CBSE Class 12

NCERT Exemplar Solutions for Class 12 Biology chapter 14 Ecosystem

These Solutions are part of NCERT Exemplar Solutions for Class 12 Biology. Here we have given NCERT Exemplar Solutions for Class 12 Biology chapter 14 Ecosystem

Multiple Choice Questions

Question 1.
Decomposers like fungi and bacteria are
(i) autotrophs
(ii) heterotrophs
(iii) saprotrophs
(iv) chemoautotrophs.
Choose the correct answer.
(a) (i) and (iii)
(b) (i) and (iv)
(c) (ii) and (iii)
(d) (i) and (ii)
Answer:
(c) : Decomposers like bacteria and fungi are heterotrophs because they are dependent on others for their food as they cannot make their own food. They are also called saprotrophs because they feed on dead and decaying organic matter.

Question 2.
The process of mineralisation by micro organisms helps in the release of
(a) inorganic nutrients from humus
(b) both organic and inorganic nutrients from detritus
(c) organic nutrients from humus
(d) inorganic nutrients from detritus and formation of humus.
Answer:
(a) : Mineralisation is the release of inorganic substances, both non-mineral (e.g., CO₂, H₂O) and minerals (e.g., Ca₂+, Mg₂+, K+, NH₄+) from organic matter. The process is slow because of trapping of these nutrients in humus and their immobilisation in decomposers/detritivores. It prevents their washing out or leaching. Nutrients get immobilised in decomposer microbes and detritivores are again exposed to humification and mineralisation after the death of these organisms. its productivity. It is measured as weight (e.g., g/m²/yr) or energy (e.g., kcal/m²/yr). Hence, only unit (iv) is correct.

Question 3.
Productivity is the rate of production of biomass expressed in terms of
(i) (kcal m^-3)yr^-1
(ii) gyr^-1
(iii) g^-1 yr^-1
(iv) (kcal m^-2) yr^-1
(a) (ii)
(b) (iii)
(c) (ii) and (iv)
(d) (i) and (iii)
Answer:
(None of the options is correct):
The rate of synthesis of energy containing organic matter or biomass by any trophic level per unit area in unit time is described as

Question 4.
An inverted pyramid of biomass can be found in which ecosystem?
(a) Forest
(b) Marine
(c) Grassland
(d) Tundra
Answer:
(b) : Biomass basically depends upon reproductive potential and age of individuals. In an aquatic ecosystem, producers have least biomass and this value gradually shows an increase towards the apex of the pyramid, thus making the pyramid inverted in shape.

Question 5.
Which of the following is not a producer?
(a) Spirogyra
(b) Agaricus
(c) Volvox
(d) Nostoc
Answer:
(b) : Spirogyra, Volvox and Nostoc are chlorophyll containing organisms and thus prepare their own food. Agaricus is a fungus (Basidiomycetes), it is a chlorophyllous and not a producer. It possesses saprotrophic mode of nutrition.

Question 6.
Which of the following ecosystems is most productive in terms of net primary production?
(a) Deserts
(b) Tropical rainforests
(c) Oceans
(d) Estuaries
Answer:
(b) : Tropical rainforests have an average net primary productivity (NPP) of 1,500 g/m²/ yr. Open oceans and deserts have average NPP of 125 and 90g/m²/yr respectively. Algal beds and reefs ecosystem have average NPP of 2,500 g/m²/yr.

Question 7.
Pyramid of numbers is
(a) always upright
(b) always inverted
(c) either upright or inverted
(d) neither upright nor inverted.
Answer:
(c) : Ecological pyramids are pictorial representation of relationship between organisms at different trophic levels, regarding energy, biomass or number. Pyramid of numbers can be either upright or inverted. For example in a grassland ecosystem, number of primary consumers are less than primary producers and that of secondary consumers are less than primary consumers and so on. On the other hand, if a single big tree ecosystem is taken into consideration, pyramid of number will be inverted.

Question 8.
Approximately how much of the solar energy that falls on the leaves of a plant is converted to chemical energy by photosynthesis?
(a) Less than 1%
(b) 2-10%
(c) 30%
(d) 50%
Answer:
(b) : 50% of the solar energy incident over earth is present in PAR (photosynthetically active radiation). About 1-5% of total incident solar energy or 2-10% of PAR is captured by the photosynthetic organisms for photosynthesis.

Question 9.
Among the following, where do you think the process of decomposition would be the fastest?
(a) Tropical rainforest
(b) Antarctic
(c) Dry arid region
(d) Alpine region
Answer:
(a) : Tropical rainforests are the richest and most productive ecosystem in the world. Consequently, the rate of decomposition is also high, as the conditions of moisture and temperature are optimum.

Question 10.
How much of the net primary productivity of a terrestrial ecosystem is eaten and digested by herbivores?
(a) 1%
(b) 10%
(c) 40%
(d) 90%
Answer:
(b) : Energy flow in the ecosystem follows the ten percent law (put forth by Lindemann in 1942).
From the level of primary producers onwards, only 10% of energy is stored at the higher trophic level and 90% is lost (as heat or in respiration etc). This energy transfer forms the basis of life.

Question 11.
During the process of ecological succession the changes that take place in communities are
(a) orderly and sequential
(b) random
(c) very quick
(d) not influenced by the physical environment.
Answer:
(a) : Biotic or ecological succession is the natural development of a series of biotic communities at the same site, one after the other till a climax community develops which does not change further because it is in perfect harmony with the environment of the area. The change is orderly and sequential. The first biotic community which develops in a bare area is called pioneer community. It has very little diversity. Climax community is a stable, self perpetuating and final biotic community that develops at the end of biotic succession.

Question 12.
Climax community is in a state of
(a) non-equilibrium
(b) equilibrium
(c) disorder
(d) constant change.
Answer:
(b) : The process of sequential establishment of various plant communities in any habitat is called plant succession. Finally that community is established which is in complete equilibrium with the prevailing environment. This stage is called climax stage and its establishment is called stabilisation.

Question 13.
Among the following biogeochemical cycles which one does not have losses due to respiration?
(a) Phosphorus
(b) Nitrogen
(c) Sulphur
(d) All of the above
Answer:
(d) : Phosphorus, nitrogen and sulphur do not have losses due to respiration because they are not particularly involved in gaseous exchange.

Question 14.
The sequence of communities of primary succession in water is
(a) phytoplankton, sedges, free-floating hydrophytes, rooted hydrophytes, grasses and trees
(b) phytoplankton, free-floating hydrophytes, rooted hydrophytes, sedges, grasses and trees
(c) free-floating hydrophytes, sedges, phyto-plankton, rooted hydrophytes, grasses and trees
(d) phytoplankton, rooted submerged hydro-phytes, floating hydrophytes, reed swamp, sedges, meadow and trees.
Answer:
(d) : Primary succession in water is also called as hydrarch, which will lead from hydric to mesic conditions. Phytoplanktons (autotrophic) are generally the first to appear. Later zooplanktons feeding on phytoplanktons also appear. Next stage is characterised by the soft mud on the bottom having organic matter favouring the growth of rooted submerged plants. They are then replaced by free floating hydrophytes (Lenina, Woljfia etc). Rapid growth of these plants build up bottom so that water becomes shallow on periphery. In this shallow water, comes the reed swamp stage (e.g. Typha). They produce abundant organic matter. Next stages are sedge and meadow stage which transpire rapidly and build up soil, on which the next stage, trees can grow.

Question 15.
The reservoir for the gaseous type of biogeo-chemical cycle exists in
(a) stratosphere
(b) atmosphere
(c) ionosphere
(d) lithosphere.
Answer:
(b) : Biogeochemical cycles can be grouped into 3 types:

1. Gaseous cycle (material involved in circulation are gases or vapours and the reservoir pool is the atmosphere or hydrosphere) e.g., nitrogen cycle.
2. Sedimentary cycle (Materials involved in circulation are non-gaseous and the reservoir pool is lithosphere) e.g. phosphorus, calcium and magnesium cycles.
3. Mixed cycle (materials involved in circulation has both gaseous and non- gaseous state), e.g., sulphur cycle.

Question 16.
If the carbon atoms fixed by producers already have passed through three species, the trophic level of the last species would be
(a) scavenger
(b) tertiary producer
(c) tertiary consumer
(d) secondary consumer.
Answer:
(c) : Length of a food chain i.e. number of trophic levels is limited by the efficiency of energy transfer i.e. 10% law.
Producers —> 1° consumers —> 2°consumers —> 3° consumers
If the carbon atoms fixed by producers already have passed through three species then the trophic level of the last species (i.e. third species) would be tertiary consumer. Scavengers (e.g. vultures) can be tertiary consumers, but they can be at other trophic levels too.

Question 17.
Which of the following type of ecosystem is expected in an area where evaporation exceeds precipitation, and mean annual rainfall is below 100mm?
(a) Grassland
(b) Shrubby forest
(c) Desert
(d) Mangrove
Answer:
(c) : Deserts have been variously classified as true deserts, having less than 120 mm annual rainfall, or extreme deserts showing less than 70 mm annual rainfall. In desert biomes, evaporation from soil always exceeds rainfall by 7 to 50 times.

Question 18.
The zone at the edge of a lake or ocean which is alternatively exposed to air and immersed in water is called
(a) Pelagic zone
(b) Benthic zone
(c) Lentic zone
(d) Littoral zone.
Answer:
(d) : Littoral zone is the shallow coastal zone. Light is available upto bottom in this zone. Therefore, producers are found throughout from surface to bottom in this zone. Rooted vegetation occurs along shores. Consumers are also available throughout i.e., from surface to the bottom in this zone.

Question 19.
Edaphic factor refers to
(a) water
(b) soil
(c) relative humidity
(d) altitude.
Answer:
(b) : Edaphic factors are classified under the abiotic factors affecting an ecosystem. Edaphic factors include factors of soil e.g. soil texture, substratum, topography, mineral composition, pH etc. These factors can influence the distribution and interrelationships of organisms, as well as rate of decomposition.

Question 20.
Which of the following is an ecosystem service provided by a natural ecosystem?
(a) Cycling of nutrients
(b) Prevention of soil erosion
(c) Pollutant absorption and reduction of the threat of global warming
(d) All of the above
Answer:
(d) : The products of ecosystem processes which have environmental, aesthetic and indirect economic value are named as ecosystem services. Soil formation and soil protection are the major ecosystem services accounting for nearly 50% of their total worth. Plant cover protects the soil from drastic changes in temperature. There is little wind or water erosion as soil particles are not exposed to them. The soil remains spongy and fertile. There are no landslides and no floods. Plant cover of natural ecosystem absorbs polluting gases, causes settling of suspended particulate matter, removes C02 and releases 0:. Purified air becomes available. There is no overall depletion of nutrients as the same are repeatedly circulated and recirculated. This keeps the fertility of soil intact.

Very Short Answer Type Questions

Question 1.
Name an organism found as secondary carnivore in an aquatic ecosystem.
Answer:
Large fish, catfish, water snake.

Question 2.
What does the base tier of the ecological pyramid represent?
Answer:
If Producers represent the base tier of the ecological pyramid.

Question 3.
Under what conditions would a particular stage in the process of succession revert back to an earlier stage?
Answer:
If At any time during primary or secondary succession, natural or human induced disturbances like fire, deforestation etc. can convert a particular seral stage of succession to an earlier stage.

Question 4.
Arrange the following as observed in vertical stratification of a forest: Grass, Shrubby plants, Teak, Amaranthus.
Answer:
Vertical stratification of forest are grass, Amaranthus, shrubby plants, teak.

Question 5.
Name an omnivore which occurs in both grazing food chain and the decomposer food chain.
Answer:
Crow is omnivore and occur in both grazing food chain and decomposer food chain.

Question 6.
Justify the pitcher plant as a producer.
Answer:
Leaf lamina in pitcher plant is modified into pitcher and consists of chlorophyll. It also undergoes photosynthesis, therefore pitcher plant is a producer.

Question 7.
Name any two organisms which can occupy more than one trophic level in an ecosystem.
Answer:
If Man and sparrow occupy more than one trophic level in an ecosystem.

Question 8.
In the North East region of India, during the process of Jhum cultivation, forests are cleared by burning and left for regrowth after a year of cultivation. How would you explain the regrowth of forest in ecological term?
Answer:
In Jhum cultivation, farmers cut down the trees of the forest and burn the plant remains. The ash is used as a fertiliser and land is used for farming or cattle grazing. After cultivation the area is left for several years so as to allow its recovery. This regrowth of forest is called secondary succession.

Question 9.
Climax stage is achieved quickly in secondary succession as compared to primary succession. Why?
Answer:
If Secondary succession occurs on a fertile land where living matter is already existing, whereas primary succession begins on a barren sterile area with no living matter present in it.

Question 10.
Among bryophytes, lichens and ferns which one is a pioneer species in a xeric succession?
Answer:
Crustose lichens are the pioneer species in xeric succession.

Question 11.
What is the ultimate source of energy for the ecosystems?
Answer:
Sun is the ultimate source of energy for the ecosystem.

Question 12.
Is the common edible mushroom an autotroph or a heterotroph?
Answer:
Edible mushroom is a heterotroph, as it is without chlorophyll and does not perform photosynthesis.

Question 13.
Why are oceans least productive?
Answer:

1. Oceans have high salinity i.c., 3.5%.
2. Low concentration of dissolved nutrients especially nitrogen.
3. Deep abyssal zone in ocean has no producers.

Question 14.
Why is the rate of assimilation of energy at the herbivore level called secondary productivity?
Answer:
Herbivores are primary consumers and depend on plants to obtain biomass. Plants are producers and herbivores gain biomass from them by primary productivity. Rate of assimilation of primary productivity at the herbivore level is therefore, called secondary productivity.

Question 15.
Why are nutrient cycles in nature called biogeochemical cycles?
Answer:
The nutrients move from living organisms to environment in a cyclic manner and similarly back to the organism. Therefore, nutrient cycles in nature are called biogeochemical cycles.

Question 16.
Give any two examples of xerach succession.
Answer:

1.  L ithosere – Succession on a rock.
2. Psammosere – Succession on a sandy area.

Question 17.
Define self sustainability.
Answer:
Self sustainability is the utilisation of natural resources in a way that their rate of consumption is equal to the rate of regeneration, so that same amount becomes available to the next generation. It is possible by judicious utilisation and avoiding wastage.

Question 18.
Given below is a figure of an ecosystem. Answer the following questions.
(1) What type of ecosystem is shown in figure.
(2) Name any plant that is characteristic of such ecosystem.

Answer:

1. It is a desert biome.
2. Xerophytic plants like cactus, Euphorbia.

Question 19.
What is common to earthworm, mushroom, soil mites and dung beetle in an ecosystem.
Answer:
All are decomposers involved in decomposition of organic remains.

Short Answer Type Questions

Question 1.
Organisms at a higher trophic level have less energy available. Comment.
Answer:
This is because of 10% (ten percent) law which was proposed by Lindemann 1942. According to this law, during transfer of energy from lower trophic level to higher trophic level 90% of energy is lost, and only 10% of energy is transferred to next trophic level. As the trophic level increases, the available energy goes on decreasing.

Question 2.
The number of trophic levels in an ecosystem are limited. Comment.
Answer:
The amount of energy made available in a trophic level goes on decreasing from lower level to higher level. The transfer of energy follows 10% law. According to this a stage comes when energy available is too less to sustain the trophic level. Therefore number of trophic levels in an ecosystem remain limited.

Question 3.
Is an aquarium a complete ecosystem?
Answer:
Yes, a balanced aquarium is a artificial ecosystem consisting of both biotic and abiotic components. Water, oxygen supply source, light source are abiotic factors, whereas aquatic plants, small animals and decomposers serve as biotic components.

Question 4.
What could be the reason for the faster rate of decomposition in the tropics?
Answer:
Tropics mostly have temperature above 25°C, along with humid conditions, which is suitable environment for growth and multiplication of decomposers. That is why rate of decomposition is faster in tropics.

Question 5.
Human activities interfere with carbon cycle. List any two such activities.
Answer:
Human activities that are adding CO₂ to the atmosphere are

1. Excessive burning of fossil fuels.
2. Deforestation.

Question 6.
Flow of energy through various trophic levels in an ecosystem is unidirectional and non- cyclic. Explain.
Answer:
Ultimate source of energy is sun. Green plants (producers) produce food by using solar energy. This food is consumed by herbivores (Primary consumers) to get energy. The energy is further transferred to next levels of consumers i.e., secondary consumers and tertiary consumers. During transfer of energy about 90% of it is wasted or consumed up in respiration and only 10% becomes part of the higher trophic level. The energy cannot be transferred from consumers to producers and even to the sun. Therefore, energy transfer is always unidirectional and non-cyclic accompanied by decrease in usable energy.

Question 7.
Apart from plants and animals, microbes form a permanent biotic component in an ecosystem. While plants have been referred to as autotrophs and animals as heterotrophs, what are microbes referred to as? How do the microbes fulfil their energy requirements?
Answer:
Microorganisms like bacteria and fungi are heterotrophs and are known as decomposers. They fulfil their energy needs by decomposition, which involves break down of complex organic matter into simple compounds. They absorb these simplified compounds and use them during their metabolism.

Question 8.
Poaching of tiger is a burning issue in today’s world. What implication would this activity have on the functioning of the ecosystem of which the tigers are integral part?
Answer:
Tigers and lions are top carnivores and play an important role in maintaining the stability of an ecosystem. Excessive poaching of tigers will lead to increase in population size of herbivores which in turn will damage the crops in abundance. This will create an imbalance in the ecosystem and will make it unstable.

