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Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Bio Botany Guide Pdf Chapter 10 Economically Useful Plants and Entrepreneurial Botany Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 12th Bio Botany Solutions Chapter 10 Economically Useful Plants and Entrepreneurial Botany

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

I. Choose the correct answer :

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:
a) Assertion is correct, Reason is wrong

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 – Lawsonia inermis
c) Timber – Cyperus papyrus
d) Pulp – Hevea brasiliensis
Answer:
b) Dye : Lawsonia inermis

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:
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) Stevecide – Natural sweetener
d) Palmyra sap – Fermented to give ethanol
Answer:
c) Stevecide – Natural sweetener

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

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:

• These are foods that are prepared and eaten as whole grain. Eg. quinoa (தினை) is a seed from Chenopodium quinoa plant. It belongs Amaranthaceae family.
• It is gluten free, whole grain carbohydrate.
• It is a whole protein with a essential amino acids.
• Taken for 6000 years in Andes hills.

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:

• Henna is the best alternative dye.
• It is in North Africa, South west Asia. It is in Gujarat, Madhya Pradesh and Rajesthan.
• Orange dye henna is from leaves and shoots of
  Lawsonia inermis.
• Principal colouring matter is ‘lacosone’
• It is harmless causing no skin irritation.
• It is u sed to dye skin, hair and finger nails.

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:

• Alternative agricultural system.
• Plants and crops are cultivated in natural ways, by using biological inputs.
• It helps to maintain soil fertility and ecological balance.
• It minimizes pollution, wastage.

Question 21.
Which is called 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:
Bio-medicines: Medicinally useful molecules obtained from plants are marketed as drugs. These are called bio-medicines.
Botanical Medicines: Medicinal plants are marketed as powders or in other modified forms. They are called Botanical medicines.

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

• Native of India
• Archaeological evidence is in Maharashtra.
• Cultivated in Madhya Pradesh, Karnataka and Tamil Nadu.

Origin and area of cultivation of Red Gram

• The only pulse native of South India.
• Grown in Maharashtra, Andhra Pradesh,
  Madhya Pradesh, Karnataka, Gujarat.

Question 24.
What are millets? What are its types? Give example for each type.
Answer:
Millet’s: Small seeds cultivated by ancient people of Africa, Asia. Gluten-free with the less glycemic index.

Finger Millet (Ragi) (Eleusine coracana)
Came to India from East Africa. It is rich in calcium.’

Uses:

• Staple food in South Indian hills.
• Made into porridge, gruel.
• Ragi malt is a nutrient drink.
• Source of fermented beverages.

Sorghum vulgare.
Native of Africa. Major millet of the world with calcium, iron
Uses:

• Feed to poultry, birds, pigs, cattle
• Alcoholic beverage source.

Fox tail Millet (Setariaitalica)
Oldest traditional millet of India. Domesticated in China about 6000 years.
Uses: Strengthens heart, eye sight, lactation.

Kodo Millet (Paspalum scrobiculatum)
From West Africa.
Uses:

• Flour for pudding
• Diuretic, cures constipation.
• Reduce obesity, blood sugar, blood pressure.

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:
Benefits of Coffee:

• Stimulates central nervous system.
• Mild diuretic
• Enhances acetyl choline release in brain.
• Enhances efficiency.
• Lower fatty liver diseases, cirrhosis, cancer.
• 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 classify and what does it focus on?
Answer:
Traditional System of Medicines (TSM)
It is classified into

1. Institutionalized (documented) system
2. Non-institutionalized (oral) system

Institutionalized system:

• It includes Siddha, Ayurvedha
• It is practiced for 2000 years.
• Text with symptoms, diagnosis, drugs, preparation of drugs, dosage, diet regimen.

Non-Institutionalized system:

• Do not have any records
• Practiced by rural, tribal people of India.
• Knowledge is in oral form.

Focus of TSM:

• Healthy lifestyle
• A healthy diet for good health
• Disease reversal.

Siddha system

• Siddha is the most popular, widely practiced and culturally accepted systm in Tamil Nadu.
• Siddha is principally based on the pancabute philosophy
• This system specializes in using minerals for preparing drugs with a 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.

Ayurveda system:

• Ayurveda supposed to have originated from Brahma.
• The core knowledge is documented by charaka, sushruta and vagbhata in compendiums written by them.
• This system uses more of herbs and few animal parts as drug sources.
• Plant sources include a good propertion of Himalayan plants.
• The Ayurvedic pharmacopoeia of India lists about 500 plants used as source of drugs.

Folk system of medicine

• Major tribal communities in Tamil Nadu who are known for their medicinal knowledge indued Irulas, Malayalis, Kurumbas, paliyans and kaanis.
• Folk system survive as oral traditions among innumerable rural and tribal communities of India.

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 of the role of Jasminum in perfuming.
Answer:
Role of Jasminum in perfuming:

• Used in India for worship, ceremonial purposes, incense, fumigants.
• For making perfumed hair oil, cosmetics and soaps.
• Essential oil for soothing relaxing, antidepressant qualities.
• Blends with other perfumes.
• Used in modern perfumery and cosmetics.
• Popular in air fresheners, antiperspirants, talcum powder, shampoo, and deodorants.

Rose:
The average oil yield is a little less than 0.5 g from lOOOg of flowers.
Uses:

• Rose oil is largely used in perfumes, scenting
  soaps, flovouring soft drinks, liqueurs, and certain types of tobacco, particularly snuff of chewing tobacco.
• In India, water is much used in eye lotions
  and eyewashes.
• Rosewater (panner) containing much of phenyl ethyl alcohol and other compounds in dissolved confectioneries syrups and soft drinks.
• In addition, it is sprinkled on guests as a ceremonial welcome.

Question 30.
Give an account of active principle and medicinal values of any two plants you have studied.
Answer:
A) Medical importance of Keezhanelli (Phyllanthus amarus):
Active principle: Phyllanthus is a major chemical component.
Medical Importance:

• Hepatoprotective.
• Used in Tamil Nadu for jaundice treatment.
• Effective against hepatitis B virus.

B) Nilavembu (Andrographis paniculata) (King of Bitters)
Active principle: Andrographolides.
Medicinal Importance:

• Potent hepatoprotective
• Treats liver disorders.
• A concoction of Andrographis + 8 herbs
  (Nilavembu Kudineer) treats malaria, dengue.

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 system of Medicine:

• It is widely practiced and culturally accepted in Tamil Nadu.
• Based on text of 18siddhars.
• Knowledge is documented as Tamil poems.
• Based on Pancabuta philosophy.
• Vatam, Pittam, Kapam are 3 humors. They are responsible for the health.
• Drug sources are plant, animal parts, marine products, minerals.
• Minerals are used for preparing drugs with long self life.
• 800 herbs are source of drugs.
• Disease prevention, health promotion, rejuvenation and cure are important.

Question 33.
What are psychoactive drugs? Add a note of Marijuana and Opium.
Answer:
Phytochemicals or drugs from some of the plants alter an individual’s 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 – tetrahydrocannabinol (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:
Queen of Spices: Cardamom (Elettaria Cardamomum)

Origin and area of cultivation:

• Indigenous to Southern India and Sri Lanka.
• Main cash crop in the Western Ghats, North-Eastern India.

Uses:

• For flavouring confectionaries, Bakery products, beverages.
• Seeds are used in curry powder, pickles and cakes.
• Medicinally, a stimulant and carminative (a drug for flatulence)
• Chewed as mouth fresheners.

King of Spices: Black Pepper (Piper nigrum)

Origin and area of cultivation:

• Indigenous to western ghats.
• Black gold of India.
• Kerala, Karnataka and Tamil Nadu are top producers in India.
• Pungency is due to alkaloid piperine.
• 2 types (Black pepper, white pepper)

Uses:

• Flavouring sauce, soup, curry, and pickles
• Aromatic stimulant for salivary gastric secretions as a stomachic.
• Pepper enhances the absorption of medicines.

Question 35.
How will you prepare an organic pesticide for your home garden with the vegetables available from your kitchen?
Answer:
Preparation of Organic Pesticide:

Step 1: Mix 120 g of hot chilies 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.

