Category: CBSE Class 12

Here we are providing Class 12 Biology Important Extra Questions and Answers Chapter 15 Biodiversity and Conservation. Important Questions for Class 12 Biology are the best resource for students which helps in Class 12 board exams.

Class 12 Biology Chapter 15 Important Extra Questions Biodiversity and Conservation

Biodiversity and Conservation Important Extra Questions Very Short Answer Type

Question 1.
Define biodiversity.
Answer:
It refers to the totality of genes, species, and ecosystem of a region, e.g. Forest.

Question 2.
Is biodiversity the same from place to place?
Answer:
No, it differs from one place to another place.

Question 3.
What is the advantage of genetically uniform crop plants?
Answer:
The monoculture of crop plants will give a high yield.

Question 4.
What is the disadvantage of genetically uniform crop plants?
Answer:
Such crop plants are highly prone to diseases.

Question 5.
What is the total number of species discovered and described presently on earth? What is the predicted number?
Answer:
1.7 million and 50 million, respectively.

Question 6.
What are the characteristics that make a community stable?
Answer:
The stability of an ecosystem is controlled by:

1. Carrying capacity,
2. Recycling of wastes,
3. Density-related self-regulation and
4. Feedback system.

Question 7.
What accounts for the greater ecological diversity of India?
Answer:
India has high ecological diversity due to a variety of topography, soil types, climates, rainfall zones, sea coasts, islands, etc. Ten well-demarcated biogeographical zones with different biota occur in India.

Question 8.
What is the difference between endemic and exotic species?
Answer:
Endemic species belong to a local area and are of limited distribution due to biotic and abiotic regulations, e.g. Lion Tailed Macaque.

Exotic species enter from outside with under distribution due to non-control of biotic and abiotic factors. They face little resistance by endemic species, e.g. Eucalyptus.

Question 9.
How does species diversity differ from ecological diversity?
Answer:
Species diversity is the occurrence of variety and abundance of species in a community while ecological diversity is the occurrence of different ecosystems and communities in a geographical area.

Question 10.
Why is genetic variation important in the plant Rauwolfia vomitoria?
Answer:
It results in differences in the potency and concentration of drug reserpine in the plant Rauwolfia vomitoria found in different regions of the Himalayas.

Question 11.
What is the advantage to a species having more genetic diversity?
Answer:
It enables the species to adjust and adapt to changed environmental conditions.

Question 12.
What are the consequences of lower genetic diversity?
Answer:
Lower genetic diversity leads to uniformity.

Question 13.
Coin a term for the following:
(i) Within community diversity
Answer:
Alpha diversity.

(ii) Between community diversity.
Answer:
Beta diversity.

Question 14.
What is the approximate drop of temperature with a corresponding increase/decrease of about 1000 m in altitude?
Answer:
There is about a 6.5° drop in temperature.

Question 15.
What is the cause of mortality of ducks, swans, and cranes?
Answer:
Lead poisoning as they take in spent gunshots that fall into lakes and marshes.

Question 16.
List two uses of the Red List.
Answer:

1. Developing awareness about the importance of threatened biodiversity.
2. Identification and documentation of endangered species.

Question 17.
The red list contains an assessment of how many species?
Answer:
18,000 species out of which 11,000 are threatened.

Question 18.
What could have triggered mass extinctions of species in the past?
Answer:
Glaciation, melting of snow, the eruption of large volcanoes, earthquakes, movement of continents, large meteorites falling on the earth, drought, etc. could have triggered mass extinctions.

Question 19.
What does ‘red’ indicate in the IUCN red list (2004)?
Answer:
‘Red’ in IUCN red list (2004) indicates the threatened species, i.e. the species under varying degrees of extinction risk categories.

Question 20.
What is the importance of cryopreservation in the conservation of biodiversity? (CBSE (Delhi) 2015)
Answer:
It serves the purpose of ex situ conservation.

Question 21.
Name the type of biodiversity represented by the following:
(a) 50,000 different strains of rice in India.
Answer:
Genetic diversity

(b) Estuaries and alpine meadows in India. (CBSE (Delhi) 2013)
Answer:
Ecological diversity.

Question 22.
Write the basis on which an organism occupies a space in its community-natural surrounding. (CBSE Outside Delhi 2013)
Answer:
Alexander von Humboldt observed that within a region species richness increased with increasing explored area up to a limit.

Question 23.
Mention the kind of biodiversity of more than a thousand varieties of mangoes in India. How is it possible? (CBSE (Delhi) 2015)
Answer:

1. Genetic diversity
2. (a) It is the measure of variation in genetic information contained in the organism.
   (b) It enables a population to adapt to the environment.

Biodiversity and Conservation Important Extra Questions Short Answer Type

Question 1.
What will be the consequences of the loss of biodiversity?
Answer:
Consequences of loss of biodiversity:

1. It would check the evolutionary capability of biota to cope up with environmental changes.
2. It would result in the extinction of species.
3. As man is dependant on food and other necessities, its loss will be hard-pressed for mankind.

Question 2.
What are the causes of the loss of biodiversity?
Answer:
Biological diversity is lost before its size is known. Causes of the toss of biodiversity:

1. Increased human population.
2. Increased consumption of resources.
3. Pollution due to human activities.

Question 3.
How many genes are present in mycoplasma, E. coli, Drosophila, Oryza sativa, and Homo sapiens?
Answer:

Name of organism	Number of genes
1. Mycoplasma	450-700
2.E.coli	3,200
3. Drosophila	13,000
4. Oryza sativa	32,000-50,000
5. Homo sapiens	31,000.

Question 4.
List three levels of biodiversity.
Answer:
Biodiversity can be studied at the following levels:

1. Genetic diversity
2. Species diversity
3. Ecological/Ecosystem diversity.

Question 5.
What is the basis of speciation?
Answer:
Basis of speciation. The amount of genetic variation is the basis of speciation (evolution of new species). It has a key role in the maintenance of diversity at species and community levels. The total genetic diversity of a community will be greater if there are many species, as compared to a situation where there are only a few species.

Genetic diversity within a species often increases with environmental variability.

Question 6.
Write a note on ecological diversity.
Answer:
Ecological diversity. It is related to species diversity and genetic diversity. India has a greater ecosystem diversity than a Scandinavian country. India has several ecosystems/biomes like alpine meadows, rain forests, deserts, wetlands, mangroves, coral reefs, etc.

Question 7.
List the Natural World Heritage sites of India.
Answer:
Natural World Heritage sites of India:

Site	Location
Kaziranga National Park	Assam
Keoladeo Ghana National Park	Rajasthan
Manas Wildlife Sanctuary	Assam
Nanda Devi National Park	Punjab
Sundarban National Park	West Bengal.

Question 8.
Depict with the help of simple sketches the representation of global biodiversity of major taxa of plants, invertebrates, and vertebrates. (CBSE 2009)
Answer:

Representation of global biodiversity of major taxa of plants, invertebrates, and vertebrates

Question 9.
Elaborate how invasion by an alien species reduces the diversity of an area.
Answer:
This may be due to any of the following reasons:

1. Fast-growing species may compete with less vigorous local species.
2. Alien species will proliferate in that area if their natural pests and predators are not present.
3. By amensalism, i.e. it may harm local species by producing chemicals.
4. Such species may grow vigorously and may form conditions unfavorable for the growth of local native species like Eichhornia.

Question 10.
Broadly classify the extinction processes.
Answer:
Classification of extinction processes:

1. Natural extinction: It is due to a change in environmental conditions. It is at a very slow rate.
2. Mass extinction: Mass extinction occurs due to catastrophes. In this case, a large number of species became extinct in millions of years.
3. Anthropogenic extinction: Extinction of species is due to man’s activities. It is occurring in a short period of time.

Question 11.
What is an endangered species? Give an example of an endangered plant and animal species each.
Answer:
An endangered species is a population of organisms, which is facing a high risk of becoming extinct because

• Its number is very low.
• It is threatened by changing environment.
• It is facing a predator threat. Endangered plant species-Venus fly trap Endangered animal species-Siberian tiger.

Question 12.
List the benefits of protected areas for the conservation of biodiversity.
Answer:
Benefits of protected areas:

1. Maintaining viable populations of all native species and subspecies.
2. Maintaining the number and distribution of communities and habitats, and conserving the genetic diversity of all the present species.
3. Preventing the human-caused introduction of alien species.
4. Making it possible for species/ habitats to shift in response to changes in the environment.

Question 13.
List the biosphere reserves of India.
Answer:
Biosphere reserves of India:

1. Nanda Devi
2. Nokrek
3. Manas
4. Dibru Saikhowa
5. Dean Debang
6. Sunderbans
7. Gulf of Mannar
8. Nilgiri
9. Great Nicobar
10. Simitipal
11. Ichanghendzonga
12. Pachmarhi
13. Agasthyamatai

Question 14.
Write a note on sacred forests.
Answer:
Sacred forests: These are forest patches protected by tribal communities in India and other Asian countries due to religious belief. These are undisturbed forests having no human intervention and frequently surrounded by highly degraded landscapes. Such forests are located in many states of India and have a number of rare, endangered, and endemic species. Likewise, Khecheopalri which contains several aquatic fauna and flora is declared as a sacred lake in Sikkim.

Question 15.
Differentiate between in situ and ex situ conservation. (CBSE Delhi 2011, Sample Paper 2020)
Answer:
Differences between in situ conservation and ex situ conservation:

In situ Conservation	Ex-situ Conservation
1. It is the conservation of endangered species in their natural habitat.	1. It is the conservation of endangered species outside their natural habitat.
2. Protection from predators is ensured.	2. Protection from all adverse factors is ensured.
3. The population recovers in its natural habitat.	3. Offsprings produced in captive breeding are released in natural habitats for acclimatization.

Question 16.
(i) Explain the ‘Ex situ’ conservation of Biodiversity. How is the In situ conservation different from it? (CBSE Delhi 2018 C)
Answer:

• It is the conservation of endangered species outside their natural habitat.
• Protection from all adverse factors is ensured.
• Offsprings produced in captive breeding are released in natural habitats for acclimatization.

(ii) Which one of the two in situ or ex situ biodiversity conservation measures helps the larger number of species to survive? Explain. (Outside Delhi 2019)
Answer:

1. In situ, biodiversity conservation measures will help the larger number of species to survive.
2. In situ is onsite conservation, i.e. it is the conservation of endangered species in their natural habitat. Whereas ex situ conservation is the conservation of endangered species outside their natural habitat.
3. To conserve species in their natural habitat, the entire ecosystem has to be conserved including all the biotic and abiotic components of the ecosystem that are associated with the target species.
4. In situ conservation helps in the restoration of degraded ecosystems and habitats that are means of conserving genetic resources species ecosystems and landscapes, without uprooting the local people.

Question 17.
Compare the ecological biodiversity existing in India and Norway. (CBSE Outside Delhi 2019)
Answer:
At the ecosystem level, India is far more diverse than the Scandinavian country Norway with its deserts, rain forests, mangroves, coral reefs, wetlands, estuaries, and alpine meadows.

Question 18.
What are Ramsar sites?
Answer:
Ramsar sites are wetlands spread over a wide area. These sites support a wide range of flora and fauna. There are about 25 Ramsar sites in India. Conservation of wetland is the main mission. World wetland day is observed on 2nd February.

Question 19.
Among the ecosystem services are control of floods and soil erosion. How is this achieved by the biotic components of the ecosystem?
Answer:

1. Earth’s rich biodiversity is vital for indirect benefits like control of floods and soil erosion. Species richness checks soil erosion by binding the soil particles thereby reducing the rate of water velocity, hence reducing the chances of floods.
2. The roots of plants make the soil porous.

Question 20.
The species diversity of plants at 22% is much less than that of animals. What could be the explanation of how animals achieved greater diversification?
Answer:

1. Since plants cannot move from their predators and harsh treatment of environmental conditions, thus have become extinct.
2. As the animals can move away from such conditions, the evolution of favorable characters has taken place in them.

Question 21.
Suggest two practices giving one example of each that helps protects rare or threatened species. (CBSE 2017)
Answer:

1. By using the cryopreservation (preservation at -196°C) technique, sperms, eggs, tissues, and embryos can be stored for long period in gene banks, seed banks, etc.
2. Plants are propagated in vitro using tissue culture methods.

Question 22.
Pollen banks are playing a very important role in promoting plant breeding programs the world over. How are pollens preserved in the pollen banks? Explain. How are such banks benefitting our farmers? Write any two ways. (CBSE Delhi 2019)
Answer:
Preservation of pollen is done in the pollen banks: Pollen grains of a large number of species can be preserved for years in liquid nitrogen at a temperature of -196°C. It is called cryopreservation. Pollen remains viable for a very long duration.

It helps to conserve a large number of species. It can prevent the complete extinction of many species and help to maintain biodiversity.

Question 23.
What is cryopreservation? Mention how it is used in the conversation of biodiversity. (CBSE Outside Delhi 2019)
Answer:
Cryopreservation or cryo conservation is a process where organelles, cells, tissues, extracellular matrix, organs, or any other biological structures are preserved by cooling to very low temperatures, i.e. 196°C in liquid nitrogen.

Role in the conservation of biodiversity: It is an ex-situ method of conservation of biodiversity. Gametes of threatened species are preserved in viable and fertile conditions for long period. They can be used as and when required.

Biodiversity and Conservation Important Extra Questions Long Answer Type

Question 1.
What is biodiversity? Why has it become important recently?
Answer:
Biodiversity: The term biodiversity was coined by W.G. Rosen in 1985. It is the occurrence of different kinds of organisms and the complete range of varieties adapted to different climates, environments, and areas being constituents of food chains and food webs of biotic interrelationships. Biodiversity refers to the totality of genes, species, and ecosystems of a region. Biodiversity differs from place to place.

Significance of biodiversity: As there is a continuous loss of biodiversity due to increasing population, resource consumption, urbanization, and pollution, it is important to conserve it. The basic reason for concern is that biodiversity is being lost even before it attains its size. Loss of biodiversity would check the evolutionary capability of biota to cope up with an environmental loss.

Question 2.
Explain what is meant by species diversity? (CBSE 2010)
Answer:
Species diversity. The diversity includes the whole range of organisms found on earth. The number of identified species worldwide is between 1.7 and 1.8 million. However, the estimates of total known species maybe 50 million. A large number of plant and animal species are yet to be identified. There are many more species present in the tropics.

The two important measures of species diversity are:

1. Species richness: It refers to the number of species per unit area.
2. Species evenness: It refers to the relative abundance with which each species is represented in an area.
3. The variety and number of individuals determine the level of diversity of an ecosystem.
4. The Western Ghats have a greater diversity of amphibian species than the Eastern Ghats.

Question 3.
What is genetic diversity? Explain. (CBSE 2010)
Answer:
Genetic diversity:

1. The greater the genetic diversity among organisms of a species, the more sustenance it has against environmental perturbations. The genetically uniform populations are highly prone to diseases and harsh environments.
2. The genetic variation shown by Rauwolfia can be in terms of the concentration and potency of the chemical reserpine.
   There are more than 50,000 genetically different strains of rice and 1,000 varieties of mango in India.

Question 4.
Describe the ecological role of biodiversity.
Answer:
The ecological role of biodiversity:

1. Biodiversity provides plant pollinators, predators, decomposers and contributes to soil fertility.
2. It helps in the purification of air and water, management of flood, drought, and other environmental disasters,
3. Ecosystems with more diversity can withstand the environmental challenges better because genetically diverse species present in the ecosystem will have different tolerance ranges for a given environmental stress, hence they cannot be easily eliminated by any single stress at a time. However, if the ecosystem contains only a few species, it will become a fragile or unstable ecosystem.
4. The species with high genetic diversity and the ecosystems with high biodiversity have a greater capacity for adaptation against environmental perturbations.

Question 5.
Write a short note on three perspectives of community and ecosystem level of diversity.
Answer:
The three perspectives of diversity at the level of community and ecosystem are:

1. Alpha diversity
2. Beta diversity and
3. Gamma diversity.

1. Alpha diversity: It refers to the diversity of organisms sharing the same community. It has been found that there is an increase in diversity with a decrease in latitude.

2. Beta diversity: The rate of replacement of species along a gradient of habitat or communities is called beta diversity.

3. Gamma diversity: It is the rate at which additional species are found as a replacement in different localities of the same habitat.

Question 6.
Give an account of global biodiversity.
Answer:
Global biodiversity:
1. According to IUCN (2004), the total number of plant and animal species described is about 1.5 million. The species inventories for taxonomic groups in temperate countries/ regions are more complete than those in tropical countries/regions.

2. A more conservative and scientifically sound estimate has been made by Robert May; it puts the global species diversity at about seven million. More than 70% of all the species recorded are animals and plants account for about 22%; 70% of the animals are insects.

These estimates do not give any figure for prokaryotes for the following reasons:

• The conventional taxonomic methods are not suitable/sufficient for identifying these microbial species,
• Many of these species cannot be cultured under laboratory conditions.
• Biochemical and molecular biology techniques would put their diversity into millions.

Question 7.
Describe the species-area relationship.
Answer:
Species-Area relationship
Alexander Von Humboldt has observed that within a region, species richness increased with the increased explored area, but only up to a limit.

The relationship between species richness and area for a number of taxa like angiosperms plants, freshwater fishes, and birds is found to be a rectangular hyperbola.

On a log scale, the relationship becomes linear (straight line) and is described by the equation.
Species area relationship

log S = log C + Z log A, where,
S = Species Richness
Z = Slope of the line (regression coefficient)
A = area and
C = y-intercept Ecologists have found out that the value of the Z-line ranges between 0.1 and 0.2 irrespective of the taxonomic group or the region.

But this analysis in very large areas like a continent, the Z value ranges between 0.6 and 1.2. The Z value for frugivorous birds and mammals in the tropical forests is found to be 1.15.

Question 8.
What kinds of threats to biodiversity may lead to its loss?
Answer:
Threats to biodiversity:

1. Habitat loss. In order to utilize the resources there occurs the destruction of habitat.
2. Disturbance and pollution. A large number of organisms are destroyed due to natural disturbances such as fire, tree fall, defoliation by insects. Man’s activities are causing pollution.
3. Introduction of exotic species. The introduction of new species into an area causes disturbances that may lead to the disappearance of native species.
4. The extinction of species is a natural process.

Question 9.
Give a brief account of the loss of biodiversity at the global level.
Answer:
The colonization of the tropical Pacific Islands by human beings has led to the extinction of more than two thousand species of native birds.

IUCN red list (2004) documents the extinction of 784 species in the last 500 years that include 359 invertebrates, 338 vertebrates, and 87 plants.

Some of the animals that have become extinct in recent times are given below:

1. Steller’s sea cow (Russia)
2. Dodo (Mauritius)
3. Thylacine (Australia)
4. Quagga (Africa)
5. Three subspecies (Bali, Javan, Caspian) of the tiger.
6. 27 species have become extinct in the last twenty years alone.
7. Amphibians are more vulnerable to extinction.

Though about 15,5000 species are facing extinction, at present following the face the threat of extinction:

• 31% of gymnosperms
• 32% of amphibians
• 12% of bird species
• 23% of mammals

Since the origin of life on earth and evolution, three have been five episodes of mass extinction, but the current rate of extinction is 100-1000 times faster than them, due to human activities.

Question 10.
Write an explanatory note on the efforts for the conservation of biodiversity in India.
Answer:
Conservation of biodiversity in India:

1. In situ conservation is carried out through biosphere reserves, national parks and wildlife sanctuaries, and other protected areas by the Ministry of Environment and Forest reserve.
2. The National Bureau of Plants, Animals, and Fish Genetic Resources collects, conserves and stores germplasms of plants and animals in seed gene banks or field gene banks.
3. Botanical Gardens and Zoological Parks have a large collection of plant and animal species.

Question 11.
Explain the following:
(a) IUCN Red List
Answer:
IUCN Red List: It is a catalog of taxa that are facing the risk of extinction.

The uses of the red list are:

• developing awareness about the threat of loss of biodiversity
• identification and documentation of endangered species
• providing a global index of the decline of biodiversity
• defining conservation.

IUCN has recognized eight red list categories of species.
They are:

1. Extinct
2. Extinct in wild
3. Critically endangered
4. Endangered
5. Vulnerable
6. Lower risk
7. Data deficient and
8. Not evaluated.

The 2000 red list contains an assessment of more than 18,000 species, 11,000 of which are threatened.

(b) Protected areas.
Answer:
Protected areas: These areas are land or sea and are dedicated to the protection and maintenance of biological diversity. They include National Parks, Sanctuaries, and Biosphere reserves. As of September 2002, India has 581 protected areas.

National parks: A national park is an area that is strictly reserved for the betterment of the wildlife and where activities like forestry, grazing, or cultivation are not permitted. In these parks, even private ownership rights are not allowed. Sanctuaries. A sanctuary is a protected area that is reserved for the conservation of only animals and human activities like harvesting of timber, collection of minor forest products and private ownership is allowed so long as they do not interfere with the well-being of animals.

A biosphere reserve is a specified area in which multiple uses of the land are permitted by dividing it into certain zones, each zone is specified for a particular activity.

Question 12.
What is a biosphere reserve? Show the zonation of the biosphere reserve.
Answer:
Biosphere Reserves. Biosphere reserves are a special category of protected areas of land and/or coastal environments, wherein people are an integral component of the system. These are representative examples of natural biomes and contain unique biological communities.

The zonation in a terrestrial Biosphere Reserve

Question 13.
Show the in situ and ex situ approaches of conserving biodiversity in India.
Answer:

The In situ and ex situ approaches of conserving biodiversity in India.

Question 14.
Write critical notes on the following:
1. Hot spots of biodiversity.
2. Ex-situ conservation.
3. India’s effort in biodiversity. (CBSE Delhi 2019)
Or
States any two criteria for determines biodiversity hotspots. Name any two hotspots of India. (CBSE Sample Paper 2020)
Answer:
1. Hot spots of biodiversity: The concept of ‘Hotspots’ was developed by Norman Myers (1988) to designate specific areas for in situ conservation. The hotspots are the richest and most threatened reservoirs of plant and animal life on earth.

The criteria for determining hotspots are:
(a) Number of endemic species.
(b) Degree of threat which is measured in terms of habitat loss.

There are 25 hotspots in the world out of which two are in India. They are the Western Ghats and Eastern Himalayas.

Hotspot of Eastern Himalayas is active centers of evolution and rich in diversity of flowering angiosperms. The Western Ghats have semi-evergreen forests.

The Western Ghats include two main centers of biodiversity:

• Aqastyamatai hills
• Silent valley.

2. Ex-situ conservation: It means maintenance of off-site collections either in gardens by farmers, botanical garden or storing seeds, genes, pollen, tissue culture, etc. The rare plants have been found to flourish in large numbers under the care and protection of gardeners and nature lovers.

The farmers have been maintaining genetic diversity (enormous varieties) of crop plants since ancient times by saving seeds or other components for the next plantings.

Collection of samples of cultivated and wild varieties of plants and storing them in botanical gardens is another method of conservation of germplasm.

In seeds, the living material remains in a metabolically suspended state. When the seeds are to be stored for longer periods, it is necessary to avoid conditions that favor respiration and enzymatic action.

Advantages of ex situ conservation:

• Threatened and endangered species can be conserved.
• Genetic strains of commercially important plants can be preserved for a long time (seed banks).
• Gametes of threatened species can be preserved in viable and fertile conditions for a longer duration by the cryopreservation technique.
• Loss of biodiversity can be reduced.
• Eggs can be fertilized in-vitro.
• Can conserve a large number of species and the aesthetic value.

3. India’s effort in biodiversity conservation: India has greatly contributed to the conservation of biodiversity owing to its great utility and need for conservation.

The following measures have been taken for this purpose:
(a) Setting up of bodies like Indian Board for Wildlife, Bombay Natural History Society, etc.
(b) Observation of the first week of October as national wildlife week.
(c) introduction of the Wildlife Protection Act in 1972.
(d) Setting up of sanctuaries, national parks, and biosphere reserves.

Question 15.
(i) Why should we conserve biodiversity?
Answer:
We should conserve biodiversity for the following reasons:
(a) Narrow utilitarian arguments Human beings draw direct economic benefits from nature like food, firewood, fiber, construction materials, industrial products, medicines, etc.

(b) Broadly utilitarian arguments. Biodiversity plays a major role in maintaining and sustaining the supply of goods and services

• Amazon forests provide 20% of the oxygen present in the atmosphere.
• The ecosystem provides pollinators.

(c) Ethical reasons. Every organism has intrinsic value even if there is no economic value. Conservation of biodiversity.

(ii) Explain the importance of biodiversity hotspots and sacred groves. (CBSE Delhi 2016)
Answer:
Importance of biodiversity hotspots and sacred groves They provide protection of species richness and a high degree of endemism in natural habitat. Hotspots are regions of the high level of species in their natural habitat. Sacred groves are forest tracts set aside and all trees and wildlife within the area are given total protection.

Question 16.
What is the ES Nino effect? Explain how it accounts for biodiversity loss. (CBSE Delhi 2011)
Answer:
1. El Nino: It is an abnormal warming of surface ocean waters in the eastern tropical Pacific. It is called the Southern Oscillation. The Southern Oscillation is the see-saw pattern of reversing surface air pressure between the eastern and western tropical Pacific.

When the surface pressure is high in the eastern tropical Pacific it is low in the western tropical Pacific, and vice versa. Because ocean warming and pressure reversals are, for the most part, simultaneous, scientists call this phenomenon the El Nino/Southern Oscillation or ENSO for short.

El Nino Effect
Unfortunately not all El Ninos are the same nor does the atmosphere always react in the same way from one El Nino to another. This is why NASA’s Earth scientists continue to take part in international efforts to understand El Nino events. Hopefully, one day scientists will be able to provide sufficient warning so that we can be better prepared to deal with the damages and changes that El Nino causes in the weather.

Question 17.
Can you think of a situation where we deliberately want to make a species extinct? How would you justify it? (CBSE Delhi 2011)
Answer:

1. Human effort to eradicate disease-causing organisms (Poliovirus).
2. This situation may appear when alien species are introduced unintentionally or deliberately into an area. Some of them may become invasive and cause damage to indigenous species.
3. Example: Nile perch introduced into lake Victoria in East Africa led to the extinction of 200 species of cichlid fish in the lake. The introduction of African catfish Clarias gariepinus for aquaculture is threatening the indigenous catfishes in rivers of India.

Question 18.
Explain biodiversity as sources of food and improved varieties.
Or
State the uses of biodiversity in modern agriculture. (CBSE 2011)
Answer:
Use of biodiversity in agriculture:

• As a source of new crops.
• As source material for breeding varieties.
• As a source of new biodegradable pesticides.

Only 20% of total plant species are cultivated to produce 85% of the world’s food.

Wheat, corn, and rice, the three major carbohydrate crops, yield nearly two-third of the food sustaining the human population. Fats, oils, fibers, etc. are other uses for which more and more new species need to be investigated.

Question 19.
(i) “India has a greater ecosystem diversity than Norway.” Do you agree with the statement? Give reasons in support of your answer.
Answer:
Yes. I agree with the statement. India has greater ecosystem diversity than Norway because it has deserts, rain forests, mangroves, coral reefs, wetlands, estuaries, and alpine meadows.

(ii) Write the difference between genetic biodiversity and species biodiversity that exists at all the levels of biological organization. (CBSE 2018)
Answer:
Genetic diversity:

1. Genetic diversity is the total number of genetic characteristics in the genetic makeup of a species.
2. A single species might show high diversity at the genetic level, e.g. Man: Chinese, Indian American, African, etc. India has more than 50,0 genetically different strains of rice and 1,000 varieties of mango.
3. Genetic diversity allows species to adapt to changing environments. This diversity aims to ensure that some species survive drastic changes and thus carry on desirable genes.

Specific diversity:

1. Specific diversity is the ratio of one species population over a total number of organisms across all species in the given biome. ‘Zero’ would be infinite diversity, and ‘one’ represents only one species present.
2. Species diversity is a measure of the diversity within an ecological community that incorporates both species richness, i.e. the number of species in a community and the evenness of species.

Question 20.
Give the approximate numbers of species that have been described and identified all over the world.
Answer:
An approximate number of species that have been described and identified from all over the World:

Group	Number of species
Higher plants	2,70,000
Algae	40,000
Fungi	72,000
Bacteria (including cyanobacteria)	4,000
Viruses	1,550
Mammals	4650
Birds	9700
Reptiles	7150
Fish	26,959
Amphibians	4780
Insects	10,25,000
Crustaceans	43,000
Molluscs	70,000
Nematodes and worms	25,000
Protozoa	40,000
Others	1,10,000

Question 21.
Give an account of latitudinal gradients of biodiversity.
Answer:
Latitudinal gradients of biodiversity:

1. Species diversity decreases from the equator towards the poles.
2. The tropics (between 23.5°N to 23.5°S) harbor more species than temperate and polar regions.
3. For example, the Western Ghats have a greater amphibian species diversity than the Eastern Ghats. There are more than 2,00,000 species in India of which several are confined to India (endemic).
4. For example, Columbia situated near the equator has about 1400 species of birds, while Hew York (41 °N) has 105 species, Greenland (70°N) has about 56 species and India (in the equator region) has 1200 species.
5. The number of species of vascular plants in the tropics is about ten times more than that of temperate forests.
6. The Amazonian rain forest in South America has the greatest biodiversity on earth; it harbors about 40,000 species of plants, 1,25,0 species of insects, 3000 fishes, 427 amphibians, 378 reptiles, 1300 birds, and 427 mammals.

Question 22.
Discuss the characteristics of India’s biodiversity.
Answer:
Characteristics of India’s biodiversity. Biodiversity is not uniformly distributed in space and time. It is rich in tropics.

India’s biodiversity is characterized by the following:
1. India contains 10 bio-geographic regions which include the Himalayan, Trans-Himalayan, the Indian desert, the Semi-arid zone, the Western Ghats, the Deccan Peninsula, the Gangetic Plain, North-East India, and the Islands and Coasts which possess different biodiversity levels.

2. India is one of the world’s 12 leading biodiversity centers of the origin of cultivated plants.

3. Though India has only 2.4% of the land area of the world, it has 8.1% of the global species biodiversity.

4. There are about 45,000 species of plants and 90,000-1,00,000 species of animals; many more species are yet to be discovered and named.

5. If we apply Robert May’s global estimate that only 22% of the total species have been recorded, India probably has more than 1,00,000 species of plants and 3,00,000 species of animals to be discovered and described.

6. India has five natural world heritage sites, 14 biosphere reserves, 89 national parks, 492 wildlife sanctuaries, and 2 hotspots. Heritage sites are places that attract tourists.

7. About 33 percent of the country’s recorded flora are endemic to India and concentrated in the North-East, Western Ghats, North-West Himalaya, and Andaman and Nicobar islands.

Question 23.
(a) List any two ways biodiversity loss affects any region.
Answer:
Effects of biodiversity loss:

1. The decline in plant production
2. Lowered resistance to environmental perturbations such as drought
3. Increased variability in certain ecosystem processes.

(b) Explain any two causes of biodiversity loss, with the help of suitable examples. (CBSE Outside Delhi 2019)
Answer:
Causes of loss of biodiversity:
(A) Habitat fragmentation:

1. Habitat loss and fragmentation create barriers that limit the potential of species to disperse and colonize new areas.
2. Species get divided into smaller populations that are unable to sustain themselves.
3. Migratory birds lose their seasonal habitats.
4. It increases edge areas thus making the species more vulnerable to predators as well as wind and fire.

Thus there is the loss of biodiversity because a large number of animals, e.g. elephants, lions, bears, and large cats require big territories to move around and live in. Likewise, some birds reproduce successfully only in deep forests.

(B) Introduction of exotic species leading to endangering the species Exotic species are having a large impact especially in the island ecosystem, which harbor much of the world’s threatened biodiversity.

A few examples are:

1. Nile perch, an exotic predatory fish introduced into Lake Victoria (South African), threatens the entire ecosystem of the lake by eliminating several native species of the small Cichlid fish species that were endemic to this freshwater aquatic system.
2. Water hyacinth clogs rivers and lakes and threatens the survival of many aquatic species in lakes and river flood plains in several tropical countries including India.

Question 24.
List the uses of biodiversity.
Answer:
Uses of biodiversity:

1. As a source of food and improved varieties.
2. As a source of drugs and medicines.
3. Aesthetic and cultural benefits. Examples of aesthetic rewards include ecotourism, bird watching, wildlife, pet-keeping, gardening, etc. Throughout human history, people have related biodiversity to the very existence of the human race through cultural and religious beliefs.
4. Biodiversity is essential for the maintenance and sustainable utilization of goods and services from the ecological systems as well as from individual species.

Question 25.
What are the uses of IUCN Red list categories?
Answer:
Uses of Red list categories:

• Developing awareness about the importance of threatened biodiversity.
• Identification and documentation of endangered species.
• Providing a global index of the decline of biodiversity.
• Defining conservation priorities at the local level and guiding conservation action.

Question 26.
How is the “sixth episode of extinction” of species on earth, now currently in progress, different from the five earlier episodes? What is it due to? Explain the various causes that have brought about this difference.
Answer:
Earth is heading for sixth extinction due to human activities.

Anthropogenic extinction. An increasing number of species are disappearing from the face of the earth due to human activities. This man-made mass extinction represents a very severe depletion of biodiversity, particularly because it is occurring within a short period of time.

It has been estimated at World Conservation Monitoring Centre that about 384 plant species (mostly phanerogams) and 533 animal species (mostly vertebrates) have become extinct since the year 1600. This rate of extinction of species is 1000 to 10,000 times higher than the earlier rate.

A few interesting points of extinction of species noticed are:

• Tropical forests are losing 14,000-40,0 species per year, i.e. at the rate of 2-5 species per hour.
• Near about 50 percent of species may become extinct at the end of the 21st century, if the present rate does not retard.
• Loss of 17,000 endemic plant species and 350,000 endemic animals may take place in the near future from 10 high diversity localities in tropical forests.

It seems that earth is heading for the sixth extinction.

Causes:

• Increase in human population and settlements.
• Hunting.
• ver-exploitation of natural resources.
• Destruction of habitat.

Question 27.
Give three hypotheses for explaining, why tropics show the greatest levels of species richness.
Answer:
The following three hypotheses explain how tropics show the greatest level of species richness:
1. Undisturbance in the tropics: Speciation is usually a function of time, unlike temperate areas subjected to frequent glaciations in the past. This type of disturbance has not occurred or remained relatively undisturbed in tropical latitudes for millions of years. Tropical regions, thus, got a long evolutionary time for species diversification.

2. Constancy in season: In tropical regions, the environment is more constant, less seasonal, and predictable. This is not so in temperate regions. Due to this stability and constancy, niche speciation takes place at a faster rate and leads to species richness.

3. Availability of more solar energy: Due to more availability of solar energy in the tropics, productivity is higher. This contributes indirectly to greater species diversity.

Question 28.
How is biodiversity important for Ecosystem functioning?
Answer:
Importance of biodiversity to the ecosystem:

1. Ecologists believe that communities with more species tend to be more stable than those with fewer species,
2. A stable community has the following attributes.
   (a) It shall not show too many variations in the year-to-year productivity.
   (b) It must be either resistant or resilient to seasonal disturbances.
   (c) It must be resistant also to invasion by alien species.
3. David Tilman had shown through his ecology experiments using outdoor plots the following features.
   (a) The plots with more species showed less year-to-year variation in the total biomass.
   (b) Plots with increased diversity showed higher productivity.
4. It is now realized that species richness and diversity are essential for ecosystem health as well as the survival of the human race on the earth.

Question 29.
What are sacred groves? What is their role in conservation? (CBSE Outside Delhi 2016)
Answer:
Sacred Groves are the secret forest patches around places of worship. They are of great religious value among tribal communities.

In such cases, nature is protected by prevailing religious and cultural traditions. Here tracts of forests are set aside and all plants and animals are venerated and provided with complete protection. Examples of sacred groves are Khasi and Jaintia hills in Meghalaya, Aravali hills in Rajasthan, Western Ghat regions of Karnataka, Maharashtra and Sargiya, Chanda and Bastar areas of M.P. In Sikkim, Khecheopalri lake is declared sacred lake by people, thus protecting the aquatic flora and fauna.

Role in conservation. Many rare and threatened plants have been protected in the sacred groves of Meghalaya. Such areas have been found to be most undisturbed and they are usually surrounded by the most degraded landscapes.

Question 30.
What is Ramsar’s mission? Explain the ‘Wise Use concept’?
Answer:
The Ramsar mission.
The Convention’s mission is “the conservation and wise use of all wetlands through local and national actions and international cooperation, as a contribution towards achieving sustainable development throughout the world”.

The Convention uses a broad definition of the types of wetlands covered in its mission, including lakes and rivers, swamps and marshes, wet grasslands and peatlands, oases, estuaries, deltas and tidal flats, near-shore marine areas, mangroves, and coral reefs, and human-made sites such as fish ponds, rice paddies, reservoirs, and salt pans. The Wise Use concept At the center of the Ramsar philosophy is the “wise use” concept.

The wise use of wetlands is defined as “the maintenance of their ecological character, achieved through the implementation of ecosystem approaches, within the context of sustainable development”. “Wise use” therefore has at its heart the conservation and sustainable use of wetlands and their resources, for the benefit of humankind.

Ramsar commitments The Ramsar Contracting Parties, or the Member States, have committed themselves to implement the “three pillars” of the Convention: to designate suitable wetlands for the List of Wetlands of International Importance (“Ramsar List”) and ensure their effective management; to work towards the wise use of all their wetlands through national land-use planning, appropriate policies and legislation, management actions, and public education; and to cooperate internationally concerning transboundary wetlands, shared wetland systems, shared species, and development projects that may affect wetlands.

Question 31.
Sanctuaries are tracts of land where animals are protected from all types of exploitation. Private ownership is permitted. Collection of minor forest products are allowed.
(i) How many sanctuaries are present in India?
Answer:
India has 551 sanctuaries.

(ii) How much land area they cover?
Answer:
3.6% of geographical area.

(iii) Name any three sanctuaries.
Answer:
(a) Keoladeo Ghana Bird Sanctuary, Bharatpur, Rajasthan
(b) Sultanpur Lake Bird Sanctuary, Gurgaon, Haryana
(c) Periyar Sanctuary, Kerala

(iv) List any three human activities which are allowed In sanctuaries.
Answer:
(a) CoLLection of forest products
(b) Harvesting of timber
(C) Tilling of Land
(d) Private ownership of Land.

Question 32.
Make a list of Ramsar Sites In India.
Answer:
A-List of Ramsar sites in India is as following:

Sr Name of Ramsar Site	Location
1. Ashtamudi Wetland	Kerala
2. Bhitarkanika Mangroves	Odisha
3. Bhoj Wetland	Madhya Pradesh
4. Chandra Taal	Himachal Pradesh
5. Chitika lake	Odisha
6. Deepor Beet	Assam
7. East Calcutta Wetlands	West Bengal
8. Harike Wetland	Punjab
9. Hokersar Wetland	Jammu and Kashmir
10. Kanji Wetland	Punjab
11. Keoladeo National Park	Rajasthan
12. Kolleru lake	Andhra Pradesh
13. Loktak lake	Manipur
14. Nalsarovar Bird Sanctuary	Gujarat
15. Point Calimere Wildlife and Bird Sanctuary	Tamil Nadu
16. Pong Dam lake	Himachal Pradesh
17. Renuka Wetland	Himachal Pradesh
18. Ropar	Punjab
19. Rudrasagar lake	Tripura
20. Sambhar lake	Rajasthan
21. Sasthamkotta lakes	Kerala
22. SurinsarMansar lakes	Jammu and Kashmir
23. Tsomoriri	Jammu and Kashmir
24. Upper Ganga River (Brljghat to Narora Stretch)	Uttar Pradesh
25. Vembanad-Kol Wetland	Kerala
26. Wular lake	Jammu and Kashmir

Here we are providing Class 12 Biology Important Extra Questions and Answers Chapter 12 Biotechnology and its Applications. Important Questions for Class 12 Biology are the best resource for students which helps in Class 12 board exams.

Class 12 Biology Chapter 12 Important Extra Questions Biotechnology and its Applications

Biotechnology and its Applications Important Extra Questions Very Short Answer Type

Question 1.
What is a transgenic organism?
Answer:
An organism that carries a foreign functional gene in its genome is termed a transgenic organism.

Question 2.
Write two applications of biotechnology.
Answer:

1. Treatment of diseases.
2. Preparation of processed fortified food.

Question 3.
What are probes?
Answer:
It is a single-stranded DNA or RNA, tagged with a radioactive molecule, which is complementary to the DNA in a clone of cells. It is used for detecting the presence of nucleotides complementary to the probe.

Question 4.
Name two diseases that can be treated by producing biological compounds in transgenic animals.
Answer:
Cystic fibrosis, rheumatoid arthritis, Alzheimer’s disease, cancer, and emphysema.

Question 5.
Name the toxin produced by Bacillus Thurinsiensis.
Answer:
Bt Toxin.

Question 6.
What is the utility of the Bt-toxin gene?
Answer:
Bt-Toxin gene provides Bt-toxin which is involved in providing resistance to cotton plants against insects.

Question 7.
Bt-toxin protein exists in which form?
Answer:
Inactive protoxins.

Question 8.
How is inactive Bt-toxin converted into active form? (CBSE Outside Delhi 2019)
Answer:
The inactive toxin is converted into active form due to the alkaline pH of the gut of insects which solubilizes the crystal converting the toxin to its active form.

Question 9.
How does Bt toxin cause the death of insects?
Answer:
Activated Bt toxin binds to the surface of midgut epithelial cells and creates pores in it that cause cell swelling and lysis. It finally leads to the death of the insect.

Question 10.
Name a few forms of cry gene.
Answer:
cry I Ac, cry II Ab, cry III Ab, and cry III Bb.

Question 11.
List the specific insects killed by:
(i) cry I Ac and
(ii) cry II Ab.
Answer:
cry I Ac, cry II Ab-both control cotton bollworm.

Question 12.
Name the insects killed by proteins coded by cry III Ab and cry III Bb.
Answer:

1. Colorado potato beetle
2. corn rootworm.

Question 13.
What is unique about transgenic animals?
Answer:
Animals that have their DNA manipulated to possess and express a foreign gene are known as transgenic animals. e.g. Rabit, sheep, cows, fish, etc.

Question 14.
How infestation of Meloidogyne incognita was prevented in the Tobacco plant?
Answer:
An infestation of Meloidogyne incognita was prevented on the basis of RNA interference. This method involves the silencing mRNA by complementary dsRNA molecule that binds to mRNA and prevents its translation.

Question 15.
What is the silencing of mRNA?
Answer:
The binding of single-stranded mRNA with complementary and double-stranded RNA to prevent translation of mRNA is called silencing of mRNA.

Question 16.
What is the source of the complementary strand in mRNA silencing?
Answer:

• Viruses having RNA genomes.
• Mobiles genetic elements (transposons).

Question 17.
How is dsRNA prepared?
Answer:
Reverse transcription.

Question 18.
Name the genetically engineered insulin.
Answer:
Humulin.

Question 19.
Write the name of the transgenic protein used to treat emphysema.
Answer:
Alpha-1-antitrypsin.

Question 20.
How is Indian basmati unique?
Answer:
It is unique for its aroma and flavor.

Question 21.
What is complementary DNA (cDNA)?
Answer:
DNA synthesized on RNA template with the help of reverse transcriptase.

Question 22.
Mention the chemical change that proinsulin undergoes, to be able to act as mature insulin. (CBSE 2018)
Answer:
The proinsulin is cleaved to remove extra stretch called the C-peptide to form mature insulin having only A-chain and B-chain joined by a disulfide bond.

Question 23.
What is the application of genetically engineered bacterium namely Pseudomonas Putidal?
Answer:
Pseudomonas putida is used for scavenging oil spills by digesting hydrocarbons of crude oil.

Question 24.
How did the first transgenic cows Rosie differ from other cows with respect to the quality of milk? (CBSE 2008)
Answer:
Rosie produced a human protein (alpha-lactalbumin) enriched milk which is nutritionally a more balanced product for human babies.

Question 25.
State the role of C-peptide in human insulin. (CBSE 2014)
Answer:
C-peptide maintains its nature as pro-hormone (pro-insulin) and during maturation, it is removed. Thus proinsulin matures into insulin.

Biotechnology and its Applications Important Extra Questions Short Answer Type

Question 1.
List three critical research areas of biotechnology.
Answer:
Three critical research areas of biotechnology are:

1. Providing the best catalyst in the form of an improved organism usually a microbe or pure enzyme.
2. Creating optimal conditions through engineering for a catalyst to act, and
3. Downstream processing technologies to purify the protein/organic compound.

Question 2.
Give the few characteristics of GMOs. (CBSE Delhi 2019)
Answer:
Genetically modified organisms:

1. They are capable of producing pharmaceutically useful proteins.
2. They are capable of producing en¬hanced, modified, or new metabolites.
3. They can be used for crop protection by control of insects, fungal diseases, frost damage, etc.
4. They degrade non-biological wastes and detoxify toxic wastes.
5. They show enhanced nitrogen fixation.

Question 3.
List a few transgenic organisms and their potential application. (CBSE Delhi 2019)
Answer:
Transgenics and their potential applications:

Transgenics	Useful Applications
Bt Cotton	Pest resistance, herbicide tolerance, and high yield
Flavr Savr Tomato	Increased shelf life (delayed ripening) and better nutrient quality.
Golden Rice	Vitamin A-rich
Cattles (cow, sheep, goat)	Therapeutic human proteins in their milk
pig	Organ transplantation without risk of rejection.

Question 4.
In view of the current food crisis, it is said that we need another green revolution. Highlight the major limitations of the earlier green revolution.
Answer:

1. Excessive use of fertilizers and pesticides which are polluting the water bodies, soil, and food items.
2. It was related to better management practices that can improve food availability to a limited extent.
3. Genetic cap for improvement in food yield.

Question 5.
Differentiate between diagnostics and therapeutics. Give one example for each category.
Answer:
Difference between diagnostics and therapeutics:

Diagnostics	Therapeutics
1. It finds out the cause and nature of the disease.	1. It treats patients to cure them of the disease.
2. It provides a logical basis for treatment. Example: ELISA test or HIV.	2. It provides relief from the disease. Example: Antibiotic for bacterial infection.

Question 6.
Gene therapy can cure important genetic disorders in humans. Comment.
Answer:
Through the “human genome project,” most human genes and sequences have been identified, genetic disorders such as Eczema, cancer, hemophilia, thalassemia, and cystic fibrosis can be cured by the insertion of correct genes into these patients.

Question 7.
What are the advantages of molecular diagnostic over conventional methods?
Answer:
Advantages of molecular diagnostics:

1. Early diagnosis is not possible using conventional methods, but by using rDNA technology and PCR, early diagnosis is possible.
2. It is also a powerful technique to identify many genetic disorders.
3. It is used to detect mutations in suspected cancer patients.

Question 8.
List two uses of cloned genes in molecular diagnostics.
Answer:

1. Cloned genes, when expressed to pro¬duce recombinant proteins, help in developing sensitive diagnostic techniques.
2. Cloned genes are used as probes to detect the presence of its complementary DNA strand; mutated genes will not hybridize with the probe and hence do not appear on the photographic film.

Question 9.
How is early detection of infectious diseases possible by molecular diagnostics?
Answer:
Molecular diagnostic of early detection of infectious diseases:

1. A low concentration of viral or bacterial DNA in a host body can be detected (much before the symptoms of the disease appear) by polymerase chain reaction (amplification).
2. Clones of genes can be used as probes to detect the presence of complementary (normal) strands of DNA in a mutant-clone. Hybridization does not occur and hence the radioactivity does not appear in the (autoradiography) photographic films.

Question 10.
How was insulin obtained before the advent of rDNA technology? What were the problems encountered?
Answer:
Before the advent of rDNA technology, insulin was obtained from slaughtered cattle and pigs.
Problems:

1. Insulin obtained from slaughtered cattle and pigs was slightly different from human insulin. It had a harmful effects over long periods.
2. The drug has been eliciting an immune response in some patients.

