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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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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 2 Sexual Reproduction,in Flowering Plants

These Solutions are part of NCERT Exemplar Solutions for Class 12 Biology. Here we have given NCERT Exemplar Solutions for Class 12 Biology chapter 2 Sexual Reproduction,in Flowering Plants

Multiple Choice Questions

Question 1.
Among the terms listed below, those that are not technically correct names for a floral whorl are
(a) androecium
(b) corolla|
(c) (i) and (iv)
(d) (ii) and (iv)
Answer:
(c) : The technically correct terms for the floral whorls are (from outermost to innermost) calyx, corolla, androecium and gynoecium. They are made up of sepals, petals, stamens and carpels, respectively.

Question 2.
Embryo sac is to ovule as________ is to an anther.
(a) stamen
(b) filament
(c) pollen grain   
(d) androecium
Answer:
(c) : A typical carpel consist of ovary which can have many ovules. Each ovule has an embryo sac, which in turn has a single egg cell. Similarly, in majority of angiosperms each stamen consists of a bilobed anther, which in turn has two pollen sacs in each lobe, consisting of pollen grains.

Question 3.
In a typical complete, bisexual and hypogynous flower the arrangement of floral whorls on the thalamus from the outermost to the innermost is
(a) calyx, corolla, androecium and gynoecium
(b) calyx, corolla, gynoecium and androecium|
(c) gynoecium, androecium, corolla and calyx
(d) androecium, gynoecium, corolla and calyx.
Answer:
(a)

Question 4.
A dicotyledonous plant bears flowers but never produces fruits and seeds. The most probable cause for the above situation is
(a) plant is dioecious and bears only pistillate flowers
(b) plant is dioecious and bears both pistillate and staminate flowers
(c) plant is monoecious
(d) plant is dioecious and bears only staminate flowers.
Answer:
(d) : Fruits can develop from a single ovary of a single flower (simple fruit) or from several free carpels of a single flower (aggregate) or from whole inflorescence (multiple). In total, fruits develop from ovaries. This is why, a dioecious plant (unisexual) bearing only staminate (male) flowers will not produce fruits, whereas monoecious plants (bisexual) or dioecious plants bearing only pistillate (female) flowers or pistillate and staminate both can bear fruits (pollination).

Question 5.
The outermost and innermost wall layers of microsporangium in an anther are respectively
(a) endothecium and tapetum
(b) epidermis and endodermis
(c) epidermis and middle layer
(d) epidermis and tapetum.
Answer:
(d) : The wall layers of a micro­sporangium from outermost to innermost are : epidermis, endothecium, middle layers and tapetum. The first three layers generally provide protection and help in dehiscence of anther. Tapetum performs nutritive function for pollen grains.

Question 6.
During microsporogenesis, meiosis occurs in
(a) endothecium
(b) microspore mother cells
(c) microspore tetrads
(d) pollen grains.
Answer:
(b) : Microsporogenesis is the formation of microspores in the form of tetrads, which later separate and are called pollen grains. Microspore mother cell (2n) undergoes meiosis for the formation of haploid pollen grains formed first in the form of spore tetrads.

Question 7.
From among the sets of terms given below, identify those that are associated with the gynoecium.
(a) Stigma, ovule, embryo sac, placenta
(b) Thalamus, pistil, style, ovule
(c) Ovule, ovary, embryo sac, tapetum
(d) Ovule, stamen, ovary, embryo sac
Answer:
(a) : Stigma,is a part of pistil on which pollen grain lands. Each ovary may have many ovules, which have embryo sacs in them. Placenta is a tissue inside ovary to which ovules are attached. Thalamus is the terminal part of the axis of flower which bears all floral appendages. Tapetum is the innermost layer of microsporangium while stamen is component of androecium.

Question 8.
Starting from the innermost part, the correct sequence of parts in an ovule are
(a) egg, nucellus, embryo sac, integument
(b) egg, embryo sac, nucellus, integument
(c) embryo sac, nucellus, integument, egg
(d) egg, integument, embryo sac, rtucellus.
Answer:
(b) : Egg cell is inside the embryo sac in the ovule. The embryo sac is further enclosed by the parenchymatous tissue, nucellus, which later provides nutrition to developing embryo. Nucellus is ultimately surrounded by integuments.

Question 9.
From the statements given below choose the option that are true for a typical female gametophyte of a flowering plant.
(1) It is 8-nudeate and 7-celled at maturity.
(2) It is free-nuclear during the development.
(3) It is situated inside the integument but outside the nucellus.
(4) It has an egg apparatus situated at the chalazal end.
(a) (i) and (iv)
(b) (ii) and (iii)
(c) (i) and (ii)   
(d) (ii) and (iv)
Answer:
(c) : Female gametophyte or embryo ‘* sac is present in the nucellus of an ovule. It is formed by free nuclear mitotic divisions of megaspore which forms 8 nucleate structure (4 nuclei each at micropylar and chalazal end). One nucleus from each side moves to the middle to form polar nuclei which later on fuse to form secondary nucleus. The remaining three nuclei at each end get surrounded by wall to form cells. Hence, female gametophyte is 8-nucleate and 7-celled at maturity because of presence of secondary nucleus. Egg apparatus is situated at the micropylar end whereas 3 antipodal s are situated at chalazal end.

Question 10.
Autogamy can occur in a chasmogamous flower if
(a) pollen matures before maturity of ovule
(b) ovules mature before maturity of pollen
(c) both pollen and ovules mature simultaneously
(d) both anther and stigma are of equal lengths.
Answer:
(c) : Autogamy is pollination within a flower, chasmogamous flowers are those in which anthers and stigma are exposed. For autogamy, in such a flower to take place, pollen and ovule should mature simultaneously and anther and stigma should lie close to each other.

Question 11.
Choose the correct statement from the following.
(a) Cleistogamous flowers always exhibit autogamy.
(b) Chasmogamous flowers always exhibit geitonogamy.
(c) Cleistogamous flowers exhibit both autogamy and geitonogamy.
(d) Chasmogamous flowers never exhibit autogamy.
Answer:
(a) : Autogamy is pollination within the same flower. Geitonogamy is pollination between different flowers of same plant. Xenogamy is pollination between flowers of different plants of same species. Cleistogamous flowers (that do not open at all) always exhibit autogamy, where as chasmogamous flowers (with exposed anthers and stigma) can exhibit autogamy, geitonogamy or xenogamy.

Question 12.
A particular species of plant produces light, non-sticky pollen in large numbers and its stigmas are long and feathery. These modifications facilitate pollination by
(a) insects
(b) water
(c) wind
(d) animals
Answer:
(c) : Light, non-sticky pollens produced in large numbers are generally traits of wind pollinated (anemophilous) plants. Insect and animal pollinated plants have sticky pollens. Long and feathery stigma is also characteristic of anemophilous plants. Maize, Cannabis and many grasses are some of the examples of this category.

Question 13.
From among the situations given below, choose the one that prevents both autogamy and geitonogamy.
(a) Monoecious plant bearing unisexual flowers.
(b) Dioecious plant bearing only male or female flowers.
(c) Monoecious plant with bisexual flowers.
(d) Dioecious plant with bisexual flowers.
Answer:
(b) : Monoecious plant (bisexual) bearing either bisexual or unisexual flowers can exhibit both autogamy as well as geitonogamy. Dioecious (unisexual) plants bearing only male or female flowers will not show autogamy or geitonogamy hence, only xenogamy is possible.

Question 14.
In a fertilised embryo sac, the haploid, diploid and triploid structures are
(a) synergid, zygote and primary endosperm nucleus
(b) synergid, antipodal and polar nuclei
(c) antipodal, synergid and primary endo­sperm nucleus
(d) synergid, polar nuclei and zygote.
Answer:
(a) : Double fertilisation is the fusion of two male gametes brought by a pollen tube with two different cells of the same female gametophyte in order to produce two different structures. It is found only in angiosperms where it was first discovered by Nawaschin in 1898 in Fritillaria and Lilium. Out of the two male gametes one fuses with egg or oosphere to perform generative fertilisation. Generative fertilisation is also called syngamy or true fertilisation. It gives rise to a diploid zygote or oospore. The second male gamete fuses with two haploid polar nuclei or diploid secondary nucleus of the central cell to form a triploid primary endosperm nucleus (PEN). This is called as vegetative fertilisation (or triple fusion).

Question 15.
In an embryo sac, the cells that degenerate after fertilisation are
(a) synergids and primary endosperm cell
(b) synergids and antipodals
(c) antipodals and primary endosperm cell
(d) egg and antipodals.
Answer:
(b)

Question 16.
While planning for an artificial hybridisation programme involving dioecious plants, which of the following steps would not be relevant?
(a) Bagging of female flower
(b) Dusting of pollen on stigma
(c) Emasculation
(d) Collection of pollen
Answer:
(c) : Artificial hybridisation is human performed crossing of two different plants having complementary good traits in order to obtain an overall superior variety. Artificial hybridisation has been used by plant breeders for crop improvement programme. Two precautionary measures in artificial hybridisation are emasculation and bagging. Emasculation is removal of stamens from the floral buds of female parent so that chances of self pollination are eliminated. In case of dioecious (unisexual) plants, emasculation is not required.

Question 17.
In the embryos of a typical dicot and a grass, true homologous structures are
(a) coleorhiza and coleoptile
(b) coleoptile and scutellum
(c) cotyledons and scutellum
(d) hypocotyl and radicle.
Answer:
(c) : During the development of dicot embryo, initially the dicot embryo is globular and undifferentiated. Early embryo with radial symmetry is called proembryo. It is transformed into embryo with the development of radicle, plumule and cotyledons. Two cotyledons differentiate from the sides with a faint plumule in the centre. At this time the embryo becomes heart-shaped. Part of embryo axis between the plumule and cotyledonary node is epicotyl (above the level of cotyledons) while the part between radicle and cotyledonary node is called hypocotyl (below the level of cotyledons). The single cotyledon of monocotyledonous seed (e.g. maize grain) is called scutellum. It occupies the major portion of the embryo regions of grain.

Question 18.
The phenomenon observed in some plants wherein parts of the sexual apparatus is used for forming embryos without fertilisation is called
(a) parthenocarpy
(b) apomixis
(c) vegetative propagation
(d) sexual reproduction.
Answer:
(b) : Apomixis is the term given to any phenomenon that leads to formation of embryo wherein parts of the sexual apparatus are used, but without fertilisation. Fertilisation is also absent in vegetative propagation, but parts of sexual apparatus are not involved. An example of apomixis is Citrus.

Question 19.
In a flower, if the megaspore mother cell forms megaspores without undergoing meiosis and if one of the megaspores develops into an embryo sac, its nuclei would be 
(a) haploid
(b) diploid
(c) a few haploid and a few diploid
(d) with varying ploidy.
Answer:
(b)

Question 20.
The phenomenon wherein, the ovary develops into a fruit without fertilisation is called
(a) parthenocarpy
(b) apomixis
(c) asexual reproduction
(d) sexual reproduction.
Answer:
(a) : Fertilised ovary is technically called fruit. But if ovary develops into fruit, without fertilisation, it is called parthenocarpic fruit. Such fruits are generally seedless. Some common examples found in nature are : Citrus, banana, etc. Parthenocarpy can also be artificially induced by the application of certain plant hormones, specially, auxin and gibberellins.

Very Short Answer Type Questions

Question 1.
Name the component cells of the ‘egg apparatus’ in an embryo sac.
Answer:
Egg apparatus consists of two synergids and one egg cell.

Question 2.
Name the part of gynoecium that determines the compatible nature of pollen grain.
Answer:
Stigma is that part of gynoecium which determines the compatible nature of pollen grain.

Question 3.
Name the common function that cotyledons and nucellus perform.
Answer:
Common function of nucellus and cotyledons is to provide nourishment.

Question 4.
Complete the following flow chart:

Answer:

Question 5.
Indicate the stages where meiosis and mitosis occur (1, 2 or 3) in the flow chart.
Megaspore mother cell ——-1—— > Megaspores –2— > Embryo sac —-3—- > Egg
Answer:
1 = Meiosis 2 = Mitosis 3 = Mitosis
In the diagram given below, show the path of a pollen tube from the pollen on the stigma into the embryo sac. Name the components of egg apparatus

Question 6.
In the diagram given below, show the path of a pollen tube from the pollen on the stigma into the embryo sac. Name the components of egg apparatus.

Answer:

Egg apparatus consists of two synergids and one egg cell.

Question 7.
Name the parts of pistil which develop into fruit and seeds.
Answer:
Fruit develops from ovary and seeds develop from ovules of a pistil.

Question 8.
In case of polyembryony, if an embryo develops from the synergid and another from the nucellus, which is haploid and which is diploid?
Answer:
In case of polyembryony, if an embryo develops from synergid, it will be haploid and if it develops from nucellus, it will be diploid.

Question 9.
Can an unfertilised, apomictic embryo sac give rise to a diploid embryo? If yes, then how?
Answer:
Apomictic embryo sac develops without the involvement of meiosis and syngamy. Here, a diploid embryo is formed from diploid egg cell or from some other diploid cell of embryo sac.

Question 10.
Which are the three cells found in a pollen grain when it is shed at the three celled stage?
Answer:
The three cells found in a pollen grain when it is shed at the three celled stage are two male gametes and one vegetative cell.

Question 11.
What is self-incompatibility?
Answer:
If a pistil carrying functional female gametes fails to set seeds following pollination with viable and fertile pollen, capable of bringing about fertilisation in another pistil, the two are said to be incompatible, and the phenomenon is known as sexual incompatibility. Sexual incompatibility may be interspecific (between individuals of different species) or intraspecific (between individuals of the same species). The latter is also called self-incompatibility.

Question 12.
Name the type of pollination in self­ incompatible plants.
Answer:
Self incompatible plants do not undergo self pollination. They can undergo only cross pollination.

Question 13.
Draw the diagram of a mature embryo sac and show its 8-nudeate, 7-celled nature. Show the following parts: antipodals, synergids, egg, central cell, polar nuclei.
Answer:

Question 14.
Which is the triploid tissue in a fertilised ovule? How is the triploid condition achieved?
Answer:
Endosperm is the triploid tissue in a fertilised ovule which is formed by the division of Primary Endosperm Nucleus (PEN). PEN is formed by the fusion of one male gamete (haploid) with secondary nucleus (diploid) hence, it is triploid.

Question 15.
Are pollination and fertilisation necessary in apomixis? Give reasons.
Answer:
In apomixis, there is no need of pollination and fertilisation. Embryo can develop directly from the nucellus or synergid or egg cell.

Question 16.
Identify the type of carpel with the help of diagrams given below:
Answer:

(a) Polycarpellary syncarpous
(b) Polycarpellary apocarpous

Question 17.
How is, pollination carried out in water plants?
Answer:
Plants which occur below the water level undergo epihydrophily or hypohydrophily,i.e pollination takes place by the agency of water. Aquatic plants which are emerged or present at the water surface can undergo entomophily, anemophily or epihydrophily.

Question 18.
What is the function of the two male gametes produced by each pollen grain in angiosperms?
Answer:
The function of the two male gametes produced by each pollen grain in angiosperms are as follows:

1. One of the male gamete fuses with egg cell to produce zygote (2n).
2. Second male gamete fuses with two polar nuclei/diploid secondary nucleus to form primary endosperm nucleus (3n)

Short Answer Type Questions

Question 1.
List three strategies that a bisexual chasmogamous.flower can evolve to prevent self pollination (autogamy).
Answer:
Strategies that a bisexual chasmogamous flower can evolve to prevent self pollination are as follows:

1. Pollen release and stigma receptivity is not synchronised. Either the pollen is released before stigma is receptive or stigma becomes receptive much before pollen release.
2. Anther and stigma are placed at different positions, so that pollen cannot come in contact with stigma.
3. When the pollen of the flower reach the stigma of the same flower, pollen grains
4. do not germinate and the phenomenon is called self-incompatibility.

Question 2.
Given below are the events that are observed in an artificial hybridisation programme. Arrange them in the correct sequential order in which they are followed in the hybridisation programme.
(a) Re-bagging
(b) Selection of parents
(c) Bagging
(d) Dusting the pollen on stigma
(e) Emasculation
(f) Collection of pollen from male parent.
Answer:
(b) —> (e) —> (c) —> (f) —> (d) —> (a)

Question 3.
Vivipary automatically limits the number of offsprings in a litter. How?
Answer:
Viviparity is a form of reproduction in animals in which the development of embryos takes place within the mother’s/ female parent’s body. The embryo obtains its nourishment directly from mother via placenta or by other means and subsequently mother gives birth to the full term young one. It is common in most mammals. Viviparity limits the number of offspring in a litter due to the following reasons:

1. As the number of eggs released during ovulation are limited during oestrous or menstrual cycles so the number of eggs fertilised during reproductive cycle of female are also limited.
2. As the entire period of development called gestation is passed within the mother’s/female parent’s body; it restricts the number of embryo that can develop together at one time.
3. During gestation no ova or egg are released.

Question 4.
Does self incompatibility impose any restrictions on autogamy? Give reasons and suggest the method of pollination in such plants.
Answer:
Self incompatibility is the phenomenon in which self pollens fail to germinate on stigma of pistil. It is a gene physiological process which is controlled by single gene S. If pollen and pistil will have the S alleles in common, pollens will not be functional on that pistil. As self pollens will have common S alleles with pistil (of the same flower), self pollination (autogamy) cannot take place in self incompatible plants. Cross pollens on the other hand, will not have common S alleles with pistil and hence, cross pollination can easily take place in such plants.

Question 5.
In the given diagram, write the names of parts shown with lines.

Answer:

Question 6.
What is polyembryony and how can it be commercially exploited?
Answer:
The phenomenon of having more than one embryo is called polyembryony, e.g., onion, groundnut.

Polyembryony is practically important because genetically uniform parental type seedlings are obtained from nucellar embryos. Nucellar embryos are superior to those obtained by vegetative propagation because nucellar embryo seedlings are disease free and maintain their superiority for long time.

Question 7.
Are parthenocarpy and apomixis different phenomena? Discuss their benefits.
Answer:
Parthenocarpy is production and development of seedless fruits. Apomixis is a type of sexual reproduction which does not involve meiosis and syngamy and produces seeds without fertilisation.
Benefits of parthenocarpy are as follows:
(1) It produces fruits which do not contain irritant seeds.
(2) Processing of fruits by food industry requires the removal of seeds which is quite difficult. Therefore, seedless fruits are preferred by food industry.
(3) Fruits can be developed inside green houses where pollinators are not available.
Benefits of apomixis are as follows:

• Production of hybrid seeds is costly and hence the cost of hybrid seeds becomes too expensive for the farmers. If these hybrids are made into apomicts, there is no segregation of characters in the hybrid progeny. Then the farmers can keep on using the hybrid seeds to raise new crop year after year and do not have to buy hybrid seeds every year.
• Adventive embryos are better clones than cuttings.
• Embryos formed through apomixis are generally free from infections.

Question 8.
Why does the zygote begin to divide only after the division of primary endosperm cell (PEC)?
Answer:
Endosperm development precedes embryo development. The primary endo¬sperm cell divides repeatedly and forms the triploid endosperm tissue. The cells of this tissue are filled with reserve food materials and are used for the nutrition of the developing embryo at the micropylar end of the embryo sac. This is an adaptation to provide assured nutrition to the developing embryo.

Question 9.
The generative cell of a two-celled pollen divides in the pollen tube but not in a three- celled pollen. Give reasons.
Answer:
In some plants pollen grains are shed at 2 celled stage whereas in others pollen grains are shed at 3 celled stage. The pollens which are shed at 2 celled stage contain vegetative cell and a generative cell. At the time of pollen germination, generative cell divides to form two male gametes in the pollen tube.
On the other hand, in case of pollens which are shed at 3 celled stage, the generative cell has already divided to form two male gametes before pollination or pollen germination, i.e., formation of pollen tube.

Question 10.
In the ,figure given below, label the following parts: .
Male gametes, egg cell, polar nuclei, synergid and pollen tube.

Answer:

Long Answer Type Questions

Question 1.
Starting with the zygote, draw the diagrams of the different stages of embryo development in a dicot.
Answer:

Question 2.
What are the possible types of pollinations in chasmogamous flowers. Give reasons.
Answer:
Chasmogamous flowers are the flowers with stamens and stigma exposed .They may undergo self pollination or cross pollination.
(1) Self pollination in chasmogamous flowers : Bisexual flowers where stigma and the stamens both mature almost at the same time can undergo autogamy or geitonogamy.

• Autogamy – Transfer of pollen grains from anther of a flower to the stigma of the same flower.
• Geitonogamy – When pollen grains from one flower are deposited on the stigma of another flower borne on the same plant.

(2) Cross pollination in chasmogamous flowers : Transfer of pollen grains from anther of a flower to the stigma of different flower growing on different plant of same species. Cross pollination is carried by various abiotic agents like wind and water or by biotic agents like insects, birds and animals etc.

Question 3.
With a neat, labelled diagram, describe the parts of a mature angiosperm embryo sac. Mention the role of synergids.
Answer:
Neat and labelled diagram of mature angiospermic embryo sac is as follows :

Ovule is an integumented megasporangium that encloses an embryo sac. Common type of ovule is anatropous. Parts of mature angiospermic ovule are:
(1) Funicle : It is the stalk of the ovule. It is attached to placenta by funicle. In anatropous ovules the funicle is fused with the body of the ovule lengthwise to form raphe. Place of union of funicle and the body of ovule is called hilum.

(2) Integuments : They are one or two cuticularised coverings of the ovule. The  place of origin of integuments is called chalaza. A pore’occurring on one side of ovule where integuments are absent is known as micropyle.

(3)  Nucellus : It is parenchymatous tissue  contained in the ovule.

(4) Embryo sac : It is female gametophyte which is covered by a thin membrane. Embryo sac has seven cells. Three cells form egg apparatus towards micropylar end. There are two synergids and one egg or oosphere in the egg apparatus. Three cells on the opposite side are called antipodal cells. The seventh cell of the embryo sac is the largest cell called central cell. Central cell has two polar nuclei which may fuse to form a diploid secondary nucleus. Synergids are short­lived (one of them degenerates long before fertilisation and second after entry of pollen tube into embryo sac).
These synergids help :

• In growth of pollen tube towards egg by secreting chemotropically active substances.
• In nutrition of embryo sac by absorption and transport of food from nucellus through their filiform apparatus.

Filiform apparatus in the form of finger like projections from cell wall is present in upper part of each synergid. The filiform apparatus is useful for the absorption and transportation of

materials from the nucellus to the embryo sac. Hook like structures help in easy penetration of pollen tube and liberation of male gamete from the pollen tube.