12th Bio Botany Guide Economically Useful Plants and Entrepreneurial Botany Additional Important Questions and Answers

I. Choose the correct answer :

Question 1.
Staple food of North India is ………………….
a) Sorghum
b) Millet
c) Paddy
d) Wheat
Answer:
d) Wheat

Question 2.
Folk system of medicine is popular in ………………
a) Nigeria
b) USA
c) India
d) UK
Answer:
c) India

Question 3.
State not growing black gram
a) Uttar Pradesh
b) Tamil Nadu
c) Chattisgarh
d) Karnataka
Answer:
b) Tamil Nadu

Question 4.
The very common rubber yielding plant of Tamilnadu is ……………………
a) Manihot esculenta
b) Ficus elastica
c) Hevea benthamiana
d) Hevea brasiliensis
Answer:
d) Hevea brasiliensis

Question 5.
Not a major cultivar mango in India
a) Alphonsa
b) Neelam
c) Malgova
d) Salem Mango
Answer:
d) Salem Mango

Question 6.
Toddy is from ……………… tree
a) Palmyra
b) Coconut
c) Mango
d) Sugar cane
Answer:
a) Palmyra

Question 7.
Chillies are a good source of:
a) Vitamin A, C and E
b) Vitamin K
c) Vitamin D
d) Vitamin B complex and Vitamin D
Answer:
a) Vitamin A, C and E

Question 8.
Gingeelly or sesame is originated in ……………………..
a) Asia
b) Africa
c) China
d) Europe
Answer:
b) Africa

Question 9.
Coffee is native of ……………….
a) Nigeria
b) Cuba
c) Ethiopia
d) Egypt
Answer:
c) Ethiopia

Question 10.
India is the largest producer of
a) Chilly
b) Tamarind
c) Turmeric
d) Pepper
Answer:
c) Turmeric

Question 11.
World’s largest turmeric market is in ………………. of Tamil Nadu
a) Coimbatore
b) Erode
c) Madurai
d) Nagercoil
Answer:
b) Erode

Question 12.
Asia contributes …………… % of latex in world production.
a) 80
b) 90
c) 70
d) 50
Answer:
b) 90

Question 13.
……………….. is the largest producer of latex
a) Kerala
b) Karnataka
c) Andhra
d) Delhi
Answer:
a) Kerala

Question 14.
…………….. is native of Sudan
a) Henna
b) Aloe
c) Jasmine
d) Turmeric
Answer:
b) Aloe

Question 15.
Thovalai of Tamil Nadu produces ………………..
a) Aloe
b) Tamarind
c) Turmeric
d) Jasmine
Answer:
d) Jasmine

Question 16.
Paste of this plant is used in bone fracture
a) Ocimum
b) Phyllanthus
c) Cissus
d) Acalypha
Answer:
c) Cissus

Question 17.
Find the Matching Pair
a) Ocimum – Antiseptic
b) Phyllanthus – Ringworm disease
c) Acalypha – Immune modulator
d) Aegle marmelos – Bone fracture
Answer:
a) Ocimum – Antiseptic

Question 18.
Capsaicin is in ……………..
a) Chilly
b) Pepper
c) tea
d) coffee
Answer:
a) Chilly

Question 19.
Veldt grape is the common name of …………….
a) Ocimum
b) I’hyllanthus
c) Acalypha
d) Cissus
Answer:
d) Cissus

Question 20.
Find the Mismatching Pair
a) Pappaver somniferum – Opium
b) Cannabis sativa – Marijuana
c) Phyllanthus amarus – Keezhanelli
d) Andrographis paniculata – Turmeric
Answer:
d) Andrographis paniculata – Turmeric

Question 21.
Match

a) A-4, B-3, C-l, D-2
b) A-l, B-2, C-3, D-4
c) A-4, B-3, C-2, D-l
d) A-2, B-l, C-4, D-3
Answer:
a) A-4, B-3, C-l, D-2

Question 22.
Vigna mungo is the botanical name of ………………..
a) Black gram
b) Red gram
c) Green gram
d) Brown gram
Answer:
a) Black gram

Question 23.
Match

a) A-4, B-3, C-2, D-l
b) A-l, B-2, C-3, D-4
c) A-2, B-l, C-4, D-3
d) A-3, B-l, C-2, D-4
Answer:
a) A-4, B-3, C-2, D-l

Question 24.
Which one of the following is an incorrect pair?
a) Turmeric – Erode
b) Cardamom – Queen of spices
c) Rubber – Kerala
d) Banana – National fruit of India
Answer:
d)Banana – National fruit of India

Question 25.
Assertion (A): Rice is the staple food for most of people in the world.
Reason (R): It is easily digestible and calorie-rich food.
a) (A) correct; (R) wrong
b) (A) wrong; (R) correct
c) (A) correct; (R) correct; but (R) does not explain (A)
d) (A) correct; (R) correct; (R) explains (A)
Answer:
d) (A) correct; (R) correct; (R) explains (A)

Question 26.
………………… is the largest consumer of coffee in India?
a) Tamil Nadu
b) Andhra
c) Kerala
d) Karnataka
Answer:
a) Tamil Nadu

Question 27.
………………… is the largest coffee producing estate in India
a) Kerala
b) Karnataka
c) Tamil Nadu
d) Andhra
Answer:
b) Karnataka

Question 28.
Curcumin is extracted from
a) Turmeric
b) Chilly
c) Cardamom
d) Tamarind
Answer:
a) Turmeric

Question 29.
Vilvum belongs to ……………………
a) Lamiaceae
b) Rutaceae
c) Vitaceae
d)Euphorbiaceae
Answer:
b) Rutaceae

Question 30.
Dr. Thyagarajan of university of Madras proved effect of Phyllanthus amarus against
a) Hepatitis-B
b) Cirrhosis
c) Cancer
d) Typhoid
Answer:
a) Hepatitis-B

Question 31.
Which one of the following is highly effective against jaundice?
a) Nilavembu
b) Opium poppy
c) Marijuana
d) Phyllanthus
Answer:
d) Phyllanthus

Question 32.
……………… Are gluten free with less Glycemic indess
a) pulses
b) gram
c) vegetables
d) millets
Answer:
d) millets

Question 33.
…………….. is native to tropical region of Africa.
a) Sugar cane
b) Palmyra
c) Peanut
d) Sesame
Answer:
b) Palmyra

Question 34.
Nuts contain ……………. Oil
a) 54%
b) 45%
c) 44%
d) 54%
Answer:
c) 44%

Question 35.
The medicinal plant commonly known as “King of Bitters” is ……………………
a) Nilavembu
b) Holy basil
c) Adathodai
d) Turmeric
Answer:
a) Nilavembu

Question 36.
Pungency of cayenne pepper is ……………….. Scoville Heat Units (SHU)
a) 30,000 to 50,000
b) 1,349,000
c) 2,200,000
d) 1,200,000
Answer:
a) 30,000 to 50,000

Question 37.
Foxtail millet is domesticated in China ……………….. years ago
a) 4000
b) 3000
c) 5000
d) 6000
Answer:
d) 6000

Question 38.
Setaria italica is the scientific name of ……………………
a) kodo millet
b) foxtail millet
c) sorghum
d) finger millet
Answer:
b) foxtail millet

Question 39.
Lady’s finger is not grown in abundance in ……………..
a) Tamil Nadu
b) Assam
c) Maharashtra
d) Gujarat
Answer:
a) Tamil Nadu

Question 40.
Which is the temperature region fruit?
a) Mango
b) Jack
c) Banana
d) plum
Answer:
c) Banana and d) plum

Question 41.
The following are the activities of entrepreneurship
a) Mushroom cultivation
b) Single cell protein production
c) Organi farming
d) Above all
Answer:
d) Above all

Question 42.
…………………. is a bio-pest repellent
a) Tamarind
b) Chilly
c) Sesame
d) Neem
Answer:
d) Neem

Question 43.
Indigenous to western ghats of India
a) Black pepper
b) Cardamom
c) Turmeric
d) Red pepper
Answer:
a) Black pepper

Question 44.
Endosperm of ………………… is a refreshing summer food
a) Coconut
b) Groundnut
c) Gingelly
d) Palmyra
Answer:
d) Palmyra

Question 45.
……………….. Enhances salivary and gastric secretions
a) Cardamom
b) Black pepper
c) Red pepper
d) Turmeric
Answer:
b) Black pepper

Question 46.
………………… is used in gerontological applications
a) Aloe
b) Turmeric
c) Jasmine
d) Phyllanthus
Answer:
a) Aloe

Question 47.
Lacosone (Colouring Matter) is in ……………………
a) Aloe
b) Jasminum
c) Henna
d) Turmeric
Answer:
c) Henna

Question 48.
Paper pulp is made from ………………
a) Eucalyphis
b) Casuarina
c) Neolamarkia
d) all the above
Answer:
d) all the above

Question 49.
Eco friendly packaging material is ……………………
a) cotton
b) latex
c) wood pulp
d) jute
Answer:
d) jute

Question 50
……………….. is a ingredient of Ponga I of Tamil Nadu
a) Green gram
b) Red gram
c) Black gram
d) Brown gram
Answer:
a) Green gram

II Two Marks

Question 1.
Name the 3 grass species of food plants?
Answer:
Rice, Wheat, Maize.

