2nd PUC

You can Download Chapter 15 Biodiversity and Conservation Questions and Answers, 2nd PUC Biology Question Bank with Answers, Karnataka State Board Solutions help you to revise complete Syllabus and score more marks in your examinations.

Karnataka 2nd PUC Biology Question Bank Chapter 15 Biodiversity and Conservation

2nd PUC Biology Biodiversity and Conservation NCERT Text Book Questions and Answers

Question 1.
Name the three important components of biodiversity.
Answer:
Genetic diversity, species diversity, and ecological diversity.

Question 2.
How do ecologists estimate the total number of species present in the world?
Answer:
Ecologists make a significant comparison of species richness of exhaustively studied groups of insects of the temperate and tropical regions and extrapolate this ratio to other groups of animals and plants to calculate a gross estimate of the total number of species existing on the earth.

Question 3.
Give three hypotheses for explaining why tropics show the greatest levels of species richness.
Answer:
a. Tropical regions have remained relatively undisturbed for millions of years and thus had a long time for species diversification.
b. Tropical environments are less seasonal, relatively more constant, and predictable.
c. Availability of more solar energy and contributes much productivity.

Question 4.
What is the significance of the slope of regression in a species-area relationship?
Answer:
When analysis of species-area relationships is done among small areas, the values of the slope of regression are remarkably similar regardless of the taxonomic group or the region. However, when such analysis is done among very large areas, i.e., continents, then the slope of regression would be much steeper. Biodiversity also changes with the change in altitude. It increases from higher to lower altitudes.

Question 5.
What are the major causes of species losses in a geographical region?
Answer:
Causes of biodiversity losses:
The extinction of species and consequent loss of biodiversity is due to mainly four major reasons. The Evil Quartet is the sobriquet used to describe them.

Question 6.
How is biodiversity important for ecosystem functioning?
Answer:
Biodiversity is useful in ecosystem services. Maintenance and sustainable utilisation of useful products and services of various ecosystems as well as individual species require the presence of biodiversity. The rich biodiversity is important for stability, productivity, resilience and ecosystem health. Increased biodiversity contributes to higher productivity. Forest and oceanic systems control climate and maintain gaseous composition of the atmosphere through photosynthesis. Biodiversity is essential for natural pest control, maintenance of populations of various species, pollination by insects and birds, nutrient cycling, conservation and purification of water, formation and protection of soil, etc.

Question 7.
What are sacred groves? What is their role in conservation?
Answer:
Sacred groves are forest patches protected by several tribal communities and religious groups due to the religious sanctity of the forest. Such sacred groves are found in Khasi and Jaintia hills in Meghalaya, Aravalli hills of Rajasthan, Western Ghat regions of Karnataka and Maharashtra. In Meghalaya, the sacred groves are the last refuge for a large number of rare and threatened plants.

Question 8.
Among the ecosystem services are control of floods and soil erosion. How is this achieved by the biotic components of the ecosystem?
Answer:
Biotic components of the ecosystem include plants and animals. Floods and soil erosion are prevented by the producers especially trees present in the ecosystem. The roots of trees bind to the soil firmly thus preventing soil erosion by wind or water. Raindrops in a forest do not hit the floor directly. The canopy layer of the forest intercepts the flow of raindrops so that the rainwater falls on the leaves of trees and then drips slowly onto the forest floor. Thus, water does not collect and stagnate on the forest floor. This prevents flooding. Moreover, trees in the forest also regulate the water cycle.

Question 9.
The species diversity of plants (22%) is much less than that of animals (72%). Give explanation.
Answer:
The great structural and resource heterogeneity provided by plants is the principal reason of high animal diversity. A single plant species can accommodate many animal species.

Question 10.
Can you think of a situation where we deliberately want to make a species extinct? How would you justify it?
Answer:
Certain species are the causal organisms of a number of human diseases which are epidemic. Mostly many disease-causing viruses like Varicella virus, Poliovirus, HIV are harmful to human beings and have no role in ecosystem stability as decomposers. But to eradicate the diseases caused by them, these viruses should be eradicated i.e., made extinct if possible.

2nd PUC Biology Biodiversity and Conservation Additional Questions and Answers

2nd PUC Biology Biodiversity and Conservation One Mark Question

Question 1.
What are endemic species?
Answer:
Those species which are confined to a particular region and are not found anywhere in the world are called endemic species.

Question 2.
Where does the Tiger project is established in Madhya Pradesh?
Answer:
Kanha national park.

Question 3.
Name some biodiversity hotspots in India
Answer:
Western Ghats, Himalayas, Indo-Burma region – Srilanka.

Question 4.
Where do great Indian bustards are found in India?
Answer:
Rajasthan, Punjab, Gujrat.

Question 5.
Some plants are used to extract chemicals used as drugs.
a. Mention the type of products extracted.
b. Give some examples of this product and its uses.
Answer:
a. Botano chemicals.
b. Papaver somniferum – Morphine
Chinchona ledgeriana – Quinine
Taxus brevifolia – Taxol

Question 6.
Which part of the world is known as lesser biodiversity?
Answer:
Poles have lesser biodiversity.

Question 7.
Name the district in Kerala
a. which has the highest forest coverage?
b. which has the lowest forest coverage?
Answer:
a. Idukki
b. Alappuzha (has no forest)

Question 8.
Name the unlabeled areas ‘a’ and ‘b’ of the pie chart representing the global diversity of invertebrates showing their proportionate number of species of major taxa (Delhi 2009)

Answer:
(a) Insects
(b) Molluscs

Question 9.
Name the unlabeled areas ‘a’ and ‘b’ of the pie chart representing the biodiversity of plants showing their proportionate number of species of major taxa. (AI2009)

Answer:
(a) Fungi
(b) Angiosperms.

Question 10.
Name the unlabeled areas ‘a’ and ‘b’ of the pie chart representing the biodiversity of vertebrates showing the proportionate number of species of major taxa. (CBSE 2009)

Answer:
(a) Fishes
(b) Amphibians.

Question 11.
Name 2 animals that have become extinct by overexploitation by humans.
Answer:
Steller’s sea cow and the passenger pigeon.

2nd PUC Biology Biodiversity and Conservation Two Marks Questions

Question 1.
Write the scientific name of the plant that yields reserpine. How is genetic variation expressed in this plant?
Answer:

• Rauwolfia vomitoria are the plant that yields reserpine.
• The genetic variation is shown in terms of the potency and concentration of the active chemical reserpine.

Question 2.
What is social forestry?
Answer:
The planning of social forestry started in India in 1976, which is related to the conservation of forests. This project is useful for local people in various ways, such as fulfills their requirements, provides work to the unemployed, uses wasteland, and help to maintain O₂ and CO₂ balance in the atmosphere, etc. Thus, the project is started by the Indian government, the chief objectives of this project are as follows:

• Plantation of useful plants in the forest.
• Development of forests on personal lands by the cooperation of government.
• To prevent the harmful effects of pollution by the development of artificial forests.
• Preservation of endangered wild animals.

Question 3.
Explain co-extinction.
Answer:
It is a phenomenon in which when a species becomes extinct, the plant and animal species associated with an obligatory manner also become extinct. eg:- When a host fish becomes extinct, its unique assemblage of parasites also faces extinction.

Question 4.
Differentiate between zoological and national parks
Answer:

Zoological park	National park
These are the areas where the threatened animals are kept under conditions very similar to their natural habitat.	These are large areas where animals and plants are protected in their natural habitat.
It is an exsitu method os conser­vation.	It is an in situ conservation method.

Question 5.
What are biosphere reserves? How many of them are present in India?
Answer:
Biosphere reserves are the ecologically unique and bio-diversity-rich regions that are legally protected. There are 14 biosphere reserves in India.

Question 6.
Why does the introduction of alien species into an ecosystem cause loss of biodiversity?
Answer:
The alien species become invasive and resources, so there is a decline in the indigenous species.

Question 7.
Name the region of the earth called the “lungs of the planet”. Mention giving reasons, the activities which are being carried out in this region now.
Answer:
Amazon forest is called the “lungs of the planet”. Human activities which are being carried out in this region are

• Grazing of beef cattle
• Cultivation of soya bean crop.

Question 8.
When and where the convention on biological diversity (earth summit) was held? While its aim.
Answer:

• The earth summit was held in Rio de Janeiro in 1992.
• The aim was to make the nation take appropriate measures for the conservation of biodiversity and sustainable utilization of its benefits.

Question 9.
What are sacred groves? Where are they found in India?
Answer:
Sacred groves are the tracts of forests that are set aside where all the trees and wildlife are venerated and given complete protection.
In India, sacred groves are found in

• Kashi and Jaintia Hills in Meghalaya
• Aravalli hills of Rajasthan
• Western Ghat regions of Karnataka and Maharashtra.
• Sarguja, Chanda, and Bastar areas of M.R
• Sarpakavu of Kerala.

Question 10.
Write any five features of Indian forests.
Answer:
Indian forests are characterized by :

1. Indian forests are mainly tropical forests.
2. Himalayan forests are characterized by the presence of coniferous trees.
3. In few parts of our country having temperate forests.
4. Our forests contain a large number of useful varieties of plants and animals.
5. Great variations are present in Indian forests.

2nd PUC Biology Biodiversity and Conservation Three Marks Questions

Question 1.
Differentiate between In situ conservation and Ex-situ conservation.
Answer:
In situ conservation:- It is the method of protecting the endangered species of plants or animals in the natural habitat either by protecting or cleaning up the habitat itself or by defending the species from predators.

• It helps in recovering populations in the surroundings where they have developed their distinct features.

Ex-situ conservation:-

• It is the method of protecting the endangered species of plants or animals by removing them from the unsafe or threatened habitat and placing them under the care of humans.
• It helps in recovering or preventing their extinction under stimulated conditions.

Question 2.
What are the characteristics of a stable community?
Answer:

• A stable community does not show much variation in productivity from year to year.
• Such a community must be either resistant or resilient to occasional disturbances both natural and man-made.
• It must be resistant to invasions by alien species.

Question 3.
Differentiate between Genetic diversity and species diversity.
Answer:

Genetic diversity	Species diversity
(a)  It is a trait of the organisms. (b)  It represents the variety of genetic information present in the organisms. (c)  It is important for adaptation to the enviro­nment and changes occurring in it.	(a) It is a trait of the biotic community. (b) It is a variety of species and their abundance found within a region. (c) It influences the stability of the ecosystem.

2nd PUC Biology Biodiversity and Conservation Five Marks Question

Question 1.
What is the influence of biodiversity on the ecosystem? Mention the 3 possible consequences that loss of biodiversity in a region can lead to.
Answer:
Influences of biodiversity

• The ecosystems with more species diversity show less year-to-year variation in total biomass production.
• The ecosystem with species diversity shows more productivity compared to those with less species diversity.

Consequences are
The loss of biodiversity can lead to

• A decline in the crop productivity
• Increased variability in a certain ecosystem
• A decreased resistance to environmental perturbates like drought.

Question 2.
Prepare a notice to be distributed to the lower classes on the importance of the ‘Conservation of Biodiversity’.
Answer:
The great biodiversity on earth is vital for the existence of mankind.
The reasons for conserving biodiversity are the following.

i. Narrowly utilitarian: This argument for conserving biodiversity are obvious i.e., humans derive countless economic benefits such as food, firewood, fiber, construction materials, industrial products (such as tannins, dyes, lubricant, resins, perfumes etc.), medicinal products etc. More than 25% of medicines are derived from plants and nearly 25,000 species of plants contribute to the traditional medicines used by the native peoples around the world. It is unknown that how many medicinal plants are waiting to be explored in the tropical rain forest.

ii. Broadly utilitarian: This argument says that biodiversity plays a vital role in all ecosystem services. Amazon rain forest alone produces 20% of the total oxygen in the atmosphere through photosynthesis. Can you estimate the economic value of this life-saving gas provided by nature? The other indirect benefits that we receive are
a. Pollination: Without pollination, the plants cannot give us fruits and seeds. The pollination is done by bees, birds, bats etc.

b. Aesthetic and cultural benefits: The aesthetic values include ecotourism, bird watching, wildlife, pet-keeping, gardening, etc. Plants like Tulsi (Ocimum sanctum), Pipal (Ficus religiosa) etc. are considered sacred and are used by many Indians for religious purposes. Nowadays we recognize plants and animals as symbols of national pride and cultural heritage. Walking through a forest, watching spring flowers in full bloom or waking up to a bulbul’s song in the morning etc. also gives pleasure and smoothness.

iii. Ethical argument: This says that we have a moral responsibility to take care of earth’s biodiversity such as millions of plants, animals, microbes etc. with whom we share this planet. We need to realize philosophically or spiritually that each and every species on the earth has its own intrinsic value.

Question 3.
“Ecology is permanent economy”. This is the Chipko slogan coined by Sundarlal Bahuguna. Write a short note on Chipko Movement.
Answer:
It is one of the People’s movements in biodiversity conservation. The Chipko Movement is the result of hundreds of decentralized and locally autonomous initiatives. Its leaders and activists are primarily village women, acting to save their means of subsistence and their communities. Men are involved too, however, and some of these have given wider leadership to the movement. Prominent Chipko figures include: Sunderlal Bahuguna, a Gandhian activist and philosopher, whose appeal to Mrs. Gandhi resulted in the green-felling ban and whose 5,000-kilometer trans-Himalaya foot march in 1981-83 was crucial in spreading the Chipko message. Bahuguna coined the Chipko slogan: ‘ecology is the permanent economy’.

Question 4.
Describe the National and International efforts prescribed for the conservation of forests.
Answer:
Forest conservation is started in India on a national level by the British government. In 1856, Lord Dalhousie had formulated a policy for the conservation of forest in Burma. In 1894, the Indian government also prepared a forest policy on a national level. The main points of this policy are:

• Forest management,
• Proper use of forest land,
• Policy for protected forests,
• Improved forest production.

The Indian government established national parks, sanctuaries, and zoological parks. The F.A.O. of the United Nations is also functioning on forest conservation on an international level. This organization also provides financial help for this purpose. In 1952, the Indian government also prepared India’s New National Forest Policy under the direction of F.A.O. Forest policy has been planned for

• Prevention of deforestation of hill plants.
• Reforestation of grazing land.
• Development of grazing land,
• Plantation of economically useful forest trees.
• Increase in the profit of government from forests.

Question 5.
Which are the different “methods of in situ conservation?
Answer:
In situ conservation is the process of protecting the whole ecosystem and its biodiversity at all levels. The endangered species of plants and animals are protected in their natural habitat without disturbing them from their own habitat. It helps in recovering the population in the surroundings where they have developed their distinctive features.

• Hotspots -These are the regions with high levels of species richness and a high degree of endemism to provide maximum protection.
• Biosphere reserves – These are ecologically unique and biodiversity-rich regions that are legally protected. In India, there are 14 biosphere reserves.
• National park and wildlife sanctuaries- These are the places where animals are in their own undisturbed habitat but protected these areas by law.
• In India, there are 90 national parks and 448 wildlife sanctuaries are there.
• Sacred groves – These are some patches of forest where all the trees and wildlife within are venerated and given total protection, eg: Sarpakavu in Kerala. Sarguja area of M.P. etc.

You can Download Chapter 12 Biotechnology and its Applications Questions and Answers, 1st PUC Biology Question Bank with Answers, Karnataka State Board Solutions help you to revise complete Syllabus and score more marks in your examinations.

Karnataka 2nd PUC Biology Question Bank Chapter 12 Biotechnology and its Applications

2nd PUC Biology Biotechnology and its Applications Ncert Text Book Questions and Answers

Question 1.
Crystals of Bt toxin produced by some bacteria do not kill the bacteria themselves because
(a) bacteria are resistant to the toxin
(b) toxin is immature;
(c) toxin is inactive;
(d) bacteria encloses toxin in a special sac.
Answer:
(c) Bacillus thuringiensis forms protein crystals during a particular phase of their growth. These crystals contain a toxic insecticidal protein. Actually, the Bt toxin protein exists as inactive protoxins but once an insect ingests the inactive toxin, it is converted into an active form of toxin due to the alkaline pH of the insect gut which solubilise the crystals. The activated toxin binds to the surface of midgut epithelial cells and creates pores that cause cell swelling and lysis and eventually cause the death of the insect, but it does not harm Bacillus itself.

Question 2.
What are transgenic bacteria? Illustrate using any one example. (CBSE – 2006)
Answer:
Bacteria carrying foreign genes are called transgenic bacteria. For example, two DNA sequences (A and B chains of human insulin) were introduced into the plasmid of bacteria E.coli. The leans genic bacteria start producing insulin chains.

Question 3.
Compare and contrast the advantages and disadvantages of the production of genetically modified crops.
Answer:
Advantages of GM crops:

• Genetic modification has made crops more tolerant to abiotic stresses (cold, drought, heat, salt.)
• Viral resistance can be introduced.
• Over ripening, losses can be reduced. Example: Flavr Savr Tomato.
• Enhanced nutritional value of food. Example: Golden Rice.
• Reduced reliance on chemical pesticides.

Disadvantages of GM crops:

• Transgenes in crop plants can endanger native species.
• Example: The gene for Bt toxin expressed in pollen may end natural pollinators such as honeybees.
• Weeds also become resistant.
• Products of transgenes may be allergic or toxic.
• They cause damage to the natural environment.

Question 4.
What are Cry proteins? Name an organism that produces it. How has man exploited this protein to his benefit?
Answer:
Cry proteins are protein responsible for killing lepidopteran insect and their larvae (also called Bt toxin). It is secreted by Bacillus thuringienesis. Man exploited gene encoding this toxin, by transferring it into cotton genome with the help of Agrobacterium TDN A as vector.

Question 5.
What is gene therapy? Illustrate using the example of adenosine deaminase (ADA) deficiency,
Answer:
It is a collection of methods which allows correction of a gene defect that has been diagnosed in a child or embryo. In gene therapy, normal genes are inserted into a person’s cells or tissues to treat a hereditary defect. Gene therapy is being tried for sickle cell anaemia and Severe Combined Immuno Deficiency (SCID).

In some children, ADA deficiency can be cured by bone marrow transplantation. In others, it can be treated by enzyme replacement therapy, in which functional ADA is given to the patient by injection. However, both of these approaches are not completely curative.

