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Thursday, 17 March 2022

Important questions boards March 17, 2022

Microbes in human welfairs important questions class 12

Microbes in human welfairs important questions

1. Conversion of milk to curd improves its nutritional value by increasing the amount of

(a) vitamin D

(b) vitamin A

(d) vitamin E

Answer:- (c) : Lactobacillus bacteria convert milk into curd. They produce acids that coagulate and partially digest the milk proteins. A small amount of curd added to the fresh milk as inoculum or starter contains millions of Lactobacillus which at suitable temperature multiply, thereby converting milk to curd this improves its nutritional value by increasing vitamin B12 content and a number of organic acids.

2. Identify the organism shown in the given diagram.

(a) Bacteria

(b) Bacteriophage

(d) HIV

Answer:- (c) : Tobacco mosaic virus is a rod like virus with a length of 300 nm and diameter of 18 mm. TMV capsids are composed of 2130 identical protein subunits which assemble around the viral ssRNA to form helical structure with a hollow central cavity of 4 mm diameter.

3. The puffed-up appearance of dough is due to fermentation by bacteria. Identify the gas liberated during the process.

(a) Hydrogen sulphide

(b) Methane

(d) Carbon dioxide

Answer:- (d) : Fermentation of glucose in dough mainly forms ethyl alcohol and carbon dioxide which causes leavening of dough.

4. Yeast is used in the production of

(a) bread and beer

(b) cheese and butter

(d) lipase and pectinase.

Answer:- (a) : The dough which is used for making bread, is fermented by baker’s yeast (Saccharomyces cerevisiae).

Besides bread, yeast fermentation is involved in the production of beer, wine, vinegar, etc. Several traditional drinks and food items are also made by fermentation of yeast.

5. Which one of the following is a bacterial biofertiliser?

(a) Anabaena

(b) Nostoc

(d) Oscillatoria

Answer:- (c) Rhizobium

6. Which one of the following is not used in organic farming?

(a) Glomus

(b) Earthworm

(d) Snail

Answer:- (d) : Organic farming is a method of farming system which primarily aimed to keep the soil alive and in good health by use of organic wastes and other biological material along with beneficial microbes (biofertilisers) to release nutrients to crops for increased sustainable production in an ecofriendly, pollution free environment. Basic components of organic farming are green manures, farm yard manure, vermicompost, crop rotation, biopesticides and biofertilisers. Glomus being a mycorrhizal component, earthworm being a vermicompost and Oscillatoria being a nitrogen fixing blue green algae can be used in organic farming. Snail cannot be a component of organic farming.

7. Anorganism used as a biofertiliser for raising soyabean crop is

(a) Azotobacter

(b) Azospirillum

(d) Nostoc.

Answer:- (c) : Rhizobium is used as a biofertiliser for raising crop. Rhizobium japonicum forms symbiotic association in the roots of the leguminous plant, soyabean.

8. The microbial biocontrol agent for butterfly caterpillar is

(a) Bacillus thuringiensis

(b) Saccharomyces

(d) Cyanobacteria

Answer:- (a) : Spores of bacterium Bacillus thuringiensis produce the insecticidal cry protein. Therefore, spores of this bacterium kill larvae of insects which eat them.

9. Which of the following is correctly matched for the product produced by them?

(a) Methanobacterium: Lactic acid

(b) Penicillium notatum : Acetic acid

(d) Acetobacter aceti: Antibiotics

Answer:- (c) : Methanobacterium is useful in the production of biogas. Penicillium notatum is used to produce penicillin, an antibiotic. Acetobacter aceti is used to obtain acetic acid.

10. A nitrogen-fixing microbe associated with Azolla in rice fields is

(a) Spirulina

(b) Anabaena

(d) Tolypothrix.

Answer:- (b) : Azolla plays a very important role in rice production. Azolla and its nitrogen-fixing partner, Anabaena, have been used as green manure to fertilise rice paddies and increase production. With the help of Azolla, rice can be grown year after year, several crops a year, with little or no decline in productivity; hence no rotation of crops is necessary.

11. Match column I with column II and select the correct option using the codes given below.

Column-I Column-II

A. Citric acid (i) Trichoderma

B. Cyclosporin A (ii) Clostridium

C. Statins (iii) Aspergillus

D. Butyric acid (iv) Monascus

(a) A-(iii), B-(i), C-(ii), D-(iv)

(b) A-(iii), B-(i), C-(iv), D-(ii)

(d) A-(iii), B-(iv), C-(i), D-(ii)

Answer:- (b)

12. Which of the following is wrongly matched in the given table?

Microbe Product Application

(a) Streptococcus Streptokinase Removal of clot from blood vessel

(b) Clostridium butylicum Lipase Removal of oil stains

(d) Monascus purpureus Statins Lowering of blood cholesterol

Answer:- (b) : Clostridium butylicum helps in the production of butyric acid. Candida lipolytica and Geotrichum candidum help in production of lipases that are added in detergents for removing oily stains from laundry.

13. Select the incorrect match.

(a) Citric acid – Aspergillus niger

(b) Streptokinase – Streptococcus

(d) Cyclosporin-A – Monascus purpureus

Answer:- (d) : Cyclosporin is used as immunosuppressant. This drug is derived from a fungus, Trichoderma polysporum and is used to prevent rejection of kidney, heart and liver transplants. Cyclosporin inhibits the activity of helper T-cells that acts against foreign bodies.

14. Which of the following microbes is used for the commercial production of ethanol?

(a) Clostridium butylicum

(b) Trichoderma polysporum

(d) Saccharomyces cerevisiae

Answer:- (b) : Yeast (Saccharomyces cerevisiae) is used for the commercial production of ethanol.

15. Viruses of the genus Nucleopolyhedrovirus are employed as

(a) gobar gas producers

(b) biological control agents

(d) atmospheric nitrogen fixing agents.

Answer:- (b) : Baculoviruses (mostly of genus Nucleopolyhedrovirus NPV) are useful in controlling many insects and other arthropods. They are species specific narrow spectrum bioinsecticides. There is no side effect on plants, mammals, birds, fish and nontarget insects. Beneficial insects are conserved. Baculoviruses are, therefore, an important component of integrated pest management (IPM) programme and are dealing with ecological sensitive areas.

16. A person admitted to hospital as he had myocardial infarction. A cardiologist injected

him 'streptokinase', why?

(a) It stimulates heart beat.

(b) It reduces hypertension.

(d) It reduces the level of blood cholesterol.

Answer:- (c) : Enzyme streptokinase is produced by the bacterium Streptococcus and modified by genetic engineering to be used as a clot buster for removing clots from blood vessels of patients who have undergone mycocardial infarction leading to heart attack.

17. Ernst Chain and Howard Florey's contribution was an

(a) establishing the potential of penicillin as effective antibiotic

(b) discovery of streptokinase

(d) discovery of DNA sequence.

Answer:- (a) : Penicillin was the first antibiotic to be discovered by Alexander Fleming (1928). He found that fungus Penicillium notatum or its extract could inhibit the growth of bacterium Staphylococcus aureus. However, its full potential as an effective antibiotic was established much later by Ernst Chain and Howard Florey. This antibiotic was extensively used to treat American soldiers wounded in the World War II. Fleming, Chain and Florey were awarded the Nobel prize in 1945, for this discovery.

18. Select the correct pair of microorganism and the product obtained from it.

(a) Monascus purpureus - produces large holes in Swiss cheese.

(b) Saccharomyces cerevisiae - used for making wine and beer after distillation.

(d) Aspergillus niger - produces citric acid and butyric acid.

Answer:- (c) : Statins produced by the yeast Monascus purpureus is used as blood-cholesterol lowering agents. Saccharomyces cerevisiae is used for making bread and ethanol. Aspergillus niger (a fungus) produces citric acid only.

19. Refer to the given figure and select the correct match.

(a) A-Fungal colony, B-Bacterial colony

(b) A-Viral colony, B-Bacterial colony

(d) A-Bacterial colony, B-Fungal colony

Answer:- (d) : The given figures (A) and (B) respectively shows bacteria and fungal colonies growing in petri dishes.

20. ‘Roquefort cheese’ is ripened by using a

(a) bacterium (b) yeast

Answer:- (d) : Adding lactic acid bacteria (LAB) to milk, results in formation of curd, because of coagulation of milk proteins. This curd is further processed to form cheese. Cheese can be classified on the basis of their texture or hardness and ripening. Roquefort cheese is a semisoft cheese. In this cheese, spores of fungus Penicillium roqueforti are added to the curd before final stages of cheese production.

21. Monascus purpureus is a yeast used commercially in the production of

(a) ethanol

(b) streptokinase for removing clots from the blood vessels

(d) blood cholesterol lowering statins.

Answer:- (d) blood cholesterol lowering statins.

22. Which among these are produced by distillation of fermented broth?

(i) Whisky (ii) Wine

(iii) Beer (iv) Rum

(v) Brandy

(a) (ii) and (iii) only (b) (i) and (ii) only

Answer:- (d) : Hard liquors such as gin, whisky, sake, arrak, brandy or rum are obtained by distillation of fermented broth.

23. Select the correct statement.

(a) Acetobacter aceti produces citric acid.

(b) Saccharomyces cerevisiae is used as clot buster.

(d) Methanogens are found in aerobic conditions.

Answer:- (c) : Acetobacter aceti produces acetic acid. Saccharomyces cerevisiae is used for commercial production of ethanol. Bacteria which produce methane are collectively called methanogens, and one such common bacterium is Methanobacterium. These bacteria are commonly found in the anaerobic sludge during sewage treatment.

24. Match the microbes in column I with their commercial/industrial products in column II and choose the correct answer.

Column I Column II

A. Aspergillus niger 1. Ethanol

B. Clostridium butylicum 2. Statins

C. Saccharomyces cerevisiae 3. Citric acid

D. Trichoderma polysporum 4. Butyric acid

E. Monascus purpureus 5. Cyclosporin A

(a) A – 4, B – 5, C – 2, D – 1, E – 3

(b) A – 5, B – 4, C – 1, D – 2, E – 3

(d) A – 3, B – 4, C – 5, D – 1, E – 2

Answer:- (c)

25. Which of the following in sewage treatment removes suspended solids?

(a) Secondary treatment

(b) Primary treatment

(d) Tertiary treatment

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.