Question 4.
Draw the diagram of a microsporangium and label its wall layers. Write briefly on the role of the endothecium.
Answer:

Question 5.
Endothecium protects the sporogenous tissue and helps in dehiscence of pollen grains.
Embryo sacs of some apomictic species appear normal but contain diploid cells. Suggest a suitable explanation for the condition.
Answer:
Embryo sacs of some apomictic species appear normal but contain diploid cells, because embryo sacs either develop from diploid nucellar cells or from diploid megaspore mother cells without undergoing meiosis. It leads to formation of apomictic diploid embryos.

We hope the NCERT Exemplar Solutions for Class 12 Biology chapter 2 Sexual Reproduction,in Flowering Plants help you. If you have any query regarding NCERT Exemplar Solutions for Class 12 Biology chapter 2 Sexual Reproduction,in Flowering Plants, drop a comment below and we will get back to you at the earliest.

NCERT Solutions for Class 10 Social Science Economics Chapter 1 Development

These Solutions are part of NCERT Solutions for Class 10 Social Science. Here we have given NCERT Solutions for Class 10 Social Science Economics Chapter 1 Development.

TEXTBOOK EXERCISES

Question 1.
How is development of a country determined ?
Answer:
The development of a country can be generally determined by average income or per capital income.

Question 2.
Which neighbouring country of India has better performance in terms of human development than India?
Answer:
Sri Lanka.

Question 3.
Assume there are four families in a country. The average or per capita income of these families is 5,000. If the income of the three families is 4,000, 7,000, and 3,000 respectively, what is the income of the fourth family?
Answer:
Rs. 6,000.

Question 4.
What is the main criterion used by the World Bank in classifying different countries? What are the limitations of this criterion, if any?
Answer:
The average income, i.e. per capita income is the main criterion used by the World Bank in classifying different countries.
According to the World Development Report 2006, published by the World Bank, countries with per capita income of $10066 per annum and above in 2004 are called rich or developed countries. On the other hand, countries with per capita income of $825 or less are called low-income countries.

Limitations: It does not tell us about how the average income is distributed among the people in the individual countries. The countries with the same per capita income might be very different with regard to income distribution. One might have equitable distribution of income, while the other might have great disparities between the rich and the poor.

Question 5.
In what respects is the criterion used by the UNDP for measuring develop­ment different from the one used by the World Bank?
Answer:
The criterion used by the UNDP for measuring development is different from the one used by the World Bank in the following respects:
The World Bank – The World Bank uses per capita income as the sole criterion for measuring development.
The UNDP – It uses the Human Development Index (HDI) based on a combination of factors such as health, education, and income as the criterion for measuring development.
Thus, the UNDP does not rely solely on per capita income, as the criterion for measuring development, as in the case with the World Bank.

Question 6.
Why do we use averages? Are there any limitations to their use? Illustrate with your own examples related to development.
Answer:
We use averages because they are useful for comparing differing quantities of the same category. For example, to compute the per capita income of a country, averages have to be used because there are differences in the incomes of diverse people. However, there are limitations to the use of averages. Even though they are useful for comparison, they may also hide disparities. For example, the infant mortality rate of a country does not differentiate between the male and female infants born in that country. Such an average tells us nothing about whether the number of children dying before the age of one are mostly boys or girls.

Question 7.
Kerala, with lower per capital income has a better human development ranking than Punjab. Hence, per capital income is not a useful criterion at all and should not be used to compare states. Do you agree? Discuss.
Answer:
No, I do not agree with the statement that per capita income is not a useful criterion at all. Kerala, with lower per capita income, has a better human development ranking than Punjab because, human development ranking is determined using a combination of factors such as health, education, and income. So, this does not imply that per capita income is not useful. Rather, per capita income is one of the development factors and can not be neglected. The World Bank uses per capita income as the criterion for measuring development and comparing states. But this criterion has certain limitations because of which determination of Human Development Index (HDI) is done using this criterion along with some other development factors like health, education etc.

Question 8.
Find out present sources of energy used by people in India. What could be possibilities fifty years from now?
Answer:
(1) The present sources of energy used by the people in India are as given below :

• Conventional sources: Coal, petroleum, natural gas, electricity.
• Non-conventional sources: Solar energy, wind energy and energy produced by using biogas, geothermal energy, tidal energy, and wave energy.

(2) Position of energy after 50 years in India: The position of energy in India after about 50 years will not be good due to the reasons mentioned below :

1.  The consumption of non-renewable resources at present is very high in comparison to production and reserves.
2. The reserves for the world as a whole would last for 43 years.
3. The countries like India depend on importing oil from abroad because they do not have enough stock of their own.
4. If price of oil increases, it becomes a burden on the country’s finances. India too has to spend a lot of foreign exchange for importing oil and petroleum and its products. It is putting a heavy strain on India’s economic development. However, India has many advantages due to its geographical features. Thus India could be in a better position to face the energy crisis, if any, as mentioned below :
   • There should be judicious utilization of the abundant renewable energy resources, such as biomass energy, solar energy, wind energy, and geothermal energy.
   • Apart from augmenting the energy supply renewable resources will help India in mitigating climate change.
   • Solar power has got the tremendous potential of energy which can be harnessed. Solar energy systems are available for industrial and domestic use with the added advantage of minimum maintenance. Solar energy could be made financially viable with government tax incentives and rebates.
   • Wind energy is one of the most efficient alternative energy sources. India now ranks as a “wind superpower” having a net potential of about 45000 MJV only from 13 identified states.
   • India has huge hydropower potential, out of which around 20% has been realized so far.
   • Biomass energy can play a major role in reducing India’s reliance on fossil fuels by making use of thermo-chemical conversion technologies.

Question 9.
Why is the issue of sustainability important for development?
Answer:
Sustainability for development or sustainable development refers to the development which is done without damaging the environment and other resources. In other words, balancing the need to use resources and also conserve them for future is known as sustainable development.

The issue of sustainability is important for development because development must happen in tandem with future. If natural resources are not sustained, it will cause stagnation of development after a point of time. Exploiting resources unethically will ultimately undo the development that a country may have achieved. This is because in the future, those resources will not be available for further progress.

Question 10.
“The Earth has enough resources to meet the needs of all but not enough to satisfy the greed of even one person.” How is this statement relevant to the discussion of development? Discuss.
Answer:
It is a fact that earth has enough resources to meet the needs of all but not enough to satisfy the greed of even one person because one person may exploit the natural resources recklessly. The reckless exploitation of resources may lead to their exhaustion and may damage the environment. In such a situation what will happen if no natural resources are available? It will hamper the development process in all the countries in the world because fossil fuels and minerals are essential for development. The reckless use of minerals may disturb the balance in nature. Thus, it is necessary to use resources judiciously for development and to adopt a strategy of economic development that is environment friendly.

Question 11.
List a few examples of environmental degradation that you may have observed around you.
Answer:
Environmental degradation manifests itself in different ways. Deforestation, falling levels of groundwater, soil erosion, water pollution, burning of fossil fuels, the hole in the ozone layer, and combustion from automobiles causing extreme air pollution especially in urban areas are some of the examples of environmental degradation.

Question 12.
For each of the items given in the following table, find out which country is at the top and which is at the bottom.
SOME DATA REGARDING INDIA AND ITS NEIGHBOURS FOR 2004

Country	Per capital income in US$	Life expectancy at birth	Literacy rate for 15+ yrs population	Gross enrollment ratio for three levels	HDI rank in the world
Sri Lanka	4390	74	91	69	93
India	3139	64	61	60	126
Myanmar	1027	61	90	48	130
Pakistan	2225	63	50	35	134
Nepal	1490	62	50	61	138
Bangladesh	1870	63	41	53	137

Answer:
For each of the items given in the above table, the country at the top and at the bottom is as given below :

Item	Country at the top	Country at the bottom
(1)  Per capita income	Sri Lanka	Myanmar
(2) Life expectancy at birth	Sri Lanka	Myanmar
(3)  Literacy rate for 15+ yrs
population	Sri Lanka	Bangladesh
(4) Gross enrolment ratio
for three levels	Sri Lanka	Pakistan
(5) HDI rank in the world	Sri Lanka	Nepal

Question 13.
The following table shows the proportion of undernourished adults in India. It is based on a survey of various states for the year 2001. Look at the table and answer the following questions :

State	Males (%)	Females (%)
Kerala	22	19
Karnataka	36	38
Madhya Pradesh	43	42
All States	37	36

1. Compare the nutritional level of people in Kerala and Madhya Pradesh.
2. Can you guess why around 40 percent of people in the country are under­nourished even though it is argued that there is enough food in the country? Describe in your own words.

Answer:

1. The undernourished adults in Kerala are males 22% and females 19% only whereas in Madhya Pradesh, it is 43% males and 42% females. It is, thus clear that the number of undernourished adults in Madhya Pradesh is higher than Kerala.
2. Around 40 percent of people in the country are undernourished even though it is argued that there is enough food in the country due to reasons as mentioned below :
   • The Public Distribution System has failed in its objects.
   • The average consumption of PDS grain at the All India level is only 1 kg per month per person.
   • The average consumption is as low as less than 300 gm per person per month in the states of Bihar, Orissa, and Uttar Pradesh.
   • Sometimes PDS dealers are found resorting to malpractices like diverting the grain to open market to get better margin, selling poor quality grains at ration shops. People generally do not buy such low-quality grains.
   • When ration shops are unable to sell, massive stock of foodgrains piles up with the FCI. Such stock gets rotten or eaten by rats. This leads to a shortage of foodgrains and to a situation as described above that there is enough food in the country but around 40 percent of people are undernourished.

We hope the NCERT Solutions for Class 10 Social Science Economics Chapter 1 Development helps you. If you have any query regarding NCERT Solutions for Class 10 Social Science Economics Chapter 1 Development, drop a comment below and we will get back to you at the earliest.

NCERT Exemplar Solutions for Class 12 Biology chapter 9 Strategies for Enhancement in Food Production

These Solutions are part of NCERT Exemplar Solutions for Class 12 Biology. Here we have given NCERT Exemplar Solutions for Class 12 Biology chapter 9 Strategies for Enhancement in Food Production

Multiple Choice Questions

Question 1.
The chances of contracting bird flu from a properly cooked (above 100°C) chicken and egg are
(a) very high
(b) high
(c) moderate
(d) none.
Answer:
(d)

Question 2.
A group of animals which are related by descentand share many similarities are referred to as
(a) breed
(b) race
(c) variety
(d) species.
Answer:
(a)

Question 3.
Inbreeding is carried out in animal husbandry because it
(a) increases vigour
(b) improves the breed
(c) increases heterozygosity
(d) increases homozygosity.
Answer:
(d) : Breeding of animals of the same breed for 4-6 generations is called inbreeding. Thus, inbreeding is necessary if we want to develop a pureline. Inbreeding, as a rule, increases homozygosity.

Question 4.
Sonalika and Kalyan Sona are varieties of
(a) wheat
(b) rice
(c) millet
(d) tobacco.
Answer:
(a) Sonalika and Kalyan Sona are high yielding and disease resistant varieties of wheat.

Question 5.
Which one of the following is not a fungal disease?
(a) Rust of wheat
(b) Smutofbajra
(c) Black rot of crucifers
(d) Red rot of sugarcane
Answer:
(c) : The pathogen responsible for black rot of crucifer is a bacterium Xanthomonas campestris. It lives on the plant residue in the soil or in the seed. The bacterium enters leaves through the hydathodes. Bacteria may also enter leaves through wounds including those made by insects. First sign of the disease appears near the leaf margin, characterised by chlorosis, which progresses towards the center of leaf blade. In the affected portions, the vein and veinlets turn brown and finally black.

Question 6.
In virus-infected plants the meristematic tissues in both apical and axillary buds are free of virus because
(a) the dividing cells are virus resistant
(b) meristems have anti viral compounds
(c) the cell division of meristems are faster than the rate of viral multiplication
(d) viruses cannot multiply within meristem cell(s).
Answer:
(c)

Question 7.
Several South Indian states raise 2-3 crops of rice annually. The agronomic feature that makes this possible is because of
(a) shorter rice plant
(b) better irrigation facilities
(c) early yielding rice variety
(d) disease resistant rice variety.
Answer:
(c)

Question 8.
Which one of the following combination would a sugarcane farmer look for in the sugarcane crop?
(a) Thick stem, long internodes, high sugar content and disease resistant.
(b) Thick stem, high sugar content and profuse flowering.
(c) Thick stem, short internodes, high sugar content, disease resistant.
(d) Thick stem, low sugar content, disease resistant.
Answer:
(a)

Question 9.
Fungicides and antibiotics are chemicals that
(a) enhance yield and disease resistance
(b) kill pathogenic fungi and bacteria, respectively
(c) kill all pathogenic microbes
(d) kill pathogenic bacteria and fungi respectively.
Answer:
(b)

Question 10.
Use of certain chemicals and radiation to change the base sequences of genes of crop plants is termed
(a) recombinant DNA technology
(b) transgenic mechanism
(c) mutation breeding
(d) gene therapy.
Answer:
(c) : Mutation is the process by which genetic variations are created through changes in the base sequence within genes resulting in the creation of a new character or trait not found in the parental type. It is possible to induce mutations artificially
through use of chemicals or radiations (like gamma radiations), and selecting and using the plants that have desirable characters as a source in breeding. This process is called mutation breeding.

Question 11.
The scientific process by which crop plants are enriched with certain desirable nutrients is called
(a) crop protection
(b) breeding
(c) biofortification
(d) bioremediation.
Answer:
(c) : Breeding of crops with higher levels of vitamins and minerals or higher protein and healthier fats is called biofortification. This is the most practical aspect to improve nutritional value of food.

Question 12.
The term ‘totipotency’ refers to the capacity of a
(a) cell to generate whole plant
(b) bud to generate whole plant
(c) seed to germinate
(d) cell to enlarge in size.
Answer:
(a) : The capacity to generate a whole plant from any cell/explant is called cellular totipotency. Infact, a whole plant can be regenerated from any plant part (referred to as explant) or cell.

Question 13.
Given below are a few statements regarding somatic hybridisation. Choose the correct statements.
(i) Protoplasts of different cells of the same plant are fused.
(ii) Protoplasts from cells of different species can be fused.
(iii) Treatment of cells with cellulase and pectinase is mandatory.
(iv) The hybrid protoplast contains characters of only one parental protoplast.
(a) (i) and (iii)
(b) (i)and(ii)
(c) (i)and(iv)
(d) (ii) and (iii)
Answer:
(d) :When a hybrid is produced by fusion of somatic cells of two varieties or species, it is known as somatic hybrid. The process of producing somatic hybrids is called somatic hybridisation. First, the cell wall of the plant cells are removed by digestion with a combination of pectinase and cellulase. The plant cells without cell wall are called protoplasts. The fusion of protoplasts not only involves the fusion of their cytoplasm. but also their nuclei. The fused protoplasts are allowed to grow on culture medium. Soon they develop their own walls and are called somatic hybrid cells. The hybrid cells give rise to callus. Callus later differentiates into new plant with characters of two plants in one.

Question 14.
An explant is
(a) dead plant
(b) part of the plant
(c) part of the plant used in tissue culture
(d) part of the plant that expresses a specific gene.
Answer:
(c) : An explant is the plant part taken out to be grown in test-tube in special nutrient medium. It has the ability to produce whole new plant.

Question 15.
The biggest constraint of plant breeding is
(a) availability of desirable gene in the crop and its wild relatives
(b) infrastructure
(c) trained manpower
(d) transfer of genes from unrelated sources.
Answer:
(a ) : Conventional plant breeding results in hybrid varieties which had a tremendous impact on agricultural productivity over the last decades. Conventional plant breeding also has its limitations. First, breeding can only be done between two plants that are sexually compatible with each other. This limits the new traits that can be added to those that already exist in a particular species. Second, when plants are crossed, many traits that have undesirable effects on yield potential are transferred along with the trait of interest.

Question 16.
Lysine and tryptophan are
(a) proteins
(b) non-essential amino acids
(c) essential amino acids
(d) aromatic amino acids.
Answer:
(c)

Question 17.
Micropropagation is
(a) propagation of microbes in vitro
(b) propagation of plants in vitro
(c) propagation of cells in vitro
(d) growing plants on smaller scale.
Answer:
(b) : Micropropagation is the in vitro propagation of plants by rapidly multiplying stock plant material to produce a large number of progeny plants (clone), using modern plant tissue culture methods.

Question 18.
Protoplast is
(a) another name for protoplasm
(b) an animal cell
(c) a plant cell without a cell wall
(d) a plant cell.
Answer:
(c) : Refer Answer 13.

Question 19.
To isolate protoplast, one needs
(a) pectinase
(b) cellulase
(c) both pectinase and cellulase
(d) chitinase.
Answer:
(c) : Refer Answer 13.

Question 20.
Which one of the following is a marine fish?
(a) Rohu
(b) Hilsa
(c) Catla
(d) Common carp.
Answer:
(b) : Rohu, Catla and common carp are freshwater fish.

Question 21.
Which one of the following products of apiculture is used in cosmetics and polishes?
(a) Honey
(b) Oil
(c) Wax
(d) Royal jelly
Answer:
(c) : Abdominal glands of worker bees produce a secretion from which bee wax is made. It is used in the manufacture of many items like cosmetics, shaving cream, face cream, plaster, pencils, electric goods, toothpaste, lotions, furniture-polishes, boot- polishes, protective coating, ink paints and candles.

Question 22.
More than 70 per cent of livestock population is in
(a) Denmark
(b) India
(c) China
(d) India and China
Answer:
(d)

Question 23.
The agriculture sector of India employs about
(a) 50 per cent of the population
(b) 70 per cent of the population
(c) 30 per cent of the population
(d) 60 per cent of the population.
Answer:
(d) : India is an agricultural country. Agriculture accounts for approximately 33% of India’s GDP and employs nearly 62% of the population.

Question 24.
33 percent of India’s (Gross Domestic Product) comes from
(a) industry
(b) agriculture
(c) export
(d) small-scale cottage industries.
Answer:
(b) : Refer answer 23.

Question 25.
A collection of all the alleles of all the genes of a crop plant is called
(a) germplasm collection
(b) protoplasm collection
(c) herbarium
(d) somaclonal collection.
Answer:
(a) : The entire collection of plants/ seeds having all the diverse alleles for all the genes in a given crop is called germplasm collection.

 Very Short Answer Type Questions

Question 1.
Millions of chicken were killed in West Bengal, Assam, Orissa and Maharashtra recently. What was the reason?
Answer:
Reason for killing of millions of chickens in West Bengal, Assam, Orissa and Maharashtra was spreading of bird flu caused by H5N1 strain of bird flu virus. This bird flu was spreading like an epidemic. Evidences showed that disease could affect many animals like cats, dogs, pigs, white tigers and also human beings.

Question 2.
Can gamma rays used for crop improvement programmes prove to be harmful for health? Discuss.
Answer:
Gamma rays are physical mutagens used to induce changes in DNA and chromosomes. ^ These are frequently employed in mutation breeding. Plants obtained by mutation breeding are high yielding and disease resistant. Useful plants with desirable mutations are selected for further breeding programmes. It has been found that such plants do not produce any harmful effect on health.

Question 3.
In animal husbandry, if two closely related animals are mated for a few generations, it results in loss of fertility and vigour. Why is this so?
Answer:
Inbreeding is the crossing of closely related animals of the same breed for 4-6 generations. Continuous inbreeding leads to
homozygosity. The recessive alleles tend to get together and express harmful effects in the progeny. Resulting in reduced survival and fertility of offspring known as inbreeding depression.

Question 4.
In the area of plant breeding, it is important not only to preserve the seeds of the variety being cultivated, but also to preserve all its wild relatives. Explain with a suitable example.
Answer:
Collection and preservation of all the different wild varieties, species and relatives of the cultivated species in plant breeding is essential for effective exploitation of natural genes a vai lable in the population. For example, in sugarcane (Saccharum officinarum) the character for disease resistance has been incorporated by breeding it with wild variety (Saccharum spontaneum).

Question 5.
Name a man-made cereal. Trace how it was developed and where is it used?
Answer:
Triticalc (6n) is a man-made cereal. It is a hexaploid variety developed by crossing wheat and rye followed by doubling of chromosomes. Triticale is the first man made crop.

Question 6.
Fill in the blanks.

Answer:

Question 7.
A few statements are given below followed by a set of terms in a box. Pick the correct term and write it against the appropriate statement.
(a) Mating of closely related individuals within the same breed.
(b) Mating of animals of same breed but having no common ancestors on either side for 4-6 generations.
(c) Mating of animals of two different species.
(d) Breeding of animals belonging to different breeds.
(i) Cross breeding,
(ii) Inter-specific hybridisation,
(iii) Outbreeding,
(iv) Outcrossing,
(v) Inbreeding.
Answer:
(a) – (v), (b) – (iv), (c) – (ii), (d) – (i)

Question 8.
What is meant by ‘hidden hunger’?
Answer:
Hidden hunger is consumption of food deficient in nutrients, particularly, in micronutrients, proteins and vitamins.

Question 9.
Why are plants obtained by protoplast culture called somatic hybrids?
Answer:
In protoplast culture, plants are obtained by fusion of somatic cells of two varieties, therefore, they are called somatic hybrids.

Question 10.
What is protoplast fusion?
Answer:
The plant cells without cell wall are called protoplasts. Protoplast fusion is the fusion of protoplasts of two plant cells by means of electro fusion or chemicals.

Question 11.
Why is it easier to culture meristems compared to permanent tissues?
Answer:
It is easier to culture meristems compared to permanent tissues because meristem is group of actively dividing and undifferentiated cells while cells of permanent tissues lose their ability to divide and have to dedifferentiate to resume their ability to divide.

Question 12.
Why are proteins synthesised from Spirulina called single cell proteins?
Answer:
Proteins synthesised from Spirulina are called single cell proteins (SCPs) because they are obtained from unicellular microorganism. The term is a misnomer as in SCP production, the biomass is produced from both unicellular and multicellular microorganisms, viz. bacteria, algae, fungi, etc.

Question 13.
A person who is allergic to pulses was advised to take a capsule of Spirulina daily? Give the reasons for the advice.
Answer:
Both Spirulina and pulses are rich sources of proteins. But if an individual is allergic to pulses, then he can take Spirulina capsule to fulfill the daily requirement of protein intake.

Question 14.
What is aquaculture? Give example of an animal that can be multiplied by aquaculture.
Answer:
Aquaculture is rearing and management of useful aquatic organisms such as aquatic plants and animals like fish, prawns, oysters etc., in both freshwater and saltwater under controlled conditions.

Question 15.
What are the duties of a veterinary doctor in management of a poultry farm?
Answer:
Duties of a veterinary doctor in management of poultry farm includes :

1. Ensuring proper and safe conditions for establishment of poultry farm.
2. Regular inspection of poultry birds to ‘protect them from diseases.
3. Vaccination of poultry birds.
4. Early detection of diseases followed by proper medication

Question 16.
Would it be wrong to call plants obtained through micropropagation as ‘clones’? Comment.
Answer:
No, it is not wrong to call plants obtained by micropropagation as clones because these plants will be genetically similar to each other and to the original plant.

