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Samacheer Kalvi 12th Bio Botany Economically Useful Plants and Entrepreneurial Botany Text Book Back Questions and Answers

Question 1.
Consider the following statements and choose the right option:
(i) Cereals are members of grass family.
(ii) Most of the food grains come from monocotyledon.
(a) (i) is correct and (ii) is wrong
(b) Both (i) and (ii) are correct
(c) (i) is wrong and (ii) is correct
(d) Both (i) and (ii) are wrong
Answer:
(b) Both (i) and (ii) are correct

Question 2.
Assertion: Vegetables are important part of healthy eating.
Reason: Vegetables are succulent structures of plants with pleasant aroma and flavours.
(a) Assertion is correct, Reason is wrong
(b) Assertion is wrong, Reason is correct
(c) Both are correct and reason is the correct explanation for assertion.
(d) Both are correct and reason is not the correct explanation for assertion.
Answer:
(d) Both are correct and reason is not the correct explanation for assertion.

Question 3.
Groundnut is native of _____________
(a) Philippines
(b) India
(c) North America
(d) Brazil
Answer:
(d) Brazil

Question 4.
Statement A: Coffee contains caffeine.
Statement B: Drinking coffee enhances cancer.
(a) A is correct, B is wrong
(b) A and B – Both are correct
(c) A is wrong, B is correct
(d) A and B – Both are wrong
Answer:
(a) A is correct, B is wrong

Question 5.
Tectona grandis is coming under family _____________
(a) Lamiaceae
(b) Fabaceae
(c) Dipterocaipaceae
(d) Ebenaceae
Answer:
(a) Lamiaceae

Question 6.
Tamarindus indica is indigenous to _____________
(a) Tropical African region
(b) South India, Sri Lanka
(c) South America, Greece
(d) India alone
Answer:
(a) Tropical African region

Question 7.
New world species of cotton _____________
(a) Gossipium arboretum
(b) G.herbaceum
(c) Both a and b
(d) G.barbadense
Answer:
(d) G.barbadense

Question 8.
Assertion: Turmeric fights various kinds of cancer.
Reason: Curcumin is an anti-oxidant present in turmeric.
(a) Assertion is correct, Reason is wrong
(b) Assertion is wrong, Reason is correct
(c) Both are correct
(d) Both are wrong
Answer:
(c) Both are correct

Question 9.
Find out the correctly matched pair:
(a) Rubber Shorea robusta
(b) Dye
(c) Timber Cyperus papyrus
(d) Pulp Hevea brasiliensis
Answer:
(b) Dye

Question 10.
Observe the following statements and pick out the right option from the following:
Statement I – Perfumes are manufactured from essential oils.
Statement II – Essential oils are formed at different parts of the plants.
(a) Statement I is correct
(b) Statement II is correct
(c) Both statements are correct
(d) Both statements are wrong
Answer:
(c) Both statements are correct

Question 11.
Observe the following statements and pick out the right option from the following:
Statement I: The drug sources of Siddha include plants, animal parts, ores and minerals.
Statement II: Minerals are used for preparing drugs with long shelf-life.
(a) Statement I is correct
(b) Statement II is correct
(c) Both statements are correct
(d) Both statements are wrong
Answer:
(c) Both statements are correct

Question 12.
The active principle trans-tetra hydro canabial is present in _____________
(a) Opium
(b) Curcuma
(c) Marijuana
(d) Andrographis
Answer:
(c) Marijuana

Question 13.
Which one of the following matches is correct?
(a) Palmyra – Native of Brazil
(b) Saccharun – Abundant in Kanyakumari
(c) Steveocide – Natural sweetener
(d) Palmyra sap – Fermented to give ethanol
Answer:
(c) Steveocide – Natural sweetener

Question 14.
The only cereal that has originated and domesticated from the New world.
(a) Oryza sativa
(b) Triticum aestivum
(c) Triticum duram
(d) Zea mays
Answer:
(c) Triticum duram

Question 15.
Write the cosmetic uses of Aloe.
Answer:
Aloe gel are used as skin tonic. It has a cooling effect and moisturizing characteristics and hence used in preparation of creams, lotions, shampoos, shaving creams, after shave lotions and allied products. It is used in gerontological applications for rejuvenation of aging skin. Products prepared from aloe leaves have multiple properties such as emollient, antibacterial, antioxidant, antifungal and antiseptic. Aloe vera gel is used in skin care cosmetics.

Question 16.
What is pseudo cereal? Give an example.
Answer:
The term pseudo-cereal is used to describe foods that are prepared and eaten as a whole grain, but are botanical outliers from grasses. Example: quinoa. It is actually a seed from the Chenopodium quinoa plant, belongs to the family Amaranthaceae.

Question 17.
Discuss which wood is better for making furniture.
Answer:
Teak wood is the ideal type of wood for making household furnitures because, it is highly durable and shows great resistance against the attack of termites and fungi. Moreover it doesnot split or crack and is a carpenter friendly wood.

Question 18.
A person got irritation while applying chemical dye. What would be your suggestion for alternative?
Answer:
If a grey haired person is allergic on using chemical dyes then he can go for natural dyes like Henna. Henna is an organic dye obtained from leaves and young shoots of Lawsonia inermis. The principal colouring matter is Tacosone’ which is harmless and causes no irritation on skin.

Question 19.
Name the humors that are responsible for the health of human beings.
Answer:
Vatam, Pittam and Kapam.

Question 20.
Give definitions for organic farming?
Answer:
Organic farming is an alternative agricultural system in which plants/crops are cultivated – in natural ways by using biological inputs to maintain soil fertility and ecological balance thereby minimizing pollution and wastage.

Question 21.
Which is called as the “King of Bitters”? Mention their medicinal importance.
Answer:
Andrographis paniculata is called as King of Bitters. Andrographis is a potent hepatoprotective agent and is widely used to treat liver disorders. Concoction of Andrographis paniculata and eight other herbs (Nilavembu Kudineer) is effectively used to treat malaria and dengue.

Question 22.
Differentiate bio-medicines and botanical medicines.
Answer:

1. Bio-medicines: Medically useful molecules obtained from plants that are marketed as drugs are called Bio-medicines.
2. Botanical medicines: Parts of medicinal plants which are modified as powers or pills or other forms and marketed. These are called botanical medicines.

Question 23.
Write the origin and area of cultivation of green gram and red gram.
Answer:

Question 24.
What are millets? What are its types? Give example for each type.
Answer:
Millet is the term applied to a variety of very small seeds originally cultivated by ancient people in Africa and Asia. They are gluten-free with less glycemic index.
Types of millet:

1. Major millets – E.g: Ragi (Eleusine coracana)
2. Minor millets – E.g: Foxtail millet (Setaria italica)

Question 25.
If a person drinks a cup of coffee daily it will help him for his health. Is this correct? If it is correct, list out the benefits.
Answer:
Yes, drinking coffee in moderation enhances the health of a person. Caffeine enhances release of acetylcholine in brain, which in turn enhances efficiency. It can lower the incidence of fatty liver diseases, cirrhosis and cancer. It may reduce the risk of type 2 diabetes.

Question 26.
Enumerate the uses of turmeric.
Answer:
Turmeric is one of the most important and ancient Indian spices and used traditionally over thousands of years for culinary, cosmetic, dyeing and for medicinal purposes. It is an important constituent of curry powders. Turmeric is used as a colouring agent in pharmacy, confectionery and food industry. Rice coloured with turmeric (yellow) is considered sacred and auspicious which is used in ceremonies. It is also used for dyeing leather, fibre, paper and toys.

Curcumin extracted from turmeric is responsible for the yellow colour. Curcumin is a very good anti-oxidant which may help fight various kinds of cancer. It has anti-inflammatory, anti- ‘ diabetic, anti-bacterial, anti-fungal and anti-viral activities. It stops platelets from clotting in arteries, which leads to heart attack.

Question 27.
What is TSM? How does it classified and what does it focuses on?
Answer:
TSM stands for Traditional Systems of Medicines India has a rich medicinal heritage. Anumber of Traditional Systems of Medicine (TSM) are practiced in India some of which come from outside India. TSM in India can be broadly classified into institutionalized or documented and non-institutionalized or oral traditions. Institutionalized Indian systems include Siddha and Ayurveda which are practiced for about two thousand years.

These systems have prescribed texts in which the symptoms, disease diagnosis, drugs to cure, preparation of drugs, dosage and diet regimes, daily and seasonal regimens. Non-institutional systems, whereas, do not have such records and or practiced by rural and tribal peoples across India. The knowledge is mostly held in oral form. The TSM focus on healthy lifestyle and healthy diet for maintaining good health and disease reversal.

Question 28.
Write the uses of nuts you have studied.
Answer:
Cashews are commonly used for garnishing sweets or curries, or ground into a paste that forms a base of sauces for curries or some sweets. Roasted and raw kernels are used as snacks.

Question 29.
Give an account on the role of Jasminum in perfuming.
Answer:
The essential jasmine oil is present in the epidermal cells of the inner and outer surfaces of both the sepals and petals. One ton of Jasmine blossom yields about 2.5 to 3 kg of essential oil, comprising 0.25 to 3% of the weight of the fresh flower. Jasmine oil is an essential oil that is valued for its soothing, relaxing and antidepressant qualities.

Question 30.
Give an account of active principle and medicinal values of any two plants you have studied
Answer:

Question 31.
Write the economic importance of rice.
Answer:
Rice is the easily digestible calorie rich cereal food which is used as a staple food in Southern and North East India. Various rice products such as Flaked rice (Aval), Puffed rice / parched rice (Pori) are used as breakfast cereal or as snack food in different parts of India. Rice bran oil obtained from the rice bran is used in culinary and industrial purposes. Husks are used as fuel, and in the manufacture of packing material and fertilizer.

Question 32.
Which TSM is widely practiced and culturally accepted in Tamil Nadu? – explain.
Answer:
Siddha is the most popular, widely practiced and culturally accepted system in Tamil Nadu. It is based on the texts written by 18 Siddhars. There are different opinions on the constitution of 18 Siddhars. The Siddhars are not only from Tamil Nadu, but have also come from other countries. The entire knowledge is documented in the form of poems in Tamil. Siddha is principally based on the Pancabuta philosophy. According to this system three humors namely Vatam, Pittam and Kapam that are responsible for the health of human beings and ‘ any disturbance in the equilibrium of these humors result in ill health.

The drug sources of Siddha include plants, animal parts, marine products and minerals. This system specializes in using minerals for preparing drugs with the long shelf-life. This system uses about 800 herbs as source of drugs. Great stress is laid on disease prevention, health promotion, rejuvenation and cure.

Question 33.
What are psychoactive drugs? Add a note Marijuana and Opium.
Answer:
Phytochemicals or drugs from some of the plants alter an individuals perceptions of mind by producing hallucination are known as psychoactive drugs.

1. Marijuana: Marijuana is obtained from Cannabis sativa. The active principle in Marijuana is trans – tetrahydrocanabinal (TCH). It is used as pain killer and reduce hypertension. It is also used in the treatment of Glaucoma, cancer radiotherapy and asthma, etc.
2. Opium: Opium is obtained from the exudates of the fruits of papaver somniferum (poppy plants). It is used to induce sleep and relieve pain. Opium yields morphine which is used as a strong analgesic in surgeries.

Question 34.
What are the King and Queen of spices? Explain about them and their uses.
Answer:

1. King of Spices: Pepper is one of the most important Indian spices referred to as the “King of Spices” and also termed as “Black Gold of India”. Kerala, Karnataka and Tamil Nadu are the top producers in India. The characteristic pungency of the pepper is due to the presence of alkaloid Pipeline. There are two types of pepper available in the market namely black and white pepper.
2. Uses: It is used for flavouring in the preparation of sauces, soups, curry powder and pickles. It is used in medicine as an aromatic stimulant for enhancing salivary and gastric secretions and also as a stomachic. Pepper also enhances the bio-absorption of medicines.
3. Queen of Spices: Cardamom is called as “Queen of Spices”. In India, it is one of the main cash crops cultivated in the Western Ghats, and North Eastern India.
4. Uses: The seeds have a pleasing aroma and a characteristic warm, slightly pungent taste. It is used for flavouring confectionaries, bakery products and beverages. The seeds are used in the preparation of curry powder, pickles and cakes. Medicinally, it is employed as a stimulant and carminative. It is also chewed as a mouth freshener.

Question 35.
How will you prepare an organic pesticide for your home garden with the vegetables available from your kitchen?
Answer:
Step 1: Mix 120 g of hot chillies with 110 g of garlic or onion. Chop them thoroughly.

Step 2: Blend the vegetables together manually or using an electric grinder until it forms a thick paste.

Step 3: Add the vegetable paste to 500 ml of warm water. Give the ingredients a stir to thoroughly mix them together.

Step 4: Pour the solution into a glass container and leave it undisturbed for 24 hours. If possible, keep the container in a sunny location. If not, at least keep the mixture in a warm place.

Step 5: Strain the mixture. Pom- the solution through a strainer, remove the vegetables and collect the vegetable-infused water and pour into another container. This filtrate is the pesticide. Either discard the vegetables or use it as a compost.

Step 6: Pour the pesticide into a squirt bottle. Make sure that the spray bottle has first been cleaned with warm water and soap to get rid it of any potential contaminants. Use a funnel to transfer the liquid into the squirt bottle and replace the nozzle.

Step 7: Spray your plants with the pesticide. Treat the infected plants every 4 to 5 days with the solution. After 3 or 4 treatments, the pest will be eliminated. If the area is thoroughly covered with the solution, this pesticide should keep bugs away for the rest of the season.

Samacheer Kalvi 12th Bio Botany Economically Useful Plants and Entrepreneurial Botany Additional Questions and Answers

1 – Mark Questions

Question 1.
Paddy, Wheat and Sorghum, etc., comes under the category of cereals. All the members of cereals belong to which of the following family?
(a) Fabaceae
(b) Poaceae
(c) Leguminoneae
(d) Caesalpinaeceae
Answer:
(b) Poaceae

Question 2.
Match the common names of the given plant species with their respective binomials

Answer:
(a) A – iii, B – iv, C – iv, C – ii and D – i

Question 3.
Given below are the plant species and their parts used. Which is the incorrect pair(s)?
(i) Cajanus cajan : Seeds
(ii) Anacardium occidentale : nuts
(iii) Borassus flabellifer : Endosperm
(iv) Capsicum annum : leaves
(a) i and ii
(b) ii and iii
(c) iii only
(d) iv only
Answer:
(d) iv only

Question 4.
Identify the tamil name for flaked rice
(a) Nel
(b) Aval
(c) Pori
(d) Umi
Answer:
(b) Aval

Question 5.
Pigeon pea is the common name for.
(a) Vigna radiata
(b) Vigna mungo
(c) Cajanus cajan
(d) Sorghum vulgare
Answer:
(c) Cajanus cajan

Question 6.
Sorghum is native to ______ continent.
Answer:
Africa

Question 7.
Match the following:

Answer:
(a) 1 – ii, 2 – iii, 3 – iv and 4 – i

Question 8.
_____ is the state tree of Tamil Nadu.
Answer:
Palmyra

Question 9.
Statement 1: Arachis hypogea belongs to Fabaceae Statement 2: It is a native of Brazil.
(a) Statement 1 is correct and Statement 2 is also correct
(b) Statement 1 is correct and Statement 2 is incorrect
(c) Statement 1 is incorrect and Statement 2 is correct
(d) Both the Statements are incorrect
Answer:
(a) Statement 1 is correct and Statement 2 is also correct

Question 10.
Statement 1: Chinese discovered the paper.
Statement 2: Eucalyptus and Casuarina are the widely used tree species for making paper pulp.
(a) Statement 1 is correct and Statement 2 is also correct
(b) Statement 1 is correct and Statement 2 is incorrect
(c) Statement 1 is incorrect and Statement 2 is correct
(d) Both the Statements are incorrect
Answer:
(a) Statement 1 is correct and Statement 2 is also correct

Question 11.
Statement 1: Andrographis paniculata is known as King of Bitters.
Statement 2: The decoction of Andrographis is used against Diabetes mellitus.
(a) Statement 1 is correct and Statement 2 is also correct
(b) Statement 1 is correct and Statement 2 is incorrect
(c) Statement 1 is incorrect and Statement 2 is correct
(d) Both the Statements are incorrect
Answer:
(b) Statement 1 is correct and Statement 2 is incorrect

Question 12.
Statement 1: Aloe vera belongs to the family Asphodelaceae.
Statement 2: Jasminum grandiflorum belongs to the family of Oleaceae.
(a) Statement 1 is correct and Statement 2 is also correct
(b) Statement 1 is correct and Statement 2 is incorrect
(c) Statement 1 is incorrect and Statement 2 is correct
(d) Both the Statements are incorrect
Answer:
(a) Statement 1 is correct and Statement 2 is also correct

Question 13.
Which district of Tamil Nadu is the World’s largest wholesale turmeric dealer?
Answer:
Erode

Question 14.
Assertion (A): Turmeric is used to treat cancer.
Reason (R): Curcumin is biomolecule present in turmeric.
(a) A is right R is wrong
(b) Both A and R are wrong
(c) A is wrong R is right
(d) A and R are right. R explains A.
Answer:
(d) A and R are right. R explains A.

Question 15.
Assertion (A): Black pepper is a spice.
Reason (R): Condiments are flavouring substances, generally added after the cooking of food.
(a) A is right R is wrong
(b) R explains A
(c) Both A and R are right. R is not correct explanation for A.
(d) Both A and R are wrong
Answer:
(c) Both A and R are right. R is not correct explanation for A.

Question 16.
Select the new world species of cotton.
(i) Gossypium hirsutum
(ii) Gossypium arboreum
(iii) Gossypium arboreum
(iv) Gossipium
(a) i and ii only
(b) i and iii only
(c) iii and iv only
(d) ii and iv only
Answer:
(a) i and ii only

Question 17.
Binomial of teak is ________
Answer:
Tectona grandis

Question 18.
The plant source of Marijuana is ________
(a) Andrographis paniculata
(b) Phyllanthus maderspatensis
(c) Cannabis sativa
(d) Papaver somniferum
Answer:
(c) Cannabis sativa

Question 19.
Identify the incorrect statements:
(a) Morphine is used as potent hepatoprotective.
(b) Phyllanthin is used as a strong analgesic in surgery.
(c) Indian Acalypha is used to cure skin diseases.
(d) Cissus quadrangularis is widely used for treating bone fractures.
(i) a and c
(ii) a and d
(iii) b and c
(iv) a and b
Answer:
(iv) a and b

Question 20.
Identify the mismatched pair:
(a) Holy basil – Ocimum sanctum
(b) Indian gooseberry – Phyllanthus amarus
(c) Vilvam – Aegle marmelos
(d) Veldt grape – Cissus quadrangularis
Answer:
(b) Indian gooseberry – Phyllanthus amarus

Question 21.
The active principle in Marijuana is _________
Answer:
Trans-tetrahydrocanabinal (THC).

Question 22.
The second GI tag for Jasmine is ‘Madurai Malli’, which jasmine type has got the first GI tag?
Answer:
‘Mysore Malli’

Question 23.
The principal coloring agent present in the leaf paste of Lawsonia inermis which give orange to dark red when applied on skin is _______
Answer:
Lacosone

Question 24.
Which state of India ranks first in coffee consumption?
Answer:
Tamil Nadu

Question 25.
Name the plant family to which the coffee plant belongs to?
Answer:
Rubiaceae

2 – Mark Questions

Question 1.
Give the binomials of paddy and wheat.
Answer:

1. Paddy : Oryza sativa.
2. Wheat: Triticum aestivum.

Question 2.
What is maida? Mention its culinary purpose.
Answer:
Processed wheat flour, that has little fibre, is called Maida which is used extensively in making parota, naan and bakery products.

Question 3.
Name any four millet varieties.
Answer:
Finger millet, Sorghum, Foxtail millet and Kodo millet.

Question 4.
Ragi rich food helps to overcome bone related ailments – justify.
Answer:
Ragi (Finger millet) is a kind of millet which has less glycemic index and rich in calcium. Hence consuming ragi food items enhances the bone strength.

Question 5.
Eating millets is good for diabetic patients. How?
Answer:
Millets are gluten free and have less glycemic index, hence it is good for diabetic patients.

Question 6.
Write the health benefits of Kodo millet.
Answer:
Kodo millet is a good diuretic and cures constipation. Helps to reduce obesity, blood sugar and blood pressure.

Question 7.
Name the family does the cereals and pulses belong to.
Answer:

1. Cereals belong to the family Poaceae.
2. Pulses belong to the family Fabaceae.

Question 8.
Write the common name for

1. Vigna radiata
2. Cajanus cajatt

Answer:

1. Vigna radiata – Green gram.
2. Cajanus cajan – Red gram or pigeon pea.

Question 9.
Why do we take vegetables in our diet?
Answer:
Vegetables are the important part of healthy eating and provide many nutrients, including potassium, fiber, folic acid and vitamins A, E and C. The nutrients in vegetables are vital for maintenance of our health.

Question 10.
Write the binomial and the family to which lady’s finger belongs to.
Answer:
Binomial of lady’s finger is Abelmoschus esculentus.
Family: Malvaceae.

Question 11.
Which is our National fruit and State tree of Tamil Nadu?
Answer:

1. National fruit of India is Mango.
2. State tree of Tamil Nadu is Palmyra.

Question 12.
Point out the uses of sugarcane.
Answer:
Sugar cane is the raw material for extracting white sugar. Sugarcane supports large number of industries like sugar mills producing refined sugars, distilleries producing liquor grade ethanol and millions of jaggery manufacturing units. Fresh sugarcane juice is a refreshing drink. Molasses is the raw material for the production of ethyl alcohol.

Question 13.
What is toddy?
Answer:
Inflorescence of palmyra is tapped for its sap which is used as health drink. Sap is processed to get palm jaggery or fermented to give toddy.

Question 14.
How coffee plants are introduced to India?
Answer:
Coffea arabica is the prime source of commercial coffee which is native to the tropical Ethiopia. An Indian Muslim saint, Baba Budan introduced coffee from Yemen to Mysore.

Question 15.
Compare Spices and Condiments.
Answer:
Spices are accessory foods mainly used for flavouring during food preparation to improve , their palatability. Spices are aromatic plant products and are characterized by sweet or bitter taste. Spices are added in minimal quantities during the cooking process. For example black pepper. Condiments, on the other hand, are flavouring substances having a sharp taste and are usually added to food after cooking. For example, curry leaves.

Question 16.
Mention any two Jute species.
Answer:

1. Corchorus capsularis
2. Corchorus olitorius

Question 17.
Name the plant species used for making paper pulp.
Answer:
Wood of Melia azadirachta, Neolamarkia chinensis, Casuarina spp and Eucalyptus spp are used for making paper pulp.

Question 18.
What does NCB stands for? Mention its role.
Answer:
The Narcotics Control Bureau (NCB) is the nodal drug law enforcement and intelligence agency of India and is responsible for fighting drug trafficking and the abuse of illegal substances.

3 – Mark Questions

Question 19.
Write a short note on foxtail millet. Foxtail Millet
Answer:

1. Botanical name: Setaria italica This is one of the oldest millet used traditionally in India. Which is domesticated first in China about 6000 years. Rich in protein, carbohydrate, vitamin B and C, Potassium and Calcium.
2. Uses:It supports in strengthening of heart and improves eye sight. Thinai porridge is given to lactating mother.

Question 20.
Write the botanical name and culinary uses of green gram.
Answer:
Botanical name: Vigna radiata Uses It can be used as roasted, cooked and sprouted pulse. Green gram is one of the ingredients of pongal, a popular breakfast dish in Tamil Nadu. Fried dehulled and broken or whole green gram is used as popular snack. The flour is traditionally used as a cosmetic, especially for the skin.

Question 21.
Mention the ways in which mangoes are used in Indian culinary.
Answer:
Mango is the major table fruit of India, which is rich in beta carotenes. It is utilized in many ways, as dessert, canned, dried and preserves in Indian cuisine. Sour, unripe mangoes are used in chutneys, pickles, side dishes, or may be eaten raw with salt and chili. Mango pulp is made into jelly. Aerated and non-aerated fruit juice is a popular soft drink.

Question 22.
Write a note on origin and area of cultivation of cotton.
Answer:
Cotton is one of the oldest cultivated crops of the world. It has been cultivated for about 8000 years both in new world and in old world. Commercial cotton comes from four cotton species: two from the new world and two from the old world.

1. Gossypium hirsutum
2. G.barbadense are the New world species and
3. G. arboretum
4. G. herbaceum are the old world species. In India, cotton is cultivated in Gujarat, Maharashtra, Andhra Pradesh and Tamil Nadu.

Question 23.
What is vulcanization? Who invented it?
Answer:
The heating of the rubber with sulphur under pressure at 150°C is called vulcanization. It was invented by Charles Goodyear.

Question 24.
Write a note on Henna.
Answer:

1. Botanical name: Lawsonia inermis.
2. Family: Lythraceae.
3. Origin and Area of cultivation: It is indigenous to North Africa and South-west Asia. It is grown mostly throughout India, especially in Gujarat, Madhya Pradesh and Rajasthan.
4. Uses: An orange dye ‘Henna’ is obtained from the leaves and young shoots of Lawsonia inermis. The principal colouring matter of leaves Tacosone” is harmless and causes no irritation to the skin. This dye has long been used to dye skin, hair and finger nails. It is used for. colouring leather, for the tails of horses and in hair-dyes.

Question 25.
What are organic pesticides?
Answer:
Pest like aphids, spider and mites can cause serious damage to flowers, fruits, and vegetables. These creatures attack the garden in swarms, and drain the life of the crop and often invite disease in the process. Many chemical pesticides prove unsafe for human and the environment. It turns fruits and vegetables unsafe for consumption. Thankfully, there are many homemade, organic options to turn to war against pests.

5 – Mark Questions

Question 26.
Write the botanical name, origin, cultivational area and uses of Black gram.
Answer:

1. Botanical name : Vigna mungo
2. Origin and Area of cultivation : Black gram is native to India. Earliest archeobotanical evidences record the presence of black gram about 3,500 years ago. It is cultivated as a rain fed crop in drier parts of India. India contributes to 80% of the global production of black gram. Important states growing black gram in India are Uttar Pradesh, Chattisgarh and Karnataka.
3. Uses : Black gram is eaten whole or split, boiled or roasted or ground into fl our. Black gram batter is a major ingredients for the preparation of popular Southern Indian breakfast dishes. Split pulse is used in seasoning Indian curries.

Question 27.
Give a detailed account on any one fibe yielding plant. Jute
Answer:

1. Botanical name : Corchorus app.
2. Family: Malvaceae
3. Origin and Area of cultivation: Jute is derived from the two cultivated species Corchorus capsularis  Colittorius is of African origin whereas C. Capsularis, is believed to Indo- Burmaese origin. It is an important cultivated commercial crop in Gangetic plains of India and Bangladesh.
4. Use : It is one of the largest exported fibre material of India. The jute industry occupies an important place in the national economy of India. Jute is used for ‘safe’ packaging in view of being natural, renewable, bio-degradable and eco-friendly product. It is used in bagging and wrapping textile. About 75% of the jute produced is used for manufacturing sacks and bags. It is also used in manufacture of blankets, rags, curtains etc. It is also being used as a textile fibre in recent years.

Question 28.
Draw a table mentioning binominals, family name, useful part and their medical uses of any five local medicinal plants.
Answer:

Question 29.
Explain the stepwise preparation of Bio-pest repellent from the leaves of Azadirachta indica
Answer:
Botanical pest repellent and insecticide made with the dried leaves of Azadirachta indica
Preparation of Bio-pest repellent

1. Pluck leaves from the neem tree and chop the leaves finely.
2. The chopped up leaves were put in a 50-liter container and fill to half with water; put the lid on and leave it for 3 days to brew.
3. Using another container, strain the mixture which has brewed for 3 days to remove the leaves, through fine mesh sieve. The filtrate can be sprayed on the plants to repel pests.
4. To make sure that the pest repellent sticks to the plants, add 100 ml of cooking oil and the same amount of soap water. (The role of the soap water is to break down the oil, and the role of the oil is to make it stick to the leaves).
5. The stewed leaves from the mixture can be used in the compost heap or around the base of the plants.

Higher Order Thinking Skills (HOTs) Questions

Question 1.
The following are the binomials of economically useful plants. Mention their respective families.
(a) Borassus flabellifer
(b) Mangifera indica
(c) Saccharum officinarum
(d) Curcuma longa
(e) Capsicum annum
(f) Tamarindus indica
Answer:

Question 2.
Name the plant source from which the following products are obtained.

1. Maida
2. Rubber
3. Morphine
4. Coffee

Answer:

1. Maida – Triticum durum
2. Rubber – Hevea brasiliensis
3. Morphine – Papaver somniferum
4. Coffee – Coffea arabica

Question 3.
Mention the plant species or their products which are popularly known as

1. Queen of species
2. Black Gold of India and
3. King of bitters.

Answer:

1. Queen of Species – Cardamom
2. Black Gold of India – Pepper
3. King of bitters – Andrographis paniculata

Question 4.
Complete the statements.

1. Alphonsa is a variety of ______
2. Cayenne pepper is a variety of ______

Answer:

1. Mango
2. Capsicum

Question 5.
Siddha medicine is an age old practice in our nation. According to this medicinal system, what is the major cause for all sort of illness in humans?
Answer:
According to Siddha medicine, three humors namely vatam, pittam and kapam are responsible for the health of human being and any disturbance in the equilibrium of these humors result in ill health.

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Tamilnadu Samacheer Kalvi 12th Bio Botany Solutions Chapter 2 Classical Genetics

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Samacheer Kalvi 12th Bio Botany Classical Genetics Text Book Back Questions and Answers

Question 1.
Extra nuclear inheritance is a consequence of presence of genes in __________
(a) Mitrochondria and chloroplasts
(b) Endoplasmic reticulum and mitrochondria
(c) Ribosomes and chloroplast
(d) Lysosomes and ribosomes
Answer:
(a) Mitrochondria and chloroplasts

Question 2.
In order to find out the different types of gametes produced by a pea plant having the genotype AaBb, it should be crossed to a plant with the genotype __________
(a) aaBB
(b) AaBB
(c) AABB
(d) aabb
Answer:
(d) aabb

Question 3.
How many different kinds of gametes will be produced by a plant having die genotype AABbCC?
(a) Three
(b) Four
(c) Nine
(d) Two
Answer:
(d) Two

Question 4.
Which one of the following is an example of polygenic inheritance?
(a) Flower colour in MirabilisJalapa
(b) Production of male honey bee
(c) Pod shape in garden pea
(d) Skin Colour in humans
Answer:
(d) Skin Colour in humans

Question 5.
In Mendel’s experiments with garden pea, round seed shape (RR) was dominant over wrinkled seeds (rr), yellow cotyledon (YY) was dominant over green cotyledon (yy). What are the expected phenotypes in the F₂ generation of the cross RRYY xrryy?
(a) Only round seeds with green cotyledons
(b) Only wrinkled seeds with yellow cotyledons
(c) Only wrinkled seeds with green cotyledons
(d) Round seeds with yellow cotyledons an wrinkled seeds with yellow cotyledons
Answer:
(d) Round seeds with yellow cotyledons an wrinkled seeds with yellow cotyledons

Question 6.
Test cross involves __________
(a) Crossing between two genotypes with recessive trait
(b) Crossing between two F₁ hybrids
(c) Crossing the F₁ hybrid with a double recessive genotype
(d) Crossing between two genotypes with dominant trait
Answer:
(c) Crossing the F₁ hybrid with a double recessive genotype

Question 7.
In pea plants, yellow seeds are dominant to green. If a heterozygous yellow seed pant is crossed with a green seeded plant, what ratio of yellow and green seeded plants would you expect in generation?
(a) 9:3
(b) 1:3
(c) 3:1
(d) 50:50
Answer:
(d) 50:50

Question 8.
The genotype of a plant showing the dominant phenotype can be determined by __________
(a) Back cross
(b) Test cross
(c) Dihybrid cross
(d) Pedigree analysis
Answer:
(b) Test cross

Question 9.
Select the correct statement from the ones given below with respect to dihybrid cross
(a) Tightly linked genes on the same chromosomes show very few combinations
(b) Tightly linked genes on the same chromosomes show higher combinations
(c) Genes far apart on the same chromosomes show very few recombinations
(d) Genes loosely linked on the same chromosomes show similar recombinations as the tightly I linked ones
Answer:
(a) Tightly linked genes on the same chromosomes show very few combinations

Question 10.
Which Mendelian idea is depicted by a cross in which the F₁ generation resembles both the parents
(a) Incomplete dominance
(b) Law of dominance
(c) Inheritance of one gene
(d) Co-dominance
Answer:
(d) Co-dominance

Question 11.
Fruit colour in squash is an example of __________
(a) Recessive epistasis
(b) Dominant epistasis
(c) Complementary genes
(d) Inhibitory genes
Answer:
(b) Dominant epistasis

Question 12.
In his classic experiments on Pea plants, Mendel did not use __________
(a) Flowering position
(b) Seed colour
(c) Pod length
(d) Seed shape
Answer:
(c) Pod length

Question 13.
The epistatic effect, in which the hybrid cross 9:3:3:1 between AaBb Aabb is modified as
(a) Dominance of one allele on another allele of both loci
(b) Interaction between two alleles of different loci
(c) Dominance of one allele to another alleles of same loci
(d) Interaction between two alleles of some loci
Answer:
(b) Interaction between two alleles of different loci

Question 14.
In a test cross involving F₁ dihybrid flies, more parental type offspring were produced than the recombination type offspring. This indicates __________
(a) The two genes are located on two different chromosomes
(b) Chromosomes failed to separate during meiosis
(c) The two genes are linked and present on the some chromosome
(d) Both of the characters are controlled by more than one gene
Answer:
(c) The two genes are linked and present on the some chromosome

Question 15.
The genes controlling the seven pea characters studied by Mendel are known to be located on h6w many different chromosomes?
(a) Seven
(b) Six
(c) Five
(d) Four
Answer:
(a) Seven

Question 16.
Which of the following explains how progeny can posses the combinations of traits that none
of the parent possessed?
(a) Law of segregation
(b) Chromosome theory
(c) Law of independent assortment
(d) Polygenic inheritance
Answer:
(d) Polygenic inheritance

Question 17.
“Gametes are never hybrid”. This is a statement of __________
(a) Law of dominance
(b) Law of independent assortment
(c) Law of segregation
(d) Law of random fertilization
Answer:
(c) Law of segregation

Question 18.
Gene which suppresses other genes activity but does not lie on the same locus is called as __________
(a) Epistatic
(b) Supplement only
(c) Hypostatic
(d) Codominant
Answer:
(c) Hypostatic

Question 19.
Pure tall plants are crossed with pure dwarf plants. In the F₁ generation, all plants were tall. These tall plants of generation were selfed and the ratio of tall to dwarf plants obtained was 3:1. This is called __________
(a) Dominance
(b) Inheritance
(c) Codominance
(d) Heredity
Answer:
(a) Dominance

Question 20.
The dominant epistatis ratio is _________
(a) 9:3:3:1
(b) 12:3:1
(c) 9:3:4
(d) 9:6:1
Answer:
(b) 12:3:1

Question 21.
Select the period for Mendel’s hybridization experiments.
(a) 1856 – 1863
(b) 1850 – 1870
(c) 1857-1869
(d) 1870 – 1877
Answer:
(a) 1856 – 1863

Question 22.
Among the following characters which one was not considered by Mendel in his experimentation pea?
(a) Stem – Tall or dwarf
(b) Trichomal glandular or non-glandular
(c) Seed – Green or yellow
(d) Pod – Inflated or constricted
Answer:
(b) Trichomal glandular or non-glandular

Question 23.
Name the seven contrasting traits of Mendel.
Answer:
Plant Height, Seed Shape, Cotyledon colour, Flower colour, Pod colour, Pod form, Flower position

Question 24.
What is meant by true breeding or pure breeding lines / strain?
Answer:
A true breeding lines (Pure-breeding strains) means it has undergone continuous self pollination having stable trait inheritance from parent to offspring. Matings within pure breeding lines produce offsprings having specific parental traits that are constant in inheritance and expression for many generations. Pure line breed refers to homozygosity only.

Question 25.
Give the names of the scientists who rediscovered Mendelism.
Answer:
Mendel’s experiments were rediscovered by three biologists, Hugo de Vries of Holland, Car Correns of Germany and Erich von Tschermak of Austria.

Question 26.
What is back cross?
Answer:
Back cross is a cross of F₁ hybrid with any one of the parental genotypes. The back cross is of two types; they are dominant back cross and recessive back cross. It involves the cross between the F₁ off spring with either of the two parents.

Question 27.
Define Genetics.
Answer:
“Genetics” is the branch of biological science which deals with the mechanism of transmission of characters from parents to offsprings. The term Genetics was introduced by W. Bateson in 1906.

Question 28.
What are multiple alleles?
Answer:
Three or more alternative forms of a gene that occupy the same locus and control the expression of a single trait.
E.g : ABO blood group

Question 29.
What are the reasons for Mendel’s successes in his breeding experiment?
Answer:
Mendel was successful because:

1. He applied mathematics and statistical methods to biology and laws of probability to his breeding experiments.
2. He followed scientific methods and kept accurate and detailed records that include quantitative data of the outcome of his crosses.
3. His experiments were carefully planned and he used large samples.
4. The pairs of contrasting characters which were controlled by factor (genes) were present on separate chromosomes.
5. The parents selected by Mendel were pure breed lines and the purity was tested by self
   crossing the progeny for many generations.

Question 30.
Explain the law of dominance in monohybrid cross.
Answer:
Law of dominance states that the offsprings of an individual with contrasting (dissimilar) traits will only express the dominant trait in F₁ generation and both the characters are expressed in F₂ generation. This law also explains the proportion of 3 : 1 ratio in F₂ generation.

Question 31.
Differentiate incomplete dominance and codominance.
Answer:
Incomplete Dominance:

1. In incomplete dominance, neither of the allele is not completely dominant to another allele rather combine and produce new trait
2. New phenotype is formed due to character blending (not alleles)
3. Example : Pink flowers of Mirabilis Jalapa

Co-dominance:

1. In co-dominance, both the alleles in heterozygote are dominant and the traits are equally expressed (joint expression)
2. No formation of new phenotype rather both dominant traits are expressed, conjointly
3. Example: Red and white flowers of camellia

Question 32.
What is meant by cytoplasmic inheritance
Answer:
DNA is the universal genetic material. Genes located in nuclear chromosomes follow Mendelian inheritance. But certain traits are governed either by the chloroplast or mitochondrial genes. This phenomenon is known as extra nuclear inheritance. It is a kind of Non-Mendelian inheritance. Since it involves cytoplasmic organelles such as chloroplast and mitochondrion that act as inheritance vectors, it is also called Cytoplasmic inheritance.

Question 33.
Describe dominant epistasis with an example.
Answer:
Dominant Epistasis – It is a gene interaction in which two alleles of a gene at one locus interfere and suppress or mask the phenotypic expression of a different pair of alleles of another gene at another locus. The gene that suppresses or masks the phenotypic expression of a gene at another locus is known as epistatic.

The gene whose expression is interfered by non-allelic genes and prevents from exhibiting its character is known as hypostatic. When both the genes are present together, the phenotype is determined by the epistatic gene and not by the hypostatic gene.

In the summer squash the fruit colour locus has a dominant allele ‘W’ for white colour and a recessive allele ‘w’ for coloured fruit. ‘W’ allele is dominant that masks the expression of any colour.

Dominant epistasis in summer squash

In another locus hypostatic allele ‘G’ is for yellow fruit and its recessive allele ‘g’ for green fruit. In the first locus the white is dominant to colour where as in the second locus yellow is dominant to green. When the white fruit with genotype WWgg is crossed with yellow fruit with genotype wwGG, the F₁ plants have white fruit and are heterozygous (WwGg). When F₁ heterozygous plants are crossed.

they give rise to F₂ with the phenotypic ratio of 12 white : 3 yellow : 1 green.Since W is epistatic to the alleles ‘G’ and ‘g’, the white which is dominant, masks the effect of yellow or green. Homozygous recessive ww genotypes only can give the coloured fruits (4/16). Double recessive ‘wwgg’ will give green fruit (1/16). The Plants having only ‘G’ in its genotype (wwGg or wwGG) will give the yellow fruit(3/l 6).

Question 34.
Explain polygenic inheritance with an example.
Answer:
Polygenic inheritance – Several genes combine to affect a single trait. A group of genes that together Dark Red determine (contribute) a characteristic of an organism is called polygenic inheritance. It gives explanations to the inheritance of continuous traits which are compatible with Mendel’s Law. The first experiment on polygenic inheritance was demonstrated by Swedish Geneticist H.

Nilsson-Ehle (1909) in wheat kernels. Kernel colour is controlled by two genes each with two alleles, one with red kernel colour was dominant to white. He crossed the two pure breeding wheat varieties dark red and a white. Dark red genotypes F₁ generation R₁R₁R₂R₂ and white genotypes are r₁r₁r₂r₂ – F₁ generation medium red were obtained with the genotype R₁r₁R₂r₂. F₁ wheat plant produces

four types of gametes R₁R₂, R₁r₂, r,₁r₂. The intensity of the red colour is determined by the number of R genes in the F₂ generation. Four R genes: A dark red kernel colour is obtained. Three R genes: Medium – dark red kernel colour is obtained. Two R genes: Medium-red kernel colour is obtained. One R gene: Light red kernel colour is obtained. Absence of R gene: Results in White kernel colour.

The R gene in an additive manner produces the red kernel colour. The number of each phenotype is plotted against the intensity of red kernel colour which produces a bell shaped curve. This represents the distribution of phenotype.

Conclusion: Finally the loci that was studied by Nilsson – Ehle were not linked and the genes assorted independently. Later, researchers discovered the third gene that also affect the kernel colour of wheat. The three independent pairs of alleles were involved in wheat kernel colour. Nilsson – Ehle found the ratio of 63 red : 1 white in F₂ generation – 1 : 6 : 15 : 20 : 15 : 6 : 1 in F₂ generation.

Question 35.
Differentiate continuous variation with discontinuous variation.
Answer:
1. Discontinuous Variation: Within a population there are some characteristics which show a limited form of variation.
Example: Style length in Primula, plant height of garden pea. In discontinuous variation, the characteristics are controlled by one or two major genes which may have two or more allelic forms.

These variations are genetically determined by inheritance factors. Individuals produced by this variation show differences without any intermediate form between them and there is no overlapping between the two phenotypes. The phenotypic expression is unaffected by environmental conditions. This is also called as qualitative inheritance

2. Continuous Variation: This variation may be due to the combining effects of environmental and genetic factors. In a population most of the characteristics exhibit a complete gradation, from one extreme to the other without any break. Inheritance of phenotype is determined by the combined effects of many genes, (polygenes) and environmental factors. This is also known as quantitative inheritance.
Example: Human height and skin color.

Question 36.
Explain with an example how single genes affect multiple traits and alleles the phenotype of an organism.
Answer:
In Pleiotropy, the single gene affects multiple traits and alter the phenotype of the organism. The Pleiotropic gene influences a number of characters simultaneously and such genes are called pleiotropic gene. Mendel noticed pleiotropy while performing breeding experiment with peas (Pisum sativum).

Peas with purple flowers, brown seeds and dark spot on the axils of the leaves were crossed with a variety of peas having white flowers, light coloured seeds and no spot on the axils of the leaves, the three traits for flower colour, seed colour and a leaf axil spot all were inherited together as a single unit. This is due to the pattern of inheritance where the three traits were controlled by a single gene with dominant and recessive alleles. Example: sickle cell anemia.

Question 37.
Bring out the inheritance of chloroplast gene with an example.
Answer:
Chloroplast Inheritance

It is found in 4 ‘O’ Clock plant (Mirabilis jalapa). In this, there are two types of variegated leaves namely dark green leaved plants and pale green leaved plants. When the pollen of dark green leaved plant (male) is transferred to the stigma of pale green leaved plant (female) and pollen of pale green leaved plant is transferred to the stigma of dark green leaved plant, the F₁ generation of both the crosses must be identical as per Mendelian inheritance. But in the reciprocal cross the F₁ plant differs from each other.

In each cross, the F plant reveals the character of the plant which is used as female plant. This inheritance is not through nuclear gene. It is due to the chloroplast gene found in the ovum of the female plant which contributes the cytoplasm during fertilization since the male gamete contribute only the nucleus but not cytoplasm.

