Class 12

Balbharti Maharashtra State Board 12th Biology Textbook Solutions Chapter 15 Biodiversity, Conservation and Environmental Issues Textbook Exercise Questions and Answers.

Maharashtra State Board 12th Biology Solutions Chapter 15 Biodiversity, Conservation and Environmental Issues

1. Multiple choice questions

Question 1.
Observe the graph and select the correct option.

(a) Line A represents, S = CA²
(b) Line B represents, log C = log A + Z log S
(c) Line A represents, S = CA^Z
(d) Line B represents, log S = log Z + C log A
Answer:
(c) Line A represents, S = CA^Z

Question 2.
Select odd one out on the basis of Ex situ conservation.
(a) Zoological park
(b) Tissue culture
(c) Sacred groves
(d) Cryopreservation
Answer:
(a) Zoological park

Question 3.
Which of the following factors will favour species diversity?
(a) Invasive species
(b) Glaciation
(c) Forest canopy
(d) Co-extinction
Answer:
(a) Invasive species

Question 4.
The term “terror of Bengal’ is used for
(a) algal bloom
(b) water hyacinth
(c) increased BOD
(d) eutrophication
Answer:
(b) water hyacinth

Question 5.
CFC are air polluting agents which are produced by
(a) Diesel trucks
(b) Jet planes
(c) Rice fields
(d) Industries
Answer:
(b) Jet planes

2. Very short answer type questions.

Question 1.
Give two examples of biodegradable materials released from sugar industry.
Answer:

1. Molasses
2. Bagasse.

Question 2.
Name any two modern techniques of protection of endangered species.
OR
Two modern methods of ex-situ conservation of species
Answer:

1. Tissue culture
2. In vitro fertilization of eggs
3. Cryopreservation.

Question 3.
Where was ozone hole discovered?
Answer:
Ozone hole was discovered in Antarctica.

Question 4.
Give one example of natural pollutant.
Answer:
Volcanic ash is a natural pollutant.

Question 5.
What do you understand by EW category of living being?
Answer:
A species which becomes extinct in the wild (EW) is called EW category, their members are seen only in captivity or as a naturalized population outside its historic range due to massive habitat loss.

3. Short answer type questions.

Question 1.
Dandiya raas is not allowed after 10.00 pm. Why?
Answer:
Dandiya rass involves blaring loudspeakers which cause noise pollution. It is undesired loud sound which could be hazardous for ears and general health. In India, the Air (Prevention and Control of Pollution) Act 1981, Amendment 1987, includes noise as an air pollutant. As per law noise after 10 pm is not allowed as many people may be resting. Therefore, Dandiya Raas is not allowed after 10 pm.

Question 2.
Tropical regions exhibit species richness as compared to polar regions. Justify.
Answer:

1. Tropical regions are bestowed by thicker vegetation and ample food due to available sunlight and humidity.
2. Polar regions are covered over with snow, with almost no vegetation.
3. Only handful species of animals can survive here due to their adaptations.
4. Species richness always shows latitudinal gradient for many plants and animal species. It is high at lower latitudes and there is a steady decline towards the poles. Therefore, tropical regions show more species richness.

Question 3.
How does genetic diversity affect sustenance of a species?
Answer:

1. Genetic diversity develops the capability of the species to adapt to the varying changes in the environment.
2. The large variation of the different gene sets allows an individual or the whole population to have the capacity to endure environmental stress in any form.
3. Some individuals have, a better capacity to endure the increasing pollution in the environment whereas some do not have it.
4. Those that do not have show infertility or even death from the same conditions.
5. Those who are able to endure and adapt to this change survive and live in a better way.
6. This is called natural selection which leads to a loss of genetic diversity in particular habitats.
7. Thus, due to genetic diversity can affect sustenance of some species.

Question 4.
Greenhouse effect is boon or bane? Give your opinion.
Answer:
(1) The natural greenhouse effect is good, it is a boon but human enhanced greenhouse effect is a bane.

(2) In the absence of an atmosphere, Earth’s surface temperature would be about -18 °C, or 0 °F, which is too cold for sustaining life.

(3) Earth is habitable because of the natural greenhouse effect. Heating of Earth’s atmosphere due to the presence of greenhouse gases such as water vapour, carbon dioxide (CO₂), methane (CH₄) and oxides of nitrogen (NO₂).

(4) Greenhouse gases have just the right molecular structure to absorb infrared radiation that the Earth emits. It re-emits most of that infrared energy in all directions, warming the atmosphere to its comfortable average temperature of 15 °C (60 °F). So, the greenhouse effect was a boon in olden days before industrialization and invention of automobiles.

(5) However, due to human impact, the proportion of greenhouse gases has increased tremendously causing global warming. Thus, now greenhouse effect has become a bane.

Question 5.
State the effects of CO in human body.
OR
How does CO cause giddiness and exhaustion?
Answer:
Effects of Carbon monoxide:

1. Carbon monoxide is tasteless, colourless and odourless gas, therefore its presence goes unnoticed.
2. It can inhibit the blood’s ability to carry oxygen to body tissues.
3. Supply of oxygen to vital organs such as.the heart and brain is affected due to presence of CO.
4. When CO is inhaled, it combines with the oxygen carrying haemoglobin of the blood to form carboxyhaemoglobin. Once combined with the haemoglobin, that haemoglobin is no longer available for transporting oxygen.
5. The symptoms of CO poisoning are headache, nausea, giddiness, etc.

Question 6.
Name two types of particulate pollutants found in air. Add a note on ill effects of the same on human health.
OR
Describe any 2 particulate and gaseous pollutants.
Answer:
I. Types of gaseous pollutants include CO₂, CO, SO₂, NO, NO₂, etc.
(1) Carbon dioxide : It is a greenhouse gas. It is produced in excess due to human activities such as burning of fossil fuels. It is also rising due to increasing deforestation. The natural cycle of Carbon dioxide is disturbed due to human interference. Otherwise, the process of photosynthesis can balance CO₂ : O₂ ratio of the air. Aeroplane traffic such as a jet plane also emits lots of CO₂.

(2) Carbon monoxide (CO) : CO is produced due to incomplete combustion of fuels. It is a toxic gas. Vehicular exhausts produce lot of CO.

II. Types of particulate pollutants are mist, dust, fume and smoke particles, smog, pesticides, heavy metals and radioactive elements, etc.
(1) Dust are fine particles which enter the respiratory passage and can cause damage to delicate tissues in the lungs. Various processes such as construction work, demolition of buildings and traffic can cause dust pollution. There are natural causes of release of dust too, through wind or volcanic eruption.

(2) Smoke and smog are worst type of particulate air pollutants which can cause many respiratory problems like emphysema or asthma.

4. Long answer type questions.

Question 1.
Montreal Protocol is an essential step. Why is it so?
Answer:

1. Montreal Protocol was an international treaty signed at Montreal in Canada in 1987.
2. Later many more efforts have been made and protocols have laid down definite roadmaps separately for developing and developed countries.
3. All these efforts were for reducing emission of CFCs and other ozone depleting chemicals.
4. All nations realized that ozone depletion can cause penetration of harmful UV radiations to the earth’s surface. This is very hazardous, for flora, fauna and for mainly human beings. Therefore, urgent action was needed to combat this effect.
5. Montreal Protocol was a very positive move because after 1987, there have been much better condition of ozone layer.

Question 2.
Name any 2 personalities who have contributed to control deforestation in our country. Elaborate on importance of their work.
Answer:
Two personalities who have contributed to control deforestation in our country are:
Saalumara Thimmakka from Karnataka and Moirangthem Loiya from Manipur.
1. Saalumara Thimmakka :

• Saalumara Thimmakka is the best example of peoples’ participation in reforestation.
• She is an Indian environmentalist from Karnataka. She has taken up work of planting and tending to 385 banyan trees along a 4 km stretch of highway between Hulikal and Kudur. Other 800 trees are also planted by her.
• She is honoured with the National Citizens Award of India and Padma Shri in 2019.

2. Moirangthem Loiya :

• Moirangthem Loiya is from Manipur who has restored Punshilok forest. For last 17 years he is planting trees after leaving his job.
• He brought the lost glory back for the 300 acres forest land. He planted a variety of trees like, bamboo, oak Ficus, teak, jackfruit and Magnolia.
• This forest now has over 250 varieties of plants including 25 varieties of bamboo along with many animals making the forest rich in biodiversity.

Question 3.
How BS emission standards changed over time? Why is it essential?
Answer:

1. BS emission standards changed over the time due to changing city life and more vehicular traffic on the road, especially in the megacities.
2. Since capital city of Delhi was declared as worst polluted city as far as its air quality is concerned, various measures were taken by the Government of India. There was new fuel policy declared, in which Bharat stage emission standards (BS) were set.
3. These norms were set to reduce sulphur and aromatic content of petrol and diesel. Also the vehicular engines were upgraded.
4. Bharat stage emission standards (BS) are standards which are equivalent to Euro norms and have evolved on similar lines as Bharat Stage II (BS II) to BS VI from 2001 to 2017.
5. Since population of Delhi was to be saved, in 2001, Bharat stage II emission norms were set for CNG and LPG vehicles.
6. This helped in reduced emission of sulphur which was controlled at 50 ppm in diesel and 150 ppm in petrol. Also aromatic hydrocarbons were reduced at 42% in concerned fuel according to norms.
7. Because, in spite of all the efforts, Delhi was declared as worst air-polluted city in the world in 2016, therefore, Government of India directly adapted BS VI in the year 2018, skipping BS V These efforts decreased the levels of CO₂ and SO₂ in Delhi.

Question 4.
During large public gatherings like Pandharpur vari, mobile toilets are deployed by the government. Explain how this organic waste is disposed.
Answer:

1. The toilets deployed at Pandharpur at the time of vari are of the Ecosan type.
2. Ecosan toilet is a closed system without water and it is an alternative to leach pit toilets.
3. When the pit of an Ecosan toilet fills up after some time, then it is closed and sealed for about 8-9 months.
4. In this time the faeces get completely composted to organic manure. In this way the organic waste can be disposed.
5. It is a practical, efficient and cost-effective solution for human waste disposal.
6. Also, open-air defecation is prohibited which can cause health problems. Therefore, during large public gatherings like Pandharpur vari mobile toilets like Ecosan are deployed by the government.

Question 5.
How Indian culture and traditions helped in bio-diversity conservation? Give importance of conservation in terms of utilitarian reasons.
Answer:
In Indian culture and traditions in different religions, biodiversity is protected and conserved. Few examples of worship of animals and plants can be given here.

1. Nagpanchami festival is towards the respect of snakes. They are worshipped on that day and the local people are aware of their role in ecosystem of control of rat population.
2. Vatapournima festival is worshipping a banyan tree.
3. Various other festivals teach the value of plants and animals surrounding us. Even the cattle are worshipped on a particular day as a tradition.
4. Jain religion strongly advocates protection of all animals through vegetarianism.

Conservation in terms of utilitarian reasons:
The conservation of biodiversity can be done in utilitarian way or for ethical reasons. Utilitarian reasons are further classified into narrowly utilitarian and broadly utilitarian reasons:

I. Narrowly utilitarian reasons:

1. Humans always reap material benefits from biodiversity in the form of resources for basic needs such as food, clothes, shelter.
2. Industrial products like resins, tannins, perfume base, etc. are also obtained through biodiversity resources.
3. For making ornaments or artefacts for aesthetic purpose, again biodiversity is sacrificed.
4. Many medicines are also obtained through biodiversity resources which shares 25% of global medicine market.
5. Around 25000 species are used for traditional medicines by tribal population worldwide.
6. Bioprospecting which is a systematic search for development of new sources of chemical compounds, genes, microorganisms, macroorganisms, and other valuable products from nature which is of economically important species is also due to biodiversity.

II. Broadly utilitarian reasons:

1. Production of oxygen done by all green plants helps human beings to thrive. Amazon forest alone gives 25% of the oxygen to the entire world.
2. Insects carry out pollination and seed dispersal.
3. If insects do not carry out pollination and seed dispersal, man would go hungry without crops and fruits.
4. Biodiversity also is useful in recreation of human beings.

III. Taking all these aspects in consideration, conservation of biodiversity becomes essential. Therefore, to protect and conserve our rich biodiversity on the planet, we have to remember all the utilitarian reasons.

Balbharti Maharashtra State Board 12th Biology Textbook Solutions Chapter 14 Ecosystems and Energy Flow Textbook Exercise Questions and Answers.

Maharashtra State Board 12th Biology Solutions Chapter 14 Ecosystems and Energy Flow

1. Multiple choice questions

Question 1.
Which one of the following has the largest population in a food chain?
(a) Producers
(b) Primary consumers
(c) Secondary consumers
(d) Decomposers
Answer:
(a) Producers

Question 2.
The second trophic level in a lake is ……………………
(a) Phytoplankton
(b) Zooplankton
(c) Benthos
(d) Fishes
Answer:
(b) Zooplankton

Question 3.
Secondary consumers are …………………….
(a) Herbivores
(b) Producers
(c) Carnivores
(d) Autotrophs
Answer:
(c) Carnivores

Question 4.
What is the % of photosynthetically active radiation in the incident solar radiation?
(a) 100%
(b) 50%
(c) 1-5%
(d) 2-10%
Answer:
(b) 50%

Question 5.
Give the term used to express a community in its final stage of succession?
(a) End community
(b) Final community
(c) Climax community
(d) Dark community
Answer:
(c) Climax community

Question 6.
After landslide which of the following type of succession occurs?
(a) Primary
(b) Secondary
(c) Tertiary
(d) Climax
Answer:
(a) Primary

Question 7.
Which of the following is most often a limiting factor of the primary productivity in any ecosystem?
(a) Carbon
(b) Nitrogen
(c) Phosphorus
(d) Sulphur
Answer:
(c) Phosphorus

2. Very short answer question.

Question 1.
Give an example of ecosystem which shows inverted pyramid of numbers.
Answer:
Number of insects dependent on a single tree, is an example of ecosystem having inverted pyramid of numbers.

Question 2.
Give an example of ecosystem which shows inverted pyramid of biomass.
Answer:
Oceanic ecosystem has inverted pyramid of biomass.

Question 3.
Which mineral acts as limiting factor for productivity in an aquatic ecosystem?
Answer:
Phosphorus acts as limiting factor for productivity in an aquatic ecosystem.

Question 4.
Name the reservoir and sink of carbon in carbon cycle.
Answer:
Atmosphere is the reservoir of carbon cycle, while fossil fuels embedded in ocean and oceanic waters are the sink of carbon in carbon cycle.

3. Short answer questions.

Question 1.
Upright and inverted pyramid of biomass.
Answer:

Upright pyramid	Inverted pyramid
1. In upright pyramid, the number and biomass of the organisms which are at first trophic level of producers is high.	1. In inverted pyramid, the number and biomass of organisms at first trophic levels of producers is lowest.
2. The biomass goes on decreasing at each trophic level.	2. The biomass foes on increasing at each trophic level.
3. The base of the pyramid is always in large number of producers.	3. The base of pyramid is always in small numbers of producers.
4. Pyramid is always upright.	4. Pyramid is always inverted.

Question 2.
Food chain and Food web.
Answer:

Food chain	Food web
1. Food chain is the linear sequence of organisms for feeding purpose.	1. Food web is interconnections between many small food chains.
2. In food chain the flow of energy is through a single straight pathway from the lower trophic level to the higher trophic level.	2. In food web, the energy flow is interconnected through numerous food chains in the ecosystem.
3. In a food chain, members present at higher trophic level feeds on only single type of organisms.	3. In a food web, one organism can feed on multiple types of organisms.
4. Energy flow can be easily calculated in food chain.	4. Energy flow is difficult to calculate in a food web.
5. In food chain there is increased instability due to increasing number of separate and confined food chains.	5. In food web there is increased stability due to the presence of the complex food chains.
6. The whole food chain gets affected even if one group of an organism is disturbed.	6. The food web does not get disturbed by the removal of one group of organisms.
7. Member of higher trophic level depends or feed upon the single type of organisms of the lower trophic level.	7. The members of higher trophic level depend or feed upon many different types of the organism of the lower trophic level.
8. Food chain consists of only 4-6 trophic levels of different species.	8. Food web contains numerous trophic levels and also of different populations of species.
9. Competition is seen in members of same trophic level.	9. Competition is seen in members of same as well as different trophic levels.
10. Food chains are of two types: 1. Grazing food chain 2. Detritus food chain.	10. In food web there are no types.

4. Long answer questions

Question 1.
Define ecological pyramids and describe with examples, pyramids of number and biomass.
Answer:
1. Ecological Pyramids : Ecological Pyramids are the representation of relationships between different components of ecosystem at successive trophic levels.

2. Pyramid of numbers:

• Pyramid of numbers is the diagrammatic representation which shows the relationship between producers, herbivores and carnivores at successive trophic levels in terms of their numbers.
• As we go up the trophic levels, the interdependent organisms keep on reducing in their numbers.
• For example, the number of grasses are more than the number of herbivores which eat them. The number of herbivores such as rabbits would be lesser than grass but greater than the carnivores that are dependent upon the population of rabbits.
• Thus, the producers would be more than primary consumers and primary consumers would be more than secondary consumers. The top level consumers would be least in their numbers. This pyramid shows upright nature.

3. Pyramid of biomass:
(1) Pyramid of biomass are constructed by taking into consideration the different biomass in every successive trophic level.
(2) Pyramid of biomass in seas in inverted as the biomass of fishes is more than the biomass of phytoplankton.

Question 2.
What is primary productivity? Give brief description of factors that affect primary productivity.
Answer:
(1) Primary Productivity : The rate of generation of biomass in an ecosystem which is expressed in units of mass per unit surface (or volume) per unit time, for instance grams per square metre per day (g/m²/day) is called primary productivity.

(2) Primary productivity is described as gross primary productivity (GPP) and net primary productivity (NPP).

(3) The rate of production of organic matter during photosynthesis is called gross primary productivity of an ecosystem. Of this the amount of energy lost through respiration of plants is called respiratory losses.

(4) Gross primary productivity minus respiratory losses gives the net primary productivity (NPP).

(5) Net primary productivity is the available biomass for the consumption to heterotrophs (herbivores, carnivores and decomposers).

(6) Factors affecting primary productivity: Gross primary productivity (GPP) depends on the following factors:

• Plant species inhabiting a particular area.
• Variety of environmental factors such as temperature, sunlight, salinity, oxygen and carbon dioxide content, etc.
• Availability of nutrients and
• Photosynthetic capacity of plants.

Question 3.
Define decomposition and describe the processes and products of decomposition.
Answer:

1. Decomposition is the process carried out by the decomposer organisms.
2. Most of the bacteria, actinomycetes and fungi are decomposers. They convert the dead and decaying organic matter into simpler compounds. These simpler inorganic substances return back to the environment.
3. Decomposition takes place through detritus food chain. It starts from the dead organic matter. Detritus eating organisms called detritivores like earthworm, etc. breakdown the detritus into smaller fragments. Therefore, this first step of decomposition is called fragmentation.
4. Water soluble inorganic nutrients seep into the soil after fragmentation. These nutrients get precipitated as salts. Therefore, this second step of decomposition is called leaching.
5. The third step of decomposition is called catabolism. In this step, fungal and bacterial enzymes degrade the detritus into simple inorganic substances.
6. The partially decomposed organic matter is called humus which is formed by the process of humification. Humus is a dark coloured amorphous substance which is the reservoir of nutrients.
7. Humus too undergoes decomposition by bacterial action at a very slow rate and ultimately releases inorganic matter. This process is therefore called mineralization.
8. Decomposition requires oxygen in greater amount. The rate of decomposition is dependent upon the temperature and the humidity of the environment.

Question 4.
Write important features of a sedimentary cycle in an ecosystem.
Answer:

1. Reservoir of sedimentary cycles is earth’s crust.
2. The nutrients such as phosphorus which show sedimentary cycle, moves through hydrosphere, lithosphere and biosphere.
3. There is no respiratory release of nutrients into the atmosphere which show sedimentary cycle.
4. Natural reservoir of such nutrients are usually in the form of rocks. The rocks upon weathering release such nutrients into circulation.
5. Sedimentary cycles are very slow in their reactions.

Question 5.
Describe carbon cycle and add a note on the impact of human activities on carbon cycle.
Answer:
I. Carbon cycle:
(1) The entire carbon cycle has following basic processes viz. Photosynthesis, Respiration, Decomposition, Sedimentation and Combustion.

(2) Carbon is an important element as it forms 49% of the dry weight of all organisms. 71% of global carbon is present in the oceans. Therefore, ocean is the major reservoir of carbon. Carbon is also present in all fossil fuels. This is long term storage places or sinks for carbon which is in the form of coal, natural gas, etc.

(3) Respiration and photosynthesis are the two events that keep the carbon in cyclic circulation. During respiration, oxygen is used for combustion of carbohydrates as a result of which carbon dioxide and water are formed with the release of energy. The process of photosynthesis utilizes carbon dioxide and water vapour liberating oxygen and producing carbohydrates at the same time.

(4) Solar energy is stored in the carbon-carbon bonds of carbohydrates during photosynthesis whereas respiration releases the same stored energy.