Question 17.
How is a somatic hybrid different from a hybrid?
Answer:
A normal sexual hybrid is produced by fusion of ordinary gametes. The major limitation in creating a sexual hybrid is the sexual compatibility of two species. However, success in some cases can be achieved by hybrid-specific modifications used to achieve fusion of gametes, and to rescue, isolate and culture the developing embryo in vitro. Sexual hybrids are diploid and cytoplasmic contribution of fusing cells is unequal. Somatic hybrids result from the fusion of somatic cells instead of ordinary gametes. This technique requires specific handling of protoplasts and is totally dependent on tissue culture technique. Sexual compatibility is not at all, a barrier to this technique and it can be performed between sexually incompatible species. Somatic hybrids are generally tetraploid and cytoplasmic contribution of two fusing cells is equal to the hybrid.

Question 18.
What is emasculation? Why and when it is done?
Answer:
Emasculation is removal of anthers from a bisexual flower before its maturity. It is done during artificial hybridisation programme to avoid self-pollination.

Question 19.
Discuss the two main limitations of plant hybridisation programme.
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The limitations of plant hybridisation programme are as follows:

1. Availability of a limited number of identified disease resistant genes that are present in various crop varieties or wild relatives.
2. Compatibility of parents.

Question 20.
Interspecific crosses are rare in nature and intergeneric crosses almost unknown. Why?
Answer:
In interspecific crosses, male and female animals of two different related species are mated. In some cases, the progeny may combine desirable features of both the parents, and may have considerable economic value, e.g., the mule. Intergeneric hybridisation is the crossing of two different animals/plants of different genera. Both interspecific and intergeneric crosses are rare in nature. Sexual incompatibility, reproductive isolation and sterility are the main barriers in crossing between non-related species. For example, plants belonging to different species do not flower at the same time and often do not use the same pollinators. Intergeneric hybrids more or less never form complete zygotes/embryos. These are sterile/ infertile.

Question 21.
Differentiate between pisciculture and aquaculture.
Answer:
Pisciculture is rearing and management of fish in water bodies. Aquaculture involves rearing and management of useful aquatic plants and animals, such as fish, prawns, crabs, molluscs etc.

Question 22.
Give two important contributions of Dr. M.S. Swaminathan.
Answer:
Dr. M. S. Swaminathan is called father of “Green revolution” in India. His contributions are as follows:

1. of high-yielding varieties of wheat.
2. Development of semi-dwarf, high- yielding rice varieties.
   These helped in increasing the crop production and meeting requirements of increasing population.

Question 23.
The term ‘desirable trait’ can mean different thingsfor different plants. Justify the statement with suitable examples.
Answer:
Different plants have different desirable traits. This can be explained with example of rice. Certain rice areas in tropical Asia and Africa are subject to prolonged and deep flooding. In such areas, farmers grow floating rice with internodes that elongate when the wafer rises. These are not very good yielding varieties. Yet they are ‘desirable’ because internode elongation prevents them from getting submerged and dying. But in Latin America, where fields flood no deeper than 40 to 100 cm, that too for shorter periods, farmers prefer hybrids obtained by crossing deep water varieties with improved dwarfs which are superior, as they contain high yield combined with dwarfism.

Short Answer Type Questions

Question 1.
You are planning to set up a Dairy Farm. Describe the various aspects you would consider before you start the venture.
Answer:
Various aspects to be considered while setting up a dairy farm include:

1. 1. Selection of good breeds containing high yielding potential (under the climatic conditions of the area) and resistance to the diseases.
   2. Ensurance of adequate ventilation, suitable temperature, sufficient light, water and proper drainage channel in cattle shed.

1. Emphasis on feeding cattle in scientific manner i.e., monitoring quantity and quality of fodder.
2. Employing people with good knowledge of cattle hygiene, handling of cattle during milking and storage and transport of milk and its products.
3. Proper arrangement of regular inspection by a veterinary doctor.

Question 2.
It is said, that diseases are spreading faster due to globalisation and increased movement of people. Justify the statement taking the example of H5N1 virus.
Answer:
bird flu virus affects the poultry birds and can turn into an epidemic. This disease can be transmitted to other animals like cats, dogs, white tigers or even human beings. It can also be transferred from one human being to another. Due to globalisation and increased movementof people, the disease spreads rapidly from one country to another. The spreading of disease also increases when affected products are imported from one country to another.

Question 3.
Explain the concept of the ‘Blue Revolution’.
Answer:
The term ‘Blue Revolution’ refers to the remarkable emergence of aquaculture as an important and highly productive activity. Aquaculture refers to all forms of active culturing of aquatic animals and plants occurring in fresh water, marine or brackish forms.
In India, it was started in 1970. ‘Blue Revolution’ has brought improvement in aquaculture by adopting new techniques of breeding, rearing, marketing and export of fish. It lead to tremendous increase in shrimp production.

Question 4.
A farmer was facing the problem of low yield from his farm. He was advised to keep a beehive in the vicinity. Why? How would the beehive help in enhancing yield?
Answer:
Bees are pollinators of many crop species such as sunflower, Brassica, apple and pear. Honey bees feed on pollen and nectar of flowers. Keeping beehive in crop fields during flowering period increases pollination efficiency and improves the yield (both crop and honey).

Question 5.
Lifestyle diseases are increasing alarmingly in India. We are also dealing with large scale malnutrition in the population, is there any method by which we can address both of these problems together?
Answer:
Increase in lifestyle diseases and malnutrition, both can be checked by biofortification. It is a technique to produce crops with high nutritional value with higher levels of vitamins, minerals, proteins and healthier fats.
A few examples of crop with improved food quality are :

1. Maize hybrids have twice the amount of essential amino acids like lysine and tryptophan.
2. Wheat variety with high protein content.
3. Iron rich wheat variety.
4. Carrots, spinach, pumpkin enriched with vitamin A.
5. Bitter gourd, mustard, tomato enriched with vitamin C.

Question 6.
How can we improve the success rate of fertilisation during artificial insemination in animal husbandry programmes?
Answer:
Multiple Ovulation Embryo Transfer Technology (MOET) is used to improve the
success rate of fertilisation during artificial insemination. In this method, hormones (with FSH-like activity) are given to the cow for inducing follicular maturation and super ovulation i.e., instead of one egg, which they usually give per cycle, they produce 6-8 eggs. The cow is either mated with a best bull or artificially inseminated. The embryos at 8-32 cell stages are recovered and transferred to surrogate mothers. The genetic mother is available for another super ovulation. MOET has been done in cattle, sheep, rabbits, buffaloes, mares etc. High milk giving breeds of females and high quality (lean meat with less lipid) meat giving bulls have been bred successfully to obtain better breed in a short time.

Question 7.
What is meant by germplasm collection? What are its benefits?
Answer:
Germplasm collection refers to collection of variability. It involves collection and preservation of all different wild varieties, species and relatives of the cultivated species. It is sum total of all the alleles of the genes present in a crop. Germplasm collection is essential for successful breeding programme as it offers, to the breeders, the entire of genes and alleles and their characteristics which a plant expresses. The breeder is exposed to the best options and he can select the most favourable traits of a particular gene.

Question 8.
Name the improved characteristics of wheat that helped India to achieve green revolution.
Answer:
The term ‘Green Revolution’ refers to increase in crop yield due to development of high yielding varieties of crop plants like wheat and rice through plant breeding technique. Sonalika and Kalyan Sona are improved wheat varieties introduced in India for ‘Green Revolution’.
They show following characteristics:

1. High yield
2. Disease resistance
3. Semi-dwarf habit

Question 9.
Suggest some of the features of plants that will prevent insect and pest infestation.
Answer:
Insect resistance in plants may be produced by morphological, biochemical or physiological features.
These are discussed as follows:

1. Presence of hair growth on aerial plant parts, e.g., hairy leaves in cotton plant protect it from jassids.
2.  Rendering the flowers nectarless, e.g., nectarless cotton varieties provide protection from cotton bollworms.
3. Low nitrogen and sugar content in stems prevent maize plants from stem borers.
4. Solid stem in wheat is not preferred by stem saw fly.

Question 10.
It is easier to culture plant cells in vitro as compared to animal cells. Why?
Answer:
Plant cells are easier to culture in vitro because they have the property of cellular totipotency, i.e., capacity to generate whole plant from any cell. But in case of animals, whole new organism cannot be produced from single cell.

Question 11.
The culture medium (nutrient medium) can be referred to as a highly enriched laboratory soil’. Justify the statement.
Answer:
The plant tissue culture medium is referred to as highly enriched laboratory soil as it consists of both micro- and macro¬elements, vitamins, carbon source, amino acids and growth regulators required for the growth of explants.

Question 12.
Is there any relationship between dedifferen-tiation and the higher degree of success achieved in plant tissue culture experiments?
Answer:
In plants, living, differentiated cells, those who have lost the capacity to divide can regain the capacity of division under certain conditions. This is called dedifferentiation. Dedifferentiated cells undergo callus formation and callus gets differentiated to form plantlets.

Question 13.
“Give me a living cell of any plant and I will give you a thousand plants of the same type” Is this only a slogan or is it scientifically possible? Write your comments and justify them.
Answer:
Living cells of plants have property of totipotency, i.e., ability to form whole new plant jfrom any cell. Large number of plants can be produced from single cell by technique called micropropagation. In this method of tissues culture, each plant produced is genetically similar to parent plant.

Question 14.
What is the difference between a breed and a species. Give an example for each category.
Answer:
A ‘breed’ is a specific group of animals which share many similarities such as homogenous appearance, behaviour etc., that distinguish it from other animals or plants of the same species. Sahiwal is an indigenous breed, Jersey, Brown Swiss are exotic breeds of cattle (Bos indica).
A ‘species’ is the largest group of individuals capable of freely interbreeding among themselves and producing fertile offspring, e.g., all cows of the whole world have only one scientific name Bos indica.

Question 15.
Plants raised through tissue cultures are clones of the ‘parent’ plant. Discuss the utility of these plants.
Answer:
The plants obtained through tissue culture are genetically identical to parent plants. This is of great use when desirable characters of the parent plant have to be maintained. Hence, plant raised through tissue culture technique are of great use.

Question 16.
Discuss the importance of testing of new plant varieties in a geographically vast country like India.
Answer:
India is a vast country with varying climatic conditions. Different hybrid varieties after growing in research fields are evaluated by growing in farmer’s field, at least for three growing seasons at several locations in the country representing all agroclimatic zones. The hybrid is evaluated for various characters like disease resistance, tolerance, yield in comparison to the best available local crop cultivar. Testing of new plant varieties before release is of great importance as it ensures that the plant varieties can grow in vivid environmental conditions present in different geographical regions of our country.

Question 17.
Define the term ‘stress’ for plants. Discuss briefly the two types of stress encountered by plants.
Answer:
External factors which reduce the growth and yield of plants are known as stress. These external factors may be abiotic or biotic. Two common types of stress encountered by plants are:

1. Water – Plants need sufficient water to grow. Less availability of water or drought is unfavourable and negatively affects growth and yield of the plant.
2. Temperature – Plants require optimum temperature for their proper growth. Very high or very low temperature negatively affects the growth rate of plants.

Question 18.
Discuss natural selection and artificial selection. What are the implications of the latter on the process of evolution?
Answer:
Selection is the process by which certain individuals with desired characters and considered to be fit are favoured. Natural selection involves selection of adapted individuals by nature. It is a continuous process and has resulted in evolution. Evolution that takes place in natural selection is a slow’ process.
In artificial selection, parent organism with desirable characters, such as high yield, disease resistance, increased tolerance etc., is selected by humans and is hybridised with superior variety to produce hybrid.
Artificial selection provides a model that helps us understand natural selection. It is a small step to envision natural conditions acting selectively on populations and causing natural changes. Artificial selection allows rapid changes in a species. This is because man has one power which nature does not have – man ensures breeding that preserves only beneficial traits of species. Hence, evolution through artificial selection is much faster as compared to evolution through natural selection.

Question 19.
Discuss briefly how purelines are created in animal husbandry.
Answer:
When breeding is between animals of the same breed, it is called inbreeding. In animal husbandry, purelines are produced by inbreeding of animals for 4-6 generations.

Question 20.
What are the physical barriers of a cell in the protoplast fusion experiment? How are the barriers overcome?
Answer:
Physical barriers in protoplast fusion involve presence of protective cell wall in plant cells and then fusion of protoplasm. Cell wall is digested by cellulase and pectinase and then protoplasts of two cells are fused in a solution of polyethylene glycol (PEG) or by high voltage pulse, known as electrofusion.

Question 21.
Give few examples of biofortified crops. What benefits do they offer to the society?
Answer:
SPinn Examples of biofortified crops are :

1. Maize variety rich in lysine and arginine.
2. High protein content wheat varieties.
3. Iron rich wheat variety.
4. Vitamin A enriched carrots, pumpkin, spinach.
5. Calcium and iron enriched spinach and bathua.
   Biofortified crops help to improve the health of people and meet their daily requirements and thus alleviate malnutrition.

Long Answer Type Questions

Question 1.
You are a botanist working in the area of plant breeding. Describe the various steps that you will undertake to release a new variety.
Answer:
The steps required for developing new varieties are as follows:
(1) Collection of germplasm : Collection and preservation of all the different wild varieties, species and relatives of the cultivated species is a pre-requisite for effective exploitation of natural genes available in the population. Germplasm is the sum total of all the alleles of the genes present in a crop and its related species.

(2) Evaluation and selection of parents : The germplasm is evaluated to identify plants with desirable combination of characaters. Seeds of plants having desirable characters are selected for multiplication.

(3) Cross-hybridisation among selected parents : Hybridisation is the most common method of creating genetic variation. Hybridisation is crossing of two or more types of plants for bringing their traits together in the progeny. It brings about useful genetic/heritable variations of two or more lines together.

(4) Selection and testing of superior recom-binants : This step comprises selecting among the progeny of the hybrids, i.e., those plants that have the desired character combination. The selection process yields plants that are superior to both of the parents. These plants are self-pollinated for several generations till they come to a state of uniformity (homozygosity).

(5) Testing, release and commercialisation of new cultivars : The newly selected lines are evaluated in comparison to best available local crop cultivar for their yield and other agronomic traits of quality, disease resistance etc. and then are released for commercialisation.

Question 2.
(a) The shift from grain to meat diets creates more demands for cereals. Why?
(b) A 250 kg cow produces 200 g of protein per day but 250 g of Methylophillus methylotrophus can produce 25 tonnes of protein. Name this emerging area of research. Explain its benefits.
Answer:
(a) The shift from grain to meat diets creates more demands on cereals as it takes 3-10 kg of grain to produce 1 kg of meat by animal farming.

(b) The emerging research area discussed in the question refers to Commercial production of proteins from micro¬organisms, called single cell protein. Advantages of SCP are as follows:

1. It provides a protein rich supplement in human diet.
2. It reduces the pressure on agricultural production systems for the supply of the required proteins.
3. SCP production is based on industrial effluents so it helps to minimise environmental pollution.

Question 3.
What are the advantages of tissue culture methods over conventional method of plant breeding in crop improvement programmes?
Answer:
Advantages of tissue culture method over conventional methods of breeding are :

1. It is fast method of growing new plants.
2. New plants can be produced from explants or even from single cell.
3. New plants produced are genetically similar to the parents.
4. Disease free plants can be produced.
5. Purelines can be created easily.

Question 4.
Modem methods of breeding animals and plants can alleviate the global food shortage’. Comment on the statement and give suitable examples.
Answer:
The population of the world is increasing day by day. To meet the food requirements of the ever increasing population, is the major concern of plant and animal breeders. Modern biological principles are applied to animal husbandry and plant breeding to enhance food production.

Modem methods in animal breeding include Multiple Ovulation Embryo Transfer Technology (MOET). In this method, hormones (with FSH-like activity) are given to the cow for inducing follicular maturation and super ovulation i.e., instead of one egg, which they usually give per cycle, they produce 6-8 eggs. The cow is either mated with a best bull or artificially inseminated. The embryos at 8-32 cell stage are recovered and transferred to surrogate mothers. The genetic mother is available for another super ovulation. MOET has been done in cattle, sheep, rabbits, buffaloes, mares, etc. High milk yielding breeds of females and high quality (lean meat with less lipid) meat-giving bulls have been bred successfully to obtain better breed in a short time.
Modern methods in plant breeding include the following:

1. Tissue culture which is an in vitro technique of regeneration of a whole plant from any part of a plant (explant) by growing it on culture medium under aseptic conditions.
2. Biofortification which is the method for developing crops with higher levels of vitamins, minerals, proteins and healthier fats to improve public health.
3. Producing single cell protein as an alternative protein source for animal and human nutrition from certain beneficial microorganisms like Spirulim.

Question 5.
Does apiculture offer multiple advantages to farmers? List its advantages if it is located near a place of commercial flower cultivation.
Answer:
Advantages of apiculture to farmers are:

1. It provides honey which has nutritional as well as medicinal advantage.
2. It provides bee wax used in cosmetics, paints, polish etc.
3. It provides bee venom used to cure gout and arthritis.
4. It provides royal jelly used as a tonic by heart patients and growing children.
5. Bees are pollinating agents. If they are kept in field during flowering season they feed on nectar and pollen to produce honey, and bring about effective pollination of crop plants thereby increasing crop yield.

Question 6.
(a) Mutations are beneficial for plant breeding.Taking an example, justify the statement,
(b) Discuss briefly the technology that made us self-sufficient in food production.
Answer:
(a) Mutation is the process by which genetic variations are created through changes in the base sequence within genes resulting in the creation of a new character or trait. It is possible to induce mutations artificially in plants through use of chemicals or radiations and then selecting and using those plants that have the desirable characters or traits as a source in breeding. This process is called mutation breeding.
Some important achievements of mutation breeding are:

1. In mung bean, resistance to yellow mosaic virus and powdery mildew were introduced by mutations.
2. Resistance to yellow mosaic virus in bhindi was transferred from a wild variety and resulted in new variety called Parbhani Kranti.
3. High yielding Mexican wheat varieties were originally red grained. The colour was not liked by the Indians. Their cultivation was adopted in India on large scale only when exposure to gamma radiations converted them to amber grained through mutations, e.g., Sharbati Sonora (from Sonora 64), Pusa Lerma (from Lerma Rojo 64A).

(b) After independence, efforts to make country self sufficient for food were not successful for two decades. It was a major task for agricultural scientists in India to produce sufficient food for increasing population with little land available. The positive results appeared only in the form of ‘Green Revolution’. Basic aims of Green Revolution were:

1. Expansion of existing farm lands.
2. Doubling cropping in existing farm lands.
3. Use of genetically improved varieties. Plant breeding as a technology helped increase food production to a great extent. By using this technology, high yielding varieties of wheat, rice, millet and corn were developed. Tremendous growth was observed in industries and it created many opportunities for new jobs. As a result, India was transformed from a starving nation to the exporter of food.

Question 7.
Discuss how the property of plant cell totipotency has been utilised for plant propagation and improvement.
Answer:
Totipotency is the ability of single cell to grow into complete new plant. Plant * cell totipotency can be utilised for plant propagation and improvement. Any part of a plant (explant) is taken out and is grown in test tube under sterile conditions in special nutrient medium. Plant hormones present in the culture medium play an important role in growth and differentation of cultured cells and tissues.”

By meristem culture, virus-free plants can be generated from an infected plant. Anther cells can be cultured in vitro to produce haploid plants. These are useful for immediate expression of mutations and quick formation of purelines. Young embryos can be cultured in vitro. Embryo culture is useful in case of orchids as orchid seeds do not have any form of stored food. Moreover in certain species, inhibitors present in the endosperm or seed coat make the seed dormant. Such embryos can escape dormancy by culturing on suitable medium.

Question 8.
What are three options to increase food production? Discuss each giving the salient features, merits and demerits.
Answer:
Traditional farming can only yield a limited biomass as food for humans and animals. Better management practices and increase in agricultural land can increase yield, but only to a limited extent.
Following are the three options to increase food production:
(a) Single cell protein : The biomass obtained from microorganisms can be treated or processed in industry to be used as food and is called single cell protein.
Merits of single cell proteins:

1. Its production reduces pollution as it uses wastes and industrial effluents.
2. It provides a protein-rich diet.
3. It fulfills the demand of protein for human diet and takes off the pressure on agriculture system.

Demerits of single cell proteins:

1. Sometimes the microbial biomass when taken as diet supplement may lead to allergic reactions.
2. Many types of microorganisms produce substances which are toxic to the humans and also to the animals.

(b) Biofortification : It is the method for developing crops with higher levels of vitamins, minerals proteins and healthier fats to improve public health.
Merits of biofortification :

1. It can potentially improve nutritional value of food and other health benefits.
2. It can help combat various lifestyle diseases and malnutrition.

Demerits of biofortification :

1. Its successful implementation into society requires safe delivery systems, stable policies, appropriate social infrastructures.
2. It requires continuous financial support from government.

(c) Tissue-culture :
It is an in vitro technique for regeneration of a whole plant from any part of a plant by growing it on culture medium under aseptic conditions.
Merits of tissue culture :

1. A large number of plants can be grown in short time.
2. Disease-free plants can be multiplied.
3. Seedless plants can be multiplied.
4. Characters of the plants where sexual reproduction is absent can be combined using somatic hybridisation.

Demerits of tissue culture :

1. It requires great expense because it needs latest techniques in the laboratory.
2. It requires acclimatisation of plants grown by tissue culture to the external environment.
3. It requires special expertise.

 We hope the NCERT Exemplar Solutions for Class 12 Biology chapter 9 Strategies for Enhancement in Food Production help you. If you have any query regarding .NCERT Exemplar Solutions for Class 12 Biology chapter 9 Strategies for Enhancement in Food Production, drop a comment below and we will get back to you at the earliest.

 

NCERT Exemplar Solutions for Class 12 Biology chapter 5 Principles of Inheritance and Variation

These Solutions are part of NCERT Exemplar Solutions for Class 12 Biology. Here we have given NCERT Exemplar Solutions for Class 12 Biology chapter 5 Principles of Inheritance and Variation

Multiple Choice Questions

Question 1.
All genes located on the same chromosome
(a) form different groups depending upon their relative distance
(b) form one linkage group
(c) will not from any linkage groups
(d) form interactive groups that affect the phenotype.
Answer:
(b) : Linked genes are the genes which occur on the same chromosome. ,A linkage group is a physical association of linked genes which are normally inherited together except for crossing over.

Question 2.
Conditions of a karyotype 2n ± 1 and 2n ± 2 are called
(a) aneuploidy
(b) polyploidy
(c) allopolyploidy
(d) monosomy.
Answer:
(a) : The chromosomal disorders are caused due to absence or excess or abnormal arrangement of one or more chromosomes. Failure of segregation of chromatids during cell division cycle results in the gain or loss of chromosome(s), is called aneuploidy. For example, Down’s syndrome results due to gain of extra copy of chromosome 21. Similarly, Turner’s syndrome results due to loss of an X chromosome in human females.