Question 9.
In relation to energy transfer in ecosystem, explain the statement “10 kg of deer’s meat is equivalent to 1 kg of lion’s flesh”.
Answer:
The statement is very true, because lion is a predator which eats deer. According to 10% energy transfer law only 10% of energy will be made available to the lion from deer, and 90% of the energy will be lost in the atmosphere during transfer.

Question 10.
Primary productivity varies from ecosystem to ecosystem. Explain?
Answer:
Primary productivity is the amount of energy accumulation in green plants as biomass or organic matter per unit area over a time period through the process of photosynthesis. Primary productivity in an ecosystem depends on number of factors like photosynthetic capacities of producers, environmental factors like temperature, sunlight intensity, rainfall and availability of nutrients. These factors are different in different ecosystems, therefore productivity varies from ecosystem to ecosystem. In tropical rainforest primary productivity is high 20 tones/hectare/year whereas in desert it is low 0.7 tones /hectare/year.

Question 11.
Sometimes due to biotic/abiotic factor the climax remain in a particular seral stage (pre climax) without reaching climax. Do you agree with this statement. If yes give a suitable example.
Answer:
Yes, this statement is right. Sometimes certain biotic or abiotic factors do not let a seral stage to reach climax, and process of succession may get arrested at pre-climax stage. This may be due to certain reasons like forest fires, land slide, change in soil characteristic, increase in herbivore population and overgrazing etc.

Question 12.
What is an incomplete ecosystem? Explain with the help of suitable example.
Answer:
lf any essential component of an ecosystem is absent in an ecosystem, it is said to be an incomplete ecosystem. E.g., an ecosystem at the bottom of fish tank or deep aphotic zone of the ocean. In both cases producers are absent, therefore both are the examples of incomplete ecosystem.

Question 13.
What are the shortcomings of ecological pyramids in the study of ecosystem?
Answer:
Shortcoming of ecological pyramids are:

1. It does not take into account the same species belonging to two or more trophic levels.
2. It assumes simple food chain, that almost never exists in nature.
3. It does not accommodate food web.
4. Saprophytes are not given any place in ecological pyramids, even though they play a vital role.

Question 14.
How do you distinguish between humification and mineralisation?
Answer:
Humification – The process by which detritus is changed into dark-coloured amorphous substance called humus, which acts as a reservoir of nutrients.
Mineralisation – The process by which inorganic substances (water, CO₂) and minerals (Ca, Mg’1″, KT NH₄+) are released in the soil.

Question 15.
Fill in the trophic levels (1, 2, 3 and 4) in the boxes provided in the figure.

Answer:
(1) is T₁ first trophic level i.e., producers
(2) is T₂ – Second Trophic level i.e., herbivore (Primary consumer)
(3) is T₃ – Third trophic level i.e., small carnivorous bird (Secondary consumers)
(4) is T₄– Fourth Trophic level i.e., large carni-vorous bird (tertiary consumers)

Question 16.
The rate of decomposition of detritus is affected by the abiotic factors like availability of oxygen, pH of the soil substratum, temperature etc. Discuss.
Answer:
Breakdown of complex organic compounds of dead bodies of plants and animals and wastes of animals by microbial action into simple substances is known as decomposition.
It depends on various factors like –

1. Temperature – Higher temperature i. e., above 25°C increases the rate of decomposition. Low temperature at high altitude and latitude decreases the metabolism of microbes.
2. Availability of oxygen – Rate of decomposition of detritus is faster in presence of oxygen i.e., in aerobic conditions, while absence of oxygen or anaerobic conditions (anaerobiosis) reduces decomposition and causes piling up of detritus.
3. pH of soil substratum – Neutral and slightly alkaline soils are rich in detritivores and favour decomposition. Increase in acidity decreases the rate of decomposition.

Long Answer Type Questions

Question 1.
A farmer harvests his crop and expresses his harvest in three different ways.
(a) I have harvested 10 quintals of wheat.
(b) I have harvested 10 quintals of wheat today in one acre of land.
(c) I have harvested 10 quintals of wheat in one acre of land, 6 months after sowing.
Do the above statements mean one and the same thing. If your answer is yes, give reasons. And if your answer is ‘no’ explain the meaning of each expression.
Answer:
NO, the above three statements do not mean one and the same thing because –

1. Statement (a) indicates total productivity of wheat but does not indicate the extent of area.
2. Statement (b) indicates the productivity of wheat/acre/on a specific day.
3. Statement (c) indicates productivity of wheat/acre/specific period.
   The third statement (c) is most accurate as it expresses the productivity of wheat per unit time and per unit area.

Question 2.
Justify the following statement in terms of ecosystem dynamics.”Nature tends to increase the gross primary productivity, while man tends to increase the net primary productivity”.
Answer:
Gross primary productivity of an ecosystem is the rate of production of organic matter during photosynthesis. Nature always tend to increase gross primary productivity. Ecological succession is the natural development of a series of biotic communities at the same site, one after the other till a climax community develops which does not change further because it is in perfect harmony with the environment of the area.

In ecological succession there is tendency to increase species diversity, complexity of organisms, and gross primary productivity as higher serai communities have higher photosynthetic efficiency.

Net primary productivity is the available biomass for the consumption to heterotrophs. It is equal to the rate of organic matter created by photosynthesis minus the rate of respiration and other losses.
NPP = GPP – R

Human always tries to increase net primary productivity by cultivating food and crops, which are important for our survival.

Question 3.
Which of the following ecosystems will be more productive in terms of primary productivity? Justify your
A young forest, a natural old forest, a shallow polluted lake, alpine meadow.
Answer:
Primary productivity is the rate of biomass or organic matter produced per unit area over a time period by green plants during photosynthesis. Primary productivity depends upon the type of ecosystem. The ecosystem which has more producers, will be more productive in terms of primary productivity. A young forest will be more productive than an old forest. It is because the rate of total photosynthesis will be highest in it, due to fully grown vegetation, more number and large surface area of the leaves and higher quantum number. On the other hand, a shallow polluted lake will have low oxygen content and an alpine area is at high altitude where low temperature inhibits productivity.

Question 4.
What are the three types of ecological pyramids. What information is conveyed by each pyramid with regard to structure, function and energy in the ecosystem.
Answer:
An ecological pyramid is a graphic representation of an ecological parameter like number of individuals present in various trophic levels of a food chain with producers forming the base and top carnivores the tip. Each trophic level represents a functional level.
There are three types of ecological pyramids.
(1) Pyramid of numbers
(2) Pyramid of biomass
(3) Pyramid of energy
(1) Pyramid of numbers : It is a graphic representation of the number of individuals per unit area of various trophic levels stepwise with producers being kept at the base and top carnivores kept at the tip. In most cases the pyramid of number is upright with members of successive higher trophic level being fewer than the previous one.

In a grassland, a larger number of grass plants or herbs support a fewer number of grasshoppers that support a still smaller number of frogs, the latter still smaller number of snakes and the snakes very few peacocks or falcons. This is, however, not applicable in all the cases.
A single large sized producer like tree can, however, provide nourishment to several herbivores (e.g., birds). The birds may support a still larger population of ectoparasites. Such a pyramid shall be inverted.

(2) Pyramid of biomass : The amount of living organic matter is called biomass. It is measured both as fresh and dry- weight. Pyramid of biomass is a graphic representation of biomass present sequence-wise per unit area of different trophic levels with producers at the base and top carnivores kept at the tip.

Maximum biomass occurs in producers. There is a progressive reduction of biomass found in herbivores, primary carnivores, secondary carnivores, etc. It is found that about 10-20% of the biomass is transferred from lower trophic level to higher trophic level.

Pyramid of biomass is upright for terrestrial habitats. Inverted or spindle¬shaped pyramids are obtained in aquatic habitats where the biomass of a trophic level depends upon reproductive potential and longevity of its members,

(3) Pyramid of energy – It is graphic representation of amount of energy trapped per unit time and area in different trophic levels of food chain. It is always upright as the amount of energy always decreases from lower to higher trophic level of food chain and follows 10% energy transfer law.

Question 5.
Write a short note on pyramid of numbers and pyramid of biomass.
Answer:
Re/er answer 4.

Question 6.
Given below is a list of autotrophs and heterotrophs. With your knowledge about food chain, establish various linkages between the organisms on the principle of’eating and being eaten’. What is the this inter-linkage established known as?
Algae, Hydrilla, grasshopper, rat, squirrel, crow, maize plant, deer, rabbit, lizard, wolf, snake, peacock, phytoplankton, crustaceans, whale, tiger, lion, sparrow, duck, crane, cockroach, spider, toad, fish, leopard, elephant, goat, Nymphaea, Spirogyra.
Answer:
A straight line sequence of ‘who eats whom’ or eating and being eaten in an ecosystem is called a food chain. A network of cross connecting food chains involving producers, consumers and decomposers are termed as a food web.

Lion, tiger, leopard, whale – Top carnivore (Top trophic level)
Spider, cockroach, lizard, wolf, snake, toad, fish, crow, sparrow, crane, duck, peacock – Secondary consumers (IIIrd trophic level).

Crustaceans, grasshopper, deer, rat, squirrel, rabbit, elephant, goat – Primary consumer (IInd trophic level). Phytoplankton, algae, Hydrilla, maize plant, Nymphaea, Spirogyra – Producers (Ist trophic level).

Question 7.
“The energy flow in the ecosystem follows the second law of thermodynamics”. Explain.
Answer:
Second law of thermodynamics, also called law of entropy, states that energy transfer or energy transformation is never cent percent. It involves degradation or dissipation of energy from a concentrated to a dispersed form as is used to maintain the metabolism. So only a part of energy is stored in the biomass. When a producer (plant) traps radiant energy for photosynthesis, only 2-10% of PAR (only 1-5% of incident solar radiations) is used for photosynthesis called GPP. About 0.2 to 1% of incident radiations is used by plants for respiration and 0.8 to 4% of incident radiations is used to produce biomass called NPP.

When a herbivore eats a producer about 90% of energy will be dissipated and only 10% of energy is available for producing biomass. It will be repeated when herbivore will be eaten by a carnivore. It is called 10% (ten percent) law which was proposed by Lindemann, 1942.

Question 8.
What will happen to an ecosystem if:
(a) All producers are removed;
(b) All organisms of herbivore level are eliminated; and
(c) All top carnivore population is removed.
Answer:
(a) In food chain plants occupy the place of producers. They are the source of food or energy for every organism. If plants producers are removed there will be reduction in primary productivity. No biomass will be available for consumption to higher levels of organisms and they all will die.

(b) If all herbivores are eliminated, there will be increase in primary productivity because of lack of consumers. Carnivores population will starve to death due to unavailability of food.

(c) ‘If all top carnivores are removed, the population of herbivores will increase. They will consume more producers, which can lead to desertification.

Question 9.
Give two examples of artificial or man made ecosystems. List the salient features by which they differ from natural ecosystems.
Answer:
Artificial or man-made ecosystem is created and maintained by human beings. Agriculture, garden, aquarium are artificial
ecosystems.

In artificial ecosystem, biotic and abiotic components are maintained artificially e.g., feeding, cleaning and supply of oxygen to fishes in aquarium.
A natural ecosystem is one which develops in nature without human support or interference i.e., forest, marine ecosystem.

In natural ecosystem, biotic and abiotic components are maintained naturally like light, nutrient cycle, self-sustainability, etc.

Question 10.
The biodiversity increases when one moves from the pioneer to the climax stage. What could be the explanation?
Answer:
The biodiversity increases in an ecological succession when one moves from pioneer community to climax stage due to following reasons –

1. Environmental conditions become more and more favourable for survival of different organisms.
2. Variety of ecological niches increases and become available to many organisms.
3. Biomass and standing crop of organic matter increases with succession. According to Odum, the increase in amount of and the change in organic structure are two of the main factors bringing about succession of species. The enlargement of organic structure is of course, related in a cause-and-effect manner to increase in species diversity.

Question 11.
What is a biogeochemical cycle. What is the role of the reservoir in a biogeochemical cycle. Give an example of a sedimentary cycle with resefvoir located in earth’s crust.
Answer:
Biogeochemical cycles are cyclic pathways through which chemical elements move from environment to organism and back to the environment. Biogeochemicals are essential elements required by the organism for their body building and metabolism, which are provided by earth and return to earth after their death and decay.

Reservoir pool is the reservoir of biogenetic nutrients from which the latter are slowly transferred to cycling pool e.g., phosphates in rocks. The function of reservoir is to meet deficient of nutrient which occurs due to differences in rate of influx and efflux. Atmosphere acts as reservoir pool for carbon and nitrogen cycles.

Phosphorus cycle is an example of sedi-mentary cycle.Phosphate present in the soil may occur in the insoluble form, which is dissolved by chemicals secreted by microbes and plant roots. The dissolved phosphate when absorbed by the plant as orthophosphate ions change to organic form and is then transferred to consumers and decomposer through food chain. Animal excretion and dead bodies when acted upon by decomposer, releases phosphorus, which is recycled/reutilised again. Leaching, erosion and mining also releases phosphate and make it available to plants. In aquatic environment, phosphate is taken from water by phytoplankton, consumed by zooplankton, which in turn excrete it into water.

1. It is a an imperfect cycle as the biological processes (teeth and bones formation and excretion) account for considerable losses of phosphorus.
2. It also shows one way flow : Phosphate rock —» land ecosystem > oceans ocean sediment.
   It, causes eutrophication and pollution when its concentration increases in natural water.

Question 12.
What will be the P/R ratio of a climax community and a pioneer community. What explanation could you offer for the changes seen in P/R ratio of a pioneer community and the climax community.
Answer:
The species that invade a bare area are called pioneer species. These are generally lichens. A pioneer community has maximum number of producers. The rate of production P is higher than rate of respiration R. Therefore P/R ratio of pioneer community is more than 1. This increases the biomass. But with the progress in succession i.e., advancement towards climax community biomass of organisms increases and P/R ratio becomes equal to 1. This shows stability of the ecosystem.

When number of organisms increases by reaching climax community, rate of respiration increases greater than rate of production and community is dominated by heterotroph, where P/R ratio become less than 1.

We hope the NCERT Exemplar Solutions for Class 12 Biology chapter 14 Ecosystem help you. If you have any query regarding .NCERT Exemplar Solutions for Class 12 Biology chapter 14 Ecosystem, drop a comment below and we will get back to you at the earliest.

NCERT Exemplar Solutions for Class 12 Biology chapter 13 Organisms and Populations

These Solutions are part of NCERT Exemplar Solutions for Class 12 Biology. Here we have given NCERT Exemplar Solutions for Class 12 Biology chapter 13 Organisms and Populations

Multiple Choice Questions

Question 1.
Autecology is the
(a) relation of heterogenous population to its environment
(b) relation of an individual to its environ-ment
(c) relation of a community to its environ-ment
(d) relation of a biome to its environment.
Answer:
(a, b) : Ecology is the study of interactions among organisms and between the organisms and their biotic and abiotic environment. The study of reciprocal relationships between every stage of development of a population/ species/individual and its environment is called autecology.

Question 2.
Ecotone is
(a) a polluted area
(b) the bottom of a lake
(c) a zone of transition between two communi¬ties
(d) a zone of developing community.
Answer:
(c) : The adjacent biotic communities do not always have sharp lines of demarcation between them. There are usually transition zones, the ecotones, between them. An ecotone often has some populations from each adjacent community and some characteristic of itself. The total number of species in ecotone is generally greater than in the adjoining communities, a phenomenon called edge effect.

Question 3.
Biosphere is
(a) a component in the ecosystem.
(b) composed of the plants present in the soil
(c) life in the outer space
(d) composed of all living organisms present on earth which interact with the physical environment.
Answer:
(d) : Biosphere or ecosphere is self-sufficient system. It includes the earth’s atmosphere (air, land, water) that sustains life. In biosphere living organisms interact with their physical environment.

Question 4.
Ecological niche is
(a) the surface area of the ocean
(b) an ecologically adapted zone
(c) the physical position and functional role of a species within the community
(d) formed of all plants and animals living at the bottom of a lake.
Answer:
(c) : Ecological niche is specific part of habitat occupied by individuals of a species which is circumscribed by its range of tolerance, range of movement, microclimate, type of food and its availability, shelter, type of predator and timing of activity. Tadpole and adult frog occupy different ecological niches as the former is herbivorous aquatic while the latter is carnivorous amphibian.

Question 5.
According to Allen’s Rule, the mammals from colder climates have
(a) shorter ears and longer limbs
(b) longer ears and shorter limbs
(c) longer ears and longer limbs
(d) shorter ears and shorter limbs.
Answer:
(d) : According to Allen’s Rule, animals in colder climates generally have smaller extremities like shorter ears and limbs. It is a mechanism to conserve heat by checking heat loss from the body. Heat loss can be minimised by decreasing the surface area to volume ratio of body. In the polar regions, small animals are rarely found because of their high surface area and less volume.

Question 6.
Salt concentration (salinity) of the sea measured in parts per thousand is
(a) 10-5
(b) 30-70
(c) 0-5
(d) 30-35
Answer:
(d) : Salinity of water bodies is generally measured in parts per thousand. It determines what kind of organisms can live in it. Salinity of the sea is 30-35 parts per thousand, while for inland waters and some hypersaline lagoons it is less than 5 and more than 100 per thousand parts, respectively. Fresh water animals generally cannot. live for long in sea water and vice versa because of osmotic problems.