(5) The main reservoirs for carbon dioxide are in the oceans and in rocks. Carbon dioxide is highly soluble in water and forms mild carbonic acid upon dissolving. This dissolved carbon dioxide precipitate as a solid rock or limestone which is calcium carbonate. This reaction in the seas is aided by corals and algae which in turn builds the coral reefs made up of limestone.

(6) Carbon moves through food chains. Autotrophic green plants on land and in water take up carbon dioxide and manufacture carbohydrates by the process of photosynthesis. The carbon stored in plants has three different fates, viz. liberation into atmosphere, consumption by animals upon feeding, storage in the plant till the plant dies.

(7) Animals get their carbon requirement through their food. When autotrophs are consumed, the heterotrophs obtain carbon. Carbon in animals also has three fates, viz. release back into the atmosphere in the process of respiration, release of stored carbon from the body by the action of decomposers or conversion into fossil fuels if buried intact.

(8) Fossil fuels such as coal, oil, natural gas, etc. can be mined and burned for energy purposes. This burning releases carbon dioxide back into the atmosphere.

(9) Carbon from limestone can also be released if pushed to the surfaces and slowly weathered away. Subducting and volcanic eruptions can also release the stored carbon from sediments.

II. Impact of human activities on carbon cycle:
(1) Excessive burning of fossils fuels for power plants, industrial processes and vehicular traffic, adds excessive carbon dioxide into atmosphere. When fossil fuels burn to run factories, power plants, motor vehicles, most of the carbon quickly enters the atmosphere as carbon dioxide gas.

(2) Each year, 5.5 billion tonnes of carbon is released through combustion of fossil fuels. Of this massive amount, 3.3 billion tonnes stays in the atmosphere.

(3) Rapid deforestation also increases carbon dioxide. Since plants absorb carbon dioxide for their photosynthesis, they always reduce the concentration of CO₂. But deforestation upsets this balance.

(4) Massive burning of fossil fuel for energy and transport, have significantly increased the rate of release of carbon dioxide into the atmosphere which is causing global warming and resultant climate change.

Balbharti Maharashtra State Board 12th Biology Textbook Solutions Chapter 13 Organisms and Populations Textbook Exercise Questions and Answers.

Maharashtra State Board 12th Biology Solutions Chapter 13 Organisms and Populations

1. Multiple choice questions

Question 1.
Which factor of an ecosystem includes plants, animals, and microorganisms?
(a) Biotic factor
(b) Abiotic factor
(c) Direct factor
(d) Indirect factor
Answer:
(a) Biotic factor

Question 2.
An assemblage of individuals of different species living in the same habitat and having functional interactions is ……………….
(a) Biotic community
(b) Ecological niche
(c) Population
(d) Ecosystem
Answer:
(a) Biotic community

Question 3.
Association between sea anemone and Hermit crab in gastropod shell is that of ………………..
(a) Mutualism
(b) Commensalism
(c) Parasitism
(d) Amensalism
Answer:
(b) Commensalism

Question 4.
Select the statement which explains best parasitism.
(a) One species is benefited.
(b) Both the species are benefited.
(c) One species is benefited, other is not affected.
(d) One species is benefited, other is harmed.
Answer:
(d) One species is benefited, other is harmed.

Question 5.
Growth of bacteria in a newly inoculated agar plate shows ………………….
(a) exponential growth
(b) logistic growth
(c) Verhulst-Pearl logistic growth
(d) zero growth
Answer:
(c) Verhulst-Pearl logistic growth

2. Very short answer questions.

Question 1.
Define the following terms
a. Commensalism
Answer:
The interaction between two species in which one species gets benefits and the other is neither harmed nor benefited is called commensalism.

b. Parasitism
Answer:
The interaction between two species in which one parasitic species derives benefit from the other host species by harming it is called parasitism.

c. Camouflage
Answer:
Camouflage is the disguising colouration or behaviour to merge with the surrounding so that prey or predator can remain hidden.

Question 2.
Give one example for each
a. Mutualism
b. Interspecific competition
Answer:
a. Lichen is composed of alga (cyanobacteria) and fungus. They cannot survive independently. Their association is mutualistic alga synthesises food by photosynthesis and fungus does the absorption of moisture.

b. Leopard and lion competing for a same prey. Sheep and cow competing for grazing in the same land.

Question 3.
Name the type of association:
a. Clown fish and sea anemone
b. Crow feeding the hatchling of Koel
c. Humming birds and host flowering plants
Answer:
a. Commensalism
b. Brood parasitism
c. Mutualism

Question 4.
What is the ecological process behind the biological control method of managing with pest insects?
Answer:

1. Pest insects act as prey to predator birds or frogs.
2. The biological control method consists of releasing the predators in the farms so that they can control the pest population in the natural way.
3. This also eliminates the use of chemical pesticides.
4. Frogs are natural predators of locust, therefore the population of this hazardous insect is controlled by frogs and the produce from agricultural farm can be saved.

Protocooperation:

1. Protocooperation is a type of population interaction where two species interact with each other.
2. Both are benefited but they have no need to interact with each other.
3. They can survive and grow even in the absence of other species.
4. Therefore this interaction is purely for the gain that they receive in such type of interaction.
5. The interaction that occurs can be between different kingdoms.

3. Short answer questions.

Question 1.
How is the dormancy of seeds different from hibernation in animals?
Answer:
In dormancy seed is not showing any metabolic activities. It can come back to life if and only if it gets suitable moisture and sunlight. Hibernation is suspended state, in which metabolic reactions do take place but at a very reduced pace. Animal arouses on its own after the winter sleep is over. This arousal is spontaneous and depends upon the ambient temperature. Dormant seed does not show such change unless it is planted or thrown in to moist place.

Question 2.
If a marine fish is placed in a fresh water aquarium, will it be able to survive? Give reason.
Answer:
Marine fish has its own osmoregulation which is different from the osmoregulation seen in fresh water fish. In marine water, the ambient salinity is more than the concentration of ions in the body. But in fresh water reverse is the case. Therefore, marine fish has different machinery to cope up with high saline environment. Therefore, it cannot survive in fresh water as its osmoregulation is not possible in less saline waters.

Question 3.
How is the dormancy of seeds different from hibernation in animals?
Answer:
In dormancy seed is not showing any metabolic activities. It can come back to life if and only if it gets suitable moisture and sunlight. Hibernation is suspended state, in which metabolic reactions do take place but at a very reduced pace. Animal arouses on its own after the winter sleep is over. This arousal is spontaneous and depends upon the ambient temperature. Dormant seed does not show such change unless it is planted or thrown into moist place.

Question 4.
An orchid plant is growing on the branch of mango tree. How do you describe this interaction between the orchid and the mango tree?
Answer:

1. Orchid is an epiphyte. It gets the support from the mango tree. But it does not cause any harm to the mango tree.
2. Mango tree does not derive any benefit from this association. Therefore, this interaction is of type of commensalism.

Question 5.
Distinguish between the following:
a. Hibernation and Aestivation
Answer:

Hibernation	Aestivation
1. Hibernation is winter sleep shown by some warm-blooded and some cold-blooded animals.	1. Aestivation is the type of summer sleep, shown by cold-blooded animals.
2. It is for the whole winter.	2. It is of short duration.
3. The animals look out for the warmer place to enter into hibernation.	3. Animals search for the moist, shady and cool place to sleep.
4. Metabolic activities of hibernators slowdown in this dormant stage.	4. Metabolic activities of aestivators remain low during aestivation period.
5. Hibernation helps in maintaining the body temperature and prevents any internal body damage due to low temperatures. E.g. Bats, birds, mammals, insects, etc. show hibernation.	5. Aestivation helps in maintaining the body temperature by avoiding the excessive water loss and thus prevents any internal body damaged due to high temperatures. E.g. Bees, snails, earthworms, salamanders, frogs, earthworms, crocodiles, tortoise, etc. show aestivation.

b. Ectotherms and Endotherms
Answer:

Ectotherms	Endotherms
1. Ectotherms do not have ability to generate heat in the body.	1. Endotherms possess the ability to generate their own body heat.
2. Ectotherms depend on the environmental sources to heat their bodies. E.g sunlight.	2. Endotherms do not depend upon outside sources to generate heat.
3. Most ectotherms are confined to warmer parts of the world.	3. Endotherms inhabit coldest parts of the earth.
4. Body temperature of ectotherms fluctuate according to ambient temperature.	4. Body temperatures of endotherms remain constant and do not show fluctuations as per ambient temperatures.
5. Metabolic rate of ectotherms is low. E.g. Amphibians and reptiles.	5. Metabolic rate of endotherms is high. E.g. Mammals and birds

c. Parasitism and Mutualism
Answer:

Parasitism	Mutualism
1. Parasitism is the relationship where only one organism receive benefits, while the other is harmed in return.	1. Mutualism is the relationship where both the organisms of distinct species are benefited.
2. Parasite cannot survive without host but if the host is overexploited then parasite too dies.	2. Both the species are dependent on each other for their benefits and survival.
3. Parasitism can be facultative or obligatory.	3. Mutualism is obligatory relationship.
4. Parasitism is a negative interaction.	4. Mutualism is a positive interaction.

Question 6.
Write a short note on
a. Adaptations of desert animals
Answer:

1. Animals which are well-adapted to live in deserts are called xerocoles. These animals show adaptations for water conservation or heat tolerance.
2. These animals show low basal metabolic rate. They obtain moisture from succulent plants and rarely drink water. E.g Gazella and Oryx.
3. Desert animals like camel produce concentrated urine and dry dung.
4. Many other hot desert animals are nocturnal, seeking out shade during the day or dwelling underground in burrows.
5. Smaller animals from desert, emerge from their burrows at night.
6. Mammals living in cold deserts have developed greater insulation through warmer body fur and insulating layers of fat beneath the skin.
7. Few adaptations to desert life are unable to cool themselves by sweating so they shelter during the heat of the day. Many desert reptiles are ambush predators and often bury themselves in the sand, waiting for prey to come within range.
8. Other animals have bodies designed to save water. Scorpions and wolf spiders have a thick outer covering which reduces moisture loss. The kidneys of desert animals concentrate urine, so that they excrete less water.

b. Adaptations of plants to water scarcity
Or
Adaptations in desert plants.
Answer:

1. Thick cuticle on their leaf surfaces
2. Stomata of desert plants is sunken that is it is in deep pits to minimize loss of water through transpiration.
3. Desert plants also have a special photosynthetic pathway (CAM -Crassulacean acid metabolism) that enables their stomata to remain closed during daytime.
4. Some desert plants like Opuntia, have their leaves reduced or they are modified to spines. Loss of leaf surface helps in prevention of transpiration.
5. Photosynthetic function is taken over by the flattened stems called as phylloclade.

c. Behavioural adaptations in animals
Answer:

1. Behavioural responses to cope with variations in their environment are shown by few animals.
2. Desert lizards manage to keep their body temperature fairly constant by behavioural adaptations. They bask in the sun and absorb heat, when their body temperature drops below the comfort zone, but move into shade, when the ambient temperature starts increasing. Even snakes also show basking during winter months.
3. Since they are ectothermic, this kind of behaviour saves them from extreme temperatures.
4. Many smaller animals show burrowing behaviour to adapt to the temperature extremes.
5. Some species burrow into the sand to hide and escape from the heat.
6. Migrations shown by the birds and mammals are also behavioural responses for adapting to severe winter temperatures.

Question 7.
Define Population and Community.
Answer:
Population:
Group of organisms belonging to same species that can potentially interbreed with each other and live together in a well-defined geographical area by sharing or competing for similar resources, is called population.

Community:
Several populations of different species in a particular area makes a community.

4. Long answer questions.

Question 1.
With the help of suitable diagram, describe the logistic population growth curve.
Answer:

1. Naturally all populations of any species always have limited resources to permit exponential growth. Due to this there is always competition between individuals for limited resources. The most fit organisms succeed by survival and reproduction.
2. A given habitat has enough resources to support a maximum possible number, but beyond a particular limit the further growth is impossible.
3. This limit is called nature’s carrying capacity (K) for that species in that habitat.
4. A population growing in a habitat with limited resources show following phases in a sequential manner, (a) A lag phase (b) Phase of acceleration (c) Phase of deceleration (d) An asymptote, when the population density reaches the carrying capacity.
5. A plot of N in relation to time (t) results in a sigmoid curve. This type of population growth is called Verhulst-Pearl Logistic Growth.
6. Since resources for growth for most animal populations are finite and become limiting sooner or later, the logistic growth model is considered as a more realistic one.
7. Logistic growth thus always shows sigmoid curve.

Question 2.
Enlist and explain the important characteristics of a population.
Answer:
Important characteristics of a population are as follows:
1. Natality:

1. Natality is the birth rate of a population. Due to increased natality the population density rises.
2. Natality is a crude birth rate or specific birth rate.
3. Crude birth rate : Number of births per 1000 population/year gives crude birth rate. Crude birth rate is helpful in calculating population size.
4. Specific birth rate : Crude birth rate is relative to a specific criterion such as age. E.g. If in a pond, there were 200 carp fish and their population rises to 800. Then, taking the current population to 1000, the birth rate becomes 800/200 = 4 offspring per carp per year. This is specific birth rate.
5. Absolute Natality : The number of births under ideal conditions when there is no competition and the resources such as food and water are abundant, then it give absolute natality.
6. Realized Natality : The number of births under different environmental pressures give realized natality. Absolute natality will be always more than realized natality.

2. Mortality:

1. Mortality is the death rate of a population. It gives a measure of the number of deaths in a particular population, in proportion to the size of that population, per unit of time.
2. Mortality rate is typically expressed in deaths per 1,000 individuals per year.
   A mortality rate of 9.5 (out of 1,000) in a population of 1,000 would mean 9.5 deaths per year in that entire population or 0.95% out of the total.
3. Absolute Mortality : The number of deaths under ideal conditions when there is no competition, and all the resources such as food and water are abundant, then it gives absolute mortality.
4. Realized Mortality : The number of deaths under environmental pressures come into play gives realized mortality.
5. It must be remembered that absolute mortality will always be less than realized mortality.

3. Density:
The density of a population in a given habitat during a given period fluctuates due to changes in four basic processes, viz.

1. Natality i.e. birth rate (The number of births during a given period in the population that are added to the initial density).
2. Mortality i.e. death rate (The number of deaths in the population during a given period).
3. Immigration i.e. number of individuals of the same species that have come into the habitat from elsewhere during the time period under consideration.
4. Emigration i.e. the number of individuals of the population who left the habitat and gone elsewhere during the time period under consideration.
5. Natality and immigration increase in population density whereas mortality and emigration decrease it.

4. Sex ratio : Ratio of the number of individuals of one sex (male) to that of the other sex (female) is called sex ratio. In nature male, female ratio is always 1 : 1. This 1 : 1 ratio is called evolutionary stable strategy of ESS for each population.

5. Age distribution and age pyramid : This parameter is important for human population. Each population is composed of individuals of different ages. The age distribution is plotted for the population, the resulting structure is called an age pyramid. For making the age pyramid, the entire population is divided into three age groups as Pre-Reproductive (age 0-14 years), Reproductive (age 15-44 years) and Post-reproductive (age 45 -85+ years).

6. Growth : Growth of a population causes rise in its density. The size and density are dynamic parameters as they keep on changing with time, and various factors including food, predation pressure and adverse weather. From the density, one comes to know if the population is flourishing or declining.

Balbharti Maharashtra State Board 12th Biology Textbook Solutions Chapter 12 Biotechnology Textbook Exercise Questions and Answers.

Maharashtra State Board 12th Biology Solutions Chapter 12 Biotechnology

1. Multiple choice questions

Question 1.
MU The bacterium which causes a plant disease called crown gall is ………………..
(a) Helicobacter pylori
(b) Agrobacterium tumifaciens
(c) Thermophilus aquaticus
(d) Bacillus thuringienesis
Answer:
(b) Agrobacterium tumtfaciens

Question 2.
The enzyme nuclease hydrolyses ……………….. of polynucleotide chain of DNA.
(a) hydrogen bonds
(b) phosphodiester bonds
(c) glycosidic bonds
(d) peptide bonds
Answer:
(b) phosphodiester bonds

Question 3.
In vitro amplification of DNA or RNA segment is known as ………………..
(a) chromatography
(b) southern blotting
(c) polymerase chain reaction
(d) gel electrophoresis
Answer:
(c) polymerase chain reaction

Question 4.
Which of the following is the correct recognition sequence of restriction enzyme hind III.
(a) 5′ —A-A-G-C-T-T— 3′
3′ —T-T-C-G-A-A—5′
(b) 5′ — G-A-A-T-T-C—3′
3′ — C-T-T-A-A-G—5′
(c) 5′ — C-G-A-T-T-C—3′
3′ — G-C-T-A-A-G—5′
(d) 5′ — G-G-C-C—3′
3′ — C-C-G-G—5′
Answer:
(a) 5’ —A-A-G-C-T-T—3’
3’ —T-T-C-G-A-A—5’

Question 5.
Recombinant protein ……………….. is used to dissolve blood clots present in the body.
(a) insulin
(b) tissue plasminogen activator
(c) relaxin
(d) erythropoietin
Answer:
(b) tissue plasminogen activator

Question 6.
Recognition sequence of restriction enzymes are generally ……………….. nucleotide long.
(a) 2 to 4
(b) 4 to 8
(c) 8 to 10
(d) 14 to 18
Answer:
(b) 4 to 8

2. Very short answer questions

Question 1.
Name the vector which is used in production of human insulin through recombinant DNA technology.
Answer:
PBR 322

Question 2.
Which cells from Langerhans of pancreas do produce a peptide hormone insulin?
Answer:
cells of islets of Langerhans of a peptide hormone insulin.

Question 3.
Give the role of Ca⁺⁺ ions in the transfer of recombinant vector into bacterial host cell.
Answer:
Ca⁺⁺ ions promotes binding of plasmid DNA to lipo polysaccharides on bacterial cell surface. Then plasmid can enter the cell on heat shock.

Question 4.
Expand the following acronyms which are used in the held of biotechnology:

1. YAC
2. RE
3. dNTP
4. PCR
5. GMO
6. MAC
7. CCMB.

Answer:

1. YAC : Yeast Artificial chromosome
2. RE : Restriction Endonuclease
3. dNTP : Deoxyribonucleoside triphosphates
4. PCR : Polymerase Chain Reaction
5. GMO : Genetically Modified Organisms
6. MAC : Mammalian Artificial Chromosome
7. CCMB : Centre for Cellular and Molecular Biology

Question 5.
Fill in the blanks and complete the chart.

GMO	Purpose
(i) Bt cotton	———–
(ii) ———-	Delay the softening of tomato during ripening
(iii) Golden rice	———–
(iv) Holstein cow	———–

Answer:

GMO	Purpose
(i) Bt cotton	Insect resistance
(ii) Flavr savr Tomato	Delay the softening of tomato during ripening
(iii) Golden rice	Rich in vitamin A
(iv) Holstein cow	High milk productivity

3. Short answer type questions.

Question 1.
Explain the properties of a good or ideal cloning vector for r-DNA technology.
Answer:
Desired characteristics of ideal cloning vector are as follows:

1. Vector should be able to replicate independenly (through ori gene), so that as vector replicates, multiple copies of the DNA insert are also produced.
2. It should be able to easily transferred into host cells.
3. It should have suitable control elements like promoter, operator, ribosomal binding sites, etc.
4. It should have marker genes for antibiotic resistance and restriction enzyme recognition sites within them.

Question 2.
A PCR machine can rise temperature up to 100 °C but after that it is not able to lower the temperature below 70 °C automatically. Which step of PCR will be hampered first in this faulty machine? Explain why?
Answer:

1. If the faulty machine is not able to lower the temperature below 70 °C, then the primer annealing step will be hampered first.
2. Each primer has a specific annealing temperature, depending upon its A, T, G, C content.
3. For most of the primers annealing temperature is about 40-60 °C.
4. Hence, if temperature is more than primers annealing temperature, it will be able to pair with its complementary sequence in ssDNA.

Question 3.
In the process of r-DNA technology, if two separate restriction enzymes are used to cut vector and donor DNA then which problem will arise in the formation of r-DNA or chimeric DNA? Explain.
Answer:
In the process of r-DNA technology, if two separate restriction enzymes are used to cut vector and donor DNA, then it will result in fragments with different sticky ends which will not be complementary to each other.

Question 4.

Recombinent protein	Its use in or for
(1) Platelet derived growth factor	(a) Anemia
(2) a-antitrypsin	(b) Cystic fibrosis
(3) Relaxin	(c) Haemophilia A
(4) Eryhthropoietin	(d) Diabetes
(5) Factor VIII	(e) Emphysema
(6) DNA ase	(f) Parturition
	(g) Atherosclerosis

Answer:

Recombinent protein	Its use in or for
(1) Platelet derived growth factor	(g) Atherosclerosis
(2) a-antitrypsin	(e) Emphysema
(3) Relaxin	(f) Parturition
(4) Eryhthropoietin	(a) Anemia
(5) Factor VIII	(c) Haemophilia A
(6) DNA ase	(b) Cystic fibrosis

4. Long answer type questions.

Question 1.
(i) Define and explain the terms Bioethics.
Answer:

1. Bioethics is the study of moral vision, decision and policies of human behaviour in relation to biological phenomena or events.
2. Bioethics deals with wide range of reactions on new developments like cloning, transgenic, gene therapy, eugenics, r-DNA technology, in vitro fertilization, sperm bank, gene therapy, euthanasia, death, maintaining those who are in comatose state, prenatal genetic selection, etc.
3. Bioethics also includes the discussion on subjects like what should and should not be done in using recombinant DNA techniques.