Question 2.
What are the nutrients provided by cereals?
Answer:
Carbohydrates, proteins, fibres, vitamins and minerals.

Question 3.
Classify cereals based on size? Give example.
Answer:

• Major Cereals. Eg. Rice, Wheat
• Minor Cereals. Eg. Millet’s, Sorghum

Question 4.
Comment on Maida?
Answer:

• Processed wheat flour is called Maida.
• It is used in making Parota, Naan and Bakery products.

Question 5.
Explain the rice products?
Answer:

• Flaked rice (Aval)
• Puffed rice (Pori) are used as breakfast cereal (or) snack food in India.

Question 6.
What are millet’s?
Answer:
A variety of very small seeds. These were originally cultivated by ancient people in Africa. It is gluten free, less glycemic index.

Question 7.
Enlist the uses of finger millet?
Answer:

• Rich in calcium
• Staple food of south hilly regions in India.
• Ragi is made into porridge and gruel.
• Ragi malt is a nutrient drink
• It is the source of fermented beverage.

Question 8.
How is Sorghum useful?
Answer:

• It is used to feed poultry, birds, pigs, cattle.
• Source of fermented alcoholic beverage.

Question 9.
Discuss the medicinal uses of Fox tail millet?
Answer:

• Strengthens heart
• Improves eye sight
• Thinai porridge is given to lactating mother.

Question 10.
Kodo Millet is medicinally useful – Discuss?
Answer:

• It is a good diuretic
• It cures constipation
• It reduces obesity, blood sugar, blood pressure.

Question 11.
Which is the only pulse native to southern India? Give it’s uses?
Answer:
Red gram (Pigeon pea) Cajanus cajan is the only pulse native to south India.
Uses:

• Major ingredient of Sambar
• Roasted, salted, unsalted seeds are snacks.
• Young pods are cooked and consumed.

Question 12.
Enlist the nutrients in vegetables?
Answer:
Potassium, fibre, folic acid, vitamin A, E, C

Question 13.
Molecular farming plants are different from natural medicinial plants. How?
Answer:

Question 14.
What are the major cultivating states of okra in Tamil Nadu?
Answer:
Coimbatore, Dharmapuri, Vellore.

Question 15.
Classify fruits based on the climatic region in which they grow?
Answer:

• Temperate Eg. Apple, Pear, Plum
• Tropical fruits Eg. Mango, Jack, Banana.

Question 16.
Which is the National fruit of India? Give its origin and area of cultivation?
Answer:
Mango (Mangifera Indica)
Origin and area of cultivation.

• Native of southern Asia, Burma and Eastern India.
• Mango producing states are Andhra Pradesh, Bihar, Gujarat and Karnataka.
• Salem, Krishnagiri, Dharmapuri are major mango producing districts of Tamil Nadu.

Question 17.
Name the Major Cultivars of Mango in India.
Answer:
Alphonsa, Banganapalli, Neelam, Malgova.

Question 18.
Which food is the source of antioxidants?
Answer:
Dry fruits with hard shell and edible kernel are nuts. They are the good source of health fat, fibre, protein, vitamin, mineral, antioxidants.

Question 19.
Name the plants ideal for the extraction of commercial sugar?
Answer:
Sugar Cane, Palmyra

Question 20.
What is sugar?
Answer:
It is the generic name for sweet tasting soluble carbohydrate. They are used in food, beverages.

Question 21.
Give the sources of sugar?
Answer:
Roots of Sugar beet, Stems of Sugar cane, Fruits of Apple, Palmyra sap.

Question 22.
How is cultivated Saccharum officinarum evolved?
Answer:
By repeated back crossing of Saccharum officinarum of new guinea with wild Saccharum Spontaneum.

Question 23.
Toddy-Comment?
Answer:
The sap from Palmyra inflorescence is fermented to get toddy.

Question 24.
Name the 2 kinds of oils?
Answer:

• Essential oil
• Vegetable, fatty oil

Question 25.
Define Essential oil?
Answer:
They evaporate or volatilize in contact with air. So, they are called volatile oils.

Question 26.
Give the sources of essential oils?
Answer:
Flowers of jasmine, fruits of orange and roots of ginger.

Question 27.
Comment on vegetable oil?
Answer:
These are non-volatile oils or fixed oils. They do no evaporate. Eg. Whole seeds or endosperm are the sources.

Question 28.
What are spices?
Answer:
Aromatic plant products. They are of sweet or bitter taste. They give flavour and improve the palatability of food.

Question 29.
Comment on condiments?
Answer:
Flavouring substances with sharp taste. They are added to food after cooking Eg. Curry leaves.

Question 30.
Dates of India – Discuss.
Answer:
Tamarindus is an Arabian word. It means dates of India (Tamar – Taste; Indus – India)

Question 31.
Write any two uses of THC.
Answer:
THC is used in treating Glaucoma a condition in which presšure develops in the eyes.

• THC is also used in reducing nausea of cancer patients under going radiation and chemotherapy.
• It is an effective pain reliever and reduces hypertension.

III. Three Marks

Question 1.
Suggest the 4 commercial cotton species?
Answer:

• G. hirsutum
• G. barbadense
• G. arboreum
• G. herbaceum

Question 2.
Give the uses of cotton?
Answer:
Manufacturing of textile, hosiery products, toys. UsedinFlospitals.

Question 3.
Name the 2 species of plants from which Jute is derived?
Answer:

• Corchorus capsularis (Indo – Burmese origin)
• Corchorus olitorius (African origin)

Question 4.
Define Vulcanization?
Answer:
Heating rubber with sulphur under pressure at 150°C. It overcomes defect in rubber articles. (Vulcan is the Roman god of fire)

Question 5.
Name the woods used for making paper pulp?
Answer:

• Wood of Melia azadirachta.
• Neolamarkia chienensis
• Cauarinaspe, Eucalyptus spe.

Question 6.
Dyeing is in use since that ancient times? Substantiate?
Answer:

• Authentic records of dyeing is in the tomb painting of ancient Egypt.
• Colouring of mummy cements (wrapping) included saffron and indigo.
• Found in rock paintings of India.

Question 7.
Give the significance of Henna?
Answer:

• Orange dye henna is from the leaves and shoots of Lawsonia intermis.
• The colouring matter Lacosone is harmless and causes no skin irritation.
• This dye is used for skin, hair and finger nails.
• It is used for colouring leather for tails of horses and in hair dyes.

Question 8.
What do the south Indian people traditionally for skin and hair care?
Answer:
People of South India use turmeric, green gram powder, henna, sigaikai and usilai for skin, hair care.

Question 9.
What does the word’perfume’mean?
Answer:

• ‘Perfume’ is a word derived from Latin.
• Per (through) and fumus (to smoke) means through smoke.
• Age old tradition of burning scented woods at religious ceremonies.

Question 10.
Give an account of NCB?
Answer:
The Narcotics Control Bureau is the drug law enforcement and intelligence agency of India. It is responsible for drug trafficking and abuse of illegal substances.

Question 11.
Can we create new business using plant resources?
Answer:
Entrepreneurial botany is the study of new business created using plant resources.