Question 11.
Why has the Indian Parliament cleared the second amendment of the country’s patents bill?
Answer:
The major change in the patent regime achieved through the second amendment is not in the area of medicine and drugs but in the area of seeds and plants, especially genetically engineered seeds. It has opened the flood gates for patenting genetically engineered seeds.

Question 12.
Give any two reasons why the patent on Basmati should not have gone to an American Company.
Answer:

1. In India, Basmati rice is being cultivated for several years.
2. American company by producing hybrids of this scented (basmati) rice cannot claim to have the patent rights.

Question 13.
What is a gene library?
Answer:
Gene Library: Several clones of cells each containing one of a few foreign genes are finally obtained, representing almost all the genes of an organism, it is termed the gene library of that organism. From that gene library, it is possible to identify a clone containing the gene of interest.

In order to obtain the gene library of an organism, its genome is first to cut into smaller DNA fragments containing one or a few genes, and such fragments can be cloned in the cell where such a cell multiplies to form a group of cells, all cells have same foreign DNA and are termed clone.

Question 14.
What is a reporter or marker gene?
Answer:
A reporter or marker gene produces a phenotype that is either easily and specifically detected or which allows a differential multiplication of the cells.

Question 15.
Why is the use of probes considered better than conventional diagnostic tools for disease diagnosis?
Answer:
Probes are better than conventional diagnostic tools because:

1. They are highly specific, relatively rapid, and much simpler.
2. They are extremely powerful especially when combined with PCR, even a single molecule in the test sample can be detected.
3. Since the culture of microbe is not required, the risk of accidental infection to laboratory personnel is eliminated.

Question 16.
Name different transfection methods.
Answer:
Calcium phosphate precipitation, direct microinjection, retrovirus infection, lipofection, particle gun delivery, and electroporation.

Question 17.
Why mice are considered the most suitable animals for transgenic production?
Answer:
The mouse is preferred for studies of gene transfers due to its many favorable features like a short estrous cycle and gestation period, production of several offspring per pregnancy, convenient in vitro fertilization, production, and maintenance of embryonic stem cell lines.

Question 18.
Define ‘Germline gene therapy.
Answer:
It is a therapy in which germ cells, i.e. sperms or eggs (even zygotes) are modified by the introduction of functional genes which are ordinarily integrated into their genomes. Therefore, the change due to therapy is heritable and passed on to later generations.

Question 19.
Write the advantages of recombinant therapeutics. How many of them have been approved the world over for human use and how many are available in the Indian market?
Answer:
Advantages of Recombinant Therapeutics:

1. The recombinant therapeutics do not induce unwanted immunological responses like similar products of non-human origin.
2. About 30 recombinant therapeutics have been approved the world over.
3. 12 of them are being marketed in India.

Question 20.
What is interest-sensitive speciesism? Is it one of the ethical issues related to transgenic animals?
Answer:
The use of animals in biotechnological research causes greater suffering to the animals. But most people seem to accept some animal suffering to serve the basic interest and welfare of mankind; this attitude has been termed as interest-sensitive speciesism. It is one of the most common ethical issues.

Question 21.
Name a few useful products obtained from animal cell lines.
Answer:
Useful products obtained from animal cell lines:

1. Production of vaccines for influenza, measles, and mumps from chick embryo fluid.
2. Production of vaccines for rabies and rubella from duck embryo fluid.

Question 22.
Can you suggest a method to remove oil (hydrocarbon) from seeds based on your understanding of DNA technology and the chemistry of oil?
Answer:
The genes for the formation of oil in the seed should be identified. The specific gene can be removed by using enzyme restriction endonucleases. Such DNA molecules should be treated with DNA ligases to seal at the broken ends. These cells when grown in a minimum nutrient medium, under aseptic conditions will differentiate into a new plant whose seeds will not have oil in them.

Question 23.
Find out from the Internet what is Golden Rice.
Answer:
Golden Rice. It is genetically engineered rice rich in Vitamin A. It was prepared by introducing three genes involved in the biosynthetic pathway for carotenoid, the precursor of vitamin A. The color of golden rice is yellow due to the synthesis of provitamin A in the entire grain.

Question 24.
Describe the responsibility of GEAC, set up by the Indian Government. (CBSE 2009)
Answer:

1. Genetic Engineering Approval Committee (GEAC) makes decisions regarding the validity of GM research.
2. It also ensures the safety of introducing GM organisms for public services.

Question 25.
Why insulin is being extracted from bacteria rather than animal sources?
Or
Name the source from which insulin was extracted earlier. (CBSE 2011)
Answer:
Insulin for the use of diabetic patients was earlier extracted from the pancreas of slaughtered cows and pigs. It caused allergy and other reactions in patients, due to foreign proteins. So these days insulin is being extracted from bacteria.

Question 26.
What are embryonic stem cells? What stages of early embryonic development are important for generating embryonic stem cells?
Answer:
Embryonic stem cells, as their name suggests, are derived from embryos. Most embryonic stem cells are derived from embryos that develop from eggs that have been fertilized in vitro-in an in vitro fertilization clinic and then donated for research purposes with the informed consent of the donors. They are not derived from eggs fertilized in a woman’s body. Embryonic stem cells are obtained from the inner cell mass of the blastocyst stage of the embryo.

Question 27.
Name the first transgenic cow developed and state the improvement in the quality of the product produced by it. (CBSE Sample Paper 2018)
Answer:

• Name of the first transgenic cow developed: Rosie
• Advantage: It produced human protein-enriched milk (2.4 grams per liter).

Question 28.
What are cry genes? In which organism are they present? (CBSE 2017)
Answer:
Cry genes code for a toxin that is poisonous to some insects thus making plant insect resistant. They are present in the bacterium Bacillus Thuriengiensis.

Question 29.
Name one toxin gene isolated from B. Thuringiensis and its target pest. (CBSE Delhi 2019 C)
Answer:
Cry I AC is a toxin isolated from B. Thuringiensis and its target pest is cotton bollworm.

Question 30.
Why does the toxin produced by B. Thuringiensis not kill the Bacillus? (CBSE Delhi 2019 C)
Answer:
The toxin produced is in inactive form as protoxin. It does not kill the bacteria and attacks only its target pest because protoxin is activated in the optimum pH medium of the gut of insect pest.

Biotechnology and its Applications Important Extra Questions Long Answer Type

Question 1.
Expand GMO. How is it different from a hybrid?
Answer:

• GMO – Genetically modified organism.
• Differences between GMO and Hybrid.

GMO	Hybrid
1. Formation of GMO does not require crossing between different organisms.	1. It is formed as a result of crossing between two different organisms.
2. One or more foreign genes are incorporated into GMOs.	2. It contains complete genomes of two different organisms.
3. A completeLy new trait has been introduced.	3. OnLy the existing traits are improved.

Question 2.
Mention any six fields of application of biotechnology for human welfare.
Answer:
Applications of Biotechnology:

1. Therapeutics
2. Genetically modified crops
3. Molecular diagnostics
4. Processed food items
5. Bioremediation
6. Biological waste treatment
7. Energy production.

Question 3.
“Specific Bt Toxin gene is incorporated into the cotton plant so as to control the infestation of Bollworm”. Mention the organism from which the gene was isolated and explain its mode of action. (CBSE Sample paper 2019-20)
Answer:

• Specific Bt toxin genes isolated from Bacillus Thuringiensis are incorporated into cotton. Cry I AC and Cry II AC control the bollworm.
• Bt gene forms protein crystals that contain a toxin insecticidal protein.
• It is in an inactive state.
• The inactive toxin is converted into active form due to the alkaline pH of the gut which solubilizes the crystal.
• Activated Bt-toxin binds to the surface of midgut epithelial cells and creates pores that cause cell swelling and lysis. It finally leads to the death of the insect.

Question 4.
How is the ELISA test carried out?
Answer:
ELISA (Enzyme-Linked Immunosorbent Assay Test):

1. It is a technique of detecting a very small amount of protein (antibody or antigen) with the help of enzyme peroxidase or alkaline phosphatase and stain-producing substrates like 5-aminosalicylic acid or orthophenylene diamine.
2. The serum is sorbed to the surface of the ELISA plate.
3. An antibody is specific to the antigen for diagnosis placed over an immobilized antigen.
4. The spot is washed to remove the free antibody.
5. Antibody bound to the enzyme is poured over the spot so as to react with com¬plex antibody.
6. The area is washed again to remove the free antibody-enzyme complex.
7. Chromagen is added. It will produce a stain showing the antigen was present.
8. ELISA is a quick method of diagnosis of pregnancy (by detection hCG in urine), AIDS, hepatitis, STDs, thyroid disorder, and Rubella virus.

Question 5.
While creating genetically modified organisms, genetic barriers are not respected. How this can be dangerous in the long run?
Answer:

1. Genetic pollution. The ecological imbalance may occur due to transferring of transgenes from one organism to another.
2. Formation of Superweeds. Due to the introduction of weedicide genes in crops, there is a danger that any such crop may itself become superweed.
3. Formation of super insecticides.
4. Foreign proteins formed with foreign genes may get attacked by the defense system of the organism leading to the formation of defective biochemicals,
5. Transgenes may exhibit changes in their expressivity after attaining certain age and change in environment.

Question 6.
Explain the structure of Insulin. How insulin is synthesized in humans (or mammals)? (CBSE Outside Delhi 2011) Answer:
(i) Insulin is made up of two short polypeptide chains; A and chain B which are linked together by disulfide bridges.

(A), (B) Conversion of proinsulin after removal of C-peptide.

(ii) Insulin is synthesized as pro-hormone (i.e. which is to be processed before becoming functional) which contains an extra stretch called C-peptide which is usually removed during the maturation of insulin.

Question 7.
Explain the social, economical, and environmental implications of genetic engineering techniques.
Answer:

1. Genetically prepared human insulin and edible vaccines will be readily available and also will be economical.
2. Transgenic crop plants for human consumption may cause concern for safety due to unwanted properties they may have.
3. Some people believe that transgenic plants and animals can solve many human problems especially hunger and disease.

Question 8.
Write a short note on:
(i) Production of human growth hormone by E. coli.
Answer:
Production of human growth hormone by E. coli. The Human growth hormone is produced commercially by transgenic Escherichia coli. The pituitary gland of humans produces growth hormones that regulate growth and development. However, in children stunted growth occurs due to deficiency of the hormone called pituitary dwarfism. For this, the hGH is now available as a recombinant protein.

The high-coding DNA sequence is linked with the bacterial signal sequence of E. coli. The hGH is secreted into the periplasmic space of bacterial cells by the signal peptides wherefrom the protein is purified.

(ii) Animals as organ donors for humans.
Answer:
Animals as organ donors for humans. Organ transplantation from animals to humans is called xenotransplantation. The first experiment was done in 1906 by French Surgeon Mathieu Jaboulay who implanted a pig’s kidney into one woman and a goat’s liver into another woman, but it was not successful.

Though now, organ transplantation from animals has been made possible in America and the United Kingdom. Of all, baboons and pigs have favored xenotransplant donors. Pig organs have been transplanted to humans several times in the last few years. Baboons are genetically close to humans, so they are most often used. Six baboon kidneys were transplanted into humans in 1964. Today, however, xenotransplantation is still experimental and there is a serious risk to the procedures.

(iii) Plant Variety Protection and Farmers’ Right Act.
Answer:
Plant Variety Protection and Farmers’ Right Act. This act provides the establishment of an effective system for the protection of plant breeder’s rights. It gives concurrent attention to the right of farmers, breeders and researchers, and the protection of public interest. Public interest is related to issues like compulsory licensing of rights and to the import of varieties incorporating Genetic Use Restriction Technology.

Question 9.
Explain the following terms in one or two sentences: intellectual property rights, humulin, and biofortified foods.
Answer:
1. Intellectual Property Rights: It is the general term covering patents, copyright, trademark, industrial designs, geographical indications, protection of layout designs of integrated circuits, and protection of undisclosed information (trade secrets).

2. Humulin: It is a crystalline suspension of human insulin. It is made using a chemical process called recombinant DNA technology and is of various types (Humulin R, U, N, L, 70/30, 50/50) depending on the percentage of insulin present in suspension. Humulin does not come from human beings, but they are synthesized in special non¬disease producing special lab strains of E. coli which are genetically altered by the addition of the gene for human insulin production. Humulin is identical in chemical structure to human insulin and is made in a factory by recombinant DNA technology.

3. Biofortified foods: They are the modified food rich in nutritional values. Biofortification is the process of breeding food crops that are rich in bioavailable micronutrients. These crops fortify themselves, they toad high levels of minerals and vitamins in their seeds and roots, which are harvested and eaten.

Question 10.
What are transgenic bacteria? Illustrate using any one example.
Answer:
Those bacteria whose DNA is manipulated to possess and express an extra (foreign) gene are known as transgenic bacteria.
Example: Human Growth hormone production by transgenic Escherichia coli.

1. The pituitary gland of humans produces growth hormones that regulate growth and development.
2. However, in children stunted growth occurs due to deficiency of the hormone which is called pituitary dwarfism.
3. For this, the hGH is now available as a recombinant protein.
4. The hGH-coding DNA sequence is linked with the bacterial signal sequence of E. coli.
5. The hGH is secreted into the periplasmic space of bacterial cells by the signal peptides wherefrom the protein is purified.

Question 11.
Write properties of stem cells. How is the population of stem cells maintained?
Answer:
Properties of stem cells:
The classical definition of a stem cell requires that it possess two properties:

1. Self-renewal: The ability to go through numerous cycles of cell division while maintaining the undifferentiated state,
2. Potency: The capacity to differentiate into specialized cell types. In the strictest sense, this requires stem cells to be either totipotent or pluripotent- to be able to give rise to any mature cell type, although multipotent or unipotent progenitor cells are sometimes referred to as stem cells. Apart from this, it is said that stem cell function is regulated in a feedback mechanism.

1. Self-renewal:
Two mechanisms exist to ensure that a stem cell population is maintained:

• Obligatory asymmetric replication: A stem cell divides into one mother cell that is identical to the original stem cell, and another daughter cell that is differentiated.
• Stochastic differentiation: When
  one stem cell develops into two differentiated daughter cells, another stem cell undergoes mitosis and produces two stem cells identical to the original.

2. Totipotency
They have the potential to develop into any cell found in the human body.

Question 12.
Show with a simple sketch the location of stem cells and their role in treatment.
Answer:
Stem cells:

Diseases and conditions where stem cell treatment is being investigated.

Question 13.
(i) Why are transgenic animals so called?
Answer:
Animals that have had their DNA manipulated to possess and express (foreign) genes are called transgenic animals.
Example: Transgenic mice, transgenic rabbits.

(ii) Explain the role of transgenic animals in (a) vaccine safety (b) biological products with the help of an example for each.
Answer:
(a) Role of transgenic animals in vaccine safety:

• Transgenic mice are being developed for use in testing the safety of vaccines before they are used on humans.
• Transgenic mice are being used to test the safety of the polio vaccine.

(b) Role of transgenic animals in biological products: In 1997, the first transgenic cow, Rosie, produced human protein-enriched milk of 2.4 gm per liter. The milk contained the human alpha-lactalbumin and was a more balanced product for human babies.

Question 14.
Name the genes responsible for making Bt cotton plants resistant to bollworm attack. How do such plants attain resistance against bollworm attacks? Explain. (CBSE 2012)
Answer:

1. Genes for making Bt cotton resistant to bollworm attack:
   1. acrylic
   2. cry IIAb
2. Specific Bt toxin genes are isolated from Bacillus Thurinsiensis and incorporated into the cotton plant.
3. The toxin is coded by a gene name cry.
4. The protein synthesized by these is insecticidal protein. It is present as an inactive protoxin.
5. Once the insect ingests the protoxin it is converted into the active form of toxin due to the alkaline medium of the gut.
6. The activated toxin binds to the surface of midgut epithelial cells and creates pores in them.
7. It causes swelling and breakdown and eventually leads to the death of the insect.

Question 15.
Explain the various steps involved in the production of artificial insulin. (CBSE 2017)
Or
Explain how Eli Lilly, an American company, produced insulin by recombinant DNA technology. (CBSE Delhi 2018C)
Answer:
Genetically engineered insulin:

1. Insulin contains two short polypeptide chains: chain A and chain Blinked together by disulfide bridges.
2. In mammals, insulin is synthesized as a pro-hormone. It contains an extra stretch called C-peptide.
3. C-peptide is absent in the mature insulin and is removed during maturation into insulin.
4. Production of insulin by rDNA techniques was achieved by an American company, Eli Lilly in 1983. It prepared two DNA sequences corresponding to A and B, chains of human insulin, and introduced them in plasmids of E. coli for production.
5. The Aand B chains produced were separated, extracted and combined, by creating disulfide bonds to form human insulin.

Question 16.
(i) What are transgenic animals?
Answer:
Animals whose DNA has been manipulated to possess and express an extra/foreign gene are known as transgenic animals.

(ii) Name the transgenic animal having the largest number amongst all the existing transgenic animals.
Answer:
Mice

(iii) Mention any three purposes for which these animals are produced. (CBSE Delhi 2018C)
Answer:
(a) Transgenic animals are designed to allow the study of how genes are regulated and how they affect the normal functions of the body and its developments, e.g. information is obtained as to how insulin has a role as a growth factor.
(b) Transgenic animals are designed to increase our understanding of how genes control the development of diseases; they serve as models for human diseases.
(c) Transgenic mice are being developed to test the safety of vaccines, e.g. polio vaccine has been tested on mice.

Question 17.
Explain the following terms in not more than 70 words.
(i) Single-cell proteins (SCP)
Answer:
Cells from different kinds of organisms such as bacteria, filamentous fungi, yeast, and algae are treated in different ways so that they are used as food or feed, are called single-cell protein. The biomass is obtained from both unicellular and multicellular microorganisms. The common substrate used for preparing such food containing SCP ranges from whey sawdust, and paddy straw. SCP provides a valuable protein-rich supplement in the human diet.

(ii) Biopatent
Answer:
A patent is a right granted by a Government to an inventor to prevent others to make commercial use of such an invention. At present patents that are granted for biological entities and the various products obtained from these organisms, are termed as biopatent.

Biopatents are being granted for the following:
(1) Strains of microorganisms
(2) Cell lines
3) Genetically modified strains of living organisms
(4) DNA sequences
(5) The proteins prepared by DNA sequences
(6) Biotechnological process
(7) Production process
(8) Products
(9) Product application.

(iii) Bioethics
Answer:
Bioethics is a set of standards that may be used to regulate our activities in relation to biological works.

The major bioethical concerns are as follows:
(a) Introduction of transgenes from one species to another violates the integrity of species.
(b) Transfer of human genes to other animals and vice versa is against ethics.
(c) Making of the clone.
(d) May cause risk to biodiversity.
(e) Suffering from animals used in biotechnology will increase.

(iv) Biopiracy
Answer:
The exploitation of patent biological resources without proper permission is called biopiracy. The collection of such material without a benefit-sharing agreement is likely to find its way into the list of criminal violations in many countries.

(v) Genetically modified food
Answer:
The food prepared from the production of genetically modified crops is called genetically modified food (GM food). It contains proteins produced by a transgene.

Question 18.
Briefly explain why are Transgenic animals produced? (CBSE Delhi 2013)
Answer:
Transgenic animals:
Transgenic animals are produced for the following purposes:

1. Transgenic animals are designed to allow the study of how genes are regulated and how they affect the normal functions of the body and its developments, e.g. information is obtained as to how insulin has a role as a growth factor.
2. Transgenic animals are designed to increase our understanding of how genes control the development of diseases; they serve as models for human diseases.
3. Transgenic animals that produce useful biological compounds are created by introducing a portion of the DNA that codes for that product, e.g. a-1 antitrypsin is produced for curing emphysema.
4. Transgenic mice are being developed to test the safety of vaccines, e.g. polio vaccine has been tested on mice.
5. Transgenic animals with more sensitivity to toxic substances are being developed to test the toxicity of drugs.

Question 19.
Describe the hazards of transgenic animals.
Answer:
Hazards of transgenic animals:

1. Proteins: Genes introduced in various organisms operate through the synthesis of polypeptide proteins and enzymes. However, foreign proteins are generally attacked by the defense system resulting in damaged biochemicals which may prove harmful and in the long term produce allergy.
2. Human organs: Replaceable human organs like kidneys, liver, heart, pancreas, etc. can be obtained only from autografts and isografts. Will it be ethical to grow the human body, human body organs for obtaining the required organs?
3. Human cloning: This can solve the problem of infertility. However, such a method of human reproduction will destroy the family system, fine human feelings, and the fabric of human society.
4. Recreation: It is not only a fancy but also the desires of numerous children, adults, and elders to see dinosaurs live.

Question 20.
Compare and contrast the advantages and disadvantages of the production of genetically modified crops.
Or
Write advantage of GM crops. (CBSE Outside Delhi 2019)
Answer:
The advantages of the production of genetically modified crops are:

1. They have proved to be extremely valuable tools in studies on plant molecular biology, regulation of gene action, identification of regulatory/ promontory sequences.
2. Genetically modified crops have improved agronomic and other features such as resistance to biotic and abiotic stresses.
3. Over-ripening losses can be reduced, e.g. flavor saves tomato.
4. Nutritional values are improved, e.g. Golden rice has high vitamin A content.
5. Viral resistance can be introduced.
6. The number of pharmaceuticals like insulin, interferon, blood clotting factors are improved.
7. Insect resistance can be introduced, e.g. cry gene can be introduced into cotton, wheat, and rice from Bacillus Thuringiensis.

The main disadvantages of the production of genetically modified crops are:

1. Many transgenes are expressed at low levels which usually limit their usefulness.
2. Sometimes, the expression of transgenes is suppressed in transgenic plants, this is called gene silencing.
3. The undesirable features are also carried along with desirable features in transgenic plants such as necrosis, reduced growth, sterility, etc.
4. Genetic pollution can be there.
5. Weeds also become resistant.
6. Bt cotton, Bt wheat also destroy pollinators and disseminators.
7. The product of transgene may be allergic or toxic.

Question 21.
Why is the introduction of genetically engineering lymphocyte into an ADA deficiency patient, not a permanent cure? Suggest a possible permanent cure. (CBSE 2010, 2011)
Or
Explain how a hereditary disease can be corrected? Give an example of the first successful attempt made towards this objective. (CBSE 2011, 2019 C)
Or
Explain enzyme replacement therapy to treat ADA deficiency. (CBSE Outside Delhi 2016, 2019 C)
Or
What is gene therapy? Illustrate using the example of adenosine deaminase (ADA) deficiency? (CBSE Delhi 2011, 2013, 2016)
Or
Two children A and B aged 4 and 5 years respectively visited a hospital with a similar genetic disorder. Girl A was provided enzyme-replacement therapy and was advised to revisit periodically for further treatment. Girl B was, however, given therapy that did not require a revisit for further treatment.
(a) Name the ailments the two girls were suffering from.
(b) Why did the treatment provided to girl A require repeated visits?
(c) How was girl B cured permanently? (CBSE Delhi 2019)
Answer:
Gene Therapy. It is defined as the introduction of a normal functional gene into cells that contain the defective allele of the concerned gene with the objective of correcting a genetic disorder or an acquired disorder.

Treatment of ADA deficiency:

1. Gene therapy was used to correct the genetic disorder called Severe Combined Immunodeficiency (SCID) syndrome produced by adenosine deaminase (ADA) deficiency.
2. In this, Normal ADA gene copies were produced by cloning.
3. Packed into a retrovirus, most of the viral genes were replaced by the ADA gene.
4. Lymphocytes were isolated from the patients.
5. Recombinant DNA of the recombinant retroviruses was used to infect the lymphocytes.
6. The infected cells expressing the ADA gene were injected back into the patients.
7. The normal ADA gene was then expressed in the patients and ADA deficiency is partially corrected. If the gene isolated from bone marrow cells producing ADA is introduced into embryonic cells at early stages, it could provide a permanent cure.

Question 22.
How is the transgenic tobacco plant protected against Meloidogyne incognita? Explain the process? (CBSE 2009)
Or
Explain the process of RNA interference. (CBSE Delhi 2011, 2016)
Or
How has the use of Agrobacterium as vectors helped in controlling Meloidogyne incognita infestation in tobacco plants? Explain in the correct sequence. (CBSE 2018, Sample paper 2020)
Answer:
Protection of tobacco plant against Nematodes, Meloidogyne incognita:

1. A nematode Meloidogyne incognita infects tobacco plants and reduces their yield.
2. The specific genes (in the form of cDNA) from the parasite are introduced into the plant using Agrobacterium as the vector.
3. The genes are introduced in such a way that both sense/coding RNA and antisense RNA (Complementary to the sense/coding RNA) are produced.
4. Since these two RNAs are complementary, they form a double-stranded RNA (ds RNA)
5. This neutralizes the specific RNA of the nematode, by a process called RNA- interference.
6. As a result, the parasite cannot live in the transgenic host, and the transgenic plant is protected from the pest.

Question 23.
What are the ethical concerns of biotechnology?
Answer:

1. Biotechnology is producing newer genotypes. Some of them can be extremely harmful due to intragenomic interactions and mutations.
2. It introduces unfamiliar proteins into transgenics which may react to form toxins and allergens.
3. The genes introduced into crops can pass into weeds through pollen transfer. It will produce superweeds.
4. It is going to cause genetic pollution which is likely to disturb natural balance in a big way.
5. Animals employed in experiments of biotechnology are made to suffer.
6. Animals being used to produce particular structures and pharmaceutical proteins are reduced to the status of factories.
7. As a gene is introduced from outside into an organism, its integrity as a species is violated.
8. Transfer of genes from human beings to specific animals or vice-versa violates the concept of humanness.
9. Biotechnology has no respect for living beings. Its only goal is to exploit them for commercial use in benefitting human society.
10. In their race to gain supremacy over others, companies and individuals are rushing for biopatents even of those products which are produced through the traditional knowledge of tribals, communities, and societies.

Question 24.
The Green Revolution succeeded in increasing the yield of crops but it is not sufficient to feed the growing human population. Thus there is a need for another green revolution.
(i) Name the technique which will help in increasing the yield of crops.
Answer:
Genetic engineering (Recombinant DNA technology).

(ii) Name any two genetically modified crops.
Answer:
(a) Bt cotton
(b) ‘Flavr Savr Tomato’

(iii)What is golden Rice’?
Answer:
Golden rice is a transgenic variety of rice (Oryza sativa) that contains good quantities of (3-carotene (provitamin A – inactive state of vitamin). Since the grains of the rice are yellow in color due to [3-carotene, the rice is called golden rice.

(iv) Name a natural genetic engineer.
Answer:
Agrobacterium tumefaciens.

Question 25.
How have transgenic animals proved to be beneficial in:
(i) Production of biological products
(ii) Chemical safety testing. (CBSE 2014)
Answer:
(i) Production of biological products:
(a) Medicines required for treating human diseases are obtained by genetic engineering.
(b) a-1-antitrypsin used to treat emphysema.
(c) Transgenic cow ‘Rosie’ produces human-protein enriched milk.
(ii) Chemical safety (Toxicity/safety testing) Transgenic animals are made that carry genes that make these more sensitive toxic substances than non-transgenic animals.

Question 26.
List the disadvantages of insulin obtained from the pancreas of slaughtered cows and pigs.
Answer:

1. A slaughtered animal produces very little hormone so that the demand was always higher than the supply.
2. It is unethical to slaughter animals for obtaining the drug.
3. Contamination was quite common.
4. The immune response is common.

Question 27.
List the advantages of recombinant insulin.
Answer:

1. Recombinant insulin is exactly similar to human insulin and is, therefore, also called humulin.
2. It is available in pure form with little chances of contamination.
3. There is no slaughtering of animals.
4. There is no immune response or any other side effect.
5. There is enough manufacturing capacity so that the chances of short supply are few.

Question 28.
Explain the process of synthesis of insulin.
Answer:
Production of human insulin: Gene transfer involves essentially the following stages:

Steps involved in gene transfer for the production of human insulin:

1. Isolation of donor or DNA segment. A useful DNA segment is isolated from the donor organism.
2. Formation of Recombined DNA (rDNA). Both the vector and donor DNA segments are cut in the presence of restriction endonuclease. In the presence of ligase DNA segments of both are joined to form rDNA.
3. Production of Multiple Copies of rDNA. The next step in the process is the production of multiple copies of this recombinant DNA.
4. Introduction of rDNA in the recipient organism. This rDNA is inserted into a recipient organism.
5. Screening of the transformed cells. The recipient (host) cells are screened for the presence of rDNA and the product of the donor gene. The transformed cells are separated and multiplied, using an economical method for its mass production.

Here we are providing Class 12 Biology Important Extra Questions and Answers Chapter 14 Ecosystem. Important Questions for Class 12 Biology are the best resource for students which helps in Class 12 board exams.

Class 12 Biology Chapter 14 Important Extra Questions Ecosystem

Ecosystem Important Extra Questions Very Short Answer Type

Question 1.
What is an ecosystem?
Answer:
Ecosystem. It is a functional unit of nature where living organisms interact among themselves and also with the physical environment.

Question 2.
Who coined the term ecosystem?
Answer:
Sir Arthur Tansley (1935).

Question 3.
Name two major kinds of ecosystems.
Answer:

1. Terrestrial ecosystem.
2. Aquatic ecosystem.

Question 4.
Write three examples of terrestrial ecosystems.
Answer:
Forests, grasslands, and deserts ecosystem.

Question 5.
Give one example of the smallest and another of large-sized ecosystems.
Answer:

1. Pond
2. Forest.

Question 6.
Give three examples of freshwater ecosystems.
Answer:
Ponds, Lakes, and Streams ecosystems.

Question 7.
Name two saltwater ecosystems.
Answer:
Marine and estuaries ecosystems.

Question 8.
Name two man-made ecosystems.
Answer:

1. Spacecrafts
2. Aquarium
3. Crop field.

Question 9.
Write two examples where man has interfered in the ecosystem.
Answer:

1. Cutting of forests
2. Construction of dams.

Question 10.
How do decomposers obtain food?
Answer:
Decomposers release their enzymes into dead and decaying plants and animal remains and absorb the simple inorganic substances.

Question 11.
Give two examples where there is the transfer of matter from terrestrial to the aquatic ecosystem or vice-versa.
Answer:

1. Dropping of leaves of riverbank trees into the water.
2. Terrestrial birds diving into the water to catch fish.

Question 12.
Write the equation that helps in deriving the net primary productivity of an ecosystem. (CBSE Delhi 2013)
Or
How will you calculate net productivity?
Answer:
GPP – R = NPP

Gross primary productivity – Respiratory processes (loss) = Net primary productivity
Or
Net primary productivity = Gross productivity – respiration loss.

Question 13.
Why are green plants not found beyond certain depths in the ocean? (CBSE 2011)
Answer:
The sunlight cannot penetrate beyond certain depths in the ocean.

Question 14.
Why are green algae not likely to be found in the deepest strata of the ocean? (CBSE Outside Delhi 2013)
Answer:
Green algae are not likely to be found in the deepest strata of the sea because the environment is perpetually dark and the sun as a source of energy is not available.

Question 15.
Name two aquatic ecosystems which have a rich diversity of macrophytes.
Answer:
Wetlands and lakes.

Question 16.
Name two structural features of the ecosystem.
Answer:

1. Species composition
2. Stratification.

Question 17.
How does the standing state of nutrients differ?
Answer:
The standing state of nutrients differs from one ecosystem to another or with the season in the same ecosystem.

Question 18.
Define detritus.
Answer:
Dead plant and animal remains are called detritus.

Question 19.
Name above ground and below ground detritus.
Answer:

1. Dried plant parts and dead animals.
2. Dead roots (root detritus).

Question 20.
What is nutrient immobilization?
Answer:
Soil nutrient gets tied up with biomass of microbes and is not available for other organisms.

Ecosystem Important Extra Questions Very Short Answer Type

Question 1.
List the abiotic components of an ecosystem.
Answer:
The abiotic components of an ecosystem are of four types:

1. Inorganic substances like C, N₂, O₂, C0₂, H₂0, etc.;
2. Organic compounds like carbohydrates, proteins lipids, etc.;
3. Climatic factors (including temperature, light, wind, gases, humidity, rain, and water) (also wave action, water currents),
4. the edaphic factor which includes soil.

Question 2.
List the differences between biotic and abiotic components of the ecosystem.
Answer:
Differences between Biotic and Abiotic Components of the ecosystem:

Biotic components	Abiotic components
1. Biotic components of an ecosystem are those Living organisms that are different members of a community.	1. Abiotic components are non-living factors.
2. Biotic components of an ecosystem are: (i) Producers (ii) Consumers (iii) Decomposers.	2. It includes water, minerals, salts, humidity, light, temperature, pH, wind, topography, and background.

Question 3.
List the kinds of ecosystems.
Answer:
An ecosystem can be as small as an aquarium in the house or as big as an ocean. The biotic community is spread everywhere. Land, water, and air show the presence of living things.

The major ecosystems of the world are:

Aquatic ecosystem	Terrestrial ecosystem	Man-made ecosystem
[A] Marine (i) Oceans (ii) Seashore (iii) Brackish[B] Fresh Water (i) Lakes and rivers (ii) Spring and ponds (iii) Swamp	[A] Forest (i) TropicaL (ii) Temperate (iii) Taiga[B] Grassland (i) Tropical (ii) Temperate[C] Desert [D] Tundra	(i) Aquarium (ii) Crop fields (iii) Spaceship (iv) Garden and parks (v) Small fishery tanks (vi) Animal husbandry (vii) Dams and reservoirs

Question 4.
Show the processes involved in the decomposition of detritus.
Or
Describe the process of decomposition of detritus under the following heads: Fragmentation, Leaching, Catabolism, Humification, and Mineralisation. (CBSE 2010)
Answer:
Process of decomposition:

Processes involved in the decomposition of detritus

1. Fragmentation: Detrivores break down detritus in small particles.
2. Leaching: Water-soluble inorganic nutrients seep inside the soil and get precipitated as unavailable salt.
3. Catabolism: Here, bacteria and fungus degrade detritus by their enzymes into simple inorganic ions.
4. Humification: Formation of hummus at a slow rate.
5. Mineralization: Release of minerals from humus by microbes.

Question 5.
What are the functions of the ecosystem?
Answer:
Functions of the ecosystem. Ecosystems possess a natural tendency to persist. This is made possible by a variety of functions (activities undertaken to ensure persistence) performed by the structural components.

The key functions of the ecosystem are:

• Productivity;
• Energy flow;
• Nutrient cycling;
• Decomposition

Question 6.
What is a food chain? List the kinds of food chains.
Answer:
Food Chain: A nutritive interaction among biotic communities (organisms) involving a producer, various levels of consumers, and a decomposer forms a food chain. Each step in a food chain is called a trophic level. Kinds of the food chain. There are 3 kinds of food chains: predator, parasitic and saprophytic chains.

Question 7.
In relation to energy transfer in the ecosystem, explain the statement “10 kg of deer’s meat is equivalent to 1 kg of lion’s flesh”.
Answer:
Passage of food energy from lower trophic level to higher trophic level follows 10% law. 10% law states that only 10% of total energy is transmitted from one trophic level to another. The rest of the energy is dissipated. A lion feeds on deer (lower trophic level). So 10 kg of meat of deer will form only 1 kg of lion’s flesh. The rest of the food energy present in deer meat is wasted and dissipated.

Question 8.
Primary productivity varies from ecosystem to ecosystem. Explain.
Answer:
Primary productivity depends upon the producers (plant species), their photosynthetic potential, soil, climate, and other environmental factors of an ecosystem. They are seldom similar in different ecosystems. Therefore, primary productivity varies from ecosystem to ecosystem.

Question 9.
What is an incomplete ecosystem? Explain with the help of a suitable example.
Answer:
An ecosystem comprises biotic and abiotic components. Abiotic components include light, air, water, temperature, humidity, etc., while biotic factor comprises all living organisms. The absence or limited availability of any component (either abiotic or biotic) makes an ecosystem incomplete like the profundal and benthic zones in an aquatic ecosystem.

Question 10.
What are the shortcomings of ecological pyramids in the study of ecosystems?
Answer:
Ecological pyramids are the graphical representation of ecological parameters. These are characterized by a pyramid of the number, the pyramid of mass, and the pyramid of energy in an ecosystem. The assumption of a simple food chain is the major shortcoming of ecological pyramids.

If we do not provide a food web, a clear position of the trophic levels of an organism cannot be given. Saprophytic organisms are not given any place in the ecological pyramid, though they are an important component in an ecosystem.

Question 11.
Write a note on secondary productivity.
Answer:
Secondary productivity. It refers to synthesized biomass at different trophic levels beginning from primary consumer level to top-level carnivores. At each consumer level, a part of the energy is used for respiration and a part of it is stored. If there is more storage than consumption, the biomass of the population would be higher at the end of a time period than at the beginning. This rate of increase in biomass of heterotrophs is called secondary productivity.

Question 12.
Sketch Pyramid of energy. Why is pyramid energy upright? (CBSE Delhi 2019 C)
Answer:
A pyramid of energy is always upright: It is a graphic representation of energy, amount of energy trapped per unit time, and area in different trophic levels of a food chain with producers forming the base and top consumer at the top.

An ideal pyramid of energy. Observe that primary producers convert only 1% of the energy from the sunlight available to them into NPP.

There is the unidirectional flow of energy in a food chain. As the energy passes from one trophic level to a higher trophic level along the food chain, its amount decreases. So, the pyramid of energy is always upright.

Question 13.
Construct a pyramid of biomass starting with phytoplankton. Label its three trophic levels. Is the pyramid upright or inverted? Justify your answer. (CBSE Sample Paper 2020)
Answer:
A pyramid of biomass is an aquatic ecosys¬tem and is always inverted. The biomass of primary producer (PP) phytoplankton is smaller than the biomass of zooplankton. The latter constitute primary consumers (PC). The biomass of carnivores, i.e. fish, a secondary consumer (SC), is more than the biomass of the primary consumer. So the pyramid of the biomass of the aquatic ecosystem is inverted.

Pyramid of biomass

Question 14.
What is eco-succession? Write its kinds and pattern.
Answer:
A continuous change in a biotic community is called biotic or eco-succession. A stage of climax is also achieved in the process of succession. Succession is of two types primary and secondary. The pattern of succession in any community is decided by pre-existing conditions of that place.

Three kinds of patterns have been observed:

1. Xerosere,
2. Hydrosere and
3. Merosere.

Question 15.
What are the causes of ecological succession?
Answer:

• Biotic and physiographic factors operating simultaneously are the causes of ecological succession.
• Biotic factors, direct succession, and physiographic factors include the climate and other physical factors such as erosion of hills, filling up of lakes and streams.

Question 16.
Give a generalized mode of ecosystem nutrient cycling.
Answer:
A generalized model of ecosystem nutrient cycling:

A generalized model of ecosystem nutrient cycling: Nutrients are brought in (input), moved out (output), and cycled internally in the ecosystem. Boxes represent ecosystem components and arrows show the pathways of nutrient transfers.

Question 17.
Where would you look for signs of secondary succession? When does secondary succession end?
Answer:
Secondary succession occurs where an early community has been damaged leaving a few organisms and considerable organic matter. A destroyed grassland, destroyed forest, or destroyed area by fire or floods will be the site for secondary succession. The development of climax forests marks the end of secondary succession. A destroyed grassland takes 50 to 100 years to recuperate and a destroyed forest takes 200 years to fully recover.

Question 18.
Name the pioneer and the climax species in a waterbody. Mention the changes observed in the biomass and biodiversity of the successive serai communities developing in the waterbody. (CBSE 2009)
Answer:

1. Pioneer communities are the small phytoplankton, which is replaced with time by rooted-submerged plants, rooted-floating angiosperms followed by free-floating plants, then reed swamp, marsh-meadow, scrub, and finally the trees.
2. The climax again is forest.

Ecosystem Important Extra Questions Long Answer Type

Question 1.
What is an ecosystem? Write its main components.
Answer:
Ecosystem. A stable, self-supporting ecological unit resulting from an interaction between a biotic community (living organisms) and its abiotic environment is called an ecosystem.

An ecosystem comprises two main components:

1. biotic including plants, animals, and microorganisms and
2. abiotic mainly including substratum, water, minerals, temperature, carbon dioxide, and oxygen. It must also receive a constant supply of energy (light).

Question 2.
Briefly describe the biotic components of an ecosystem.
Answer:
Biotic components: Of an ecosystem’s biotic components, the plants are producers as they introduce food materials and energy into the living world. The animals are consumers because they get food and energy by consuming plants directly thus called primary consumers (herbivores); secondary/ tertiary consumers (carnivores) obtain energy and food indirectly from plants, and microorganisms are decomposers for they flourish by breaking dead organic matter to simple substances that are returned to environment for reuse by plants.

In an ecosystem, nutrients are used again and again in a cyclic manner, whereas energy trapped from sunlight is lost as heat.

Question 3.
Give an account of factors affecting the rate of decomposition.
Answer:
Factors affecting decomposition:

1. The upper layer of soil is the main site of decomposition processes in the ecosystem.
2. The rate of decomposition of detritus is affected by climatic factors and the chemical quality of detritus.
3. Temperature and soil moisture affect the activities of root microbes.
4. The chemical quality of detritus is determined by the relative proportion of water-soluble substances, polyphenols, lignin, and nitrogen.

Question 4.
List the important differences between producers and decomposers.
Answer:
Differences between producers and decomposers:

Producers	Decomposers
(i) These are organisms that synthesize their own food by the process of photosynthesis. These are also called autotrophs.	(i) These organisms feed on the dead bodies of plants and animals.
(ii) They convert the raw materials in the earth and water into carbohydrates which give them food.	(ii) They enrich the earth with raw materials trapped in dead bodies of plants and animals.
(iii) They are dependent on decomposers for soil nutrients.	(iii) They are dependent on plants and animals for their food.

Question 5.
Explain the terms standing crop, biomass, and standing state.
Or
State what does a standing crop of a trophic level represent? (CBSE Outside Delhi 2013)
Answer:
Standing crop: Each trophic level has a certain mass of living material at a particular time called the standing crop. Biomass. The standing crop is measured as the mass of living organisms (biomass) or the number in a unit area. The biomass of a species is expressed in terms of fresh or dry weight. Measurement of biomass in terms of dry weight is more accurate as the moisture content of biomass fluctuates greatly.

Standing State: Organisms need a constant supply of nutrients to grow, reproduce and regulate various body functions. The amount of nutrients, such as carbon, nitrogen, phosphorus, calcium, etc. present in the soil at any given time, is referred to as the standing state. It varies in different kinds of ecosystems and also on a seasonal basis.

Question 6.
Give a diagrammatic representation of trophic levels in an ecosystem.
Answer:

Diagrammatic representation of trophic levels in an ecosystem.

Question 7.
Sometimes due to biotic/abiotic factors, the climax remains in a particular serai stage (pre-climax) without reaching climax. Do you agree with this statement? If yes give a suitable example.
Answer:
Sometimes pre-climax stage remains in a particular serai stage without reaching the climax because during ecological succession any change in abiotic and biotic components may affect the particular serai stage, leading to the pre-climax stage before the climax is achieved.

This type of condition occurs in the presence of seeds and other propagules. This secondarily based area may be invaded by moss or exotic weeds thus exhibiting succession seriously and the climax community is never regenerated.

Question 8.
Explain the meaning of the food web and illustrate with a ray diagram.
Answer:
Food web. In nature, the food chains are not strictly linear, but are interrelated and interconnected with one another. Generally, the various food chains in a community are so interlinked as to form a sort of web. As a result, one animal may be a link in more than one food chain.

A network of food chains in a community is referred to as a food web. A food web may have all or some of the three types of food chains, i.e. detritus, predator, and parasitic. The food webs become more complicated because of the variability of taste and preference, availability and compulsion, and several other factors at each level. For example, tigers normally do not eat fish or crab, but they are forced to feed on them in the Sundarbans.

Question 9.
Starting from a bare rock or a site of volcanic eruption, trace the organisms that participate in the process of succession. (CBSE Delhi 2011)
Or
Describe the process of succession on a base rock. (CBSE 2012)
Answer:

1. Simple organisms appear first of all on such an exposed site as lichens. Lichens make conditions suitable for mosses (bryophytes).
2. Gradually a variety of complex organisms join the community.
   Stages of biotic succession.
3. Finally, large plants, trees, etc. appeared. It can be illustrated that lichens are pioneers, followed by mosses, annual grasses, perennial herbs, shrubs, and finally trees along with their characteristic animal populations.

Question 10.
How does succession differ in terrestrial and aquatic systems? Give salient points. (CBSE Delhi 2019)
Answer:
Differences between terrestrial and aquatic succession:

Succession on land/rock	Succession In Water
1. lichens and mosses are the pioneer community.	1. Phytoplankton is a pioneer community.
2. Soil is formed by the action of lichens.	2. Waterbodies are prone to silting due to soil erosion.
3. There is a deficiency of water.	3. Water is abundant.
4. The various stages are crustose lichen stage, foliose lichen stage, moss stage, herb stage, shrub stage, and forest stage.	4. The various stages are the plankton stage, floating stage, rooted stage, swamp stage, woodland stage, and forest stage.

Question 11.
Explain the difference between the serai stage and climax community during succession.
Answer:
Change during succession:

Characteristics	Seral	Climax
Community structure: Size of individuals Ecological niches	Small Few, generalized	large Many specialized
Community organization:	Simple	Complex
Community functions: Food chains and food Efficiency of energy use Nutrient conservation	Simple low low	Complex High High

Question 12.
(a) Write the pioneer species each of xerarch and Hydr-arch successions. Which type of climax community is attained by both these successions?
Answer:

• Pioneer species of xerarch succession: Wind-borne lichen propagules settle on wet rock soon after rain or heavy dew. Pioneer lichens are crustose lichens. Examples: Graphis, Rhizocarpon.
• Climax stages of xerarch succession: Trees growing in areas occupied by shrubs.
• Pioneer species of Hydr-arch succession: Phytoplankton examples are diatoms, green flagellates, single-celled or filamentous green algae as well as blue-green algae.
• Climax stages of hydra succession: Climax forest of trees

(b) Why is secondary succession faster than primary succession? Explain. (CBSE 2019 C)
Answer:
Secondary succession is faster than primary succession:

• It has a secondarily based area built-in soil organic matter.
• It is biologically fertile.
• Underground parts, some seeds, and remnant species quickly give rise to a new community.

Secondary succession takes 50-100 years to complete in grassland and 100—200 years for the development of the forest. It takes short time as com¬pared to primary succession.

Question 13.
Outline the salient features of the ecosystem nitrogen cycling.
Answer:
Salient features of the ecosystem nitrogen cycle are listed below.

1. The ultimate source of nitrogen is an atmosphere that cannot be directly metabolized by living organisms.
2. Nitrogen-fixing bacterial activities facilitate the entry of nitrogen into biological pathways.
3. Azotobacter and Clostridium are the major free nitrogen-fixing bacteria in the soil.
4. Ammonification is done by many bacteria.
5. Ammonia is converted into nitrate by a group of chemoautotrophic bacteria through a two-step process called nitrification.
6. Denitrifying bacteria transform nitrate nitrogen to nitrous and nitric oxides ultimately to gaseous nitrogen.
7. Most plants absorb nitrate from soil.

Question 14.
List the features of the phosphorus cycle.
Answer:
Features of the phosphorus cycle are listed below.

1. The natural reservoir of phosphorus is a rock in the form of phosphates.
2. Minute quantities of phosphates get dissolved in the soil solution during weathering of rocks.
3. Phosphates enter the plants through their roots and then the food chain.
4. The organic wastes and dead organisms are decomposed by phosphate-solubilizing bacteria, which release phosphorus back into the soil.
5. The atmospheric input of phosphorus through rainfall or gaseous exchange of phosphorus between organisms and the environment is negligible.