Ethical aspects pertaining to the use of biotechnology are:

1. Use of animals cause great sufferings to them.
2. Violation of integration of species caused due to transgenosis.
3. Transfer of human genes into animals and vice versa.
4. Indiscriminate use of biotechnology pose risk to the environment, health and biodiversity.
5. The effects of GMO on non-target organisms, insect resistance crops, gene flow, the loss of diversity.
6. Modification process disrupting the natural process of biological entities.

(ii) Define and explain the term Biopiracy.
Answer:

1. Biopiracy is defined as ‘theft of various natural products and then selling them by getting patent without giving any benefits or compensation back to the host country’.
2. It is unauthorized misappropriation of any biological resource and traditional knowledge.
3. It is bio-patenting of bio-resource or traditional knowledge of another nation without proper permission of the concerned nation or unlawful exploitation and use of bioresources without giving compensation.

Following are the examples of biopiracy:
(a) Patenting of Neem (Azadirachta indica):

1. Pirating India’s traditional knowledge about the properties and uses of neem, the USDA and an American MNC W.R. Grace sought a patent from the European Patent Office (EPO) on the “method for controlling on plants by the aid of hydrophobic extracted neem oil,” in the early 90s.
2. The patenting of the fungicidal properties of Neem, was an example of biopiracy.

(b) Patenting of Basmati:

1. Texmati is a trade name of “Basmati rice line and grains” for which Texas based American company Rice Tec Inc was awarded a patent by the US Patent and Trademark Office (USPTO) in 1997.
2. This is a case of biopiracy as Basmati is a long-grained, aromatic variety of rice indigenous to the Indian subcontinent.
3. Very broad claims about “Inventing” the said rice was the basis of patent application.
4. The UPSTO has rejected all the claims due to people movement against Rice Tec in March 2001.

(c) Haldi (Turmeric) Biopiracy:

1. A patent claim about the healing properties of Haldi was made by two American researchers of Indian origin of the University of Mississippi Medical Center, to the US Patent and Trademark Office.
2. They were granted a patent in March 1995.
3. This is an example of biopiracy because healing properties of Haldi is not a new discovery, but it is a traditional knowledge in ayurvedas for centuries.
4. The Council of Scientific and Industrial Research (CSIR) applied to the US Patent Office for a reexamination and they realized the mistake and cancelled the patent.

(iii) Define and explain the term Biopatent.
Answer:

1. Biopatent is a biological patent awarded for strains of microorganisms, cell lines, genetically modified strains, DNA sequences, biotechnological processes, product processes, product and product applications.
2. It allows the patent holder to exclude others from making, using, selling or importing protected invention for a limited period of time.
3. Duration of biopatentis five years from the date of the grant or seven years from the date of filing the patent application, whichever is less.
4. Awarding biopatents provides encouragement to innovations and promote development of scientific culture in society. It also emphasizes the role of biology in shaping human society.
5. First biopatent was awarded for genetically engineered bacterium ‘Pseudomonas’ used for clearing oils spills.
6. Patent jointly issued by Delta and Pineland company and US department of agriculture having title ‘control of plant gene expression’, is based on a gene that produces a protein toxic to plant and thus prevents seed germination.

This patent was not granted by Indian government. Such a patent is considered morally unacceptable and fundamentally unequitable. Such patents would pose a threat to global food security as financially powerful corporations would acquire monopoly over biotechnological process.

Question 2.
Explain the steps in process of r-DNA technology with suitable diagrams.
Answer:

The steps involved in gene cloning are as follows:
(1) Isolation of DNA (gene) from the donor organism:

• To obtain the desired gene to be cloned, the cells of the donor organism are sheared with the blender and treated with suitable detergent. Genetic material is then isolated and purified.
• Isolated purified DNA is then cleaved using restriction Endonucleases.
• Restriction fragment containing desired gene is isolated and selected for cloning. This is now called foreign DNA or passanger DNA.
• A desired gene can also be obtained directly from genomic library or c-DNA library.

(2) Insertion of desired foreign gene into a cloning vector (vehicle DNA):

• The foreign DNA or passanger DNA is inserted into a cloning vector (vehicle DNA) like bacterial plasmids and the bacteriophages like lamda phage and M13. The most commonly used plasmid is pBR 322.
• Plasmids are isolated from the bacteria and are cleaved by using same RE which is used in the isolation of the desired gene from the donor.
• Enzyme DNA ligase is used to join foreign DNA and the plasmid DNA.
• Plasmid DNA containing foreign DNA is called recombinant DNA (r-DNA) or chimeric DNA.

(3) Transfer of r-DNA into suitable competent host or cloning organism:

• The r-DNA is introduced into a competent host cell, which is mostly a bacterium.
• Host cell takes up naked r-DNA by process of ‘transformation’ and incorporates it into its own chromosomal DNA which finally expresses the trait controlled by passenger DNA.
• The transfer of r-DNA into a bacterial cell is assisted by divalent Ca⁺⁺.
• The cloning organisms are E.coli and Agrobacterium tumifaciens.
• The competent host cells which have taken up r-DNA are called transformed cells.
• By using techniques like electroporation, microinjection, lipofection, shot gun, ultrasonification, biolistic method, etc. Foreign DNA can also be transferred directly into the naked cell or protoplast of the competent host cell, without using vector.
• In plant biotechnology the transformation is through Ti plasmids of A. tumifaciens.

(4) Selection of the transformed host cell:

• For isolation of recombinant cell from non-recombinant cell, marker gene of plasmid vector is employed.
• For example, pBR322 plasmid vector contains different marker genes like ampicillin resistant gene and tetracycline resistant gene. When pstl RE is used, it knocks out ampicillin resistant gene from the plasmid, so that the recombinant cells become sensitive to ampicillin.

(5) Multiplication of transformed host cell:

• The transformed host cells are introduced into fresh culture media where they divide.
• The recombinant DNA carried by them also multiplies.

(6) Expression of gene to obtain desired product. Then desired products like enzymes, antibiotiocs etc. separated and purified through down stream processing using bioreactors.

Question 3.
Explain the gene therapy. Give two types of it.
Answer:
Gene therapy is the treatment of genetic disorders by replacing, altering or supplementing a gene that is absent or abnormal and whose absence or abnormality is responsible for the disease.
Types of gene therapy:
(a) Germ line gene therapy:

1. In this germ cells are modified genetically to correct a genetic defect.
2. Normal gene is introduced into germ cells like sperms, eggs, early embryos.
3. It allows transmission of the modified genetic information to the next generation.
4. Although it is highly effective in treatment of the genetic disorders, its use is not preferred in human beings because of various technical and ethical reasons.

(b) Somatic cell gene therapy:

1. In this somatic cells are modified genetically to correct a genetic defect.
2. Healthy genes are introduced in somatic cells like bone marrow cells, hepatic cells, fibroblasts endothelium and pulmonary epithelial cells, central nervous system, endocrine cells and smooth muscle cells of blood vessel walls.
3. Modification of somatic cells only affects the person being treated and the modified chromosomes cannot be passed on the future generations.
4. Somatic cell gene therapy is the only feasible option and the clinical trials have already employed for the treatment of disorders like cancer, rheumatoid arthritis, SCID, Gaucher’s disease, familial hypercholesterolemia, haemophilia, phenylketonuria, cystic fibrosis, sickle-cell anaemia, Duchenne muscular dystrophy, emphysema, thalassemia, etc.

Question 4.
How are the transgenic mice used in cancer research?
Answer:

1. Transgenic mice are used in various research areas of cancer research.
2. Transgenic mice containing a particular oncogene (cancer causing gene) develop specific cancer.
3. They are used to study the relationship between oncogenes and cancer development, cancer treatment and prevention of malignancy.
4. The transgenic mouse model for the investigation of the breast cancer was developed in the laboratory of Philip Leder in Harvard (USA).
5. Transgenic mice containing oncogenes myc and ras were analyzed to find out role of these genes in the development of breast cancer.

Question 5.
Give the steps in PCR or polymerase chain reaction with suitable diagrams.
Answer:

(1) The DNA segment and excess of two primer molecules, four types of dNTPs, the thermostable DNA polymerase are mixed together in ‘eppendorf tube’.

(2) One PCR cycle is of 3-4 minutes duration and it involves following steps:

• Denaturation : The reaction mixture is heated at 90-98°C. Due to this hydrogen bonds in the DNA break and two strands of DNA separate. This is called denaturation.
• Annealing of primer : When the reaction mixture is cooled to 40-60°C, the primer pairs with its complementary sequences in ssDNA. This is called annealing.
• Extension of primer : In this step, the temperature is increased to 70-75°C. At this temperature thermostable Taq DNA polymerase adds nucleotides to 3’end of primer using single-stranded DNA as template. This is called primer extension. Duration of this step is about two minutes.

(3) In an automatic thermal cycler, the above three steps are automatically repeated 20-30 times.
(4) Thus, at the end of ‘n’ cycles 2n copies of DNA segments, get synthesized.

Question 6.
What is a vaccine? Give advantages of oral vaccines or edible vaccines.
Answer:

1. A vaccine is a biological preparation that provides active acquired immunity against a certain disease.
2. Vaccine is often made from a weakened or killed form of the microorganism, its toxins or one of its surface protein antigens.
3. Edible vaccine is an edible plant part engineered to produce an immunogenic protein, which when consumed gets recognized by immune system.
4. Immunogenic protein of certain pathogens are active when’administered orally.
5. When animals or mainly humans consume these plant parts, they get vaccinated against certain pathogen.
6. Oral or edible vaccines have low cost, they are easy to administer and store.

Question 7.
Enlist different types of restriction enzymes commonly used in r-DNA technology? Write on their role.
Answer:

1. Different restriction enzymes commonly used in r-DNA technology are Alu I, Bam HI, Eco RI, Hind II, Hind III, Pst I, Sal I, Taq I, Mbo II, Hpa I, Bgl I, Not I, Kpn I, etc.
2. They are the molecular scissors which recognize and cut the phosphodiester back bone of DNA on both strands, at highly specific sequences.
3. The sites recognized by them are called recognition sequences or recognition sites.
4. Different restriction enzymes found in different organisms recognize different nucleotide sequences and therefore cut DNA at different sites.
5. Restriction cutting may result in DNA fragments with blunt ends or cohesive or sticky ends or staggered ends (having short, single stranded projections).
6. Restriction endonucleases like Bam HI and EcoRI produce fragments with sticky ends.
7. Restriction endonucleases like Alu I, Hind III produce fragments with blunt ends.
8. Type I restriction endonucleases fuction simultaneously as endonuclease and methylase e.g. EcoK.
9. Type II restriction endonucleases have separate cleaving and methylation activities. They are more stable and are used in r-DNA technology e.g. EcoRI, Bgll. They cut DNA at specific sites within the palindrome.
10. Type III restriction endonucleases cut DNA at specific non-palindromic sequences e.g. Hpal, MboII.
11. In bacterial cells, REs destroy various viral DNAs that might enter the cell, thus restricting the potential growth of the virus.

Question 8.
Enlist and write in brief about the different biological tools required in r-DNA technology.
Answer:
The biological tools used in r-DNA technology are various enzymes, cloning vectors and competent hosts.
(1) Enzymes:

• Enzymes like lysozymes, nucleases (exonucleases and endonucleases), DNA ligase, reverse transcriptase, DNA polymerase, alkaline phosphatases, etc. are used in r-DNA technology.
• The restriction endonucleases are used as biological or molecular scissors. They are able to cut a DNA molecule at a specific recognition site.

(2) Vectors:

• Vectors are DNA molecules which carry foreign DNA segment and replicate inside the host cell.
• Vectors may be plasmids, bacteriophages (M13, lambda virus), cosmid, phagemids, BAC (bacterial artificial chromosome), YAC (yeast artificial chromosome), transposons, baculoviruses and mammalian artificial chromosomes (MACs).
• Most commonly used vectors are plasmid vectors (pBR 322, pUC, Ti plasmid) and bacteriophages (lamda phage, M13 phage).

(3) Competent host cells:

1. They are bacteria like Bacillus haemophilus, Helicobacter pyroliand E. coli.
2. Mostly E. coli is used for the transformation with recombinant DNA.

Balbharti Maharashtra State Board 12th Biology Textbook Solutions Chapter 11 Enhancement of Food Production Textbook Exercise Questions and Answers.

Maharashtra State Board 12th Biology Solutions Chapter 11 Enhancement of Food Production

1. Multiple choice questions

Question 1.
Antibiotic Chloromycetin is obtained from ………………….
(a) Streptomyces erythreus
(b) Penicillium chrysogenum
(c) Streptomyces venezuelae
(d) Streptomyces griseus
Answer:
(c) Streptomyces venezuelae

Question 2.
Removal of large pieces of floating debris, oily substances, etc. during sewage treatment is called ………………….
(a) primary treatment
(b) secondary treatment
(c) final treatment
(d) amplification
Answer:
(a) primary treatment

Question 3.
Which one of the following is free living bacterial biofertilizer?
(a) Azotobacter
(b) Rhizobium
(c) Nostoc
(d) Bacillus thuringiensis
Answer:
(a) Azotobacter

Question 4.
Most commonly used substrate for industrial production of beer is ………………….
(a) barley
(b) wheat
(c) corn
(d) sugar cane molasses
Answer:
(a) barley

Question 5.
Ethanol is commercially produced through a particular species of ………………….
(a) Aspergillus
(b) Saccharomyces
(c) Clostridium
(d) Trichoderma
Answer:
(b) Saccharomyces

Question 6.
One of the free-living anaerobic nitrogen- fixers is ………………….
(a) Azotobacter
(b) Beijerinckia
(c) Rhodospirillum
(d) Rhizobium
Answer:
(c) Rhodospirillum

Question 7.
Microorganisms also help in production of food like ………………….
(a) bread
(b) alcoholic beverages
(c) vegetables
(d) pulses
Answer:
(a) bread

Question 8.
MOET technique is used for ………………….
(a) production of hybrids
(b) inbreeding
(c) outbreeding
(d) outcrossing
Answer:
(a) production of hybrids

Question 9.
Mule is the outcome of ………………….
(a) inbreeding
(b) artificial insemination
(c) interspecific hybridization
(d) outbreeding
Answer:
(c) interspecific hybridization

2. Very Short Answer Questions

Question 1.
What makes idlis puffy?
Answer:
During preparation of idlis, rice and black gram flour is fermented by air borne Leuconostoc and Streptococcus bacteria. CO₂ produced during fermentation makes them puffy.

Question 2.
Bacterial biofertilizers.
Answer:
Rhizobium, Frankia, Pseudomonas striata, Bacillus polymyxa, Agrobacterium, Microccocus, Azotobacter, Costridium, Beijerinkia, Klebsiella.

Question 3.
What is the microbial source of vitamin B₁₂?
Answer:
The microbial source of vitamin B₁₂ is Pseudomonas denitrificans.

Question 4.
What is the microbial source of enzyme invertase?
Answer:
The microbial source of enzyme invertase is Saccharomyces cerevisiae.

Question 5.
Milk starts to coagulate when Lactic Acid Bacteria (LAB) are added to warm milk as a starter. Mention any two other benefits of LAB.
Answer:
Lactic Acid Bacteria (LAB) check the growth of disease causing microbes and produce vitamin B.

Question 6.
Name the enzyme produced by Streptococcus bacterium. Explain its importance in medical sciences.
Answer:
The enzyme produced by Streptococcus spp. is streptokinase. It is used as a ‘clot buster’ for clearing blood clots in the blood vessels in heart patients.

Question 7.
What is breed?
Answer:
Breed is a group of animals related by descent and similar in most characters like general appearance, features, size, configuration, etc.

Question 8.
Estuary
Answer:
Estuary is a place where river meets the sea.

Question 9.
What is shellac?
Answer:
Shellac is the pure form of lac obtained by washing and filtering.

3. Short Answer Questions.

Question 1.
Many microbes are used at home during preparation of food items. Comment on such useful ones with examples.
Answer:

1. Many food preparations made at home involves the use of microorganisms.
2. The microbes Lactobacilli are used in the preparation of dhokla from gram flour and buttermilk by the process of fermentation.
3. Dosa and idlis are prepared by using batter of rice and black gram which is fermented by air borne Leuconostoc and Streptococcus bacteria.
4. Large, fleshy fruiting bodies of some mushrooms and truffles are directly used as food. It is sugar free, fat free food rich in proteins, vitamins, minerals and amino acids. It is food with low calories.
5. Curd is prepared by inoculating milk with Lactobacillus acidophilus. Lactic acid produced during fermentation causes coagulation and partial digestion of milk protein casein and milk turns into curd.
6. Buttermilk is the acidulated liquid left after churning of butter from curd, is called buttermilk.

Question 2.
What is biogas? Write in brief about the production process.
Answer:
Biogas is a mixture of methane CH₄ (50-60%), CO₂ (30-40%), H₂S (0-3%) and other gases (CO, N₂, H₂) in traces.

Biogas production process:
a. A typical biogas plant consists of digester (made up of concrete bricks and cement or steel and is partly buried in the soil) and gas holder (a cylindrical gas tank to collect gases).
b. Raw materials like cow dung is mixed with water in equal proportion to make slurry which is fed into the digester’ through a side opening (charge pit).

Anaerobic digestion involves following processes:
i. Hydrolysis or solubilization:
Anaerobic hydrolyzing bacteria like Clostridium and Pseudomonas hydrolyse carbohydrates into simple sugars, proteins into amino acids and lipids into fatty acids.

ii. Acidogenesis:
Facultative and obligate anaerobic, acidogenic bacteria convert simple organic substances into acids like formic acid, acetic acid, H₂ and CO₂

iii. Methanogenesis:
Anaerobic methanogenic bacteria like Methanobacterium, Methanococcus convert acetate, H₂ and CO₂ into Methane, CO₂ and H₂O and other products.
12 mol CH₃COOH → 12CH₄ + 12CO₂ 4mol H.COOH → CH₄ + 3CO₂ + 2H₂O CO₂ + 4H₂ → CH₄ + 2H₂O

Question 3.
Biocontrol agents.
Answer:
(1) Biocontrol agents are the organisms like (bacteria, fungi, viruses and protozoans) act which are employed for controlling pathogens, pests and weeds.

(2) They cause the disease to the pest or compete or kill them.

(3) The use of biocontrol measures greatly reduces use of toxic chemicals and pesticides that are harmful to human beings and also pollute our environment.

(4) Biocontrol agents and their hosts.

• Bacteria (Bacillus thuringiensis, B. papilliae and B. lentimorbus Hosts : Caterpillars, cabbage worms, adult beetles
• Fungi (Beauveria bassiana, Entomophothora, pallidaroseum, Zoophthora radicans) Host : Aphid crocci, A. unguicilata, mealy bugs, mites, white flies, etc.
• Protozoans (Nosema locustae) Host: Grasshoppers, caterpillars, crickets
• Viruses (Nucleopolyhedro virus-NPV, Granulovirus-GV) Host : Caterpillars, Gypsy moth, ants and beetles.

(5) Some examples:

• Bacillus thuringiensis (Bt) is a microbiai pesticide used to get rid of butterfly, caterpillars.
• Trichoderma fungus is an effective biocontrol agent against soil borne fungal plant pathogens which infect roots and rhizomes.
• Phytophthora palmiuora is a mycoherbicide that controls milk weed in orchards.
• Pseudomonas spp. is a bacterial herbicide that attacks several weeds.
• Tyrea moth controls the weed Senecio jacobeac.

Question 4.
Name any two enzymes and antibiotics with their microbial source.
Answer:

1. Microbial source of Chloromycetin. – Streptomyces venezuelae
2. Microbial source of Erythromycin. – Streptomyces erythreus
3. Microbial source of Penicillin. – Penicillium chrysogenum
4. Microbial source of Streptomycin. – Streptomyces griseus
5. Microbial source of Griseofulvin. – Penicillium griseojulvum
6. Microbial source of Bacitracin. – Bacillus licheniformis
7. Microbial source of Oxytetracyclin / Terramycin. – Streptomyces aurifaciens
8. The enzyme produced by Streptococcus bacterium. – Streptokinase
9. Microbial source of Invertase. – Saccharomyces cerevisiae
10. Microbial source of Pectinase. – Sclerotinia libertine, Aspergillus niger
11. Microbial source of Lipase. – Candida lipolytica
12. Microbial source of Cellulase. – Trichoderma konigii

Question 5.
Write the principles of farm management.
Answer:
The principles of farm management are as follows:

1. Selection of high-yielding breeds.
2. Understanding the feed requirements of farm animals.
3. Supply of adequate nutritional sources for the animals.
4. Maintaining the cleanliness of environment.
5. Maintenance of health with the help of veterinary supervision.
6. Undertaking vaccination programmes.
7. Development of high-yielding cross-bred varieties.
8. Making various products and their preservation.
9. Distribution and marketing of the farm produce.