Question 12.
What is entrepreneurship?
Answer:
Developing ideas to create new ventures among young people.

Question 13.
What is the role of an Entrepreneur?
Answer:

• One who works to create a product or service that people will buy
• He builds an organization to support the sales.

Question 14.
Discuss about ‘Capsaicin’
Answer:

• It is an active component of chillies.
• It has pain relieving properties.
• It gives pungency or spicy taste to chillies.
• Pungency is measured in Scoville Heat Unit (SHU)
• Eg. Naga piper is the hottest chilly of India with 1,349,000 SHU.

Question 15.
Name some plants used in making paper pulp?
Answer:

• Melia azadirachta
• Neolamarkia chinensis
• Casuarina spe
• Eucalyptus spe

Question 16.
Give the uses of purified dissolving pulp?
Answer:
It helps to manufacture rayon, artificial silk, fabrics, transparent films (cellophane, cellulose, acetate films), plastic. Viscose process of making rayon is common.

Question 17.
Which the second geographical Indication tag after Mysore Malli? How?
Answer:

• Madurai Malli is the second GI Tag.
• It has thick petals, long stalk.
• Distinct fragrance is due to chemicals like jasmine, alpha terpineol.

Question 18.
Name the major tribal communities in Tamil Nadu known for their medicinal knowledge?
Answer:
Irulas, Malayalis, Kurumbas, Paliyans and Kaanis

Question 19.
Discuss the origin and area of cultivation of black gram?
Answer:

• Archeo botanical evidence show the presence of black gram 3500 years ago.
• India gives 80 % of global production.
• Black gram is grown in Uttar Pradesh, Chattisgarh and Karnataka.

Question 20.
Suggest the nutrients in fruits?
Answer:
Potassium, dietary fibre, folic acid, vitamins.

Question 21.
What is rayon?
Answer:

• Rayon is purified dis solving pulp is used as a basic material in the manufacture of rayon or artificial silk, fabrics, transport films (cellophane), cellulose, acetate films), plastics.
• This viscose process of making rayon is the most common process.

Question 22.
What is known as sustainable development of agriculture?
Answer:

• Use of biofertilizers is one of the important components of integrated organic farm management, as they are cost effective and renewable source of plant nutrients to supplement the chemical fertilizers for sustainable agriculture.

IV . Five Marks

Question 1.
What kind of cereal can be eaten as a whole grain? Discuss?
Answer:

• Pseudocereal can be eaten as wholegrain.
• These are botanical outliers from grasses.
• Eg. Seed from the Chenopodium quinoa (Family: Amaranthaceae)
• Gluten-free, whole grain carbohydrate, whole protein with all essential amino acids.
• Eaten for 6000 years in Andes hill region.

Question 2.
Suggest the attributes of cereals as food plants?
Answer:

• Adaptability and colonisation on every type of habitat.
• Ease of cultivation.
• Tillering property gives high yield per unit area.
• Compact dry grains are easily handled, transported, stored without spoilage.
• High-calorie value provides energy.

Question 3.
How will you prepare a Bio-pest repellent?
Answer:

• Neem tree leaves are plucked.
• Put chopped leaves in 50 litre container half-filled with water. Leave it for 3 days to brow.
• Strain the mixture and spray on plants.
• 100 ml of cooking oil is added to make the repellent stick to the plants.
• Soap water is added to break down the oil.
• A stewed leaf mixture can be composted around the base of the plant.

Question 4.
Tabulate the uses of common medicinal plants?
Answer:

Question 5.
Give a detailed account on the ‘National Fruit of India’?
Answer:
Mango (Mangifera indica) belongs to the family Anacardiaceae
Origin and area of cultivation.

• Native of southern Asia, Burma and Eastern India.
• Andhra, Bihar, Gujarat and Karnataka are mango producing states.
• Salem, Krishnagiri, Dharmapuri are mango producing districts of Tamil Nadu.
• Major cultivars of Mango are Alphonsa, Banganapalli, Neelam and Malgova.

Uses:

• Major Indian table fruit.
• Rich in beta carotenes.
• Used as dessert, canned, dried, preserves in Indian cuisine.
• Unripe mangoes are used in chutney, pickle, side dishes, eaten raw with salt, chilli.
• Pulp is made as jelly
• Aerated, non aerated soft drinks are prepared.

Question 6.
Enlist the uses of Sugar cane.
Answer:
Botanical Name : Saccharum officinarum of
Poaceae family
Uses:

• Raw material for white sugar
• Industries supported are
• Sugar mills producing refined sugar
• Distilleries producing liquor grade ethanol.
• Jaggery manufacturing unit.
• Refreshing drink can be extracted.
• Gives molasses. It is the raw material for ethyl alcohol.

Question 7.
Detail on the State Tree of Tamil Nadu.
Answer:
Botanical Name : Borassus flabellifer of Arecaceae family
Origin and area of cultivation

• Native of tropical Africa, Asia, New Guinea.
• All over Tamil Nadu especially in coastal districts.

Uses:

• Exudate from inflorescence gives palm sugar
• Sap of inflorescence is a healthy drink
• Processed sap gives palm sugar
• Fermented sap gives toddy
• Endosperm is a refreshing summer food
• Elongated embryo of germinated seeds is edible.

Question 8.
Enlist the uses of Chilly / Red pepper?
Answer:
Botanical Name : Capsicum annuum of
Solanaceae family
Uses:

• Capsicum annuum is less pungent
• Capsicum annuum includes large sweet bell peppers.
• Long fruit cultivars called ‘Cayenne Pepper’ are crushed, powdered and used as condiment.
• Sauce, Curry powder, pickle can be prepared.
• Capsaicin has pain relieving property. Good source of vitamins A, C, E.

Question 9.
Which plant contributes to 90% of world production by Asia? Detail the uses?
Answer:
Rubber (Hevea brasiliensis) of Euphorbiaceae.
Origin and Cultivation

• Native of Brazil
• Kerala is the largest Indian producers

Uses:

• Tyre, automobile parts consume 70% of rubber.
• To manufacture footwear, wire, cable insulation, rain coats, household, hospital goods, shock absorbers, belts, sports goods, erasers, adhesives, rubber band
• Hard rubber is used in electrical and radio engineering
• Latex makes gloves, balloons, condoms.
• Foamed latex used for the manufacture of cushion, pillow, life belts.

Question 10.
Which system of medicine originated from Brahma? Explain?
Answer:

• Ayurveda system of Medicine
• Core knowledge is documented in compendium of Charaka, Sushruta and Vagbhata.
• It is based on 3 humor principles Vatha, Pitha, Kapha.
• Herbs, few animal parts are drug sources.
• Himalayan plants are plant sources.
• Ayurvedic pharmacopoeia of India list 500 plant sources.

Question 11.
Which system of medicine survives as oral tradition? Explain?
Answer:
Folk system of Medicine:

• It is a oral tradition in rural, tribal communities.
• Document of plants used by ethnic communities was launched by Ministry of Environment and Forest, Government of India.
• The document is All India Co-ordinated Research Project on Ethnobiology.
• 8000 species of medical plants are documented.
• Major tribal communities with medicinal knowledge are Irulas, Malayalis, Kurumbas, Paliyans and Kaanis.

Question 12.
Jute Industry occupies an important place in the national economy of India. Explain?
Answer:

• One of the largest exported fibre of India.
• Used for safe packaging of natural, renewable, bio degradable, Eco-friendly products.
• Used in bagging, wrapping textile.
• 75% is used to prepare sack, bag.
• Manufacture of blanket, rag, curtain
• Used in textiles recently.

Question 13.
Organic farming is considered as the movement towards the philosophy of Back to Nature. Explain
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
• Indians were organic farmers by default until the green revolution came in to practice.
• Use of bio-fertilizers is one of the important components of integrated organic farm management, as they are cost effective and renewable source of plant nutrients to supplement the chemical fertilizers for sustainable agriculture.
• Several microorganisms and their association with crop plants are being exploited in the production of bio-fertilizers.
• Organic farming is thus considered as the movement directed towards the philosophy of Back to Nature.

Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Bio Botany Guide Pdf Chapter 3 Chromosomal Basis of Inheritance Text Book Back Questions and Answers, Notes.

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

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

I. Choose the correct answer

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:
c) Two 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 test crossed 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) 10 AB,10 ab
c) 45 AB, 45 ab
d) 45 AB, 45 ab, 5 Ab, 5aB
Answer:
c) 45 AB, 45 ab

Question 3.
Match list I with list II

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. Crossing 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 L and G isT5%, S and L is 50%, H and S are 20%. The correct order of genes is
a) GHSL
b) SHGL
c) SGHL
d) HSLG
Answer:
c) SGHL

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 38
c) 37 and 35
d) 17 and 19
Answer:
a) 35 and 37

Question 10.
Changing the codon AGC to AGA represents
a) missense mutation
b) nonsense mutation
c) frameshift mutation
d) deletion mutation
Answer:
a) missense mutation

Question 11.
Assertion (A): Gamma rays are generally use 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:
c) 9 cM

Question 13.
When two different genes came from same parent they tend to remain together.
i) What is the name of this phenomenon?
ii) Draw the cross with suitable example.
iii) Write the observed phenotypic ratio.
Answer:
i) The name of this phenomenon is known as Linkage.
This is reported in Sweet pea Lathyrus odoratus by Willium Bateson & Reginald C Punnet in 1906.

Genes for Purple colour and Long pollen grain were found close together in the same homologous pair of chromosomes – They do not assort independently and this condition is known as linkage.

Question 14.
If you cross dominant genotype PV/PV male Drosophila with double recessive female and obtain FI hybrid. Now you cross FI male with double recessive female,
i) What type of linkage is seen?
ii) Draw the cross with correct genotype.
iii) What is the possible genotype in F₂ generation?

Question 15.

i) What is the name of this test cross?
ii) How will you construct gene mapping from the above given data?
iii) Find out the correct order of genes.
Answer:
i) It is three point test cross – It refers to analysing the inheritance, patterns of three alleles by crossing a triple recessive herterozygote with a triple recessive homozygote.
ii) The relative distance between the three alleles & the order in which they are located can be determined with the help of frequency of recombination between them.

All the loci are linked because all the RF values are considerable less then 50%. In AC loci show highest RF value, they must be farthest apart. There fore the B locus must lie between them. The order of genes should be abc. A genetic map can be drawn.

A final point note that two small map distances. 19.9 m.u and 21.75 m.u is add up to 41.95 m.u which is greater then 40.16 m.u the stance calculated for 1 and g. We must identify the two least number of progenius (totalling 8) in relation to recombination of AC . These two least progenius are double cross over. The two least progenies not only counted once should have count each of them twice because each represents a double recombinant progeny. Hence, We can correct the value adding the number 114 + 125 + 116 + 128 + 5 + 14 + 4 = 500 of the total 1200 this number exactly 41.65% which is identical with the same of two component values.
The test cross parental combination can be rewritten as follows.

Gene order showing double recombinant.

Question 16.
What is the difference between missense 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 the codon for one amino acid is changed into a termination or stop codon is called Nonsense mutation.

Question 17.

From the above figure identify the type of mutation and explain it.
Answer:

• It is a change in the arrangement of gene loci,
• Here the duplicated segment is located immediately aftear the normal segment but the gene sepuence order will be reversed – (Paracentric inversion)

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 – it is a very significant biological process
It is a precise one with several stages

i) Synapsis:
During zygotene – of prophase. I of meiosis I the homologous chromosomes come and align side by side known as – bivalents.
This pairing – is known as synapsis or syndesis.
Types of synapsis

ii) Tetrad Formation:
Each homologous chromosome of – a bivalent begin to form two identical sister chromatids – held together by a centromere.
Each bivalent has 4 chromatids – (tetrad stage),

iii) Cross Over:
At pachytene stage cross over occur. The points of contact at one or more points between non-sister chromatids is called Chiasmata.
Crossing over is exchange of corresponding segments occur, in the chiasma region.

Synaptonemal Complex (SC)
The highly organised structure of filaments called SC – facilitate chiasma formation.

SC formation & chiasma formation – is absent in Drosophila

Terminalisation:
After crossing over, chiasma starts to moving towards the terminal end of chromatids is known as terminalisation. Complete separation of homologous chromosomes occurs after terminalization.

Question 20.
Write the steps involved in molecular mechanism of DNA recombination with diagram.
Answer:
Proposed by Robin Holliday in 1964

Steps:

• Homologous DNA molecules are paired side by side with their duplicated copies of DMAs
• One strand of both DNAs cut in one place by the enzyme endonuclease.
• The cut strands cross and join the homologous strands – Holliday junction.
• Holliday junction migrates away from the original site, a process called branch migration, as a result heteroduplex region is formed.
• DNA strands may cut along through the vertical (V) line or horizontal (H) line.
• The vertical cut will result in heteroduplexes with recombinants.
• The horizontal cut will result in heteroduplex with non recombinants.

Question 21.
How is Nicotiana exhibit self¬incompatibility? Explain its mechanism.
Answer:
In Nicotiana self sterility or self-incompatibility is due to multiple alleles.
The pollen from a plant is unable to germinate on its own stigma – and no fertilization.
The gene for self incompatibility can be – ‘S’ which has allelic series S₁, S₂, S₃, S₄ & S₅.
Cross-fertilizing tobacco – were not always homozygous as S₁S₁ or S₂S₂, but heterozygous
Crosses between different S₁S₂ plants, pollen tube did not develop normally.
But effective – development observed when cross was made with other than S₁S₂ Eg. S₃S₄.

Question 22.
How is sex determined in monoecious plants. Write the genes involved in it.
Answer:
Zeamays (maize) – monoecious plant
Made & Female flowers are present on the same plant.

• Terminal inflorescence – arise from tassel bear staminate flowers
• Lateral inflorescence – arise from ear or cob bear pistillate flowers.
• Unisexvality in maize – occurs through selective abortion of ear florets and pistils in tassel florets.
• The allele for barren plant (ba)- when homozygous makes the stalk staminate (eliminating silk and ears)
• The allele for tassel seed (ts) – transforms tassel into a pistillate structure (no pollen produced)
• Most of these mutations are shown to be defects in Gibberellins biosynthesis.
• Gibbercilins 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 the 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 a 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 the results of dihybrid and trihybrid crosses.
• It allows the geneticists to understand the overall genetic complexity of the 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:

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

12th Bio Botany Guide Chromosomal Basis of Inheritance Additional Important Questions and Answers

I. Fill in the blanks

1. The scientists who independently rediscovered mendelian works were
Answer:
De Vries, Correns & Tschermak

2. The worm-shaped cells formed during cell division are called in the earlier period as
Answer:
Chromosomes

3. Who postulated that the chromosomes of a cell are responsible for transferring heredity
Answer:
Wilhelm Roux (1883)

4. ……………………… was the first to find out physical mutagen in Drosophila
Answer:
Muller (1927)

5. ………………………used X-rays for the first time to induce mutation in the fruit fly
Answer:
H.J. Muller

6. Induced mutations are planted was reported for the first time by
L.J. Stadler

7. Chemical mutagenesis was first reported by
Answer:
Auerback (1944)

8. Double nullisomy is
Answer:
2n-2-2

9. Trisomis were first reported by Blakeslee in
Datura Stramonium

10. All possible tetrasomics are available in ……………………… plant
Answer:
Wheat

11. The kind of Aneuploid are usually lethal are
Answer:
Nullisomy

12. The alkaloid used to induce polyploidy is
Answer:
Colchicine

13. Raphano brassicas the sterile hybrid of Radish & Cabbage was produced by
Answer:
G.D. Karpechenko (1927)

14. The cross between hexaploid wheat Triticum aestivum and rye produced is a
Answer:
Octoploidy

15. Colchicine is extracted from the root and corms of
Answer:
Colchicum autumnale

16. Who first reported duplication in drosophila
Answer:
Bridges (1919)

17. In which types of cells chromosomal aberration is commonly found?
Answer:
Cancer cells

18. Recombination frequencies are the same for
Answer:
CIS and trans heterozygotes

19. The map distance between gene A and B is 3 units between B & C is 10 units and between C & A is 7 units – the order of genes in a linkage map constructed on the about would perhaps be
Answer:
B-A-C