Samacheer Kalvi 12th Bio Botany Classical Genetics Additional Questions and Answers

1 – Mark Questions

Question 1.
The term ‘Genetics’ was introduced by __________
(a) Gregor Mendel
(b) Bateson
(c) Hugo de vries
(d) Carl Correns
Answer:
(b) Bateson

Question 2.
Which is not a correct statements?
(A) Variations are the raw materials for evolution
(B) Variations provide genetic material for natural selection
(C) It helps the individual to adapt to changing environment
(D) Variations allow breeders to improve the crop field
(a) A and D
(b) B only
(c) C and D
(d) nono of he above
Answer:
(d) nono of he above

Question 3.
The process of removal of anthers from the flower is called __________
Answer:
Emasculation

Question 4.
An allede is __________
(a) another word for a gene
(b) alternate forms of a gene
(c) morphological expression of a gene
(d) genitic
Answer:
(b) alternate forms of a gene

Question 5.
Gregor Mendel __________
(i) was born in Czechoslovakia
(ii) did his experiments in Pisum fulvum
(iii) was the first systemic researcher in genetics
(iv) Published his results in the paper “Experiments on Plant Hybrids”
(a) All are correct
(b) (ii),(iii), (iv) are correct
(c) (i), (iii),(iv) are correct
(d) (i), (iii),(iv) are correct
Answer:
(c) (i), (iii),(iv) are correct

Question 6.

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

Question 7.
How many characters studied by Mendel in pisum sativum
(a) Three
(b) Five
(c) Seven
(d) Nine
Answer:
(c) Seven

Question 8.
Mendel’s work were rediscovered by __________
(a) Hugo de Vries
(b) Tschermak
(c) Carl Correns
(d) All the above
Answer:
(d) All the above

Question 9.
Crossing of F₁, to any one of the parent refers to __________
(a) selling
(b) back cross
(c) test cross
(d) all of the above
Answer:
(b) back cross

Question 10.
Match the following

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

Question 11.
In an intergenic interaction, the gene that suppresses the pherotype of a gene is said to Crossing of F, to any one of the parent refers to __________
(a) Dominant
(b) Inhibitory
(c) Epistatic
(d) Hypostatic
Answer:
(c) Epistatic

Question 12.
Assertion (A) : Test cross is done between F₂ hybrid with F₁ recessive
Reason (R) : It helps to identify the homozygosity of hybrids
(a) A and R are correct R explains A
(b) A and R are incorrect
(c) A is correct R is incorrect
(d) A is incorrect R is correct
Answer:
(b) A and R are incorrect

Question 13.
Assertion (A) : Codominance is an example for intragenic interaction
Reason (R) : Interaction take place between the alleles of same gene
(a) A and R are correct R explains A
(b) A and R are incorrect
(c) A is correct R is incorrect
(d) A is incorrect R is correct
Answer:
(a) A and R are correct R explains A

Question 14.
Assertion (A) : Pleiotropic gene affects multiple traits
Reason (R) : ABO blood group is an example for Pleiotropism
(a) A and R are correct R explains A
(b) A and R are incorrect
(c) A is correct R is incorrect
(d) A is incorrect R is correct
Answer:
(c) A is correct R is incorrect

Question 15.
Assertion (A) : Cytoplasmic male sterility is a Mendelian inheritance
Reason (R) : The genes for cytoplasmic male sterility in peal maize is located at mitochondrial DNA
(a) A and R are correct R explains A
(b) A and R are incorrect
(c) A is correct R is incorrect
(d) A is incorrect R is correct
Answer:
(d) A is incorrect R is correct

Question 16.
What is the phenotypic ratio in case of incomplete dominance
(a) 9 : 7
(b) 3 : 1
(c) 1 : 2 : 1
(d) 1 : 1 : 1 : 1
Answer:
(c) 1 : 2 : 1

Question 17.
Identify the mismatched pair
(a) Chloroplast inheritance – Gregor Mendel
(b) Polygenic inheritance – H. Nilsson
(c) Lethal genes – E. Baur
(d) Incomplete dominance – Carl Correns
Answer:
(a) Chloroplast inheritance – Gregor Mendel

Question 18.
Statement 1 : Intergenic gene interaction occurs between alleles at same locus
Statement 2 : Co-dominance is an example for intergenic gene interaction
(a) Statement 1 is correct & Statement 2 is incorrect
(b) Statement 1 is incorrect & Statement 2 is correct
(c) Both Statements 1 & 2 are correct
(d) Both Statements 1 & 2 are incorrect
Answer:
(c) Both Statements 1 & 2 are correct

Question 19.
Statement 1 : Test cross is done between F₁ individual with homozygous recessive
Statement 2 : If F₁ individual is homozygous, the rate of a monohybrid cross will be 1:1
(a) Statement 1 is correct & Statement 2 is incorrect
(b) Statement 1 is incorrect & Statement 2 is correct
(c) Both Statements 1 & 2 are correct
(d) Both Statements 1 & 2 are incorrect
Answer:
(a) Statement 1 is correct & Statement 2 is incorrect

Question 20.
Identify the incorrect statement
Answer:
(a) In incomplete dominance, the traits are blended not the genes
(b) Incomplete dominance is noticed in Mirabilis jalapa by Carl Correns
(c) It is a type of Intragenic gene interaction
(d) Incomplete dominance F₂ ratio is 1 : 3 : 1
Answer:
(d) Incomplete dominance F₁ ratio is 1 : 3 :1

Question 21.
In case of co-dominance, monohybrid F_{1 __________} is 1 : 2 : 1
(a) Genotype ratio
(b) Phenotype ratio
(c) Both genotype & Phenotype ratio
(d) Ratio is wrong
Answer:
(c) Both genotype & Phenotype ratio

Question 22.
Identify the wrong statement (s)
(i) Monohybrid cross involve the inhertance of teo alleles of a gene
(ii) The dwarf traits reappeared in F₂
(iii) Law of dominance was proved by monohybrid cross
(iv) F₁ monohybrid was an hererozygous
(a) i and ii
(b) iii and iv
(c) i only
(d) none of the above
Answer:
(d) none of the above

Question 23.
Result of incomplete dominance is __________
(а) Intermediate genotype
(b) Intermediate phenotype
(c) Recessive phenotype
(d) Epistasis
Answer:
(b) Intermediate phenotype

Question 24.
Heterozygous Tall mono hybrid is cross with homozygous dwarf. What will be characteristic of offspring?
(a) 25 % recessive 75% dominant
(b) 75 % recessive 25% dominant
(c) 50 % recessive 50% dominant
(d) All are dominance
Answer:
(c) 50 % recessive 50% dominant

Question 25.
ABO blood group is a classical example for __________
(a) Polygenic inheritance
(b) Incomplete Dominance
(c) Epistasis
(d) Dominance
Answer:
(d) Dominance

Question 26.
RR (Red) flower of Mirabilis is crossed with White (WW) flowers. Resultant offspring are pink RW. This is an example of __________
(a) Epistasis
(b) Co-dominance
(c) Incomplete dominance
(d) Pleiotropism
Answer:
(c) Incomplete dominance

Question 27.
How many genetically different gametes are produced by a plant have genotype TtYyRr?
(a) 2
(b) 4
(c) 6
(d) 8
Answer:
(d) 8

Question 28.
When a single gene influences multiple traits then the phenomenon is called __________
(a) Pleiotropy
(b) Polygenic inheritance
(c) Epistasis
(d) Atavism
Answer:
(a) Pleiotropy

Question 29.
According to Mendel which character shown dominance.
(a) Yellow flower color
(b) Yellow cotyledon color
(c) Wrinkled seeds
(d) Inflated pod
Answer:
(d) Inflated pod

Question 30.
Ratio of recessive epistasis is __________
(a) 12 : 3 : 1
(b) 9 : 7
(c) 9 : 3 : 4
(d) 9 : 6 : 1
Answer:
(c) 9 : 3 : 4

Question 31.
According to Mendel, which is not a dominant trait?
(a) Wrinkled seeds
(b) Purple flower
(c) Inflated pod form
(d) Axial flower portion
Answer:
(a) Wrinkled seeds

Question 32.
Identify the allelic interaction.
(a) Domination epistasis
(b) Co – dominance
(c) Recessive epistasis
(d) Duplicate genes
Answer:
(b) Co – dominance

Question 33.
Gametes are never hybrid’ is concluded by __________
(a) Law of dominance
(b) Law of segregation
(c) Law of independent environment
(d) Law of lethality
Answer:
(b) Law of segregation

Question 34.
Factor hypothesis was proposed by __________
(a) Reginald Punnett
(b) W. Bateson
(c) Gregor Mende
(d) Carl Correns
Answer:
(b) W. Bateson

Question 35.
The 1:2:1 ratio of co-dominance process Mendel’s __________
(a) Law of dominance
(b) Law of recessiveness
(c) Law of segregation
(d) Law of independent assortment
Answer:
(b) Law of recessiveness

Question 36.
Match the following:
Epistatic interaction Example
(A) Complementary genes (i) Seed capsule in xxxxx
(B) Supplementary genes (ii) Leaf color in rice plant
(C) Inhibitory genes (iii) Grain color in maize
(D) Duplicate genes (iv) Flower color in sweet peas
Answer:
A – (iv)
B – (iii)
C – (ii)
D – (i)

2 – Mark Questions

Question 1.
Who coined the term genetics? Also define it.
Answer:
“Genetics” is the branch of biological science which deals with the mechanism of transmission of characters from parents to off springs. The term Genetics was introduced by W. Bateson in 1906.

Question 2.
Name the four major subdisciplines of genetics.
Answer:
(a) Classical genetics
(b) Molecular genetics
(c) Population genetics
(d) Quantitative genetics

Question 3.
Define Heredity and variations.
Answer:
Heredity : Heredity is the transmission of characters from parents to off springs.
Variations : The organisms belonging to the same natural population or species that shows a difference in the characteristics is called variation.

Question 4.
Mendel’s theory is a particulate theory – justify.
Answer:
Mendel’s theory of inheritance, known as the Particulate theory, establishes the existence of minute particles or hereditary units or factors, which are now called as genes.

Question 5.
Which organism was studied by Gregor Mendel? How many traits does he considered on his experiments?
Answer:
Gregor Mendel selected seven pairs of characters in Pisum sativum (garden pea)

Question 6.
Name any four characters of pisum sativum that was studied by Mendel.
Answer:
Seed shape, flower color, flower position & pod color.

Question 7.
Define the terms

1. Emasculation
2. Alleles.

Answer:

1. Emasculation : Removal of anthers from the flower
2. Alleles : Alternate forms of a gene

Question 8.
Name the first and second law of Mendel.
Answer:

1. The Law of Dominance
2. The Law of Segregation

Question 9.
What is genotype & phenotype?
Answer:
genotype & phenotype

1. The term genotype is the genetic constitution of an individual.
2. The term phenotype refers to the observable characteristic of an organism.

Question 10.
Write the phenotypic and genotypic ratio of monohybrid cross.
Answer:
(a) Phenotypic ratio = 3:1.
(b) Genotypic ratio =1 : 2 : 1

Question 11.
What is test cross? Why it is done?
Answer:

1. Test cross is crossing an individual of unknown genotype with a homozygous recessive.
2. Test cross is used to identify whether an individual is homozygous or heterozygous for dominant character.

Question 12.
State the law of independent assortment.
Answer:
When two pairs of traits are combined in a hybrid, segregation of one pair of characters is independent to the other pair of characters. Genes that are located in different chromosomes assort independently during meiosis.

Question 13.
Give the phenotypic ratio of
(a) Dihybrid cross
(b) Dihybrid test cross
Answer:
(a) Dihybrid cross ratio = 9 : 3 : 3 : 1
(b) Dihybrid test cross ratio = 1 : 1 : 1 : 1

Question 14.
RrYyf (F₁ hybrid)  rryy (recessive parent). Name the type of cross. Mention its ratio.
Answer:
Dihybrid test cross and the ratio is 1 : 1 : 1 : 1

Question 15.
How many types of gametes are produced by a dihybrid plant. If the same plant is self fertilized, how many second generation offsprings are developed?
Answer:
Four different gametes are produced by a dihybrid plant and on selfing, it yield 16 off springs.

Question 16.
Write the phenotypic ratio of trihybrid cross.
Answer:
27 : 9 : 9 : 9 : 3 : 3 : 3 : 1

Question 17.
Define gene interaction.
Answer:
A single phenotype is controlled by more than one set of genes, each of which has two or more alleles. This phenomenon is called Gene Interaction.

Question 18.
Classify gene interactions with an example.
The gene interactions may be
(a) Intragenic gene interaction. E.g.: Codominance
(b) Intergenic gene interaction. E.g.: Epistasis

Question 19.
Provide any four intergenic gene interactions.
Answer:
(a) Incomplete dominance
(b) Codominance
(c) Multiple alleles
(d) Pleiotropic genes are common examples for intragenic interaction.

Question 20.
Define intragenic interaction
Answer:
Interactions take place between the alleles of the same gene i.e., alleles at the same locus is called intragenic or intralocus gene interaction.

Question 21.
In which plant does the incomplete dominance was studied by Carl Correns? Write the ratio of the cross.
Answer:
Mirabilis Jalapa (4 o’ clock plant). Incomplete dominance ratio is 1 : 2 : 1

Question 22.
What are lethal alleles? Give example.
Answer:
An allele which has the potential to cause the death of an organism is called a Lethal Allele.
E.g : Recessive lethality in Antirrhinum species.

Question 23.
Give the proper terminologies for the following statement
(a) Single gene affecting multiple traits
(b) Single trait affected by many genes.
Answer:
(a) Pleiotropism
(b) Poly genic inheritance

Question 24.
What is intergenic gene interactions? Give example
Answer:
Interlocus interactions take place between the alleles at different loci i.e. between alleles of different genes.
Eg: Dominant Epistasis

Question 25.
Name any two extranuclear inheritance.
Answer:
(a) Chloroplast inheritance
(b) Mitrochondrial inheritance

Question 26.
What are plasmogenes?
Answer:
Plasmogenes are independent, self-replicating, extra-chromosomal units located in cytoplasmic organelles, chloroplast and mitochondrion

Question 27.
What are extra nuclear inheritance?
Answer:
Certain characters/traits are governed and inherited by genes located in cytoplasmic organelles (chloroplast or mitochondrion) other than nucleus. This is called extra nuclear inheritance.

Question 28.
Why extranuclear inheritance is called as cytoplasmic inheritance.
Answer:
Extra nuclear inheritance is due to genes located on the cytoplasmic organelles such as chloroplast and mitochondrion hence it is called cytoplasmic inheritance.

Question 29.
What is cytoplasmic male sterility?
Answer:
In Sorghum vulgare (Pearl maize), the gene located for the sterility pollens are located in the mitochondrial DNA. This phenomenon is called as cytoplasmic male sterility.

3 – Mark Questions

Question 30.
Point out any three importance of variations.
Answer:

1. They help the individuals to adapt themselves to the changing environment.
2. Variations allow breeders to improve better yield, quicker growth, increased resistance and lesser input.
3. They constitute the raw materials for evolution.

Question 31.
Why Mendel selected pea plants for his experiments.
Answer:
He choose pea plant because,

1. It is an annual plant and has clear contrasting characters that are controlled by a single gene separately.
2. Self-fertilization occurred under normal conditions in garden pea plants. Mendel used both self-fertilization and cross-fertilization.
3. The flowers are large hence emasculation and pollination are very easy for hybridization.

Question 32.
State the law of segregation.
Answer:
The Law of Segregation (Law of Purity of gametes): Alleles do not show any blending. During the formation of gametes, the factors or alleles of a pair separate and segregate from each other such that each gamete receives only one of the two factors. A homozygous parent produces similar gametes and a heterozygous parent produces two kinds of gametes each having one allele with equal proportion. Gametes are never hybrid.

Question 33.
How many types of gametes are produced by heterozygous dihybrid plant with a genotypeRrYy? Write them.
Answer:
Four gametes – RY, Ry, rY, ry

Question 34.
Define trihybrid cross. Mention its F₂ phenotypic ratio.
Answer:
A cross between homozygous parents that differ in three gene pairs (i.e. producing trihybrids) is called trihybrid cross, F₂ Phenotypic ratio -27 : 9 : 9 : 9 : 3 : 3 : 3 : 1

Question 35.
Define co-dominance. How it is proved by using Gossypium species?
Answer:
The phenomenon in which two alleles are both expressed in the heterozygous individual is known as codominance. The codominance was demonstrated in plants with the help of electrophoresis or chromatography for protein or flavonoid substance.

Example: Gossypium hirsutum and Gossypium sturtianum, their F₁ hybrid (amphiploid) was tested for seed proteins i by electrophoresis. Both the parents have different banding patterns for their seed proteins. In hybrids, additive banding pattern was noticed. Their hybrid shows the presence of both the types of proteins similar to their parents.

Question 36.
Give an account on cytoplasmic male sterility.
Answer:
Male sterility found in pearl maize (Sorgum vulgare) is the best example for mitochondrial cytoplasmic inheritance. So it is called cytoplasmic male sterility. In this, male sterility is inherited maternally. The gene for cytoplasmic male sterility is found in the mitochondrial DNA.

Question 37.
Write a short note on Atavism.
Answer:
Atavism is a modification of a biological structure whereby an ancestral trait reappears after having been lost through evolutionary changes in the previous generations. Evolutionary traits that have disappeared phenotypically do not necessarily disappear from an organism’s DNA. The gene sequence often remains, but is inactive.

Such an unused gene may remain in the genome for many generations. As long as the gene remains intact, a fault in the genetic control suppressing the gene can lead to the reappearance of that character again. Reemergence of sexual reproduction in the flowering plant Hieracium pilosella is the best example for Atavism in plants.

5 – Mark Questions

Question 38.
Explain Dihybrid cross in pea plant.
Answer:
The crossing of two plants differing in two pairs of contrasting traits is called dihybrid cross. In dihybrid cross, two characters (colour and shape) are considered at a time. Mendel considered the seed shape (round and wrinkled) and cotyledon colour (yellow & green) as the two characters. In seed shape round (R) is dominant over wrinkled (r); in cotyledon colour yellow (Y) is dominant over green (y).

Hence the pure breeding round yellow parent is represented by the genotype RRYY and the pure breeding green wrinkled parent is represented by the genotype rryy. During gamete formation the paired genes of a character assort out ‘ independently of the other pair. During the F₁ x F, fertilization each zygote with an equal probability receives one of the four combinations from each parent. The resultant gametes thus will be genetically different and they are of the following four types:

(1) Yellow round (YR) – 9/16
(2) Yellow wrinkled (Yr) – 3/16
(3) Green round (yR) – 3/16
(4) Green wrinkled (yr) -1/16

These four types of gametes of F₁ dihybrids unite randomly in the process of fertilization and produce sixteen types of individuals in F₂ in the ratio of 9:3:3:1 as shown in the figure. Mendel’s 9:3:3:1 dihybrid ratio is an ideal ratio based on the probability including segregation, independent assortment and random fertilization. In sexually reproducing organism / plants from the garden peas to human beings, Mendel’s findings laid the foundation for understanding inheritance and revolutionized the field of biology. The dihybrid cross and its result led Mendel to propose a second set of generalisations that we called Mendel’s Law of independent assortment.

Question 39.
How does the wrinkled gene make Mendel’s peas wrinkled? Find out the molecular explanation.
Answer:
The protein called starch branching enzyme (SBEI) is encoded by the wild-type allele of the gene (RR) which is dominant. When the seed matures, this enzyme SBEI catalyzes the formation of highly branched starch molecules. Normal gene (R) has become interrupted by the insertion of extra piece of DNA (0.8 kb) into the gene, resulting in allele. In the homozygous mutant form of the gene (R) which is recessive, the activity of the enzyme SBEI is lost resulting in wrinkled peas.

The wrinkled seed accumulates more sucrose and high water content. Hence Ore osmotic pressure inside the seed rises. As a result, the seed absorbs more water and when it matures it loses water as it dries. So it becomes wrinkled at maturation. When the seed has at least one copy of normal dominant gene heterozygous, the dominant allele helps to synthesize starch, amylopectin an insoluble carbohydrate, with the osmotic balance which minimises the loss of water resulting in smooth structured round seed.

Question 40.
Describe incomplete dominance exhibited by Mirabilis jalapa.
Answer:
The German Botanist Carl Correns’s (1905) Experiment – In 4 O’ clock plant, Mirabilis jalapa when the pure breeding homozygous red (R¹R¹) parent is crossed with homozygous white (R²R²), the phenotype of the F₁ hybrid is heterozygous pink (R¹R²). The F₁ heterozygous phenotype differs from both the parental homozygous phenotype. This cross did not exhibit the character of the dominant parent but an intermediate colour pink. When one allele is not completely dominant to another allele it shows incomplete dominance. Such allelic interaction is known as incomplete dominance. F₁ generation produces intermediate phenotype pink coloured flower.

When pink coloured plants of F₁ generation were interbred in F₂ both phenotypic and genotypic ratios were found to be identical as 1 : 2 :1(1 red: 2 pink: 1 white). Genotypic ratio is 1 R¹ R¹ : 2 R¹R² : 1 R²R². From this we conclude that the alleles themselves remain discrete and unaltered proving the Mendel’s Law of Segregation. The phenotypic and genotypic ratios are the same. There is no blending of genes. In the F¹ generation R¹ and R² genes segregate and recombine to produce red, pink and white in the ratio of 1 : 2 : 1. R¹ allele codes for an enzyme responsible for the formation of red pigment. R² allele codes for defective enzyme.

R¹ and R² genotypes produce only enough red pigments to make the flower pink. Two R¹ R² are needed for producing red flowers. Two R²R² genes are needed for white flowers. If blending had taken place, the original pure traits would not have appeared and all F₂ plants would have pink flowers. It is very clear that Mendel’s particulate inheritance takes place in this cross which is confirmed by the reappearance of original phenotype in F₂.

Higher Order Thinking Skills (HOTs) Questions

Question 1.
A yellow colour flower plant indicated by YY is crossed with white color flower plant denoted by yy.
(a) following the Mendelian inheritance pattern, what would be the flower color is first filial generation?
(b) Which Mendelian principle is illustrated in this cross?
(c) Derive the cross and state the phenotypic ratio of yellow flowers to white flowers in F₂ generation?
Answer:
(a) F₁ plants produce yellow colour flower plants.
(b) Law of dominance and Law of segregation
(c)

Question 2.
Mala is a genetic research student. She was given a plant to identify whether it is a homozygous or heterozygous for a particular trait. How will she proceed further?
Answer:
To identify the plant genotype whether homozygous or heterozygous Mala can perform test cross, where the individual is crossed with homozygous recessive for the trait. If the plant is heterozygous then the resultant progenies would be in the ratio 50:50

Question 3.
In the chart given below, ‘AA’ are the genes located in a chromosome of Pisum sativum.
Answer:

Observe the chart and mention the genetic phenomenon does it indicates.
Pleitrophy – A single gene affecting many traits. Here the single gene AA controls the traits – for flower colour, seed colour and leaf axil spot.

Question 4.
Give the F₂ phenotypic ratio of
(a) Supplementary genes
(b) Complementary genes
(c) Dominant epistasis
Answer:
(a) Supplementary genes – 9 : 3 : 4
(b) Complementary genes – 9 : 7
(c) Dominant epistasis -12 : 3 : 1

Question 5.
Name the respective pattern of inheritance where F₁ phenotype
(a) resembles any one of the two parents
(b) is an intermediate between two parental traits.
Answer:
(a) Dominance
(b) Incomplete dominance

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Samacheer Kalvi 12th Bio Botany Principles and Processes of Biotechnology Text Book Back Questions and Answers

Question 1.
Restriction enzymes are ___________
(a) Not always required in genetic engineering
(b) Essential tools in genetic engineering
(c) Nucleases that cleave DNA at specific sites
(d) both b and c
Answer:
(d) both b and c

Question 2.
Plasmids are ___________
(a) circular protein molecules
(b) required by bacteria
(c) tiny bacteria
(d) confer resistance to antibiotics
Answer:
(d) confer resistance to antibiotics

Question 3.
EcoRI cleaves DNA at
(a) AGGGTT
(b) GTATATC
(c) GAATTC
(d) TATAGC
Answer:
(c) GAATTC

Question 4.
Genetic engineering is ___________
(a) making artificial genes
(b) hybridization of DNA of one organism to that of the others.
(c) production of alcohol by using micro organisms.
(d) making artificial limbs, diagnostic instruments such as ECG and EEG, etc.
Answer:
(b) hybridization of DNA of one organism to that of the others.

Question 5.
Consider the following statements:
i. Recombinant DNA technology is popularly known as genetic engineering is a stream of,biotechnology which deals with the manipulation of genetic materials by man invitro
ii. pBR322 is the first artificial cloning vector developed in 1977 by Boliver and Rodriguez from E.coli plasmid.
iii. Restriction enzymes belongs to a class of enzymes called nucleases. Choose the correct option regarding above statements
(a) i and ii
(b) i and iii
(c) ii and iii
(d) i,ii and iii
Answer:
(d) i,ii and iii

Question 6.
The process of recombinant DNA technology has the following steps
i. Amplication of the gene.
ii. Insertion of recombinant DNA into the host cells.
iii. Cutting of DNA at specific location using restriction enzyme.
iv. Isolation of genetic material (DNA).
Pick out the correct sequence of step for recombinant DNA technology.
(a) ii, iii, iv, and i
(b) iv, ii, iii, and i
(c) i, ii, iii and iv
(d) iv, iii, i, and ii
Answer:
(d) iv, iii, i, and ii

Question 7.
Which one of the following palindromic base sequence in DNA can be easily cut at about the middle by some particular restriction enzymes?
(a) 5′ CGTTCG 3′ ATCGTA5′
(b) 5′ GATATG 3′ CTACTA5′
(c) 5′ GAATTC 3′ CTTAAG 5′
(d) 5′ CACGTA 3′ CTCAGT 5′
Answer:
(c) 5′ GAATTC 3′ CTTAAG 5′

Question 8.
pBR 322, BR stands for
(a) Plasmid Bacterial Recombination
(b) Plasmid Bacterial Replications
(c) Plasmid Boliver and Rodriguez
(d) Plasmid Baltimore and Rodriguez
Answer:
(c) Plasmid Boliver and Rodriguez

Question 9.
Which of the following one is used as a Biosensors?
(a) Electrophoresis
(b) Bioreactors
(c) Vectors
(d) Electroporation
Answer:
(b) Bioreactors

Question 10.
Match the following

Column A	Column B
1. Exonuclease	a. add or remove phosphate
2. Endonuclease	b. binding the DNA fragments
3. Alkaline Phosphatase	c. cut the DNA at terminus
4. Ligase	d. cut the DNA at middle

(A) a b c d
(B) c d b a
(C) a c b d
(D) c d a b
Answer:
(D) c d a b

Question 11.
In which techniques Ethidium Bromide is used?
(a) Southern Blotting techniques
(b) Western Blotting techniques
(c) Polymerase Chain Reaction
(d) Agarose Gel Electrophoresis
Answer:
(d) Agarose Gel Electrophoresis

Question 12.
Assertion: Agrobacterium tumifaciens is popular in genetic engineering because this bacteriumis associated with the root nodules of all cereals and pulse crops.
Reason: A gene incorporated in the bacterial chromosomal genome gets automatically transferred to the cross with which bacterium is associated.
(a) Both assertion and reason are true. But reason is correct explanation of assertion.
(b) Both assertion and reason are true. But reason is not correct explanation of assertion.
(c) Assertion is true, but reason is false.
(d) Assertion is false, but reason is true.
(e) Both assertion and reason are false.
Answer:
(a) Both assertion and reason are true. But reason is correct explanation of assertion.

Question 13.
Which one of the following is not correct statement?
(a) Ti plasmid causes the bunchy top disease
(b) Multiple cloning site is known as Polylinker
(c) Non-viral method of transfection of Nucleic acid in cell
(d) Polylactic acid is a kind of biodegradable and bioactive thermoplastic.
Answer:
(a) Ti plasmid causes the bunchy top disease

Question 14.
An analysis of chromosomal DNA using the southern hybridisation technique does not use
(a) Electrophoresis
(b) Blotting
(c) Autoradiography
(d) Polymerase Chain Reaction
Answer:
(a) Electrophoresis

Question 15.
An antibiotic gene in a vector usually helps in the selection of
(a) Competent cells
(b) Transformed cells
(c) Recombinant cells
(d) None of the above
Answer:
(a) Competent cells

Question 16.
Some of the characteristics of Bt cotton are
(a) Long fibre and resistant to aphids
(b) Medium yield, long fibre and resistant to beetle pests
(c) high yield and production of toxic protein crystals which kill dipteran pests.
(d) High yield and resistant to ball worms
Answer:
(b) Medium yield, long fibre and resistant to beetle pests

Question 17.
How do you use the biotechnology in modern practice?
Answer:
In modem practice, biotechnology is used in the development of herbicide resistance plants, improved crop varieties, producing pharma products like insulin, developing vaccines, diagnosing genetic diseases and designing drgus etc.

Question 18.
What are the materials used to grow microorganism like Spirulinal
Answer:
Spirulina can be grown easily on materials like waste water from potato processing plants (containing starch), straw, molasses, animal manure and even sewage, to produce large quantities.

Question 19.
You are working in a biotechnology lab with a bacterium namely E.coli. How will you cut the nucleotide sequence? explain it.
Answer:
The DNA nucleotide sequence can be cut using Restriction endonucleases (RE). Restriction endonucleases – EcoRI cuts the DNA at GAATTC seqUence, producing sticky ends.CTTAAG

Question 20.
What are the enzymes you can use to cut terminal end and internal phospho diester bond of nucleotide sequence?
Answer:
Restriction exonuclease are the restriction enzyme used to cut nucleotides from the terminal end of DNA. Whereas, restriction endonucleases cut the internal phospho diester bond with DNA molecule.

Question 21.
Name the chemicals used in gene transfer.
Answer:
Polyethylene Glycol (PEG) and Dextran Sulphate.

Question 22.
What do you know about the word pBR332?
Answer:
pBR 322 plasmid is a reconstructed plasmid and most widely used as cloning vector; it contains 4361 base pairs. In pBR, p denotes plasmid, B and R respectively the names of scientist Roliver and/fodriguez who developed this plasmid. The number is the number of plasmid developed from their laboratory.

It contains ampR and tetR two different antibiotic resistance genes and recognition sites for several restriction enzymes. (Hind III, EcoRI, BamH I, Sal I, Pvu II, Pst I and Cla I), ori and antibiotic resistance genes. Rop codes for the proteins involved in the replication of the plasmid.

Question 23.
Mention the application of biotechnology.
Answer:

1. Biotechnology is one of the most important applied interdisciplinary sciences of the 21st century. It is the trusted area that enables us to find the beneficial way of life.
2. Biotechnology has wide applications in various sectors like agriculture, medicine,environment and commercial industries.
3. This science has an invaluable outcome like transgenic varieties of plants e.g. transgenic cotton (Bt-cotton), rice, tomato, tobacco, cauliflower, potato and banana.
4. The development of transgenics as pesticide resistant, stress resistant and disease resistant varieties of agricultural crops is the immense outcome of biotechnology.
5. The synthesis of human insulin and blood protein in E.coli and utilized for insulin deficiency disorder in human is a breakthrough in biotech industries in medicine.
6. The synthesis of vaccines, enzymes, antibiotics, dairy products and beverages are the products of biotech industries.
7. Biochip based biological computer is one of the successes of biotechnology.
8. Genetic engineering involves genetic manipulation, tissue culture involves aseptic cultivation of totipotent plant cell into plant clones under controlled atmospheric conditions.
9. Single cell protein from Spirulina is utilized in food industries.
10. Production of secondary metabolites, biofertilizers, biopesticides and enzymes.
11. Biomass energy, biofuel, bioremediation and phytoremediation for environmental biotechnology.

Question 24.
What are restriction enzyme. Mention their type with role in biotechnology.
Answer:
Restriction enzymes are the enzymes of bacterial origin which cleaves DNA into fragments at or near specific recognition sites within DNA molecules. This principle is used in biotechnology to cut and insert the desired gene (gene of interest) thereby generating an rDNA with desirable characters.

Question 25.
Is there any possibilities to transfer a suitable desirable gene to host plant without vector? Justify your answer.
Answer:
Yes, it is possible to transfer a suitable desired gene to a host plant using certain chemicals, microinjection method, electroporation or by biolistics.

Question 26.
How will you identify a vector?
Answer:

1. Vectors are able to replicate autonomously to produce multiple copies of them along with their DNA insert in the host cell.
2. It should be small in size and of low molecular weight, less than 10 Kb (kilo base pair) in size so that entry/transfer into host cell is easy.
3. Vector must contain an origin of replication so that it can independently replicate within the host.
4. It should contain a suitable marker such as antibiotic resistance, to permit its detection in transformed host cell.
5. Vector should have unique target sites for integration with DNA insert and should have the ability to integrate with DNA insert it carries into the genome of the host cell. Most of the commonly used cloning vectors have more than one restriction site. These are Multiple Cloning Site (MCS) or polylinker. Presence of MCS facilitates the use of restriction enzyme of choice.

Question 27.
Compare the various types of Blotting techniques.
Answer:

Question 28.
Write the advantages of herbicide tolerant crops.
Answer:
Advantages of Herbicide Tolerant Crops:

• Weed control improves higher crop yields;
• Reduces spray of herbicide;
• Reduces competition between crop plant and weed;
• Use of low toxicity compounds which do not remain active in the soil; and
• The ability to conserve soil structure and microbes.

Question 29.
Write the advantages and disadvantages of Bt cotton.
Answer:
The advantages of Bt cotton are:

1. Yield of cotton is increased due to effective control of bollworms.
2. Reduction in insecticide use in the cultivation of Bt cotton
3. Potential reduction in the cost of cultivation.
4. Cost of Bt cotton seed is high.
5. Effectiveness up to 120 days after that efficiency is reduced.
6. Ineffective against sucking pests like jassids, aphids and whitefly.
7. Affects pollinating insects and thus yield.

Question 30.
What is bioremediation? Give some examples of bioremediation.
Bioremediation:
It is defined as the use of microorganisms or plants to clean up environmental pollution. It is an approach used to treat wastes including wastewater, industrial waste and solid waste. Bioremediation process is applied to the removal of oil, petrochemical residues, pesticides or heavy metals from soil or ground water.

In many cases, bioremediation is less expensive and more sustainable than other physical and chemical methods of remediation. Bioremediation process is a cheaper and eco-friendly approach and can deal with lower concentrations of contaminants more effectively. The strategies for bioremediation in soil and water can be as follows:

1. Use of indigenous microbial population as indicator species for bioremediation process.
2. Bioremediation with the addition of adapted or designed microbial inoculants.
3. Use of plants for bioremediation – green technology.

Question 31.
Write the benefits and risk of Genetically Modified Foods.
Answer:
GM Food – Benefits:

1. High yield without pest.
2. 70% reduction of pesticide usage.
3. Reduce soil pollution problem.
4. Conserve microbial population in soil.

Risks – believed to:

1. Affect liver, kidney function and cancer.
2. Hormonal imbalance and physical disorder.
3. Anaphylactic shock (sudden hypersensitive reaction) and allergies.
4. Adverse effect in immune system because of bacterial protein.
5. Loss of viability of seeds show in terminator seed technology of GM crops.

Samacheer Kalvi 12th Bio Botany Principles and Processes of Biotechnology Additional Questions and Answers

Question 1.
Which of the following person coined the term biotechnology?
(a) Ernst Hoppe
(b) Stanley Cohen
(c) Ian Wilmet
(d) Karl Ereky
Answer:
(d) Karl Ereky

Question 2.
Zymology deals with
(a) Study of yeast fungus and its practical applications.
(b) Study of fermentation and its uses.
(c) Study of Bioreactors and their construction methodology.
(d) Study of zymase producing microbes and its benefits.
Answer:
(b) Study of fermentation and its uses.

Question 3.
Match column I with column II

Column I	Column II
A. One gene one enzyme hypothesis	i. Kohler and Milstein
B. Monoclonal antibodies	ii. Kary Mullis
C. First transgenic animal	iii. Beadle and Tatum
D. Development of PCR technology	iv. Ian Wilmet

(a) A – iii, B – i, C – iv, D – ii
(b) A – i, B – iv, C – ii, D – iii
(c) A – iv, B – iii, C – ii D – i
(d) A – ii, B – iv C – i, D – iii
Answer:
(a) A – iii, B -1, C – iv, D – ii

Question 4.
Identify the incorrect statement:
(a) French chemist Louis Pasteur demonstrated the fermentation.
(b) Fermentor is a vessel providing optimal condition for microbial action.
(c) Solvent extraction is an upstream process of fermentation.
(d) Distillation and filtration comes under down stream process.
Answer:
(c) Solvent extraction is an upstream process of fermentation.

Question 5.
Pick out the mismatched pair(s):
(i) Amphotericin-B – Streptomyces notatum
(ii) Penicillin – Penicillum nodosus
(iii) Streptomycin – Streptomyces grises
(iv) Tetracycline – Streptomyces aureofacins
(a) i and ii
(b) ii and iii
(c) iii and iv
(d) i only
Answer:
(a) i and ii

Question 6.
Identify the non-fungal species used in SCP production.
(i) Candida
(ii) Chlorella
(iii) Chlamydomonas
(iv) Cellulomonas
(a) i and ii
(b) ii and iii
(c) ii, iii and iv
(d) All the above
Answer:
(c) ii, iii and iv

Question 7.
Select the correct restriction enzyme which breaks the phosphodiester bond within a DNA
molecule.
(i) Bal 31
(ii) Hind II
(iii) Bam HI
(iv) Pvul
(a) i and iii
(b) i, ii and iii
(c) ii, iii and iv
(d) i only
Answer:
(c) ii, iii and iv

Question 8.
Cohesive ends are _______
(a) Blunt ends
(b) Flush ends
(c) Sticky ends
(d) Symmetric cuts
Answer:
(c) Sticky ends

Question 9.
Self-ligation is prevented by __________
(a) DNA Polymerase
(b) Helicase
(c) Alkaline phosphate
(d) DNA lipase
Answer:
(c) Alkaline phosphate

Question 10.
Observe the diagram and name A and B.

(a) A – Plasmid – B – Vector
(b) A – Nucleoid – B – Plasmid
(c) A – Bacterial chromosome B – Vector
(d) A – Nucleoid – B – x phage DNA
Answer:
(b) A – Nucleoid – B – Plasmid

Question 11.
A vector should __________
(i) contain suitable marker
(ii) contain ori site
(iii) have poly linkess
(iv) be small in size
(a) i, ii and iii
(b) ii, iii and iv
(c) i, ii and iv
(d) all the above
Answer:
(d) all the above

Question 12.
Number of base pairs does pBR 322 plasmid contains __________
(a) 322
(b) 4322
(c) 4361
(d) 3264
Answer:
(c) 4361

Question 13.
__________ is the plasmid present in Agrobacterium.
Answer:
Ti plasmid

Question 14.
ptlC 19 is an example for.
(a) Shuttle vector
(b) Expression vector
(c) Cosmid
(d) Phagemid vector
Answer:
(b) Expression vector

Question 15.
Statement 1: YAC plasmid behaves like a yeast chromosome.
Statement 2: Circular YAC multiplies in bacteria.
(a) Statement 1 is correct and Statement 2 is also correct.
(b) Statement 1 is correct and Statement 2 is incorrect.
(c) Both the statements are incorrect.
(d) Statement 1 is incorrect and Statement 2 is correct.
Answer:
(a) Statement 1 is correct and Statement 2 is also correct.

Question 16.
Statement 1: Liposomes are the artificial lipoprotein vesicles. Statement 2: Liposomes are highly used in gene transfer.
(a) Statement 1 is correct and Statement 2 is also correct.
(b) Statement 1 is correct and Statement 2 is incorrect.
(c) Both the statements are incorrect.
(d) Statement 1 is incorrect and Statement 2 is correct.
Answer:
(d) Statement 1 is incorrect and Statement 2 is correct.

Question 17.
Statement 1: DNA is a hydrophobic molecule.
Statement 2: T-DNA is a part of E-coli plasmid.
(a) Statement 1 is correct and Statement 2 is also correct.
(b) Statement 1 is correct and Statement 2 is incorrect.
(c) Both the statements are incorrect.
(d) Statement 1 is incorrect and Statement 2 is correct.
Answer:
(c) Both the statements are incorrect.

Question 18.
Statement 1: Bioventing procedure increases 02 flow to accelerate degradation of pollutants.
Statement 2: Bioaugmentation uses microbes to recover metal pollutants from contaminated sites.
(a) Statement 1 is correct and Statement 2 is also correct.
(b) Statement 1 is correct and Statement 2 is incorrect.
(c) Both the statements are incorrect.
(d) Statement 1 is incorrect and Statement 2 is correct.
Answer:
(b) Statement 1 is correct and Statement 2 is incorrect.

Question 19.
Assertion (A) : Golden rice helps to overcome childhood blindness.
Reason (R) : It is rich in P carotene.
(a) Both A and R are wrong.
(b) A is right R is wrong.
(c) R explains A.
(d) A and R are right, R does not explain A.
Answer:
(c) R explains A.

Question 20.
Assertion (A): Expression vectors are suitable for expressing foreign proteins.
Reason (R): pBR 322 is an expression vectors.
(а) Both A and R are wrong.
(b) A is right R is wrong.
(c) R explains A.
(d) A and R are right, R does not explain A.
Answer:
(b) A is right R is wrong.

Question 21.
Assertion (A) : Pseudomonas putida is utilized in the production of Biological hydrogen.
Reason (R): During photosynthesis, it releases oxygen.
(a) Both A and R are wrong.
(b) A is right R is wrong.
(c) R explains A.
(d) A and R are right, R does not explain A.
Answer:
(a) Both A and R are wrong.

Question 22.
Assertion (A): DMH -11 is a transgenic mustard.
Reason (R): It is developed by using bamase/ barstar technology.
(a) Both A and R are wrong.
(b) A is right R is wrong.
(c) R explains A.
(d) A and R are right, R does not explain A.
Answer:
(c) R explains A.

Question 23.
Green fluorescent protein (GFP) was isolated from
(a) Aequorea victoria
(b) Arabidopsis thaliana
(c) Agrobacterium tumifaciens
(d) Escherichia coli
Answer:
(a) Aequorea victoria

Question 24.
Tetracycline is obtained from
(a) S.nodosus
(b) S.aureofacins
(c) S.grises
(d) P. chryosogenum
Answer:
(a) S.aureofacins

Question 25.
Today more than restriction enzymes have been isolated.
(a) 800
(b) 900
(c) 1000
(d) 870
Answer:
(6) 900

2. Mark Questions

Question 1.
How modern biotechnology differs from conventional biotechnology?
Answer:
There are two main features of modem biotechnology, that differentiated it from the conventional technology are its:
(i) ability to change the genetic material for getting new products with specific requirement through recombinant DNA technology
(ii) ownership of the newly developed technology and its social impact.

Question 2.
What is a fermentor?
Answer:
Bioreactor (Fermentor) is a vessel or a container that is designed in such a way that it can provide an optimum environment in which microorganisms or their enzymes interact with a substrate to produce the required product. In the bioreactor, aeration, agitation, temperature and pH are controlled.

Question 3.
Define fermentation.
Answer:
Fermentation refers to the metabolic process in which organic molecules (normally glucose) are converted into acids, gases, or alcohol in the absence of oxygen or any electron transport chain.

Question 4.
What are primary metabolites? Give example.
Answer:
Metabolites produced for the maintenance of life process of microbes are known as primary metabolites.
E.g. Ethanol, citric acid, lactic acid and acetic acid.

Question 5.
How microbial enzymes are produced? Mention its significance.
Answer:
When microbes are cultured, they secrete some enzymes into the growth media. These enzymes are industrially used in detergents, food processing, brewing and pharmaceuticals.
E.g. Protease, amylase, isomerase, and lipase.

Question 6.
Mention any two bacterial species used as SCP.
Answer:

1. Cellulomonas
2. Alcaligenes

Question 7.
Name any two fungal species used as SCP.
Answer:

1. Agaricus campestris
2. Saccharomyces cerevisiae

Question 8.
Expand PCR and mention its use.
Answer:
PCR stands for Polymerase Chain Reaction.
PCR is a common lab technique used to make copies of particular region of DNA.

Question 9.
What is Restriction Endonuclease?
Answer:
A restriction enzyme or restriction endonuclease is an enzyme that cleaves DNA into fragments at or near specific recognition sites within the molecule known as restriction sites.