Question 7.
Formation of tropical forests needs mean annual temperature and mean annual precipitation as
(a) 18-25°C and 150-400 cm
(b) 5-15°C and 50-100 cm
(c) 30-50°C and 100-150 cm
(d) 5-15°C and 100-200 cm.
Answer:
(a) : Temperature and precipitation are generally most important climatic abiotic factors that influence the geographical distribution of plants. Average temperature decreases from equator to the poles. Precipitation includes rainfall, snow, dew, etc. Tropical forests, which are very productive, need 18 to 25°C mean annual temperature and 150-400 cm mean annual rainfall.

Question 8.
Which of the following forest plants controls the light conditions at the ground?
(a) Lianas and climbers
(b) Shrubs
(c) Tall trees
(d) Herbs
Answer:
(c) : In a forest, plants get arranged in various strata according to their shade tolerance. Tall trees form the canopy of the forest i.e., roof of the forest, thus, controlling the amount of light reaching the ground.

Question 9.
What will happen to a well growing herbaceous plant in the forest if it is transplanted outside the forest in a park?
(a) It will grow normally.
(b) It will grow well because it is planted in the same locality.
(c) It may not survive because of change in its microclimate.
(d) It grows very well because the plant gets more sunlight.
Answer:
(c) : Herbaceous plants are of small height. Since, this herbaceous plant is growing in forest, it must have been adapted to the light intensity reaching it, moisture in surrounding air and soil, soil characteristics, etc. This constitutes the microclimate of this plant. If this plant is transplanted to a park outside forest, this microclimate might get lost, partially or completely, because of which plant might not be able to survive.

Question 10.
If a population of 50 Paramecium present in pool increases to 150 after an hour, what would be the growth rate of population?
(a) 50 per hour
(b) 200 per hour
(c) 5 per hour
(d) 100 per hour
Answer:
(d). Population of Paramecium at time t = 50
Population of Paramecium 1 hour after t= 150
Growth rate of population = 150 – 50 per
hour = 100 per hour

Question 11.
What would be the percent growth or birth rate per individual per hour for the same population mentioned in the previous question (Question 10)?
(a) 100
(b) 200
(c) 50
(d) 150
Answer:
(b) : Initial number of Paramecium = 50
Number of Paramecium after 1 hour = 150 Birth rate

Question 12.
A population has more young individuals compared to the older individuals. What would be the status of the population after some years?
(a) It will decline.
(b) It will stabilise.
(c) It will increase.
(d) It will first decline and then stabilise.
Answer:
(c) : If in a population more young individuals are present as compared to older individuals, population will increase after some years. This is because of the reason that number of individuals in pre-reproductive age is high and thus, more number of individuals will enter the reproductive age in coming years. Such a population can be represented by an age structure forming an upright pyramid.

Question 13.
What parameters are used for tiger census in our country’s national parks and sanctuaries?
(a) Pug marks only
(b) Pug marks and faecal pellets
(c) Faecal pellets only
(d) Actual head counts
Answer:
(b) : In our country, parameters for tiger census in national parks and sanctuaries include both pug marks and faecal pellets. They can be used to determine the number as well as distribution of tiger population. New techniques of excreta DNA analysis and camera trappings are considered more accurate for tiger census.

Question 14.
Which of the following would necessarily decrease the density of a population in a given habitat?
(a) Natality > mortality
(b) Immigration > emigration
(c) Mortality and emigration
(d) Natality and immigration
Answer:
(c) : Population density in a given habitat is influenced by four processes :

• Natality : It is the number of births in a population during a given period of time.
• Mortality : It is the number of deaths in a population during a given period of time.
• Immigration : Number of individuals coming from other habitats in a given period of time
• Emigration : Number of individuals exiting from a given pop-.Tation in a given period of time.
• Mortality and emigration will, therefore decrease the population density while natality and immigration will increase it.
  

Question 15.
A protozoan reproduces by binary fission. What will be the number of protozoans in its population after six generations?
(a) 128
(b) 24
(c) 64
(d) 32
Answer:
(c) : By binary fission, an individual protozoan will divide in two. In this way after six generations their number will be 64.

Question 16.
In 2005, for each of the 14 million people present in a country, 0.028 were born and 0. 008 died during the year. Using exponential equation, the number of people present in 2015 is predicted as
(a) 25 millions
(b) 17 millions
(c) 20 millions
(d) 18 millions.
Answer:

Question 17.
Amensalism is an association between two species where
(a) one species is harmed and other is benefited
(b) one species is harmed and other is unaffected
(c) one species is benefited and other is , unaffected
(d) both the species are harmed.
Answer:
(b) : Amensalism is the interaction between two populations in which oneis adversely affected, whereas other is apparently, neither harmed nor benefited. For example, Penicilliitm does not allow the growth of Staphylococcus bacterium by secreting certain chemicals.

Question 18.
Lichens are the associations of
(a) bacteria and fungus
(b) algae and bacterium
(c) fungus and algae
(d) fungus and virus.
Answer:
(c) : Lichens are association (mutually beneficial) between fungus and alga. The fungal partner is mycobiont and algal partner is phycobiont. Lichens can grow in extremely inhospitable conditions. In many ecosystems they are the pioneer species. The role of mycobiont is to provide body structure and anchorage and absorption of minerals and water. The role of phycobiont is to manufacture food through photosynthesis for itself and also for fungus.

Question 19.
Which of the following is a partial root parasite?
(a) Sandal wood
(b) Mistletoe
(c) Orobanche
(d) Ganoderma
Answer:
(a) : Parasites can be divided into : holo- parasitesandhemiparasites. Holoparasitesare those which are dependent on their host for all of their requirements, while hemiparasites are those, which receive only a part of their nourishment from host. Holoparasites and hemiparasites are also known as complete and partial parasites, respectively. Sandal wood is partial root parasite, which synthesises its own food but is dependent on host’s roots for water and inorganic nutrients. Mistletoe is partial stem parasite. Orobanche (Broomrope) is complete root parasite. Ganoderma, a fungus, is parasitic on hardwood.

Question 20.
Which one of the following organisms reproduces sexually only once in its lifetime?
(a) Banana plant
(b) Mango
(c) Tomato
(d) Eucalyptus
Answer:
(c) : Since tomato is an annual plant, so it reproduces sexually only once in its life time.

Very Short Answer Type Questions

Question 1.
Species that can tolerate narrow range of temperature are called …………
Answer:
stenothermal organisms

Question 2.
What are eurythermic species?
Answer:
Eurythermal organisms are those which can tolerate wide range of temperature variations, e.g., most mammals and birds.

Question 3.
Species that can tolerate wide range of salinity are called
Answer:
eurvhaline species

Question 4.
Define stenohaline species.
Answer:
Species which can tolerate only a narrow range of salinity are called stenohaline species e.g., sharks.

Question 5.
What is the interaction between two species called?
Answer:
The interaction between two species is called interspecific interaction, it can be positive, negative or neutral.

Question 6.
What is commensalism?
Answer:
Commensalism isanassociation between two organisms, in which one is benefited and second is neither harmed nor benefited. E.g., interaction between sucker fish and shark.

Question 7.
Name the association in which one species produces poisonous substance or a change in environmental conditions that is harmful to another species.
Answer:
In amensalism one organism inhibits the growth of the other. This inhibition is done by secreting chemicals called allochemics.

Question 8.
What is mycorrhiza?
Answer:
Symbiotic association between fungus and the roots of higher plants is called mycorrhiza. The fungus helps the plant in absorption of essential nutrients from the soil, while the plant in turn provides the fungus with energy yielding carbohydrates.

Question 9.
Emergent land plants that can tolerate the salinities of the sea are called .
Answer:
halophytes

Question 10.
Why do high altitude areas have brighter sunlight and lower temperatures as compared to the plains?
Answer:
Brighter sunlight at high altitudes is due to thinner air and less pollution. High altitudes have lower temperature because as we go higher, air pressure decreases that causes the temperature to be colder at high altitudes.

Question 11.
What is homeostasis?
Answer:
Homeostasis is the ability to maintain constant body temperature and osmotic concentration irrespective of the environmental conditions.

Question 12.
Define aestivation.
Answer:
The period of dormancy or suspended metabolism during summer months to escape from heat and unfavourable conditions is called aestivation or summer sleep, c.g., snails undergo aestivation to avoid heat and dessication.

Question 13.
What is diapause and its significance?
Answer:
Diapause is the stage of suspended development. During stress period, many zooplanktons and larvae of insects suspend their development and enter a stage of dormancy. They resume their growth during favourable conditions. This helps organisms to escape harsh, extreme conditions.

Question 14.
What would be the growth rate pattern, when the resources are unlimited?
Answer:
When resources like food, space, etc., are unlimited for a population, it grows in an exponential or geometric ratio resulting in a J-shaped growth curve, c.g., growth of algal bloom.

Question 15.
What are the organisms that feed on plant sap and other plant parts called?
Answer:
Organisms feeding on plant sap and other plant parts are called phytophagous.

Question 16.
What is high altitude sickness? Write its symptoms.
Answer:
High altitude sickness is experienced at high altitudes (> 3500 m) like Leh, Rohtang Pass, etc. It is due to low atmospheric pressure at high altitude due to which the body does not get enough oxygen. Its symptoms include nausea, fatigue, heart palpitation.

Question 17.
Give a suitable example for commensalism.
Answer:
The cattle egret and grazing cattle show an example of commensalism. The egrets always forage close to where cattle are grazing, because the cattle, as they move, stir up and flush out insects from the vegetation which are otherwise difficult for egrets to find and catch.

Question 18.
Define ectoparasite and endoparasite and give suitable examples.
Answer:
Ectoparasites live on the surface of the host. They suck blood (in animals), or juice (in plants), c.g., lice on humans and ticks on dogs.

Endoparasites live inside the body of host Most of them spend a part of their life cycle in another host. Their life cycle is more complex because of their extreme specialisation, c.g., Plasmodium in RBCs and Taenia solium in intestine.

Question 19.
What is brood parasitism? Explain with the help of an example.
Answer:
Brood parasitism is an example of parasitism in which the parasitic bird lays its eggs in the nest of its host and the host incubates them. During the course of evolution, the eggs of the parasitic bird have evolved to resemble the host’s eggs in size and colour to reduce the chances of the host bird detecting the foreign eggs and ejecting them from the nest. Most common example is koel whose female lays its eggs in the crow’s nest.

Short Answer Type Questions

Question 1.
Why are coral reefs not found in the regions from West Bengal to Andhra Pradesh but are found in Tamil Nadu and on the east coast of India?
Answer:
Coral reefs are found in zone with high salt concentration (salinity), optimal temperature and with a less siltation condition which fairly facilitate the corals to colonise. In case of high siltation and very high freshwater inflow, coral reefs do not colonise.

Question 2.
If a freshwater fish is placed in an aquarium containing sea water, will the fish be able to survive? Explain giving reasons.
Answer:
freshwater fish is placed in aquarium having sea water, then it would not be able to survive due to osmoregulatory problems. Outer hypertonic environment having high salt concentration would cause exosmosis and fish would die.

Question 3.
Why do all the freshwater organisms have contractile vacuoles whereas majority of marine organisms lack them?
Answer:
Osmotic concentration of freshwater organisms is hypertonic to the surrounding water. The water enters their body by endosmosis. Excess water is expelled out of the body through contractile vacuole. Osmotic concentration of most of the marine organisms is almost equal to surrounding water, i.e., isotonic to surroundings, therefore, they need pot expel extra water out of their body and hence lack contractile vacuole.

Question 4.
Define heliophytes and sciophytes. Name a plant from your locality that is either heliophyte or sciophyte.
Answer:
Plants adapted to bright light are known as heliophytes and are commonly called sun plants. E.g., sunflower.
Plants growing in partial shade or low intensity of light are called sciophytes or shade plants. E.g., money plant.

Question 5.
Why do submerged plants receive weaker illumination than exposed floating plants in a lake?
Answer:
There is a light zonation in deep lakes and oceans. Floating plants are found in upper part of lake and receive more light, however less light penetrates into deep lake and thus submerged plants receive weaker illumination of light.

Question 6.
In a sea shore, the benthic animals live in sandy, muddy and rocky substrata and accordingly developed the following adaptations.
(a) Burrowing
(b) Building cubes
(c) Holdfasts/peduncle
Find the suitable substratum against each adaptation.
Answer:
ln a sea shore, water current restrict distribution of organisms. In streamed areas of ocean, animals are strong swimmers or possess attaching organs such as pedunde, or live under stone, in burrows etc. Burrowing animals like tubeworm, Nereis are strong swimmers. Burrowing, building cubes and hold fast or penduncle are found in sandy muddy and rocky substratum respectively.

Question 7.
Categorise the following plants into hydrophytes, halophytes, mesophytes and xerophytes. Give reasons for your answer.
(a) Salvinia
(b) Opuntia
(c) Rhizophora
(d) Mangifera
Answer:
(a) Salvinia – Hydrophyte – Free floating aquatic plant
(b) Opuntia – Xerophyte – Grows in hot, dry conditions with less rainfall.
(c) Rhizophora – Halophyte – Grows in saline soil.
(d) Mangifera – Mesophyte – Grows in mesic (moderate) conditions with high moisture.

Question 8.
In a pond, we see plants which are free-floating; rooted-submerged; rooted emergent; rooted with floating leaves. Write the type of plants against each of them.”

Answer:

Question 9.
The density of a population in a habitat per unit area is measured in different units. Write the unit of measurement against the following:
(a) Bacteria …………
(b) Banyan …………
(c) Deer ………….
(d) Fish …………..
Answer:
(a) Bacteria – Number per unit volume
(b) Banyan tree – Biomass per unit area
(c) Deer – Number per unit area
(d) Fish – Weight per unit area

Question 10.

(a) Label the three tiers 1, 2, 3 given in the above age pyramid.
(b) What type of population growth is represented by the above age pyramid?
Answer:
(a)
1 – Pre-reproductive individuals
2 – Reproductive individuals
3 – Post-reproductive individuals

(b) Given age pyramid has broad base or triangular structure which indicates a rapidly expanding population with a high percentage of young individuals and only few old individuals.

Question 11.
In an association of two animal species, one is a termite which feeds on wood and the other is a protozoan Trichonympha present in the gut of the termite. What type of association they establish?
Answer:
Termites and Trichonympha show mutualistic relationship. Termites feed on wood though they do not possess enzymes for digesting the same. Termites harbour cellulose digesting flagellates (e.g., Trichonympha campanula) for this purpose. Flagellates are unable to live independently. Termites would die of starvation in the absence of flagellates.

Question 12.
Lianas are vascular plants rooted in the ground and maintain erectness of their stem by making use of other trees for support. They do not maintain direct relation with those trees. Discuss the type of association the lianas have with the trees.
Answer:
Lianas are woody vascular climbers. They maintain their erectness of stem by using the support of the tree without any direct relation with it. In this relation, lianas are benefited, but the tree neither gets benefit nor is harmed. It represents commensalism, in which one is benefited and other remains unaffected.

Question 13.
Give the scientific names of any two microorganisms inhabiting the human intestine.
Answer:
Escherichia coli and Lactobacillus are two microorganisms which inhabit human intestine.

Question 14.
What is a tree line?
Answer:
The tree line is the edge of the habitat till which trees are capable of growing. Beyond the tree line, trees cannot tolerate the environmental conditions, like, temperature, humidity. Beyond this limit trees are not found.

Question 15.
Define ‘zero population growth rate’. Draw a age pyramid for the same.
Answer:
When birth rate and death rate become equal and there is no increase in population, it is called zero population growth. It is obtained when the number of pre-reproductive and reproductive individuals is almost equal and post-reproductive individuals are comparatively

Question 16.
List any four characters that are employed in human population census
Answer:
Census is an official counting of population and preparing data considering various population characteristics, such as :

1. Birth rate
2. Sex ratio
3. Death rate
4. Age distribution

Question 17.
Give one example for each of the following types.
(a) Migratory animal
(b) Camouflaged animal
(c) Predator animal
(d) Biological control agent
(e) Phytophagous animal
(f) Chemical defence agent
Answer:
(a) Migratory animal – Arctic tern
(b) Camouflaged animal – Leaf insect
(c) Predator animal – Lion
(d) Biological control agent – Gambusia
(e) Phytophagous – Butterfly
(f) Chemical defense agent – Cardiac glycosides produced by Calotropis

18.
Fill in the blanks:

Species  A	Species B	Type of Interaction	Example
+	–
+	+
+		Commensalism

Answer:

Species A	Species B	Type of Interaction	Example
+	–	Predation	Tiger and Deer
+	+	Mutualism	Lichen
+	O	Commensalism	Sucker fish and shark

Question 19.
Observe the set of 4 figures A, B, C and D and, answer the following questions.

1. Which one of the figures shows mutualism?
2. What kind of association is shown in D?
3. Name the organisms and the association in
4. What role is the insect performing in B?
   

Answer:

1. Fig. A shows mutualism, in plant- animal relationship. Plants need the help of animals for pollinating their flowers. Animals get rewards in the form of pollen and nectar.
2. Fig. D shows predation where, tiger is the predator and deer is a prey.
3. Fig. C shows commensalism between the cattle egret and grazing cattle. The cattle egret and grazing cattle show an example of commensalism. The egrets always forage close to where cattle are grazing, because the cattle, as they move, stir up and flush out insects from the vegetation which are otherwise difficult for egrets to find and catch.
4. In fig. B, insect is scavenging, i.c., it is ‘ feeding on dead and decaying matter.