Question 3.
Distance between the genes and percentage of recombination shows
(a) a direct relationship
(b) an inverse relationship
(c) a parallel relationship
(d) no relationship.
Answer:
(a) : Linkage is the phenomenon of physical association of genes on a chromosome and recombination is the generation of non- parental gene combinations. Strength of the linkage between two genes is inversely proportional to the distance between the two, i.e., two linked genes show higher frequency of crossing over (recombination) if the distance between them is higher and lower frequency if the distance is small. But distance between gene has a direct relationship with percentage of recombination because with increase in distance between genes, percentage of recombination increases.

Question 4.
If a genetic disease is transferred from a phenotypically normal but carrier female to only some of the male progeny, the disease is
(a) autosomal dominant
(b) autosomal recessive
(c) sex-linked dominant
(d) sex-linked recessive.
Answer:
(d) : The sex linked recessive disease shows its transmission from unaffected carrier female to some of the male progeny e.g., haemophilia. The heterozyous female (carrier) of haemophilia may transmit the disease to sons. The possibility of a female becoming a haemophilic is extremely rare because mother of such a female has to be at least carrier and the father should be haemophilic.

Question 5.
In sickle cell anaemia glutamic acid is replaced by valine. Which one of the following triplets codes for valine?
(a) GGG
(b) A AG
(c) GAA
(d) GUG
Answer:
(d) :The sickle-cell anaemia is caused by the substitution of glutamic acid (Glu) by valine (Val) at the sixth position of the beta globin chain of the haemoglobin molecule. The substitution of amino acid in the globin protein results due to the single base substitution at the sixth codon of the beta globin gene from GAG to GUG. The mutant haemoglobin molecule undergoes polymerisation under low oxygen tension causing the change in the shape of the RBC from biconcave disc to elongated sickle like structure.

Question 6.
Person having genotype /^A /^B would show the blood group as AB.This is because of
(a) pleiotropy
(b) codominance
(c) segregation
(d) incomplete dominance.
Answer:
(b) : ABO blood groups are controlled by the gene I. The plasma membrane of the red blood cells has sugar polymers that protrude from its surface and the kind of sugar is controlled by the gene. The gene (I) has three alleles I^A, I^B and i. The alleles I^A and /^B produce a slightly different form of the sugar while allele i does not produce any sugar. Because humans are diploid organisms, each person possesses any
two of the three I gene alleles. I^A and /^B are completely dominant over /, in other words when I^A and i are present then only I^A expresses (because i does not produce any sugar), and when /^B and i are present I^B expresses. But when I^A and /^B are present together they both express their own types of sugars, this is because of codominance. Hence, in this case red blood cells have both A and B types of sugars.

Question 7.
ZZ/ZW type of sex determination is seen in
(a) platypus
(b) snails
(c) cockroach
(d) peacock.
Answer:
(d) : In birds and some reptiles both the sexes possess two sex chromosomes but unlike human beings the females contain heteromorphic sex chromosomes (AA + ZW) while the males have homomorphic sex chromosomes (AA + ZZ). Because of having heteromorphic sex chromosomes, the females are heterogametic (female heterogamety) and produce two types of eggs, (A + Z) and (A + W). The male gametes or sperms are of the one type (A + Z). 1:1 sex ratio is produced in the offspring.\

Question 8.
A cross between two tall plants resulted in offspring having few dwarf plants. What would be the genotypes of both the parents?
(a) TTandTt
(b) TtandTt
(c) TT andTT
(d) Tt and tt
Answer:
(b) : The genotypes of both the parents are :

Question 9.
In a dihybrid cross, if you get 9:3 :3 :1 ratio it denotes that
(a) the alleles of two genes are interacting with each other
(b) it is a multigenic inheritance
(c) it is a case of multiple allelism
(d) the alleles of two genes are segregating independently.
Answer:
(d) : Cross involving two contrasting characters is called dihybrid cross or a two factor cross. The two factors of each trait assort at random and independent of the factors of other traits at the time of meiosis (gametogenesis) and get randomly as well as independently rearranged in the offspring producing both parental and new combinations of traits. This explained the law of independent assortment given by Mendel.

Question 10.
Which of the following will not result in variations among siblings?
(a) Independent assortment of genes
(b) Crossing over
(c) Linkage
(d) Mutation
Answer:
(c) : Linkage will not result in variations among siblings because linked genes occur on the same chromosome and are transmitted together. In linkage there is a tendency to maintain the parental gene combinations except for occasional crossovers.

Question 11.
Mendel’s law of independent assortment holds good for genes situated on the
(a) non-homologous chromosomes
(b) homologous chromosomes
(c) extra nuclear genetic element
(d) same chromosome.
Answer:
(b) : According to the Mendel’s law of independent assortment, the pairs of “factors” segregate independently of each other when germ cells are formed. Homologous chromosomes synapse during meiosis and then separate to segregate independently into different cells which establishes the quantitative basis for segregation and independent assortment of hereditary factors.

Question 12.
Occasionally, a single gene may express more than one effect. The phenomenon is called
(a) multiple allelism
(b) mosaicism
(c) pleiotropy
(d) polygeny
Answer:
(c) : Pleiotropic allele has more than one effect in an organism. For example, the allele that causes the erythrocytes to have a distorted form in sickle-cell anemia also causes these blood cell to rupture easily.

Question 13.
In a certain taxon of insects some have 17 chromosomes and the others have 18 chromosomes. The 17 and 18 chromosome­bearing organisms are
(a) males and females, respectively
(b) females and males, respectively
(c) all males
(d) all females.
Answer:
(a) : In certain insects XX – XO type of sex determination is observed where females have autosomes + i pair of sex chromosomes while males have autosomes + 1 sex chromosome.

Question 14.
The inheritance pattern of a gene over generations among humans is studied by the pedigree analysis. Character studied in the pedigree analysis is equivalent to
(a) quantitative trait
(b) Mendelian trait
(c) polygenic trait
(d) maternal trait
Answer:
(b) : Pedigree analysis is study of pedigree for the transmission of particular trait and finding the possibility of absence or presence of that trait in homozygous or heterozygous state in a particular individual.  It indicates that Mendel’s principles are also applicable to human genetics with some modifications like quantitative inheritance, sex linked characters and other linkages.

Question 15.
It is said that Mendel proposed that the factor controlling any character is discrete and independent. His proposition was based on the
(a) results of F₃ generation of a cross
(b) observations that the offspring of a cross made between the plants having two contrasting characters shows only one character without any blending
(c) self pollination of F₁ offsprings
(d) cross pollination of F! generation with recessive parent.
Answer:
(b) : When plants having two contrasting characters are crossed, F, generation shows dominant phenotype and the recessive phenotype is not lost but appears in F₂ generation. This suggested that there is no blending of Mendelian factors in Fj generation but they stay together and only one of them is expressed. At the time of formation of gametes, these two factors obviously separate or segregate, otherwise recessive type will not appear in F₂ generation. The gametes which are formed are always pure for a particular character. That is why it is called as ‘principle of segregation’ or Taw of purity of gametes’.

Question 16.
In the F₂ generation of a Mendelian dihybrid cross the number of phenotypes and genotypes are
(a) phenotypes-4;genotypes-16
(b) phenotypes-9; genotypes-4
(c) phenotypes-4; genotypes-8
(d) phenotypes-4; genotypes-9.
Answer:
(d) : Mendel performed crosses involving two characters called as dihybrid cross. Results of the dihybrid cross where the two parents different in two pairs of contrasting traits i.e. seed colour and seed

Question 17.
Mother and father of a person with ‘O’ blood group have’A’and’B’blood group respectively. What would be the genotype of both mother and father?
(a) Mother is homozygous for ‘A’ blood group and father is heterozygous for’B’.
(b) Mother is heterozygous for’A’blood group and father is homozygous for’B’.
(c) Both mother and father are heterozygous for’A’and’B’blood group, respectively.
(d) Both mother and father are homozygous for ‘A’ and ‘B’ blood group, respectively.
Answer:
(c) : If mother and father of a person with ‘O’ blood group have ‘A’ and ‘B’ blood group then both mother and father will be heterozygous for ‘A’ and ‘B’ blood group respectively.

Very Short Answer Type Questions

Question 1.
What is the cross between the progeny of F, and the homozygous recessive parent called? How is it useful?
Answer:
The cross between the progeny of F, and the homozygous recessive plant is known as test cross.-This tells about unknown genotype,  i.e., homozygous or heterozygous dominant nature of genotype.

Question 2.
Do you think Mendel’s laws of inheritance would have been different if the characters that he chose were located on the same chromosome?
Answer:
Mendel’s laws of dominance and segregation would have remained the same. Mendel’s law of independent assortment would have been different if the characters that he chose were located on the same chromosome. If the characters are present on the same chromosome they would not assort independently and show linkage.”

Question 3.
Enlist the steps of controlled cross pollination. Would emasculation be needed in a cucurbit plant? Give reasons for your answer.
Answer:
Steps of controlled cross pollination are:

1. Selection of parent
2. Selfing
3. Emasculation
4. Bagging
5. Tagging
6. Artificial pollination
7. Seed setting and harvesting.

Cucurbit plants are generally unisexual plants. So, emasculation cannot be needed. Emasculation is done in bisexual plants to prevent self pollination.

Question 4.
A person has to perform crosses for the purpose of studying inheritance of a few traits/ characters. What should be the criteria for selecting the organisms?
Answer:
The criteria for selecting the organism to study inheritance of characters are as follows:

1. It should be easily available.
2. Shorter life span.
3. Breeding should be cheap and can be done throughout the year.
4. A large number of progeny can be produced.
5. Easily identifiable contrasting characters.

Question 5.
The pedigree chart given below shows a particular trait which is absent in parents but present in the next generation irrespective of sexes. Draw your conclusion on the basis of the pedigree.

Answer:
Given pedigree chart shows autosomal recessive trait. Both parents of generation I are heterozygous (i.e., Hb^A Hb^s) and carrier for the gene. The second and fourth child developed the trait so the genotype is homozygous recessive (i.e., Hb^s Hb^s). The first, third and fifth are carrier for the gene and genotype is heterozygous (i.e., Hb^AHb^s).

Question 6.
In order to obtain the F, generation Mendel pollinated a pure breeding tall plant with a pure breeding dwarf plant. But for getting the F₂ generation, he simply self-pollinated the tall Ft Why?
Answer:
Mendel cross pollinated plants of two different traits of the character of height in order to study their mixing in F1, generation. Only one trait appeared in F1 plants. For knowing the fate of other trait, he allowed the F1 plants to self-pollinate. F₂ generation showed both the traits indicating the recessive trait of the dwarfness remains in F1 generation, but without expresison.

Question 7.
“Genes contain the information that is required to express a particular trait.” Explain.
Answer:
Gene is a unit of inheritance which consists of linear chromosome situated at a specific locus, and carries coded information associated with particular trait. The process of gene expression involves central dogma of genetics e.g., transcription and translation.

Question 8.
How are alleles of particular gene differ from each other? Explain its significance.
Answer:
Allele has a unique nucleotide sequence, it is one of the alternative forms of a gene. In a diploid cell there are usually two alleles of any one gene which occupy the same relative position on homologous chromosomes. These allele may be similar or one allele may be dominant to the other (known as the recessive). It controls the expression of a character.

Question 9.
In a monohybrid cross of plants with red and white flowered plants, Mendel got only red flowered plants. On self-pollinating these F, plants got both red and white flowered plants in 3 : 1 ratio. Explain the basis of using RR and rr symbols to represent the genotype of plants of parental generation.
Answer:

‘R’ represent the dominant character gene whereas V represent the recessive character gene. In F, plants (Rr) both the factors for red and white colour are present. However, the factors for white colour is unable to express itself in the presence of red colour. Hence, the factor for red colour is dominant over the factor for white colour. The factor for white colour is recessive.

Question 10.
For the expression of traits genes provide only the potentiality and the environment provides the opportunity. Comment on the veracity of the statement.
Answer:
The observable characteristics of an organism are determined by its genes, the relationships between the alleles and by the interaction of genes with the environment. Phenotype = Genotype + Environment (Trait) (Potentiality) (Opportunity) Genes can provide only the potentiality and the environment provides the opportunity for the expression of traits. For example a genetically tall plant can become tall only if it receives proper amount of sunlight, minerals and water.

Question 11.
A, B, D are three independently assorting genes with their recessive alleles a, b, d respectively. A cross was made between individuals of Aa bb DD genotype with aa bb dd. Find out the type of genotypes of the offspring produced.
Answer:
The cross between AabbDD and aabbdd and the type of offsprings produced would be:

So, genotypes of offspring produced will be AAbbDd and aabbDd.

Question 12.
In our society a woman is often blamed for not bearing male child. Do you think it is right? Justify.
Answer:
In humans, sex of the child is determined at the time of fertilisation. The female parent produces only one type of egg with X-chromosome. The male gametes are of two types with X-chromosome and Y-chromosome. Fertilisation of the egg with sperm carrying X-chromosome produces a female child while fertilisation with sperm carrying Y-chromosome give rise to male child. Thus sex of the child is determined by male (father) and not by the female (mother).

Question 13.
Discuss the geneticbasisof wrinkled phenotype of pea seed.
Answer:
Seed shape is determined by a single gene, with the allele (R) for round peas domipant over the allele (r) for wrinkled peas (recessive trait). If the alleles for the gene controlling the seed shape are homozygous in a plant, it will show the character or phenotype of the same alleles i.e., RR-Round seed, rr-wrinkled seed. On the other hand, if the alleles of gene are heterozygous, they will express the phenotype of dominant allele. Rr – Round seed (r-wrinkled is recessive). This is the genetic basis of wrinkled phenotype of pea seed.

Question 14.
Even if a character shows multiple allelism, an individual will only have two alleles for that character. Why?
Answer:
Despite multiple allelism, an individual will have only two alleles because an individual develops from a zygote which is the result of fusion of sperm (carrying father set of (n) haploid chromosomes) and an egg (carrying mother set of haploid chromosomes). Sperm and an egg have only one gene (allele) for each trait. Hence an individual will have two alleles despite the occurrence of several alleles in the population.

Question 15.
How does a mutagen induce mutation? Explain with example.
Answer:
Mutagen is an agent that causes an increase in the number of mutants in a population. There are basically two types of mutagens – physical mutagens like UV-rays, gamma rays and chemical mutagens like hydroxyl radicals. Mutagens operate either by causing changes in the DNA of the genes, so interfering with the coding system, or by causing chromosome damage. A mutagen can induce mutation by inducing a change in the base sequence by insertion, deletion or substitution.

Short Answer Type Questions

Question 1.
In a Mendelian monohybrid cross, the F₂ generation shows identical genotypic and phenotypic ratios. What does it tell us about the nature of alleles involved? Justify your answer.
Answer:
In Mirabilis jalapa (Four o’clock) and Antirrhinum majus (Snapdragon or dog flower), there are two types of flower colour in pure state.When the two types of plants are crossed, the hybrid or plants of F1 generation have pink flowers, lithe latter are selfed, the plants of F2 generation are of three types—red, pink and white flowered in the ratio of 1 : 2 : 1 both phenotypically and genotypically. The pink colour apparently appears either due to mixing of red and white colours (incomplete dom inance) oi expression of a single gene for pigmented flower which produces only pink colour (quantitative inheritance).

Question 2.
Can a child have blood group O if his parents have blood group ‘A’ and ‘B’. Explain.
Answer:
A child can have blood group O, if both the parents are heterozygous for their blood groups

Question 3.
What is Down’s syndrome? Give its symptoms and cause. Why is it that the chances of having a child with Down’s syndrome increases if the age of the mother exceeds forty years?
Answer:
Down’s syndrome is a congenital form of mental retardation due to a chromosome defect in which there are three copies of chromosome no. 21 instead of the usual two (i.e., trisomy). The affected individual has a short broad face and slanted eyes (as in the  Mongolian races), short fingers, and weak muscles. Down’s syndrome is caused when the chromosomes of the pair 21 pass into a single egg due to nondisjunction during oogenesis. Then, the egg possesses 24 chromosomes, instead of 23 and offspring has 47 chromosomes instead of 46. The frequency of Down’s syndrome increases, if the age of the mother exceeds forty years. It is due to factors that adversely affect meiotic chromosome behaviour with advancement of women age. In human females, meiosis starts in the foetus to produce egg cells, but it is not completed until after the egg is fertilised. During the long time prior to fertilisation, egg cells are arrested in prophase I. In this suspended state, chromosome may become unpaired. The longer the time in prophase I, the greater the chances for unpairing and chromosome nondisjunction.

Question 4.
How was it concluded that genes are located on chromosomes?
Answer:
The chromosomal theory of inheritance states that the Mendelian factors or genes are located at specific loci on the chromosomes. Occurence of genes over chromosomes was proved by Morgan (1910) during study of sex- linked inheritance of eye colour in Drosophila. It shows criss-cross inheritance, in which female fly passes its X-chromosome and eye colour to male offspring and male fly passes its X-chromosome and eye colour to female offspring.

Question 5.
A plant with red flowers was crossed with another plant with yellow flowers. If F, showed all flowers orange in colour, explain the inheritance.
Answer:
It is the phenomenon of neither of the two alleles being dominant so that expression in the hybrid is intermediate between the expressions of the two alleles in homozygous state. It is called incomplete dominance or intermediate inheritance. There are two types of flower colour in pure state: red and yellow. When the two types of plants are crossed, the hybrid or plants of F, generation have orange flowers. If the latter are selfed, the plants of F₂ generation are of three types – red, orange and yellow flowered in the ratio of 1 : 2 : 1. The orange colour apparently appears due to mixing of red and yellow colours (incomplete dominance).

Question 6.
What are the characteristic features of a true- breeding line?
Answer:
A true breeding line is a pure line that shows same traits, generation after generation due to presence of homozygous characters. They are produced by repeated self fertilisation or breeding between identical homozygous ancestors. They are not superior to high yielding varieties, but are maintained for cross breeding and formation of new varieties.

Question 7.
In peas, tallness is dominant over dwarfness, and red colour of flowers is dominant over the white colour. When a tall plant bearing red flowers was pollinated with a dwarf plant bearing white flowers, the different phenotypic groups were obtained in the progeny in numbers mentioned against them.
Tall, red = 138
Tall, white = 132
Dwarf, red = 136
Dwarf, white = 128
Mention the genotypes of the two parents and of the four offspring types.
Answer:
The result shows four types of offspring with almost similar in number with ratio 1 : 1 : 1 : 1. This kind of result is possible in case of dihybrid test cross. One parent is double hybrid (TtRr) and the other one is double recessive (ttrr).
The test cross can be shown as :

Question 8.
Why is the frequency of red-green colour blindness is many times higher in males than that in the females?
Answer:
Red and green colour blindness is due to X-linked recessive alleles. Males have XY sex chromosomes and presence of single recessive allele expresses itself. Females have XX sex chromosomes. The possibility of female being colour blind is very less because in females both X chromosome should carry the recessive trait, which is very rare due to large scale heterozygosity in human population.

Question 9.
If a father and son are both defective in red- green colour vision, is it likely that the son inherited the trait from his father? Comment.
Answer:
Red and green colour blindness is X-linked character. Colour blindness like any sex-linked trait, shows criss-cross inheritance, i.e., mother transfers traits to her granddaughter through their son, while the father passes traits to her grandson through his daughter. Males have XY type of sex chromosomes, and they get X chromosome from their mother. In this case it is possible that the mother of that male (son) must be a carrier for the character but not affected. Male to male inheritance of X-linked character is not possible.

Question 10.
Discuss why Drosophila has been used extensively for genetical studies.
Answer:
Drosophila is used extensively for genetical studies because of following advantages:

1. It is easily available hovering over ripe mango/banana fruits where it feeds over yeast cells present over the fruit surface.
2. The flies can be reared inside bottles having yeast culture over medium ^:         containing cream of wheat, molasses and agar.
3. A new generation can be raised within 2 weeks with single mating producing hundreds of individuals.
4. The animals can be temporarily inactivated with ether and examined by hand lens/dissection microscope.
5. Female is distinguishable from male by its larger size and ovipositor at the rear end.
6. The animals possess four pairs of chromosomes of different sizes. The male fly possesses XY sex chromosomes while the female has XX chromosomes. Y chromosome is hooked and easily distinguished.
7. Polytene chromosomes occur in salivary glands of larva which can indicate any type of abnormality.
8. Breeding Drosophila is quite cheap. Further, it can be done throughout the year.

Question 11.
How do genes and chromosomes share similarity from the point of view of genetical studies?
Answer:
Similarities of behaviour of chromo-somes with Mendel’s factors / genes.

1. Both pass from generation to generation in altered form.
2. Both are present singly in gametes.
3. An organism receives two genes and two chromosomes of each type from its two parents.
4. Both follows mendelian principles.
5. Each gene and each chromosome replicates during S-phase of cell cycle.

Question 12.
What is recombination? Discuss the applica­tions of recombination from the point of view of genetic engineering?
Answer:
Recombination is the rearrangement of genes that occurs when reproductive cells (gametes) are formed. It results from the I  independent assortment of parental sets of chromosomes and exchange of chromosomal material that occur during meiosis. Useful recombinations produced by crossing over are picked up by breeders to produce useful new varieties of crop plants and animals. Green revolution has been achieved in India due to this selective picking up of useful recombinations operation flood or white revolution is also being carried out on the similar lines. In genetic engineering, the desired genes can be taken from one source and introduce them in the cells of another organism in order to improve and change the same.

Question 13.
What is artificial selection? Do you think it affects the process of natural selection? How?
Answer:
Artificial selection is the modification of species by selective breeding. Animals or plants with desirable characteristics are interbred with the aim of altering the genotype and producing a new strain of the organism for a specific purpose. For example, sheep are bred by means of artificial selection in order to improve wool quality.Yes, it affects the process of natural selection. Natural selection selects trait based on their effect on the fitness of the organism. In artificial selection traits are selected based on human preference for improving traits.The process of natural selection leads to evolutionary change in the expression of the trait in the populations whereas the artificial selection being same process involves the traits preferred by humans for its own benefit. It is much faster than the natural selection, and it impose threat on diversity in long run making it unfit to the natural selected organism.

Question 14.
With the help of an example differentiate between incomplete dominance and co­dominance.
Answer:
Differences between incomplete dominance and codominance are as follows:

Incomplete dominance	Codominance
(1) Effect of one of the two alleles is more conspicuous	The effect of both the alleles is equally conspicuous.
(2) It produces a fine mixture of the expression of two alleles.	There is no mixing of the effect of the two alleles
(3) The effect in hybrid is intermediate of the expression of the two alleles, e.g., flower colour in Mirabilis jalapa Red, Pink, and White	Both the alleles produce their effect independently, e.g., AB blood group.
(4) Alleles show quantitative effect. One dominant allele produces half and two dominant alleles produce full phenotype	There is no quantitative effect of the alleles.