26. The primary treatment of sewage water involves

(a) anaerobic bacterial activity

(b) sludge digestion

(d) aerobic bacterial activity.

Answer:- (c)

27. The given figure represents a typical biogas plant. Select the correct option for A, B and C respectively.

(a) A is the inlet for cattle dung.

(b) C is the outlet for the release of biogas.

(d) All of these

Answer:- (a) : The given figure is of a biogas plant, in which A is Inlet for cattle dung and water; B is Gas holder and C is Outlet for leftover slurry.

28. What gases are produced in anaerobic sludge digesters?

(a) Methane and CO2 only

(b) Methane, Hydrogen sulphide and CO2

(d) Hydrogen sulphide and CO2

Answer:- (b) : The type of gases produced by microbial activity depend upon the microbes and the organic substrates they utilise. Certain bacteria called methanogens grow anaerobically on cellulosic material and produce large amount of methane along with carbon dioxide. These bacteria are commonly found in the anaerobic sludge during sewage treatment.

Other anaerobic bacteria, involved in the process of anaerobic digestion produce other gases like ammonia and hydrogen sulphide.

29. The domestic sewage in large cities

(a) has a high BOD as it contains both aerobic and anaerobic bacteria

(b) is processed by aerobic and then anaerobic bacteria in the secondary treatment in Sewage Treatment Plants (STPs)

(d) has very high amount of suspended solids and dissolved salts.

Answer:- (b)

30. When domestic sewage mixes with river water

(a) small animals like rats will die after drinking river water

(b) the increased microbial activity releases micronutrients such as iron

(d) the river water is still suitable for drinking as impurities are only about 0.1%.

Answer:- (c) : When sewage mixes with water body, microorganisms present in it biodegrade organic matter of sewage using oxygen. This results into a sharp decline in dissolved oxygen which may cause mortality of aquatic creatures.

31. In biogas plant, the role of acidogenic bacteria is to

(a) convert monomers to organic acids

(b) convert monomers to complex polymers

(d) digest fungi in the sludge.

Answer: (a) convert monomers to organic acids

32. Refer to the given diagram and choose the correct option for it.

(a) Bacteriophage - Tobacco mosaic disease

(b) Adenovirus - Respiratory disease

(d) Prion - Alzheimer’s disease

Answer:- (b) : The picture shown in the question is of adenovirus. It most commonly causes respiratory illness which ranges from common cold to pneumonia, croup and bronchitis.

33. In gobar gas, the maximum amount is that of

(a) butane (b) methane

Answer:- (b) methane

34. Select the correct statement from the following.

(a) Biogas is produced by the activity of aerobic bacteria on animal waste.

(b) Methanobacterium is an aerobic bacterium found in rumen of cattle.

(d) Activated sludge-sediment in settlement tanks of sewage treatment plant is a rich source of aerobic bacteria.

Answer:- (d) Activated sludge-sediment in settlement tanks of sewage treatment plant is a rich source of aerobic bacteria.

35. Which of the following can be used as a biocontrol agent in the treatment of plant disease?

(a) Lactobacillus (b) Trichoderma (c) Chlorella (d) Anabaena

Answer:- (b) : Trichoderma species are free-living fungi that are very common in the root ecosystems. They are effective biocontrol agents of several plant pathogens.

36. Select the correct group of biocontrol agents.

(a) Nostoc, Azospirillum, Nucleopolyhedrovirus

(b) Bacillus thuringiensis, Tobacco mosaic virus, Aphids

(d) Oscillatoria, Rhizobium, Trichoderma

Answer:- (c) Trichoderma, Baculovirus, Bacillus thuringiensis

37. A biocontrol agent to be a part of an integrated pest management should be

(a) species-specific and symbiotic

(b) free living and broad spectrum

(d) species-specific and inactive on non-target organisms.

Answer:- (d) species-specific and inactive on non-target organisms.

38. Match the following organisms with the products they produce.

(A) Lactobacillus (i) Cheese

(B) Saccharomyces cerevisiae (ii) Curd

(D) Acetobacter aceti (iv) Bread

(v) Acetic acid

Select the correct option.

(A) (B) (C) (D)

(a) (ii) (i) (iii) (v)

(b) (ii) (iv) (v) (iii)

(d) (iii) (iv) (v) (i)

Answer:- (c)

39. Microbe used for biocontrol of pest butterfly caterpillars is

(a) Saccharomyces cerevisiae

(b) Bacillus thuringiensis

(d) Trichoderma sp.

Answer:- (b) Bacillus thuringiensis

40. Select the mismatch.

(a) Rhodospirillum – Mycorrhiza

(b) Anabaena – Nitrogen fixer

(d) Frankia – Alnus

Answer:- (a) : Rhodospirillum is a free-living nitrogen fixing bacteria. Mycorrhiza is the symbiotic association between fungi and roots of higher plants. The most common fungal partners of mycorrhiza are Glomus species.

Case based MCQs

Case I : Read the following passage and answer the questions from 41 to 45 given below.

Yeast and certain bacteria play a key role fermentation to breakdown carbohydrates into ethanol and carbon dioxide which then further used to prepare acetic acid with the help of bacterium Acctobacter aceti. Alcoholic fermentation in anaerobic process, but the conversion of alcohol to acetic acid is aerobic one. This process can be represented by following equation :

Zymase

41. The rate of alcohol production is measured

on the basis of

(a) amount of sugar present in the medium

(b) amount of CO2 produced per unit time

(d) all of these.

Answer:- (b) : The fermentation rate of the yeast can be calculated by measuring the volume of CO2 and dividing it by the amount of time it took for that volume to form.

42. A number of chemicals are produced at the time of alcoholic fermentation with the change of nutrient media, pH and aeration. Select such by-product from the following.

(a) Butanol

(b) Succinic acid

(d) All of these

Answer:- (d) All of these

43. During alcoholic fermentation of cereals and potato, the crushed food mixed with hot water for obtaining malt is called

(a) juice

(b) mash

(d) none of these.

Answer:- (b) : The crushed food mixed with hot water for obtaining malt is called mash. The sweetened nutrient medium prior to fermentation is called wort. Wort is added into the bioreactor tank, sterilised and then inoculated with yeast.

44. Distilled alcohol with 95% ethanol content is called

(a) absolute alcohol

(b) rectified spirit

(d) brandy.

Answer:- (b) : Alcoholic content in gin is 40%, brandy in 60-70%. These are hard liquors. Rectified spirit contains 95% alcohol and absolute alcohol is 100% alcohol. Rectified spirit and absolute alcohol are extensively used in laboratory works.

45. _____ is used in the preparation of vinegar.

(a) toddy (b) acetic acid

Answer:- (d)

Case II : Read the following passage and answer the questions from 46 to 50 given below.

Villagers in a place near Chambur started planning to make power supply for agricultural purposes from cow dung. They have started a biogas plant for the purpose. Study the flow chart for biogas production given below and answer the following questions.

46. Biogas is composed of majorly

(a) methane, CO2 and O2

(b) CO2, H2S and H2O

(d) H2S, H and O2

Answer:- (c) : Biogas is composed of methane (50-70%), carbon dioxide (30-40%) with traces of nitrogen, hydrogen sulphide and hydrogen.

47. In the given flow chart, ‘A’ denotes

(a) aerobic bacteria

(b) methanogenic bacteria

(d) yeast and protozoa.

Answer:- (b) methanogenic bacteria

48. What is represented by ‘B’ in the flow chart?

(a) Carbohydrates

(b) Protein polymers

(d) Fat globules

Answer:- (c) : In the second stage of biogas production, the simple soluble compounds or monomers are acted upon by fermentation causing microbes and converted into organic acids especially acetic acid.

49. ‘C’ in the given flow chart causes

(a) aerobic breakdown of complex organic compounds

(b) anaerobic digestion of complex organic compounds

(d) fermentation of monomers.

Answer:- (b) anaerobic digestion of complex organic compounds

50. If ‘A’ is not added in the procedure

(a) methane will not be formed

(b) CO2 will not be formed

(d) O2 will not be form

Answer:- (a) methane will not be formed

Assertion & Reasoning Based MCQs

For question numbers 51-60, two statements are given-one labelled Assertion and the other labelled Reason. Select the correct answer to these questions from the codes (a), (b), (c) and (d) as given below.

(a) Both assertion and reason are true and reason is the correct explanation of assertion.

(b) Both assertion and reason are true but reason is not the correct explanation of assertion.

(d) Assertion is false but reason is true.

51. Assertion : Azotobacter fixes nitrogen in symbiotic form.

Reason : Rhizobium form root nodules in the roots of leguminous plants.

Answer:- (d) : Azotobacter is a non-symbiotic nitrogen fixing bacteria i.e., it lives freely in the soil and perform nitrogen fixation. It is saprotrophic living on organic remains. Root nodules are formed by Rhizobium, a symbiotic nitrogen fixing bacteria.

52. Assertion : Azolla is used as a biofertiliser in rice fields.

Reason : Azolla shows the presence of N2 - fixing bacteria in its leaf cavities.

Answer:- (a) : Azolla can co-exist with rice plants because it does not interfere with their growth. In some South East Asian countries, especially China, the rice fields are regularly provided with Azolla. The fern is stocked in a corner of rice field and provided with manure, ash and straw before transplantation. When the field is flooded at the time of transplantation, the fern spreads rapidly and covers the space in between the rice plants. Anabaena, a blue green algae 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 utilisation by the rice plants.

53. Assertion : Whisky develops colour during the aging process.

Reason : Vodka is colourless.

Answer:- (b) : The whiskies are aged in charred oak containers. At first the whisky is colourless, the colour develops during the aging process. The vodka is not aged and bottled immediately after distillation and therefore it remains colourless.