Long Answer Type Questions

Question 1.
Comment on the following figures 1,2 and 3. A, B, C, D, G, P, Q, R and S are species.
Answer:
In fig. 1, all the individuals are of same species and show intraspecific competition. Here, all the members of species have similar requirements of food, light, water, space, shelter and mate.
In Fig. 2, there are individuals of three different species (A, B and C) and have interspecific competition. Here two or more populations usually belonging to same trophic level or feeding habit compete with one another for natural resources. E.g., tigers and leopards may compete for same prey in forest, trees, shrubs, herbs and vines compete for water, sunlight, nutrients, pollinators, etc.
In Fig. 3, three different biotic communities interact with one another and with their physical environment and exchange energy and material with surroundings. It represents a biome.

Question 2.
An individual and a population has certain characteristics. Name these attributes with definitions.
Answer:
An individual or an organism is the smallest unit of ecological hierarchy.Characteristics of an individual are :

1. It performs all body functions indepen-dently, but cannot live in isolation.
2.  It depends on both biotic and abiotic environment.
3. It reproduces to give rise to young ones.
4. Organism is well adapted to the environment and has a definite life span.

Population is the total number of individuals of a species in a specific geogra¬phical area.
Characteristics of population are:

1. Natality : It refers to the number of births during a given period in the population that are added to the initial density.
2. Mortality : It is the number of deaths in the population during a given period.
3. Immigration : It is the number of individuals of the same species that have come into the habitat from elsewhere during the time period.
4. Emigration : It is the number of individuals of the population who left the habitat and gone elsewhere during the time period.
5. Sex ratio : Number of females produced per 1000 males.

Question 3.
The following diagrams are the age pyramids of different populations. Comment on the status of these populations.

Answer:
Fig. A shows a triangular age pyramid. It indicates that number of individuals in pre-reproductive age group are much more than those in reproductive age group, which in turn are more than the individuals in post reproductive age group. This type of population shows rapid growth and is known as expanding population. Fig. B is a bell shaped age pyramid. Here, individuals in pre-reproductive age group and reproductive age group are almost same. Such population show the zero growth rate because death rate and birth rate are equal. The population is termed as stable or stationary population.
Fig. C shows an urn shaped age pyramid.
Here, . number of individuals in pre- reproductive age group are less than number of individuals in reproductive age group. Number of post-reproductive individuals is also sizeable. Death rate is more than the birth rate and this pyramid shows declining population.

Question 4.
Comment on the growth curve given below.

Answer:
The given growth model shows logistic growth, where the resources are limited and a given habitat can support certain number of individuals. The limit beyond which population cannot grow further is known as carrying capacity. Population shows sigmoid growth curve, (S-shaped curve) and has following phases – lag phase, log phase, exponential phase and stationary phase.
This population growth is called Verhulst – Pearl logistic growth, it has following equation :

where,
N = Population density at a time
K = Carrying capacity
r = Intrinsic rate of natural increase.

Question 5.
A population of Paramecium caudatum was grown in a culture medium. After 5 days the culture medium became overcrowded with Paramecium and had depleted nutrients. What will happen to the population and what type of growth curve will the population attain? Draw the growth curve.
Answer:
Paramecium caudatum when grown in a culture medium will show logistic or sigmoid growth curve. Such type of growth occurs when a population is growing in a habitat with ‘limited resources. The population of Paramecium initially shows a lag phase followed by phase of exponential growth as the nutrients and space will be abundant in nutrient medium. When the resources get depleted, the population density starts decreasing and finally the population density reaches carrying capacity beyond which no further growth is possible.

Question 6.
Discuss the various types of positive interactions between species.
Answer:
In positive population interactions, both members involved are either benefited or one is benefited and other remains unaffected. Types of positive interactions are as follows :
(1) Mutualism – An interaction between two organisms of different species where both the partners are benefited. Examples of symbiosis or mutualism are lichens, symbiotic nitrogen fixation, mycorrhizae etc.
Lichen is a composite entity which is formed jointly by an alga (phycobiont) and a fungus (mycobiont). The main body of the lichen is formed of fungus. The fungus also provides, water, minerals and shelter to the alga. The alga manufactures food not only for itself but also for the fungus. This interaction or relationship allows the lichen to grow in highly hostile environment like bare rock,

(2) Commensalism – It is the relationship between two living individuals of different species in which one is benefited while the other is neither harmed nor benefited except to a negligible extent. E.g., sucker fish attaches itself to the under-surface of shark with the help of its dorsal fin which is modified into holdfast. Sucker fish gets a free ride. It is widely dispersed and remains protected from its predators. Sucker fish detaches itself from shark w’hen the latter is feeding to obtain smaller pieces of food.

Question 7.
In an aquarium two herbivorous species of ‘fish are living together and feeding on phytoplanktons. As per the Gause’s principle, one of the species is to be eliminated in due course of time, but both are surviving well in the aquarium. Give possible reasons.
Answer:
According to Gause’s ‘Competitive Exclusion Principle’ two closely related species competing for same resources cannot co-exist indefinitely and the competitively inferior one will be eliminated eventually. If two herbivorous species of fish are co-existing in same ecological niche, then it might be possible that the two herbivorous species of fish may have developed different specialisations, i.e., they may have evolved mechanisms to encourage co-existence, e.g., resource partitioning.

Question 8.
While living in and on the host species, the animal parasite has evolved certain adaptations. Describe these adaptations with examples.
Answer:
Parasites living inside or outside the bodies of host species evolve certain special adaptations to have better survival. The few parasitic adaptations are as follow’s:

1. Parasites living in intestine usually develop a covering called cuticle to resist digestive enzvmes of the host, e.g., Ascnris.
2. Endoparasites undergo anaerobic respi-ration.
3. Some develop adhesive structures to avoid expulsion, e.g., hooks and suckers in tapeworm.
4. Loss of locomotorv organs and absence of joined appendages, e.g., Sacculimi.
5. High rate of reproduction to compensate loss during transfer from one host to another.
6. Simplified sense organs and nervous system.

Question 9.
Do you agree that regional and local variations exist within each biome? Substantiate your answer with suitable example.
Answer:
Abiome is a major ecological communitv or complex of communities characterised by a major vegetation type and distinct landscape
characters. Different types of flora and fauna exist in a biome. Climate is the main factor that determines the type of soil which in turn determines the type of vegetation. The type of vegetation and climate together determine the type of animal population inhabiting the area. The organisms of a biome are adapted to the climatic conditions associated with the region. There are no distinct boundaries between adjacent biomes which merge gradually with each other.

World major biomes include tropical forests, savanna, deserts, polar regions, chaparral, temperate grasslands, temperate deciduous forests, coniferous forests and tundra. Each biome has regional and local variations so that there is wide variety of habitats and ecological niches. For example, the temperate broad leaf and mixed forests biome include a number of distinct ecoregions with characteristic species. In the south, the middle Atlantic coastal forest occupies the flat Atlantic coastal plain, from the eastern shore of Maryland and Delaware to just south of Georgia. The river swamp or bottom land forests in this ecoregion are dominated by majestic bald cypress and swamp tupelo. Farther north, the north eastern forest type is typified by white and red oak.

Question 10.
Which element is responsible for causing soil salinity? At what concentration does the soil become saline?
Answer:
The most important element responsible for causing soil salinity is Na present as NaCl in the soil. Chlorides, nitrates, sulphates and carbonates of potassium, magnesium and ^sodium; chlorides and nitrates of calcium are important elements to make soil more saline.

Question 11.
Does light factor affect the distribution of organisms? Write a brief note giving suitable examples of either plants or animals.
Answer:
Light is an important abiotic factor responsible for growth, development and distribution of plants. There are three parameters of light.

1. Light intensity,
2. Light quality
3. Light duration

(1) Decreasing light intensity at various depths in ocean determines distribution of plants.
Green algae – found along the shoreline Brown algae – deeper level Red algae – Very deep in water

(2) Terrestrial communities are regulated by light. The dominant species occupy highest canopy level to utilise maximum light. The quantity and quality of light result in different types of plants growing in different areas in various states.

(3) Day length varies, when we move from equator to poles, it plays important role in plant distribution. According to duration of light available to the plants, they can be long day plants, short day plants or day neutral plants. The flowering period varies depending on the availability and duration of light.

Question 12.
Give one example for each of the following:

1. Eurythermal plant species
2. A hot water spring organism
3. An organism seen in deep ocean trenches
4. An organism seen in compost pit
5. A parasitic angiosperm
6. A stenothermal plant species
7. Soil organism
8. A benthic animal
9. Antifreeze compound seen in antarctic fish
10. An organism which can conform

Answer:

1. Eurvthermal plant species – Artemcsin.
2. A hot water spring organism – Thermits aquatints
3. An organism seen in deep ocean trenches – Euplectclla
4. An organism seen in compost pit – Pseudomonas
5. A parasitic angiosperm – Cuscuta
6. A stenothermal plant species – Coconut
7. Soil organism – Earthworm
8. A benthic animal – Sponge
9. Antifreeze compound seen in antarctic fish – Glycoprotein
10. An organism which can conform – Asterias

We hope the NCERT Exemplar Solutions for Class 12 Biology chapter 13 Organisms and Populations help you. If you have any query regarding .NCERT Exemplar Solutions for Class 12 Biology chapter 13 Organisms and Populations, drop a comment below and we will get back to you at the earliest.

NCERT Exemplar Solutions for Class 12 Biology chapter 12 Biotechnology and Its Applications

These Solutions are part of NCERT Exemplar Solutions for Class 12 Biology. Here we have given NCERT Exemplar Solutions for Class 12 Biology chapter 12 Biotechnology and Its Applications

Multiple Choice Questions

Question 1.
Bt cotton is not
(a) a GM plant
(b) insect resistant
(c) a bacterial gene expressing system
(d) resistant to all pesticides.
Answer:
(d) : Several transgenic plants have been developed. One such plant is Bt cotton in which the Bt toxin from the bacterium Bacillus thuringiensis has been used as a biological insecticide. The choice of genes depends upon the crop and targeted pest. Bt cotton is resistant to insects but is not pesticide resistant.

Question 2.
C-peptide of human insulin is
(a) a part of mature insulin molecule
(b) responsible for formation of disulphide bridges
(c) removed during maturation of pro-insulin to insulin
(d) responsible for its biological activity.
Answer:
(c) : In mammals, including humans, insulin is synthesized as a pro-hormone which needs to be processed before it becomes a fully mature and functional hormone. It contains an extra, stretch called the C peptide which is removed during maturation into insulin.

Question 3.
GEAC stands for
(a) Genome Engineering Action Committee
(b) Ground Environment Action Committee
(c) Genetic Engineering Approval Committee
(d) Genetic and Environment Approval Committee.
Answer:
(c) : Genetic modification of organisms can have unpredictable results when such organisms are introduced into the ecosystem. Therefore, the Indian Government has set up organizations such as GEAC (Genetic Engineering Approval Committee), which makes decisions regarding the validity of GM research and the safety of introducing GM- organisms for public services.

Question 4.
a-1 antitrypsin is
(a) an antacid
(b) an enzyme
(c) used to treat arthritis
(d) used to treat emphysema.
Answer:
(d) : Disorders of a-l-antitrypsin protein include a-l-antitrypsin deficiency, an autosomal, codominant hereditary disorder in which deficiency of a-l-antitrypsin leads to a chronic uninhibited tissue breakdown. This causes the degradation of lung tissue, and eventually leads to characteristic manifestations of pulmonary emphysema.

Question 5.
A probe which is a molecule used to locate specific sequences in a mixture of DNA or RNA molecules could be
(a) a single stranded RNA
(b) a single stranded DNA
(c) either RNA or DNA
(d) can be ss DNA but not ss RNA.
Answer:
(a, b) : The molecular probes are usually single stranded pieces of DNAs (sometimes RNAs) labelled with radio isotopes such phosphorus-32. Molecular probes are available for many genetic disorders such as Duchenne muscular dystrophy, cystic fibrosis, Tay-Sachs disease.

Question 6.
Choose the correct option regarding retrovirus.
(a) An RNA virus that can synthesise DNA during infection.
(b) A DNA virus that can synthesise RNA during infection.
(c) A ssDNA virus.
(d) A dsRNA virus.
Answer:
(a) : A retrovirus is a ssRNA virus that stores its nucleic acid in the form of an mRNA genome and targets a host cell as an obligate parasite. Once inside the host cell cytoplasm, the virus uses its own reverse transcriptase enzyme to produce DNA from its RNA genome (the reverse of usual pattern, thus retro).

Question 7.
The site of production of ADA in the body is
(a) erythrocytes
(b) lymphocytes
(c) blood plasma
(d) osteocytes.
Answer:
(b) : Lymphocytes are a kind of white blood cells present in bone marrow. ADA (adenosine deaminase) is an enzyme that is present in lymphocytes and is very important for the immune system to function.

Question 8.
A protoxin is
(a) a primitive toxin
(b) a denatured toxin
(c) toxin produced by protozoa
(d) inactive toxin.
Answer:
(d)

Question 9.
Pathophysiology is the
(a) study of physiology of pathogen
(b) study of normal physiology of host
(c) study of altered physiology of host
(d) none of the above.
Answer:
(c) : Pathophysiology means the functional changes in the affected person associated with or resulting from a disease or injury or a syndrome.

Question 10.
The trigger for activation of toxin of Bacillus thuringiensis is
(a) acidic pH of stomach
(b) high temperature
(c) alkaline pH of gut
(d) mechanical action in the insect gut.
Answer:
(c) : The Bt toxin proteins exist as inactive protoxins but once an insect ingests the inactive toxin it is converted into an active form of toxin due to the alkaline pH of the
alimentary canal that solubilises the crystals. The activated toxin binds to the surface of midgut epithelial cells and create pores which causes cell swelling and lysis and finally cause death of the insect.

Question 11.
Golden rice is
(a) a variety of rice grown along the yellow river in China
(b) long stored rice having yellow colour tint
(c) a transgenic rice having gene for (3-carotene)
(d) wild variety of rice with yellow coloured grains.
Answer:
(c) : Golden rice is a transgenic variety of rice (Oryza sativa) which contains good quantities of (3-carotene (provitamin A- inactive state of vitamin A). (3-carotene is a principle source of vitamin A. Since the grains (seeds) of the rice are yellow in colour due to (3-carotene, the rice is commonly called golden rice.

Question 12.
In RNAi, genes are silenced using
(a) ssDNA
(b) dsDNA
(c) dsRNA
(d) ssRNA.
Answer:
(c) : RNAi takes place in all eukaryotic organisms as a method of cellular defense. This method involves silencing of a specific mRNA due to a complementary dsRNA molecule that binds to and prevents translation of the mRN A (silencing). The source of this complementary RNA could be from an infection by viruses having RNA genomes or mobile genetic elements (transposons) that replicate via an RNA intermediate.

Question 13.
The first clinical gene therapy was done for the treatment of
(a) AIDS
(b) cancer
(c) cystic fibrosis
(d) SCID (Severe Combined Immuno Defici-ency resulting form deficiency of ADA).
Answer:
(d) : The first clinical gene therapy was given in 1990 to a 4-year old girl with adenosine deaminase (ADA) deficiency. This enzyme is very important for the immune system to function. ADA deficiency can lead to Severe Combined Immuno Deficiency (SCID).

Question 14.
ADA is an enzyme which is deficient in a genetic disorder SCID. What is the full form of ADA?
(a) Adenosine deoxyaminase
(b) Adenosine deaminase
(c) Aspartate deaminase
(d) Arginine deaminase
Answer:
(b) Refer answer 13.

Question 15.
Silencing of a gene could be achieved through the use of
(a) RNAi only
(b) antisense RNA only
(c) both RNAi and antisense RNA
(d) none of the above.
Answer:
(c) : For RNAi refer answer 12.
Antisense RNA is an RNA molecule whose base sequence is complementary to that of the sense RNA. It can undergo base pairing with its comlementary mRNA and block the gene expression either by preventing access for ribosomes to translate the mRNA or by triggering degradation of the dsRNA by ribonucleases. Flavr savr variety of tomato is produced using this antisense technology in which an artificial gene inserted for antisense RNA prevented expression of genes that cause ripening. It has been found that RNA interference is more effective than antisense RNA.

Very Short Answer Type Questions

Question 1.
In view of the current food crisis, it is said, that we need another Green Revolution. Highlight the major limitations of the earlier Green Revolution.
Answer:
BID Limitations of Green Revolution are:

1. Increased crop yield is not sufficient to feed the increasing human population.
2. Fertilisers and pesticides which are used to increase the food production are not good for environment.
3. Green revolution was agrochemical based, and as agrochemicals are expensive so, small farmers cannot afford them.

Question 2.
Expand GMO. How is it different from a hybrid?
Answer:
GMO is Genetically Modified Organism.
Differences between GMO and hybrid are as follows:

GMO	Hybrid
(1) GMO is obtained by inserting foreign gene into an organism.	Hybrid is produced by crossing two superior individuals
(2) Change in genotype is precisely controlled.	Change in genotype depends on chance segregation or aggregation during fertilisation.
(3) New traits are introduced	Existing traits of parents are reshuffled.