Question 6.
Give the economic importance of fisheries.
Answer:
Economic importance of fisheries is as follows:

1. Fish is a nutritious food and thus is a source of many vitamins, minerals and nutrients.
2. Commercial products such as fish oil, fish meal and fertilizers, fish guano, fish glue, isinglass are prepared from fish.
3. These by-products are used in paints, soaps, oils and medicines.
4. Some organisms like prawns and lobsters have high export value and market price.
5. Fish farming and other fishery trades provide job opportunity and self-employment
6. Productivity and national economy is improved through fishery practices.

Question 7.
Enlist the species of honey bee mentioning their specific uses.
Answer:
(1) The four species of honey bees commonly found in India : Apis dorsata (rock bee, or wild bee), Apis jlorea (little bee), Apis mellifera (European bee) and Apis indica (Indian bee).

(2) Uses:

• Rock bee : They produce 36 kg of honey per comb per year. They produce bee wax.
• Little bee : They produce half kg of honey per hive per year.
• European bee : The average production per colony per year is 25 to 40 kg.
• Indian bee : The average production per colony per year is 6 to 8 kg.

Question 8.
What are A, B, C, D in the table given below.

Types of microbe	Name	Commercial Product
Fungus	A	Penicillin
Bacterium	Acetobacter aceti	B
C	Aspergillus niger	Citric acid
Yeast	D	Ethanol

Answer:
A : Penicillium chrysogenum
B : Vinegar (Acetic acid)
C : Fungus
D : Sachharomyces cerevisiae var. ellipsoidis

4. Long Answer Questions.

Question 1.
Explain the process of sewage water treatment before it can be discharged into natural bodies. Why is this treatment essential?
Answer:
Sewage treatment includes following steps:
(1) Preliminary Treatment:

• Screening: The larger suspended or floating objects are filtered and removed in screening chambers by passing the sewage through screens or net in the chambers.
• Grit Chamber : Filtered sewage is passed into series of grit chambers which contain large stones (pebbles) and brick-ballast. Coarse particles which settle down by gravity are removed.

(2) Primary treatment (physical treatment):

• The sewage water is pumped into the primary sedimentation tank where 50-70% of the suspended solid or organic matter get sedimented and about 30-40% (in number) of coliform organisms are removed.
• The organic matter which is settled down is called primary sludge.
• Primary sludge is removed by mechanically operated devices.
• Dissolved organic matter and micro-organisms in the supernatant (effluent) are then removed by the secondary treatment.

(3) Secondary treatment (biological treatment):

• The primary effluent is passed into large aeration tanks where it is constantly agitated mechanically and air is pumped into it.
• The mesh like masses of aerobic bacteria, slime and fungal hyphae, known as floes are formed.
• Aerobic microbes consume most of the organic matter and this reduces BOD (Biochemical Oxygen Demand) of the effluent.

(4) Tertiary treatment:

• Once the BOD is sufficiently reduced, waste water is passed into a settling tank where the floes are allowed to sediment.
• The sediment is called activated sludge.
• Small part of activated sludge is transferred to aeration tank and the major part is pumped in to large anaerobic sludge digesters.
• In these tanks, anaerobic bacteria grow and digest the bacteria and fungi in the sludge and gases like methane, hydrogen sulphide, CO₂, etc. are released.
• Effluents from these digesters are released in natural water bodies like rivers and streams after chlorination which kills pathogenic bacteria.
• Digested sludge is then disposed.

Question 2.
Lac culture.
Answer:

1. Lac is a pink coloured resin secreted by dermal glands of female lac insect (Trachardia lacca) that hardens on coming in contact with air forming lac.
2. Lac is a complex substance having resin, sugar, water, minerals and alkaline substances.
3. Lac insect is colonial in habit and it feeds on succulent twigs like ber, peepal, palas, kusum, babool,
4. These plants are artificially inoculated in order to get better and regular supply of good quality and quantity of lac.
5. Natural lac is always contaminated and pure form of lac obtained by washing and filtering is called as shellac.
6. Lac is used to make bangles, toys, woodwork, inks, mirrors, etc.
7. India’s share is 85% of total lac produced in the world.

Question 3.
Describe various methods of fish preservation.
Answer:

1. Fish is a highly perishable commodity.
2. After catching the fish it immediately starts spoilage process.
3. In order to prevent this process, the fish preservation is done.

The different methods of fish preservation are as follows:

1. Chilling : This involves covering the fish with layers of ice. Ice is effective for short term preservation. It inhibits the activity of autolytic enzymes.
2. Freezing : It is a long duration preservation method. Fish are freezed at 0°C to -20°C. This also inhibit autolytic enzyme activities and slows down bacterial growth.
3. Freeze drying : The deep frozen -fish at -20°C are dried by direct sublimation of ice to water vapour with any melting into liquid water. This is achieved by exposing the frozen fish to 140°C in a vacuum chamber. The fish is then packed or canned in dried condition.
4. Sun drying : This inhibits the growth of microorganisms that spoil the fish.
5. Smoke drying : Smoke is prepared by burning woods with less resinous matter. Bacteria are destroyed by the acid content of the smoke. Smoking also give the characteristic colour, taste and odour to fish.
6. Salting : Salt removes the moisture from the fish tissues by osmosis. High salt concentration destroys autolytic enzymes and halts bacterial activity.
7. Canning : Canning involves sealing the food in a container, heat ‘sterilising’ the sealed unit and cooling it to ambient temperature for subsequent storage.

Question 4.
Give an account of poultry diseases.
Answer:
Various poultry diseases are as follows:

1. Viral diseases : Ranikhet, Bronchitis, Avian influenza (bird flu), etc. Bird flu had serious impact on poultry farming and also caused human infection.
2. Bacterial diseases : Pullorum, Cholera, Typhoid, TB, CRD (chronic respiratory disease), Enteritis, etc.
3. Fungal diseases : Aspergillosis, Favus and Thrush.
4. Parasitic diseases : Lice infection, round worm, caecal worm infections, etc.
5. Protozoan diseases : Coccidiosis.

Question 5.
Give an account of mutation breeding with examples.
Answer:

1. Mutations are sudden heritable changes in the genotype.
2. Natural mutations occur at a very slow rate.
3. Natural physical mutagens include exposure to high temperature, high concentration of C02, X-rays, UV rays.
4. Mutations can be induced by using various mutagens.
5. Mutagens cause gene mutations and chromosomal aberrations.
6. Chemical mutagens include nitrous acid, EMS (Ethyl – Methyl – Sulphonate), mustard gas, colchicine, etc.
7. Seedlings or seeds are irradiated by using CO60 or UV bulbs or X-ray machines.
8. The mutated seedlings are then screened for resistance to diseases/pests, high yield, etc.
9. Examples of mutant varieties in different crops are Jagannath (rice), NP 836 (rust resistant wheat variety), Indore-2 (cotton variety resistant to bollworm), Regina-II (cabbage variety resistant to bacterial rot).

Question 6.
Describe briefly various steps of plant breeding methods.
Answer:
The main steps of the plant breeding program (Hybridization) are as follows:

(1) Collection of variability:

• Germplasm collection is the entire collection of all the diverse alleles for all genes in a given crop.
• Wild species and relatives of the cultivated species having desired traits are collected and preserved.
• Forests and natural reserves are the means of in situ conservation of germplasm.
• Botanical gardens, seed banks, etc. are means of ex situ conservation of germplasm.

(2) Evaluation and selection of parents:

• The collected germplasm is evaluated to identify healthy and vigorous plants with desirable and complementary characters.
• Selected parents are selfed for three to four generations to increase homozygosity.
• Only pure lines are selected, multiplied and used in the hybridization.

(3) Hybridization:

• The variety showing maximum desirable features is selected as female (recurrent) parent and the other variety which lacks good characters found in recurrent parent is selected as male parent (donor).
• The pollen grains from anthers of male parent are artificially dusted over stigmas of emasculated flowers of female parent.
• Hybrid seeds are collected and sown to grow F₁ geneartion.

(4) Selection and Testing of Superior Recombinants:

• The F₁ hybrid plants which are superior to both the parents and having high hybrid vigour, are selected and selfed for few generations to make them homozygous for the said desirable characters.
• This ensures that there is no further segregation of the characters.

(5) Testing, release and commercialization of new cultivars:

• The newly selected lines are evaluated for the productivity and desirable features like disease resistance, pest resistance, quality, etc.
• They are initially grown under controlled conditions of water, fertilizers, etc. and their performance is recorded.
• The selected lines are then grown for at least three generations in natural field, in different agroclimatic zones.
• Finally variety is released as new variety for use by the farmers.

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

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

1. Multiple Choice Questions

Question 1.
Which of the following is NOT caused by unsterilized needles?
(a) Elephantiasis
(b) AIDS
(e) Malaria
(d) Hepatitis
Answer:
(a) Elephantiasis

Question 2.
Opium derivative is …………………
(a) Codeine
(b) Caffeine
(c) Heroin
(d) Psilocybin
Answer:
(c) Heroin

Question 3.
The stimulant present in tea is …………………
(a) tannin
(b) cocaine
(C) caffeine
(d) crack
Answer:
(c) caffeine

Question 4.
WhIch of the following Is caused by smoking?
(a) Liver cirrhosis
(b) Pulmonary tuberculosis
(c) Emphysema
(d) Malaria
Answer:
(c) Emphysema

Question 5.
An antibody is …………………
(a) molecuic that binds specifically an antigen
(b) WBC which invades bacteria
(c) secretion of mammalian RBC
(d) cellular component of blood
Answer:
(a) molecule that binds specifically an antigen

Question 6.
The antiviral proteins released by a virus-infected cell are called …………………
(a) histamines
(b) interferons
(c) pyrogens
(d) allergens
Answer:
(b) interferons

Question 7.
Both B-cells and T-cells are derived from …………………
(a) lymph nodes
(b) thymus glands
(c) liver
(d) stem cells in bone marrow
Answer:
(b) thymus glands

Question 8.
Which of the following diseases can be contracted by droplet infection?
(a) Malaria
(b) Chicken pox
(c) Pneumonia
(d) Rabies
Answer:
(c) Pneumonia

Question 9.
Confirmatory test used for detecting HIV infection is …………………
(a) ELISA
(b) Western blot
(c) Widal test
(d) Eastern blot
Answer:
(b) Western blot

Question 10.
Elephantiasis is caused by …………………
(a) W. barterofti
(b) P. vivax
(c) Bedbug
(d) Elephant
Answer:
(a) W. bancrofti

Question 11.
Innate immunity is provided by …………………
(a) phagocytes
(b) antibody
(c) T-lymphocytes
(d) B-lymphocytes
Answer:
(c) T-lymphocytes

2. Short Answer Questions

Question 1.
What is the source of cocaine?
Answer:
Source of cocaine is coca plant – Erythroxylum coca.

Question 2.
Name one disease caused by smoking.
Answer:
Emphysema. (Damaged and enlarged lungs causing breathlessness)

Question 3.
Which cells stimulate B-cells to form antibodies ?
Answer:
Helper T-cells stimulate B-cells to form antibodies.

Question 4.
What does the abbreviation AIDS stand for?
Answer:
AIDS stands for Acquired Immuno Deficiency Syndrome.

Question 5.
Name the causative agent of typhoid fever.
Answer:
Salmonella typhi

Question 6.
What is Rh factor?
Answer:
Antigen ‘D’ present on the surface of RBCs is known as Rh factor.

Question 7
What is schizont?
Answer:
Schizont is a ring-like form produced from merozoites inside the erythrocytes of human beings, infected by Plasmodium, which again forms new merozoites.

Question 8.
Name the addicting component found in tobacco.
Answer:
Nicotine

Question 9.
Name the pathogen causing Malaria.
Answer:
Plasmodium vivax

Question 10.
Name the vector of Filariasis.
Answer:
Female Culex mosquito

Question 11.
Name of the causative agent of ringworm.
Answer:
Trichophyton

Question 12.
Health
Answer:
Health is defined as the state of complete physical, mental and social well¬being and not merely the absence of disease or infirmity.

3. Short Answer Questions

Question 1.
What are acquired diseases?
Answer:
Diseases which are developed after the birth of an individual are called acquired diseases. These are of two types, viz. (a) Communicable or infectious diseases and (b) Non- communicable or Non-infectious diseases. Communicable or infectious diseases are transmitted from infected person to another healthy person either directly or indirectly. They are caused due to pathogens like viruses, bacteria, fungi, helminth worms, etc. Non-communicable or Non-infectious diseases cannot be transmitted from infected person to another healthy one either directly or indirectly.

Question 2.
Antigen and antibody.
Answer:

Antigen	Antibody
1. Antigens are foreign proteins which are capable of producing infection.	1. Antibodies are immunoglobulins produced by the body to act against the antigens.
2. The structure of antigens is variable dependent upon the type of pathogen.	2. The structure of antibody is Y-shaped.
3. The antigen is the ‘non-self’ molecule.	3. The antibody is ‘self’ molecule.
4. The antigens have epitope sites which bind with the antibody molecule.	4. The antibodies have paratope sites which bind with the antigen molecule.

Question 3.
Name the infective stage of Plasmodium. Give Symptoms of malaria
Answer:
Sporozoite
I. Symptoms of malaria:

1. Fever accompanied by shivering.
2. Joint pain or arthralgia.
3. Vomiting.
4. Anaemia caused due to rupture of RBCs or haemolysis.
5. Haemoglobinuria.
6. Retinal damage.
7. Convulsions.
8. Cyclical occurrence of sudden coldness followed by rigor and then fever and sweating lasting for four to six hours. This is called a classic symptom of malaria.
9. Splenomegaly or enlarged spleen, severe headache, cerebral ischemia, hepatomegaly, i. e. enlarged liver, hypoglycaemia and haemoglobinuria with renal failure may occur in severe infections.

II. Spread / Transmission of malaria:

1. Malaria parasite is transmitted through the female Anopheles mosquito and hence it is known as mosquito-borne disease. Mosquito acts as a vector.
2. There are four species of Plasmodium, viz., P. vivax, P. falciparum, P. ovale and P. malariae which transmit malaria.

Question 4.
Explain the mode of infection and cause of elephantiasis.
Answer:
Mode of infection, i.e. transmission:

1. The parasite Wuchereria bancrofti is transmitted from a patient to other normal human being by female Culex mosquito.
2. The filarial larvae leave mosquito body and arrive on the human skin where they penetrate the skin and enter inside.
3. They undergo two moultings to become adults. Later they settle in the lymphatic system. They incubate for about 8-16 months.
4. When they settle in lymphatic system, this infection is called lymphatic filariasis.
5. The worms start infecting lymphatic circulation resulting into enlargement of lymph vessels and lymph nodes. The extremities like legs or limbs become swollen which resembles elephant legs. Therefore it is called elephantiasis.
6. This condition is lymphoedema, i.e. accumulation of lymph fluid in tissue causing swelling.

Question 5.
Why is smoking a bad habit?
Answer:

1. Smoking involves inhaling the cigarette smoke which contains nicotine and other toxic substances like N-nitrosodimethlene. There is some amount of carbon monoxide.
   All these substances affect the normal respiratory health.
2. Smoking invites problems like asthma, hypertension, heart disease, stroke, lung damage.
3. The worst impact is that these substances are carcinogenic and hence can cause cancer of larynx, trachea, lung, etc.
4. Smoking not only affects the smokers but also has bad effect on others due to passive smokers.
5. In women, smoking is still hazardous as their ovaries can undergo mutations due to mutagenic chemicals found in smoke.
6. Therefore, smoking is a very bad habit.

Question 6.
What do the abbreviations AMIS and CMIS denote?
Answer:
AMIS is Antibody-mediated immune system or humoral immunity and CMIS is cell- mediated immune system.

Question 7.
What is a carcinogen? Name one chemical carcinogen with its target tissue.
Answer:

1. Carcinogen is the substance or agent that causes cancer.
2. Urinary bladder cancer caused by 2-naphthylamine and 4-aminobiphenyl.

Question 8.
Active immunity and passive immunity.
Answer:

Active immunity	Passive immunity
1. Active immunity is produced in response to entry of pathogens and their antigenic stimuli.	1. Passive immunity is produced due to antibodies that are transferred to the body.
2. Active immunity is the long lasting immunity.	2. Passive immunity is short-lived immunity.
3. In active immunity, the body produces its own antibodies.	3. In passive immunity, antibodies are given to the body from outside.
4. Natural acquired active immunity is obtained due to infections by pathogens.	4. Natural acquired passive immunity is obtained through antibodies of mother transmitted- to baby by placenta or colostrum.
5. Artificial acquired active immunity is obtained through vaccinations. These vaccines contain dead or live but attenuated pathogens.	5. Artificial acquired passive immunity is also obtained through vaccinations, but here the vaccines contain the readymade antibodies which are prepared with the help of other animals such as horses.

4. Short Answer Questions

Question 1.
B-cells and T-cells.
Answer:

B-cells	T-cells.
1. B-cells are type of lymphocyte whose origin is in bone marrow but maturation is in blood.	1. T-cells are type of lymphocytes which originate in bone marrow but maturation occurs in thymus.
2. B-cells Eire type of lymphocytes which are involved in humoral mediated immunity.	2. T-cells are type of lymphocytes which are involved in cell-mediated immunity.
3. 20% of lymphocytes present in the blood are B-cells.	3. 80% of lymphocytes present in the blood are T-cells.
4. Two types of B-cells are Memory cells and Plasma cells.	4. T-cells are of following subtypes : Cytotoxic T-cells, helper T-cells, suppressor T-cells.
5. They are involved in antibody mediated immunity. (AMI)	5. They are involved in cell-mediated immunity (CMI).
6. B-cells produced antibodies with which they fight against pathogens.	6. T-cells do not produce antibodies.
7. B-cells have membrane bound immunoglobulins located on the surface.	7. There is a presence of T cell receptors on the T-cell surface.

Question 2.
What are the symptoms of malaria? How does malaria spread?
Answer:
Symptoms of malaria:

1. Fever accompanied by shivering.
2. Joint pain or arthralgia.
3. Vomiting.
4. Anaemia caused due to rupture of RBCs or haemolysis.
5. Haemoglobinuria.

Question 3.
AIDS.
Answer:
(1) AIDS or the acquired immuno deficiency syndrome, is fatal viral disease caused by a retrovirus (ss RNA) known as the human immuno deficiency virus (HIV) which weakens the body’s immune system. It is called a modern pandemic.

(2) The HIV attacks the immune system which in turn causes many opportunistic infections, neurological disorders and unusual malignancies ultimately leading to death.

(3) AIDS was first noticed in USA in 1981 whereas in India, first confirmed case of AIDS was in April 1986 from Tamil Nadu.

(4) HIV is transmitted through body fluids such as saliva, tears, nervous system tissue, spinal fluid, blood, semen, vaginal fluid and breast milk. However, only blood, semen, vaginal secretions and breast milk generally transmit infection to others.

(5) The transmission of HIV occurs by sexual contact, through blood and blood products and by contaminated syringes, needles, etc. There is also transplacental transmission or through breast milk at the time of nursing.

(6) Accidental needle injury, artificial insemination with infected donated semen and transplantation with infected organs are some of the rare occasions of transmission of HIV.

(7) HIV infection is not spread by casual contact such as hugging, bite of mosquitoes or using other objects touched by a patient.

(8) Acute HIV infection progresses over time to asymptomatic HIV infection and then to early symptomatic HIV infection. Later, it progresses to full blown AIDS when patient shows advanced HIV infection with CD4 T-cell count below 200 cells/mm.

Question 4.
Give the symptoms of cancer.
Answer:
Symptoms of cancer:

1. Presence of lump or tumour.
2. White patches in the mouth.
3. Change in a wart or mole on the skin.
4. Swollen or enlarged lymph nodes.
5. Vertigo, headaches or seizures if cancer affect the brain.
6. Coughing and shortness of breath if lungs are affected due to cancer.

Question 5.
Antigens on blood cells.
Answer:

1. There are about 30 known antigens on the surface of human red blood cells. They decide the type of blood group such as ABO, Rh, Duffy, Kidd, Lewis, P MNS, Bombay.
2. The different blood groups are determined genetically due to presence of a particular antigen.
3. Landsteiner found two antigens or agglutinogens on the surface of human red blood cells which are named as antigen A and antigen B.
4. There is another antigen called Antigen D which decides the Rh status of the blood. If Antigen D is present, the person is said to be RH positive and when it is lacking, the person is Rh negative.
5. These antigens are responsible for types of blood group and the specific transfusions.
6. Antigens present on the RBCs and antibodies present in the serum can cause agglutination reactions if they are non-compatible. Therefore, at the time of transfusion blood groups are checked properly.

Question 6.
Antigen-antibody complex:
Img 1
Answer:

1. Between antigen and antibody there is specificity.
2. Each antibody is specific for a particular antigen.
3. On the antigens there are combining sites which are called antigenic determinants or epitopes.
4. Epitopes react with the corresponding antigen binding sites of antibodies which are called paratopes.
5. The antigen binding sites are located on the variable regions of the antibody. Variable regions have small variations which make each antibody highly specific for a particular antigen.
6. Owing to variable region the antibody can recognize the specific antigen.
7. Antibody thus binds to specific antigen in a lock and key manner, forming an antigen- antibody complex.