54. Assertion : An organism which acts as herbicide is called bioherbicide.

Reason : Phytophthora palmivora is a mycoherbicide.

Answer:- (b) : An organism which controls or destroys unwanted plant growth without harming the useful plant is called bioherbicide. The first bioherbicide happened to be mycoherbicide (fungal herbicide). It was elaborated in 1981. This herbicide is Phytophthora palmivora. The fungus does not allow the Milkweed vine to grow in citrus orchards.

55. Assertion : Saccharomyces ellipsoidens is Brewer’s yeast.

Reason : Brewer’s yeast produces beer not wine.

Answer:- (d)

56. Assertion : Curdling is required in the manufacture of cheese.

Reason : Lactic acid bacteria and rennet is used for the purpose.

Answer:- (b) : The manufacture of cheese requires two main steps curdling and the ripening. Curdling the milk proteins forms a solid material from which the liquid is drained away. The curdling process may be exclusively microbiological, since acid production of lactic acid bacteria is sufficient to coagulate milk proteins. However, an enzyme known as rennin (extracted from the stomachs of calves or rennet) which curdles milk is often used for this purpose.

57. Assertion : Acetic acid is prepared by acetic acid bacteria.

Reason : Alcoholic fermentation and the conversion of alcohol to acetic acid are aerobic processes.

Answer:- (c) : Acetic acid is prepared from fermented alcohols with the help of acetic acid bacteria, Acetobacter aceti. Alcoholic fermentation is anaerobic process but the conversion of alcohol to acetic acid is aerobic one.

58. Assertion : Aspergillus niger produces lactic acid.

Reason : Aspergillus niger carry out fermentation.

Answer:- (d) : Citric acid is obtained through the fermentation carried out by Aspergillus niger and Mucor species on sugary syrups. Lactic acid was the first organic acid to be produced from the microbial fermentation. Lactic acid fermentation is carried out by both bacteria (e.g. Streptococcus lactis, Lactobacillus species and fungi (e.g. Rhizopus).

59. Assertion : The alcoholic content of fortified wines are high.

Reason : The fermentation is stopped before all the sugars are being converted.

Answer:- (b) : In fortified wines, the fermentation is stopped before all the sugar is being converted and at least 1 percent is still present. The still wines have a higher alcoholic content due to the addition of wine, brandy or alcohol.

60. Assertion : Champagne gives off bubbles.

Reason : Alcoholic content is 12 - 16% in champagne.

Answer:- (b) : Champagne wines are bottled before fermentation is complete. These wines give off bubbles of gas. Alcohol content is 12 - 16%.

Short Answer Type Questions (SA-I)

1. What is the economic value of Spirulina?

Answer:- Spirulina is single cell protein (SCP), which is rich in high quality protein and is used for consumption as human food and animal feed.

2. BOD of two samples of water A and B were 120 mg/L and 400 mg/L respectively. Which sample is more polluted?

Answer:- Water body having high BOD is more polluted as compared to water body having low BOD. Hence water sample having BOD 400 mg/L is more polluted as compared to water sample having BOD 120 mg/L.

3. Name the enzyme produced by Streptococcus bacterium. Explain its importance in medical sciences.

Answer:- Streptokinase is produced by bacterium Streptococcus. Streptokinase is modified by genetic engineering which is further used as a ‘clot bluster’ for removing clots from the blood vessels of the patients who have undergone myocardial infarction leading to heart attack.

4. Name the metabolic pathway associated with the rising of dough in making bread. What makes the dough rise?

Answer:- When Saccharomyces cerevisiae (baker’s yeast) is added to dough it causes its fermentation and releases CO2 gas which is responsible for puffed up appearance of dough.

5. Write the commercial use of lipases.

Answer:- Lipases are used in detergents formulations and are helpful in removing oily stains from the laundry.

6. Name the nutrient that gets enhanced while curdling of milk by Lactobacillus.

Answer:- The curdling of milk by Lactobacillus changed milk into curd and its nutritional quality is enhanced due to increase in vitamin B12 content.

7. Name the group of organisms and the substrate they act on to produce biogas.

Answer:- Methanogenic bacteria or methanogens are the group of anaerobic microbes which digest organic mass as well as aerobic microbes of the sludge to produce a mixture of gases containing methane, H2S and CO2 called biogas.

8. List two advantages that a mycorrhizal association provides the plant.

Answer:- Mycorrhiza perform following functions for the plant :

(i) Absorption of water and minerals like phosphorus from the soil and passing it to the plants.

(ii) Solubilisation of organic matter of the soil humus and their transfer to roots.

9. Write the biochemical reaction of yeast fermentation of molasses for alcohol production.

Answer:- The biochemical reaction for anaerobic fermentation is

10. Which of the following is a free living bacteria that can fix nitrogen in the soil?

Spirulina, Azospirillum, Sonalika

Answer:- Azospirillum is a free living nitrogen fixing bacteria.

11. Name the bacterium responsible for the large holes seen in “Swiss Cheese”.What are these holes due to?

Answer:- Bacterium responsible for large holes in Swiss cheese is Propionibacterium shermanii. Large holes in Swiss cheese are due to CO2 gas produced by these bacteria.

12. Why are some molecules called bioactive molecules? Give two examples of such molecules.

Answer:- Bioactive compounds are those compounds that have an effect on living organisms tissues or cells. Bioactive compounds are found in both plant and animal products or can be synthetically produced. Two examples of bioactive compounds are cyclosporin A and statins.

13. What makes the Nucleopolyhedrovirus a desirable biological control agent?

Answer:- Nucleopolyhedrovirus a genus of baculoviruses are useful in controlling many insects and other arthropods. They are species specific narrow spectrum bioinsecticides with no side effects on plants, mammals, birds, fish and non-target insects. Therefore, they serve as an important component of integrated pest management programme in dealing with ecological sensitive areas. These properties are useful in organic farming.

14. Name the blank spaces a, b, c and d in the table given below.

Answer:- a = Lactobacillus bulgaricus b = Trichoderma polysporum c = Yeast (Fungus) d = Penicillin

15. How has the fungus, Trichoderma polysporum proved to be very essential to organ-

transplant patients?

Answer:- Trichoderma polysporum produces a bioactive molecule known as cyclosporin–A which has antifungal, antiinflammatory and immunosuppressive properties. It inhibits activation of T-cells and therefore, prevents rejection reactions in organ transplantation.

16. How does the application of the fungal genus, Glomus, to the agricultural farm increase the farm output?

(a) Recommend two microbes that can enrich the soil with nitrogen.

(b) Why do leguminous crops not require such enrichment of the soil?

Answer:- Many members of the genus Glomus form symbiotic associations with plants to form mycorrhiza. Glomus helps to absorb phosphorus from soil and passes it to the plant. Plants having such associations show other benefits also, such as resistance to root-borne pathogens, tolerance to salinity and drought and an overall increase in plant growth and development. Therefore, Glomus increases the farm yield.

17. Name the source of streptokinase. How does this bioreactor molecule function in our body?

18. Your advice is sought to improve the nitrogen content of the soil to be used for cultivation of a non-leguminous terrestrial crop.

Answer:- (a) Azotobacter and Azospirillum.

(b) Leguminous crops have symbiotic nitrogen fixing bacteria such as Rhizobium that live in the root nodules of these plants. These bacteria obtain food and shelter from the plant and in return they trap nitrogen directly from the atmosphere which they provide to the plant.

19. State one reason for adding blue-green algae to the agricultural soil.

Answer:- A number of free living and symbiotic blue green algae or cyanobacteria have the property of nitrogen fixation and are photosynthetic. Therefore, they add organic matter as well as extra nitrogen to the soil. Hence, blue green algae serve as biofertilisers and are added to agricultural fields such as cotton, maize, jowar, rice, etc.

20. Name the organism that fixes nitrogen in symbiotic association with a water fern. Where does it live in such plant?

Answer:- Anabaena azollae (blue green alga) fixes nitrogen in symbiotic association with a water fern. It resides in the leaf cavities of the fern.

Short Answer Type Questions (SA-II)

21. Why biogas is more suitable and advantageous in rural areas?

Answer:- Biogas is more suitable and advantageous in rural areas as :

(i) It provides both energy and manure.

(ii) Biogas has wider applications than the direct burning of organic wastes.

(iii) The energy value of biogas is lower than that of organic matter but due to more efficient handling, the net energy output is roughly equal to the output in direct burning of organic wastes.

(iv) It minimises the chances of spread of faecal pathogens.

(v) The fertiliser value of the manure produced in biogas plants is similar to that of manure formed directly from organic wastes.

(vi) Biogas use does not add to pollution.

22. (a) Why are the fruit juices bought from market clearer as compared to those made at home?

(b) Name the bioactive molecules produced by Trichoderma polysporum and Monascus purpureus.

Answer:- (a) The fruit juices sold in market or bottled juices are treated with pectinases and proteases which makes them clearer than those made at home.

(b) Cyclosporin A is obtained from fungus Trichoderma polysporum whereas statin is obtained from yeast Monascus purpureus.

Cyclosporin A has immunosuppressive properties. It inhibits activation of T cells and therefore prevents rejection of transplants. Statin inhibits cholesterol synthesis and is therefore used in lowering blood cholesterol.

25. Distinguish between the roles of flocs and anaerobic sludge digesters in sewage treatments.

Answer:- Aerobic bacteria and fungi constitute ‘flocs’. Flocs are masses of aerobic bacteria held together by slime and fungal filaments to form mesh like structures. These microbes digest a lot of organic matter converting it into microbial biomass and releasing a lot of minerals. This reduces biochemical oxygen demand or BOD.

In anaerobic sludge digesters, aerobic microbes present in the sludge get killed. Anaerobic microbes digest the organic mass as well as aerobic microbes of the sludge. During this digestion, bacteria produce a mixture of gases such as methane, hydrogen sulphide and carbon dioxide. These gases form biogas which can be used as source of energy as it is inflammable. The spent sludge of anaerobic sludge digester can be used as manure or part of compost.