In gene therapy, lymphocytes from the blood of the patient are grown in a culture outside the body. A functional ADA cDNA (using a retroviral vector) is then introduced into these lymphocytes, which are subsequently returned to the patient. Because these cells are not immortal, the patient requires a periodic infusion of such genetically engineered lymphocytes. However, if the gene isolated from marrow cells producing ADA is introduced into cells at early embryonic stages, the disease could be cured permanently.

Question 6.
Diagrammatically represent the experimental steps in cloning and expressing a human gene (say the gene for growth hormone) into a bacterium like E. coli?
Answer:
The given diagram represents the experimental steps in cloning and expressing a human gene for growth hormone into a bacterium E. coli.

Question 7.
Can you suggest a method to remove oil (hydrocarbon) from seeds based on your understanding of rDNA technology and the chemistry of oil?
Answer:
To remove oil from seeds using recombinant DNA technology would involve:

• Identifying the genes that code for oil production.
• Deleting these genes from the seed genome.
• Splicing back together with the remaining DNA.
• Putting it back into the cell.

It will not be very easy because the oils are made up of fatty acids and glycerol. Since fatty acids are important components of the cell membrane system, deleting or switching off of its genes might affect the cell structure itself.

Question 8.
Find out from the internet what is golden rice.
Answer:
Golden rice is transgenic rice having gene coding for vitamin A synthesis enzyme. Golden rice was developed by the Swiss Federal Institute of Technology, rich in vitamin A (beta carotene). The rice grains are golden yellow in colour due to colour it gets from the beta carotene.

Question 9.
Does our blood have proteases and nucleases?
Answer:
No, our blood does not contain enzyme proteases and nucleases. If these two enzymes were there in the blood, it causes the degeneration of blood cells and lining cells of blood cells.

Question 10.
Consult the internet and find out how to make orally active protein pharmaceuticals. What is the major problem to be encountered?
Answer:
Proteinaceous drugs cannot be taken orally because they can be degraded by the proteases of our alimentary canal. To counteract this problem or to make an orally active protein pharmaceutical, it must be coated by a film that is resistant to protein degrading enzymes.

2nd PUC Biology Biotechnology and its Applications Additional Questions and Answers

2nd PUC Biology Biotechnology and its Applications One Mark Questions

Question 1.
Name the organism whose genetic material has been altered using genetic engineering techniques.
Answer:
Genetically Modified Organisms (GMO).

Question 2.
Genetically modified cotton is named Bt cotton. What does the prefix ‘Bt’ mean?
Answer:
Cotton containing a toxin gene from Bacillus thuringiensis

Question 3.
Through whom Bt toxin protein originates?
Answer:
By Bacillus thuringiensis.

Question 4.
Biopiracy affects developing countries like India more than industrialized nations because our country is rich in ………………. and …………………. related to bio-resources.
(Bioetics / Biodiversity/Biopatent, Traditional resources/Traditional cultivation/Traditional knowledge)
Answer:
Biodiversity and Traditional knowledge

Question 5.
Full form of RNAi.
Answer:
RNA interference (RNAi).

Question 6.
What GEAC?
Answer:
Genetic engineering Approval committee.

Question 7.
Name the scientific name of bacteria in which be form organism toxin.
Answer:
Bacillus thuringiensis.

Question 8.
A multinational company outside India tried to sell new varieties of turmeric without proper patent rights. What is such an act referred to as?
Answer:
Biopiracy

Question 9.
What is Hirudin?
Answer:
It is a protein which prevent blood dotting.

2nd PUC Biology Biotechnology and its Applications Two Marks Questions

Question 1.
In case of Bt Cotton how does the toxic insecticide protein produced by the bacterium kill the insect pest but not the cells of Bacillus thuringiensis where the toxic protein is generated?
Answer:
The toxin produced by Bacillus thuringiensis is an endotoxin called cry protein. It is crystalline and nontoxic when formed being in the protoxin stage. As it reaches the gut of insects, the cry protein is converted into toxic and soluble state. It attaches the receptors present on the epithelial cells of the gut produces pores and kills the cells resulting in the death of the insects.

Question 2.
What are the 4 main objectives of genetically modified crop plants? (CBSE 2008)
Answer:

• Higher nutritional value eg: Vitamin. A in golden rice
• Abiotic stresses- Increased tolerance to drought etc.
• Post Harvest losses- Prevention of over-ripening and other post-harvest losses.
• Insect and Pest resistance. eg: Bt-cotton.

Question 3.
Define transgenic organisms (CBSE 2006)
Answer:
They are organisms which have been modified genetically through the introduction of genes of another organism artificially by the technique of genetic engineering instead of conventional hybridisation.

Question 4.
How is early detection of diseases possible using molecular diagnostics?
Answer:

• Low concentration of viral or bacterial DNA in the host cell/body can be detected much before the symptoms of the disease appear i.e. early detection of disease is possible
• Clones of genes can be used to as probe to detect the presence of mutual alleles in cancer suspected patients.

Question 5.
Name any two biological products that are produced in transgenic animals and mention their uses.
Answer:

• a-1- antitrypsin is used to treat emphysema.
• a-lactalbumin is produced in the milk of the transgenic cow, Rosie, this is a nutritionally more balanced product for human babies than normal cow milk.

Question 6.
Bacillus thuringiensis makes our environment pesticide-free. Comment.
Answer:
Bacillus thuringiensis produces a toxin called Bt toxin. Gene for Bt toxin has been cloned from bacteria and been expressed in plants to provide resistance to pests. In this manner, many plants are produced. So in the future, this makes our environment pesticide-free.

Question 7.
What is the cause of adenosine deaminase deficiency in a person?
Answer:
It is due to the deletion of the gene coding for the enzyme adenosine deaminase, this enzyme is crucial for the functioning of the immune system.

Question 8.
Briefly describe the RNA interference process for preventing nematode infestation of plants.
Answer:
RNA interference (RNAi) takes place in all eukaryotic organisms as a method of cellular defense. This method involves the silencing of a specific mRNA due to a complementary double-stranded RNA molecule that binds to it and prevents translation of mRNA. The source of this complementary RNA could be from injection by viruses having RNA genomes or mobile genetic elements.

Question 9.
Why does Bt toxin not kill the bacillus? How does it kill the insect larvae?
Answer:

• When Bt toxin is ingested by an insect, it is converted into its active form when exposed to the alkaline pH in the gut.
• The activated toxin binds to the surface of the epithelial cells of the midgut creates pores.
• Water enters the cells and causes their swelling and lysis.

Question 10.
Write the advantages of recombinant therapeutics?
Answer:
The recombinant therapeutics do not induce any unwanted immunological response like the similar products of non-human origin such therapeutics are highly effective.

Question 11.
How is ELISA used to detect pathogens in the body?
Answer:

• Pathogens are detected by the presence of antigens, which may be a protein or glycoprotein.
• Pathogens can be also be detected by the presence of antibodies synthesized against the pathogens.

2nd PUC Biology Biotechnology and its Applications Three Marks Questions

Question 1.
What is genetically modified (GM) food? Give two examples.
Answer:
Genetically modified food (GM food): The food substances produced from genetically modified crops or transgenic crops is called GM food. This food differs from conventionally developed varieties in the following aspects :

• GM food contains an antibiotic resistance gene itself.
• It contains protein produced by transgene, e.g. Cry protein in insect resistance varieties.
• These GM foods contain enzyme produced by the antibiotic resistance gene that was used during gene transfer by recombinant DNA technology.

Examples of GM Crops, Food, and Fruits:

1. Flavr Savr Tomato: It is the first food containing genetically engineered DNA. . These tomatoes contain genes for antibiotic resistance for kanamycin.

2. Maize: GM maize has a bacterial gene that increases its resistance to pests and
diseases. It also has a gene for ampicillin resistance which is harmful for us, therefore the introduction of GM maize is opposed by many European countries. ,

3. Rape oilseed: It is a new type of plant that contains genes for resistance to the herbicide Basta. It has more potential, dangers and can become a weed and would be impossible to control with Basta. It could cross-fertilize with relatives such as wild mustard, thus, spreading the resistance to wild plants. Such type of environmental risks could occur with genetically modified rapeseed crop. They might also affect food chains in unpredictable ways.

Question 2.
Mention three reasons for the success of the green revolution in India.
Answer:

• Use of improved verities of crops
• Employing better management practice
• Use of agrochemicals such as fertilizers and pesticides.

Question 3.
Enumerate the fields of application of biotechnology.
Answer:
The applications of biotechnology include

• Molecular diagnostics
• Bioremediation
• Waste treatment
• Energy production
• Therapeutics.

Question 4.
Explain the steps involved in the production of genetically engineered insulin.
Answer:
Insulin used for diabetes was earlier extracted from the pancreas of slaughtered cattle and pigs. Insulin consists of two short polypeptide chains, chain A and chain B that are linked together by disulphide bridge.

In mammals, including humans, insulin is synthesised as a prohormone which contains an extra stretch called the C peptide. This C peptide is not present in the mature insulin and is removed during maturation into insulin. The main challenge for the production of insulin using rDNA techniques was getting insulin assembled into a mature form.

In 1993 Eily Lilly an American company prepared two DNA sequences corresponding to A and B, chain of human insulin and introduced them in the plasmid of E.coli to produce insulin chains, Chains A and B were produced separately, extracted, and combined by creating disulphide bonds to form human insulin.

Question 5.
Name any 6 plants where Bt toxin-producing genes have been included.
Answer:

• Cotton
• Tomato
• Rice
• Potato
• Soybean
• Com.

2nd PUC Biology Biotechnology and its Applications Five Marks Questions

Question 1.
Two of the steps involved in producing nematode-resistant tobacco plants base on the process of RNA I are mentioned below. Write the missing steps in the proper sequence.
Answer:

• Nematode specific genes
• Production of both sense and antisense RNAs
• Double strand RNA
• Silencing specific mRNAs of nematode
• Death of nematodes
• Protection of transgenic plants from nematodes.

Question 2.
Describe the application of genetic engineering in the field of Agriculture and Medicine.
Answer:
(A) Application of Genetic engineering or Biotechnology in Agriculture: Genetic engineering is found to be very beneficial in agriculture. Its important use in agriculture are:
1. Increase in photosynthetic efficiency: An increase in photosynthetic efficiency of crop plants can be achieved by introducing suitable Carbon dioxide Fixation Gene (cfx) from any plant into the crop plants.

2. Transfer of nitrogen-fixing ability: Number of symbiotic and non-symbiotic micro-organisms have a capacity of fixing atmospheric nitrogen. Nitrogen fixers are found to possess nitrogen-fixing gene (nif genes) which are located on chromosomes or plasmids. Introduction of nif gene in crop plants results inability in crop plants to fix atmospheric nitrogen and reduction in the use of chemical nitrogen fertilizers.

3. Disease resistance in crop plants: Plant breeders at present are developing high-yield varieties by transferring genes for disease resistance through conventional breeding.

4. Plant tissue in crop improvement: Some of the areas of plant improvement where tissue culture has been applied with success are as follows :

• Rescuing hybrids through embryo culture.
• Multiplication of germplasm.
• Production of disease-free plants.
• Production of haploid through another culture.
• Somaclonal variation.
• Somatic hybridization.
• Cryopreservation of germplasm.

5. VAM (Vesicular-Arbuscular Mycorrhiza) fungi with Rhizobium can boost the yields: Recently there has been a new dimension to this farm practice by the way of increasing Rhizobium inoculation effect by simultaneous inoculating seeds with VAM as well as Rhizobium culture. VAM is a structural modification of hyphae helping in absorption and storage of phosphorus.

(B) Application of Genetic engineering in the Medical field:

1. Hereditary diseases like color-blindness, haemophilia which are caused by recessive genes and also many inborn metabolic disorders due to defective genes as alkaptonuria, phenylketonuria can be cured with gene therapy.
2. Substances like vitamins, hormones, amino acids and antibodies can be synthesized in bacteria by introducing the genes which code these substances. In this way, bacteria can be used as biofactories for the synthesis of these substances.
3. Production of insulin: Insulin is a medicine used for the treatment of diabetes. Initially, it is derived from animals (pig and cows) but today it is produced by gene splicing.
4. Hepatitis-B vaccine: Hepatitis-B is a viral disease of the liver. Today this vaccine is prepared with the help of genetic engineering.

You can Download Chapter 11 Biotechnology: Principles and Processes Questions and Answers, 1st PUC Biology Question Bank with Answers, Karnataka State Board Solutions help you to revise complete Syllabus and score more marks in your examinations.

Karnataka 2nd PUC Biology Question Bank Chapter 11 Biotechnology: Principles and Processes

2nd PUC Biology Biotechnology: Principles and Processes NCERT Text Book Questions and Answers

Question 1.
Can you list 10 recombinant proteins which are used in medical practice? Find out where they are used as therapeutics (use the internet).
Answer:
Recombinant proteins used in medical practice as therapeutics are as follows:

• OKT-3, a therapeutic antibody is used for the reversal of acute kidney transplantation rejection.
• ReoPro is for the prevention of blood clots.
• Tissue plasminogen activator (TPA) is for acute myocardial infarction.
• Asparaginase is for the treatment of some types of cancer.
• DNase is for the treatment of cystic fibrosis.
• Insulin is used in diabetes mellitus.
• Blood clotting factor VIII is used for the treatment of haemophilia A.
• Blood clotting factor IX is for the treatment of haemophilia B.
• Hepatitis B vaccine is for the prevention of hepatitis B.
• Platelet-derived growth factor has been approved for diabetic/skin ulcers. It also stimulates wound healing.

Question 2.
Make a chart (with diagrammatic representation) showing a restriction enzyme, the substrate. DNA on which it acts, the site at which it cuts DNA, and the product it produces.
Answer:

Question 3.
From what you have learned, can you tell whether enzymes are bigger or DNA is bigger in molecular size? How did you know?
Answer:
Human cells vary in size and volume. DNA is fixed in its size. That makes it kind of hard (like impossible) to state a concentration for DNA. Only a range can be given, and that information has little to no practical value and even less meaning.

Question 4.
What would be the molar concentration of human DNA in a human cell? Consult your teacher.
Answer:
Molar concentration is the ratio of the number of moles of solute in a solution divided by the volume of the solution expressed in liters.
The average weight of a DNA basepair = 650 daltons (1 dalton equals the mass of a single hydrogen atom or 1.67 x 10^-24 grams)
The molecular weight of a double-stranded
DNA molecule = Total no. of basepairs x 650 daltons
The human genome is 3.3 x 10⁹ bp in length.
Hence, the weight of human genome,
= 3.3 x 10⁹ bp x 650 Da
= 3.59 x 10^-12 gm.
The molar concentration of DNA can be calculated accordingly.

Question 5.
Do eukaryotic cells have restriction endonucleases? Justify your answer.
Answer:
The stirred tank bioreactor facilitates the mixing and oxygen availability. It controls the temperature and pH inside the bioreactor.

Question 6.
Besides better aeration and mixing properties, what other advantages do stirred tank bioreactors to have over shake flasks?
Answer:
Shake flasks are used for growing and mixing the desired materials on a small scale in the laboratory. Bioreactors are vessels in which raw materials are biologically converted into specific products by microbes, plant and animal cells, and their enzymes. Bioreactors are used for large-scale production of biomass or sell products under aseptic conditions. Here large volumes (100-1000 litres) of culture can be processed. A bioreactor provides the optimal conditions for achieving the desired product by providing optimum growth conditions (temperature, pH, substrate, salts, vitamins, oxygen). The most commonly used bioreactors are of stirring type.

A bioreactor is more advantageous than shake flasks. It has an agitator system to mix the contents properly, an oxygen delivery system to make available of oxygen, a foam control system, a temperature control system, a pH control system, and a sampling port to withdraw the small volumes of the culture periodically.

Question 7.
Collect 4 examples of palindromic DNA sequences by consulting your teacher. Better try to create a palindromic sequence by following base-pair rules.
Answer:

Question 8.
Can you recall meiosis and indicate at what stage recombinant DNA is made?
Answer:
Meiosis I – Pachytene – When recombination nodule appears after synaptonemal complex formation.

Question 9.
Can you think and answer how a reporter enzyme can be used to monitor the transformation of host cells by foreign DNA in addition to a selectable marker? Ans: A selectable marker helps to identify transformed host cells and non transformed cells will be eliminated and only transformed cells will grow. Whereas reporter gene is the one whose phenotypic expression can be monitored and thus it reports about activity or change in advance of the effect of the modification, in addition to eliminating non transformed cells by selectable markers. HLDescribe briefly the followings:
(a) Origin of replication
(b) Bioreactors
(c) Downstream processing
Answer:
(a) Origin of replication (on): One of the major components of a plasmid is a sequence of bases where replication starts. It is called the origin of replication (on). This is a specific portion of the plasmid genome that serves as a start signal for self-replication (to make another copy of itself). Any piece of DNA when linked to this sequence can be made to replicate within the host cells. This property is used to make a number of copies of linked DNA (or DNA insert).

(b) Bioreactors: These are vessels in which raw materials are biologically converted into specific products by microbes, plant and animal cells, and their enzymes. They are allowed to synthesise the desired proteins which are finally extracted and purified from cultures. Small volume cultures are usually employed in laboratories for research and production of fewer quantities of products. However, large-scale production of the products is carried out in bioreactors’. The most commonly used bioreactors is a stirring type bioreactor (fermenter) that has a provision for batch culture or continuous culture.

(c) Downstream processing: After the formation of the product in the bioreactors, it undergoes some processes before a finished product is ready for marketing. These processes include separation and purification of products which are collectively called downstream processing. The product is then subjected to quality control testing and kept in suitable preservatives. The downstream process and quality control test are different for different products

Question 11.
Explain briefly
(a) PCR
(b) Restriction enzymes and DNA
(c) Chitinase
Answer:
a. The in vitro amplification of DNA by repeated cycles of strand separation and polymerisation is PCR.
b. The nuclease enzyme that cuts the DNA at a unique sequence is called restriction endonuclease. They are also known as molecular knives, molecular scissors, or molecular scalpels.
c. Chitinase is digestive enzymes that break down glycosidic bones in chi tin.