20. The percentage of crossing over will be more if
Answer:
Linked genes are located apart from each other

21. A point mutation that changes an amino acid coding codon into a stop codon, prematurely terminating synthesis of the encoded protein ………………………
Answer:
Nonsense mutation

22. Single base change in DNA is known as ………………………
Answer:
Point mutation

23. Genetic change in a non-sex cell is known as
Answer:
Somatic mulalion

24. A duplicated DNA sequence next to the original sequence is known as
Answer:
Tandem duplication

25. A missing sequence of DNA or part of a chromosome
Answer:
Deletion mutation

26. Mutation that alters the genes reading frame is known as
Answer:
Frame shift mutation

27. A single base change mutation that alters and amino acid ………………………
Answer:
Missense

28. A substance that changes, adds, or deletes a DNA base
Answer:
Multagen

29. The mutation that introduces a section of aminoacids not normally found is known as ………………………
Answer:
Frame shift mutation

30. A mutation that changes an adenine to guanine is an example of a ………………………
Answer:
transition

31. A point mutation that has no obvious effect at all on the phenotype is called a ……………………… mutation
Answer:
silent

32. A point mutation that changes a codon specifying an amino acid into a stop codon is called a ………………………
Answer:
Non sense mutation

33. Changing the codon AGC to AGA represents ……………………… of a point mutation
Answer:
missense

34. A point multation that alters a codon so that the encoded aminoacid is substituted with another is called a ………………………mutation
Answer:
missense

35. A ………………………mutation occurs during the DNA replication that precedes meiosis. while a ……………………… mutation occurs during the DNA replication that preceeds mitosis.
Answer:
germline, somatic

36. A mutation that introduction of section of aminoacids not normally found is ………………………
Answer:
Frame shift mutation.

37. A point mutation altering a purine to pyrimidine or vice versa is ………………………
Answer:
transversion

38. A spontaneous mutation usually originates as an error in ………………………
Answer:
DNA replication

39. The codon for leucine is CUC. How many different aminoacids could possibly result from a single base substitution
Answer:
7

40. How may map units separate two alleles if the recombination frequency is o.o7?
Answer:
7cM

41. In a population of 1000 individuals 360 belong to genotype AA. 480 to Aa and the remaining 160 to aa – Based on the data, the frequency of allela A in the population is
Answer:
0.7

II. Find out the incorrect statement

Question 43.
Which one of the following is incorrect regarding chromosomal behaviour during cell division?
a) The alleles of a genotype are found in the some locus of a homologous chromosome
b) In the S phase of meiotic interphase each chromosome replicates forming two copies of each allele, one on each chromatid.
c) The Homologus chromosomes segregate in metaphase I, thereby separating two different alleles.
d) In anaphase II of meiosis separation of sister chromatid of homologous chromosomes takes place.
Answer:
c) The Homologus chromosomes segregate in metaphase I, thereby separating two different alleles.

Question 44.
a) 94% of all flowering plants are sexually monomorphic
b) When three or more allelic forms of a gense occupy the same locus in given pair of homologous chromosome they are known as Multiple alleles
c) The mutation that result in the change of one
codon of an aminoacid changed into codon of another amino acid is known as Frame shift mutation.
d) Muller (1928) first time used x-rays to induce mutation in Drosophila
Answer:
c) The mutation that result in the change of one codon of an aminoacid changed into codon of another amino acid is known as Frame shift mutation.

Question 45.
Which of the following statement is not correct of two genes that show 50% recombination frequency?
a) The genes may be on different chromosomes
b) The genes are tightly linked
c) The genes show independent assortment
d) If the genes are present on the same chromosome, they undergo more than one crossover in every meiosis.
Answer:
d) If the genes are present on the same chromosome, they undergo more than one crossover in every meiosis.

Question 46.
a) Selfing of monosomic plants produce nullisomics.
b) In a true diploid both the monoploid and haploid chromosome number are same.
c) An Auto triploids can be produced artificially by crossing between haploid and a diploid.
d) An increase in the number of chromosome
sets has been an important factor in the origin of new plant species.
Answer:
c) An Auto triploids can be produced artificially by crossing between haploid and a diploid.

III. Match the following

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

IV. Choose the correct statement

Question 48.
When red eyed female Drosophila is crossed with white eyed male, the FI offsprings would be
a) Females are with white eye and males are with red eye.
b) Males are with red eye and females are with yellow eye.
c) Both males and females are with red eye
d) Both males and females are with white eye.
Answer:
c) Both males and females are with red eye

V. Find the Odd man out with reference to Allopolyploidy

Question 49.
a) All organisms which possess two or more basic sets of chromosomes derived from two
different specie’s.
b) They have four or six copies of its own genome – induced by doubling of the diploid species.
c) They can be developed by inter-specific crosses and fertility is restored by chromosome doubling with colchicine treatment.
d) They are formed between closely related species only..
Answer:
b) They have four or six copies of its own genome – induced by doubling of the diploid species.

VI. Find the Odd man out with reference to Altotriploidy

Question 50.
a) Thev can be produced artificially bv crossing between autotetraploid & diploid.
b) They are highly fertile due to large number of gametes.
c) Cultivated triploid bananas are seedless having larger fruits than diploid.
d) Common doob cross is a natural autotriploid.
Answer:
b) They are highly fertile due to large number of gametes.

VII. Find the Odd man out regarding crossing over

Question 51.
a) It occur in germinal cells during gametogenesis.
b) Take place during Pachytene state of prophase I of meiosis..
c) It is directly proportional to the frequency of recombination between them.
d) It has universal occurrence has great significance.
Answer:
c) It is directly proportional to the frequency of recombination between them.

VIII. Choose the wrongly matched pair

Question 52.

Answer:
d

Question 53.

Answer:
d

Question 54.

Answer:
c

IX. Choose the incorrect statement with reference to Deletion

Question 55.
a) Deletions occur due to chemicals, drugs & radiation.
b) On the basis of location of breakage on chromosome it is divided in to Terminal deletion & inter calary deletion
c) Larger deletions have evolutionary significance.
d) Deletions are recorded in Drosophila & Maize
Answer:
c) Larger deletions have evolutionary significance.

Question 56.
How can we reverse the sterility of FI hybrid?
a) Genetic Engineering
b) Protoplasmic fusion
c) Induced Mutation
d) Induced chromosomal aberration
Answer:
d) Induced chromosomal aberration

Question 57.
If haploid number in a cell is 23. The double monosomic and pentasomy number will be
a) 44 and 49
b) 17 and 34
c) 47 and 46
d) 45 and 48
Answer:
a) 44 and 49

Question 58.
Genes located close together on the same chromosome and inherited together represented as
a) linked genes
b) unlinked gene
c) syntenic genes
d) trans genes
Answer:
a) linked genes

XI. Assertion (A) & Reason (R)

Question 59.
Assertion (A): Arabidopsis plant chromosomes have more repeats of TTT nucleotide sequences in the telomeres.
Reason (R) : Restriction endonuclease enzyme is used in the formation of nucleotide sequence (Telomeres) mui
a) (A) is incorrect, (R) is correct
b) (A) is correct, (R) is the correct explanation (A)
c) (A) is correct, (R) is the incorrect explanation (A)
d) (A) and (R) are wrong.
Answer:
b) (A) is correct, (R) is the correct explanation (A)

Question 60.
Assertion (A) : Linkage and crossing over are two processes that have opposite effects. Reason (R) : Linkage keeps particular genes together but crossing over mixes them.
Answer:
a) If both the Assertion (A) & Reason (R) are true and the reason is a correct explanation of the Assertion. .

Question 61.
Assertion (A) : Increase in temperature increases the rate of mutation.
Reason (R) : While rise in temperature hydrolyses DNA by the restriction endonuclease which degrade Nucleotides.
Answer:
c) Assertion (A) is true but Reason (R) is false

XII. Two Marks

Question 1.
Define chromosome theory of inheritance.
Answer:
It states the Mendelian factors (genes) have specific locus on chromosomes & they carry information from one generation to the next generation.