Question 10.
What is a palindrome sequence?
Answer:
Palindrome is a sequence of nucleotide in DNA strands at the site which reads the same in 5′-3′ direction and in the 3′-5′ direction.

Question 11.
Write an palindrome sequence of DNA.
Answer:
5′ … CATTATATAATG … 3′
3′ … GTAATATATTAC … 5′

Question 12.
Differentiate between flush end and cohesive end of DNA.
Answer:
Flush End:
Some restriction enzymes cut the strands of DNA through the centre resulting in blunt end or flush end or symmetric cuts.

Cohesive End:
Some restriction enzymes cut the strands in a way producing protruding and recessed ends known as cohesive end or sticky end or asymmetric cuts.

Question 13.
What is the role of DNA ligase in genetic engineering?
Answer:
DNA ligase enzyme joins the sugar and phosphate molecules of double stranded DNA (dsDNA) with 5’-P0₄ and a 3’-OH in an Adenosine Triphosphate (ATP) dependent reaction. This is isolated from T4 phage.

Question 14.
Define plasmids.
Plasmids are extrachromosomal, self replicating ds circular DNA molecules, found in the bacterial cells in addition to the bacterial chromosome. Plasmids contain Genetic information for their own replication.

Question 15.
Classify vectors and explain them.
Answer:
Vectors are of two types:

1. Cloning Vector
2. Expression Vector.

Cloning vector is used for the cloning of DNA insert inside the suitable host cell. Expression vector is used to express the DNA insert for producing specific protein inside the host.

Question 16.
What are expression vectors?
Answer:
Vectors which are suitable for expressing foreign proteins are called expression vectors. This vector consists of signals necessary for transcription and translation of proteins in the host. This helps the host to produce foreign protein in large amounts.
Example: pUC 19.

Question 17.
Name any four vectors that you know?
Answer:
Cosmid, plasmid, Bacteriophage and Phagemids.

Question 18.
Write a brief note on BAC vector.
Answer:
Bacterial Artificial Chromosome (BAC) Vector is a shuttle plasmid vector, created for cloning large-sized foreign DNA. BAC vector is one of the most useful cloning vector in r-DNA technology they can clone DNA inserts of upto 300 Kb and they are stable and more user-friendly.

Question 19.
What does Blotting refers to?
Answer:
Blotting refers to the process of immobilization of sample nucleic acids on solid support (nitrocellulose or nylon membranes). The blotted nucleic acids are then used as target in the hybridization experiments for their specific detection.

Question 20.
Point out any two disadvantages of Bt cotton?
Answer:
Bt cotton has some limitations:

• Cost of Bt cotton seed is high.
• Effectiveness up to 120 days after that efficiency is reduced.

Question 21.
What are the benefits of Genetically Modified plants?
Answer:
GM Food – Benefits:

1. High yield without pest.
2. 70% reduction of pesticide usage.
3. Reduce soil pollution problem.
4. Conserve microbial population in soil.

Question 22.
Expand:

1. PEG
2. PHB

Answer:

1. PEG – Poly Ethylene Glycol
2. PHB – Poly Hydroxy Butyrate

Question 23.
What is Biopharming?
Answer:
Biopharming also known as molecular pharming is the production and use of transgenic plants genetically engineered to produce pharmaceutical substances for use of human beings. This is also called “molecular farming or pharming”. These plants are different from medicinal plants which are naturally available.

Question 24.
Define the terms

1. Bioventing
2. Bioaugmentation

Answer:

1. Bioventing is the process that increases the oxygen or air flow to accelerate the degradation of environmental pollutants.
2. Bioaugmentation is the addition of selected microbes to speed up degradation process.

Question 25.
How hydrogen is biologically synthsized?
Answer:
The biological hydrogen production with algae is a method of photo biological water splitting. In normal photosynthesis the alga, Chlamydomonas reinhardtii releases oxygen. When it is deprived of sulfur, it switches to the production of hydrogen during photosynthesis and the electrons are transported to ferredoxins. [Fe]-hydrogenase enzymes combine them into the production of hydrogen gas.

Question 26.
Define Biopiracy.
Answer:
Biopiracy can be defined as the manipulation of intellectual property rights laws by corporations to gain exclusive control over national genetic resources, without giving adequate recognition or remuneration to the original possessors of those resources.

Question 27.
What are polylinkers?
Answer:
Mostly cloning vectors have more than one restriction sites. These are called as Multiple Cloning Site (MCS) or polylinkers. Presence of MCS facilitates the use of restriction enzyme of choice.

3-Mark Questions

Question 28.
Mention any three historical events which took place in the 21st century for the development of biotechnology.
Answer:
2002 – First crop plant genome sequenced in Oryza sativa.
2003 – Human genome project is completed, providing information on the locations and 1 sequence of human genes on all 46 chromosomes.
2016 – Stem cells injected into stroke patients re-enable patient to walk – Stem cell therapy.

Question 29.
In the fermentation process, what does upstream and downstream refers to? Explain.
Answer:
Upstream process:
All the process before starting of the fermenter such as sterilization of the fermenter, preparation and sterilization of culture medium and growth of the suitable inoculum are called upstream process.

Downstream process:
All the process after the fermentation process is known as the downstream process. This process includes distillation, centrifuging, filtration and solvent extraction. Mostly this process involves the purification of the desired product.

Question 30.
Provide a stepwise procedure of fermentation process.
Procedure of Fermentation
Answer:

1. Depending upon the type of product, bioreactor is selected.
2. A suitable substrate in liquid media is added at a specific temperature, pH and then diluted.
3. The organism (microbe, animal/plant cell, sub-cellular organelle or enzyme) is added to it.
4. Then it is incubated at a specific temperature for the specified time.
5. The incubation may either be aerobic or anaerobic.
6. Withdrawal of product using downstream processing methods.

Question 31.
What are secondary metabolites? Give two examples.
Answer:
Secondary metabolites are those which are not required for the vital life process of microbes, but have value added nature, this includes antibiotics
e.g. -Amphotericin-B (Streptomyces nodosus) and Penicillin (Penicillium chryosogenum).

Question 32.
What is SCP? Mention its nutritional value.
Answer:
Single Cell Protein (SCP) are dried cells of microorganisms that are used as protein supplement in human foods or animal feeds. SCP are rich in proteins, aminoacids, vitamins, carbohydrates, fats and minerals. It is used as food source by Astronauts and Antarctica expedition scientists.

Question 33.
Mention any three algal species used for SCP production.
Answer:
Spirulina, Chlorella and Chlamydomonas.

Question 34.
Though SCP is a rich protein source, it has not been widely used as food supplement. Point a reason to support this statement.
Answer:
Yes, although SCP is a rich protein source it is not widely used by most of the people in countries due to its higher nucleic acid content and slow digestibility.

Question 35.
Point out few advantages of single cell protein.
Answer:
Applications of Single-Cell Protein:

1. It is used as protein supplement.
2. It is used in cosmetics, products for healthy hair and skin.
3. It is used in poultry as the excellent source of proteins and other nutrients, it is widely used for feeding cattle, birds and fishes, etc.
4. It is used in food industry as aroma carriers, vitamin carrier, emulsifying agents to improve the nutritive value of baked products, in soups, in ready-to-serve-meals, in diet recipes.
5. It is used in industries like paper processing and leather processing as foam stabilizers.

Question 36.
Classify restriction enzymes based on their mode of action.
Answer:
Based on their mode of action restriction enzymes are classified into Exonucleases and Endonucleases.

a. Exonucleases are enzymes which remove nucleotides one at a time from the end of a DNA molecule,
e.g. Bal 31 and Exomiclease III.

b. Endonucleases are enzymes which break the internal phosphodiester bonds within a DNA molecule,
e.g. Hind II, EcoRI, Pvul, BamHI and TaqI.

Question 37.
Which type of restriction enzyme is widely used in rDNA technology? Why?
Answer:
Type II enzyme is preferred for use in recombinant DNA technology as they recognise and cut DNA within a specific sequence typically consisting of 4-8 bp.

Question 38.
Explain the procedure behind the naming of Restriction Enzymes by citing an example.
Answer:
Restriction endonucleases are named by a standard procedure. The first letter of the enzymes indicates the genus name, followed by the first two letters of the species, then comes the strain of the organism and finally a roman numeral indicating the order of discovery. For example, EcoRI is from Escherichia (E) coli (co), strain RY 13 (R) and first endonuclease (I) to be discovered.

Question 39.
Give a short note on Alkaline phosphate.
Answer:
Alkaline phosphate is a DNA modifying enzymes and adds or removes specific phosphate group at 5’ terminus of double stranded DNA (dsDNA) or single stranded DNA (ssDNA) or RNA. Thus it prevents self ligation. This enzyme is purified from bacteria and calf intestine.

Question 40.
What are the features that a vector must possess to facilitate cloning?
Answer:
The following are the features that are required to facilitate cloning into a vector.

1. Origin of replication (ori): This is a sequence from where replication starts and piece of DNA when linked to this sequence can be made to replicate within the host cells.
2. Selectable marker: In addition to ori the vector requires a selectable marker, which helps in identifying and eliminating non-transformants and selectively permitting the growth of the transformants,
3. Cloning sites: In order to link the alien DNA, the vector needs to have very few, preferably single, recognition sites for the commonly used restriction enzymes.

Question 41.
Draw and label Ti plasmid
Answer:

Question 42.
What do you mean by the term “walking genes”? Could you explain?
Answer:
Transposons (Transposable elements or mobile elements) are DNA sequence able to insert itself at a new location in the genome without having any sequence relationship with the target locus and hence transposons are called walking genes or jumping genes. They are used as genetic tools for analysis of gene and protein functions, that produce new phenotype on host cell. The use of transposons is well studied in Arabidopsis thaliana and bacteria such as Escherichia coli.

Question 43.
How does shuttle vectors differ from other types of vectors?
Answer:
The shuttle vectors are plasmids designed to replicate in cells of two different species. These vectors are created by recombinant techniques. The shuttle vectors can propagate in one host and then move into another host without any extra manipulation. Most of the Eukaryotic
vectors are Shuttle Vectors.

Question 44.
Given below are the three different DNA palindrome sequences. Name the respective restriction enzymes which cleaves those sequences and also mention the microbial sources of the enzymes.
Answer:
(a) 5 AGCT3′
(b) 5 GGCC3′
(c) 5 GAATTC3′
Answer:

Question 45.
Why is it difficult for DNA to pass through cell membrane? How the bacterial cells can be made competent to take up DNA?
Answer:
Since the DNA is a hydrophilic molecule, it cannot pass through cell membranes, In order to force bacteria to take up the plasmid, the bacterial cells must first be made competent to take up DNA. This is done by treating them with a specific concentration of a divalent cation such as calcium.

Question 46.
Write a brief note on Biolistics.
Answer:
The foreign DNA is coated onto the surface of minute gold or tungsten particles (1-3 pm) and bombarded onto the target tissue Or cells using a particle gun (also called as gene gun/micro projectile gun/shotgun). Then the bombarded cells or tissues are cultured on selected medium to regenerate plants from the transformed cells.

Question 47.
Agrobacterium – a natural genetic engineer of plants. Justify the statement.
Answer:
Among the various vectors used for plant transformation, the Ti-plasmid from Agrobacterium tumefaciens has been used extensively. This bacterium has a large size plasmid, known as Ti plasmid (Tumor inducing) and a portion of it referred as T-DNA (transfer DNA) is transferred to plant genome in the infected cells and cause plant tumors (crown gall). Since this bacterium has the natural ability to transfer T-DNA region of its plasmid into plant genome, upon infection of cells at the wound site, it is also known as the natural genetic engineer of plants.

Question 48.
Give a brief account on antibiotic resistant marker.
Answer:
An antibiotic resistance marker is a gene that produces a protein that provides cells with resistance to an antibiotic. Bacteria with transformed DNA can be identified by growing on a medium containing an antibiotic. Recombinants will grow on these medium as they contain genes encoding resistance to antibiotics such as amphicillin, chloroamphenicol, tetracycline or kanamycin, etc., while others may not be able to grow in these media, hence it is considered useful selectable marker.

Question 49.
Mention the types of Blotting techniques.
Answer:
Types of Blotting Techniques:

1. Southern Blotting: The transfer of DNA from agarose gels to nitrocellulose membrane.
2. Northern Blotting: The transfer of RNA to nitrocellulose membrane.
3. Western Blotting: Electrophoretic transfer of proteins to nitrocellulose membrane:

Question 50.
Expand CRISPR – Cas9.
Answer:
Clustered Regularly Interspaced Short Palindromic Repeats and CRISPR-associated protein 9.

Question 51.
What is RNA interference (RNAi)? How it is carried out?
Answer:
RNA interference is a biological process in which RNA molecules inhibit gene expression or translation. This is done by neutralising targeted mRNA molecules.

A simplified model for the RNAi pathway is based on two steps, each involving ribonuclease enzyme. In the first step, the trigger RNA (either dsRNA or miRNA primary transcript) is processed into a short interfering RNA (siRNA) by the RNase II enzymes called Dicer and Drosha. In the second step, siRNAs are loaded into the effector complex RNA-induced silencing complex (RISC). The siRNA is unwound during RISC assembly and the single- stranded RNA hybridizes with mRNA target. This RNAi is seen in plant feeding nematodes.

Question 52.
Prepare a protocol for Glyphosphate resistant potato plant development.
Answer:

Question 53.
How Bt cotton crops are developed?
Answer:
Bt cotton is a genetically modified organism (GMO) or genetically modified pest resistant plant cotton variety, which produces an insecticide activity to bollworm. Strains of the bacterium Bacillus thuringiensis produce over 200 different Bt toxins, each harmful to different insects. Most Bt toxins are insecticidal to the larvae of moths and butterflies, beetles, cotton bollworms and gatflies but are harmless to other forms of life.The genes are encoded for toxic crystals in the Cry group of endotoxin.

When insects attack and eat the cotton plant the Cry toxins are dissolved in the insect’s stomach.The epithelial membranes of the gut block certain vital nutrients thereby sufficient regulation of potassium ions are lost in the insects and results in the death of epithelial cells in the intestine .membrane which leads to the death of the larvae.

Question 54.
Comment on Golden rice.
Answer:
Golden rice is a variety of Oryza sativa (rice) produced through genetic engineering of biosynthesized beta-carotene, a precursor ofVitamin-A in the edible parts of rice developed by – Ingo Potrykus and his group. The aim is to produce a fortified food to be grown and consumed in areas with a shortage of dietary Vitamin-A, which kills so many children under five year age.

Golden rice differs from its parental strain by the addition of three beta-carotene biosynthesis , genes namely ‘psy’ (phytoene synthase) from daffodil plant Narcissus pseudonarcissus and ‘crt-1’ gene from the soil bacterium Erwinia auredorora and ‘lyc’ (lycopene cyclase) gene from wild-type rice endosperm. The endosperm of normal rice, does not contain beta-carotene. Golden-rice has been genetically altered so that the endosperm now accumulates Beta-carotene. This has been done using Recombinant DNA technology. Golden rice can control childhood blindness – Xerophthalmia.

Question 55.
Name any 3 bacterial species used to generate polyhydroxybutyrates (PHB).
Answer:
Bacillus megaterium.
Corynebacterium glutamicum.
Alcaligenes eutrophus.

Question 56.
Write short note on Green fluorescent protein.
Answer:
The green fluorescent protein (GFP) is a protein containing 238 amino acid residues of 26.9 kDa that exhibits bright green fluorescence when exposed to blue to ultraviolet range (395 nm). GFP refers to the protein first isolated from the jellyfish Aequorea victoria. GFP is an excellent tool in biology due to its ability to form internal chromophore without requiring any accessory cofactors, gene products, enzymes or substrates other than molecular oxygen. In cell and molecular biology, the GFP gene is frequently used as a reporter of expression. It has been used in modified forms to make biosensors.

Question 57.
How turmeric biopiracy is prevented by Indian Government?
Answer:
The United States Patent and Trademark Office, in the year 1995 granted patent to the method of use of turmeric as an antiseptic agent. Turmeric has been used by the Indians as a home remedy for the quick healing of the wounds and also for purpose of healing rashes. The journal article published by the Indian Medical Association, in the year 1953 wherein this remedy was mentioned.

Therefore, in this way it was proved that the use of turmeric as an antiseptic is not new to the world and is not a new invention, but formed a part of the traditional knowledge of the Indians. The objection in this case US patent and trademark office was upheld and traditional knowledge of the Indians was protected.

5 – Mark Questions

Question 58.
Describe the steps involved in recombinant DNA technology.
Answer:
The steps involved in recombinant DNA technology are:

1. Isolation of a DNA fragment containing a gene of interest that needs to be cloned. This is called an insert.
2. Generation of recombinant DNA (rDNA) molecule by insertion of the DNA fragment into a carrier molecule called a vector that can self-replicate within the host cell.
3. Selection of the transformed host cells that is carrying the rDNA and allowing them to multiply thereby multiplying the rDNA molecule. The
4. entire process thus generates either a large amount of rDNA or a large amount of protein expressed by the insert.
5. Wherever vectors are not involved the desired gene is multiplied by PCR technique. The multiple copies are. injected into the host cell protoplast or it is shot into the host cell protoplast by shot gun method.

Question 59.
Explain in detail about various ty pes of direct gene transfer method,
Answer:
a. Chemical mediated gene transfer: Certain chemicals like polyethylene glycol (PEG) and dextran sulphate induce DNA uptake into plant protoplasts.

b. Microinjection: The DNA is directly injected into the Electric field induces a voltage across cell membrane nucleus using fine tipped glass needle or micro pipette Electroporation Methods of Gene Transfer to transform plant cells. The protoplasts are immobilised on a solid support (agarose on a microscopic slide) or held with a holding pipette under suction.

c. Electroporation Methods of Gene Transfer: Apulse of high voltage is applied to protoplasts, cells or tissues which makes transient pores in the plasma membrane through which uptake of foreign DNA occurs.

d. Liposome mediated method of Gene Transfer: Liposomes the artificial phospholipid vesicles are useful in gene transfer. The gene or DNA is transferred from liposome into vacuole of plant cells. It is carried out by encapsulated DNA into the vacuole. This technique is advantageous because the liposome protects the introduced DNA from being damaged by the acidic pH and protease enzymes present in the vacuole. Liposome and tonoplast of vacuole fusion resulted in gene transfer. This process is called lipofection.

e. Biolistics: The foreign DNA is coated onto the surface of minute gold or tungsten particles (1-3 pm) and bombarded onto the target tissue or cells using a particle gun (also called as gene gun/micro projectile gun/shotgun). Then the bombarded cells or tissues are cultured on selected medium to regenerate plants from the transformed cells.

Question 60.
Describe the procedure of Blue-White colony selection methods.
Answer:
Blue- White Colony Selection Method is a powerful method used for screening of recombinant plasmid. In this method, a reporter gene lacZ is inserted in the vector. The lacZ encodes the enzyme P-galactosidase and contains several recognition sites for restriction enzyme.P-galactosidase breaks a synthetic substrates called X-gal (5-bromo-4-chloroindolyl- P-D- galacto-pyranoside) into an insoluble blue coloured product. If a foreign gene is inserted into lacZ, this gene will be inactivated.

Therefore, no-blue colour will develop (white) because P-galactosidase is not synthesized due to inactivation of lacZ. Therefore, the host cell containing r-DNA form white coloured colonies on the medium contain X-gal, whereas the other cells containing non-recombinant DNA will develop the blue coloured colonies. On the basis of colony colour, the recombinants can be selected.

Question 61.
Write a note on Replica plating technique.
Answer:

A technique in which the pattern of colonies growing on a culture plate is copied. A sterile filter plate is pressed against the culture plate and then lifted. Then the filter is pressed against a second sterile culture plate. This results in the new plate being infected with cell in the same
– relative positions as the colonies in the original plate. Usually, the medium used in the second plate will differ from that used in the first. It may include an antibiotic or without a growth factor. In this way, transformed cells can be selected. Replica plating technique

Question 62.
How Agarose Gel Electrophoresis is performed?
Answer:
1. Agarose Gel Electrophoresis is used mainly for the purification of specific DNA fragments. Agarose is convenient for separating DNA fragments ranging in size from a few hundred to about 20000 base pairs. Polyacrylamide is preferred for the purification of smaller DNA fragments.

2. The gel is complex network of polymeric molecules. DNA molecule is negatively charged molecule – under an electric field DNA molecule migrates through the gel. The electrophoresis is frequently performed with marker DNA fragments of known size which allow accurate size

3. determination of an unknown DNA molecule by interpolation. The advantages of agarose gel electrophoresis are that the DNA bands can be readily detected at high sensitivity. The bands of DNA in the gel are stained with the dye Ethidium Bromide and DNA can be detected

4. as visible fluorescence illuminated in UV light will give orange fluorescence, which can be photographed.

Question 63.
Explain the procedure of Southern Blotting Technique. Southern Blotting Techniques – DNA
Answer:
The transfer of denatured DNA from Agarose gel to Nitrocellulose Blotting or Filter Paper technique was introduced by Southern in 1975 and this technique is called Southern Blotting Technique.

Steps:
The transfer of DNA from agarose gel to nitrocellulose filter paper is achieved by Capillary Action.
A buffer Sodium Saline Citrate (SSC) is used, in which DNA is highly soluble, it can be drawn up through the gel into the Nitrocellulose membrane. By this process ss-DNA becomes ‘Trapped’ in the membrane matrix.

This DNA is hybridized with a nucleic acid and can be detected by autoradiography. Autoradiography – A technique that captures the image formed in a photographic emulsion due to emission of light or radioactivity from a labelled component placed together with unexposed
film.

Higher Order Thinking Skills (HOTs) Questions 

Question 1.
Give the technical terminologies for the following statements.
(а) Autonomous, self-replicating, circular DNA
(b) Molecular scissors
(c) Symmetrical repreated sequence in DNA strands
(d) Mobile genetic elements
Answer:
(a) Plasmid
(b) Restriction Enzymes
(c) Palindrome sequence
(d) Transposons

Question 2.
Observe the given flow chart and complete it.

Answer:
A. Plasmid
B. rDNA/chimeric DNA

Question 3.
Name the products of the following combinations.
(a) Bacterial plasmid + cos – site = ______
(b) Bacterial plasmid + phage DNA = ______
Answer:
(a) Cosmid
(b) Phagemid

Question 4.
Golden rice is a bio-fortified rice developed by rDNA technology. It differes from its parental strain by possessing ‘psy’ gene, ‘crt-1’ gene and ‘lye’ gene which are responsible for beta – carotene synthesis.
(a) Name the sources of the above mentioned genes.
(b) Which disease can be controlled / prevented if a person’s diet has golden rice?
Answer:
(a) ‘psy’ gene is obtained from Daffodil plant (Narcissus pseudonarcissus).
(b) ‘crt-1’ gene is from Erwinia auredorora bacterium.
(c) Tyc gene is from wild-type rice endosperm.
(b) Golden rice can control Xerophthalmia.

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Samacheer Kalvi 12th Bio Botany Plant Breeding Text Book Back Questions and Answers

Question 1.
Assertion: Genetic variation provides the raw material for selection.
Reason: Genetic variations are differences in genotypes of the individuals.
(a) Assertion is right and reason is wrong.
(b) Assertion is wrong and reason is right.
(c) Both reason and assertion is right.
(d) Both reason and assertion is wrong.
Answer:
(c) Both reason and assertion is right.

Question 2.
While studying the history of domestication of various cultivated plants recognized earlier.
(a) Centres of origin
(b) Centres of domestication
(c) Centres of hybrid
(d) Centres of variation
Answer:
(a) Centres of origin

Question 3.
Pick out the odd pair.
(a) Mass selection – Morphological characters
(b) Purline selection – Repeated self pollination
(c) Clonal selection – Sexually propagated
(d) Natural selection – Involves nature
Answer:
(a) Mass selection – Morphological characters

Question 4.

Answer:
(b) i – III, ii – I, iii – IV, iv – II

Question 5.
The quickest method of plant breeding is __________
(a) Introduction
(b) Selection
(c) Hybridization
(d) Mutation breeding
Answer:
(d) Mutation breeding

Question 6.
Desired improved variety of economically useful crops are raised by __________
(a) natural selection
(b) hybridization
(c) mutation
(d) biofertilisers
Answer:
(b) hybridization

Question 7.
Plants having similar genotypes produced by plant breeding are called __________
(a) clone
(b) haploid
(c) autopolyploid
(d) genome
Answer:
(a) clone

Question 8.
Importing better varieties and plants from outside and acclimatising them to local environment is called __________
(a) cloning
(b) heterosis
(c) selection
(d) introduction
Answer:
(d) introduction

Question 9.
Dwarfing gene of wheat is __________
(a) pall
(b) Atomita 1
(c) Norin 10
(d) pelita 2
Answer:
(c) Norin 10

Question 10.
Crosses between the plants of the same variety are called __________
(a) interspecific
(b) intervarietal
(c) intravarietal
(d) intergeneric
Answer:
(c) intravarietal

Question 11.
Progeny obtained as a result of repeat self pollination a cross pollinated crop to called __________
(a) pure line
(b) pedigree line
(c) inbreed line
(d) heterosis
Answer:
(a) pure line

Question 12.
Jaya and Ratna are the semi dwarf varieties of __________
(a) wheat
(b) rice
(c) cowpea
(d) mustard
Answer:
(b) rice

Question 13.
Which one of the following are the species that are crossed to give sugarcane varieties with high sugar, high yield, thick stems and ability to grow in the sugarcane belt of North India?
(a) Saccharum robustum and Saccharum officinarum
(b) Saccharum barberi and Saccharum officinarum
(c) Saccharum sinense and Saccharum officinarum
(d) Saccharum barberi and Saccharum robustum
Answer:
(b) Saccharum barberi and Saccharum officinarum

Question 14.
Match column I (crop) with column II (Corresponding disease resistant variety) and select the correct option from the given codes.

Answer:
(b) ii, i, iii, iv

Question 15.
A wheat variety, Atlas 66 which has been used as a donor for improving cultivated wheat, which is rich in __________
(a) iron
(b) carbohydrates
(c) proteins
(d) vitamins
Answer:
(c) proteins

Question 16.
Which one of the following crop varieties correct matches with its resistance to a disease?

Answer:
(a) Pusa Komal – Bacterial blight

Question 17.
Which of the following is incorrectly paired?
(a) Wheat – Himgiri
(b) Milch breed – Sahiwal
(c) Rice – Ratna
(d) Pusa Komal – Brassica
Answer:
(d) Pusa Komal – Brassica

Question 18.
Match list I with list II:

Answer:
(b) i-d, ii-c, iii-a, iv-b

Question 19.
Differentiate primary introduction from secondary introduction
Answer:

1. Primary Introduction: Primary introduction – When the introduced variety is well adapted to the new environment without any alternation to the original genotype.
2. Secondary Introduction: Secondary introduction – When the introduced variety is subjected to selection to isolate a superior variety and hybridized with a local variety to transfer one or a few characters to them.

Question 20.
How are microbial innocnlants used to increase the soil fertility?
Answer:
Biofertilizers or microbial innoculants are defined as preparations containing living cells or latent cells of efficient strains of microorganisms that help crop plants uptake of nutrients by their interactions in the rhizosphere when applied through seed or soil.

They are efficient in fixing nitrogen, solubilising phosphate and decomposing cellulose. They are designed to improve the soil fertility, plant growth, and also the number and biological activity of beneficial microorganisms in the soil. They are ecofriendly organic agro inputs and are more efficient and cost effective than chemical fertilizers.

Question 21.
What are the different types of hybridization?
Answer:
Types of Hybridization
According to the relationship between plants, the hybridization is divided into.

1. Intravarietal hybridization – The cross between the plants of same variety. Such crosses are useful only in the self-pollinated crops.
2. Intervarietal hybridization – The cross between the plants belonging to two different varieties of the same species and is also known as intraspecific hybridization. This technique has been the basis of improving self-pollinated as well as cross pollinated crops.
3. Interspecific hybridization – The cross between the plants belonging to different species belonging to the same genus is also called intragenic hybridization. It is commonly used for transferring the genes of disease, insect, pest and drought resistance from one species to another.

Question 22.
Explain the best suited type followed by plant breeders at present?
Answer:
Mutation breeding represents a new method of conventional breeding procedures as they have the advantage of improving the defect without losing agronomic and quality character in agriculture and crop improvement. Mutation means the sudden heritable changes in the genotype or phenotype of an organism. Gene mutations are of considerable importance in plant breeding as they provide essential inputs for evolution as well as for re-combination and selection. It is the only method for improving seedless crops.

Question 23.
Write a note on heterosis.
Answer:
The superiority of the F₁ hybrid in performance over its parents is called heterosis or hybrid vigour. Vigour refers to increase in growth, yield, greater adaptability of resistance to diseases, pest and drought. Vegetative propagation is the best suited measure for maintaining hybrid vigour, since the desired characters are not lost and can persist over a period of time. Many breeders believe that its magnitude of heterosis is directly related to the degree of genetic diversity between the two parents. Depending on the nature, origin, adaptability and reproducing ability heterosis can be classified as:

1. Euheterosis- This is the true heterosis which is inherited and is further classified as
2. Mutational Euheterosis – Simplest type of euheterosis and results from the sheltering or eliminating of the deleterious, unfavourable often lethal, recessive, mutant genes by their adaptively superior dominant alleles in cross pollinated crops.
3. Balanced Euheterosis – Well balanced gene combinations which is more adaptive to environmental conditions and agricultural usefulness.
4. Pseudoheterosis – Also termed as luxuriance. Progeny possess superiority over parents in vegetative growth but not in yield and adaptation, usually sterile or poorly fertile.

Question 24.
List out the new breeding techniques involved in developing new traits in plant breeding.
Answer:
New Breeding Techniques (NBT) are a collection of methods that could increase and accelerate the development of new traits in plant breeding. These techniques often involve genome editing, to modify DNA at specific locations within the plants to produce new traits in crop plants. The various methods of achieving these changes in traits include the following.

• Cutting and modifying the genome during the repair process by tools like CRISPR /Cas.
• Genome editing to introduce changes in few base pairs using a technique called Oligonucleotide-directed mutagenesis (ODM).
• Transferring a gene from an identical or closely related species (cisgenesis).
• Organising processes that alter gene activity without altering the DNA itself (epigenetic methods)

Samacheer Kalvi 12th Bio Botany Plant Breeding Additional Questions and Answers

1 – Mark Questions

Question 1.
____________ is the process of bringing a plant species under human control.
(a) Emasculation
(b) Hybridization
(c) Domestication
(d) Acclimatization
Answer:
(c) Domestication

Question 2.
Which of the following scientist developed world’s first cotton hybrid?
(a) Dr. B.P. Pal
(b) C.T. Patel
(c) Dr. K. Ramiah
(d) N.G.P. Rao
Answer:
(b) C.T. Patel

Question 3.
Identify the incorrect statement:
(a) Bio-inoculants are efficient in solubilising phosphate
(A) Bio-inoculants are ecofriendly organic agro outputs
(c) Bio-inoculants are obtained from dead organic matters
(d) Bio-inoculants are designed to improve soil fertility
Answer:
(c) Bio-inoculants are obtained from dead organic matters

Question 4.
Which is not a free-living nitrogen fixing species?
(a) Azotobacter
(b) Clostridium
(c) Nostop
(d) Anabaena
Answer:
(d) Anabaena

Question 5.
Arbuscular mycorrhizae is a symbiotic association between ________
(a) Algae and fungi
(b) Angiosperm roots and fungi
(c) Blue green algae and Azolla fern
(d) Cyanobacteria and corolloid root
Answer:
(b) Angiosperm roots and fungi

Question 6.
Azolla is best suited biofertilizer for ____________
(a) Sugar cane cultivation
(b) Paddy cultivation
(c) Wheat cultivation
(d) Cotton cultivation
Answer:
(A) Paddy cultivation

Question 7.
Assertion (A): SLF promotes vigorous growth and provide resistance against diseases.
Reason (R): SLF is made from kelp containing more than 70 minerals.
(a) Both A and R are true. R explains A.
(A) A is true R is false
(c) A is false R is true
(d) Both A and R are false
Answer:
(a) Both A and R are true. R explains A.

Question 8.
Assertion (A): Pure line varieties show homozygosity.
Reason (R): Pure line species are obtained through cross pollination.
(a) Both A and R are true. R explains A.
(b) A is true R is false
(c) A is false R is true
(d) Both A and R are false
Answer:
(b) A is true R is false

Question 9.
Assertion (A): Hybrids show increased growth and elevated yield.
Reason (R): F₁ hybrids show Heterosis.
(a) Both A and R are true. R explains A.
(b) A is true R is false
(c) A is false R is true
(d) Both A and R are false
Answer:
(a) Both A and R are true. R explains A.

Question 10.
Statement (1): Trichoderma species is a free-living bacteria.
Statement (2): It acts as a potent bio-control agent
(a) Statement 1 is correct and Statement 2 is incorrect
(b) Statement 1 is incorrect and Statement 2 is correct
(c) Both the statements are correct
(d) Both the statements are incorrect
Answer:
(b) Statement 1 is incorrect and Statement 2 is correct

Question 11.
Statement (1): Clonal selection is carried out in asexually propagating plants.
Statement (2): Clones show similar genotypes.
(a) Statement 1 is correct and Statement 2 is incorrect
(b) Statement 1 is incorrect and Statement 2 is correct
(c) Both the statements are correct
(d) Both the statements are incorrect
Answer:
(c) Both statements are correct

Question 12.
Removal of anthers from a flower to overcome self-pollination and the phenomenon is.
Answer:
Emasculation

Question 13.
Identify the proper sequence of hybridisation technique.
(a) Emasculation → Selection → Bagging → Crossing → Harvesting
(b) Harvesting → Selection → Crossing → Emasculation → Bagging
(c) Selection → Harvesting → Crossing → Emasculation → Bagging
(d) Selection → Emasculation → Bagging → Crossing → Harvesting
Answer:
(d) Selection → Emasculation → Bagging → Crossing → Harvesting

Question 14.
Intra specific hybridization is also termed as
(a) Intravarietal hybridization (b) Intervarietal hybridization
(c) Interspecific hybridization (d) Intergeneric hybridization
Answer:
(b) Intervarietal hybridization

Question 15.
The period of opening of a flower is ________
Answer:
Anthesis

Question 16.
Superiority of hybrids over parents only in vegetative growth not in yield. This phenomenon is termed as ________
(a) Euheterosis
(b) Balanced euheterosis
(c) Luxuriance
(d) Mutational heterosis
Answer:
(c) Luxuriance

Question 17.
The term green revolution was coined by ________
(a) William S Gaud
(b) M.S. Swaminathan
(c) Dr. B.P. Pal
(d) Dr. N.E. Borlaug
Answer:
(a) William S Gaud

Question 18.
Which is the second gamma garden in India?
Answer:
Indian Agricultural Research Institute (IARI)

Question 19.
Match the following:

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

Question 20.
Who is popularly called as the “father of green revolution in India”?
(a) Nel Jeyaraman
(b) Dr. M.S. Swaminathan
(c) Dr. Nammalvar
(d) N.G.P. Rao
Answer:
(a) Dr. M. S. Swaminathan

Question 21.
Pusa swamim variety of Brassica species show resistance to ________
(a) White rust
(b) Leaf curl
(c) Black rot
(d) Hill bunt
Answer:
(a) White rust

Question 22.
The first established Atomic Garden in India was ________
(a) Bhabha Atomic Research Institute
(b) Indira Gandhi Centre for Atomic Research
(c) Indian Agricultural Research Institute
(d) Bose Research Institute
Answer:
(d) Bose Research Institute

Question 23.
Triticale is polyploid breed of ________
(a) Triticum cereale x Secale sativus
(b) Triticum durum x Secale cereale
(c) Triticum cereale x Secale sativus
(d) Triticum sativus x Secale cereale
Answer:
(b) Triticum durum x Secale cereale

Question 24.
Raphanobrassica is an example for ________
(a) Autopolyploid
(b) Allopolyploid
(c) Polyploid
(d) Polysomy
Answer:
(b) Allopolyploid

Question 25.
Atlas 66 is a improved variety of ________
(a) Rice
(b) Maize
(c) Wheat
(d) Spinach
Answer:
(c) Wheat

Question 26.
Pusa Sawani variety of okra is resistant against ________
(a) Aphids
(b) Fruit borers
(c) Shoot and fruit borers
(d) Jassids
Answer:
(c) Shoot and fruit borers

Question 27.
________ was the first person to develop world’s first hybrid of sorghum.
Answer:
N.G.P. Rao

Question 28.
Identify the incorrect statement:
(i) SLF is obtained from Kelps – a brown algae
(ii) Azolla is a fern
(iii) Rhizobium is found in association with root nodules
(iv) AM forms symbiotic relation with angiospermic roots
(a) i only
(b) ii, iii and iv only
(c) none of the above
(d) all the above
Answer:
(c) none of the above

Question 29.
Damping off of tomato is controlled by ________
(a) Beauveria species
(b) Trichoderma species
(c) Acacia species
(d) Pseudomonas species
Answer:
(a) Beauveria species

Question 30.
Match the following:

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

Question 31.
Identify the correct symbiotic association
(a) Rhizobium × Corolloid roots
(b) Arbuscular Mycorrhizae × Angiospermic roots
(c) Azolla × Amantia
(d) Azospirillum × Azolla
(b) Arbuscular Mycorrhizae × Angiospermic roots
Answer:
(b) Arbuscular Mycorrhizae × Angiospermic roots

Question 32.
Atomita 2 – rice is a product by.
(a) Polyploid breeding
(b) Hybridization
(c) Mutation breeding
(d) Clonal selection
Answer:
(c) Mutation breeding

Question 33.
Luxuriance is the term used on par with ________
(a) Heterosis
(b) Anthesis
(c) Hybrids
(d) Mutant breeds
Answer:
(a) Heterosis

Question 34.
The term pureline was coined by.
Answer:
Johannsen

2 – Mark Questions

Question 1.
What is meant by domestication of plants?
Answer:
Domestication is the process of bringing a plant species under the control of humans and gradually changing it through careful selection,, genetic alteration and handling so that it is more useful to people.

Question 2.
Mention any two free – living nitrogen fixing bacteria.
Answer:
Azotobacter and Clostridium.

Question 3.
What are bio-pesticides? Why they are considered better than synthetic pesticides?
Answer:
Bio-pesticides are biologically based agents used for the control of plant pests. They are in high use due to their non-toxic, cheaper and eco-friendly characteristics as compared to chemical or synthetic pesticides.

Question 4.
Name any four plants used in Green leaf manuring.
Answer:

1. Cassia fistula
2. Sesbania grandiflora
3. Azadirachta indica
4. Pongamia pinnata

Question 5.
What is plant breeding?
Answer:
Plant breeding is the science of improvement of crop varieties with higher yield, better quality, resistance to diseases and shorter durations which are suitable to particular environment.

Question 6.
Define acclimatization.
Answer:
The newly introduced plant has to adapt itself to the new environment. This adjustment or adaptation of the introduced plant in the changed environment is called acclimatization.

Question 7.
Who coined the term “pureline”? Define it.
Answer:
Johannsen in 1903 coined the word pureline. It is a collection of plants obtained as a result of repeated self-pollination from a single homozygous individual.

Question 8.
Define the following terms:
Answer:
(a) Emasculation
(b) Bagging
Emasculation: It is a process of removal of anthers to prevent self pollination before anthesis (period of opening of a flower).
Bagging: The stigma of the flower is protected against any undesirable pollen grains, by covering it with a bag.

Question 9.
What is Heterosis?
Answer:
The superiority of the F₁ hybrid in performance over its parents is called heterosis or hybrid vigour. Vigour refers to increase in growth, yield, greater adaptability of resistance to diseases, pest and drought.

Question 10.
What does the term ‘luxuriance’ stands for in plant breeding? Explain.
Answer:
Pseudoheterosis – Also termed as luxuriance. Progeny possess superiority over parents in vegetative growth but not in yield and adaptation, usually sterile or poorly fertile.

Question 11.
Mutagens are the substances that induces mutation. Name any two physical and chemical mutagens.
Answer:

1. UV short waves, X-rays – Physical mutagens.
2. Nitromethyl, Urea – Chemical mutagens.

Question 12.
Write in brief about Atomic Garden.
Answer:
Atomic Garden: Is a form of mutation breeding where plants are exposed to radioactive sources typically cobalt-60 or caesium-137 in order to generate desirable mutation in crop plants.

Question 13.
State any one advantage and one disadvantage of polyploid breeding.
Answer:

1. Advantage: Polyploidy often exhibit hybrid vigour and increased tolerance to biotic and abiotic stresses.
2. Disadvantage: Polyploidy results in reduced fertility due to meiotic error resulting in seedless varieties.

Question 14.
What are polyploids? Mention its nature.
Answer:
Majority of flowering plants are diploid (2n). The plants which possess more than two sets of chromosome are called polyploids. Polyploidy often exhibit increased hybrid vigour, increased heterozygosity, increase the tolerance to both biotic and abiotic stresses and buffering of deleterious mutations.

Question 15.
Name any two allopolyploid plant species.
Answer:

1. Triticale (Triticum durum x Secale cereale).
2. Raphanobrassica (Brassica oleraceae x Raphanus sativus).

Question 16.
What is Biofortification?
Answer:
Biofortification is breeding crops with higher levels of vitamins and minerals or higher protein and healthier fats. It is the most practical means to improve public health.

Question 17.
Name the insect resistant varieties developed in the following crops.

1. Okra
2. Rapeseed mustard.

Answer:

1. Okra: Pusa Sawani and Pusa A-4 varieties of Okra are resistant to shoot and fruit borers.
2. Rapeseed mustard: Pusa gaurav variety of rapeseed mustard shown

Question 18.
In which plant, and by whom the first natural hybridization was performed?
Answer:
The first natural hybridization was done by Cotton Mather in maize.

3 – Mark Questions

Question 19.
In 1926, Vavilov initially proposed eight main geographic centres of crop origin. Mention any six of them.
Answer:
China, India, South America, Mediterranean, Mesoamerica and South East Asia.

Question 20.
Name any three eminent plant breeders of Indian origin.
Answer:

1. Dr. M.S. Swaminathan
2. C.T. Patel
3. Dr. B.P. Pal

Question 21.
How Rhizobium acts as a efficient bio-fertilizer?
Answer:
Bio-fertilizers containing rhizome bacteria are called rhizome bio-fertilizer culture. Symbiotic bacteria that reside inside the root nodules convert the atmospheric nitrogen into a bio available form to the plants. This nitrogen fixing bacterium when applied to the soil undergoes multiplication in billions and fixes the atmospheric nitrogen in the soil. Rhizobium is best suited for the paddy fields which increase the yield by 15 – 40%.

Question 22.
Azolla increases the yield of paddy crop – support your answer.
Answer:
Azolla is a free-floating water fern that fixes the atmospheric nitrogen in association with nitrogen fixing blue green alga Anabaena azolla. It is used as a bio-fertilizer for wetland rice cultivation and is known to contribute 40 – 60 kg/ha/crop. The agronomic potential of Azolla is quite significant particularly for increasing the yield of rice crop, as it quickly decompose in soil.

Question 23.
Mention any three advantages of Arbuscular mycorrhizal association.
Answer:

1. Arbuscular mycorrhizae have the ability to dissolve the phosphate found in soil.
2. Provides resistance against diseases, germs and unfavourable weather conditions.
3. It assures water availability.

Question 24.
What makes the Trichoderma an effective bio-pesticide?
Answer:
Trichoderma species are free-living fungi that are common in soil and root ecosystem. They have been recognized as bio-control agent for

1. the control of plant disease
2. ability to enhance root growth development
3. crop productivity
4. resistance to abiotic stress and
5. uptake and use of nutrients.

Question 25.
Write a note on Green manuring.
Answer:
Green manuring is defined as the growing of green manure crops and use of these crops directly in the field by ploughing. One of the main objectives of the green manuring is to increase the content of nitrogen in the soil. Also it helps in improving the structure and physical properties of the soil. The most important green manure crops are Crotalaria juncea, Tephrosia purpurea and Indigofera tinctoria.