Question 12.
Discuss with your teacher and find out how to distinguish between
(a) Plasmid DNA and Chromosomal DNA
(b) RNA and DNA
(c) Exonuclease and Endonuclease
Answer:

(a) Plasmid DNA and Chromosomal DNA

Plasmid DNA	Chromosomal DNA
(i) It is extranuclear DNA	It is nuclear DNA
(ii) It caries nonvital genes	It possesses vital genes
(iii) A bacterial cell may carry one to several plasmid DNAs.	A bacterial cell carries only one chromosome DNA

(b) RNA and DNA

RNA	DNA
(i) It is ribonucleic acid	Deqxy ribonucleic acid
(ii) It is single strandred	Double standard
(iii) RNA produced from DNA template	Parental DNA acts as a DNA template
(iv) Nitrogen base Uracil is present instead of thymine, which pairs with adenine.	Thymine pairs with adenine

(c) Exonuclease and Endonuclease

Exonuclease	Endonuclease
(i) It breaks DNA from ends	It cuts DNA from inside
(ii) The separated fragments are small nucleotides	These separated fragments  are generally large sized
(iii) The separated fragments cannot be used in genetic engineering	The desirable separated fragments are used in genetic engineering

2nd PUC Biology Biotechnology: Principles and Processes Additional Questions and Answers

2nd PUC Biology Biotechnology: Principles and Processes One Mark Question

Question 1.
By observing the given pair fill-up the blanks

1. Cutting DNA : Restriction endonuclease :: Joining DNA : …………………….
2. Chitinase : Fungus :: Cellulose : ………………………..
3. Protein : Protease :: RNA : …………………………

Answer:

1. DNA ligase
2. Plant cells
3. Ribonuclease

Question. 2.
Which DNA polymerase is active in high temperatures?
Answer:
Taq DNA polymerase is active at high temperatures.

Question 3.
Restriction endonucleases are used to cut DNA at specific sites.. Name the first endonuclease isolated from Escherichia coli.
Answer:
EcoRI

Question. 4.
Write the full form of PCR. Which enzyme is used in?
Answer:
Polymerase Chain Reaction (PCR).
Taq DNA is used in PCR.

Question 5.
Name the technique used for separating DNA fragments in the laboratory
Answer:
Electrophoresis. (Dehli 2005)

Question. 6.
First recombinant DNA was formed in.
Answer:
In Bacteria Salmonella Typhimurium.

Question 7.
Write the function of the restriction enzyme in a bacterial cell?
Answer:
Responsible for restricting the growth of bacteriophage.

Question. 8.
Where does Hind II cut the DNA molecule?
Answer:
It cuts the DNA molecule at a specific 6 base pair seQuestionuence.

Question 9.
Why are plasmids and bacteriophages commonly used as cloning vectors?
Answer:
Plasmids and bacteriophages have the ability to replicate within bacterial cells independently of the chromosomal DNA.

Question. 10.
Which type of charge found in DNA?
Answer:
Negative charge.

Question 11.
What is downstream processing?
Answer:
Downstream processing is the recovery of product from the fully grown genetically modified cells, its purification and preservation

Question. 12.
What is electrophoresis?
Answer:
Electrophoresis is a techniQuestionue used in laboratories in order to separate macromolecules based on size.

Question 13.
What is gene therapy?
Answer:
It is the replacement of a defective gene by normal healthy and functional gene. This method helps to overcome the effect of various disorders like sickle cell anaemia, alkaptonuria, SCID, colour blindness etc.

Question. 14.
Name the technique in which we should be isolated the DNA segment.
Answer:
Electrophoresis.

Question 15.
What is amplification?
Answer:
It is the process of making multiple copies of gene/DNA segments of interest.

Question 16.
What is recombinant protein?
Answer:
It is a biochemical compound or useful protein produced inside the heterologous host cell by recombinant biotechnology method.

Question 17.
What is a bioreactor or fermenter?
Answer:
It is a container in which the biochemical process is carried out by using living cells and their growth medium.

Question 18.
What meant by bioconversion?
Answer:
It is the process by which raw materials are biologically converted into specific products using microbes, plant or animal cells and or their enzymes.

Question 19.
Why are antibiotic resistance genes used as selectable markers for E.Coli?
Answer:
Since E.Coli doesn’t have any of antibiotic resistance genes, antibiotic resistance genes are used from outside as selectable marker.

2nd PUC Biology Biotechnology: Principles and Processes Two Marks Questions

Question 1.
Name the scientists who constructed recombinant DNA. Name the bacterium from which they isolated the gene.
Answer:
Stanley Cohen and Herbert Boyer were the first to construct recombinant DNA. They isolated the gene from the bacterium. Salmonella typhimurium.

Question 2.
Few gaps have been left in the following table showing certain terms and their meanings, fill up the gaps.
Term Meanings
(i) ………….. Non-coding sequence in eukaryotic DNA
(ii) ………….. The technique used in solving paternity disputes
(iii) Restriction endonuclease …………..
(iv) Plasmids …………..
(v) Transgenics …………..
(vi) Nucleotide sequences with single base deficiencies …………….

Question 3.
Name the particular technique in biotechnology whose steps are shown in the figure use the figure to summarize the technique in three steps.
Answer:

• Template stand
• DNA fingerprinting
• Extranuclear DNAs
• Organisms having genes of other organisms obtained through genetic engineering.
• Single nucleotide polymorphism. (SNPs)

Question 3.
Name the particular technique in biotechnology whose steps are shown in the figure use the figure to summarize the technique in three steps.
Answer:

(a) Recombinant technology
(b)

• Cutting and isolation of human gene
• Incorporation of human gene into the plasmid to produce recombinant DNA or plasmid
• Incorporation of recombinant plasmids into bacterium to obtain gene product

Question 4.
Refer to the diagram and answer the following:
(i) From what T₁– plasmid is obtained?
(ii) Name the enzyme which is involved in step I
(iii) What happens in step II
(iv) The plant produced is called hybrid or transgenic
(v) Will the plant produced have other genes along with desired genes? Yes or No explain.
Answer:

(i) Agrobacterium tumefaciens
(ii) Restriction endonuclease
(iii) Incorporation of genes in T₁ plasmid in the region of T-DNA.
(iv) Transgenic
(v) Yes. Selectable marker gene which is often an antibiotic resistance gene

Question 5.

Study the linking of DNA fragments shown above
(i) Name “a” DNA and “b” DNA
(ii) Name the restriction enzymes that recognize this palindrome
(iii) Name the enzyme that can link these two DNA fragments (CBSE 2008)
Answer:
(i) (a) – vector DNA
(b) – foreign DNA

(ii) EcoRI
(iii) DNA ligase

Question 6.
Explain the importance of
(a) Ori
(b) amp R and
(c) rop in E. Coli vector shown below (CBSE 2008)
Answer:

• Ori – Origin of replication
• amp R – ampicillin antibiotic resistance gene
• rop – gene that produces proteins involved in the replication of plasmid.

Question 7.
An interesting property of restriction enzymes is molecular cutting and pasting Restriction enzymes typically recognize a symmetrical

Notice that the top strand is the same as the bottom strand but reads backward. When the enzyme cut the strand between G and A, it leaves overhanging chains

(A) What is the symmetrical sequence of DNA known as?
(B) What is the significance of these overhanging chains?
(C) Name the restriction enzyme that cuts the strand between G and A.
Answer:
(A) Palindromic sequence
(B) sticky ends
(C) Eco RI

Question. 8.
What is DNA ligase?
Answer:
DNA ligase is a specific type of enzyme, a ligase that facilitates the joining of DNA strands together by catalyzing the formation of a phosphodiester bond.

Question 9.
What are cloning vectors? What functions do these vectors perform?
Answer:
Cloning vectors are those organisms on their DNAs which can multiply independently of the host DNA and increase their copy number along with the alien DNA attached to them. Functions are:

• They help in linking the foreign/alien DNA with that of the host
• They also help in the selection of recombinants from non-recombinants

Question. 10.
What is Ti-plasmid?
Answer:
Ti or tumor-inducing plasmid is a plasmid that is a part of the genetic equipment that Agrobacterium tumefacient use to transduce their genetic material to plants.

Question 11.
Palindromic nucleotide sequences have significance in recombinant DNA technology. Explain. Give example for a palindromic DNA sequence.
Answer:
Palindrome in a DNA is a sequence of base pairs that reads same on the two strands when orientation of reading is kept the same. Each restriction endonuclease recognises a specific palindromic nucleotide sequence and cuts the strand of DNA a little away from the centre of the palindrome site, but between the two bases on the opposite strands.
Example of palindrome DNA is
5′ – GAATTC – 3′
3′ – CTTAAG – 5′

Question. 12.
Name the scientist who discovered the artificial DNA synthesizing method.
Answer:
The Nobel prize in physiology in 1968 was awarded jointly to Robert W. Holley, Hargobind Khorana, and M. Nirenberg for their interpretation of the genetic code.

2nd PUC Biology Biotechnology: Principles and Processes Three Marks Question

Question 1.
Given below are the different steps in recombinant DNA technology. Arrange them according to the sequence of occurrence.
a. Transferring the recombinant DNA into the host.
b. Extraction of the product.
c. Fragmentation of DNA by restriction endonucleases
d. Ligation of DNA fragment into a vector.
e. Isolation of DNA.
f. Culturing the host ceils in a medium at large scale.
g. Isolation of the desired DNA fragment.
Answer:
a. Isolation of DNA.
b. Fragmentation of DNA by restriction endonucleases
c. Isolation of the desired DNA fragment.
d. Ligation of DNA fragment into a vector.
e. Transferring the recombinant DNA into the host.
f. Culturing the host cells in a medium at large scale.
g. Extraction of the product.

Question 2.
Represent diagrammatically the E Coli cloning vector pBR 322 showing the restriction site.
Answer:

Question 3.
Draw a labelled diagram of a sparged stirred tank bioreactor.
Answer:

Question 4.
Read the following base sequence of a certain DNA strand and answer the questions that follow.

(i) What is called “Palindromic sequence” in a bNA?
(ii) Write the Palindromic nucleotide sequence shown in the DNA strand given and mention the enzyme that will recognise such a sequence.
(iii) State the significance of enzymes that identify palindromic nucleotide sequences.
(AI – 2008)
Answer:
(i) A palindromic sequence of DNA is a sequence of base pairs that reads the same on the two strands, when orientation of reading is kept the same, i.e. in the 5′ → 3′ direction.
(ii) 5′ GAA TT C – 3′
3′ CTTA AG – 5′
This sequence is restricted by the restriction enzyme Eco RI
(iii) The enzyme that identifies the palindromic nucleotide sequence cuts the strands between the same 2 bases, more often producing sticky ends; hence they are useful in the formation of recombinant DNA.

2nd PUC Biology Biotechnology: Principles and Processes Five Marks Question

Question 1.
Describe in detail the components of a simple stirred tank bioreactor along with a labeled diagram.
Answer:
A stiered tank bioreactor is usually a cylindrical vessel with a curved base to facilitate the mixing of the content. The stirrer facilitates even mixing and oxygen availability throughout the bioreactor.

The bioreactor has an agitator system, an oxygen delivery system along with a foam control system, a temperature control system, pH control system and sampling ports to remove small volumes of culture periodically. It provides optimum growth conditions of pH, temperatures, substrate, oxygen etc. for achieving the desired products.

Question. 2.
What is a clone? Give its preparation, extracted and purified.
Answer:
An organism or cell or group of organisms, produced asexually from an ancestor, to which they are genetical.

1. Gene cloning: Following steps are used by gene cloning:

1. Preparation of gene: DNA extracted from an organism, with the gene of interest is cut into gene size pieces with a restriction enzyme.

2. Insertion into a vector: Bacterial plasmids are cut with the same restriction enzyme. Plasmids are small circles of DNA in bacterial cells that are naturally present in addition to the bacterial other DNA.

3. Transformation of host cells: The recombinant plasmids are then transferred into bacteria using either electrophOration. The plasmid is small enough to pass through the holes into the cells. However, rather than using electricity to create holes in the bacterium, it is done by alternating the temperature between hot and cold. The bacteria are grown on a culture dish and allowed to grow into colonies. All the colonies on all the plates are called a gene library.

2. Plant cloning: Plant tissue culture is a method of propagation that has been sprouting in popularity as an alternative to cloning.

The plant can be cloned artificially using tissue culture. Vegetative propagation works because the end of the cutting forms a mass of nonspecialized cells called a callus, the callus will grow divide and form various specialized cells eventually forming a new plant.

Question 3.
(a) If the restriction enzyme has to cut a DNA, the DNA must be in pure form, i.e. free from the associated RNA and proteins. How is it achieved?
(b) Represent only diagrammatically the steps in the recombinant DNA (r DNA) technology.
Answer:
(a) The RNAs are removed by using enzymes called ribonucleases (RNases) The proteins are removed by using enzyme proteases.
(b)

Question 4.
(a) Show only diagrammatically the three steps in the polymerase chain reaction,
(b) How is repeated amplification achieved using this method?
Answer:

(b) Repeated amplification is achieved by the use of a thermostable DNA polymerase, it is isolated from the bacterium Thermus aquaticus. It remains active during the high temperature used for denaturation of the double-stranded DNA.

Question 5.
Make a diagrammatic representation of showing a restriction enzyme, the substrate DNA on u which it acts the site at which it cuts DNA and the product it produces.
Answer:

You can Download Chapter 6 Molecular Basis of Inheritance Questions and Answers, 2nd PUC Biology Question Bank with Answers, Karnataka State Board Solutions help you to revise complete Syllabus and score more marks in your examinations.

Karnataka 2nd PUC Biology Question Bank Chapter 6 Molecular Basis of Inheritance

2nd PUC Biology Molecular Basis of Inheritance Ncert Text Book Questions and Answers

Question 1.
Group of the following as nitrogenous bases and nucleosides:
Adenine, Cytidine, Thymine, Guanosine, Uracil, and Cytosine.
Answer:
Nitrogenous bases: Adenine, Thymine, Uracil and Cytosine.
Nucleosides: Cytidine and Guanosine.

Question 2.
If a double-stranded DNA has 20 percent of cytosine, calculate the percent of adenine in the DNA.
Answer:

Question 3.
If the sequence of one strand of DNA is written as follows:
5ATGCATGCATGC ATGCATGCATGC ATGC-3′
Write down the sequence of the complementary strand in 5′ → 3’ direction.
Answer:
In 3’→ 5′ direction, 3′ –
TACGTACGTACGTACGTACGTACGTACG-5′
In 5’→ 3′ 5′ direction, 5-
GCATGCATGCATGCATGCATGCATGCAT-3′.

Question 4.
If coding strand in a transcription unit is written as follows:
5′-ATGCATGCATGCATGCATGCATGCATGC – 3′ Write down the sequence of mRNA.
Answer:
UACGUACGUACGUACGUACGUACGUACG

Question 5.
Which property of DNA double helix led Watson and Crick to hypothesize a semi-conservative model of DNA replication? Explain.
Answer:
The complementary base pairing property of DNA double helix led Watson and Crick to hypothesise a semi-conservative mode of DNA replication. Watson & Crick observed that the nitrogenous bases are in complementary pairing in two strands of the double helix of DNA molecule. Such an arrangement of DNA molecules led them to hypothesize the semi-conservative mode of replication of DNA.

Question 6.
Depending upon the chemical nature of the template (DNA or RNA) and the nature of nucleic acids synthesized from it (DNA or RNA), list the types of nucleic acid polymerases.
Answer:
The types of nucleic acid polymerases required for the synthesis of DNA and RNA are :

1. DNA polymerase I, II, and III – They help in the replication of DNA.
2. RNA dependent DNA polymerase – It helps in the synthesis of DNA from RNA (reverse transcription).
3. DNA dependent RNA polymerase – It helps in the synthesis of RNA from DNA (transcription).

In eukaryotes, there are at least three RNA polymerases in addition to those found in cell organelles.

• RNA polymerase I transcribes rRNA (28S, 18S, and 5.8 S).
• RNA polymerase II transcribes the precursor of mRNA called heterogeneous nuclear RNA (hnRNA).
• RNA polymerase III transcribes tRNA, 5SrRNA, and snRNAs.

Question 7.
How did Hershey and Chase differentiate between DNA and protein in their experiment while proving that DNA is the genetic material?
Answer:
A. D. Hershey and Martha Chase performed an experiment to prove that DNA is the genetic material not the protein. They used T₂ bacteriophages which infect the Escherichia coil. To differentiate between DNA and protein during the experiment, they used radioactive isotopes of phosphorus (³²P) and sulphur (³⁵S) in the culture medium of bacterial hosts. They could not use radioactive isotnpes of carbon (¹⁴C) etc., because they are present in both molecules and would not distinguish the two. They grew two cultures of Eschcrichia cou.

One culture was supplied with radioactive sulphur, ³⁵S. The other culture was provided with radioactive phosphorus, ³²P. Radioactive sulphur gets incorporated into sulphur containing amino acids and therefore, becomes part of bacterial proteins. Radioactive phosphorus gets incorporated into nucleotides which form nucleic acids, mostly DNA. Therefore, bacteria of both cultures became labelled.

They then introduced bacteriophage T₂ in both bacterial cultures. The virus entered the bacteria where it multiplied. The viral progeny was tested in both cases. It was labelled, one type with radioactive protein and the other types with radioactive DNA. These labelled phages were introduced in new bacterial cells. It was found that phages with labelled protein did not make the bacterial host labelled, while those with labelled DNA made the host labelled. It indicated that viral DNA is the genetic material (not proteins) which is transferred to the infected host.

Question 8.
Differentiate between the followings:
(a) Repetitive DNA and Satellite DNA
Answer:

Repetitive DNA	Satellite DNA
(i) the sequence of N2 bases present in more than 1 copy in a genome.	(i) part of DNA  having repeated short sequences of N2 bases
(ii) Repeated DNA sequence may or may not be present in the text.	(ii) Repeated sequence occur in tandem
(iii) Variability may or may not be present	(ii) Variability occurs

(b) mRNA and tRNA
Answer:

mRNA	tRNA
(i) Large-sized RNA with cap and tail. (ii) Carries codon information	(i) Small-sized with 3-4 loops and a limb (ii) Carries infor­mation for the association of AA with anti-Condon for incor­poration

(c) Template strand and Coding strand
Answer:

Template strand	Coding strand
(i) The strand of DNA takes part in transcription.	(i) It does not take part in transcription.

Question 9.
List two essential roles of ribosomes during translation.
Answer:
Two essential roles of the ribosome during translation are as follows:

• They provide a surface for binding of mRNA in the groove of the smaller subunit of the ribosome.
• As a larger subunit of the ribosome has peptidyl transferase on its ‘P’ site, therefore it helps in joining amino acids by forming peptide bonds.