Question 15.
It is said, that the harmful alleles get eliminated from population over a period of time, yet sickle cell anaemia is persisting in human population. Why?
Answer:
Harmful alleles become lethal, whenever they occur in homozygous state. The affected individual does not transfer them to the next generation. Over a period of time, these alleles get eliminated from the population. This is not so in the case of sickle cell anaemia. Sickle cell anaemia is an autosomal recessive disorder in which the erythrocytes become sickle shaped under oxygen deficiency as during strenuous exercise and at high altitude.
Despite having harmful effect, the allele for sickle cell anaemia continues to persist in «human population because it has survival value in malaria infested areas like tropical Africa. Malarial parasite is unable to penetrate the erythrocyte membrane and cause any harm. Further, the sickle cell heterozygotes do not always suffer from syndrome. Their erythrocytes appear normal till there is oxygen deficiency when some sickle-shaped erythrocytes may be observed.

Long Answer Type Questions

Question 1.
In a plant tallness is dominant over dwarfness and red flower is dominant over white. Starting with the parents work out a dihybrid cross.What is’standard dihybrid ratio? Do you think the values would deviate if the two genes in question are interacting with each other?
Answer:

Tall red = 9
Tall white = 3 Phenotypic ratio = 9 : 3 : 3 : 1 dwarf red = 3 dwarf white = 1
This is a standard dihybrid phenotypic ratio. Alleles of two or more independent genes interact to produce a phenotypic expression different from the normal expression. They show⁷ dominant epistasis (12 : 3 : 1), recessive epistasis (9:3:4), dominant recessive epistasis (13 : 3) etc.

Question 2.
(a) In humans, males are heterogametic and females are homogametic. Explain. Are there any examples where males are homogametic and females heterogametic?
(b) Also describe as to, who determines the sex of an unborn child? Mention whether temperature has a role in sex determination.
Answer:
(a) Human males are said to be heterogametic because they produce two types of gametes as they have two different sex chromosomes X and Y in addition to autosomes e.g. (A + X) and (A + Y) type of gametes. Whereas human females are homogametic because they produce only one type of gametes, as they have XX sex chromosomes in addition to autosomes (A + X), (A+X). In birds females are heterogametic and have ZW sex chromosomes and males are homogametic with ZZ sex chromosomes.

(b) Sex of unborn child in human is determined by male. Males produce two types of gametes (A + X) and (A + Y). Females produce only one type of gametes (A + X). If (A + X) x (A + X) – female child If (A + X) x (A + Y) – male child Therefore sex of the unborn child completely depends on males. In crocodile and some lizards temperature has a role in sex determination, high temperature induces maleness and low temperature induces femaleness.

Question 3.
A normal visioned woman, whose father is colourblind, marries a normal visioned man. What would be the probability of her sons and daughters to be colour blind? Explain with the help of a pedigree chart.
Answer:
Colour blindness is a recessive sex- linked trait in which eye fails to distinguish red and green colours. The gene for normal vision is dominant. The normal gene and its recessive allele are carried by X-chromosomes. In females, colour blindness appears only when both the sex chromosomes, carry the recessive gene (X^CX^C). The females have normal vision but function as carrier if a single recessive gene for colour blindness is present (XX^C). In males, the defect appears in the presence of a single recessive gene (X^CY) because Y-chromosome of male does not carry any gene for colour vision.Flere, woman is normal visioned, whose father is colour blind, so woman act as carrier because she has got one gene for colour blindness from her father. She marries a normal man, then her 50% daughters will be carrier and 50% will be completely normal. Her 50% sons will be colour blind and 50% will be normal.

Question 4.
Discuss in detail the contributions of Morgan and Sturtevant in the area of genetics.
Answer:
Morgan carried out many dihybrid crosses in Drosophila to study the genes which were sex linked. He hybridised yellow bodied white eyed female to brown bodied red eyed males and intercrossed their F[ progeny. He observed that two genes did not segregate independently and deviated normal F,
ratio 9 : 3 : 3 : 1.
According to him if two genes in a dihybrid cross are situated on same chromosome, the parental gene combinations are much higher than non-parental ones. He gave the term linkage to physical association of genes and recombination to non-parental characters.His students Alfred Sturtevant used the frequency of recombination between gene pairs on the same chromosomes as a measure of distance between genes and mapped their position on the chromosome. The first chromosome maps were prepared by Sturtevant in 1911 for two chromosomes in Drosophila. Genetic maps are extensively used these days as a starting point in sequencing of whole genome in Human Genome Project (HGP).

Question 5.
Define aneuploidy. How is it different from polyploidy? Describe the individuals having following chromosomal abnormalities.
(a) Trisomy of 21^st chromosome
(b) XXY
(c) XO
Answer:
Aneuploidy is the phenomenon of gain or loss of one or more chromosomes that results due to failure of separation of the members of the homologous pairs of chromosomes during meiosis. Polyploidy is a increase of entire set of chromosomes whereas aneuploidy is the addition or deletion of one or few chromosomes from the original genome. Polyploid occurs in nature due to the failure of chromosomes to separate at the time of anaphase either due to non-disjunction or due to nonformation of spindle. Aneuploidy commonly arises due to nondisjunction of the two chromosomes of homologous pair so that one gamete comes to have an extra chromosome (N + 1) while the other becomes deficient in one chromosome (N – 1).
(1) Trisomy of 21^st chromosome :
Trisomy of 21^st chromosome is commonly known as Down’s syndrome. Down’s syndrome is a relatively common birth defect caused by the presence of an extra chromosome number 21 (three instead of two number 21 chromosomes, or, trisomy 21). Both the chromosomes of the pair 21 pass into a single egg due to nondisjunction during oogenesis. This chromosome abnormality adversely affects both the physical and intellectual development of the individual.
The affected individuals have a very different but characteristic external appearance. They display prominent folding at the comer of eyes and have short stature. They have small round head; protruding furrowed tongue that cause the mouth to remain partially open; and short, broad hands with fingers showing characteristic fingerprint patterns. Physical, psychomotor and mental development is retarded and the life expectancy is shortened.

(2) XXY:
An abnormal condition of male sexual characteristics in which the body cells contain one or more extra X-chromosomes. It is caused due to trisomy of sex (X) chromosome. It is known as Klinefelter’s syndrome. Characteristics include undeveloped testicles; long legs and female like breasts. The severity of the abnormalities increases with greater numbers of X-chromosomes. About 30-90 percent of all patients with Klinefelter’s syndrome have the karyotype 47 (2A+XXY). The man may appear generally normal, although infertile.

(3) XO:
It is known as Turner’s syndrome. Individuals having a single X chromosome 2A+XO (45) have female sexual differentiation but ovaries are rudimentary. Other associated phenotypes of this condition are short stature, webbed-neck, broad chest, lack of secondary sexual characteristics and sterility. Thus, any imbalance in the copies of the sex chromosomes may disrupt the genetic information necessary for normal sexual development.

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NCERT Exemplar Solutions for Class 12 Biology chapter 1 Reproduction in Organisms

These Solutions are part of NCERT Exemplar Solutions for Class 12 Biology. Here we have given NCERT Exemplar Solutions for Class 12 Biology chapter 1 Reproduction in Organisms

Multiple Choice Questions

Question 1.
A few statements describing certain features of reproduction are given below.
(i) Gametic fusion takes place.
(ii) Transfer of genetic material takes place.
(iii) Reduction division takes place.
(iv) Progeny have some resemblance with parents.
Select the options that are true for both asexual and sexual reproduction from the options given below.
(a) (i) and (ii)
(b) (ii) and (iii)
(c) (ii) and (iv)
(d) (i) and (iii)
Answer:
(c) : Reproduction is a biological process in which-an organism produces young ones (offspring) similar to itself. In reproduction, offsprings have some resemblance with parents. Both sexual and asexual reproduction involve transfer of genetic material.

Question 2.
The term ‘clone’ cannot be applied to offspring formed by sexual reproduction because
(a) offspring do not possess exact copies of parental DNA
(b) DNA of only one parent is copied and passed on to the offspring
(c) offspring are formed at different times
(d) DNA of parent and offspring are completely different.
Answer:
(a) : In sexual reproduction, there is fusion of male gametes and female gametes, the offspring produced are not identical to their parents. This genetic recombination leads to variations, which play an important role in evolution.

Question 3.
A sexual method of reproduction by binary fission is common to which of the following?
(i) Some eukaryotes
(ii) All eukaryotes
(iii) Some prokaryotes
(iv) All prokaryotes
(a) (i) and (ii)
(b) (ii) and (iii)
(c) (i) and (iii)
(d) (iii) and (iv)
Answer:
(c)

Question 4.
A few statements with regard to sexual reproduction are given below.
(i) Sexual reproduction does not always require two individuals.
(ii) Sexual reproduction generally involves gametic fusion.
(iii) Meiosis never occurs during sexual reproduction.
(iv) External fertilisation is a rule during sexual reproduction.
Choose the correct statements from the options below.
(a) (i) and (iii)
(b) (i) and (ii)
(c) (ii) and (iii)
(d) (i) and (iv)
Answer:
(b) : Meiosis is required for the production of haploid gametes during sexual reproduction. External fertilisation is not a rule during sexual reproduction, it can occur internally also.

Question 5.
A multicellular, filamentous alga exhibits a type of sexual life cycle in which the meiotic division occurs after the formation of zygote. The adult filament of this alga has
(a) haploid vegetative cells  and   diploid gametangia
(b) diploid vegetative cells  and   diploid gametangia
(c) diploid vegetative cells  and  haploid gametangia
(d) haploid vegetative cells and  haploid gametangia.
Answer:
(d)

Question 6.
The male gametes of rice plant have 12 chromosomes in their nucleus. The chromo­some number in the female gamete, zygote and the cells of the seedling will be, respectively
(a) 12,24,12
(b) 24,12,12
(c) 12,24,24  
(d) 24,12,24.
Answer:
(c) : Chromosome number in male gamete of rice plant is n = 12 therefore chromosome number in female gamete would also be 12. Zygote is diploid as it is the product of fertilisation and the cells of the seedling would be meiocytes and other diploid cells. Hence, the chromosome number in both zygote and cells of seedling will be 2n = 24.

Question 7.
Given below are a few statements related to external fertilisation. Choose the correct statements.
(i) The male and female gametes are formed and released simultaneously.
(ii) Only a few gametes are released into the medium.
(iii) Water is the medium in a majority of organisms exhibiting external fertilisa­tion.
(iv) Offspring formed as a result of external fertilisation have better chance of survival than those formed inside an organism.
(a) (iii) and (iv)
(b) (i) and (iii)
(c) (ii) and (iv)
(d) (i) and (iv)
Answer:
(b) : A large number of gametes are released into the medium to increase the chances of fertilisation. The chances of survival of offsprings from external fertilisation are lesser than those of internal fertilisation as they face more risk from predators.

Question 8.
The statements given below describe certain features that are observed in the pistil of flowers.
(i) Pistil may have many carpels.
(ii) Each carpel may have more than one ovule.
(iii) Each carpel has only one ovule.
(iv) Pistil have only one carpel.
Choose the statements that are true from the options below.
(a) (i) and (ii)
(b) (i) and (iii)
(c) (ii) and (iv)
(d) (iii) and (iv)
Answer:
(a) : Gynoecium is the female part of the flower, a unit of which is called pistil. A pistil may have one or more than one carpels (monocarpellary, bicarpellary etc). Each carpel may have more than one ovules.

Question 9.
Which of the following situations correctly describe the similarity between an angiosperm egg and a human egg?
(i) Eggs of both are formed only once in a lifetime.
(ii) Both the angiosperm egg and human egg are stationary.
(iii) Both the angiosperm egg and human egg are motile transported.
(iv) Syngamy in both results in the formation of zygote.
Choose the correct answer from the options given below.
(a) (ii) and (iv)
(b) (iv) only
(c) (iii) and (iv)
(d) (i) and (iv)
Answer:
(b) : Syngamy is the complete and permanent fusion of male and female gametes to form a zygote.

Question 10.
Appearance of vegetative propagules from the nodes of plants such as sugarcane and ginger is mainly because
(a) godes are shorter than internodes
(b) nodes have meristematic cells
(c) nodes are located near the soil
(d) nodes have non-photosynthetic cells.
Answer:
(b) : Vegetative propagation is asexual reproduction from ‘various parts in plants. In plants, such as, sugarcane and ginger, appearance of vegetative propagules from nodes is because of presence of meristematic cells in them. Meristematic cells have the ability to divide to form new cells which can differentiate and give rise to permanent tissues.

Question 11.
Which of the following statements, support the view that elaborate sexual reproductive process appeared much later in the organic evolution?
(i) Lower groups of organisms have simpler body design.
(ii) Asexual reproduction is common in lower groups.
(iii) Asexual reproduction is common in higher groups of organisms.
(iv) The high incidence of sexual reproduction in angiosperms and vertebrates.
Choose the correct answer from the options given below.
(a) (i), (ii) and (iii)
(b) (i), (iii) and (iv)
(c) (i), (ii) and (iv) 
(d) (ii), (iii) and (iv)
Answer:
(c)

Question 12.
Offspring formed by sexual reproduction exhibit more variation than those formed by asexual reproduction because
(a) sexual reproduction is a lengthy process
(b) gametes of parents have qualitatively different genetic composition
(c) genetic material comes from parents of two different species.
(d) greater amount of DNA is involved in sexual reproduction.
Answer:
(b) : Sexual reproduction occurs in almost all types of animals and mostly in higher plants. It is usually biparental. Daughter organisms genetically differ from the parents. Since there are variations, so it contributes to evolution of the species.

Question 13.
Choose the correct statement from amongst the following.
(a) Dioecious organisms are seen only in animals.
(b) Dioecious organisms are seen only in plants.
(c) Dioecious organisms are seen in both plants and animals.
(d) Dioecious organisms are seen only in vertebrates.
Answer:
(c) : Dioecious organisms are those in which male and female sex organs are present in different organisms, where as monoecious organisms are those in which male and female sex organs are present in the same organism. Monoecious organisms are also called as hermaphrodite. Dioecious organisms are seen in both plants and animals. Papaya, date palm and most of the animals are dioecious.

Question 14.
There is no natural death in single celled organisms like Amoeba and bacteria because
(a) they cannot reproduce sexually
(b) they reproduce by binary fission
(c) parental body is distributed among the offspring
(d) they are microscopic.
Answer:
(c) : These are no natural death in single celled organisms like Amoeba and bacteria. It is so, because of asexual reproduction, the body of parent is divided into daughter cells. So, in effect, there is no practical death in Amoeba and bacteria.

Question 15.
There are various types of reproduction. The type of reproduction adopted by an organism depends on
(a) the habitat and morphology of the organism
(b) morphology of the organism
(c) morphology and physiology of the organism
(d) the organism’s habitat, physiology and genetic makeup.
Answer:
(d) : There are various types of reproduction, both asexual (fission, budding, etc.) and sexual (internal and external). The type of reproduction, an organism undergoes depends ultimately on its genetic makeup which influences its physiology. Habitat also influences the type of reproduction, that organism undergoes.

Question 16.
Identify the incorrect statement.
(a) In asexual reproduction, the offspring produced are morphologically and genetically identical to the parent.
(b) Zoospores are sexual reproductive structures.
(c) In asexual reproduction, a single parent produces offspring with or without the formation of gametes.
(d) Conidia are asexual structures in
Answer:
(b) : Spores formation is also a type of asexual reproduction. Zoospores, conidia, oidia, etc. are all asexually reproducing structures. There is generally no gamete formation in asexual reproduction and the offsprings produced are called clones.

Question 17.
Which of the following is a post-fertilisation event in flowering plants?
(a) Transfer of pollen grains
(b) Embryo development
(c) Formation of flower
(d) Formation of pollen grains
Answer:
(b) : Events in sexual reproduction after the fertilisation are called post-fertilisation events. After fertilisation, a diploid zygote is formed in all sexually reproducing organisms. The process of development of embryo from the zygote is called embryogenesis.

Question 18.
The number of chromosomes in the shoot tip cells of a maize plant is 20. The number of chromosomes in the microspore mother cells of the same plant shall be
(a) 20 
(b) 10
(c) 40
(d) 15.
Answer:
(a) : Shoot tip cells and microspore mother cells both are diploid in maize plant. If number of chromosomes in shoot tip cell (2n) = 20, then number of chromosomes in microspore mother cell will be (2n) = 20.

Very Short Answer Type Questions

Question 1.
Mention two inherent characteristics of Amoeba and yeast that enable them to reproduce asexually.
Answer:
The two inherent characteristics for asexual reproduction in Amoeba and yeast are :

1. Gametes are not formed in Amoeba and yeast.
2. Uniparental condition (i.e., single parent) is involved in reproduction of both Amoeba and yeast.

Question 2.
Why do we refer to offspring formed by asexual method of reproduction as clones?
Answer:
Offsprings produced by asexual reproduction are morphologically and genetically similar to their parents. Hence, they are known as clones.

Question 3.
Although potato tuber is an underground part, it is considered as a stem. Give two reasons.
Answer:
Reasons for considering potato as a stem are:

1. Differentiation into nodes and inter­nodes.
2. The nodes bear buds which can grow to from leaf shoots or plantlets.

Question 4.
Between an annual and a perennial plant, which one has a shorter juvenile phase? Give one reason.
Answer:
Annual plant have shorter juvenile phase as they complete their life cycle in single season i.e., a few weeks to a few months, e.g., wheat, maize, pea, gram.

Question 5.
Rearrange the following events of sexual reproduction in the sequence in which they occur in a flowering plant : embryogenesis, fertilisation, gametogenesis, pollination.
Answer:
The correct sequence of events of sexual reproduction in flowering plants are as follows : gametogenesis, pollination, fertilisation, embryogenesis.

Question 6.
The probability of fruit set in a self-pollinated bisexual flower of a plant is far greater than a dioecious plant. Explain.
Answer:
Presence of male and female repro­ductive organs on same plant provides more chances of self pollination and therefore increases chances of fruit set. However in dioecious plants agents of pollination are needed which may therefore decrease the chances of a fruit set.

Question 7.
Is the presence of large number of chromosomes in an organism a hindrance to sexual reproduction? Justify your answer by giving suitable reasons.
Answer:
The presence of a large number of chromosomes in an organism is not a hindrance to sexual reproduction.

Question 8.
Is there a relationship between the size of an organism and its life span? Give two examples in support of your answer.
Answer:
There is no direct correlation between the life span of organisms and their sizes.

1. The crow and parrot are nearly of the same size, but life span of crow and parrot are 15 years and 140 years respectively.
2. The mango and the peepal trees are nearly of the same size but the life span of mango and peepal are 100 years and 1000 years respectively.

Question 9.
In the figure given below the plant bears two different types of flowers marked ‘A’ and ‘B’. Identify the types of flowers and state the type of pollination that will occur in them.

Answer:
The plant shown in figure has two types of flowers:
Flower A – chasmogamous type with stigma and anthers exposed. Such type of flowers may undergo self or cross pollination.
Flower B – Cleistogamous type – closed type of flowers, where stigma and anthers are not exposed. Such type of flowers undergo only self pollination.

Question 10.
Give reasons as to why cell division cannot be a type of reproduction in multicellular organisms.
Answer:
In multicellular organisms cell division does not divide the whole body into daughter cells as in unicellular organisms. But multicellular organisms have a well developed reproductive organs which help in reproduction.

Question 11.
In the figure given below, mark the ovule and pericarp.

Answer:

Question 12.
Why do gametes produced in large numbers in organisms exhibit external fertilisation?
Answer:
In external fertilisation, there are great chances that the sperm and the eggs released by the organisms can be affected by desiccation, predators, etc. So, to make up for the high fatality rate of the gametes, the organisms produces a lot of gametes.

Question 13.
Which of the followings are monoecious and dioecious organisms.
(a) Earthworm____________
(b) Chara ______________
(c) Marchantia____________
(d) Cockroach ___________
Answer:
(a) Earthworm – Monoecious animal
(b) Chara – Monoecious plant
(c) Marchantia- Dioecious plant
(d) Cockroach – Dioecious plant

Question 14.
Match the organisms given in Column ‘A’ with the vegetative propagules given in Column ‘B’

Column A		Column B
i	Bryophyllum	a.	Offset
ii	Agave	b.	Eyes
iii.	Potato	c.	Leaf buds
iv.	Water hyacinth	d.	Bulbils

1. Rrunphyllum. – (c) Leaf buds
2. Agave –  (d)   Bulbils
3. Potato –  (b)   Eyes
4. Water hyacinth –  (a) Offset

Question 15.
What do the following parts of a flower develop into after fertilisation?
(a) Ovary__________
(b) Ovules__________
Answer:
(a) Ovary – fruit
(b) Ovules-seed

Short Answer Type Questions

Question 1.
In haploid organisms that undergo sexual reproduction, name the stage in the life cycle when meiosis Give reasons for your answer.
Answer:
In zygospore (formed by zygote) meiosis occurs. Because after meiosis it can form meiospores which can develop into haloid organisms.

Question 2.
The number of taxa exhibiting asexual reproduction is drastically reduced in higher plants (angiosperms) and higher animals (vertebrates) as compared with lower groups of plants and animals. Analyse the possible reasons for this situation.
Answer:
Both angiosperms and vertebrates have more complex structural organisation. They have evolved very efficient mechanism of sexual reproduction.

1. The offspring produced due to sexual reproduction adapt better to the changing environmental conditions.
2. Genetic recombination, interaction, etc. during sexual reproduction provide vigour and vitality to the offspring.

Question 3.
Honeybees produce their young ones only by sexual reproduction. Inspite of this, in a colony of bees we find both haploid and diploid  individuals Name the haploid and diploid individuals in the colony and analyse the reasons behind their formation.
Answer:
Honeybees produce their young ones by sexual reproduction but parthenogenesis also occurs alongwith sexual reproduction. Fertilised eggs and parthenogenetically developed eggs give rise to different castes. In honey bees, fertilised eggs (zygotes) which is diploid give rise to queens and workers (both are females) and unfertilised eggs (ova) which is haploid develop into drones (males).

Question 4.
With which type of reproduction do we associate the reduction division? Analyse the reasons for it.
Answer:
We can associate the reduction division (meiosis) with sexual reproduction. Sexual reproduction is the process of development of new organisms through the formation and fusion of male and female gametes. It involves meiosis or reduction division i.e., halving the parental chromosomes inside the male and female gamete and their subsequent fusion resulting in the restoration of the original parental number of chromosomes, causing an increased genetic diversity.