Question 26.
Point out the objectives of plant breeding.
Answer:

1. To increase yield, vigour and fertility of the crop.
2. To increase tolerance to environmental condition, salinity, temperature and drought.
3. To prevent the premature falling of buds and fruits, etc.
4. To improve synchronous maturity.
5. To develop resistance to pathogens and pests.
6. To develop photosensitive and thermos-sensitive varieties

Question 27.
Give an account on clonal selection.
Answer:
In asexually propagated crop, progenies derived from a plant resemble in genetic constitution with the parent plant as they are mitotically divided. Based on their phenotypic appearance, clonal selection is employed to select improved variety from a mixed population (clones). The selected plants are multiplied through vegetative propagation to give rise to a clone. The genotype of a clone remains unchanged for a long period of time.

Question 28.
Write a short note on autopolyploidy with an example.
Answer:
When chromosome number is doubled by itself in the same plant, is called autopolyploidy.
Example: A triploid condition in sugarbeets, apples and pear has resulted in the increase in vigour and fruit size, large root size, large leaves, flower, more seeds and sugar content in them. It also resulted in seedless tomato, apple, watermelon and orange.

5-Mark Questions

Question 29.
Give a comparative account on Seaweed liquid fertilizer.
Answer:
Seaweed liquid fertilizer (SLF) contains cytokinin, gibberellins and auxin apart from macro and micro nutrients. Most seaweed based fertilizers are made from kelp (brown algae) which grows to length of 150 metres. Liquid seaweed fertilizer is not only organic but also eco-friendly. The alginates in the seaweed that reacts with metals in the soil and form long, cross- linked polymers in the soil. These polymers improve the crumbing in the soil, swell up when they get wet and retain moisture for a long time.

They are especially useful in organic gardening which provides carbohydrates for plants. Seaweed has more than 70 minerals, vitamins and enzymes. It promotes vigorous growth. Improves resistance of plants to frost and disease. Seeds soaked in seaweed extract germinate much rapidly and develop a better root system.

Question 30.
Explain the steps involved in hybridization.
Answer:
Steps involved in hybridization are as follows:

1. Selection of Parents: Male and female plants of the desired characters are selected. It should be tested for their homozygosity.
2. Emasculation: It is a process of removal of anthers to prevent self pollination before anthesis (period of opening of a flower).
3. Bagging: The stigma of the flower is protected against any undesirable pollen grains, by covering it with a bag.
4. Crossing: Transfer of pollen grains from selected male flower to the stigma of the female emasculated flower.
5. Harvesting seeds and raising plants: The pollination leads to fertilization and finally seed formation takes place.
   The seeds are grown into new generation which are called hybrid.

Higher Order Thinking Skills (HOTs) Questions

Question 1.
Given below are the examples of symbiotic association in which one partner was mentioned. Write the name of the mutual co-partners.
(a) Azollafem + ______________
(b) Root nodules of legume plants + ______________
(c) phycomycetous fungi + ______________
(a) Anabaena azolla (blue green alga)
(b) Rhizobium
Answer:
(c) Angiosperm roots

Question 2.
Provide an example for each of the following agricultural components.

1. Bio pesticide
2. Green manure crop
3. Bio-Fertilzier

Answer:

1. Bio pesticide – E.g. Trichoderma species
2. Green manure – E.g. Crotalariajuncea
3. Bio fertilizer – E.g. Azolla

Question 3.
State the objective of using green manuring.
Answer:
Green manuring helps to increase the content of nitrogen soil and also improves the structure and physical property of soil.

Question 4.
Why plant Breeding is carried out by farmers and scientists?
Answer:
Plant breeding is a proposal manipulation of plant species in order to develop improved crop varieties with higher yield, better quality, resistance to disease and quick maturation.

Question 5.
Yesterday, Ramu visited his friend’s Orchard, where in he noticed few flowers of certain guava trees are covered using thin paper bags.

1. Name the process carried out there.
2. Why it was done so?

Answer:

1. The process is referred as bagging in plant hybridization.
2. It is done after emasculation to protect the contact of the stigma of the flower against any other pollen grain.

Question 6.
Who am I?

1. Father of Green Revolution.
2. Father of Indian Green Revolution.

Answer:

1. Dr. Norman E. Borlaug.
2. Dr. M.S. Swaminathan.

Question 7.
A plant breeder developed a hybrid sugarcane by grafting two different varieties with desirable characters. The resultant hybrid showed as excellent growth and productivity with increased sucrose content compared to its parental forms.

1. What does this phenomenon refers to?
2. How this condition can be maintained through further generation?

Answer:

1. Heterosis or Hybrid vigour
2. Vegetative propogation is the best suited measure to maintain the vigourisity of hybrid.

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Tamilnadu Samacheer Kalvi 12th Bio Botany Solutions Chapter 3 Chromosomal Basis of Inheritance

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Samacheer Kalvi 12th Bio Botany Chromosomal Basis of Inheritance Text Book Back Questions and Answers

Question 1.
An allohexaploidy contains ___________
(a) Six different genomes
(b) Six copies of three different genomes
(c) Two copies of three different genomes
(d) Six copies of one genome
Answer:
(b) Six copies of three different genomes

Question 2.
The A and B genes are 10 cM apart on a chromosome. If an AB/ab heterozygote is testcrossed to ab/ab, how many of each progeny class would you expect out of 100 total progeny?
(a) 25 AB, 25 ab, 25 Ab, 25 aB
(b) 10AB, 10ab
(c) 45 AB, 45 ab
(d) 45 AB, 45 ab, 5 Ab, 5aB
Answer:
(b) 10AB, 10ab

Question 3.
Match list I with list II

List I	List II
A. A pair of chromosomes extra with diploid	(i) monosomy
B. One chromosome extra to the diploid	(ii) tetrasomy
C. One chromosome loses from diploid	(iii) trisomy
D. Two individual chromosomes lose from diploid	(iv) double monosomy

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

Question 4.
Which of the following sentences are correct?
1. The offspring exhibit only parental combinations due to incomplete linkage
2. The linked genes exhibit some crossing over in complete linkage
3. The separation of two linked genes are possible in incomplete linkage
4. grossing over is absent in complete linkage
(a) 1 and 2
(b) 2 and 3
(c) 3 and 4
(d) 1 and 4
Answer:
(c) 3 and 4

Question 5.
Accurate mapping of genes can be done by three point test cross because increases ___________
(a) Possibility of single cross over
(b) Possibility of double cross over
(c) Possibility of multiple cross over
(d) Possibility of recombination frequency
Answer:
(b) Possibility of double cross over

Question 6.
Due to incomplete linkage in maize, the ratio of parental and recombinants are ___________
(a) 50 : 50
(b) 7 : 1 : 1 : 7
(c) 96.4 : 3.6
(d) 1 : 7 : 7 : 1
Answer:
(b) 7 : 1 : 1 : 7

Question 7.
Genes G S L H are located on same chromosome. The recombination percentage is between and G is 15%, S and L is 50% and H and S are 20%. The correct order of genes is ___________
(a) GHSL
(b) SHGL
(c) SGHL
(d) HSLG
Answer:
(b) SHGL

Question 8.
The point mutation sequence for transition, transition, transversion and transversion in DNA are ________
(a) A to T, T to A, C to G and G to C
(b) A to G, C to T, C to G and T to A
(c) C to G, A to G, T to A and G to A
(d) G to C, A to T, T to A and C to G
Answer:
(b) A to G, C to T, C to G and T to A

Question 9.
If haploid number in a cell is 18. The double monosomic and trisomic number will be ___________
(a) 35 and 37
(b) 34 and 37
(c) 37 and 35
(d) 17 and 19
Answer:
(b) 34 and 37

Question 10.
Changing the codon AGC to AGA represents ___________
(a) mis-sense mutation
(b) non-sense mutation
(c) frameshift mutation
(d) deletion mutation
Answer:
(a) mis-sense mutation

Question 11.
Assertion (A): Gamma rays are generally used to induce mutation in wheat varieties.
Reason (R): Because they carry lower energy to non-ionize electrons from atom
(a) A is correct. R is correct explanation of A
(b) A is correct. R is not correct explanation of A
(c) A is correct. R is wrong explanation of A
(d) A and R is wrong
Answer:
(c) A is correct. R is wrong explanation of A

Question 12.
How many map units separate two alleles A and B, if the recombination frequency is 0.09?
(a) 900 cM
(b) 90 cM
(c) 9 cM
(d) 0.9 cM
Answer:
(d) 0.9 cM

Question 13.
When two different genes came from same parent they tend to remain together.

1. What is the name of this phenomenon?
2. Draw the cross with suitable example.
3. Write the observed phenotypic ratio.

Answer:
(i) Linkage
(ii)

(iii) Observed Pherotypic ratio 7 : 1 : 1 : 7

Question 14.
If you cross dominant genotype PV/PV male Drosophila with double recessive female and obtain F₁ hybrid. Now you cross F₁ male with double recessive female.

1. What type of linkage is seen?
2. Draw the cross with correct genotype.
3. What is the possible genotype in F₂ generation?

Answer:
(i) Incomplete linkage
(ii) Parent Garnets

F₁ hybrid

Question 15.

1. What is the name of this test cross?
2. How will you construct gene mapping from the above given data?
3. Find out the correct order of genes.

Answer:
(i) Three point test cross.
(ii) Construction of gene map:
To construct the gene map, the recombinant frequency (RF) of the alleles has to be calculated.
From the given data it is clear that ABC and abc are parental (P) types and the others (Abe, abC, AbC, aBc, ABc) are recombinant (R) type.

Lets analyse the loci of two alleles at a time starting with A and B. Since the genes AB and ab are parental type, the recombinants will be Ab and aB.
Therefore
Recombinant frequency of alleles Ab and aB = \(\frac { No.of recombinant }{ Total progenies }\) × 100
\(\frac { 114+5+4+116 }{ 1200 }\) × 100 = 19.91%

Recombinant frequency for the loci B and C
The parental form are Be and bC and the recombinant are Be and bC.

Recombinant frequency of alleles Be and bC = \(\frac { 4+128+124+5 }{ 1200 }\) × 100 + 21.75%
Since the recombinant frequency of the alleles A and C shown highest frequency, they must be the farthest apart and alleles B must lie in between A and C. So the gene map can be constructed as follows

(iii) The correct gene order is ABC/abc.

Question 16.
What is the difference between mis-sense and nonsense mutation?
Answer:
Mis-sense Mutation:
The mutation where the codon for one amino acid is changed into a codon for another amino acid is called Missense or non-synonymous mutations.

Non-sense Mutation:
The mutations where codon for one amino acid is changed into a termination or stop codon is called Nonsense mutation.

Question 17.
A B C C B D E F G H I
From the above figure identify the type of mutation and explain it.
Answer:
In reverse tandem duplication, the duplicated segment is located immediately after the normal segment but the gene sequence other will be reversed.

Question 18.
Write the salient features of Sutton and Boveri concept.
Answer:
Salient features of the chromosomal theory of inheritance:

1. Somatic cells of organisms are derived from the zygote by repeated cell division (mitosis). These consist of two identical sets of chromosomes. One set is received from female parent (maternal) and the other from male parent (paternal). These two chromosomes constitute the homologous pair.
2. Chromosomes retain their structural uniqueness and individuality throughout the life cycle of an organism.
3. Each chromosome carries specific determiners or Mendelian factors which are now termed as genes.
4. The behaviour of chromosomes during the gamete formation (meiosis) provides evidence to the fact that genes or factors are located on chromosomes.

Question 19.
“Explain the mechanism of crossing over.
Answer:
Crossing over is a precise process that includes stages like synapsis, tetrad formation, cross over and terminalization.

(i) Synapsis: Intimate pairing between two homologous chromosomes is initiated during zygotene stage of prophase I of meiosis I. Homologous chromosomes are aligned side by side resulting in a pair of homologous chromosomes called bivalents. This pairing phenomenon is called synapsis or syndesis. It is of three types:

1. Procentric synapsis: Pairing starts from middle of the chromosome.
2. Proterminal synapsis: Pairing starts from the telomeres.
3. Random synapsis: Pairing may start from anywhere.

(ii) Tetrad Formation: Each homologous chromosome of a bivalent begin to form two identical sister chromatids, which remain held together by a centromere. At this stage each bivalent has four chromatids. This stage is called tetrad stage.

(iii) Cross Over: After tetrad formation, crossing over occurs in pachytene stage. The non-sister chromatids of homologous pair make a contact at one or more points. These points of contact between non¬sister chromatids of homologous chromosomes are called Chiasmata (singular-Chiasma).

At chiasma, cross-shaped or X-shaped structures are formed, where breaking and rejoining of two chromatids occur. This results in reciprocal exchange of equal and corresponding segments between them. A recent study reveals that synapsis and chiasma formation are facilitated by a highly organised structure of filaments called Synaptonemal Complex (SC). This synaptonemal complex formation is absent in some species of male Drosophila, hence crossing over does not takes place.

(iv) Terminalisation: After crossing over, chiasma starts to move towards the terminal end of chromatids. This is known as terminalisation. As a result, complete separation of homologous chromosomes occurs.

Question 20.
Write the steps involved in molecular mechanism of DNA recombination with diagram.
Answer:

1. Homologous DNA molecules are paired side by side with their duplicated copies of DNAs
2. one stand place by the enzyme endonuclease.
3. The cut strands cross and join the homologous strands forming the Holliday structure or Holliday junction.
4. The Holliday junction migrates away from the original site, a process called branch migration, as a result heteroduplex region is formed.
5. DNA strands may cut along through the vertical (V) line or horizontal (H) line.
6. The vertical cut will result in heteroduplexes with recombinants.
7. The horizontal cut will result in heteroduplex with non recombinants.

Question 21.
How is Nicotiana exhibit self-incompatibility. Explain its mechanism.
Answer:
Self-sterility means that the pollen from a plant is unable to germinate on its own stigma and will not be able to bring about fertilization in the ovules of the same plant. East (1925) observed multiple alleles in Nicotiana which are responsible for self-incompatibility or self-sterility. The gene for self-incompatibility can be designated as S, which has allelic series S₁ S₂, S₃, S₄ and S₅. The cross-fertilizing tobacco plants were not always homozygous as S₁ S₁ or S₂ S₂, but all plants were heterozygous as S₁ S₂, S₃ S₄ and S₅  S₆. When crosses were made between different S₁ S₂ plants, the pollen tube did not develop normally. But effective pollen tube development was observed when crossing was made with other than S₁S₂ for example S₃S4.

When crosses were made between seed parents with S₁S₂ and pollen parents with S₂S₃, two kinds of pollen tubes were distinguished. Pollen grains carrying S₂ were not effective, but the pollen grains carrying S₃ were capable of fertilization. Thus, from the cross S₁ S₂ X S₃ S₄, all the pollens were effective and four kinds of progeny resulted: S₁S₃, S₁ S₄, S₂ S₃ and S₂ S₄.

Question 22.
How sex is determined in monoecious plants? Write their genes involved in it.
Answer:
Zea mays (maize) is an example for monoecious, which means male and female flowers are present on the same plant. There are two types of inflorescence. The terminal inflorescence which bears staminate florets that develops from shoot apical meristem called tassel. The lateral inflorescence which develop pistillate florets from axillary bud is called ear or cob. Unisexuality in maize occurs through the selective abortion of stamens in ear florets and pistils in tassel florets. A substitution of two single gene pairs ‘ba’ for barren plant and ‘ts’ for tassel
seed makes the difference between monoecious and dioecious (rare)

maize plants. The allele for barren plant (ba) when homozygous makes the stalk staminate by eliminating silk and ears. The allele for tassel seed
(ts) transforms tassel into a pistillate structure that produce no pollen. The table is the resultant sex expression ‘ based on the combination of these
alleles. Most of these mutations are shown to be defects in gibberellin biosynthesis. Gibberellins play an important role in the suppression of stamens in florets on the ears.

Question 23.
What is gene mapping? Write its uses.
Answer:
The diagrammatic representation of position of genes and related distances between the adjacent genes is called genetic mapping. It is directly proportional to the frequency of recombination between them. It is also called as linkage map.
Uses of genetic mapping:

• It is used to determine gene order, identify the locus of a gene and calculate the distances between genes.
• They are useful in predicting results of dihybrid and trihybrid crosses.
• It allows the geneticists to understand the overall genetic complexity of particular organism.

Question 24.
Draw the diagram of different types of aneuploidy.
Answer:

Question 25.
Mention the name of man-made cereal. How it is formed?
Answer:
Triticale, the successful first man made cereal. Depending on the ploidy level Triticale can be divided into three main groups:

1. Tetraploidy: Crosses between diploid wheat and rye.
2. Hexaploidy: Crosses between tetraploid wheat Triticum durum (macaroni wheat) and rye.
3. Octoploidy: Crosses between hexaploid wheat T. aestivum (bread wheat) and rye. Hexaploidy Triticale hybrid plants

demonstrate characteristics of both macaroni wheat and rye. For example, they combine the high-protein content of wheat with rye’s high content of the amino acid lysine, which is low in wheat.

Samacheer Kalvi 12th Bio Botany Chromosomal Basis of Inheritance Additional Questions and Answers

1 – Mark Questions

Question 1.
Name the scientist(s) who rediscovered the Mendelian work?
(i) Hugo de Vries
(ii) Carl Correns
(iii) Tschermak
(a) i and iv
(b) i, ii and iv
(c) (iii) i, ii and iii
(d) ii, iii and iv
Answer:
(c) i, ii and iii

Question 2.
Which is not a feature of the chromosomal theory of inheritance?
(a) Somatic cells of organisms are derived from zygote by repeated meiosis.
(b) Chromosomes retain their structural uniqueness throughout the life of an organism.
(c) Mendelian factors are located in chromosomes
(d) Sutton and Boveri independently proposed the theory.
Answer:
(a) Somatic cells of organisms are derived from zygote by repeated meiosis.

Question 3.
The following sequence represents the location of genes in a chromosome. A-B-C-M-R – S – Y – Z. Which of the gene pairs will have least chance of getting inherited together?
(a) A and M
(b) S and Y
(c) M and Z
(d) A and Y
Answer:
(d) A and Y

Question 4.
Match the column I with column II:

(a) 1 – iii, 2 – i, 3 – iv, 4 – ii
(b) 1 – ii, 2 – i, 3 – iv, 4 – iii
(c) 1 – iv, 2 – iii, 3 – ii, 4 – i
(d) 1 – ii, 2 – iii, 3 – i, 4 – iv
Answer:
(a) 1 – iii, 2 – i, 3 – iv, 4 – ii

Question 5.
Number of chromosomes (2n) in Ophioglossum is ________
(a) 1226
(b) 1622
(c) 1262
(d) 2126
Answer:
(c) 1262

Question 6.
Identify the syntenic gene from the given genes sequence of a chromosome G-H-I-J-K-L-M-A-B
(a) G and H
(b) J, K and L
(c) G and B
(d) A and B
Answer:
(c) G and B

Question 7.
Incomplete linkage was reported by Hutchinson in
(a) Drosophila
(b) Maize
(c) Neurospora
(d) Lathyrus odoratus
Answer:
(b) Maize

Question 8.
Mechanism of crossing over involves the following stages. Select the correct sequence.
(а) Tetrad stage → Synapsis → Bivalent stage → cross over
(b) Syndesis → Tetrad → Crossing over → Terminalisation
(c) Terminalisation → Tetrad → Bivalent → Cross over
(d) Cross over → Bivalent → Tetrad → Terminalisation
Answer:
(b) Syndesis → Tetrad → Cross over → Terminalisation

Question 9.
During cross over, chiasma occurs between
(a) Sister chromatids of non-homologous chromosomes
(b) Non-sister chromatids of non- homologous chromosomes
(c) Non-sister chromatids of homologous chromosomes
(d) Sister chromatids of homologous chromosomes
Answer:
(c) Non-sister chromatids of homologous chromosomes

Question 10.
The term crossing over was coined by _________
Answer:
T.H. Morgan

Question 11.
At which stage of meiosis, does the chromosomes undergo recombination process?
(a) Leptotene stage of prophase I
(b) Zygotene stage of prophase I
(c) Diplotene stage of prophase I
(d) Pachytene stage of prophase I
Answer:
(d) Pachytene stage of prophase I

Question 12.
Which of the following statement(s) is/are wrong with respect to Recombination process?
(i) Mitotic crossing over occurs rarely in somatic cells.
(ii) Syndesis refers to pairing of non-homologous chromosome.
(iii) Procentric synapsis starts from telomeres.
(iv) A Bivalent has four chromatids.
(a) i and iv
(b) ii and i
(c) ii and iii
(d) All the above
Answer:
(c) ii and iii

Question 13.
Recombination frequency (RF) is equal to
(a) \(\frac { No. of recombinants }{ No. of recombinants }\) × 100
(b) \(\frac { No. of recombinants }{ No. of patental strains }\) × 100
(c) \(\frac { No. of recombinants }{ No. of offsprings }\) × 100
(d) \(\frac { No. of recombinants }{ No. of Parental strains }\) × 100
Answer:
(c) \(\frac { No. of recombinants }{ No. of offsprings }\) × 100

Question 14.
In a population of 250 progenies produced, only 120 resemble the parental forms. Calculate the recombinant frequency.
(a) 66%
(b) 52%
(c) 59%
(d) 49%
Answer:
(b) 52%

Question 15.
The unit of distance in genetic map is called as __________
Answer:
Map unit or Centimorgan

Question 16.
The “2n” condition of Carica papaya is __________
Answer:
36

Question 17.
Mutation theory was proposed by. __________
(a) T. H. Morgan
(b) Hugo de Vries
(c) Alfred variety
(d) Erectiferm
Answer:
(b) Hugo de Vries

Question 18.
Identity the mutant variety of castor.
(a) Sharbathi Sonora variety
(b) Aruna variety
(c) Reimei variety
(d) Erectiferm variety
Answer:
(b) Aruna variety

Question 19.
Which is not a non-ionizing radiation?
(a) X-rays
(b) Gamma rays
(c) Alpha rays
(d) UV rays
Answer:
(d) UV rays

Question 20.
Transition type of gene mutation is caused when __________
(a) AC is replaced by GT
(b) AG is replaced by TC
(c) AC is replaced by TG
(d) TC is replaced by AG
Answer:
(a) AC is replaced by GT

Question 21.
Pick out the co-mutagen from the following:
(a) Eosin
(b) Mustard gas
(c) Ascorbic acid
(d) Nitrous acid
Answer:
(c) Ascorbic acid

Question 22.
Sharbati Sonara is a mutant wheat variety which is developed by irradiating the seeds with __________
(a) Thermal neutrons
(b) Gamma radiation
(c) X-rays
(d) UV radiations
Answer:
(b) Gamma radiation

Question 23.
Which one of the following ploidy is irrelevant to others?
(a) Monosomy
(b) Trisomy
(c) Tetrasomy
(d) Pentasomy
Answer:
(a) Monosomy

Question 24.
Match with correct

(a) 1 – ii, 2 – iii, 3 – iv, 4 – i
(b) 1 – ii, 2 – i, 3 – iv, 4 – iii
(c) 1 – iv, 2 – iii, 3 – ii, 4 – i
(d) 1 – ii, 2 – iii, 3 – i, 4 – iv
Answer:
(a) 1 – ii, 2 – iii, 3 – iv, 4 – i

Question 25.
Statement 1: Euploidy involves entire sets of chromosomes
Statement 2: Aneuploidy involves individual chromosomes within a diploid net.
(a) Statement 1 is correct and Statement 2 is incorrect
(b) Statement 1 is incorrect and Statement 2 is correct
(c) Both the statements are correct
(d) Both the statements are incorrect
Answer:
(c) Both the statements are correct

Question 26.
Statement 1: In transversion mutation, single purine is changed to pyrimidine.
Statement 2: In transition mutation, a purine replaced by another purine.
(a) Statement 1 is correct and Statement 2 is incorrect
(b) Statement 1 is incorrect and Statement 2 is correct
(c) Both the statements are correct
(d) Both the statements are incorrect
Answer:
(a) Statement 1 is correct and Statement 2 is incorrect

Question 27.
Statement 1: Pairing of homologous chromosome is called as syndesis.
Statement 2: Proterminal synapsis occurs from telomeres.
(a) Statement 1 is correct and Statement 2 is incorrect
(b) Statement 1 is incorrect and Statement 2 is correct
(c) Both the statements are correct
(d) Both the statements are incorrect
Answer:
(c) Both the statements are correct

Question 28.
Statement 1: The widely accepted DNA replication model is Holliday’s hybrid DNA model.
Statement 2: The vertical cut in the DNA results in heteroduplex with non-recombinants.
(a) Statement 1 is correct and Statement 2 is incorrect
(b) Statement 1 is incorrect and Statement 2 is correct
(c) Both the statements are correct
(d) Both the statements are incorrect
Answer:
(d) Both the statements are incorrect.

Question 29.
Statement 1: Self-sterility in Nicotiana is controlled by multiple alleles.
Statement 2: Multiple alleles are always responsible for the same character.
(a) Statement 1 is correct and Statement 2 is incorrect
(b) Statement 1 is incorrect and Statement 2 is correct
(c) Both the statements are correct
(d) Both the statements are incorrect
Answer:
(c) Both the statements are correct

Question 30.
Sex determination in plants was first discovered by.
Answer:
C.E. Allen

Question 31.
One of the following is not the kind of euploidy
(a) Diploidy
(b) Polyploidy
(c) Hyperploidy
(d) Autoploidy
Answer:
(c) Hyperploidy

Question 32.
The chromosomal condition 2n -2 represents __________
(a) Monosomy
(b) Nullisomy
(b) Nullisomy
(c) Trisomy
Answer:
(d) Tetrasomy

Question 33.
(a) Potato
(b) Coffee
(c) Groundnut
(d) Apple
Answer:
(d) Apple

Question 34.
Assertion (A): Polyploidy is common in plants.
Reason (R): Polyploids possess more than 2 basic sets of chromosomes.
(a) A is true R is false
(b) Both A and R are false
(c) A is true, R is not correct explanation for A
(d) R explains A
Answer:
(d) R explains A

Question 35.
Assertion (A): Complete linkage is noticed in male species of Drosophila.
Reason (R): Completely linked genes show some crossing over.
(a) A is true R is false
(b) Both A and R are false
(c) A is true, R is not correct explanation for A
(d) R explains A
Answer:
(a) A is true R is false

Question 36.
Assertion (A): Self-sterility is observed in Nicotiana species.
Reason (R): Because the genes are located on chromosome.
(a) A is true R is false
(b) Both A and R are false
(c) A is true, R is not correct explanation for A
(d) R explains A
Answer:
(c) A is true, R is not correct explanation for A

Question 37.
The first man made cereal is _______
Answer:
Triticale

Question 38.
Observe the gene sequence and identify the types of aberration A B C B C D E F?
(a) Tandem duplication
(b) Simple duplication
(c) Reverse tandem duplication
(d) Displaced tandem duplication
Answer:
(a) Tandem duplication

2 – Mark Questions

Question 1.
Write any one contribution of the following scientists to the field of molecular biology,
(a) Hugo de Vries
(b) Sutton and Boveri
Answer:
(a) Hugo de Vries proposed Mutation theory.
(b) Sutton and Boveri proposed chromosomal theory of inheritance.

Question 2.
Define Linkage. Mention its types.
Answer:
Tendency of genes to remain together during separation of chromosomes is called linkage. Linkage are of 2 types – complete linkage and incomplete linkage.

Question 3.
What does the condition synteny refers to?
Answer:
The two genes that are sufficiently far apart on the same chromosome are called unlinked genes or syntenic genes. Such condition is known as synteny.

Question 4.
What are linked genes?
Answer:
Genes located close together on the same chromosome and inherited together are called linked genes.

Question 5.
Who coined the term crossing over? When does the crossing over occurs in a cell?
Answer:
The term ‘crossing over ’ was coined by Morgan (1912). It takes place during pachytene stage of prophase I of meiosis.

Question 6.
Crossing over occurs only in germinal cells. Yes or no. Support your answer.
Answer:
No. Though crossing over is a common process in germinal cells rarely it also occurs in somatic cells during mitosis. Such crossing over is called mitotic crossing over or somatic crossing over.

Question 7.
Mention the major stages involved in crossing over.
Answer:
Crossing over is a precise process that includes stages like synapsis, tetrad formation, cross over and terminalization.

Question 8.
What are bivalents? When does this condition is noticed in a cell?
Answer:
During zygotene stage of prophase I of meiosis I, homologous chromosomes are aligned side by side resulting in a pair of homologous chromosomes called bivalents.

Question 9.
What is meant by synaptonemal complex?
Answer:
Synaptonemal complex is a highly organised structure or filaments that facilitates the synapsis and chiasma formation during crossing over mechanism.

Question 10.
What will be the result if there is a failure in the formation of synaptonemal complex. Give one example of organism where such condition is noticed?
Answer:
Failure in the formation of synaptonemal complex leads to the absence of crossing over to take place. It is noticed in certain male species of Drosophila.

Question 11.
Define terminalization.
Answer:
After crossing over, chiasma starts to move towards the terminal end of chromatids. This is known as terminalization. As a result, complete separation of homologous chromosomes occurs.

Question 12.
Write the formula to calculate recombination frequency.
Answer:
\(\frac { No. of recombinants }{ No. of offsprings }\) × 100

Question 13.
What is genetic mapping?
Answer:
The diagrammatic representation of position of genes and related distances between the adjacent genes is called genetic mapping. It is directly proportional to the frequency of recombination between them. It is also called as linkage map.

Question 14.
Define the terms
(a) locus
(b) centimorgan.
Answer:
(a) Locus: A specific location of genes on a chromosome.
(b) Centimorgan: Unit of distance in a genetic map.

Question 15.
What are multiple alleles?
Answer:
When any of the three or more allelic forms of a gene occupy the same locus in a given pair of homologous chromosomes, they are said to be called multiple alleles.

Question 16.
Does environment determines the sex of a plant? Explain in brief with an example.
Answer:
Yes, the equisetum (horsetail plant), which grow under good conditions develop as female plants those grown under stressed condition develops as male then supporting the fact environment can also determines the sex of plant in few species.

Question 17.
Why do we call papaya a dimorphic plant.
Answer:
Papaya is a dimorphic plant since their sexes are separate i.e., male plant produces flowers with stamens and female plants produces flowers with carpels.

Question 18.
Define the term mutation. Who coined the term?
Answer:
A sudden change in the genetic material of an organisms is called mutation. The term mutation was introduced by Hugo de Vries.

Question 19.
Compare point mutation with chromosomal mutation.
Answer:
Mutational events that take place within individual genes are called gene mutations or point mutation, whereas the changes occur in structure and number of chromosomes is called chromosomal mutation.

Question 20.
Based on the effect on translation, classify mutations.
Answer:
(a) Silent mutation
(b) Mis-sense mutation
(c) Non-sense mutation
(d) Frameshift mutation

Question 21.
What does indel mutation refers to?
Answer:
Addition or deletion mutations are actually additions or deletions of nucleotide pairs and also called base pair addition or deletions. Collectively, they are termed as indel mutations.

Question 22.
Define mutagens and mention its types.
Answer:
The factors which cause genetic mutation are called mutagenic agents or mutagens. Mutagens are of two types, physical mutagen and chemical mutagen.

Question 23.
Point out any four physical mutagens.
Answer:
Temperature, X-rays, Gamma rays and UV rays.

Question 24.
Write a brief note on Castor Aruna variety.
Answer:
Castor Aruna is mutant variety of castor which is developed by treatment of seeds with thermal neutrons in order to induce very early maturity (120 days instead of 270 days as original variety).

Question 25.
How Sharbati Sonora was developed by Dr. M.S. Swaminathan etal?
Answer:
Sharbati Sonora is a mutant variety of wheat, which is developed from Mexican variety (Sonora 64) by irradiating of gamma rays.

Question 26.
Name any four chemical mutagens.
Answer:
(a) Ethyl methane sulphonate (EMS)
(b) Mustard gas
(c) Magnous salt
(d) Formaldehyde

Question 27.
Nitrous oxide is a potent mutagen – comment.
Answer:
Nitrous oxide alters the nitrogen bases of DNA and disturb the replication and transcription that leads to die formation of incomplete and defective polypeptide during translation.

Question 28.
What is ploidy?
Answer:
The chromosome number of somatic cells changes due to addition or elimination of individual chromosome or basic set of chromosomes. This condition in known as numerical chromosomal aberration or ploidy.

Question 29.
Differentiate Aneuploidy from Euploidy.
Answer:
Aneuploidy:

1. Ploidy involving individual chromosomes within a diploid set.
2. E.g: Trisomy.

Euploidy:

1. Ploidy involving entire sets of chromosomes.
2. E.g: Polyploidy.

Question 30.
Comment on the chromosomal condition; 2n – 2.
Answer:
Loss of pair of homologous chromosome from the diploid set is called Nullisomy. Selling of monosomic plants produces Nullisomics. They are generally lethal.

Question 31.
Name any 3 auto tetraploids and one natural autotriploid plant species.
Answer:
(a) Autotetraploids : Rye, potato and coffee.
(b) Natural autotriploid : Cyanodon dactylon (common doob grass).

Question 32.
What are deficiency loops?
Answer:
Deletions are observable during meiotic pachytene stage and polytene chromosome. The unpaired loop formed in the normal chromosomal part at the time of chromosomal pairing. Such loops are called as deficiency loops and it can be seen in meiotic prophase.

Question 33.
Given below are the gene sequences on the chromosome. Compare them with the normal chromosome and identify the type of structural chromosomal aberrations.
Normal Chromosome : A-B-C-D-E-F-G-H-I.
Chromosome 1 : A-B-C-B-C-D-E-F-G-H-I.
Chromosome 2 : A-B-C-D-F-G-H-I
Answer:
Chromosome 1 : Tandem duplication.
Chromosome 2 : Intercalary deletion.

3 – Mark Questions

Question 34.
Point out any three salient features of chromosomal theory of inheritance.
Answer:
Salient features of the chromosomal theory of inheritance:

1. Somatic cells of organisms are derived from die zygote by repeated cell division (mitosis). These consist of two identical sets of chromosomes. One set is received from female parent (maternal) and the other from male parent (paternal). These two chromosomes constitute the homologous pair.
2. Chromosomes retain their structural uniqueness and individuality throughout the life cycle of an organism.
3. Each chromosome carries specific determiners or Mendelian factors which are now termed as genes.

Question 35.
Compare Mendelian factors with chromosome
Answer:
Mendelian factors:

1. Alleles of a factor occur in pair.
2. Similar or dissimilar alleles of a factor separate during the gamete formation.
3. Mendelian factors can assort independently.

Chromosomes behaviour:

1. Chromosomes occur in pairs.
2. The homologous chromosomes separate during meiosis.
3.  The paired chromosomes can separate independently during meiosis but the linked genes in the same chromosome normally do not assort independently.

Question 36.
State Coupling and Repulsion theory.
Answer:
The two dominant alleles or recessive alleles occur in the same homologous chromosomes, tend to inherit together into same gamete are called coupling or configuration. If dominant or recessive alleles are present on two different, but homologous chromosomes they inherit apart into different gamete are called repulsion or trans configuration.

Question 37.
Give a short note on incomplete linkage.
Answer:
If two linked genes are sufficiently apart, the chances of their separation are possible. As a result, parental and non-parental combinations are observed. The linked genes exhibit some crossing over. This phenomenon is called incomplete linkage. This was observed in maize. It was reported by Hutchinson.

Question 38.
How crossing over differs from linkage?
Answer:
Linkage:

1. The genes present on chromosome stay close together.
2. It involves same chromosome of homologous chromosome.
3. It reduces new gene combinations.

Crossing over:

1. It leads to separation of linked genes.
2. It involves exchange of segments between non-sister chromatids of homologous chromosome.
3. It increases variability by forming new gene combinations. Leading to formation of new organism.

Question 39.
What is Synapsis? Explain its types.
Answer:
Pairing of homologous chromosomes during zygotene stage of prophase I of meiosis I is called synapsis. Synapsis is of three types:

1. Procentric synapsis: Pairing starts from middle of the chromosome.
2. Proterminal synapsis: Pairing starts from the telomeres.
3. Random synapsis: Pairing may start from anywhere.

Question 40.
How and where chiasma is formed?
Answer:
After tetrad formation, crossing over occurs in pachytene stage. The non-sister chromatids of homologous pair make a contact at one or more points. These points of contact between non sister chromatids of homologous chromosomes are called Chiasmata (singular-Chiasma). At chiasma, cross-shaped or X-shaped structures are formed, where breaking and rejoining of two chromatids occur. This results in reciprocal exchange of equal and corresponding segments between them.

Question 41.
Classify cross over.
Answer:

1. Single cross over: Formation of single chiasma and involves only two chromatids out of four.
2. Double cross over: Formation of two chiasmata and involves two or three or all four strands.
3. Multiple cross over: Formation of more than two chiasmata and crossing over frequency is extremely low.

Question 42.
What is recombination? Which is the widely accepted model of DNA recombination?
Answer:
Crossing over results in the formation of new combination of characters in an organism called recombinants. In this, segments of DNA are broken and recombined to produce new combinations of alleles. This process is called recombination.
The widely accepted model of DNA recombination during crossing over is Holliday’s hybrid DNA model.

Question 43.
Which type of test cross provides the data to construct an efficient genetic map? Explain.
Answer:
A more efficient mapping technique is to construct based on the results of three-point test cross. It refers to analyzing the inheritance patterns of three alleles by test crossing a triple recessive heterozygote with a triple recessive homozygote. It enables to determine the distance between the three alleles and the order in which they are located on the chromosome. Double cross overs can be detected which will provide more accurate map distances.

Question 44.
Enumerate the uses of Genetic mapping.
Answer:

1. It is used to determine gene order, identify the locus of a gene and calculate the distances between genes.
2. They are useful in predicting results of dihybrid and trihybrid crosses.
3. It allows the geneticists to understand the overall genetic complexity of particular organism.

Question 45.
List any three characteristic features of multiple allele.
Answer:

1. Multiple alleles of a series always occupy the same locus in the homologous chromosome. Therefore, no crossing over occurs within the alleles of a series.
2. Multiple alleles are always responsible for the same character.
3. The wild type alleles of a series exhibit dominant character whereas mutant type will influence dominance or an intermediate phenotypic effect.

Question 46.
Explain the sex determination mechanism in Carica papaya.
Answer:
Carica papaya, 2n=36 (Papaya) has 17 pairs of autosomes and one pair of sex chromosomes. Male papaya plants have XY and female plants have XX. Unlike human sex chromosomes, papaya sex chromosomes look like autosomes and it is evolved from autosome. The sex chromosomes are functionally distinct because the Y chromosome carries the genes for male organ development and X bears the female organ developmental genes. In papaya sex determination is controlled by three alleles. They are m, Mj and M2 of a single gene.

Sex chromosome of papaya

Question 47.
Differentiate Mis-sense mutation from Non-sense mutation.
Answer:
Mis-sense Mutation:
The mutation where the codon for one amino acid is changed into a codon for another amino acid is called Mis-sense or non-synonymous mutations.

Non-sense Mutation:
The mutations where codon for one amino acid is changed into a termination or stop codon is called Non-sense mutation.

Question 48.
What is frameshift mutation?
Answer:
Mutations that result in the addition or deletion of a single base pair of DNA that changes the reading frame for the translation process as a result of which there is complete loss of normal protein structure and function are called Frameshift mutations.

Question 49.
How temperature induces mutation?
Answer:
Increase in temperature increases the rate of mutation. While rise in temperature, breaks the hydrogen bonds between two DNA nucleotides which affects the process of replication and transcription.

Question 50.
Explain co-mutagens with examples.
Answer:
The compounds which are not having own mutagenic properties but can enhance the effects of known mutagens are called co-mutagens.
Example: Ascorbic acid increase the damage caused by hydrogen peroxide. Caffeine increase the toxicity of methotrexate.

Question 51.
Name the following chromosomal conditions.
(a) 2n + 2 + 2
(b) 2n – 1 – 1
(c) x
(e) 2n + n + n
(f) 2n + 1
Answer:
(a) Double tetrasomy
(b) Double monosomy
(c) Monoploidy
(e) Polyploidy
(f) Trisomy

Question 52.
Give an account on colchicine.
Answer:
Colchicine, an alkaloid is extracted from root and conns of Colchicum autumnale, when applied in low concentration to the growing tips of the plants it will induce polyploidy. Surprisingly it does not affect the source plant Colchicum, due to presence of anticolchicine.

Question 53.
Explain the three types of duplication with diagram.
Answer:

1. Tandem duplication: The duplicated segment is located immediately after the normal segment of the chromosome in the same order.
2. Reverse tandem duplication The duplicated segment is located immediately after the normal segment but the gene sequence order will be reversed.
3. Displaced duplication: The duplicated segment is located in the same chromosome, but away from the normal segment.

5 – Mark Questions

Question 54.
Why crossing over is important?
Answer:

1. Exchange of segments leads to new gene combinations which plays an important role in evolution.
2. Studies of crossing over reveal that genes are arranged linearly on the chromosomes.
3. Genetic maps are made based on the frequency of crossing over.
4. Crossing over helps to understand the nature and mechanism of gene action.
5. If a useful new combination is formed it can be used in plant breeding.

Question 55.
Draw a flow chart depicting the various types of ploidy.
Answer:

Question 56.
Explain hyperploidy with its types.
Answer:
Hyperploidy
Addition of one or more chromosomes to diploid sets are called hyperploidy. Diploid set of chromosomes represented as Disomy. Hyperploidy can be divided into three types. They are as follows:

(a) Trisomy: Addition of single chromosome to diploid set is called Simple trisomy (2n+l). Trisomics were first reported by Blackeslee (1910) in Datura stramonium (Jimson weed). But later it was reported in Nicotiana, Pisum and Oenothera. Sometimes addition of two individual chromosome from different chromosomal pairs to normal diploid sets are called Double trisomy (2n+l+l).

(b) Tetrasomy: Addition of a pair or two individual pairs of chromosomes to diploid set is called tetrasomy (2n+2) and Double tetrasomy (2n+2+2) respectively. All possible tetrasomics are available in Wheat.

(c) Pentasomy: Addition of three individual chromosome from different chromosomal pairs to normal diploid set are called pentasomy (2n+3).

Question 57.
List out the significance of ploidy.
Answer:

1. Many polyploids are more vigorous and more adaptable than diploids.
2. Many ornamental plants are autotetraploids and have larger flower and longer flowering duration than diploids.
3. Autopolyploids usually have increase in fresh weight due to more water content.
4. Aneuploids are useful to determine the phenotypic effects of loss or gain of different chromosomes.

Question 58.
Explain the Translocation type of chromosomal aberration. Translocation
Answer:
The transfer of a segment of chromosome to a non- homologous chromosome is called translocation. Translocation should not be confused with crossing over, in which an exchange of genetic material between homologous chromosome takes place. Translocation occurs as a result of interchange of chromosome segments in non-homologous cliromosomes.
There are three types:

1. Simple translocation
2. Shift translocation
3. Reciprocal translocation

i. Simple translocation : A single break is made in only one chromosome. The broken segment gets attached to one end of a non-homologous chromosome. It occurs very rarely in nature.

ii. Shift translocation : Broken segment of one chromosome gets inserted interstitially in a non-homologous chromosome.

iii. Reciprocal translocations : It involves mutual exchange of chromosomal segments between two non-homologous chromosomes. It is also called illegitimate crossing over. It is further divided into two types.

a. Homozygous translocation: Both the chromosomes of two pairs are involved in translocation. Two homologous of each translocated chromosomes are identical.

b. Heterozygous translocation: Only one of the chromosome from each pair of two homologous are involved in translocation, while the remaining chromosome is normal. Translocations play a major role in the formation of species.

Higher Order Thinking (HOTs) Questions

Question 1.
Given below is a sequence of alphabets representing the genes of chromosome. Observe it and answer the questions.
A-B-C-D-E-F-G-H-I-J-K.
(a) Write the sequence of genes after the chromosome undergoes terminal deletion of single gene.
(b) What will be the gene sequence, if the genes E and F undergoes tandem duplication?
(c) Consider the centromere is located between the genes F and G and write a gene sequence, after paracentric inversion occurs in between the genes C, D and E.
Answer:
(a) B – C – D – E – F – G – H -I – J – K (or) A- B – C – D – E – F – G – H -I – J.
(b) A – B – C – D – E – F – E – F – G – H – I – J – K.
(c) A – B – E – D – C – F – G – H -1 – J – K.

Question 2.
In Drosophila melanogaster, there are four pairs of chromosomes. If there occurs chromosomal aberrations resulting in trisomic and monosomic condition, what will be the chromosomal count? Write the correct chromosomal count against respective chromosomal aberration.
Answer:
Normal chromosome of Drosophila melanogaster (2n) = 8
Frisomic condition (2n+1) = 9
Monosomic condition (2n – 1) = 7

Question 3.
Study the figures given below and answer the questions.
Answer:

Which type of cross produces higher recombinant percentage? Give reason in support of your answer.
(a) Cross B produces more recombinants.
(b) The gene e and f are located sufficiently apart leading to crossing over resulting in recombinant varieties.