26. Describe the functions of anaerobic sludge digester in a sewage treatment plant.

Answer:- In secondary sewage treatment, the sediment of settling tank is called activated sludge. A part of it is used as inoculum in aeration tank. The remaining is passed into a large tank called anaerobic sludge digester. It is designed for continuous operation. The aerobic microbes present in the sludge get killed. Anaerobic microbes digest the organic mass as well as aerobic microbes of the sludge. They are of two types, nonmethanogenic and methanogenic. Methanogenic bacteria produce a mixture of gases containing methane, H2S and CO2. The mixture called biogas is inflammable and is a source of energy. The spent sludge can be used as manure or part of compost.

27. (a) Give the metabolic pathway involved in the puffing up of idli dough.

(b) Name the two different categories of microbes naturally occurring in sewage water. Explain their role in cleaning sewage water into usable water.

Answer:- (a) The metabolic pathway involved in the puffing up of idli dough is fermentation in which incomplete oxidation of glucose is achieved under anaerobic conditions by sets of reactions that produces CO2. Carbon dioxide produced during fermentation causes puffing up of idli dough. (b) Aerobic heterotrophs like bacteria and fungi occur in sewage water. They are natural decomposers and digest a lot of organic matter present in the polluted water thereby releasing minerals and reducing organic waste. Hence, they play an important role in cleaning water and making it fit for various domestic uses.

28. Name the genus to which baculoviruses belong. Describe their role in the integrated pest management programmes.

29. Choose any three microbes from the following which are suited for organic farming which is in great demand these days for various reasons. Mention one application of each one chosen. Mycorrhiza; Monascus; Anabaena; Rhizobium; Methanobacterium; Trichoderma.

Answer:- Among the given microbes, the ones which are in great demand for organic farming are: Mycorrhiza, Anabaena and Rhizobium. 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. Plants having mycorrhizal associations show resistance to root-borne pathogens. Anabaena is free living and symbiotic nitrogen fixing cyanobacteria. Cyanobacteria are photosynthetic and have the property of nitrogen fixation. They add organic matter as well as extra nitrogen to the soil. Cyanobacteria are an extremely low cost biofertilisers. Rhizobium is symbiotic nitrogen fixing bacteria. They form a mutually beneficial association with the plants. The bacteria obtain food and shelter from plants. In return, they give a part of their fixed nitrogen to the plants, thus enhancing the availability of nutrient to crops. It forms nodules on the roots of legume plants. They develop the ability to fix nitrogen only when they are present inside the root nodules.

30. Explain the role of baculoviruses as biological control agents. Mention their importance in organic farming.

Answer:- Baculoviruses (a group of viruses) are known to infect the larval stages of many harmful insects such as ants, wasps, gnats and beetles. These biological weapons are not only effective as potential biological control of these insects, but are also harmless to non-target organisms (plants, mammals, birds, fish, or even non-target insects). Majority of baculoviruses belong to the genus Nucleopolyhedrovirus and are used as biopesticides during organic farming. Organic farming is a technique of raising crops through the use of manure, fertilizers and pesticides of biological origin.

31. (a) What are the properties of an antibiotic?

(b) Explain the scientific reason for growing Azolla pinnata in the rice field.

Answer:- (a) Antibiotics are chemical substances produced by some microbes which in small concentration can kill or retard the growth of harmful microbes without adversely affecting the host. Broad spectrum antibiotic can kill or destroy a number of pathogens that belong to different groups with different structures and wall compositions. Specific antibiotics are effective only against one type of pathogen. Good antibiotics should be harmless to host with no side effects. They should be harmless to beneficial microorganism of alimentary canal and should be effective against all strains of pathogen. They should also be quick in action.

(b) Azolla pinnata supplies nitrogen, increases physicochemical properties of soils such as soil structure, texture, water holding capacity, cation exchange capacity and pH by providing several nutrients and sufficient organic matter. So, Azolla pinnata grows in rice field to increase productivity.

32. (a) Organic farmers prefer biological control of diseases and pests to the use of chemicals for the same purpose. Justify.

(b) Give an example of a bacterium, a fungus and an insect that are used as biocontrol agents.

Answer:- (a) Chemical pesticides used in agricultural fields are toxic and they kill even useful organisms along with harmful ones, harm human beings and animals, pollute soil, water and crop plants. It is estimated that despite the use of chemical pesticides 30% of the agricultural produce is lost to pathogens and pests because these continue to develop resistance against various pesticides. Now, organic farmers prefer biological control of diseases and pests to the use of chemicals. Biopesticides are the biological agents that control the growth of weeds, insects and pathogens in an agricultural field. They have targeted actions and are harmless to the crop plants, other beneficial field animals and humans. In organic farming, pests and pathogens are not eradicated but kept at manageable levels by a system of checks and balances as operating in ecosystem. An organic farmer holds the view that eradicating pests is undesirable because without them the beneficial predatory and parasitic organisms which depend upon them for food would also be annihilated. (b) Bacterium as a biocontrol agent : Bacillus thuringiensis is effective against the cabbage looper.

Fungi as a biocontrol agent : Trichoderma found in root ecosystem exerts biocontrol over several plant pathogens.

Insect as a biocontrol agent : Lady bird beetle and dragonflies feeds on aphids and prey upon mosquitoes, respectively.

33. Define the term biofertilizers? How does each of the following serve as a biofertilizer

(i) Rhizobium (ii) Ectomycorrhiza

Answer:- Biofertilizers are microorganisms which bring about nutrient enrichment of soil by enhancing the availability of nutrients like nitrogen (N) and phosphorus (P) to crops. The microorganisms which act as biofertilizers are bacteria, cyanobacteria (blue green algae) and mycorrhizal fungi.

(i) Rhizobium is one of the most important symbiotic nitrogen fixing bacteria. It forms nodules on the roots of legume plants. They develop the ability to fix nitrogen only when they are present inside the root nodules. In the nodule cells, bacteria (bacteroids) lie in groups surrounded by membrane of the host which is lined by a pink-red pigment called leghaemoglobin. Leghaemoglobin in root nodules rapidly combines with oxygen and protects the enzyme nitrogenase by any inhibitory effect of oxygen.

(ii) In ectotrophic mycorrhiza, the fungal mycelium completely encloses the feeder rootlets forming sheath or mantle. The mantle of fungal hyphae increases the absorptive surface of roots and hence serve better intake of nutrients such as nitrogen, phosphorus, calcium and potassium from the surrounding soil. The major functions which ectomycorrhiza perform are (a) absorption of water, (b) solubilisation of complex organic molecules into simple inorganic nutrients, their absorption and transfer to the roots, and (c) protection of plants from attack of disease-inciting pathogens by secreting antimicrobial substances. Ectomycorrhizae are known to occur in Pinus, Quercus, Betula, Eucalyptus, peach, etc.

Long Answer Type Questions (LA)

34. Name the site of nitrogen fixation in legumes. List and describe the biochemical components at this site.

Answer:- Symbiotic nitrogen fixation is carried out by bacteria frequently found in the root nodules of leguminous plants. The different components and their role in symbiotic nitrogen fixation are as follows:-

(i) Rhizobium : They are symbiotic nitrogen fixing bacteria commonly present in root nodules of leguminous plants and carry out conversion of atmospheric nitrogen into nitrogenous salts to make it available for absorption by plants. (ii) Lectin : Lectins are the key proteins involved in the

Rhizobium-legume symbiotic association. Bauer (1981) proposed that these lectins interact selectively with microbial cell carbohydrates (or glycoproteins) found in the capsule of bacteria and serve as determinants of recognition or host specificity.

(iii) Tryptophan : Leguminous plants release tryptophan in the soil which is absorbed by Rhizobium and is metabolised to produce IAA.

(iv) Root hair curling factor : The rhizobia produce another characteristic substance called root hair curling factor that causes deformation and twisting of root hairs.

(v) Leghaemoglobin : The nodule contains a pink coloured pigment leghaemoglobin which is like true haemoglobin combines with oxygen and CO2 gets readily oxidised into brown form with a trivalent iron.

(vi) Nitrogenase : Process of nitrogen fixation, involves reduction of atmospheric nitrogen to ammonia (NH3) by the enzyme nitrogenase. Nitrogenase is made up of two protein components, one containing iron and molybdenum, known as Mo-Fe protein or molybdo-ferredoxin (component I) and the other containing only iron called Fe-protein or azoferredoxin. Nitrogenase is extremely sensitive to oxygen. The enzyme remains active under anaerobic conditions. The leghaemoglobin binds with oxygen and protect nitrogenase from O2 inactivation. At the same time it is able to make O2 available to bacteroids for ATP production, required for nitrogen fixation.

35. What are biofertilisers? Describe their role in agriculture. Why are they preferred over chemical fertilisers?

Answer:- Biofertilisers are microorganisms which bring about nutrient enrichment of soil by enhancing the availability of nutrients like nitrogen (N) and phosphorus (P) to crops. Biofertilisers includes–nitrogen fixing bacteria, nitrogen fixing cyanobacteria and mycorrhiza. Azotobacter occurring in fields of cotton, maize, jowar and rice, not only increases yield but also saves nitrogen fertiliser upto 10–25 kg/ha.

A number of free living cyanobacteria or blue-green algae have the property of nitrogen fixation, e.g., Anabaena, Nostoc, Oscillatoria, Tolypothrix. Cyanobacteria are extremely low cost biofertilizers.

The most important of the symbiotic nitrogen fixing bacteria is Rhizobium. It forms nodules on the roots of legume plants. There are about a dozen species of Rhizobium which form association with different legume roots, e.g., R.

leguminosarum, R. lupini, R. trifolii, R. meliloti, R. phaseoli. Nitrogen fixing cyanobacteria (blue–green algae) form symbiotic association with several plants, e.g., cycad roots, lichens, liverworts, Azolla (fern). Azolla–Anabaena association is of great importance to agriculture. Azolla pinnata is a free floating fresh water fern which multiplies rapidly, doubling every 5–7 days. The fern can coexist with rice plants because it does not interfere with their growth. In some South-East Asian countries, especially China, the rice fields are regularly provided with Azolla.