Question 10.
In the medium where E. coli was growing, lactose was added, which induced the lac operon. Then, why does lac operon shut down sometime after the addition of lactose in the medium?
Answer:
Lac operon is switched on adding lactose in the medium, as lactose acts as inducer and make repressor inactive. Due to this switch on of lac operon system, (3-galactosidase is formed which converts lactose into glucose and galactose. As soon as lactose is consumed, the repressor again becomes active and causes a switch off (shut down) of the system.

Question 11.
Explain (in one or two lines) the function of the followings:

(a) Promoter
Answer:
It is a gene that lies near the operator which functions as binding site structural genes if the operator allows.

(b) tRNA
Answer:
Functions as adapter molecule that picks  up a particular amino – acid from the cellular pool and takes same over A site, m – RNA for incorporation into a polypeptide

(c) Exons
Answer:
Coding segments present in the primary transcript which on splicing by snRNP’s get joined to form functional m-RNA.

Question 12.
Why is the Human Genome project called a megaproject?
Answer:
HGP (Human Genome Project) is called a megaproject because:

• it involved many countries (USA, UK, Japan, France, Germany, China) for determining the nucleotide sequences of genes
• it involved sequencing 3 x 10⁹ base pairs costing 9 billion US dollars
• it required bioinformatics databasing and other high-speed computational devices for analysis, storage, and retrieval of information.

Question 13.
What is DNA fingerprinting? Mention its applications.
Answer:
DNA Fingerprinting: Every human individual is characterized by unique print at the fingertips. The study of fingers, palm, and sole print is called dermatologyphics’.

Like prints of the fingertips, each individual has a unique DNA fingerprint. Unlike the prints of a finger, the DNA fingerprints can not be altered by surgery. The latter is exactly similar in all the cells and tissues of an individual. It can not be changed by medical treatment. The distinction of individuals on the basis of DNA fingerprint is due to the sequence of nucleotides in whole genomic DNA.

The technique to identify a person on the basis of his/her DNA specificity is called DNA fingerprinting. This was invented by Sir Alec Jeffreys in 1984 at Leicester University U.K. In India, Dr. V. K. Kashyap and Dr. Lalji Singh started this technique at CCMB, Hyderabad.

Question 14.
Briefly describe the following:
(a) Transcription
Answer:
Transcription – Formation of-RNA over the template of DNA.

The snRNA formed has codon information similar to the sense or coding strand of DNA with just U replacing T.

• The DNA strand which functions as a template is a template or an anti-sense strand.
• Only one of the DNA strands is transcribed
• The main enzyme taking part in transcription is RNA polymerase.

(b) Polymorphism
Answer:
Polymorphism It is the occurrence of more than one form of genetic material
Types:

• Allelic polymorphism – multiple alleles in a gene. So it alters the structure and function of the protein formed.
• SNP or single nucleotide Polymorphism – Unique in every human being. Useful in locating specific alleles, disease-associated sequences.
• RFLP or Restriction Fragment Length polymorphism – Different sized fragments are formed by cleavage with same enzyme most RFPL have no effect on phenotypes.

(c) Translation
Answer:
Translation (Biosynthesis of proteins) Coded genetic message brought by m – RNA from DNA is charged into a polypeptide chain (proteins).

DNA transcription mRNA translation Proteins Materials required for translation are ribosomes, AA, t RNA’s, aninoacylt – RNA synthetase, m – RNA and same factors Step in Translation. Activation of Amino Acids → charging of t RNA→ Initiation → Elongation → Termination → Modification.

(d) Bioinformatics
Answer:
Science of handling, storing as databases, analysing, modeling and providing access to various aspects of biological information, especially molecules connected with genomies and proteomics
Applications

• Organisation of Biological Data
• Functional Genomics
• Proteomics, Pharmacogenomics
• Medical informatics, chemoinformatics
• Faster drey research.

2nd PUC Biology Molecular Basis of Inheritance Additional Questions and Answers

2nd PUC Biology Molecular Basis of Inheritance One Mark Questions

Question 1.
Structure formed by regulation + structural + operator+ promoter gene.
Answer:
Operon.

Question 2.
Capping and tailing are seen during the transcription of RNA.
a. How is this process done?
b. What is the use of this process?
Answer:
a. Capping is done by adding methylated guanosine at the 5′ end. Tailing is done by adding 200-300 adenylate residues at the 3′ end.
b. Capping and tailing are used to process the hnRNA. The fully processed hnRNA is called mRNA.

Question 3.
What are the group of cells or organisms which have the same hereditary characters?
Answer:
Clone.

Question 4.
What is a codon?
Answer:
It is the triplet nitrogenous base sequence which codes for one amino acid. It lies on the m- RNA.

Question 5.
What is an anticodon?
Answer:
It is the triplet nitrogenous bases in the t-RNA complementary to an mRNA codon. It identifies a particular codon on mRNA.

Question 6.
In the lac operon, there is no apo-repressor and co-repressor. But it is present in tryptophan operon.
a. Who proposed tryptophan operon?
b. What is the significance of the two types of repressors given above?
Answer:
a. Francois Jacob and Jacques Monod
b. Apo-repressor: This is a protein produced by the regulator gene which is unable to bind the operator gene.
Co-repressor: When tryptophan binds with the apo-repressor, it is known as co-repressor.

Question 7.
Name the enzyme which takes part in transcription.
Answer:
RNA Polymerase.

Question 8.
Name the enzyme that catalyzes
(a) Replication of DNA and
(b) Formation of RNA. (CBSE 1995)
Answer:
(a) Topoisomerase.
(b) RNA polymerase.

Question 9.
Which bond is made in DNA when joining the sugar and phosphoric acid?
Answer:
Phosphodiester bond.

Question 10.
Name the bond present between two adjacent nucleotides
Answer:
Phosphodiester bond.

Question 11.
Due to a mistake during transcription, ATG forms UAG in m-RNA. What change would occur in the polypeptide chain translated by this mRNA? (Cbse 1996)
Answer:
UAG is a termination codon and so at point protein synthesis will get stopped.

Question 12.
Name three different non – sense codons.
Answer:
UAA, UAG and UGA.

Question 13.
Write the full name of Sn RNP.
Answer:
Small nuclear Ribonucleo Proteins.

Question 14.
Name the scientist who proposed one gene-one enzyme hypothesis. (CBSE 1997)
Answer:
Beadle and Tatum.

Question 15.
Name the enzyme that joints the short pieces in the lagging strand during synthesis of DNA.
Answer:
DNA legase. (CBSE 1998)

Question 16.
In which direction 5′ – 3′ or 3′ -5′ are the new strands of DNA formed during replication ? (CBSE 1992, 2K)
Answer:
5′ – 3′ direction

Question 17.
Give the present-day representation of central dogma.
Answer:

Question 18.
State Chargaff s base complementary rule.
Answer:
The total molar amount of adenine in any specimen of DNA is always equal to that of thymine. In a given DNA A = T and G = C.

Question 19.
Define mutation
Answer:
It is an abrupt and distinct change in the structure of base pair. It is a discontinuous inheritable and sudden change in an organism.

Question 20.
What do you call the kind of mutation in which a single base is added to a base strand : (CBSE 2K)
Answer:
Frameshift mutation.

Question 21.
Sickle cell anaemia is caused due to abnormal haemoglobin. Which chain of haemoglobin is responsible for this disease?
Answer:
β – chain of haemoglobin

Question 22.
What do the triplets AUG and UGA respectively code for during proteins synthesis?
Answer:
AUG – Methionine UGA – Termination codon (Nonsense codon)

Question 23.
Name the technique used by Watson and Crick to propose the double-helical structure of DNA molecule?
Answer:

• X-ray crystallography
• X-ray diffraction method.

Question 24.
Who discovered nucleic acid DNA? What was it called them?
Answer:
Fredrich Meischer. It was called nuclein.

Question 25.
(a) What is the length of the pitch of helix?
(b) What is the distance between 2 base pairs is a stand of DNA?
Answer:
(a) 3.4 nm
(b) 34 nm.

Question 26.
Why is the distance between the 2 nucleotide chains in a DNA maintained almost constant?
Answer:
The complementary base pairing between the 2 strands i.e. adenine and thymine (double bond) and Guanine and cytosine (triple bond) is responsible for the uniform distance between the 2 strands.

Question 27.
Name the process which occurs in virus where the formation of DNA occurs from RNA.
Answer:
Reverse transcription.

Question 28.
Name the components ‘a’ and ‘b’ in the nucleotides with a parent given below. (CBSE, Delhi 2008)
Answer:

Question 29.
What are histones?
Answer:
Histones are positively charged proteins found in association with DNA in a eukaryotic cell.

Question 30.
Name 2 organisms where RNA is the genetic material.
Answer:
QB Bacteriophage, Tobacco mosaic virus.

Question 31.
Name the main enzyme involved in the replication of DNA.
Answer:
DNA dependent DNA polymerase enzyme.

Question 32.
Name the types of synthesis ‘a’ and ‘b’ occurring in the replication of DNA as shown below (CBSE 2008)
Answer:

• a – continuous synthesis
• b – discontinuous synthesis.

Question 33.
What is a replication fork in DNA?
Answer:
During DNA replication the unwinding of DNA leads to the formation of a ‘Y’ shaped structure to the 2 strands of DNA duplex. This is known as the replication fork.

Question 34.
Define a cistron
Answer:
A cistron is defined as the length of mRNA, that codes for a polypeptide.

Question 35.
What is meant by hnRNA?
Answer:
The hnRNA is the precursor of mRNA transcribed by RNA polymerase II in eukaryotic cells hn represents heteronuclear RNA.

Question 36.
When and at what end does the ‘tailing’ of hnRNA take place? (AI 2009)
Answer:
After splicing, tailing occurs at the 3′ end of hn RNA.

Question 37.
Why is hn RNA required to undergo splicing?
Answer:
Since hn RNA contains both the coding sequences (exon) and the non-coding sequences (introns), hn RNA has to undergo splicing for the removal of introns.

Question 38.
Who proposed the operon concept?
Answer:
Francois Jacob and Jacque Monod proposed the operon concept.

Question 39.
Name the induce of lac operon in E. Coli?
Answer:
Lactose.

Question 40.
Given below is a schematic representation of a lac operon in the absence of an inducer. Identify ‘a’ and ‘b’ in it
Answer:

a – Repressor
b – operator.

Question 41.
What term is given to a single base DNA difference?
Answer:
When the repressor binds to the operator, the operon is switched off and transcription is stopped.

Question 42.
Expand VNTR.
Answer:
Variable Number of Tandem Repeats.

Question 43.
Regulation of lac operon by a repressor is referred to as negative why is it so?
Answer:
When the repressor binds to the operator, the operon is switched off and transcription is stopped.

Question 44.
Name 2 plants whose genome has been sequenced.
Answer:
Rice and Arabidopsis.

Question 45.
Define DNA polymorphism
Answer:
Inheritable mutations at high frequency in a population. It refers to the variation at the genetic level.

Question 46.
Who discovered the techniques of DNA fingerprinting
Answer:
Alec Jeffreys.

Question 47.
What is a probe in DNA-finger printing?
Answer:
A probe is a short stretch of DNA, with the nucleotide sequence that is complementary to that of the VNTR sequence.

Question 48.
Name the branch of science that HGP is closely associated with.
Answer:
Bioinformatics.

Question 49.
Given below is the sequence of steps of transcription in eukaryotic cells.
Fill up the blank 1, 2, 3, 4 left in the sequence.
Answer:

Single nucleotide polymorphism (SNPs).

1. RNA polymerase
2. hn
3. processed RNA
4. tail.

Question 50.
Write any 3 unusual bases present in Yeast’s alanine tRNA with their sources.
(CBSE 2004)
Answer:
All tRNA have 2 unusual bases – dihydro uridine (derived from uracil) and pseudouridine (from uracil. The third common unusual base is hypoxanthine (from adenine).

2nd PUC Biology Molecular Basis of Inheritance Two Marks Questions

Question 1.
What is genetic code? What do you know about the discovery of genetic code?
Answer:
Genetic code is that sequence of three nitrogenous bases of mRNA in which genetic information for the synthesis of one amino acid is coded.

The triplet codons of the genetic codes are discovered for the first time by M.W. Nirenberg in 1950. He synthesized an RNA by the use of a repetitive sequence of Uracil which is called polyuracil (UUUUU …………). They added synthesized mRNA to a cell-free extract containing protein-synthesizing enzymes and ribosomes from E. coli together with a mixture of 20 amino acids. The only molecules synthesized in a polypeptide chain were phenylalanine and their number in the chain was one-third of the Uracil base on poly-U-m-RNA. This confirmed the triplet nature of the genetic code.

Question 2.
Label the diagram 1, 2, 3, 4, 5, 6. (CBSE 2004)
Answer:

• 5′ end
• Ribosome binding site
• Start
• Stop signal
• Open reading frames (ORFs)
• 3′ end.

Question 3.
Name the components (parts) A and B of the transcription unit ‘given below.
Answer:

A – Promoter
B – Coding strand.

Question 4.
(a) Who first proposed a semi-conservative mode of replication of DNA.
(b) Which organisms are used in this experiment?
(c) Name the techniques used
(d) What is the result of first, second, and 3rd generations?
Answer:
(a) Watson and Crick (1953)
Escherichia Coli

• Use heavy isotope ¹⁵N instead of normal ¹⁴
  
• Use of Cesium chloride-based density gradient configuration with ethidium bromide as a fluorochrome.
• First-generation DNA is hybrid as with intermediate heaviness due to the presence of both ¹⁴N and ¹⁵N strands. The parent generation is the heaviest.

Second generation 50% light (¹⁴N/¹⁴N) and 50% intermediate heavy (¹⁴N/¹⁵N). Third generation75% light (¹⁴N/¹⁴N) and 25% of intermediate heaviness (¹⁴N/¹⁵N). This is possible only if DNA is double-stranded with semi-conservative replication. (one parent strand and other new strands) UACGAG AGAUUUi

Question 5.

Study the messenger RNA segment given above which is complete to be translated into a polypeptide chain

• Write the codons and ‘a’ and ‘b’
• What do they code for?
• How is a peptide bond formed between 2 amino acids in the ribosome (CBSE 2008)

Answer:
(i) a – AUG
b – UAA | UAG | UGA

(ii) AUG codes for methionine.
UAA/ UAG/ UGA is stop / nonsense codon.

(iii) Peptide bond is formed between -COOH group of P- site amino acid and NH₂ – group of A – site amino acid with the help of ribozine peptidyl transferase provided by ribosome.

Question 6.
State any one reason to explain why RNA virus mutate and evolve faster than other viruses.
Answer:
RNA is an unstable highly reactive molecule due to its single-stranded structure and exposure of its nitrogen bases. But DNA is stable, the molecule, as its, nitrogen bases are not exposed, became its double-helical nature.

Repressor binds to the operator region
(o)  and prevent RNA polymerase from transcribing the given.
Look at the figure above depicting lac operon in E-Coii.
(a) What could be a series of events when an inducer is present in the medium in which E. Coli is growing.
(b) Name the inducer.
Answer:
(a) When the inducer is present in the medium it is absorbed at first slowly into the bacterium. The inducer binds with the repressor attached to the operator gene the repressor leaves the operator gene and allows the RNA polymerase to pass from the promotor to the structural genes for transcription or formation of a polycistronic mRNA.
(b) Lactose or galactoside.

Question 7.
Draw schematically a single polynucleotide strand (with at least three nucleotides)
Answer:

Question 8.
Differentiate between euchromatin and heterochromatin.
Answer:
Euchromatin

• These are the regions where chromatin is loosely packed.
• Euchromatin stains lighter
• This is transcriptionally more active.

Hetero chromatin

• These are the regions where chromatin is tightly packed
• Hetero chromatin stains darker
• This is transcriptionally less active or inert.

Question 9.
Write notes on the structural gene, regulator gene, and operator gene.
Answer:
Structural gene – It is a type of gene containing the code for the synthesis of enzymes necessary for lactose catabolism.
Regulator gene – The regulator gene is a type of control gene which is involved in lactose catabolism.
Operator gene – It is also a type of control gene involved in lactose metabolism.

Question 10.
Compare the roles of the enzymes DNA polymerase and DNA ligase in the replication fork of DNA.
Answer:
DNA polymerases the nucleotides in the 5’ → 3′ direction, as a continuous stretch on the template strand with 3′ → 5′ polarity and short stretches on the template strand with 5′ → 3′ polarity. → DNA ligase joins the short stretches of DNA formed on the template strand with 5’→ 3′ polarity.

Question 11.
Write the differences between mono-cistronic and polycistronic mRNAs.
Answer:

Monocistromic mRNA	Polycistromic mRNA
(1) It is the mRNA that can code for only one polypeptide i.e. it has one cistron. (2) it is normally found in eukaryotic cells.	(1) It is the mRNA that can code for more than one polypeptide i.e. it has more than one cistron. (2) It is found in prokaryotic cells.

Question 12.
Differentiate between Exons and Introns.
Answer:
Exons (HOTS):

• They are the coding sequence of DNA/RNA transcript, that form parts of mRNA and code for different regions of the polypeptide.
• They are joined together during splicing to make the information continuous.

Introns:

• Introns are non-coding sequences of DNA/ RNA transcript that do not become part of mRNA.
• They are removed during splicing.

Question 13.
Amino acids are the building blocks of protein. How is a protein synthesised from aminoacids?
Answer:
By linking the aminoacids with the help of peptide bonds.

Question 14.
Why is that transcription and translation can be coupled in the prokaryotic cell but not in eukaryotic cells?
Answer:
(a) In prokaryotes, mRNA does not require any processing to become active

(b) Transcription and translation occur in the same compartment cytosol, as there is no well-defined nuclear membrane. Therefore it can be coupled.

In eukaryotic mRNA has to be processed (splicing) before it becomes active. Since RNA is synthesized inside the nucleus and translation occurs in the cytoplasm, coupling of transcription and translation is not possible.

Question 13.
What is transcription? Name the enzyme catalyzing it.
Answer:
Transcription: The formation of mRNA from DNA in the presence of an enzyme is called transcription. It is the first stage of protein synthesis which is catalysed, by an RNA polymerase enzyme. The process of transcription involves the following steps:

1. Exposing of the bases of DNA: The two strands of DNA are separated due to the presence of an unwinding protein and thus, their bases are exposed. The exposed chain of DNA functions as a template for the synthesis of mRNA in the presence of RNA polymerase enzyme.