Question 7.
What are the various public health measures, which you would suggest as safeguard against infectious diseases?
Answer:
Infectious diseases spread through pathogens, therefore, it is an important duty of each person to decrease the risk of infecting our own self or others. This can be achieved by

1. Washing hands often, especially whenever, we are in contact with food and water. Before and after preparing food, before eating and after using the toilet, hand wash is a must.
2. Vaccinations : Immunization helps us to protect against contracting many diseases. Therefore, timely vaccination should be taken. Especially at the time of epidemic, one must keep distance from infected area or get vaccinated.
3. One must be at home if there are signs and symptoms of an infection. By going out, we may infect other healthy persons.
4. Proper diet and exercise should be followed to improve one’s own immunity.
5. Hygiene should be utmost in the kitchen and dining area. One must take care while eating uncovered and leftover food.
6. Bathroom and toilet should be cleaned daily as there can be a high concentration of bacteria or other infectious agents in these areas.
7. One should have responsible sexual behaviour to avoid sexually transmitted diseases.
8. Personal items such as toothbrush, comb, towel, undergarments or razor blade should never be shared.
9. Travelling should be avoided because we may infect other passengers during travel. Moreover, our illness can be aggravated. Some special immunizations are needed during certain travels, such as anti-cholera vaccine while going to Pandharpur during Ashadhi.

Question 8.
How does the transmission of each of the following diseases take place?
(a) Amoebiasis:
Answer:
Amoebiasis is usually transmitted by the following ways:

1. The faecal-oral route.
2. Through contact with dirty hands or objects.
3. By anal-oral contact.
4. Through contaminated food and water.

(b) Malaria:
Answer:
Symptoms of malaria:

1. Fever accompanied by shivering.
2. Joint pain or arthralgia.
3. Vomiting.
4. Anaemia caused due to rupture of RBCs or haemolysis.
5. Haemoglobinuria.

(c) Ascariasis:
Answer:

1. Unsafe and unhygienic food and drinks contaminated with the eggs of Ascaris are the main mode of transmission.
2. Eggs hatch inside the intestine of the new host.
3. The larvae pass through various organs and settle as adults in the digestive system.

(d) Pneumonia:
Answer:

1. Pneumonia usually spreads by direct person to person contact.
2. It is also spread via droplet infection, i.e. droplets released by infected person.
3. Using clothes and utensils of the patient.

Question 9.
What measures would you take to prevent water-borne diseases?
Answer:

1. To prevent water-borne diseases, use of safe, clean and potable water is a must. Water should be filtered, then boiled and stored in covered container. If possible water purifier systems should be installed at home.
2. One should preferably use bottled water or carry our own water container while travelling.
3. Cleaning of water containers and maintaining personal hygiene near water storage is a must.
4. Megacities offer chlorinated and purified water for citizens. But villages and smaller rural set ups use river water which may be highly contaminated with pathogens. Such water should be purified before consumption to prevent water-borne diseases.

Question 10.
Typhoid.
Answer:
Typhoid is an infective disease caused by Gram-ve bacterium, Salmonella typhi.
(1) It is food and water-borne infection. In the intestinal lumen of infected person this bacteria is found.

(2) The bacterium has “O” – antigen, which is a lipopolysaccharide (LPS), present on surface coat and its flagella has “H” – antigen. Thus it becomes pathogenic.

(3) Signs and Symptoms of typhoid are as follows:
Prolonged and high fever with nausea, fatigue, headache.
Abdominal pain, constipation or diarrhoea. In severe cases rose-coloured rash is seen on skin. Tongue shows white coating and there is cough. Anorexia or loss of appetite is seen. In chronic cases there is breathlessness, irregular heartbeats and haemorrhage.

(4) Poor hygiene habits and poor sanitation and insects like houseflies and cockroaches spread typhoid.

(5) Typhoid is diagnosed by Widal test.

(6) Antibiotics like Chloromycetin can cure typhoid. Preventive vaccines such as oral Ty21a vaccine and injectable typhim vi and typherix against typhoid are also available. Chronic cases need surgical removal of gall bladder.

5. Match the following.

Column I	Column II
(a) AIDS	(i) Antibody production
(b) Lysozyme	(ii) Activation of B-cells
(c) B-cells	(iii) Immunoglobulin
(d) T-helper cells	(iv) Tears
(e) Antibody	(v) Immuno deficiency

Answer:

Column I	Column II
(a) AIDS	(v) Immuno deficiency
(b) Lysozyme	(iv) Tears
(c) B-cells	(i) Antibody production
(d) T-helper cells	(ii) Activation of B-cells
(e) Antibody	(iii) Immunoglobulin

6. Long Answer Questions

Question 1.
Describe the structure of antibody.
Answer:
Img 2

1. Antibodies are highly specific to specific antigens. They are glycoprotein called immunoglobulins (Igs.).
2. They are produced by plasma cells. Plasma cells are in turn formed by B-lymphocytes.
3. About 2000 molecules of antibodies are formed per second by the plasma cells.
4. Antibody is a ‘Y’-shaped molecule. It has four polypeptide chains, two heavy or H-chains and two light or L-chains.
5. Disulfide bonds (-s-s-) hold the polypeptide chains together to form a ‘Y’-shaped structure.
6. The region holding arms and stem of antibody is termed as hinge. Each chain of the antibody has two distinct regions, the variable region and the constant region.
7. Variable regions have a paratope which is an antigen-binding site. This part of antibody recognizes and binds to the specific antigen forming an antigen-antibody complex.
8. Antibodies are called bivalent as they carry two antigen binding sites.

Question 2.
Vaccination.
Answer:

1. Vaccines are prepared from inactivated pathogen, in the form of protein or sugar from pathogen or dead form of pathogen or toxoid from pathogens or attenuated pathogen.
2. These when they are administered to a person to protect against a particular pathogen, it is called vaccination.
3. Vaccination ’teaches’ the immune system to recognize and eliminate pathogenic organism. Because, already in the body the vaccine is injected and body has made antibodies in response to it. Thus, body is prepared before the attack, if at all it is exposed to pathogen.
4. Thus, it is an important form of primary prevention, which reduces the chances of illness by protecting people. It works by exposing the pathogen in a safe form.
5. Vaccinations control spread of diseases like measles, polio, tetanus and whooping cough that once threatened many lives.
6. Vaccination controls the epidemic outbreak of diseases, if all the people Eire pre-vaccinated.
7. Some hazardous diseases like small, pox and polio have been completely eradicated by the vaccination.

Question 3.
What is cancer? Differentiate between benign tumour and malignant tumour. The main five types of cancer
Answer:
I. Cancer : Cancer is a disease caused by uncontrolled cell division due to disturbed cell cycle.

II. Difference between benign tumour and malignant tumour:

Benign tumour	malignant tumour
1. Benign tumour is localized and it does not spread to neighbouring areas.	1. Malignant tumour starts as local but spreads rapidly to neighbouring areas.
2. Benign tumour is enclosed in connective tissue sheath.	2. Malignant tumour is not enclosed in connective tissue sheath.
3. Benign tumour compresses the surrounding normal tissue.	3. Malignant tumour invades and destroys the surrounding tissue.
4. Benign tumours can be removed surgically.	4. Malignant tumours need further treatment after removal.
5. Except for brain tumour, benign tumours are usually not fatal.	5. Malignant tumours are fatal.
6. Benign tumours do not show metastasis.	6. Malignant tumours show metastasis.
7. Benign tumours are well differentiated.	7. Malignant tumours are poorly differentiated.
8. Benign tumours show slow and progressive growth.	8. Malignant tumours show rapid and erratic growth.

III. The main five types of cancer:

Types of Cancer : According to the tissue affected, the cancers are classified into five main types. These are as follows:

1. Carcinoma : Cancer of epithelial tissue covering or lining the body organs is known as carcinoma. E.g. breast cancer, lung cancer, cancer of stomach, skin cancer, etc.
2. Sarcoma : Cancer of connective tissue is called sarcoma. Following are the types of sarcoma osteosarcoma (bone cancer), myosarcoma (muscle cancer),
   chondrosarcoma (cancer of cartilage) and liposarcoma (cancer of adipose tissue).
3. Lymphoma : Cancer of lymphatic tissue is called lymphoma. Lymphatic nodes, spleen and tissues of immune system are affected due to lymphoma.
4. Leukaemia : Leukaemia is blood cancer. In this condition, excessive formation of leucocytes take place in the bone marrow. There are millions of abnormal immature leucocytes which cannot fight infections. Monocytic leukaemia, lymphoblastic leukaemia, etc. are the types of leukaemia.
5. Adenocarcinoma : Cancer of glandular tissues such as thyroid, pituitary, adrenal, etc. is called adenocarcinoma.

Question 4.
Describe the different type of immunity.
Answer: There are two basic types of immunity, viz. innate immunity and acquired immunity.
(A) Innate immunity:

1. Innate immunity is natural, inborn immunity, which helps the body to fight against the invasion of microorganisms.
2. Innate immunity is non-specific because it does not depend on previous exposure to foreign substances.
3. Innate immunity mechanisms consist of various types of barriers such as anatomical barriers, physiological barriers, phagocytic barriers and inflammatory barriers. They prevent entry of foreign agents into the body.

(B) Acquired immunity:

1. The immunity that an individual acquires during his life is called acquired immunity or adaptive immunity or specific immunity. It helps the body to adapt by fighting against specific antigens hence it is called adaptive immunity. Since it is produced specifically against an antigen, it is called specific immunity.
2. Acquired immunity takes long time for its activation.
3. This type of immunity is seen only in vertebrates.
4. Due to acquired immunity, the body is able to defend against any invading foreign agent.

Question 5.
Describe the ill-effects of alcoholism on health.
Answer:

1. Alcohol in any form is toxic for the body. Hence as soon as alcohol is consumed, the liver tries to detoxify it.
2. In low doses it acts as a stimulant but in high dose, it acts on central nervous system, especially the cerebrum and cerebellum. Still higher dose can induce a comatose condition.
3. Alcohol affect the gastrointestinal tract by causing inflammation and damage to gastric 4 mucosa. Ulceration and painful condition arises in alcoholics.
4. Excessive doses of alcohol induce vomiting.
5. The worst effect of alcohol is on liver causing diseases like cirrhosis.
6. Alcohol induces hypertension and cardiac problems.
7. Apart from physical effect, it causes deterioration of mental health and emotional well-being.
8. Alcoholic person cannot think due to numbness in his/her cerebrum.
9. The social health is greatly affected as the alcoholic can cause problems to his family, friends and society in general.

Question 6.
In your view, what motivates the youngsters to take to alcohol or drugs and how can this be avoided?
Answer:
I. Taking drugs or alcohol:

1. Youngsters are at the vulnerable age, where they lack the planning about their future.
2. If they fall into bad company or are facing parental neglect, they get hooked on to alcohol or drugs.
3. Some common causes for addiction among youngsters are insufficient parental supervision and monitoring or excessive pressure and expectations from them. Lack of communication between an adolescent and parents.
4. Poorly defined rules for the family. Continuous family conflicts.
5. Favourable parental attitudes towards alcohol and drug uses. Many a times, at home children are exposed to such habits.
6. Inability to cope up with present and hence switching to the addictions. Risk taking behaviour which is common among youngsters.

II. Methods/measures to avoid drug abuse:

1. There should be complete acceptance for the child, because the adolescent phase is the most crucial phase when the children should be treated with love, care and respect.
2. Many physical, hormonal and psychological transformations are taking place in this phase. Therefore child suffers from stressful situation.
3. Wrong company and bad influence of peer group can trap the child in bad addictive habits. Thus, family should be supportive and communicative to help such children.
4. The sexual thoughts should be sublimed by channelizing energy into healthy pursuits like sports, reading, music, yoga and other extracurricular activities.
5. Ill-effects of drugs or alcohol should be told to youngsters.
6. Education and counselling can control the children from getting hooked on to the addictions.

Question 7.
Do you think that friends can influence one to take alcohol/drugs? If yes, how may one protect himself/herself from such an influence?
Answer:
Friends can influence one to take alcohol and drugs, if a boy or girl is timid and non-communicative with his or her parents and teachers. It also depends on the personality of the indtvidual. In the adolescent age, many fall in trap due to such peer pressure. The confusion in the mind and role of hormones playing on the psyche and thought process makes one unable to understand the hazards of such habits. Also there is curiosity to do these experimentations due to bad influence of media.

If there is complete trust and friendship with sensible parents, then such influence does not work. One should protect himself or herself by a strong denial. Communicating such incidents to an elder in whom a boy or girl can confide, is very important. One should tell his or her friends about the ill-effects of alcohol and drugs. He should be made aware of these aspects that he or she has learnt in this lesson.

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

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

1. Multiple choice questions

Question 1.
The nervous system of mammals uses both electrical and chemical means to send signals via neurons. Which part of the neuron receives impulse?
(a) Axon
(b) Dendron
(c) Nodes of Ranvier
(d) Neurilemma
Answer:
(b) Dendron

Question 2.
……………. is a neurotransmitter.
(a) ADH
(b) Acetyl CoA
(c) Acetyl choline
(d) Inositol
Answer:
(c) Acetyl choline

Question 3.
The supporting cells that produce myelin sheath in the PNS are …………….
(a) Oligodendrocytes
(b) Satellite cells
(c) Astrocytes
(d) Schwann cells
Answer:
(d) Schwann cells

Question 4.
A collection of neuron cell bodies located outside the CNS is called …………….
(a) tract
(b) nucleus
(c) nerve
(d) ganglion
Answer:
(d) ganglion

Question 5.
Receptors for protein hormones are located …………….
(a) in cytoplasm
(b) on cell surface
(c) in nucleus
(d) on Golgi complex
Answer:
(b) on cell surface

Question 6.
If parathyroid gland of man Eire removed, the specific result will be …………….
(a) onset of aging
(b) disturbance of Ca⁺⁺
(c) onset of myxoedema
(d) elevation of blood pressure
Answer:
(b) disturbance of Ca⁺⁺

Question 7.
Hormone thyroxine, adrenaline and non¬adrenaline are formed from ……………
(a) Glycine
(b) Arginine
(c) Ornithine
(d) Tyrosine
Answer:
(d) Tyrosine

Question 8.
Pheromones are chemical messengers produced by animals and released outside the body. The odour of these substance affects …………….
(a) skin colour
(b) excretion
(c) digestion
(d) behaviour
Answer:
(d) behaviour

Question 9.
Which one of the following is a set of discrete endocrine gland?
(a) Salivary glands, thyroid, adrenal, ovary
(b) Adrenal, testis, ovary, liver
(c) Pituitary, thyroid, adrenal, thymus
(d) Pituitary, pancreas, adrenal, thymus
Answer:
(c) Pituitary, thyroid, adrenal, thymus

Question 10.
After ovulation, Graafian follicle changes into …………….
(a) corpus luteum
(b) corpus albicans
(c) corpus spongiosum
(d) corpus callosum
Answer:
(a) corpus luteum

Question 11.
Which one of the following pairs correctly matches a hormone with a disease resulting from its deficiency?
(a) Parathyroid hormone – Diabetes insipidus
(b) Luteinising hormone – Diabetes mellitus
(c) Insulin – Hyperglycaemia
(d) Thyroxine – Tetany
Answer:
(c) Insulin – Hyperglycaemia

Question 12.
……………. is in direct contact of brain in humans.
(a) Cranium
(b) Dura mater
(c) Arachnoid
(d) Pia mater
Answer:
(d) Pia mater

2. Very very short answer questions.

Question 1.
What is the function of red nucleus?
Answer:
Red nucleus plays an important role in controlling posture and muscle tone, modifying some motor activities and motor coordination.

Question 2.
What is the importance of corpora quadrigemina?
Answer:
Corpora quadrigemina consists of 4 solid rounded structures, viz. superior and inferior colliculi. Superior colliculi control visual reflexes while inferior colliculi control auditory reflexes.

Question 3.
What does the cerebellum of brain control?
Answer:
Cerebellum of brain is an important centre which maintains equilibrium of body, posture, balancing orientation, moderation of voluntary movements and maintenance of muscle tone.

Question 4.
Name the three ear ossicles.
Answer:
Malleus [hammer], incus [anvil] and stapes [stirrup].

Question 5.
Name the anti abortion hormone.
Answer:
Progesterone.

Question 6.
Name an organ which acts as temporary endocrine gland.
Answer:
Placenta. Corpus luteum in ovary.

Question 7.
Name the type of hormones which bind to the DNA and alter the gene expression.
Answer:
Steroid hormones.

Question 8.
What is the cause of abnormal elongation of long bones of arms and legs and of lower jaw.
Answer:
Hypersecretion of growth hormones in adults causes abnormal elongation of long bones of arms and legs and of lower jaw i.e. acromegaly.

Question 9.
Name the hormone secreted by the pineal gland.
Answer:
Melatonin.

Question 10.
Which endocrine gland plays important, role in improving immunity?
Answer:
The endocrine gland, thymus plays an important role in improving immunity.

3. Match the organism with the type of nervous system found in them.

Column A	Column B
(1) Neurons	(a) Earthworm
(2) Ladder type	(b) Hydra
(3) Ganglion	(c) Flatworm
(4) Nerve net	(d) Human

Answer:

Column A	Column B
(1) Neurons	(d) Human
(2) Ladder type	(c) Flatworm
(3) Ganglion	(a) Earthworm
(4) Nerve net	(b) Hydra

4. Very short answer questions.

Question 1.
Describe the endocrine role of islets of Langerhans.
OR
Islets of Langerhans.
Answer:
Endocrine cells of pancreas form groups of cells called Islets of Langerhans. There are four kinds of cells in islets of Langerhans which secrete hormones.

1. Alpha (α) cells : They are 20% and secrete glucagon. Glucagon is a hyperglycemic hormone. It stimulates liver for glucogenolysis and increases the blood glucose level.
2. Beta (β) cells : They are 70% and secrete insulin. Insulin is a hypoglycemic hormone. It stimulates liver and muscles for glycogenesis. This lowers blood glucose level.
3. Delta (δ) cells : They are 5% and secrete somatostatin. Somatostatin inhibits the secretion of glucagon and insulin. It also decreases the gastric secretions, motility and absorption in digestive tract. In general it is a growth inhibiting factor.
4. PP cells or F cells : They form 5%. They secrete pancreatic polypeptide (PP) which inhibits the release of pancreatic juice.

Question 2.
Mention the function of testosterone?
Answer:
Testosterone is a steroid sex hormone secreted by testes and cortex of adrenal glands. It controls the secondary sexual characters in males.

Question 3.
Give symptoms of the disease caused by hyposecretion of ADH.
Answer:
Polydipsia, i.e. frequent thirst and polyuria, i.e. frequent urination are the symptoms of the disease caused by hyposecretion of ADH.

5. Short answer questions

Question 1.
Rakesh got hurt on his head when he fell down from his motorbike. Which inner membranes must have protected his brain? What other roles do they have to play
Answer:

1. When Rakesh fell down from his motorbike, the inner membranes that protected his brain were meninges, viz. dura mater, arachnoid membrane and pia mater. Morevover, CSF must have also acted as a shock absorber.
2. Dura mater : It is the outer tough membrane protective in function.
3. Arachnoid membrane : It is the middle web-like membrane which communicates with fluids of upper sub dural space and lower sub arachnoid space.
4. Pia mater : It is the innermost highly vascularised nutritive membrane in close contact with brain and spinal cord.

Question 2.
Injury to medulla oblongata may prove fatal.
OR
Injury to medulla oblongata causes sudden death. Explain.
Answer:

1. Medulla oblongata is the region of the brain that controls all the involuntary activities.
2. Vital activities such as heartbeats, respiration, vasomotor activities, peristalsis, etc. are under the control of medulla oblongata.
3. When medulla oblongata is injured, all these vital functions are instantly stopped.
4. Therefore, injury to medulla oblongata causes sudden death.

Question 3.
Distinguish between the sympathetic and parasympathetic nervous system on the basis of the effect they have on:
Heartbeat andUrinary Bladder.
Answer:

Sympathetic Nervous System	Parasympathetic	Nervous System
(1) Heartbeat	Increases	Decreases
(2) Urinary bladder	Relaxes and stores urine	Contracts causing micturition

Question 4.
While holding a tea cup Mr. Kothari’s hands rattle. Which disorder he may be suffering from and what is the reason for this?
Answer:

1. This condition is due to Parkinson’s disease.
2. It is due to degeneration of dopamine- producing neurons in the CNS.
3. 80% of the patients develop this condition along with stiffness, difficulty in walking, balance and coordination.

Question 5.
List the properties of the nerve fibres.
Answer:

1. Excitability / irritability
2. Conductivity
3. Stimulus
4. Summation
5. All or none
6. Refractory period
7. Synaptic delay
8. Synaptic fatigue
9. Velocity.

Question 6.
How does tongue detect the sensation of taste?
Answer:

1. The surface of tongue is with gustatoreceptors.
2. These receptors are sensitive to the chemicals [sweet, salt, sour, bitter and umami (savory)] present in the food.
3. The receptor cells get stimulated, generate the impulse which is given to the sensory neuron.

Question 7.
State the site of production and function of Secretin, Gastrin and Cholecystokinin.
Answer:

Hormone	Site of production	Functions
1. Secretin	Duodenal mucosa	Stimulates secretion of pancreatic juice and bile from pancreas and liver respectively.
2. Gastrin	Gastric mucosa	Stimulates gastric glands to secrete gastric juice.
3. Cholecystokinin	Duodenal mucosa	Stimulates pancreas and gall bladder to release pancreatic enzymes and bile respectively.