Question 15.
“In a food-chain, a trophic level represents a functional level, not a species. Explain.” (CBSE Delhi 2016)
Answer:
In a food chain, a trophic level represents a functional level:

1. Trophic level is a step or division of the food chain. It is characterized by the method of obtaining food.
2. The number of trophic levels is equal to the number of steps in the food chain.
3. The two fundamental trophic levels are producers and consumers.
4. Producers are autotrophic organisms found in the ecosystem which synthesize food from raw materials. There are many such organisms at this level that comprise the first trophic level and not just one species, e.g. green algae plant, phytoplankton, etc.
5. Consumers are heterotrophic organisms. Herbivores are primary consumers. In a pond ecosystem many crustaceans, larvae of insects constitute this trophic level.

Similarly, at the next trophic level, a number of small fishes, not just one species consist of the secondary consumer.

Question 16.
Explain the impact of human activities on the carbon cycle in nature and list its harmful effects. (CBSE Delhi 2019 C)
Answer:
Impact of human activities on the carbon cycle: Carbon is a component of all living organic compounds of protoplasm. It constitutes 49% of dry weight. C0₂ is present in the air. It is dissolved in water as carbonic acid and bicarbonates. It is also present in fossil fuels and graphite in rocks.

There is a regulation of the carbon cycle in nature. Rapid deforestation and the massive burning of fossil fuel for energy and transport have significantly increased the rate of releasing C0₂ into the atmosphere. Harmful effect. Greenhouse effect and global warming.

Question 17.
What are ecosystem services? Briefly explain. (CBSE Delhi 2019)
Answer:
Ecosystem Services. The products of ecosystem processes are named ecosystem services.
Examples:
The following services are provided by the forest ecosystem.
They are:

• purify air,
• mitigate droughts and floods,
• help in the cycling of nutrients,
• provide habitat to a number of the wildlife,
• act as a storehouse of carbon,
• influence the hydrological cycle and
• maintain biodiversity.

The value of services of biodiversity is difficult to determine. Robert Constanza et al. have tried to put price tags on nature’s life-support services. Researchers have estimated them to be 33 trillion US dollars a year, while our global gross production is only 18 trillion US dollars.

Question 18.
What are the two main components of an ecosystem? Describe the physical factors which affect the distribution of organisms in different habitats.
Answer:
Abiotic (physical) and biotic components are the two main components of an ecosystem.
Abiotic components or physical factors:
1. Temperature: The physiological and behavioral adaptations of most animals depend upon the changes in the environmental temperature. The rates of photosynthesis and respiration in plants also fluctuate depending upon the change in temperature.

2. Water: The extent to which an organism is dependent on an abundant water supply depends on its requirements and its ability to conserve it in adverse conditions. Organisms living in dry habitats generally have good water conservation such as in cacti and camels.

3. Light: This is essential for all green plants and photosynthetic bacteria, and for all the animals dependent on the plants.

4. Humidity: This is important because it can affect the rate at which water evaporates from the surface of an organism, which in turn influences its ability to withstand drought.

5. Wind and air currents: This particularly applies to plants. Only plants with strong root systems and tough stems can live in exposed places where winds are fierce. The wind is also instrumental in the dispersal of spores and seeds.

6. pH: This influences the distribution of plants in soil and fresh-water ponds. Some plants thrive in acidic conditions others in neutral or alkaline conditions. Most are highly sensitive to changes in pH.

7. Soil nutrients: These particularly affect the distribution of plants in the soil.

8. Water currents: Particularly in rivers and streams. Only organisms capable of swimming or avoiding strong currents can survive.

9. Topography. Minor topographical differences may be just as important in influencing the distribution of organisms as wide geographical separation.

10. Background. The distribution of organisms whose shape or coloration is such that they are camouflaged when viewed against a particular background is related to the general texture and pattern of the environment.

Question 19.
Name the two fundamental trophic levels and describe the general makeup of each.
Answer:
The two fundamental trophic levels include the following:
1. Producers (Autotrophic organisms): Green plants are the producers in any ecosystem. They also include photosynthetic bacteria. The producers use radiant energy of the Sun during photosynthesis whereby carbon dioxide is assimilated and the light energy is covered into chemical energy.

This energy is locked up into the energy-rich carbon compounds i.e. carbohydrates. The oxygen that is evolved as a by-product in photosynthesis is used in respiration by all living organisms.

2. Consumers (Heterotrophic organisms): They are the living members of the ecosystem which consume the food synthesized by the producers. All living animals are thought to be consumers.

The consumers may be of the following types:
(a) Primary consumers (also called first-order consumers) which are purely herbivorous and depend upon green plants i.e. on producers for their food e.g., Cow, Goat, Rabbit, Deer, Grasshopper, and other insects.
(b) Secondary consumers (also called second-order consumers) are carnivorous animals and eat the flesh of herbivorous animals e.g., Tiger, Lion, Dog, Cat, Frog, etc.
(c) Tertiary consumers are the carnivorous animals that eat other carnivores e.g. Snake eats a frog, birds eat fishes
(d) Top consumers are carnivores of an ecosystem that are not killed and eaten by other animals e.g. Lions, vultures, etc.

Question 20.
Explain the meaning of biotic succession taking an example of succession in a hydrosere.
Answer:
Biotic Succession: The organisms interact among themselves. They not only influence their community but also change their physical or abiotic environment. The alteration in the physical environment is such as to continually favor another set of organisms till a stable community is formed.

Such a biologically controlled modification in the composition of a community of a particular area is called biotic succession or ecological succession. Biotic succession is also known as a successive development of different communities in a particular area till a climax community is formed.

The following seven stages of the hydrosere pattern of succession can be observed:

1. Phytoplanktonic stage (Free-floating angiosperm)
2. Rooted submerged stage
3. Rooted floating stage
4. Amphibian or Reed Swamp stage
5. Sedge meadow stage
6. Woodland stage
7. Forest stage.

Question 21.
Give the graphic representation of the nitrogen cycle.
Answer:

Nitrogen Cycle in Nature.

Question 22.
Distinguish between the following:
(i) Grazing food chain and detritus food chain (CBSE 2008)
Answer:
Differences between grazing food chain and detritus food chain:

grazing food chain	detritus food chain
1. Primary source of energy is solar radiation.	1. It is detritus.
2. First trophic level is formed of all producers (plants).	2. Detritus Food Chain (DFC) begins with detritus or dead organic matter. Detrivores or decomposers feed over it.
3. long-sized food chains.	3. Small-sized food chains.
4. Examples: Predatory food chains on land and in water.	4. Examples: Fallen leaves of mangroves in the brackish zone of South Florida.

(ii) Production and decomposition
Answer:
Differences between production and decomposition:

Production	Decomposition
1. It is the process of formation of organic food material by the process of photosynthesis.	1. It is the process by which complex organic compounds are broken into simpler inorganic substances.
2. It is done by green plants (producers).	2. It is done by bacteria and fungi (decomposers).

(iii) Upright and inverted pyramid
Answer:
Differences between Upright pyramid and Inverted pyramid:

Upright Pyramid	Inverted Pyramid
In a terrestrial habitat, the pyramid of biomass is maximum at the level of producers and there is a progressive decrease in biomass from lower to higher trophic levels.	In aquatic habitats, the pyramid of biomass is inverted or spindle-shaped whereas the biomass of a trophic level depends upon the reproductive potential and longevity of the members because the biomass of phytoplankton is less than that of zooplanktons.

(iv) Food chain and food web
Answer:
Differences between food chain and food web:

food chain	food web
1. A single energy path where energy is transferred from producer to successive order of consumers.	1. It is the network of various food chains which are interconnected with each other like an interlocking pattern.
2. All food chain starts from plants that are the original source of food.	2. It has many linkages and intercrossing among the producers and consumers.

(v) Litter and detritus
Answer:
Difference between Litter and detritus:

Litter	Detritus
1. It refers to discarded paper and plastics.	1. Dead plant remains like leaves, bark, flowers, and remains of dead animals including fecal matter is catted detritus.
2. It cannot be decomposed.	2. It can be decomposed.

(vi) Primary and secondary productivity.
Answer:
Differences between primary productivity and secondary productivity:

primary productivity	secondary productivity
The amount of biomass or organic matter produced per unit area over a time period by plants during photosynthesis is catted primary productivity.	The rate of formation of new organic matter by consumers is called secondary productivity.

Question 23.
Define ecological pyramid and describe pyramids of number and biomass.
Or
Construct a pyramid of biomass starting with phytoplankton. Label three trophic levels. Is the pyramid upright or inverted? Why? (CBSE 2009)
Answer:
Definition: An ecological pyramid is a graphical/mathematical representation of an ecological parameter, like a number or biomass or accumulated energy at different trophic levels in a food chain in an ecosystem.
1. Pyramid of number: The pyramid is the graphical/mathematical representation of a number of organisms present at each trophic level in a food chain of a particular ecosystem.

When the number of organisms at successive levels are calculated mathematically and plotted, they assume the shape of a pyramid. This is called a pyramid of numbers. The base represents the number of producers whereas the tip is represented by top consumers; inverted pyramids can also be formed.

In the given pyramid of the number, the number of producers (here grass and trees are the producers) will be far more than any other level. The number of consumers will be comparatively less than herbivores. You can guess if the number of carnivores increases and exceeds the herbivores then what the result will be?

The figure shows an inverted pyramid and there you can see producer is one tree and dependents like primary and secondary consumers are many. Parasite insects of birds will be definitely more than the number of birds.

Pyramid of Numbers:
A. Straight—Forest and Water ecosystem
B. Inverted—Tree Ecosystem.
C. SpindLe shaped

2. Pyramid of Biomass: Biomass is defined as the total weight of dry matter present in an ecosystem at a given time. Graphical measurement of biomass is called the pyramid of biomass. Trophic levels at the base represent biomass as numbers in a pyramid of numbers. Here also you can obtain an upright and inverted pyramid.

Pyramid of biomass Upright-Tree and forest ecosys¬tem. Inverted-Aquatic ecosystem.

The aquatic pyramid of biomass is always inverted because the biomass of fishes far exceeds the biomass of phytoplankton.

Question 24.
What is primary productivity? Give a brief description of factors that affect primary productivity. (CBSE Delhi 2011)
Answer:
The amount of biomass or organic matter produced per unit area over a time period by plants during photosynthesis is called primary productivity.

It is expressed in terms of gm^-2yr^-1 or kcal m^-2yr^-1. Primary productivity depends upon a number of environmental factors like:

1. Availability of nutrients which varies in different types of ecosystem.
2. Photosynthetic capacity of plants.
3. The plant species inhabiting a particular area.
4. Environmental factors.

Question 25.
Define decomposition and describe the processes and products of decomposition.
Answer:
Decomposition is defined as the process by which complex organic compounds are broken down into simpler inorganic substances that can be reutilized by plants for their growth.

Decomposition involves the following processes:

Decomposition process	End products
1. Fragmentation of detritus by the detritivorous invertebrates.	1. Increases the surface area of detritus for the action of microbes.
2. lead, Ingor Eluviation.	2. Carrying away soluble nutrients from the A-horizon of soil by downward-moving gravitational water.
3. CatabolIsm involves the breakdown of detritus in the presence of extracellular enzymes released by the decomposers.	3. Simple organic compounds and Inorganic substances are formed.
4. Humlfication involves the transformation of simplified detritus into fully decomposed, dark-colored, and amorphous humus.	4. Humus acts as a reservoir of nutrients.
5. Minerailsatlon	5. Release of inorganic substances like CO2, water, etc., and minerals like K+, Mg++, Ca++, NH+4, etc. In the soil by the action of microbes.

Question 26.
Give an account of energy flow in an ecosystem.
Answer:
Different components for a universal model of energy flow are shown below:
A generated energy flow model of ecosystem-Boxes represent biotic components and the arrows show the pathways of energy transfer

When a herbivore eats a plant, then it digests and oxidizes the ingested food to liberate energy which is equal to that used in synthesizing the organic biomass by the plant. Some of the released energy is lost as heat while only a part of the energy is used in building the biomass of the herbivore, which is called gross secondary productivity.

The same is repeated when the herbivore is eaten by a primary carnivore and so on. At each transfer, about 80-90% of potential energy is dissipated as heat while only 10-20% of energy is available to the next trophic level.

Thus, there is a decline in the amount of energy passing from one trophic level to the next trophic level. The study of energy transfer is called bioenergetics. Secondary productivity tends to be about 10 percent at the herbivore level, although efficiency may be higher, 20 percent at the carnivore level.

So regarding the energy flow, an ecosystem is characterized by the following:

• Unidirectional flow of energy.
• The decrease in useful energy.
• Return of radiant energy of the Sun to the non-living system as heat.

Question 27.
Outline salient features of carbon cycling in an ecosystem. (CBSE Delhi 2012)
Answer:
The reservoirs of carbon are listed below:

1. Carbon dioxide present in the air.
2. Carbon dioxide dissolved in water.
3. Carbonates in the earth’s crust.
4. Fossil fuels like coal and petroleum,
5. Bicarbonates in oceans.

The carbon cycle is the simplest of all nutrient cycles. Its salient features are listed below:
1. CO₂ utilization: Carbon dioxide is used by green plants for the process of photosynthesis, and oxygen is released as a by-product. The fixed carbon enters the food chain and is passed to herbivores, carnivores, and decomposers. About 4 x 1013 kg of carbon is fixed in the biosphere through photosynthesis annually.

Carbon cycle in nature.

2. C0₂ production and return to the atmosphere:
(a) Carbon dioxide is released into the atmosphere by the respiration of producers and consumers.
(b) It is also released by the decomposition of organic wastes and dead bodies by decomposers by the action of bacteria and fungi of decay.
(c) Burning of wood and fossil fuels also releases C0₂ into the atmosphere.
(d) Volcanic eruptions and hot springs also release C02 into the atmosphere.
(e) Weathering of carbonate-containing rocks by the action of acids also adds C0₂ to the atmosphere.

Question 28.
Name the pioneer species on a bare rock. How do they help in establishing the next type of vegetation? Mention the type of climax community that will ultimately get established. (CBSE 2009)
Answer:

1. Lichens and mosses are the pioneer community. Lichens secrete acids which dissolve the rocks and helps in weathering rocks to form soil.
2. Soil is formed by the action of lichens.
3. Invasion is the successful establishment of species.
4. The reaction is most important in succession. It is the mechanism of modification of the environment due to the influence of living organisms. Changes take place; annual grasses, perennial herbs, shrubs, and finally trees appear along with animal communities.
5. Angiosperms trees forming forests will form climax communities.

Question 29.
(i) What is an ecological pyramid? Compare the pyramids of energy, biomass, and numbers.
Answer:
Ecological pyramid: An ecological pyramid is a graphical representation of an ecological parameter, like a number or biomass or accumulated energy at different trophic levels in a food chain in an ecosystem.

Ecological pyramids are of three types:

1. Pyramid of number
2. Pyramid of biomass
3. Pyramid of energy

Pyramid of energy	Pyramid of biomass	Pyramid of number
(i) The relationship between producers and consumers in an ecosystem represented in the form of a pyramid in terms of the flow of energy is called the pyramid of energy.	(i) The relationship between producers and consumers in an ecosystem represented in the form of a pyramid in terms of biomass is called the pyramid of biomass.	(i) The relationship between producers and consumers in an ecosystem represented in the form of a pyramid in terms of number is called the pyramid of number.
(ii) It is always upright.	(ii) It can be upright or inverted.	(ii) It can be upright or inverted.

(ii) Write any two limitations of ecological pyramids. (CBSE 2017)
Answer:
Limitations of ecological pyramids:
(a) Ecological pyramid never takes into account the same species belonging to two or more trophic levels.
(b) It assumes a simple food chain, which never exists in nature.

Here we are providing Class 12 Biology Important Extra Questions and Answers Chapter 11 Biotechnology: Principles and Processes. Important Questions for Class 12 Biology are the best resource for students which helps in Class 12 board exams.

Class 12 Biology Chapter 11 Important Extra Questions Biotechnology: Principles and Processes

Biotechnology: Principles and Processes Important Extra Questions Very Short Answer Type

Question 1.
What is genetic engineering?
Answer:
Genetic engineering. It is a technique for artificially and deliberately modifying DNA (genes) to suit human needs. It is also called recombinant DNA technology or DNA splicing. It is a kind of biotechnology.

Question 2.
Define recombinant DNA.
Answer:
Recombinant DNA. They are molecules of DNA that are formed through genetic recombination methods.

Question 3.
What is the role of restriction endonuclease?
Answer:
Restriction endonucleases are specific enzymes which can cleave double-stranded DNA at the specific site.

Question 4.
What are BACs and YACs? (CBSE 2016)
Answer:
BACs and YACs are artificial chromosomes from bacteria and yeast efficient for gene transfer. They are vectors.

Question 5.
Name the soil bacterium which contains the gene for production of endotoxins.
Answer:
Agrobacterium tumefaciens.

Question 6.
Name a technique by which DNA fragments can be separated. (CBSE Delhi 2008)
Answer:
Gel electrophoresis.

Question 7.
What is the principle of Gel electrophoresis?
Answer:
DNA fragments are negatively charged so they move to anode under electric field through the matrix (usually agarose). This matrix gel acts as sieve and DNA fragments resolve according to their size.

Question 8.
Name the compound used for staining DNA to be used in Recombinant Technology. What is the colour of such stained DNA?
Answer:
The compound used for staining DNA is ethidium bromide. Stained DNA becomes orange.

Question 9.
Name the technique for vector less direct gene transfer.
Answer:
Gene gun.

Question 10.
What is the role of ‘Ori’ in any plasmid?
Answer:
The plasmid is prokaryotic circular DNA which has a sequence of nucleotides from where the replication starts. This is called the origin of replication.

Role of Ori is to start replication. Also, the copy number of linked DNA is controlled by Ori.

Question 11.
Do normal £. coli cells have any gene resistance against antibiotics?
Answer:
No.

Question 12.
What is the function of TPA?
Answer:
TPA (Tissue plasminogen activator) dissolves blood clots after a heart attack and stroke.

Question 13.
Give an example in which recombinant DNA technology has provided a broad range of tool in the diagnosis of diseases.
Answer:
Construction of probes, which are short segments of single-stranded DNA attached to a radioactive or fluorescent marker.

Question 14.
Give the full form of PCR. Who developed it? (CBSE Delhi 2013)
Answer:
PCR is a polymerase chain reaction. It was developed by Kary Mullis in 1985.

Question 15.
What is the source of DNA polymerase, i.e. Taq polymerase? (CBSE Outside Delhi 2013)
Answer:
Taq polymerase is isolated from the bacterium Thermus Aquaticus.

Question 16.
Define “melting of target DNA”.
Answer:
The target DNA containing the sequence to be amplified is heat-denatured (around 94° C for 15 seconds) to separate its complementary strands. This process is called melting of target DNA.

Question 17.
Expand ELISA. Write one application. (CBSE Delhi 2013)
Answer:
ELISA-Enzyme Linked ImmunoSorbent Essay. Importance-lt is used for the diagnosis of AIDS.

Question 18.
What are transgenic animals? Give one example. (CBSE Outside Delhi 2016)
Answer:
Transgenic animals: The animals obtained by genetic engineering containing transgenes are known as transgenic animals.
Example. Transgenic cow ‘Rosie’.

Question 19.
How many PCR cycles are adequate for proper amplification of DNA segment?
Answer:
20-30 cycles.

Question 20.
Define gene therapy.
Answer:
Gene Therapy: It is the replacement of a faulty gene by normal healthy functional gene.

Question 21.
What is the importance of gene bank?
Answer:
It provides a stock from which genes can be obtained for improving the varieties or used in genetic engineering.

Question 22.
What can be the source of thermostable DNA?
Answer:
Thermostable DNA is obtained from a bacterium Thermus Aquaticus.

Question 23.
What are selectable markers?
Answer:
Genes which are able to select transformed cell from the non-recombinant cells are called selectable markers.

Question 24.
Why is enzyme cellulase used for isolating genetic material from plant cells and not from animal cells? (CBSE 2010)
Answer:
Cellulase is used for breaking the cell wall of plant cells whereas animal cells lack a cell wall. The cell wall is made of cellulose which can be broken down by cellulase.

Question 25.
Give one example each of transgenic plant and transgenic animal.
Answer:

1. Transgenic tomato plant called Flavor Savr.
2. A transgenic mouse called Supermouse.

Question 26.
What would be the molar concentration of human DNA in a human cell?
Answer:
Humans have 3 M of DNA per cell, i.e. the molar concentration is 3.

Question 27.
Do eukaryotic cells have restriction endonucleases?
Answer:
Yes, eukaryotic cells possess restriction endonucleases. They are involved in editing (Proofreading) and DNA repairs during DNA replication.

Question 28.
Name a technique by which DNA fragments can be separated. (CBSE Delhi 2008)
Answer:
Get electrophoresis.

Question 29.
Biotechnological techniques can help to diagnose the pathogen much before the symptoms of the disease appear in the patient. Suggest any two such techniques. (CBSE Outside Delhi 2019)
Answer:
PCR – Polymerase chain reaction
ELISA – Enzyme-linked immunosorbent assay

Question 30.
Why is it not possible for an alien DNA to become part of chromosome anywhere along its length and replicate? (CBSE 2014)
Answer:
For multiplication of any alien DNA, it needs to be a part of a chromosome which has a specific sequence known as the origin of replication.

Question 31.
Mention the type of host cells suitable for the gene guns to introduce an alien DNA. (CBSE Delhi 2014)
Answer:
Plant cells.

Question 32.
Name the enzymes that are used for isolation of DNA from bacterial and fungal cells for rDNA technology. (CBSE 2014)
Answer:
Lysozyme for bacterial cell and chitinase for the fungal cell.

Question 33.
What is EcoRI? How does EcoRI differ from an exonuclease? (CBSE Outside Delhi 2015)
Answer:
EcoRI is an endonuclease restriction enzyme which cut both the stands of palindromic DNA at a specific position of nitrogen base 5′ (GAATTC) 3′ while exonuclease removes nucleotides from terminals of DNA strands.

Biotechnology: Principles and Processes Important Extra Questions Short Answer Type

Question 1.
(i) While cloning vectors, which of the two will be preferred by biotechnologists, bacteriophages or plasmids. Justify with reason.
Answer:
Biotechnologists prefer bacteriophages for cloning over plasmids because they have very high copy numbers of their genome within the bacterial cells whereas some plasmids may have only one or two copies per cell and others may have 15-100 copies per cell. Phage vectors are more efficient than plasmids for cloning large DNA fragments.

(ii) Name the first transgenic cow developed and state the Improvement In the quality of the product produced by it. (CBSE Sample paper 2018-19)
Answer:
Transgenic cow Rosie produced human protein-enriched milk (2.4 grams per litre).

Question 2.
What are the two core techniques that enabled the birth of biotechnology?
Answer:
The two core techniques that enabled the birth of modern biotechnology are:

1. Genetic engineering techniques to alter the chemistry of genetic material (DNA and RNA), to introduce these into host organisms and thus change the phenotype of the host organism.
2. Maintenance of sterile (microbial contamination-free) ambience in chemical engineering processes to enable the growth of only the desired microbe/eukaryotic cell in large quantities for the manufacture of biotechnological products like antibiotics, vaccines, enzymes, etc.

Question 3.
Make a list of tools of recombinant DNA technology. (CBSE Delhi, 2011)
Answer:
Key tools of recombinant DNA technology:

1. Restriction enzymes
2. Polymerase enzymes
3. Ligases
4. Vectors
5. Host organism.

Question 4.
What does EcoRI signify? How its name is derived?
Answer:
EcoRI signifies the name of restriction endonuclease:

1. First capital letter of the name that comes from the genus Escherichia is ‘E’.
2. Second two small letters come from the species Coli of prokaryotic cells from which they are isolated, i.e. ‘co’.
3. Letter R is derived from the name of the strain, i.e. Escherichia coli Ry 13.
4. The Roman number indicates the order in which enzymes were isolated from that strain of bacteria.

Question 5.
What are recognition sequences or recognition sites?
Answer:
The sites recognised by restriction endonucleases are called recognition sites. The recognition sequences are different and specific for the different restriction endonucleases. These sequences are palindromic in nature.

Question 6.
Define vector. Give the properties of a “Good Vector”.
Answer:
A vector is a DNA molecule that has the ability to replicate in an appropriate host cell, and into which the DNA fragment to be cloned is integrated for cloning.

A good vector must have the following properties:

• It should have an origin of replication so that it is able to replicate autonomously.
• It should be easy to isolate and purify.
• It should get easily introduced into the host cells.

Question 7.
What is the difference between cloning and expression vectors?
Answer:
All vectors that are used for propagation of DNA inserts in a suitable host are called cloning vectors. When a vector is designed for the expression of, i.e. production of the protein specified by, the DNA insert, it is termed as an expression vector.

Question 8.
What do you understand by the term selectable marker?
Answer:
Selectable marker:

1. A marker is a gene which helps in selecting those host cells which contain the vector (transformant) and eliminating the non-transformants. It selectively permits the growth of transformants.
2. Common selectable markers for E. coli include the genes encoding resistance to antibiotics such as ampicillin. chloramphenicol tetracycline and kanamycin or the gene for (i-galactosidase which can be identified by a colour reaction. Normal E.Coli do not carry resistance against any of these antibodies.

Question 9.
Explain the principle that helps in separation of DNA fragments in Gel electrophoresis. (CBSE Delhi 2009 C)
Answer:
Get electrophoresis is a technique of molecules such as DNA/RNA/protein on the basis of their size under the influence of the electric field so that they migrate in the direction of electrode bearing the opposite charge. Positively charged molecules move towards cathode (-ve electrode) and vice versa. These molecules move through a medium or matrix and can be separated on the basis of their size.

Question 10.
Give the applications of PCR technology. (CBSE, Delhi 2013)
Answer:

1. Amplification of DNA and RNA.
2. Determination of orientation and location of restriction fragments relative to one another.
3. Detection of genetic diseases such as sickle cell anaemia, phenylketonuria and muscular dystrophy.

Question 11.
Why is “Agrobacterium-mediated genetic transformation” in plants described as natural genetic engineer of plants? (CBSE Delhi 2011)
Answer:
Agrobacterium tumefaciens is a plant pathogenic bacterium which can transfer part of its plasmid DNA because it infects host plants. Agrobacterium produces crown gall in most of the dicotyledonous plants. These bacteria contain large tumours inducing plasmid (Ti-plasmids) which pass on their tumour causing gene into the genome of the host plant. Thus gene transfer is happening in nature without human involvement hence Agrobacterium-mediated genetic transformation is described as natural genetic engineering in plants.

Question 12.
Differentiate gene therapy and gene cloning.
Answer:
Differences between gene therapy and gene cloning:

Gene therapy	Gene cloning
It is the replacement and/ or alteration of defective genes responsible for hereditary diseases by normal genes.	It is the technique of obtaining identical copies of a particular segment of DNA or a gene.

Question 13.
From what you have learnt, can you tell whether enzymes are bigger or DNA is bigger in molecular size? How did you know?
Answer:
DNA molecules are bigger in size as compared to the molecular size of enzymes. Enzymes are proteins. Protein synthesis occurs from a small portion of DNA called genes.

Question 14.
How does restriction endonuclease work? (CBSE Delhi 2013, 2014, Outside Delhi 2019)
Or
What are molecular scissors? Explain their role. (CBSE 2009)
Answer:
Restriction endonuclease enzymes are called molecular scissors which can cut double-stranded DNA at specific sites.

Role of restriction endonuclease:

• Restriction endonuclease inspects the length of DNA sequence.
• It finds specific recognition sequence, i.e. palindromic nucleotide sequence in DNA.
• These enzymes cut the strand of DNA a little away from the centre of palindromic sites.
• Thus restriction endonucleases leave overhanging stretches called sticky ends on each strand.

Question 15.
How and why is the bacterium Thermus Aquaticus employed in recombinant DNA technology? Explain. (CBSE Delhi 2009)
Answer:
Thermus Aquaticus bacterium is employed in recombinant DNA technology because it has thermostable DNA polymerase (Taq. Polymerase) that remains active during high temperature-induced denaturation of a step of PCR.

This enzyme is employed during amplification of gene using PCR (Polymerase Chain Reaction). The amplified fragment can be used to ligate with a vector for further cloning.

Question 16.
Name the source of Taq polymerase. Explain its advantages. (CBSE Outside Delhi 2009)
Answer:
Taq Polymerase is extracted from Thermostable bacteria, namely Thermus Aquaticus. It remains active at a higher temperature and is used for denaturation of DNA during PCR.

Question 17.
What are recombinant proteins? How do bioreactors help in their production? (CBSE Outside Delhi 2009, 2015)
Answer:
Recombinant proteins. When any protein-encoding gene is expressed in a heterologous host, it is called recombinant protein. Bioreactors help in the production of recombinant proteins on large scale. A bioreactor provides optimal conditions for achieving the desired recombinant protein by biological methods.

Question 18.
What is meant by gene cloning?
Answer:
Formation of multiple copies of a particular gene is called gene cloning. A gene is separated and ligated to a vector-like plasmid. The recombinant plasmid is introduced into a plasmid-free bacterium through transformation. The transformed bacterium is made to multiply and form a colony. Each and every bacterium of the colony has a copy of the gene.

Question 19.
Both the wine marker and a molecular biologist who has developed a recombinant vaccine claim to be biotechnologist. Who in your opinion is correct?
Answer:
Both are considered biotechnologists. Wine marker utilises a strain of yeast which produces wine by fermentation. The molecular biologist uses a cloned gene for the antigen. The antigen is used as a vaccine. This permits the formation of antigen in huge quantity. Both generate products and services using living organisms useful to mankind.

Question 20.
You have created a recombinant DNA molecule by ligating a gene to a plasmid vector. By mistake, your friend adds an exonuclease enzyme to the tube containing the recombinant DNA. How will your experiment get affected as you plan to go to the transformation now?
Answer:
The experiment is not likely to be affected as the recombinant DNA molecule is circular and closed, with no free ends. Hence, it will not be a substrate for exonuclease enzyme which removes nucleotides from the free ends of DNA.

Question 21.
Explain the work carried out by Cohen and Boyer that contributed immensely to biotechnology. (CBSE2012)
Answer:
Work of Cohen and Boyer:

1. Discovery of restriction endonuclease, an enzyme of E.coli which cut DNA at palindromic sequence.
2. Preparation of recombinant DNA (Plasmid and DNA of interest.)
3. Their work established recombinant DNA (rDNA) technology also called genetic engineering.

Question 22.
How are ‘Sticky ends’ formed on a DNA strand? Why are these so-called? (CBSE Delhi 2014)
Answer:
1. Restriction enzymes cut the strand of DNA a little away from the centre of palindrome site, but between the same two bases on opposite strands. As a result, single-stranded portions are left at each end. These overhanging stretches of DNA are called ‘Sticky ends’.

2. The Sticky ends are named so because they form hydrogen bonds with their complementary cut counterparts. The stickiness helps in the action of DNA ligase.

Question 23.
How is a continuous culture system maintained in bioreactors and why? (CBSE Delhi 2019)
Answer:
In order to maintain a continuous culture system, the used medium is drained out from one side of the bioreactor and the fresh medium is added from one side. This type of culturing method produces larger biomass leading to higher yields of the desired product.

Question 24.
Galactosidase enzyme is considered a better selectable marker. Justify the statement. (CBSE Delhi 2019)
Answer:
Recombinant strains can be differentiated from the non-recombinants ones easily by using this selectable marker. The selection is done on the basis of the colour change. All are grown on a chromogenic substance. Non-recombinants will change from colourless to blue while in recombinants insertional inactivation of galactosidase gene occurs.

Hence, recombinants showed no colour change. This is a single step, an easy method for selection.

Biotechnology: Principles and Processes Important Extra Questions Long Answer Type

Question 1.
What is genetic engineering? Explain briefly the distinct steps common to all genetic engineering technology.
Or
With the help of diagrams show the different steps in the formation of recombinant DNA by the action of restriction endonuclease. (CBSE 2011)
Answer:
Genetic engineering: It is a technique for artificially and deliberately modifying DNA (genes) to suit human needs. It is also called recombinant DNA technology or DNA splicing.

It is a kind of biotechnology:

1. Isolation of genetic material which has the gene of interest.
2. Cutting of gene of interest from genome and vector with the same restriction endonuclease enzyme. Amplifying gene of interest (PCR).
3. Ligating gene of interest and vector using DNA ligase forming rDNA.
4. Transformation of rDNA into the host cell.
5. Multiplying host cell to create clones.

Diagram showing various steps Involved in DNA recombinant technology for the production of a recombinant protein.

Question 2.
List three important features necessary for preparing a genetically modifying organism.
Answer:
Conditions necessary for preparing:

1. Identification of DNA with desirable genes.
2. Introduction of the identified DNA into the host.
3. Maintenance of introduced DNA in the host and transfer of the DNA to its progeny.

Question 3.
How are restriction endonuclease enzymes named? Write examples. (CBSE 2014
Answer:
The naming of restriction enzymes is as follows:

1. The first letter of the name comes from the genus and the next two letters from the name of the species of the prokaryotic cell from which they are isolated.
2. The next letter comes from the strain of the prokaryote.
3. The roman numbers following these four letters indicate the order in which the enzymes were isolated from that strain of the bacterium.

Examples:

1. EcoR I is isolated from Escherichia coli RY 13.
2. Hind II is from Haemophilus influenza.
3. Bam H I is from Bacillus amylotiquefaciens.
4. Sal I is from Streptomyces Albus
5. Pst I is from Providencia stuartii.

Question 4.
Explain any three methods of vector less gene transfer. (CBSE Outside Delhi 2013)
Answer:
Vectors of gene transfer. Following are common methods of vectors gene transfer.

1. Microinjection: Microinjection is the process/technique of introducing foreign genes into a host cell by injecting the DNA directly into the nucleus by using microneedle or micropipette.
2. Electroporation: Electroporation is the process by which transient holes are produced in the plasma membrane of the (host) cell to facilitate entry of foreign DNA.
3. Gene Gun: Gene gun is the technique of bombarding microprojectiles (gold or tungsten particles) coated with foreign DNA with great velocity into the target cell.

Question 5.
Write a note on the cloning vector.
Answer:
Cloning vectors:

1. Plasmids and bacteriophages are the commonly used vectors
2. Presently genetically engineered/ synthetic vectors are also used for easily linking the foreign DNA and selection of recombinants from non-recombinants.
3. The following features are required to facilitate cloning in a vector:
   (a) Origin of replication (Ori)
   (b) Selectable marker
   (c) Cloning (Recognition) site
   (d) Small size of the vector.

Question 6.
What is PCR? List the three main steps. Show the steps with a diagrammatic sketch.
Answer:

1. PCR. Polymerase Chain Reaction.
2. Three steps of PCR.
   (a) Denaturation
   (b) Primer annealing and
   (c) Extension of primers.

The three steps of PCR

Question 7.
Name the various cloning vectors and explain how a plasmid can be used for genetic engineering.
Answer:
Cloning vectors:

• Plasmids
• Bacteriophages
• Plant and animal vectors
• Jumping genes (Transposons)
• Artificial chromosomes of bacteria, yeast and mammals (BAC, YAC).

Use of plasmid as genetic material Plasmids are obtained from bacteria. They are treated with a restriction endonuclease enzyme to obtain the fragments of the desired genome. They are allowed to fuse with the help of a DNA ligase enzyme. The recombinant plasmids thus formed are used as genetic material.

Question 8.
Give various means by which a competent host is formed for recombinant DNA technology. Why and how bacteria can be made ‘competent’? (CBSE Delhi 2013)
Answer:
A host cell should be competent enough to take the DNA molecule for the transformation as the following methods can be used.

1. Using divalent cations: Bacteria are treated with Ca²⁺, etc. so that DNA enters the bacterium through pores in its cell wall.
2. Heat shock: Cells can be incubated on ice and then at 42°C for a heat shock and then again put on ice.
3. Microinjection: Recombinant DNA is directly injected into the nucleus of an animal cell.
4. Biolistic. Cells bombarded with high- velocity micro-particles of gold or tungsten coated with DNA is known as a gene gun.

Question 9.
How is recombinant DNA transferred to host?
Answer:
Transfer of recombinant DNA into the host:

1. The bacterial cells must be made competent to take up DNA; this is done by treating them with a specific concentration of calcium, that increases the efficiency with which DNA enters the cell through the pores in its cell wall.
2. Recombinant DNA can then be forced into such cells by incubating the cells with recombinant DNA on ice followed by placing them at 42°C and then putting them back on ice (heat shock treatment),
3. Microinjection is a method in which the recombinant DNA is directly injected into the nucleus of the animal cell with the help of microneedles or micropipettes.
4. Gene gun or biolistics is a method suitable for plant cells, where cells are bombarded with high-velocity microparticles of gold or tungsten coated with DNA.
5. Disarmed pathogens are used as vectors; when they are allowed to infect the cell, they transfer the recombinant DNA into the host.

Question 10.
Why DNA cannot pass through the cell membrane? How can the bacteria be made competent to take up a plasmid? Explain a method for the introduction of alien DNA into a plant host cell. Name a pathogen that is used as a disarmed vector. (CBSE Outside Delhi 2019)
Answer:
DNA is a hydrophilic molecule thus it cannot pass through the cell membrane.

Bacterial cells are made ‘competent’ by treating them with a specific concentration of divalent cation such as calcium in order to take up the plasmid. The divalent cation increases the efficiency with which DNA enters the bacterium through the pores of the cell wall.

Procedure: Recombinant DNA is forced into ‘competent’ bacterial host cells by incubating them on the ice. It is followed by placing them briefly at 42°C. It is termed ‘heat stock’ treatment. Again they are placed back on ice. This process allows bacteria to take up the recombinant DNA.

Gene gun or biolistic method is used for the introduction of alien DNA into a plant host cell. Here, the plant cells are bombarded with high-velocity micro-particles of gold or tungsten coated with DNA. Agrobacterium tumefaciens or Retroviruses can be used as a disarmed vector.

Question 11.
Write a note on vectors used during recombinant DNA technology. (CBSE Delhi 2008)
Answer:
A vector or vehicle DNA is used as a carrier for transferring selected DNA into cells. A plasmid with its small DNA from a bacterium is a good choice for indirect gene transfer because it can move from one cell to another and make several copies of itself. However, artificial chromosomes from bacteria and yeast called BACs and YACs respectively are more efficient for eukaryotic gene transfers.

Plasmid and Yeast Artificial Chromosome

Question 12.
(i) Identify A and B illustrations in the following:
(a)
(b)
Answer:
A= 5′ GAATTC 3′
B marks for ORI (origin of replication).

(ii) Write the term given to A and C and why?
Answer:
A represents a nucleotide palindromic sequence. C-sticky end.

(iii) Expand PCR. Mention its importance in biotechnology. (CBSE Delhi 2011)
Answer:
PCR, Polymerase chain reaction. It helps in gene amplification.

Question 13.
Write the role of the following sites in pBR322 cloning vector:
(a) rop
Answer:
Role of rop, ori and selectable marker in pBR322 cloning vector.

Role of rop: Rop gene regulates copy number. Rop process is involved in stabilising the interaction between RNA I and RNA II which in turn prevents replication of pBR322.

(b) ori
Answer:
Origin of replication (Ori):

• It is a specific sequence of DNA bases, which is responsible for initiating replication.
• An alien DNA for replication should be linked to the origin of replication.
• A prokaryotic DNA has normally a single origin of replication, while eukaryotic DNA may have more than one origin of replication.
• The sequence is responsible for controlling the copy number of linked DNA.

(c) selectable marker (CBSE Delhi 2019 C)
Answer:
Selectable marker:

• A marker is a gene which helps in selecting those host cells which contain the vector (transformant) and eliminating the non-transformants.
• Common selectable markers for E. coli include the genes encoding resistance to antibiotics such as ampicillin. Chloramphenicol, tetracycline and kanamycin or the gene for B-galactosidase can be identified by a colour reaction.

Question 14.
(i) Explain the significance of palindromic nucleotide sequence in the formation of recombinant DNA.
Answer:
The palindromic sequences, i.e. the sequence of base pairs read the same on both the DNA strands when the orientation of reading is kept the same, e.g.
5’ — GAATTC — 3’
3’ — CTTAAG — 5’

Every endonuclease inspects the entire DNA sequence for palindromic recognition sequence.

(ii) Write the use of restriction endonuclease in the above process. (CBSE 2017)
Answer:
On finding the palindrome, the endonuclease binds to the DNA. It cuts the opposite strands of DNA, but between the same bases on both the strands and forms sticky ends. This sticky ends facilitate the action of enzyme DNA ligase and help in the formation of recombination DNA.

Question 15.
Describe the roles of heat, primers and the bacterium Thermus Aquaticus in the process of PCR. (CBSE 2017)
Answer:
Role of heat: Heat helps in the denaturation process in PCR. The double-stranded DNA is heated in this process at very high temperature (95°C) so that both the strands separate.

Role of primers: Primers are chemically synthesised small oligonucleotides of about 10-18 nucleotides. These are complementary to a region of template DNA and helps in the extension of the new chain. Rote of Bacterium Thermus

Aquaticus: A thermostable Taq DNA polymerase is isolated from this bacterium, which can tolerate high temperatures and forms new strand.

Question 16.
How has the use of Agrobacterium as vectors helped in controlling Meloidogyne incognita infestation in tobacco plants? Explain in the correct sequence. (CBSE 2018, Outside Delhi 2019)
Or
(a) Write the mechanism that enables Agrobacterium tumefaciens to develop tumours in their host dicot plant.
(b) State how Agrobacterium tumefaciens and some retroviruses have been modified as useful cloning vectors. (CBSE Delhi 2019 C)
Answer:
(a) Cloning
(b) A nematode Meloidogyne incognita infects the roots of tobacco plants and causes a great reduction in yield.

To prevent this infestation a novel strategy was adopted which was based on the process of RNA interference (RNAi).

Nematode-specific genes were introduced into the host plants using Agrobacterium vectors. The introduction of DNA was such that it produced both sense and anti-sense RNA in the host cells. These two RNAs, being complementary to each other, formed a double-stranded RNA (dsRNA) that initiated RNAi and thus, silenced specific mRNA of the nematode. Due to this the parasite could not survive in a transgenic host by expressing specific interfering RNA. The transgenic plant, therefore, got itself protected from the parasite.

Question 17.
Explain the roles of the following with the help of an example each in recombinant DNA technology:
(i) Restriction Enzymes
Answer:
Restriction enzymes :
(a) Restriction enzymes belong to nucleases class of enzymes which breaks nucleic acids by cleaving their phosphodiester bonds.
(b) Since restriction endonucleases cut DNA at a specific recognition site, they are used to cut the donor DNA to isolate the desired gene.
(c) The desired gene has sticky ends which can be easily ligated to cloning vector cut by same restriction enzymes having complementary sticky ends to form recombinant DNA.
(d) An example is EcoR1 which is obtained from E.coli bacteria “R” strain which cuts DNA at specific palindromic recognition site.
5‘ GAATTC 3‘
3‘ CTTAAG 5‘

(ii) Plasmids (CBSE 2018)
Answer:
Plasmids: Plasmids are autonomous, extrachromosomal circular double-stranded DNA of bacteria. They are used as cloning vectors in genetic engineering because they are small and self-replicating. Some plasmids have antibiotic resistance genes which can be used as marker genes to identify recombinant plasmids from non-recombinant ones.

To obtain the desired products, plasmids are cut and ligated with desired genes and transformed into a host cell for amplification. An example of artificially modified plasmids is pBR322 (constructed by Bolivar and Rodriguez) or pUC (constructed at University at California).

Question 18.
When the gene product is required in large amounts, the transformed bacteria with the plasmid inside the bacteria are cultured on a large scale in an industrial fermenter which then synthesises the desired protein. This product is extracted from the fermenter for commercial use.
(a) Why is the used medium drained out from one side while the fresh medium is added from the other? Explain.
Answer:
In the bioreactor used medium is drained out and the fresh medium is added to maintain the cells in their physiologically most active log / experimental phase,

(b) List any four optimum conditions for achieving the desired product in a bioreactor. (CBSE Sample Paper 2020)
Answer:
Condition for obtaining the desired product in a bioreactor:

• Temperature
• Substrates
• SaLts
• pH
• Oxygen

Question 19.
List the steps in the formation of rDNA.
Answer:
Steps in formation of rDNA:
Recombined DNA technology involves the following steps:

1. Isolation of DNA.
2. Fragmentation of DNA by restriction endonucleases.
3. Isolation of the desired DNA fragment.
4. Amplification of the gene of interest.
5. Ligation of the DNA fragment into a vector using DNA ligase.
6. Transfer of DNA fragment into the vector using DNA ligase.

Question 20.
How is the isolated gene of interest amplified? (CBSE Delhi 2019, 2019 C)
Answer:
Amplification of the DNA/gene of interest:

1. Amplification refers to the process of making multiple copies of the DNA segment in vitro.
2. It employs the polymerase chain reaction (PCR).
3. The process was designed by K. Mullis,
4. This technique involves three main steps:
   (a) Denaturation
   (b) Primer annealing and
   (c) Extension of primers.
5. The double-stranded DNA is denatured by subjecting it to high temperatures.
6. Two sets of primers are used; primers are the chemically synthesised short segments of DNA (oligonucleotides), that are complementary to the segment of DNA (of interest).
7. DNA polymerase enzyme (Taq polymerase) is used to make copies of DNA making use of genomic template DNA and primer.

Question 21.
List the features required to facilitate cloning into a vector. Show with a sketch the E. coli cloning vector showing restriction sites.
Or
Sketch pBR322. (CBSE 2012, Outside Delhi 2019)
Answer:
Features required to facilitate cloning vector.

1. Origin of replication (Ori)
2. Selectable marker
3. Cloning sites
4. Vectors for cloning genes in plants and animals
   Sites of cloning vector

E. coli Cloning Vector pBr322 showing restriction sites (Hindlll, EcoRI, BamHI, Sal I, Pvu II, Pst I, ClaI), oriV and antibiotic resistance genes (ampR and tetR). Rop codes for the proteins involved in the replication of the plasmid.

Question 22.
With the help of simple sketch show the action of restriction enzyme (EcoR1).
Answer:
The action of restriction enzyme.

Ecol cuts the DNA between bases G and A only when the sequence GAATTC is present in the DNA.

Question 23.
Explain the importance of (a) ori, (b) ampR and (c) rop in the E. colt vector. (CBSE Outside Delhi 2009, Outside Delhi 2019)
Answer:

1. Importance of ori: This is a sequence from where replication starts and any piece of DNA, when linked to this sequence, can be made to replicate within the host cells, it allows multiple copies per cell.
2. Importance of ampR: It is the antibiotic resistance gene for ampicillin. It helps in the selection of transformer cells.
3. Importance of rop: It codes for the proteins involved in the replication of plasmid.

Question 24.
Name any two cloning vectors. Describe the features required to facilitate cloning into a vector. (CBSE Sample Paper)
Answer:
Plasmids and bacteriophages are two examples of the cloning vector. A vector is a DNA molecule that has the ability to replicate in an appropriate host cell and into which the DNA fragment to be cloned is integrated for cloning.

A good vector must have the following properties:

• It should be able to replicate autonomously.
• it should be easy to isolate and purify,
• It should be easily introduced into the host cells.

Cloning vectors: All vectors that are used for propagation of DNA inserts in a suitable host are called cloning vectors. When a vector is designed for the expression of, i.e. production of the protein specified by, the DNA insert, it is termed as an expression vector.

Question 25.
What are bioreactors? Sketch the two types of bioreactors. What is the utility? Which is the common type of bioreactors? (CBSE Delhi 2013)
Or
How do bioreactors help in the production of recombinant proteins? (CBSE Outside Delhi 2009)
Or
(i) How has the development of bioreactor helped in biotechnology?
(ii) Name the most commonly used bioreactor and describe its working. (CBSE Delhi 2018, 2019 C)
Answer:
Small volume cultures cannot yield appreciable quantities of products. To produce these products in large quantities the development of ‘bioreactors’ was required where large volumes (100-1000 litres) of culture can be processed. Thus bioreactors can be thought of as vessels in which raw materials are biologically converted into specific products, using microbial, plant, animal or human cells or individual enzymes.