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Tamilnadu Samacheer Kalvi 12th Bio Botany Solutions Chapter 7 Ecosystem

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Samacheer Kalvi 12th Bio Botany Ecosystem Text Book Back Questions and Answers

Question 1.
Which of the following is not a abiotic component of the ecosystem?
(a) Bacteria
(b) Humus
(c) Organic compounds
(d) Inorganic compounds
Answer:
(b) Humus

Question 2.
Which of the following is / are not a natural ecosystem?
(a) Forest ecosystem
(b) Rice field
(c) Grassland ecosystem
(d) Desert ecosystem Pond is a type
Answer:
(b) Rice field

Question 3.
Pond is a type of __________
(a) Forest ecosystem
(b) grassland ecosystem
(c) marine ecosystem
(d) fresh water ecosystem Pond ecosystem
Answer:
(d) fresh water ecosystem Pond ecosystem

Question 4.
Pond ecosystem is __________
(a) not self sufficient and self regulating
(b) partially self sufficient and self regulating
(c) self sufficient and not self regulating
(d) self sufficient and self regulating
Answer:
(d) self sufficient and self regulating

Question 5.
Profundal zone is predominated by heterotrophs in a pond ecosystem, because of __________
(a) with effective light penetration
(b) no effective light penetration
(c) complete absence of light
(d) a and b
Answer:
(d) a and b

Question 6.
Solar energy used by green plants for photosynthesis is only __________
(a) 2 -8%
(b) 2-10%
(c) 3-10%
(d) 2-9%
Answer:
(b) 2-10%

Question 7.
Which of the following ecosystem has the highest primary productivity?
(a) Pond ecosystem
(b) Lake ecosystem
(c) Grassland ecosystem
(d) Forest ecosystem
Answer:
(c) Grassland ecosystem

Question 8.
Ecosystem consists of __________
(a) decomposers
(b) producers
(c) consumers
(d) all of the above
Answer:
(d) all of the above

Question 9.
Which one is in descending order of a food chain?
(a) Producers → Secondary consumers → Primary consumers → Tertiary consumers
(b) Tertiary consumers → Primary consumers → Secondary consumers → Producers
(c) Tertiary consumers → Secondary consumers → Primary consumers → Producers
(d) Tertiary consumers → Producers → Primary consumers → Secondary consumers Significance of food web is / are
Answer:
(c) Tertiary consumers → Secondary consumers → Primary consumers → Producers

Question 10.
Significance of food web is / are __________
(a) it does not maintain stability in nature
(b) it shows patterns of energy transfer
(c) it explains species interaction
(d) b and c
Answer:
(d) b and c

Question 11.
The following diagram represents __________

(a) pyramid of number in a grassland ecosystem
(b) pyramid of number in a pond ecosystem
(c) pyramid of number in a forest ecosystem
(d) pyramid of biomass in a pond ecosystem
Answer:
(c) pyramid of number in a forest ecosystem

Question 12.
Which of the following is / are not the mechanism of decomposition?
(a) Eluviation
(b) Catabolism
(c) Anabolism
(d) Fragmentation
Answer:
(c) Anabolism

Question 13.
Which of the following is not a sedimentary cycle?
(a) Nitrogen cycle
(b) Phosphorous cycle
(c) Sulphur cycle
(d) Calcium cycle
Answer:
(a) Nitrogen cycle

Question 14.
Which of the following are not regulating services of ecosystem services?
(i) Genetic resources
(ii) Recreation and aesthetic values
(iii) Invasion resistance
(iv) Climatic regulation
(a) i and iii
(b) ii and iv
(c) i and ii
(d) i and iv
Answer:
(c) i and ii

Question 15.
Productivity of profundal zone will be low. Why?
Answer:
The producers of the pond ecosystem depends on phytoplankton through photosynthesis. Profundal zone lies below the limnetic zone with no effective light penetration, hence productivity rate is very low.

Question 16.
Discuss the gross primary productivity is more efficient than net primary productivity.
Answer:
Gross primary productivity:

1. It refers to the total amount of food energy or organic matter produced in an ecosystem by autotrophs.
2. GPP = NPP + Respiration

Net primary productivity:

1. It refers to the amount of energy that remain in autotrophs after respiration loss.
2. NPP = GPP – Respiration

Question 17.
Pyramid of energy is always upright. Give reasons.
Answer:
The energy pyramid represents a successive energy flow at each trophic level in an ecosystem. There is a gradual decrease in energy transfer at successive tropic levels from producers to higher levels, hence the pyramid of energy is always upright.

Question 18.
What will happen if all producers are removed from ecosystem?
Answer:
Producers are the autotrophs which occupy the first tropic level in an ecosystem. The energy produced by them is utilized by the herbivores and then by carnivores, thereby maintaining the stability of ecosystem. If producers are removed from an ecosystem, it would lead to starvation and death of herbivores and subsequently the carnivores, thus terminating the entire food web.

Question 19.
Construct the food chain with the following data.
Hawk, plants, frog, snake, grasshopper.
Answer:
Plants → Grasshopper → Frog → Snake → Hawk

Question 20.
Name of the food chain which is generally present in all type of ecosystem. Explain and write their significance.
Answer:
Detritus food chain is common in all type of ecosystem. In detritus food chain, the dead remains of plant and animals or their excreta are broken down by detrivores and the organic and inorganic substances are returned back to environment. Thus maintaining die company of various biogeochemical cycles. Also Microbes growing on detritus makes the soil nutritious for consumers.

Question 21.
Shape of pyramid in a particular ecosystem is always different in shape. Explain with example.
Answer:
In a forest ecosystem the pyramid of number is spindle in shape, it is because the base (T₁) of the pyramid occupies large sized trees (Producer) which are lesser in number. Herbivores (T₂) (Fruit eating birds, elephant and deer) occupying second trophic level, are more in number than the producers. In final trophic level (T₄), .tertiary consumers (lion) are lesser in number than the secondary consumer (T₃) (fox and snake).

Question 22.
Generally human activities are against to the ecosystem, where as you a student how will you help to protect ecosystem?
Answer:

1. Buying and using only ecoffiendly products and recycle them.
2. Growing more trees.
3. Choosing sustained farm products (vegetables, fruits and greens, etc.)
4. Reducing the use of natural resources.
5. Recycling the waste and reduce the amount of waste you produce.
6. Reducing consumption of water and electricity.
7. Reducing or eliminating the use of house-hold chemicals and pesticides.
8. Maintaining your cars and vehicles properly to reduce carbon emission.
9. Creating awareness and educate about ecosystem protection among your friends and family members and ask them to find out solution to minimise this problem.

Question 23.
Generally in summer the forest are affected by natural fire. Over a period of time it recovers itself by the process of successions. Find out the types of succession and explain. Secondary succession.
Answer:
The development of a plant community in an area where an already developed community has been destroyed by natural causes is known as secondary succession. This type of succession takes less time to occur.

Question 24.
Draw a pyramid from following details and explain in brief.
Answer:
Quantities of organisms are given-Hawks-50, plants-1000, rabbit and mouse-250 +250, pythons and lizard – 100 + 50 respectively

T₄ (50) T₃ (100) T₂ (550) T₁ (1000)
The pyramid produced is an upright pyramid of numbers where there is a gradual decrease in number of organisms at each trophic level from T₁ to T₄. This is an example for grassland ecosystem.

Question 25.
Various stages of succession are given bellow. From that rearrange them accordingly. Find out the type of succession and explain in detail.
Answer:
Reed-swamp stage, phytoplankton stage, shrub stage, submerged plant stage, forest stage, submergedfree floating stage and marsh medow stage.

(1) Phytoplankton stage – It is the first stage of succession consisting of the pioneer community like blue green algae, green algae, diatoms, bacteria, etc., The colonization of these organisms enrich the amount of organic matter and nutrients of pond due to their life activities and death. This favors the development of the next serai stages.

(2) Submerged plant stage – As the result of death and decomposition of planktons, silt brought from land by rain water, lead to a loose mud formation at the bottom of the pond. Hence, the rooted submerged hydrophytes begin to appear on the new substratum.
Example: Vallisneria and Hydrilla etc. The death and decay of these plants will build up the substratum of pond to become shallow.

(3) Submerged free floating stage – During this stage, the depth of the pond will become almost 2-5 feet Hence, the rooted hydrophytic plants and with floating large leaves start colonising the pond.
Example: Rooted floating plants like Nelumbo, Nymphaea and some free floating species like Azolla, and Pistia are also present in this stage. By death and decomposition of these plants, further the pond becomes more shallow.

(4) Reed-swamp stage – It is also called an amphibious stage. During this stage, rooted floating plants are replaced by plants which can live successfully in aquatic as well as aerial environment.
Example: Typha, Phragmites, Sagittaria and Scirpus etc. At the end of this stage, water level is very much reduced, making it unsuitable for the continuous growth of amphibious plants.

(5) Marsh meadow stage – When the pond becomes swallowed due to decreasing water level, species of Cyperaceae and Poaceae colonise the area. They form a mat-like vegetation with the help of their much branched root system. This leads to an absorption and loss of large quantity of water. At the end of this stage, the soil becomes dry and the marshy vegetation disappears gradually and leads to shurb stage.

(6) Shrub stage – Here areas are invaded by terrestrial plants like shrubs (Salix and Comus) and trees (Populus and Alnus). These plants absorb large quantity of water and make the habitat dry. Further, the accumulation of humus with a rich flora of microorganisms produce minerals in the soil, ultimately favouring the arrival of new tree species in the area.

(7) Forest stage – It is the climax community of hydrosere. A variety of trees invade the area and develop any one of the diverse type of vegetation.
Example.Temperate mixed forest (Ulmus,Acer and Quercus), Tropical rain forest (Artocaipus and Cinnamomum ) and Tropical deciduous forest (Bamboo and Tectona).

Samacheer Kalvi 12th Bio Botany Ecosystem Additional Questions and Answers

1 – Mark Question

Question 1.
Ecosystem is the structural and functional unit of ecology. This statement was given by ___________
(a) Tansley
(b) Odum
(c) Charles Elton
(d) Edwin
Answer:
(b) Odum

Question 2.
Identify the incorrect option among the following component sequence.
(a) air, water, sunlight and temperature
(b) latitude, altitude, direction of mountain and aptitude
(c) soil air, pH of soil, saltwater and soil moisture
(d) carbohydrate, protein, lipids and humic substances
Answer:
(b) latitude, altitude, direction of mountain and aptitude

Question 3.
Pick out the edaphic factor among the following.
(a) Rainfall
(b) Temperature
(c) Soil pH
(d) Latitude
Answer:
(c) Soil pH

Question 4.

Answer:
a -ii, b – iii, c – iv, d-i

Question 5.
Which is not a macro consumer?
(a) Herbivore
(b) Carnivore
(c) Ominivore
(d) Decomposer
Answer:
(d) Decomposer

Question 6.
Photosynthetically Active Radiation ranges between the wavelength of.
(a) 400 – 600 nm
(b) 600 – 700 nm
(c) 400 – 500 nm
(d) 400 – 700 nm
Answer:
(d) 400 – 700 nm

Question 7.
Who coined the term Ecosystem?
Answer:
A.G. Tansley

Question 8.
Identify the incorrect statement.
(a) Carbon stored in oil is referred as Grey carbon
(b) Carbon stored in atmosphere is referred as Blue carbon
(c) Carbon stored in industrialized forests is referred as Green carbon
(d) Carbon emitted from gas, died engine is referred as Black carbon
Answer:
(c) Carbon stored in industrialized forests is referred as Green carbon

Question 9.
Which group of organism occupies the third tropic level in an ecosystem?
(a) Primary consumers
(b) Secondary consumers
(c) Secondary carnivores
(d) Omnivores
Answer:
(b) Secondary consumers

Question 10.
Which is irrelevant to the first law of thermodynamics?
(a) Energy can be transmitted from one system to other in many forms.
(b) Energy transformation results in reduction of free energy.
(c) Energy can neither be created nor destroyed.
(d) Energy in the universe is constant.
Answer:
(b) Energy transformation results in reduction of free energy.

Question 11.
If 1200 Joules of solar energy is trapped by producers, how much of Joules of energy does the organism in the third trophic level will receive?
(a) 120 Joules
(b) 12 Joules
(c) 1.2 Joules
(d) 0.12 Joules
Answer:
(c) 1.2 Joules

Question 12.
Which of the following food chain is in improper sequence?
(a) Plants → snake → rabbit → lizard → eagle
(b) Plants → grasshopper → lizard → snake → hawk
(c) Plants → lizard → rabbit → snake → eagle
(d) Plants → rabbit → lizard → hawk → eagle
Answer:
(b) Plants → grasshopper → lizard → snake → hawk

Question 13.
Which one of the following is not a functional unit of an ecosystem?
(a) Productivity
(b) Conductivity
(c) Energy flow
(d) Decomposition
Answer:
(b) Conductivity

Question 14.
The upright pyramid is not a feature of. __________
(a) Pond ecosystem
(b) Grassland ecosystem
(c) Forest ecosystem
(d) Terminal ecosystem
Answer:
(a) Pond ecosystem

Question 15.
The type of ecosystem with maximum net primary productivity is __________
(a) Desert ecosystem
(b) Deciduous forest ecosystem
(c) Tropical rain forest ecosystem
(d) Grassland ecosystem
Answer:
(c) Tropical rain forest ecosystem

Question 16.
Pyramid of numbers with broad base indicates __________
(a) High population of old individuals
(b) Low population of young individuals
(c) High population of young individuals
(d) Low population of old individuals
Answer:
(b) Low population of young individuals

Question 17.
Spindle shaped pyramid is a character of __________
(a) Pond ecosystem
(b) Grassland ecosystem
(c) Parasite ecosystem
(d) Forest ecosystem
Answer:
(d) Forest ecosystem

Question 18.
Read the statement and select the correct terminology for the same:
“Carrying away of inorganic compounds of soil by water”.
(a) Eluviation
(b) Fragmentation
(c) Humification
(d) Mineralisation
Answer:
(a) Eluviation

Question 19.
Complete the food chain by filling the link X:
Paddy → Grassopper → Frog → X → Hawk
(a) King cobra
(b) Gorilla
(c) Rabbit
(d) Tasmanial wolf
Answer:
(a) King cobra

Question 20.
Which of the following is abundant in rock deposits and guano?
(a) Nitrogen
(b) Phosphorous
(c) Oxygen
(d) Calcium
Answer:
(b) Phosphorous

Question 21.
The bottom most zone of a pond is termed as
(a) Limnetic zone
(b) Littoral zone
(c) Benthic zone
(d) Profundal zone
Answer:
(c) Benthic zone

Question 22.
Observe the figures and select the correct type of pyramid of numbers

Answer:
(a) (A) Grassland ecosystem (B) Forest Echosystem (c) Parasite ecosystem

Question 23.
Lotic ecosystem refers to ___________
(a) Open water ecosystem
(b)Running water ecosystem
(c) Standing water
(d) Ocean water ecosystem
Answer:
(b)Running water ecosystem

Question 24.
Identify the correct sequence of various zones from surface to depth in a pond ecosystem.
(a) Profundal, limnetic, littoral and benthic
(b) Benthic, littoral, profundal and limnetic
(c) Limnetic, profundal, littoral and benthic
(d) Littoral, limnetic, profundal and benthic
Answer:
(d) Littoral, limnetic, profundal and benthic

Question 25.
Which type of ecosystem service does the genetic resources comes under?
(a) Provisioning services
(b) Supporting services
(c) Regulating services
(d) Cultural services
Answer:
(a) Provisioning services

Question 26.
Assertion (A): Pyramid of energy is upright.
Reason (R): During the energy transfer at successive trophic levels from producers there will be a gradual decrease
(a) Both A and R are wrong
(b) A is right R is wrong
(c) R explains A
(d) A is right R is not the correct explanation for A
Answer:
(c) R explains A

Question 27.
Assertion (A): In forest ecosystem, the pyramid of number is spindle shaped.
Reason (R): Tropical level (T₁) of the pyramid occupies large trees which are maxium in number
(a) Both A and R are wrong
(b) A is right R is wrong
(c) R explains A
(d) A is right R is not the correct explanation for A
Answer:
(b) A is right R is wrong

Question 28.
Species that indicate the health of the ecosystem are called as __________
Answer:
Flagship species

Question 29.
Succession initiating on a sand referred as
(a) Hydrosere
(b) Psammosere
(c) Halosere
(d) Lithosere
Answer:
(b) Psammosere

Question 30.
Match the column I with column II

Answer:
(a) a – iii, b – iv, c – i, d – ii

Question 31.
Statement (I): Allogenic succession occurs as a result of abiotic factors.
Statement (II): Autogenic succession occurs as result of biotic factors.
(a) Statement I is correct; Statement II is incorrect.
(b) Statement I is incorrect; Statement II is correct.
(c) Both Statements I and II are correct.
(d) Both Statements I and II are incorrect.
Answer:
(c) Both Statements I and II are correct.

Question 32.
Statement (I): The first invaded plants in a barren area are called as pioneers.
Statement (II): Marsh meadow stage of hydrosere succession is also called as amphibious stage.
(a) Statement I is correct; Statement II is incorrect.
(b) Statement I is incorrect; Statement II is correct.
(c) Both Statements I and II are correct.
(d) Both Statements I and II are incorrect.
Answer:
(a) Statement I is correct; Statement II is incorrect.

Question 33.
_______ is the climax community of hydrosere.
(a) Reed swamp stage
(b) Marsh medow stage
(c) Shrub stage
(d) Forest stage
Answer:
(d) Forest stage

2-Mark Questions 

Question 1.
According to A.G. Tansley, what is an ecosystem?
Answer:
A.G. Tansley (1935), who defined ecosystem as ‘the system resulting from the integration of all the living and non-living factors of the environment.

Question 2.
Mention any two climatic factors and edaphic factors of an ecosystem.
Answer:

1. Climatic factors: Light and Air.
2. Edaphic factors: Soilwaer and pH of soil.

Question 3.
Define ‘Standing state’ with regard to ecosystem.
Answer:
The total inorganic substances present in any ecosystem at a given time is called standing quality (or) standing state.

Question 4.
What are biotic components?
Answer:
Biotic (living) components includes all living organisms like plants, animals, fungi and bacteria. They form the trophic structures of any ecosystem.

Question 5.
Name the macro consumers and micro consumers.
Answer:

1. Macro consumers are herbivores, carnivores and omnivores.
2. Micro consumers are decomposers.

Question 6.
How will you define decomposers?
Answer:
Decomposers are organisms that decompose the dead plants and animals to release organic and inorganic nutrients into the environment which are again reused by plants.
Example: Bacteria, Actinomycetes and Fungi.

Question 7.
What is standing crop?
Answer:
The amount of living materials present in a population at any given time is known as standing crop, which may be expressed in terms of number or biomass per unit area.

Question 8.
What do you mean by PAR? Mention its significance.
Answer:
The amount of light available for photosynthesis of plants is called Photosynthetically Active Radiation (PAR) which is between the range of 400-700 mm wave length.

Question 9.
Pointout the factors that affects the photosynthetically active radiation.
Answer:
PAR is not always constant because of clouds, tree shades, air, dust particles, seasons, latitudes and length of the daylight availability.

Question 10.
Define Grey carbon and Black carbon.
Answer:
Grey carbon – carbon stored in fossil fuel (coal, oil and biogas deposits in the lithosphere). Black carbon – carbon emitted from gas, diesel engine and coal fired power plants.

Question 11.
What is meant by productivity of an ecosystem?
Answer:
The rate of biomass production per unit area in a unit time is called productivity. It can be expressed in terms of gm /m²/year or Kcal/m²/ year.

Question 12.
How Net Primary Productivity can be derived?
Answer:
Net Primary Productivity (NPP) is derived by the difference between Gross primary productivity (GPP) and respiration.
NPP = GPP – Respiration

Question 13.
Expand GPP and define it.
Answer:
Gross Primary Productivity (GPP) is the total amount of food energy or organic matter or biomass produced in an ecosystem by autotrophs through the process of photosynthesis is called gross primary productivity.

Question 14.
Write the name of four important aspects of ecosystem.
Answer:

1. Productivity
2. Energy flow
3. Decomposition
4. Nutrient cycling

Question 15.
State the role of herbivores and microconsumers in a terrestrial ecosystem.
Answer:

1. Herbivores acts as primary consumers of producers (Plants).
2. Microconsumers decomposes the dead remains and excreta of plants and animals.

Question 16.
Name the category of organisms that occupy the first tropic level (T₁) and fourth tropic level (T₄) in an ecosystem.
Answer:
Producers (autotrophs) occupy the first tropic level (T₄) whereas the fourth tropic level (T₄) is occupied by Tertiary Consumers (Secondary carnivore).

Question 17.
What is energy flow?
Answer:
The transfer of energy in an ecosystem between trophic levels can be termed as energy flow. It is the key function in an ecosystem. Energy flow is always unidirectional in an ecosystem.

Question 18.
State the first law of thermodynamics.
Answer:
First law of thermodynamics states that energy can be transmitted from one system to another in various forms. Energy cannot be destroyed or created. But it can be transformed from one form to another. As a result, the quantity of energy present in the universe is constant.

Question 19.
State the ten percent law.
Answer:
Ten percent law states that during transfer of food energy from one trophic level to other, only about 10% stored at every level and rest of them (90%) is lost in respiration, decomposition and in the form of heat.

Question 20.
Define food chain.
Answer:
The movement of energy from producers upto top carnivores is known as food chain, i.e., in any food chain, energy flows from producers to primary consumers, then from primary consumers to secondary consumers, and finally secondary consumers to tertiary consumers. Hence, it shows linear network links.

Question 21.
Name the two types of food chain.
Answer:

1. Grazing food chain
2. Detritus food chain

Question 22.
Which is the first link in a grazing food chain and a detritus food chain?
Answer:

1. The first link of grazing food chain is plants (Producers).
2. The first link of detritus food chain is dead remains and excreta of plants and animals.

Question 23.
Rearrange the components of the ecosystem and frame a food chain. Also mention the type of food chain.
Hawk, Earthworm, Animal excreta and Black bird.
Answer:
Animal excreta → Earthworm → Black bird → Hawk. It is a detritus food chain.

Question 24.
Define food web.
Answer:
The inter-locking pattern of a number of food chain form a web like arrangement called food web. It is the basic unit of an ecosystem, to maintain its stability in nature.

Question 25.
What is Eltoian pyramid?
Answer:
Eltonian pyramid or Ecological pyramid is a graphic representation of the trophic structure and function at successive trophic levels of an ecosystem.

Question 26.
Why do we obtain an inverted pyramid in a parasite ecosystem?
Answer:
The pyramid of number in a parasite ecosystem is always inverted, because it starts with a single tree. Therefore there is gradual increase in the number of organisms in successive tropic levels from producer to tertiary consumers.

Question 27.
What is pyramid of biomass?
Answer:
A graphical representation of the amount of organic material (biomass) present at each successive trophic level in an ecosystem is called pyramid of biomass.

Question 28.
Biogeochemical cycle comprises both gaseous cycle and sedimentary cycle. How they differ from one another?
Answer:

1. The components of gaseous cycle are placed in atmosphere
   E.g: Oxygen in air
2. whereas in sedimentary cycle, the components are present in/on Earth.
   E.g: Phosphorous in rocks.

Question 29.
Name the type of ecosystem that exhibits the following types of pyramid.

1. Inverted pyramid of biomass
2. Spindle shaped pyramid of number

Answer:

1. Pond ecosystem
2. Forest ecosystem.

Question 30.
Cite few examples of biomolecules that contain phosphorus.
Answer:
DNA, RNA, ATP and NADP.

Question 31.
What are Blue carbon ecosystems?
Answer:
Sea grasses and mangroves of Estuarine and coastal ecosystems are the most efficient in carbon sequestration. Hence, these ecosystems are called as “Blue carbon ecosystems”.

Question 32.
Mention the four categories of ecosystem services.
Answer:

1. Provisioning services
2. Cultural services
3. Supporting services
4. Regulating services

Question 33.
What is meant by Ecosystem resilience?
Answer:
Ecosystem is damaged by disturbances from fire, flood, predation, infection and drought, etc. removing a great amount of biomass. However, ecosystem is endowed with the ability to resist the damage and recover quickly. This ability of ecosystem is called ecosystem resilience or ecosystem robustness.

Question 34.
Ecosystem management – comment on the statement.
Answer:
Ecosystem management is a process that integrates ecological, socio economic and institutional factors into a comprehensive strategy in order to sustain and enhance the quality of the ecosystem to meet current and future needs

Question 35.
Which kind of organisms constitute the pioneer community and climax community of a Hydrosere succession?
Answer:
Phytoplanktons from the poineer community and a variety of trees makes the climaxe community of a Hydrosere

Question 36.
Define plant succession.
Answer:
Successive replacement of one type of plant community by the other of the same area/ place is known as plant succession.

Question 37.
Define

1. Hydrosere
2. Xerosere

Answer:

1. Hydrosere: Succession of plants in a freshwater ecosystem.
2. Xerosere: Succession of plants in areas with minimal amount of water.

3 – Mark Questions

Question 38.
What is secondary productivity? Explain its types.
Answer:
The amount of energy stored in the tissues of heterotrophs or consumers is called secondary productivity.

1. Gross secondary productivity: It is equivalent to the total amount of plant material ingested by the herbivores minus the materials lost as faces.
2. Net secondary productivity : Storage of energy or biomass by consumers per unit area per unit time, after respiratory loss is called net secondary productivity.

Question 39.
List the factors affecting primary productivity.
Answer:
Primary productivity depends upon the plant species of an area, their photosynthetic capacity, availability of nutrients, solar radiation, precipitation, soil type, topographic factors (altitude, latitude and direction), and other environmental factors.

Question 40.
Give an account on the concept of trophic level in an ecosystem.
Answer:
A trophic level refers to the position of an organism in the food chain. The number of trophic levels is equal to the number of steps in the food chain. The green plants (producers) occupying the first trophic level (T₁) are called producers. The energy produced by the producers is utilized by the plant eaters (herbivores) they are called primary consumers and occupies the second trophic level (T₂).

Herbivores are eaten by carnivores, which occupy the third trophic level (T₃). They are also called secondary consumers or primary carnivores. Carnivores are eaten by the other carnivores, which occupy the fourth trophic level (T₄). They are called the tertiary consumers or secondary carnivores. Some organisms which eat both plants and animals are called as omnivores (Crow). Such organisms may occupy more than one trophic level in the food chain.

Question 41.
State the second law of thermodynamics.
Answer:
Second law of thermodynamics states that energy transformation results in the reduction of the free energy of the system. Usually energy transformation cannot be 100% efficient. As energy is transferred from one organism to another in the form of food, a portion of it is stored as energy in living tissue, whereas a large part of energy is dissipated as heat through respiration. The transfer of energy is irreversible natural process.

Question 42.
Explain Grazing food chain with example.
Answer:
Main source of energy for the grazing food chain is the Sun. It begins with the first link, producers (plants). The second link in the food chain is primary consumers (mouse) which get their food from producers. The third link in the food chain is secondary consumers (snake) which get their food from primary consumers. Fourth link in the food chain is tertiary consumers (eagle) which get their food from secondary consumers.
Grass → Mouse → Snake → Eagle
Producers Primary consumers Secondary Consumers Tertiary consumers

Question 43.
Write a brief note on Detritus food chain.
Answer:
Detritus food chain is a type of food chain which begins with dead organic matter which is an important source of energy. A large amount of organic matter is derived from the dead plants, animals and their excreta. This type of food chain is present in all ecosystems. The transfer of energy from the dead organic matter, is transferred through a series of organisms called detritus consumers (detritivores)- small carnivores – large (top) carnivores with repeated eating and being eaten respectively. This is called the detritus food chain.
Fallen leaves → Earthworm → Blackbird → Hawk
Detritus Detritivores Small carnivores Top carnivores

Question 44.
Enumerate the significance of food webs.
Answer:
Significance of food web

1. Food web is constructed to describe species interaction called direct interaction.
2. It can be used to illustrate indirect interactions among different species.
3. It can be used to study bottom-up or top-down control of community structure.
4. It can be used to reveal different patterns of energy transfer in terrestrial and aquatic ecosystems.

Question 45.
Name the three types of ecological pyramids.
Answer:

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

Question 46.
Spindle shaped pyramid of number is noticed in forest ecosystem. Give reasons.
Answer:
In a forest ecosystem the pyramid of number is spindle in shape, it is because the base (T₁) of the pyramid occupies large sized trees (Producer) which are lesser in number. Herbivores (T₂) (Fruit eating birds, elephant and deer) occupying second trophic level, are more in number than the producers. In final trophic level (T₄), tertiary consumers (lion) are lesser in number than the secondary consumer (T₃) (fox and snake).

Question 47.
Pyramid of energy is always upright – Justify.
Answer:
A graphical representation of energy flow at each successive trophic level in an ecosystem is called pyramids of energy. The bottom of the pyramid of energy is occupied by the producers. There is a gradual decrease in energy transfer at successive tropic levels from producers to the upper levels. Therefore, the pyramid of energy is always upright.

Question 48.
What does the term ‘Eluviation’ stands for?
Answer:
The movement of decomposed, water soluble organic and inorganic compounds from the surface to the lower layer of soil or the carrying away of the same by water is called leaching or eluviation.

Question 49.
What is biogeochemical cycle? Explain its types.
Answer:
Circulation of nutrients within the ecosystem or biosphere is known as biogeochemical cycles and also called as cycling of materials.
There are two basic types:

1. Gaseous cycle – It includes atmospheric Oxygen, Carbon and Nitrogen cycles.
2. Sedimentary cycle – It includes the cycles of Phosphorus, Sulphur and Calcium – Which are present as sediments of Earth.

Question 50.
Explain the cycling of phosphorus in an ecosystem.
Answer:
Phosphorus cycle is a type of sedimentary cycle in which phosphorus is found in the biomolecules like DNA, RNA, ATP, NADP and phospholipid molecules of living organisms. Phosphorus is not abundant in the biosphere, whereas a bulk quantity of phosphorus is present in rock deposits, marine sediments and guano. It is released from these deposits by weathering process.

After that, it circulates in lithosphere as well as hydrosphere. The producers absorb phosphorus in the form of phosphate ions, and then it is transferred to each trophic level of food chain through food. Again death of the organisms and degradation by the action of decomposers, the phosphorus is released back into the lithosphere and hydrosphere to maintain phosphorus cycle.

Question 51.
Discuss the three zones of a lentic ecosystem.
Answer:
There are three zones, littoral, limnetic and profundal. The littoral zone, which is closest to the shore with shallow water region, allows easy penetration of light. It is warm and occupied by rooted plant species. The limnetic zone refers the open water of the pond with an effective penetration of light and domination of planktons.

The deeper region of a pond below the limnetic zone is called profundal zone with no effective light penetration and predominance of heterotrophs. The bottom zone of a pond is termed benthic and is occupied by a community of organisms called benthos (usually decomposers).

Question 52.
What is ecosystem services? Why it is of much importance?
Answer:
Ecosystem services are defined as the benefits that people derive from nature. Study on ecosystem services acts as an effective tool for gaining knowledge on ecosystem benefits and their sustained use. Without such knowledge gain, the fate of any ecosystem will be at stake and the benefits they provide to us in future will become bleak.

Question 53.
Point out any three mangrove ecosystem services.
Answer:

1. Act as bridge between sea and rivers by balancing sedimentation and soil erosion.
2. Help to reduce water force during cyclones, tsunamis and high tide periods.
3. Help in wind break, O₂ production, carbon sequestration and prevents salt spray from waves.

Question 54.
What are the human activities that disturb an ecosystem?
Answer:

1. Habitat destruction.
2. Deforestation and over grazing.
3. Erosion of soils.
4. Introduction of non-native species.
5. Over harvesting of plant material.
6. Pollution of land, water and air.
7. Rim off pesticides, fertilizers and animal wastes.

Question 55.
What is primary succession?
Answer:
The development of plant community in a barren area where no community existed before is called primary succession. The plants which colonize first in a barren area is called pioneer species or primary community or primary colonies. Generally, Primary succession takes a very long time for the occurrence in any region.
Example: Microbes, Lichen and Mosses.

5 – Mark Questions

Question 56.
Describe the various stages of decomposition process.
Answer:

1. Fragmentation – The breaking down of detritus into smaller particles by detritivores like bacteria, fungi and earth worm is known as fragmentation. These detritivores secrete certain substances to enhance the fragmentation process and increase the surface area of detritus particles.
2. Catabolism – The decomposers produce some extracellular enzymes in their surroundings to break down complex organic and inorganic compounds in to simpler ones. This is called catabolism
3. Leaching or Eluviation – The movement of decomposed, water soluble organic and inorganic compounds from the surface to the lower layer of soil or the carrying away of the same by water is called leaching or eluviation.
4. Humification – It is a process by which simplified detritus is changed into dark coloured amorphous substance called humus. It is highly resistant to microbial action, therefore
   decomposition is very slow. It is the reservoir of nutrients.
5. Mineralisation – Some microbes are involved in the release of inorganic nutrients from the humus of the soil, such process is called mineralisation.

Question 57.
Give a detailed account of Biotic and abiotic components of a pond ecosystem. Abiotic components
Answer:
A pond ecosystem consists of dissolved inorganic (CO₂, O₂, Ca, N and Phosphate) and organic substances (amino acids and humic acid) formed from the dead organic matter. The function of pond ecosystem is regulated by few factors like the amount of light, temperature, pH value of water and other climatic conditions.

Biotic components:
They constitute the producers, variety of consumers and decomposers (microorganisms).

(a) Producers: A variety of phytoplanktons like Oscillatoria, Anabaena, Eudorina, Volvox and Diatoms. Filamentous algae such as Ulothrix, Spirogyra, Cladophora and Oedogonium; floating plants Azolla, Salvia, Pistia, Wolffia and Eichhornia; sub-merged plants Potamogeton and Phragmitis; rooted floating plants Nymphaea and Nelumbo; macrophytes like Typha and Ipomoea, constitute the major producers of a pond ecosystem.

(b) Consumers: The animals represent the consumers of a pond ecosystem include zooplanktons like Paramoecium and Daphnia (primary consumers); benthos (bottom living animals) like molluscs and annelids; secondary consumers like water beetles and frogs; and tertiary consumers (carnivores) like duck, crane and some top carnivores which include large fish, hawk and man, etc.

(c) Decomposers: They are also called as microconsumers. They help to recycle the nutrients in the ecosystem. These are present in mud water and bottom of the ponds. Example: Bacteria and Fungi. Decomposers perform the process of decomposition in order to enrich the nutrients in the pond ecosystem.

Question 58.
What are the strategies of eco system management?
Answer:
Strategy of ecosystem management

1. It is used to maintain biodiversity of ecosystems.
2. It helps in indicating the damaged ecosystem (Some species indicate the health of the ecosystem: such species are called a flagship species).
3. It is used to recognize the inevitability of ecosystem change and plan accordingly.
4. It is one of the tools used for achieving sustainability of ecosystem through sustainable development programme (or projects).
5. It is also helpful in identifying ecosystems which are in need of rehabilitation.
6. It involves collaborative management with government agencies, local population, communities and NGO’s.
7. It is used to build the capacity of local institutions and community groups to assume responsibility for long term implementation of ecosystem management activities even after, the completion of the project.

Question 59.
List the characteristics of ecological succession.
Answer:

1. It is a systematic process which causes changes in specific structure of plant community.
2. It is resultant of changes of abiotic and biotic factors.
3. It transforms unstable community into a stable community.
4. Gradual progression in species diversity, total biomass, niche specialisation, and humus content of soil takes place.
5. It progresses from simple food chain to complex food web.
6. It modifies the lower and simple life form to the higher life forms.
7. It creates inter-dependence of plants and animals.

Question 60.
Differentiate Primary succession and Secondary succession
Answer:
Primary succession:

1. Developing in an barren area.
2. Initiated due to a biological or any other external factors.
3. No soil, while primary succession starts
4. Pioneer species come from outside environment.
5. It takes more time to complete.

Secondary succession:

1. Developing in disturbed area.
2. Starts due to external factors only.
3. It starts where soil covers is already present.
4. Pioneer species develop from existing environment.
5. It takes comparatively less time to complete.

Question 61.
Write in detail about Autogenic succession and Allogenic succession.
Answer:
Autogenic succession
Autogenic succession occurs as a result of biotic factors. The vegetation reacts with its environment and modifies its own environment causing its own replacement by new communities. This is known as autogenic succession.
Example: In forest ecosystem, the larger trees produce broader leaves providing shade to the forest floor area. It affects the shrubs and herbs which require more light (heliophytes) but supports the shade tolerant species (sciophytes) to grow well.

Allogenic succession:
Allogenic succession occurs as a result of abiotic factors. The replacement of existing community is caused by other external factors (soil erosion and leaching, etc) and not by existing organisms.
Example: In a forest ecosystem soil erosion and leaching alter the nutrient value of the soil leading to the change of vegetation in that area.

Question 62.
What are the significance of plant succession?
Significance of Plant Succession
Answer:

1. Succession is a dynamic process. Hence an ecologist can access and study the serai stages of a plant community found in a particular area.
2. The knowledge of ecological, succession helps to understand the controlled growth of one or more species in a forest.
3. Utilizing the knowledge of succession, even dams can be protected by preventing siltation.
4. It gives information about the techniques to be used during reforestation and afforestation.
5. It helps in the maintenance of pastures.
6. Plant succession helps to maintain species diversity in an ecosystem.
7. Patterns of diversity during succession are influenced by resource availability and disturbance by various factors.
8. Primary succession involves the colonization of habitat of an area devoid of life.
9. Secondary succession involves the re-establishment of a plant community in disturbed area or habitat.
10. Forests and vegetation that we come across all over the world are the result of plant succession.

Higher Order Thinking Skills (HOTs) Questions

Question 1.
Nutrients are retained in the ecosystem. Discuss how and name the process through which it is achieved?
Answer:
Nutrients will retain in the ecosystem through biogeo chemical cycles. Various nutrients present in soil and atmosphere enter the organism at various trophic levels through food and other process and are cycled back to their origin. Thus the nutrients are not lost but retained in the ecosystem.

Question 2.
According to ten percent law, how many Joules of energy does the individuals at the fourth tropic level will receive; if the individuals at first trophic level receives 1000 Joules of energy?
Answer:
Organism at T₄ level will receive 0.1 Joule of energy as per ten percent law.

Question 3.
Frame any two food chain patterns from the given organisms. Each chain must contain a minimum of four organisms.
Diatoms, Hawk, Rabbit, Vallisneria Stoke, guppies, grass, snake, large fishes, grasshopper, crane.
Answer:

1. Diatom → Guppies → Large fishes → Crane
2. Grass → Rabbit → Snake → Hawk

Question 4.
Mention any two vital biomolecules that requires phosphorous for their biosynthesis.
Answer:
DNA and ATP

Question 5.
The below diagramatic sketch shows the stratisfication of pond ecosystem. Considering it, name X and Y.
Answer:

Diagrammatic sketch shows strarification of Pond ecosystem
X – Limnetic Zone
Y – Benthic Zone

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Samacheer Kalvi 12th Bio Botany Plant Tissue Culture Text Book Back Questions and Answers

Choose the correct answer from the given option

Question 1.
Totipotency refers to
(a) capacity to generate genetically identical plants
(b) capacity to generate a whole plant from any plant cell / explant
(c) capacity to generate hybrid protoplasts
(d) recovery of healthy plants from diseased plants
Answer:
(b) capacity to generate a whole plant from any plant cell / explant

Question 2.
Micro propagation involves
(a) vegetative multiplication of plants by using micro-organisms
(b) vegetative multiplication of plants by using small explants
(c) vegetative multiplication of plants by using microspores
(d) Non-vegetative multiplication of plants by using microspores and megaspores
Answer:
(b) vegetative multiplication of plants by using small explants

Question 3.
Match the following:

(a) C A D B
(b) A C B D
(c) B A D C
(d) D B C A
Answer:
(a) C A D B

Question 4.
The time duration for sterilization process by using autoclave is ___________ minutes and the temperature is ___________
(a) 10 to 30 minutes and 125° C
(b) 15 to 30 minutes and 121° C
(c) 15 to 20 minutes and 125° C
(d) 10 to 20 minutes and 121° C
Answer:
(b) 15 to 30 minutes and 121° C

Question 5.
Which of the following statement is correct?
(a) Agar is not extracted from marine algae such as seaweeds
(b) Callus undergoes differentiation and produces somatic embryoids
(c) Surface sterilization of explants is done by using mercuric bromide
(d) PH of the culture medium is 5.0 to 6.0
Answer:
(d) PH of the culture medium is 5.0 to 6.0

Question 6.
Select the incorrect statement from given statements:
(a) A tonic used for cardiac arrest is obtained from Digitalis purpuria
(b) Medicine used to treat Rheumatic pain is extracted from Capsicum annum
(c) An anti-malarial drug is isolated from Cinchona officinalis
(d) Anti-carcinogenic property is not seen in Catharanthus roseus.
(e) All the above are correct
Answer:
(e) All the above are correct

Question 7.
Vims free plants are developed from
(a) Organ culture
(b) Meristem culture
(c) Protoplast culture
(d) Cell suspension culture
Answer:
(b) Meristem culture

Question 8.
The prevention of large scale loss of biological integrity
(a) Biopatent
(b) Bioethics
(c) Biosafety
(d) Biofuel
Answer:
(c) Biosafety

Question 9.
Cryopreservation means it is a process to preserve plant cells, tissues or organs
(a) at very low temperature by using ether
(b) at very high temperature by using liquid nitrogen
(c) at very low temperature of-196 by using liquid nitrogen
(d) at very low temperature by using liquid nitrogen
Answer:
(c) at very low temperature of-196 by using liquid nitrogen

Question 10.
Solidifying agent used in plant tissue culture is
(a) Nicotinic acid
(b) Cobaltous chloride
(c) EDTA
(d) Agar
Answer:
(d) Agar

Question 11.
What is the name of the process given below? Write its 4 types.
Answer:

Question 12.
How will you avoid the growing of microbes in nutrient medium during culture process? What are the techniques used to remove the microbes?
Answer:
The microbial growth in culture medium can be overcome by autoclaving the medium at Plant Tissue Culture II 121°C (15 psi) for 15 to 30 minutes.

Chemical sterilization using chemicals, sterilizing using UV radiation. Alcoholic sterilization using ethanol, autoclaving and filtration etc., are the various techniques used to remove microbes.

Question 13.
Write the various steps involved in cell suspension culture.
Answer:
Step 1: Growing of cells/callus in medium (Single or aggregates).
Step 2 : Transfer of callus to a liquid medium.
Step 3: Agitation of callus using rotary shaker.
Step 4: Filtration and separation of cells.

Question 14.
What do you mean by Embryoids? Write its application.
Answer:
Somatic embryogenesis is the formation of embryos from the callus tissue directly and these embryos are called Embryoids or from the in vitro cells directly form pre-embryonic cells which differentiate into embryoids.
Applications:

1. Somatic embryogenesis provides potential plantlets which after hardening period can establish into plants.
2. Somatic embryoids can be used for the production of synthetic seeds.
3. Somatic embryogenesis is now reported in many plants such as Allium sativum, Hordeum vulgare, Oryza sativa, Zea mays and this is possible in any plant.

Question 15.
Give the examples for micro propagation performed plants.
Answer:
Pineapple, banana, strawberry and potato.

Question 16.
Explain the basic concepts involved in plant tissue culture.
Answer:
Basic concepts of plant tissue culture are totipotency, differentiation, dedifferentiation and redifferentiation.

1. Totipotency: The property of live plant cells that they have the genetic potential when cultured in nutrient medium to give rise to a complete individual plant.

2.Differentiation: The process of biochemical and structural changes by which cells become specialized in form and function.

3. Redifferentiation: The further differentiation of already differentiated cell into another type of cell. For example, when the component cells of callus have the ability to form a whole plant in a nutrient medium, the phenomenon is called redifferentiation.

4. Dedifferentiation: The phenomenon of the reversion of mature cells to the meristematic state leading to the formation of callus is called dedifferentiation. These two phenomena of redifferentiation and dedifferentiation are the inherent capacities of living plant cells or tissue. This is described as totipotency.