2. Base pairing: The ribonucleotides are jointed in a definite fashion on the exposed strand of DNA. G is bonded with ‘C’, ‘C’ bonded with ‘G’, ‘T’ bonded with ‘A’, and ‘A’ bonded with‘‘T’respectively.

3. Synthesis of RNA chain: The new ribonucleotide bonded on DNA template are jointed with the help of RNA polymerase and thus, forming a new chain of RNA. Then this mRNA is separated from DNA and reaches the cytoplasm. Where it combines with ribosomes and thus, initiating the synthesis of protein.

Question 16.
What is an operon? How does lactose act as an inducer in the lac operon? (Al 2008)
Answer:
All the genes controlling a metabolic pathway, collectively constitute an operon. Lactose binds to the repressor and inactivates it, and prevents it from binding to the operator. As a result, the RNA polymerase gets access to the promoter and transcription proceeds, ie., the operon is induced to function.

Question 17.
Explain VNTR as the basis of DNA fingerprinting.
Answer:

• Variable number of tandem repeats belong to the class of satellite DNA referred to as minisatellite.
• The number of repeats shows very high degree of polymorphism as a result the size of VNTR varies.
• After hybridization of DNA sample with VNTR probe, an autoradiogram developed which gives many bands give the characteristic pattern of an individual DNA.

Question 18.
Give the applications of DNA fingerprinting?
Answer:

• To identify criminals in the forensic labs.
• To determine the biological parent in case of dispute.
• To verify whether an immigrant is really a close relative of the mentioned resident.
• To identify racial groups to rewrite the biological evolution.

Question 19.
How is the nucleosome formed? Draw a diagram of the nucleosome.
Answer:

In eukaryotes, histones which are positively charged proteins, become organised as a unit of 8 molecules, called histone octamer. The negatively charged DNA is wrapped around the positively charged histone octamer, to form the structure, called a nucleosome.

Question 20.
What is meant by R – cells and S – a cell with which Frederick Griffith carried out his experiments on Diplococcus pneumonia? What did he prove from these experiments?
Answer:
R – cells are those bacterial cells that form rough colonies, without a capsule and are non-virulent. S – cells are those cells, that form smooth colonies, with a capsule and are virulent. He proved that virulence of S – cells had somehow been transferred into R – cells, which became transformed into S- cells, it is the genetic material that had affected transformation.

Question 21.
Draw the schematic representation to show the continuous or discontinuous synthesis of DNA (replication of DNA) and label it.
Answer:

Question 22.
Explain frames shift mutation.
Answer:

• It is the type of mutation where insertion or deletion of one or two bases changes the reacting frame from the point of insertion or deletion.
• When three or multiples of three bases are added there is the addition of one or more amino acids and the reacting of the frame remains unaltered after that. This provides genetic proof that codons are triplets.

Question 23.
A tRNA is charged with the amino acid phenylalanine.
(a) At what end of the tRNA is the amino acid attached?
(b) Give the mRNA codon that codes for phenylalanine.
(c) Which enzyme is responsible for this attachment?
Answer:
(a) Amino acid is attached to the 3′ end of tRNA.
(b) UUC or UUU are the codes for phenylalanine.
(c) Amino Acyl tRNA synthetase enzyme is responsible for this attachment.

Question 24.
(a) Draw the schematic diagram of tRNA showing the following.
(i) Methionine attached to the amino acid accept the site.
(ii) Correct base sequence at the anticodon loop.
(b) Write the role of “untranslated regions” on mRNA segment play in protein synthesis?
Answer:
(a)

(b) They are needed for efficient translation

Question 25.
Draw a labeled diagram depicting schematically the process of elongation in translation.
Answer:

Question 26.
Describe the goals of Human genoine project.
Answer:
The major goals of Human genome project are

• Determine the sequence of the 3 billion base pairs present in Human DNA.
• Identify the genes in the human DNA.
• Store the information in databases.
• Improve the tool for data analyses.
• Address ELSI (ethical, legal, social issues) that arise from the project.
• Transfer the technologies to other sectors.

2nd PUC Biology Molecular Basis of Inheritance Five Marks Questions

Question 1.
Describe the features of the double-helical model of DNA.
Answer:
The features of double-helical DNA are

• It is made up of 2 polynucleotides of sugar-phosphate and the nitrogen bases inside.
• The two chains have antiparallel polarity ie., one has 5′ → 3’ polarity and the other with 3′ → 5′ polarity.
• The bases of 2 strands are joined by a double hydrogen bond between adenine and thymine and a triple hydrogen bond between guanine-cytosine.
• The distance between 2 base pairs is 0.34nm and the distance between each turn is 3.4nm. Each turn consists of 10 base pairs.
• The plane of one base pair stacks over the other in double Helen this gives the stability of the double-helical structure.

Question 2.
(1) Represent diagrammatically the Watson Crick model for semi conservation replication of DNA.
(2) Differentiate between continuous and discontinuous synthesis of DNA.
Answer:

(2)

Continuous	Discontinuous
(a) One strand of DNA is synthesised as a continuous stretch in the 5′ → 3′ direction.	(a) Short stretches  are synthesised in the 5’→ 3′ direction from the replication fork
(b) The template strand of the DNA strand is with 3’→ 5′ polarity.	(b) The DNA strand with 5′ → 3′ polarity is the template strand for this.
(c) No need for enzyme ligase (for joining)	(c) DNA ligase enzyme is required for joining short stretches.
(d) There is no need for primers.	(d) There is a need for primers.

Question 3.
Write short notes on different types of RNAs.
Answer:
The RNAs are of 3 types

• Messenger RNA (m – RNA)
• Transfer RNA (t – RNA)
• Ribosomal RNA (r – RNA)

m – RNA It provides the template for polypeptide synthesis, ie., it decides the sequence of amino acids in the polypeptide through the sequence of bases on it. t – RNA It has the shape of a cloverleaf in a two-dimensional structure It transports the amino acids to the site of protein synthesis it recognizes the codon on m RNA. r – RNA It plays the structural and catalytic role during translation.

Question 4.
Define genetic code and write its salient features.
Answer:
Genetic code is the relationship between the sequence of nucleotides on mRNA and the sequence of amino acids in the polypeptide.
Its features are

• The genetic code is universal ie., the codons code for one amino acid is the same in all organisms.
• Codons are triplet codons and there are 64 codons. 61 codons code for twenty different amino acids while the other three (UAA, UAG, UGA) are termination codons that do not code for any amino acids.
• Each codon codes for only one particular amino acid. Therefore it is unambiguous.
• Some amino acids are coded by more than one codon there for it is said to be degenerate.
• The codons are read in a continuous manner, without any punctuation ie., codons are comma less.
• AUG has dual functions of coding for methionine as well as functioning as initiation codon.

Question 5.
(1) Represent schematically the process of transcription in eukaryotic cells.
(2) How does DNA polymerase function in the replication fork of DNA?
Answer:

(2) The DNA polymerase can catalase the polymerisation of nucleotides only in one direction is 5′ → 3′ direction.
On the template stand with 3′ → 5′ polarity the polymerisation occurs continuously and on the template strand with 5′ → 3′ polarity, polymerisation occurs in short stretches (discontinuous synthesis).

Question 6.
(a) Explain with the help of schematic representation, the lac operon in E-Coli.
(b) Mention the role of lactose in this operon.
Answer:
(a) Jacob and Monod explained that lactose inducer the expression of genes leading to its catabolism.

(b) Role of lactose

• Lactose is a substrate for the enzyme (3- galactosidase.
• Its functions as the inducer and regulates the switching on and off of the operon.
• When lactose is present, it combines with the repressor protein which otherwise has a high affinity for the operator.
• This inactivates the repressor from binding to the operator and hence transcription continuous ie., the operon is switched on.

You can Download Chapter 4 Reproductive Health Questions and Answers, 2nd PUC Biology Question Bank with Answers, Karnataka State Board Solutions help you to revise complete Syllabus and score more marks in your examinations.

Karnataka 2nd PUC Biology Question Bank Chapter 7 Evolution

2nd PUC Biology Evolution  NCERT Text Book Questions and Answers

Question 1.
Explain antibiotic resistance observed in bacteria in light of Darwinian selection theory.
Answer:
When a bacterial population encounters a particular antibiotic, those sensitive to it, die. Sometimes a bacterial population happens to contain a few bacteria having mutations which make them resistant to the antibiotic. Such resistant bacteria survive and multiply quickly. Soon the resistance providing genes become widespread and entire bacterial population becomes resistant. This type of sorting out of the organisms with useful variations has been called as ‘natural selection’ by Darwin.

Question 2.
Find out from newspapers and popular science articles any new fossil discoveries or controversies about evolution.
Answer:

• Fossil of small terrestrial dinosaur with feathers covering limbs and body. (Archaeopteryx lithographic)
• Mesohippus – intermediate horse size of goat with 3 toes on each foot and molar teeth had serration.

Question 3.
Attempt giving a clear definition of the term species.
Answer:
Species (used both as singular and plural) is a natural population of individuals or group of populations which resemble one another in all essential morphological and reproductive characters so that they are able to interbreed freely and produce fertile offspring.

Question 4.
Try to trace the various components of human evolution (hint: brain size and function, skeletal structure, dietary preference, etc.)
Answer:

Question 5.
Find out through the internet and popular science articles whether animals other than man have self-consciousness.
Answer:
Yes, chimpanzees are most near to man than any other living animal and have self-consciousness.

Question 6.
List 10 modern-day animals and using the internet resources link it to a corresponding
ancient fossil. Name both.
Answer:
Please do survey in internet.

Question 7.
Practice drawing various animals and plants.
Answer:
Please practice drawing.

Question 8.
Describe one example of adaptive radiation.
Answer:
The process of evolution of different species in a given geographical area starting from a point and literally radiating to other areas of geography (habitats) is called adaptive radiation. Adaptive radiation gave rise to Australian marsupials. A number of marsupials, each different from the other evolved from an ancestral stock, but all within the Australian island continent.

Question 9.
Can we call human evolution adaptive radiation?
Answer:
No, we cannot call human evolution adaptive radiation

Question 10.
Using various resources such as your school library or the internet and discussions with your teacher, trace the evolutionary stages of any one animal say horse.
Answer:
E.g.: Horse – See textbook diagram

2nd PUC Biology Evolution Additional Questions and Answers

2nd PUC Biology Evolution One Mark Questions 

Question 1.
Name the non-cellular organism.
Answer:
Virus.

Question 2.
It is said the earth has formed about 4.5 billion years back. How did the earth crust and atmosphere formed according to Big Bang theory?
Answer:
It is said there was no atmosphere on early earth and the earth was a hot ball of gases. With the passage of years, the gases condensed into a molten core and different elements got stratified according to their density. Heavy elements such as iron, nickel etc. sank deep into centre and formed the core. Lighter elements like silicon, aluminum etc. formed the middle layer, while the lighter elements like hydrogen, helium, nitrogen etc. formed the gaseous part.

Question 3.
Releasing of 02 by the evolution of photosynthesis an ancient time is called?
Answer:
Oxygen revolution.

Question 4.
Define analogous organs. (CBSE 93,94,2007)
Answer:
The organs which perform the same function but differ in their origin and structures are called analogous organs.

Question 5.
Where did life originate?
Answer:
In water.

Question 6.
What is paleontology?
Answer:
It is the branch of science that deals with fossils.

Question 7.
Name the method of the changing of the structure of a gene resulting in a variant form that may be transmitted to.
Answer:
Mutation.

Question 8.
Name any two vertebrate body parts that are homologous to human forelimbs.
(CBSE 2008)
Answer:
Wings of birds, flippers of whale forelimbs of cheetah, wings of bats.

Question 9.
Who wrote the origin of species?
Answer:
Charles Darwin.

Question 10.
Mention the type of evolution that has brought similarity as seen in potato tuber and sweet potato.
Answer:
Convergent evolution.

Question 11.
Who proposed the Recapitulation theory or Biogenetic law?
Answer:
Haeckel.

Question 12.
Thorns of Bougainvillea and tendrils of cucurbit are analogous or homologous.
Or
Wings of bird and forelimbs of horse are homologons or analogous
Or
Flippers of penguin and dolphin are analogous or Homologies.
What type of evolution has brought similarity in these cases? (CBSE 1992, Delhi 2009, Foreign 2009)
Answer:

• All these are homologous.
• Divergent type evolution has brought this similarity.

Question 13.
Name the theory by which earth is said to originate.
Answer:
The big bang theory.

Question 14.
What kind of evidence is afforded by Darwin’s finches in support of organic evolution? (CBSE 1991)
Answer:
Adaptive radiation i.e. All these finches evolved from a common ancestor but they diverge in various directions.

Question 15.
What is a mutation?
Answer:
New species originate due to changes of hereditary characters is called a mutation.

Question 16.
Define ontogenetic law.
Answer:
Ontogeny recapitulates phytogeny.

Question 17.
Which era is called the golden period of Dinosaurs?
Answer:
Mesozoic period is called the golden period of Dinosaurs.

Question 18.
As per Hugo deVries, what is the cause of speciation? (Delhi 2008, CBSE 1995)
Answer:
Mutation. (Single-step large mutation)

Question 19.
What is meant by genetic equilibrium?
Answer:
When allele frequencies in a population are stable, the allele frequency of a population remains constant. It is called as genetic equilibrium, i.e. the sum total of all the allelic frequency is one.

Question 20.
Define evolution as per Hardy Weinberg.
Answer:
As per Hardy Weinberg, change of frequency of alleles in a population would be considered as evolution i.e. disturbance in the genetic equilibrium.

Question 21.
Who is the early man of the modern human?
Answer:
Cro-Magnon peoples are early human of modern human.

Question 22.
What is meant by gene flow?
Answer:
Changes in the gene pool of population when there is continuous migration of organisms between them i.e. it refers to the addition or loss of genes.

Question 23.
Which human form first started to walk on two legs?
Answer:
Australopithecus form first started to walk on two legs.

Question 24.
Define natural selection?
Answer:
Natural selection is the process in which heritable variations that enable better survival, are enabled to produce and leave behind a greater number of progeny.

Question 25.
On the basis of evolution which human had a brain size of 1400cc?
Answer:
Neanderthal.

Question 26.
Define the term “reproductive isolation”.
Answer:
If the population of 2 different species whether they are isolated or not, cannot enter bread to produce offspring.

Question 27.
Differentiate between Dryopithecus and Ramapithecus.
Answer:
Dryopithecus was apelike but Ramapithecus was mostly human-like.

Question 28.
Name the group of animals that evolved into amphibians.
Answer:
Lobe-fins evolved into amphibians.

Question 29.
Mention the key concepts about the mechanism of biological evolution/ speciation according to
(a) Devries and
(b) Darwin (Delhi 2008)
Answer:
(a) De Vries – Mutation
(b) Darwin – Natural selection and branching descent.

2nd PUC Biology Evolution Two Marks Questions

Question 1.
What is a virus? Why is it treated as a link between living and non-living?
Answer:
Viruses are the simplest organisms of the earth, which consists of nucleic acid (DNA or RNA) surrounded by protein cover. It shows characteristics of living as well as non-living organisms.
(A) Living characters of virus:

• The virus shows structural differentiations.
• They contain hereditary material.
• They exhibit mutation.
• They spread plant and animal diseases.
• Growth and development present.
• They exhibit adaptation.
• They possess sensitivity.

(B) Non-living characters of virus:

• Lack protoplasm and cell organelles.
• Can be crystallized.
• No metabolic activities are seen.
• Cannot reproduce outside living cells.
• They lack enzymes.

Due to the above reason, viruses are considered as a link between the living and non-living organisms, thus, it is the first life that originated on the earth.

Question 2.
Life has originated in seawater. Justify the statement.
Answer:
According to Oparin and Haldane, the first cell protobionts arose in the broth (not dilute soup). So it is considered that life has originated in water.

Question 3.
Explain the origin of the earth.
Answer:
Origin of the earth:

• Earth was formed 4-5 billion years back.
• Initially, die surface was covered with water vapOur, methane, C02 and NH3.
• The UV rays of the sun broke water into hydrogen and oxygen.
• Hydrogen escaped and oxygen combined with NH3 and CH4 to form water, C02 and other gases, also forming die ozone layer.
• Cooling of water vapour led to rain which filled the depressions on earth’s surface, forming water bodies.

Question 4.
Ontogeny recapitulates phylogeny. Discuss.
Answer:
In the higher animals their development passes through stages which are similar to adult stages of lower animals which were their ancestors. For e.g. The tadpole larva of frog represents its fish like ancestors.

Question 5.
Given below are the names of 2 pairs of limbs. Categorize them into homologous and analogous organs, give reason.
(i) Human arm and fore leg of cow
(ii) Bat’s wing and Grass hopper’s wing. (CBSE 1999)
Answer:
(i) Human arm and foreleg of cow are homologous organs because they are built upon the same fundamental plan (pentadactyl pattern) but they perform different functions as grasping in man and locomotion in cow.

(ii) Bat’s wing and grass hopper’s wing are analogous organs because they perform the same function bat differ in their origin and structure. Wings of bats are modified forelimbs whereas grasshoppers wings are modified outgrowth of the body wall.

Question 6.
Give 2 examples each of analogy and homology in plants. (Hots)
Answer:
Analogy:

• Tubers of sweet potato and potato
• Tendrils of pea and cucurbits.

Homology:

• Tendrils of cucurbit and lemon thorn
• Tendrils of Pea and Spines of opuntia

Question 7.
What do you mean by analogous?
Answer:
Analogous organs: Organs which are different in origin and structure but per¬forming similar functions are known as analogous organs and the phenomenon is called an analogy. Analogous organs do not indicate phylogeny.
Examples: Wings of butterflies are made up of chitin, wings of birds made by production of feathers on forelimbs, and skin present between the fingers of bat are the examples of analogous organs.

Question 8.
Define adaptive radiation with 2 eggs. (Hots)
Answer:
Adaptive radiation is defined as the process of evolution of different species in a given geographical area starting from single species and radiated to other habitats.
Example:

• Australian marsupials, each different from the other, have evolved from ancestral stock.
• Darwin’s finches – from the original seed-eating stock insectivores and vegetarian birds have evolved.