Question 3.
Differentiate between diagnostics and therapeutics. Give one example and for each category.
Answer:
Diagnosis refers to the detection of disease and understanding its pathophysiology (symptoms, causes etc.) so that it can be cured. It involves ELISA, PCR, blood tests etc. Whereas, therapeutics is treatment of disease by providing proper medication or replacing defective genes by normal ones. E.g., Vaccines, production of humulin, antibiotics etc.

Question 4.
Give the full form of ELISA. Which disease can be detected using it? Discuss the principle underlying the test.
Answer:
HELISA is Enzyme Linked Immuno- sorbant Assay. ELISA is used for detecting HIV, Hepatitis-B virus etc. diseases. ELISA f is based on principle of antigen-antibody interaction. An antibody (Ab) reacts with the concerned antigen (Ag) in a highly specific manner and forms Ag-Ab complex. A second antibody conjugated with enzyme specific to a second site on the test protein is added. The enzyme causes colour change and this colour change is analysed to identify the condition.

Question 5.
Can a disease be detected before its symptoms appear? Explain the principle involved.
Answer:
Yes, a disease can be detected when symptoms are not yet visible due to very low
coqnt of pathogens, by the technique called Polymerase Chain Reaction (PCR). PCR multiplies pathogen nucleic acids and by analysing them, disease can be diagnosed. By PCR, even very low amounts of DNA can be detected and amplified. It is used to detect HIV and many other genetic disorders etc.

Question 6.
Write a short note on biopiracy highlighting the exploitation of developing countries by the developed countries.
Answer:
Biopiracy is exploitation of bioresources of a country by organisations and multi-nationals for commercial exploitation with or without patent but without any Access and Benefit Sharing Agreement (ABA). Bioresources or biological resources are all those organisms which can provide commercial benefits. They are abundant in developing countries which are poor in technology though the countries have traditional knowledge related to bioresources. On the other hand, developed countries are poor in bioresources but are rich in technology. Traditional knowledge helps in saving time, effort and expenditure in developing refined product for commercialisation. Therefore, based on traditional knowledge, institutions and companies of industrialised nations are collecting and exploiting bioresources of other nations by getting them patented.

Question 7.
Many proteins are secreted in their inactive form. This is also true of many toxic proteins produced by microorganisms. Explain how the mechanism is useful for the organism producing the toxin?
Answer:
Bacillus thuringiensis is one such organism which produces a toxic protein in its inactivated form (protoxin). The benefit of doing so is that the bacterium itself remains unaffected from the toxic effects but once an insect ingests it, the alkaline pH of insect’s gut changes it into active form. The activated toxin binds to the surface of midgut epithelial cells and creates pores which cause cell swelling and lysis and death of an insect, but the bacterium remains unaffected.

Question 8.
While creating genetically modified organisms, genetic barriers are not respected. How can this be dangerous in the long run?
Answer:
Genetically modified organisms can be dangerous in long run due to following reasons :

1. It is possible that insects might become resistant to Bt or other crops that have been genetically modified to produce their own pesticides.
2. Genetically engineered herbicide crop plants and weeds may cross breed, resulting in transfer of herbicide resistance genes from crops into weeds. Therefore, weeds would also be herbicide tolerant.
3. The enzyme produced by antibiotic resistance gene can cause allergies, because it is a foreign protein.
4. Genetically engineered microbes may be uncontrollable if get released in ecosystems. They may also transfer virulence to other microbial populations.
5. GMOs can be dangerous to other organisms in the ecosystem. Such as honeybees show toxicity to pollens of insect resistant crops.

Question 9.
Why has the Indian Parliament cleared the second amendment of the country’s patents bill?
Answer:
The Indian Parliament has recently also cleared second amendment of the Indian Patents Bill because there has been growing realisation of injustice, inadequate compensation and benefit sharing between developed and developing countries. Such patent bills prevent unauthorised exploitation of bioresources and traditional knowledge of a country and consider emergency provisions and development initiative.

Question 10.
Give any two reasons why the patent on basmati should not have gone to an American Company.
Answer:
Patents on basmati should not have gone to an American Company because:

1. Presence of rice goes back thousands of years in Asia’s agricultural history. There are over 2,00,000 varieties of rice in India alone.
2. The “new” variety of basmati having the US patent had actually been derived from Indian farmer’s varieties by crossing Indian basmati with semi-dwarf varieties.

Question 11.
How was insulin obtained before the advent of rDNA technology? What were the problems encountered?
Answer:
Before the advent of rDNA technology, insulin was extracted from pancreas of slaughtered pigs and cattle, as insulin is secreted by (3-cells of islets of Langerhans of pancreas. Problem encountered was that this insulin was different from human insulin and caused some undiserable side effects such as allergy.

Question 12.
With respect to understanding diseases, discuss the importance of transgenic animal models.
Answer:
Animals having foreign gene inserted in their genome by rDNA technology are called transgenic animals. Many transgenic animals have been developed to increase our understanding of how genes contribute to the development of disease so that investigation of new treatments for diseases is made possible. Now transgenic models exist for many human diseases such as cancer, cystic fibrosis, rheumatoid arthritis, Alzheimer’s disease, haemophilia, thalessaemia, etc. Such transgenic animals have been created which code for particular products of therapeutic value such as human protein (a-l-antitrypsin) which is used to treat emphysema (sheep), tissue plasmogen activator (goat), blood clotting factors VIII and IX (sheep) and lactoferrin (cow).

Question 13.
Name the first transgenic cow. Which gene was introduced in this cow?
Answer:
First transgenic cow was ‘Rosie’. The gene introduced was the gene for human a lactalbumin. It resulted in production of a lactalbumin (human protein) rich milk which is for more nutritious than normal milk.

Question 14.
PCR is a useful tool for early diagnosis of an infectious disease. Elaborate.
Answer:
Early detection of disease refers to diagnosis before the symptoms are seen. In this case, pathogen (bacteria or virus) are present in very less number, i.e., not capable of producing visible symptoms. Here, PCR is used to multiply their genetic material to get enough quantities to be analysed. E.g., it is used to detect HIV virus in AIDS patient, gene mutations in suspected cancer patients, etc.

Question 15.
What is GEAC and what are its objectives?
Answer:
GEAC is Genetic Engineering Approval Committee, which has been set up by Indian government.
Objectives of GEAC are as follows:

1. To ensure safety of Genetically Modified Organisms (GMO) for public services, i. c., safely using GM organisms for food, medicine etc.
2. To take decisions regarding validity of GM research.

Question 16.
For which variety of Indian rice, the patent was filed by USA company?
Answer:
USA company got patent rights on basmati rice of India. They claimed to develop “new” variety by crossing basmati rice with semi-dwarf varieties.

Question 17.
Discuss the advantages of GMO.
Answer:
USAdvantages of Genetically Modified Organisms (GMO) are:

1. GM crops are more tolerant to abiotic stresses.
2. They are disease resistant.
3. GM plants have increased nutritional value.
4. Transgenic mice are being used for testing vaccine safety.
5. Transgenic animals have been used + as model for studying many human diseases.

Short Answer Type Questions

Question 1.
Gene expression can be controlled with the help of RNA. Explain the method with an example.
Answer:
Silencing of gene expression using a dsRNA is called RNA interference (RNAi). It involves silencing of specific mRNA due to complementary dsRNA molecule that binds to and prevent translation of n;RNA. Using Agrobacterium vectors, nematode specific genes have been introduced into the host plant (tobacco plant). The introduction of DNA was such that it produced both sense and anti¬sense RNA in the host cells. These two RNAs being complementary to each other formed a dsRNA (double stranded RNA) that initiated RNAi.
Different steps involved in making tobacco plant resistant to nematode are briefly described below:

1. RNase enzyme called ‘dicer’ cuts all dsRNA molecules into small interfering RNAs (siRNAs) (21-23 nucleotides long).
2. Each siRN A complexes with ribonucleases (distinct from dicer) to form an RNA- induced silencing complex (RISC).
3. The siRNA unwinds and RISC is activated.
4. The activated RISC targets complementary mRNA molecules. The siRNA strands act as guides where the RISCs cut the transcripts in an area where the siRNA binds to the wRNA. This destroys the 77! RNA.
5. When 77/RNA of the parasite is destroyed no parasite proteins are synthesised. It results in the death of the parasite (nematode) in the transgenic host. Thus the transgenic plant gets itself protected from the parasite.

Question 2.
Ignoring our traditional knowledge can prove costly in the area of biological patenting. Justify.
Answer:
Ignoring traditional knowledge is harmful for developing countries as developed nations take advantage of this. Developing nations are rich in traditional knowledge and biodiversity but are poor in finances and thus are not able to exploit modern applications for commercialisation of bioresources. E.g., basmati rice of India was patented by US company which was later challenged by Indian government.

Question 3.
Highlight any four areas where genetic modification of plants has been useful.
Answer:
Four areas where genetic modification of plants has been successful are :

1. Tolerance of abiotic stress like drought and salinity.
2. Better nutritional value.
3. Increased efficiency of mineral utilisation.
4.  Reduced post harvest losses.

Question 4.
What is a recombinant DNA vaccine? Give two examples.
Answer:
Recombinant DNA vaccines are vaccines that are produced by genetic modification technique. For production of such vaccines, microbes are programmed to produce desired antigenic fragment. E.g., vaccine against hepatitis B virus consists of a portion of the viral protein coat produced by genetically modified yeast.

Question 5.
Why is it that the line of treatment for a genetic disease is different from infectious diseases?
Answer:
Infectious diseases are due to infection with a pathogen. They are treated by killing or inhibiting the growth of pathogen by taking various drugs or by strengthening our immune system. Genetic disorders are due to defect in genes, and subsequent defective protein and enzyme formation and subsequent can be cured by gene therapy, which involves replacement of faulty genes by healthy genes.

Question 6.
Discuss briefly how a probe is used in molecular diagnostics.
Answer:
The molecular probes are usually pieces of ssDNA (or RNA) labelled with radio isotopes such as 32P. These are used for molecular diagnosis of various diseases such as Duchenne muscular dystrophy, cystic fibrosis, Tay-Sachs disease etc.

In molecular diagnosis, a single stranded DNA or RNA joined with a radioactive molecule (probe) is allowed to hybridise with its complementary strand followed by detection using autoradiography.

Question 7.
Who was the first patient who was given gene therapy? Why was the given treatment recurrent in nature?
Answer:
The first clinical gene therapy was given in 1990 to a 4 – year old girl named Ashanti de Silva with adenosine deaminase (ADA) deficiency. This enzyme is very important for the immune system to function as its deficiency causes SCID. In some children, ADA deficiency can be cured by bone marrow transplantation. However, in others it can be treated by enzyme replacement therapy, in which functional ADA is given to the patient by injection. But in both approaches the patients are not completely cured.

As a first step towards gene therapy, lymphocytes, a kind of white blood cells, are extracted from the bone marrow of the patient and are grown in a culture outside the body. A functional ADA cDNA (using a retroviral vector) is then introduced into these lymphocytes, which are reinjected to the patient’s bone marrow. But as these cells do not always remain alive, the patient requires periodic infusion of such genetically engineered lymphocytes. However, if the isolated gene from bone marrow cells producing ADA is introduced into cells at early embryonic stages, it can be a permanent cure.

Question 8.
Taking examples under each category, discuss upstream and downstream processing.
Answer:
A biotechnological manufacturing process can be separated into upstream processes and downstream processes. Upstream processing refers to the entire process of selecting and isolating cells to be used, their cultivation, cell banking and culture expansion of the cells until final harvest (termination of the culture and collection of the live cell batch) is achieved. For example, in production of humulin, upstream processing includes cell line preparation by obtaining human insulin gene, and inserting it into E. coli cells and media and equipment preparation.

Downstream processing is the part where the final harvest is purified and quality checked to make it suitable for commercial applications. Suitable preservatives may also be added. For example, in production of humulin, downstream processing includes lysis of the bacterial cells, separation of cell components from the products, synthesis of active insulin by joining A and B chains to produce mature insulin and its purification to obtain highly purified insulin, fit for medicinal use.

Question 9.
Answer:
Antigens (Ag) are substances which when introduced into the body, stimulate the production of antibodies. Antibodies (Ab) are immunoglobulins which are produced in the body in response to antigens or foreign bodies.
Two diagnostic kits based on Ag and Ab are :

1. ELISA which is used for detection of diseases like HIV.
2. Pregnancy test kit.

Question 10.
ELISA technique is based on the principles of antigen-antibody interaction. Can this technique be used in the molecular diagnosis of a genetic disorder, such as phenylketonuria?
Answer:
Yes, ELISA can be used in diagnosis of a genetic disorder, such as phenylketonuria. In this process antibody against the enzyme which is responsible for the metabolism of phenylalanine i.e. phenylalanine hydroxylase, is used to develop the ELISA based diagnostic technique. The patient has deficiency of this enzyme, thus will test negative while a normal individual which has this enzyme in its body, will test positive.

Question 11.
How is a mature, functional insulin hormone different from its prohormone form?
Answer:
The mature insulin consists of two polypeptide chains – chain A and chain B. The pro-hormone insulin contains an extra C-peptide which is removed during maturation and is not present in mature insulin hormone.

Question 12.
Gene therapy is an attempt to correct a genetic defect by providing a normal gene into the individual. By this, the normal function can be restored. An alternate method would be to provide the gene product (protein/enzyme) known as enzyme replacement therapy, which would also restore the function. Which in your opinion is a better option? Give reason for your answer.
Answer:
Gene therapy is better to cure genetic defect than enzyme replacement therapy because in enzyme replacement therapy a functional ADA is given to the patient by injection and patients do not have functional T-lymphocytes to provide immune response against pathogens and thus patient is not completely cured. In gene therapy, the isolated gene from bone marrow cells producing ADA can be introduced into cells at early embryonic stages and therefore can provide permanent cure.

Question 13.
Transgenic animals are the animals in which a foreign gene is expressed. Such animals can be used to study the fundamental biological process, phenomenon as well as for producing products useful for mankind. Give one example for each type.
Answer:
Transgenic animals used to study fundamental biological processes include transgenic mice in which some specific genes are deleted or replaced with nonfunctional genes thought to be associated with fundamental processes like ageing. Then, the effect of this gene’s absence is studied.

A transgenic animal used to produce products useful for mankind may be exemplified by rosie, a transgenic cow that has gene for human lactalbumin inserted in its genome which is expressed in its milk making it more nutritious.

Question 14.
When a foreign DNA is introduced into an organism, how is it maintained in the host and how is it transferred to the progeny of the organism?
Answer:
The foreign genes are inserted into the genome of an organism using recombinant DNA technology to produce a transgenic organism. This foreign gene is maintained in the host organism because as its cells undergo division, rDNA also gets replicated alongwith the cell chromosomes and daughter cells receive copies of rDNA. Similarly, during reproduction, reproductive cells get the rDNA along ,with the other chromosomes and as a result, the transgenic trait is transferred to the progeny. Chances are 100%, if the organism reproduces asexually.

Question 15.
Bt cotton is resistant to pests, such as iepidopteran, dipterans and coleopterans. is Bt cotton also resistant to other pests as well?
Answer:
Bt cotton is made resistant against specific pests only. Soil bacterium Bacillus thuringiensis produces proteins that kills certain insects like lepidopterans (tobacco budworm), coleopterans (beetles) and dipterans (flies) etc. It does not have gene which is effective against all type of insect pests. Therefore, other pests may attack cotton plant.

Long Answer Type Questions

Question 1.
A patient is suffering from ADA deficiency. Can he be cured? How?
Answer:
Gene therapy is the technique of genetic engineering used to replace a faulty gene by a normal healthy functional gene. The first clinical gene therapy was given in 1990 to a 4 years old girl with adenosine deaminase deficiency (ADA deficiency). This enzyme is very important for the immune system to function. Severe combined immunodeficiency (SCID) is caused due to defect in the gene for the enzyme adenosine deaminase. SCID patient lacks functional T-lymphocytes and, therefore, fails to fight the infecting pathogens.

To perform gene therapy, lymphocytes are extracted from the patient’s bone marrow and a normal functional copy of human gene coding for ADA is introduced into these lymphocytes with the help of retroviral vector. The cells so treated are reintroduced into the patient’s bone marrow. The lymphocytes produced by these cells contain functional ADA gene which reactivates the victim’s immune system.

But, as these lymphocytes do not divide and are short lived, so periodic infusion of genetically engineered lymphocytes is required. This problem can be overcome, if stem cells are modified at early embryonic stage.

Question 2.
Define transgenic animals. Explain in detail any four areas where they can be utilised.
Answer:
The organisms which have their DNA manipulated to possess and express a foreign or extra gene are known as transgenic animals. Various areas where transgenic animals can be used are as follows:

1. Transgenic animals produce useful products, such as human protein (a-1- antitrypsin) used to treat emphysema. Attempts are being made for treatment of phenylketonuria and cystic fibrosis.
2. Transgenic mice are being used for testing the safety of vaccines before they are used for human beings. E.g., they are being used to test the safety of polio vaccine.
3. Transgenic animals act as models for human diseases like cancer, cystic fibrosis, rheumatoid arthritis, Alzheimer’s disease etc., and their possible new methods of their treatment.
4. Transgenic animals are specifically developed to study, how genes are regulated and how they affect normal functioning of the body and its development. E.g. study of complex factors involved in growth, such as insulin like growth-factor (IGF).