Role. A bioreactor provides the optimal conditions for achieving the desired product by providing optimum growth conditions (temperature, pH, substrate, salts, vitamins, oxygen).

One of the most commonly used bioreactors is of stirring type.

A stirred tank reactor is cylindrical or a container with a curved base which facilitate the mixing of the reactor contents. The stirrer facilitates even mixing and oxygen availability throughout the bioreactor. Alternatively, air can be passed through the reactor. It consists of agitator system, an oxygen delivery system, a foam control system, a temperature control system, pH control system and sampling ports so that small volumes of the culture can be withdrawn periodically.

(a) Simple stirred-tank bioreactor (b) Sparged stirred-tank bioreactor through which sterile air bubbles are sparged

Question 26.
Describe briefly the following:
(i) Origin of replication (Ori).
Answer:
(a) It is a specific sequence of DNA bases, which is responsible for initiating replication.
(b) An alien DNA for replication should be linked to the origin of replication.
(c) A prokaryotic DNA has normally a single origin of replication, while eukaryotic DNA may have more than one origin of replication.
(d) The sequence is responsible for controlling the copy number of linked DNA.

(ii) Bioreactor.
Answer:
(a) They are vessels in which raw materials are biologically converted into specific products using microbial, plant or human cells.
(b) A bioreactor provides optimal conditions for achieving the desired product by providing optimum growth conditions, pH, substrate salts, vitamins, oxygen, etc.
(c) The commonly used bioreactors are of stirring type.
(d) A stirred-tank reactor is usually cylindrical or with a curved base to facilitate the mixing of the contents.
(e) The stirrer facilitates the even-mixing and oxygen availability throughout the bioreactor.
(f) The bioreactor has the following components:

• An agitator system.
• An oxygen delivery system.
• A foam control system.
• A temperature control system.
• pH control system and
• Sampling ports.

(iii) Downstream processing.
Answer:
(a) It refers to the series of processes, to which a genetically modified product has to be subjected before it is ready to be marketed.
(b) The processes include two processes:

1. separation and
2. purification.

(c) The product has to be formulated with suitable preservatives.
(d) Such formulation has to undergo thorough clinical trials in the case of drugs. Strict quality control testing is also required.
(e) A proper quality controlled testing of each product is also required.

Question 27.
Besides better aeration and mixing properties, what other advantages do stirred tank bioreactors to have over shake flasks?
Answer:
Shake flasks are the conventional flasks for fermentation studies during secondary screening or laboratory process development. So, stirred-tank bioreactors are used to produce the product in large quantities.

Besides aeration and mixing:

1. it also helps in providing optimum growth conditions (temperature, pH, substrate, salts, vitamins, oxygen) to achieve the desired product.
2. cost-effective
3. due to baffles, the oxygen transfer rate is very high
4. the capacity of fermenters is more.

Question 28.
Explain briefly the following
(i) PCR
(ii) Restriction enzymes and DNA
(iii) Chitinase. (CBSE 2012)
Or
Explain the three steps involved in a polymerase chain reaction. (CBSE Delhi 2018C)
Answer:
(i) PCR-Polymerase Chain Reaction; It is the process in which multiple copies of the gene or segment of DNA of interest are synthesised in vitro using primers and DNA polymerase.

Working Mechanism of PCR: A single PCR amplification cycle involves three basic steps: denaturation, annealing and extension (polymerisation).
(a) Denaturation. In the denaturation step, the target DNA is heated to a high temperature (usually 94°C), resulting in the separation of the two strands. Every single strand of the target DNA then acts as a template for DNA synthesis.

(b) Annealing (Anneal = Join). In this step, the two oligonucleotide primers anneal (hybridize) to each of the single-stranded template DNA since the sequence of the primers is complementary to the 3’ ends of the template DNA. This step is carried out at a lower temperature depending on the length and sequence of the primers.

(c) Primer Extension (Polymerisation): The final step is an extension, wherein Taq DNA polymerase (of a thermophilic bacterium Thermus aquatics) causes synthesis of the DNA region between the primers, using dNTPs (deoxynucleoside triphosphates) and Mg²⁺. It means the primers are extended towards each other so that the DNA segment lying between the two primers is copied.

The optimum temperature for this polymerisation step is 72°C. To begin the second cycle, the DNA is again heated to convert all the newly synthesised DNA into single strands, each of which can now serve as a template for synthesis of more new DNA. Thus the extension product of one cycle can serve as a template for subsequent cycles and each cycle essentially doubles the amount of DNA from the previous cycle. As a result, from a single template molecule, it is possible to generate 2ⁿ molecules after n number of cycles.

Applications of PCR:

• Diagnosis of pathogen
• Diagnosis of the specific mutation
• DNA fingerprinting
• Detection of plant pathogens
• Cloning of DNA fragments from mummified remains of humans and extinct animals.

(ii) Restriction enzymes:
(a) They are called “molecular scissors” or chemical scalpels.
(b) Restriction enzymes, synthesised by micro-organisms as a defence mechanism, are specific endonucleases, which can cleave double-stranded DNA.
(c) Restriction enzymes belong to a class of enzymes called nucleases.
(d) They are of two kinds:

• Exonucleases, which remove nucleotides from the ends of DNA.
• Endonucleases, which cut the DNA at specific positions anywhere in its length (within).

(e) The recognition sequence is a palindrome, where the sequence of base pairs reads the same on both the DNA strands when the orientation of reading is kept the same, i.e. 5′ → 3′ direction or 3′ → 5′ direction.
e.g. 5′ – GAATTC – 3′
3′ – CTTAAG – 5′

(f) Each restriction endonuclease functions by inspecting the length of a DNA sequence and binds to the DNA at the recognition sequence.
(g) It cuts the two strands of the double helix at specific points in their sugar-phosphate backbones, a little away from the centre of the palindrome sites, but between the same two bases on both the strands.
(h) As a result, single-stranded portions called sticky ends are produced at the ends of the DNA; this stickiness of the end facilitates the action of enzyme DNA ligase.

1. When cut by the same restriction endonuclease, the DNA fragments (of the donor as well as the host/ recipient) yield the same kind of ‘sticky ends’ which can be joined end-to-end by DNA ligases.
2. Chitinase. This cell wall in fungi is made of chitin. The enzyme is used in fungi to break open the cell to release DNA along with their macromolecules like RNA proteins, lipids and polysaccharides.

Question 29.
Discuss with your teacher and find out how to distinguish between
(i) Plasmid DNA and chromosomal DNA
Answer:
Differences between plasmid DNA and chromosomal DNA:

Plasmid DNA	Chromosomal DNA
1. It is self-replicating, DNA molecule found naturally in many bacteria and yeast.	1. Chromosomal DNA present in chromosomes of all organisms.
2. It is not essential for normal growth and division.	2. It Is essential for growth and division.
3. It contains information for a few traits.	3. It contains information for all traits.

(ii) Exonuclease and Endonuclease (CBSE, Delhi 2013)
Answer:
Differences between Exonuclease and Endonuclease:

Endonuclease	Exonuclease
It cuts the DNA at a specific position of nitrogen bases anywhere within the length of DNA except the ends.	This enzyme removes nucleotides from the terminals from 5’ or 3’ ends of DNA molecules.

Question 30.
Collect the examples of palindromic sequences by consulting your teacher. Better try to create a palindromic sequence by following base pair rules.
Answer:

Question 31.
Can you list 10 recombinant proteins which are used In medical practice? Find where they are used as therapeutics (use the Internet).
Answer:

Recombinant Protein	Therapeutic Use
1. Insulin	For the treatment of diabetes Mellitus
2. Human Growth Hormone	For the treatment of dwarfism
3. Interferons	For the treatment of viral diseases, cancer and AIDS.
4. Streptokinase	For treating thrombosis.
5. Tumour Necrosis factor	For treating sepsis and cancer
6. Interleukins	For treating various cancers.
7. Hepatitis-B Surface Antigen	The vaccine against Hepatitis- B
8. Granulocyte Colony-stimulating factor	For treating cancer and AIDS and in bone marrow transplantation
9. Granulocyte-macrophage Colony-stimulating factor	For treating cancer and AIDS
10. Bovine growth hormone	For increasing milk yield.

Question 32.
How is DNA isolated in a purified form? (CBSE Outside Delhi 2009)
Answer:
Isolation of DNA in the purified form:

1. DNA has to be isolated in pure form for the action of restriction enzymes.
2. DNA can be released from the cells by digesting the cell envelope by the use of enzymes like lysozyme for bacterial cells, chitinase for fungal cells and cellulase for plants cells.
3. Since DNA is intertwined with histone proteins and RNAs, proteins are removed by treatment with proteases and RNAs by ribonucleases.
4. Other impurities are removed by employing suitable treatments.
5. The purified DNA is precipitated by the addition of chilled ethanol. It is seen as fine threads in suspension.

Question 33.
How is isolation and Fragmentation of DNA of interest carried out in recombinant DNA technology? (CBSE Outside Delhi 2009, 2019)
Answer:
1. Fragmentation DNA: Fragmentation of DNA is carried out by incubating the purified DNA molecules with suitable restriction enzymes at optimal conditions of temperature and pH.

2. Isolation of DNA (gene) of Interest:
(a) The fragments of DNA are separated by a technique called gel electrophoresis.
(b) The DNA is cut into fragments by restriction endonucleases.
(c) These fragments are separated by a technique called gel electrophoresis.
(d) Agarose, the natural polymer obtained from seaweeds, is used as the matrix.
(e) DNA fragments being negatively charged are separated by forcing them to move through the matrix towards the anode under an electric field.
(f) The DNA fragments separate/ resolve according to their size.
(g) The separated molecules are stained by ethidium bromide and visualised by exposure to UV radiation, as bright orange coloured bands.
(h) The separated bands of DNA (on the gel) are cut from the gel and extracted from the gel piece (elution).
(i) Such DNA fragments are purified and used for constructing recombinant DNA by joining them with cloning vectors.

Here we are providing Class 12 Biology Important Extra Questions and Answers Chapter 6 Molecular Basis of Inheritance. Important Questions for Class 12 Biology are the best resource for students which helps in Class 12 board exams.

Class 12 Biology Chapter 6 Important Extra Questions Molecular Basis of Inheritance

Molecular Basis of Inheritance Important Extra Questions Very Short Answer Type

Question 1.
Name the genetic material for the majority of organisms.
Answer:
DNA (Deoxyribose nucleic acid)

Question 2.
List the function of RNA.
Answer:
RNA acts as genetic material in viruses and also functions as an adapter, structural, and in some cases as a catalytic molecule.

Question 3.
How many nucleotides are present in a bacteriophage Φ × 174?
Answer:
5386.

Question 4.
List the number of base pairs in:
(i) lambda bacteriophage
Answer:
48502 bp

(ii) E.coli and
Answer:
4.6 × 10⁶bp

(iii) haploid content of human DNA.
Answer:
3.3 × 10⁹bp.

Question 5.
Comment two chains of DNA have antiparallel polarity.
Answer:
If one chain has 5′ → 3′ polarity, the other chain has 3′ → 5′ polarity.

Question 6.
What is the difference between DNA and DNAase?
Answer:
DNAs are the number of molecules of DNA and DNAase is an enzyme that digests DNA.

Question 7.
What made DNA as genetic material?
Answer:
As RNA was unstable, it evolved further with certain chemical modifications and formed DNA. DNA became more stable.

Question 8.
What is the average rate of polymerization?
Answer:
2000 bp per second.

Question 9.
List three components of the transcription unit.
Answer:

1. A promoter,
2. The structural gene,
3. A terminator.

Question 10.
What is the term used for fully processed hn RNA?
Answer:
A messenger RNA (mRNA).

Question 11.
What is splicing?
Answer:
It is the removal of introns and the joining of exons in a definite manner.

Question 12.
Where are UTRs present in mRNA strand?
Answer:
UTRs (Untranslated region) are present at both 5′ end (before start codon) and at the 3′ end (after stop signal).

Question 13.
Write the significance of UTRs.
Answer:
They are required for the regulation of an efficient translation process.

Question 14.
Name any two non-sense codons (stop signal).
Answer:

1. UGA (Opal)
2. UAA (Ochre).

Question 15.
What are the exceptions to the general rule that DNA is the genetic material in all organisms? Give evidence that supports these exceptions.
Answer:
Some animal viruses and all plant viruses contain RNA as their genetic material.

Question 16.
What is a replication fork? (CBSE, Delhi 2011)
Answer:
When replication starts the two strands of DNA unwind to form a Y-shaped structure called the replication fork.

Question 17.
Name the technique by which Gene expression can be controlled with the help of RNA molecule. (CBSE Sample Paper 2018-19)
Answer:
Northern blotting

Question 18.
Name the parts ‘A’ and ‘B’ of the transcription unit given below. (C.B.S.E. Delhi 2008)
Answer:

A-Promoter
B-Coding strand.

Question 19.
Mention the role of the codons AUG and UGA during protein synthesis. (CBSE 2011, 2016)
Or
Mention two functions of codon AUG. (CBSE 2010)
Answer:

1. AUG acts as a start signal and also code for methionine amino acid.
2. UGA acts as a stop signal.

Question 20.
How do histones acquire positive charge? (CBSE Delhi 2011)
Answer:
A histone protein acquires a positive charge because they are rich in basic amino acid residues such as lysine, arginine, and histidine. All are positively charged in their side chains.

Question 21.
Why do DNA fragments move towards the anode during gel electrophoresis? (CBSE Delhi 2018C)
Answer:
DNA molecule is negatively charged. When placed in an electric field it moves towards a positively charged anode.

Question 22.
Name one amino acid which is coded by only one codon. (CBSE Delhi 2018 C)
Answer:
Methionine – AUG/ Tryptophan – UGG

Molecular Basis of Inheritance Important Extra Questions Short Answer Type

Question 1.
If the sequence of coding strand In a transcription unit Is written as follows: (CBSE Delhi 2011)
5′- A T G C A T G C A T G C A T G C A T G C A T G C A T G C – 3′
Write down the sequence of mRNA.
Answer:
The sequence of mRNA shall be:
5′ – U A C G U A C G U A C G U A C G U A C G U A C G U A C G – 3′.

Question 2.
What is Chargaff rule? (CBSE Delhi 2011)
Answer:
Chargaff Rules:

1. In DNA molecule, A — T base pairs equals in number to G — C base pairs.
2. A + G = T + C, i.e. purines and pyrimidines equal in amount.
3. A = T and C = G (Amount).
4. The base ratio A + T/G + C may vary from one species to another but is constant for each species. It helps in identifying the source of DNA.
5. The deoxyribose sugar and phosphate component occur in equal proportions.

Question 3.
(a) Name the component of a nucleotide responsible for giving 5’— 3′ polarity to a polynucleotide.
Answer:
A polymer has at one end a free phosphate moiety at 5′- end of ribose sugar which is referred to as 5′-end of a polynucleotide chain. Similarly, at the other end of the polymer, the ribose has a free 3-OH group which is referred to as the 3′-end of a polynucleotide chain.

(b) Where in a nucleotide is the glycosidic bond present? (CBSE Delhi 2019 C)
Answer:
A nitrogenous base is linked to pentose sugar through an N-glycosidic linkage to form nucleoside.

Question 4.
Which property of DNA double helix led Watson and Crick to hypothesize a semi-conservative model of DNA replication? Explain.
Or
Write a note on the semi-conservative mode of DNA replication. (CBSE Delhi 2008; Delhi 2011, 2014)
Answer:
The semi-conservative model of DNA replication hypothesized by Watson and Crick was based upon the property that during replication, the two strands would separate and act as a template for the synthesis of new complementary strands. After the completion of replication, each DNA molecule would have one parental and one newly synthesized strand. This scheme was termed as ‘Semi-conservative’ DNA replication.

Question 5.
RNA was the first genetic material, DNA evolved later on. Explain.
Or
Why is RNA considered the first genetic material? (CBSE, 2009 2012)
Answer:
RNA was the first genetic material. There is now enough evidence to suggest that essential life processes (such as metabolism, translation, splicing, etc.) evolved around RNA. RNA used to act as a genetic material as well as a catalyst (there are some important biochemical reactions in living systems that are catalyzed by RNA catalysts and not by protein enzymes). But, RNA being a catalyst was reactive and hence unstable. Therefore, DNA has evolved from RNA with chemical modifications that make it more stable. DNA being double-stranded and having complementary strands further resists changes by evolving a process of repair.

Question 6.
Discuss the significance of the heavy isotope of nitrogen in Meselson and Stahl’s experiment.
Answer:

1. By using the heavy isotope of nitrogen, they could find out the semiconservative nature of DNA replication as the densities of DNA having ¹⁵N in both the strands ¹⁵N/¹⁴N DNA and ¹⁴N/¹⁴N DNA were all different.
2. The hybrid DNA (¹⁵N/¹⁴N DNA) had a density intermediate between that of heavy DNA (¹⁵N/¹⁵N DNA) and that of light/normal DNA (¹⁴N/¹⁴N DNA).

Question 7.
Differentiate polycistronic mRNA and monocistronic mRNA.
Answer:
Differences between polycistronic mRNA and monocistronic mRNA:

Polycistronic mRNA	Monocistronic mRNA
1. It is the mRNA that can code for only one polypeptide, i.e. it has one cistron	1. It is the mRNA that can code for more than one polypeptide, i.e. it has more than one cistron.
2. It is normally found in eukaryotic cells.	2. It is found in prokaryotic cells.

Question 8.
You are repeating the Hershey-Chase experiment and are provided with two isotopes ³²p and ¹⁵N (in place of ³⁵S in the original experiment). How do you expect your results to be different?
Answer:

1. The use of ¹⁵N will not give any conclusive result because it is only a heavy isotope of nitrogen.
2. In the original experiment, ³⁵S was detected only in the supernatant as it was incorporated in protein only, while 15N will be incorporated into proteins as well as in DNA and hence it would appear both in the supernatant and in the sediment as well.

Question 9.
What is DNA fingerprinting? Mention its applications.
Or
List any two applications of the DNA fingerprinting technique. (CBSE Delhi 2018)
Answer:
DNA fingerprinting: The technique of using DNA fragments, resulting from restriction endonuclease enzyme cleavage to identify particular individuals, is called DNA fingerprinting.

Applications of DNA Fingerprinting:

1. Paternity disputes can be solved by DNA fingerprinting.
2. It can solve the problems of evolution.
3. It can be used to study the breeding patterns of animals facing the danger of extinction.
4. It is useful in restoring the health of the patients suffering from leukemia (blood cancer).
5. It is very useful in the detection of crime and legal pursuits.

Question 10.
What are the major enzymes of DNA replication? (CBSE 2009)
Answer:
Enzymes of DNA Replication:

1. DNA-dependent DNA polymerase. It catalyzes the polymerization of deoxynucleotides,
2. Okazaki fragments, which are then quickly joined together by an enzyme known as DNA ligase.
3. The discovery of helicases and topoisomerase enzymes explains the unwinding of the DNA helix.

Question 11.
One of the codons on mRNA is AUG. Draw the structure of the tRNA adapter molecule for this codon. Explain the uniqueness of this tRNA. (CBSE Delhi and Outside Delhi, 2008)
Answer:
Transfer RNA or soluble RNA or Adapter RNA (tRNA or sRNA). It constitutes 15% of total RNA and is the smallest out of three with only 70-85 nucleotides having a sedimentation coefficient of 45.

It is unique because it has double specificity, one for codon on mRNA strand and the other for the corresponding amino acid.

Structure of tRNA adapter

Question 12.
Briefly describe the termination of a polypeptide chain. (C8SE 2009)
Answer:
Termination of polypeptide synthesis:

1. When one of the termination codons (UAA, UAG, UGA) comes at the A-site, it does not code for any amino acid and there is no tRNA molecule for it.
2. As a result, the polypeptide synthesis (or elongation of the polypeptide) stops.
3. The polypeptide synthesized is released from the ribosome, catalyzed by a ‘release factor’.

Question 13.
Write the dual purpose served by Deoxyribonucleoside triphosphates in polymerization. (CBSE Delhi 2018)
Answer:
Deoxyribonucleoside triphosphates serve as substrates, i.e. nucleotides during replication, and also provide energy for polymerization reaction by cleavage of high energy terminal phosphates bond. Thus they serve dual purposes.

Question 14.
Although a prokaryotic cell has no defined nucleus, yet DNA is not scattered throughout the cell. Explain. (CBSE Delhi 2018)
Answer:
In the nucleoid region, DNA that is negatively charged is held with some proteins that are positively charged. The DNA in the nucleoid is organized in large loops held by proteins. And the DNA in the form of single chromosomes is attached to the mesosome at a point. 15

Question 15.
Differentiate between the genetic codes given below:
(i) Unambiguous and Universal
Answer:

• Unambiguous: The code is specific, i.e. one Condon code for onLy one amino acid.
• Universal: The code is the same in aLL organisms.

(ii) Degenerate and Initiator (CBSE Delhi 2017)
Answer:

• Degenerate: When an amino acid is coded by more than one codon, it is said to be degenerate.
• Initiator: AUG is an initiator codon, i.e. it initiates the translation process and also codes for methionine.

Question 16.
Why does the lac operon shut down sometime after the addition of lactose in the medium where E.coti was growing? Why low-level expression of the lac operon is always required? (CBSE Sample Paper 2018-19)
Answer:
After the addition of lactose, a complete breakdown of lactose to glucose and galactose takes place. Therefore, there is no more lactose to bind to the repressor protein and the lac operon shuts down.

A very low level of expression of lac operon has to be present in the cell all the time, otherwise, lactose cannot enter the cells.

Question 17.
Carefully examine structures A and B of pentose sugar given below. Which one of the two is more reactive? Give reasons. (CBSE Sample Paper 2019-20)

 

 

 

 

 

 

Answer:
The pentose sugar of figure A is more reactive.
Reasons:

 

1. There are two -OH groups present in the pentose sugar.
2. It makes it more labile.
3. It is easily degradable.

Molecular Basis of Inheritance Important Extra Questions Long Answer Type

Question 1.
(i) How are the following formed and involved in DNA packaging in a nucleus of a cell?
(a) Histone octamer
(b) Nucleosome
(c) Chromatin
Answer:
Packaging of DNA.
(a) Histone octamer: Five types of histone proteins (H₁, H₂A, H₂B, H_3, and H₄) are involved. Out of these, four of them H₁ A, H₂ B, H_3, and H₄ occur in pairs to produce histone octamer also called the nu body. Histones are organized in a form of a compact unit formed of 8 molecules hence called histone octamer.

(b) Nucleosome: The unit of compaction of DNA is the nucleosome. About 146 bp of DMA is wrapped
over histone octamer for 1 \(\frac{3}{4}\) turn to form nucleosome of the size 110 × 60 A.

(c) Chromatin: Linker DNA connects two adjacent nucleosomes. It bears H₁ protein. As a result, a chain is formed called chromatic. Nucleosome chain gives ‘beads on string’ appearance under the electron microscope. Chromatin are repeating units of a structure located on the nucleosome.

(ii) Differentiate between Euchromatin: 1 and Heterochromatin. (CBSE Delhi 2016)
Answer:

Heterochromatin	Euchromatin
1. Darkly stained.	1. lightly stained.
2. Condensed regions of chromatin fibers.	2. less tightly coiled regions of chromatin fibers.
3. Transcriptionatty inactive or less active.	3. Transcnptionally active.
4. less affected by temperature, sex, or age. it is not acetylated.	4. More affected by temperature, sex, or age. It is acetylated during interphase.

Question 2.
Distinguish between heterochromatin and euchromatin. Which of the two is transcriptionally active?
Answer:
Differences between heterochromatin and euchromatin:

Heterochromatin	Euchromatin
1. Darkly stained.	1. lightly stained.
2. Condensed regions of chromatin fibers.	2. less tightly coiled regions of chromatin fibers.
3. Transcriptionatty inactive or less active.	3. Transcnptionally active.
4. less affected by temperature, sex, or age. it is not acetylated.	4. More affected by temperature, sex, or age. It is acetylated during interphase.

Euchromatin is more active transcriptionally.

Question 3.
Write a note on messenger RNA.
Answer:
Messenger RNA (mRNA).
It forms only 5% of total RNA but is the longest of all. It brings instructions from DNA for the formation of a particular polypeptide. The instructions are coded in the form of a base sequence called genetic code. Three adjacent nitrogen bases specify a particular amino acid. The formation of polypeptides occurs over the ribosomes. mRNA gets attached to ribosomes.

mRNA.

It starts as a cap for attachment with the ribosome. It is followed by an initiation codon (AUG) either immediately or after a small non-coding region. It is followed by the coding region followed by the termination codon (UAA, UAG, and UGA). Then there is a small non-coding region and poly-A area at 3 termini. The mRNA may specify only a single polypeptide or a number of them called monocistronic and polycistronic respectively.

The life span of mRNA maybe a few minutes to an hour or even days in the case of RBC.

Question 4.
What is genetic code? List the properties of genetic code.
Answer:
The code language of DNA and mRNA is complementary. So, genetic code is the sequence of nucleotides in DNA and RNA that determines the amino acid sequence in proteins. Some amino acids are specified by more than one codon. The sequence of nucleotide on the tRNA molecule which complements the codon is called anticodon, e.g. one of the codons for the amino acid leucine is CUG and the anticodon is GAC. Similarly, the codon for phenylalanine is UUU, while the anticodon is AAA.

Properties of genetic code:
The following properties of genetic code have now been proved by experimental evidence.

1. The code is a triplet.
2. The code is degenerate.
3. The code is non¬overlapping.
4. The code is commaless.
5. The code is non-ambiguous.
6. The code is universal,
7. Collinearity.

Both polypeptide and DNA or mRNA have a linear arrangement of their components.

The term code letter stands for a nucleotide A, T, G, or C in DNA and A, U, G, or C in RNA. The sequence of three does not code for any amino acid, such codons are called a non-sense codon, e.g. UGA.

Question 5.
What is the role of ribosomes during translation? Ribosomes move along mRNA molecules and catalyze the assembly of amino acids into protein chambers. (CBSE Outside Delhi 2019)
Answer:
Role of ribosomes: Ribosomes usually form linear or helical groups during active protein synthesis called polyribosomes or polysomes. The mRNA strand having coded information joins along with smaller subunits of ribosomes. The adjacent ribosomes are 360 A apart.

The different parts of the ribosome connected with protein synthesis are:

1. A tunnel for mRNA.
2. A groove for the passage of newly synthesized polypeptide (larger subunit).
3. Two active sites (P-site-peptidyl transfer or donor site and A-site or aminoacyl or acceptor site).
4. A binding site for tRNA near A-site.
5. Presence of enzyme peptidyl transferase.
6. Recognition point of smaller subunit for mRNA.
7. Presence of GTP-ase, binding sites for elongation factors, and translocases.

Question 6.
Describe the steps in the sequencing of the human genome.
Answer:
Sequencing of a genome.
The method involved two major approaches:

1. Expressed Sequence Tags (ESTs): It is focused on identifying all the genes that are expressed as RNAs.
2. Sequence Annotation: It involves simply sequence the whole set of the genome that included all the coding and non-coding sequences and then assigning functions to different regions in the sequence.

HGP followed the second technique:

1. The total DNA from the cell is isolated and converted into random fragments of relatively smaller sizes.
2. These fragments are then cloned in suitable hosts using specialized vectors; the commonly used hosts are bacteria and yeast and the vectors are bacterial artificial chromosomes (BAC) and yeast artificial chromosomes (YAC).
3. The fragments are then sequenced using automated DNA sequences.
4. The sequences were then arranged on the basis of certain overlapping regions present in them; this required the generation of overlapping fragments for sequencing.
5. These sequences are annotated and assigned to the respective chromosomes.

Question 7.
(i) Why is DNA molecule a more stable genetic material than RNA? Explain.
Answer:
There is now enough evidence to suggest that essential life processes (such as metabolism, translation, splicing, etc.) revolved around RNA. RNA used to act as a genetic material as well as a catalyst (there are some important biochemical reactions in living systems that are catalyzed by RNA catalyst and not by protein enzymes). But, RNA being a catalyst was reactive and hence unstable. Therefore, DNA has evolved from RNA with chemical modifications that make it more stable. DNA being double-stranded and having complementary strands further resists changes by evolving a process of repair.

(ii) ‘Unambiguous’, ‘degenerate’ and ‘universal’ are some of the salient features of genetic code. Explain. (CBSE 2008, 2011, Delhi 2014, 2016)
Answer:
Unambiguous: Each codon codes for one amino acid, none for more than one.

Degenerate: One amino acid often has more than one triplet codon; only methionine and tryptophan have a single triplet codon.

Universal: The genetic code is universal. A given codon in DNA and m RNA specifies the same amino acid in all organisms from viruses, bacteria to human beings.

Question 8.
Explain the role of tRNA in the initiation of protein synthesis. (CBSE 2012)
Answer:
Role of tRNA’ in the initiation of protein synthesis:

1. tRNA is an adapter molecule that on one hand would bind to a specific amino acid.
2. The tRNA then called sRNA (soluble RNA).
3. tRNA has an anticodon loop that has bases complementary to the code.
4. It has an amino acid acceptor end to which it binds to an amino acid.
5. tRNAs are specific for each amino acid.
6. For initiation, there is another tRNA, that is referred to as initiator tRNA.
7. Amino acids are activated in the presence of ATP and joined to their corresponding tRNA it is called charging of tRNA or aminoacylation of tRNA.
8. Two such charged tRNAs are brought close and the formation of peptide bond occurs.

Question 9.
List the criteria that can act as genetic material must fulfill. Which one of the criteria is best fulfilled by DNA or RNA thus making one of them a better genetic material? Explain. (CBSE (Delhi) 2016)
Answer:
Essential requirements of genetic material:

1. A genetic material should be able to store and express information to confer the heritable characters of living organisms.
2. It should be capable to make its own replica.
3. Genetic material should also have the mechanism to undergo mutations that will generate variations.

Question 10.
A small stretch of DNA strand that codes for a polypeptide are shown below:
3’— — — — CAT CAT AGA TGA AAC — — — — 5’
(a) Which type of mutation could have occurred in each type resulting in the following mistakes during rep¬lication of the above original se¬quence?
(i) 3’ … … … … CAT CAT AGA TGA ATC … … … 5’
Answer:
A point mutation (single base substitution)

(ii) 3’ … … … … CAT ATA GAT GAA AC … … … 5’
Answer:
A point mutation (single base deletion)

(b) How many amino acids will be translated from each of the above strands (i) and (ii)? (CBSE Sample Paper 2019-20)
Answer:
(i) 4 amino acids
(ii) 4 amino acids

Question 11.
(i) In the human genome which one of the chromosomes has the most genes and which one has the fewest?
Answer:
Chromosome 1 has the most genes (2968) and Y-chromosome has the least gene (231).

(ii) Scientists have identified about 1.4 million single nucleotide polymorphs in the human genome. How is the information of their existence going to help the scientists? (CBSE2009)
Answer:
This information promises to revolutionize the processes of finding the chromosomal location for disease-associated sequences and tracing human history.

Question 12.
Describe the structure of an RNA polynucleotide chain having four different types of nucleotides. (CBSE Delhi 2013, 2019)
Answer:

RNA polynucleotide

RNA polynucleotide:

1. RNA is formed of polynucleotides of ribose series.
2. Each nucleotide is formed of a nitrogen base, ribose sugar (pentose), and phosphoric acid.
3. There are two types of nitrogen bases-purines and pyrimidines.
4. Adenine (A) and Guanine (G) are purines.
5. Cytosine (C) and Uracil (U) are pyrimidines.
6. The nitrogen base is linked to pentose sugar by N-glycosidic linkage to form nucleoside.
7. When a phosphate group is linked to 5′ -OH of a nucleoside through phosphodiester linkage thus forms nucleotide.
8. Nucleotides of ribose series present are AMP, GMP, CMP, and UMP.
9. Two nucleotides are linked through 3′ – 5′ phosphodiester linkage to form dinucleotides.
10. More nucleotides can be joined in such a manner to form a polynucleotide chain.

Question 13.
Describe the initiation process of transcription in bacteria. (CBSE 2010, 2016, 2019)
Answer:
Initiation of transcription in bacteria:

1. The process of copying genetic information from antisense or template strands of DNA into RNA is called transcription.
2. The segment of DNA that takes part in transcription is called the transcription unit. It has three components
   (a) a promoter,
   (b) the structural gene and
   (c) a terminator.
3. The structural gene is composed of that strand of DNA that has 3′ → 5′ polarity as transcription can occur only in the 5′ → 3′ direction.
4. Transcription requires a DNA-dependent-RNA polymerase and initiation factor.
5. Bacteria have only one type of RNA polymerase which transcribes all three types of RNAs.
6. Ribonucleotides of ribose series are activated through phosphorylation (ATP, GTP, CTP, UTP).
7. Transcription begins at the initiation site. A promotor has an RNA polymerase recognition site and it binds to the specific site.
8. Enzymes required for the unwinding of the chain are unwound and single-stranded binding proteins.
9. Nucleotides are added as per the base-pairing rule.

Question 14.
State the role of VNTRs in DNA fingerprinting. (CBSE Outside Delhi 2013)
Answer:
Role of Variable Number of Tandem Repeats (VNTRs) Short nucleotide repeats in the DNA are very specific in each individual and vary in number from person to person but are inherited. These are the ‘Variable Number Tandem Repeats’ (VNTRs). These are also called ‘minisatellites. Each individual inherits these repeats from his/her parents which are used as genetic markers in a personal identity test.

For example, a child might inherit a chromosome with six tandem repeats from the mother and the same tandem repeated four times in the homologous chromosome inherited from the father. The half of VNTR alleles of the child resemble that of the mother and half that of the father.

Question 15.
(i) List the two methodologies which were involved in the human genome project. Mention how they were used.
Answer:
(a) Expressed Sequence Tags (ESTs): This method focuses on identifying all the genes that are expressed as RNA.
(b) Sequence Annotation: It is a method of simply sequencing the whole set of the genome that contains all the coding and non-coding sequences, and then assigning different regions in the sequence with functions.

(ii) Expand ‘YAC’ and mention what it was used for. (CBSE Delhi 2017)
Answer:
‘YAC’ is an abbreviated form of ‘Yeast Artificial Chromosome’. It is used as a cloning vector for cloning DNA fragments in a suitable host so that DNA sequencing can be done.

Question 16.
Summarise the process by which the sequence of DNA bases in the Human Genome Project was determined using the method developed by Frederick Sanger. Name a free-living non-pathogenic nematode whose DNA has been completely sequenced. (CBSE Sample Paper 2019-20)
Answer:
(a) The HGP strategy:

(b) Chromosome 1
(c) Caenorhabditis elegans

Question 17.
Why is a DNA molecule considered a better hereditary material than an RNA molecule? (CBSE Delhi 2018C
Answer:
DNA is the better hereditary material compared to RNA because of the following features:

1. DNA molecule is more stable tha> RNA (as thymine is present here instead of uracil found in RNA).
2. DNA is less reactive than RNA (a- does not have 2’ OH group wl makes RNA very reactive).
3. Since DNA is less reactive so it is t easily degradable.
4. Chances of mutation are less. compared to RNA.

Question 18.
Name the three RNA polymerases found in eukaryotic cells and mention their functions. (CBSE Delhi 2018C)
Answer:

• RNA polymerase I – It transcribes ribosomal RNAs (28S, 18S, 5.8S rRNAs)
• RNA polymerase II – It transcribes precursors of mRNA – heterogeneous nuclear RNA (hnRNA)
• RNA polymerase III – It transcribes transfer RNA (tRNA), 5SrRNAand small nuclear RNAs (snRNAs)

Question 19.
Explain the post-transcriptional modifications the hn-RNA undergoes in eukaryotic cells. (CBSE Delhi 2018C)
Answer:
In eukaryotes, the primary transcript is not functional due to the presence of exons and introns. Therefore it requires certain modifications to make it functional. First of all, introns are removed and the remaining exons are joined together in order by the process called Splicing.

Then the hnRNA undergoes capping followed by tailing. In capping, methyl guanosine triphosphate is added to the 5’ end of hnRNA. In tailing 200 -300 adenylate residues are added to 3’ end in a template-independent manner. Now the hnRNA has been processed into mRNA.

Question 20.
What are the aims of bioinformatics?
Answer:
Aims of bioinformatics:

1. To spread scientifically investigated knowledge for the benefit of the research community.
2. To transform the biological polymeric sequences into sequences of digital symbols and to store them as databases.
3. To develop a variety of methods and tools of software for data analysis.

Question 21.
Describe Griffith’s experiment to demonstrate that DNA is the basic genetic material. What explanation for Griffith’s observation was given by Avery, McCarty, and MacCleod? (CBSE Delhi 2008, 2009, 2012)
Answer:
Griffith’s experiment demonstrates DNA as genetic material. Transformation experiments were initially conducted by F. Griffith in 1928.

1. He injected a mixture of two strains of Pneumococcus (Diplococcus pneumoniae) into mice. One of these two strains S III was virulent and the other strain All was non-virulent.
2. The S III type bacteria when injected into mice cause pneumonia and ultimately to death. The R II type bacteria when injected, no pneumonia occurred.
3. The S III bacteria, prior to injection are killed by heating. It is then injected into the mice. It did not cause the disease.
4. Heat killed S III and RII (non-virulent) bacteria, when injected into mice, causing pneumonia. Finally, death occurred.

Griffith’s experiment demonstration transformation in Pneumococcus.

This proved that the DNA of S III type bacteria has transformed the DNA of R II type bacteria into virulent type S III. This phenomenon of transferring characters of one strain to another by using a DNA extract of the former is called transformation. Conclusion. Griffith concluded that virulence was transferred from S-Type dead cells to R-Type living cells in the form of a capsule (some component of a cell) rather than the whole cells.

Contribution of Avery, MacCleod, and MacCarty: Avery, MacCleod, and McCarty gave the proof that the “transforming agent” is DNA. They carried out the experiments with Diplococcus and showed the transformation of type R-ll to type S-III. They found that proteases and RNases did not affect transformation but DNAses affect it. Thus they gave proof that the active component was DNA and not the RNA, proteins, or polysaccharides.

Question 22.
How did Hershey and Chase differentiate between DNA and protein in their experiment while proving that DNA is the genetic material? (CBSE 20015)
Or
Why did Hershey and Chase use 35S and 32P in their experiment? Explain. State the importance of blending and centrifugation in their experiment. Write the conclusion they arrived at after completing their experiment. (CBSE 20019 C)
Or
Hershey and Chase carried out their experiment under three steps: (a) Infection, (b) Blending, and (c) Centrifugation.
Or
Explain each one of these steps that helped them to prove that DNA is the hereditary material. (CBSE (Outside Delhi) 2019)
Answer:
1. Alfred Hershey and Martha Chase (1952) worked with viruses that infect bacteria called bacteriophages.

2. They used radioactive sulfur (35S) to identify protein and radioactive phosphorus (32P) to identify the components of nucleic acid.

3. The tadpole-shaped bacteriophage attaches to the bacteria. Its genetic material enters the bacterial cell by dissolving the cell wall of bacteria. The bacterial cell treats the viral genetic material as if it was it’s own and subsequently manufactures more virus particles.

4. Infection: They grew some viruses on a medium that contained radioactive phosphorus and others on a medium that contained radioactive sulfur.

The Hershey—Chase Experiment

5. Viruses grown in the presence of radioactive phosphorus contained radioactive DNA but not radioactive protein because DNA contains phosphorus but protein does not. Similarly, viruses grown on radioactive sulfur contained radioactive protein but not radioactive DNA because DNA does not contain sulfur.

6. Radioactive phages were allowed to attach to E. co/i bacteria. Then as the infection proceeded, the viral coats were removed from the bacteria by agitating them in a blender. The virus particles were separated from the bacteria by spinning them in a centrifuge.

7. Bacteria that were infected with viruses that had radioactive DNA were radioactive, indicating that DNA was the material that passed from the virus to the bacteria.

8. Bacteria that were infected with viruses that had radioactive proteins were not radioactive.

9. This indicates that proteins did not enter the bacteria from the viruses.

10. DNA is, therefore, the genetic material that is passed from virus to bacteria.

Question 23.
List the characteristics of DNA molecules.
Answer:
Characteristics of DNA molecule:

1. The two chains are spirally coiled about around a common axis to form a regular, right-handed double helix.
2. The double helix has a major groove and minor groove alternately.
3. The helix is 20Å wide; it’s one complete turn is 34Å long, and has 10 base pairs, and the successive base pairs are 3.4Å apart.
4. The two chains are complementary to each other with respect to base sequence.
5. The two strands are hydrogen-bonded: A on one chain is joined to T on the other chain by 2 hydrogen
   bonds; C on one chain is linked to G on the other chain by 3 hydrogen bonds.
6. The two strands run in an antiparallel direction.
7. The amount of A + G = the amount of T + C; the amount of A = the amount of T: the amount of G = the amount of C. Sugar and phosphate groups occur in equal proportion.
8. The DNA molecule is remarkably stable due to hydrogen bonding and hydrophobic interactions.
9. The DNA molecule undergoes denaturation and renaturation easily.
10. The denatured DNA strands are hyperchromic.
11. The DNA molecule can replicate and repair itself, and can also transcribe RNAs.
12. The base sequence of one chain serves as the genetic code.
13. The DNA can function in vitro.
14. The amount of DNA per nucleus is constant in all the body cells of a given species.

Question 24.
How is a long DNA molecule adjusted in a nucleus? (CBSE Delhi 2008)
Or
(i) What is this diagram representing?
(ii) Name the parts A, B, and C.
(iii) In the eukaryotes, the DNA molecules are organized within the nucleus. How is the DNA molecule organized in a bacterial cell in the ab¬sence of a nucleus? (CBSE 2009)

Answer:
The distance between two consecutive base pairs is 0.34 nm (0.34 × 10-9 m). If the length of DNA double helix in a typical mammalian cell is calculated (simply by multiplying the total number of bp with the distance between two consecutive bp, that is, 6.6 × 109 bp × 0.34 × 1(T9 m/bp), it comes out to be approximately 2.2 meters. A length that is far greater than a dimension of a typical nucleus (approx 10-6 m).

In eukaryotes, this organization is much more complex. There is a set of positively charged, basic proteins called ‘histones’. A protein acquires charge depending upon the abundance of amino acid residues with charged side chains. Histones are rich in amino acid residues lysines and arginines. Both the amino acid residues carry positive charges in their side chains. Histone organized to form a unit of eight histone molecules called ‘histone octamer’. The negatively charged DNA is wrapped around positively charged histone-octamer to form a structure called ‘Nucleosome’.

A typical nucleosome contains 200 bp of DNA helix. Nucleosomes constitute the repeating unit of a structure in the nucleus called ‘Chromatin’, thread-like stained (colored) bodies seen in the nucleus. The nucleosomes in chromatin are seen as a “beads-on-string” structure when viewed under an electron microscope (EM).
Or
(i) Nucleosome

(ii)

(iii) Organization of DNA molecule in bacteria. In prokaryotes like bacteria which lacks a nucleus, DNA is not scattered. The negatively charged DNA binds with positively charged proteins in a region called a nucleoid.

Question 25.
Describe briefly the mechanism of DNA replication. (CBSE (Delhi) 2019)
Or
Draw a labeled diagram of the replication fork. (CBSE 2011)
Or
Draw a neat labeled sketch of replicating fork of DNA.
Or
(a) Why does DNA replication occur within a replication fork and not in its entire length simultaneously?
(b) “DNA replication is continuous and discontinues on the two strands within the replication fork.” Give reasons. (CBSE (Outside Delhi) 2019)
Answer:
Mechanism of DNA replication. Watson and Crick, while explaining their model of DNA structure, suggested the semi-conservative mechanism of its replication. The quality and quantity of DNA in parent and daughter cells must be the same.

Replication is one of the most important properties of DNA and forms the very basis of life:
1.  Replication takes place during the S-phase of the interphase between two mitotic cycles.

2. Replication is a semi-conservative process in which each of the two double helices formed from the parent double strand has one old and one new strand. Repair replication is non-conservative.

3. DNA replication requires a DNA template, a primer, deoxyribonucleoside triphosphates (dATP, dGTP, dTTP, dCTP), Mg++, DNA unwinding protein, superhelix relaxing protein, a modified RNA polymerase to synthesize the RNA primer, a joining polynucleotide ligase, enzyme.

4. Watson and Crick suggested that the two strands of DNA molecules uncoil and separate, and each strand serves as a template for the synthesis of a new strand alongside it.

5. The sequence of bases which should be present in the new strands can be easily predicted because these would be complementary to the bases present in the old strands. A will pair with T, T with A, C with G, and G with C.

6. Thus, two daughter molecules are formed from the parent molecule and these are identical to the parent molecule.

7. Each daughter DNA molecule consists of one old (parent) strand and one new strand.

Continuous and discontinuous synthesis of DNA.

8. Since only one parent strand is conserved in each daughter molecule, this mode of replication is said to be semiconservative.

9. In one strand replication takes place discontinuously and short pieces called Okazaki fragments are synthesized. One strand may synthesize a continuous strand and the other Okazaki fragments. Both new strands are synthesized in the 5’ → 3’ direction. Thus one strand is synthesized forward and the other backward.

10. The Okazaki pieces are joined by polynucleotide ligase, a joining enzyme, to form continuous strands.

Question 26.
Provide experimental evidence for the semi-conservative mode of replication of DNA. (CBSE 2008 (Outside Delhi) 2012, 2016, 2019 C)
Answer:
Meselson and Stahl (1958) experimentally proved that DNA replication is semi-conservative.

1. E. coli bacterium was grown for many generations in a culture medium in which the nitrogen source contained heavy isotope N¹⁵ thus the labeling of bacterial DNA was done.
2. Later on, these bacteria were cultured in N¹⁴ non-radioactive isotope.
3. DNA was analyzed to determine the distribution of radioactivity.
4. The experiment showed that one strand of each daughter DNA molecule was radioactive whereas the other was non-radioactive.
5. During the second replication in the N¹⁴ medium, the radioactive and non-radioactive strands separated and served as the template for the synthesis of non-radioactive strands.
6. Out of four DNA molecules, two are completely non-radioactive and the other two have half of the molecule as non-radioactive.
7. This evidence shows that DNA replication is semi-conservative.
8. This biochemical evidence was supported by the direct cytological observation of duplicating DNA of E. coli.

Question 27.
(i) State the ‘Central dogma’ as proposed by Francis Crick. Are there any exceptions to it? Support your answer with a reason and an example.
Answer:
Francis Crick proposed the Central dogma in molecular biology, which states that the genetic information

Central dogma
In some viruses, central dogma is seen in the reverse direction, that is from RNA to DNA as RNA is the main genetic material. Example: Retrovirus (HIV) and the process is reverse transcription.

(ii) Explain how the biochemical characterization (nature) of the ‘Transforming Principle’ was determined, which was not defined from Griffith’s experiments. (CBSE Delhi 2018)
Answer:
Transforming principle: In 1928, Frederick Griffith, in a series of experiments with Streptococcus pneumonia which is a bacterium responsible for pneumonia, witnessed a miraculous transformation in the bacteria. During the experiment, a living organism (bacteria) changed its physical form. He concluded that the R strain bacteria had somehow been transformed by the heat-killed S strain bacteria.

Some ‘transforming principle’, transferred from the heat-killed S strain, had enabled the R strain to synthesize a smooth polysaccharide coat and become virulent. This must be due to the transfer of the genetic material. However, the biochemical nature of genetic material was not defined from his experiments.

Oswald Avery, Colin MacLeod, and Maclyn McCarty worked to determine the biochemical nature of the ‘transforming principle’ in Griffith’s experiment. They purified biochemicals (proteins, DNA, RNA, etc.) from the heat-killed S cells to see which ones could transform live R cells into S cells. They discovered that DNA alone from S bacteria caused R bacteria to become transformed.

They also discovered that protein-digesting enzymes (proteases) and RNA-digesting enzymes (RNases) did not affect transformation, so the transforming substance was not a protein or RNA. Digestion with DNase did inhibit transformation, suggesting that the DNA caused the transformation. They concluded that DNA is the hereditary material, but not all biologists were convinced.