Question 8.
An adult patient suffers from low heart rate, low metabolic rate and low body temperature. He also lacks alertness, intelligence and initiative. What can be this disease? What can be its cause and cure ?
Answer:

1. The above symptoms indicate that the person is suffering from Myxoedema.
2. Myxoedema is condition caused due to hypothyroidism.
3. Hypothyroidism causes deficiency of thyroid hormones like T₃ and T₄ (thyroxine). This results in low BMR.
4. This condition can be cured by giving injections of thyroxine or tablets containing hormone preparation.

Question 9.
Where is the pituitary gland located? Enlist the hormones secreted by anterior pituitary.
Answer:
The pituitary gland is attached to hypothalamus on the ventral surface of brain. It is lodged in a bony depression called sella turcica of sphenoid bone.
For names of hormones:

1. GH : [Growth Hormone/STH : Somatotropic Hormone]
2. TSH/TTH – [Thyroid Stimulating Hormone/ Thyrotropic Hormone]
3. ACTH – [adrenocorticotropic hormone]
4. PRL – [prolactin]

Gonadotropins-

1. FSH [follicle stimulating hormone]
2. LH/ICSH – [leutinizing hormone/ interstitial cells stimulating hormone]

Question 10.
Explain how the adrenal medulla and sympathetic nervous system function as a closely integrated system.
Answer:

1. Adrenal medulla originates from embryonic neuro – ectoderm.
2. It consists of rounded group of large granular cells called chromaffin cells. They are modified post-ganglionic cells of sympathetic nervous system which have lost normal processes and acquired glandular function.
3. These cells are connected with pre-ganglionic fibres of sympathetic nervous system.
4. Hence adrenal medulla is an extension of sympathetic nervous system.
5. Thus adrenal medulla and sympathetic nervous system functions as a closely integrated system.

Question 11.
Name the secretion of alpha, beta and delta cells of islets of Langerhans. Explain their role.
Answer:

Pancreatic islet cells	Secretion	Functions
1. Alpha cells	Glucagon	Stimulates glycogenolysis in the liver
2. Beta cells	Insulin	Stimulates glycogenesis in the liver and muscles
3. Delta cells	Somatostatin	Inhibits the secretion of glucagon and insulin. It also decreases the gastric secretions, motility and absorption in digestive tract.

Question 12.
Which are the two types of goitre? What are their causes?
Answer:
(1) Goitre is the enlargement of thyroid gland. It is easily visible at the base of neck when a person is suffering from it.

(2) Goitre is of two types.

1. Simple goitre : It is also called endemic goitre. This is due to iodine deficiency in the food. This causes iodine deficit in blood. In an attempt to take more iodine from blood, the blood supply to the gland increases. This results in swelling on the thyroid.
2. Exophthalmic goitre : It is also called toxic goitre. This is due to hyperactive thyroid gland. This can happen if there is overstimulation of thyroid due to excess of ACTH. This disorder is also called Grave’s disease or hyperthyroidism.

Question 13.
Name the ovarian hormone and give their functions.
Answer:

Hormone	Functions
Oestrogen	It is responsible for secondary sexual characters in female.
Progesterone	Essential for thickening of uterine endometrium, thus preparing the uterus for implantation of fertilized ovum. It is responsible for development of mammary glands during pregnancy. It inhibits uterine contractions during pregnancy.
Relaxin	It relaxes the cervix of the pregnant female and ligaments of pelvic girdle during parturition.
Inhibin	It inhibits the FSH and GnRH production.

6. Answer the following.

Complete the table.

Location	Cell type	Function
PNS	————-	Produce myelin sheath.
PNS	Satellite cells	————-
————	Oligodendrocytes	Form myelin sheath around central axon.
CNS	————	Pathogens are destroyed by phagocytosis. (Phagocytose)
CNS	————	Form the epithelial lining of brain cavities and central canal.

Answer:

Location	Cell type	Function
PNS	Schwann cells	Produce myelin sheath.
PNS	Satellite cells	Supply nutrients to surrounding neurons, protect and cushion nearby neurons.
CNS	Oligodendrocytes	Form myelin sheath around central axon.
CNS	Microglia	Pathogens are destroyed by phagocytosis. (Phagocytose)
CNS	Ependyma	Form the epithelial lining of brain cavities and central canal.

7. Long answer questions.

Question 1.
Explain the process of conduction of nerve impulses up to development of action potential.
Answer:

1. The origin and maintenance of resting potential depends on the original state of no stimulation.
2. Any stimulus or disturbance to the membrane will make the membrane permeable to Na⁺ ions. This causes rapid influx of Na⁺ ions.
3. The voltage gated Na⁺/K⁺ channels are unique. They can change the potential difference of the membrane as per the stimulus received and also the gates operate separately and are self closing.
4. During resting potential, both gates are closed and resting potential is maintained.
5. However during depolarization, the Na⁺ channels open but not the K⁺ channels. This causes Na⁺ to rush into the axon and bring about a depolarisation. This condition is called action potential.
6. Extra cellular fluid (ECF) becomes electronegative with respect to the inner membrane which becomes electropositive.

Question 2.
Draw the neat labelled diagrams.
a. Human ear.
Answer:

b. Sectional view of human eye.
Answer:

c. Draw the neat labelled diagram of sagittal section or L.S. of human brain
Answer:

d. Draw the neat labelled diagram of Multipolar Neuron.
Answer:

Question 3.
Answer the questions after observing the diagram given below.

a. What do the synaptic vesicles contain?
Answer:
Synaptic vesicles contain a neurotransmitter – acetyl choline.

b. What process is used to release the neurotransmitter ?
Answer:
Exocytosis.

c. What should be the reason for the next impulse to be conducted?
Answer:
Removal of neurotransmitter by the action of acetyl cholinesterase.

d. Will the impulse be carried by post synaptic membrane even if one pre-synaptic neuron is there?
Answer:
As far as impulse is transmitted by pre-synaptic neuron, it will be received by post-synaptic neuron.

e. Can you name the channel responsible for their transmission?
Answer:
Ca⁺⁺ channel

Question 4.
Explain the Reflex Pathway with the help of a neat labelled diagram.
OR
With the help of a neat and labelled diagram, describe reflex arc.
Answer:

I. Reflex action : Reflex action Is defined as a quick, automatic involuntary and often unconscious action brought about when the receptors are stimulated by external or internal stimuli.

II. Reflex arc : Reflex actions are controlled by CNS. Reflex arc is the structural or functional unit of reflex action. Simple reflex arc is formed of the following five components.
(1) Receptor organ : The sensory part that receives the stimulus is called receptor organ. It can be any sense organ that receives the stimulus and converts it into the impulse, e.g. skin, eye, ear, tongue, nasal epithelium, etc.

(2) Sensory neuron or afferent neuron:
Sensory part carrying impulse from receptor organ to CNS is called sensory neuron. Its cyton is located in dorsal root ganglion. Its dendron is long and connected to receptor while the axon enters in the grey matter of spinal cord to form a synapse.

(3) Association, adjustor or intermediate neuron : It is present in the grey matter of spinal cord. Receiving impulse from sensory neuron, interpreting it and generating motor impulse are done by association neuron.
(4) Motor neuron (effector) : The cyton of motor neuron is present in the ventral horn of grey matter and axon travels through ventral root. It conducts motor impulse from spinal cord to effector organ.

(5) Effector organ : Effector organ is a specialized part of the body which is excited by receiving the motor impulse. It gives proper response to the stimulus, e.g. muscles or glands. The path of reflex action is followed by the unidirectional impulse. It originates in the receptor organ and ends in effector organ through CNS.

Question 5.
Krishna was going to school and on the way he saw a major bus accident. His heartbeat increased and hands and feet become cold. Name the part of the nervous system that had a role to play in this reaction.
Answer:

1. The symptoms observed in Krishna were due to sympathetic nervous system. Emergency conditions trigger sympathetic nervous system to stimulate adrenal medulla.
2. The cells of adrenal medulla secrete catecholamines like adrenaline and nor¬adrenaline.
3. These hormones have direct effect on the pacemaker of the heart which causes increase in the heart rate and other associated symptoms.
4. This is a typical fright reaction caused by intervention of sympathetic nervous system.

Question 6.
What will be the effect of thyroid gland atrophy on the human body?
Answer:

1. Atrophy means degeneration. Atrophy of thyroid gland will result in deficient secretion of thyroid hormones leading to hypothyroidism. Deficiency of thyroid hormones [T₃ and T₄] and thyrocalcitonin will cause following effects on the body.
2. Decrease in BMR i.e. basal metabolic rate, decrease in the blood pressure, heart beat, body temperature, etc.
3. Occurrence of myxoedema in which there is abnormal deposition of fats under the skin giving puffy appearance in adults.
4. Irregularities in menstrual cycle in case of female patients.
5. Hair become brittle and fall.
6. Calcium metabolism also disturbs due to lack of thyrocalcitonin.

Question 7.
Write the names of hormones and the glands secreting them for the regulation of following functions
(a) Growth of thyroid and secretion of thyroxine.
Answer:
TSH by adenohypophysis.

(b) Helps in relaxing pubic ligaments to facilitate easy birth of young ones.
Answer:
Relaxin by degenerating corpus luteum of the ovary.

(c) Stimulate intestinal glands to secrete intestinal juice.
Answer:
Secretin by duodenal mucosa.

(d) Controls calcium level in the blood.
Answer:
Calcitonin [hypocalcemic hormone] by thyroid and parathormone [ hypercalcemic hormone] by parathyroid glands.

(e) Controls tubular absorption of water in kidneys.
Answer:
ADH by hypothalamus.

(f) Urinary elimination of water.
Answer:
Atrial natriuretic factor by atria of heart.

(g) Sodium and potassium ion metabolism.
Answer:
Aldosterone by adrenal cortex.

(h) Basal Metabolic rate.
Answer:
T₃ and T₄ by thyroid gland.

(I) Uterine contraction.
Answer:
Oxytocin by hypothalamus.

(j) Heartbeat and blood pressure.
Answer:
Adrenaline, non-adrenaline [stimulation] and acetylcholine [inhibition] by adrenal medulla.

(k) Secretion of growth hormone.
Answer:
GHRF by hypothalamus.

(l) Maturation of Graafian follicle.
Answer:
FSH by anterior pituitary.

Question 8.
Explain the role of hypothalamus and pituitary as a coordinated unit in maintaining homeostasis.
Answer:

1. Homeostasis is maintenance of constant internal environment of the body.
2. When certain hormones from any endocrine glands are secreted in excess quantity, the : inhibiting factors from hypothalamus, automatically exert negative feedback and stop the production of stimulating hormones from pituitary.
3. Similarly, if any hormone is in deficit, then j the concerned gland is given message through releasing factor. This way the hormone production remains in a balanced state or homeostasis.
4. E.g. If thyroxine from thyroid gland is secreted in excess, the secretion of TSH from pituitary is stopped by stopping the production of TRF from hypothalamus.
5. Though most of the endocrine glands are under the influence of pituitary gland, it is in turn controlled by hypothalamus.
6. Hypothalamus secretes releasing factors and inhibiting factors and hence regulate the secretions of pituitary (hypophysis).
7. There is negative feedback mechanism in controlling the secretions of the endocrine glands.
8. Hypothalamus forms the hypothalamo- hypophyseal axis through which transportation of neurohormones take place.

Following are the releasing and inhibiting factors produced by hypothalamus:

1. Somatotropin/GHRF : It stimulates release of growth hormone.
2. Somatostatin/GHRIF : It inhibits the release of growth hormone.
3. Adrenocorticotropin Releasing Hormone / CRF : It stimulates the release of ACTH by the anterior pituitary gland.
4. Thyrotropin Releasing Hormone /TRF : It stimulates the release of TSH by anterior pituitary gland.
5. Gonadotropin Releasing Hormone (GnRH) : It stimulates pituitary to secrete gonadotropins.
6. Prolactin Inhibiting Hormone (Prolactostatin) : It inhibits prolactin released by anterior pituitary gland.
7. Gastrin Releasing Peptide (GRP).
8. Gastric Inhibitory Polypeptide (GIP).

Question 9.
What is adenohypophysis ? Name the hormones secreted by it.
Answer:

1. Adenohypophysis is the large anterior lobe of pituitary gland.
2. It is derived from embryonic ectoderm in the form of Rathke’s pouch which is a small outgrowth from the roof of embryonic stomodaeum.
3. It is made up of epitheloid secretory cells.

It secretes following hormones:

1. GH : [Growth Hormone/STH : Somatotropic Hormone]
2. TSH/TTH – [Thyroid Stimulating Hormone/ Thyrotropic Hormone]
3. ACTH – [adrenocorticotropic hormone]
4. PRL – [prolactin]

Gonadotropins-

1. FSH [follicle stimulating hormone]
2. LH/ICSH – [leutinizing hormone/ interstitial cells stimulating hormone]

Question 10.
Describe, in brief, an account of disorders of adrenal gland.
Answer:
(1) Disorders of adrenal cortical secretions are caused due to hyposecretion and hypersecretion of adrenal corcoid hormones.

(2) Hyposecretion of corticosteroids causes Addison’s disease.

(3) The symptoms of Addison’s disease are low blood sugar, low body temperature, feeble heart action, low BR acidosis, low Na⁺ and K⁺ concentration in plasma, excessive loss of Na⁺ and water in urine, impaired kidney functioning and kidney failure, etc. it leads to weight loss, general weakness, nausea, vomiting and diarrhoea.

(4) Hypersecretion of corticoids causes Cushing’s disease.

(5) The symptoms of Cushing’s disease are high blood sugar level, glucosuria, alkalosis, enhancement of total quantity of electrolytes in extracellular fluid, polydipsia, increased BR muscle paralysis, obesity, wasting of limb muscles, etc.

Question 11.
Explain action of steroid hormones and proteinous hormones.
OR
Explain the mode of action of steroid hormones.
Answer:
The hormones always act on their target organs or tissues to induce their effects. The target tissues have specific binding sites or receptor sites which contain hormone receptors.
I. Steroid hormones:

1. The steroid hormones are lipid soluble and can easily cross the lipoproteinous plasma membrane.
2. The hormone receptors for steroid hormones are present in cytoplasm or in nucleus.
3. Hormone-receptor complex formed in cytoplasm enters the nucleus and regulate the gene expression or chromosome function.
4. In some cases the receptors are present inside the nucleus where hormone receptor complex is formed.
5. These complexes interact with the genome to evoke biochemical changes that result in physiological and developmental functions.

II. Protein hormones:

1. The hormone receptors for protein hormones are present on the cell membrane (i.e. membrane bound receptors).
2. When the hormone binds to its receptor, it forms hormone-receptor complex. Each receptor is specific to a specific hormone.
3. The hormones which interact with membrane bound receptors normally do not enter the target cell but generate second messengers. Such as cyclic AMP Ca⁺⁺ or IP (Inositol triphosphate), etc.
4. This leads to certain biochemical changes : in the target tissue.
5. Thus, the tissue metabolism and consequently the physiological functions are regulated by hormones.

Question 12.
Describe in brief an account of disorders of the thyroid.
OR
What are the functional disorders of thyroid gland? Describe in brief.
Answer:
Disorders of thyroid gland are of three types, viz. hypothyroidism, hyperthyroidism and simple goitre.
(1) Hypothyroidism : Hypothyroidism is deficient secretion of thyroxine. This hyposecretion causes two types of disorders, viz. cretinism in children and myoxedema in adults.
(i) Cretinism : Hyposecretion of thyroxine in childhood causes cretinism. The symptoms of cretinism are retardation of physical and mental growth.

(ii) Myxoedema : Deficiency of thyroxine in adults causes this disorder. It is also referred to as Gull’s disease. Symptoms are thickening and puffiness of the skin and subcutaneous tissue particularly of face and extremities. Patients with low BMR. It also causes mental dullness, loss of memory, slow action.

(2) Hyperthyroidism : Excessive secretion of thyroxine causes exophthalmic goitre or Grave’s disease. There is slight enlargement of thyroid gland. It increases BMR, heart rate, pulse rate and BE Reduction in body weight due to rapid oxidation, nervousness, irritability. Peculiar symptom is exophthalmos, i.e. bulging of eyeballs with staring look and less blinking. This is caused by deposition of fats behind the eye balls in eye sockets. There is muscular weakness and loss of weight.

(3) Simple goitre (Iodine deficiency goitre) : Simple goitre occurs due to deficiency of iodine in diet or drinking water. Simple goitre causes enlargement of thyroid gland. Thyroid gland in an attempt to get more iodine from the blood, swells due to increased blood supply. Prevention of goitre can be done by administering iodized table salt. It is also called endemic goitre as it is common in hilly areas.

Balbharti Maharashtra State Board 12th Biology Textbook Solutions Chapter 8 Respiration and Circulation Textbook Exercise Questions and Answers.

Maharashtra State Board 12th Biology Solutions Chapter 8 Respiration and Circulation

1. Multiple choice questions

Question 1.
The muscular structure that separates the thoracic and abdominal cavity is …………………..
(a) pleura
(b) diaphragm
(c) trachea
(d) epithelium
Answer:
(b) diaphragm

Question 2.
What is the minimum number of plasma membrane that oxygen has to diffuse across to pass from air in the alveolus to haemoglobin inside a R.B.C.?
(a) two
(b) three
(c) four
(d) five
Answer:
(a) two

Question 3.
…………………. is a sound producing organ.
(a) Larynx
(b) Pharynx
(c) Tonsils
(d) Trachea
Answer:
(a) Larynx

Question 4.
The maximum volume of gas that is inhaled during breathing in addition to T.V. is …………………..
(a) residual volume
(b) IRV
(c) GRV.
(d) vital capacity
Answer:
(b) IRV

Question 5.
………………….. muscles contract when the external intercostals muscles contract.
(a) Internal abdominal
(b) Jaw
(c) Muscles in bronchial walls
(d) Diaphragm
Answer:
(d) Diaphragm

Question 6.
Movement of cytoplasm in unicellular organisms is called …………………..
(a) diffusion
(b) cyclosis
(c) circulation
(d) thrombosis
Answer:
(b) cyclosis

Question 7.
Which of the following animals do not have closed circulation?
(a) Earthworm
(b) Rabbit
(c) Butterfly
(d) Shark
Answer:
(c) Butterfly

Question 8.
Diapedesis is performed by …………………..
(a) erythrocytes
(b) thrombocytes
(c) adipocytes
(d) leucocytes
Answer:
(d) leucocytes

Question 9.
Pacemaker of heart is …………………..
(a) SA node
(b) AV node
(c) His bundle
(d) Purkinje fibers
Answer:
(a) SA node

Question 10.
Which of the following is without nucleus?
(a) Red blood corpuscle
(b) Neutrophil
(c) Basophil
(d) Lymphocyte
Answer:
(a) Red blood corpuscle

Question 11.
Cockroach shows which kind of circulatory system?
(a) Open
(b) Closed
(c) Lymphatic
(d) Double
Answer:
(a) Open

Question 12.
Diapedesis can be seen in …………………..
(a) RBC
(b) WBC
(c) Platelet
(d) neuron
Answer:
(b) WBC

Question 13.
Opening of inferior vena cava is guarded by …………………..
(a) bicuspid valve
(b) tricuspid valve
(c) Eustachian valve
(d) Thebesian valve
Answer:
(c) Eustachian valve

Question 14.
…………………. wave in ECG represent atrial depolarization.
(a) P
(b) QRS complex
(c) Q
(d) T
Answer:
(a) P

Question 15.
The fluid seen in the intercellular spaces in Human is …………………..
(a) blood
(b) lymph
(c) interstitial fluid
(d) water
Answer:
(b) lymph

2. Match the columns

Question 1.
Respiratory surface Organism

Respiratory surface	Organism
(1) Plasma membrane	(a) Insect
(2) Lungs	(b) Salamander
(3) External gills	(c) Bird
(4) Internal gills	(d) Amoeba
(5) Trachea	(e) Fish

Answer:

Respiratory surface	Organism
(1) Plasma membrane	(d) Amoeba
(2) Lungs	(c) Bird
(3) External gills	(b) Salamander
(4) Internal gills	(e) Fish
(5) Trachea	(a) Insect

3. Very Short Answer Questions

Question 1.
Why does trachea have ‘C’-shaped rings of cartilage?
Answer:
Trachea is supported by ‘C’-shaped rings of J cartilage which prevent it from collapsing and always keep it open.

Question 2.
Why is respiration in insect called direct respiration?
Answer:
Respiration in insect is called direct because tracheal tubes exchange O₂ and CO₂ directly with the haemocoel which then exchange them with tissues.

Question 3.
Why is gas exchange very rapid at alveolar level?
OR
Why does gas exchange in the alveolar region very rapid?
Answer:
Gas exchange is very rapid at alveolar level because numerous alveoli (about 700 millions) in the lungs provide large surface area for gaseous exchange.

Question 4.
Name the organ which prevents the entry of food into the trachea while eating.
Answer:
Epiglottis prevents the entry of food into trachea while eating.