Question 3.
You have identified a useful gene in bacteria. Make a flow chart of the steps that you would follow to transfer this gene to a plant.
Answer:

Question 4.
Highlight five areas where biotechnology has influenced our lives.
Answer:
Various areas where biotechnology has influenced our lives are :

1. Gene therapy – A collection of methods that allows correction of a gene defect diagnosed in a person, child or an embryo.
2. Molecular diagnosis – Early detection and treatment of diseases before their symptoms start appearing.
3. Production of proteins using rDNA technique – Several proteins have been produced in abundance for curing certain diseases. They include insulin, growth hormone, interferons, vaccines etc.
4. Agricultural applications – rDNA technology has developed transgenic plants which can tolerate drought, various diseases and have increased productivity.
5. Industrial applications – Enzymes are synthesised and used to produce sugar, cheese and detergents etc.

Question 5.
What are the various advantages of using genetically modified plants to increase the overall yield of the crop?
Answer:
Advantages of genetically modified plants are as follows :

1. Tolerance – GM plants are resistant to abiotic stresses (drought, cold, salinity, heat).
2. Pest resistance – GM plants are pest resistant and can reduce the utilisation of chemical insecticides or pesticides e.g. Bt cotton.
3. Disease resistance – Genetically modified plants are resistant to various diseases caused by bacteria, virus, fungus etc.
4. Reduced post harvest losses – GM plants have helped to reduce post harvest losses, e.g. flavr savr transgenic tomatoes.
5. Increased efficiency of mineral usage – GM plants can more efficiently utilise soil minerals and thus prevent early exhaustion of fertility of soil.
6. Increased nutritional value of food – GM plants have enhanced nutritional value of food, e.g. golden rice is rich in vitamin A.

Question 6.
Explain with the help of one example how genetically modified plants can
(a) reduce usage of chemical pesticides.
(b) enhance nutritional value of food crops.
Answer:
(a) GM crops are pest resistant crops. E.g., a nematode Meloidogyne incognita infects the roots of tobacco plant and causes lots of damage by reducing the yield. Using Agrobacterium vectors nematode specific genes are introduced into the host plant, as a result they produce both sense and antisense RNA. This initiates RNAi i.e. a specific segment of RNA is made silent and is unable to produce the protein required by the nematode. The nematode dies in such a transgenic host. By this way, using biotechnological technique the transgenic plant gets protected by itself without using chemical pesticides,

(b) GM plants have increased nutritional vdlue of food. E.g., golden rice is a transgenic variety of rice, which contains good quantities of (3 carotene (provitamin A). Since the contents of vitamin A are very low in rice, so genetically engineered rice have been produced by introducing three genes associated with synthesis of carotene. The grains of transgenic rice are rich in provitamin and reduce the occurrence of vitamin A deficiency diseases.

Question 7.
List the disadvantages of insulin obtained from the pancreas of slaughtered cows and pigs.
Answer:
Disadvantages of insulin obtained from slaughtered cows and pigs are as follows :

1. As insulin from slaughtered animals is produced in vary small amounts, so it requires killing of large number of animals.
2. Insulin produced by animals is slightly different from human insulin, therefore it sometimes does not respond well and causes allergy.
3. It is unethical to kill so many animals to obtain a drug.
4. Due to infection by microorganisms, some animals might produce contaminated insulin.
5. Its production is a more time taking process and the supply is also limited.

Question 8.
List the advantages of recombinant insulin.
Answer:
Advantages of recombinant insulin are as follows:

1. Recombinant insulin exactly resembles human insulin in structure and is commonly called humulin.
2. It is available in pure form, therefore, chances of contamination are very little.
3. Its production does not involve killing of animals.
4. There is no immune response or allergy or any other side effects.
5. There is no shortage of supply.

Question 9.
What is meant by the term biopesticide? Name and explain the mode of action of a popular biopesticide.
Answer:
A biopesticide is a living organism and not a chemical substance. Its product or gene can kill the pest, thus it is used for pest control and is thus named so. It is highly specific and safe for the environment. It does not cause any pollution in the environment. Bt toxin is a biopesticide produced from Bacillus thuringiensis.

Bt toxins are proteins that kill certain insects like lepidopterans, coleopterans and dipterans. This protein exists in inactive form in bacteria and does not cause any harm to it. Once an insect ingests the inactive toxin, it gets converted into active form of toxin due to alkaline pH of alimentary canal of the insect, that solubilises the crystals.

The activated toxin binds to the surface of the midgut epi thelial cells of the insect and create pores which cause swelling and lysis and finally the death of insect. Bt toxin crystals are used as pesticide in agriculture. The protein crystals are available commercially as liquids which are sprayed on the leaves. When a insect pest ingests them, the toxins get inactivated inside the insect and kill it. With advances in the field of biotechnology, genetic engineering techniques have been used to incorporate pest resistance in crop plants. The Bt toxin genes are isolated from Bacillus thuringiensis and incorporated into several crop plants like cotton. Bt cotton farming has shown good results in various parts of the country.

Question 10.
Name the five key tools for accomplishing the tasks of recombinant DNA technology. Also mention the functions of each tool.
Answer:
Five key tools for recombinant DNA technology are :

1. Cleaving enzymes – Restriction endonucleases cleave DNA at specific points and exonucleases cut DNA at the terminal end.
2. DNA ligase – Enzyme used for sealing gaps in DNA fragment and for joining foreign gene with plasmid DNA.
3. Vectors – Vectors are DNA molecule that carry foreign DNA segment and replicate inside host cell, e.g., plasmid.
4. Competent host – Cell capable of transformation, i.e., able to take foreign DNA.
5. Eysing enzymes – To open the cell to isolate the DNA.

We hope the NCERT Exemplar Solutions for Class 12 Biology chapter 12 Biotechnology and Its Applications help you. If you have any query regarding .NCERT Exemplar Solutions for Class 12 Biology chapter 12 Biotechnology and Its Applications, drop a comment below and we will get back to you at the earliest.

NCERT Exemplar Solutions for Class 12 Biology chapter 11 Biotechnology: Principles and Processes

These Solutions are part of NCERT Exemplar Solutions for Class 12 Biology. Here we have given NCERT Exemplar Solutions for Class 12 Biology chapter 11 Biotechnology: Principles and Processes

Multiple Choice Questions

Question 1.
Rising of dough is due to
(a) multiplication of yeast
(b) production of CO2
(c) emulsification
(d) hydrolysis of wheat flour starch into sugars.
Answer:
(b)

Question 2.
An enzyme catalysing the removal of nucleotides from the ends of DNA is
(a) endonuclease
(b) exonuclease
(c) DNA ligase
(d) HindII
Answer:
(b) : Exonucleases remove nucleotides from the terminal ends (either 5′ or 3′) of DNA of one strand of duplex. Endonucleases make cut at specific position within the DNA. DNA ligases are known as joining or sealing enzymes. Ligases form phosphodiester bonds between adjacent nucleosides and covalently link two individual fragments of DNA. Hindll is restriction endonuclease, it cuts the DNA and produces blunt end.

Question 3.
The transfer of genetic material from one bacterium to another through the mediation of a vector like virus is termed as
(a) transduction
(b) conjugation
(c) transformation
(d) translation.
Answer:
(a)

Question 4.
Which of the given statements is correct in the context of observing DNA separated by agarose gel electrophoresis? 
(a) DNA can be seen in visible light.
(b) DNA can be seen without staining in visible light.
(c) Ethidium bromide stained DNA can be seen in visible light.
(d) Ethidium bromide stained DNA can be seen under exposure to UV light.
Answer:
(d) : Electrophoresis is a technique of separation of molecules such as DNA, RNA or protein, under the influence of an electric field, so that they migrate in the direction of electrode bearing the opposite charge, viz, positively charged molecules move towards cathode (-ve electrode) and negatively charged molecules travel towards anode (+ve electrode) through a medium/ matrix. Since DNA fragments are negatively charged molecules, they can be separated by allowing them to move towards the anode (+ve electrode) under an electric field through a matrix of agarose gel.

The DNA fragments separate’according to their size through the agarose gel, with smaller fragments moving farther away as compared to larger ones. The DNA fragments can be visualised by staining them with ethidium bromide followed by exposure to UV radiations. Bright orange coloured bands of DNA can be observed. The separated DNA bands are then cut out from the agarose gel and extracted from the gel piece, this step is known as elution.

Question 5.
‘Restriction’ in restriction enzyme refers to
(a) cleaving of phosphodiester bond in DNA by the enzyme.
(b) cutting of DNA at specific position only.
(c) prevention of the multiplication of bacteriophage in bacteria.
(d) all of the above.
Answer:
(c) : A restriction enzyme is an enzyme that cuts DNA by hydrolysing phosphodiester bonds at or near a specific recognition nucleotide sequences known as restriction sites. The term ‘restriction’ refers to the function of an enzyme restricting the propagation of foreign DNA of bacteriophage in the host bacterium. These enzymes are found in bacteria and provide a defense mechanism against invading virus. The host DNA is protected by a modification enzyme that modifies the prokaryotic DNA and blocks cleavage. Together, these processes form restriction modification system.

Question 6.
Which of the following is not required in the preparation of a recombinant DNA molecule?
(a) Restriction endonuclease
(b) DNA ligase
(c) DNA fragments
(d) E.coli
Answer:
(d) : To produce a recombinant DNA following procedure is followed :

• Genetic material is isolated by using the enzyme, lysozyme or cellulase or chitinase for animal, plant or fungal cell respectively.
• Purified DNA is cut at specific sites by restriction enzymes.
• DNA fragments are separated using agarose gel electrophoresis and the gene or fragment of interest is amplified using PCR.
• After the cutting of the source DNA and the vector DNA with specific restriction enzyme, the cut out ‘gene of interest’ from the source DNA and the cut vector with space are mixed and ligase enzyme is added. This results in the formation of a rDNA or hybrid DNA or chimeric DNA.
• The ultimate aim of recombinant DNA technology is to produce a desirable protein. The foreign gene gets expressed under appropriate condition, culturing methods are used to produce higher yields of the desired protein. Therefore, E. coli is not required for preparation of a recombinant DNA molecule. we have to break the cell open to release DNA and other macromolecules

Question 7.
In agarose gel electrophoresis, DNA molecules are separated on the basis of their
(a) charge only
(b) size only
(c) charge to size ratio
(d) all of the above.
Answer:
(b) : Electrophoresis is a technique of separation of molecules, such as DNA, RNA or proteins on the basis of their size, under the influence of an electric field. So they migrate towards electrode of opposite charge. DNA fragments separate according to the size through pores of agarose gel.

Question 8.
The most important feature in a plasmid to be used as a vector is
(a) origin of replication (or/)
(b) presence of a selectable marker
(c) presence of sites for restriction endonuclease
(d) its size.
Answer:
(a,b,c & d) : Origin of replication (On), a selectable marker, sites for restriction 1 endonuclease and its size, all are importantfeatures required to facilitate cloning into a vector. A good DNA vector should be able to replicate autonomously in the host cell, for which it needs to have an origin of replication site (Ori). This is important for the replication of inserted gene. A prokaryotic DNA has a single Ori while eukaryotic DNA may have more than one Ori. Selectable marker helps in identifying and eliminating non-transformants and selectively permitting the growth of transformants. Cloning sites (recognition sites) are the sites where the DNA is cut by a restriction endonuclease. A vector should be ideally less than 10 kb in size because large DNA molecules can break down during purification procedure.

Question 9.
While isolating DNA from bacteria, which of the following enzymes is not used?
(a) Lysozyme
(b) Ribonuclease
(c) Deoxyribonuclease
(d) Protease
Answer:
(c) : In order to cut the DNA with restriction enzymes, it needs to be in pure form, free from other macromolecules. Since the DNA is enclosed by the membranes,we haive to break the cell open to release DNA and other macromolecules like RNA, proteins, polysaccharides and lipids. It is obtained by treating the bacterial cells/ plant or animal tissue with enzymes such as lysozyme (bacteria), cellulase (plant cells) and chitinase (fungus). DNA is interwined with proteins like histones. RNA can be removed by treatment with ribonuclease while proteins can be removed by treatment with protease. Other molecules are removed by appropriate treatments and purified DNA ultimately precipitates out after the addition of chilled ethanol.

Question 10.
Which of the following has popularised the PCR (polymerase chain reaction)?
(a) Easy availability of DNA template.
(b) Availability of synthetic primers.
(c) Availability of cheap deoxyribonudeo- tides.
(d) Availability of ‘thermostable’ DNA poly-merase.
Answer:
(d) : The final step of PCR is extension, wherein Taq DNA polymerase (isolated from a thermophilic bacterium Thermits aquaticus) synthesises the DNA region between the primers, using dNTPs (deoxynucleoside triphosphates) and Mg²⁺. The primers are extended towards each other so that the DNA segment lying between the two primers is copied. The optimum temperature for this polymerisation step is 72CC. Taq polymerase remains active during high temperature induced denaturation of double stranded DNA.

Question 11.
An antibiotic resistance gene in a vector usually helps in the selection of
(a) competent cells
(b) transformed cells
(c) recombinant cells
(d) none of the above.
Answer:
(b) :The ligation of alien DNA is carried out at a restriction site present in one of the two antibiotic resistance genes. For example, you can ligate a foreign DNA at the BamHl site of tetracycline resistance gene in the vector pBR322. The recombinant plasmids will lose tetracycline resistance other macromolecules like RNA, proteins, polysaccharides and lipids. It is obtained by treating the bacterial cells/ plant or animal tissue with enzymes such as lysozyme (bacteria), cellulase (plant cells) and chitinase (fungus). DNA is interwined with proteins like histones. RNA can be removed by treatment with ribonuclease while proteins can be removed by treatment with protease. Other molecules are removed by appropriate treatments and purified DNA ultimately precipitates out after the addition of chilled ethanol.

Question 12.
Significance of ‘heat shock’ method in bacterial transformation is to facilitate
(a) binding of DNA to the cell wall
(b) uptake of DNA through membrane transport proteins
(c) uptake of DNA through transient pores in the bacterial cell wall
(d) expression of antibiotic resistance gene.
Answer:
(c) : Transformation is a process by which a cell takes up naked DNA fragment from the environment, incorporates it into its own chromosomal DNA and finally expresses the trait controlled by the incoming DNA. Since DNA is a hydrophilic molecule, it cannot pass through membranes, so the bacterial cells must be made competent to take up DNA. This is done by treating them with a specific concentration of a divalent cation, such as calcium (Ca²⁺) which increases the efficiency with which DNA enters the bacterium through pores in its cell wall. Recombinant DNA (rDNA) can then be forced into such cells by incubating the cell with recombinant DNA on ice, followed by placing them briefly at 42°C (heat shock), and then putting them back on ice. This enables the bacteria to take up the recombinant DNA.

Question 13.
The role of DNA ligase in the construction of a recombinant DNA molecule is
(a) formation of phosphodiester bond between two DNA fragments
(b) formation of hydrogen bonds between sticky ends of DNA fragments
(c) ligation of all purine and pyrimidine bases
(d) none of the above.
Answer:
(a) : DNA ligases (joining or sealing enzymes) are also called genetic gum. They join two individual fragments of double  straqded DNA by forming phosphodiester bonds between them. Thus, they help in sealing gaps in DNA fragments. Therefore, they act as a molecular glue.

Question 14.
Which of the following is not a source of restriction endonuclease?
(a) Haemophilus influenzae
(b) Escherichia coli
(c) Entamoeba coli
(d) Bacillus amyloliquifaciens
Answer:
(c)

Source	Restriction endonuclease
(1) Haemophilus influenzae	Hindll and HmdIII
(2) Escherichia coli	EcoRI and EcoRII
(3)Bacillus amyloliquefaciens	Bam HI

Question 15.
Which of the following steps are catalysed by Taq polymerase in a PCR reaction?
(a) Denaturation of template DNA.
(b) Annealing of primers to template DNA.
(c) Extension of primer end on the template DNA.
(d) All of the above.
Answer:
(c) : Refer answer 10.

Question 16.
A bacterial cell was transformed with a recombinant DNA that was generated using a human gene. However, the transformed cells did not produce the desired protein. Reasons could be
(a) human gene may have intron which bacteria cannot process
(b) amino acid codonsfor humans and bacteria are different
(c) human protein is formed but degraded by bacteria
(d) all of the above.
Answer:
(a) : Eukaryotic genes do not function properly when transferred into bacterial cell because introns are present in eukaryotic cells but are absent in prokaryotic cells. Hence, when bacterial cell is transformed with recombinant DNA which is generated using human gene, it could not process it. As a result, no desired protein will be produced.

Question 17.
Which of the following should be chosen for best yield if one were to produce a recombinant protein in large amounts?
(a) Laboratory flask of largest capacity
(b) A stirred-tank bioreactor without in-lets and out-lets
(c) A continuous culture system
(d) Any of the above
Answer:
(c) : The cells having cloned genes of interest can be grown on a small scale in the laboratory. The cultures may be used tor extracting and purifying the desired protein. The cells can also be multiplied in a continuous culture system where the used medium is passed out from one side and fresh medium is added from the other side to maintain the cells in their physiologically most active log/ exponential phase – rapid multiplication of the cells. This type of culturing method produces a larger biomass to get higher yields of desired protein.

Question 18.
Who among the following was awarded the Nobel Prize for the development of PCR technique?
(a) Herbert Boyer
(b) Hargovind Khurana
(c) KaryMullis 
(d) Arthur Kornberg
Answer:
(c)

Question 20.
Which of the following statements does not hold true for restriction enzyme?
(a) It recognises a palindromic nucleotide sequence.
(b) It is an endonuclease.
(c) It is isolated from viruses.
(d) It produces the same kind of sticky ends in , different DNA molecules.
Answer:
(c) : More than 900 restriction enzymes have been isolated from over 230 strains of bacteria each of which recognise different recognition sequences. No restriction enzyme has been isolated from viruses.