Question 9.
What do you mean by vestigial organs?
Or
Write the two names of vestigial organs of man.
Or
What are vestigial organs? Explain. Write four vestigial organs of the human body.
Answer:
Vestigial organs: Organs that are reduced and have become functionless in an organism but were functional in their ancestors are called vestigial organs:
Examples:

• Vermiform appendix
• Coccygeal vertebrae
• The nictitating membrane in the eyes of human
• Muscles of external ear (Pinna).

2nd PUC Biology Evolution Three Marks Questions

Question 1.
Whose theory was put to test by Miller Urey and what was the theory? How did their experiment prove the abiotic origin of life on earth?
Answer:
Oparin and Haldane proposed that the first life form could have come from the „ nonliving organic molecules like RNA, protein etc.

• The orgaine molecules must have been produced by chemical evolution, i.e. formation of divers organic molecules from inorganic constituents.
• The condition, of the earth that favoured chemical origin were

(i) very high temperature
(ii) volcanic storms
(iii) Reducing atmosphere that contained methane, ammonia and water vapour.

• Energy must have been produced by U-V radiation and lightning.
• Analysis of the products of their experiment showed the presence of amino acids which help to form proteins.

Question 2.
Give a brief account as how evolution has taken pleace from the time the non-cellular aggregate of giant molecules turned into cells.
Answer:
The first formed cells were anaerobic heterotrophs. But slowly some of these cells developed coloured proteins, that could release oxygen, in a process that could have been similar to the light reactions of photosynthesis. As oxygen started coming into the atmospheres is as the atmosphere started becoming an oxidising one, new formes of life could not arise from nonliving organic molecules.

The organisms started becoming aerobic and autotrophic. The single celled organism slowly became multicellular form, where some became autotrophic while many remained heterotrophic. Plants like bryophytes were the first to invade land, followed by reptiles among animals.

Question 3.
Differentiate between natural selection and artificial selection.
Answer:

Natural selection	Artificial selection
(a) It is the process occurring in nature over a number of generations to increase the number of fit individuals in a population. (b) The characters/ adaptations are advantageous to the organism.	(a)  It is the process practised by man over a number of generations, to select organisms with better qualities. (b) The characters are advantageous human.

Question 4.
Fossils are the written documents of evolution. Comment on it.
Answer:
Palaeontological evidence:
The direct and concrete evidence of the process of evolution can be obtained from the study of fossils. The word fossil is derived from the Latin word fossilium meaning something dugout. The term fossil refers to the remnants of the previously existed animals and plants preserved in the earth’s crust. In other words, the fossil is the dead remains of the past. The remnants include bones, teeth, shells, and other hard parts and also the impressions or imprints left by some previous organisms.

The process of preservation of organisms or their parts in the form of fossils is known as fossilization. The fossils are regarded as the written documents of evolution. The study of fossils is called paleontology. It includes paleozoology (fossil animals) and paleobotany (fossil plants). Leonardo da Vinci is considered the Father of Palaeontology.

Question 5.
How is genetic drift differ from gene migration?
Answer:

Genetic drift	Gene migration
Random changes in the allele frequencies of a population occurring only by chance events constitute genetic drift.	It refers to the change in allele frequencies of a given population when individuals ‘ migrate into the population or leave the population.

Question 6
(a) Name the largest dinosaurs and mention any two characteristic features.
(b) How did Darwin explain the existence of different varieties of finches on the Galapagos Islands?
Answer:
(a) Tyrannosaurus rex was the largest dinosaur it was about 20 feet in height. It had huge fearsome dagger-like teeth

(b) Darwin explained that all the verities evolved on the island itself from the original seed-eating birds, many other forms with altered beaks arose some others became insectivorous while remained the vegetarian flinches. Such process of evolution of different species in a given geographical area, starting from a point and literally radiating to other habitats, is called adaptive radiation.

Question 7.
Explain Landmark’s theory of evolution with an example.
Answer:
According to Lamarck, the evolution of life forms had occurred, drives by the use and disease of organs. This theory is as known as the theory of inheritance of acquired characters. He gave the examples of giraffes, who in an attempt to forage leaves on tall trees had to stretch their neck. As they passed this acquired character to the next generations giraffe slowly over the years, came to acquire tons of necks.

Question 8.
Stanley Miller and Harold C Urey performed an experiment by recreating in the laboratory the probable condition of the atmosphere of the primitive earth.
i. What was the purpose of the experiment?
ii. In what forms was the energy supplied for the chemical reactions to occur?
Answer:
i. Stanley Miller gave an experiment based on the theory of the origin of life on earth.
ii. In primitive times at the time of formation of the earth, there was a non-oxygenated atmosphere of CH₄, NH₃, CO₂, CO, NO₂, H₂O (water vapours), etc. Due to cooling and compression caused by increased pressure, the chemicals reacted, compressed against each other, and cooled (liquefied) to form the first life on earth in water.

You can Download Chapter 5 Principles of Inheritance and Variation Questions and Answers, 2nd PUC Biology Question Bank with Answers, Karnataka State Board Solutions help you to revise complete Syllabus and score more marks in your examinations.

Karnataka 2nd PUC Biology Question Bank Chapter 5 Principles of Inheritance and Variation

2nd PUC Biology Principles of Inheritance and Variation NCERT Text Book Questions and Answers

Question 1.
Mention the advantages of selecting pea plant for the experiment by Mendel.
Answer:
Mendel selected garden pea for his experimental work because it had the following advantages:

• The pea plants showed a number of well-defined contrasting characters.
• It has perfect bisexual flowers containing both male and female parts. The flowers are predominantly self-pollinating.
• Because of self fertilisation, plants are homozygous. It is, therefore, easy to get pure lines for several generations.
• It is an annual plant. Its short life cycle made it possible to study several generations within a short period.
• It is easy to cultivate.
• It is easy to cross because pollen from one plant can be introduced to the stigma of another plant.

Question 2.
Differentiate between:-
(a) Dominance and Recessive.
Answer:

Dominance	Recessive
1. Able to express even in presence of contrasting alleles. 2. Does not require another similar allele to produce its phenotype. 3. Produces complete polypeptide, protein, or enzyme. 4.  It is usually wild type allele.	1. Cannot express in presence of contrasting alleles. 2. Can produce its phenotype only along with a similar allele. 3. Forms incomplete products. 4.  Usually a mutant allele.

(b) Homozygous and Heterozygous.
Answer:

Homozygous	Heterozygous
1. Possesses similar alleles. 2.  2 types – Homozygous dominant and recessive. 3. Individual is pure for the trait. 4. On self breeding, a similar types of offspring are formed. 5. Only one type of gamete is formed	1. Possesses different alleles. 2.  Is of one type 3. Individual is seldom pure for the trait. 4. On breeding, 3 types of offspring are formed. 5. 2 types of gametes produced.

(c) Monohybrid and Dihybrid
Answer:

Monohybrid	Dihybrid
1. Cross made between individuals having contrasting traits in order to study the inheritance of a pair of alleles. 2.  Phenotypic monohybrid ratio in F2 generation is 3:1. 3. Genotypic monohybrid ratio in F2 generation is 2:1.	1. Cross made between individuals having contrasting traits in order to study the inheritance of 2 pairs of alleles. 2. Phenotypic dihybrid ratio is 9:3:3:1 3. Genotypic dihybrid ratio is 1:2:1:2:4:2:1:2:1

Question 3.
A diploid organism is heterozygous for 4 loci, how many types of gametes can be produced?
Answer:
A diploid organism heterozygous for 4 loci will have the supported genetic constitution YyRr for two characters. The alleles Y-y and R-r will be present on different 4 loci. Each parent will produce four types of gametes – YR, Yr, yR, yr.

Question 4.
Explain the law of Dominance using a monohybrid cross.
Answer:
Law of Dominance:

• Characters are controlled by discrete units called factors.
• Factors occur in pairs.
• In a dissimilar pair of factors, one member of the pair dominates (dominant) the other (recessive) suppresses.

In a monohybrid cross:

Question 5.
Define and design a test cross.
Answer:
Test cross: An organism showing dominant phenotype (genotype is to be determined.) from F₂ is crossed with a recessive parent instead of self-crossing. Progenies of such a cross can be analysed to predict the genotype of the test organism. Eg: A tall and short plant taken into the experiment. From F₂ generation, TT or Tt could be tall  I plant. The recessive parent will be tt. So, if the test plant was TT, the resulting plant will only I tall. If the test plant was Tt, the resulting plant will be tall and short.

Question 6.
Using a Punnett square, workout distribution of phenotypic features in the F₁ generation after crossing between homozygous female and heterozygous male for a single locus.
Answer:

Question 7.
When the cross is made between a hybrid tall plant with yellow seeds (Tt Yy) and tall plant with green seeds (Tt yy), What proportion phenotype in offspring could be expected to be
(a) Tall and Green
(b) Dwarf and Green
Answer:

Question 8.
Two heterozygous plants are crossed. If 2 loci are linked what would the distribution of phenotypic features in F₁ generation in a dihybrid cross.
Answer:

Question 9.
Mention the contributions of TH Morgan in Genetics.
Answer:
Thomas Hunt Morgan (1866-1945) is called father of experimental genetics. He was an American scientists, famous for his experimental research with the fruit fly (Drosophila) by which he established the chromosome theory of heredity. He discovered presence of gene over chromosomes, chromosome theory of linkage, chromosome mapping, crossing over, criss-cross inheritance & mutability of genes. Morgan’s work played a key role in establishing the field of genetics. He received the nobel prize for physiology or Medicine in 1933.

Question 10.
What is pedigree analysis? Suggest how such analysis can be useful.
Answer:
Pedigree analysis: Study of transmission of particular traits graphically over the present and the last few generations for finding out the possibility of their occurrence in future generations. So, analysis of traits in several generations of a family is called Pedigree Analysis.
Importance:

• To know the possibility of a recessive allele which may create disorder.
• Can indicate the origin of a trait in the ancestors.
• Analysis is used for genetic counseling.
• Extensively used in medical research.

Question 11.
A child has blood group O. If the father has blood group A and mother B, work out genotypes of parent and the possible genotypes of other offsprings.
Answer:
Blood group O can appear with 1° 1° only (2 recessive alleles).

Question 12.
How is sex determined in human beings?
Answer:
Chromosomal determination of sex in human beings is of XX-XY type. Human beings have 22 pairs of autosomes and one pair of sex chromosomes. The female possess two homomorphic (= isomorphic) sex chromosomes, named XX. The males contain two heteromorphic sex chromosomes, i.e., XY. All the ova formed by female are similar in their chromosome type (22 + X). Therefore, females are homogametic. The male gametes or sperms produced by human males are of two types, gymnosperms (22 + X) and androsperms (22 + Y). Human males are, therefore, heterogametic.

Sex of the offspring is determined at the time of fertilisation. Fertilisation of the egg (22 + X) with a gynosperm (22 + X) will produce a female child (44 + XX) while fertilisation with an androsperm (22 + Y) gives rise to male child (44 + XY). As the two types of sperms are produced in equal proportions, there are equal chances of getting a male or female child in a particular mating. As Y-chromosomes determine the male sex of the individual, it is also called androsome.

Question 13.
Explain the following terms with an example
(a) Co-dominance
(b) Incomplete dominance
Answer:
a. Codominance is a condition in which the F₁ generation resembles both parents, (e.g. Blood groups).
b. The type of inheritance in which the F₁ does not show the dominant character but shows an intermediate character between the dominant and recessive is called incomplete dominance.

Question 14.
What is point mutation? Give one example.
Answer:
When heritable alterations occur in a very small segment of DNA molecule i.e., a single nucleotide or nucleotide pair, then these types of mutations are called point mutations (also called gene mutations). The point mutations may occur due to inversion, substitution (transition and transversion), and frameshift (insertion and deletion) type of nucleotide change in the DNA or RNA. Phenylketonuria (PKU; Foiling 1934) is an inborn, autosomal, recessive metabolic disorder in which the homozygous recessive individual lacks the enzyme phenylalanine hydroxylase needed to change phenylalanine (amino acid) to tyrosine (amino acid) in the liver. It results in hyperphenylalaninemia which is characterized by the accumulation and excretion of phenylalanine, phenyl pyruvic acid, and related compounds.

The lack of the enzyme is due to the abnormal autosomal recessive gene on chromosome 12. This defective gene is due to substitution. Affected babies are normal at birth but within a few weeks, there is a rise (30 – 50 times) in plasma phenylalanine level which impairs brain development. Other symptoms are mental retardation, decreased pigmentation of hair and skin, and eczema.

Question 15.
Who had proposed the chromosomal theory of inheritance?
Answer:
Sutton and Boveri proposed the chromosomal theory of inheritance. The theory believes that chromosomes are vehicles of hereditary information that possess mendelian factors or genes and it is the chromosomes which segregate and assort independently during transmission from one generation to the next.

Question 16.
Mention any two autosomal genetic disorders with their symptoms.
Answer:
Phenylketonuria: This inborn error of metabolism is also inherited as the autosomal recessive trait. The affected individual lacks an enzyme that converts the amino acid phenylalanine into tyrosine. As a result, this phenylalanine is accumulated and converted into phenyl pyruvic acid and other derivatives. Accumulation of these in the brain results in mental retardation.

Sickle cell anemia: This is an autosome liked recessive trait that can be transmitted from parents to the offspring when both the partners are carriers for the gene of heterozygous.

2nd PUC Biology Principles of Inheritance and Variation Additional Questions and Answers

2nd PUC Biology Principles of Inheritance and Variation One Mark Questions

Question 1.
Give the meaning of the term allele.
Answer:
It represents a pair of genes of 2 alter-natives of the same character and occupies the same position (locus) on the homologous chromosome. OR The various forms of genes are called alleles.

Question 2.
What is known as the alternative forms of a gene that can occupy the same locus on a particular chromosome and that controls the same character?
Answer:
Allelomorph.

Question 3.
Differentiate between mutants and mutons.
Answer:
Mutants: The organism showing mutation.
Mutons: The smallest segment of DNA which can undergo mutation.

Question 4.
What do you mean by heredity material which is found in the outer part of the chromosome?
Answer:
Plasmogene.

Question 5.
What term did Mendel use for what we now call the genes? [Hots]
Answer:
Factors or determination.

Question 6.
What do you say when a gene shows a dominant effect?
Answer:
Dominant gene.

Question 7.
Define Phenotype.
Answer:
The observable or external characteristics of an organism, constitute its phenotype.

Question 8.
What is meant by true breeding?
Answer:
True breeding means that the parents will also pass-down a specific phenotypic trait to their offspring.

Question 9.
Mendel studied seven traits in garden pea. Which one or more of the following were recessive?
Wrinkled seed, axial flower, yellow colour of pod, tall, purple flower.
Answer:
Wrinkled seed

Question 10.
What do you mean by Fj in mendels law?
Answer:
First fertile generation.

Question 11.
Name the plant in which Mendel performed his experiment.
Answer:
Pea plant – Pisum sativum.

Question 12.
Who gave the name ’X’ body? Which genes are select sex determination in human beings?
Answer:
Herman Henking suggested the name and occur.es the sex determination in humans by XX and XY chromosome.

Question 13.
Mendel observed 2 kinds of ratios, i.e., 3:1 and 1:2:1 in F₂ generation in his experiments on the garden pea. Name these two kinds of ratios respectively.
Answer:

• 3:1 – Phenotypic ratio
• 1:2:1 – Genotypic ratio.

Question 14.
Give one example of Co-dominancy.
Answer:
‘A’, ‘B’ and ‘O’ genes are co-dominant which found in blood group.

Question 15.
What is epistasis?
Answer:
It is the interaction between two different genes (non-allelic genes) where one gene masks the effect of another gene.

Question 16.
What do you mean by the diagrammatic representation of hereditary characters which shows specific characters from generation-to-generation?
Answer:
Pedigree analysis.

Question 17.
In man, four types of blood groups A, B, AB, and O are controlled by three alleles of a gene. Name the type of inheritance involved in the blood group.
Answer:
Multiple allelic inheritances.

Question 18.
Give the name of any allopolyploid species which is originated in a natural way.
Answer:
Wheat (Triticum astivum).

Question 19.
What is the phenotypic and genotypic ratio of incomplete dominance?
Answer:
The phenotypic and genotypic ratios are same i.e. 1AA : 2Aa : laa

Question 20.
When are individuals exhibiting morphological characteristics of both sexes in drosophila called this method?
Answer:
Gynandromorphy.

Question 21.
Name the pigment that gives colour to the skin of man.
Answer:
Melanin

Question 22.
Names a specific DNA segment which is functioned as a unit of heredity.
Answer:
Gene.

Question 23.
What are complementary genes?
Answer:
These are the two independent pairs of genes that interact to produce a trait together but each dominant gene alone does not show its effect.

Question 24.
What is the name of the arms of a chromosome?
Answer:
Chromatid.

Question 25.
Name the pigments that control skin colouration in man. [Hots]
Answer:
Melanin.

Question 26.
Who first observed the ‘X’ chromosome?
Answer:
Henking. [Hots]

Question 27.
Why ‘X’ chromosome is called a sex chromosome?
Answer:
It involves the sex determination of an individual.

Question 28.
Name the disease caused by inadequate or defective nutrition.
Answer:
Dystrophy.

Question 29.
Define heterogamety. Give an example.
Answer:
It is a phenomenon in which organisms produce 2 types of (more than one type) gametes.
Eg: Human male, male drosophila / female fowl.

Question 30.
What is the exception of Mendel’s Independent assortment theory?
Answer:
Linkage.

Question 31.
Give the term for the factors which cause mutation.
Answer:
Mutagens.

Question 32.
Mention 2 sex-linked Mendelian disorders.
Answer:
Haemophilia and colour blindness

Question 33.
Name some diseases which can be avoided in the progeny through pedigree analysis of parents.
Answer:
Colour blindness, Tuberculosis, Turner’s syndrome.

Question 34.
Give the reason for Down’s syndrome.
Answer:
Trisomy of 21st chromosome.

Question 35.
Define trisomic condition.
Answer:
When a particular chromosome is present in 3 copies in a diploid cell, the condition is called trisomic condition.

36.
Define monosomic condition.
Answer:
When a particular chromosome is present in a single copy in a diploid cell, the condition is called a monosomic condition.

2nd PUC Biology Principles of Inheritance and Variation Two Marks Questions

Question 1.
Define test cross. What is its significance?
Answer:
It is the cross between an individual with a dominant phenotype with an individual homozygous recessive for the trait. It is used to determine the genotype of an individual for any character trait.