4. Short Answer Questions

Question 1.
Why is it advantageous to breathe through the nose than through the mouth?
Answer:
Breathing through nose is better than breathing through the mouth because of the following reasons:

1. The nostrils are smaller than the mouth so air exhaled through the nose creates a backflow of air into the lungs.
2. As we exhale more slowly through the nose than we do through the mouth, the lungs have more time to extract oxygen from the air that we have already taken in.
3. The hairs inside nostrils filter any dust particles and microbes in the air and it only lets the clean air pass through.
4. The air gets warm and humidified in nostrils as it passes into our bodies.
5. Moreover breathing through the mouth can dry the oral cavity and lead to bad breath, gum disease and tooth decay.

Question 2.
Identity the incorrect statement and correct it.
(a) A respiratory surface area should have a. large surface area.
(b) A respiratory surface area should be kept dry.
(c) A respiratory surface area should be thin, may be 1 mm or less.
Answer:
Statement (a) and statement (c) are correct whereas statement (b) is incorrect. A respiratory surface area should be kept moist, is the correct statement.

Question 3.
Given below are the characteristics of some modified respiratory movement. Identify them.
a. Spasmodic contraction of muscles of expiration and forceful expulsion of air through nose and mouth.
Answer:
Sneezing

b. An inspiration followed by many short convulsive expiration accompanied by facial expression.
Answer:
Laughing, Crying.

Question 4.
Blood plasma.
Answer:

1. Plasma is a straw coloured, slightly alkaline viscous fluid part of the blood, having 90-92% water and 8-10% soluble proteins.
2. Serum albumin, serum globulin, heparin, fibrinogen and prothrombin are the plasma proteins which form 7% of the plasma.
3. Glucose, amino acids, fatty acids and glycerol are the nutrients dissolved in plasma.
4. Nitrogenous wastes (urea, uric acid, . ammonia and creatinine) and respiratory gases (oxygen and carbon dioxide) is present in plasma.
5. Enzymes and hormones too are transported Ada plasma.
6. Inorganic minerals are also present in plasma such as bicarbonates, chlorides, phosphates and sulphates of sodium, potassium, calcium and magnesium.

Question 5.
Blood clotting/Coagulation of blood.
OR
Explain blood clotting in short.
Answer:

1. The process of converting the liquid blood into a semisolid form is called blood clotting or coagulation.
2. The process of clotting may be initiated by contact of blood with any foreign surface (intrinsic process) or with damaged tissue (extrinsic process).
3. Intrinsic and extrinsic processes involve interaction of various substances called clotting factors by a step wise or cascade mechanism.
4. There are in all twelve clotting factors numbered as I to XII (factor VI is not in active use).
5. Interaction of these factors in a cascade manner leads to formation of enzyme, Thromboplastin which helps in the formation of enzyme prothrombinase.
6. Prothrombinase inactivates heparin and also converts inactive prothrombin into active thrombin.
7. Thrombin converts soluble blood protein- fibrinogen into insoluble fibrin. Fibrin forms a mesh in which platelets and other blood cells are trapped to form the clot.
8. These reactions occur in 2 to 8 minutes. Therefore, clotting time is said to be 2 to 8 minutes.

Question 6.
Describe pericardium.
Answer:

1. Pericardium is the double layered peritoneum that encloses the heart. It consists of two layers, viz. fibrous pericardium and serous pericardium.
2. Fibrous pericardium is the outer layer having tough, inelastic fibrous connective tissue whereas serous pericardium is the v inner double layered membrane. It has in turn an outer parietal layer and inner visceral layer.
3. Parietal layer of serous pericardium lies on the inner side of fibrous pericardium.
4. Visceral layer also known as epicardium adheres to heart and thus forms outer covering over the heart.
5. There is a pericardial fluid in the pericardial space which is present in between the parietal and visceral layers of serous pericardium.

Question 7.
Describe valves in the human heart.
Answer:
Human heart has following main valves:

1. Tricuspid valve : Tricuspid valve is present between the right atrium and right ventricle. It has three cusps or flaps. It prevents the backflow of blood into right atrium.
2. Bicuspid valve : Bicuspid valve, also called mitral valve is present between the left atrium and left ventricle. It has two flaps. It prevents the backflow of blood in left atrium. Both tricuspid and bicuspid valves are attached to papillary muscles with tendinous chords or chordate tendinae to prevent valves from turning back into atria at the time of systole.
3. Semilunar valve : These are present at the opening of pulmonary artery and systemic aorta. They prevent the back flow of blood when ventricles undergo systole.
4. Thebesian valve : Thebesian valve is present at the opening of coronary sinus.
5. Eustachian valve : Eustachian valve is present at the opening of inferior vena cava.

Question 8.
What is the role of papillary muscles and chordae tendinae in human heart?
Answer:

1. Papillary muscles are large and well- developed muscular ridges present along the inner surface of the ventricles.
2. Bicuspid and tricuspid valves are attached to papillary muscles of ventricles by chordae tendinae.
3. Chordae tendinae are inelastic fibres present in the lumen of ventricles.
4. The chordae tendinae prevent the valves from turning back into the atria during the contraction of ventricles and regulate the opening and closing of bicuspid and tricuspid valves.

Question 9.
Explain in brief the factors affecting blood pressure.
Answer:

1. Cardiac output : Normal cardiac output is 5 lit/min. Increase in cardiac output increases systolic pressure.
2. Peripheral resistance : Peripheral resistance depends upon the diameter of blood vessels. Decrease in diameter of arterioles and capillaries under the effect of vasopressin cause increase in peripheral resistance and thereby increase in blood pressure.
3. Blood volume : Loss of blood in accidents decreases blood volume and thus cause decrease in blood pressure.
4. Viscosity of blood : Blood pressure is directly proportional to viscosity of blood.
5. Age : Blood pressure increases with age due to increase in inelasticity of blood vessels.
6. Venous return : Amount of blood brought to the heart via the veins per unit time is called the venous return and it is directly proportional to blood pressure.
7. Length and diameter of blood vessels : Blood pressure is directly proportional to the total length of the blood vessel. Blood pressure can also be affected by vasoconstriction or vasodilation.
8. Gender : Females have slightly lower BP than males of her age before menopause. However, the risk of high B. P increases in the females after menopause sets in.

5. Give Scientific Reason

Question 1.
Closed circulation is more efficient than open circulation.
Answer:

1. Closed circulation considerably enhances the speed, precision and efficiency of circulation.
2. The blood flows more rapidly, it takes less time to circulate through the closed system and return to the heart.
3. This fastens the supply and removed of materials to and from the tissues by the blood as compared to open circulation.
4. In open circulation, there are no blood vessels such as arteries or veins, to pump the blood.
5. Therefore, the blood pressure is very low.
6. Organisms with an open circulatory system typically have a relatively high volume of hemolymph and low blood pressure. Closed circulation is thus more efficient than open circulation.

Question 2.
Human heart is called as myogenic and autorhythmic?
Answer:

1. The heart shows auto rhythmicity because the impulse for its rhythmic movement develops inside the heart. Such heart is called myogenic.
2. Some of the cardiac muscle fibres become auto rhythmic (self-excitable) and start generating impulse during development.
3. These autorhythmic fibres perform two important function, viz. acting as a pacemaker and setting the rhythm for heart.
4. They also form conducting system for conduction of nerve impulses throughout the heart muscles.

Question 3.
In human heart, the blood flows only in one direction.
Answer:

1. In veins there are valves, which prevent the back flow of the blood.
2. In arteries, blood flows with unidirectional pressure.
3. Hence the circulation takes place only in one direction.

Question 4.
Arteries are thicker than veins.
Answer:

1. Arteries have relatively thick walls to enable them to withstand the high pressure of blood ejected from the heart.
2. Arteries expand when the pressure increases as the heart pushes blood out but then recoil (shrink) Wn the pressure decreases when the heart relaxes between heartbeats.
3. This expansion and recoiling occurs to maintain a smooth blood flow.
4. Veins, on the other hand, have thinner walls and larger lumen veins have no need for thick walls as then need not have to withstand high pressure like arteries.
5. Moreover, as veins transport relatively low pressure blood, they are commonly equipped with valves to promote the unidirectional flow of blood towards the heart.

Question 5.
Left ventricle is thick than all other chambers of heart.
OR
Left ventricle has thicker wall than the right ventricle.
Answer:

1. Left ventricle pumps oxygenated blood to all parts of the body. Therefore, there is greater pressure from the blood in left ventricle.
2. Right ventricle sends deoxygenated blood to lungs for oxygenation. This does not put more pressure and lungs are in vicinity of the heart.
3. Due to these functional differences between the two ventricles, left ventricle has thicker wall than that of the right ventricle.

6. Distinguish Between

Question 1.
Open circulation and Closed circulation
Answer:

Open circulation	Closed circulation
1. In open circulation, blood flows through large open spaces and channels called lacunae and haemocoels among the tissues.	1. In closed circulation, blood flows through a network of blood vessels all over the body.
2. Tissues are in direct contact with the blood.	2. Blood does not come in direct contact with tissue.
3. Blood flows with low pressure and usually does not contain any respiratory pigment like haemoglobin.	3. Blood flows with high pressure and contains respiratory pigment like haemoglobin.
4. Exchange of material takes place directly between blood and cells or tissues of the body.	4. Exchange of material takes place between blood and body tissues through an intermediate fluid called lymph.
5. Volume of blood flowing through a tissue cannot be controlled as blood flows out in open space.	5. Volume of blood can be regulated by the contraction and relaxation of the smooth muscles of the blood vessels.
6. Open circulatory system is found in arthropods and some molluscs.	6. Closed circulatory system is found in annelids, echinoderms and all vertebrates.

Question 2.
Arteries and veins.
Answer:

Arteries	Veins
1. The blood vessels that arise from the heart and carry blood away from heart are called arteries.	1. The blood vessels that bring blood to the heart are called veins.
2. Arteries are thick walled blood vessels, situated in deep layers in the body.	2. Veins cure thin walled blood vessels, situated superficially in the body.
3. Arteries do not have valves.	3. Veins have valves.
4. Tunica adventitia, the outermost layer of arteries is thick and elastic.	4. Tunica externa, the outermost layer of veins is thin.
5. Tunica media is very thick and contain elastic fibres.	5. Tunica media is thin layer and contain involuntary muscle fibres.
6. The lumen of arteries is small.	6. The lumen of the veins is very spacious.
7. With the exception of pulmonary arteries, all other arteries carry oxygenated blood.	7. With the exception of pulmonary veins, all other veins carry deoxygenated blood.
8. Blood in the arteries show high blood pressure.	8. Blood in the veins show lesser blood pressure.

Question 3.
Blood and Lymph.
Answer:

Blood	Lymph
1. Contains blood plasma with proteins and all three types of blood cells namely RBCs, WBCs and blood platelets.	1. Contains blood plasma without blood proteins, RBCs and platelets and contains lymphocytes.
2. Red in colour due to presence of RBCs.	2. Light yellow in colour and does not contain RBCs.
3. Carries oxygen in the body.	3. Does not carry oxygen.
4. The flow of blood in blood vessels is fast.	4. The flow of lymph in lymph capillaries is slow.
5. Lymphocytes are present.	5. Lymphocytes are present, more in number than those present in the blood.

Question 4.
Blood capillary and Lymph capillary.
Answer:

Blood capillary	Lymph capillary
1. Reddish, easy to observe.	1. Colourless, difficult to observe.
2. Joined to arterioles at one end and to venules at another end.	2. Blind (closed at the tip).
3. Narrower than lymph capillaries.	3. Wider than blood capillaries.
4. Wall consists of normal endothelium and basement membrane.	4. Wall consists of thin endothelium and poorly developed basement membrane.
5. Contains red blood.	5. Contains colourless lymph.
6. Have relatively high pressure.	6. Have relatively low pressure.

Question 5.
Intrinsic and Extrinsic process of clotting.
Answer:

Intrinsic process	Extrinsic process
1. The intrinsic pathway requires only clotting factors found within the blood itself – in particular, clotting factor XII (Hageman factor) from the platelets.	1. The extrinsic pathway is initiated by factors external to the blood, in the tissues adjacent to damaged blood vessel – in particular, it is initiated by clotting factor III, thromoboplastin from the damaged tissues.
2. It is a longer, multistep process and it takes a little longer for the blood to clot by this mechanism.	2. It involves fewer chemical reaction steps and produce a clot a little more quickly than the intrinsic pathway.

7. Long Answer Questions

Question 1.
Smita was working in a garage with the doors closed and automobiles engine running. After some time she felt breathless and fainted. What would be the reason? How can she be treated
OR
While working with the car engine in a closed garage, John suddenly felt dizzy and fainted what is the possible reason?
Answer:

1. As Smita and John were working with the car engine running in a closed garage, they must be suffering from carbon monoxide poisoning.
2. Carbon monoxide (CO) is a highly toxic gas produced when fuels burn incompletely from automobile engines.
3. Because of strong affinity of haemoglobin with carbon monoxide, it readily combines with carbon monoxide to from a stable compound, carboxyhaemoglobin. Thus, less haemoglobin is available for oxygen transport depriving the cells of oxygen.
4. Exposure to carbon monoxide can usually leads to throbbing headache, drowsiness, breathlessness and often person gets fainted. In extreme cases carbon monoxide poisoning usually leads to unconsciousness, convulsions, cardiovascular failure, coma and eventually death.

The breathless persons can be treated by following method:

1. Oxygen treatment : The best way to treat carbon monoxide poisoning is to breathe in pure oxygen (high-dose oxygen treatment)
2. Oxygen chamber : Doctor may temporarily place her in a pressurized oxygen chamber (also known as a hyperbaric oxygen chamber)

Question 2.
Shreyas went to a garden on a wintry morning. When he came back, he found it difficult to breath and stated wheezing. What could be the possible condition and how can he be treated?
Answer:
(1) It indicates that Shreyas might be suffering from allergic reactions. He may have come in contact with allergens such as pollen, dust, pet dander or other environmental substances on his way in the garden. Or Shreyas may be already a patient of Asthma and his symptoms may have aggrevated due to wintry climate.

(2) If a person is allergic to a substance, such as pollen, his immune system reacts to the substance as if it was foreign and harmful, and tries to destroy it.

(3) The body reacts to these allergens by making and releasing substances known as IgE antibodies. These IgE antibodies attach to most cells in the body which release histamine. Histamine is the main substance responsible for pollen allergy symptoms such as difficulty in breathing, wheezing, sneezing, itchy throat, etc.

(4) Treatment : There are several drugs to treat the allergic reactions:

• Antihistamines such as cetirizine or diphenhydramine.
• Decongestants, such as pseudoephedrine or oxymetazoline.
• Medications that combine an antihistamine and decongestant such as Actifed and Claritin-D.

Question 3.
Why can you feel a pulse when you keep a finger on the wrist or neck but not when you keep them on a vein?
Answer:
(1) When the heart contracts, it creates pressure that pushes blood out of heart. This pressure acts like a wave. This “wave” of pressure is the pulse you feel. But this pressure is not constant.

(2) When the heart pumps the blood out of it at the time of systole, there is maximum pressure in the arteries. This pressure weakens considerably when it reaches capillaries, and so the veins which are away from the heart are under less pressure. Due to low pressure veins have valves to prevent backflow of blood.

(3) The pressure in the arteries can be felt every time the heart beats, especially in arteries which come to surface of the body like that of the wrist and neck but not in veins.

(4) The pressure in veins is always weaker than in arteries, resulting in a weaker pulse to the point that it is undetectable by touch
alone.

(5) Owing to this, when we keep finger on the arteries of wrist or neck, we feel a pulse but not when we keep it on a vein.

Question 4.
A man’s pulse rate is 68 and cardiac output is 5500 cm3. Find the stroke volume.
Answer:
Cardiac output is the volume of blood pumped out per min for a normal adult human being it is calculated as follows:
Cardiac output = Heart rate × Stroke volume
Given : Cardiac output = 5500 cm³
Pulse rate = Heart rate = 68
By using these values stroke volume of is calculated as follows:
∴ Cardiac output = Heart rate × Stroke volume
∴ Stroke volume = Cardiac output/Heart rate
= 5500/68
= Approx. 80. ∴ Stroke volume is 80 ml.

Question 5.
Which blood vessel leaving from the heart will have the maximum content of oxygen and why?
Answer:

1. The Aorta leaving the heart from left ventricle carry the maximum content of oxygen.
2. Deoxygenated blood becomes oxygenated in the pulmonary capillaries surrounding the alveoli of lungs. The oxygenated blood from lungs is collected by the four pulmonary veins.
3. These pulmonary veins carry that oxygenated blood to left atrium of heart. During atrial systole that blood is carried to left ventricle.
4. Left ventricle then pumps that oxygenated blood to Aorta during ventricular systole. Therefore, aorta has the maximum content of oxygen.

Question 6.
If the duration of the atrial ‘systole is 0.1 second and that of complete diastole is 0.4 second, then how does one cardiac cycle complete in 0.8 second?
Answer:

1. The time duration required to complete one cardiac cycle is 0.8 second.
2. Cardiac cycle is divided into three important phases, viz, atrial systole, ventricular systole and joint diastole.
3. Atrial systole in normal condition lasts for 0.1 second, ventricular systole follows atrial systole and lasts for 0.3 second whereas joint diastole or complete diastole lasts for about 0.4 second.
4. In this way one cardiac cycle is completed in 0.8 second.

Question 7.
How is blood kept moving in the large veins of the legs?
Answer:
1. When heart undergoes systole, it pushes the blood with pressure in aorta. This pressure moves the entire circulation of the blood throughout the body. Aorta gives rise to dorsal aorta after supplying to upper parts of body. Then it divides into two arteries which enter two legs. The blood is forced to move in the legs due to blood pressure and also aided by gravity.

2. In addition, the muscles in legs help transport blood back to our heart. As the muscles of our body contract and relax to move our limbs, they squeeze the blood in veins and the blood is then pushed towards the heart.

3. The veins in legs also have valves to keep this process going and prevent blood from flowing back down towards the feet.

4. In this way blood is kept moving in the large veins of the legs.

Question 8.
Describe histological structure of artery, vein and capillary.

Answer:
Histological structure of artery and vein.

1. Artery is a thick walled blood vessel that carries oxygenated blood. (Exception is pulmonary artery which carries deoxygenated blood from heart to lungs for oxygenation.)
2. All the arteries arise from heart and carry blood away from the heart.
3. Each artery is made up of three layers, viz. tunica externa, tunica media and tunica interna.
4. Tunica externa or adventitia is the thickest layer of all. It is the outermost coat made up of connective tissue with elastic and collagen fibres.
5. Tunica media is the middle coat made up of smooth muscle fibres and elastic fibres. It withstands high blood pressure during ventricular systole. It is also thick.
6. Tunica interna or intima is the innermost coat made of endothelium and elastic layer.

Histology of Capillaries:

1. Capillaries are the smallest and thinnest blood vessels. Capillaries are formed by the division and re-division of the arterioles.
2. The wall of the capillary is made up of endothelium or squamous epithelium.
3. The capillary wall is permeable to water and dissolved substances.
4. Exchange of respiratory gases, nutrients, excretory products, etc. takes place through the capillary wall.
5. Capillaries unite to form venules.

Question 9.
What is blood pressure? How is it measured? Explain factors affecting blood pressure.
Answer:
1. Blood pressure:

1. The pressure exerted by blood on the wall of the blood vessels is called blood pressure. Pressure exerted by blood on the wall of arterial wall is arterial blood pressure. Blood pressure is described in two terms viz. systolic blood pressure and diastolic blood pressure.
2. Systolic blood pressure is the pressure exerted on arterial wall during ventricular contraction (systole). For a normal healthy adult the average value is 120 mmHg.
3. Diastolic blood pressure is the pressure on arterial wall during ventricular relaxation (diastole). For a normal healthy adult it is 80 mmHg.
4. B. E = SP/DP = 120/80 mmHg. Blood pressure is normally written as 120/80 mmHg. Difference between systolic and diastolic pressure is called pulse pressure normally, it is 40 mmHg.

2. Measurement of blood pressure:

1. Blood pressure is measured with the help of an instrument called sphygmomanometer.
2. The instrument consists of inflatable rubber bag cuff covered by a cotton cloth. It is connected with the help of tubes to a mercury manometer on one side and a rubber bulb on the other side.
3. During measurement, the person is asked to lie in a sleeping position. The instrument is placed at the level of heart and the cuff is tightly wrapped around upper arm.
4. The cuff is inflated till the brachial artery is blocked due to external pressure. Then pressure in the cuff is slowly lowered till the first pulsatile sound is produced. At this moment, pressure indicated in manometer is systolic pressure. Sounds heard during this measurement of blood pressure are called as Korotkoff sounds.
5. Pressure in the cuff is further lowered till any pulsatile sound cannot be heard due to smooth blood flow. At this moment, pressure indicated in manometer is diastolic pressure an optimal blood pressure (normal) level reads 120/80 mmHg.