Question 17.
Based on the material used, how will you classify the culture technology? Explain it.
Answer:
Based on explants used culture technology are of following types:

1. Organ culture – Embryos, anthers, root and shoot part are used.
2. Meristem culture – Meristematic tissues are used.
3. Protoplast culture – Protoplasts are used.
4. Cell culture – Single cells or aggregate of cells from callus are used.

Question 18.
Give an account on Cryopreservation.
Answer:
Cryopreservation, also known as Cryo-conservation, is a process by which protoplasts, cells, tissues, organelles, organs, extracellular matrix, enzymes or any other biological materials are subjected to preservation by cooling to very low temperature of-196°C using liquid nitrogen. At this extreme low temperature any enzymatic or chemical activity of the biological material will be totally stopped and this leads to preservation of material in dormant status.

Later these materials can be activated by bringing to room temperature slowly for any experimental work. Protective agents like dimethyl sulphoxide, glycerol or sucrose are added before cryopreservation process. These protective agents are called cryoprotectants, since they protect the cells, or tissues frofn the stress of freezing temperature.

Question 19.
What do you know about Germplasm conservation? Describe it.
Answer:
Germplasm conservation refers to the conservation of living genetic resources like pollen, seeds or tissue of plant material maintained for the purpose of selective plant breeding, preservation in live condition and used for many research works.

Germplasm conservation resources is a part of collection of seeds and pollen that are stored in seed or pollen banks, so as to maintain their viability and fertility for any later use such as hybridization and crop improvement. Germplasm conservation may also involve a gene bank, DNAbank of elite breeding lines of plant resources for the maintenance of biological diversity and also for food security.

Question 20.
Write the protocol for artificial seed preparation.
Answer:

Later these seeds are grown in vitro medium and converted into plantlets. These plantlets require a hardening period (either green house or hardening chamber) and then shifted to normal environment condition.

Samacheer Kalvi 12th Bio Botany Plant Tissue Culture Additional Questions and Answers

1 – Mark Questions

Question 1.
_________ is regarded as the Father of tissue culture.
Answer:
Gottlieb Haberland

Question 2.
Identify the group of scientists who developed the intergenic hybrid – the pomato.
(a) Yamada et al.
(b) Horsh et al.
(c) Takebe et al.
(d) Melchers et al.
Answer:
(d) Melchers et al.

Questiom 3.
The production of secondary metabolites require the use of ___________
(a) Protoplast culture
(b) Organ culture
(c) Cell suspension culture
(d) Virus free germ culture
Answer:
(c) Cell suspension culture

Question 4.
Which of the following condition favours callus induction?
(a) Temperature of 25°C ± 5°C with 12 hours of photoperiod
(b) Temperature of 25°C ± 2°C with 18 hours of photoperiod
(c) Temperature of 25°C + 5°C with 14 hours of photoperiod
(d) Temperature of 25°C ± 2°C with 16 hours of photoperiod
Answer:
(d) Temperature of 25°C ± 2°C with 16 hours of photoperiod

Question 5.
Protoplast are the cells devoid of ___________
(a) Cell wall
(b) Cell membrane
(c) Plasma membrane
(d) both A and B
Answer:
(a) Cell wall

Question 6.
A widely used fusogen in protoplast culture is ___________
(a) Polymethyl glycol
(b) Polyethylene glycol
(c) Polyethylene chloride
(d) Polyvinyl chloride
Answer:
(b) Polyethylene glycol

Question 7.
Source of agar is ___________
Answer:
Marine algae (Sea weeds)

Question 8.
Synseeds are developed by encapsulating embryoids with ___________
(a) Sodium chloride
(b) Potassium iodide
(c) Sodium alginate
(d) Potassium dichromate
Answer:
(c) Sodium alginate

Question 9.
The optimal pH of culture medium is generally ___________
(a) Acidic
(b) Basic
(c) Neutral
(d) Slightly basic
Answer:
(a) Acidic

Question 10.
Identity the correct sequence regarding steps involved in PTC
(a) Sterilization → Incubation → Inoculation → Embryogenesis → Hardening
(b) Inoculation → Induction →Sterilization → Hardening → Embryogenesis
(c) Induction → Incubation → Inoculation → Hardening → Sterilization
(d) Sterilization → Inoculation → Incubation → Embryogenesis → Hardening
Answer:
(d) Sterilization → Inoculation → Incubation → Embryogenesis → Hardening

Question 11.
Dimethyl sulphoxide is a ___________
(a) Solidifying agent
(b) Cryoprotectant
(c) Fusogenic agent
(d) Stimulant
Answer:
(b) Cryoprotectant

Question 12.
Assertion (A) : Incubation is followed by Inoculation.
Reason (R) : Explant is inoculated to media.
(a) Both A and R are correct but R is not a correct explanation to A
(b) R explains A
(c) A is correct R is incorrect
(d) Both A and R are incorrect
Answer:
(b) R explains A

Question 13.
Assertion (A) : Sterilization helps to overcome microbes.
Reason (R) : Explants are autoclaved.
(a) Both A and R are correct but R is not a correct explanation to A
(b) R explains A
(c) A is correct R is incorrect
(d) Both A and R incorrect
Answer:
(c) A is correct R is incorrect

Question 14.
Assertion (A) : Protoplasts are cells devoid of cell wall.
Reason (R) : Secondary metabolites are synthesized by protoplasmic fusion.
(a) Both A and R are correct but R is not a correct explanation to A
(b) R explains A
(c) A is correct R is incorrect
(d) Both A and R are incorrect
Answer:
(c) A is correct R is incorrect

Question 15.
Assertion (A) : Development of root from callus is called caulogenesis.
Reason (R) : Caulogenesis is the final step of protoplasmic fusion.
(a) Both A and R are correct but R is not a correct explanation to A
(b) R explains A
(c) A is correct R is incorrect
(d) Both A and R are incorrect
Answer:
(d) Both A and R are incorrect

Question 16.
Assertion (A) : Liquid nitrogen is used in cryopreservation techniques.
Reason (R) : Gene bank DNA bank are the parts of germplasm conservation.
(a) Both A and R are correct but R is not a correct explanation to A
(b) R explains A
(c) A is correct R is incorrect
(d) Both A and R are incorrect
Answer:
(a) Both A and R are correct but R is not a correct explanation to A

Question 17.
Identify the cryoprotectant
(a) Dimethyl formamide
(b) Fructose
(c) Glycerol
(d) Sodium alginate
Answer:
(c) Glycerol

Question 18.
Identify the wrong statement:
(a) Artificial seeds are stored for long time under cryopreservation
(b) Somatic embryos are used for artificial seed production
(c) Period of dormancy of artificial seeds is greatly reduced
(d) Encapsulation of embryoids is done using cryoprotectant
Answer:
(d) Encapsulation of embryoids is done using cryoprotectant

Question 19.
Identify the plant tissue used for virus free germplasm
(a) Apical meristem
(b) Intercalary meristem
(c) Lateral meristem
(d) Plate meristem
Answer:
(a) Apical meristem

Question 20.
Match the following:
(A) Solidifying agent (i) Sucrose
(B) Cryoprotectant (ii) PEG
(C) Growth hormone (iii) Agar
(D) Fusogen (iv) IAA
(1) A – iii B-i C – iv D – ii
(2) A – ii B – iv C – iii D – i
(3) A – iv B – ii C-i D – iii
(4) A – i B – iii C -ii D – iv
Answer:
(1) A – iii B – i C – iv D – ii

Question 21.
Identify the incorrect statement:
(a) Explants are surface sterilized
(b) Nutrient media are autoclaved
(c) Culture rooms are UV radiated for 15 minutes
(d) Glasswares and accessories are autoclaved
(a) a only (b) b and c (c) d only (d) none of the avove
Answer:
(d) none of the above

Question 22.
The enzymatic mixture for chemical isolation of protoplast is
(a) 0.5% macrozyme, 2% onozuka cellulase, 13% mannitol
(b) 1.5% macrozyme, 0.5% onozuka cellulase, 12% sorbitol
(c) 2% macrozyme, 0.5% onozuka cellulase, 13% sorbitol
(d) 0.1% macrozyme, 2% onozuka cellulase, 15% mannitol
Answer:
(a) 0.5% macrozyme, 2% onozuka cellulase, 13% mannitol

Question 23.
The term used to define the ability of a cell to generate entire individual is
(a) Pleuripotent
(b) Totipotent
(c) Multipotent
(d) Unipotent
Answer:
(b) Totipotent

Question 24.
The phenomenon of reversion of mature cells to meristematic state leading to callus ___________
formation is
(a) Redifferentiation
(b) Dedifferentiation
(c) either (a) or (b)
(d) none of these
Answer:
(b) Dedifferentiation

Question 25.
Somatic hybridization is achieved through ___________
(a) Protoplast fusion
(b) r-DNA technology
(c) Transformation
(d) Grafting
Answer:
(a) Protoplast fusion

Question 26.
Identify the mismatched pair:
(a) Digoxin – Digitalis purpuria
(b) Codeine – Capsicum annum
(c) Vincristine – Catharanthus roseus
(d) Quinine – Cinchona officinalis
Answer:
(b) Codeine – Capsicum annum

2 – Marks Questions

Question 1.
Define tissue culture.
Answer:
Growing plant protoplasts, cells, tissues or organs away from their natural or normal environment, under artificial condition, is known as Tissue Culture.

Question 2.
Name the four basic concepts of plant tissue culture.
Answer:
Basic concepts of plant tissue culture are totipotency, differentiation, dedifferentiation and redifferentiation.

Question 3.
What is the term totipotency refers to?
Answer:
The property of live plant cells that they have the genetic potential when cultured in nutrient medium to give rise to a complete individual plant.

Question 4.
Define sterilization.
Answer:
Sterilization is the technique employed to get rid of microbes such as bacteria and fungi in the culture medium, vessels and explants.

Question 5.
Mention the way by which culture media and explants are sterilized.
Answer:

• Culture media are sterilized by Autoclaving.
• Explants are surface sterilized by chemicals.

Question 6.
Name any four culture media used in plant tissue culture,
Answer:

1. Murashige and Skoog medium
2. Gamborg medium
3. White medium
4. Nitsch’s medium

Question 7.
What is Agar? Mention its role in plant tissue culture.
Answer:
Agar is a complex mucilaginous polysaccharide obtained from marine algae (sea weeds), used: as solidifying agent of culture media.

Question 8.
Write the composition of vitamins used in MS medium.
Answer:
Glycine 2.0 mg/1, Nicotinic acid 0.5 mg/1, Pyridoxin HC10.5 mg/1 and Thiamine HC10.1 mg/1.

Question 9.
Define
(a) Callus
(b) Embryoids
Answer:
(a) Callus is a mass of unorganized growth of plant cells or tissues in in vitro culture medium,
(b) The callus cells undergoes differentiation and produces somatic embryos, known as Embryoids.

Question 10.
What do you mean by “Hardening” in plant tissue culture technique?
Answer:
Hardening is the gradual exposure of in vitro developed plantlets in humid chambers in r diffused light for acclimatization so as to enable them to grow under normal field conditions.

Question 11.
Classify plant tissue culture based on types of explants used.
Answer:
Based on the explanin some other plant tissue culture
types are:

1. Organ culture
2. Meristem culture
3. Protoplast culture
4. Cell culture

Question 12.
What is cell suspension culture?
Answer:
The growing of cells including the culture of single cells or small aggregates of cells in vitro in liquid medium is known as cell suspension culture.

Question 13.
What is a protoplast? Which chemical stain is used to test its viability?
Answer:
Protoplasts are the cells without a cell wall, bounded by cell membrane. Fluorescein diacetate (FDA) is used to test the viability of protoplast.

Question 14.
What is a cybrid?
Answer:
Cybrid is a cytoplasmic hybrid obtained by the fusion of cytoplasm of cells of different parental sources a term applied to the fusion of cytoplasms of two different protoplasts.

Question 15.
Given below are the secondary metabolites. Mention their plant source.

1. Digoxin
2. Vincristine.

Answer:

1. Digoxin is obtained from Digitalis purpuria.
2. Vincristine is obtained from Catharanthus roseus.

Question 16.
Define Organogenesis.
Answer:
The morphological changes occur in the callus leading to the formation of shoot and roots is called organogenesis.

Question 17.
How virus free plants are developed?
Answer:
Shoot meristem tip culture is the method to produce virus-free plants, because the shoot meristem tip is always free from viruses.

Question 18.
State the role of cryoprotectants in conservation of plant resources.
Answer:
Cryoprotectants are the protective agents that are used to protect the cells or tissues from the stress of freezing temperature.
E.g: Sucrose.

Question 19.
Name any two widely used cryoprotectants.
Answer:

1. Dimethyl sulphoxide
2. Glycerol

Question 20.
Expand and define IPR.
Answer:
Intellectual property right (IPR) is a category of property that includes intangible creation of the human intellect, and primarily consists of copyrights, patents, and trademarks.

Question 21.
Define the patent type – Grant.
Answer:
Grant is a signed document, actually the agreement that grants patent right to the inventor. It is filled at patent office and not published.

Question 22.
Point any four ways by which IPR is protected in India.
Answer:
Patents, Copyrights, trade secrets and geographical indications.

Question 23.
What does ELSI represents to?
Answer:
ELSI which represents ethical, legal and social implications of biotechnology broadly covers the relationship between biotechnology and society with particular reference to ethical and legal aspects.

Question 24.
Mention any two competent national authorities that implement Bio safety guidelines.
Answer:

1. Review Committee on Genetic Manipulation (RCGM)
2. Genetic Engineering Approval Committee (GEAC)

Question 25.
What does the term ‘Bioethics’ refers to?
Answer:
Bioethics refers to the study of ethical issues emerging from advances in biology and medicine. It is also a moral discernment as it relates to medical policy and practice.

Question 26.
State the mission of ELSI program.
Answer:
The mission of the ELSI program was to identify and address issues raised by genomic research that would affect individuals, families and society.

Question 27.
Name the two inherent capacity responsible for cellular totipotency.
Answer:
Redifferentiation and Dedifferentiation.

3 – Mark Questions

Question 28.
Compare Redifferentiation with Dedifferentiation.
Answer:

Question 29.
How the autoclaving is done for culture media?
Answer:
Culture media are dispensed in glass containers, plugged with non-absorbent cotton or sealed with plastic closures and then sterilized using autoclave at 15 psi (121°C) for 15 to 30 minutes.

Question 30.
Briefly explain the surface sterilization of explants.
Answer:
The plant materials to be used for tissue culture should be surface sterilized by first exposing the material in running tap water and then treating it in surface sterilization agents like 0.1% mercuric chloride, 70% ethanol under .aseptic condition inside the Laminar Air Flow Chamber.

Question 31.
Point out the factors that determine success rate of tissue culturing.
Answer:
The success of tissue culture lies in the composition of the growth medium, plant growth [ regulators and culture conditions such as temperature, pH, light and humidity.

Question 32.
Mention any three macronutrients and micronutrients used in MS medium.
Answer:
Macronutrients:

1. Ammonium nitrate
2. Potassium nitrate
3. Calcium chloride

Micronutrients:

1. Manganese sulphate
2. Zinc sulphate
3. Potassium iodide

Question 33.
What are the optimal conditions that favours the induction of callus from nutrient medium?
Answer:
After the inoculation of explant in the nutrient medium supplemented with auxins and incubated 1 at 25°C ± 2°C in an alternate light and dark period of 12 hours with light intensity of 1000 lux, induces cell division leading to the development of callus from the surface of explant.

Question 34.
How cell suspension is prepared?
Answer:
The cell suspension is prepared by transferring a portion of callus to the liquid medium and agitated using rotary shaker instrument. The cells are separated from the callus tissue and used for cell suspension culture.

Question 35.
What are Secondary metabolites? Give example.
Answer:
Secondary metabolites are chemical compounds that are not required by the plant for normal growth and development but are produced in the plant as ‘byproducts’ of cell metabolism.
Example: Biosynthesis and isolation of indole alkaloids from Catharanthus roseus plant cell culture.

Question 36.
Name any three secondary metabolites obtained from plants and mention their medicinal aspects.
Answer:
Secondary Metabolites:

1. Capsaicin
2. Vincristine
3. Quinine

Medicinal use:

1. Rheumatic pain treatment
2. Anti-carcinogenic
3. Anti-malarial

Question 37.
What is somatic embryogenesis? Give any two of its applications.
Answer:
Somatic embryogenesis is the formation of embryos from the callus tissue directly and these embryos are called Embryoids.
Applications:

1. Somatic embryogenesis provides potential plantlets which after hardening period can establish into plants.
2. Somatic embryoids can be used for the production of synthetic seeds.

Question 38.
Differentiate between Somaclonal and Gametoclonal variations.
Answer:

1. Somaclonal variations: Somatic variations found in plants regenerated in vitro
   (i.e. variations found in leaf, stem, root, tuber or propagule).
2. Gametoclonal variations: Gametophytic variations found in plants regenerated in vitro gametic origin
   (i.e. variations found in gametes and gametophytes)

Question 39.
How synthetic seeds are developed?
Answer:
Artificial seeds or synthetic seeds (synseeds) are produced by using embryoids (somatic embryos) obtained through in vitro culture. They may even be derived from single cells from any part of the plant that later divide to form cell mass containing dense cytoplasm, large nucleus, starch grains, proteins, and oils, etc. To prepare the artificial seeds different inert materials are used for coating the somatic embryoids like agrose and sodium alginate.

Question 40.
Give an account on germplasm conservation.
Answer:
Germplasm conservation refers to the conservation of living genetic resources like pollen, seeds or tissue of plant material maintained for the purpose of selective plant breeding, preservation in live condition and used for many research works.

Germplasm conservation resources is a part of collection of seeds and pollen that are stored in seed or pollen banks, so as to maintain their viability and fertility for any later use such as hybridization and crop improvement. Germplasm conservation may also involve a gene bank and DNA bank of elite breeding lines of plant resources for the maintenance of biological diversity and also for food security.

Question 41.
How cryopreservation works?
Answer:
Cryopreservation, also known as Cryo-conservation, is a process by which protoplasts, cells, tissues, organelles, organs, extracellular matrix, enzymes or any other biological materials are subjected to preservation by cooling to very low temperature of-196°C using liquid nitrogen. At this extreme low temperature any enzymatic or chemical activity of the biological material will be totally stopped and this leads to preservation of material in dormant status. Later these materials can be activated by bringing to room temperature slowly for any experimental work.

Question 42.
What does the terms specification and claim refers with respect to patents?
Answer:

1. The specification part is narrative in which the subject matter of invention is described as how the invention was carried out.
2. The claim specifically defines the scope of the invention to be protected by the patent which the others may not practice.

Question 43.
Comment on Biosafety.
Answer:
Biosafety is the prevention of large-scale loss of biological integrity, focusing both on ecology and human health. These prevention mechanisms include conduction of regular reviews of the biosafety in laboratory settings, as well as strict guidelines to follow. Biosafety is used to protect from harmful incidents.

Many laboratories handling biohazards employ an ongoing risk management assessment and enforcement process for biosafety. Failures to follow such protocols can lead to increased risk of exposure to biohazards or pathogens. Human error and poor techniques contribute to unnecessary exposure to hazards and compromise the best safeguards set into place for protection.

Question 44.
Write any three points that you know about Genetic Engineering Appraisal Committee (GEAC).
Answer:
GEAC is an apex body under Ministry of Environment, Forests and Climate change for regulating manufacturing, use, import, export and storage of hazardous microbes or genetically modified organisms (GMOs) and cells in the country. It was established as an apex body to accord approval of activities involving large scale use of hazardous microorganisms and recombinants in research and industrial production.

The GEAC is also responsible for approval of proposals relating to release of genetically engineered organisms and products into the environment including experimental field trials (Biosafety Research Level trial-I and II known as BRL-I and BRL-II).

Question 45.
What is cybrid?
Answer:
The fusion product of protoplasts without nucleus of different cells is called a cybrid.

5 – Mark Questions

Question 46.
Explain the steps involved in protoplast culture.
Answer:
Protoplasts are cells without a cell wall, but bounded by a cell membrane or plasma membrane. Using protoplasts, it is possible to regenerate whole plants from single cells and also develop somatic hybrids.
The steps involved in protoplast culture are:

(i) Isolation of protoplast: Small bits of plant tissue like leaf tissue are used for isolation of protoplast. The leaf tissue is immersed in 0.5% Macrozyme and 2% Onozuka cellulase enzymes dissolved in 13% sorbitol or mannitol at pH 5.4. It is then incubated over night at 25°C. After a gentle teasing of cells, protoplasts are obtained, and these are then transferred to 20% sucrose solution to retain their viability. They are then centrifuged to get pure protoplasts as different from debris of cell walls.

(ii) Fusion of protoplast: It is done through the use of a suitable fusogen. This is normally PEG (Polyethylene Glycol). The isolated protoplast are incubated in 25 to 30% concentration of PEG with Ca++ ions and the protoplast shows agglutination (the formation of clumps , of cells) and fusion.

(iii) Culture of protoplast: MS liquid medium is used with some modification in droplet, plating or micro-drop array techniques. Protoplast viability is tested with fluorescein diacetate before the culture. The cultures are incubated in continuous light 1000-2000 lux at 25°C. The cell wall formation occurs within 24-48 hours and the first division of new cells occurs between 2-7 days of culture.

(iv) Selection of somatic hybrid cells: The fusion product of protoplasts without nucleus of different cells is called a cybrid. Following this nuclear fusion happen. This process is called somatic hybridization.

Question 47.
Point out the applications of plant tissue culture.
Answer:
Plant tissue culture techniques have several applications such as:

1. Improved hybrids production through somatic hybridization.
2. Somatic embryoids canbe encapsulated into synthetic seeds (synseeds). These encapsulated seeds or synthetic seeds help in conservation of plant biodiversity.
3. Production of disease resistant plants through meristem and shoot tip culture.
4. Production of stress resistant plants like herbicide tolerant, heat tolerant plants.
5. Micropropagation technique to obtain large numbers of plantlets of both crop and tree species useful in forestry within a short span of time and all through the year.
6. Production of secondary metabolites from cell culture utilized in pharmaceutical,

Question 48.
Discuss the protocol for micropropagation of banana.
Answer:

Question 49.
Enumerate the advantages of Artificial seeds.
Answer:
Advantages of Artificial seeds:
Artificial seeds have many advantages over the true seeds as follows:

1. Millions of artificial seeds can be produced at any time at low cost.
2. They provide an easy method to produce genetically engineered plants with desirable traits.
3. It is easy to test the genotype of plants.
4. They can potentially stored for long time under cryopreservation method.
5. Artificial seeds produce identical plants.
6. The period of dormancy of artificial seeds is greatly reduced, hence growth is faster with a shortened life cycle.

Question 50.
Prepare a protocol for virus free meristem tip culture.
Answer:

Higher Order Thinking Skills (HOTs) Questions

Question 1.
Transferred to field condition
Study the process given below and mention the phenomena A and B

1. Meristematic tissue → Permanent tissue
2. Callus → Embryoid

Answer:

1. A – Differentiation
2. B – Redifferentiation

Question 2.
Given below are the list of components and accessories used in PTC technique. Sort them out according to their mode of sterilization.

1. Glass wares
2. Laminar air flow chamber
3. Nutrient medium
4. Explain

Answer:

1. Glass wares are sterilized by auto claving.
2. Nutrient medium is sterilized by auto claving.
3. Laminar air flow chamber is sterilized by UV radiation.
4. Explant is surface sterilized using chemicals.

Question 3.
Name the explant through which virus free plantlets can be generated using tissue culturing technique.
Answer:
Shoot – tip meristem

Question 4.
Geographical Indication refers to the products confined to a specific geographical origin. Name any three Tamil Nadu products of your knowledge that hold GI tag.
Answer:
Kanchipuram Silk, Tanjavore dancing doll, Madurai malli.

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Tamilnadu Samacheer Kalvi 12th Bio Botany Solutions Chapter 8 Environmental Issues

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Samacheer Kalvi 12th Bio Botany Environmental Issues Text Book Back Questions and Answers

Question 1.
Which of the following would most likely help to slow down the greenhouse effect?
(a) Converting tropical forests into grazing land for cattle.
(b) Ensuring that all excess paper packaging is buried to ashes.
(c) Redesigning landfill dumps to allow methane to be collected.
(d) Promoting the use of private rather than public transport.
Answer:
(d) Promoting the use of private rather than public transport.

Question 2.
With respect to Eichhornia
Statement A: It drains off oxygen from water and is seen growing in standing water.
Statement B: It is an indigenous species of our country.
(a) Statement A is correct and Statement B is wrong.
(b) Both Statements A and B are correct.
(c) Statement A is correct and Statement B is wrong.
(d) Both statements A and B are wrong.
Answer:
(a) Statement A is correct and Statement B is wrong.

Question 3.
Find the wrongly matched pair.
(a) Endemism – Species confined to a region and not found anywhere else.
(b) Hotspots – Western ghats
(c) Ex-situ Conservation – Zoological parks
(d) Sacred groves – Saintri hills of Rajasthan
(e) Alien sp Of India – Water hyacinth
Answer:
(d) Sacred groves – Saintri hills of Rajasthan

Question 4.
Depletion of which gas in the atmosphere can lead to an increased incidence of skin cancer?
(a) Ammonia
(b) Methane
(c) Nitrous oxide
(d) Ozone
Answer:
(d) Ozone

Question 5.
One green house gas contributes 14% of total global warming and another contributes 6%. These are respectively identified as ________
(a) N₂O and CO₂
(b) CFCs and N₂O
(c) CH₄ and CO₂
(d) CH₄ and CFCS
Answer:
(b) CFCs and N₂O

Question 6.
One of the chief reasons among the following for the depletion in the number of species making endangered is ______
(a) over hunting and poaching
(b) green house effect
(c) competition and predation
(d) habitat destruction
Answer:
(d) habitat destruction

Question 7.
Deforestation means ______
(a) growing plants and trees in an area where there is no forest
(b) growing plants and trees in an area where the forest is removed
(c) growing plants and trees in a pond
(d) removal of plants and trees
Answer:
(d) removal of plants and trees

Question 8.
Deforestation does not lead to _______
(a) Quick nutrient cycling
(b) Soil erosion
(c) alternation of local weather conditions
(d) Destruction of natural habitat weather conditions
Answer:
(a) Quick nutrient cycling

Question 9.
The unit for measuring ozone thickness ______
(a) Joule
(b) Kilos
(c) Dobson
(d) Watt
Answer:
(c) Dobson

Question 10.
People’s movement for the protection of environment in Sirsi of Karnataka is ________
(a) Chipko movement
(b) Amirtha Devi Bishwas movement
(c) Appiko movement
(d) None of the above
Answer:
(c) Appiko movement

Question 11.
The plants which are grown in silivpasture system are ________
(a) Sesbania and Acacia
(b) Solenum and Crotalaria
(c) Clitoria and Begonia
(d) Teak and sandal
Answer:
(a) Sesbania and Acacia

Question 12.
What is ozone hole?
Answer:
The decline in the thickness of the ozone layer over restricted area is called Ozone hole.

Question 13.
Give four examples of plants cultivated in commercial agroforestry.
Answer:
Casuarina, Eucalyptus, Teak, Malai vembu.

Question 14.
Expand CCS.
Answer:
Carbon Capture and Storage.

Question 15.
Hpw do forests help in maintaining the climate?
Answer:
Forests play a major role in regulating the CO₂ level in the atmosphere. As the result global warming in highly reduced.

Question 16.
How do sacred groves help in the conservation of biodiversity?
Answer:
These are the patches or grove of cultivated trees which are community protected and are based on strong religious belief systems which usually have a significant religious connotation for protecting community. Each grove is an abode of a deity mostly village God Or Goddesses like Aiyanar or Amman.

448 grooves were documented throughout Tamil Nadu, of which 6 groves (Banagudi shola, Thirukurungudi and Udaiyankudikadu, Sittannnavasal, Puthupet and Devadanam) were taken up for detailed floristic and faunistic studies. These groves provide a number of ecosystem services to the neighbourhood like protecting watershed, fodder, medicinal plants and micro climate control.

Question 17.
Which one gas is most abundant out of the four commonest greenhouse gases? Discuss the effect of this gas on the growth of plants?
Answer:
Carbondioxide is the most abundant greenhouse gas. Increase in CO₂ level in the air decreases the uptake of nitrogen components leading to protein deficiency and chlorophyll formation.

Question 18.
Suggest a solution to water crisis and explain its advantages.
Answer:
Rainwater harvesting is the accumulation and storage of rain water for reuse in-site rather than allowing it to run off. Rainwater can be collected from rivers, roof tops and the water collected is directed to a deep pit. The water percolates and gets stored in the pit. RWH is a sustainable water management practice implemented not only in urban area but also in agricultural fields, which is an important economical cost effective method for the future. Environmental benefits of Rain Water Harvesting:

1. Promotes adequacy of underground water and water conservation.
2. Mitigates the effect of drought.
3. Reduces soil erosion as surface run-off is reduced.
4. Reduces flood hazards.
5. Improves groundwater quality and water table / decreases salinity.
6. No land is wasted for storage purpose and no population displacement is involved.
7. Storing water underground is an eco-friendly measure and a part of sustainable water storage strategy for local communities.

Question 19.
Explain afforestation with case studies.
Answer:
Afforestation is planting of trees where there was no previous tree coverage and the conversion of non-forested lands into forests by planting suitable trees to retrieve the vegetation. Example: Slopes of dams afforesed to reduce water run-off, erosion and siltation. It can also provide a range of environmental sendees including carbon sequestration, water retention. The Man who Single Handedly Created a Dense Forest

Jadav “Molai” Payeng (bom 1963) is an environmental activist has single-handedly planted a forest in the middle of a barren wasteland. This Forest Man of India has transformed the world’s largest river island, Majuli, located on one of India’s major rivers, the Brahmaputra, into a dense forest, home to rhinos, deers, elephants, tigers and birds. And today his forest is larger than Central Park.

Former vice-chancellor of Jawahar Lai Nehru University, Sudhir Kumar Sopory named Jadav Payeng as Forest Man of India, in the month of October 2013. He was honoured at the Indian Institute of Forest Management during their annual event Coalescence. In 2015, he was honoured with Padma Shri, the fourth highest civilian award in India. He received honorary doctorate degree from Assam Agricultural University and Kaziranga University for his contributions.

Question 20.
What are the effects of deforestation and benefits of agroforesty?
Answer:
Effects of deforestation:

1. Burning of forest wood release stored carbon, a negative impact just opposite of carbon sequestration.
2. Trees and plants bind the soil particles. The removal of forest cover increases soil erosion and decreases soil fertility. Deforestation in dry areas leads to the formation of deserts.
3. The amount of runoff water increases soil erosion and also creates flash flooding, thus reducing moisture and humidity.
4. The alteration of local precipitation patterns leading to drought conditions in many regions. It triggers adverse climatic conditions and alters water cycle in ecosystem.
5. It decreases the bio-diversity significantly as their habitats are disturbed and disruption of natural cycles.
6. Loss of livelihood for forest dwellers and rural people.
7. Increased global warming and account for one-third of total CO₂ emission.
8. Loss of life support resources, fuel, medicinal herbs and wild edible fruits.

Benefits of agroforestry:

1. It is an answer to the problem of soil and water conservation and also to stabilise the soil (salinity and water table) reduce landslide and water run-off problem.
2. Nutrient cycling between species improves and organic matter is maintained.
3. Trees provide micro climate for crops and maintain CO₂ balanced, atmospheric temperature and relative humidity.
4. Suitable for dry land where rainfall is minimum and hence it is a good system for alternate land use pattern.
5. Multipurpose tree varieties like Acacia are used for wood pulp, tanning, paper and firewood – industries.
6. Agro-forestry is recommended for the following purposes. It can be used as Farm Forestry for the extension of forests, mixed forestry, shelter belts and linear strip plantation.

Samacheer Kalvi 12th Bio Botany Environmental Issues Additional Questions and Answers

1 – Mark Questions

Question 1.
Which is not a green house gas?
(a) CO₂
(b) N₂O
(c) O₃
(d) CFC
Answer:
(C) O₃

Question 2.
Identify the incorrect statement with regard to Global warming.
(a) Leads to species enrichment
(b) Decreases irrigation
(c) Increases vector population
(d) Frequent heat waves
Answer:
(a) Leads to species enrichment

Question 3.
__________ is the unit of measurement of total ozone.
Answer:
Dobson unit

Question 4.
The total ozone layer over the earth surface is __________
(a) 30 DU
(b) 300 DU
(c) 3000 DU
(d) 0.3 DU
Answer:
(b) 300 DU

Question 5.
Methane is __________ times as effective as CO₂ at trapping heat.
(a) 5
(b) 10
(c) 20
(d) 100
Answer:
(b) 20

Question 6.
Which is not a beneficial aspect of Agroforestry?
(a) Nutrient cycling is improved
(b) Balance in O₂ – CO₂ composition
(c) Suitable for wet land where rainfall is maximum
(d) Reduces water run-off problem
Answer:
(c) Suitable for wet land where rainfall is maximum

Question 7.
The production of woody plants combined with pasture is referred to system.
Answer:
Silvopasture

Question 8.
Assertion (A): CO₂ is a main cause for global warming
Reason (R) : Greenhouse gases trap the radiant heat from sun
(a) A is correct R is incorrect.
(b) A is incorrect R is correct.
(c) R explains A.
(d) Both A and R are incorrect.
Answer:
(c) R explains A.

Question 9.
Assertion (A): Ozone acts as a natural sun block.
Reason (R): UV rays reaching the earth are deviated from earth.
(a) A is correct R is incorrect.
(b) A is incorrect R is correct.
(c) R explains A.
(d) Both A and R are incorrect.
Answer:
(c) R explains A.

Question 10.
Assertion (A): Social forestry refers to management of forests and afforestation on barren lands.
Reason (R): Afforestation involves the cutting of trees.
(a) A is correct R is incorrect.
(b) A is incorrect R is correct.
(c) R explains A.
(d) Both A and R are incorrect.
Answer:
(a) A is correct R is incorrect.

Question 11.
Assertion (A): Prosopis juliflora is native to Afganisthan.
Reason (R): Alien species refers to non-native species.
(a) A is correct R is incorrect.
(b) A is incorrect R is correct.
(c) R explains A.
(d) Both A and R are incorrect.
Answer:
(b) A is incorrect R is correct.

Question 12.
Assertion (A): In zoological parks, the animals are maintained in their natural habitat.
Reason (R): Ex-situ conservation refers to protecting species in their natural habitat.
(a) A is correct R is incorrect.
(b) A is incorrect R is correct.
(c) R explains A.
(d) Both A and R are incorrect.
Answer:
(d) Both A and R are incorrect.

Question 13.
Which is not reduced by deforestation?
(a) Amount of habitat
(b) Amount of animal population
(c) Amount of biodiversity
(d) Amount of agricultural land
Answer:
(d) Amount of agricultural land

Question 14.
Identify the potent cause for deforestation.
(a) Agriculture
(b) Soil erosion
(c) Afforestation
(d) Forest fire
Answer:
(a) Agriculture

Question 15.
Total number of forestry extension centres in Tamil Nadu is __________
(a) 16
(b) 32
(c) 18
(d) 51
Answer:
(b) 32

Question 16.
Who is celebrated as Forest Man of India?
(a) Anand Mohan Chakrabarthy
(b) Dr. M.S Swaminathan
(c) Jadav Molai Payeng
(d) Choudary Ram Dhan
Answer:
(c) Jadav Molai Payeng

Question 17.
Invasive species __________
(a) alter the soil system
(b) are more adapted
(c) are fast growing
(d) all the above
Answer:
(d) all the above

Question 18.
Pick out the odd one out __________
(a) Biosphere reserve
(b) National Park
(c) Wild life sanctuaries
(d) Botanical gardens
Answer:
(d) Botanical gardens

Question 19.
Which is not true with respect to prosopis juliflora?
(a) Invasive species native to Mexico
(b) Arrest wind erosion
(c) Absorb hazardous chemical from soil
(d) Decreases O₂ content of water bodies
Answer:
(d) Decreases O₂ content of water bodies

Question 20.
How many numbers of scared grooves were documented in tamil nadu?
(a) 484
(b) 844
(c) 488
(d) 448
Answer:
(c) 488

Question 21.
Match the following:

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

Question 22.
Process of heating biomass in low oxygen environment is called as ______
Answer:
Pyrolysis

Question 23.
Biochar is __________
(i) a kind of char coal used as a soil amendment
(ii) a potent way of sequestring carbon
(iii) made from biomass via pyrolysis
(iv) a notable solid, rich in carbon.
(a) (i) and (iii) is correct
(b) (ii) and (iv) is correct
(c) (i) and (ii) is correct
(d) all the above is correct.
Answer:
(d) all the above is correct.

Question 24.
Which is not a true statement regarding rain water harvesting?
(a) Mitigates groundwater quality
(b) Reduces soil erosion
(c) Decreases soil salinity
(d) No wastage of land for storing
Answer:
(a) Mitigaes groundwater quality

Question 25.
EIA stands for __________
(a) Ecological Information Analysis
(b) Environmental Information Assessment
(c) Environmental Impact Analysis
(d) Environmental Impact Assessment
Answer:
(d) Environmental Impact Assessment

Question 26.
__________ is the 100th Satellite launched to watch border surveillance.
(a) GSAT-6A
(b) SCAT SAT-I
(c) INSAT 3DR
(d) CARTOSAT-2
Answer:
(d) CARTOSAT-2

Question 27.
The ozone layer of __________ is called bad ozone.
(a) Stratosphere
(b) Mesosphere
(c) Troposphere
(d) Exosphere
Answer:
(c) Troposhere

Question 28.
When does World Ozone Day is observed?
(a) June 17th
(b) December 1st
(c) October 12th
(d) September 16th
Answer:
(d) September 16th

Question 29.
Clean Development Mechanism (CDM) is defined in __________
(a) Copenhagen Acord
(b) Montreal Protocol
(c) Paris Agreement
(d) Kyoto Protocol
Answer:
(d) Kyoto Protocol

Question 30.
__________ is a plant species which acts as an indicator of Nitrate pollution.
(a) Petunia
(b) Lichens
(c) Gladiolus
(d) Pinus
Answer:
(a) Petunia

Question 31.
Identify the plant species that is not used as a live fence.
(a) Sesbania grandiflora
(b) Acacia species
(c) Petunia species
(d) Erythrina species
Answer:
(c) Petunia species

Question 32.
Match the following:

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

2 – Mark Questions

Question 1.
What are greenhouse gases? Give an example.
Answer:
The gases that capture heat are called Green House Gases.
E.g: Carbon dioxide (CO₂)

Question 2.
Name any four green house gases.
Answer:
CO₂, CH₄, N₂O and CFC

Question 3.
Define global warming.
Answer:
The increase in mean global temperature due to increased concentration of green house gases is called global warming.

Question 4.
Methane is one of a potent green house gas. Point out few sources from where methane is generated.
Answer:
Paddy field, fossil fuel production, bacteria in water bodies, cattle rearing, non-wetland soils and forest & wild fires.

Question 5.
List out the anthropogenic sources of Nitrous oxide.
Answer:
Man-made sources include nylon and nitric acid production, use of fertilizers in agriculture, manures cars with catalytic converter and burning of organic matter.

Question 6.
What is Dobson unit?
Answer:
Dobson Unit is the unit of measurement for total ozone. One DU is the number of molecules of ozone that would be required to create a layer of pure ozone 0.01 millimetre thick at a temperature of 0° C and a pressure of 1 atmosphere.

Question 7.
What do you mean by good ozone and bad ozone?
Answer:

1. The ozone layer of troposphere is called bad ozone.
2. The ozone layer of stratosphere is called good ozone.

Question 8.
What is ozone layer? Why it is essential?
Answer:
Ozone layer is a region of Earth’s stratosphere that absorbs most of the Sim’s ultra violet radiation. The ozone layer is also called as the ozone shield and it acts as a protective shield, cutting the ultra-violet radiation emitted by the sun.

Question 9.
Define ozone hole. Name any one potent chemical that is responsible for the effect.
Answer:
The decline in the thickness of the ozone layer over restricted area is called Ozone hole. Chlorofluorocarbons (CFC) damages the ozone layer to a great extent.

Question 10.
Where did Montreal Protocol was held? State its objectives.
Answer:
The International treaty called the Montreal Protocol (1987) was held in Canada on substances that deplete ozone layer and the main goal of it is gradually eliminating the production and consumption of ozone depleting substances and to limit their damage on the Earth’s ozone layer.

Question 11.
What is meant by biological indicators? Give example.
Answer:
The plant species or plant community acts as a measure of environmental conditions, it is referred as biological indicators or phytoindicators or plant indicators.
E.g: Lichens as SO₂ pollution indicator.

Question 12.
Name the plant species which act as indicators for Nitrate pollution & Flouride pollution.
Answer:

1. Petunia species acts as indicators for Nitrate pollution.
2. Gladiolus species acts as indicators for Flouride pollution.

Question 13.
Define the term Agroforestry. Name any two major tree species cultivated in Agroforestry.
Answer:
Agroforestry is an integration of trees, crops and livestock on the name plot of land. The commercial Agroforestry trees are casuarina and Eucalyptus.

Question 14.
What is silvopasture system? How it helps economy?
Answer:
The production of woody plants combined with pasture is referred to silvopasture system. The trees and shrubs may be used primarily to produce fodder for livestock or they may be grown for timber, fuel wood and fruit or to improve the soil.

Question 15.
Mention the name of any four plant species that are widely used as live fences in agricultural practicer.
Answer:

1. Gliricidia sepium
2. Sesbaniagrandiflora
3. Erythrina species
4. Acacia species

Question 16.
Define social forestry.
Answer:
Social forestry refers to the management of forests and afforestation on barren lands with the purpose of helping the environmental, social and rural development and benefits.

Question 17.
Compare Deforestation and Afforestation
Answer:

1. Deforestation: The conversion of forested area into a non- forested area is known as deforestation.
2. Afforestation: Afforestation is planting of trees where there was no previous tree coverage and the conversion of non-forested lands into forests by planting suitable trees to retrieve the vegetation.

Question 18.
What is invasive species?
Answer:
A non-native species to the ecosystem or country under consideration that spreads naturally, interferes with the biology and existence of native species, poses a serious threat to the ecosystem and causes economic loss

Question 19.
Point out the adverse effect caused by invasive species.
Answer:
Invasive species are fast growing and are more adapted. They alter the soil system by changing litter quality thereby affecting the soil community, soil fauna and the ecosystem processes. It has a negative impact on decomposition in the soils by causing stress to the neighbouring native species.

Question 20.
Distinguish between In-situ conservation and Ex-situ conservation.
Answer:
In-situ conservation:

1. Conservation of species in their natural habitat.
2. E.g: Biosphere reserves

Ex-situ conservation:

1. Conservation of species outside their natural habitat.
2. E.g: Zoological parks

Question 21.
Mention any two historical community level conservation movements held for the protection of environment.
Answer:
Chipko Movement and Appiko Movement

Question 22.
Give the names of any four famous sacred grooves in Tamil Nadu.
Answer:

1. Banagudi Shola
2. Thirukurungudi
3. Udaiyankudikadu
4. Sittannavasal

Question 23.
What does the term ‘Endemic’ refers to?
Answer:
Any species found restricted to a specified geographical area is referred to as ENDEMIC.

Question 24.
Name any two endemic trees of Peninsular India.
Answer:

1. Agasthiyamalaia pauciflora
2. Harawickia binata

Question 25.
What is carbon sequestration?
Answer:
Carbon sequestration is the process of capturing and storing CO₂ which reduces the amount of CO₂ in the atmosphere with a goal of reducing global climate change.

Question 26.
Carbon sequestration occurs naturally by plants and oceans. Name any four microalgal species involved in the process.
Answer:

1. Chlorella
2. Scenedesmus
3. Chroococcus
4. Chlamydomonas

Question 27.
Explain the term Carbon sink.
Answer:
Any system having the capacity to accumulate more atmospheric carbon during a given time interval than releasing CO₂.
Example: forest, soil, ocean are natural sinks. Landfills are artificial sinks.

Question 28.
How EIA is beneficial to a society?
The benefits of EIA to society
Answer:

1. A healthier environment
2. Maintenance of biodiversity
3. Decreased resource usage
4. Reduction in gas emission and environment damage

Question 29.
What is GIS?
Answer:
Geographic Information System (GIS) is a computer system for capturing, storing, checking and displaying data related to positions on Earth’s surface. Also to manipulate, analyse, manage and present spacial or geographic data.