Question 5.
Is it possibletoconsider vegetative propagation observed in certain plants like Bryophyllum, water hyacinth, ginger etc., as a type of asexual reproduction? Give two/ three reasons.
Answer:
Yes, it is possible to consider vegetative propagation as a type of asexual reproduction because of the following reasons:

1. Production of plantlet occurs by a single parent plant without the formation and fusion of gametes.
2. Plantlets receives all genes from their one parent plant.
3. Vegetative reproduction involves only mitotic cell division.
4. There is no gamete formation.

Question 6.
Fertilisation is not an obligatory event for fruit production in certains plants’ Explain the statement.
Answer:
No, fertilisation is not an obligatory event for fruit production in certain plants, as fruits can also develop from unfertilised ovary also. Such fruits are called parthenocarpic  fruits. The formation of this type of fruit takes place without prior’fertilisation of the flower by pollen. The resulting fruits are seedless and therefore do not contribute to the reproduction of the plant, examples are bananas and pineapples. Plant growth substances may have a role in this phenomenon, which can be induced by auxins in the commercial production of tomatoes and other fruits.

Question 7.
In a developing embryo, analyse the consequences if cell divisions are not followed by cell differentiation.
Answer:
Cell division increase the number of cells in the developing embryo while cell differentiation helps the groups of cells to undergo certain modifications and form specialised tissues and organs.

Initially all the cells of the developing embryo are alike with same genetic information. Later however, due to the phenomenon of differentiation depending on their location and internal cellular mechanism, different cells of the embryo develop differently forming embryonal axis, plumule and radicle. However, if cell differentiation does not occur the embryo will remain just a mass of undifferentiated cells or callus. There would not any plumule, radicle, cotyledons or embryo axis. A new plant will not be formed from such an embryo.

Question 8.
List the changes observed in an angiosperm flower subsequent to pollination and fertilisation.
Answer:
In an angiosperm flower, the changes that occur subsequent to pollination and fertilisation are called post-fertilisation changes. Pollen grain germinates over the stigma and forms a pollen tube carrying two gametes. Pollen tube reaches ovary and enters an ovule through one of its synergids. Fertilisation produces a diploid zygote and triploid primary endosperm cell. Zygote produces embryo and primary endosperm cell forms endosperm. Transformation takes place and ovule transforms into seed, ovary transforms into fruit and ovary wall into pericarp. The petals, stamens and style wither away.

Question 9.
Suggest a possible explanation why the seeds in a pea pod are arranged in a row, whereas those in tomato are scattered in the juicy pulp.
Answer:
Arrangement of seeds inside a fruit depends upon the type of placentation and the growth of placental axis. In pea ovary, the ovules are attached to the ventral suture i.e., marginal placentation and carpel is mono- carpellary. So, seeds are arranged in a row. In tomatoes, the gynoecium is tricarpellary with axile placentation. The placentae grow and become pulpy during fruit formation. As a result seeds get scattered in the pulpy mass in tomato.

Question 10.
Draw the sketches of a zoospore and a conidium. Mention two dissimilarties between them and atleast one feature common to both structures.
Answer:

Dissimilarities : Zoospores are motile and flagellated whereas conidia are nonmotile. Zoospore are formed inside a zoosporangium and conidia are formed (exogenously) outside on a conidiophore.
Similarity : Both are reproductive structures for asexual reproduction.

Question 11.
Justify the statement’Vegetative reproduction is also a type of asexual reproduction’.
Answer:
Refer answer 5.

Long Answer Type Questions

Question 1.
Enumerate the differences between asexual and sexual reproduction. Describe the types of asexual reproduction exhibited by unicellular organisms.
Answer:

A sexual reproduction	Sexual reproduction
(1) Asexual reproduction involves the participation of single individual parent.	Sexual reproduction involves participation of two separate parents.
(2) It generally occurs without the involvement of sex organs.	It usually involves the sex organs.
(3) It does not involve meiosis or reduction division.	It involves meiosis which occurs at the time of gamete formation.
(4) Asexual reproduction does not involve sexual fusion or fusion of two gametes. Zygotes are not formed.	The sexual reproduction requires fertilisation to take place between two opposite gametes leading to the production of a zygote.
(5) Since asexual reproduction does not involve meiosis and fusion of gametes, the offsprings are genetically similar to parents and they do not show variation.	The individuals produced as a result of meiosis and gametic fusion exhibit genetic variation and differ from either of the two parents.
(6) It is a very quick method of multiplication and, therefore, used by plant breeders for cloning.	It is very slow method of multiplication of individuals.

Asexual reproduction occurs usually in some organisms such as monerans, protists, in plants and in some animals.
Following are the methods of asexual reproduction in unicellular organisms :

(1) Binary fission : In this type of asexual reproduction, the parent organism divides into two halves, each half forms an independent daughter organism e.g.,

(2) Multiple fission : In this process, the parent body divides into many similar daughter individuals. Multiple fission occurs in Amoeba during unfavourable condition.

(3) Bidding : In budding, a daughter individual is formed from a small part or bud, arising from the parent body g., yeast.
Spore formation or sporulation : Spores are minute, single celled, thin or thick walled propagules. In this type of asexual reproduction dispersive structures called spores are released from parent body that germinate under favourable conditions to form new individuals. Motile spores called zoospores are formed in unicellular alga like

Question 2.
Do all the gametes formed from a parent organism have the same genetic composition (identical DNA copies of the parental genome)? Analyse the situation with the background of gametogenesis and provide or give suitable explanation.
Answer:
No, All the gametes formed from a parent organism don’t have the same genetic combination. Formation of two types of gametes-male and female, inside the gametangia, is called gametogenesis. The reproductive units in sexual reproduction are specialised cells called gametes. The gametes are generally of two kinds : male and female. The gametes of all the organisms are usually haploid cells, i.e., possess single set (or n number) of chromosomes. The gametes are usually formed by meiotic divisions. Therefore, they are haploid, i.e., have halved or reduced (n) number of chromosomes. During this processes, random segregation of chromosomes (independent assortment) and exchange of genetic material between homologous chromosomes (crossing over) result in new combinations of genes in the gametes, and this reshuffling increases genetic diversity. The coming together of two unique sets of chromosomes in the zygote forms the genetic basis of variation within the species.

Question 3.
Although sexual reproduction is a long drawn, energy-intensive complex form of reproduction, many groups of organisms in Kingdom Animalia and Plantae prefer this mode of reproduction. Give atleast three reasons for this.
Answer:
Reasons for preference of sexual reproduction in higher groups of organisms are as follows:

1. During sexual reproduction, there is fusion of gametes, hence genetic recombination takes place causing variations.
2. The offspring produced due to sexual reproduction adapt better to the changing environmental conditions.
3. Genetic recombination, interaction, etc. during sexual reproduction provide vigour and vitality to the offspring.
4. Variation being a major factor of natural selection, therefore, it plays an important role in evolution.

Question 4.
Differentiate between
(a) oestrous and menstrual cycles;
(b) ovipary and vivipary. Cite an example for each type.
Answer:
(a) Differences between menstrual and oestrous cycles are as follows:

Menstrual cycle	Oestrous cycle
(1) This cycle consists of menstrual phase, proliferative phase and the secretory phase.	It consists of a short period of oestrous or heat (e.g., 12-24 hours in cow) followed by anoestrous or passive period
(2) Blood flows in the last few days of this cycle.	Blood does not flow in this cycle.
(3) The broken endometrium is passed out during menstruation	The broken endometrium is reabsorbed.
(4) Sex urge is not increased during menstruation.	Sex urge is increased during oestrous period.
(5) Female does not permit copulation during menstrual phase of the cycle.	Female permits copulation only during oestrous period.
(6) It occurs in primates (monkeys, apes and human beings) only.	It occurs in nonprimates such as cows, dogs, etc

 (b) Differences between ovipary and vivipary are as follows:

Ovipary	Vivipary
(1) The process in which females lay fertilised/ unfertilised eggs.	The process in which females give birth to young ones.
(2) The development of zygote takes place outside the female’s body.	The development of zygote takes place inside the female’s body.
(3) Females lay eggs in a safe place in the environment but the chances of survival are less.	Females deliver young ones and the chances of survival are more.
(5) Example. All birds, most of reptiles and egg­laying mammals.	Example. Mammals except egg-laying mammals

Question 5.
Rose plants produce large, attractive bisexual flowers but they seldom produce fruits. On the other hand a tomato plant produces plenty of fruits though they have small flowers. Analyse the reasons for failure of fruit formation in rose.
Answer:
Both rose and tomato plants are selected by human beings for different characteristics, the rose for its flower and tomato for its fruit. Rose plant is vegetatively propagated and does not produce seeds. Fruit formation fails in rose plant due to following reasons:

1. Viable pollen may not be produced, resulting in failure of fertilisation.
2. Functional egg may not be produced.
3. The ovule produced may be defective or non functional.
4. There may be pollen-pistil incom­patibility.
5. The plant may be a hybrid with abnormal segregation of chromosomes resulting in non-viable gametes.
6. The plant may be a hybrid and sterile.
7. There may be internal barriers for growth of pollen tube and fertilisation

We hope the NCERT Exemplar Solutions for Class 12 Biology chapter 1 Reproduction in Organisms help you. If you have any query regarding NCERT Exemplar Solutions for Class 12 Biology chapter 1 Reproduction in Organisms, drop a comment below and we will get back to you at the earliest.

NCERT Solutions for Class 10 Social Science Civics Chapter 8 Challenges to Democracy

These Solutions are part of NCERT Solutions for Class 10 Social Science. Here we have given NCERT Solutions for Class 10 Social Science Civics Chapter 8 Challenges to Democracy.

TEXTBOOK EXERCISES

Different contexts, different challenges

Question 1.
Each of these cartoons represents a challenge to democracy. Please describe what that challenge is. Also place it in one of the three categories mentioned in the first section.

Answer:

1. Rigging of elections – Challenge of deepening of democracy.
2. Use of muscle power in elections – Challenge of deepening of democracy.
3. Not adequate representation for women or discrimination against women – Challenge of expansion.
4. Use of money power in elections or too much expenditure on election campaign – Challenge of deepening of democracy.

Question 2.
In the following cases and context (as given in the Textbook page 104) give description of the challenges for democracy in that situation.
Answer:

Case and context	Your description of the challenges for democracy in that situation
Chile : General Pinochet’s government defeated, but military still in control of many institutions	Foundational challenges i.e., Establishing civilian control over all governmental institutions, holding the first multi-party elections, recalling all political leaders from exile.
Poland : After the first success of Solidarity, the government imposed martial law and banned solidarity.	Foundational challenges i.e., to bring down the existing non-democratic regime, to remove martial law and grant freedom of association.
Ghana : Just attained independence, Nkrumah elected president.	Challenge of expansion – to apply basic principles of democracy in all regions.
Myanmar : Suu Kyi under \ house arrest for more than 15 years, army rulers getting global acceptance.	Foundational challenge to keep military away from controlling government and establishing a sovereign and functional state.
International Organisations : US as the only super power disregards the UN and takes	Challenge of expansion – to ensure equal powers and cooperation among members to solve international problems.
Mexico : Second free election after the defeat of PRI in 2000; defeated candidate alleges rigging.	Challenge of deepening of democracy – to strengthen the institutions and practices of democracy; to conduct free and fair elections.
China : Communist Party adopts economic reforms but maintains monopoly over political power.	Challenge of deepening of democracy – no division of power or public participation. To have free and fair elections.
Pakistan : General Musharraf holds referendum, allegations of fraud in voters’ list.	Foundational challenge – to make transition to democracy – and establish civilian government on the basis of free and fair elections.
Iraq : Widespread sectarian violence as the new government fails to establish its authority.	Foundational challenge to democracy – to set up democratic government on the basis of free and fair elections
South Africa : Mandela  retires from active politics, pressure on his successor Mbeki to withdraw some concessions given to White minority.	Challenge of deepening of democracy – to safeguard the interests of White minority.
US, Guantanamo Bay : UN Secretary General calls this a violation of international law, US refused to respond.	Foundational challenge – to safeguard various rights of people including freedom. To refrain the USA from following unjust policies and compel them to obey international law.
Saudi Arabia : Women not allowed to take part in public activities, no freedom of religion for minorities.	Challenge of expansion – to grant equal rights to ‘ women without any discrimination. Interests of the minorities to be protected.
Yugoslavia : Ethnic tension between Serbs and Albanians on the rise in the province of Kosovo. Yugoslavia disintegrated.	Challenge of expansion of democracy – basic principles . of democracy should have been applied in Yugoslavia i.e., extension of federalism, to protect the interests of minorities etc.
Belgium : One round of constitutional change taken place, but the Dutch speakers not satisfied; they want more autonomy.	Challenge of deepening of democracy – to strengthen the institutions of democracy to realise the expectations of the people. More powers to local bodies.
Sri Lanka : The peace talks between the government and the LTTE break down, renewed violence.	Challenge of expansion of democracy – to adopt federal principles – to avoid majoritarianism. To accommodate minorities and protect their interests.
US Civil Rights : Blacks have won equal rights, but are still poor, less educated and marginalised.	Challenge of deepeni ng of democracy – equal opportunities in economic field, education to be provided such as reservation for SCs / STs / OBCs in India.
Northern Ireland : The civil war has ended but Catholics and Protestants yet to develop trust.	Challenge of expansion of democracy – extension of federal principle to all the units, women and minorities.
Nepal : Constituent Assembly about to be elected, unrest in Terai areas, Maoists have not surrendered arms.	Foundational challenge of making the transition – to democracy and setting up democratic government.To establish a sovereign functional state.

Question 3.
Different types of challenges :
Now that you have noted down all these challenges, let us group these together into some broad categories. Given below are some spheres or sites of democratic politics. You may place against each of these the specific challenges that you noted for one or more countries or cartoons in the previous section. In ease you find that some challenges do not fit into any of the categories given below, you can create new categories and put some items under that.
Answer:

Constitutional design	Nepal
Democratic rights	Poland, Myanmar, Pakistan, US and Guantanamo Bay, Saudi Arabia and US Civil Rights
Working of institutions	International organisations, Mexico and China
Elections	Mexico
Federalism, decentralisation	Yugosl avia an d Belgium
Accommodati on of diversity	Iraq and Northern Ireland
Political organisations .	Ghana and South Africa
Globalisation	Bolivia

Question 4.
Let us group these again, this time by the nature of these challenges as per the classification suggested in the first section. For each of these categories, find at least one example from India as well.
Answer:

Foundational challenges	Poland, Myanmar, Pakistan, Iraq, US, Guantanamo Bay, Nepal, India (Naxalites problems, insurgency in North-Eastern states)
Challenge of expansion	Ghana, International organisations, Saudi Arabia, Yugoslavia, Sri Lanka, Northern Ireland, India (More power to local governments)
Challenge of deepening	Mexico, South Africa, Belgium, US, Civil Rights, Bolivia, India (corruption, less public participation)

Question 5.
Now let us think only about India. Think of all the challenges that democracy faces in contemporary India. List those five that should be addressed first of all. The listing should be in order of priority, i.e., the challenge you find most important or pressing should be mentioned at number 1, and so on. Give one example of that challenge and your reasons for assigning it the priority.
Answer:

Priority	Challenges to democracy	Example	Reasons for preference
1.	Challenge of deepening	Use of money muscle power Free and fair elections Decentralisation – more powers to local governments to increase public participation.	To make India democratic in practice at all levels – national, state and local.
2.	Challenge of expansion	5. Representation of women to elected bodies.	Women consist of half of the population. They should have adequate representation.

Question 6.
Here are some challenges that require political reforms. Discuss these chal­lenges in detail, study the reform options offered here and give your preferred solution with reasons. Remember that none of the options offered here is ‘right’ or ‘wrong’. You can opt for a mix of more than one option, or come up with something that is not offered here. But you must give your solution’in details and offer reasons for your choice.

Doctors absenteeism	Political funding
Challenge: Uttar Pradesh government got a survey done and found out that most of the doctors posted in the rural primary health centres are simply not there. They live in a town, carry out private practice and visit the village where they are posted only once or twice in the entire month. Villagers have to travel to towns and pay very high fee to private doctors even for common ailments.	Challenge:    On an average, every candidate who contested the last Lok Sabha elections owned a property of more than ? 1 crore. There is a fear that only wealthy people or those with their support can afford to fight elections. Most of the political parties are dependent on money given by big business houses. The worry is that the role of money in politics will reduce whatever little voice the poor have in our democracy.
Reform proposals: Tire government should make it compulsory for the doctors to live in the village where they are posted, otherwise their service should be terminated. They should be given some monetary incentives.	Reform proposals: The financial accounts of every political party should be made public. These accounts should be examined by government auditors
District administration and police should carry out surprise checks to ensure the attendance of the doctors.	There should be state funding of elections. Parties should be given some money by the government to meet their election expenditure.
Village panchayat, should be given the power to write the annual report of the doctor which should be read out in the gram sabha meeting.	Citizens should be encouraged to give more donations to parties and to political workers. Such donations should be exempt from income tax.
Problems like this can be solved only if Uttar Pradesh is split into several smaller states which can be administered more efficiently.

Answer:
(1) Doctor’s absenteeism :

• Doctor’s absenteeism can be controlled by making it compulsory for the doctors to live in the village of their posting,
• Secondly, they should be debarred from doing private practice.
• They should be given some monetary incentives.
• These rules should be part of their terms and conditions for appointment.
• These conditions must be strictly followed,
• Surprise checks by the department should be made.
• The erring doctors should be punished by the department.

(2) Political funding :

1. Role of money in elections in India has been increasing for the last few years,
2. It is necessary to curb the role of money to make democracy more successful.
3. The first, two proposals e., auditing of accounts of the political parties and state funding are good because these reforms will bring transparency in the finances of the political parties.
4. Secondly, state funding will reduce the expenditure on elections. Role of money will be less and the poor candidates may also be able to fight elections.

Question 7.
Write your own definition of good democracy and its features.
Answer:

1. Definition of good democracy : A government that is run and elected by the people to look after the interests of all people – rich and poor, rural, adivasis and others without any discrimination. Basic needs of all should be-fulfilled.
2. Features :
   1. Government elected by the people.
   2. Run by the people directly or indirectly with maximum public participation at all levels.
   3. Public opinion should be respected as was done in Ram Rajya
   4. There should be no poverty, illiteracy, social divisions, unemployment in the state
   5. Citizens should be satisfied in different spheres of life. There should be no tension in the society.

We hope the NCERT Solutions for Class 10 Social Science Civics Chapter 8 Challenges to Democracy help you. If you have any query regarding NCERT Solutions for Class 10 Social Science Civics Chapter 8 Challenges to Democracy, drop a comment below and we will get back to you at the earliest.

NCERT Exemplar Solutions for Class 12 Biology chapter 8 Human Health and Diseases

These Solutions are part of NCERT Exemplar Solutions for Class 12 Biology. Here we have given NCERT Exemplar Solutions for Class 12 Biology chapter 8 Human Health and Diseases

Multiple Choice Questions

Question 1.
The term ‘Health’ is defined in many ways. The most accurate definition of the health would be
(a) health is the state of body and mind in a balanced condition
(b) health is the reflection of a smiling face
(c) health is a state of complete physical, mental and social well-being
(d) health is the symbol of economic prosperity.
Answer:
(c) : Health could be defined as a state of complete physical, mental and social well¬being. When people are healthy, they are more efficient at work. This increases productivity and brings economic prosperity. Health also increases longevity of people and reduces infant and maternal mortality.

Question 2.
The organisms which cause diseases in plants and animals are called
(a) pathogens
(b) vectors
(c) insects
(d) worms.
Answer:
(a) : A wide range of organisms belonging to bacteria, viruses, fungi, protozoans, helminths, etc. could cause diseases in pants and animals including humans. Such disease causing organisms are called pathogens. Most parasites are therefore pathogens as they cause harm to the host by Jiving in (or on) them. The pathogens can enter our body by various means, multiply and interfere with normal vital activities, resulting in morphological and functional damage.

Question 3.
The chemical test that is used for diagnosis of typhoid is
(a) ELISA-Test
(b) ESR – Test
(c) PCR – Test
(d) Widal-Test.
Answer:
(d) :Salmonella typhi is a pathogenic bacterium which causes typhoid fever in human beings. These pathogens generally enter the small intestine through food and water contaminated with them and migrate to other organs through blood. Sustained high fever (39 to 40°C), weakness, stomach pain, constipation, headache and loss of appetite are some of the common symptoms of the disease. Intestinal perforation and death many occur in severe cases. Typhoid fever could be confirmed by Widal test. Widal test is an agglutination test for the presence of antibodies against the Salmonella organisms that cause typhoid fever.

Question 4.
Diseases are broadly grouped into infectious and non-infectious diseases. In the list given below, identify the infectious diseases.
(1) Cancer
(2) Influenza
(3) Allergy
(4) Small pox
(a) (i) and (ii)
(b) (ii) and (iii)
(c) (iii) and (iv)
(d) (ii) and (iv)
Answer:
(d) : Diseases can be broadly grouped into infectious and non-infectious. Diseases which are easily transmitted from one person to another, are called infectious diseases. Influenza and small pox are infectious diseases which are spreaded by direct contact, inhalation and droplet infections. Cancer and allergy are non-communicable diseases, as these diseases remain confined to the person who suffer from them. They are not transmitted from infected person to a healthy person.

Question 5.
The sporozoites that cause infection, when a female Anopheles mosquito bites a person, are formed in
(a) liver of the person
(b) RBCs of mosquito
(c) salivary glands of mosquito
(d) intestine of mosquito.
Answer:
(d) : When a female Anopheles mosquito bites an infected person, gametocyte of Plasmodium enter the mosquito’s gut along with blood meal and fuse to form zygote which elongates and becomes motile, called ookinets. Ookinetes bore through the wall of gut and change to oocysts. Inside oocyst, sporozoites are formed. Mature infective stage(sporozoite) are released in the body cavity of mosquito and migrate to salivary gland of mosquito. When these mosquitoes bite a human, the sporozoites are introduced into his/her body, thereby initiating the disease in the host’s body.

Question 6.
The disease chikungunya is transmitted by
(a) houseflies
(b) Aedes mosquitoes
(c) cockroach
(d) female Anopheles.
Answer:
(b) : Chikungunya is caused by chikungunya virus. The virus was first isolated from human patients and Aedes aegypti mosquitoes from Tanzania in 1952. The disease is transmitted by the bite of Aedes aegypti mosquito.

Question 7.
Many diseases can be diagnosed by observing the symptoms in the patient. Which group of symptoms are indicative of pneumonia?
(a) Difficulty in respiration, fever, chills, cough, headache.
(b) Constipation, abdominal pain, cramps, blood clots.
(c) Nasal congestion and discharge, cough, sore throat, headache.
(d) High fever, weakness, stomach pain, loss of appetite and constipation.
Answer:
(a) : Bacteria like Streptococcus pneumoniae and Haemophilus influenzae are responsible for the disease pneumonia in humans which infects the alveoli (air filled sacs) of the lungs. As a result of the infection, the alveoli get filled with fluid leading to severe problems in respiration. The symptoms of pneumonia include fever, chills, cough and headache.

