alcohol and constitutes about 60% of starch
(d) in soluble in alcohol and constitutes about 60% of starch
Answer:
(b) insoluble in water and constitutes about 80% of starch

48. Insulin contains
(a) 51 amino acids
(b) 151 amino acids
(c) 15 amino acids
(d) 115 amino acids
Answer:
(a) 51 amino acids

49. Pyranose structure of glucose is
(a) an open chain structure of glucose
(b) a structure of reduction product of glucose
(c) a cyclic six-membered structure of glucose
(d) a four-membered cyclic form of glucose
Answer:
(c) a cyclic six-membered structure of glucose

50. The number of – OH groups present in ribulose is
(a) 3
(b) 4
(c) 6
(d) 5
Answer:
(b) 4

51. Peptide linkage is
Maharashtra Board Class 12 Chemistry Solutions Chapter 14 Biomolecules 109
Answer:
(d)

52. Stachyose is an example of
(a) monosaccharides
(b) disaccharides
(c) trisaccharides
(d) tetrasaccharides
Answer:
(d) tetrasaccharides

53. How many moles of (CH₃CO)₂O will be required to form glucose pentaacetate form 2 moles of glucose?
(a) 2
(b) 5
(c) 10
(d) 2.5
Answer:
(c) 10

54. Which of the following NOT present in DNA?
(a) Adenine
(b) Guanine
(c) Thymine
(d) Uracil
Answer:
(d) Uracil

55. Maltose is a
(a) polysaccharide
(b) disaccharide
(c) trisaccharide
(d) monosaccharide
Answer:
(b) disaccharide