Chemical fertilisers cause pollution of water bodies as well as ground water, besides getting stored in crop plants. Therefore, farmers are pressing for switch over to organic farming which includes the use of manures biofertilisers, biopesticides. Biofertilisers are microorganisms which bring about nutrient enrichment of soil by enhancing the availability of nutrients to crops. The microorganisms which act as biofertilisers are bacteria, cyanobacteria (blue green algae) and mycorrhizal fungi. Bacteria and cyanobacteria have the property of nitrogen fixation while mycorrhizal fungi preferentially withdraw minerals from organic matter for the plant with which they are associated. Phosphate is also solubilised by some bacteria and by some fungi that form association with plant roots.

36. (a) Explain the process of sewage water treatment before it can be discharged into natural water bodies.

(b) Why is this treatment essential?

Answer:- (a) Sewage water can be purified by passing it through sewage treatment plants with the action of microorganisms. A sewage treatment plant separates solids from liquids by physical processes and purifies the liquid by biological processes. There are three stages of this treatment; primary, secondary and tertiary. Primary treatment is physical, secondary biological and tertiary chemical. Primary treatment phase of sewage treatment removes floating and suspended solids from sewage through two processes of filtration and sedimentation. First floating matter is removed through sequential filtration. The filtrate is kept in large open settling tanks where grit settles down. Aluminium or iron sulphate is added in certain places to flocculation and settling down of solids. The sediment is called primary sludge while the supernatant is called effluent. The primary sludge traps a lot of microbes and debris. It is subjected to composting or land fill where anaerobic digestion removes the organic matter.

During secondary treatment, the primary effluent is taken to aeration tanks. A large number of aerobic heterotrophic microbes grow in the aeration tank. They form flocs which are masses of bacteria held together by slime and fungal filaments to form mesh like structures. The microbes digest a lot of organic matter, converting it into microbial biomass and releasing a lot of minerals. As a result the BOD of the waste matter is reduced to 10-15% of raw sewage, which is then passed into settling tank. In settling tank, the bacterial flocs are allowed to undergo sedimentation. The effluent or supernatant is generally passed into natural water bodies and sediment of settling tank is called activated sludge.

(b) This treatment prevents water pollution and water borne diseases. So, it is essential in order to protect the natural water bodies from sewage pollution.

37. Prior to sowing rice, a legume crop was cultivated and ploughed back in the field, why?

Answer:- Prior to sowing rice, a leguminous crop was cultivated and ploughed back in the field. The root nodules of leguminous plants contain Rhizobium bacteria. Rhizobium is one of the most important source of biofertilizer. It forms an efficient symbiotic relationship with leguminous plants and can fix upto 100 – 500 kg nitrogen per hectare of land. The fixed nitrogen is used up by the leguminous plants. However, a sizeable amount of fixed nitrogen is left behind in the soil in the form of residue which can be utilized by the succeeding crop.

StudyPur

Tuesday, 8 March 2022

Important questions boards March 08, 2022

Organisms and populations important questions

1. Niche is

(a) all the biological factors in the organism’s environment

(b) the physical space where an organism lives

(d) the functional role played by the organism where it lives.

Answer (d) : Niche is specific part of habitat occupied by individuals of a species which is circumscribed by its range of tolerance, range of movement, microclimate, etc. Hence, niche refers to the functional role played by the organism where it lives.

2. Presence of plants arranged into well defined vertical layers depending on their height can be seen best in

(a) tropical rainforest (b) grassland (c) temperate forest (d) tropical savannah.

Answer:- (a) tropical rainforest

3. Match mean annual precipitation in column I with the biome in column II and choose the right option.

Column I Column II

(i) 0-50 cm A. Tropical forest

(ii) 50-100 cm B. Coniferous forest

(iii) 150-400 cm C. Grassland

(iv) 50-250 cm D. Desert

(a) (i)-D (ii)-C (iii)-A (iv)-B

(b) (i)-C (ii)-A (iii)-B (iv)-D

(d) (i)-B (ii)-D (iii)-A (iv)-C

Answer :- (a) (i)-D (ii)-C (iii)-A (iv)-B

4. Plants requiring low light intensity for optimum photosynthesis are called

(a) heliophytes (b) pteridophytes (c) sciophytes (d) bryophytes.

Answer (c) : Plants growing in partial shade or low intensity light are called shade plants or sciophytes. In shade plants, the stems are soft, slender with large internodes. Leaves are thin and large sized. They have bright green colour. Cuticle is thin. Stomata are present on both the surfaces. Palisade parenchyma is less developed. There is more vegetative growth as compared to flowering and fruiting. In low light intensity they acquire optimum photosynthesis.

5. –1°C to 13°C annual variations in the intensity and duration of temperature and 50 to 250 cm annual variation in precipitation, account for the formation of a major biome as

(a) temperate forest (b) coniferous forest (c) tropical forest (d) grassland.

Answer (b) coniferous forest

6. Pneumatophores occur in

(a) halophytes

(b) free-floating hydrophytes

(d) submerged hydrophytes.

Answer (a) : Pneumatophores are breathing or respiratory roots which are found in plants growing in mangroves or water logged soil or saline swamps. Such plants are called halophytes.

7. Organisms possessing strong and stout forelimbs with clawed digits show _______ adaptations.

(a) arboreal (b) cursorial

Answer (c) : The fossorial adaptation is adaptation for digging. Organisms showing this type of adaptation have strong and stout forelimbs. They are provided with claws which are useful in digging. They may have a pointed, elongated snout which is also used in digging.

8. Some desert beetles can survive on “metabolic water”, without ever drinking liquid water which

(a) is a breakdown product of pyruvate inside the mitochondria, along with carbon dioxide

(b) was produced as water in the organisms they eat

(d) is absorbed from the air along with respiratory oxygen.

Answer:- (a) is a breakdown product of pyruvate inside the mitochondria, along with carbon dioxide

9. Which one of the following is the specific xerophytic adaptation?

(a) Presence of spines

(b) Absence of stomata

(d) Presence of stipular leaves

Answer:- (c) : Xerophytes are the plants living in xeric (dry) habitats. Deserts are the typical xerophytic habitat. Plants growing in xerophytic habitats develop various morphological and physiological adaptations. Roots of these plants are deep tap roots penetrating the soil to great depths so as to absorb water to the maximum. Roots can also be shallow but extensive and spreading so as to collect and hold rain water. In many xerophytes, leaves are reduced to form spines to reduce the transpiration losses as in cacti.

10. Which of the following statements regarding responses of organisms to abiotic factors is false?

(a) All birds and mammals are capable of thermoregulation.

(b) Majority of animals and nearly all plants cannot maintain a constant internal environment.

(d) Very small animals are commonly found in polar regions as they have to spend less energy to generate body heat.

Answer:- (d) : Heat loss or heat gain is a function of surface area. Since small animals have a larger surface area relative to their volume they tend to lose body heat very fast, when it is cold outside and they have to expend much more energy to generate body heat through metabolism. Hence, very small animals are rarely found in polar areas.

11. The success of mammals on earth is largely because

(a) they have the ability to maintain constant body temperature

(b) they can conform to the changes in the environment

(d) they can reduce metabolic activity and go into a state of dormancy during unfavourable conditions in the environment.

Answer:- (a) : Mammals are endothermic. Endothermy (maintenance of internal temperature) enable these animals to live at temperatures that other land vertebrates cannot bear.

12. Which one of the following refers to Allen’s rule?

(a) An organism can move from a stressful habitat to a more hospitable area and return when the stressful period is over.

(b) If the stressful conditions are localised or remain only for a short duration, an organism either migrates or suspends itself.

(d) Mammals from colder climates have shorter ears and limbs to minimize heat loss.

Answer:- (d) Mammals from colder climates have shorter ears and limbs to minimize heat loss.

13. Which one of the following is not a parasitic adaptation?

(a) Development of adhesive organs

(b) Loss of digestive organs

(d) Loss of unnecessary sense organs

Answer (c) : Parasitism is a relationship between two living organisms of different species in which one organism called parasite obtains its food directly from another living organism called host. The parasite spends a part or whole of its life either on or inside the body of the host. The general parasitic adaptations are

(i) anaerobic respiration in internal parasites,
(ii) loss of certain organs,

(iii) presence of adhesive organs,
(iv) excessive multiplication,

(v) resistant cysts and eggs for safe transfer of their progeny to new hosts and

(vi) well developed and complicated reproductive organs.

14. The stage of suspended development shown by zooplanktons is called

(a) desiccation (b) diapause (c) hibernation (d) homeostasis

Answer (b) : Environment keeps changing with time and these changes may be unfavourable for an organism e.g., extreme heat or extreme cold. Some organisms migrate from the unfavourable place to the place with favourable conditions e.g., birds. Some organisms, unable to migrate, might avoid the stress by escaping in time. They do so by reducing their metabolic activity and going into a state of ‘dormancy’. Under unfavourable conditions many zooplankton species in lakes and ponds are known to enter diapause, a stage of suspended development.

16. Select the incorrect statement.

(a) An overwhelming majority of animals and nearly all plants maintain a constant internal temperature.

(b) An orchid growing as an epiphyte on a mango branch is an example of commensalism.

(d) Very small animals are rarely found in polar regions.

Answer:- (a) An overwhelming majority of animals and nearly all plants maintain a constant internal temperature.

17. Match the following and select the correct option.

Column I Column II

A. Bears 1. Diapause

B. Snail 2. Hibernation

C. Zooplanktons3. Dormancy

D. Seeds 4. Aestivation

(a) A -3, B-4, C-1, D-2 (b) A-2, B-1, C-4, D-3

Answer:- (d) A-2, B-4, C-1, D-3

18. In a growing population of a country,

(a) pre-reproductive individuals are more than the reproductive individuals

(b) reproductive individuals are less than the post-reproductive individuals

(d) pre-reproductive individuals are less than the reproductive individuals.

Answer:- (a) : In a population where the number of prereproductive individuals or the younger individuals is higher than the reproductive individuals, the population will increase.