Question 28.
(a) Write the contributions of the following scientists in deciphering the genetic code: George Gamow; Hargobind Khorana; Marshall Nirenberg; Severo Ochoa.
Answer:
Contributions of scientists:

1. George Gamow: He suggested that in order to code for all the 20 Amino acids, the code should be made up of three nucleotides. He proved that the codon is a triplet.
2. Hargobind Khorana: He provided experimental proof that genetic codon is always triplet. He was able to synthesize RNA molecules with a defined combination of bases (homopolymers and copolymers)
3. Marshall Nirenberg: He prepared a cell-free system for protein synthesis and finally deciphered the genetic code.
4. Severe Ochoa: He established that enzyme polynucleotide phosphorylase was helpful in RNA synthesis with a defined sequence in a template-independent manner.

(b) State the importance of a Genetic code in protein biosynthesis. (CBSE Delhi 2019)
Answer:
Importance of Genetic code: Genetic code codes for a specific amino acid that is required for protein synthesis. (CBSE Outside Delhi 2018)

Question 29.
Explain translation in detail.
Or
Explain the process of charging tRNA.
Answer:
Translation: During the translation process, proteins are made by the ribosomes on the mRNA strand:
The main steps are:

Continuous and discontinuous synthesis of DNA:

1. Activation of amino acid.
2. Transfer of activated amino acid to tRNA.
3. Initiation of synthesis.
4. Elongation of a polypeptide chain.
5. Termination of a chain.

Question 30.
Explain the steps involved in DNA fingerprinting. (CBSE 2009)
Answer:
Steps of DNA fingerprinting:
The steps/procedure in DNA fingerprinting include the following:

1. Extraction: DNA is extracted from the cells in a high-speed, refrigerated centrifuge.
2. Amplification: Many copies of the extracted DNA are made by a polymerase chain reaction.
3. Restriction Digestion: DNA is cut into fragments with restriction enzymes into precise reproducible sequences.
4. Separation of DNA sequences/ restriction fragments: The cut DNA fragments are introduced and passed through an electrophoresis setup containing agarose polymer gel; the separated fragments can be visualized by staining them with a dye that shows fluorescence under ultraviolet radiation.
   
   Lac operon
5. Southern Blotting: The separated DNA sequences are transferred onto a nitrocellulose or nylon membrane.
6. Hybridization: The nylon membrane is immersed in a bath and radioactive probes (DNA segments of known sequence) are added: these probes target a specific nucleotide sequence that is complementary to them.
7. Autoradiography: The nylon membrane is pressed on an X-ray film and dark bands develop at the probe sites.

Question 31.
What is the inducer in the lac operon? How does it ensure the “switching on” of genes?
(i) Draw a schematic representation of the Lac operon.
Answer:
Inducer. It is a chemical that may be a substrate, hormone, or some other metabolite which after coming in contact with the repressor, changes the latter into the non-DNA binding state so as to free the operator gene. Thus the “switch on” occurs.

(ii) Explain how does this operon gets switched ‘on’ or ‘off’.
Answer:
The expression of the genes is usually controlled to achieve maximum cellular economy. This means that the gene will be turned on or off as per requirement. A set of genes will be switched on when there is the necessity to handle and metabolize a new substrate. When these genes are turned on enzymes are produced, which metabolize the new substrate. The phenomenon is known as induction.

(iii) What is the role of a regulatory gene? (CBSE Delhi 2012, 2019 C)
Answer:
Role of the regulatory gene in a lac operon: It synthesizes a biochemical or regulator protein that can act positively as an activator and negatively as a repressor. It controls the activity of the operator gene.

Question 32.
(i) Describe the structure and function of a tRNA molecule. Why is it referred to as an adapter molecule?
Answer:
t-RNA (transfer RNA) reads the genetic code on one hand and transfers amino acids on the other hand. So it was called adapter molecule by Francis Crick. It is also called soluble RNA (SRNA).

Note: For the figure, please see the answer of Q. no. 11 short answer type question. The secondary structure of tRNA is clover leaf-like but the 3-D structure is inverted L-shaped. t-RNA has five arms or loops:

• Anticodon loop: It has bases complementary to the code.
• Amino acid acceptor end: Amino acid binds to it.
• T-loop: It helps in binding to ribosomes.
• D-loop: It helps in binding aminoacyl synthetase.
• Variable loop: Its function is not known.

(ii) Explain the process of splicing of hn-RNA in a eukaryotic cell. (CBSE Delhi 2017)
Answer:
The primary transcript formed in eukaryotes is non-functional, containing both the coding region exon and non-coding region intron in RNA and are called heterogeneous RNA or hn-RNA. hn-RNA undergoes a process where the introns are removed and exons are joined to form mRNA by the process called splicing.

Question 33.
(i) Why does replication occur in small replication forks and not in entire lengths?
Answer:
Replication occurs in a small opening of a DNA helix called replication forks for very long DNA molecules. The reason is in the case of long DNA molecules, the two strands of DNA cannot be separated in their entire length due to very high energy demand.

(ii) Why is DNA replication continuous and discontinuous in a replication fork?
Answer:
DNA-dependent DNA polymerase catalyzes DNA polymerization only in one direction, that is, 5′ → 3′. DNA strands are anti-parallel and have opposite polarity. So on template strand with polarity 3′ → 5′ DNA replication is continuous white on the template strand with polarity 5′ → 3′ replication is discontinuous.

(iii) State the importance of the origin of replication in a replication fork. (CBSE Delhi 2018C)
Answer:
Replication cannot start randomly at any point in DNA. It requires a definite region of sequences where the replication originates. This site is called the origin of replication.

Question 34.
Compare the processes of DNA replication and transcription in prokaryotes. (CBSE Delhi 2019C)
Answer:

Replication	Transcription
(i) It is a synthesis of DNA from DNA.	(i) It is a synthesis of RNA from DNA.
(ii) Both strands take part in replication.	(ii) Only one strand functions as a template.
(iii) It forms double-stranded DNA.	(iii) It forms single-stranded RNA.
(iv) RNA primer is essential for initiation.	(iv) No primer Is required.
(v) Occurs in the S phase of the cell cycle.	(v) Occurs in the G1 and G2 phases of the cell cycle.
(vi) Catalysed by DNA polymerase enzymes.	(vi) Catalysed by RNA polymerase enzymes.
(vii) Deoxyribonucteoside triphosphates (dATP, dGTP, dCTP, dTTP) serve as raw materials.	(vii) Ribonucleoside triphosphates (ATP, GTP, CTP, UTP) serve as raw materials.
(viii)Involves unwinding and splitting of the entire DNA molecule (chromosome).	(viii) Involves unwinding and splitting of only those genes which are to be transcribed.
(ix) Two double-stranded DNA molecules are formed from one DNA molecule.	(ix) A single one-strand RNA molecule is formed from a segment of one DNA strand.

Question 35.
Write the different components of a lac operon in E. coil. Explain its expression while in an ‘open’ state. (CBSE Delhi 2017, 2019 C)
Answer:
The Lac operon (Inducible operon): The concept of the operon was first proposed in 1961 by Jacob and Monod.
Components of an operon:

1. Structural genes: It Is the fragment of DNA that transcribes mRNA for polypeptide synthesis.
2. Promoter: It Is the sequence of DNA where RNA poLymerase binds and initiates transcription.
3. Operator: It is the sequence of DNA adjacent to the promoter.
4. Regulator gene: It is the gene that codes for repressor protein which binds to the operator due to which operon Is switched “off”.
5. Inducer: Lactose Is an Inducer that helps in switching “on” of an operon. Lac operon consists of three structural genes (z, y, a), operator (o), promoter (p), regulatory gene (r).

(a)

(b)

• Gene z codes for p-galactosidase
• Gene y codes for permease.
• Gene codes for enzymes transacetylase. When lactose is absent:

When lactose is absent, i.e. gene produces repressor protein.

This repressor protein binds to the operator and as a result, prevents RNA polymerase to bind to the operon, and the operon is switched off.

When lactose is present:

• Lactose acts as an inducer that binds to the repressor and forms an inactive repressor.
• The repressor cannot bind to the operator.
• Now the RNA polymerase binds to the operator and transcribes lac mRNA.
• Lac mRNA is polycistronic, i.e. produces all three enzymes p-galactosidase, permease, and transacetylase.
• The lac operon is switched on.

Question 36.
What is the human genome project (HGP)? Write salient features of the human genome project.
Answer:
Human Genome Project. It is a mega project involving a lot of money, the most advanced techniques, numerous computers, and scientists at work. The magnitude of the project can be imagined that if the cost of sequencing a bp is 3 dollars, sequencing of 3 × 109 bp would be a billion dollars. If the data is to be stored in books, with each book having 1000 pages and each page with 1000 letters, some 3300 books will be required. Here bioinformatics databasing and other high-speed computational devices have helped in the analysis, storage, and retrieval of information.

Salient features of the human genome:

1. The human genome contains 3164.7 million nucleotides (base pairs).
2. The size of the genes varies; an average gene consists of 3000 bases, while the largest gene, dystrophin, consists of 2.4 million bases.
3. The total number of genes is estimated to be 30000 and 99.9% of the nucleotides are the same in humans.
4. The functions of over 50% of the discovered genes are not known.
5. Only less than 2% of the genome codes for proteins.
6. Repetitive segments made up a large portion of the human genome.
7. Repetitive sequences throw light on chromosome structure and dynamics and evolution, though they are thought to have no direct coding functions.
8. (Chromosome 1 has 2968 genes and Y-chromosome has the least number (231 genes).
9. Scientists have identified about 1.4 million locations, where DNA differs in a single base in human beings. These are called single nucleotide polymorphisms (SNPs).
10. Repeated sequences make up a large portion of the human genome.

Very Important Figures:

Unidirectional repLication

BidirectionaL replication

The flow of genetic information.

Transcription in eukaryotes

Here we are providing Class 12 Biology Important Extra Questions and Answers Chapter 1 Reproduction in Organisms. Important Questions for Class 12 Biology are the best resource for students which helps in Class 12 board exams.

Class 12 Biology Chapter 1 Important Extra Questions Reproduction in Organisms

Reproduction in Organisms Important Extra Questions Very Short Answer Type

Question 1.
What is the life span?
Answer:
The period from birth to natural death of an organism is termed its life span.

Question 2.
Why is reproduction essential for organisms?
Answer:
Reproduction is a process by which an organism produces young ones of its own kind to maintain the continuity of the species. It enables the species to live generation after generation.

Question 3.
What type of modification are ginger, potato, onion and Samarkand?
Answer:
Underground modification of stem.

Question 4.
Name sub-aerial stems which help in multiplication.
Answer:
Pistia, Chrysanthemum Eichhornia, Pineapple.

Question 5.
Name artificial methods of vegetative propagation.
Answer:

1. Cutting of stem, root and leaf.
2. Grafting
3. Layering
4. Gootee.

Question 6.
Which type of division is involved in asexual reproduction?
Answer:
Only mitotic division occurs during asexual reproduction.

Question 7.
Male honeybee has 16 chromosomes whereas its female has 32 chromosomes. Give one reason. (CBSE Outside Delhi 2016)
Answer:
Male honey bee develops from haploid unfertilised egg (Ovum), whereas female develops from the diploid fertilized zygote.

Question 8.
The diploid number of chromosomes in an angiospermous plant is 16. What will be the number of chromosomes in its endosperm and antipodal cells? (CBSE Outside Delhi 2019)
Answer:
Number of chromosomes in endosperm = 24 (3N)
Number of chromosomes in antipodal cells = 8(N)

Question 9.
Banana is a true fruit and also a parthenocarpic fruit. Justify. (CBSE Foreign 2008)
Answer:
Banana develops from the ovary (true fruit) and develops without fertilisation (parthenocarpic fruit).

Question 10.
Pick out the ancestral line of angiosperms from the list given below: Conifers, seed ferns, cycads, ferns. (CBSE 2008)
Answer:
Seed ferns.

Question 11.
Why is apple referred to as false fruit? (CBSE 2010)
Answer:
In the case of apple, thalamus contributes to fruit formation, while most of the plant’s fruit develops from the ovary.

Question 12.
Name the type of cell division that takes place in the zygote of an organism exhibiting haplontic life cycle. (CBSE 2011)
Answer:
Meiosis

Question 13.
Mention the unique flowering phenomenon exhibited by Strobilanthus Ludhiana (Neelakuranji). (CBSE 2012)
Answer:
It is a monocarpic flowering plant. It flowers once in 12 years.

Question 14.
Some flowers, selected for artificial hybridisation, do not require emasculation but bagging is essential for them. Give a reason. (CBSE Delhi 2019 C)
Answer:
Bagging is the covering of flower by butter paper on polythene. The emasculated flower buds of the female parent and floral buds or male parent are bagged in order to protect them from contamination with unwanted pollen grains.

Question 15.
Cucurbits and papaya plants bear staminate and pistillate flowers. Mention the categories they are put under separately on the basis of the type of flowers they bear. |HOTSj (CBSE 2012)
Answer:
Cucurbits-Monoecious plants Papaya-Dioecious plants.

Question 16.
Why is banana considered a good example of parthenocarpy? (CBSE 2011)
Answer:
It is propagated vegetatively because there is no seed formation.

Question 17.
Name an alga that reproduces asexually through zoospores. Why are these reproductive units called so? (CBSE Outside Delhi 2013)
Answer:

1. Chlamydomonas
2. They are called so because they are microscopic motile structures.

Question 18.
Name the phenomenon and one bird where the female gamete directly develops into a new organism. (CBSE Outside Delhi 2013)
Answer:

1. Parthenogenesis
2. Turkey bird

Question 19.
Name the vegetative propagules in the following: (CBSE 2014)
(a) Agave
Answer:
Bulbils

(b) Bryophyllum.
Answer:
Leaf bud

Question 20.
Mention a characteristic and a function of zoospores in some algae. (CBSE 2010)
Answer:

• Zoospores are microscopic and flagellated motile spores.
• They are reproductive structures.

Reproduction in Organisms Important Extra Questions Short Answer Type

Question 1.
Is there a relationship between the size of an organism and its life span? Give two examples in support of your answer.
Answer:
No, there is no relationship between size and life span of organisms. Large-sized tiger and small-sized dog both live for about 20 years. The very large-sized elephant has a life span of up to 90 years. On the other hand, small-sized tortoise lives for 200 years. Similarly, the mango tree has a much shorter life span as compared to peep at the tree.

Question 2.
Offspring formed due to sexual reproduction have better chances of survival. Why?
Answer:
The offspring formed due to sexual reproduction show variations due to crossing over during gametogenesis, random segregation of gametes or random fertilisation. These useful variations produced in offspring help the organisms to adapt and survive.

Question 3.
How are the progeny formed from asexual reproduction different from those formed by sexual reproduction?
Answer:
The progeny formed from asexual reproduction is genetically similar to the parent, while those formed by sexual reproduction are genetically different from the parents due to new gene combinations formed during crossing over, random segregation and fertilisation.

Question 4.
Mention two inherent characteristics of Amoeba and yeast that enable them to reproduce asexually.
Answer:

1. Amoeba and yeast are unicellular organisms.
2. Both have a very simple body structure.
3. Both reproduce by fission.

Question 5.
Why do we refer to offspring formed by the asexual method of reproduction as clones?
Answer:
Offspring formed by asexual reproduction are called clones because they are morphologically and genetically similar to the parent.

Question 6.
Higher organisms have resorted to sexual reproduction in spite of its complexity. Why?
Answer:
Higher organisms have resorted to sexual reproduction in spite of its complexity because, at the same time, sexual reproduction provides two-fold advantages:

1. Here genetic recombination, interaction, etc. take place which causes variations in the offspring thus also form raw materials for evolution.
2. The offspring adapt more comfortably and quickly to the changes in the environmental conditions.

Question 7.
What are gemmules and conidia? Name one organism each in which these are formed. (CBSE Sample Paper 2019, 20)
Answer:
Gemmules: These are internal buds. They consist of a small group of archaeocytes, enclosed by a protective coat. They are formed in freshwater sponges e.g. Spongilla.

Conidia: They are formed in Penicillium. They are non-motile spores produced single or in the chain by a constriction at the tip of special hyphal branches called conidiophores.

Question 8.
Give examples of plants which are propagated vegetatively from underground stems and creeping stems.
Answer:
Underground stems. Mint and Chrysanthemum, Banana, Turmeric, Ginger, Aspidium, Adiantum. Creeping stems. Runners (mint, grass), stolons (strawberry) and offset (Eichhornia).

Question 9.
Differentiate between a zoospore and a zygote.
Answer:
Difference between a zoospore and a zygote:

Zoospore	Zygote
1. It is an asexual spore produced by algae and some fungi and Is capable of moving about by means of flagella.	1. It is a non-mottle cell produced by the certain union of male and female gametes. It Lacks flagella.
2. It Is haploid or diploid in nature.	2. It is diploid in nature.

Question 10.
List the pre-fertilisation events.
Answer:
Pre-fertilisation events. These include all the events of sexual reproduction prior to the fusion of gametes. The two main pre-fertilisation events are gametogenesis and gamete transfer.

Question 11.
Why does the zygote in angiosperms start developing into embryo only after some endosperm is formed?
Answer:
Zygote in angiosperms starts developing into embryo only after some endosperm is formed because endosperm is nutritive in function. It provides nutrients to the zygote for further growth and development.

Question 12.
Why is the offspring formed by asexual reproduction referred to as clone?
Answer:
The offspring formed by asexual reproduction is referred to as clone because the offspring is morphologically and genetically similar to the parent.

Question 13.
Mention the site where syngamy occurs in amphibians and reptiles respectively. (CBSE 2010)
Answer:

1. In amphibians, syngamy occurs in the external medium, i.e. water.
2. In reptiles, syngamy occurs in the body of an organism.

Question 14.
Why do internodal segments of sugarcane fail to propagate vegetatively even when they are in contact with damp soil? (HOTS) (CBSE Sample Paper)
Answer:
Sugarcane plants propagate vegetatively only when nodes are in contact with damp soil. Adventitious roots emerge from nodes and not from internodes because nodes bear buds.

Question 15.
Why do algae and fungi shift to a sexual mode of reproduction just before the onset of adverse conditions? (CBSE Delhi 2014, 2015)
Answer:
The organisms produced through asexual reproduction have low adaptability to the changing environment. Thus algae and fungi shift to a sexual mode of reproduction during the onset of adverse conditions.

Question 16.
A moss plant produces a large number of antherozoids but relatively only a few egg cells. Why? (CBSE 2010)
Answer:
In a moss plant, an antheridium produces many sperms while one archegonium produces only one egg cell. That is why there are a large number of antherozoids and a few egg cells.

Question 17.
Mention the reasons for the difference in ploidy of zygote and primary endosperm nucleus in an angiosperm. (CBSE 2010)
Answer:
A zygote is diploid (2n) as one male gamete fuses with egg or oosphere, while primary endosperm nucleus is triploid as one male gamete fuses with a secondary nucleus which is already diploid.

Question 18.
In haploid organisms that undergo sexual reproduction, name the stage when meiosis occurs. Give reasons for your answer.
Answer:
Haploid organisms form gametes without meiosis. Mate and female gametes fuse to form a diploid zygote. Zygote being diploid undergoes meiosis to form haploid organisms e.g. Ulothrix, Chlamydomonas.

Question 19.
Describe the importance of syngamy and meiosis in the life cycle of an organism. (CBSE Delhi 2016)
Answer:
Syngamy is a fusion of haploid gametes. It restores diploid nature in the zygote. Meiosis occurs during gametogenesis, thus produces haploid gametes. Both are important for maintaining chromosome number (ploidy) in an organism.

Question 20.
Angiosperms bearing unisexual flowers are said to be either monoecious or dioecious. Explain with the help of one example each. (CBSE Delhi 2016)
Answer:
In dioecious plants, male flowers termed a staminate flower, and female flowers, termed as pistillate flowers, are borne on different; plants. Thus plants are either male or female.

Examples: Papaya, date palm, etc.
In monoecious plants, male and female flowers are present on the same plants. Example: Maize, coconut, cucurbits, etc.

Question 21.
Write the significance of meiocytes. (CBSE (Delhi) 2016)
Answer:
Significance of meiocytes. Meiocytes are gamete-producing cells which undergo meiosis. They are diploid. As a result of meiosis, they produce haploid gametes. During fertilisation, a fusion of haploid gametes restores diploid nature of zygote. It undergoes mitosis to form complete new young one.

Question 22.
Why do organisms like algae and fungi shift from asexual mode of reproduction to sexual mode? (CBSE Delhi 2018C)
Answer:
During favourable conditions, organisms opt for asexual reproduction but when the conditions are adverse or unfavourable, organisms undergo sexual reproduction.

Question 23.
What is a juvenile phase in organisms? (CBSE Delhi 2018C)
Answer:
It is the stage of growth and attaining maturity in their life before they can reproduce sexually. It is also called the vegetative phase.

Question 24.
(i) State the difference between meiocyte and gamete with respect to chromosome number.
Answer:
Meiocytes are diploid (2n) and gametes are haploid.

(ii) Why is a whiptail lizard referred to as parthenogenetic? (CBSE 2012)
Answer:
Whiptail lizard eggs develop without fertilisation

Reproduction in Organisms Important Extra Questions Long Answer Type

Question 1.
Define:
(i) juvenile phase,
Answer:
Juvenile phase. The period of growth in the life of organisms before they start reproducing sexually and attain a level of maturity is called juvenile phase. It is followed by the reproductive phase.

(ii) reproductive phase
Answer:
Reproductive phase. The period of active reproductive behaviour, when the organisms show marked morphological and physiological changes is called reproductive phase. It is followed by senescence phase.

(iii) senescence phase.
Answer:
Senescence phase. The period when the reproductive phase ends and concomitant changes occur in the body such as slowing of metabolism is called senescence phase. It is followed by death.

Question 2.
Distinguish between asexual and sexual reproduction. Why is vegetative reproduction also considered as a type of asexual reproduction?
Answer:
1. Differences between asexual reproduction and sexual reproduction.

Asexual Reproduction	Sexual Reproduction
1. The process involves only one cell or one parent.	1. This process involves two cells or gametes belonging to either the same or different parents.
2. The whole body of the parent may act as a reproductive unit or it can be a single cell or a bud.	2. The reproductive unit is called gamete which is unicellular and haploid.
3. The offspring are genetically similar to the parent.	3. The offspring differ from the parents.
4. Only mitotic division takes place.	4. Meiosis and mitosis both take place.
5. No formation of sex organs.	5. Formation of sex organs is essential.
6. No evolutionary significance.	6. It introduces variation; hence it is of evolutionary significance.

2. Vegetative reproduction is also considered a type of asexual reproduction because it does not involve meiotic division and there is no formation and fusion of gametes.

Question 3.
How does an encysted Amoeba reproduce on the return of favourable conditions? (CBSE Sample Paper 2019-20)
Answer:
Multiple fission in encysted Amoeba:

• Amoeba withdraws pseudopodia and secretes a cyst wall around itself. This phenomenon is called encystation.
• Amoeba divides by multiple fission.
• It produces a large number of pseudo- conidiospores.
• The cyst wall breakdown.
• The spores are liberated and settle down on suitable substrates and grow as amoebae. This process is also called sporulation.

Question 4.
Discuss the advantages and disadvantages of asexual reproduction.
Answer:
Advantages of asexual reproduction:

1. It involves simple mitotic division in single-parent and it may produce a large number of young ones.
2. Young ones produced by asexual methods are genetically similar to the parent.
3. It helps in the dispersal of offspring to far off places.

Disadvantages of asexual reproduction.

1. The young ones thus produced do not possess much capacity to adapt rapidly to the environmental changes taking place in quick succession.
2. No genetic recombination occurs; thus no variation occurs.

Question 5.
Discuss the advantages and disadvantages of sexual reproduction.
Answer:
Advantages of sexual reproduction:

1. Genetic recombination, interaction, etc. take place which causes variations in the offspring, thus also form raw materials for evolution.
2. The offspring adapt more comfortably and quickly to the change in environmental conditions and have better chances of survival.

Disadvantages of sexual reproduction. Usually, two parents of opposite sexes are required (except in hermaphrodite).

Question 6.
List various methods of natural vegetative propagation. Give examples:
Answer:

1. Vegetative propagation by stems, e.g.Grasses, Turmeric, Onion, Colocasia, Potato, Gladiolus and Crocus.
2. Vegetative propagation by roots, e.g. Murraya sp., Albizzia Lebbac, Dalbergia sissoo, Tuberous roots of sweet potato, Asparagus, Tapioca, Dahlia and Yams (Dioscorea).
3. Vegetative propagation from reproductive organs. Flower buds of century plant (Agave sp.) develop into bulbils.

Question 7.
Define external fertilisation. Mention its disadvantages:
Answer:
The fertilisation in which the fusion of gametes occurs outside the body of the female in an external medium, i.e. water, is called external fertilisation.

Examples. Bony fishes, amphibians, etc. Organisms that exhibit external fertilisation show great synchrony between the sexes in order to liberate the gametes at the same time.

Disadvantages of external fertilisation:

1. A large number of gametes are produced to ensure fertilisation, thus there is wastage.
2. The offspring formed are extremely vulnerable to predators, thus threatening their survival up to adulthood.

Question 8.
Explain the process of budding in yeast. (CBSE 2010)
Answer:
Budding in yeast. It is a common type of vegetative reproduction. In a medium which is abundantly supplied with sugar, yeast cytoplasm forms a bud-like outgrowth. The growth soon enlarges and a part of the nucleus protrudes into the bud and breaks off. The bud then begins to grow and then separates from the mother cell. Often it will itself form a bud before it breaks away, and straight or branched chains are produced.

Thus, as a result, branched or unbranched chains of cells called pseudo my cilium are produced. The cells are loosely held together. Sooner or later they become independent.

Question 9.
Describe the importance of vegetative propagation.
Answer:
Merits of vegetative propagation:

1. Plants produced by vegetative propagation are genetically similar and constitute a uniform population called a clone.
2. Plants with reduced power of sexual reproduction, long dormant period of seed, poor viability, etc. are multiplied by vegetative methods.
3. Some fruit trees like banana and pineapple do not produce viable seeds. So these are propagated by only vegetative methods.
4. It is a more rapid and easier method of propagation.
5. Good characters are preserved by vegetative propagation.
6. Some plants such as doob grass (Cynodon dactylon) which produce only a small quantity of seed are mostly propagated by vegetative propagation.
7. Grafting helps in getting an economically important plant having useful characteristics of two different individuals in a short time.

Question 10.
Describe the post-fertilisation changes in a flower.
Answer:
Post-fertilisation changes in a flower.

Question 11.
Write a note on sexuality in plants.
Or
Coconut palm is monoecious while date palm is dioecious. Why are they called so?
Answer:
Sexuality in organisms: Sexual reproduction in organisms generally involves the coming together of gametes from two different individuals. But this is not always true.

Sexuality in Plants: Plants may have both male and female reproductive structures in the same plant (bisexual) or on different plants (unisexual). In several fungi and plants, terms such as homothallic and monoecious are used to denote the bisexual condition, and heterothallic and dioecious are used to describe the unisexual condition.

In flowering plants, the unisexual male flower is staminate, i.e. bearing stamens, while the female is pistillate or bearing pistils. In some flowering plants, both male and female flowers may be present on the same individual (monoecious) or on separate individuals (dioecious). Some examples of monoecious plants are cucurbits and coconuts and dioecious plants are papaya and date palm.

Very Important Figures:

Here we are providing Class 12 Biology Important Extra Questions and Answers Chapter 4 Reproductive Health. Important Questions for Class 12 Biology are the best resource for students which helps in Class 12 board exams.

Class 12 Biology Chapter 4 Important Extra Questions Reproductive Health

Reproductive Health Important Extra Questions Very Short Answer Type

Question 1.
Expand STD and CDRI:
Answer:
STD: Sexually transmitted diseases.
CDRI: Central Drug Research Institute.

Question 2.
Comment on the RCH program of the government to improve the reproductive health of the people:
Answer:
Creating public awareness regarding reproduction-related aspects and providing facilities to build up a healthy society with an added emphasis on the health of mother and child are the basic aims of the RCH programs.

Question 3.
What is the significance of progesterone- estrogen combination as a contraceptive measure?
Answer:
Birth control pills like Mala-D and Saheli are commonly called combined pills as they have progesterone-estrogen combinations. These prevent ovulation from the ovaries. These also retard the entry of sperms in the uterus.

Question 4.
Males in whom testes fail to descend to the scrotum are generally infertile. Why?
Answer:
If the testes fail to descend to the scrotum, gametogenesis could be inhibited as the process of spermatogenesis requires a marginally lesser ambient temperature (2°C less) than that in the abdominal cavity.

Question 5.
Mention one positive and one negative application of amniocentesis. (CBSE 2010)
Answer:

• Positive application: To detect congenital defects or genetic disorders.
• Negative application: Sex determination leads to female foeticide.

Question 6.
Which age group has the highest infection rates of sexually transmitted infections (STIs) including HIV?
Answer:
Young people aged between 15 and 24 years.

Question 7.
Mention any two events that are inhibited by the intake of oral contraceptive pills to prevent pregnancy in humans: (CBSE 2009 (S))
Answer:

1. Inhibition of ovulation.
2. Inhibition of motility and secretory activity of oviducts.

Question 8.
After a successful in vitro fertilization, the fertilized egg begins to divide. Where is this egg transferred before it reaches the 8-cell stage and what is this technique named? (CBSE Sample Paper)
Answer:

1. Fallopian tube.
2. Zygote intrafallopian transfer (ZIFT).

Question 9.
Name two STDs that can be transmitted through contaminated blood. (CBSE 2009 (S))
Answer:

1. AIDS
2. Hepatitis-B.

Question 10.
List the main methods of birth control:
Answer:

1. Mechanical methods
2. Surgical methods
3. Chemical and hormonal methods
4. Natural methods of birth control.

Question 11.
What are the two factors which have raised life expectancy in developing countries?
Answer:

1. The decline in death rate.
2. Increase in longevity.

Question 12.
What are the causes of death of women between ages 15 and 19 years?
Answer:
Complications of pregnancy, childbirth, unsafe abortions.

Reproductive Health Important Extra Questions Short Answer Type

Question 1.
Define reproductive health. How does this affect society?
Answer:
Reproductive health means total well-being in all aspects of reproduction, i.e. physical, emotional, behavioral, and social. A society with people having physically and functionally normal reproductive organs and normal emotional and behavioral interactions among them in all sex-related aspects are called a reproductively healthy society.

Question 2.
What is the significance of reproductive health in a society?
Answer:
A reproductively healthy society survives in a normal way. Reproductive health ensures that all individuals in the reproductive age group are fertile and able to carry on the human race further. Individuals should be educated about reproduction-related aspects like controlled population, absence of sex-abuses, and sex-related crimes in the society and this will enable people to think and take up necessary steps to build up a society that is reproductively healthy.

Question 3.
Is sex education necessary in schools? If so why?
Answer:
Yes, sex education should be encouraged in schools to give the right information to the young minds to save them from myths and misconceptions about sex-related aspects. Proper information about reproductive organs, adolescence and related changes, safe and hygienic sexual practices, sexually transmitted diseases like AIDS, etc. would help people to lead a reproductively healthy life. Thus sex education is necessary to make society reproductively healthy.

Question 4.
What are the suggested reasons for the population explosion?
Answer:
The following are the suggested reasons for population explosion:

1. the rapid decline in death rate.
2. the decline in maternal mortality rate (MMR).
3. the decline in infant mortality rate (IMR).
4. increase in the number of people in reproducible age.

Question 5.
The present population growth rate in India is alarming. Suggest ways to check it.
Answer:

1. Widespread information and materials for birth control.
2. The spread of education facilities.
3. Making young people career conscious.
4. Providing vocational training for gainful employment.
5. Raising marriage age.

Question 6.
Suggest the reproduction-related aspects in which counseling should be provided at the school level.
Answer:
Counseling should be provided to students about reproduction-related problems as:

1. Students are experiencing physical, physiological, and psychological changes during adolescence.
2. Students should be guided about the harms of early sex, hygiene of reproductive organs, and STDs.

Question 7.
Is the use of contraceptives justified? Give reasons.
Answer:
The use of contraceptives is justified because these methods prevent or delay pregnancy and in the long run, help to check the uncontrolled growth of the population and thereby reduce the ill effects of overpopulation on the reproductive health of the society.

Question 8.
Amniocentesis, the fetal sex determination test, is banned in our country. Is it necessary? Comment.
Answer:
Amniocentesis is a fetal sex determination test based upon the chromosomal pattern in the amniotic fluid surrounding the developing embryo. It should be legally banned throughout the country as such a ban shall check increasing female foeticide cases and maintain a normal sex ratio in the country.

Question 9.
Name the hormone composition of oral contraceptives used by a human female. Explain how does it act as a contraceptive? (CBSE 2009)
Answer:

1. Progesterone or progesterone- estrogen combination is used as an oral contraceptive by human females.
2. These pills inhibit ovulation as well as implantation. They also alter the quality of cervical mucus to prevent or retard entry of sperms.

Question 10.
What are the conditions in which medical termination of pregnancy is advised?
Answer:

1. If pregnancy is likely to produce a congenitally malformed child.
2. In case of rape.
3. Contraceptive failure.
4. Pregnancy is likely to harm the mother.

Question 11.
Comment on the essential features required for an ideal contraceptive.
Answer:

1. User friendly, i.e. comfortable and easy to use.
2. Without any side effects.
3. Reversible.
4. Completely effective against pregnancy.

Question 12.
All reproductive tract infections (RTIs) are STDs, but all STDs are not RTIs. Justify with an example:
Answer:
Among the common STDs gonorrhea, syphilis, genital herpes, chlamydiosis, hepatitis-B, AIDS, etc., hepatitis-B and AIDS are not infections of the reproductive organs, though their mode of transmission could be through sexual contact also. All other diseases are transmitted through sexual contact and are also infections of the reproductive tract.

Reproductive Health Important Extra Questions Long Answer Type

Question 1.
Suggest the aspects of reproductive health that need to be given special attention in the present scenario:
Answer:
Conferring upon the demands of the present situation of our country, the following aspects of reproductive health should be given special attention:

1. Control of the human population.
2. Creation of awareness about reproduction-related aspects among people including sex education to the students.
3. Implementation of various action plans to attain reproductive health and building a reproductively healthy society.
4. Continued research on reproduction-related areas.
5. Providing medical assistance and care to people especially during pregnancy, delivery, STDs, abortions, contraception, menstrual problems, infertility, etc.
6. Measures of birth control.

Question 2.
Briefly give an account of various intra-uterine contraceptive (IUD) measures. What are their advantages? How do they function?
Answer:
Different types of lUDs are presently available such as the non-medicated lUDs (e.g. Lippes loop), copper releasing IUDs (CUT, CU 7, Multiload 375), and the hormone-releasing ones (Progestasert, LNG-20). Functions. lUDs increase phagocytosis of sperms within the uterus and the Cu ions released by some suppress sperm motility and the fertilizing capacity of the sperms. The hormone-releasing lUDs, in addition, make the uterus unsuitable for implantation and the cervix hostile to the sperms. Advantages. LEDs are ideal contraceptives for females who want to delay pregnancy and/or space children. It is one of the most widely accepted methods of contraception in India.

Question 3.
What are the measures one has to take to prevent contracting STDs?
Answer:
STDs are sexually transmitted diseases that are transmitted from an infected person to a normal person through sexual intercourse. STDs are a major threat to a healthy society and can be prevented by adopting the following practices:

1. Avoidance of sex with multiple partners.
2. Avoidance of sex with unknown partners.
3. Complete abstinence from sex with infected individuals.
4. Use of condoms during sexual intercourse.
5. In case of doubt, a qualified doctor should be consulted.

Question 4.
People of which age group are more vulnerable to the incidence of infection of STDs? Suggest three ways of preventing STDs.
Answer:
Though all persons are vulnerable to these infections, their incidence is reported to be very high among persons in the age group of 15-24 years.

Preventive measures:

1. Avoid sex with unknown partners/ multiple partners.
2. Always use condoms during coitus.
3. In case of doubt, go to a qualified doctor for early detection and get it completely cured.

Question 5.
Removal of gonads cannot be considered a contraceptive option. Why?
Answer:
Removal of gonads cannot be considered as a contraception option because of the following reasons:

1. It should not have side effects. Gonads also secrete hormones. Thus the other functions will be harmed.
2. It should not interfere with sexual desire.
3. It is an irreversible process. It will lead to infertility.

Question 6.
What are the various oral contraceptives used? How do they function? What is the advantage of Saheli?
Or
Why do some women use ‘Saheli’ pills? (CBSE2009)
Or
Name an oral pill used as a contraceptive by a human female. Explain how does it prevent pregnancy? (CBSE 2011)
Answer:

1. Few oral contraceptives used are Mala D, Saheli, etc.
2. Oral administration of small doses of either progestagens or progestagen- estrogen combinations is another contraceptive method used by females.
3. They are used in the form of tablets and hence are popularly called the ‘pills’.
4. Pills have to be taken daily for a period of 21 days starting preferably within the first five days of the menstrual cycle. After a gap of 7 days (during which menstruation occurs), it has to be repeated in the same pattern till the desired period of contraception.
5. Functions. They inhibit ovulation and implantation as well as alter the quality of cervical mucus to prevent/ retard entry of sperms.
6. Pills are very effective with lesser side effects and are well accepted by females.

Question 7.
A mother of a one-year-old daughter wanted to space her second child. Her doctor suggested Cu-T. Explain its contraceptive actions. (CBSE Delhi 2008)
Or
How does Cu-T act as an effective contraceptive for human females? (CBSE 2009, 2011)
Answer:
Role of Cu-T as the contraceptive:

1. Copper-T increases phagocytosis of spermatozoa within the uterus.
2. Copper ions released also suppress the motility of sperms as well as their fertilizing ability.
3. It is a reversible technique and can be removed as and when required.

Question 8.
Name two hormones that are constituents of contraceptive pills. Why do they have high and effective contraceptive value? Name a commonly prescribed non¬steroidal oral pill: (CBSE Outside Delhi 2016)
Answer:

1. Oral contraceptive pills contain progesterone alone or a combination of progesterone and estrogen hormones.
2. Effective contraceptive pills:
   (a) Inhibition of ovulation
   (b) Inhibition of motility and secretory activity of fallopian tubes.
   (c) Changes in the endometrium layer of the uterus make it unsuitable for implantation.
3. ‘Saheli’ contains a non-steroidal preparation called centchroman.

Question 9.
Suggest some methods to assist infertile couples to have children. (CBSE Delhi 2008 (S) Outside Delhi 2013)
Answer:
There are numerous assisted reproductive technologies (ART) available that can bless infertile couples with children.

They are:

• IVF: In Vitro Fertilisation (Test-tube babies).
• ET: Embryo Transfer
• ZIFT: Zygote Intra Fallopian transfer.
• GIFT: Gamete Intra Fallopian Transfer.
• ICSI: Intracytoplasmic sperm injection.
• IUI: Intrauterine Insemination.

Question 10.
(i) Name any two copper-releasing lUDs:
Answer:
Copper releasing lUDs.
(a) CuT,
(b) Cu7 and
(c) Multiload 375.

(ii) Explain how do they act as effective contraceptives in human females? (CBSE 2014, Sample Paper 2019-20)
Answer:
The action of copper-releasing lUDs.
(a) Increases phagocytosis of sperms within the uterus.
(b) Suppress sperm motility.
(c) Then lUDs also suppress fertilizing capacity of sperms.

Question 11.
Suggest and explain the assisted reproductive techniques which will help a couple to have children, where the female had a blockage in the fallopian tube and the male partner had a low sperm count. (CBSE Sample paper 2018-19) Answer:
1. Since the female partner is having a blockage in the fallopian tube, In Vitro fertilization, followed by embryo transfer (ET) will help her conceive. In this case, sperms from the male partner will be collected and injected into the ovum of the female partner to form a zygote under simulated conditions in the laboratory, and an embryo with more than 8 blastomeres will be transferred into the uterus IUT – intrauterine transfer, to complete its further development.

2. Since the male partner is suffering from low sperm count, and intracytoplasmic sperm injection technique should be used to directly inject sperm into the ovum.

Question 12.
List the objectives of Reproductive and Child Health Care Programmes (RCH):
Answer:
Objectives of RCH are as follows:

1. Creating awareness about various reproduction-related problems.
2. Providing facilities and support for building up a reproductively healthy society.
3. Providing audiovisual and print media support to various government and non-government organizations.
4. Educating the people and providing the right information to save them from myths and misconceptions.
5. Providing proper education regarding reproductive organs, adolescence and related changes, safe and hygienic sexual practices.
6. Providing information regarding the danger of sexually transmitted diseases, AIDS, etc.

Question 13.
Define population. What are the aims of the population study?
Answer:
The population is defined as the total number of individuals of a species present in a particular area. The members of a population have some common characteristics, share a common gene pool, and are capable of interbreeding among themselves to produce fertile offsprings. Aims of Population Study. An alarming rise in the human population has created many serious problems. Therefore, population education has been introduced into the school and college curricula.

Population education is aimed at making the students aware of the:

• consequences of uncontrolled population growth such as environmental pollution, depletion of natural resources, extinction of species, etc ;
• benefits of lowering population growth rate to the biosphere ;
• advantages of a small family to humans ;
• growth, distribution, and density of population ;
• relation of population to the standards of life.

Question 14.
Define birth rate, death rate, and fertility rate.
Answer:
1. Birth or natality rate: It is generally expressed as the number of births per 1,000 individuals of a population per year. It increases the population size (total number of individuals of a population) and population density.

The national average birth rate in India is about 28.6 per 1,000 per year. Among Indian states, Kerala has the lowest birth rate of 18 per 1,000, while U.P. has the highest of 34.8 per 1,000.

2. Death or mortality rate: It is the opposite of the natality rate. It is commonly expressed as the number of deaths per 1,000 individuals of a population per year.

3. Fertility rate: It is the number of live births per unit time per unit number of fertile females. Fertility Rate

Question 15.
What is family planning? List the ways of family planning:
Answer:
Family planning: The main objective of family planning or family welfare program is to prevent the fertilization of the ovum by the male sperm and stop the increase in population growth by various methods, such as contraceptives, intrauterine devices, vasectomy, and tubectomy. The contraceptives (Anirudh) for males and intra-uterine devices, loop for females are used to avoid pregnancy.

Vasectomy is the method of sterilizing males by surgical operation of sperm duct or vas deferens. Tubectomy is the method of sterilizing females by the surgical operation of fallopian tubes. Whatever the method employed, it must take care of the health of the persons concerned.

Because of the family planning methods, the birth rate in India is reduced to some extent. The government gives incentives to those who adopt family planning.

Ways of family planning:

• Late marriage for young persons.
• Increase in the sources of recreation so as to divert the attention from sex.
• The couple should not mate between 8-18th days from the start of the menstrual cycle.
• Use of contraceptives.
• Sterilization.
• Use of drugs.
• Abortion.
• Restrict the family to two children.

Question 16.
Suggest the various measures of population control:
Answer:
Population control: Population explosion can be checked by two methods-population education and birth control.

A. Population education: The knowledge about the relationship of population size and the availability of resources for the welfare of the society is called population education.

1. The students should be convinced about the relationship between overpopulation and unemployment.
2. The citizens should be told how the large size of the population is eating away the resources of the state and the reasons for the limited availability of healthcare, education facilities, and other welfare schemes.
3. People should be made aware of how a large number of children eat away the meager resources of the family with nothing left for bad days, how large families rely on indebtedness to meet emergencies, how child bread earners do not improve the conditions of the family, how uneducated children remain a burden on the society, etc. They should be convinced that a small family can live comfortably even with meager resources.

B. Birth control:

1. Mass media of communication. Radio, television, newspapers, magazines, hoardings, and posters should be employed to spread the message of family planning and birth control and its advantages. The future of mankind depends on the stabilization of the human population at a level that ensures basic necessities of life, employment, and happiness,
2. The law about marriageable age should be widely published and strictly enforced (21 years for boys and 18 years for girls). In developed countries, women marry at the age of 25-35 years.
3. As far as possible, stress should be laid on raising the social status of women. Women having higher social status prefer smaller families. Such women generally marry late.
4. Remove the superstitions and wrong beliefs in the society about a higher number of children being God’s gift connected with earthly or heavenly prosperity.

Question 17.
What is amniocentesis? Write its procedure and significance:
Answer:
Amniocentesis is a fetal sex determination test based upon the chromosomal pattern in the amniotic fluid surrounding the developing embryo. It should be legally banned throughout the country as such a ban shall check increasing female foeticide cases and maintain a normal sex ratio in the country.

Procedure:

1. The fetus bathes in the amniotic fluid that fills the amniotic cavity. At an early stage of pregnancy (14th or 15th week), the location of the fetus and placenta is determined by sonography (use of high-frequency sound waves).
2. Then a small amount of amniotic fluid is drawn by passing a special surgical syringe needle through the abdominal wall and uterine wall into the amniotic sac containing the amniotic fluid.
3. Celts that have sloughed from the fetus’s skin or respiratory tract into the fluid are thus sucked into the syringe.

Significance:

1. These cells can be examined for chromosomal abnormalities, such as Down’s syndrome, Klinefelter’s syndrome, Turner’s syndrome, etc resulting from non-disjunction during cell division.
2. The cells can also be cultured and in about a fortnight enough cells become available for test. The cells and fluid are also tested for metabolic disorders such as phenylketonuria, sickle-cell anemia, etc.

Question 18.
Write a note on test-tube babies: (CBSE 2014)
Answer:
Test-Tube Babies: In some women normal conception is not possible because of blocked oviducts or spermicidal secretions in the vagina or the low sperm count of the husband. In such cases, her ovum is removed, fertilized by her husband’s sperm in a laboratory dish, checked that development has begun, and a morula (up to 32 cell stage) replaced or implanted in her uterus.

The entire operation is carried out under sterilized conditions. With proper medical care, she will give birth to a normal child on the completion of gestation. The baby produced in this manner (conceived out of and nursed in the uterus) is called a test-tube baby. The baby is not reared in the test tube. A scientific term for this procedure is in vitro (“in glass”) fertilization.

The success rate of the technique is less than 20%. To increase the chances of success, the prospective mother is given fertility drugs which cause many ovarian follicles to mature at the same time. This releases many eggs simultaneously, thereby increasing the chances of success.

A developing embryo can be inserted into the uterus of another female, called a surrogate mother, provided her hormones are in the proper phase of the reproductive cycle for implantation to occur.

Question 19.
Correct the following statements:
1. Surgical methods of contraception prevent gamete formation.
Answer:
Surgical methods of contraception block gamete transport and thereby prevent conception.

2. All sexually transmitted diseases are completely curable.
Answer:
Few sexually transmitted diseases are curable if detected at an early stage and some sexually transmitted diseases are not curable, e.g. AIDS.

3. Oral pills are very popular contraceptives among rural women.
Answer:
Oral pills are very popular contraceptives among educated urban women.

4. In E.T. techniques, embryos are always transferred into the uterus.
Answer:
In E.T. techniques, 8-celled embryos may be transferred into the fallopian tubes and more than 8-celled embryos are transferred into the uterus.

Question 20.
Reproductive and Child Healthcare (RCH) programs are currently in operation. One of the major tasks of these programs is to create awareness amongst people about the wide range of reproduction-related aspects. This is important and essential for building a reproductive health society.
1. “Providing sex education in schools is one of the ways to meet this goal.” Give four points in support of your opinion regarding this statement. (CBSE Delhi 2016)
Answer:
Sex education is important in schools:
(a) to provide the right information about myths and misconceptions.
(b) to create awareness about reproduction.
(c) to provide knowledge about the growth of reproductive organs and sexually transmitted diseases (STDs)
(d) to guide the students about social evils such as sex abuse, sex-related crimes, etc.

(ii) List any two indicators that indicate a reproductively healthy society.
Answer:
Indicators about a reproductively healthy society.
(a) Low infant mortality rate (IMR)
(b) Low maternal mortality rate (MMR)

• Increased number of couples with small families.
• Better detection and cure of STDs.