Question 8.
The genes causing cancer are
(a) structural genes
(b) expressor genes
(c) oncogenes
(d) regulatory genes.
Answer:
(c) : Cancer is one of the most serious medical problems caused by the abnormal cell growth and proliferation due to the loss of regulation. Cancer-causing genes are oncogenes. Oncogene results from mutation of a normal gene. It is capable of both initiation and continuation of malignant transformation of normal cells.

Question 9.
In malignant tumours, the cells proliferate, grow rapidly and move to other parts of the body to form new tumours. This stage of disease is called
(a) metagenesis
(b) metastasis
(c) teratogenesis
(d) mitosis.
Answer:
(b) : There are two major types of tumours with respect to overall form or growth pattern. If the tumour cells remain in place to form a compact mass, the tumour is benign. In contrast, cells from malignant or cancerous tumours can actively spread, throughout the body in a process known as metastasis, often by floating in the blood and establishing secondary tumours.

Question 10.
When an apparently healthy person is diagnosed as unhealthy by a psychiatrist, the reason could be that
(a) the patient was not efficient at his work
(b) the patient was not economically pros-perous
(c) the patient shows behavioural and social maladjustment
(d) he does not take interest in sports.
Answer:
(c) : If a patient shows behavioural and social maladjustment then, he cannot be considered as healthy. Health is the condition in which the organism (and all of its parts) performs its vital functions normally. It is a state of physical, social and mental well-being and not merely the absence of disease.

Question 11.
Which of the following are the reason(s) for rheumatoid arthritis? Choose the correct option.
(i) The ability to differentiate pathogens or foreign molecules from self cells increases.
(ii) Body attacks self cells.
(iii) More antibodies are produced in the body.
(iv) The ability to differentiate pathogens or foreign molecules from self cells is lost.
(a) (i)and(ii)
(b) (ii) and (iv)
(c) (iii) and (iv)
(iv) (d) (i) and (iii)
Answer:
(b) : Rheumatoid arthritis is a disorder of the body’s defence mechanisms in which an immune response is elicited against its own tissues, which are thereby damaged or destroyed i.e., autoimmune disease.
Rheumatoid arthritis is the second most common form of arthritis (after osteoarthritis). It typically involves the joints of the fingers,
wrists, feet, and ankles, with later involvement of the hips, knees, shoulders, and neck. It is a disease of the synovial lining of joints; the joints are initially painful, swollen, and stiff and are usually affected symmetrically. As the disease progresses the ligaments supporting the joints are damaged and there is erosion of the bone, leading to deformity of the joints. Tendon sheath can be affected, leading to tendon rupture.

Question 12.
AIDS is caused by HIV. Among the following, which one is not a mode of transmission of HIV?
(a) Transfusion of contaminated blood.
(b) Sharing the infected needles.
(c) Shaking hands with infected persons.
(d) Sexual contact with infected persons.
Answer:
(c) : AIDS is a result of an infection by the human immunodeficiency virus (HIV) :
HIV can be transmitted by the following ways.

• Transfusion of infected blood.
• Use of contaminated needles and syringes to inject drugs or vaccines.
• Use of contaminated razors.
• Use of contaminated needles for boring pinnae.
• Sexual intercourse with an infected partner without a condom.
• From infected mother to child through placenta.
• Artificial insemination.
• Organ transplantation.

Question 13.
‘Smack’is a drug obtained from the
(a) latex of Papaver somniferum
(b) leaves of Cannabis sativa
(c) flowers of Datura
(d) fruits of Erythroxylum coca.
Answer:
(a) : Heroin commonly called smack is chemically diacetylmorphine which is a white, odourless, bitter crystalline compound. This is obtained by acetylation of morphine, which is extracted from the latex of poppy plant (Papaver somniferum).

Question 14.
The substance produced by a cell in viral infection that can protect other cells from further infection is
(a) serotonin
(b) colostrum
(c) interferon
(d) histamine.
Answer:
(c) : Interferons are glycoproteins released by living cells in response to viral attack, and make the surrounding cells resistant to viral infection by inhibiting multiplication of viral particles.
Interferons are divided into three groups based on the cell of origin, namely leucocyte (alpha interferon), fibroblast (beta interferon) and lymphocyte (gamma interferon). They are cytokine (chemical mediators) barriers.

Question 15.
Transplantation of tissues/organs to save certain patients often fails due to rejection of such tissues/organs by the patient. Which type of immune response is responsible for such rejections?
(a) Auto immune response
(b) Humoral immune response
(c) Physiological immune response
(d) Cell-mediated immune response
Answer:
(d) : Cell mediated immune response is responsible for graft rejection. Tissue matching and blood group matching are essential before undertaking any graft/transplant and even after this the patient has to take immuno¬suppressants throughout his/her life as body is able to differentiate ‘self’ from ‘non-self’.

Question 16.
Antibodies present in colostrum which protect the new born from certain diseases is of
(a) IgGtype
(b) IgA type
(c) IgDtype
(d) IgEtype.
Answer:
(b) : IgA antibody is the major antibody present in the colostrum. It protects the infant from inhaled and ingested pathogens. It is the second most abundant class of immunoglobulins, constituting about 13% of serum immunoglobulins.

Question 17.
Tobacco consumption is known to stimulate secretion of adrenaline and nor-adrenaline. The component causing this could be
(a) nicotine
(b) tannic acid
(c) curaimin
(d) catechin.
Answer:
(a) : Tobacco contains a large number of chemical substances including nicotine, an alkaloid. Nicotine stimulates adrenal gland to release adrenaline and nor-adrenaline into blood circulation, both of which raise blood pressure and increase heart rate.

Question 18.
Anti venom against snake poison contains
(a) antigens
(b) antigen-antibody complexes
(c) antibodies
(d) enzymes.
Answer:
(c) : In case of snakebite, the injection which is given to a patient, contains preformed antibodies against the snakevenom. This type of immunisation is called passive immunisation.

Question 19.
Which of the following is not a lymphoid tissue?
(a) Spleen
(b) Tonsils
(c) Pancreas
(d) Thymus
Answer:
(c) : Lymphoid organs are the organs where origin and/or maturation
and proliferation of lymphocytes occur. The primary lymphoid organs are bone marrow and thymus, where immature lymphocytes differentiate into antigen – sensitive lymphocytes. After maturation the lymphocytes migrate to secondary lymphoid organs like spleen, lymph nodes, tonsils, Peyer’s patches of small intestine and mucosa associated lymphoid tissue (MALT). The secondary lymphoid organs provide the sites for interaction of lymphocytes with the antigen, which then proliferate to become effector cells. Pancreas is not a lymphoid tissue.

Question 20.
Which of the following glands is large sized at birth but reduces in size with ageing?
(a) Pineal
(b) Pituitary
(c) Thymus
(d) Thyroid _____
Answer:
(c) : The thymus is a lobed organ located near the heart and beneath the breastbone. The thymus is quite large at the time of birth but keeps reducing in size with age and by the time puberty is attained, it reduces to a very small size.

Question 21.
Haemozoin is a
(a) precursor of haemoglobin
(b) toxin released from Streptococcus infected cells.
(c) toxin released from Plasmodium infected cells.
(d) toxin released from Haemophilus infected ‘cells.
Answer:
(c) : Plasmodium enters the human body as sporozoites (infectious form) through the bite of infected female Anopheles mosquito. The parasites initially multiply within the liver cells and then attack the red blood cell (RBCs) resulting in their rupture. The rupture of RBCs is associated with release of a toxic substance, haemozoin, which is responsible for the chill and high fever recurring every three to four days.

Question 22.
One of the following is not the causal organism for ringworm.
(a) Microsporum
(b) Trichophyton
(c) Epidermophyton
(d) Macrosporum
Answer:
(d) : Many fungi belonging to the genera Microsporum, Trichophyton and Epidermophyton are responsible for ringworms which is one of the most common infectious diseases in man. Appearance of dry, scaly lesions on various parts of the body such as skin, nails and scalp are the main symptoms of the disease.

Question 23.
A person with sickle cell anaemia is
(a) more prone to malaria
(b) more prone to typhoid
(c) less prone to malaria
(d) less prone to typhoid.
Answer:
(c) : Sickle cell trait protects against malaria. Malarial parasite is unable to penetrate the erythrocyte membrane of sickle shaped erythrocytes, reducing the number of parasites that actually infect the host thus conferring some protection against the disease.

Very Short Answer Type Questions

Question 1.
Certain pathogens are tissue/organ specific. Justify the statement with suitable examples.
Answer:
Some pathogens attack specific organs or tissues, such as :

1. Vibrio cholerae – It is a bacterium which causes cholera. It attacks digestive tracts and intestine and spreads through contaminated food and drinks.
2. Streptococcus pneumoniae – It is a bacterium which causes pneumonia. It is a serious disease of lungs characterised by
   accumulation of mucus/fluid in alveoli and bronchioles hence, breathing becomes difficult.

Question 2.
The immune system of a person is suppressed. In the ELISA test, he was found positive to a pathogen.
(a) Name the disease the patient is suffering from.
(b) What is the causative organism?
(c) Which cells of body are affected by the pathogen?
Answer:
(a) The patient must be suffering from Acquired Immune Deficiency Syndrome (AIDS). .
(b) The causative agent is Human Immuno-deficiency Virus (HIV).
(c) Virus affects helper T cells of the body.

Question 3.
Where are B-cells and T-cells formed? How do they differ from each other?
Answer:
B-cells and T-cells are lymphocytes which are responsible for immune response in the body. Both B-cells and T-cells are formed in bone marrow.
Differences between B-cells and T-cells are

B-cells	T-cells
(1) Bone marrow is the site of both formation and maturation.	They are produced in bone marrow but mature in thymus.
(2) B-cells are responsible for humoral or antibody mediated immunity.	T-cells are responsible for cell mediated immunity.

Question 4.
Given below are the pairs of pathogens and the diseases caused by them. Which out of these is not a matching pair and why?
(a) Virus common cold
(b) Salmonella typhoid
(c) Microsporum filariasis
(d) Plasmodium malaria
Answer:
(c) : Microsporum is a fungus and causes ringworm. It attacks the hair and skin.Filariasis is caused by Wuchereria bancrofti, a helmirith, which results in swelling of limbs, scrotal sacs and breasts.

Question 5.
What would happen to immune system, if thymus gland is removed from the body of a person?
Answer:
T-lymphocytes get differentiated in thymus gland and are responsible for cell mediated immunity (CMI). If thymus gland is removed, then CMI will weaken and person will become more susceptible to infectious diseases.

Question 6.
Many microbial pathogens enter the gut of humans along with food. What are the preventive barriers to protect the body from such pathogens? What type of immunity do you observe in this case?
Answer:
Physiological barriers of innate immunity protect the body against the pathogens which enter the gut along with food. These barriers are as follows:

1. Saliva contains lysozyme which kills the microorganisms that are not the normal inhabitants of the buccal cavity and come with food and drinks.
2. Bile, a bitter alkaline secretion of the liver, checks the growth of foreign bacteria on semidigested food, the chyme, in the intestine.
3. Acidity of gastric juice kills most of the microorganisms entering the body through digestive tract.

Question 7.
Why is mother’s milk considered the most appropriate food for a new born infant?
Answer:
Mother’s milk (colostrum) contains all the necessary nutrients needed by the newborn and has abundant antibodies (IgA) which provide passive immunity and protect the infant for few months after birth.

Question 8.
What are interferons? How do interferons check infection of new cells?
Answer:
Interferons are cytokine barriers of innate immunity. These are proteins secreted by virus infected cells which protect non- infected cells from further viral infections.

Question 9.
In the figure, structure of an antibody molecule is shown. Name the parts A, B and C

Answer:
A – Variable region of heayy chain
B – Constant region of light chain .
C – Disulphide bond

Question 10.
If a regular dose of drug or alcohol is not provided to an addicted person, he shows some withdrawal symptoms. List any four such withdrawal symptoms.
Answer:
Withdrawal symptoms shown by an addicted person when not provided with drug or alcohol includes anxiety, nausea, muscle twitch, sweating, nervousness and epilepsy etc.

Question 11.
Why is it that during changing weather, one is advised to avoid closed, crowded and airconditioned places like cinema halls etc.?
Answer:
It is advised to avoid overcrowded places during changing weather because chances of getting infections are very high during this time. This is because changing seasons are the time when infectious agents are more prevalent as moist condition favours pathogen growth. Moreover, people are more vulnerable as their body system is busy in adapting to the changing environmental conditions.

Question 12.
The harmful allele of sickle cell anaemia has not been eliminated from human population. Such afflicted people derive some other benefit. Discuss.
Answer:
Sickle cell anaemia is caused due to a recessive allele in haemoglobin. It is lethal
in homozygous (HbsHbs) condition and the person dies in early age but in heterozygous condition (HbAHbs) person is carrier of disease. Despite having harmful effect, the allele for sickle-cell anaemia continues to persist in human population because it has survival value in malaria infested areas like tropical Africa. Heterozygous individuals show resistance to malarial infection, because malarial parasite is unable to penetrate the membrane of sickle shaped erythrocytes.

Question 13.
Lymph nodes are secondary lymphoid organs. Explain the role of lymph nodes in our immune response.
Answer:
The lymph nodes are small solid structures located at different points along the lymphatic system. Lymph nodes serve to trap the microorganisms or other antigens, which happen to get into the lymph and tissue fluid. Antigens trapped in the lymph nodes are responsible for the activation of lymphocytes present there and cause the immune response.

Question 14.
Why is an antibody molecule represented as H2L2?
Answer:
Antibody molecule is made up of four polypeptide chains-2 Heavy (H) and 2 Light (L), therefore it is represented as H:L2.

Question 15.
What does the term ‘memory’ of the immune system mean?
Answer:
Memory response is a feature of acquired immunity. It develops during first encounter between specific foreign agent and body’s immune system. Second encounter with the same pathogen generates quicker and heightened immune response due to activated memory cells.

Question 16.
lf a patient is advised anti retroviral therapy, which infection is he suffering from? Name the causative organism.
Answer:
The patient must be suffering from Required Immune Deficiency Syndrome (AIDS) caused by Human Immunodeficiency Virus (HIV), which is a member of retrovirus group

short Answer Type Questions

Question 1.
Differentiate between active immunity and passive immunity.
Answer:
Differences between active immunity and passive immunity are as follows:

Active immunity	Passive immunity
(1) It develops in response to vaccine or infection.	It develops when readymade antibodies are injected from outside.
(2) It is long lasting.	It lasts for short period only.
(3) Antibodies are produced within the body.	Antibodies are injected from outside.
(4) It has no side effects.	It may cause reaction.
(5) Immunity is not immediate. A time lapse occurs for its development.	Immunity develops immediately

Question 2.
Differentiate between benign tumour and malignant tumour.
Answer:
Difference between benign tumour and malignant tumour are as follows:

Benign tumour	Malignant
(1) It is non-cancerous tumour	It is cancerous tumour.
(2) It does not show metastasis and remains confined to the affected organ.	It shows metastasis and spreads to other organs of the body.
(3) Rate of growth is slow. It shows definite growth.	Rate of growth is rapid. It shows indefinite growth.

Question 3.
Do you consider passive smoking is more dangerous than active smoking? Why?
Answer:
Passive smoking is inhalation of air containing tobacco smoke exhaled by an active smoker. Passive smoker is at higher risk as compared to an active smoker, because active smoker might inhale 10% of the smoke, but passive smoker inhales 90% of the smoke.

Question 4.
“Prevention is better than cure”. Comment.
Answer:
Prevention is always better than cure.This is because there are some diseases which have detrimental effects on our body and lives. Certain diseases like cancer and AIDS etc., are fatal. They can be avoided if our health systems focus on prevention of such diseases rather than concentrating on treatment methods. Many people can be saved from contracting diseases. Similar conditions apply for communicable, infectious diseases. If preventive measures are undertaken, their spread, can be checked and controlled easily.

Question 5.
Explain any three preventive measures to control microbial infections.
Answer:
Maintenance of personal and public hygiene is important to control microbial infections. Some of the measures that can be undertaken are as follows:

1. Eradication of vectors and destroying their breeding sites
2. Use of mosquito nets and repellents.
3. Proper disposal of waste.
4. Periodic cleaning of water reservoirs.
5. Checking water stagnation and garbage accumulation.
6. Implementation of vaccination and immunisation programmes for diseases.

Question 6.
In the given flow diagram, the replication of retrovirus in a host is shown. Observe and answer the following questions.
a. Fill in (1) and (2).
b. Why is the virus called retrovirus?
c. Can the infected cell survive while viruses are being replicated and released?

Answer:
(a)
(1) – Viral DN A is produced by reverse transcriptase.
(2) – New viral RNA is produced by the infected cell.
(b) HIV is called retrovirus because viral DNA is produced from viral RNA by reverse transcription.
(c) Infected cell survives but T-lymphocytes decrease in number due to replication and release of virus.

Question 7.
“Maintenance of personal and public hygiene is necessary for prevention and control of many infectious diseases”. Justify the statement giving suitable examples.
Answer:
Maintenance of personal and public hygiene is very important for prevention and control of many infectious diseases because diseases spread through contaminated food, water, air and through vectors. Measures for personal hygiene include keeping the body clean, consumption of clean drinking water, food, etc.
Public hygiene includes proper disposal of waste and excreta, periodic cleaning and disinfection of water reservoirs, pools, and tanks and observing standard practices of hygiene in public catering.
In case of air-borne diseases such as pneumonia and common cold, close contact with the infected persons or their belongings should be avoided. Vector borne diseases can be controlled by eliminating the vectors and their breeding places. This can be achieved by avoiding stagnation of water in and around residential areas, regular cleaning of household coolers, use of mosquito nets, etc.

Question 8.
The following table shows certain diseases, their causative organisms and symptoms. Fill the gaps.”

Name of the Disease	Causative organism	Symptoms
(1) Ascariasis	Ascaris	—
(2) —	Trichophyton	Appearance of dry, scaly lesions on various parts of the body
(3) Typhoid		High fever, weakness, headache, stomach pain, constipation
(4) Pneumonia	Streptococcus pneumoniae	—
(5)	Rhino viruses	Nasal congestion and discharge, sorethroat, cough, headache
(6) Filariasis	—	Inflammation in lower limbs

Answer:
(i) Abdominal pain, indigestion, nausea, vomiting etc.
(ii) Ringworm
(iii) Salmonella typhi
(iv) Difficulty in breathing, fever, cough, headache, chills, lips and finger nails may turn grey to bluish in colour
(v) Common cold
(vi) Wuchereria bancrofti

Question 9.
The outline structure of a drug is given below.
a. Which group of drugs does this represent?
b. What are the modes of consumption of these drugs?
c. Name the organ of the body which is affected by consumption of these drugs.

Answer:
(a) Cannabinoids
(b) Cannabinoids are generally taken by inhalation and oral ingestion.
(c) They affect cardiovascular system of the body.

Question 10.
Give the full form of CT and MRI. How are they different from each other?
Answer:
CT (Computed Tomography) is recon-struction of three dimensional image made by X rays directly on a computer instead of a photographic film. It is radiological invasive technique. It is used to diagnose diseases of brain, spinal cord, chest and abdomen and detection of tumours. MRI (Magnetic Resonance Imaging) is a non invasive, non radiological technique in which the image is reconstructed on computer by detecting MRI signals generated by nuclei of hydrogen atoms in a magnetic field using non-ionising radiations. It is used to obtain multiplanar imaging of soft living tissues like in case of tumours, muscular disorders, cardiovascular disorders, haemorrhage with superior resolution.

Question 11.
Many secondary metabolites of plants have medicinal properties. It is their misuse that creates problems. Justify the statement with an example.
Answer:
Drugs like barbiturates, amphetamines, benzodiazepines, lysergic acid diethylamides (LSD) and other similar drugs, that are normally used as medicines to help patients coping with mental illnesses like depression and insomnia are secondary metabolites of plants. Misuse of plant metabolites can impair one’s .physical, physiological or functional behaviour creating problem for the society. For example, Cocaine is natural alkaloid obtained from leaves of plant Erythroxylum coca, native to South America. Cocaine has vasoconstrictor properties and therefore, is a good local anaesthetic. It is chewed, eaten or sniffed in its powdered form or taken as a drink. It is a powerful CNS stimulant (interferes with the transport of the neurotransmitter dopamine) and induces a sense of well being and pleasure and delays fatigue. Its overdose causes hallucinations, headache, insomnia, convulsions, etc.

Question 12.
Why cannabinoids are banned in sports and games?
Answer:
Cannabinoids may enhance the performance of some athletes. Smoking cannabis may decrease anxiety, fear, depression, tension, etc., and improve self confidence, relaxation, well-being and sleep, etc. These effects may allow the athletes to perform better, especially under pressure and may reduce stress, experienced before and during competition. The increase in ‘risk taking’ associated with cannabinoids may possibly improve training and performance giving athlete a competitive edge. Moreover, cannabinoids increase bronchodilation (which may improve oxygenation of tissue) and have analgesic (pain relieving) properties which could permit athletes to work during injury. Moreover, long term use of cannabinoids causes various hazards to the health of abuser. Therefore, use of cannabinoids by sportsperson is not justified, as it will lead to unfair competition. Hence, their use is banned in sports and games.

Question 13.
What is secondary metabolism?
Answer:
Secondary metabolism consists of metabolic pathways and products of metabolism that seem to have no direct function in growth and development of plants and are not absolutely required for their survival. Secondary metabolism is responsible for many bioactive compounds used medicinally.

Question 14.
Drugs and alcohol give short-term ‘high’ and long-term’damages’. Discuss.
Answer:
Curiosity, need for adventure and excitement, peer pressure, etc., constitute some common causes which motivate youngsters towards drug and alcohol abuse. Short term effects of drugs/alcohols are generally pleasing and relaxing which habituate the abuser for using these again and again. But long term effects of drugs or alcohol are devastating and detrimental. For example, alcohol when taken occasionally in small quantities, acts as sedative, analgesic and anaesthetic. It stimulates the secretion of gastric juices. But, if a person becomes addicted to it, he consumes high concentration alcohol frequently. Heavy drinking causes depression. Suicide attempt is much common in alcoholics than in the rest of society.

Sexual relationship is usually deteriorated because of erectile dysfunction or rejection by the partner.
Caffeine is a mild stimulant and taken as beverages- tea, coffee, cocoa and cola drinks. Caffeine is CNS stimulant and thus increases alertness and thought. It also acts as a cardiac and respiratory stimulant. It is a mild diuretic and also increases basal metabolic rate. Excessive use may cause anxiety, irritability, diarrhoea, irregular heart beat (arrhythmia) and decreased concentration. It also causes indigestion and disturbs pancreatic and renal functions. If taken empty stomach, it can cause stomach ulcers.