19. Asymptote in a logistic growth curve is obtained when

(a) K = N

(b) K > N

(d) the value of ‘r’ approaches zero.

Answer (a) : Asymptote in a logistic growth curve is obtained when population density (N) reaches the carrying capacity (K), i.e., N = K.

20. Select the correct statement.

(a) In a population, number of births is different from birth rate

(b) A sigmoid growth curve is depiction of exponential growth

(d) ‘r’ is equal to the difference between number of births and number of deaths in a population.

Answer:- (a) : Birth, i.e., production of new offspring is an attribute of an individual whereas birth rate i.e., production of new individuals per unit population per unit time is an attribute of population. A sigmoid growth curve is depiction of logistic growth, a more realistic growth model where individuals compete for limited resources and the fittest individual survives and reproduces. In an exponential growth curve (J-shape curve) the asymptote, is beyond carrying capacity as resources are unlimited. ‘r’ is intrinsic rate of natural increase.

21. A biologist studied the population of rats in a barn. He found that the average natality was 250, average mortality 240, immigration 20 and emigration 30. The net increase in population is

(a) 05 (b) zero (c) 10 (d) 15.

Answer:- (b) : Natality and immigration positively contribute to the population growth while mortality and emigration are negative factors. In the given question,

The net increase in population is
natality + immigration = 250 + 20 = 270
The net decrease in population is
mortality + emigration = 240 + 30 = 270
Thus, net increase in population = 270 – 270 = 0

22. Which of the following is correct?

(a) Population change = (Birth + immigration) – (death + emigration)

(b) Population change = (Birth + immigration) + (death + emigration)

(d) Population change = (Birth – immigration) – (death + emigration)

Answer:- (a) Population change = (Birth + immigration) – (death + emigration)

23. Which is correctly labelled with respect to the given diagram?

(a) B : Logistic curve

(b) C : Carrying capacity

(d) A : Carrying capacity

Answer:- (d) : The given figure shows population growth curve, in which A is carrying capacity, B is exponential growth curve and C is logistic growth curve.

25. The formula of growth rate for population in given time is

(a) dt/ dN = rN (b) dt/rN = dN

Answer:- (d) : The population grows in an exponential or geometric ratio. If in a population of size N, the birth rates (not total number but per capita births) are represented as b and death rates (per capita death rates) as d, the increase or decrease in N (growth rate) during a unit time period

t (dN/dt) will be
dN/dt = (b – d) × N
Let (b – d) = r, then
dN/dt = rN.
The r in this equation is called intrinsic rate of natural increase.

26. Between which among the following, the relationship is not an example of commensalism?

(a) Orchid and the tree on which it grows

(b) Cattle Egret and grazing cattle

(d) Female wasp and fig species

Answer:- (d) : Female wasp and fig species show mutualism.

27.If ‘+’ sign is assigned to beneficial interaction, ‘–’ sign to detrimental and ‘O’ sign to neutral interaction, then the population interaction represented by ‘+’ ‘–’ refers to

(a) mutualism (b) amensalism

Answer:- (d) : Parasitism is an association in which one organism (the parasite) lives on (ectoparasitism) or in (endoparasitism) the body of another organism (host), from which it obtains its nutrients. This association is beneficial for the parasites as they get continuous supply of nutrients from their host and are able to rapidly multiply their numbers. But it is detrimental for the host organism as parasitic infection leads to various complications and diseases in the host body may also be fatal to him under certain circumstances.

28. Match the following.

Population Example

Interaction

1. Predation A. Cuscuta and hedge plants

2. Commensalism B. Balanus and Chthamalus

3. Parasitism C. Cactus and moth

4. Competition D. Orchid and mango

(a) 1-C, 2-D, 3-A, 4-B (b) 1-D, 2-C, 3-B, 4-A

Answer:- (a) 1-C, 2-D, 3-A, 4-B

29. Connel’s field experiment on the rocky sea coast of Scotland, where larger barnacle Balanus dominates the intertidal area and removes the smaller barnacle Chthamalus. This happened due to

(a) parasitism (b) predation

Answer (d) : Competition is a rivalry between two or more organisms for obtaining the same resources.

30. All the following interactions are mutualism, except

(a) plant and animal relation for pollination

(b) association of algae and fungi in lichens

(d) association of fungi and roots of higher plants in mycorrhiza.

Ans:- (c) association of cattle egret and grazing cattle

31. Gause’s principle of competitive exclusion states that

(a) no two species can occupy the same niche indefinitely for the same limiting resources

(b) larger organisms exclude smaller ones through competition

(d) competition for the same resources exclude species having different food preferences.

Answer:- (a) : Gause’s ‘competitive exclusion principle’ states that if resources are limited, then two closely related species competing for the same resource cannot co-exist indefinitely and the competitively inferior one will be eliminated eventually.

32. Match column I with column II and select the correct option from the codes given below.

Column I Column II

A. Commensalism (i) One inhibited, other unaffected

B. Parasitism (ii) One benefitted, other unaffected

C. Mutualism (iii) Both are benefitted

D. Amensalism (iv) One benefitted, other harmed

(a) A-(iv), B-(ii), C-(iii), D-(i)

(b) A-(iii), B-(iv), C-(ii), D-(i)

(d) A-(ii), B-(iv), C-(i), D-(iii)

Answer:- (c) A-(ii), B-(iv), C-(iii), D-(i)

33. Carnivorous animals lions and leopards, occupy the same niche but lions predate mostly larger animals and leopards take smaller ones. This mechanism of competition is referred to as

(a) character displacement

(b) altruism

(d) competitive exclusion.

Answer (c) resource partitioning

34. A biologist studied the population of rats in a barn. He found that the average natality was 250, average mortality 240, immigration 20 and emigration 30. The net increase in population is

(a) 05 (b) zero

Answer:- (b) : Natality and immigration positively contribute to the population growth while mortality and emigration are negative factors. In the given question,

The net increase in population is
natality + immigration = 250 + 20 = 270
The net decrease in population is
mortality + emigration = 240 + 30 = 270
Thus, net increase in population = 270 – 270 = 0

35. Select the correct option w.r.t. age pyramids.

(a) A-Expanding, B-Stable, C-Declining

(b) A-Stable, B-Expanding, C-Declining

(d) A-Declining, B-Stable, C-Expanding

Answer:- (a) : In a growing or expanding population, the number of pre-reproductive individuals is very large, number of reproductive individuals is moderate while post reproductive individuals are fewer. In a stable population, the number of pre reproductive and reproductive individuals is almost equal whereas post reproductive individuals are fewer. In declining population, proportion of reproductive age group is higher than pre-reproductive age group. Number of post reproductive individual is also sizeable.

36. Which one of the following population interactions is widely used in medical science for the production of antibiotics?

(a) Commensalism

(b) Mutualism

(d) Amensalism

Answer:- (d) : Amensalism is a relationship between organisms of different species in which an organism does not allow other organism to grow or live near it. Inhibition is achieved through the secretion of chemicals called allochemics. Antibiotics are a kind of allochemics produced by some microbes which in small concentration can kill or retard growth of harmful microbes without adversely affecting the host.

37. Which of the following statements is false regarding predators?

(a) Predators keep prey populations under control.

(b) Predators help in maintaining species diversity in a community.

(d) Herbivores (predators) have a greater advantage since the plants cannot run away to avoid predation.

Answer:- (c) : Predators are those organisms that devour other organisms to obtain energy. They may be carnivores (eating other animals) or herbivores (eating plants). Besides acting as ‘conduits’ for energy transfer across trophic levels, predators play other important roles. They keep prey populations under control. Predators also help in maintaining species diversity in a community, by reducing the intensity of competition among competing prey species. But if a predator is too efficient and overexploits its prey, then the prey might become extinct and following it, the predator will also become extinct for lack of food.

38. Which one of the following causes population explosion?

(a) Decrease in infant mortality rate and increase in death rate

(b) Decrease in death rate, maternal mortality rate and infant mortality rate

(d) Decrease in death rate and increase in maternal mortality rate

Answer:- (b) : Death rate is defined as the number of deaths per 1,000 individuals in a population per year. It decreases population size and population density. Maternal mortality rate is the rate of maternal deaths. Maternal death is the death of a woman while pregnant or within 42 days of termination of pregnancy, irrespective of the duration of pregnancy, from any cause related to or aggravated by the pregnancy or its management, but not from accidental or incidental causes.

MMR = Maternal deaths
Live births
Infant mortality rate is the number
of deaths per 1,000 infants who die between birth and 1 year of age.
A decrease in all three of these factors can result in population explosion.

39. Which one of the following is most appropriately defined?

(a) Host is an organism which provides food to another organism.

(b) Amensalism is a relationship in which one species is benefitted whereas the other is unaffected.

(d) Parasite is an organism which always lives inside the body of other organism and may kill it.

Answer:- (c) : Predation is an interaction between members of two species in which members of one species capture, kill and eat up members of other species ‘Host’ is a term which is specifically related to parasitism. Amensalism is an interspecies interaction in which one species is harmed whereas the other one is unaffected. Parasitic organism can live both over the surface of their host or inside their body.

40. Which one of the following microbes forms symbiotic association with plants and helps them in their nutrition?

(a) Azotobacter (b) Aspergillus

Answer (c) : Azotobacter, Aspergillus and Trichoderma all are free living microbes that help plants in their nutrition. Glomus is a fungus that symbiotically forms endomycorrhiza that helps in absorption of nutrition specially phosphorus from soil.

Case Based MCQs

Case I : Read the following passage and answer the questions from 41 to 45 given below.

Organism P has thick lips and tongue so that it can easily feed on the commonly available spiny plants. Organism Q has thick layer of insulating fat under the skin. It was strong hooves to walk steadily on steep surfaces and lives in burrows during winters. Organism R has bright colours and sticky pads on its fingers and toes. It lives on trees.

41. Which of the following is correct habitat for organisms P regarding its adaptation?

(a) Grassland biome

(b) Desert biome

(d) Tropical deciduous forest

Answer:- (b) : P is a camel adapted to desert conditions as it has thick lips and tongue so that it can easily feed on the commonly available spiny plants.