Question 21.
Give a brief account of Assisted Reproductive Technologies (ART). (CBSE Outside Delhi 2013, 2019)
Answer:
Where corrective treatments are not available, there are special techniques called Assisted Reproductive Technologies (ART) to help the couple produce children; they are as follows:
1. Test-Tube baby programs:
(a) In this method, ovum from the wife or a donor female and the sperms from the husband or a donor is allowed to fuse under simulated conditions (as that of the body) in the laboratory; it is called in vitro fertilization (IVF).

2. The zygote or early embryo is transferred into the uterus or fallopian tube for further development; this process is called Embryo Transfer (ET) and can be done in the following ways:
(a) The zygote or embryo up to eight blastomeres is transferred into the fallopian tube; it is called Zygote Intra Fallopian Transfer (ZIFT).
(b) Embryos with more than eight blastomeres are transferred into the uterus. It is called Intrauterine Transfer (IUT).

3. Gamete Intra Fallopian Transfer (GIFT): This method involves the transfer of an ovum collected from a donor female into another female, who cannot produce ova, but can provide suitable conditions for fertilization and further development of the fetus up to parturition.

4. Intra Cytoplasmic Sperm Injection (ICSI): In this method, the sperm is directly injected into the ovum to form an embryo in the laboratory, and then embryo transfer is carried out.

5. Artificial insemination:
(a) In this method, the semen collected from the husband or a healthy donor is artificially introduced into the vagina or into the uterus (intrauterine insemination).
(b) This method is used in cases where infertility is due to the inability of the male partner to inseminate the female or due to very low sperm counts in the ejaculates.

Question 22.
It is commonly observed that parents feel embarrassed to discuss freely with their adolescent children about sexuality and reproduction. The result of this parental inhibition is that the children go astray sometimes.
(i) Explain the reasons that you feel are behind such embarrassment amongst some parents to freely discuss such issues with their growing children.
Answer:
Parents feel embarrassed because of the following reasons:
(a) Indian society is not that broad-minded. So parents feel shy talking openly about these matters to their children.
(b) Improper communication and age gap are the reasons behind such embarrassment.

(ii) By taking one example of a local plant and animal, how would you help these parents to overcome such inhibitions about reproduction and sexuality? (CBSE Delhi 2017)
Answer:
Parents can take the example of China rose to explain the process of sexual reproduction. They can also take an example of the male honeybee and orchid Ophrys flower.

It is evident that sexual attraction is a natural phenomenon. The honeybee is attracted to an Ophrys flower and assumes its one petal as its female partner and pseudo copulate with it. So it is a natural phenomenon and parents should talk regarding this matter to their children.

Question 23.
(a) Explain one application of each one of the following:
(A) Amniocentesis:
Answer:

• Detection of a genetic disorder
• Detection of chromosomal disorder
• Sex determination
• Karyotyping (used for detecting chromosomal aberrations)

(B) Lactational amenorrhea:
Answer:
It is a kind of natural contraception to prevent pregnancy. When the women breastfeed regularly her menstrual cycle stops for some period and thus can’t have a baby.

(C) ZIFT:
Answer:
Application of ZIFT (Zygote Intrafallopian Transfer)- In vitro fertilization, the zygote or early embryos at eight blastomeres stage are transferred to the fallopian tale to complete its further development inside the body of the mother. Hence this method is very helpful for infertile couples.

(b) Prepare a poster for the school program depicting the objectives of the “Reproductive and Child Health Care Programme”. (CBSE Delhi 2019)
Answer:
Reproductive and Child Health Care:

Objectives of RCH:

• Creating awareness about various reproduction-related problems.
• Providing facilities and support for building up a reproductively healthy society.
• Providing audio-visual and print, media support, to various government and non-government organizations.
• Educating the people and providing the right information to save them from myths and misconceptions.
• Providing proper education regarding reproductive organs, adolescence and related changes, safe and hygienic sexual practices.
• Providing information regarding the danger of sexually transmitted diseases, AIDS, etc.
• Awareness regarding that gender selection and detection is punishable.

Example – Hum do hamare do, Beti bachao beti padhao, Do boond zindasi ke etc.

Here we are providing Class 12 Biology Important Extra Questions and Answers Chapter 3 Human Reproduction. Important Questions for Class 12 Biology are the best resource for students which helps in Class 12 board exams.

Class 12 Biology Chapter 3 Important Extra Questions Human Reproduction

Human Reproduction Important Extra Questions Very Short Answer Type

Question 1.
Given below are the events in human reproduction. Write them in correct sequential order? Insemination, gametogenesis, fertilization, parturition, gestation, implantation.
Answer:
Gametogenesis, insemination, fertilization, implantation, gestation, parturition.

Question 2.
The path of sperm transport is given below. Provide the missing steps in blank boxes:

Answer:
A = Vasa efferentia; B = Vas deferens.

Question 3.
What is the role of the cervix in the human female reproductive system?
Answer:
The cavity of the cervix called the cervical canal, and the vagina collectively forms the birth canal for parturition. It also regulates the entry of sperms into the uterus.

Question 4.
Why are menstrual cycles absent during pregnancy?
Answer:
The high levels of progesterone and estrogens during pregnancy suppress the gonadotropins which are essential for the transformation of the primary follicle into Graafian follicle and ovulation.

Question 5.
Give the scientific term for the following:
1. Release of the ovum from the ovary.
2. Onset of the menstrual cycle in females.
3. The structures that pick up ova from the body cavity.
Answer:

1. Ovulation
2. Menarche
3. Fimbriae.

Question 6.
What is the location of Cowper’s glands in the body?
Answer:
Cowper’s glands are situated laterally attached to the urethra beneath the urinary bladder at the base of spongy tissue.

Question 7.
Where are sperms stored in males?
Answer:
They are stored in the epididymis.

Question 8.
What is follicular atresia?
Answer:
Follicles in the ovaries which undergo regression and disappear due to death and disposal by phagocytes during the reproductive years of the female are referred to as follicular atresia.

Question 9.
Which particular part of mammalian sperm secretes enzymes to facilitate penetration of sperm?
Answer:
Acrosome.

Question 10.
Name the hormones involved in the regulation of spermatogenesis.
Answer:
GnRH, LH, FSH, Androgen Binding Protein (ABP), Inhibin, androgens.

Question 11.
How many eggs do you think were released by the ovary of a female dog which gave birth to six puppies? HOTS
Answer:
Six eggs.

Question 12.
What will happen if the fallopian tubes are partially blocked and the ovulated eggs are prevented from reaching the uterus?
Answer:
Fertilization may take place but the zygote may develop in the tube instead of the uterus.

Question 13.
Name the cells which produce testosterone. What is the function of this hormone?
Answer:
Interstitial cells (Leydig’s cells) of testis secrete the testosterone hormone. Function. They control secondary sexual characters.

Question 14.
How does the ovum which is released in the body cavity enter the fallopian tube?
Answer:
Ciliary movements of epithelial cells lining the lumen of nearby fallopian tubes induce the ovum to pass into the open fimbriated funnel-shaped end called ostium.

Question 15.
At what stage of life is oogenesis initiated in a human female? When does the oocyte complete oogenesis? (CBSE Sample Paper)
Answer:
Oogenesis is initiated during the embryonic development stage. It is completed at the time of ovulation.

Question 16.
When do morphogenetic cell movements take place?
Answer:
As blastocyst undergoes gastrulation.

Question 17.
Define spermiogenesis. Where does it occur? (CBSE Delhi 2008 (S))
Answer:
The transformation of non-motile spermatids into motile spermatozoa is called spermiogenesis. It occurs inside seminiferous tubules of testes.

Question 18.
How does colostrum provide initial protection against diseases to newborn infants? Give one reason. (CBSE 2009)
Answer:
The colostrum provides antibodies that are essential to developing resistance for newborn babies.

Human Reproduction Important Extra Questions Short Answer Type

Question 1.
Why scrotal sacs are present outside the body? (CBSE Outside Delhi 2014)
Answer:
The scrotal sac is a pouch of deeply pigmented skin divided into two separate sacs. Each sac contains one testis. The normal temperature of the testes in the scrotum is about 2°C lower than the internal body temperature, the ideal temperature for developing sperms.

Question 2.
Write the location and functions of Sertoli cells in humans? (CBSE 2012, 2014)
Answer:
Sertoli cells. These are present in the seminiferous tubules of the testis. They provide nutrition to germ cells. They play a vital role in the maturation of spermatids into motile sperms.

Question 3.
Name the muscular and glandular layer of the human uterus. Which one of these layers undergoes cyclic changes during the menstrual cycle? Name the hormone essential for the maintenance of this layer. (CBSE 2009)
Answer:

1. Muscular layer – Myometrium
   Glandular layer – Endometrium.
2. Endometrium undergoes cyclic changes during the menstrual cycle.
3. LH hormone maintains the lining of the uterus.

Question 4.
Study the figure given below and answer the questions that follow:

(i) Name the stage of the human embryo the figure represents.
Answer:
Blastocyst (Blastula).

(ii) Identify ‘a’ in the figure and mention its function.
Answer:
Trophoblast

(iii) Mention the fate of the inner cell mass after implantation in the uterus,
Answer:
Inner cell mass gets differentiated as embryo after implantation.

(iv) Where are the stem cells located in this embryo? (CBSE 2009)
Answer:
Stem cells are located in the inner cell mass.

Question 5.
Write two major functions each of testis and ovary?
Answer:
Functions of Testis:

• Production of sperms and
• Secretion of male sex hormones, androgens (e.g. testosterone).

Functions of Ovary:

• Production of ova and
• Secretion of female sex hormones, e.g. estrogens, progesterone.

Question 6.
What is the function of the acrosome and head?
Answer:
Functions of acrosome: It secretes sperm lysins which dissolve the membrane around the egg and thus facilitate the penetration of sperm into the ovum.

The function of the head: Head contains a nucleus that contains hereditary information. The fusion of nuclei is involved during fertilization.

Question 7.
Point out the differences in male and female urethra?
Answer:
In the case of the male urethra, there is a single opening for the elimination of urine and for the ejaculation of sperms. In females, there are two separate openings, one for the elimination of urine and the other is a vaginal orifice. The urethra in males is continued into the penis and in the case of females, it has not been found.

Question 8.
Draw a labeled diagram of a part of the seminiferous tubule showing spermatogenesis. (CBSE Delhi 2010, 2014, Outside Delhi 2019)
Answer:
Section of the testis:

Question 9.
During reproduction, the chromosome number (2n) reduces to half (n) in the gametes and again the original number (2n) is restored in the offspring. What are the processes through which these events take place?
Answer:

1. Chromosome number is reduced to half or (n) during gametogenesis.
2. Restoration of chromosome number to diploid or (2n) stage occurs during fertilization.

Question 10.
What is the significance of ampullary- isthmic junction in the female reproductive tract?
Answer:
The ampullary isthmic junction is the site where fertilization of the ovum takes place.

Question 11.
Draw a labeled diagram of the section through the ovary.
Or
Draw a labeled diagram of a section through the ovary showing various stages of follicles growing in it. (CBSE Delhi 2014, 2019)
Answer:
T.S. of Ovary:

Fig. T.S. of Ovary

Question 12.
Differentiate between Graafian follicle and Corpus luteum.
Answer:

Graafian follicle	Corpus luteum
1. SphericaL or at the sac-Like mass of cells.	1. Large mass of big conicaL yellow cells.
2. Each Graaflan follicle contains a Large, centrally placed ovum surrounded by many Layers of granuLar cells.	2. It is the structure formed after the release of the ovum.
3. Estrogen hormone is secreted.	3. Progesterone hormone is secreted.

Question 13.
Differentiate between spermiogenesis and speciation. (CBSE Outside Delhi 2019)
Answer:
Difference between spermiogenesis and Speramiation: Spermiogenesis is the process of transformation of non-motile spermatids into mature motile sperms (male-gametes) whereas speciation is the release of sperms from Sertoli cells of seminiferous tubules.

Question 14.
Define ooplasm and germinal vesicle.
Answer:

• Ooplasm: The cytoplasm of the haploid ovum is called the ooplasm.
• Germinal vesicle: The nucleus of the ovum at the time of fertilization is termed a germinal vesicle.

Question 15.
Explain the significance of fertilization.
Answer:
Significance of fertilization.

1. The fusion of male and female pronuclei in fertilization restores the diploid number of chromosomes.
2. The activation of secondary oocyte undergoes maturation to form ovum.
3. Fertilization initiates cleavage or segmentation.
4. The combination of the chromatin material from two different parents forms the physical basis of biparental inheritance and variation.

Question 16.
What is the cortical reaction?
Answer:
It is a reaction occurred during fertilization that presents polyspermy, i.e. fusion of multiple sperm with one egg. In this reaction, the cortical granules present beneath the secondary oocyte’s plasma membrane fuse with the plasma membrane and release their contents (enzymes) between the plasma membrane and zona pellucida. These enzymes harden the zona pellucida.

Question 17.
What is a fetus?
Answer:
It is the unborn young one of a viviparous animal after it has taken form in the uterus. In human beings, it represents the product of conception from the end of the eighth week to the moment of birth.

Question 18.
List the characteristics of the morula stage.
Answer:
Characteristics of morula stage:

1. The embryo consists of a solid ball of cells called blastomeres. They are arranged like a mulberry.
2. Morula consists of an outer layer of smaller cells and an inner mass of larger cells.
3. Zona pellucida layer persists during morula.

Question 19.
What is the placenta?
Answer:
It is the structure formed by the union of the fetal and uterine tissue for purpose of nutrition, respiration, and excretion of the embryo. Although the blood vessels of the embryo and the mother come close but are kept separated by some barriers between them. The useful substances pass from maternal blood to fetal blood while the wastes (excretory products and C02) are passed from the fetal blood to maternal blood.

Question 20.
Give the names and functions of the hormones involved in the process of spermatogenesis. Write the names of the endocrine glands from where they are released.
Answer:

Hormone	EndocrIne gland	Functions
(1) FSH	Antenor pituitary	StimuLates Sertoli cells to secrete some factors which help in spermatogenesis.
(2) ICSH/LH	-do-	StimuLates Leydig’s cells of testes to secrete androgens (especially testosterone) which regulate spermatogenesis.

Question 21.
How does colostrum provide initial protection against diseases to newborn infants? Give one reason. (CBSE 2009)
Answer:
The colostrum provides antibodies that are essential to developing resistance for newborn babies.

Question 22.
Diagrammatically shows the development of the human embryo in the female reproductive tract.
Answer:
Events of development:

Fig. Development of the human embryo in the female reproductive tract (Fertilisation-implantation)

Question 23.
When and where do chorionic villi appear in humans? State their function. (CBSE Delhi 2013, 2019)
Answer:

1. After implantation, finger-like projections appear on the trophoblast called chorionic villi. They are surrounded by uterine tissue and maternal blood.
2. The chorionic villi and uterine tissue become interdigitated with each other and jointly form structural and functional units between the developing embryos and maternal bodies called the placenta.

Question 24.
State the fate of the trophoblast of a human blastocyst at the time of implantation and that of the inner cell mass immediately after implantation. (CBSE Outside Delhi 2019)
Answer:
The trophoblast layer of the human blastocyst gets attached to the endometrium and the inner cell mass gets differentiated into an embryo. After attachment, the uterine cells divide rapidly and cover the blastocyst. As a result, the blastocyst becomes embedded in the endometrium of the uterus. It is termed Implantation.

Human Reproduction Important Extra Questions Long Answer Type

Question 1.
Briefly explain the primary male sex organs of man.
Answer:
The testes, male gonads, produce sperms that are suspended outside the abdominal cavity in a sac of skin called the scrotum. It results in maintenance of the temperature of the testis which is lower than the rest of the body. It is a condition favorable to sperm production.

Each testis is an oval-shaped structure and is composed of a large number of seminiferous tubules surrounded by connective tissue in which occurs numerous cells called Interstitial cells or Leydig cells. These cells produce a male sex hormone named testosterone. Seminiferous tubules are lined by a layer of germinal epithelial cells. In between the germinal cells, certain large cells called Sertoli cells are present. They are nutritive in function. The germinal epithelial cells produce sperms by spermatogenesis.

Question 2.
Draw well-labeled sketches of the front view and sagittal section of the male reproductive system of man. (CBSE Delhi 2019)
Answer:
The male reproductive system of man:

Question 3.
Explain the events taking place at the time of fertilization of an ovum in a human female. (CBSE 2010, 2014, (Delhi) 2016)
Answer:
Fertilization is the fusion of two gametic nuclei to form a diploid zygote. It involves a series of chemical and physical steps as follows:

The cortical cytoplasm of the ovum shows the physicochemical reactions called the cortical reactions:

1. Sperm lysins dissolve the membranes around the egg.
2. The Head of sperm containing a nucleus and proximal centriole physically passes into the ovum.
3. Normally these reactions result in the formation of a fertilization membrane outside the egg plasma membrane.
4. Cortical granules burst and release their contents between the egg plasma membrane and zona pellucida, i.e. perivitelline space (no fertilization membrane formation). The plasma membrane shows increased permeability for water, phosphate, and potassium.
5. The electrical potentiality of plasma membrane changes from positive to negative, NAD kinase enzyme becomes activated after fertilization for the oxidation and reduction reaction of the cell.
6. The rate of DNA synthesis increases with great pace after fertilization. Hence the ovum is now ready for mitosis (cleavage).
7. Cleavage results in multicellular individuals.

Question 4.
Give the hormonal control of the male reproductive system. (CBSE Delhi 2009)
Answer:
Hormonal control of the male reproductive system:
1. Hypothalamus: It releases gonadotropin-releasing hormones (GnRH) which bring the release of gonadotropins from the pituitary.

2. Anterior pituitary secretes two gonadotropins:
(a) Follicle Stimulating Hormone (FSH): It stimulates spermato¬genesis in the germ cells of seminiferous tubules of the testis to produce haploid and motile male gametes, called spermatozoa, so is also called games kinetic hormone.

Fig. Hormonal control of the male reproductive system

(b) Interstitial Cells Stimulating Hormone (ICSH): It stimulates interstitial cells to secrete male sex hormone called testosterone.

3. Sertoli cells secrete androgen binding protein (ABP) which concentrates testosterone in the seminiferous tubules.

4. Testosterone hormone secreted by interstitial or Leydig’s cells of testes. It regulates secondary sex organs and secondary sexual characters. It regulates the growth, maintenance, and functioning of secondary sex organs like epididymis, vasa deferentia, prostate gland, seminal vesicles, penis, etc. Testosterone also stimulates the development of mate secondary sexual characters like the appearance of facial hair, deepening of the voice, broadening of shoulders, elongation of bones, and increase in height.

Question 5.
Give an account of the secondary sex organs of a human male. (CBSE Delhi 2011)
Answer:
Secondary Sex OrgAnswer: The tubules meet and join at one end of the seminiferous to form ducts called rete testes that form vasa efferentia leading to the epididymis. It is a narrow tube about 6 m long lying in a coiled mass on the outside of the testis and store sperms. The epididymis in turn leads to the muscular sperm duct, the two sperm ducts open into the top of the urethra, just below the point where it leaves the bladder.

In a short coiled tube, the seminal vesicle branches from each sperm duct just above its opening into the urethra. Reproductive glands. They open into the urethra, the prostate gland, which surrounds it at the point where it leaves the bladder and Cowper’s gland a little lower down.

The urethra in males traverses through the corpus Spongiosum of the penis. Corpora Spongiosum consists of spongy and highly vascular connective tissue containing numerous small spaces that are normally empty but filled with blood when the penis is erect during copulation. The penis becomes erect and firm to facilitate the ejaculation of semen (spermatic fluid) into the vagina of the female.

Question 6.
Differentiate between vasa deferentia and vasa efferentia.
Answer:
Differences between vasa deferentia and vasa efferentia:

Vas deferens	Vasa efferentia
1. Vas deferens is the main duct that carries the sperms from the epididymis to the urethra.	1. They are 10-20 smaLL tubuLes which transfer sperms from rete testes to the epididymis.
2. It conveys mature sperms.	2. It conveys immature sperms.
3. Accessory reproductive glands pour their secretions into it.	3. No secretions are added to it.

Question 7.
Describe the structure of the seminiferous tubule.
Answer:
Each seminiferous tubule is lined by germinal epithelium. The majority of cells in this epithelium are cuboidal spermatogenic cells but a few are large, pyramidal, supporting Sertoli or nurse cells. The cuboidal cells, by mitotic divisions, produce spermatogonia into the lumen of the seminiferous tubule.

The spermatogonia grow into primary spermatocytes, which undergo meiosis, producing haploid cells, first secondary spermatocytes, and then spermatids. The latter metamorphose into spermatozoa. Differentiation of spermatozoa from spermatogonia is called spermatogenesis. The spermatozoa are nourished during development by nurse cells. Mature spermatozoa lie free in the cavity of the seminiferous tubules. Scattered in the connective tissue are Leydig’s cells which secrete a hormone called testosterone.

Question 8.
The mother germ cells are transformed into a mature follicle through series of steps. Provide the missing steps in the blank boxes.

Answer:
A = Primary oocyte; B = Secondary follicle, C = Tertiary follicle.

Question 9.
What is the difference between a primary oocyte and a secondary oocyte?
Answer:

primary oocyte	secondary oocyte
1. It is a diploid structure.	1. It is a haploid structure.
2. It is formed from oogonium through mitosis and differentiation.	2. It is formed from the primary oocyte after it undergoes the first meiotic division.
3. No polar body is formed during its development.	3. A polar body is extruded during its formation.

Question 10.
What is spermiogenesis? List the changes that take place during spermiogenesis.
Answer:
Spermiogenesis. The process of differentiation of motile sperm from the spermatid is called spermiogenesis.

Changes during spermiogenesis:

1. The nucleus becomes compact and forms the head.
2. Distal centriole forms the tail of the sperm.
3. The acrosome is formed at the tip of the head.
4. Mitochondria get arranged in the form of a spiral.
5. The cytoplasm is changed to a sheath.

Question 11.
Draw a mammalian sperm and label its four major parts. (CBSE Delhi 2008, 2011, 2013)
Answer:

Question 12.
Give the functions of the following:
1. corpus luteum
2. endometrium
3. acrosome
4. sperm tail
5. fimbriae.
Answer:
1. Functions of corpus luteum:
(a) It secretes a small amount of estradiol hormone and a significant amount of progesterone.
(b) It also secretes the relaxin hormone.

2. Functions of the endometrium:
(a) It is an internal layer of the uterus. It undergoes cyclic changes during the menstrual cycle.
(b) It provides pits for implantation of the blastocyst.
(c) It takes part in the formation of the placenta.

3. Function of acrosome: It contains hydrolytic enzymes that are used to contact and penetrate the egg at the time of fertilization.

4. Function of sperm tail: It provides motility to the sperm.

5. Function of fimbriae: They help in the collection of ovum after ovulation.

Question 13.
Draw a labeled diagram of the human female reproductive system. (CBSE Delhi 2011, 2013)
Answer:
The female reproductive system of human:

Question 14.
Write a short note on the structure and functions of the vagina.
Answer:
Structure of vagina: The vagina is a muscular tube lined with a membrane. It is composed of a special type of stratified epithelium which is supplied with blood vessels and nerves. It extends from the uterus to the vestibule. The anterior surface of the vagina is in relation to the base of the bladder and urethra. The vagina receives the male sex organ during copulation.

Functions of the vagina:

1. It receives the penis of males during copulation.
2. It serves as the receptacle for the sperms.
3. It acts as a birth canal.

Question 15.
Draw a labeled diagram of the Graafian follicle and ovum. (CBSE 2011, 2012)
Answer:
Graafian follicle:

Question 16.
What is a menstrual cycle? Give an account of a simplified 28-day menstrual cycle. (CBSE 2019 C, Outside Delhi 2019)
Answer:
Menstrual cycle: The cycLic changes that occur in the reproductive organs of primate femaLes (monkeys, apes, and human beings) constitute the menstruaL cycLe. The cycLe of events starts from one menstruation tiLt the onset of the next and Lasts for about 28/29 days (a menses).

Simplified menstrual cycle (28-day Cycle):

Phase	Days	Events
Menstrual phase	1-5	Endometrium breaks down, menstruation begins. The cells of en domain urn, secretions, blood, and the unfertilized ovum constitute the menstrual flow.
Follicular phase (ProLiferative phase)	6-13	Progesterone production is reduced. In fact, menstrual flow is associated with withdrawaL of progesterone.
Ovulatory phase	14	Endometrium rebuilds, FSH secretion, and estrogen secretion increase.
Luteal phase (Secretory phase)	15-28	Both LH and FSH attain a peak level. The concentration of estrogen in the blood is also high and reaches its peak. Ovulation occurs. Corpus luteum secretes progesterone. The endometrium thickens and uterine glands become secretory.

Question 17.
Show with graphic sketch the hormonal control over the menstrual cycle.

Fig. MenstruaL cycLe showing hormonal relations: (A) Gonadotropin, (B) Ovarian cycle, (C) Ovarian hormones, (D) Uterine cycle.

Question 18.
DifferentIate between proliferative and secretory phases.
Answer:
Differences between proliferative and secretory phases:

Proliferative phases	Secretory phases
1. Growth of Graafian follicle and maturation and release of ovum occurs.	1. Prepares the uterus for receiving the fertilized ovum and in case, there is no fertilization, endometrium breaks resulting in bleeding.
2. It is the first phase that follows menstruation.	2. It is the second phase that folLows after the release of an ovum.
3. It starts on the 5th day of menstruaL cycLe and Lasts up to 14 days.	3. It starts after the 14th day when the corpus Luteum is formed.
4. It is caused by estrogen and controLled by FSH.	4. Corpus Luteum is formed under the impact of LH.
5. Contraction of uterine muscLes increases.	5. Contraction of uterine muscles decreases.

Question 19.
1. What is ovulation?
2. What happens to the Graafian follicle after ovulation?
Answer:
1. Ovulation: The release of eggs (at secondary oocyte stage) after rupturing of Graafian follicle is called ovulation.

2. After the ovulation, the granulosa cells as well as the stroma cells from theca Interna rapidly multiply to fill the cavity of the Graafian follicle which becomes the corpus luteum. If fertilization occurs, the corpus luteum grows further and secretes hormone. If fertilization does not take place, the corpus luteum regresses and forms a yellow body.

Question 20.
Differentiate between zona pellucida and corona Radiata.
Answer:
Differences between zona pellucida and corona Radiata:

Zona Pellucida	Corona Radiata
1. The non-ceLLuLar, secretory layer around the ovum is termed zona Pellucidar (or Coleman).	1. The ceLLular layer around zona petLucida is termed Corona Radiata.
2. This Layer is thick and transparent.	2. It is a very thick layer of follicular cells with granular cytoplasm.
3. It encLoses perivitelline space around the pLasma membrane.	3. It does not enclose any space.

Question 21.
Compare mature mammalian sperm and ovum.
Answer:
Similarities:

1. Both are gametes and take part in fertilization,
2. Both are haploid.
3. Both are a carrier of the heredity characters,
4. Both are unicellular structures.

Differences:

Mammalian sperm	Mammalian ovum
1. Sperm is motiLe and encLosed by an only singLe plasma membrane.	1. Ovum Is sphericaL, non-motile, and surrounded by zona pesticide and corona radiate.
2. It is differentiated into four parts, i.e. head, neck, middle piece, and tait.	2. It is a spherical, aLecithaL egg with acentric nucLeus.
3. SmaLL amount of cytopLasm present.	3. A large amount of cytoplasm.
4. Centriotes and acrosome present.	4. CentnoLe and acrosome absent.
5. Mitochondria arranged in the form of a spiral.	5. Mitochondria scattered in the cytoplasm.

Question 22.
Briefly explain the steps of fertilization. (CBSE Delhi 2019)
Answer:
Fertilization is the process of fusion of motile sperm and non-motile ovum which takes place in the fallopian tube (oviduct).

• Fertilization is internal in human females.
• Fertilization is a Physico-chemical process.
• Acrosome in sperm’s head releases enzymes that dissolve the membrane of the egg to facilitate entry of sperm into the ovum.

Steps involved during fertilization:

1. Acrosomal reactions and penetration of sperm.
2. Cortical granule reaction.
3. Formation of pronuclei.
4. The fusion of pronuclei.

Question 23.
In our society, women are often blamed for giving birth to their daughters. Can you explain why this is not correct? (CBSE Delhi 2014)
Answer:
Sex chromosomes determine sex in human beings. In males, there are 44+XY chromosomes, whereas, in females, there are 44+XX chromosomes. Here X and Y chromosomes determine sex in human beings.

Two types of gametes are formed in males, one type is having 50% X-chromosome, whereas another type is having Y-chromosome. In females, gametes are of one type and contain X-chromosome. Thus females are homogametic. If a male gamete having Y-chromosome (endosperm) undergoes fusion with a female gamete having an X-chromosome, the zygote will have an XY chromosome and this gives rise to a male child.

If a male gamete having an X-chromosome (gymnosperm) undergoes fusion with a female gamete having an X-chromosome, the zygote will be having XX-chromosome and this gives rise to a female child.

Question 24.
What is fetal ejection reflex? Explain how it leads to parturition?
Answer:
Fetal ejection reflex is a neuroendocrine mechanism initiated by a fully formed fetus and placenta which helps in parturition.

It operates as below:

1. Foetus secretes hormones from its adrenal glands. These hormones stimulate the posterior pituitary of the mother to release oxytocin.
2. Oxytocin causes forceful contraction of smooth muscles of myometrium called labor pains, which further stimulates more secretion of oxytocin.
3. The stimulatory reflex between uterine contraction and oxytocin secretion continues which causes more forceful contraction of uterine muscles.
4. Uterine contraction pushes the young gradually out through the dilated cervix (caused by relaxin) and vagina.

Question 25.
Give an account of the histology of testis.
Answer:
Each testis is composed of a large number of seminiferous tubules surrounded by connective tissue in which there are numerous cells present called interstitial cells or Leydig cells. The testis is covered by a layer of dense connective tissue called tunica albuginea. From it, various septa radiate inward into the cavity of the testis dividing it into many lobules. The seminiferous tubules are coiled structures.

Each tubule is lined by a layer of germinal epithelial cells. In between the germinal cells, certain large cells called Sertoli cells are present. They are nutritive in function. The germinal epithelial cells give rise to sperms. They are in various stages of development like spermatogonia, primary spermatocytes, secondary spermatocytes, and spermatids. The Leydig cells which lie in connective tissue produce a male sex hormone called testosterone.

Question 26.
Explain the histological structure of the mammalian ovary.
Answer:
The ovary is a female gonad: It produces female sex cells called ova and the female sex hormones. Histologically the ovary is composed of fibrous connective tissue.

Its central part is called the stroma. It contains blood capillaries and nerve fibers. The outer part is called the cortex. A large number of Graafian follicles are embedded in a cortical layer. Each follicle is sac-shaped and contains female germ cells. It develops from the germinal layer. In a sexually mature female, the follicles are all at different stages of development and of different sizes. The cells of follicles are all of different sizes.

The cells of the follicle secrete a hormone called estrogen. In the connective tissue are present corpus luteum. It is formed at the site of a ruptured Graafian follicle. It secretes another hormone called progesterone.

Question 27.
What is spermatogenesis? Briefly describe the process of spermatogenesis.
Or
Give a schematic view of spermatogenesis in humans. (CBSE Delhi 2008, 2010)
Answer:
Spermatogenesis. The process of sperm formation from the sperm mother cells of testis (male gonad) is called spermatogenesis. It is completed in four phases, viz. spermatocytogenesis, meiosis I, meiosis II, and spermiogenesis. Spermatocytogenesis, Meiosis I and Meiosis II. The sperms are formed from the sperm mother cells present in the germinal layer of seminiferous tubules of the testis. Some of the mother cells enlarge to divide mitotically to form spermatogonia.

Growth phase: Some of them enter a period of growth and are called primary spermatocytes which are diploid.

Maturation phase: These cells divide meiotically to form two haploid secondary spermatocytes. Each secondary spermatocyte again divides mitotically. Thus one primary spermatocyte forms haploid spermatids.

Spermiogenesis: These develop into complete spermatozoan. These possess head which is embedded in the nourishing cells called Sertoli cells. The process of conversion of spermatid into spermatozoan is called spermiogenesis or spermine peliosis.

Since each primary spermatocyte divides meiotically to form 4 sperms thus 400 sperms or spermatozoa will be formed from 100 primary spermatocytes. The sperm consists of the acrosome (head), nucleus, and tail.

When sperm production starts, this age is called sexual maturity. It is generally 14-16 years. Spermatogenesis is controlled by the hormone (FSH) of the anterior lobe of the pituitary gland.

Question 28.
What role do pituitary gonadotropins play during the follicular and ovulatory phases of the menstrual cycle? Explain the shifts in steroidal secretions.
Answer:
Gonadotropin-releasing hormone (GnRH) also called gonadotropin-releasing factor (GnRH) is secreted by the hypothalamus of the brain, which stimulates the release of follicle-stimulating hormone (FSH) and luteinizing hormone (LH). FSH stimulates the ovarian follicles to produce estrogens during the proliferative phase. LH stimulates the corpus luteum of the ovary to secrete progesterone.

1. The menstrual phase is caused by the reduction of progesterone and estrogens.
2. The proliferative phase is caused by the increased production of estrogens.
3. LH causes ovulation. LH also causes a transformation of the empty Graafian follicle into the corpus luteum by luteinization inside the ovary. LH also stimulates the corpus luteum to secrete progesterone hormone to help in implantation, placentation, and maintenance of pregnancy.
4. A secretory phase is caused by the increased production of progesterone.

Question 29.
A woman has conceived and implantation has occurred. Discuss the sequence of changes up to parturition which will take place within her body under the influence of hormones.
Answer:

1. Under the impact of progesterone and estrogen the size of the uterus and birth canal increases. Relaxation of pelvic ligament takes place. The placenta is developed between chorionic and uterine tissues.
2. Breast size increases to full maturity under the impact of progesterone. Corpus luteum grows in the early month and then regresses in the later month.
3. Menstrual cycle and ovulation do not occur.
4. Uterine changes: Sex hormones, the estrogenic and placental gonadotropins are excreted with urine.
5. The volume of blood increases.

Some of the endocrine glands such as the adrenal cortex, anterior pituitary, and thyroid show enlargement. Relaxin hormone is essential in relaxing the pelvic ligament during parturition.

Oxytocin controls uterine contraction during childbirth.

Question 30.
The zygote passes through several developmental stages till implantation. Describe each stage briefly with suitable diagrams. (CBSE 2010)
Answer:
Cleavage is a holoblastic type but there is a tendency to show size differences of the blastomeres from the very start. Cleavage

Fig. CLeavage in mammalian ovum (pig) Fri sectional view A. zygote; B. two-ceU stage; C. four-cell stage: D. sixteen-celL stage; E. morula stage; F and G. Blastocyst (diagrammatic)

The first cleavage takes place in the cytoplasm of the ovum on an imaginary axis (animal pole to vegetal pole). The first division results in two blastomeres. One blastomere is slightly larger than the other. The larger cell divides first and thus forms the three blastomeres. Then the smaller blastomere divides and thus four blastomeres are formed. It is a 4-celled stage.

2nd cleavage is at a right angle to the 1st cleavage. One member of the large blastomeres of 4-cell stage divides forming 5-cell condition followed by 6-7 stage ultimately 8-celled stage. Successive cleavage divisions result in the formation of the solid mass of cells. The thus solid ball-like stage is called the morula stage.

Conversion of morula into blastula is initiated by dynamic rearrangement of small blastomeres. A central fluid-filled cavity called blastocoel (segmentation cavity) appears. Externally placed blastomeres lose their rounded form and become flat. This trophoblast along with the lining of the uterus forms extra-embryonic membranes. This provides protection and nourishment to the developing embryo.

Fig. Development of a blastodermic vesicle (blastocyst) in pig (diagrammatic sectional view).

Simultaneously zona pellucida disappears and the embryo gets implanted into the uterus. Below the blastodermic vesicle, there appears formative tissue, i.e. inner cell mass. From this inner cell mass, the proper embryo is formed.

Question 31.
Explain the following:
(i) Failure of testes to descend into the scrotum produces sterility.
Answer:
Failure of testes to descend into scrotum is called cryptorchidism. It leads to sterility because the scrotum provides a lower temperature (2°C) than the body temperature necessary for the spermatogenic tissue of testes to produce sperms.

(ii) Spermatids possess a haploid chromosome number.
Answer:
Spermatids possess haploid chromosome numbers because they are produced as a result of spermatogenesis (a kind of reductional or meiotic division). Moreover, it is essential because during fertilization diploid number of chromosomes will be restored in the fertilized egg. The haploid number of chromosomes in the spermatids helps in the maintenance of a specific number of chromosomes in the species.

(iii) The first half of the menstrual cycle is called the proliferative phase as well as the follicular phase.
Answer:
The first half of the menstrual cycle is called the follicular phase because estrogens secreted by the cells of maturing Graafian follicles control the changes in the secondary sex organs. It is also called the proliferative phase because the growth and proliferation of tissue on the wall of the uterus, vagina take place.

(iv) The second half of the menstrual cycle is called the luteal phase as well as the secretory phase.
Answer:
The second half of the menstrual cycle is called the secretory phase as well as a luteal phase because during this phase secretion occurs due to the luteinizing hormone (LH) which stimulates ovulation, the formation of corpus luteum and progesterone.

(v) Primary sex organs control the growth, function, and maintenance of secondary sex organs.
Answer:
Primary sex organs control the growth, function, and maintenance of secondary sex organs because they secrete hormones such as testosterone in males and estrogen and progesterone in females.

(vi) Why are the human testes located outside the abdominal cavity? Name the pouch in which they are present. (CBSE 2014)
Answer:
(a) Testes are located in the scrotal sac (scrotum).
(b) It helps in maintaining the low temperature (2-2.5°C) lower than the normal body temperature required for maturation of sperms.

Question 32.
Describe the formation of three germ layers in a mammalian embryo.
Answer:
The blastodermic vesicle is surrounded by the outer cellular layer (trophectoderm). The blastodermic vesicle shows an internal cluster of cells due to differentiation. This cluster of cells is called inner cell mass. Now the morphogenetic movements (epiboly, emboly, etc.) of the cells in small masses or sheets take place. As a result, three germinal layers (i.e. endoderm, mesoderm, and ectoderm) are formed. Formation of endoderm.

Some cells from the inner cell mass detach. Cells move in sheets or masses in the blastocoel. These are the potential endodermal cells. These cells arrange themselves as second layer inner to the outer layer of a blastodermic vesicle. The blastocoel disappears and a new cavity appears. This new cavity is called archenteron or primitive gut. This archenteron will give rise to the gut tract.

Formation of mesoderm: At the margin of the embryonic disc, cells multiply at an increased rate. The thickness of the embryonic disc increases. These cells detach from the embryonic disc and give rise to the mesoderm.

Formation of ectoderm: After the formation of mesoderm, the cells of the embryonic disc arrange themselves so as to form the ectoderm.

Question 33.
Write functions of the placenta.
Answer:
The placenta serves primarily as an organ that permits the interchange of materials carried in the blood of mother and fetus. Its main functions are:

• Nutrition: Supply of nutrient materials to the fetus.
• Respiration: Supply of 02 to the fetus and receives C02 back from it.
• Excretion: Fluid nitrogenous waste products escape through the placenta.
• Barrier: The placenta is a barrier-like semipermeable membrane.
• Storage: The placenta stores fat, glycogen, and iron for the embryo before the formation of the liver.
• Hormonal function: The placenta secretes extra ovarian hormones estrogen and progesterone in females during pregnancy that serves to maintain the fetus.

Question 34.
Briefly describe the process of oogenesis. (CBSE 2004)
Or
Give a schematic representation of oogenesis in humans. Mention the number of chromosomes at each stage. Correlate the life phases of the individual with the stages of the process. (CBSE (Delhi) 2008, 2009 (S), Outside Delhi 2013)
Or
When and where are primary oocytes formed in a human female? Trace the development of these oocytes till ovulation (in the menstrual cycle). How do gonadotropins influence this developmental process? (CBSE 2010)
Answer:
Oogenesis:
1. The phenomenon of the formation of haploid ova (egg) from the diploid egg mother cell of Ovary (female gonad) is called oogenesis. It occurs in three phases, i.e. multiplication, growth, and maturation.

Fig. Oogenesis

2. One ovum (n) is formed from one primary oocyte (2n) as a result of oogenesis.
3. The oogenesis is also controlled by the FSH hormone.

Question 35.
How does parturition take place? Which hormones are involved in the induction of parturition? (CBSE 2019 C)
Or
‘Parturition is induced by a complex neuroendocrine mechanism’. Justify. (CBSE Sample Paper 2018-2019)
Answer:
The onset of labor is termed parturition. The human gestation period (duration of pregnancy) is normally 280 days from the time of the last menstrual period to the birth of the baby. The process of uterine contractions that expel the baby and placenta is called ‘labor pains’.

It can be divided into three phases. The first stage is dilation, which usually lasts from 2 to 20 hours and ends up with the cervix of the uterus fully open or dilated. The second state, expulsion which lasts from about 2 to 100 minutes begins with full crowning, the appearance of the baby’s head in the cervix and continues while the baby is pushed, head first, down through the vagina into outside where it draws its first breath.

During this phase, placentation which lasts for 10-15 minutes after the birth of the child, the placenta and the fetal membranes are loosened from the lining of the uterus by another series of contractions and expelled. They are collectively called after birth or decidua.

Hormonal control:

• Oxytocin (birth hormone) promotes the contraction of the uterine muscles.
• Relaxin allows the pubic symphysis and ligaments to dilate.
• Corticotropin-releasing hormone (CRH) takes part in the ‘Clock’ that establishes the timing of birth.

Question 36.
A mother is ready to feed her newborn baby just after parturition by nature.
(i) Name the cells which secrete milk.
Answer:
Lacteriferous cells of alveoli of mammary glands. They are present in the breast.

(ii) Name the process of producing the milk.
Answer:
Lactation.

(iii) Name the milk produced during the first few days, after parturition.
Answer:
Colostrum.

(iv) Name the hormone meant for the release of milk.
Answer:
Oxytocin hormone released by the posterior lobe of the pituitary body.

(v) Why the doctors recommend breastfeeding during the initial period of infant growth? (CBSE Delhi 2016)
Answer:
Mother’s milk is low in fat but rich in proteins such as lactalbumin and lactoprotein. Colostrum also contains major immunoglobin IgA. It provides passive immunity to the newborn baby.

Question 37.
Describe the structure of an ovum.
Or
1. Draw a diagram of the structure of a human ovum surrounded by corona Radiata. Label the following parts:
(a) Ovum
(b) Plasma Membrane
(c) Zona Pellucida.
2. State the function of Zona Pellucida. (CBSE Delhi 2013)

Answer:
1. Structure of ovum: Most animaL eggs are spherical or oval, non-motiLe but on close examination, it is noted that one pole is different from the other. The pole from which polar bodies are given off is called an animal pole while the opposite is termed as a vegetal pole. Thus it is said to have polarity. Thus various cytoplasmic substances are distributed along the axis in an unequal manner.

The nucleus also called (germinal vesicle) having a chromatin network is surrounded by the nuclear membrane. It also contains prominent nucleolus. All animal eggs contain some reserve material to provide food called yolk. The cytoplasm of an ovum is called the ooplasm. It lacks a centrosome but contains cortical granules derived from Golgi bodies in its outer region termed cortex.

2. Function of Zona pellucida: During fertilization as a result of the cortical reaction, zona pellucida hardens and prevents the entry of additional sperms (polyspermy).

Question 38.
(i) Explain the following phases in the menstrual cycle of a human female:
(a) Menstrual phase
(b) Follicular phase
(c) Luteal phase
Answer:
Menstrual cycle: The cyclic changes that occur in the reproductive organs of primate femaLes (monkeys, apes, and human beings) constitute the menstruaL cycLe. The cycLe of events starts from one menstruation tiLt the onset of the next and Lasts for about 28/29 days (a menses).

Simplified menstrual cycle (28-day Cycle):

Phase	Days	Events
Menstrual phase	1-5	Endometrium breaks down, menstruation begins. The cells of en domain urn, secretions, blood, and the unfertilized ovum constitute the menstrual flow.
Follicular phase (ProLiferative phase)	6-13	Progesterone production is reduced. In fact, menstrual flow is associated with withdrawaL of progesterone.
Ovulatory phase	14	Endometrium rebuilds, FSH secretion, and estrogen secretion increase.
Luteal phase (Secretory phase)	15-28	Both LH and FSH attain a peak level. The concentration of estrogen in the blood is also high and reaches its peak. Ovulation occurs. Corpus luteum secretes progesterone. The endometrium thickens and uterine glands become secretory.

(ii) A proper understanding of the menstrual cycle can help immensely in family planning. Do you agree with the statement? Provide reasons for your answer. (CBSE Delhi 2017)
Answer:
Yes, a proper understanding of the menstrual cycle can help immensely in family planning. Day 10 to 17 of the menstrual cycle is called the fertile period of a human female. If coitus is done during this period, there is an increased chance of conception. But if coitus abstains during this period, pregnancy can be prevented. Pills used by females are also dependent on the menstrual cycle.

The pills have to be taken daily for a period of 21 days starting preferably within the first five days of the menstrual cycle. It is repeated again after a period of 7 days. These inhibit ovulation and implantation as well.

Question 39.
(i) Explain the menstrual cycle in human females.
Answer:
Menstrual cycle: The cycLic changes that occur in the reproductive organs of primate femaLes (monkeys, apes, and human beings) constitute the menstruaL cycLe. The cycLe of events starts from one menstruation tiLt the onset of the next and Lasts for about 28/29 days (a menses).

(ii) How can the scientific understanding of the menstrual cycle of human females help as a contraceptive measure? (CBSE Delhi 2018)
Answer:
Scientific understanding of the menstrual cycle can be helpful in human females to adopt any contraceptive measure as it can help in demarking the dates during which the chances of contraception are higher. For example, periodical abstinence is one such method in which couples avoid coitus during the 10-17 day of the menstrual cycle when ovulation is expected. Coitus can also be avoided during the time span in which the chances of fertilization are higher if there is a scientific understanding of the menstrual cycle.

Question 40.
(i) Draw a diagram of the adult human female reproductive system and label the different:
(a) parts of the fallopian tube
(b) layers of the uterus wall
Answer:
The female reproductive system of human:

(ii) Explain the events during the fertilization of an ovum in humans. (CBSE Delhi 2018C)
Answer:
When sperm comes in contact with zona pellucida of the ovum, this induces changes in the membrane that blocks the entry of additional sperm.

Thus, it ensures that only one sperm can fertilize an ovum. Secretion of acrosome helps sperm enter into the cytoplasm of the ovum through zona pellucida and plasma membrane. This induces completion of the meiotic division of secondary oocyte and the formation of a haploid ovum and a second polar body. The haploid nucleus of the ovum fuses with the sperm nucleus to form a diploid zygote, i.e. fertilization occurs.

Question 41.
(a) Name the hormones secreted and write their functions:
(i) by corpus luteum and placenta (any two).
Answer:
The Corpus luteum and placenta secrete progesterone. It helps in maintaining the pregnancy as a rising level of progesterone inhibits the release of GnRH which in turn inhibits the production of FSH, LH, and progesterone.

(ii) during the Follicular phase and parturition.
Answer:
FSH is secreted during the follicular phase. It stimulates the growth of the ovarian follicles and also stimulates the formation of estrogen. Relaxin is secreted during parturition. It softens the connective tissue of public symphysis that facilitates parturition.

(b) Name the stages in a human female where:
(i) Corpus luteum and placenta co-exist.
Answer:
The Corpus luteum and placenta co-exist up to blastocyst formation.

(ii) Corpus luteum temporarily ceases to exist. (CBSE (Delhi) 2017)
Answer:
During the menstruation phase corpus luteum ceases to exist.

Very Important Figures:

Here we are providing Class 12 Biology Important Extra Questions and Answers Chapter 9 Strategies for Enhancement in Food Production. Important Questions for Class 12 Biology are the best resource for students which helps in Class 12 board exams.