Question 2.
What are multiple alleles? Give an example.
Answer:
When a gene exists in more than 2 allelic forms, the alleles are called multiple alleles.
Eg: The gene ‘I’ controls the human blood group. It exist in 3 different alleleic form i.e., I^A I^B and i.

Question 3.
Write T.H.Morgan’s contribution to genetics.
Answer:

• Morgan conducted experiments in Drosophila melanogaster and discovered sex linkage.
• He also found that linked genes may show the phenomenon called linkage where the recombinants in a test than 50%. cross progeny are less.

Question 4.
Differentiate between monohybrid and dihybrid.
Answer:

Monohybrid	Dihybrid
It is a cross, where 2 forms of the single trait are hybridised.	It is a cross where 2 forms of 2 different traits are hybridised.

Question 5.
Differentiate between genotypes and phenotype.
Answer:

Genotype	Phenotype
(a) It is the total genetic constitution of an individual.	(a) It is the external appearance of an individual.
(b) It is the expression of the genome or more specifically the alleles present at one locus.	(b) It is the expression of genotype produced under the influence of an environment.

Question 6.
Differentiate between test cross and reciprocal cross.
Answer:

Test cross	Reciprocal cross
It is a cross between, F1 hybrid and recessive parent. It confirms the purity of the F1 hybrid whether it is homozygous or heterozygous.	It is the second cross involving the same strains carried by sexes opposite to those in the first cross. It is able to distinguish between nuclear chromosomal and sex-linked inheritance.

Question 7.
Is not possible to study the inheritance of traits in humans in the same way as in peas? Give 2 main reasons for it and the alternative method used for such study.[Hots]
Answer:
Mendel’s laws are not applicable for human beings because:

• In human the generation time is too long and produces a small progeny. It creates difficulty in the statistical computation of any generation of individuals.
• Human cannot be crossed at will. Pedigree analysis is another method used to study the family instances and the transmission of particular trait generation after generation.

Question 8.
What is back cross?
Answer:
When an intercross done between 2 genetically different parents, a hybrid is produced which may be homozygous or heterozygous. To determine the purity of parents and to test genotypes of F₁ hybrid, a cross is made between F₁, hybrid and of the parent. Such cross is known as backcross.

Question 9.
How would you find the genotype of an organism exhibiting a dominant phenotypic trait?[Delhi 2008]
Answer:
First a test cross will be done between the dominant phenotypic individual (F₁ hybrid) with the recessive phenotypic individual (parent). If the individuals are homozygous dominant, all the individuals in the progeny

If the individuals are heterozygous, the progeny will show dominant phenotype and recessive phenotype in the ratio 1:1.

Question 10.
Give the structure of E. coli chromosome.
Answer:
Isolated E. coli chromosomes display many individually supercoiled loops emanating from a central region. These supercoiled loops were hypothesized to be topological domains. These domains are sufficient to generate the observed precision of E. coli chromosome structure. Its chromosomal DNA has been completely sequenced by lab researchers. E. coli has a single chromosome with about 4,600 kb, about 4,300 potential coding sequences, and only about 1,800 known E. coli proteins.

Question 11.
What is meant by chromosomal abberation? In which type of cells it commonly occurs?
Answer:
Alternation in chromosomes is due to the deletion/duplication/addition of a segment of DNA is called chromosomal abberation. Commonly occurs in cancerous cells.

Question 12.
What is a modifier gene?
Answer:
Modifier genes, instead of making the effects of another gene. A gene can modify the expression of a second gene. In mice, coat colour is controlled by the B gene. THeB allele conditions black coat colour and is dominant to the b allele that produces a brown coat.

Question 13.
Classify the following into chromosomal and Mendelian disorders.
(a) Cystic fibrosis
(b) Turner’s syndrome
(c) Klinefelter’s syndrome
(d) Hemophilia
Answer:
(a) Cystic fibrosis – Mendelian disorder
(b) Turner’s syndrome – Chromosomal disorder
(c) Klinefelter’s syndrome – Chromosomal disorder
(d) Haemophilia – Mendelian disorder

Question 14.
Describe the causes of hereditary variation.
Or
Why do hereditary variations originate?
Answer:
Any change in the structure of the gene of an organism will produce heritable variations. The chief causes of these variations are as follows:

(i) Genetic recombination: The exchange of chromosomal segments of genes takes place during reduction division which causes recombination of genes. These recombinant organisms fuse to form a zygote. The genetic structure of these zygotes is different from their parents. Adults formed from this zygote will show variations.

(ii) Changes in the number of chromosomes: Changes in the number of chromosomes will also cause genetic variations, e.g., modern varieties of wheat contain 16,21 and 42 chromosomes. These varieties are produced from their parents having only 7 chromosomes.

(iii) Changes in the structure of genes: Changes in the structure of genes will also cause genetic variations, e.g., 30 varieties of wheat is the best example of genetic variations. Like this, if any changes in the genes of human beings which are related to pigment production, will inhibit the production of these pigments and thus man become albino.

Question 15.
Explain how an XXY individual can arise in humans.
Answer:
An XXY individual can arise if an abnormal egg with XX chromosomes fertilizes with a normal sperm carrying Y’ chromosome.

Question 16.
Why are women normally carriers of sex-linked diseases? Write with an example.
Or
Men are generally suffering from color-blindness but women are only carriers of this disease. Explain the reason.
Answer:
The gene for color-blindness is recessive and carried by X-chromosomes. A color-blind man has a single recessive gene (X^CY). The gene for normal vision is dominant. When sex chromosomes (X) of a man contain genes for color-blindness (X^C), they express themselves and produce color-blindness in them. A carrier woman contains one recessive gene which remains suppressed due to the presence of the dominant gene of normal vision on the other sex chromosomes (X^CX). These women play important role in the conduction of genes of color-blindness. Hence, they are carrier (X^CX). Women become color-blind only when both chromosomes contain the genes of color-blindness (X^CX^C).

Question 17.
Differentiate between haemophilia and sickle cell anemia.
Answer:

Haemophilia	Sickle cell anaemia
1. It is due to recessive defective allele present on X – chromosome. 2.  It is a sex-linked ‘disorder. 3.  A protein necessary for clotting of blood is deficient.	1. It is due to point mutation i.e., a single base pair change leading to a change in amino acid. 2. It is an autosomal disorder. 3. The defective haemoglobin leads to a change in the shape of RBCs, which become a sickle cell.

Question 18.
Differentiate between Turners and Klinefelter’s syndrome.
Answer:

Turner’s Syndrome	Klinefelter’s Syndrome
1. The individual is female.	1. The individual is male.
2. The individual has one X chromosome less i.e., she has 45 chromosomes.	2. The individual has an extra X chromosome i.e., 47 chromosomes.
3.The individual has a short stature with an underdeveloped feminine character.	3. The individual has a tall stature with a feminised character.

Question 19.
Describe Mendel’s observation on the hybridization experiments on the garden pea.
Answer:

• The F hybrids always showed one of the parental forms of the trait and there was no blending of traits.
• Both the parental form of traits appeared without any change / blending in the F₂ generation.
• The form of the trait that appeared in the F₁ generation is called the dominant character and it appeared in the F₂ generation about 3 times in frequency as that of its alternate form.

Question 20.
Why Mendel selected pea plants for his experiments?
Answer:
Mendel selected pea plants on the basis of the following characters:

• The Pea plant exhibited a number of contrasting characters.
• It normally undergoes self-pollination but can be cross-pollinated manually.
• It is an annual plant and yield’s results in a year’s time, so it can be observed for many generations.
• Many varieties were available with observable alternative forms for a trait.
• Pure varieties of pea were available which always bred true.
• A large number of seeds are produced per plant which can be easily handled and cultivated.

Question 21.
List any 7 traits in garden pea which Mendel studied in his breeding
Answer:

Question 22.
What is Klinefelter’s syndrome? Also, describe their symptoms.
Answer:
Klinefelter’s syndrome: It is a sex chromosomal syndrome. It is found only in men. They possess 47 chromosomes in their cells (22 pairs autosomes + XXY sex chromosomes). One extra X chromosome is found in them, thus, there is Trisomy in the sex-chromosome.

Question 23.
A pea plant with Purple flowers was crossed with a plant with white flowers producing 40 plants with only purple flowers. On selfing, these plants produced 470 plants with purple flowers 162 with white flowers. What genetic mechanism accounts for these results?
Answer:

F₂ generation 470 purple: 162 white. In the F₁ generation only purple flower-producing plants appeared. This means the purple colour is dominant. Which doesn’t allow the white colour to express itself.

In F₂ generation, purple and white coloured flowers were produced in the ratio 3:1. Here the parental character of white again reappeared in about one-fourth of the progeny. This occurs due to the segregation of genes in the gamete formation. This represents the law of segregation and the monohybrid ratio.

Question 24.
In human beings, blue eye colour is recessive to brown eye colour. A brown-eyed man has a blue-eyed mother
(a) What is the genotype of the man and his mother?
(b) What are the possible genotype of his father?
(c) If a man marries a blue-eyed woman, what are the possible genotypes of their offsprings?
Answer:
As per the given condition, the brown eye colour is dominant over the blue eye colour.

(a) The mother’s genotype must be ‘bb’ as she is recessive for blue coloured eye. The man is brown-eyed (dominant). Its possible genotype must be ‘Bb’ as he is procuring one of the recessive gene from his mother.

(b) As the genotype of the man is ‘Bb’ so the possible genotypes of his father may be BB or Bb

Question 25.
How do you relate dominance, codominance, and incomplete dominance in the inheritance of character?
Answer:

• Dominance: It is the phenomenon in which one of the alleles of a gene expresses itself in the hybrid / heterozygous condition. While the other is suppressed (recessive) in the presence of the other.
• Codominance: It is the phenomenon in which 2 alleles of a gene are equally dominant and express themselves in the presence of the other
• Incomplete dominance: It is the phenomenon in which neither of two alleles of a gene is completely dominant over the other and the hybrid is intermediate between the two parents.

Question 26.
Describe the XO- type of sex determination. [HOTS]
Answer:
XO – type of sex determination occurs in certain insects like grasshoppers. The males have only one X – chromosome and hence they have one chromosome less than females. All the ova contains autosomes and one X chromosome. 50% of the sperm contains one X chromosome besides the autosomes, while 50% of sperm don’t have X – chromosomes. Sex of the individual is determined by the type of sperm fertilizing ovum. Egg fertilized by sperm having an X- chromosome developed into female. Egg fertilized by sperm having no chromosome develop into male.

Question 27.
Describe the different types of mutation.
Answer:

• A mutation is a phenomenon arising from the alteration of DNA sequences or structure or number of chromosomes.
• Loss (deletion) or gain (insertion/duplication) of a segment of DNA results in the alteration in chromosome structure, which is called a chromosomal aberration.
• When a mutation arises due to a change in single base pair of DNA, it is known as a point mutation. Eg: Sickle cell anaemia. Deletion or addition of base pairs of DNA causes frameshift mutation.

2nd PUC Biology Principles of Inheritance and Variation Five Marks Questions

Question 1.
What is incomplete dominance? Describe with one example.
Answer:
When 2 parents are intercrossed with each other, the hybrid that produced doesn’t resemble either of the parents, but mid-way between 2 parents. The phenomenon of incomplete dominance occurs in the Four O’clock plant (Mirabilis Jalopa) and Snapdragon (Antirrhinum Majus). A cross between red-flowered (RR) and a white-flowered (rr) plant yields the hybrid flowered (Rr) one which is of pink colour in F, generation.

The pink hybrids on crossing with each other give the usual Mendelian ratio 1 : 2 : 1 (one red : 2 pink : 1 white). In this, the gene “R is not completely dominant over the allele ‘r’. The heterozygous Rr synthesizes only half pigment, so they are pink in colour.

Question 2.
In Snapdragons, tall (DD) is dominant over dwarf (dd) and Red (RR) is incompletely dominant over white (rr), the hybrid being pink. A pure tall white is crossed to a pure dwarf red and F₂ is self-fertilized. Give the expected genotype and phenotype in F₁ and F₂ generations.
Answer:

Question 3.
Sex determination is based on particular chromosomes in both birds and humans. State two points of difference between their mechanisms of sex determination.
Answer:
i. In birds female has dissimilar chromosomes (ZW) and male has two similar ZZ chromosomes.
But in humans female has XX (homogametic) and male has XY (heterogametic)
ii. In humans the sperm is responsible for the sex of child but in birds egg is responsible for the sex of the chicks.

Question 4.
Study the Pedigree analysis chart and answer the question given below: [CBSE 2008]
Answer:

(a) Is the trait recessive or dominant?
(b) Is the trait sex-linked or autosomal?
(c) Give the genotypes of the parents in Generation I and of their 3rd and 4th child in Generation II.
Answer:
(a) The trait is recessive
(b) The trait is autosomal
(c) Parent Aa and Aa
3rd child is II generation – aa 4th child is II generation – Aa

Question 5.
Study the given pedigree chart to show the inheritance pattern of a human trait and answer questions given below.
(a) Give the genotype of a parent in 1st generation and of son and daughter were shown in 2nd generation.
(b) give the genotype of daughters shown in 3rd generation.
(c) Is the trait sex-linked or autosomal. Justify your answer.
Answer:
(a) Genotypes of parents in Generation 1. Male – Aa Female – Aa Son (Generation II) – Aa Daughter (Generation II) – aa.
(b) Genotype of the daughter in Generation III – Aa
(c) It is an autosomal trait and not sex-linked because if it is sex-linked, the daughter in generation II cannot have it.

Question 6.
A girl baby has been reported to suffer from haemophilia. How is it possible? Explain with the help of a cross.
Answer:
(a) Haemophilia is due to the presence of a recessive defective allele on the X – chromosome.
(b) A female with XX – chromosome must be homozygous for the disease to appear.
(c) She must receive one of the alleles from her haemophilie father (X^hY) and the other from her mother who is also an haemophilie or at least a carrier i.e. the heterozygous for the disease XX^h.

Question 7.
How are Pedigree charts prepared?
Answer:

• Squares denote males.
• Circles denote females.
• Mating is shown by a horizontal line connecting a male symbol with a female symbol.
• Offsprings symbols are rearranged from left to right in the order of their birth and connected by a horizontal line below the parents and this line is connected to the parent’s marriage line by a vertical line.
• A solid or blackened symbol represents the individual with the trait under study.
• An open/clear symbol represents the absence of a trait under study.

 

You can Download Chapter 10 Microbes in Human Welfare Questions and Answers, 2nd PUC Biology Question Bank with Answers, Karnataka State Board Solutions help you to revise complete Syllabus and score more marks in your examinations.

Karnataka 2nd PUC Biology Question Bank Chapter 10 Microbes in Human Welfare

2nd PUC Biology Microbes in Human Welfare NCERT Text Book Questions and Answers

Question 1.
Bacteria cannot be seen with the naked eyes, but these can be seen with the help of a microscope. If you have to carry a sample from your home to your biology laboratory to demonstrate the presence of microbes under a microscope, which sample would you carry and why?
Answer:
Microbes or microorganisms like bacteria, fungi, and protozoans are omnipresent (i.e., found everywhere) but are not visible to the naked eye because they have a size of 0.1 mm or less. The common household product that shows the presence of bacteria is curd. Milk is converted to curd by Lactobacillus bulgaricus and Streptococcus thermophilous.

Question 2.
Give examples to prove that microbes release gases during metabolism.
Answer:
The examples of bacteria that release gases during metabolism are:

• Bacteria and fungi carry out the process of fermentation and during this process, they release carbon dioxide.
• The dough used for making idli and dosa gives a puffed appearance. This is because of the action of bacteria which releases carbon dioxide. This CO₂ released from the dough gets trapped in the dough, thereby giving it a puffed appearance.

Question 3.
In which food would you find lactic acid bacteria? Mention some of their useful applications.
Answer:
Microorganisms such as Lactobacillus and others commonly called lactic acid bacteria (LAB) grow in milk. They convert lactose sugar of milk into lactic acid. Lactic acid causes coagulation of milk protein casein. Milk is changed into curd, yogurt and cheese etc.

• Curd – Indian curd is prepared by inoculating cream and skimmed milk with Lactobacillus acidophilus at a temperature of about 40°C or less.
• Yogurt – It is produced by curdling milk with the help of Streptococcus thermophilus and Lactobacillus bulgaricus. It has a flavor of lactic acid and acetaldehyde. It is often sweetened and flavored with fruit.
• Cheese – It consists of milk curd separated from liquid part. In preparation for raw cheese milk is curdled with the help of lactic acid bacteria.
  In our stomach too, the LAB play very beneficial role in checking disease-causing microbes.

Question 4.
Name some traditional Indian foods made of wheat, rice and Bengal gram (or their products) which involve the use of microbes.
Answer:

• Wheat product: Bread, Cake etc.
• Rice product: Idli, dosa.
• Bengal gram Product: Dhokla, Khandvi.

Question 5.
In which way have microbes played a major role in controlling diseases caused by harmful bacteria?
Answer:
Microbes are very useful to combat disease-causing harmful bacteria. A number of antibiotics have been isolated from microorganisms. An antibiotic is a substance which in low concentration inhibits the growth and metabolic activity of pathogenic organisms without harming the host. Penicillin was the first antibiotic to be discovered by Alexander Fleming from fungus Penicillium notation. Antibiotics are obtained from lichens, fungi, eubacteria, and actinomycetes. Some common antibiotics and their sources are as follows :

1. Polymyxin – Bacillus polymyxa
2. Chloramphenicol – Streptomyces venezuelae
3. Neomycin – Streptomyces fradiae
4. Tetracycline (Terramycin) – Streptomyces rimosus
5. Cephalosporin – Cephalosporium acremonium

Question 6.
Name any two species of fungus, which are used in the production of antibiotics.
Answer:

• Penicillin – Penicillin chrysogenum
• Griseofulvin – Penicillin griseofulvin
• Fumagillin – Aspergillus puniyatus
• Cephalosporin – Cephalosporium acremonium.