42. Which of the following is correct match regarding organism Q and its habitat?

(a) Tundra - Polar bear

(b) Tropical rain forest - Deer

(d) Desert - Camel

Answer:- (a) : Q is polar bear as it has thick insulating fat layer under the skin.

43. Which of the following is incorrect regarding organisms R’s habitat?

(a) The vegetation shows stratification

(b) Epiphytic growth is rich

(d) Deep rooted shrubs are common due to abundant sunlight

Answer:- (d) : R could be poison dart frog as it has bright colours and sticky pads on its fingers and toes and its habitat is tropical rainforest. Deep rooted shrubs are not found in tropical rainforest. These are found in deserts. 44. (a) : Habitat of P is desert, so, the dominant plant is Opuntia.

44. The dominant plants in habitat where P lives could be

(a) Opuntia

(b) Nymphaea

(d) both (a) and (c).

45. Organisms P, Q and R respectively most likely occur in

(a) F, B and A (b) C, A, E

Answer (c) A, F and C

Assertion & Reasoning Based MCQs

(a) Both assertion and reason are true and reason is the correct explanation of assertion.

(b) Both assertion and reason are true but reason is not the correct explanation of assertion.

(d) Assertion is false but reason is true.

51. Assertion : Microclimate generally differs from the prevailing regional climatic conditions.

Reason : Microclimate represents the climatic conditions that prevail at local scale or in areas of limited size.

Answer:- (a) : Microclimate generally differs from the prevailing regional climatic conditions, because the microclimate represents the climatic conditions that prevail at local scale or in areas of limited size, such as the immediate surroundings of plants and animals.

52. Assertion : Aerenchyma is present in the leaves and petioles of hydrophytes.

Reason : Aerenchyma imparts buoyancy to the hydrophytes.

Answer (a) : Plants which remain permanently immersed in water are called hydrophytes. They may be submerged or partly submerged and show the presence of aerenchyma (large air space) in the leaves and petioles. Aerenchyma helps to transport oxygen produced during photosynthesis and permits its free diffusion to other parts, including roots located in anaerobic soils. These tissues also impart buoyancy to the plants.

53. Assertion : Ectotherms are able to remain active under cold conditions.

Reason : Ectotherms are unable to maintain a constant internal temperature.

Answer:- (d) : Endotherms (warm - blooded animals) can regulate their body temperature by physiological means and are able to maintain a more or less constant internal temperature, even when the temperature outside fluctuates (for example - birds and mammals). They have physiological mechanisms for keeping body temperature constant or within tolerance limits. Ectotherms, (cold-blooded animals) cannot regulate their body temperature and are unable to maintain a constant internal temperature. Their body temperature tends to match with the environmental temperature in which they live, (for example-frogs and snakes).

54. Assertion : Many mangrove plants possess high levels of organic solutes.

Reason : This is an adaptation to cope with the conditions of high salt concentration and osmotic potential.

Answer (a) : Mangroves are found in marshy conditions of tropical deltas and along ocean edges. For coping with conditions of high salt concentration and osmotic potential, many mangrove plants have high levels of organic solutes, such as proline and sorbitol. Dunaliella species (green and halophytic algae found in hyper saline lakes) can tolerate saline conditions by accumulating glycerol in the cells, which helps in osmoregulation. Some species of mangroves can excrete salts through the salt glands on the leaves. Some mangroves can exclude salts from the roots by pumping excess salts back into soil.

55. Assertion : With increase in population size, environmental resistance tends to increase.

Reason : This is a nature’s way to check the expression of biotic potential.

Answer:- (a) : The inherent maximum capacity of an organism to reproduce or increase in number is termed as biotic potential (designated by the symbol ‘r). Biotic potential is realised only when the environmental conditions are non-limiting, so that natality rate (birth- rate) is maximum and mortality rate (death - rate) is minimum. Under these conditions, population size increases at the maximum rate. However, nature keeps a check on the expression of biotic potential. For example, if a pair of flies is allowed to reproduce unchecked, the fly population may outweigh the earth in a few years. The environmental check on population size, or its biotic potential is called environmental resistance. With increase in population size, the environmental resistance (against the population) tends to increase. The environmental resistance represents the limiting effect of abiotic (e.g., water, space) and biotic factors (e.g., food, competition) that do not allow organisms to attain their biotic potential and keep the population size at a much lower level.

56. Assertion : The soil profiles of grassland, forest and desert biomes differ from each other.

Reason : Soil profile develops due to weathering process, accumulation of organic matter and leaching of mineral matter.

Answer:- (b)

57. Assertion : Heliophytes, generally have low photosynthetic, respiratory and metabolic activities.

Reason : Heliophytes are the sun adapted plants which are adapted to high intensity of light.

Answer (d) : Plants have special traits that help them to enlarge their tolerance limits to light regimes. Individual plants, as well as plant communities, adapt to different light intensities by becoming shade tolerant or sun adapted. Heliophytes are the sun adapted plants which are adapted to high intensity of light, and have higher temperature optima for photosynthesis,as well as have high rate of respiration. On the other hand, shade adapted plants also called as sciophytes generally have low photosynthetic, respiratory and metabolic activities. Plants such as ferns and several herbaceous plants growing on the ground under the dense canopy of trees, are shade tolerant plants.

58. Assertion : Mycorrhizal relation exists between Boletus and Pinus.

Reason : Mycorrhizal association is a symbiotic interaction.

Answer:- (b) : Mycorrhiza is a mutualistic or symbiotic interaction in which a fungus (e.g. Boletus) and a root of plant (e.g. Pinus) are involved. The root provides food and shelter to the fungus. The fungus helps the plant in solubilisation and absorption of minerals, water uptake and protection against pathogenic fungi.

59. Assertion : Mimicry is the resemblance of one organism to another.

Reason : Mimicry may be protective or aggressive.

Answer:- (b) : Mimicry is defined as the resemblance of one organism to another or to any natural object for the purpose of concealment, protection or for some other advantage. Mimicry employed by a prey is known as protective mimicry while the other used by a predator is termed as aggressive mimicry.

60. Assertion:Phytoplanktons grow in abundance in the profundal zone of lake.

Reason : Profundal zone is the dark zone where light does not reach.

Answer:- (d) : Life activity under water is often controlled by the availability of light. In aquatic systems the presence of light determines where producers and consumers are to live in water. For example, the phytoplanktons (phyto : plants; plankton : small) live in the illuminated surface layer of water, whereas benthic organisms live in, or at, the segments of a lake. Profundal zone is the dark zone where light does not reach.

SUBJECTIVE TYPE QUESTIONS

Very Short Answer Type Questions (VSA)

1. Soil horizons A and B represent solum. Why?

Answer:- Only A and B horizons of soil represents solum or true soil because they have weathered products of the parent rock.

2. Plants like Calotropis have evolved adaptations for defence against grazers. Explain.

Answer:- In order to keep predators or herbivores away from grazing, certain plants like Calotropis produce highly poisonous cardiac glycosides that can make herbivores sick when eaten.

Hence, grazers like cattle, goats and sheep do not graze it.

3. Name the type of interaction that exists between barnacles and whale.

Answer:- Commensalism

4. When and why do some animals like frogs hibernate?

Answer:- When animals are exposed to low temperatures, hibernation is necessary for cold-blooded animals like frogs to prevent their metabolic rate from getting slow down.

5. In a population, per capita birth rate is 0.025 and per capita death rate is 0.008 during a unit time period. What is the value of intrinsic rate of natural increase, ‘r’ for the population?

Answer:- Intrinsic rate of natural increase,

r = Per capita birth rate – Per capita death rate

= 0.025 – 0.008 = 0.017.

6. When and why do some animals like snails go into aestivation?

Answer:- When animals like snails are exposed to lethal high temperatures, they go into aestivation to avoid the heat of summer.

7. Give one example where population estimation of an organism is done indirectly without actually counting the organism.

Answer:- Sometimes population size is indirectly estimated without actually counting them, for example, tiger census in our National parks and tiger reserves is often based on pug (animal’s foot print) marks and faecal pellets.

8. What is an interaction called when an orchid grows on a mango plant?

Answer:- Commensalism

9. Give example of an organism that enters ‘diapause’ and why.

Answer:- Bombyx mori (silk moth) is an insect that enters diapause due to some adverse environmental conditions such as drought, extreme temperature, reduced food availability which, in turn, delays the overall development. The physiological and metabolic activities diminish at this particular time. 10. Emphemerals are drought escapers xerophytic plants, for example: Tribulus terrestris.

Short Answer Type Questions (SA-I)

11. What type of interaction is seen when koel lays eggs in crow’s nest?

Answer:- Koel or cuckoo laying its eggs in crow’s nest is an example of brood parasitism. In brood parasitism, the parasitic bird lays its eggs in the nest of its host, and the host incubates them. Here, the eggs of the parasitic bird have evolved to resemble the host’s egg (in size and colour) so as to avoid being detected.

13. Draw and explain expanding age pyramids of human population. Why is it so called?

Answer:- An age pyramid for expanding human population is as follows:

Pyramid with broad base or triangular shape indicates a rapidly expanding population with a high percentage of prereproductive individuals followed by reproductive then postreproductive individuals. Thus, in rapidly growing population, birth rate is high and population keeps growing.

14. Differentiate between the mutualism and competition.

Answer

15. What does the given age pyramid signify about the status of a population? (The bar at the base represents pre-reproductive individuals.)

Answer:- Given is a bell-shaped age pyramid which signifies that the population is stable. Such age pyramid is formed when the number of pre-reproductive and reproductive individuals is almost equal and the post-reproductive individuals are comparatively fewer. It implies that the population is neither decreasing nor increasing, instead is maintained at a stable level.

16. Shark is eurythermal while polar bear is stenothermal. What is the advantage the former has and what is the constraint the later has?

Answer:- Sharks being eurythermal can tolerate wide range of temperature variations and thus have wider distribution on earth, on the other hand, polar bear being stenothermal can tolerate only narrow range of temperature and is restricted to specific regions only.