Class 12 Biology Chapter 9 Important Extra Questions Strategies for Enhancement in Food Production

Strategies for Enhancement in Food Production Important Extra Questions Very Short Answer Type

Question 1.
Define the livestock.
Answer:
Domesticated animals reared to provide milk, hide, flesh, etc.

Question 2.
What do you mean by the white revolution?
Answer:
Increased production of milk.

Question 3.
Who is called the Father of the White revolution?
Answer:
Dr. V. Kurien.

Question 4.
Give one example of each of the indigenous and exotic milch breeds of cow.
Answer:
Indigenous breed: Sahiwal; Exotic breed: Brown swiss.

Question 5.
Name one bacterial and one viral disease of cattle.
Answer:
Bacterial disease: Anthrax ;
Viral disease: foot and mouth disease.

Question 6.
What are the main advantages of cross-breeds of fowls over indigenous breeds?
Answer:
These consume less feed and produce more eggs.

Question 7.
Name two high-yielding cross-breeds of fowls.
Answer:
“B-77” and “HH-260”.

Question 8.
What is meant by ‘hidden hunger’?
Answer:
It is the consumption of food that is deficient in some essential nutrients like micronutrients, proteins, and vitamins.

Question 9.
Why are plants obtained by protoplast culture called somatic hybrids?
Answer:
Plants obtained from protoplast culture are called somatic hybrids because they are formed from hybrid cells developed through the fusion of genetically different somatic cells.

Question 10.
A person who is allergic to pulses was advised to take a capsule of Spirulina daily. Give the reasons for the advice.
Answer:
Spirulina is rich in proteins. Thus it overcomes the protein deficiency due to non-eating of pulses.

Question 11.
Give examples of some microorganisms working for SCP.
Answer:
Spirulina maxima, Methylophilus methylotrophs, Candida utilize, Paecilomyces variety.

Question 12.
What is the importance of male sterility in plant breeding?
Answer:
Plant breeders use male-sterile plants in artificial hybridization to avoid self-pollination.

Question 13.
Give the significance of superovulation.
Answer:
It increases the production of eggs (6-8) per cycle. Thus it helps in increasing herd size in a short time.

Question 14.
What is the economic value of Spirulina? (CBSE 2008, 2009)
Answer:
Spirulina can be grown on easily available cheap materials like wastewater from potato processing plants, molasses, and even sewage. They provide a large quantity of protein-rich food. Hence it also minimizes pollution.

Question 15.
List the products obtained from bee-keeping. (CBSE Delhi 2008)
Answer:
Honey, wax, royal jelly, and bee venom.

Question 16.
What is the importance of MOET? (CBSE Delhi 2013)
Answer:
MOET is multiple ovulation embryo transfer technologies.

Importance:

1. It is used to increase herd size in short term.
2. It is also meant for the successful production of hybrids.

Question 17.
Name any two diseases the ‘Himgiri’ variety of wheat is resistant to. (CBSE Outside Delhi 2013)
Answer:

1. Leaf and stripe rust
2. Hill bunt.

Question 18.
Name the following:
(a) The semi-dwarf variety of wheat is high-yielding and disease-resistant.
Answer:

1. Sonalika and
2. Kalyan Sona

(b) Anyone inter-specific hybrid mammal. (CBSE 2012)
Answer:
Mule.

Strategies for Enhancement in Food Production Important Extra Questions Short Answer Type

Question 1.
What is animal breeding? Name two main methods.
Answer:
Animal breeding: The main aim of animal breeding is to produce more milk-yielding cows with longer lactation periods and sturdier work animals. So animal breeding is an important aspect of animal husbandry. Two main methods are inbreeding and outbreeding.

Question 2.
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:
The honeybee is the major pollinator of several crop plants. Keeping a beehive near the farm will ensure proper pollination of crop plants. Therefore, crop yield will be ensured.

Question 3.
List three scientific approaches to obtain a high yield of crops.
Answer:

1. Crop production management
2. Crop improvement for higher yield through genetic manipulation
3. Crop protection management.

Question 4.
List the main steps of plant breeding. (CBSE2014)
Answer:

1. Collection of variability.
2. Evaluation and selection of parents.
3. Cross-hybridisation among selected parents.
4. Selection and testing of superior recombinations.
5. Testing, release, and commercialization of new cultivars.

Question 5.
Suggest two features of plants that will prevent insect and pest infestation.
Answer:

1. Increasing hair growth on aerial parts of plants.
2. Rendering the flowers nectar less.
3. Enabling plants to secrete insect-killing chemicals (toxins).

Question 6.
Briefly explain hybridization. Give one example.
Answer:
For the process of hybridization, selected parents are crossed to combine useful characters in the progeny. Heterosis is exploited in augmenting yields in several commercial crops. A high-yielding variety of wheat is a classical example of hybridization in crop evolution.

Question 7.
How is a disease-resistant plant selected for successful breeding? (CBSE Delhi 2013)
Answer:
Successful breeding for disease resistance depends mainly on two factors:

1. A good source of resistance.
2. A dependable disease test.

Question 8.
Define germplasm. How is it maintained? (CBSE Delhi 2014)
Answer:
Germplasm is the sum total of all the alleles of the genes present in a crop and its related species. Germplasm collections are usually maintained at low temperatures in the form of seeds. In the case of fruit trees, the germplasm is maintained as trees grown in the field.

Question 9.
Name any five varieties of crop plants that have been developed in India.
Answer:

1. Himgiri variety of wheat.
2. Pusa Swarnim variety of Brassica.
3. Pusa Shubhra variety of Cauliflower.
4. Pusa Komal variety of Cowpea.
5. Pusa Sadabahar variety of Chilli.

Question 10.
What is inbreeding depression and how is it caused in organisms? Write any two advantages of inbreeding. (CBSE Delhi 2011)
Or
Explain inbreeding depression. (CBSE Delhi 2011)
Answer:
Inbreeding depression. Continued inbreeding especially close inbreeding, usually reduces fertility and even productivity. It is termed inbreeding depression.

Advantages of inbreeding:

1. Evolve and maintain a pure line in any animal.
2. It helps in the accumulation of superior genes and the elimination of less desirable genes.

Question 11.
List the four objectives with which bio-fortification has been carried out to improve public health. (CBSE Delhi 2019 C)
Answer:
The method of breeding crops with higher levels of vitamins and minerals or higher protein and healthier fats is called biofortification. It is one of the most important methods for improving public health.

Objective:

• In the recent past, many varieties of crop plants such as rice, carrots, spinach, pumpkin, and bathua have been developed which possess several times more nutritive value and vitamins as compared to existing varieties.
• High protein content and quality
• High oil content and quality
• Micronutrient and minerals content.

Question 12.
Why is animal husbandry essential?
Or
What are the objectives of animal breeding? (CBSE 2014)
Answer:

• To increase milk production.
• To increase egg production.
• To increase meat production.
• To increase fish production.
• For proper utilization of animal wastes.

Question 13.
(a) What is the breeding of crops for enhancing their nutritional value called? Why is the need felt for enhancing the nutritional value of the crops?
Or
How have these grains improved in their nutritional value in comparison to their conventional varieties? (CBSE Outside Delhi 2019)
Answer:
Iron-fortified rice variety contains over five times as much iron as compared to commonly consumed rice.

(b) Rice, wheat, and maize are the most commonly used food grains the world over.
Answer:

1. Iron-fortified rice variety contains over five times as much iron as compared to commonly consumed rice.
2. Wheat variety Atlas-66 has huge protein content.
3. Maize hybrids have twice the amount of lysine and tryptophan amino acid.

Question 14.
Name any two common Indian millet crops. State one characteristic of millets that has been improved as a result of hybrid breeding to produce high-yielding millet crops. (CBSE Delhi 2015)
Answer:
Hybrid varieties of maize, Jower, and Bajra have been developed in India by plant breeders. Hybrid breeding has led to the development of several high-yielding varieties resistant to water stress.

Question 15.
Enumerate four objectives for improving the nutritional quality of different crops for the health benefits of the human population by the process of “Biofortification.” (CBSE Delhi 2015)
Answer:
Objectives:

• Protein content and quality.
• Oil content and quality.
• Vitamin content.
• Micronutrient and mineral content.

Question 16.
(i) Why are the plants raised through
micropropagation termed as some clones 1(CBSE Outside Delhi 2015)
Answer:
Plants raised by the tissue culture method are genetically identical to the original plant from which they were grown hence called some clones.

(ii) Mention two advantages of their technique.
Answer:
(a) Plants can be produced on a commercial scale.
(b) Recovery of healthy plants from diseased plants.

Question 17.
(i) Name any two fowls other than chicken reared in a poultry farm.
Answer:
Fowls other than chickens :

1. Turkeys,
2. Geese,
3. Swans,
4. Guinea fowls,
5. Peafowls.

(ii) Enlist four important components of poultry farm management. (CBSE Delhi 2016)
Answer:
Important components of poultry farm management:
(a) Selection of disease-free and suitable breeds.
(b) Brood house should be crowd-free.
(c) Sanitation and hygiene. The brooding house should be cleaned, disinfected, and with the proper drainage system.
(d) Care of chicken during brooding.

Question 18.
Explain the process of artificial hybridization to get improved crop variety in
(i) plants bearing bisexual flowers and
Answer:
In plants bearing bisexual flowers, the anthers are removed from the flower before they dehisce. This is called emasculation. The emasculated flowers are covered with a bag of butter paper to prevent contamination of stigma with unwanted pollen. This process is called bagging. When this stigma attains receptivity, mature pollen grains are dusted on the stigma and the flowers are rebagged to allow the fruits to develop.

(ii) female parent producing unisexual flowers. (CBSE Outside Delhi 2010)
Answer:
If the female parent produces unisexual flowers, emasculation is not done. The flower buds are bagged before the flowers open. When the stigma becomes receptive, pollen is dusted on the stigma and the flower is rebagged.

Question 19.
(i) Mention the property that enables the explants to regenerate into a new plant.
Answer:
Totipotency is the ability of a cell to grow or generate the whole plant.

(ii) A banana herb is virus-infected. Describe the method that will help in obtaining healthy banana plants from this diseased plant. (CBSE Outside Delhi 2010, Delhi 2019)
Answer:
Healthy banana plants can be obtained from diseased plants by meristem culture. Although the plant is virus-infected, the apical and axillary meristems are free of virus. The meristem is removed from the plant and is grown in vitro by micropropagation. The plants produced are virus-free.

Question 20.
You have obtained a high-yielding variety of tomatoes. Name and explain the procedure that ensures retention of the desired characteristics repeatedly in large populations of future generations of the tomato crop. (CBSE Delhi 2018)
Answer:
The retention of the desired characteristics in a large population of crops can be done by micropropagation, which is a type of vegetative propagation.

In this process, a small part of the plant is excised and grown under a sterile condition in a special nutrient medium to obtain many such plants that would be genetically identical to the original plants.

Question 21.
A herd of cattle is showing reduced fertility and productivity. Provide one reason and one suggestion to overcome this problem. (CBSE Delhi 2017)
Answer:
The cattle’s productivity is decreased due to inbreeding depression. A single outcross, i.e. breeding with animals of the same breed but should not have a common ancestor on either side up to 4-6 generations, can restore its fertility.

Question 22.
Why are microbes like Spirulina being produced on a commercial scale? Mention its two advantages. (CBSE Delhi 2018C)
Answer:
25% of the human population suffers from hunger and malnutrition. Single-cell protein (SCP) is the solution to this problem. Spirulina is a rich source of protein, so it is grown on a commercial scale for SCP production.

Advantages:

1. It utilizes cheap substrate like raw material like animal manure, wastewater from potato processing plants, straw molasses, and even sewage, therefore incidentally cleaning the environment and reducing pollution.
2. A small amount of microbe can produce a huge quantity of protein.

Question 23.
“Modern methods of breeding animals and plants can alleviate the global food shortage”. Comment on the statement and give suitable examples.
Answer:
Several new techniques like inbreeding, outbreeding, artificial insemination, MOET, somatic hybridization, and tissue culture are playing important role in enhancing global food production. Because all these techniques aim at increasing productivity.

Strategies for Enhancement in Food Production Important Extra Questions  Long Answer Type

Question 1.
What is apiculture? How is it important in our lives? (CBSE Delhi 2019 C)
Or
Beekeeping practice is a good income-generating industry. Write the different points to be kept in mind for successful beekeeping. Write the scientific name of the most common Indian species used for the purpose. (CBSE Delhi 2019)
Answer:
Apiculture is the maintenance of hives of honeybees for commercial purposes. It is also called Beekeeping. Apiculture is very important in our lives because it produces products of high economic and nutritive value:

1. The main product of beekeeping is honey which is a highly nutritious edible product. It replaces the use of sugar in many industries.
2. Honey is used as a part of many indigenous medicines.
3. Another product called beeswax is also produced which is used in the cosmetic and polish industry.
4. Honeybees are chief pollinating agents so helpful in improving the plant yield.
5. The most commonly used species in the wild state is Apis Indica and in the domestic state Apis mellifera.

Question 2.
What is plant breeding? (CBSE Delhi 2013)
Answer:
Plant breeding is the genetic improvement of the crop in order to create desired plant types that are better suited for cultivation, give better yields, and are disease resistant. Conventional plant breeding is in practice from 9,000 to 11,000 years ago. Most of our major food crops are derived from domesticated varieties.

But now due to advancements in genetics, molecular biology, and tissue culture, plant breeding is being carried out by using molecular genetic tools. Classical plant breeding includes hybridization (crossing) of pure lines, artificial selection to produce plants with desirable characters of higher yield, nutrition, and resistance to diseases.

Question 3.
What is a hybrid? Explain the procedure of obtaining a hybrid.
Answer:
Hybrid: A progeny obtained by crossing two varieties or species having desired genes thus showing required characters.

Process of hybridization:
This technique involves the following steps:

1. Removal of undehisced anthers from the bisexual flower of a plant to be used as female. This is called emasculation.
2. The emasculated flowers are covered by butter paper to avoid pollination by an undesirable pollen grain. It is also termed bagging.
3. Pollen grains from known seeds of desirable plants are used to pollinate these emasculated flowers.
4. They are collected, multiplied and their desirable characters are determined.

Question 4.
(i) Give the scientific name of the soil bacterium which produces crystal (Cry) proteins.
Answer:
A soil bacterium Bacillus thuringiensis (Bt) produces a crystal (Cry) protein.

(ii) How are these proteins useful in agriculture?
Answer:
A crop expressing cry gene is usually resistant to a group of insects for which the concerned Cry protein is toxic.

(iii)What do the different written terms “Cry” and “cry” represent respectively?
Answer:
The first letter of the protein symbol is always capital and always written in roman letters, i.e. Cry. However, the gene symbol usually has small letters and written in italics, i.e. cry. So Cry represents protein and cry represents a gene.

Question 5.
Differentiate between inbreeding and outbreeding in cattle. State one advantage and one disadvantage for each one of them. (CBSE Delhi 2013)
Or
How is inbreeding advantageous as well as disadvantageous in cattle breeding programs? (Mention any two advantages and two disadvantages.) (CBSE Sample Paper 2019-20)
Answer:
Differences between inbreeding and outbreeding:

Inbreeding	Outbreeding
1. Breeding between animals of the same breed for 4-6 generations.	1. Breeding between unrelated animats.
2. Continuous inbreeding leads to inbreeding depression, i.e. reduced fertility and productivity.	2. It leads to heterosis or hybrid vigor.
3. Advantages of inbreeding: It helps in restoring fertility and yield.	3. Advantage of outbreeding: It causes hybrid vigor.
4. Disadvantages of inbreeding: Continued inbreeding reduces fertility and even productivity. It is called inbreeding depression.	4. Disadvantage of outbreeding: It is costly and a difficult procedure.

Question 6.
What are superovulation and embryo transfer?
Or
Explain the strategy used for herd improvement in cattle. (CBSE Delhi 2019 C)
Or
Expand MOET. Explain the procedure of this technology in cattle improvement. (CBSE Delhi 2008, 2012, Outside Delhi 2019)
Answer:
MOET (Multiple Ovulation and Embryo Transplantation Technique). It is the latest technique to produce super milch cows.

1. Superovulation: It involves the stimulation of high milk-producing cows to ovulate more eggs by hormonal stimulation. Fertilization is done by artificial insemination. In this way, 4 to 10 embryos are developed in such cows.
2. Embryo transfer: In this, embryos are collected from superovulated cows. Each embryo is transplanted in the uterus of surrogate cows of inferior quality to produce a large number of calves with good germplasm. It can also be done in sheep, goats, and other livestock animals as well.

Question 7.
Why does a beekeeper keep beehives in crop fields during the flowering periods? State any two advantages. (CBSE Sample Paper)
Or
Honey collection improves when beehives are kept in crop fields during the flowering season. Explain. (CBSE 2010)
Answer:

1. Bees act as pollinating agents.
2. Bees bring about cross-pollination. Thus the productivity of crops increases. This in turn is beneficial to bees too as they get nectar to make honey.

Question 8.
Give a brief account of the cultivation of sugarcane and millets. (CBSE Delhi 2011, 2015)
Or
Write the scientific name of the sugarcane variety that was originally grown in North India. Why was this variety hybridized with the tropical variety of sugarcane grown in South India? (CBSE Delhi 2019 C)
Answer:
Cultivation of sugarcane:

1. Saccharum Barberi, originally grown in North India, had poor sugar content.
2. Saccharum officinarum, grown in South India, has thicker stems and higher sugar content but could not grow properly in North India.
3. A cross has been made between these species and the hybrid variety, combining the desirable qualities like thick stem, high sugar content, and higher yield is being grown in North India.

Cultivation of millets:

1. Several hybrid varieties of maize, bajra, and Jowar have been developed in India.
2. These breeding programs have resulted in the development of high-yielding varieties that are resistant to water stress.

Question 9.
Explain out-breeding, out-crossing, and cross-breeding practices in animal husbandry. (CBSE Delhi 2018)
Or
Differentiate between out-crossing and cross breeding. (CBSE Delhi 2018C)
Answer:
Out-breeding: Out-breeding is the breeding of unrelated animals, which may be between individuals of the same breed but having no common ancestors for 4-6 generations or between different breeds or different species.

Out-crossing: Out-crossing is the practice of mating animals within the same breed, but having no common ancestors on either side of their pedigree, up to 4-6 generations. It is the best breeding method for animals that are below average in milk productivity.

Cross-breeding: Cross-breeding allows the desirable qualities of two different breeds to be combined. In this method, superior males of one breed are mated with superior females of another breed. The progeny hybrid animals may be used for commercial production or they may be subjected to some form of inbreeding and selection to develop new stable breeds that may be superior to the existing breeds. Many new animal breeds have been developed by this approach.

Question 10.
(i) Write the desirable characters a farmer looks for in his sugarcane crop,
Answer:
The desirable characters that should be present in the sugarcane crop are:
(a) High yield,
(b) thick stem,
(c) high sugar content and
(d) ability to grow in North India.

(ii) How did plant breeding techniques help north Indian farmers to develop cane with desired characters? (CBSE Delhi 2017)
Answer:
Cultivation of sugarcane:

1. Saccharum Barberi, originally grown in North India, had poor sugar content.
2. Saccharum officinarum, grown in South India, has thicker stems and higher sugar content but could not grow properly in North India.
3. A cross has been made between these species and the hybrid variety, combining the desirable qualities like thick stem, high sugar content, and higher yield is being grown in North India.

Cultivation of millets:

1. Several hybrid varieties of maize, Bajra, and Jowar have been developed in India.
2. These breeding programs have resulted in the development of high-yielding varieties that are resistant to water stress.

Question 11.
(i) Lifestyle diseases are increasing alarmingly in India. We are also dealing with large-scale malnutrition in the population. Suggest a process by which we can address both these problems.
Answer:
These problems can be addressed by the process termed biofortification,

(ii) Give any three examples to support your answer. (CBSE Sample Paper 2018-19)
Answer:
Biofortification can be done:
(a) by enhancing food quality with respect to protein, e.g. improved wheat, maize varieties, etc.
(b) by cultivating vitamin-enriched vegetables, e.g. vitamin A enriched carrots, spinach, pumpkin, etc
(c) by enriching micronutrient and mineral content in vegetables, e.g. iron and calcium-enriched spinach and bathua

Question 12.
What is biofortification? Mention the contribution of the Indian Agricultural Research Institute towards it with the help of any two examples. (CBSE Delhi 2018C)
Answer:
It is an improvement in the nutritional quality of food crops by breeding. Through this, levels of vitamins, minerals, proteins, and healthier fats are enriched in crops. Indian Agricultural Research Institute released several vitamin and mineral enriched vegetable crops like vitamin A enriched carrots and spinach; Vitamin C enriched bitter guard, mustard, tomato; iron and calcium-enriched spinach and bathua; protein-enriched broad lablab, French and garden peas.

Question 13.
(i) What is artificial insemination? Give its significance.
Answer:
Artificial insemination:

• It is the process in which the semen collected from a superior quality male is injected into the reproductive tract of the selected female by the breeder.
• The advantages of artificial insemination are as follows:
• Semen can be used immediately or stored/frozen and used at a later date when the female is in the right reproductive phase.
• Semen can be transported in the frozen form to a distant place where the selected female animals are present.
• Semen from one selected male animal can be used on a number of female animals.
• The disadvantage is that the success rate is fairly low.

(ii) Write a note on MOET. (CBSE Delhi 2008)
Answer:
Multiple Ovulation Embryo Transfer Technology (MOET):

1. It is a method to improve the herds and their size.
2. The steps in the method are as follows:
3. A cow is administered hormones (like FSH) to induce follicular maturation and superovulation, i.e. production of 6-8 ova in one cycle.
4. The cow is mated with the selected bull or artificially inseminated.
5. The fertilized eggs at 8-32 celled stages are recovered and transferred to surrogate mothers.
6. This technology has been used for cattle, rabbits, mares, etc.
7. High milk-yielding breeds of females and high-quality meat-yielding bulls have been bred successfully to increase the herd size in a short time.

Significance:

• Several cows (up to 3,000) can be inseminated by the semen of a single pedigree bull of good quality.
• Avoid the transportation of animals,
• The quality and quantity of progeny can be improved.
• New characters can be introduced in the progeny.
• It is economical.

Question 14.
What are the practices adopted to improve crop production?
Answer:
The practices adopted to improve crop production are as follows:

1. Addition of fertilizers to the soil.
2. Selective breeding.
3. Weed control.
4. Control of plant diseases.
   (a) Fertilisers: These are the chemical compounds that are added to the soil to increase fertility. They make up for the deficiency of the required nutrients and help in increasing crop production.
   (b) Selective breeding: Disease-resistant seeds are produced by selective breeding. Regular use of high yield variety results in better crop production.
   (c) Weed control: The unwanted plants or weeds are controlled by using certain chemicals called weedicides.
   (d) Control of plant diseases: Crops should be protected from insects, fungi, animals, and other diseases. It is very useful for increasing crop production. Insects are very harmful to crops. So insecticides should be used to kill insects.

Question 15.
Discuss the role of plant tissue culture in increasing food production. (CBSE Delhi 2011)
Answer:
Applications of tissue culture technique:

1. This technique is applied for the rapid multiplication of desirable and rare plants.
2. By this technique, an indefinite number of plants can be produced.
3. From the culturing of virus-free tissues of the shoot apex of an infected plant, it becomes possible to obtain virus-free plant in sufficient stock. The tissue culture technique has been used to obtain virus-free potatoes and sugarcane.
4. The technique (Embryo culture) is useful in overcoming seed dormancy, but also in producing viable plants from the crosses which normally fail due to the death of immature embryos.
5. The technique has been applied for obtaining a large number of haploid and homozygous diploids.
6. Somatic hybridization helps the fusion of cells belonging to different families.
7. This technique is also useful for the genetic improvement of useful plants.

Question 16.
“The benefits of a new variety can be achieved only if farmers grow the variety”. Explain.
Answer:
The seed of new variety must be multiplied and made available to the farmers. In-plant breeding, seed means any plant part that is used to grow a crop. Thus ‘seed’ would include grains of wheat, rice, etc. tubers of the potato, stems of sugarcane, etc., provided they are used for producing new plants.

Therefore, wheat grains used as food cannot be termed as seeds, whereas those used for raising a crop are called seeds. A seed of a variety with superior traits is called an improved seed, which must be of high purity and have a high germination percentage. It must also be free from weed seeds and from diseases.

Question 17.
(a) What is mutation breeding? Give an example of a crop and disease to which resistance was induced by this method.
Answer:
(a) Mutation breeding involves the following steps:

1. Inducing mutation(s) through various methods/ mutagens.
2. Screening the plant materials for disease resistance.
3. Multiplication of these selected plants for direct use or for use inbreeding.
4. Hybridization of the selected plant materials.
5. Selection for disease resistance, testing, and release as a variety.
6. Through mutation breeding, varieties of mung bean have been developed that are resistant to yellow mosaic virus and powdery mildew.

(b) Differentiate between pisciculture and aquaculture. (CBSE Sample Paper 2020)
Answer:

Aquaculture	Pisciculture
It involves the rearing all types of aquatic organisms to obtain products of economic value.	Pisciculture involves the rearing of fishes for obtaining food and fishery by-products such as fish oil, fish glue, fish manure, shagreen, leather, etc.

Question 18.
How can crop varieties be made disease resistant to overcome the food crisis in India? Explain. Name one disease-resistant variety in India of:
(i) Wheat to leaf and stripe rust
Answer:
Mutation breeding involves the following steps:
(a) Inducing mutation(s) through various methods/mutagens.
(b) Screening the plant materials for disease-resistance.
(c) Multiplication of these selected plants for direct use or for use inbreeding.
(d) Hybridisation of the selected plant materials.
(e) Selection for disease resistance, testing, and release as a variety.
(e) Through mutation breeding, varieties of mung bean have been developed that are resistant to yellow mosaic virus and powdery mildew.

(ii) Brassica to white rust (CBSE 2011)
Answer:
(a) Himgiri variety of wheat
(b) Pusa swarm of Brassica

Question 19.
(i) Name the technology that has helped the scientists to propagate on a large scale the desired crops in a short duration. List the steps carried out to propagate the crops by the said technique.
(ii) How are somatic hybrids obtained?
Or
Scientists tried to develop a single plant exhibiting the characteristic of tomato and potato by using cells from tomato and potato plants respectively. Name the procedure and list the steps to achieve this. (CBSE Delhi 2014, Outside Delhi 2019, Sample Paper 2020)
Answer:
1. Tissue culture:
(a) It is the technique of regeneration of whole plants from any part of a plant by growing it on a suitable culture/nutrient medium under aseptic/sterile conditions in vitro.
(b) The culture medium must supply the energy, inorganic nutrients, vitamins, amino acids, and growth regulators like cytokinins and auxins.
(c) By this method, called micropropagation, thousands of plants can be grown in a short period of time.

The advantages of micropropagation are:
(a) The plants produced are genetically identical and constitute some clones.
(b) A number of plants can be grown in a short period of time.
(c) Healthy, disease-free plants can be grown by meristem culture.
(d) Somatic hybrids can be raised by tissue culture, where sexual hybridization is not possible.

2. Somatic Hybridization: Fusion of isolated protoplasts from two different plant varieties each having desirable characters is called somatic hybridization. The resultant hybrid is called somatic hybrids. This somatic hybrid can be grown into a new plant carrying all desirable qualities.

Here we are providing Class 12 Biology Important Extra Questions and Answers Chapter 8 Human Health and Disease. Important Questions for Class 12 Biology are the best resource for students which helps in Class 12 board exams.

Class 12 Biology Chapter 8 Important Extra Questions Human Health and Disease

Human Health and Disease Important Extra Questions Very Short Answer Type

Question 1.
What is health?
Answer:
It is a state of complete physical, mental and social well-being of a person.

Question 2.
What is a disease?
Answer:
Malfunctioning of one or more organs characterised by signs and symptoms is called disease.

Question 3.
Write the role of interferons. (CBSE Delhi 2019(C))
Answer:
Role of interferon. These are glycoprotein released by virus-infected cells. They protect the adjoining cells from the attack of the virus.

Question 4.
Define autoimmunity.
Answer:
It is an abnormality which sometimes occurs in the immune system and instead of destroying foreign molecules, it attacks the body’s own cells.

Question 5.
A boy of ten years had chickenpox. He is not expected to have the same disease for the rest of his life. Mention how it is possible. (CBSE 2009)
Answer:
The body will acquire active immunity as the antibodies formed will protect him from the attack of microbes of chickenpox.

Question 6.
What is it that prevents a child to suffer from a disease he/she is vaccinated against? Give one reason. (CBSE 2010)
Answer:

1. Antibodies
2. These antibodies neutralise the action of antigens.

Question 7.
Name the pathogen which causes Typhoid. Name the test that confirms the disease. (CBSE Delhi 2019 (C))
Answer:
Causative pathogen: Salmonella typhi Test: Widal test

Question 8.
Name two types of cells in which the HIV multiplies after gaining entry into the human body. (CBSE 2008)
Answer:

1. T-Lymphocytes
2. White Blood Corpuscles (Macrophages).

Question 9.
Why is a secondary immune response more intense than the primary immune response in human? (CBSE 2014)
Answer:
It is because the human body appears to have more memory of the first encounter.

Question 10.
When does a human body elicit an anamnestic response? (CBSE (Outside Delhi) 2013)
Answer:
The primary immune response is of low intensity; a subsequent encounter with the same pathogen elicits a highly intensified anamnestic or secondary response.

Question 11.
Name the two intermediate hosts which the human liver fluke depends on to complete its life cycle so as to facilitate parasitisation of its primary host. (CBSE Delhi 2014)
Answer:

1. Freshwater Snail
2. Fish

Question 12.
Indiscriminate use of X-rays for diagnoses should be avoided. Give reason. (CBSE (Delhi) 2015)
Answer:
X-rays cause mutation thus may lead to cancer.

Question 13.
Give the scientific name of the source organism from which the first antibiotic was produced. (CBSE Sample paper 2018-19)
Answer:
Penicillium Notatum

Question 14.
Name two diseases whose spread can be controlled by the eradication of Aedes mosquitoes. (CBSE 2018)
Answer:
Dengue and chikungunya

Question 15.
How do cytokine barriers provide innate immunity in humans? (CBSE 2018)
Answer:
Cytokine barriers: Cytokines inhibit viral replication. Virus-infected cells secrete proteins called interferons which protect non-infected cells from virus.

Question 16.
Name two recent incidences of wide-spread diseases caused by Aedes mosquitoes. (CBSE Delhi 2008)
Answer:

1. Dengue
2. Chikungunya

Question 17.
How does the human body respond when haemozoin produced by Plasmodium is released in its blood? (CBSE Delhi 2019 (C))
Answer:
As haemozoin is released in the blood, the patient shows symptoms of malaria such as restlessness, sleeplessness, muscular pain and chilliness. In response to chill, the body temperature rises.

Question 18.
How does saliva act in body defence? (CBSE Delhi 2004)
Answer:
Human saliva contains lysozyme, a lytic enzyme, which kills the germs in the food.

Question 19.
Name the type of cells that produce antibodies. (CBSE 2004)
Answer:
Lymphocytes which is a form of leucocytes (white blood cells) produce antibodies.

Question 20.
Why sharing injection needles between two individuals is not recommended? (CBSE Delhi 2013)
Answer:
Sharing of injection needles between two individuals may cause the transmission of AIDS and Hepatitis B.

Question 21.
How do monocytes act as a cellular barrier in humans to provide innate immunity? (CBSE Delhi 2018C)
Answer:
Monocytes kill bacteria by the process of phagocytosis.

Human Health and Disease Important Extra Questions Short Answer Type

Question 1.
Make a list of common infectious diseases.
Answer:
Common infectious diseases: Depending on the pathogen, infectious diseases are as follows:

Question 2.
Given below are pairs of pathogens and diseases caused by them. Which of these is not a matching pair and why?
(i) Virus Common Cold
(ii) Salmonella Typhoid
(iii) Microsporum Filariasis
(iv) Plasmodium Malaria
Answer:

1. (iii) is not matching.
2. Microsporum is a fungus which causes ringworm disease. Filariasis is caused by Wuchereria bancrofti and W. Malayi (roundworm).

Question 3.
Differentiate between antibodies and interferons.
Answer:
Differences between antibodies and interferons:

Antibodies	Interferons
1. They are slow-acting and long-lasting.	1. They are quick acting and temporary.
2. They act outside the cells.	2. They act inside the cells.
3. They act against bacterial and viral infections.	3. They act against the virus only.

Question 4.
(i) Name the source plants of heroin drug. How is it obtained from the plants?
Answer:
Papaver somniferum is the source plant of heroin drug. It is obtained by acetylation of morphine, which is extracted from the latex of poppy plant (Papaver somniferum).

(ii) Write the effects of heroin on the human body. (CBSE 2018)
Answer:
Heroin is depressant and slows down body functions.

Question 5.
Mention one application for each of the following:
(i) Passive immunisation
Answer:
(i) Passive Immunisation: When ready¬made antibodies are introduced into the body, it is called passive immunisation. Passive immunisation provides a quick immune response in the body.

(ii) Antihistamine
Answer:
Anti-Histamines: Anti-Histamines are the chemicals which are given against allergic reactions.

(iii) Colostrum (CBSE 2017, 2019)
Answer:
Colostrum: Colostrum is the yellow fluid produced during the initial days of lactation. It is rich in antibodies and is essential to develop resistance in a newborn baby.

Question 6.
What is a vaccine? (CBSE Delhi 2019 C)
Answer:
Vaccine: It is a preparation of dead or altered (weakened) germs of a disease which on entry into the body of a healthy person provide temporary or permanent active/passive immunity by inducing antibody formation. Thus antibody provoking agents are called vaccines. The vaccine provides artificial active immunity.

Question 7.
Name the primary and secondary lymphoid organs. (CBSE Delhi 2019 C)
Or
State the function of primary and secondary lymphoid organs. (CBSE Delhi 2011)
Answer:
Primary lymphoid organs:

• Bone Marrow
• Thymus.

Secondary lymphoid organs:

• Spleen
• Lymph nodes
• Tonsils
• Peyer’s Patches of the small intestine.

Or

Primary lymphoid organs are the sites where immature lymphocytes differentiate and become antigen-sensitive mature lymphocytes.

However, secondary lymphoid organs provide site/location for mature lymphocyte & antigen interaction.

Question 8.
Explain what is meant by metastasis. (CBSE 2009)
Answer:
Metastasis. Small pieces of primary tumour break off and are carried to other body parts by the blood or lymph where these form the secondary tumours. This process is called metastasis. So metastasis is the process of transference of cancerous cells from the site of origin to distant parts of the body. The most frequent sites of metastasis are lymph nodes, lungs, long bones, liver, skin and brain. Metastasis is the most feared property of malignant tumours.

Question 9.
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:
The condition is called allergy. Mast cells are responsible for such reactions. To avoid such reactions, the following precautions must be taken:

• Use of drugs like antihistamine, adrenalin and steroids quickly reduces the symptoms.
• Avoid contact with substances to which a person is hypersensitive.

Question 10.
What are the symptoms of allergic reactions?
Answer:
Symptoms of allergic reactions.
The following are the symptoms of allergies:

• The person may suffer from high fever
• The mucous membrane of the lower part of the respiratory tract gets affected which leads to cough and asthma.
• Reddening of the skin, the appearance of blisters on the skin.
• Accumulation of tissue fluid below the skin.
• Watering of eyes and inability to breathe.
• Sneezing, running nose, etc.

Question 11.
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:

1. Allergy is the exaggerated (hypersensitive) response of the immune system to certain antigens present in the environment. These certain antigens are called allergens. Our immune system responds to it by releasing histamines and serotonin from mast cells. Common allergens are mites in the dust, pollens and animal dander (material shed from animals),
2. Lifestyle in metro cities is making them sensitive to allergens.
3. The symptoms are sneezing, watery eyes, running nose and difficulty in breathing.

Question 12.
Drugs and alcohol give short-term ‘high’ and long-term ‘damages’. Discuss.
Answer:
Mostly stimulant drugs (caffeine) and alcohol (depressant) give a feeling of intoxication and euphoria for only a brief period soon after use. However, prolonged use for long-term causes permanent damage to vital body parts like liver, kidneys, lungs, cardiovascular system, etc.

Question 13.
List any two adaptive features evolved in parasites enabling them to live successfully on their hosts. (CBSE Delhi 2008)
Answer:
Adaptations of parasites:

1. Presence of adhesive organs or suckers to cling to the host.
2. Loss of unnecessary sense organs.

Question 14.
What is dengue fever? List two symptoms.
Answer:
Dengue fever: Dengue fever is caused by an RNA containing arbovirus of flavivirus group which also causes yellow fever (not found in India). Thus, the virus which causes dengue fever is a mosquito-borne flavi-ribo virus. The virus of dengue fever is transmitted by the bite of Aedes aegypti (mosquito). The incubation period is 3-8 days.

Symptoms:

• Abrupt onset of high fever.
• Severe frontal headache and pain behind eyes which worsens with eye movement.

Question 15.
Why is the structure of an antibody molecule represented as H₂L₂? Name any two types of antibodies produced in humans. (CBSE Delhi 2018C)
Answer:

• The antibody molecule is made up of four peptide chains-two small chains are called light chains and two longer chains are called heavy chains. Hence it is represented as H₂L₂.
• Ig G, Ig A, Ig M. and Ig E are the antibodies produced in humans.

Question 16.
What are the preventive measures of dengue fever? Is there any vaccine available?
Answer:
Prevention and treatment:

1. Mosquitoes and their eggs should be eliminated. Put wire mesh on doors and windows.
2. No specific treatment is available.
3. Symptomatic care including bed rest, intake of adequate fluid and pain killer medicines are recommended.
4. Do not take Aspirin and Aspirin. Give plenty of liquids to the patient.
5. No vaccine for Dengue fever is available.

Question 17.
Differentiate between the roles of B-lymphocytes and T-lymphocytes in generating immune responses. (CBSE Delhi 2019)
Answer:
Role of B-lymphocytes and T-lymphocytes in the immune response:

1. B-cells (B-lymphocytes) and T-cells (T lymphocytes) comprising the immune system are produced in the bone marrow. T-cells differentiate in the thymus.
2. B-lymphocytes produce antibodies in response to foreign substances (antigens) such as pathogens and pollen. Antibodies are immunoglobulins. They are specific for each antigen. There is more than one antibody for an antigen. Antibodies bind antigens but do not destroy them. This is attacked through other mechanisms. Allergens which are weak antigens cause allergy.
3. T-cells respond to pathogens by producing three types of cells: killer T-cells, helper T-cell and suppressor T-cells. T lymphocytes either help B-lymphocytes to produce antibodies or kill the pathogen directly (killer T-cells). Both B- and T-cells produce memory cells when stimulated. These have long lives and form the basis of acquired immunity.

Question 18.
Why is tobacco smoking associated with rising in blood pressure and emphysema? Explain. (CBSE Outside Delhi 2019)
Answer:
Nicotine, an alkaloid present in tobacco, stimulates the adrenal gland to release adrenaline and nor-adrenaline into blood circulation. Both these hormones increase blood pressure and heart rate.

Smoking causes emphysema. Tobacco smoke damages the air sacs (alveoli) of the lungs. Thus surface area for exchange of gases becomes less and disorder emphysema is caused.

Question 19.
Write the scientific names of the causal organisms of elephantiasis and ringworm in humans. Mention the body parts affected by them. (CBSE DeLhi 2012)
Answer:

Name of disease	Causative organism	Organ affected
1. Elephantiasis	(i) Wuchereria bancrofti (ii) Wuchereria malayi Fungi namely	Genital organs, swelling of lower limbs.
2. Ringworm	(i) Microsporum (ii) Trichophyton (iii) Epidermophyton	Skin, nails and scalps.

Human Health and Disease Important Extra Questions Long Answer Type

Question 1.
(i) How and at what stage does Plasmodium enter a human body?
Answer:
Sporozoite stage enters human body aLong with saLiva of female anopheLes mosquito as ii
bites to suck bLood.

(ii) With the help of a flow chart only shows the stages of asexual reproduction in the life cycle of the parasite in the infected human.
Answer:
Asexual phases of the life history of plasmodium in the body of a human

(iii) Why does the victim show symptoms of high fever? (CBSE Delhi 2008, 2013)
Answer:
When the parasite attacks red blood cells, it leads to its rupture with the release of haemozoin, which is a toxin. As the haemozoin is released into blood, symptoms (high fever) of malaria appear.

Question 2.
What is Immune system? Mention the two types of the immune system. (CBSE Delhi 2011)
Answer:
The system which protects our body from pathogens and other foreign invaders is called the immune system. It is of two types.

1. Innate
2. Acquired

Innate immunity is non-specific and is present by birth. It includes physical barriers, physiological barriers, cellular and cytokinin barriers.

Acquired immunity is pathogen-specific and is obtained with experience. It is of two types- Humoral and cell-mediated

Question 3.
Distinguish between B-cells and T-cells.
Answer:
Differences between B-cells and T-cells:

B-cells	T-cells
1. They are produced in cells of bone marrow and remain there and later migrate to lymphoid tissues.	1. They are produced in cells of bone marrow and migrate to the thymus and differentiate under the influence of thymus.
2. These crafts produce plasma cells, once triggered off by the antigens.	2. These cells are responsible for recognising a specific antigen and attack it by releasing chemicals.
3. They are part of a humoral system.	3. They are part of the cell-mediated immune system.
4. They act against viruses and bacteria and do not react against transplants and cancer cells.	4. They act against pathogenic microorganism, organ transplants and cancer cells.
5. No inhibitory effect on the immune system.	5. Suppressor cells inhibit the immune system.

Question 4.
(i) Name the infective stage of Plasmodium which Anopheles mosquito takes in along with the blood meal from an infected human.
Answer:
Gametocytes

(ii) Why does the infection cause fever in humans?
Answer:
Due to the release of haemozoin toxin in the blood.

(iii) Give a flow chart of the part of the life cycle of this parasite passed in the insect. (CBSE (Delhi) 2008, 2011)
Answer:

Question 5.
(i) Why do the symptoms of malaria not appear immediately after the entry of sporozoites into the human body when bitten by female Anopheles? Explain.
Answer:
As the sporozoites enter the human body along with saliva of female anopheles mosquito, these parasites pass through hepatic schizogony in liver cells and erythrocytic schizogony in RBCs. Haemozoin present in unused cytoplasm of RBC is released, followed by the appearance of malarial symptoms. This period is also called the incubation period.

(ii) Give the scientific name of the malarial parasite that causes malignant malaria in humans. (CBSE 2009)
Answer:
Plasmodium falciparum. Causes malignant malaria in human.

Question 6.
Give the scientific name of the parasite that causes malignant malaria in humans. At what stage does this parasite enter the human body? Trace its life cycle in the human body. (CBSE 2009, 2012)
Answer:

• Plasmodium falciparum
• Sporozoites enter the human body along with saliva of the female anopheles mosquito.
• The life cycle of Plasmodium in the human body

Life Cycle of Plasmodium

Question 7.
A 17-year-old boy is suffering from high fever with profuse sweating and chills. Choose the correct option from the following diseases which explains these symptoms and rule out the rest with adequate reasons.
(i) Typhoid
Answer:
If the boy is suffering from typhoid, then he should have sustained high fever (39° to40°C), weakness, stomach pain, constipation and headache. So it cannot be typhoid.

(ii) Viral Fever
Answer:
If the boy is suffering from viral fever, he will suffer from high fever, joint pain, weakness and headache. So it cannot be a viral fever.

(iii) Malaria (CBSE Sample paper 2018-19)
Answer:
If the boy is suffering from malaria, he should have high fever recurring with profuse sweating every three to four days associated with chills and headache. There is a possibility that he is suffering from malaria because high fever associated with chills is possible with malaria.

Question 8.
Medically it is advised to all young mothers that breastfeeding is the best for their newborn babies. Do you agree? Give reasons in support of your answer. (CBSE 2018)
Answer:
Yes, I do agree with the fact that breastfeeding is the best for newborn babies. Mammary glands start producing milk at the end of pregnancy. The milk produced during the initial few days of lactation is called colostrum which contains several antibodies. It helps in developing resistance for newborn baby against diseases. It helps the baby fight off viruses and bacteria. Thus breast milk is packed with a disease-fighting substance that protects newborn babies from illness. Breast milk also naturally contains many of the vitamins and minerals that a newborn requires. It is easily digested as well. There is no constipation, diarrhoea and upset stomach.

Question 9.
Name a human disease, its causal organism, symptoms (any three) and vector, spread by intake of water and food contaminated by human faecal matter. (CBSE 2017)
Answer:

• Amoebic dysentery [Amoebiasis]
• Causal Organism: Entamoeba historlytica, protozoa.

Symptoms:

• Abdominal pain
• Constipation
• Cramps.

Vector: Housefly.

Question 10.
(i) Why is there a fear amongst the guardians that their adolescent wards may get trapped in drug/alcohol abuse?
Answer:
Reasons for alcohol abuse in adolescents:
(a) Curiosity for adventure, excitement and experiment
(b) Social pressure
(c) To escape from stress, depression and frustration
(d) To overcome hardships of life
(e) Unstable or unsupportive family structure, etc.

(ii) Explain ‘addiction’ and dependence’ in respect of drug/alcohol abuse in youth. (CBSE 2017)
Answer:
Addiction is the psychological attachment to certain effects such as euphoria and a temporary feeling of well-being, associated with drugs and alcohol. The addicted person cannot manage him/herself without drug or alcohol.

Dependence: Dependence is the tendency of the body to manifest a characteristic and unpleasant withdrawal syndrome on abrupt discontinuation of a regular dose of drug/alcohol.

Question 11.
What is the basic principle of vaccination? How do vaccines prevent microbial infections? Name the organism from which the hepatitis B vaccine is produced.
Or
Principle of vaccination is based on the property of “memory” of the immune system. Taking one suitable example, justify the statement. (CBSE Delhi 2019)
Answer:
Principle of vaccination is based on the property of ‘memory’ of the immune system.

In vaccination, a preparation of antigenic proteins of pathogens or inactivated/live but weakened pathogens is introduced into the body. The antigens generate a primary immune response by producing antibodies along with forming memory B-cells and T-cells. When the vaccinated person is attacked by the same pathogens, the existing memory B-cells and T-cells recognise the antigen and overwhelm the invaders with massive production of lymphocytes and antibodies. The hepatitis-B vaccine is produced from yeast.

Question 12.
Prior to a sports event blood and urine samples of sportspersons are collected for drug tests.
(i) Why is there a need to conduct such tests?
Answer:
Drugs are consumed by sportspersons to enhance their performance. It is necessary to test the blood and urine of sportspersons to analyse the presence of any performance-enhancing drug.

(ii) Name the drugs the authorities usually look for.
Answer:
Narcotic analgesics, anabolic steroids, diuretics.

(iii) Write the generic names of two plants from which these drugs are obtained. (CBSE Delhi 2016)
Answer:
(a) Cocaine is obtained from Erythroxylum coca
(b) Caffeine is obtained from Coffea arabica and narcotics from Papaver somniferum.

Question 13.
Explain the following terms:
(i) Benign tumour
Answer:
A benign tumour (Non-malignant tumour). Such tumours grow slowly but become quite large. It remains restricted to the place of origin and does not spread to other areas of the body. Most tumours are of this type and do not give rise to cancer.

(ii) Cancerous tumour
Answer:
A cancerous tumour (Malignant tumour): It begins as a small tumour growth at first, grows slowly in the starting and more rapidly later on. The tumour ultimately spreads to the neighbouring tissue like the roots of a tree. Later on, cancerous cells separate off from the original site and migrate through the blood to the other sites and they divide and redivide to form a secondary tumour.

(iii) Metastasis.
Answer:
Metastasis: The stage when the secondary tumour is formed and accumulated by repeated division, is called metastasis. This stage is fatal and causes death sooner or later.