Very Short Answer Type Questions

Question 1.
How is copy number of the plasmid vector related to yield of recombinant protein?
Answer:
The copy number is the number of copies of plasmids compared with the amount of chromosomal DNA found in single bacterial cell. The copy number influences the stability of plasmid. The rDNA can multiply as many times as the copy number of vector plasmid. So, higher the copy number of plasmid, higher would be the copy number of gene and protein yield would be high.

Question 2.
Would you choose an exonuclease while producing a recombinant DNA molecule?
Answer:
For producing recombinant DNA, exonuclease is not added because it removes nucleotides from terminal ends (either 5′ or 3′) and it acts on single strand of DNA. It does not recognise any specific sequence.

Question 3.
What does ‘H’, ‘in’, ‘d’ and ‘III’ refer to in the enzyme Hindlll?
Answer:
ln naming of restriction endonuclease, the first letter of enzyme is the first letter of bacterium’s, generic name, “H”-Haemophilus, “in” represents two letters of species name- influenzae, “d” is for strain “Rd”, and “HI” indicates the order in which enzyme is synthesised.

Question 4.
Restriction enzymes should not have more than one site of action in the cloning site of a vector. Comment.
Answer:
Restriction enzymes should preferably have single recognition site because presence of more than one recognition sites within the vector will generate several fragments, which will complicate the gene cloning.

Question 5.
What does ‘competent’ refer to in competent cells used in transformation experiments?
Answer:
Competent cells are the cells that have altered cell wall so that they can take up foreign DNA easily. Competent cell is the one that is able to undergo genetic transformation.

Question 6.
What is the significance of adding proteases at the time of isolation of genetic material (DNA)?
Answer:
Genes located on the DNA are interwined with proteins such as histones, which are removed by proteases. It is important to add protease during DNA isolation because DNA has to be free from any other .macromolecule such as RNA, proteins to get cut with restriction endonuclease.

Question 7.
While doing a PCR/denaturation’step is missed. What will be its effect on the process?
Answer:
Denaturation is the first step in PCR reaction. If denaturation does not take place, annealing and extension will also not take place. Therefore, there will be no amplification.

Question 8.
Name a recombinant vaccine that is currently being used in vaccination program.
Answer:
HepaH tis-R surface antigen (HBsAg) is currently used as vaccine in vaccination programme against hepatitis virus.

Question 9.
Do biomolecules (DNA, protein) exhibit biological activity in anhydrous conditions?
Answer:
DNA and protein do not show biological activity in anhydrous conditions because in non-aqueous conditions rigidity of biomolecules increases due to weakening of hydrogen bond strength.

Question 10.
What modification is done on the Ti plasmid Agrobacterium tumefaciens to convert it into a cloning vector?
Answer:
Agrobacterium tumefaciens, a pathogen of several dicot plants is able to deliver T-DNA to transform normal cell into tumour and direct tumour cells to produce chemicals required by pathogen. The Ti (tumor inducing) plasmid has been modified (disarmed) by removing gene responsible for causing tumour and inserting gene to be used as selectable marker. Modified plasmid is non-pathogenic to plants and delivers the gene of interest.

Short Answer Type Questions

Question 1.
What is meant by gene cloning?
Answer:
Gene cloning is formation of multiple identical copies of any template DNA. For cloning, an alien DNA is linked with origin of replication of a vector, so that it can replicate and multiply itself in the host organism.

Question 2.
Both a wine maker and a molecular biologist who had developed a recombinant vaccine claim to be biotechnologists. Who in your opinion is correct?
Answer:
Biotechnology is technological employment of biological entities and processes to generate products and services useful to man. In this case a wine maker has utilised a strain of yeast, which is commonly used for making wine by fermentation. The molecular biologist has used cloned gene for an antigen and antigen is used as vaccine. Thus, both can be considered as biotechnologists.

Question 3.
A recombinant DNA molecule was created by ligating a gene to a plasmid vector. By mistake, an exonuclease was added to the tube containing the recombinant DNA. How does this affect the next step in the experiment i.e. bacterial transformation?
Answer:
Recombinant DNA is formed by ligating an alien DNA with plasmid DNA, a circular, self-replicating, extra-chromosomal, dsDNA. Therefore, rDNA is also circular DNA and exonucleases cleave DNA at terminal ends (either 5′ or 3′) in a linear DNA segment. Thus, addition of exonuclease to a tube containing rDNA would not affect bacterial transformation.

Question 4.
Restriction enzymes that are used in the construction of recombinant DNA are endonucleases which cut the DNA at ‘specific- recognition sequence’. What would be the disadvantage if they do not cut the DNA at specific recognition sequence?
Answer:
If endonucleases used in the recombinant DNA technology do not cut DNA at specific locations, they may cut the DNA at random location. The sticky ends will not be generated to ligate the DNA segments and recombinant DNA would not be formed.

Question 5.
A plasmid DNA and a linear DNA (both are of the same size) have one site for a restriction endonuclease. When cut and separated on agarose gel electrophoresis, plasmid shows one DNA band while linear DNA shows two fragments. Explain.
Answer:
A plasmid DNA is a circular DNA. When this circular DNA having one site for restriction endonuclease is cleaved with an enzyme, it would form a strand of linear DN? and would be visible as single band on agarose gel. On the other hand, when linear DNA with one restriction site is cleaved, it would produce two fragments and thus two bands would be seen.

Question 6.
How does one visualise DNA on an agarose gel?
Answer:
The separated DNA fragments on an agarose gel can be visualised only after staining the DNA with ethidium bromide followed by exposure to UV radiations as bright orange coloured bands.

Question 7.
A plasmid without a selectable marker was chosen as vector for cloning a gene. How does this affect the experiment?
Answer:
If a cloning vector does not have a selectable marker, then it would not be possible to distinguish between transformants (host bacterium having rDNA) and non-transformants. Therefore, an ideal cloning vector should have selectable markers for the selection of transformants.

Question 8.
A mixture of fragmented DNA was electrophoresed in an agarose gel. After staining the gef with ethidium bromide, no DNA bands were observed. What could be the reason?
Answer:
There could be various reasons for DNA bands not visible after electrophoresis, such as:

1. DNA might be impure, i.e., have not been isolated properly and is associated with RNA and proteins.
2. Restriction endonucleases have not been added in appropriate amount.
3. Electrodes might not have been connected correctly.
4. Stained bands have not been visualised under UV rays.
5. Electrodes were put in opposite orientation in the gel assembly, that is anodes towards the well (where DNA is loaded).
6. Ethidium bromide has not been added to stain DNA fragments.

Question 9.
Describe the role of CaCI₂ in the preparation of competent cells.
Answer:
Competent cell is one which can take up naked DNA fragment and incorporate it into its own chromosomal DNA. DNA is a hydrophilic molecule and. cannot pass through membranes, so bacterial cell can be made competent by CaCH method. In this method, bacterial cell is treated with divalent cation Ca²⁺ which increases the efficiency with which DNA enters the bacterium through pores in its cell wall.

Question 10.
What would happen when one grows a recombinant bacterium in a bioreactor but forgets to add antibiotic to the medium in which the recombinant is growing?
Answer:
Antibiotics do not allow other bacteria to grow in the medium. In the absence of antibiotics, the desirable bacteria may not be able to grow to its optimum level. The plasmids may also be lost, since maintaining a high copy number of plasmids is a metabolic burden on the bacteria.

Question 11.
Identify and explain steps A, B and C in the PCR diagram given below.

Answer:
Polymerase Chain Reaction (PCR) involves amplification of gene of interest.
A single PCR amplification cycle involves three basic steps:

1. A- Denaturation : It is separation of two strands of DNA by heating the DNA strand at a temperature of 94 – 96 °C. Each single strand of target DNA is used as template DNA.
2. B-Annealing : Hybridisation of two oligo-nucleotide primers to each of the ssDNA template. It is carried out at a temperature of 40- 60 °C.
3. C-Extension : It is the final step of PCR. In this, enzyme Taq DNA polymerase synthesises the DNA region between the primers, using dNTPs and Mg²⁺.

Question 12.
Name the regions marked A, B and C.

Answer:
A – BarnHI
B – PtI
C – ampR

Long Answer Type Questions

Question 1.
For selection of recombinants, insertional inactivation of antibiotic marker has been superceded by insertional inactivation of a marker gene coding for a chormogenic substrate. Give reasons.
Answer:
Selection of recombinants by insertional inactivation of an antibiotic has been superceded by insertional inactivation of selectable marker because selection of recombinants due to inactivation of antibiotics is a cumbersome procedure, because it requires simultaneous plating on two plates having different antibiotics (ampicillin and tetracyiine). Therefore, alternative selectable markers have been developed which differentiate recombinants from non-recombinants on the basis of their ability to produce colour in the presence of chromogenic substrate. In this, a recombinant DNA is inserted within the coding sequence of an enzyme p-galactosidase.

This results in the inactivation of the enzyme (insertional inactivation). The presence of chromogenic substrate gives blue coloured colonies if the plasmid in the bacteria does not have an insert. Presence of insert results into insertional inactivation of the (1-galactosidase and the colonies do not produce any colour, and these are identified as recombinant colonies.

Question 2.
Describe the role of Agiobacterium tumefaciens in transforming a plant cell.
Answer:
Aemhactprium tumefaciens is a bacterium that causes tumours in plants. It has ability to transform plant cells. For this reason it has become an important tool in plant improvement by genetic engineering.
The plasmid is disarmed by deletion of the tumour inducing gene. These modified bacteria can still transform plant cells. The part of Ti plasmid transferred into plant cell DNA, is called T-DNA. This T-DNA with desired DNA spliced into it, is inserted into the chromosomes of the host plant where it produces copies of itself. But it no longer produces tumours. Such plant cells are then cultured, induced to multiply and differentiate to form plantlets. Thus the ability of Agrobactcrium to transfer genes to the plants has been exploited in genetic engineering for plant improvement programme.

Question 3.
Illustrate the design of a bioreactor. Highlight the difference between a flask in your laboratory and a bioreactor which allows cells to grow in a continuous culture system.
Answer:
Bioreactors are vessels in which raw materials are biologically converted into specific products by microbes, plant and anirhal cells and their enzymes. They are allowed to synthesise the desired proteins which are finally extracted and purified from cultures.

Small volume of cultures are usually employed in laboratories for research and production of less quantities of products. Large scale production of products is carried out in bioreactors. The most commonly used bioreactors are of stirring type, that have provision for batch culture or continuous culture. In continuous culture, the culture medium is added and the product is taken out continuously.

Flask is used in laboratory for testing and provides batch type of culture. With flask no gadget can be connected. Bioreactors are used commercially and provide continuous culture. With a bioreactor, gadgets can be attached to give better functioning. A bioreactor provides optimal conditions for optimal growth (temperature pH, substrate, salts, vitamins, oxygen).

We hope the NCERT Exemplar Solutions for Class 12 Biology chapter 11Biotechnology: Principles and Processes help you. If you have any query regarding .NCERT Exemplar Solutions for Class 12 Biology chapter 11Biotechnology: Principles and Processes, drop a comment below and we will get back to you at the earliest.

NCERT Exemplar Solutions for Class 12 Biology chapter 10 Microbes in Human Welfare

These Solutions are part of NCERT Exemplar Solutions for Class 12 Biology. Here we have given NCERT Exemplar Solutions for Class 12 Biology chapter 7 Evolution

Multiple Choice Questions

Question 1.
The vitamin whose content increases following the conversion of milk into curd by lactic acid bacteria is
(a) vitamin C
(b) vitamin D
(c) vitamin B12
(d) vitamin E.
Answer:
(c) : Curd is more nutritious than milk as it contains a number of organic acids and vitamins including B₁₂ .

Question 2.
Wastewater treatment generates a large quantity of sludge, which can be treated by
(a) anaerobic digesters
(b) floe
(c) chemicals
(d) oxidation pond.
Answer:
(a) : The sediment of settling tank is r called activated sludge. A part of it is used as inoculum in aeration tanks. The remaining is passed into a large tank called anaerobic sludge digesters.

Question 3.
Methanogenic bacteria are not found in
(a) rumen of cattle
(b) gobar gas plant
(c) bottom of water-logged paddy fields
(d) activated sludge.
Answer:
(d) : Methanogens are strict anaerobes. Nutritionally they are “autotrophs” which obtain both energy and carbon from decomposition products. They occur in marshy areas where they convert formic acid and carbon dioxide into methane with the help of hydrogen. This capability is commercially exploited in the production of methane and fuel gas inside gobar gas plants c.g., Methanobacterium, Methanococcus. Some of the methanogen archaebacteria live as symbionts (e.g., Methanobacterium) inside rumen or first chamber in the stomach of herbivorous animals that chew their cud (ruminants, e.g., cow, buffalo).

These archaebacteria are helpful to the ruminants in fermentation of cellulose. The warm, water logged soil of paddy fields provide ideal condition for methanogenesis and though some of the methane produced is usually oxidised by methanotrophs in the shallow overlying area, vast majority is released into atmosphere. Methanogenic bacteria are not found in activated sludge instead activated sludge contains aerobic microorganisms.

Question 4.
Match the following list of bacteria and their commercially important products.
Bacterium Product

Choose the correct match.
(a) i-(B), ii-(C), iii-(D), iv-(A)
(b) f-(B), ii-(D), iii-(C), iv-(A)
(c) i-(D), ii-(C), iii-(B), iv-(A) 
(d) i-(D), ii-(A), iii-(C), iv-(B)
Answer:
(c)

Question 5.
Match the following list of bioactive substances and their roles. Role

Choose the correct match
(a) i-(B), ii-(C), iii-(A), iv-(D)
(b) i-(D), ii-(B), iii-(A), iv-(C)
(c) i-(D), ii-(A), iii-(B), iv-(C)
(d) i-(C), ii-(D), iii-(B), iv-(A)
Answer:
(d)

Question 6.
The primary treatment of waste water involves the removal of
(a) dissolved impurities
(b) stable particles
(c) toxic substances
(d) harmful bacteria.
Answer:
(b) : Primary or physical treatment is the process of removal of small and large, floating and suspended solids from sewage through two processes of filtration and sedimentation. First floating and suspended matter is removed through sequential filtration with progressively smaller pore filters. The filtrate is then kept in large open settling tanks where grit (sand, silt, small pebbles) settles down. Aluminium or iron sulphate is added in certain places for flocculation and settling down of solids. The sediment is called primary sludge, while the supernatant v is called effluent.

Question 7.
BOD of waste water is estimated by measuring the amount of
(a) total organic matter
(b) biodegradable organic matter
(c) oxygen evolution
(d) oxygen consumption.
Answer:
(d) : Degree of impurity of water due to organic matter is measured in terms of BOD. It is the oxygen in milligrams required for five days in one liter of water at 20°C for the microorganisms to metabolise organic waste.

Question 8.
Which one of the following alcoholic drinks is produced without distillation?
(a) Wine
(b) Whisky
(c) Rum
(d) Brandy
Answer:
(a) : Wine and beer are produced without distillation whereas whisky, brandy and rum are produced by distillation of the fermented broth.

Question 9.
The technology of bio gas production from cow dung was developed in India largely due to the efforts of
(a) Gas Authority of India
(b) Oil and Natural Gas Commission
(c) Indian Agricultural Research Institute and Khadi & Village Industries Commission
(d) Indian Oil Corporation.
Answer:
(c)

Question 10.
The free-living fungus Trichoderma can be used for
(a) killing insects
(b) biological control of plant diseases
(c) controlling butterfly caterpillars
(d) producing antibiotics.
Answer:
(b) : Abiological control being developed for use in the treatment of plant disease is the fungus Trichoderma. Trichoderma species are free-living fungi that are very common in the root ecosystems. They are effective biocontrol agents of several plant pathogens.

Question 11.
What would happen if oxygen availability to activated sludge floes is reduced?
(a) It will slow down the rate of degradation of organic matter.
(b) The center of floes will become anoxic, which would cause death of bacteria and eventually breakage of floes.
(c) Floes would increase in size as anaerobic bacteria would grow around floes.
(d) Protozoa would grow in large numbers.
Answer:
(a,b) : Floes are masses of bacteria associated with fungal filaments to form mesh like structures. If oxygen availability to activated sludge floes is reduced, their rate of decomposition of organic matter will decrease. And as the center of floes will become anoxic, the bacterial cells will die, thus causing breakage of floes.

Question 12.
Mycorrhiza does not help the host plant in
(a) enhancing its phosphorus uptake capacity
(b) increasing its tolerance to drought
(c) enhancing its resistance to root pathogens
(d) increasing its resistance to insects.
Answer:
(d) : Fungi are also known to form symbiotic associations with plants (mycorrhiza). Many members of the genus Glomus form mycorrhiza. The fungal symbionts in these associations absorb phosphorus from soil and passes it to the plant. Plants having such associations show other benefits also, such as resistance to root- 1 borne pathogens, tolerance to salinity and drought, and an overall increase in plant growth and development.