Question 15.
Diseases like dysentery, cholera, typhoid etc., are more common in over crowded human settlements. Why?
Answer:
Overcrowded areas are generally unhygienic. This is because public hygeine is not maintained properly. Water bodies in these areas are generally contaminated with disease causing microbes. Water stagnation and garbage accumulation is often not checked.
Diseases like dysentery, cholera, typhoid etc., are water borne infectious diseases and may easily spread in overcrowded human dwellings.

Question 16.
From which plant cannabinoids are obtained? Name any two cannabinoids. Which part of the body is effected by consuming these substances?
Answer:
Cannabinoids are obtained from leaves, resins and flowers of hemp plants e.g., Cannabis sativa. Examples of cannabinoids are bhang, ganga, charas etc. They interact with receptors present in the brain and affect cardiovascular system of the body.

Question 17.
In the metropolitan cities of India, many children are suffering from allergy/asthma. What are the main causes of this problem. Give some symptoms of allergic reactions.
Answer:
Children living in metropolitan cities are more prone to allergies/asthma, mainly due to life style and increased level of pollution. Suspended particulate matter released by vehicles add more to the increasing pollution levels. Moreover children living in metropolitan cities are overprotected in their early childhood days and are more sensitive to allergens.

Allergy is hypersensitive response of the body against foreign particles (allergens). Common symptoms are sneezing, watery eyes, running nose, difficulty in breathing, irritation of throat, trachea and skin etc. Asthma related problems worsen due to changing weather conditions and increased pollution levels.

Question 18.
What is the basic principle of vaccination?How do vaccines prevent microbial infections? Name the organism from which hepatitis B vaccine is produced.
Answer:
Vaccine is a preparation of antigenic proteins or inactivated/live but weakened germs of a disease which on inoculation into a healthy person provides temporary or permanent, active immunity by inducing antibody formation.

Vaccination is based on the property of ‘memory’ of the immune system. When a vaccine is injected, it generates primary immune response of low intensity and also produces B cells and T cells. When vaccinated person is again attacked by same pathogen, the existing memory T and B cells recognise the antigen quickly and attack the invaders with a massive production of lymphocytes and antibodies. Hepatitis B vaccine is second generation vaccine that has been produced from yeast by recombinant DNA technology.

Question 19.
What is cancer? How is a cancer cell different from the normal cell? How do normal cells attain cancerous nature?
Answer:
Cancer is an abnormal and uncontrolled division of cells, that invade and destroy surrounding tissues. Cancer cells differ from normal cells in the following ways:

1. Cancer cells multiply in an uncontrolled manner.
2. They do not exhibit property of contact inhibition.
3. They show metastasis.
4. They have indefinite life span.

Cancer causing agents known as carcinogens include physical, chemical and biological agents which activate the proto-oncogenes or cellular oncogenes to oncogenes and cause deviation in growth pattern leading to uncontrolled growth of cells. They alter the behaviour of normal cells and make them cancerous.

Question 20.
A person shows strong unusual hypersensitive reactions when exposed to certain substances present in the air. Identify the condition. Name the cells responsible for such reactions. What precaution should be taken to avoid such reactions.
Answer:
Hypersensitive reaction to foreign substances is known as allergy. Substances which cause allergic reactions are called allergens. The common allergens are dust, pollen, feathers, paint. In allergic chemicals called histamine and serotonin are released from mast cells. Allergy can be avoided by reducing an exposure to allergen and by taking drugs like antihistamines, adrenaline and steroids.

Question 21.
For an organ transplant, it is an advantage to have an identical twin. Why?
Answer:
For an organ transplantation, tissue matching(histocompatibility)andblood group compatibility of donor and recipient are very important. If they do not match, organ may be rejected. This is because, immune system recognises the protein in the transplanted tissue or organ as foreign and initiates cellular immunity. Chances of matching of tissue as well as blood group are very high if donor and recipient are identical twins, because of genetic similarity. Transplantation between identical twins is known as isograft.

Question 22.
What are lifestyle diseases? How are they caused? Name any two such diseases.
Answer:
Diseases which occur due to improper changes in the life style are known as lifestyle diseases. These diseases may be caused due to over-eating or crash dieting, lack of exercise, sedentary lifestyle, smoking, alcoholism, drug addiction etc. Hypertension and obesity are common lifestyle diseases.

Question 23.
If there are two pathogenic viruses, one with DNA and other with RNA, which would mutate faster? And why?
Answer:
Pathogenic virus with RNA as genetic material will mutate faster than the one with DNA as genetic material, because

1. RNA is chemically and structurally unstable. Uracil present in RNA is less stable than thymine in DNA.
2. RNA is highly reactive, labile and easily degradable.

Long Answer Type Questions

Question 1.
Represent schematically the life cycle of a malarial parasite.
Answer:

Question 2.
Compare the life style of people living in the urban areas with those of rural areas and briefly describe how the lifestyle affects their health.
Answer:
Lifestyle affects human health in various ways which are as follows:
(1) Food habits – People living in urban areas usually prefer fast food like pasta, pizza, burger, noodles etc., as compared to fresh, green leafy vegetables, fruits which are more consumed by villagers, because of this they suffer from obesity, hypertension and some cardiovascular problems.

(2) Residential areas — Due to lack of space in urban areas, residential areas are generally overcrowded. Sometimes there is no facility for proper disposal of garbage, no proper ventilation etc., therefore people become prone to food, water, and air borne diseases, like cholera, tuberculosis, typhoid, pneumonia etc. On the other hand, villages have clean air, human dwellings are not crowded and the environment is free of pollution. But due to lack of personal and public hygiene and malnutrition, villagers frequently suffer from various diseases.
Moreover, medical facilities in rural areas are not up to the mark. People of urban areas are exposed to good education, proper medical facilities etc.

(3) Exercise and sleep — Because of busy lifestyle, people in urban areas do not get enough rest and have no time for exercise. Whereas people in rural areas do a lot of physical work and have enough time to rest as well.

(4) Due to overprotected environment provided during early life, people in urban areas sometimes have low immunity and are susceptible to certain diseases and are affected by allergens. However, people in rural areas are exposed to harsh environmental conditions therefore develop better immunity.

Question 3.
Why do some adolescents start taking drugs? How can this be avoided?
Answer:
Reasons for drug and alcohol abuse among adolescents are :
(1) Curiosity : Adolescents want to have personal experience of smoking cigarette and taking alcohol and drugs.

(2) Adventure and excitement : A child may go in for use of drug, smoking and alcoholic drink for the sake of adventure and excitement.

(3) Family set up : In certain families, use of alcohol, tobacco, sleeping pills and pain killers are common. It induces the youngsters to taste the
same.

(4) Group or peer pressure : Friends and peer groups often motivate some adolescents to take drugs, alcohol as a defiance of authority and feeling of independence.

(5) Progressiveness : There is a false perception that taking of drugs, alcohol or smoking is a sign of progressiveness in society.

Various measures which help to avoid drug abuse are:

(1) Discipline : Good nurturance with consistent discipline but without suffocating strictness reduces the risk of addictions.

(2) Communication : The child must be able to communicate with the parents seeking clarification of all doubts and discussing problems that arise in studies or develop in the class, with friends, siblings and others.

(3) Independent working : Give responsibility to the child for small tasks and allow him/ her to perform independently. Of course, provide guidance where required.

(4) Education and counselling : Stresses, failures, disappointments and problems are part of life. A child has to be trained, educated and counselled to face them as and when they come.

(5) Looking for danger signs : Teachers and parents should always be careful to look for and identify danger signs that can indicate tendency to go in for addiction.

Question 4.
In your locality, if a person is addicted to alcohol, what kind of behavioural changes do you observe in that person? Suggest measures to overcome the problem.
Answer:
Behavioural changes which can be observed in an alcoholic are as follows:
(1) Reckless behaviour, vandalism and violence.
(2) Drop in academic performance and unexplained absence from school/ college.
(3) Lack of interest in personal hygiene, isolation, depression, fatigue, aggressive-ness.
(4) Change in eating and sleeping habits. Measures to overcome the problem of alcohol addiction are as follows:

• Seeking professional advice : Highly qualified psychologists, psychiatrists and de-addiction and rehabilitation programmes can help individuals who are suffering from alcohol abuse.
• Avoid undue peer pressure : Every person has his/her own choice and personality, which should be kept in mind. So he/ she should not be pressed unduly to do beyond his/her capacities, in work condition and other in social get together or activities.
• Education and counselling : Helps to overcome the problems, like stresses, disappointments and failure in life. One should utilise one’s energy in some beneficial activities like sports, music, reading, yoga and other extra curricular activities.
• Seeking help from parents and peers : In case of minors, whenever, there is any problem, one should seek help and guidance from parents and peers. Help should be taken from close and trusted friends. This would help young to share their feelings of anxiety and wrong doings.

Question 5.
What are the methods of cancer detection? Describe the common approaches for treatment of cancer. Answer:
Detection and diagnosis of cancer depends upon histological features of malignant structure. Few methods of cancer detection are as follows:
(1) Bone marrow biopsy and abnormal count of WBCs in leukemia.
(2) Biopsy of tissue, direct or through endoscopy. Pap test (cytological staining) is used for detecting cancer of cervix and other parts of genital tract.
(3) Techniques such as radiography (use of X-rays), CT Scan (computed tomography), MRI Scan (magnetic resonance imaging)
are very useful to detect cancers of the internal organs. Mammography is radiographic examination of breasts for possible cancer.
(4) Monoclonal antibodies coupled to appropriate radioisotopes can detect cancer specific antigens and hence cancer.
Treatment of cancer includes following :
(1) Chemotherapy : In chemotherapy a variety of anti-cancer drugs are used that produce more injury to cancer cells than to normal cells. These drugs interfere with the cell division and growth and affect both normal and cancerous cells. Vincristine and vinblastine from Catharanthus roseus (Vinca rosea) are effective in leukaemia control. Taxol is another anti-cancer drug obtained from Taxus baccata. Tetrathiomolybdate is the new anti-cancer drug. It arrests the tumour growth by starving cancer cells of copper.

(2) Radiotherapy : It is used in addition to chemotherapy. The basic priniciple there is to bombard cancer cells with rays that damage or destroy the ability of cancer cell to grow and divide by damaging the DNA within the tumour cells, but produce minimum damage to the surrounding normal tissue.

(3) Surgery : It is removal of the cancerous cells surgically and has only limited usefulness. In breast tumour and uterine tumour, it is most effective but other treatments are also given to kill any cells that may have been left.

(4) Immunotherapy : Immunotherapy is a form of treatment that enhances the body’s ability to recognise cancer cells and destroy them. It can be given intravenously or by subcutaneous injection.

(5) Blood and marrow transplant : High dose chemotherapy or radiation therapy can destroy bone marrow’s ability to make blood cells. A blood or marrow transplant can be used to replace marrow stem cells which produce blood cells.

Question 6.
Drugs like LSD, barbiturates, amphetamines, etc., are used as medicines to help patients with mental illness. However, excessive doses and abusive usage are harmful. Enumerate the major adverse effects of such drugs in humans.
Answer:
LSD is a psychedelic drug or hallucinogen which induces behavioural abnormalities, they cause optical or auditory hallucinations, horrible dreams, emotional outburst and severe damage to central nervous system.

Barbiturates are sedatives which depress CNS activity, give feeling of calmness, relaxation and drowsiness but high doses induces deep sleep.

Amphetamines are synthetic drugs and are CNS stimulants. They cause alertness, self-confidence, talkativeness, wakefulness and increased work capacity. High doses cause euphoria, marked excitement and sleeplessness which may lead to mental confusion. Their use may produce after effects like nausea and vomiting. Amphetamine is one of the drugs included in the ‘dope test’ for athletes.

Question 7.
What is Pulse Polio Programme of Government of India? What is OPV? Why is it that India is yet to eradicate Polio?
Answer:
Pulse polio programme is a programme that involves vaccination of children in the age group of 0-5 years with polio vaccine. It began in 1995 with the aim to make India a polio-free country. Oral Polio Vaccine (OPV) is a live-attenuated vaccine, produced by the passage of the virus through non-human cells at a sub-physiological temperature, which produces spontaneous mutations in the viral genome. OPV also proved to be superior in administration, eliminating the need for sterile syringes and making the vaccine more suitable for mass vaccination campaigns. OPV also provides long lasting immunity than the Salk vaccine. One dose of OPV produces immunity to all three poliovirus serotypes in approximately 50% of recipients. WHO declared India and the entire South-East Asia region polio free in March, 2014. Last case of polio in India was reported in January, 2011.

Question 8.
What are recombinant DNA vaccines? Give two examples of such vaccines. Discuss their advantages.
Answer:
Recombinant vaccines are vaccines having a gene encoding the protein of pathogen that causes immunogenic reactions in the host so as to produce antibody against the disease. These vaccines are produced from yeast through recombinant DNA technology. Examples are Hepatitis B vaccine and herpes virus vaccine.

Advantages – By rDNA technology the vaccines can be produced on larger scale so providing greater availabilitv for immunisation. These vaccines are highly specific, pure and elicit strong immune response.

We hope the NCERT Exemplar Solutions for Class 12 Biology chapter 8 Human Health and Diseases help you. If you have any query regarding.NCERT Exemplar Solutions for Class 12 Biology chapter 8 Human Health and Diseases, drop a comment below and we will get back to you at the earliest.

NCERT Solutions for Class 10 Social Science Economics Chapter 5 Consumer Rights

These Solutions are part of NCERT Solutions for Class 10 Social Science. Here we have given NCERT Solutions for Class 10 Social Science Economics Chapter 5 Consumer Rights.

TEXTBOOK EXERCISES

Question 1.
Why are rules and regulations required in the marketplace ? Illustrate with a few examples.
Answer:
Rules and regulations are required in the marketplace to protect consumers. Sellers often abdicate responsibility for a low-quality product, cheat in weighing out goods, add extra charges over the retail price, and sell adulterated/ defective goods. Hence, rules and regulations are needed to protect the scattered buyers from powerful and fewer producers who monopolize markets. For example, a grocery shop owner might sell expired products and then blame the customer for not checking the date of expiry before buying the items.

Question 2.
What factors gave birth to the consumer movement in India ? Trace its evolution.
Answer:
(1) The following factors gave birth to the consumer movement in India :

1. There was the dissatisfaction of the consumers as many sellers indulged in various unfair practices such as less weight and measurement, more prices, and defective goods.
2. There was no legal system or law available to consumers to protect themselves from unethical and unfair trade practices and exploitation by the sellers.
3. In case of any defective product or deficiency of services, it was considered to be the responsibility of the consumer to be careful while buying a commodity or service.

(2) Evolution :

1. It took many years for organisations in India and around the world, to create awareness amongst people. This has also shifted the responsibility of ensuring quality of goods and services on the sellers.
2. In India, the consumer movement as a ‘social force’ originated with the necessity of protecting and promoting the interests of consumers against unethical and unfair trade practices.
3. Rampant food shortages, hoarding, black-marketing, adulteration of food and edible oil were also responsible for the rise of consumer movement in an organised form in the 1960s.
4. Till 1970s, consumer organisations wrote articles and held exhibitions.
5. Consumer groups were formed to look into the malpractices in ration shops and over­crowding in the road passenger transport.
6. In 1986, the Government of India enacted Consumer Protection Act 1986, popularly known as COPRA to protect the interests of the consumers.

Question 3.
Explain the need for consumer consciousness by giving two examples.
Answer:
There is a need for consumer consciousness so that the buyers themselves can take action against cheating traders. The ISI and Agmark logos are certifications of good quality. Consumers must look for such certifications while buying goods and services. Secondly, to be able to discriminate and make informed choices, a consumer needs to have adequate knowledge of the goods or services purchased by him/her.

Question 4.
Mention a few factors which cause exploitation of consumers.
Answer:
A few factors which cause exploitation of consumers are as mentioned below :

1. Individual consumers often find themselves in a weak position whenever there is a complaint regarding a good or service that had been bought, the seller tries to shift all the responsibility on to the buyers.
2. Consumers purchase in small amounts and are scattered. They do not bother about products of small value even if they are cheated.
3. Consumers do not take receipt for products of small value. Sometimes the shopkeepers too don’t give receipt for products of small value due to rush of customers as they don’t find time to issue receipts to each and every customer.
4. Limited supplies : Limited supplies ‘Of goods and services are made to exploit the consumers who are compelled to pay more than the actual price.
5. Limited competition : When one or a group of producers control the production of any product, they exploit the consumers by manipulating the prices. For example in the housing sector, where the producers and consumers are few, the sellers exploit the consumers.
6. Low literacy : Illiteracy too leads to exploitation of the consumers because they are unable to have complete knowledge about the products.
7. Life of the people is so busy that they do not find time for such matters. People have become habitual and exploitation by traders does not make any difference in their lives unless they suffer a substantial loss.

Question 5.
What is the rationale behind the enactment of Consumer Protection Act, 1986?
Answer:
The rationale behind the enactment of Consumer Protection Act of 1986 is to protect the consumer against unethical and unfair trade practices. Also, it recognises the consumer’s right to be informed, right to choose, right to seek redressal and right to represent himself/herself in consumer courts.

Question 6.
Describe some of your duties as consumers if you visit a shopping complex in your locality.
Answer:
Some of our duties as consumers are as given below :

1. While purchasing the goods, consumers should look at the quality of the products as well as on the guarantee of the products and services.
2. Consumers should ask for cash memo for the products purchased.
3. Exercise your right to choose a product of your liking without any conditions.
4. Wherever possible, consumers should insist for the warrantee card.
5. Consumers should preferably purchase products with ISI, Agmark etc.
6. Consumers should ask for expiry date about any eatables/biscuits etc. before buying the product.
7. Consumer should ask about the MRP of the product and should not pay more than MRP. On the other hand, it is his duty to ask the shopkeeper to reduce the rate.

Question 7.
Suppose you buy a bottle of honey and a biscuit packet. Which logo or mark will you have to look for and why?
Answer:
Agmark, because it is meant for agricultural products.

Question 8.
What legal measures were taken by the government to empower the consum­ers in India ?
Answer:
Legal measures taken by the government to empower consumers in India are plenty. First and foremost being the COPRA in 1986. Then, in October 2005, the Right to Information Act was passed, ensuring citizens all information about the functioning of government departments. Also, under COPRA, a consumer can appeal in state and national courts, even if his case has been dismissed at the district level. Thus, consumers even have the right to represent themselves in consumer courts now.

Question 9.
Mention some of the rights of consumers and write a few sentences on each.
Answer:
Consumer right is the right to have information about the quality, potency, quantity, purity, price, and standard of goods or services as it may be the case, but the consumer is to be protected against any unfair practices of the trade. It is therefore very essential for the consumers to know their rights which are given below:
(1) Right to safety:

1. According to this right the consumers have the right to be protected against the marketing of goods and services which are hazardous to life and property.
2. This right is important for safe and secure life.
3. This right includes concern for consumer’s long term interest as well as for their present requirement.
4. Sometimes the manufacturing defects in pressure cookers, gas cylinders and other electrical appliances may cause loss to life, health and property of customers.
5. Thus right to safety protects the consumers from sale of such hazardous goods or services.

(2) Right to information :

1. According to this right the consumer has the right to get information about the quality, quantity, purity, standard^– and price of goods or services as to protect himself against the abusive and unfair practices.
2. The producer must supply all the relevant information at a suitable place preferably on the product itself.
3. Since October 2005, the Right to Information Act ensures its citizens all the information about the functions of government departments.

(3) Right to choice :

1. A consumer has the right to choose the goods or services of his/her likings.
2. The right to choice means consumer’s access to variety of goods and services at a competitive price.
3. A consumer cannot be forced to buy things he may not wish to and is left with no choice.

(4) Right to be heard or right to representation :

1. In case a consumer has been exploited or has any complaint against the product or service then he has the right to be heard and be assured that his/her interest would receive due consideration.
2. This right includes the right to representation in the government and in other policy making’
   
3. Under this right the companies must have complaint cells to attend the complaints of customers.

(5) Right to seek redressal :

1. A consumer has the right to get compensation or seek redressal against unfair trade practices or any other exploitation.
2. This right assures justice to consumers against exploitation.
3. The right includes compensation in the form of money or replacement of goods or repair of a defect in the goods as per the satisfaction of consumer.
4. Various redressal forum e., three-tier quasi-judicial bodies under the Consumer Protection Act 1986 has been set at district, state and national level.

(6) Right to consumer education :

1. It is right of the consumer to adquire the knowledge and skills to be informed so that even the illiterate consumer may seek information about the existing acts, and agencies.
2. The government of India has included consumer education in the school curriculum and in various university courses.
3. Government is also making use of media to make the consumers aware of their rights and make wise use of their money.

Question 10.
By what means can the consumers express their solidarity?
Answer:
Consumers can express their solidarity by forming consumer groups that write articles or hold exhibitions against traders’ exploitation. These groups guide individuals on how to approach a consumer court, and they even fight cases for consumers. Such groups receive financial aid from the government to create public awareness. Participation of one and all will further strengthen consumer solidarity.

Question 11..
Critically examine the progress of consumer movement in India.
Answer:

1. It was on 24 December 1986 that the Indian Parliament enacted the Consumer Protection Act. This day i.e., 24th December is celebrated as National Consumers’ Day in the country.
2. The consumer movement has made progress in terms of numbers of organised groups and activities. There are more than 700 consumer groups in the country. Out of these groups 20­25 are well organised and recognised for their work.
3. However, the progress of the consumer movement in India has been slow. Even after twenty seven years of the enactment of COPRA in 1986, consumers are being exploited by the shopkeepers or traders. Defective or substandard home appliances are sold in the market. Duplicate articles are also being sold. There is adulteration and impurity in edible items. Incomplete information is found printed on various products. LPG gas cylinders with less weight are supplied to the housewives.
   But the main reason for this state of affairs is that the consumers have not realised their role and importance. Most of the people do not make a complaint to redress their grievances. To make consumer movement effective, it is necessary that every complaint, even for a small amount of money, should be made. People should actively get involved in the movement to protect their interests.

Question 12.

Match the following :

Answer:
(i) (e), (ii) (c), (iii) (a), (iv) (b), (v) (f), (vi) (d).

Question 13.
Say True or False :

1. COPRA applies only to goods.
   False
2. India is one of the many countries in the world which has exclusive courts for consumer redressal.
   True    
3. When a consumer feels that he has been exploited, he must file a case in the District Consumer Court.
   True,
4. It is worthwhile to move to consumer courts only if the damages incurred are of high value.
   False
5. Hallmark is the certification maintained for the standardization of jewellery.
   True
6. The consumer redressal process is very simple and quick.
   True
7. A consumer has the right to get compensation depending on the degree of the damage.
   True

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