3 – Mark Questions

Question 30.
List out the effects of global warming.
Answer:

1. Rise in global temperature which causes sea levels to rise as polar ice caps and glaciers begin to melt causing submergence of many coastal cities in many parts of the world.
2. There will be a drastic change in weather patterns bringing more floods or droughts in some areas.
3. Biological diversity may get modified, some species ranges get redefined. Tropics and sub-tropics may face the problem of decreased food production.

Question 31.
Mention any three sources of carbon dioxide emission.
Answer:

1. Coal based power plants, by the burning of fossil fuels for electricity generation.
2. Combustion of fuels in the engines of automobiles, commercial vehicles and air planes contribute the most of global warming.
3. Agricultural practices like stubble burning result in emission of CO₂.

Question 32.
What are the adverse effects of global warming on plants?
Answer:

1. Low agricultural productivity in tropics
2. Frequent heat waves (Weeds, pests, fungi need warmer temperature)
3. Increase of vectors and epidemics
4. Strong storms and intense flood damage
5. Water crisis and decreased irrigation
6. Change in flowering seasons and pollinators
7. Change in Species distributional ranges
8. Species extinction

Question 33.
Suggest few ways to overcome global warming.
Answer:

1. Increasing the vegetation cover, grow more trees
2. Reducing the use of fossil fuels and green house gases
3. Developing alternate renewable sources of energy
4. Minimising uses of nitrogeneous fertilizers, and aerosols.

Question 34.
Ozone acts a a natural sun screen – Justify.
Answer:
Ozone depletion in the stratosphere results in more UV radiations especially UV B radiations (shortwaves). UV B radiation destroys biomolecules (skin ageing) and damages living tissues. UV – C is the most damaging type of UV radiation, but it is completely filtered by the atmosphere (ozone layer). UV – a contribute 95% of UV radiation which causes tanning burning of skin and enhancing skin cancer. Hence the uniform ozone layer is critical for the wellbeing of life on earth.

Question 35.
Give a detailed account on clean development mechanism.
Answer:
Clean Development Mechanism (CDM) is defined in the Kyoto protocol (2007) which provides project based mechanisms with two objectives to prevent dangerous climate change and to reduce green house gas emissions. CDM projects helps the countries to reduce or limit emission and stimulate sustainable development.

An example for CDM project activity, is replacement of conventional electrification projectswith solar panels or other energy efficient boilers. Such projects can earn Certified Emission I Reduction (CER) with credits / scores, each equivalent to one tonne of CO₂, which can be counted towards meeting Kyoto targets.

Question 36.
What are the major activities carried out by forestry extension centres?
Answer:

1. Training on tree growing methods
2. Publicity and propaganda regarding tree growing
3. Formation of demonstration plots
4. Raising and supply of seedlings on subsidy
5. Awareness creation among school children and youth about the importance of forests through training and camps.

Question 37.
Prosopis juliflora, though an alien invasive species to India, it is beneficial to certain extent. Give reason.
Answer:
Prosopis juliflora is used to arrest wind erosion and stabilize sand dunes on coastal and desert areas. It can absorb hazardous chemicals from soil and it is the main source of charcoal.

Question 38.
Write a short note on Chipko Movement.
Chipko Movement:
Answer:
The tribal women of Himalayas protested against the exploitation of forests in 1972. Later on it transformed into Chipko Movement by Sundarlal Bahuguna in Mandal village of Chamoli district in 1974. People protested by hugging trees together which were felled by a sports goods company. Main features of Chipko movement were,

1. This movement remained non political
2. It was a voluntary movement based on Gandhian thought.
3. It was concerned with the ecological balance of nature
4. Main aim of Chipko movement was to give a slogan of five Fs – Food, Fodder, Fuel, Fibre and Fertilizer, to make the communities self sufficient in all their basic needs.

Question 39.
Give a brief account on In-situ conservation.
Answer:
In-situble conservation means conservation and management of genetic resources in their natural habitats. Here the plant or animal species are protected within the existing habitat. Forest trees, medicinal and aromatic plants under threat are conserved by this method. This is carried out by the community or by the State conservation which include wildlife, National park and Biosphere reserve. The ecologically unique and biodiversity rich regions are legally protected as wildlife sanctuaries, National parks and Biosphere reserves. Megamalai, Sathyamangalam wildlife, Guindy and Periyar National park, and Western ghats, Nilgiris, Agasthyamalai and Gulf of Mannar are the biosphere reserves of Tamil Nadu.

Question 40.
What do you mean by Biochar? How it helps the environment?
Answer:
Biochar is a long term method to store carbon. To increase plants ability to store more carbon, plants are partly burnt such as crop waste, waste woods to become carbon rich slow decomposing substances of material called Biochar. It is a kind of charcoal used as a soil amendment. Biochar is a stable solid, rich in carbon and can endure in soil for thousands of years. Like most charcoal, biochar is made from biomass via pyrolysis. (Heating biomas in low oxygen environment) which arrests wood from complete burning.

Biochar thus has the potential to help mitigate climate change via carbon sequestration. Independently, biochar when added to soil can increase soil fertility of acidic soils, increase agricultural productivity, and provide protection against some foliar and soil borne diseases. It is a good method of preventing waste woods and logs getting decayed instead we can convert them into biochar thus converting them to carbon storage material.

Question 41.
Explain the role of lakes in an ecosystem.
Answer:
Lakes as a storage of rain water provides drinking water, improves ground water level and preserve the fresh water bio-diversity and habitat of the area where in occurs.

In terms of services lakes offer sustainable solutions to key issues of water management and climatic influences and benefits like nutrient retention, influencing local rainfall, removal of pollutants, phosphorous and nitrogen and carbon sequestration.

5 – Mark Questions

Question 42.
List out the major effects of Ozone depletion.
Answer:
The main ozone depletion effects are:

1. Increases the incidence of cataract, throat and lung irritation and aggravation of asthma or emphysema, skin cancer and diminishing the functioning of immune system in human beings.
2. Juvenile mortality of animals.
3. Increased incidence of mutations.
4. In plants, photosynthetic chemicals will be affected and therefore photosynthesis will be inhibited. Decreased photosynthesis will result in increased atmospheric CO₂ resulting in global warming and also shortage of food leading for food crisis.
5. Increase in temperature changes the climate and rainfall pattern which may result in flood / drought, sea water rise, imbalance in ecosystems affecting flora and fauna.

Question 43.
What are the objectives of Afforestation programme?
Answer:
Afforestation Objectives:

1. To increase forest cover, planting more trees, increases CO₂ production and air quality.
2. Rehabilitation of degraded forests to increase carbon fixation and reducing CO₂ from atmosphere.
3. Raising bamboo plantations.
4. Mixed plantations of minor forest produce and medicinal plants.
5. Regeneration of indigenous herbs / shrubs.
6. Awareness creation, monitoring and evaluation.
7. To increase the level and availability of water table or ground water and also to reduce nitrogen leaching in soil and nitrogen contamination of drinking water, thus making it pure not polluted with nitrogen.
8. Nature aided artificial regeneration.

Question 44.
How Eichhornia crassiper spoils the Indian ecosystem?
Answer:
Eichhornia crassipes is an invasive weed native to South America. It was introduced as aquatic ornamental plant, which grows faster throughout the year. Its widespread growth is a major cause of biodiversity loss worldwide. It affects the growth of phytoplanktons and finally changing the aquatic ecosystem.

It also decreases the oxygen content of the waterbodies which leads to eutrophication. It poses a threat to human health because it creates a breeding habitat for disease causing mosquitoes (particularly Anopheles) and snails with its free floating dense roots and semi submerged leaves. It also blocks sunlight entering deep and the waer ways hampering agriculture, fisheries, recreation and hydropower.

Question 45.
Write a comparative note on Carbon Foot Print (CFP).
Answer:
Every human activity leaves a mark just like our footprint. This Carbon foot print is the total amount of green house gases produced by human activities such as agriculture, industries, deforestation, waste disposal, buring fossil fuels directly or indirectly. It can be measured for an individual, family, organisation like industries, state level or national level. It is usually estimated and expressed in equivalent tons of CO₂ per year.

The burning of fossil fuels releases CO₂ and other green house gases. In turn these emissions trap solar energy and thus increase the global temperature resulting in ice melting, submerging of low lying areas and inbalance in nature like cyclones, tsunamis and extreme weather conditions.
To reduce the carbon foot print we can follow some practices like

1. Eating indigenous fruits and products
2. Reduce use of your electronic devices
3. Reduce travelling
4. Do not buy fast and preserved, processed, packed foods
5. Plant a garden
6. Less consumption of meat and sea food. Poultry requires little space, nutrients and less pollution comparing cattle farming
7. reduce use of Laptops (when used for 8 hours, it releases nearly 2 kg. of CO₂ annually)
8. Line dry your clothes.
   (Example: If you buy imported fruit like kiwi, indirectly it increases CFP. How? The fruit has travelled a long distance in shipping or airliner thus emitting tons of CO₂)

Question 46.
Write in detail about Remote sensing and its uses.
Answer:
Remote Sensing is the process of detecting and monitoring the physical characteristics of an area by measuring its reflected and emitted radiation at a distance from the targeted area. It is an tool used in conservation practices by giving exact picture and data on identification of even a single tree to large area of vegetation and wild life for classification of land use patterns and studies, identification of biodiversity rich or less areas for futuristic works on conservation and maintenance of various species including commercial crop, medicinal plants and threatened plants.
Specific uses

1. Helps predicting favourable climate, for the study of spreading of disease and controlling it.
2. Mapping of forest fire and species distribution.
3. Tracking the patterns of urban area development and the changes in Farmland or forests over several years.
4. Mapping ocean bottom and its resources.

Higher Order Thinking Skills (HOTs) Questions

Question 1.
Name the movement launched by the people of Mandal village to protect the trees by hugging them.
Answer:
Chipko movement.

Question 2.
Recently in January 2019, our Tamil Nadu Government had imposed a ban on using 14 different products of plastic origin. Mention any six plastic products that you know.
Answer:

1. Water Packets
2. Plastic straws
3. Plastic carry bags
4. Thermocol cups
5. Plastic coated paper plates
6. Plastic flags.

Question 3.
Due to plastic ban scheme in our state, people are gradually switching over to other optionals for daily activities. As a biology student, suggest few eco-friendly alternatives for this issue.
Answer:

1. Cloth / Jute bags
2. Plantain leave, palmyra plates as plates
3. Metal cups
4. Earthen pots or ceramic wares
5. Paper made flags

Question 4.
Deforestation is creating various adverse effect on environment. Enlist the consequences of deforestation.
Answer:

1. Increased global warming.
2. Loss of livelihood for forest dwellers.
3. Loss of bio-diversity
4. Decrease in soil fertility.
5. Decline in annual rainfall.

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Tamilnadu Samacheer Kalvi 12th Bio Zoology Solutions Chapter 6 Evolution

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Samacheer Kalvi 12th Bio Zoology Evolution Text Book Back Questions and Answers

Question 1.
The first life on Earth originated __________
(a) in air
(b) on land
(c) in water
(d) on mountain
Answer:
(c) in water

Question 2.
Who published the book “Origin of species by Natural Selection” in 1859?
(a) Charles Darwin
(b) Lamarck
(c) Weismann
(d) Hugo de Vries
Answer:
(a) Charles Darwin

Question 3.
Which of the following was the contribution of Hugo de Vries?
(a) Theory of mutation
(b) Theory of natural Selection
(c) Theory of inheritance of acquired characters
(d) Germplasm theory
Answer:
(a) Theory of mutation

Question 4.
The wings of birds and butterflies is an example of __________
(a) Adaptive radiation
(b) convergent evolution
(c) divergent evolution
(d) variation
Answer:
(b) convergent evolution

Question 5.
The phenomenon of “ Industrial Melanism” demonstrates __________
(a) Natural selection
(b) induced mutation
(c) reproductive isolation
(d) geographical isolation
Answer:
(a) Natural selection

Question 6.
Darwin’s finches are an excellent example of __________
(a) connecting links
(b) seasonal migration
(c) adaptive radiation
(d) parasitism
Answer:
(c) adaptive radiation

Question 7.
Who proposed the Germplasm theory?
(a) Darwin
(b) August Weismann
(c) Lamarck
(d) analysis of bones
Answer:
(b) August Weismann

Question 8.
The age of fossils can be determined by __________
(a) electron microscope
(b) weighing the fossils
(c) carbon dating
(d) analysis of bones
Answer:
(c) carbon dating

Question 9.
Fossils are generally found in __________
(a) igneous rocks
(b) metamorphics
(c) volcanic rocks
(d) sedimentary rocks
Answer:
(d) sedimentary rocks

Question 10.
Evolutionary history of an organism is called __________
(a) ancestry
(b) ontogeny
(c) phylogeny
(d) paleontology
Answer:
(c) phylogeny

Question 11.
The golden age of reptiles was __________
(a) Mesozoic era
(b) Cenozoic era
(c) Paleozoic era
(d) Proteroic era
Answer:
(a) Mesozoic era

Question 12.
Which period was called “Age of fishes”?
(a) Permian
(b) Triassic
(c) Devonian
(d) Ordovician
Answer:
(c) Devonian

Question 13.
Modem man belongs to which period?
(a) Quaternary
(b) Cretaceous
(c) Silurian
(d) Cambrian
Answer:
(a) Quaternary

Question 14.
The Neanderthal man had the brain capacity of __________
(a) 650 – 800cc
(b) 1200cc
(c) 900cc
(d) 1400c
Answer:
(d) 1400c

Question 15.
List out the major gases seems to fie found in the primitive Earth.
Answer:
C0₂, NH₃, UV and Water vapour

Question 16.
Explain the three major categories in which fossilization occur.
Answer:
(i) Actual remains is the most common method of fossilization. When marine animals die, their hard parts such as bones and shells, etc. are covered with sediments and are protected from further deterioration. They get preserved as such as they are preserved in vast ocean

the salinity in them prevents decay. The sediments become hardened to form definite layers or strata. For example, Woolly Mammoth that lived 22 thousand years ago were preserved in the frozen coast of Siberia as such. Several human beings and animals living in die ancient city of Pompeii were preserved intact by volcanic ash which gushed out from Mount Vesuvius.

(ii) Petrifaction – When animals die the original portion of their body may be replaced molecule for molecule by minerals and the original substance being lost through disintegration. This method of fossilization is called petrifaction. The principle minerals involved in this type fossilization are iron pyrites, silica, calcium carbonate and bicarbonates of calcium and magnesium.

(iii) Natural moulds and casts – Even after disintegration, the body of an animal might leave indelible impression on the soft mud which later becomes hardened into stones. Such impressions are called moulds. The cavities of the moulds may get filled up by hard minerals and get fossilized, which are called casts. Hardened faecal matter termed as coprolites, occur as tiny pellets. Analysis of the coprolites enables us to understand the nature of diet, the prehistoric animals thrived.

Question 17.
Differentiate between divergent evolution and convergent evolution with one example for each.
Answer:
Divergent Evolution:

1. Divergent evolution is a result of homology.
2. Eg: The wings of bird and the forelimbs of human both are homologous structures modified according to functions. In birds, it is used for flight and in humans used for writing and other purposes.

Convergent Evolution:

1. Convergent evolution is a result of analogy,
2. E.g: Root modification in sweet potato, and stem modification in potato are analogous structures both performing same function i.e., storage,

Question 18.
How does Hardy-Weinberg’s expression (p² + 2pq + q² = 1) explain that genetic equilibrium is maintained in a population? List any four factors that can disturb the genetic equilibrium.
Answer:
The allele frequencies in a population are stable and are constant from generation to generation in the absence of gene flow, genetic drift, mutation, recombination and natural selection. If a population is in a state of Hardy Weinberg equilibrium, the frequencies of alleles and genotypes or sets of alleles in that population will remain same over generations. Evolution is a change in the allele frequencies in a population over time. Hence population in Hardy Weinberg is not evolving.

Suppose we have a large population of beetles, (infinitely large) and appear in two colours ’ dark grey (black) and light grey, and their colour is determined by ‘A’ gene. ‘AA’ and ‘Aa’ beetles are dark grey and ‘aa’ beetles are light grey. In a population let’s say that ‘ A’ allele has frequency (p) of 0.3 and ‘a’ allele has a frequency (q ) of 0.7. Then p+q= 1.

If a population is in Hardy Weinberg equilibrium the genotype frequencycan be estimated by Hardy Weinberg equation.
(p + q)² = p² + 2pq + q²
p² = frequency of AA
2pq = frequency of Aa
q² = frequency of aa
p = 0.3, q = 0.7 then,
p² = (0.3)² = 0.09 = 9 %AA
2pq = 2(0.3) (0.7) = 0.42 = 42 % Aa
q2 = (0.7)² 0.49 = 49 % aa
Hence the beetle population appears to be in Hardy- Weinberg equilibrium. When the beetles in Hardy- Weinberg equilibrium reproduce the allele and genotype frequency in the next generation would be: Let’s assume that the frequency of ‘A’ and ‘a’ allele in the pool of gametes that make the next generation would be the same, then there would be no variation in the progeny. The genotype frequencies of the parent appears in the next generation.
(i.e. 9% AA, 42% Aa and 49% aa).

If we assume that the beetles mate randomly (selection of male gamete and female gamete in the pool of gametes), the probability of getting the offspring genotype depends on the genotype of the combining parental gametes.

Question 19.
Explain how mutations, naturalxelection and genetic drift affect Hardy Weinberg equilibrium.
Answer:
Natural selection occurs when one allele (or combination of alleles of differences) makes an organism more or less fit to survive and reproduce in a given environment. If an allele reduces fitness, its frequencies tend to drop from one generation to the next.

The evolutionary path of a given gene (i.e) how its allele’s change in frequency in the population across generation, may result from several evolutionary mechanisms acting at once. For example, one gene’s allele frequencies might be modified by both gene flow and genetic drift, for another gene, mutation may produce a new allele, that is favoured by natural selection.

Genetic drift / Sewall Wright Effect is a mechanism of evolution in which allele frequencies of a population change over generation due to chance (sampling error). Genetic drift occurs in all population sizes, but its effects are strong in a small population. It may result in a loss of some alleles (including beneficial ones) and fixation of other alleles. Genetic drift can have major effects, when the population is reduced in size by natural disaster due to bottle neck effect or when a small group of population splits from the main population to form a new colony due to founder’s effect.

Although mutation is the original source of all genetic variation, mutation rate for most organisms is low. Hence new mutations on an allele frequencies from one generation to the next is usually not large.

Question 20.
How did Darwin explain fitness of organisms?
Answer:
Organisms struggle for food, space and mate. As these become a limiting factor, competition exists among the members of the population. Darwin denoted struggle for existence in three ways

Intra specific struggle between the same species for food, space and mate Inter specific struggle with different species for food and space.
Struggle with the environment to cope with the climatic variations, flood, earthquakes and drought, etc.

According to Darwin, nature is the most powerful selective force. He compared origin of species by natural selection to a small isolated group. Darwin believed that the struggle for existence resulted in the survival of the fittest. Such organisms become better adapted to the changed environment.

Question 21.
Mention the main objections to Darwinism.
Answer:
Some objections raised against Darwinism were Darwin failed to explain the mechanism of variation.

1. Darwinism explains the survival of the fittest but not the arrival of the fittest.
2. He focused on small fluctuating variations that are mostly non-heritable.
3. He did not distinguish between somatic and germinal variations.
4. He could not explain the occurrence of vestigial organs, over specialization of some organs like large tusks in extinct mammoths and over sized antlers in the extinct Irish deer, etc.

Question 22.
Taking the example of Peppered moth, explain the action of natural selection. What do you call the above phenomenon?
Answer:
Natural selection can be explained clearly through industrial melanism. Industrial melanism is a classical case of Natural selection exhibited by the peppered moth, Bistort betularia. These were available in two colours, white and black. Before industrialization peppered moth both white and black coloured were common in England. Pre-industrialization witnessed white colpured background of the wall of the buildings hence the white coloured moths escaped from their predators. Post industrialization, the tree trunks became dark due to smoke and soot let out from the industries.

The black moths camouflaged on the dark bark of the trees and the white moths were easily identified by their predators. Hence the dark coloured moth population was selected and their number increased when compared to the white moths. Nature offered positive selection pressure to the black coloured moths. The above proof shows that in a population, organisms that can adapt will survive and produce more progenies resulting in increase in population through natural selection.

Question 23.
Darwin’s finches and Australian marsupials are suitable examples of adaptive radiation – Justify the statement.
Answer:
Darwin’s finches are the birds whose common ancestor arrived on the Galapagos about 2 million years ago. During that time, Darwin’s finches have evolved into 14 recognized species differing in body size, beak shape and feeding behavior. Changes in the size and form of the beak have enabled different species to utilize different food resources such as insects, seeds, nectar from cactus flowers and blood from iguanas, all driven by Natural selection. Genetic variation in the ALX1 gene in the DNA of Darwin finches is associated with variation in the beak shape. Mild mutation in the ALX1 gene leads to phenotypic change in the shape of the beak of the Darwin finches.

Marsupials in Australia and placental mammals in North America are two subclasses of mammals they have adapted in similar way to a particular food resource, locomotory skill or climate. They were separated from the common ancestor more than 100 million years ago and each lineage continued to evolve independently.

Despite temporal and geographical separation, marsupials in Australia and placental mammals in North America have produced varieties of species living in similar habitats with similar ways of life. Their overall resemblance in shape, locomotory mode, feeding and foraging are superimposed upon different modes of reproduction. This feature reflects their distinctive evolutionary relationships.

Over 200 species of marsupials live in Australia along with many fewer species of placental mammals. The marsupials have undergone adaptive radiation to occupy the diverse habitats in Australia, just as the placental mammals have radiated across North America.

Question 24.
Who disproved Lamarck’s Theory of acquired characters? How?
Answer:
Lamarck’s “Theory of Acquired characters” was disproved by August Weismann who conducted experiments on mice for twenty generations by cutting their tails and breeding them. All mice bom were with tail. Weismann proved that change in the somatoplasm will not be transferred to the next generation but changes in the germplasm will be inherited.

Question 25.
How does Mutation theory of De Vries differ from Lamarck and Darwin’s view in the origin of new species.
Answer:
According to de Vries, sudden and large variations were responsible for the origin of new species, whereas Lamarck and Darwin believed in gradual accumulation of all variations as the causative factors in the origin of new species.

Question 26.
Explain stabilizing, directional and disruptive selection with examples.
Answer:
i. Stabilising selection (centipetal selection): This type of selection operates in a stable environment as shown in fig. The organisms with average phenotypes survive whereas the extreme individuals from both the ends are eliminated. There is no speciation but the phenotypic stability is maintained within the population over generation. For example, measurements of sparrows that survived the storm clustered around the mean, and the sparrows that failed to survive the storm clustered around the extremes of the variation showing stabilizing selection.

ii. Directional Selection: The environment” which undergoes gradual change is subjected to directional selection, as shown in fig. This type of selection removes the individuals from one end towards the other end of phenotypic distribution. For example, size differences between male and female sparrows. Both male and female look alike externally but differ in body weight. Females show directional selection in relation to body weight.

iii. Disruptive selection: (centrifugal selection) When homogenous environment changes into heterogenous environment this type of selection is operational as shown in fig. The organisms of both the extreme phenotypes are selected, whereas individuals with average phenotype are eliminated. This results in splitting of the population into sub population/species. This is a rare form of selection but leads to formation of two or more different species. It is also (called adaptive radiation. (E.g:) Darwin’s finches beak size in relation to seed size inhabiting Galapagos islands. Group selection and sexual selection are other types of selection. The two major group selections are Altrusim and Kin selection.

Question 27.
Rearrange the descent in human evolution.
Answer:
Australopithecus → Homo erectus → Homo sapiens → Ramapithecus → Homo habilis
Ramapithecus → Australopithecus → Homo habilis → Homo erectus → Homo sapiens

Question 28.
Differentiate between the eating habit and brain size of Australopithecus and Ramapithecus.
Answer:

	Australopithecus	Ramapithecus
Eating Habit	Herbivores	Omnivores
Brain Size	350- 450 cc	200 – 300 cc

Question 29.
How does Neanderthal man differ from the modern man in appearance?
Answer:
Neanderthal man differ from the modem human in having semierect posture, flat cranium, sloping forehead, thin large orbits, heavy brow ridges, protruding jaws and no chin.

Question 30.
Mention any three similarities found common in Neanderthal man and Homo sapiens. Common characters showed by Neanderthal man and Homo sapiens are:
Answer:

1. Usage of Fire
2. Burying of deadbodies
3. Protecting themselves from predators

Question 31.
According to Darwin, the organic evolution is due to
(а) Intraspecific competition
(b) Interspecific competition
(c) Competition within closely related species.
(d) Reduced feeding efficiency in one species due to the presence of interfering species.
Answer:
(d) Reduced feeding efficiency in one species due to the presence of interfering species.

Question 32.
A population will not exist in Hardy – Weinberg equilibrium if
(a) Individuals mate selectively
(b) There are no mutations
(c) There is no migration
(d) The population is large
Answer:
(a) Individuals mate selectively

Samacheer Kalvi 12th Bio Zoology Evolution Additional Questions and Answers

1 – Mark Questions

Question 1.
Identify the incorrect statement in concern with Neanderthals.
(a) Neanderthal human were found in Germany.
(b) They possessed flat cranium.
(c) They used to bury their dead.
(d) Their brain size is of 650 – 800 cc
Answer:
(d) Their brain size is of 650 – 800 cc

Question 2.
Which of the following statement does not satisfy Hardy Weinberg principle?
(a) A population undergoing random mating
(b) Small sized population
(c) Population where there is no mutation or gene flow
(d) Absence of natural selection
Answer:
(b) Small sized population

Question 3.
Match column I with column II

Answer:
(a) a – iii b – ii c – iv d – i

Question 4.
Placental mammals develop during _______
(a) Eocene
(b) Oligocene
(c) Pliocene
(d) Paleocene
Answer:
(d) Paleocene

Question 5.
Identify the correct sequence from oldest to youngest
(а) Cambrian → Permian → Devonian → Silurian → Ordovician
(b) Permian → Silurian → Devonian → Ordovician → Cambrian
(c) Permian → Devonian → Silurian → Cambrian → Ordovician
(d) Cambrian → Ordovician → Silurian → Devonian → Permian
Answer:
(d) Cambrian → Ordovician → Silurian Devonian → Permian

Question 6.
Match the scientists with their terminologies used
(a) Biogenesis (i) Oparin
(b) Prebiotic soup (ii) Henry Bastin
(c) Coacervates (iii) Thomas Huxley
(d) Abiogenesis (iv) Haldane
(a) a – iii b – iv c – ii d – i
(b) a – ii b – iv c – i d – iii
(c) a – iii b – i c – iv d – ii
(d) a – i b – iv c – iii d – ii
Answer:
(b) a – ii b – iv c – i d – iii

Question 7.
Anatomical structures that have similar functions but not similar structures are called
(a) Homologous structures
(b) Vestigial structures
(c) Analogus structures
(d) Generalized structures
Answer:
(c) Analogous structures

Question 8.
Who propounded the theory of recapitulation?
(a) Ernst Von Haeckel
(b) Charles Darwin
(c) Thomas Huxley
(d) Oparin
Answer:
(c) Ernst Von Haeckel

Question 9.
Mammal in human male is
(a) Atavistic organ
(b) Rudimentary Organ
(c) Vestigial organ
(d) Homologous structure
Answer:
(c) Vestigial organ

Question 10.
Which of the following is/are not examples of analogous structure
(a) Wings of Birds and Bats
(b) Wings of Birds and Insects
(c) Thom of Bougainvillea and Tendril of cururbita
(d) Flippers of Penguins and Dolphins
(i) a, b, c
(ii) a and c
(iii) b andd
(iv) All the above
Answer:
(ii) a and c

Question 11.
identify the mismatched pairs
(a) Thom of Bougainvillea and Tenrdril of – Analogy
(b) Forelimbs of whale and cat – Analogy
(c) Octopus eye & Mammalian eye – Homology
(d) Root of sweet potato & stem of potato – Homology
Answer:
(a) Thorn of Bougainvillea & Terdril of crucurbita – Analogy

Question 12.
Witnesses for evolution are found in
(a) Rocks
(b) Ocean beds
(c) Fossils
(d) Desert
Answer:
(c) Fossils

Question 13.
Assertion (A): Oparin used the term coacervases
Reason (R): Coacervates are colloidal particles in aqueous environment
(a) Both A and Rare incorrect
(b) Both A and R are correct
(c) Both A and R are correct. R explains A.
(d) A is correct R is incorrect
Answer:
(c) Both A and R are correct. R explains A.

Question 14.
According to the theory of spontaneous generation, life originated from
(a) Cosmic particles
(b) Non-living materials
(c) Coacervates
(d) Sea
Answer:
(b) Non-living materials

Question 15.
Assertion (A): Hardy – Weinberg principle states that allelic frequency of a population remain constant
Reason (R) : Constancy is maintained through natural selection and mutation
(a) A is true R is false
(b) A is false R is true
(c) Both A and R are true
(d) R explains
Answer:
(a) A is true R is false

Question 16.
Calculate the allelic frequency of Aa. frequency of 0.7
(a) 0.67
(b) 0.42
(c) 0.36
Answer:
(b) 0.42

Question 17.
Match the following

Answer:
(a) a – iv b – i c – ii d – iii

Question 18.
Genetic drift leads to
(a) Mutation
(b) Bottle neck effect
(c) Immigration
(d) Isolation
Answer:
(b) Bottle neck effect

Question 19.
Atavism refers to
(a) Inheritance of triat by mother
(b) Inheritance of triat by father
(c) Criss-cross inheritance
(d) Inheritance of characters not shown by parents
Answer:
(d) Inheritance of characters not shown by parents

2 – Mark Questions

Question 1.
State the theory of spontaneous generation.
Answer:
According to the theory of spontaneous generation or Abiogenesis, living organisms originated from non-living materials and occurred through stepwise chemical and molecular evolution over millions of years. Thomas Huxley coined the term abiogeneis.

Question 2.
List the four eras of geological time scale.
Answer:

1. Precambrian era
2. Paleozoic era
3. Mesozoic era
4. Cenozoic era

Question 3.
Which periods of paleozoic era are referred as

1. Age of fishes
2. Invertebrates

Answer:

1. Age of fishes – Devonian period
2. Age of invertebrates – Cambrian period

Question 4.
Point out the epochs of carboniferous period.
Answer:

1. Pennsylvanian
2. Mississippian

Question 5.
Compare relative dating with absolute dating.
Answer:
Relative dating is used to determine a fossil by comparing it to similar rocks and fossils of known age. Absolute dating is used to determine the precise age of a fossil by using radiometric dating to measure the decay of isotopes

Question 6.
Wing of a cockroach and the wing of parrot. What do you infer from this statement with reference to evolution?
Answer:
Both the wings of cockroach and bird are different in structure but similar in their function. Thus, they are analogous structure that brings about convergent evolution.

Question 7.
Name the scientists who propounded the following theories.

1. Mutation theory
2. Chemical theory of evolution

Answer:

1. Mutuation theory was propounded by Hugo de Vries.
2. Chemical theory of evolution was propounded by Oparin and Haldane

Question 8.
Define fossilization and mention its types.
Answer:
Fossilization is the process by which plant and animal remains are preserved in sedimentary rocks. It is of three major types,

1. Actual remains
2. Petrifaction
3. Natural moulds and casts.

Question 9.
Name the principle minerals involved in petrifaction.
Answer:
Iron pyrites, silica, calcium carbonate and bicarbonates of calcium and magnesium.

Question 10.
What is meant by petrifaction?
Answer:
When animals die the original portion of their body may be replaced molecule for molecule by minerals and the original substance being lost through disintegration. This method of fossilization is called petrifaction. The principle minerals involved in this type fossilization are iron pyrites, silica, calcium carbonate and bicarbonates of calcium and magnesium.

Question 11.
Define analogous organ with an example.
Answer:
Organisms having different structural patterns but similar function are termed as analogous structures. For example, the wings of birds and insects are different structurally but perform the same function of flight that brings about convergent evolution.

Question 12.
Mention any four organs homologous to human hand.
Answer:
Flippers of whale, wings of bat, wings of bird and forelimb of horse.

Question 13.
Thorn of Bougainvillea and tendrils of Pisum sativum represent homology. How?
Answer:
The thorn of Bougainvillea and the tendrils of Curcurbita and Pisum sativum represent homology. The thorn in former is used as a defence mechanism from grazing animals and the tendrils of latter is used as a support for climbing.

Question 14.
Which type of evolution is brought out by homologous structures and analogous structures?
Answer:
Homologous structures brings about divergent evolution. Analogous structures brings about convergent evolution.

Question 15.
What are vestigial organs? Give example.
Answer:
Structures that are of no use to the possessor, and are not necessary for their existence are called vestigial organs. Vestigial organs may be considered as remnants of structures which were well developed and functional in the ancestors, but disappeared in course of evolution due to their non-utilization.
E.g: Human appendix.

Question 16.
Human appendix is a vestige. Give reason.
Answer:
Human appendix is the remnant of caecum which is functional in the digestive tract of herbivorous animals like rabbit. Cellulose digestion takes place in the caecum of these animals. Due to change in the diet containing less cellulose, caecum in human became functionless and is reduced to a vermiform appendix, which is vestigial.

Question 17.
What are connecting link? Give example.
Answer:
The organisms which possess the characters of two different groups (transitional stage) are called connecting links. Example Peripatus (connecting link between Annelida and Arthropoda) Archaeopteryx (connecting link between Reptiles and Aves).

Question 18.
Name one fossilised connecting link between reptiles and Aves also one living connecting link between Annelida and Arthropoda.
Answer:
Archaeopteryx – connecting link between Reptiles and Aves.
Peripatus – Connecting link between Annelida and Arthropoda.

Question 19.
Why it is considered as a connecting link?
Answer:
Peripatus is a worm that shown the characters of both Annelidia and Arthropoda. Hence it is a connecting link between Annelida and Arthropoda.

Question 20.
Atavistic organs – comment.
Answer:
Sudden appearance of vestigial organs in highly evolved organisms is called atavistic organs. Example, presence of tail in human baby is an atavistic organ.

Question 21.
Define Ontogeny and Phytogeny.
Answer:

1. Ontogeny refers to the life history of an individual.
2. Phytogeny refers to the evolutionary history of a race.

Question 22.
Who proposed the theory of recapitulation? State the theory.
Answer:
Ernst Von Haeckel proposed the theory of recapitulation, which states that life history of an individual briefly repeats the evolutionary history of the race.

Question 23.
Name few Neo – Lamarckists.
Answer:
Cope, Osborn, Packard and Spencer.

Question 24.
Who proposed the theory of acquired characters? Also mention the scientist who disproved it.
Answer:
The theory of acquired characters was proposed by Jean Baptise de Lamarck and it was disproved by August Weismann.

Question 25.
Point out the basic principles of Darwin’s theory of evolution.
Answer:
Over production, struggle for existence, Universal occurence of variation, Survival of fittest and Natural selection.

Question 26.
Name any four Neo – Darwinists.
Answer:
Gregor Mendel, August Weismann, Russel Wallace and Heinrich.

Question 27.
Enumerate the salient features of mutation theory.
Answer:

1. Mutations or discontinuous variation are transmitted to other generations.
2. In naturally breeding populations, mutations occur from time to time.
3. There are no intermediate forms, as they are fully fledged.
4. They are strictly subjected to natural selection.

Question 28.
Who proposed Mutation theory? Name the organism on which the experiment was carried out.
Answer:
Mutation theory was put forth by Hugo de Vries. Based on the experiments in Oenothera lamarckiana (The evening primrose plant).

Question 29.
What are the basic factors of modern synthetic theory that leads to evolution?
Answer:
Gene mutation, Chromosomal mutation, Genetic recombination, Natural selection and Reproductive isolation.

Question 30.
Name the scientists who supported modern synthetic theory.
Answer:
Sewell Wright, Dobzhansky, Huxley and Simpson.

Question 31.
Define point mutation.
Answer:
Gene mutation refers to the changes in the structure of the gene. It is also called gene / point mutation. It alters the phenotype of an organism and produces variations in their offsprings.

Question 32.
Point out the factors that alters allelic frequency of a population.
Answer:
Natural selection, Genetic drift, Mutation and Geneflow

Question 33.
Mention any two differences between Homo habilis and Homo erectus
Answer:

1. Homo habilis: The brain capacity was between 650-800 cc. They were probably vegetarians.
2. Homo erectus: The brain capacity was around 900 cc. They probably ate meat.

Question 34.
Write a brief note on Homo sapiens with respect to evolution.
Answer:
Homo sapiens or modem human arose in Africa some 25,000 years ago and moved to other continents and developed into distinct races. They had a brain capacity of 1300 – 1600 cc. “They started cultivating crops and domesticating animals.

Question 35.
Define evolution.
Answer:
The term evolution describes heritable changes in one or more characteristics of a population of species from one generation to the other.

3 – Mark Question

Question 36.
Write a short note on Big Bang theory.
Answer:
Big bang theory explains the origin of universe as a singular huge explosion in physical terms. The primitive Earth had no proper atmosphere, but consisted of ammonia, methane, hydrogen and water vapour. The climate of the Earth was extremely high. UV rays from the Sun split up water molecules into hydrogen and oxygen. Gradually the temperature cooled and the water vapour condensed to form rain. Rain water filled all the depressions to form water bodies. Ammonia and methane in the atmosphere combined with oxygen to form carbon dioxide and other gases.

Question 37.
Theory of chemical evolution states that organisms have evolved from inorganic substances. If so, what was the atmospheric condition that favoured evolution?
Answer:
The atmosphere was devoid of O₂, and with high level of CO₂, NH₃ and UV radiations.

Question 38.
Name the periods of Mesozoic era. Also mention the flora and fauna dominates during that periods.
Answer:

1. Mesozoic era is divided into three periods namely Triassic, Jurassic and Cretaceous.
2. Dominating Fauna : Reptiles and Dinosaurs Dominating
3. Flora : Conifers, Ferns and Ginkgon.

Question 39.
Which era is referred as Age of Mammals? What are the periods of that era? And also mention the fauna during the periods.
Answer:
Cenozoic era is called as Age of Mammals. Tertiary and Quaternary are the two periods of Cenozoic era. Tertiary periods marks the abundance of mammalian fauna. Quaternary period marks the beginning of human social life.

Question 40.
Write a short note on Cenozoic era.
Answer:
Cenozoic era (Age of mammals) is subdivided into two periods namely Tertiary and Quaternary. Tertiary period is characterized by abundant mammalian fauna. This period is subdivided into five epochs namely, Paleocene (placental mammals, Eocene (Monotremes except duck billed Platypus and Echidna, hoofed mammals and carnivores), Oligocene (higher placental mammals appeared), Miocene (origin of first man like apes) and Pliocene (origin of man from man like apes). Quaternary period witnesses decline of mammals and beginning of human social life.

Question 41.
Name the gaseous mixture used in Urey – Miller’s experiment. Which type of physical force is applied to generate amino acids?
Answer:
Ammonia, Methane, Hydrogen, Water vapour are the gaseous mixture allowed to circulate over electric discharge from a tungsten electrode.

Question 42.
Which is the most common methods of fossilization? Explain how it occurs.
Answer:
Actual remains – The original hard parts such as bones, teeth or shells are preserved as such in the Earth’s atmosphere. This is the most commpn method of fossilization. When marine animals die, their hard parts such as bones and shells, etc., are covered with sediments and are protected from further deterioration.

They get preserved as such as they are preserved in vast ocean; the salinity in them prevents decay. The sediments become hardened to form definite layers or strata. For example, Woolly Mammoth that lived 22 thousand years ago were preserved in the frozen coast of Siberia as such. Several human beings and animals living in the ancient city of Pompeii were preserved intact by volcanic ash which gushed out from Mount Vesuvius.

Question 43.
What are coprolites? Mention its role in phytogeny.
Answer:
Coprolites are the hardened faecal matters occurs as small pieces. Analysing the coprolites helps to understand the nature of diet of pre-historic animals.

Question 44.
What are moulds and casts?
Answer:
Even after disintegration, the body of an animal might leave indelible impression on the soft mud which later becomes hardened into stones. Such impressions are called moulds. The cavities of the moulds may get filled up by hard minerals and get fossilized, which are called casts.

Question 45.
How will you compute the age of fossil?
Answer:
The age of fossils can be determined using two methods namely, relative dating and absolute dating. Relative dating is used to determine a fossil by comparing it to similar rocks and fossils of known age. Absolute dating is used to determine the precise age of a fossil by using radiometric dating to measure the decay of isotopes.

Question 46.
“Ontogeny recapitulates phylogeny” – comment on the statement with example.
Answer:
The embryonic stages of a higher animal resemble the adult stage of its ancestors. Appearance of pharyngeal gill slits, yolk sac and the appearances of tail in human embryos are some of the examples.

Question 47.
Biogenetic law is not universal – justify.
Answer:
The biogenetic law is not universal and it is now thought that animals do not recapitulate the adult stage of any ancestors. The human embryo recapitulates the embryonic history and not the adult history of the organisms.

Question 48.
How macro molecules like DNA and RNA play their crucial role in evolutionary history?
Answer:
Molecular evolution is the process of change in the sequence composition of molecules such as DNA, RNA and proteins across generations. It uses principles of evolutionary biology and population genetics to explain patterns in the changes of molecules.

One of the most useful advancement in the development of molecular biology is proteins and other molecules that control life processes are conserved among species. A slight change that occurs over time in these conserved molecules (DNA, RNA and protein) are often called molecular clocks. Molecules that have been used to study evolution are cytochrome (respiratory pathway) and rRNA (protein synthesis).

Question 49.
Explain the principles of Lamarckian theory.
Answer:

1. The theory of use and disuse – Organs that are used often will increase in size and those that are not used will degenerate. Neck in giraffe is an example of use and absence of limbs in snakes is an example for disuse theory.
2. The theory of inheritance of acquired characters – Characters that are developed during ’ the life time of an organism are called acquired characters and these are then inherited.

Question 50.
Write a note on Mutation theory.
Answer:
Hugo de Vries put forth the Mutation theory. Mutations are sudden random changes that occur in an organism that is not heritable. De Vries carried out his experiments in the Evening Primrose plant (Oenothera lamarckiana) and observed variations in them due to mutation. According to de Vries, sudden and large variations were responsible for the origin of new species whereas Lamarck and Darwin believed in gradual accumulation of all variations as the causative factors in the origin of new species.

Question 51.
What do you mean by “adaptive radiation”? Give example.
Answer:
The evolutionary process which produces new species diverged from a single ancestral form becomes adapted to newly invaded habitats is called adaptive radiation. Adaptive radiations are best exemplified in closely related groups that have evolved in relatively short time. Darwin’s finches and Australian marsupials are best examples for adaptive radiation.

Question 52.
Darwins finches are the classical examples studied for adaptive radation. Explain.
Answer:
Darwin’s finches are the birds whose common ancestor arrived on the Galapagos about 2 million year ago. During that time, Darwin’s finches have evolved into 14 recognized species differing in body size, beak shape and feeding behavior. Changes in the size and form of the beak have enabled different species to utilize different food resources such as insects, seeds and nectar from cactus flowers and blood from iguanas, all driven by Natural selection. Genetic variation in the ALX1 gene in the DNA of Darwin finches is associated with variation in the beak shape. Mild mutation in the ALX1 gene leads to phenotypic change in the shape of the beak of the Darwin finches.

Question 53.
What is micro evolution?
Answer:
Microevolution (evolution on a small scale) refers to the changes in allele frequencies within a population. Allele frequencies in a population may change due to four fundamental forces of evolution such as natural selection, genetic drift, mutation and gene flow.

Question 54.
Name the major types of Natural Selection.
Answer:

1. Stabilising Selection
2. Directional Selection
3. Disruptive Selection

Question 55.
What do you mean by gene flow?
Answer:
Movement of genes through gametes or movement of individuals in (immigration) and out (emigration) of a population is referred to as gene flow. Organisms and gametes that enter the population may have new alleles or may bring in existing alleles but in different proportions than those already in the population. Gene flow can be a strong agent of evolution.

Question 56.
Give an account on Genetic drift. Mention its impact over a population.
Answer:
Genetic drift is a mechanism of evolution in which allele frequencies of a population change over generation due to chance (sampling error). Genetic drift occurs in all population sizes, but its effects are strong in a small population. It may result in a loss of some alleles (including beneficial ones) and fixation of other alleles. Genetic drift can have major effects, when the population is reduced in size by natural disaster due to bottle neck effect or when a small group of population splits from the main population to form a new colony due to founder’s effect.