17. Explain why very small animals are rarely found in polar region.

Answer:- Very small animals have large surface area to body volume ratio. It results in excessive heat loss from exposed body surface. Such a great extent of heat loss makes it impossible for very small animals to survive in cold polar regions. Large animals have small surface area to body volume ratio, reducing heat loss and making temperature maintenance easier for them. This effect of temperature on the absolute size of an animal and the relative proportions of various body parts is also known as Bergmann’s rule.

18. Explain parasitism and coevolution with the help of one example of each.

Answer:- Parasitism is the interspecific interaction where one of species (called parasite) depends on the other species (host) for food and shelter and damages the host. E.g., malarial parasite in blood cells of humans. Coevolution in parasitism refers to the process in which parasite evolves mechanism to interact and neutralise the mechanism evolved by the host to reject or resist parasite.

19. Why do clown fish and sea anemone pair up? What is this relationship called?

Answer:- Mutualism is the interaction between clown fish and sea anemone. The clownfish lives among the stinging tentacles of sea anemone and gets protection from its predators. Clownfish has a slimy mucus covering that protects it from the poisonous tentacles of sea anemone. Also clown fish makes its meals from the anemone’s leftover. In return clown fish helps anemone in catching its prey by luring other fish towards it. It also eats the dead tentacles keeping the anemone and the area around it clean.

21. Why the plants that inhabit a desert are not found in a mangrove? Give reasons.

Answer:- Plants inhabiting desert (xerophytes) are not found in mangroves, because xerophytic plants are adapted to dry and hot environment. They possess various physical modifications to tolerate extreme water scarcity and heat, like extensive root system, succulent organs, leaf reduced to spine, etc. Mangrove swamp is a region of vegetation where soil is highly saline and water logged. Only halophytes can survive in such regions as they possess aerial roots called pneumatophores through which gaseous exchange occurs. Roots of xerophytes are positively geotropic and will suffocate and die in such badly aerated soil ultimately leading the whole plant to death.

Short Answer Type Questions (SA-II)

22. Refer to the given graph and answer the following questions.

(a) What does the graph represent? Identify A, and C.

(b) Differentiate between A and B.

Answer:- (a) The given graph represents various ways of organismic response i.e., possibilities of living organisms to cope with stressful conditions. A represents the conformers, B represents regulators and C represents partial regulators.

(b) Differences between A (conformers) and B (regulators) are as follows:

(c) C i.e., partials regulators have the ability to regulate body functions to a limited extent. Beyond that limit they become conformers.

23. Define the following:

(a) Ecological niche

(b) Gause’s competitive exclusion principle

Answer :- (a) Ecological niche (Grinnel, 1917) is a 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.

(b) Gause’s competitive exclusion principle states that two or more species with similar niche requirements cannot coexist indefinitely in the same area and one of the two gets eliminated.

(c) Animals develop strategies to live better in their environment. Mimicry is the resemblance of one species with another in order to obtain advantage, especially against predation.

24. (a) How are herbs able to grow on forest floors?

Answer:- (a) Sciophytes or shade plants like herbs and shrubs survive under the shadow of big canopied trees or sun plants or heliophytes in forests as they are perfect shade tolerant plants which show better growth in lower level of light intensity. They grow in a manner, that they are arranged in different strata according to their shade tolerance.

(b) What are osmoconformers?

Answer:- (b) Osmoconformers are those organisms which cannot maintain constant osmolarity of their body fluids and it varies according to their surrounding medium.

25. Predation is usually referred to as detrimental association. State any three positive roles that a predator plays in an ecosystem.

Answer:- Predators play important role in ecosystem. These are discussed as follows:

(i) Maintaining prey population : In nature, the population of predator is quite small as compared to that of the prey. The prey has high reproductive potential. If, for some time, the prey population is allowed to grow without predation, then it would grow beyond the carrying capacity of the environment. The predator keeps the population of the prey under check so that an equilibrium is maintained. Example, the prickly pear cactus introduced in Australia in the early 1920’s caused havoc by spreading rapidly into millions of hectares of rangeland. Finally, the invasive cactus was brought under control only after a cactus-feeding predator (a moth) from its natural habitat was introduced into the country.

(ii) Maintaining species diversity : Predators also help in maintaining species diversity in a community, by reducing the intensity of competition among competing prey species. Example, in the rocky intertidal communities of the American

Pacific Coast, the starfish Pisaster is an important predator. When all the starfish were removed from an enclosed intertidal area, more than 10 species of invertebrates became extinct within a year because of interspecific competition.

(iii) Vegetation : Predation helps in growth of vegetation all over the globe by restricting population of herbivores.

27. Name and explain the type of interaction that exists in mycorrhizae and between cattle egret and cattle.

Answer:- Mycorrhiza is a mutualistic interaction between fungus and roots of higher plants. The root provides food and shelter to the fungus. The fungus helps the plant in solublisation and absorption of minerals, water uptake and protection against pathogenic fungi.

The egret and grazing cattle in close association is an example of commensalism. Commensalism is the interaction in which one organism is benefitted and other organism is neither harmed nor benefitted. The egrets always forage close to where the cattle are grazing because the cattle, as they move, stir up and flush out from the vegetation insects that otherwise might be difficult for the egrets to find and catch.

29. When you go for a trek/trip to any high altitude places, you are advised to take it easy and rest for the first two days. Comment, giving reasons.

Answer:- Atmospheric pressure is low at higher altitudes as compared to plains. When we go for a trek/trip on high altitude, then due to low atmospheric pressure our body does not get enough oxygen, as a result of which we experience nausea, fatigue and heart palpitation (altitude sickness). But by taking rest for first two days, body gets acclimatised to high altitude conditions. The body compensates low oxygen availability by increasing red blood cell production, decreasing binding capacity of haemoglobin and increasing breathing rate. Hence, we automatically stop experiencing altitude sickness.

30. During a school trip to ‘Rohtang Pass’, one of your classmates suddenly developed ‘altitude sickness’. But, she recovered after sometime.

(a) Mention one symptom to diagnose the sickness.

(b) What caused the sickness?

Answer:- (a) Heart palpitation

(b) Sickness is due to low atmospheric pressure of high altitudes, as body does not get enough oxygen.

(c) After sometimes, body compensates for low oxygen availability by increasing red blood cell production, decreasing binding capacity of haemoglobin and by increasing breathing rate.

31. Water is very essential for life. Write any three features both for plants and animals which enable them to survive in water scarce environment.

Answer:- Water is very essential for life. Plants and animals show modifications according to availability of water in the area. Some of the adaptations seen in plants which enable them to survive water scarce environment are as follows:

– Plants of hot deserts are adapted to survive in dry conditions of soil and high temperatures. The plant which evade dry conditions are known as ephemerals. Some plants have deep tap root which are capable of absorbing water from deep soil e.g., Prosopis, Acacia, etc.

– In case of cacti and succulents, the presence of fleshy leaves and stems to store water (succulence) is an adaptation to dry environment. In cacti, leaves are reduced to spines, where stems are modified to fleshy structures.

– Many tropical plants, particularly grasses which grow in hot and arid climates possess C4 pathway of photosynthesis. So, these plants perform better in low soil water environments. Such plants, use less water to achieve higher rates of photosynthesis.

Some of the adaptations seen in animals which enable them to survive water scarce environment are as follows:

– Desert lizards lack the physiological ability that mammals have to deal with the high temperature. They keep their body temperature fairly constant by behavioural means. They enjoy in the sun and absorb heat when their body temperature drops below the comfort zone but move into shade when the surrounding temperature starts increasing.

– The Kangaroo rats conserves water by excreting solid urine and can live from birth to death without even drinking water. Loss of water is minimised by producing nearly solid urine and faeces.

– The camels show tolerance to wide fluctuations in body temperature and are able to maintain blood stream moisture even during extreme heat stress.

32. Differentiate between hibernation and aestivation. Give one example of each.

Answer:- The differences between hibernation and aestivation are as follows:

33. Some organisms suspend their metabolic activities to survive in unfavourable conditions. Explain with the help of any four examples.

Answer:- To tide over unfavourable conditions, some organisms suspend their metabolic activities. These are discussed as follows :

(i) Bacteria, fungi and lower plants develop thick walled spores, which germinate during suitable conditions.

(ii) Polar bears go into hibernation during winter season to escape cold.

(iii) Some snails and fish undergo aestivation to avoid summer related problems like heat and dessication.

(iv) During unfavourable conditions, zooplanktons in lakes and ponds are known to enter diapause, i.e., stage of suspended development.

34. (a) List any three ways of measuring population density of a habitat.

(b) Mention the essential information that can be obtained by studying the population of an organism.

Answer:- (a) The different methods to study population size are as follows:

– Quadrat method : It is a method which involves the use of square of particular dimension to measure number of organisms. For example the number of Parthenium plants in a given area can be measured using the quadrat method.

– Direct observation: It involves counting of organisms. For example, in order to determine the number of bacteria growing in a petri dish, their colonies are counted.

– Indirect method : The number of fishes caught per trap gives the measure of their total density in a given water body.

(b) Whether competition of survival exists or not, whether the population is increasing or decreasing, natality, mortality, emigration and immigration.

35. Name and explain the kind of interaction in the following:

(i) Algae and fungi in lichens

(ii) Hermit crab and sea anemone

(iii) Head louse and humans

Answer (i) Mutualism is found between algae and fungi in lichen. Lichen is a composite entity which is formed jointly by an alga and a fungus. The fungus provides water, minerals and shelter to the alga. In return alga provides food to the fungus.

(ii) Interaction between sea anemone and hermit crab is considered as an example of mutualism (or as protocooperation by some). Sea anemone uses hermit crab as a portable home and is able to find more food. Hermit crab in turn gets protection from its enemies. Thus it is a mutually beneficial association. Recent studies reveal that it is obligate mutualism not proto-cooperation.

(iii) Parasitism is found between head louse and humans. Head louse is an ectoparasite that sucks the blood of man or feed on living tissues of head’s skin.