Question 13.
Which one of the following is not a nitrogen fixing organism?
(a) Anabaena
(b) Nostoc
(c) Azotobacter
(d) Pseudomonas
Answer:
(d) : Pseudomonas is not a nitrogen fixing bacteria. Pseudomonas are used in antibiotic formation and biodegradation of organic pollutant like petroleum spillage. Azotobacter is a free living nitrogen fixing bacteria. Anabaena and Nostoc are free living nitrogen fixing cyanobacteria.

Question 14.
Big holes in Swiss cheese are made by a
(a) a machine
(b) a bacterium that produces methane gas
(c) a bacterium producing a large amount of carbon dioxide
(d)a fungus that releases a lot of gases during its metabolic activities.
Answer:
(c) : Ripened cheese is prepared from unripened cheese by first dipping in brine, wiping and then maturation with different strain of bacteria and fungi. It takes 1-16 months for ripening. Large holed Swiss cheese is ripened with the help of CO² producing (causing holes) bacterium called Propionibacterium sharmanii.

Question 15.
The residue left after methane production from cattle dung is
(a) burnt
(b) burried in landfills
(c) Used as manure
(d) used in civil construction.
Answer:
(c) : Biogas is a methane rich fuel gas produced by anaerobic breakdown or digestion of biomass with the help of methanogenic bacteria. Biogas is made up of methane (50-70%), carbon dioxide (30-40%) with traces of nitrogen, hydrogen sulphide and hydrogen. The effluent and residue left after the fermentative generation of biogas is rich in minerals, lignin and a part of cellulose. It is an ideal manure.

Question 16.
Methanogens do not produce
(a) oxygen
(b) methane
(c) hydrogen sulphide
(d) carbon dioxide.
Answer:
(a) : Refer answer 15.

Question 17.
Activated sludge should have the ability to settle quickly so that it can
(a) be rapidly pumped back from sedimentation tank to aeration tank
(b) absorb pathogenic bacteria present in waste water while sinking to the bottom of the settling tank
(c) be discarded and anaerobically digested
(d) absorb colloidal organic matter.
Answer:
(a) : Bacterial floes are allowed to sediment into the settling tank, this sediment is called activated sludge. A small part of activated sludge is pumped back into aeration tank to serve as inoculum. So, it must have ability to settle quickly.

Question 18.
Match the items in Column ‘A’ and Column ‘B’ and choose correct answer.

The correct answer is
(a) (i)-Q, (ii)-S, (iii)-R, (iv)-P
(b) (i)-p, (ii)-S, (iii)-Q, (iv)-P
(c) (i)-S, (ii)-P, (iii)-Q, (iv)-R
(d) (i)-R, (ii)-Q, (iii)-P, (iv)-S
Answer:
(b)

Very Short Answer Type Questions

Question 1.
Why does ‘Swiss cheese’ have big holes?
Answer:
cheese is prepared by bacteria Propionibncterium shanmnii. It produces large quantity of CO, which causes large holes in cheese.

Question 2.
What are fermentors?
Answer:
Fermentors are the containers where fermentation is carried out, also known as bioreactors.

Question 3.
Name a microbe used for statin production. How do statins lower blood cholesterol level?
Answer:
Statin is produced by yeast Monascus purpureus. It is used as blood cholesterol lowering agent by acting as a competitive inhibitor of enzyme for cholesterol synthesis.

Question 4.
Why do we prefer to call secondary waste water treatment as biological treatment?
Answer:
Secondary waste water treatment is also called biological treatment because in this step microbes digest organic matter and convert it into microbial biomass and release minerals.

Question 5.
What for nucleo poly hedroviruses are being used now a days?
Answer:
Nucleo poly hedroviruses are baculo- virus. These are biological control agents that attack insects and other arthropods. TheY are species-specific, narrow spectrum insecticides and do not harm plants, mammals, birds and fish.

Question 6.
How has the discovery of antibiotics helped mankind in the field of medicine?
Answer:
Antibiotics are the chemical substances produced by microorganisms which are used for treatment of pathogenic or infectious diseases. Antibiotics have greatly improved our capacity to treat deadly diseases such as plague, whooping cough (kali khansi), diphtheria (gal ghotu).

Question 7.
Why is distillation required for producing certain alcoholic drinks?
Answer:
Production of hard liquors require distillation of fermented broth to increase the alcoholic content. Beer and wine are formed without distillation and have 3-6% alcoholic content while rum, brandy and gin are formed by distillation and have 40% alcohol content.

Question 8.
Write the most important characteristic that Aspergillus niger, Clostridium butylicum, and Lactobacillus share.
Answer:
the three microbes, Aspergillus niger, Clostridium butylicum and Lactobacillus are used for commercial and industrial production of organic acids through fermentation. Organic acids produced are citric acid, butyric acid and lactic acid respectively.

Question 9.
What would happen if our intestine harbours microbial flora exactly similar to that found in the rumen of cattle?
Answer:
If humans also had microbes in their intestine like ruminants, then they would also have been capable of digesting cellulose presentin food itemsbecause microbes present in the rumen of cattle, called methanogens are capable of digesting cellulose as they have cellulase enzyme.

Question 10.
Give any two microbes that are useful in biotechnology.
Answer:
Two microbes that are used in bio-technology are:-

1. Escherichia coil
2. Bacillus thuringiensis

Question 11.
What is the source organism for EcoRI, restriction endonuclease?
Answer:
Bacterium Escherichia coil of strain RY13.

Question 12.
Name any genetically modified crop.
Answer:
cotton is a genetically modified crop that has been modified to resist attack by insect pests.

Question 13.
Why are blue-green algae not popular as biofertilisers?
Answer:
Blue -green algae add organic matter to the soil and increase its fertility but still these are not popular as biofertilisers. This is due to several constraints that limit the application or implementation of the biofertiliser technology. The constraints may be environmental, technological, infrastructural, financial, unawareness, quality, marketing etc.

Question 14.
Which species of Penicillium produces Roquefort cheese?
Answer:
Penicillium roqueforti

Question 15.
Name the states involved in Ganga action plan.
Answer:
Uttaranchal, Uttar Pradesh, Jharkhand, West Bengal, Bihar etc.

Question 16.
Name any two industrially important enzymes.
Answer:

1. Lipase – lipid dissolving enzymes, added in detergents for removing oil stains from laundry.
2. Pectinase – used along with protease in clearing fruit juices.

Question 17.
Name an immune immunosuppressive agent.
Answer:
Cyclosporin-A is used as immuno-suppressive agent in organ transplantation.

Question 18.
Give an example of a rod shaped virus.
Answer:
Tobacco mosaic virus (TMV).

Question 19.
What is the group of bacteria found in both the rumen of cattle and sludge of sewage treatment?
Answer:
Methanogens or Methanobacterium are found in the rumen of cattle and anaerobic sludge of sewage treatment.

Question 20.
Name a microbe used for the production of Swiss cheese.
Answer:
Propionibacterium sharmanii

Short Answer Type Questions

Question 1.
Why are floes important in biological treatment of waste water?
Answer:
Floes are masses of bacteria held together by fungal filament in mesh like structure in aeration tank for secondary sewage treatment. They digest organic matter and convert it into microbial biomass.

Question 2.
How has the bacterium Bacillus thuringiensis helped us in controlling caterpillars of insect pests?
Answer:
Bacteria Bacillus thuringiensis is used to control butterfly caterpillars. These are available in sachets as dried spores which are mixed with water and sprayed onto vulnerable plants such as brassicas and fruit trees, where they are eaten by the insect larvae. In the gut of larvae, the toxin is released and the larvae get killed.

Question 3.
How. do mycorrhizal fungi help the plants harbouring them?
Answer:
Mycorrhiza is a mutually beneficial or symbiotic association of a fungus with the roots of a higher plant. Mycorrhizal roots show a sparse or dense wooly growth of fungal hyphae on their surface.
Mycorrhiza perform several functions for the plant:

• Absorption of water.
• Solubilisation of organic matter of the soil humus, release of inorganic nutrients, absorption and their transfer to root.
• Direct absorption of minerals (e.g., phosphorus) from the soil over a large area and bending over the same to the root. Plants having mycorrhizal associations show resistance to root-borne pathogens, tolerance to salinity and drought, and overall increase in plant growth and development.

Question 4.
Why are cyanobacteria considered useful in paddy fields?
Answer:
Cyanobacteria are used as biofertilisers. They fix atmospheric nitrogen in specialised cells called heterocysts. These organism form symbiotic association with plant, add organic matter and extra nitrogen to the soil and do not interfere with plant growth, e.g., Azolla- Anabaena association is of great importance to paddy fields. Anabaena azollae resides in the leaf cavities of the fern. It fixes nitrogen. A part of the fixed nitrogen is excreted in the cavities and becomes available to the fern. The decaying fern plants release the same for the utilisation of rice plants.

Question 5.
How was penicillin discovered?
Answer:
Discovery of an antibiotic penicillin was serendipitous. It was discovered by Alexander Fleming, he found that petri dish containing culture of Staphylococcus got contaminated by mould. Mould inhibited the bacterial growth. Fleming isolated the mould and proved that filtrate of broth culture Penicillium notatum has antibacterial properties and discovered penicillin.

Question 6.
Name the scientists who were credited for showing the role of penicillin as an antibiotic.
Answer:
Penicillin was discovered by Alexander Fleming, while working on staphylococci bacteria. However its full potential as an effective antibiotic were established much later by Ernest Chain and Howard Florey and used it for treating wounded soldiers in World War II. Fleming, Chain and Florey were awarded the Noble prize, in 1945 for this discovery.

Question 7.
How do bioactive molecules of fungal origin help in restoring good health of humans?
Answer:
Bioactive molecules of fungal origin which help in restoring good health are :

1. Cyclosporin A is obtained from fungus Trichotderma pohjsporum. It has anti-fungal, anti-inflammatory and immunosuppres¬sive properties. It inhibits activation of T-cells and prevent graft rejection during organ transplantation.
2. Statin is produced from yeast Monascus purpureas, which helps in lowering blood cholesterol by inhibiting the enzyme responsible for cholesterol synthesis.

Question 8.
What roles do enzymes play in detergents that we use for washing clothes? Are these enzymes produced from some unique microorganisms?
Answer:
ipases are lipid dissolving enzymes that are added in detergents for removing oily stains from laundry. They are obtained from Candida lipoh/tica and Gcotrichum candiduiu.

Question 9.
What is the chemical nature of biogas? Name an organism which is involved in biogas production.
Answer:
Biogas is a methane rich fuel gas produced by anaerobic breakdown or digestion of biomass with the help of methanogenic bacteria. Biogas is made up of methane (50-70%), carbon dioxide (30-40%) with traces of nitrogen, hydrogen sulphide and hydrogen. Mcthanobactcrium is a common methanogenic bacteria involved in biogas production.

Question 10.
How do microbes reduce the environmental degradation caused by chemicals?
Answer:
Chemicals from fertilisers and pesticides are highly toxic to human beings and animals alike, and have been polluting our environment. To reduce the environmental degradation caused by chemicals, microbes can be used both as fertilisers and pesticides and are called biofertiliser and biopesticides, respectively. E.g., Rhizobium acts asbiofertiliser, as it can fix atmospheric nitrogen and Bacillus thuringiensis acts as biopesticide to control growth of insect pest.

Question 11.
What is a broad spectrum antibiotic? Name one such antibiotic.
Answer:
Broad spectrum antibiotic is an antibiotic which can kill or destroy a number of pathogens that belong to different groups with different structure and wall composition. e.g., Streptomycin.

Question 12.
What are viruses parasitising bacteria called? Draw a well labelled diagram of the same.
Answer:
Viruses parasitising bacteria are called bacteriophages. These viruses do not eat bacteria, but they infect and replicate within the bacteria.

Question 13.
Which bacterium has been used as clot buster? What is its mode of action?
Answer:
BacteriumBacterium Streptococcus has been modified genetically to function as clot buster. It produces enzvme streptokinase and has fibrinolytic effect. It helps in clearing blood dots inside blood vessels through dissolution of intravascular fibrin.

Question 14.
What are biofertilisers? Give two examples.
Answer:
Biofertilisers are microorganisms which bring about nutrient enrichment of the soil by enhancing the availability7 of nutrients like nitrogen and phosphorus to the crops. e.g.,

1. Nostoc, a free-living nitrogen fixing cyanobacteria,
2. Rhizobium, a symbiotic nitrogen-fixing bacteria which form nodules on roots of leguminous plants. They develop the ability to fix nitrogen only, when the) are present inside the root nodules, so in this way give fixed nitrogen to the plant.

Long Answer Type Questions

Question 1.
Why is aerobic degradation more important than anaerobic degradation for the treatment of large volumes of waste waters rich in organic matter. Discuss.
Answer:
During aerobic conditions bacteria gets associated with fungal hyphae and form floes,
They multiply very, rapidly and decompose large amount of organic waste matter present in waste water in the presence of oxygen. But in absence of oxygen i.e. anaerobic conditions, some toxic gases are also produced which kill many bacteria and fungi. That is why aerobic degradation of waste water treatment is more important than anaerobic degradation.

Question 2.
(a) Discuss about the major programs that
the Ministry of Environment and Forests, Government of India, has initiated for saving major Indian rivers from pollution.
(b) Ganga has recently been declared the National river. Discuss the implication with respect to pollution of this river.
Answer:
(a) Before 1985, very few cities and towns had sewage treatment plants. The municipal wastewater was discharged directly into rivers resulting in their pollution and high incidence of water borne diseases. The Ministry of Environment and Forests has initiated Yamuna Action Plan and Ganga Action Plan, to save these major rivers of our country from pollution. Under these plans it is proposed to build large number of sewage treatment plants so that only treated sewage is discharged into the rivers.

(b) Ganga Action Plan – Ganga, along with its tributaries, is the largest and very important I river of the country. It has been a symbol of purity, but today it is highly polluted with solid, biological and chemical pollutants. Major causes of pollution in Ganga are:

1. Urban liquid and solid wastes
2. Dead bodies of animals and humans
3. Wallowing of cattle Ganga action plan was started in 1986 by Late Sh. Rajiv Gandhi, the Prime Minister of India. by interception, diversion and treatment of domestic sewage and industrial wastes. Under the Ganga Action Plan JI programme, which started in 1993, the Pollution Control Research Institute (PCRI) of Bharat Heavy Electricals Ltd. at Haridwar has been conducting monthly studies to analyse the quality of Ganga river water. It is still under implementation.

Question 3.
Draw a diagrammatic sketch of biogas plant, and label its various components given below: Gas holder. Sludge chamber, Digester, Dung + Water chamber.
Answer:

Question 4.
Describe the main ideas behind the biological control of pests and diseases.
Answer:
The natural method of pest and pathogen control involving the use of viruses, bacteria and other insects is called biocontrol or biological control. Main ideas behind biological control of pests and diseases were to replace or supplement the use of chemical pesticides in order to reduce their ill effects. The bio-control agents are non-toxic, non persistent and biodegradable.
A few examples are as follows:

1. Lady bird beetles – Natural predator of aphids
2. Dragonflies – Control mosquitoes
3. Baculoviruses – Species specific, narrow spectrum insecticidal applications. They dre virus based bioinsecticides.
4. Bt cotton – Bacillus thuringiensis, a bacterium produces protein toxin which when ingested by larvae of insects, kills them.Main aim of the plan was pollution abatement to improve the water quality.
   These spores are available in sachet. They are dissolved in water and sprayed on vulnerable plants to kill insect larvae.
5. Trichoderma species are effective biocontrol agents of several plant pathogens.
   This is how biocontrol agents have decreased the use of chemical insecticides and controlled diseases biologically and have played very important role in regulating environmental pollution and eco-degradation.

Question 5.
(a) What would happen if a large volume of untreated sewage is discharged into a river?
(b) In what way anaerobic sludge digestion is important in sewage treatments?
Answer:
(a) Untreated sewage consists of large amount of organic matter and various microorganisms. If untreated sewage is discharged directly into river, it will lead to pollution of water with organic matter and microorganisms. These microbes are mostly disease causing bacteria and fungi (pathogenic). If such water is consumed by humans or animals, they may get number of diseases like cholera, typhoid, diarrhorea, etc. as their pathogens are found in contaminated water. At the same time high amount of organic waste increases biological oxygen demand and decrease the amount of dissolved oxygen in water which become toxic for animals and plants living in water and causes death of aquatic organisms like fish etc.

(b) In anaerobic sludge digestion, major part of activated sludge is pumped into anaerobic sludge digester. These anaerobic bacteria digest the bacteria and fungi and produce a mixture of gases like methane, carbon dioxide and H₂S, which are used as biogas. It also produces manure at same time.

Question 6.
Which type of food would have lactic acid bacteria? Discuss their useful application.
Answer:
Dairy products such as curd, butter milk, cheese etc. have lactic acid bacteria (LAB).

1. Lactic acid bacteria (LAB) like Lactobacillus are added to milk. It converts lactose sugar of milk into lactic acid. Lactic acid causes coagulation and partial digestion of milk protein casein. Milk is changed into curd, yoghurt and cheese.
2. Indian curd is prepared by inoculating skimmed and cream milk with Lactobacillus acidophilus at a temperature about 40°C or less. Curd is more nutritious than milk as it contains a number of organic acids and vitamins including B₁₂. LAB present in curd also checks growth of disease causing microbes in stomach and other parts of digestive tract.
3.  Yoghurt is produced by curdling milk with the help of Lactobacillus bulgaricus and Streptococcus thermophilus.
4.  Cottage cheese is prepared by single step fermentation which involves inoculation of skimmed milk with cheese culture (e.g., Lactobacillus and addition

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