3. Factors affecting blood pressure:

1. Cardiac output : Normal cardiac output is 5 lit/min. Increase in cardiac output increases systolic pressure.
2. Peripheral resistance : Peripheral resistance depends upon the diameter of blood vessels. Decrease in diameter of arterioles and capillaries under the effect of vasopressin cause increase in peripheral resistance and thereby increase in blood pressure.
3. Blood volume : Loss of blood in accidents decreases blood volume and thus cause decrease in blood pressure.
4. Viscosity of blood : Blood pressure is directly proportional to viscosity of blood.
5. Age : Blood pressure increases with age due to increase in inelasticity of blood vessels.
6. Venous return : Amount of blood brought to the heart via the veins per unit time is called the venous return and it is directly proportional to blood pressure.
7. Length and diameter of blood vessels : Blood pressure is directly proportional to the total length of the blood vessel. Blood pressure can also be affected by vasoconstriction or vasodilation.
8. Gender : Females have slightly lower BP than males of her age before menopause. However, the risk of high B. P increases in the females after menopause sets in.

Question 10.
Describe human blood and give its functions.
Answer:
Blood Composition:

1. Blood is a red coloured fluid connective tissue derived from embryonic mesoderm.
2. It has two components – the fluid plasma (55%) and the formed elements i.e. blood cells (44%).
3. Plasma is a straw coloured, slightly alkaline and viscous fluid having 90% water and 10% solutes such as proteins, nutrients, nitrogenous wastes, salts, hormones, etc.
4. Blood corpuscles are of three types, viz. erythrocytes (RBCs), white blood corpuscles (WBCs) and thrombocytes (platelets).
   

(5) Red blood corpuscles or Erythrocytes:

1. Erythrocytes or red blood corpuscles. They are circular, biconcave, enucleated cells.
2. The RBC size : 7 pm in diameter and 2.5 pm in thickness.
3. The RBC count : 5.1 to 5.8 million RBCs/ cu mm of blood in an adult male and 4.3 to 5.2 million/cu mm in an adult female.
4. The average life span of RBC : 120 days.
5. RBCs are formed by the process of erythropoiesis. In foetus, RBC formation takes place in liver and spleen whereas in adults it occurs in red bone marrow.
6. The old and worn out RBCs are destroyed in liver and spleen.
7. Polycythemia is an increase in number of RBCs while erythrocytopenia is decrease in their (RBCs) number.

Functions of RBCs:

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

(6) White blood corpuscles / Leucocytes:

1. Leucocytes or White Blood Corpuscles (WBCs) are colourless, nucleated, amoeboid and phagocytic cells.

2. Their size ranges between 8 to 15 pm. Total WBC count is 5000 to 9000 WBCs/cu mm of blood. The average life span of a WBC is about 3 to 4 days.

3. They are formed by leucopoiesis in red bone marrow, spleen, lymph nodes, tonsils, thymus and Payer’s patches, whereas the dead WBCs are destroyed by phagocytosis in blood, liver and lymph nodes.

4. Leucocytes are mainly divided into two types, viz., granulocytes and agranulocytes.

5. Granulocytes : Granulocytes are cells with granular cytoplasm and lobed nucleus. Based on their staining properties and shape of nucleus, they are of three types, viz. neutrophils, eosinophils and basophils.

(I) Neutrophils:

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

(II) Eosinophils or acidophils:

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

(III) Basophils:

1. The cytoplasmic granules of basophils take up basic stains such as methylene blue.
2. They have twisted nucleus.
3. In size, they are smallest and constitute about 0.5% of total WBCs.
4. They too are non-phagocytic.
5. Their function is to release heparin which acts as an anticoagulant and histamine that is involved in inflammatory and allergic reaction.

6. Agranulocytes : There are two types of agranulocytes, viz. monocytes and lymphocytes. Agranulocytes do not show cytoplasmic granules and their nucleus is not lobed. They are of two types, viz. lymphocytes and monocytes.
(I) Lymphocytes:

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

(II) Monocytes:

1. Largest of all WBCs having large kidney shaped nucleus are monocytes. They are about 5% of total WBCs.
2. They are phagocytic in function.
3. They can differentiate into macrophages for engulfing microorganisms and removing cell debris. Hence they are also called scavengers.
4. At the site of infections they are seen in more enlarged form.

(7) Thrombocytes/Platelets:

1. Thrombocytes or platelets are non- nucleated, round and biconvex blood corpuscles.
2. They are smallest corpuscles measuring about 2.5 to 5 mm in diameter with a count of about 2.5 lakhs/cu mm of blood.
3. Their life span is about 5 to 10 days.
4. Thrombocytes are formed from megakaryocytes of bone marrow. They break from these cells as fragments during the process of thrombopoiesis.
5. Thrombocytosis is the increase in platelet count while thrombocytopenia is decrease in platelet count.
6. Thrombocytes possess thromboplastin which helps in clotting of blood.
7. Therefore, at the site of injury platelets aggregate and form a platelet plug. Here they release thromboplastin due to which further blood clotting reactions take place.

(8) Functions of blood:

1. Transport of oxygen and carbon dioxide
2. Transport of food
3. Transport of waste product
4. Transport of hormones
5. Maintenance of pH
6. Water balance
7. Transport of heat
8. Defence against infection
9. Temperature regulation
10. Blood clotting/coagulation
11. Helps in healing

Balbharti Maharashtra State Board Hindi Yuvakbharati 12th Digest परिशिष मुहावरे Notes, Questions and Answers.

Maharashtra State Board 12th Hindi परिशिष मुहावरे

मुहावरा वह वाक्यांश जो सामान्य अर्थ को छोड़कर किसी विशेष अर्थ में प्रयुक्त होता है; मुहावरे में उसके लाक्षणिक और व्यंजनात्मक अर्थ को ही स्वीकार किया जाता है। वाक्य में प्रयुक्त किए जाने पर ही मुहावरा सार्थक प्रतीत होता है।

• अपना उल्लू सीधा करना – अपना स्वार्थ सिद्ध करना।
• दिन दूना रात चौगुना बढ़ना – दिन–प्रतिदिन अधिक उन्नति करना।
• अक्ल पर पत्थर पड़ना – बुद्धि काम न करना।
• आँखों में धूल झोंकना – धोखा देना।
• आँखें बिछाना – अति उत्साह से स्वागत करना।
• कान में कौड़ी डालना – गुलाम बनाना।
• कंगाली में आटा गीला होना – विपत्ति में और अधिक विपत्ति आना।
• कुएँ में बाँस डालना – जगह–जगह खोज करना।
• गुड़ गोबर करना – बने काम को बिगाड़ देना।
• गड़े मुर्दे उखाड़ना – पुरानी कटु बातों को याद करना।
• कटे पर नमक छिड़कना – दुखी को और दुखी बनाना।
• एक और एक ग्यारह – एकता में शक्ति होना
• घर फूंक तमाशा देखना – अपनी ही हानि करके प्रसन्न होना।
• घाट-घाट का पानी पिया होना – हर प्रकार के अनुभव से परिपूर्ण होना।
• चाँदी काटना – बहुत लाभ कमाना।
• जहर का चूंट पीना – अपमान को चुपचाप सह लेना।
• जी-जान से काम करना – पूरी क्षमता के साथ काम करना।
• तिल का ताड़ बनाना – छोटी बात को बढ़ा–चढ़ाकर कहना।
• पत्थर की लकीर होना – पक्की बात।
• पेट में दाढ़ी होना – छोटी आयु में बुद्धिमान होना।
• फूंक-फूंककर पाँव रखना – अति सावधानी बरतना।
• मुट्ठी गर्म करना – रिश्वत देना।
• रंग में भंग होना – प्रसन्नता के वातावरण में विघ्न पड़ना।
• शक्ल पर बारह बजना – बड़ा उदास रहना।
• सितारा चमकना – भाग्योदय होना
• आठ-आठ आँसू रोना – बहुत अधिक रोना।
• आँखें चार होना – प्रेम होना।
• अगर-मगर करना – टाल–मटोल करना।
• अपना ही राग अलापना – अपनी ही बातें करते रहना।
• आसमान पर थूकना – अशोभनीय कार्य करना।
• उल्टी गंगा बहाना – उल्टा काम करना।
• उगल देना – भेद बता देना।
• ओखली में सिर देना – जान–बूझकर जोखिम उठाना।
• एक लाठी से हाँकना – सबके साथ समान व्यवहार करना।
• चार चाँद लगाना – शोभा बढ़ाना।
• पापड़ बेलना – कड़ी मेहनत करना।
• कान भरना – चुगली करना।
• कोल्हू का बैल – लगातार काम में लगे रहने वाला। बहुत परिश्रम करने वाला।
• कब्र में पैर लटकना – मरने के समीप होना।
• कागजी घोड़े दौड़ाना – लिखा–पढ़ी करना।
• कौड़ी-कौड़ी का मोहताज – अत्यंत निर्धन होना।
• खाला का घर – आसान काम।
• खाल मोटी होना – बेशर्म होना।
• गिरगिट की तरह रंग बदलना – अवसरवादी होना।
• घोड़े बेचकर सोना – गहरी नींद सोना।
• हाथ खींचना – निश्चिंत होकर सोना।
• चोली-दामन का साथ होना – साथ न देना।
• चोर की दाढ़ी में तिनका – घनिष्ठ संबंध होना।
• जली-कटी सुनाना – अपराधी का भयभीत और सशंकित रहना।
• डकार तक न लेना – कटु–चुभती बातें करना।
• डूबती नाव पार लगाना – सब कुछ हजम कर लेना।
• तलवे चाटना – कष्टों से छुटकारा देना।
• दाल न गलना – खुशामद करना।
• पेट काटना – काम न बनना।
• पाँचों उँगलियाँ घी में होना – चतुराई काम न आना।
• पोंगा होना – भूखा रहना।
• बात का धनी – चहुँ तरफ लाभ होना।
• मरने की फुरसत न होना – नासमझ होना।
• मूंछ उखाड़ना – वचन का पक्का कामों में बहुत व्यस्त होना।
• रोटियाँ तोड़ना – घमंड चूर-चूर कर देना।
• वीरगति को प्राप्त होना-मुफ्त में खाना।
• स्वांग भरना – युद्ध में वीरतापूर्वक मृत्यु पाना।
• हवा लगना – विचित्र वेश बनाना, किसी की नकल उतारना।
• हवाई किले बनाना – असर पड़ना/होना।
• दाई से पेट छिपाना – बहुत अधिक कल्पना करना।
• सिर खपाना – भेद जानने वाले से सच्ची बात छिपाना।
• खबर गरम होना – कठोर परिश्रम करना। चर्चा-ही-चर्चा होना।
• चिराग तले अँधेरा – गुणवान व्यक्ति में भी दोष होना।

Balbharti Maharashtra State Board 12th Biology Textbook Solutions Chapter 7 Plant Growth and Mineral Nutrition Textbook Exercise Questions and Answers.

Maharashtra State Board 12th Biology Solutions Chapter 7 Plant Growth and Mineral Nutrition

1. Multiple choice questions

Question 1.
Which of the hormone can replace vernalization?
(a) Auxin
(b) Cytokinin
(c) Gibberellins
(d) Ethylene
Answer:
(c) Gibberellins

Question 2.
The principle pathway of water translocation in angiosperms is ………………..
(a) Sieve cells
(b) Sieve tube elements
(c) Xylem
(d) Xylem and phloem
Answer:
(c) Xylem

Question 3.
Abscissic acid controls ………………..
(a) cell division
(b) leaf fall and dormancy
(c) shoot elongation
(d) cell elongation and wall formation
Answer:
(b) leaf fall and dormancy

Question 4.
Which is employed for artificial ripening of banana fruits?
(a) Auxin
(b) Ethylene
(c) Cytokinin
(d) Gibberellin
Answer:
(b) Ethylene

Question 5.
Which of the following is required for stimulation of flowering in plants?
(a) Adequate oxygen
(b) Definite photoperiod
(c) Adequate water
(d) Water and minerals
Answer:
(b) Definite photoperiod

Question 6.
For short day plants, the critical period is ………………..
(a) light
(b) dark/night
(c) UV rays
(d) Both (a) and (c)
Answer:
(b) dark/night

Question 7.
Which of the following is NOT day neutral plant?
(a) Tomato
(b) Cotton
(c) Sunflower
(d) Soybean
Answer:
(d) Soybean

Question 8.
Essential macro elements are ………………..
(a) manufactured during photosynthesis
(b) produced by enzymes
(c) obtained from soil
(d) produced by growth hormones
Answer:
(c) obtained from soil

Question 9.
Function of Zinc is ………………..
(a) closing of stomata
(b) biosynthesis of 3-IAA
(c) synthesis of chlorophyll
(d) oxidation of carbohydrates
Answer:
(b) biosynthesis of 3-LAA

Question 10.
Necrosis means ………………..
(a) yellow spot on the leaves
(b) death of tissue
(c) darkening of green colour in leaves
(d) wilting of leaves
Answer:
(b) death of tissue

Question 11.
Conversion of nitrates to nitrogen is called ………………..
(a) ammonification
(b) nitrification
(c) nitrogen fixation
(d) denitrification
Answer:
(d) denitrification

Question 12.
How many molecules of ATP are required to fix one molecule of nitrogen?
(a) 12
(b) 20
(c) 6
(d) 16
Answer:
(d) 16

2. Very short answer questions

Question 1.
Enlist the phases of growth in plants.
Answer:
The three phases of growth are phase of cell division, phase of cell enlargement and phase of cell maturation.

Question 2.
Give full form of IAA.
Answer:
Full form is Indole Acetic Acid.

Question 3.
What does it mean by ‘open growth’?
Answer:
In plants the growth is indeterminate and takes place throughout the life at specific regions having meristems.

Question 4.
Plant stress hormone.
Answer:
Abscissic acid.

Question 5.
What is denitrification?
Answer:
Anaerobic bacteria can convert nitrates of soil back into nitrogen gas. That process performed by denitrifying bacteria is denitrification.

Question 6.
Bacteria responsible for conversion of nitrite to nitrate.
Answer:
Nitrobacter.

Question 7.
What is the role of gibberellins in rosette plants?
Answer:
In rosette plants like beet and cabbage, bolting, i.e. elongation of internodes before flowering is observed due to effect of gibberellins.

Question 8.
Vernalization
Answer:
The response of plant to the influence of low temperature on flowering in plants is called vernalization.

Question 9.
Photoperiodism
Answer:
The response of plant to the influence of light for initiation of flowering is known as photoperiodism.

Question 10.
What is grand period of growth?
Answer:
There are three phases of growth and the total time required for all phases to occur is called grand period of growth.

3. Short Answer Questions

Question 1.
(i) Differentiation
Answer:

1. It is a process of maturation of cells derived from apical meristems.
2. Differentiation is a permanent change in structure and function of cells that leads to its maturation.
3. Cell undergoes major anatomical and physiological change during differentiation process.
4. In hydrophytic plants parenchyma cells develop large schizogenous cavities which help them in aeration, buoyancy and mechanical support.

(ii) Redifferentiation
Answer:

1. It is a process in which cells produced by de-differentiation lose their capacity of division and become mature.
2. The cells mature to perform specific function.
3. Interfascicular cambium is formed by process of dedifferentiation loses its capacity to divide.
4. Secondary xylem and secondary phloem is formed form this cambium in vascular cylinder.

Question 2.
Arithmetic growth and Geometric growth
Answer:

Arithmetic growth	Geometric growth
1. In arithmetic growth only one daughter cell continues to divide, while the other undergoes differentiation and maturation.	1. In geometric growth both the daughter cells continue to divide and redivide again and again.
2. Rate of growth is constant.	2 Rate growth is initially slow but later on rapid rate.
3. Linear curve is obtained.	3. Exponential curve is obtained.
4. Mathematical expression is Lt = Lo + rt whereLt = length of time ‘t’ Lo = Length at time zero rt = growth rate, t = time of growth	4. Mathematical expression is Wt = Woe rt where, Wt = final size, Wo = initial size, r = growth rate, t = time of growth E = base of natural logarithm
5. e.g. Elongation of root	5. e.g. Divisions of zygote during embryo development.

Question 3.
Enlist the role and deficiency symptoms of: (a) nitrogen (b) phosphorus (c) potassium.
Answer:
(a) Nitrogen:
Role : Constituent of proteins as amino acids, nucleic acids, vitamins, hormones, coenzymes, ATP and chlorophyll molecule.
Deficiency symptoms : stunted growth and chlorosis.

(b) Phosphorus:
Role : Constituent of cell membrane, certain proteins, nucleic acids and nucleotides, required for all phosphorylation reactions.
Deficiency symptoms : Poor growth, leaves dull green

(c) Potassium :
Role : Determination of anion – cation balance in cell, necessary for protein synthesis, involved in formation of cell membrane, opening and closing of stomata, activates enzymes, helps in maintenance of turgidity of cells.
Deficiency symptom : Yellow edges in leaves, premature death.

Question 4.
What is short day plant? Give any two examples.
Answer:
The plants which flower when the day length or light period is shorter than the critical photoperiod are called short day plants or SDP
SDPs usually flower during winter and late summer.
Examples – Dahlia, Aster, Tobacco, Chrysanthemum, Soybean (Glycine max) and Cocklebur (Xanthium).

Question 5.
What is vernalization? Give its significance.
Answer:
A low temperature or chilling treatment that induces early flowering in plants is known as vernalization.

Significance:

1. Due to chilling treatment crops can be produced earlier.
2. Crops can be grown in areas where they do not grow naturally.

4. Long answer questions

Question 1.
Explain sigmoid growth curve with the help of diagram.
Answer:

1. When growth occurs in plants three distinct phases of growth are noticed.
2. Phase of cell formation is first phase where meristematic cells divide and new cells added.
3. In phase of cell enlargement newly formed cells elongate and with turgidity there is cell enlargement.
4. In phase of cell maturation cells get differentiated.
5. When we compare the growth rate it differs in these three phases.
6. In first phase or lag phase it is slow, while in log phase or exponential phase, growth rate accelerates and it reaches maximum.
7. In stationary phase of maturation growth rate slows down and comes to steady state.
8. When this changing rate of growth is plotted against time duration in a graph a sigmoid or S-shaped growth curve is obtained.

Question 2.
Describe the types of plants on the basis of photoperiod required, with the help of suitable examples.
Answer:

1. Effect of light duration on flowering of plants is known as photoperiodism.
2. Depending on photoperiodic response, plants are categorised into three types – Short day plants, long day plants and day neutral plants.

1. Short day plants : Plants that flower under short day length conditions are called short day plants. Plants such as Dahlia, Xanthium, Soybean, Aster, Tobacco and Chrysanthemum are short day plants or SDR. Short day plants require a long uninterrupted dark period for flowering. Therefore, they are also called long night plants.

2. Long day plants : Plants that flower only when they are exposed to light period longer than their critical photoperiod are called long day plants or LDP Long day plants require a short dark or night period for flowering. Hence, they are also called short night plants. Plants such as radish, spinach, wheat, poppy, cabbage, pea, sugar beet, etc. are long day plants.

3. Day neutral plants : Plants in which the flowering is not affected by the day length period are called day neutral plants or DNP or photoneutral plants. Plants such as cucumber, sunflower, cotton, balsam, maize, tomato, etc. are day neutral plants.

Question 3.
Explain biological nitrogen fixation with example.
Answer:

1. Conversion of atmospheric nitrogen into nitrogenous salts to make it available to plants for its update is described as nitrogen fixation.
2. When living organisms are involved in nitrogen fixation process it is known as biological nitrogen fixation.
3. The process is mainly carried out by prokaryotic organisms, i.e. different kinds of bacteria present in soil.
4. The nitrogen fixing organisms are known as diazotrophs or nitrogen fixers and about 70% nitrogen is fixed by them.
5. The nitrogen fixers are either free living bacteria or symbiotic associated with other higher plants e.g. Rhizobium.
6. The cyanobacteria have specialized cells heterocysts which help in process of nitrogen fixation.
7. Nitrogen fixation is high energy requiring process and 16 ATP molecules are needed for fixation of one molecule of nitrogen to ammonia.
8. Soil bacteria like Nitrosomonas, Nitrosocyccus convert ammonia to nitrate and the Nitrobacter convert nitrite to nitrate. This is known as nitrification, biological oxidation.
9. These bacteria are chemoautotrophic and utilize these processes for their metabolism.
10. Fabaceae plants like pea, bean have root nodules which harbour symbiotic bacterium Rhizobium which fixes nitrogen. It is host specific, soil bacterium, Nitrogen is made available to host plant.

Question 4.
Write on macro and micro nutrients required for plant growth.
Answer:

1. Plants absorb mineral nutrients from their surroundings.
2. For a proper growth of plants about 35 to 40 different elements are required.
3. Plants absorb these nutrients in ionic or dissolved form from soil with their root system e.g. Phosphorus as PO₄^–, Sulphur as SO₄^2- etc.
4. Based on their requirement in quantity, they are classified as major nutrients or macronutrients and those needed in small amounts Eire minor or micronutrients.
5. Macroelements are required in large amounts, as they play nutritive and structural roles e.g. C, H, O, R Mg, N, K, S and Ca. – Ca pectate cell wall component, Mg component of chlorophyll.
6. C, H, O are non-mineral major elements obtained from air and water e.g. CO₂ is source of carbon, Hydrogen from water.
7. Microelements are required in traces as they mainly have catalytic role as co-factors or activators of enzymes.
8. Microelements may be needed for certain activity in life cycle of plant e.g. B for pollen germination, Si has protective role during stress conditions and fungal attacks, Al enhances availability of phosphorus.
9. The important micronutrients for plant growth are Mn, B, Cu, Zn, Cl.