Question 2.
State the number of chromosomes of the given organism.
Answer:
1) Ophioglossum 2) Arabiodopsis 3) Sugarcane 4) Rice 5) Potato 6) Maize
Answer:
1) -1262;
2) -10;
3) 80;
4) 24;
5) 48;
6) 20

Question 3.
What are Fossil Genes?
Answer:

• Some junk DNA is made up of pseudogenes, once working but have lost their ability to make proteins.
• They are fossilized parts act as evidence for evolution.

Question 4.
State the works of T.H. Morgan
Answer:

• His works on Drosophila melanogaster – Sex linkage – helped to confirm chromosome theory of heredity.
• He received Nobel prize in Physiology of medicine in 1933 fot it.
• He coined the term crossing over.

Question 5.
What are co-mutagens ?
Answer:
Compounds which are not having own mutagenic properties – but enhance the effects of known mutagens.
Eg. Ascorbic acid – increase the damage caused by hydrogen peroxide.
Caffeine – increase the toxicity of methotrexate.

Question 6.
Differentiate between Euploidy & Aneuploidy
Answer:
Eupoidy :

1. Ploidy involving entire sets of chromosomes is known as euploidy
2. Triploidy (3x); Tetraploidy (4x); Poly ploidy ( ∞n)

Aneuploidy :

1. Here the diploid number is altered either by addition or deletion of one or more chromosomes
2. Trisomy; Tetrasomv; Monosomy; Nullisomy (2n+1)(2n+2)(2n-1)(2n-2)

Question 7.
Distinguish between Monoploidy & Haploidy
Answer:
Monoploidy
In Monoploidy the chromosome number is referred as x .
Eg.
Hexaploidy wheat
(2n) = 6 x = 72
haplaid = (n) 36
Monoploidy = x = 12

Haploidy :
Half the number of somatic chromosomes is referred as gametic chromosome number called haploidy (n) Human of haploid = 23 (n) Wheat of haploid = 36 (n)

Question 8.
Independent assortment & Linkage are alternatives of each other – Discuss
Answer:

Question 9.
How does the strength and weakness of linkage depend on linked genes?
Answer:

• The strength of linkage increases as the distance between linked genes decreases.
• The linkage becomes weaker with the increase in the distance between genes.

Question 10.
Distinguish between crossing over & Reciprocal Translocation.
Crossing over

1. It is legitimate & natural
2. Occurs between nonsister chromatids of homologous chromosomes
3. Occurs between nonsister chromatids of homologous chromosomes

Reciprocal Translocation :

1. It is illegitimate & chromosomal abnormality
2. Occurs between non sister chromatids of non homologous chromosomes
3. Also play major rome in formation of species

Question 11.
Distinguish between tetrad & bivalent Tetrad:
Answer:

• During Synapsis homologous chromosomes come together side by side resulting in bivalents
• As the stage during which each bivalent has 4 chromatids & the stage is known as tetrad stage.

Question 12.
Define Recombination.
Answer:
In this segments of DNA one broken and recombined to produce new combination of alleles – known as Recombination.

Question 13.
What is RF (Recombination Frequency)
Answer:
The frequency with which recombination occur is a certain condition

Question 14.
A diploid organism is heterozygous for 4 loci. How many types of gametes cars be produced.
Answer:
The formula 2n is applied – Organism hetr oizy gous f or 4 loci = n = 4.
So 2n = 24 = 2 x 2 x 2 x 2 = 16.
The organism produces 16 types of gametes.

Question 15.
Notes on Colchicine.
Answer:

• Alkaloid, extracted from – root and corms of colchicum autumnale
• In low concentration to the growing lips it induce polyploidy
• It does not affect the source plant due to the presence of Anticolchicine

Question 16.
Write down the significance of ploidy.
Answer:

• Polyploids – More vigorous & more adaptive
• Ornamental flowers – (Autotetraploids) larger flowers – longer flowering duration
• Increase in fresh weight (due to more water content)
• Aneuploids – help to determine the phenotypic effects (loss or gain of different chromosomes
• Allopolyploids of angiosperms play a role in an evolution of plants.

Question 17.
Distinguish between Mendelian disorder & Chromosomal disorder.
Answer:
Mendelian disorder:
Occur due to mutation of single gene & follow the well known Mendelian pattern of inheritance.
Eg. Sickle cell anaemia

Chromosomal disorder :
Chromosomal disorders are produced due to alteration in the number of chromosomes.
Eg. Down syndrome

Question 18.

Answer:
a) Which type of crossing over is mentioned in the
above diagram? I h M I
Single cross over
b) Mention the percentage of Recombination Frequency (RF)
RF = 2/4 x 100 = 50%

Question 19.
This is a type of Numerical chromosomal abnormality find it out give a note on it.

Answer:

• This numerical chromosomal abnormality is known as double monosomy (2n-l-l)
• From a diploid set of chromosome if one chromosomes is lost, the condition is known as monosomy (2n-l)
• If another chromosome is also lost it is known as double monosomy (2n-l-l)

Question 20.
Bring out the difference between Linkage & Crossingover in inheritance
Answer:
Linkage

1. It is the tendency of genes in a chromosome to stay close together
2. It involves same chromosome of homologous chromosome
3. It reduces new gene combinations
4. Not very significant in evolution

Crossing over :

1. It leads to separation of linked gene
2. It involves exchange of segments between nonsister chromatids of homologous chromosome.
3. It increases – variability by forming new combinations → lead to formation of new organism
4. play important role in evolution

Question 21.
What is chiasmata?
Answer:

• 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.

Question 22.
What is 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 be called multiple alleles.

Question 23.
What is monomorphic?
Answer:

• About 94% of all flowering plants have only one type of individual, which produces flowers with male organs (the stamens) and female organs (the carpels).
• Such plants are termed as sexually monomorphic.

Question 24.
What is Dimorphic?
Answer:
Some 6% of flowering plants which have two separate sexes are called dimorphic.

XIII. Three Marks

Question 1.
Differentiate tetrasomy from tetraploidy
Answer:

Question 2.
Give a tabulation comparing the behaviour of gene & Chromosome
Answer:

Question 3.
The important aspects about the chromosome behaviour during cell devision.
Answer:

• Alleles of a genotype – found in the same locus of a homologous chromosome (A/a)
• In ‘S’ – Phase of meiotic interphase – the replication of chromosome occur – (two copies of each allele (AA/aa) one on each chromatid
• Anaphase II of meiosis, separation of sister chromatids of homologous chromosomes. So each daughter cell (gamete) carries only a single allele of a character (A), (A), (a) and (a)

Question 4.
Write the differences between coupling and Repulsion
Answer:
Coupling

1. The two dominant alleles or recessive alleles called repulsion or trans configuration
2. It tend to inherit together into same gametes

Repulsion :

1. If dominant or recessive alleles are present on occur in the same homologous chromosomes two different but homologous chromosomes.
2. If they inherit apart in to different game es are

Question 5.
Define synapsis.
What are the types of Synapsis
Answer:

• During cygotene stage of prophase I of meiosis I – homologous chromosomes are aligned side by side resulting in a pair called (bivalents).
• This pairing phenomenon is called synapsis or syndesis.
  Based on the starting poiring of pairing there are 3 types of synapsis

  Procentric	Proterminal	Random
  Starts from middle	Starts from the telomeres	Starts from any where

Question 6.
Distinguish between sharbati sonora & Castor Aruna.
Answer:
Sharbati sonora :

1. Multant variety of Wheat – approved in 1967
2. Sonora 64 (Mexican variety subjected to gamma rays to produce sharbati sonora
3. Developed by Dr. M.D. Swaminathan
4. Early maturing & high protein content high kneading quality

Castor Aruna :

1. Mutant castor variety
2. Seeds treated with thermal neutrons
3. Early maturing – (120 days instead of 270 dyas) & High yielding.

Question 7.
How do increase in temperature cause mutation?
Answer:
Rise temperature breaks the hydrogen bonds between two DNA nucleotides – & affects the process of replication & transcription.

Question 8.
Distinguish between the impact of ionizing & non ionizing radiation in causing mutation. Ionizing radiation Non Ionizing radiation
Answer:
Ionizing radiation :

• Short wave length and carry enough higher energy to ionize electrons from atoms. They breaks the chromosome & chromatids. Ex. x-rays – gamma rays, alfa rays, beta rays & cosmic rays.