Question 14.
Why cannabinoids are banned in sports and games?
Answer:
Cannabinoids are hallucinogenic chemicals obtained from leaves, resins and inflorescence of Hemp plant, Cannabis sativa. They are used by sportspersons to increase their athletic performance. Intake of cannabinoids results in rapid heartbeat decreased vital capacity of the lung. But their misuse is associated with a number of problems in both sportsmen and sportswomen, e.g. these cause masculinisation, increased aggressiveness, mood swings, abnormal menstrual cycles, enlarged clitoris in sportswomen, while their misuse in sportsmen is known to cause acne, mood swings, reduced testicular size, decreased spermatogenesis, enlarged breasts and prostate gland, dysfunctioning of liver and kidney, etc.

Question 15.
The outline structure of a drug is given below:
1. Which group of drugs does this represent?
2. What are the modes of consumption of these drugs?
3. Name the organ of the body which is affected by the consumption of this drug.

Answer:

1. Cannabinoids.
2. By smoking or oral ingestion.
3. Cannabinoids generally affect the cardiovascular system of the body.

Question 16.
What is cannabis? List its main derivatives.
Answer:
Cannabis: It is the most ancient drug and is obtained from hemp plants.
The following three kinds of drugs are obtained from these plants (Derivative of Cannabis indica):

1. Hashish or Charas is obtained from flowering tops of female plants.
2. Bhang is obtained from dry leaves.
3. Ganja is obtained from small leaves and bracts of inflorescence.

Marijuana is another drug obtained from Cannabis sativa. The common reaction of these drugs is relaxation, euphoria, laughing tendency and rise in blood sugar level.

Question 17.
Why is using tobacco in any form injurious to the health? Explain. (CBSE Delhi 2008, Outside Delhi 2011)
Answer:
Tobacco is injurious to health:

1. Nicotine present in tobacco is toxic and addictive. It causes coronary diseases.
2. Heat irritants and carcinogens cause mouth cancer and lung cancer.
3. Tobacco leads to male infertility.
4. in pregnant women, nicotine causes decreased foetal growth and development.
5. Tobacco addiction often leads to gastric and duodenal ulcers.
6. It is an expensive habit causing staining of teeth and fingers and making breath unpleasant.
7. Swelling of respiratory tract leads to chronic bronchitis.

Question 18.
Give reasons for the following:
(a) Antibody-mediated immunity is called humoral immunity.
Answer:
(a) Antibodies produced by plasma cells are present in the blood, the response is called humoral, immunity response. Thus it is termed humoral immunity.

(b) How is a child protected from a disease for which he/she is vaccinated?
Answer:
The principle of vaccination is based on the property of the ‘memory’ of the immune system. As during vaccination, antigens are introduced in the body. In response to antigens, antibodies are produced in the body against them. They neutralise the pathogen during actual infection.

(c) Name the type of cells the AIDS virus enters after getting into the human body. (CBSE Outside Delhi 2019)
Answer:
HIV enters macrophages. Simultaneously HIV enters T helper- Lymphocytes.

Question 19.
(a) Write the scientific names of the source plants from where opioids and cannabinoids are extracted.
Answer:
Opioids are obtained from the opium plant Papaver somniferous. Cannabinoids are extracted from Cannabis sativa.

(b) Write their receptor sites in the human body. How do these drugs affect human beings? (CBSE Outside Delhi 2019)
Answer:
Receptors of opioids are present on the central nervous system and gastrointestinal tract. They are a depressant and slow down the body functions. Receptors of cannabinoids are present in the brain. They affect the cardiovascular system of the body.

Question 20.
Briefly describe the life history of the malarial parasite.
Answer:

1. Malarial parasite (Plasmodium) completes its life cycle in two hosts., i.e. female anopheles mosquito and humans.
2. Sporozoites are the infective stage.
3. The sporozoites enter the human body, reach the liver through blood and multiply within the liver cells.
4. Such liver cells burst and release the parasites (Cryptomerozoites) into the blood.
5. Then they attack RBCs, multiply and cause their rupture.
6. The rupture of RBCs is associated with the release of a toxin called haemozoin, which is responsible for the high recurring fever and the chill/ shivering and causing malaria.
7. Sexual stages (gametocytes) develop in the red blood cells.
8. The parasite then enters the female Anopheles mosquito along with the blood when it bites the infected person.
9. Further development occurs in the stomach wall of the mosquito.
10. The gametes fuse to form a zygote. It takes the worm-like shape called ookinete as it pierces the wall of the stomach.
11. The zygote undergoes further development in the body of the mosquito to form sporozoites.
12. Sporozoites are transported to and stored in the salivary glands of mosquitoes and are transferred to a human body during the bite of the mosquito.
13. Female mosquito sucks human blood because it requires blood proteins for the development of its eggs.

Question 21.
Describe the effects of drug and alcohol abuse.
Answer:
Effects of drug/alcohol abuse:

1. The immediate effects of drugs/ alcohol abuse are manifested as reckless behaviour, vandalism and violence.
2. Excess doses can lead to coma and death due to cerebral haemorrhage, respiratory and heart failure.
3. A combination of drugs or their intake of alcohol leads to death.
4. The most common warning signals of drug/alcohol abuse include:
   (a) Drop in academic performance.
   (b) Lack of interest in personal hygiene.
   (c) Withdrawal and isolation from family and friends.
   (d) Aggressive and rebellious behaviour.
   (e) Lack of interest in hobbies.
   (f) Change in sleeping and eating habits.
   (g) Fluctuations in weight, etc.

Question 22.
Name the type of immunity that is present at the time of birth in humans. Explain any two ways by which it is accomplished. (CBSE 2008)
Answer:
Innate immunity: It is also called inborn or non-specific immunity. It is the first line of defence. It is composed of the following steps:
1. Anatomic barriers: The skin and mucous membranes secrete certain chemicals which dispose of pathogens. Specific cases of this defence are cited below: The oil and sweat secreted by sebaceous and sudoriferous glands contain lactic acid and fatty acids, which make the skin surface acidic (pH 3 to 5). This does not allow the microorganisms to establish.

2. Physiological barrier: Body temperature, pH and various body secretions like saliva prevent the growth of many pathogenic microorganisms. Pyrogens and interferons aid in fighting infections.

Question 23.
On what basis diagnosis of cancer is made?
Answer:
Diagnosis of cancer:

1. Blood and bone marrow tests are done for increased cell counts in case of leukaemia.
2. Histopathological study or biopsy: In a biopsy, a piece of the suspected tissue cut into thin sections is stained and examined under a microscope by a pathologist.
3. Radiography: X-rays are used to detect cancer of the internal organs
4. Computed tomography: It uses X-rays to generate a three-dimensional internal image of an object.
5. Resonance imaging: Non-ionising radiation and strong magnetic field are used in MRI to accurately detect pathological and physiological changes in the living tissue.
6. Monoclonal antibodies: Antibodies against cancer-specific antigens are also used for the detection of certain cancers.

Question 24.
Explain with the help of sketch the action of HIV in the body. (CBSE Delhi 2011)
Or
Name the cells HIV (Human Immuno Deficiency Virus) gains entry into after infecting the human body. Explain the events that occur in these cells. (CBSE Outside Delhi 2016)
Or
Trace the events that occur in the human body to cause immunodeficiency when HIV gains entry into the body. (CBSE 2011, 2014)
Answer:
The action of HIV in the body. After getting into the body of the person, the virus enters into macrophages where RNA genome of the virus replicates to form viral DNA with the help of the enzyme reverse transcriptase. This viral DNA gets incorporated into the host cell’s DNA and directs the infected cells to produce virus particles. The macrophages continue to produce virus and in this way acts as an HIV factory.

Simultaneously, HIV enters into helper (Th) T-lymphocytes (a type or subset of T-lymphocytes about which you have read above, in the immune system), replicates and produces progeny viruses. The progeny viruses released in blood attack other helper T-lymphocytes. This is repeated leading to a progressive decrease in the number of helper T-lymphocytes in the body of the infected person.

During this period, the person suffers from bouts of fever, diarrhoea and weight loss. Due to the decrease in the number of helper T-lymphocytes, the person starts suffering from an infection due to bacteria such as Mycobacterium, viruses, fungi and even parasite Toxoplasma. The patient becomes so much immunologically deficient and unable to fight against such infections.

The action of HIV in the body cells.

Question 25.
(a) If a patient is advised anti-retroviral drug, name the possible infection he/ she is likely to be suffering from. Name the causative organism.
Answer:
The person may be suffering from AIDS. It is caused by HIV (Human Immunodeficiency Virus).

(b) How do vaccines prevent subsequent microbial infection by the same pathogen?
Answer:
Antibody-provoking agents are called a vaccine. They prevent microbial infection by initiating production of antibodies to act against antigens to neutralise the pathogenic agents during later actual infection.
The vaccine also generates memory B-cells and T-cells that actually recognise the pathogen quickly in case of infection at later stages of life.

(c) How does a cancerous cell differ from a normal cell?
Answer:
Differences between the cancerous cell and normal cell:

Cancerous cells	Normal cells
1. These cells divide in an uncontrolled manner and rate of division is high.	1. These cells divide in a regulated and controlled manner.
2. These cells do not show contact inhibition.	2. These cells show contact inhibition.
3. These cells do not undergo differentiation.	3. These cells undergo programmed differentiation.
4. They have the ability to move in the body fluid and settle at different sites and divide thus show metastasis.	4. These cells don’t show metastasis.

(d) Many microbial pathogens enter the gut of humans along with food. Name the physiological barrier that protects the body from such pathogens. (CBSE Sample Paper 2020)
Answer:
The acid present in the stomach and saliva kills the microbial pathogens that enter along with food.

Question 26.
Write the source and the effect on the human body of the following drugs:
(i) Morphine
(ii) Cocaine
(iii) Marijuana (CBSE Delhi 2011)
Answer:

Name of drug	Source	Effects on the human body
1. Morphine	Opium plant	1. Depresses respiratory centre. 2. Causes fall in blood pressure. 3. Slow heartbeat. 4. Mild hyperglycaemia
2. Cocaine	Natural coca alkaloid obtained from Erythroxylon coca	1. It is a powerful stimulant of the central nervous system (CNS) 2. Increases heartbeat, blood pressure and body temperature.
3. Marijuana	Dried flowers and top leaves of female plant of Cannabis Sativa	1. It causes psychosis 2. Raises blood sugar and increases the frequency of urination

 

Here we are providing Class 12 Biology Important Extra Questions and Answers Chapter 5 Principles of Inheritance and Variation. Important Questions for Class 12 Biology are the best resource for students which helps in Class 12 board exams.

Class 12 Biology Chapter 5 Important Extra Questions Principles of Inheritance and Variation

Principles of Inheritance and Variation Important Extra Questions Very Short Answer Type

Question 1.
Name one trait that does not blend.
Answer:
Sex does not blend.

Question 2.
Give one example of a genetic trait for each of the following in humans:
1. Lethality
Answer:
Lethality: Homozygous sickle cell anaemia.

2. Multiple allelism.
Answer:
Example of Multiple allelism: ABO blood groups.

Question 3.
What for symbols AA and Aa stand?
Answer:

• AA: Homozygous dominant,
• Aa: Heterozygous dominant.

Question 4.
Name the plant that shows incomplete dominance in respect to the colour of its flower.
Answer:
Mirabilis jalapa.

Question 5.
Write the genotypes of a man with blood group A.
Answer:
l^Al^A, l^Al⁰

Question 6.
What is Mendel’s monohybrid ratio for phenotypes?
Answer:
3:1.

Question 7.
Write down Mendel’s dihybrid ratio for phenotypes.
Answer:
9: 3: 3: 1.

Question 8.
Who were the discoverers of Mendelism?
Answer:
Hugo de Vries, Karl Correns, Erich von Tschermak were the rediscoverers of Mendelism.

Question 9.
What are the real determinants of what an organism will become?
Answer:
The complex interaction between genes and their environment really determine what an organism will become.

Question 10.
Name the disorder in humans with the following karyotype:
(a) 22 pairs of autosomes + XO
(b) 22 pairs of autosomes + 21 st chromosome + XY (CBSE Outside Delhi 2019)
Answer:

Karyotype	Name of Disorder
(a) 22 pairs of autosomes + XO	Turner’s syndrome
(b) 22 pairs of autosomes + 21st chromosome + XY	Down’s syndrome (MongoLism inmate)

Question 11.
What wilt is the genetic makeup of an organism which suffers from sickle cell anaemia?
Answer:
Homozygous (Hb^S Hb^S).

Question 12.
Name the type of cross that would help to find the genotype of a pea plant bearing violet flowers. (CBSE Delhi 2017)
Answer:
Test cross.

Question 13.
What term is used for the two chromatids resulting from the interchange of segments during crossing over?
Answer:
Recombinants (cross overs).

Question 14.
Write one example of each of organisms exhibiting
(i) male heterogamety
Answer:
Male heterogamety: Drosophila and humans

(ii) female heterogamety. (CBS£ Delhi 2019 C)
Answer:
Female heterogamety: Birds and some reptiles

Question 15.
A geneticist is interested in study variations and pattern in living being preferred to choose an organism with the short life cycle. Provide a reason. (CBSE Delhi 2015)
Answer:
An organism with a shorter life cycle is helpful in rapid study and analysis of hereditary pattern in many generations, e.g. Drosophila, Neurospora.

Question 16.
Give an example of a human disorder that is caused due to a single gene mutation. (CBSE Delhi 2016)
Answer:
Phenylketonuria.

Question 17.
Write the sex of human having XXY chromosomes with 22 pairs of autosomes. Name the disorder this human suffers from. (CBSE Delhi 2016)
Answer:
The human is male (as Y chromosome is present). He is suffering from Klinefelter’s syndrome

Principles of Inheritance and Variation Important Extra Questions Short Answer Type

Question 1.
State the Mendelian principle which can be derived from a dihybrid cross and not from monohybrid cross.
Or
State Mendel’s Law of Independent Assortment. CBSE Sample Paper 2018-19)
Answer:
From the dihybrid cross, the law of independent assortment can be derived which states that, when two pairs of traits are combined in a hybrid, segregation of one pair of characters is independent of the other pair of characters.

Question 2.
In a cross between two tall pea plants, some of the offsprings produced pure dwarf. Show with the help of Punnett square how this is possible. (CBSE (Delhi) 2013)
Answer:

Question 3.
What is a dihybrid cross?
Answer:
Dihybrid cross. A cross in which two characters are taken into consideration during experimentation, such a cross is called dihybrid cross. A cross between a pea plant with yellow smooth seeds and a pea plant with green, wrinkled seeds is a dihybrid cross.

From the dihybrid cross, it can be derived that each gene is assorted independently of the other during its passage from one generation to the other or Law of independent assortment.

Question 4.
In order to obtain the E, generation, Mendel pollinated a pure breeding tall plant with a pure breeding dwarf plant. But forgetting the F₂ generation, he simply self-pollinated the tall F₁ plants. Why?
Answer:

1. He made crosses to study the pattern of inheritance of a few characters over generations.
2. Initially, he made pure lines.
3. To create a heterozygote or hybrid, he had to cross two different plants (pure lines).
4. To study the inheritance pattern; it is enough if the hybrids are self – pollinated, thus the segregation of factors can be studied.

Question 5.
A cross between a red flower-bearing plant and a white flower-bearing plant of Antirrhinum majus produced all plants having pink flowers. Work out across, to explain how is this possible? (CBSE Outside Delhi 2013)
Answer:

Question 6.
Differentiate gene and allele.
Answer:
Difference between gene and allele:
Allele (allelomorphs) refers to the alternate form of a gene pair present on the same loci in the homologous chromosome, whereas gene is the smallest unit of an organism capable of transmitting genetic information and expressing the same.

Question 7.
In Snapdragon, a cross between true-breeding red-flowered (RR) plants and true-breeding white-flowered (RR) plants showed a progeny of plants with all pink flowers.
(i) The appearance of pink flowers is not known as blending. Why?
Answer:
It is not a case of blending. In this case, R gene was not completely dominant over r gene and this made genotype Rr to distinguish as pink.

(ii) What is this phenomenon known as? (CBSE 2014)
Answer:
Incomplete dominance.

Question 8.
With the help of one example, explain the phenomena of co-dominance and multiple allelism in the human population. (CBSE 2014)
Answer:
In the case of co-dominance, two alleles for a trait are equally expressed.

Example: ABO blood groups are controlled by the gene I. The gene I have three alleles l^A, l^B and l^O. These alleles determine the type of sugar on the RBC surface. Alleles lA and lB are co-dominant and express the AB blood group.

Since there are three different alleles and express themselves on the basis of dominance recessiveness and co-dominance, it is a case of multiple allelism.

Question 9.
The child has a blood group of O. If the father has blood group A and mother has blood group B, work out the genotypes of the parents and the possible genotypes of the other offsprings. (CBSE Outside Delhi, 2015, 2019)
Answer:

1. Genotypes. Man (l^A l^O) Mother l^B l^O and child l^O l^O.
2. The blood group of the future offspring. A type, B type, 0 types and AB type. It is based on the following cross:

Inheritance of blood groups A, B, O, AB

Question 10.
Give examples of sex-linked inheritance in Drosophila. During his studies on genes in Drosophila that were sex-linked T.H. Morgan found F₂ population phenotypic ratios deviated from expected 9: 3: 3: 1. Explain the conclusion he arrived at. (CBSE 2010)
Answer:
Examples of sex-linked inheritance in Drosophila (Morgan’s conclusion).

1. Genes for white eye colour is located in the X-chromosome and Y-chromosome is empty carrying no normal allele for white eye colour.
2. The white-eyed female possesses a gene for white eye colour (W) on both of its X-chromosomes.
3. The white-eyed males receive X-chromosome with (W) gene from mother and (Y) from father with no gene.
4. The daughter receives one X-chromosome with (W) gene from mother and one X-chromosome with dominant (W+).

Question 11.
Briefly explain XX-XO (a type of sex determination).
Answer:
In the case of roundworms, true bugs, grasshoppers and cockroaches the females have two sex chromosomes XX, whereas the males have only one X-chromosome. The male has no second chromosome thus designated as XO. The sex ratio of 1: 1 is produced as shown in the figure below.

XX-XO determination of sex in the cockroach.

Question 12.
Explain the mechanism of ‘sex determination’ in birds. How does it differ from that of human beings? (CBSE Delhi 2018)
Answer:
Sex determination is of ZW-ZZ type in birds.
In this type, the males are homogametic and have ZZ sex chromosomes, and females are heterogametic with ZW pair of sex chromosomes.

In human beings, the chromosomal mechanism of sex determination is of XX- XY type. The human male is heterogametic and has XY sex chromosomes, whereas the human female is homogametic with XX sex chromosomes.

Question 13.
Write a note on ZO-ZZ type of sex determination.
Answer:
In case of ZO-ZZ type of sex determination the female produces two types of eggs. The one-half of eggs is with Z-chromosome and the other half without Z-chromosome. The male has homomorphic sex chromosomes and is homogametic. It forms only one kind of sperms each with Z-chromosome. On fertilisation by a sperm with Z-chromosome, the Z-containing egg gives rise to male offspring ZZ and Z- lacking egg produces female offspring ZO. Such type of sex determination is found in the case of butterflies and moths.

Question 14.
Give an example of an autosomal recessive trait in humans. Explain its pattern of inheritance with the help of a cross. (CBSE Delhi 2016)
Answer:
Autosomal recessive trait. Sickle-cell anaemia is caused by autosomal recessive trait. The disease is controlled by a single pair of alleles Hb^A and Hb^S. Only the homozygous individuals for HbsHbs show the disease. The heterozygous individuals are carriers (Hb^A Hb^s)

Question 15.
How would you find the genotype of a tall pea plant bearing white flowers? Explain with the help of a cross. Name the types of the cross you would use. (CBSE Delhi 201i 5)
Answer:
In order to find the genotype of a given plant, one has to breed it with plenty of oic recessive individual. It is called test CRC is. Tall and white plant TTww or Ttww / is crossed with dwarf white plant ttww TTww x ttww

(All tall white plants indicate that the genotype is TTww)

This ratio indicates that the genotype is Ttww

Depending on the resuLt we can determine the genotype of the flower.

Question 16.
How is polygenic inheritance different from pleiotropy? Give one example of each.
Answer:

polygenic inheritance	pleiotropy
It is a type of inheritance in which. a single dominant gene contributes a part of the trait. Thus dominant alleles have a cumulative effect.	It is a condition in which a single gene influence more than one trait.
Example: 1. Cob length in maize 2. Skin colour in human	Example: 1. In Drosophila, white eye mutant causes depigmentation in many parts of the body. 2. Sickle cell anaemia

Question 17.
Why is it not possible to study the pattern of inheritance of traits in human beings, the same way as it is done in pea plant? Name the alternate method employed for such an analysis of human traits.
Answer:

1. Control cross cannot be performed in human as in other organisms.
2. The generation time is long. Pedigree analysis is the alternative method to study the inheritance of human traits in several generations.

Question 18.
A man with blood group A married a woman with a B group. They have a son with AB blood group and a daughter with blood group 0. Work out the cross and show the possibility of such inheritance. (CBSE Delhi and Outside Delhi 2008)
Answer:

Blood groups of progeny A, B, AB and O.

Question 19.
A haemophilic father can never pass the gene for haemophilia to his son. Explain. (CBSE Delhi 2016)
Answer:
Haemophilia is a sex-linked recessive disorder where X chromosome carries the defective haemophilic gene and Y chromosome is healthy. And son inherits only the Y chromosome from his father which is not carrying the gene for haemophilia. Therefore, the haemophilic father can never pass haemophilia to his son.

Principles of Inheritance and Variation Important Extra Questions Long Answer Type

Question 1.
Study the given pedigree chart and answer the questions that follow:
1. Is the trait recessive or dominant?
Answer:
Dominant.

2. Is the trait sex-linked or autosomal?
Answer:
Autosomal.

3. Give the genotypes of the parents shown in generation I and their third child is shown in generation II and the first grandchild shown in generation III. (CBSE Sample Paper 2018-19)

Answer:
The genotype of parents in generation I – Female: aa and Male: Aa
The genotype of a third child in generation II-Aa Genotype of the first grandchild in generation III – Aa

Question 2.
Haemophilia is a sex-linked recessive disorder of humans. The pedigree chart given below shows the inheritance of Haemophilia in one family.

Study the pattern of inheritance and answer the questions given.
1.Give all the possible genotypes of the members 4, 5 and 6 in the pedigree chart.
2. A blood test shows that the individual 14 is a carrier of haemophilia. The member numbered 15 has recently married the member numbered 14. What is the probability that their first child will be a haemophilic male?
(CBSE 2009, CBSE Sample Paper 2018-19)

Answer:

1. Genotypes of member 4 – XX or XX^h Genotypes of member 5 – X^hY Genotypes of member 6 – XY
2. The probability of first child to be a haemophilic male is 50%.

Question 3.
Mention the advantages of selecting a pea plant for the experiment by Mendel.
Answer:
Advantages of selecting pea plant as experimental material:

Mendel selected pea plant (Pisum sativum) because:

1. Many varieties were available with observable alternative forms for a trait or a characteristic.
2. Peas normally self-pollinate; as their corolla completely encloses the reproductive organs until pollination is complete.
3. It was easily available.
4. It has pure lines for experimental purpose, i.e. they always breed true.
5. It has contrasting characters. The traits were seed colour, pod colour, pod shape, flower shape, the position of flower, seed shape and plant height.
6. Its life cycle was short and produced a large number of offsprings.
7. The plant can be grown easily and does not require care except at the time of pollination.

Question 4.
Differentiate between the following:
(i) Dominance and recessiveness
Answer:
Differences between dominance and recessiveness:

Dominance	Recessiveness
(1) Dominant gene or factor is able to express itself even in the presence of its recessive allele.	(1) Recessive gene or factor is unable to express itself in the presence of dominant allele.
(2) It expresses itself because it forms complete polypeptide or enzyme for expressing its effect.	(2) The recessive gene forms an incomplete or defective polypeptide or enzyme thus fails to express its effect.

(ii) Homozygous and heterozygous
Answer:
Differences between homozygous individual and a heterozygous individual:

homozygous individual	heterozygous individual
(1) In a homozygous individual (homo-zygote) the two genes for a particular character are identical (TT) or (tt).	(1) The heterozygous individual (hetero zygote) possesses contrasting genes of a pair çrt).
(2) They form identical gametes for a particular character.	(2) They form dissimilar gametes for a particular character.
(3) They breed true for a specific trait.	(3) They do not breed true.

Question 5.
Explain the law of dominance using a monohybrid cross.
Answer:
Law of Dominance. According to this law, when two factors of a character are unlike, one of them will manifest in the body and is called dominant while the other remains hidden and is termed recessive factor.

The law can be well explained by the monohybrid cross by studying the following crosses:
(i) Pure tall = TT, Hybrid tall = Tt

Gametes of TT parent = \(\frac { 1 }{ 2 }\)T + \(\frac { 1 }{ 2 }\)T
Gametes of Tt parent = \(\frac { 1 }{ 2 }\)T + \(\frac { 1 }{ 2 }\)t

The 50% are pure tall and 50% hybrid tall. Then pure tall plants will produce 100% tall in F₂ generation and hybrid plants will produce in the ratio of 1: 2: 1 in the F2 generation.

(ii) When the cross is made between pure tall and pure dwarf, we get results as follows (Fig.).

A Punnett square used to understand a typical monohybrid cross conducted by Mendes between true-breeding tall plants and true-breeding dwarf plants

Question 6.
Define and design a test cross.
Answer:
Test Cross: It is a cross between an organism of an unknown genotype and a homozygous recessive organism.

Results of a Test Cross: If the test cross yields offspring of which 50% show the dominant character and 50% show the recessive character, i.e. F1 ratio is 1: 1, the individual under test is heterozygous (see fig.). This is so because the individual showing the recessive trait (say white coat colour in the guinea pig, dwarf size in pea plant) must have received one recessive allele (b in a guinea pig, t in pea plant) from each parent.

Genetics of a test cross

If all the offspring of the test cross show the dominant trait, the individual being tested is homozygous dominant with genotype BB for a guinea pig and TT for pea plant (fig.).

Question 7.
When a cross is made between a tall plant with yellow seeds (TtYy) and tall plant with green seeds (Ttyy), what proportions of phenotype In the offspring could be expected to be
(i) tall and green
(ii) dwarf and green?
Answer:

Cross between a tall plant with yellow seeds

Question 8.
Differentiate back cross and test cross.
Answer:
Differences between the back cross and test cross:

Back cross	Test cross
1. It is the hybridisation process between F1 progeny and one of its parents.	1. It represents the hybridisation between F1 progeny with the homozygous recessive parent.
2. The ratio of F2 progeny maybe 1: 1.	2. The ratio of F2 progeny is 1:1.
3. It is used to develop a good hybrid particularly through recurring back cross.	3. It is used to testify the genotype of the F1 progeny.

Question 9.
Differentiate incomplete dominance and codominance.
or
Explain the following terms with an example:
(i) Codominance
(ii) Incomplete dominance (CBSE Outside Delhi 2008; Delhi 2011, 2019)
Answer:
Differences between incomplete dominance and codominance:

Incomplete Dominance	Codominance
(1) Effect of one of the two alleles is more prominent.	(1) The effect of both the alleles is equally prominent.
(2) It produces a fine mixture of the expression of two alleles.	(2) 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. pink coloured snapdragon obtained as a result of cross-pollination between red and white snapdragon flowers.	(3) Both the alleles produce their effect independently, e.g. IA and lB, HbS and HbA.

Question 10.
(i) Why is the human ABO blood group gene considered a good example of multiple alleles?
Answer:
A B O blood groups are controlled by a single gene: (I) The plasma membrane of the red blood cells has sugar polymers that protrude from its surface and is controlled by the gene. The gene (I) has three alleles l^A, l^Band l^O. Presence of more than two types of alleles at the same locus governing the same character is called multiple alleles.

(ii) Work out across up to F₂ generation only, between a mother with blood group A (Homozygous) and the father with blood group B (Homozygous). Explain the pattern of inheritance exhibited. (CBSE (Delhi) 2013)
Or
Describe the mechanism of a pattern of inheritance of ABO blood groups in human. (CBSE 2011)
Answer:
Patterns of inheritance of the ABO blood group.

F₁ generation. All with blood group AB. It is a case of co-dominance.

Question 11.
(i) What is polygenic inheritance? Explain with the help of a suitable example, (ii) How are pleiotropy and Mendelian pattern of inheritance different from the polygenic pattern of inheritance? (CBSE Outside Delhi 2016)
Or
Certain phenotypes in the human population are spread over a gradient and reflect the contribution of more than two genes. Mention the term used for the type of inheritance. Describe it with the help of an example in the human population. (CBSE Sample Paper 2019-20)
Answer:
(i) Polygenic inheritance. It is a type of inheritance controlled by three or more genes in which the dominant alleles have a cumulative effect. Each dominant allele expresses a part or unit of a trait. It is also called quantitative inheritance or multiple factor inheritance. The genes involved in such kind of inheritance are termed polygenes.

Examples:

1. Kernel colour in wheat
2. Cob length in maize
3. Skin colour in human
4. Human intelligence Human Skin Colour. It is caused by the pigment melanin.

The quantity of melanin is due to three pairs of polygenes (A, B and C). The genotype of black will be (AA BB CC) and white will have (aa bb cc). Marriage between two such persons will show variations. In progeny (Aa Bb CC) there will be 7 types of phenotypes i.e. very dark, 6 dark, 15 fairly dark, 20 intermediates, 15 family light, 6 light and have very light.

(ii) How are pleiotropy and Mendelian pattern of inheritance different from the polygenic pattern of inheritance? (CBSE Outside Delhi 2016)
Answer:

1. Pleiotropy and Mendelian pattern of inheritance: In the case of pleiotropy one gene has an effect on two on more traits. One effect is more evident in the case of one trait (major effect) and less evident in the case of others (secondary effect). The mendelian pattern of inheritance is monogenic.

2. In pleiotropism and monogenic inheritance no intermediates are produced and show discontinuous variations in the expression of a trait. Intermediates are quite common in polygenic inheritance and produce continuous variations in the expression of a trait.

Question 12.
(i) Write the scientific name of the organism Thomas Hunt Morgan and his colleagues worked with for their experiments. Explain the correlation between linkage and recombination with respect to genes as studied by them.
Answer:
Thomas Hunt Morgan and his colleagues worked with Drosophila melanogaster. They carried out several dihybrid crosses in Drosophila to study gens that were sex-linked.

Morgan and his group knew that the genes were located on the X chromosome and noticed that when the two genes in a dihybrid cross were situated on the same chromosome, the proportion of parental gene combinations were much higher than the non-parental type.

Morgan attributed this due to the physical association or linkage of th< two genes. He coined the term links to describe this physical associate of genes on a chromosome and t/ term recombination to describe t generation of non-parental age combination. Morgan and his also found that even when genes w grouped on the same chromos or some genes were very tightly linked, they showed very low recombine while others were loosely linked.

(ii) How did Sturtevant explain gene mapping while working with Morgan? (CBSE Delhi 2018)
Answer:
Alfred Sturtevant was Morgan’s student. He used the frequency of recombination between gene pairs on the same chromosome as a measure of the distance between genes and ‘mapped’ their position on the chromosome. Today genetic mappings are extensively used as a starting point in the sequencing of the whole genomes.

Question 13.
What is recombination? Discuss the applications of recombination from the point of view of genetic engineering.
Answer:
Recombination refers to the generation of a new combination of genes which is different from the parental types. It is produced due to crossing over that occurs during meiosis prior to gamete formation.

Applications of recombination:

1. It is a means of introducing new combinations of genes and hence new traits.
2. It increases variability which is useful for natural selection under changing environment.
3. It is used for preparing linkage chromosome maps.
4. It has proved that genes lie in a linear fashion in the chromosome.
5. Breeders have to select small or large population for obtaining the required cross-overs. For obtaining cross-overs between closely linked genes, a very large population is required.
6. Useful recombinations produced by crossing over are picked up by breeders to produce useful new varieties of crop plants and animals. Green revolution and white revolution were implemented using the selective recombination technique.

Question 14.
How is sex determined? (CBSE Delhi 2015)
Answer:
Determination of the sex of the child. Sex chromosomes determine sex in human beings. In males, there are 44+XY chromosomes, whereas in female there are 44+XX chromosomes. Here X and Y chromosomes determine sex in human beings.

Two types of gametes are formed in male, one type is having 50% X-chromosome, whereas another type is having Y-chromosome. In a female, gametes are of one type and contain X-chromosome. Thus females are homogametic and males are heterogametic. If male gamete having Y-chromosome (endosperm) undergoes fusion with female gamete having X-chromosome, the zygote will have XY chromosome and this gives rise to a male child.

If male gamete having X-chromosome (gymnosperm) undergoes fusion with female gamete having X-chromosome, the zygote will be having XX-chromosome and this gives rise to the female child.

Genetics of sex in human beings. The letter A represents autosomes.

Question 15.
Both haemophilia and thalassemia are blood-related disorders in human. Write their causes and the difference between the two. Name the category of genetic disorder they both come under. (CBSE Delhi 2017)
Answer:
Both haemophilia and thalassemia are Mendelian disorders:

• Haemophilia is a sex-linked recessive disorder. The gene for haemophilia is located on X-chromosome. The gene passes from a carrier female to her son.
• Thalassemia is an autosomal-linked recessive disease.
• It occurs due to either mutation or deletion resulting in the reduced rate of synthesis of one of the globin chains of haemoglobin.
• Difference between Haemophilia and Thalassemia. In haemophilia, clotting is affected, i.e. there can be a non-stop bleeding even after a minor cut.
• In Thalassemia anaemia is the characteristic of this disease. It is caused by faulty haemoglobin synthesis.

Question 16.
Differentiate male and female heterogamety. (CBSE Delhi 2015, 2019 C)
Answer:
Differences between male and female heterogamety:

Male heterogamety	female heterogamety
(1) Mate heterogamety refers to the phenomenon, where mates produce two (more than one) types of sperms.	(1) Female heterogamety refers to the phenomenon, where females produce two (more than one) types of ova.
(2) Sex of the individual is determined by the type of sperm fertilising the ovum.	(2) Sex of the individual is determined by the type of ovum that is fertilisers.
(3) XX female XY male	(3) ZZ male and ZW female.

Question 17.
Why Drosophila has been used extensively for genetical studies? (CBSE 2014, 2019 C)
Answer:
Advantages of using Drosophila as genetic material:
Drosophila is a very useful organism for genetical experiments because:

1. A very large number of offsprings are produced after each mating.
2. It can be cultured in large number in laboratory and animals can be easily examined under a hand lens.
3. Its life cycle is very short and is completed in 10-12 days. A new generation can be obtained every two weeks.
4. It has four pairs of chromosomes all different in size and easily distinguishable.
5. They produce numerous variants.
6. It has heteromorphic (XY) chromosomes in the male.
7. Female Drosophila flies can be easily differentiated from the males by the large body size and presence of ovipositor in the abdomen.

Question 18.
Mendel published his work on the inheritance of characters in 1865, but it remained unrecognised till 1900. Give three reasons for the delay in accepting his work. (CBSE Delhi 2014)
Answer:
Mendel’s work published as “Experiments on plant hybridisation” remained unnoticed and unappreciated for some 34 years due to:

1. Limited circulation of the “Proceedings of Brunn Natural Science Society” in which it was published.
2. He could not convince himself about his conclusions being universal since Mendel failed to reproduce the results on Hawkweed (Hieracium) undertaken on the suggestion of Naegeli. It was due to non-availability of pure lines.
3. Absence of aggressiveness in his personality.
4. The scientific world was being rocket^ at that time by Darwin’s theory c}f Natural Selection.

Question 19.
Compare in any three ways the chromosomal theory of inheritance as proposed by Sutton and Boveri with that of experimental results on pea pie int presented by Mendel. (CBSE Delhi 2019)
Answer:

Sutton and Boveri	Mendel
(1) Chromosomes occur in pairs.	(1) Factors occur in pairs.
(2) Chromosomes segregate during gamete formation such that only one of each pair is transmitted to a gamete.	(2) Factors segregate during gamete formation stage and only one of each pair is transmitted to a gamete.
(3) Independent pairs of chromosomes segregate independently of each other.	(3) One pair of factors segregate independently of another pair.

Question 20.
(a) Explain linkage and recombination as put forth by T.H. Morgan based on his observations with Drosophila melanogaster crossing experiment.
(b) Write the basis on which Alfred Sturtevant explained gene mapping. (CBSE Delhi 2019)
Answer:
(a) Linkage and recombination:

• Morgan is called the father of experimental genetics.
• Morgan used Drosophila for experiments of genetics.
• Linkage: It is the phenomenon of certain genes staying together during inheritance through several generations without any change or separation of these being present on the same chromosome. The two genes do not segregate independently of each other. So, F₂ generation deviates significantly from 9:3:3:1.
• Recombination: Loosely linked genes show a higher frequency of recombinant frequency which is around 37.2%. Tightly linked genes tend to show fewer recombinant frequency which is around 1.3%.

(b) Morgan’s student Alfred Sturtevant used the frequency of recombination between gene pairs on the same chromosome as a measure of the distance between genes and mapped their position on a chromosome.

Question 21.
ExplaIn how a test cross can be conducted to distinguish between a homozygous and heterozygous dominant genotype. What is the test cross? How can it decipher heterozygosity of a plant? (CBSE Delhi 2016)
Or
How will you find out whether a given plant is homozygous dominant? (CBSE 2008)
Or
You are given a tall pea plant and asked to find its genotype. How will you find it? (CBSE Outside Delhi 2019)
Answer:
Test Cross. When an individual is crossed to recessive parent it is called a test cross. The results can be easily analysed. If you follow the monohybrid cross where the FT is test crossed, a ratio of 1:1 will be obtained. On the same basis, you can work out that in a dihybrid case, the test cross ratio will be 1:1:1:1. Test cross can also be used for another purpose.

You must have understood by now that the homo and heterozygous genotypes for a dominant trait cannot be differentiated because they show the same phenotype. If we put them through a test cross, you will see that all homozygous dominant combinations will breed true but heterozygous genotypes will follow the segregation.

A test cross can be conducted to differentiate between a homozygous and heterozygous dominant genotype.

Question 22.
Explain the law of independent assortment with a dihybrid cross. (CBSE Outside Delhi, 2013, 2014)
Answer:
Law of independent assortment: According to this law, the factors of different pairs of contrasting characters do not influence each other. They are independent of one another in their assortment to form a new combination during gamete formation. Dihybrid cross. A cross in which two characters are taken into consideration during experimentation, such a cross is called dihybrid cross.

A cross between a pea plant with yellow smooth and a pea plant with green, wrinkled seeds is considered. Explanation. When a cross is made between pea plant having yellow smooth seeds (YYSS) and a pea plant with green wrinkled seeds (yyss). At the time of cross-pollination, yellow smooth (YYSS) produces gametes with genes (YS) and green wrinkled will produce gametes with gene (ys). Gametes unite at random. The seeds obtained when placed in the soil will grow to form plants and produce seeds which are yellow smooth (YySs) because yellow and smooth characters are dominant over green and wrinkled. These are called plants of F₁ generation.

When plants of F₁ generation are allowed to self-pollinate gametes formed YS, Ys, yS and ys by meiosis, they unite at random forming seeds. The plants thus obtained are called F₂ generation. They are Yellow smooth (YYSS, YySS, YySs, YYSs); yellow wrinkled (YYss, Yyss), green smooth (yySS, yySs) and green wrinkled (yyss) in the ratio of 9: 3: 3: 1. The result of a dihybrid cross can be shown in Fig. on the chequerboard.

Result of a dihybrid cross.

From the above dihybrid cross, it can be derived that each gene is assorted independently of the other during its passage from one generation to the other or law of independent assortment is justified.

Question 23.
In four o’clock plants, red colour (R) is incompletely dominant over white (r), the heterozygous having pink colour. What will be the offspring in a cross between a red flower and a pink flower? (CBSE Outside Delhi, 2013)
Answer:

1. In the monohybrid cross, red is incompletely dominant over white.
2. Red flowered plants have genotype RR and white-flowered plants have genotype rr.
3. Pink flowers have a genotype Rr.
4. Red flowering plants will form gametes with R genes and pink flowers will produce two types of gametes with R gene and r gene.
5. Arrangement of gametes in chequerboard.

Question 24.
Explain the pattern of inheritance of haemophilia in humans. Why is the possibility of a human female becoming a haemophilic extremely rare? Explain. (CBSE Delhi, 2008; Outside Delhi 2011)
Or
Why is human female rarely haemophilic? Explain how do haemophilic patients suffer? (CBSE Outside Delhi 2013)
Answer:
A pattern of inheritance of haemophilia:

1. It is the sex-linked recessive trait which is known as bleeder’s disease because the exposed blood does not readily clot due to deficiency of plasma thromboplastin (haemophilia B/ Christmas disease) or Antihaemophilia globulin (haemophilia A)
2. The defect has been inherited in the family of British Crown through Queen Victoria.
3. In females, haemophilia appears when both the sex chromosomes carry its recessive gene, X^h X^h. Such females die before birth.
4. A woman having a single allele of the trait appears normal but is a carrier of the disease XX^h
   
5. For sex-linked genes, human males are hemizygous. Therefore, X^h Y is haemophilic.
6. Marriage between haemophilic male and carrier female produces haemophilic sons (X^hY, 50%), normal sons (XY, 50%), carrier daughters (XX^h, 50%) and haemophilic daughters (X^hX^h, 50%, die before birth).
7. Haemophilic man (X^hY) and normal woman (XX) produce carrier girls (XX^h) and normal boys (XY).
   
8. Marriage between carrier woman and normal man produce 50% carrier girls (XX^h), 50% normal girls (XX), 50% normal boys (XY) and 50% haemophilic boys (X^hY).
   
   Sons- 50% normaL 50% heamophitic
   Daughters- 50% normaL 50% camer
9. The possibility of a female becoming haemophilic is very rare because the mother of such a female has to be at least a carrier and father should be haemophilic.

Question 25.
A colourblind child is born to a normal couple. Work out a cross to show how is it possible. Mention the sex of this child. (CBSE Delhi 2014, 2016)
Answer:
(a) Colourblindness is an X-linked recessive disease

So, the sex of the child is male.

Question 26.
1. How does a chromosomal disorder differ from a Mendelian disorder?
2. Name any two chromosomal aberra¬tion associated disorders.
3. List the characteristics of the disorders mentioned above that help in their diagnosis. (CBSE 2010)
Or
How does gain or loss of chromosome(s) take place in humans? Describe one example each of chromosomal disorder along with the symptoms involving an autosome and a sex chromosome. (CBSE Sample Paper 2019-20)
Answer:
1. Mendelian disorders are mainly determined by alteration or mutation in a single gene. These disorders are transmitted to the offspring on the basis of Mendelian inheritance, e.g. haemophilia, sickle cell anaemia. Chromosomal disorders are caused due to absence or excess or abnormal arrangement of one or more chromosomes. They are caused due to failure of segregation of chromatids during cell division or due to polyploidy. e.g. Down’s syndrome, Klinefelter syndrome.

2. Chromosomal aberration associated disorders.
(a) Down’s syndrome
(b) Klinefelter syndrome.

3. (a) Down’s syndrome. It is caused due to an additional copy of chromosome number 21 (Trisomy).
Symptoms: Short statured body, small rounded head, furrowed tongue, partially open mouth.

(b) Klinefelter syndrome. It is caused due to an additional copy of X-chromosomes (47 chromosome XXY).
Symptoms. Overall masculine development but the development of breast also occurs. These individuals are sterile.

Question 27.
A true-breeding pea plant, homozygous for inflated green pods (FFGG) is crossed with another pea plant with constricted yellow pods (ffgg). What would be the phenotype and genotype  F₁ and F₂ genotype? Give the phenotype ratio of F₂ generation. (CBSE Delhi 2008)
Answer:

Question 28.
A true-breeding pea plant homozygous for axial violet flowers (AAW) crossed with another pea plant with terminal white flowers (aaw).
(i) What would be phenotype and genotype of F₁ and F₂ generations
Answer:

(ii) Give the phenotype ratio of F₂ generations. (CBSE Delhi. 2008)
Answer:

Question 29.
A child suffering from Thalassemia is born to a normal couple. But the mother is being blamed by the family for delivering a sick baby.
(i) What is Thalassemia?
Answer:
Thalassemia: It is an autosomal recessive blood disease that appears in children of two unaffected carriers, heterozygote parents. The defect occurs due to mutation or deletion of the genes controlling the formation of globin chain (commonly a and P) of haemoglobin. Imbalanced synthesis of globin chains of haemoglobin causes anaemia. Thalassemia is of three types a, p, and 8.

(ii) How would you counsel the family not to blame the mother for delivering a child suffering from this disease? Explain.
Answer:
I would explain to the people around that this disease can be caused due to the presence of a defective gene in both the parents or it may be caused due to certain changes with the genetic setup.

(iii) List the values your counselling can propagate in the families. (CBSE Delhi 2013)
Answer:
People had a good understanding and had to realize the situation. They become supportive and made joint efforts to help the patients.

Question 30.
In a dihybrid cross, white-eyed, yellow-bodied female Drosophila was crossed with red-eyed, brown-bodied male Drosophila. The cross produced 1.3 per cent recombinants and 98.7 progeny with parental type combinations in the F2 generation. Analyse the above observation and compare with the Mendelian dihybrid cross. (CBSE Sample Paper 2018-19)
Answer:
Morgan observed that the two genes did not segregate independently of each other and the F2 ratio deviated vary significantly from the 9:3:3:1 ratio.

He attributed this to physical association or linkage of two genes and coined the term linkage and the term recombination to describe the generation of non-parental gene combinations.

Morgan and his group found that even when the genes are grouped on the same chromosome, some genes are very tightly linked (show very low recombination) while others were loosely linked (showed higher recombination). in the Mendelian dihybrid cross, the phenotypes round, yellow; wrinkled, yellow; round, green and wrinkled, green appeared in the ratio 9:3:3:1.

Wrinkled, yellow and round, green is possible because the distance between two genes is more. Therefore, recombination of parental type is possible.

Question 31.
Aneuploidy of chromosomes in human beings results in certain disorders. Draw out the possibilities of the karyotype in common disorders of this kind in human beings and its consequences in individuals. (CBSE Sample Paper 2018-19)
Or
A doctor after conducting certain tests on a pregnant woman advised her to undergo M.T.P., as the foetus she was carrying showed trisomy of 21st chromosome.
(a) State the cause of trisomy of the 21 st chromosome.
Answer:
(a) Down’s syndrome, Turner’s syndrome, Klinefelter’s syndrome are common examples of Aneuploidy of chromosomes in human beings.

• Down’s syndrome results in the gain of the extra copy of chromosome 21- trisomy.
• Turner’s syndrome results due to the loss of an X chromosome in human females- XO monosomy.
• Klinefelter’s syndrome is caused due to the presence of an additional copy of X- chromosome resulting in XXY condition.

(b) Why was the pregnant woman advised to undergo M.T.P. and not to complete the full term of her pregnancy? Explain. (CBSE Delhi 2019 C)
Answer:
Down’s Syndrome: The affected individual is

• short statured with small round head furrowed tongue and partially open mouth
• Palm is broad with characteristic palm crease
• Physical, psychomotor, and mental development is retarded.

Klinefelter’s Syndrome: The affected individual is

• a male with development of breast, i.e. Gynecomastia
• Such individuals are sterile.

Turner’s Syndrome: The affected individual shows the following characters:

• Females are sterile as ovaries are rudimentary
• lack of other secondary sexual characters

Very Important Figures:

XX-XO determination of sex in the cockroach.

Formation of recombinant as well as non-recombinant (parental type) gametes.
Forms of chromosomal mutations.

Answer:

S. No Blood groups of  parents	Blood groups possible in children	Blood groups not possible in children
1. O × O	O	A, B, AB
2. O × A	O, A	B, AB
3. O × B	O, B	A, AB
4. O × AB	A, B	O, AB
5. A × A	O, A	B, AB
6. A ×  B	O, A, B, AB	None
7. A  ×  AB	A, B, AB	O
8. B ×  B	O, B	A, AB
9. B  ×  AB	A, B, AB	O
10. AB  × AB	A, B, AB	O