Question 57.
State Hardy – Weinberg equilibrium.
Answer:
The allele frequencies in a population are stable and are constant from generation to generation in the absence of gene flow, genetic drift, mutation, recombination and natural selection.

Question 58.
Write in brief about the characters of Australian ape man.
Answer:
Australopithecus lived in East African grasslands about 5 mya and was called the Australian ape man. He was about 1.5 meters tall with bipedal locomotion, omnivorous, semi erect, and lived in caves. Low forehead, brow ridges over the eyes, protruding face, lack of chin, low brain capacity of about 350 – 450 cc, human like dentition, lumbar curve in the vertebral column were his distinguishing features.

Question 59.
Who is Cro-Magnon?
Answer:
Cro-Magnon was one of the most talked forms of modem human found from the rocks of Cro-Magnon, France and is considered as the ancestor of modem Europeans. They were not only adapted to various environmental conditions, but were also known for their cave paintings, figures on floors and walls.

5 – Mark Questions

Question 60.
Explain Oparin – Haldane hypothesis on evolution.
Answer:
According to the theory of chemical evolution primitive organisms in the primordial environment of the Earth evolved spontaneously from inorganic substances and physical forces such as lightning, UV radiations, volcanic activities, etc. Oparin (1924) suggested that the organic compounds could have undergone a series of reactions leading to more molecules. He proposed that the molecules formed colloidal aggregates or ‘coacervates’ in an aqueous environment. The coacervates were able to absorb and assimilate organic compounds from the environment. Haldane (1929) proposed that the primordial sea served as a vast chemical laboratory powered by solar energy.

The atmosphere was oxygen free and the combination of CO₂, NH₂ and UV radiations gave rise to organic compounds. The sea became a ‘hot’ dilute soup containing large populations of organic monomers and polymers. They envisaged that groups of monomers and polymers acquired lipid membranes and further developed into the first living cell. Haldane coined the term prebiotic soup and this became the powerful symbol of the Oparin-Haldane view on the origin of life (1924-1929). Oparin and Haldane independently suggested that if the primitive atmosphere was reducing – and if there was appropriate supply of energy such as lightning or UV light then a wide range of organic compounds can be synthesized.

Question 61.
How Urey – Miller’s experiment supports the origin of life?
Answer:
Urey and Miller (1953), paved way for understanding the possible synthesis of organic compounds that led to the appearance of living organisms is depicted in the Figure In their experiment, a mixture of gases was allowed to circulate over electric discharge from an tungsten electrode. A small flask was kept boiling and the steam emanating from it was made to mix with the mixture of gases (ammonia, methane and hydrogen) in the large chamber that was connected condensed to form water which ran down the ‘U’ tube.

Experiment was conducted continuously for a week and the liquid was analysed. Glycine, alanine, beta alanine and aspartic acid were identified. Thus Miller’s experiments had an insight as to the possibility of abiogenetic synthesis of large amount of variety of organic compounds in nature from a mixture of sample gases in which the only source of carbon was methane. Later in similar experiments, formation of all types of amino acids, and nitrogen bases were noticed.

Question 62.
Give a detailed account of Modern Synthetic Theory.
Answer:
Sewell Wright, Fisher, Mayer, Huxley, Dobzhansky, Simpson and Haeckel explained Natural Selection in the light of Post-Darwinian discoveries. According to this theory gene mutations, chromosomal mutations, genetic recombinations, natural selection and reproductive isolation are the five basic factors involved in the process of organic evolution.

1. Gene mutation refers to the changes in the structure of the gene. It is also called gene/ point mutation. It alters the phenotype of an organism and produces variations in their off +springs.
2. Chromosomal mutation refers to the changes in the structure of chromosomes due to deletion, addition, duplication, inversion or translocation. This too alters the phenotype of an organism and produces variations in their offspring.
3. Genetic recombination is due to crossing over of genes during meiosis. This brings about genetic variations in the individuals of the same species and leads to heritable variations.
4. Natural selection does not produce any genetic variations but once such variations occur it favours some genetic changes while rejecting others (driving force of evolution).
5. Reproductive isolation helps in preventing interbreeding between related organisms

Higher Order Thinking Skills (HO’ts) Questions

Question 1.
Name the connecting link for the following groups of organisms.

1. Annelida and Arthropoda
2. Reptiles and Aves
3. Pisces and Amphibians
4. Reptiles and Mammals

Answer:

1. Peripatus
2. Archeopteryx
3. Lung fish
4. Platypus

Question 2.
Point out any four condition under which Hardy Weinberg’s equilibrium is not attained.
Answer:

1. Selected mating
2. Flow of genes (either by immigration or emmigration)
3. Occurance of mutation
4. Definite population size

Question 3.
Why are analogous structures a result of convergent evolution?
Answer:
Analogous structures are not anatomically similar though they perform same function.

Question 4.
Organs which are of no use to the organism is called as vestige. Name any four vestigal organs that can be noticed in your body.
Answer:
Wisdom teeth, Mammae in male, Body hair and Coccyx.

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Samacheer Kalvi 12th Bio Zoology Applications of Biotechnology Text Book Back Questions and Answers

Question 1.
The first clinical gene therapy was done for the treatment of ______
(a) AIDS
(b) Cancer
(c) Cystic fibrosis
(d) SCID
Answer:
(d) SCID

Question 2.
Dolly, the sheep was obtained by a technique known as _______
(a) Cloning by gene transfer
(b) Cloning without the help of gametes
(c) Cloning by tissue culture of somatic cells
(d) Cloning by nuclear transfer
Answer:
(d) Cloning by nuclear transfer

Question 3.
The genetic defect adenosine deaminase deficiency may be cured permanently by ______
(a) Enzyme replacement therapy
(b) periodic infusion of genetically engineered lymphocytes having ADA cDNA
(c) administering adenosine deaminase activators
(d) introducing bone marrow cells producing ADA into embryo at an early stage of development.
Answer:
(d) introducing bone marrow cells producing ADA into embryo at an early stage of development.

Question 4.
How many amino acids are arranged in the two chains of Insulin?
(a) Chain A has 12 and Chain B has 13
(b) Chain A has 21 and Chain B has 30 amino acids
(c) Chain A has 20 and chain B has 30 amino acids
(d) Chain A has 12 and chain B has 20 amino acids
Answer:
(b) Chain A has 21 and Chain B has 30 amino acids

Question 5.
PCR proceeds in three distinct steps governed by temperature, they are in order of ______
(a) Denaturation, Annealing, Synthesis
(b) Synthesis, Annealing, Denaturation
(c) Annealing, Synthesis, Denaturation
(d) Denaturation, Synthesis, Annealing
Answer:
(a) Denaturation, Annealing, Synthesis

Question 6.
Which one of the following statements is true regarding DNA polymerase used in PCR?
(a) It is used to ligate introduced DNA in recipient cells
(b) It serves as a selectable marker
(c) It is isolated from a Virus
(d) It remains active at a high temperature
Answer:
(d) It remains active at a high temperature

Question 7.
ELISA is mainly used for ______
(a) Detection of mutations
(b) Detection of pathogens
(c) Selecting animals having desired traits
(d) Selecting plants having desired traits
Answer:
(b) Detection of pathogens

Question 8.
Transgenic animals are those which have
(a) Foreign DNA in some of their cells
(b) Foreign DNA in all their cells
(c) Foreign RNA in some of their cells
(d) Foreign RNA in all their cells
Answer:
(b) Foreign DNA in all their cells

Question 9.
Recombinant Factor VIII is produced in the ______ cells of the Chinese Hamster
(a) Liver cells
(b) blood cells
(c) ovarian cells
(d) brain cells
Answer:
(c) ovarian cells

Question 10.
Vaccines that use components of a pathogenic organism rather than the whole organism are called ______
(a) Subunit recombinant vaccines
(b) attenuated recombinant vaccines
(c) DNA vaccines
(d) conventional vaccines
Answer:
(a) Subunit recombinant vaccines

Question 11.
Mention the number of primers required in each cycle of PCR. Write the role of primers and DNA polymerase in PCR. Name the source organism of the DNA polymerase used in PCR.
Answer:

1. For each cycle of PCR two primers are required.
2. Primers are the small fragments of single stranded DNA or RNA which serves as template for initiating DNA polymerization.
3. DNA polymerase is an enzyme that synthesize DNA molecules by pairing the Deoxyribo Nucleotides leading to formation of new strands.
4. DNA polymerase used in PCR is Taq polymerase which is isolated from a thermophilic bacteria called Thermus aquatics. Taq polymerase will remain active ever at very high temperature (80°C) and hence used in PCR amplification technique.

Question 12.
How is the amplification of a gene sample of interest carried out using PCR?
Answer:
Denaturation, renaturation or primer annealing and synthesis or primer extension, are the three steps involved in PCR. The double stranded DNA of interest is denatured to separate into two individual strands by high temperature . This is called denaturation. Each strand is allowed to hybridize with a primer (renaturation or primerannealing). The primer template is used to synthesize DNA by using Taq – DNA polymerase.During denaturation the reaction mixture is heated to 95 °C for a short time to denature the target DNA into single strands that will act as a template for DNA synthesis.

Annealing is done by rapid cooling of the mixture, allowing the primers to bind to the sequences on each of the two strands flanking the target DNA. During primer extension or synthesis the temperature of the mixture is increased to 75°C for a sufficient period of time to allow Taq DNA polymerase extend each primer by copying the single stranded template.

At the end of incubation both single template strands will be made partially double stranded. The new strand of each double stranded DNA extends to a variable distance downstream. These steps are repeated again and again to generate multiple forms of the desired DNA. This process is also called DNA amplification.

Question 13.
What is genetically engineered Insulin?
Answer:
The insulin synthesized by recombinant DNA technology is called genetically engineered Insulin. It was the first ever pharmaceutical product of DNA technology. In 1986, human insulin was marked under the trade name Humulin.

Question 14.
Explain how “Rosie” is different from a normal cow.
Answer:
Rosie was the first transgenic cow. It produced human protein enriched milk, which contained the human alpha lactalbumin (2.4 gm/litre). This milk was a nutritionally balanced food for infants than the normal milk of cows.

Question 15.
How was Insulin obtained before the advent of rDNA technology? What were the problems encountered?
Answer:
Conventionally, Insulin was isolated and refined from the pancreas of pigs and cows to treat diabetic patients. Though it is effective, due to minor structural changes, the animal insulin caused allergic reaction in few patients.

Question 16.
ELISA is a technique based on the principles of antigen-antibody reactions. Can this technique be used in the molecular diagnosis of a genetic disorder such as Phenylketonuria?
Answer:
Yes, ELISA test can be done to diagnose phenylketonuria. The affected person does not produce the enzyme phenylalanine hydroxylase. If specific antibodies are developed against the enzyme and ELISA is performed, the unaffected person will show positive result due to antigen and antibody reaction, whereas the affected individual produces negative result. [Note: phenylketonuria is an inherited metabolic disorder that causes the accumulation of Phenylalanine (an amino acid) in body cells due to defect in the synthesizing of an enzyme phenylalanine hydroxylase]

Question 17.
Gene therapy is an attempt to correct a Genetic defect by providing a normal gene into the individual. By this the function can be restored. An alternate method would be to provide gene product known as enzyme replacement therapy, which would also restore the function. Which in your opinion is a better option? Give reasons for your answer.
Answer:
Though both Gene therapy and Enzyme replacement therapy helps to restore the genetic defects, Gene therapy is much better than Enzyme replacement therapy. Because, in Gene therapy once the defective gene is repaired using normal gene, the affected individual gains complete recovery whereas, in Enzyme replacement therapy, the respective enzyme or protein has to be provided periodically and does not offer permanent cure. Moreover when compared to Gene therapy, the Enzyme replacement therapy is highly expensive.

Question 18.
What are transgenic animals? Give examples.
Answer:
Transgenesis is the process of introduction of extra (foreign/exogenous) DNA into the genome of the animals to create and maintain stable heritable characters. The foreign DNA that is introduced is called the transgene and the animals that are produced by DNA manipulations are called transgenic animals or the genetically engineered or genetically modified organisms.
Example: Mice, Cow

Question 19.
If a person thinks he is infected with HIV, due to unprotected sex, and goes for a blood test. Do you think a test such as ELISA will help? If so why? If not, why?
Answer:
Yes, ELISA is a highly sensitive and precise procedure and can detect antigens even in the range of a nanogram. So, it can be used to detect HIV in blood.

Question 20.
Explain how ADA deficiency can be corrected?
Answer:
The right approach for SCID treatment would be to give the patient a functioning ADA which breaks down toxic biological products. In some children ADA deficiency could be cured by bone marrow transplantation, where defective immune cells could be replaced with healthy immune cells from a donor. In some patients it can be treated by enzyme replacement therapy, in which functional ADA is injected into the patient.

During gene therapy the lymphocytes from the blood of the patient are removed and grown in a nutrient culture medium. A healthy and functional human gene, ADA cDNA encoding this enzyme is introduced into the lymphocytes using a retrovirus. The genetically engineered lymphocytes are subsequently returned to the patient. Since these cells are not immortal, the patient requires periodic infusion of such genetically engineered lymphocytes. The disease could be cured permanently if the gene for ADA isolated from bone marrow cells are introduced into the cells of the early embryonic stages.

Question 21.
What are DNA vaccines?
Answer:
Genetic immunisation by using DNA vaccines is a novel approach that came into being in 1990. The immune response of the body is stimulated by a DNA molecule. A DNA vaccine consists of a gene encoding an antigenic protein, inserted onto a plasmid, and then incorporated into the cells in a target animal. DNA instructs the cells to make antigenic molecules which are displayed on its surfaces. This would evoke an antibody response to the free floating antigen secreted by the cells. The DNA vaccine cannot cause the disease as it contains only copies of a few of its genes. DNA vaccines are relatively easy and inexpensive to design and produce.

Question 22.
Differentiate between Somatic cell gene therapy and Germline gene therapy.
Answer:
Somatic Cell Gene Therapy:

1. Therapeutic genes transferred into the somatic cells.
2. Introduction of genes into bone marrow cells, blood cells, skin cells etc.
3. Will not be inherited in later generations.

Germ Line Gene Therapy:

1. Therapeutic genes transferred into the germ cells.
2. Genes introduced into eggs and sperms.
3. Heritable and passed on to later generations.

Question 23.
What are stem cells? Explain its role in the field of medicine.
Answer:
Stem cells are undifferentiated cells found in most of the multi cellular animals. These cells maintain their undifferentiated state even after undergoing numerous mitotic divisions.

Stem cell research has the potential to revolutionize the future of medicine with the ability to regenerate damaged and diseased organs. Stem cells are capable of self renewal and exhibit ‘cellular potency’. Stem cells can differentiate into all types of cells that are derived from any of the three germ layers ectoderm, endoderm and mesoderm.

Question 24.
One of the applications of biotechnology is ‘gene therapy” to treat a person born with a hereditary disease

1. What does “gene therapy” mean?
2. Name the hereditary disease for which the first clinical gene therapy was used.
3. Mention the steps involved in gene therapy to treat this disease.
4. Gene therapy is the process in which the defective genes are replaced with normal genes leading to the expression of proper phenotype.

Answer:

1. SCID (Severe Combined Immuno Deficiency) disease was the first disease treated by using gene therapy.
2. There are two strategies involved in gene therapy namely Gene augmentation therapy, which involves insertion of DNA into the genome to replace the missing gene product and Gene inhibition therapy, which involves insertion of the anti sense gene which inhibits the expression of the dominant gene.

Question 25.
PCR is a useful tool for early diagnosis of an Infectious disease. Elaborate.
Answer:
The specificity and sensitivity of PCR is useful for the diagnosis of inherited disorders (genetic diseases), viral diseases, bacterial diseases, etc., The diagnosis and treatment of a particular disease often requires identifying a particular pathogen. Traditional methods of identification involve culturing these organisms from clinical specimens and performing metabolic and other tests to identify them. The concept behind PCR based diagnosis of infectious diseases is simple – if the pathogen is present in a clinical specimen its DNA will be present.

Its DNA has unique sequences that can be detected by PCR, often using the clinical specimen (for example, blood, stool, spinal fluid, or sputum) in the PCR mixture.

Question 26.
What are recombinant vaccines? Explain the types.
Answer:
Vaccines developed by using recombinant DNA technology are called recombinant vaccines. Subunit recombinant vaccines, attenuated recombinant vaccines, DNA vaccines are the types of recombinant vaccines.

Question 27.
Explain why cloning of Dolly, the sheep was such a major scientific breakthrough?
Answer:
The development of Dolly was a remarkable achievement in scientific field and it demonstrates thatthe DNA from differentiated adult cells can also be used to develop into an entire organism.

Question 28.
Mention the advantages and disadvantages of cloning.
Answer:

1. Offers benefits for clinical trials and medical research. It can help in the production of proteins and drugs in the field of medicine.
2. Aids stem cell research.
3. Animal cloning could help to save endangered species.
4. Animal and human activists see it as a threat to biodiversity saying that this alters evolution which will have an impact on populations and the ecosystem.
5. The process is tedious and very expensive.
6. It can cause animals to suffer.
7. Reports show that animal surrogates were manifesting adverse outcomes and cloned animals were affected with disease and have high mortality rate.
8. It might compromise human health through consumption of cloned animal meat.
9. Cloned animals age faster than normal animals and are less healthy than the parent organism as discovered in Dolly
10. Cloning can lead to occurrence of genetic disorders in animals.
11. More than 90% of cloning attempts fail to produce a viable offspring.

Question 29.
Explain how recombinant Insulin can be produced.
Answer:
Production of insulin by recombinant DNA technology started in the late 1970s. This technique involved the insertion of human insulin gene on the plasmids of E. coli. The polypeptide chains are synthesized as a precursor called pre-pro insulin, which contains A and B segments linked by a third chain (C) and preceded by a leader sequence. The leader sequence is removed after translation and the C chain is excised, leaving the A and B polypeptide chains Explain the steps involved in the production of recombinant hGH.

Using recombinant DNA technology hGH can be produced. The gene for hGH is isolated from the human pituitary gland cells. The isolated gene is inserted into a plasmid vector and then is transferred into E. coli. The recombinant E. coli then starts producing human growth hormone. The recombinant E. coli are isolated from the culture and mass production of hGH is carried out by fermentation technology.

Samacheer Kalvi 12th Bio Zoology Applications of Biotechnology Additional Questions and Answers

1 – Mark Questions

Question 1.
Statement 1: Human Insulin is a polypeptide
Statement 2: It is composed of 52 amino acids
(a) Statement 1 is true. Statement 2 is false.
(b) Statement 1 is false. Statement 2 is true.
(c) Both statements 1 and 2 are true.
(d) Both statements 1 and 2 are false.
Answer:
(a) Statement 1 is true. Statement 2 is false.

Question 2.
Statement 1: Rosie was the first transgenic goat.
Statement 2: Meat is enriched with human protein.
(a) Statement 1 is true. Statement 2 is false.
(b) Statement 1 is false. Statement 2 is true.
(c) Both statements 1 and 2 are true.
(d) Both statements 1 and 2 are false.
Answer:
(d) Both statements 1 and 2 are false.

Question 3.
Statement 1: Recombinant Hepatitis B vaccine is a live vaccine.
Statement 2: It is obtained by cloning HB antigen gene in yeast.
(a) Statement 1 is true. Statement 2 is false.
(b) Statement 1 is false. Statement 2 is true.
(c) Both statements 1 and 2 are true.
(d) Both statements 1 and 2 are false.
Answer:
(b) Statement 1 is false. Statement 2 is true.

Question 4.
Statement 1: ADA deficiency was the first disease treated by gene therapy.
Statement 2: ADA is an autosomal recessive metabolic disorder.
(a) Statement 4 is true. Statement 2 is false.
(b) Statement 1 is false. Statement 2 is true.
(c) Both statements 1 and 2 are true.
(d) Both statements 1 and 2 are false.
Answer:
(c) Both statements 1 and 2 are true.

Question 5.
Statement 1: Attenuated recombinant vaccines are live vaccines.
Statement 2: Polio is a live vaccine.
(a) Statement 1 is true. Statement 2 is false.
(b) Statement 1 is false. Statement 2 is true.
(c) Both statements 1 and 2 are true.
(d) Both statements 1 and 2 are false.
Answer:
(c) Both statements 1 and 2 are true.

Question 6.
Assertion (A): Interferons are used to treat herpes zoster.
Reason (R): Interferons are antiviral protein.
(a) R explains A.
(b) Both A and Rare incorrect.
(c) A is correct. R is incorrect.
(d) A and R are correct. R does not explains A.
Answer:
(a) R explains A.

Question 7.
Assertion (A): PCR is a amplification technique used in biotechnology.
Reason (R): Using PCR multiple copies of DNA can be generated.
(a) R explains A.
(b) Both A and Rare incorrect.
(c) A is correct. R is incorrect.
(d) A and R are correct. R does not explains A.
Answer:
(a) R explains A.

Question 8.
The B-chain of Insulin is composed of aminoacids __________
(a) 70
(b) 30
(c) 45
(d) 60
Answer:
(b) 30

Question 9.
The gene for the formation of factor VIII is located in __________
(a) 20th Chromosome
(b) 12th Chromosome
(c) X-chromosome
(d) Y-chromosome
Answer:
(c) X-chromosome

Question 10.
The genetic defect in the synthesis of factor VIII results in __________
(a) Polycythemia
(b) Anaemia
(c) Thalassemia
(d) Haemophilia
Answer:
(c) Haemophilia

Question 11.
Name the scientists who discovered Interferons?
Answer:
Alick Issac and Jean Lindemann

Question 12.
Which is the first synthetic vaccine produced?
(a) Polio Vaccine
(b) Hepatitis B Vaccine
(c) BCG Vaccine
(d) MMR Vaccine
Answer:
(b) Hepatitis B Vaccine

Question 13.
Identify the incorrect statement.
(i) The first clinical gene therapy was given by French Anderson.
(ii) For a four year old boy with ADA deficiency.
(iii) ACD is a autosomal dominant metabolic disorder.
(iv) Where patients have non-functioning B – lymphocytes.
(a) i and iv only
(b) ii, iii and iv
(c) i, ii and iv
(d) all the above
Answer:
(b) ii, iii and iv

Question 14.
Identify the correct statement(s).
(i) Totipotency is the ability of single cell to produce a whole organism.
(ii) Pluripotency refers to ability of stem cell with apotential to differentiate into any kind of germ layers.
(iii) Unipotency refers to ability of stem cell to differentiate into one cell type.
(iv) Oligopotency refers to stem cells to differentiate into few cell types.
(a) i and iii
(b) ii and iv
(c) i and iv
(d) all the above
Answer:
(d) all the above

Question 15.
Identify those proper sequence of ELISA testing.
(a) Coating → Blocking → Detection → Read out
(b) Detection → Read out → Coating → Blocking
(c) Read out → Coating → Detection → Blocking
(d) Blocking → Detection → Read out → Coating
Answer:
(a) Coating → Blocking → Detection → Read out

Question 16.
PCR technique was developed by
(a) Eva Engvall
(b) Peter Perlmanin
(c) Kary Mullis
(d) Wilmut
Answer:
(c) Kary Mullis

Question 17.
Arrange the steps of PCR in proper sequence.
(a) Denaturation, Primer extension, Renaturation
(b) Renaturation, Denaturation, Primer extension
(c) Primer extension, Denaturation, Renaturation
(d) Denaturation, Renaturation, Primer extension
Answer:
(d) Denaturation, Renaturation, Primer extension

Question 18.
The first cloned organism was.
(a) Goat
(b) Cow
(c) Sheep
(d) Pig
Answer:
(c) Sheep

Question 19.
The first transgenic clone of sheep was called as
(a) Rosie
(b) Dolly
(c) Sameera
(d) Joel
Answer:
(b) Dolly

Question 20.
In cloning process of Dolly, how many embryos were implemented by Ian Wilmut and Campbell, out of which one successful Dolly was developed?
(a) 267
(b) 211
(c) 287
(d) 307
Answer:
(b) 277

Question 21.
The term Biotechnology was coined by
Answer:
Karl Ereky

2 – Mark Questions

Question 1.
How insulin controls blood sugar level?
Answer:
Insulin controls the blood sugar level by facilitating the cellular uptake and utilisation of glucose for the release of energy.

Question 2.
State the role of Somatostatin and Somatotropin in human beings.
Answer:
Both somatostatin and somatotropin are peptide hormones which helps in growth and development by increasing the uptake of amino acids and promoting protein synthesis.

Question 3.
Mention the manifestation of the disease – Haemophilia-A
Answer:
Haemophilia A is a X-linked disease which is characterised by prolonged clotting time and internal bleeding.

Question 4.
Define Interferons.
Answer:
Interferons are proteinaceous, antiviral, species specific substances produced by mammalian cells when infected with viruses. They stimulate the cellular DNA to produce antiviral enzymes which inhibit viral replication and protect the cells.

Question 5.
Who discovered Interferons? On which basis it was classified?
Answer:
Interferons were discovered by Alick Isaacs and Jean Lindemann in 1957. It is classified as P and y interferons based on the structure.

Question 6.
Name the disease that are treated by using interferons.
Answer:
Interferons are used for the treatment of various diseases like cancer, AIDS, multiple sclerosis, hepatitis C and herpes zoster.

Question 7.
Recombinant vaccines are better than conventional ones – Justify.
Answer:
The recombinant vaccines are generally of uniform quality and produce less side effects as compared to the vaccines produced by conventional methods.

Question 8.
Point out four types of recombinant vaccines.
Answer:

1. Subunit recombinant vaccines
2. Attenuated recombinant vaccines
3. Edible vaccines
4. DNA vaccines

Question 9.
What are subunit recombinant vaccines? Mention its advantages.
Answer:
Vaccines that use components of a pathogenic organism rather than the whole organism are called subunit vaccines. The advantages of these vaccines include their purity in preparation, stability and safe use.

Question 10.
Define Attenuated recombinant vaccines.
Answer:
Attenuated recombinant vaccines includes genetically modified pathogenic organisms (bacteria or viruses) that are made nonpathogenic and are used as vaccines. Such vaccines are referred to as attenuated recombinant vaccines.

Question 11.
List out the benefits of recombinant vaccines.
Answer:
Vaccines produced by recombinant techniques have definite advantages like producing target proteins, long lasting immunity and trigger immune response only against specific pathogens with less toxic effects.

Question 12.
Name the two strategies involved in gene therapy
Answer:

1. Gene augmentation therapy.
2. Gene inhibition therapy.

Question 13.
Comment on SCID.
Answer:
ADA deficiency or SCID (Severe Combined Immuno Deficiency) is an autosomal recessive metabolic disorder. It is caused by the deletion or dysfunction of the gene coding for ADA enzyme. In these patients the nonfunctioning T-Lymphocytes cannot elicit immune responses against invading pathogens.

Question 14.
Differentiate between Gene augmentation therapy and gene inhibition therapy.
Answer:
Gene augmentation therapy which involves insertion of DNA into the genome to replace the missing gene product and Gene inhibition therapy which involves insertion of the anti sense gene which inhibits the expression of the dominant gene.

Question 15.
Define the terms

1. Totipotency
2. Unipotency

Answer:

1. Totipotency is the ability of a single cell to divide and produce all of the differentiated cells in an
2. organism.Unipotency refers to the ability of the stem cells to differentiate into only one cell type.

Question 16.
What are the best sources of stem cells in mammals?
Answer:
Placenta, Umbilical cord, amniotic sac, amniotic fluid.

Question 17.
Write the names of any two molecular diagnostic techniques used for early diagnosis of diseases?
Answer:

1. Polymerase Chain Reaction (PCR) technique.
2. Enzyme Linked Immuno Sorbent Assay (ELISA)

Question 18.
What does ELISA stands for? Who invented this technique?
Answer:
Enzyme Linked Immuno Sorbent Assay (ELISA). It was invented by Eva Engvall and Peter Perlmanin.

Question 19.
Name the various kinds of ELISA.
Answer:
There are four kinds of ELISA namely, Direct ELISA, Indirect ELISA, sandwich ELISA and competitive ELISA.

Question 20.
Simply define the PCR technique. Also mention its inventor.
Answer:
The Polymerase Chain Reaction (PCR) is an invitro amplification technique used for synthesising multiple identical copies (billions) of DNA of interest. The technique was developed by Kary Mullis in the year 1983.

Question 21.
Expand PCR and name the steps involved in the process.
Answer:

1. PCR – Polymerase Chain Reaction.
2. Denaturation, Renaturation or Primer annealing and Primer extension are the three steps in PCR technique.

Question 22.
For which disease does the first clinical gene therapy was done? Who accomplished it?
Answer:
The first clinical gene therapy was done for SCID. Severe Combined Immuno Deficiency disease is caused by ADA deficiency. It was done by French Anderson in 1990.

Question 23.
Define Transgenesis.
Answer:
Transgenesis is the process of introduction of foreign DNA (exogenous DNA) into the genome of the other organism to create and maintain stable heritable characters.

Question 24.
What are the Genetically Modified Organisms?
Answer:
Transgenesis is the process of introduction of extra (foreign/ exogenous) DNA into the genome of the animals to create and maintain stable heritable characters. The foreign DNA that is introduced is called the transgene and the animals that are produced by DNA manipulations are called transgenic animals or the genetically engineered or genetically modified organisms.

Question 25.
What does Biological Product refers to?
Answer:
A biological product is a substance derived from a living organism and used for the prevention or treatment of disease. These products include antitoxins, bacterial and viral vaccines, blood products and hormone extracts.

Question 26.
Define cloning. Name the first organism developed by cloning.
Answer:
Cloning is the process of producing genetically identical individuals of an organisms either naturally or artificially. The first cloned organism is a sheep named Dolly.

Question 27.
Who developed Dolly? How many embryos were aborted to develop single Dolly?
Answer:
Dolly- The first cloned organism (sheep) was developed by Iam Wilmut and Campbell. Out of 29 embryos implanted only one Dolly was developed.

Question 28.
Define Biotechnology.
Answer:
Biotechnology is defined as “any technological application that uses biological systems, living organisms or derivatives thereof, to make or modify products or processes for specific use”.

3 – Mark Questions

Question 29.
Briefly explain the structure of insulin.
Answer:
The Human insulin is synthesized by the (5 cells of Islets of Langerhans in the pancreas. It is formed of 51 aminoacids which are arranged in two polypeptide chains, A and B. The polypeptide chain A has 21 amino acids while the polypeptide chain B has 30 amino acids. Both A and B chains are attached together by disulphide bonds.

Question 30.
Who was the first to discover the role of insulin against diabetes? From which organism does was insulin isolated?
Answer:
Best and Banting in 1921, isolated insulin from the pancreatic islets of a dog and demonstrated its effectiveness against diabetes.

Question 31.
How “Rosie” differs from normal cow? Explain.
Answer:
Rosie, the first transgenic cow produced human protein enriched milk, which contained the human alpha lactalbumin. The protein rich milk (2.4 gm/litre) was a nutritionally balanced food for new born babies than the normal milk produced by the cows.

Question 32.
Point out any two microbes that play crucial role in recombinant DNA technology.
Answer:

1. Saccharomyces cerevisiae
2. Escherichia coli

Question 33.
What are Edible vaccines?
Answer:
Edible vaccines are prepared by molecular pharming using the science of genetic engineering. Selected genes are introduced into plants and the transgenic plants are induced to manufacture the encoded protein. Edible vaccines are mucosal targeted vaccines which cause stimulation of both systemic and mucosal immune response. At present edible vaccines are produced for human and animal diseases like measles, cholera, foot and mouth disease and hepatitis.

Question 34.
How recombinant hepatitis B vaccine is produced in laboratory?
Answer:
Recombinant hepatitis B vaccine as a subunit vaccine is produced by cloning hepatitis B surface antigen (HbsAg) gene in the yeast, Saccharomyces cerevisiae.

Question 35.
Suggest few methods to treat SCID.
Answer:
SCID caused by ADA deficiency could be cured by bone marrow transplantation where defective immune cells could be replaced with healthy immune cells from donor. It can also be treated by enzyme replacement therapy in which functional ADA is injected into patient’s body where it breaks down toxic biological product.

Question 36.
How gene therapy is done to treat ADA deficiency?
Answer:
During gene therapy the lymphocytes from the blood of the patient are removed and grown in a nutrient culture medium. A healthy and functional human gene, ADA cDNA encoding this enzyme is introduced into the lymphocytes using a retrovirus. The genetically engineered lymphocytes are subsequently returned to the patient. Since these cells are not immortal, the patient requires periodic infusion of such genetically engineered lymphocytes. The disease could be cured permanently if the gene for ADA isolated from bone marrow cells are introduced into the cells of the early embryonic stages.

Question 37.
How does Somatic cell therapy differ from germ line gene therapy?
Answer:
Somatic cell therapy involves the insertion of a fully functional and expressible gene into a target somatic cell to correct a genetic disease permanently whereas Germ line gene therapy involves the introduction of DNA into germ cells which is passed on to the successive generations. Gene therapy involves isolation of a specific gene and making its copies and inserting them into target cells ’to make the desired proteins.

Question 38.
Differentiate between Pluripotency and Multipotency.
Answer:

1. Pluripotency refers to a stem cell that has the potential to differentiate into any of the three germ layers-ectoderm, endoderm and mesoderm.
2. Multipotency refers to the stem cells that can differentiate into various types of cells that are related. For example blood stem cells can differentiate into lymphocytes, monocytes, neutrophils etc.

Question 39.
Write a short note on stem cell banks.
Answer:
Stem cell banking is the extraction, processing and storage of stem cells, so that they may be used for treatment in the future, when required. Amniotic cell bank is a facility that stores stem cells derived from amniotic fluid for future use. Stem cells are stored in banks specifically for use by the individual from whom such cells have been collected and the banking costs are paid.

Cord Blood Banking is the extraction of stem cells from the umbilical cord during childbirth. While the umbilical cord and cord blood are the most popular sources of stem cells, the placenta, amniotic sac and amniotic fluid are also rich sources in terms of both quantity and quality.

Question 40.
State any two uniqueness of ELISA test.
Answer:

1. ELISA is highly sensitive and can detect antigen even in nanograms.
2. ELISA test does not require radio isotopes or radiation counting apparatus.

Question 41.
What is ELISA test?
Answer:
ELISA – Enzyme Linked Immuno Sorbent Assay is a biochemical procedure done to detect the presence of specific antibodies or antigens or hormones in a sample of serum, urine etc.

Question 42.
Elucidate the methodology of ELISA test.
Answer:
During diagnosis the sample suspected to contain the antigen is immobilized on the surface of an ELISA plate. The antibody specific to this antigen is added and allowed to react with the immobilized antigen. The anti-antibody is linked to an appropriate enzyme like peroxidase. The unreacted anti-antibody is washed away and the substrate of the enzyme (hydrogen peroxidase) is added with certain reagents such as 4-chloronaphthol. The activity of the enzyme yields a coloured product indicating the presence of the antigen.

Question 43.
Whether PCR is applicable for RNA molecules? Explain.
Answer:
The PCR technique can also be used for amplifications of RNA in which case it is referred to as reverse transcription PCR (RT-PCR). In this process the RNA molecules (mRNA) must be converted to complementary DNA by the enzyme reverse transcriptase. The cDNA then
serves as the template for PCR.

Question 44.
How PCR helps forensic personnel?
Answer:
PCR technique can also be used in the field of forensic medicine . A single molecule of DNA from blood stains, hair, semen of an individual is adequate for amplification by PCR. The amplified DNA is used to develop DNA fingerprint which is used as an important tool in forensic science.Thus, PCR is very useful for identification of criminals. PCR is also used in amplification of specific DNA segment to be used in gene therapy.

Question 45.
Role of PCR in phylogenetics. Explain.
Answer:
The differences in the genomes of two different organisms can be studied by PCR. PCR is very important in the study of evolutions, more specifically phylogenetics. As a technique which can amplify even minute quantities of DNA from any source, like hair, mummified tissues, bones or any fossilized materials.

Question 46.
Enumerate the use of biological products.
Answer:
Antibodies are substances that react against the disease causing antigens and these can be produced using transgenic animals as bioreactors. Monoclonal antibodies, which are used to treat cancer, heart disease and transplant rejection are produced by this technology. Natural protein adhesives are non toxic, biodegradable and rarely trigger an immune response, hence could be used to reattach tendons and tissues, fill cavities in teeth, and repair broken bones.

Question 47.
Name the principles underlying cloning technique.
Answer:

1. Nuclear transfer
2. Totipotency (ability of a cell to develop into entire organism)

5 – Mark Questions

Question 48.
Explain in detail about stem cell therapy.
Answer:
Stem cells are undifferentiated cells found in most of the multi cellular animals. These cells maintain their undifferentiated state even after undergoing numerous mitotic divisions.

Stem cell research has the potential to revolutionize the future of medicine with the ability to regenerate damaged and diseased organs. Stem cells are capable of self renewal and exhibit ‘cellular potency’ Stem cells can differentiate into all types of cells that are derived from any of the three germ layers ectoderm, endoderm arid mesoderm.

In mammals there are two main types of stem cells – embryonic stem cells (ES cells) and adult stem cells. ES cells are pluripotent and can produce the three primary germ layers ectoderm, mesoderm and endoderm. Embryonic stem cells are multipotent stem cells that can differentiate into a number of types of cells. ES cells are isolated from the epiblast tissue of the inner cell mass of a blastocyst. When stimulated ES can develop into more than 200 cells types of the adult body. ES cells are immortal they can proliferate in a sterile culture medium and maintain their undifferentiated state.

Adult stem cells are found in various tissues of children as well as adults. An adult stem cell or somatic stem cell can divide and create another cell similar to it. Most of the adult stem cells are multipotent and can act as a repair system of the body, replenishing adult tissues.The red bone marrow is a rich source of adult stem cells.

The most important and potential application of human stem cells is the generation of cells and tissues that could be used for cell based therapies. Human stem cells could be used to test new drugs.

Question 49.
Describe the role of PCR in clinical field.
Answer:
PCR In Clinical Diagnosis: The specificity and sensitivity of PCR is useful for the diagnosis of inherited disorders (genetic diseases), viral diseases, bacterial diseases, etc., The diagnosis and treatment of a particular disease often requires identifying a particular pathogen. Traditional methods of identification involve culturing these organisms from clinical specimens and performing metabolic and other tests to identify them.

The concept behind PCR based diagnosis of infectious diseases is simple – if the pathogen is present in a clinical specimen its DNA will be present. Its DNA has unique sequences that can be detected by PCR, often using the clinical specimen (for example, blood, stool, spinal fluid, or sputum) in the PCR mixture. PCR is also employed in the prenatal diagnosis of inherited diseases by using chorionic villi samples or cells from amniocentesis. Diseases like sickle cell anemia, P-thalassemia and phenylketonuria can be detected by PCR in these samples. cDNA from PCR is a valuable tool for diagnosis and monitoring retroviral infections – eg. Tuberculosis by Mycobacterium tuberculosis.

Several virally induced cancers, like cervical cancer caused by Papilloma virus can be detected by PCR. Sex of human beings and live stocks, embryos fertilized invitro can be determined by PCR by using primers and DNA probes specific for sex chromosomes. PCR technique is also used to detect sexlinked disorders in fertilized embryos.

Question 50.
Enumerate the steps involved in producing transgenic animals.
Answer:
The various steps involved in the production of transgenic organisms are,

1. Identification and separation of desired gene.
2. Selection of a vector (generally a virus) or direct transmission.
3. Combining the desired gene with the vector.
4. Introduction of transferred vector into cells, tissues, embryo or mature individual.
5. Demonstration of integration and expression of foreign gene in transgenic tissue or animals. Transgenic animals such as mice, rat, rabbit, pig, cow, goat, sheep and fish have been produced.

Question 51.
List out the uses of Transgenesis.
Answer:

1. Transgenesis is a powerful tool to study gene expression and developmental processes in higher organisms.
2. Transgenesis helps in the improvement of genetic characters in animals.Transgenic animals serve as good models for understanding human diseases which help in the investigation of new treatments for diseases.Transgenic models exist for many human diseases such as cancer, Alzheimer’s, cystic fibrosis, rheumatoid arthritis and sickle cell anemia.
3. Transgenic animals are used to produce proteins which are important for medical and pharmaceutical applications.
4. Transgenic mice are used for testing the safety of vaccines.
5. Transgenic animals are used for testing toxicity in animals that carry genes which make them sensitive to toxic substances than non-transgenic animals exposed to toxic substances and their effects are studied.
6. Transgenesis is important for improving the quality and quantity of milk, meat, eggs and wool production in addition to testing drug resistance.

Question 52.
Describe the procedure by which Dolly was developed.
Answer:
Dolly was the first mammal (Sheep) clone developed by Ian Wilmut and Campbell in 1997. Dolly, the transgenic clone was developed by the nuclear transfer technique and the phenomenon of totipotency. Totipotency refers to the potential of a cell to develop different cells, tissues, organs and finally an organism.

The mammary gland udder cells (somatic cells) from a donor sheep (ewe) were isolated and subjected to starvation for 5 days. The udder cells could not undergo normal growth cycle, entered a dormant stage and became totipotent. An ovum (egg cell) was taken from another sheep (ewe) and its nucleus was removed to form an enucleated ovum. The dormant mammary gland cell/udder cell and the enucleated ovum were fused. The outer membrane of the mammary cell was ruptured allowing the ovum to envelope the nucleus.

The fused cell was implanted into another ewe which served as a surrogate mother. Five months later dolly was bom. Dolly was the first animal to be cloned from a differentiated somatic cell taken from an adult animal without the process of fertilization.

Question 53.
What are the ethical issues about cloning.
Answer:
Biotechnology has given to the soceity cheap drugs, better fruits and vegetables, pest resistant crops, indigenious cure to diseases and lot of controversy. This is mainly because the major part

of the modem biotechnology deals with genetic manipulations. People fear that these genetic manipulations may lead to unknown consequences. The major apprehension of recombinant DNA technology is that unique microorganisms either inadvertently or deliberately for the purpose of war may be developed that could cause epidemics or environmental catastrophies. Although many are concerned about the possible risk of genetic engineering, the risks are in fact slight and the potential benefits are substantial.

Higher Order Thinking (HOTs) Questions

Question 1.
The immune system of a person is suppressed,. In ELISA test, the result is positive

1. Name the disease associated with this.
2. Why did he loose his immunity?

Answer:

1. AIDS caused by Human Immuno Vims.
2. In AIDS, the pathogen destroys the T-lymphocytes which forms the major immune resouce of our body.

Question 2.
Why do children cured by enzyme replacement therapy for ADA deficiency need periodic treatment? Suggest a permanent solution for this issue.
Answer:
During gene therapy the lymphocytes from the blood of the patient are removed and grown in a nutrient culture medium. A healthy and functional human gene, ADA cDNA encoding this enzyme is introduced into the lymphocytes using a retrovirus. The genetically engineered lymphocytes are subsequently returned to the patient. Since these cells are not immortal, the patient requires periodic infusion of such genetically engineered lymphocytes. The disease could be cured permanently if the gene for ADA isolated from bone marrow cells are introduced into the cells of the early embryonic stages.

Question 3.
Saccharomyces cerevisiae, acts as a best host than Encherichia coli for the production of recombinant interferons. Yes or No? Support your answer.
Answer:
Yes. The Saccharomyces cerevisiae, is the best source of recombinant interferon than E-coli. Since E-coli does not possess the machinery for glycolysation of protein.

Question 4.
Isolation of blood to treat Haemophilia A is practically impossible. Give reason.
Answer:

1. Requirement of large quantity of blood.
2. Risk of transmission of blood related diseases like AIDS.

Question 5.
Functional Insulin differs from its pre-hormonal form. How?
Answer:
Pro-Insulin contains A and B segments linked by a C – chain and preceded by a leader sequence. Whereas the functional Insulin contains only A and B chain formed by the excision of C-chain and leaders sequence after translation.

Question 6.
Whether PCR can be done for RNA molecules? Explain.
Answer:
The PCR technique can also be used for amplifications of RNA in which case it is referred to as reverse transcription PCR (RT-PCR). In this process the RNA molecules (mRNA) must be converted to complementary DNA by the enzyme reverse transcriptase. The cDNA then serves as the template for PCR.

Question 7.
Suggest any two techniques for early diagnosis of bacterial / viral human diseases.
Answer:
PCR and ELISA

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