Question 7.
What is sewage? In which way can sewage be harmful to us?
Answer:
Sewage is a collective term used to represent municipal wastewater (both liquid and solid wastes) generated in cities and towns which is carried off in sewers. Chemically, the sewage consists of approximately 99% water and 1% solid waste including inorganic and organic matter. The microorganisms present in sewage include bacteria (coliforms, streptococci, clostridia, lactobacilli), microfungi, protozoa, and microalgae. Proper sewage disposal is of prime importance, because disposal of untreated sewage in the river and other water bodies may be harmful in the following ways:

• It results in the dissemination of water-borne diseases caused by microorganisms.
• It may cause depletion of dissolved oxygen (DO) in water.
• Reduction in oxygen availability may kill aerobic aquatic microorganisms.
• Untreated sewage produces an offensive odor.

Question 8.
What is the key difference between primary and secondary sewage treatment?
Answer:
Primary treatment is physical and removes grit and large prices of organic matter while secondary treatment is Biological causing digestion of organic matter by microbes.

Question 9.
Do you think microbes can also be used as source of energy? If yes, how?
Answer:
Microbes can be used as a source of energy. Biogas is a mixture of gases produced from degradable organic matter by the activity of various anaerobic microorganisms and it may be used as fuel.

The microorganisms involved in biogas production are mainly facultative and strictly anaerobic bacteria. The most important among them are methanogenic archaebacteria, represented by Methanobacterium. The major component of biogas is methane (about 50-68%) which is highly inflammable. The other gases are CO₂ (25-35%), hydrogen (1-5%), nitrogen (2-7%), oxygen (0-0.1%) and H₂S (traces).

Biogas is commercially produced inside the biogas plant. The plant is fed with a mixture of dung and water (1 : 1 ratio). Dung is an excreta of cattle, commonly called ‘gobar’, it is because of this, the plant is commonly called ‘gobar gas plant’. Cattle dung is a rich source of cellulosic material from plants. Biogas is used as fuel for heating, cooking, lighting, power for irrigation, and other purposes as an alternative to firewood, kerosene, dung cakes, or even electricity and LPG. It is considered an eco-friendly and pollution-free source of energy.

Question 10.
Microbes can be used to decrease the use of chemical fertilisers and pesticides. Explain how this can be accomplished.
Answer:
Microbes play an important role in organic farming which is done without the use of chemical fertilizers and pesticides. Bio-fertilizers are living organisms that help increase the fertility of soil. It involves the selection of beneficial microorganisms that help in improving plant growth through the supply of plant nutrients. Bio-fertilizers are introduced in seeds, roots or soil to mobilize the availability of nutrients. Thus, they are extremely beneficial in enriching the soil with organic nutrients.

Many species of bacteria and cyanobacteria have the ability to fix free atmospheric nitrogen, Rhizobium is a symbiotic bacteria found in the root nodules of leguminous plants. Azospirillium and Azotobocter are free-living nitrogen-fixing bacteria, whereas Anabena, Nostoc and Oscillitoria are examples of nitrogen-fixing cyanobacteria. Bio-fertilizers are cost-effective and eco-friendly.

Microbes can also act as bio-pesticides to control insect pests in plants. An example of bio-pesticides is Bacillus thuringiensis, which produces a toxin that kills insect pests. Dried bacterial spores are mixed in water and sprayed in agricultural fields. When larvae of insects feed on crops, these bacterial spores enter the gut of the larvae and release toxins, thereby it. Similarly, Trichoderma is free-living fungi. They live in the roots of higher plants and protect them from various pathogens. Baculoviruses are another bio-pesticides that is used as a biological control agent against insects and other arthropods.

Question 11.
Three water samples namely river water, untreated sewage water and secondary effluent discharged from a sewage treatment plant were subjected to BOD test. The samples were labelled A, B and C; but the laboratory attendant did not note which was which. The BOD values of the three samples A, B and C were recorded as 20mg/L, 8mg/L and 400mg/ L, respectively. Which sample of the water is most polluted? Can you assign the correct label to each assuming the river water is relatively clean?
Answer:
BOD (Biological Oxygen Demand) refers to the amount of oxygen that would be consumed if all the organic matter in one litre of water were oxidised by bacteria. BOD is a measure of the organic matter present in the water. The greater the BOD of wastewater, the more are the pollutants. BOD value of clean water is generally between 1 and 2 mg/L. As the amount of pollution increases, BOD is also increased and grossly polluted waters may have the BOD around 20 mg/L.

In the given problem BOD values of the three samples A, B and C are 20 mg/L, 8 mg/L and 400 mg/L, respectively. Here sample C has the greatest BOD value hence it is the most polluted. If we correctly label the three samples, then sample A should be secondary effluent discharged from a sewage treatment plant (20 mg/L), sample B should be river water (8 mg/L) and sample C should be untreated sewage water (400 mg/L).

Question 12.
Find out the name of the microbes from which Cyclosporin A (an immunosuppressive drug) and Statins (blood cholesterol-lowering agents) are obtained.
Answer:

• Cyclosporin A is produced by the fungus Trichoderma polysporum.
• Statins are produced by die yeast Monascus purpureus which acts as a blood cholesterol-lowering agent.

Question 13.
Find out the role of microbes in the following and discuss it with your teacher.
(a) Single-cell protein (SCP)
(b) Soil
Answer:
(a) Single-cell protein (SCP) – Microorganisms (e.g., bacteria, yeast, filamentous fungi, algae etc.) can be cultured on a commercial scale in a fermenter, treated in various ways, dried or used as food source or as animal feed are called single-cell protein. The term SCP, however, is misleading as it sounds as if the protein is obtained from single cell. In fact, the biomass obtained from uni and the multicellular organism is considered as SCP.

Some common microorganisms, used in the production of SCP are as follows:

• Bacteria (e.g., Methylophilus, Brevibacterium, etc.)
• Cyanobacteria (e.g., Spirulina)
• Yeasts (e.g., Saccharomyces cerevisiae, Candida utilis, etc.)
• Filamentous fungi (e.g., Fusarium graminearum)
• Algae (e.g., Chlorella)

There are several advantages of SCP using as food. Some of them are listed below :

• SCP is rich in high-quality protein and poor in fat content.
• The SCP can be produced in laboratories all year round. Its production is not dependent on climatic factors.
• The microorganisms, used in the production of SCP, are very fast-growing and produce a large amount of SCP from a relatively very small area of land.

(b) Soil – Microbes are very useful to maintain and restore soil fertility. The fertility of soil depends not only on its chemical composition but also on the quantity and quality of useful microbes present in it. Moreover, if the composition of the soil is not upto the mark and poor in fertility, materials of biological origin are added to it to improve and maintain its fertility. These materials are grouped under the two broad categories manures and’ biofertilisers. Manures are of three types, farmyard manure, compost and green manure.

Farmyard manure is the oldest manure known to mankind which is made up of dung of farm animals and plant remains etc, which are allowed to partially decay with the help of soil microorganisms. These microorganisms decompose complex organic debris into a dark amorphous substance (humus) and degradation products are easily assimilated by plants. The manure loosens the soil, increases its aeration, and makes the soil more fertile. Biofertilisers are the microorganisms which bring about soil nutrient enrichment, maximize the ecological benefits and minimize environmental hazards.

Question 14.
Arrange the following in the decreasing order (most important first) of their importance, for the welfare of human society. Give reasons for your answer. Biogas, Citric acid, Penicillin and Curd.
Answer:
The order of arrangement of products according to their decreasing importance is:
Penicillin—Biogas—Citric acid—Curd

Penicillin is the most important product for the welfare of human society. It is an antibiotic, which is used for controlling various bacterial diseases. The second most important product is biogas. It is an eco-friendly source of energy. The next important product is citric acid, which is used as a food preservative. The least important product is curd, a food item obtained by the action of lactobacillus bacteria on milk Hence, the products in the decreasing order of their importance are as follows:

Penicillin—Biogas—Citricacid—Curd.

Question 15.
How do biofertilizers enrich the fertility of the soil?
Answer:
Bio fertilizers are the microorganisms which bring about soil nutrient enrichment, maximize the ecological benefits and minimize environmental hazards. The main sources of biofertilizers are bacteria, fungi, and cyanobacteria. Some of the biofertilizers and their importance in maintaining soil fertility are as follows:

1. Free-living nitrogen-fixing bacteria fix atmospheric nitrogen in the soil and make it available for the higher plants. The best example is Azolobncter. Apart from this Clostridium, Bacillus polymyxa, Derxia, etc., are also known to fix atmospheric nitrogen.
2. Symbiotic nitrogen-fixing bacteria, Rhizobium, form an efficient symbiotic relationship with leguminous plants and can fix upto 500 kg nitrogen per hectare of land. Rhizobium forms nodules on the roots of legume plants.
3. Free-living nitrogen-fixing cyanobacteria ‘include Anabaena, Nostoc, Aulosira, Stigonema, etc. Aulosira fertilissimma is considered to be the most active nitrogen fixer of rice fields in India.
4. Nitrogen-fixing cyanobacteria form a symbiotic association with several plants, e.g., cycad roots, lichens, liverworts, Azolla (fern). Out of these, the Azollci-Amibaena association is of great importance to agriculture. Anabaena azollae resides in the leaf cavities of fern. It fixes nitrogen.
5. Fungi are also known to form symbiotic associations with plants (mycorrhiza). The fungal symbiont in these associations absorbs phosphorus from soil and passes it to the plant. Plants having such associations show other benefits also, such as resistance to root-borne pathogens, tolerance to salinity and drought, and an overall increase in plant growth and development.

2nd PUC Biology Microbes in Human Welfare Additional Questions And Answers

2nd PUC Biology Microbes in Human Welfare One Mark Questions

Question 1.
What is LAB?
Answer:
The microorganisms which commonly grow in milk and converted into curd are called Lactic Acid Bacteria or LAB.

Question 2.
Name the fungus which gave gibberellins hormone.
Answer:
Gibberellin is found by Gibberella fujikuroi.

Question 3.
Who discovered penicillin?
Answer:
Alexander Fleming.

Question 4.
Which microbes are given antibiotics?
Answer:
Generally, bacteria and fungus give us antibiotics.

Question 5.
What are fermentors?
Answer:
Fermentors are very large vessels used for growing microbes for the production of microbial products on an industrial scale.

Question 6.
What are antibiotics?
Answer:
The chemical substances which are produced by certain microbes can kill or retard the growth of other microbes.

Question 7.
What is BOD?
Answer:
The amount of oxygen that would be consumed for the oxidation or decomposition of the biodegradable organic waste present in the water is called BOD or biological oxygen demand.

Question 8.
Why does the river Ganga water not spoil even kept for a long-time?
Answer:
Bacteriophages are present in the Ganga river’s water, they feed bacteria which spoil water. So, this water does not spoil in long-time.

Question 9.
Name the enzyme used “clot buster” is to remove blood clots from blood vessels and which organism produces this enzyme.
Answer:
Streptokinase is known as clot buster. It is produced by Streptococcus bacteria.

Question 10.
Name the microorganism/fungi which gave penicillin.
Answer:
Penicillium notatum.

Question 11.
What is STP stands for?
Answer:
Sewage Treatment Plant.

Question 12.
Write the name of methanogen bacteria.
Answer:
Methano bacterium species.

Question 13.
Which of the following is a Cyanobacteria that can fix atmospheric nitrogen? Spirulina, Azospirillum, Sonalika.
Answer:
Azospirillum.

Question 14.
Milk starts to coagulate when Lactic Acid Bacteria is added to warm milk as a starter. Mention any other two benefits LAB provides. (AI CBSE – 2009)
Answer:
Increase vitamin B₁₂ and also check disease-causing microbes in stomach.

Question 15.
Name the group of organisms and the substrate they act on to produce biogas.
(CBSE 2009)
Answer:
Methanogens, Cellulose present in cattle excreta (cow dung)

Question 16.
What are floes?
Answer:
The masses of bacteria associated with fungal filaments forming the mesh-like structures are the floes.

Question 16.
Give the name of autotrophic nitrogen-fixing microbes.
Answer:
Anabaena.

Question 18.
Name the fungus that is being developed as biocontrol.
Answer:
Trichoderma.

Question 19.
What is the key differences between primary and secondary sewage treatment?
Answer:
Primary treatment of sewage is a physical process, while secondary treatment is a biological process.

Question 20.
Name a genus of a fungus that forms mycorrhiza.
Answer:
Glomus is the fungus that forms mycorrhiza.

Question 21.
List 2 advantages that a mycorrhizal association provides to the plant. (AI2008)
Answer:

• The fungal partner in mycorrhiza absorbs
• Phosphorus from the soil and passes it on to the plant.
• The plants having mycorrhizal association show resistance to root borne pathogens and tolerance to salinity

2nd PUC Biology Microbes in Human Welfare Two Marks Questions

Question 1
(a) Name an eco-friendly hloherhicide that interferes with amino add synthesis and resistance to which has been obtained through transgenic culture.
(b) Name the first biopesticide. (CBSE 2004)
Answer:
Basta or phosphinothricin (ppt)
(a) It obtained from a product of Streptomyces species. Crop plants have been made resistant to it by transferring bar gene into them (e.g. wheat)
(b) Divine and college.

Question 2.
Match the columns

Answer:

Question 3.
Name the blank spaces ‘a’, ‘b’, ‘c’ and ‘d’ give in the following label.

Type of microbe	Scientific Name	Commercial or Product
Bacterium	a	Lactic acid
Fungus	b	Cyclosporin A stains
C	Monascus Purpur
Fungus	Penicillin Nolaltisin	d

Answer:

• a – Laetobaellhts
• b – Triehodmna polysptmtm
• c – Yeast (fungus)
• d – Penicillin

Question 4.
Name the blank spaces a, b, c, and d given in the following label. (CBSE – 2008)

Type of microbe	Scientific Name	Commercial or Product
Bacterium	a	Clot buster
b	Aspergillus Trichoderma	Enzyme citric acid
Fungus	Polysporum	c
Bacterium	d	Enzyme citric acid

Answer:
a – Streptococcus
b – Fungus
c – Cyclosporin A
d – Clostridium butylicum.

Question 5.
Fill in the blanks from Cephalosporin, Cyclosporin, Cycas, Soybean, Nepiatode, Fungus, Rhizobium. (CBSE 2008)
(a) Potent immunosuppressant drug is ………….. which is obtained from a …………..
(b) Roots pines and ………….. are associated will Amanita and …………..respectively.
Answer:
(a) Cyclosporin A, Fungus
(b) Soybean, Rhizobium.

Question 5.
“BOD must be low in freshwater”
a. Expand BOD.
b. Define BOD.
c. Write the importance of BOD.
Answer:
a. Biological Oxygen Demand
b. This is the amount of oxygen that would be consumed if all the organic matter in one litre of water were oxidised by the microbes.
c. BOD is the measure of the organic matter present in water because BOD test measures the rate of uptake of oxygen by microbes in sample water. The greater the BOD in sewage, more is its polluting potential.

Question 7.
Name the blank spaces a, b, c and d from the table given below (CBSE 2007)

Type of Microbes	Scientific Name	Product	Medical Application
Fungus	a	Cyclosporin	b
c	Monascus	statin	d
	purpureus

Answer:
(a) Trichoderma polysporum
(b) Organs transplant patients
(c) Yeast (fungus)
(d) lowering blood cholesterol level.

Question 8.
What is baker’s yeast?
Answer:
Saccharomyces cerevisiae

Question 9.
What are harmful effects of using chemical pesticides?
Answer:

• The chemical pesticides are harmful to many organisms (maybe useful) other than the pests for which they used.
• They enter the food chain and cause diseases/disorders in various organism
• They are also highly toxic to man
• They cause pollution of our soil, water and air.

Question 10.
How do bacteria function as biofertilizers? Name 2 free-living soil bacteria that are biofertilizers.
Answer:

• Bacteria can fix atmospheric nitrogen into those nitrogen compounds which are used by the plants as their nutrients.
• Soil bacteria that are biofertilizers includes – Azotobacter and azospirillum.

2nd PUC Biology Microbes in Human Welfare Three Marks Questions

Question 1.
What are methanogens? Where are they generally found? Give examples.
Answer:
Methanogens are a group of anaerobic bacteria, which produce large quantities of methane from cellulosic materials.
E.g.: Methano bacterium Methanogens are found in

• The sewage wager
• Marshy place
• The rumen of cattle.

Question 2.
Describe the functions of anaerobic sludge in a sewage treatment plant.
Answer:

• A portion of the activated sludge from anaerobic sludge digester is pumped back to the aeration tank to serve as inoculum.
• The anaerobic bacteria digest the bacteria and fungi of the floes.
• The anaerobic bacteria in the sludge release CO₂, methane etc. during decomposition; these gases from the biogas is used as a source of energy, as it is inflammable.

Question 3.
What are mycorrhizae? How do they serve as biofertilizers or how are they useful to plants?
Answer:
Mycorrhizae are the symbiotic association between certain fungi and roots of higher plants.

• The fungus absorbs phosphorous from the soil and passes it to the plants
• Plants with mycorrhiza show resistance to root borne pathogens
• They show increased tolerance to salinity and drought
• There is an overall increase in plant growth and development.

Question 4.
Biocontrol is the best alternative for chemical control. Justify your answer.
Answer:
The use of biological methods for controlling plant diseases and pests is called biocontrol. This method has been replaced by the indiscriminate use of chemicals used in the chemical control of pests. The chemicals used for killing pests are toxic and extremely harmful to men and domestic animals. They also pollute the environment and our crop plants.

2nd PUC Biology Microbes in Human Welfare Five Marks Questions

Question 1.
Write seven useful activities of Bacteria.
Answer:
Useful activities of bacteria:

1. N₂ fixation: Some bacterias play an important role in nitrogen fixation e.g., Azotobacter, Clostridium, Rhizobium. These bacterias increase the fertility of soil by the fixation of atmospheric nitrogen.
2. Lactic acid synthesis: Lactobacillus lacti converts the milk sugar into lactic acid.
3. Acetic acid synthesis: Acetobacter aceti takes part in the synthesis of acetic acid or vinegar.
4. Rating of fibers: Isolation of wood fibres from the stem of plants is called rating. Clostridium butyricum is used in the rating of fibres.
5. Tobacco and Tea industry: Some bacteria like Micococcus canadiens is used to increase the flavour of the leaves of tobacco and tea. This process is called seasoning.
6. Medicine production: Bacteria are the chief source of antibiotics, hence, they are used to extract antibiotics, e.g., Streptomyces gresius (Streptomycin).
7. As symbionts: Bacteria present in our body, helps in the various metabolic reactions e.g, E.coli.