Non Ionizing radiation :

• Longer wave lengths and carry lower energy so they have lover penetrating power used to treat unicellular microbes – spores pollengrains – which have nuclei – near surface membrance. Eg. UV rays

Question 9.
What is significance of ploidy?
Answer:

• Many polyploids are more vigorous and more adaptable than diploids.
• Many ornamental plants are autotetraploids and have large flower and longer flowering duration than diploids.
• Auto polyploids usually have increase in fresh weight due to more water content.
• Aneuploids are useful to determine the phenotypic effects of loss or gain of different chromosome.
• Many angiosperms are allopolyploids and they play a role in an evolution of plants.

Question 10.
What is chemical mutagens? Give an example?
Answer:
Chemical which include mutation are called mutagens.
Example:
Nitrous oxide alters the nitrogen bases of DNA and disturb the replication and transcription that leads to the formation of incomplete and defective polypeptide during translation.

Question 11.
What is cis configuration (or) coupling?
Answer:
The two dominant alleles or recessive alleles occur in the same homologus chromosomes, tend to inherit together into same gamete are called coupling (or) cis configuration

XIV. Five Marks

Question 1.
Whose works supported the chromosomal theory of heredity? Explain.
Answer:

• T.H. Morgan works on fruit fly supported the chromosomal theory of inheritance.
• The alleles for red or white eye colour are present on the X – chromosome but there is no counter part for this gene on the Y chromosome.
• The genes for yellow body colour and miniature wings are also carried on the X – chromosome.
• By understanding the sex linked inheritance of these characters it is proved that genes are located on the chromosomes.
• Thus T.H. Morgan’s works on Drophila came as a support to the chromosomal theory of inheritance.

Question 2.
Write down the steps in the Holliday’s hybrid DNA model.
Answer:

• Homologous DNA molearles are paired side by side with their duplicated copies of DMAs.
• One strand of both DNAs cut in one place by the enzyme endonuclease.
• The cut strands cross & join the homologous strands forming the Holliday junction
• Holliday junction – migrates away from the original site, by branch migration – as a result
  heteroduplex region is formed.
• DNA strands may cut along the vertical (V) or horizontal (H) line.
• The vertical cut will result in heteroduplexes with recombinants & the Horizontal with non recombinants.

Question 3.
Explain sex determination is Silene latifolia (Melandrium album)
Answer:

• C.E Allen (1917) discovered sex determination in plants.
• Complex precess determined by
  1. genes
  2. environment
  3. hormones

Sex determination silene latifolia – is controlled by 3 distinct regions in a sex chromosome

• Y – Chromosome – determines maleners
• X – Chromosome – specify femaleness
• X & Y – show different segments (I, II, III, IV, & V)

Question 4.
How do Hawaii explain the sex determination in Papaya
Answer:

• Carica papaya 2n = 36
  
• The sex chromosomes look like autosomes
• Developed from autosomes
• Y- chromosome carries the genes for male organ
• X- chromosomes bear the gene for female organ development.

Question 5.
Explain sex determination in Sphaerocarpos donnelli. It is also known as Bottle liverwort (Bryophyta)
Answer:

• gametophyte – haploid with 8 chromosome (n).
• The sporophyte – diploid & heterogametic
• Male sfemale gameto phyte – seven autosomes are similar.
• In female 8th chromosome is X – Larger than the seven autosomes.
• In male 8th chromosome is Y – Smaller than the autosomes.
• In sporophyte – contain XY – combinations produces two types of meiospores
• Meiospore with X – produce – female gemetophyte

Question 6.
What are the various types of crossing over.
Answer:

Question 7.
The two loci A/a and D/d are so tightly linked that no recombination is ever observed. If AA dd is crossed to aa DD what phonotypes will be seen in the F2 and in what proportions.
Answer:
Genotypes of the parents are Ad/ Ad x aD x aD – If genes are so tightly linked, the only possible types of gametes produced by parents are
Ad and aD respectively (parental or nonrecombinant gametes)
FI will be all Ad / aD
only types of gametes from each FI can be Ad (50%)oraD(50%)
F2 frequencies can be calculated from these F2 will be Ad/ Ad (1/4i) Ad/ aD (1/2) aD / aD (1/4)

Question 8.
Classify maj or types of mutations.
Answer:

Question 9.
Define point mutation & explain it’s types
Answer:
Definition:
Mutation affecting single base or base pair of DNA
Types:

• Indel mutation : (Base pair insertions or . addition. Addition or deletions of nucleotide
  pairs.
• Substitution : one base pair is replaced by another

Types – (Two)

1. (Purine replaced by Purine)
2. Pyrimidine replaced by Pyrimidine
3. Transversion purine replaced by pyrimidin or pyridine replaced

Synonymous or silent mutations:
Here change in one codon for an amino acid into another codon for that same amino acid

Missense or Non synonymous mutations
Here the codon for one amino acid is changed in to -a termination or stop codon.

Frameshift mutations.
Additions or deletions of a single base pair of DNA, – changed the reading frame for translation – so there is complete loss of normal protein structure & function.

Question 10.
Explain how translocation chromosomal aberration is different from crossing over?
Answer:

Question 11.
Explain structural changes in chromosome with reference to changed to changes in the number of gene loci
Answer:

• There are 2 types
  1. Deletion
  2. Duplication

Deletion or Deficiency :

• (loss of a portion of chromosome)
• 2 types
  1. Terminal deletion (break in any one end
  2. Intercalary deletion (two breaks & reunion of terminal parts leaving the middle.

• > Unpaired loops some times formed known as deficiency loops (during meiotic prophase)
• > Larger deletions may have lethal effect Duplication or Repeat
• > Same order of genes repeated more than once in the same chromosome.
  Eg. Drosophila

Duplication
3 types

1. Tandem duplication
2. Reverse tandem
3. Displaced duplication

i) Tandem duplication
Duplicated segment is located immediately after the normal segment in the same order.

ii) Reverse tandem
Duplicated segment, immediately after the normal segment but gene sequence order will be reversed.

ii) Reverse tandem
Duplicated segment away from the normal segment.
Duplication play a maj or role in evolution.

Question 12.
Explain translocation.
Definition : The transfer of a segment of chromosome to – a non homologous chromosome is called translocation.
Types-3
i) Simple translocation:
Single break in only one chromosome the broken segment gets attached to one end of a non homologous chromosome – (Very rare occurrence)

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

iii) Reciprocal translocation.
Mutual exchange of chromosomal segments between two non homologous chromosomes – Illegitimate crossing over)

a) Homozygous translocation.

• Both the chromosomes of two pairs are involved.
• Two homologous of each translocated chromosomes are identical.

b) Heterozygous translocation.
Only me of the chromosome from each pair of two homologous are involved others remain normal.

Question 13.
Consider two hypothetical recessive auto¬somal genes a and b, where a heterozygote is testcrossed to a double homozygous mutant. Predict the phenotypic ratios under the following conditions:
a) a and b are located on separate autosomes.
b) a and b are linked on the same autosome but are so far apart that a crossover occurs between them.
c) a and b are linked on the same autosome but are so close together that a crossover almost never occurs.
Answer:
a) The problem involves an understanding of linkage, crossing over & independent assortment.
F₂geno & phenotypic ratio =

b) When crossing over occurs the result in same as the question (a)
F₂ geno & phenotypic ratio =

c) When a & b linked with out crossing on the heterozygolic parent can be AB / ab – (cis) or Ab / ab – (tr ans)
However there won’t be any recombinant gametes because no. crossing over occur.
It will produce Ab & aB (50 % of each)
The progeny of test cross will be Ab/ ab&aB/ab

Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Bio Botany Guide Pdf Chapter 2 Classical Genetics Text Book Back Questions and Answers, Notes.

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

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

I. Choose the correct answer from the given option

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) Lysososmes 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 the genotype AABbCC?
a) Three
b) Four
c) Nine
d) Two
Answer:
b) Four

Question 4.
Which one of the following is an example of polygenic inheritance?
a) Flower colour in Mirabilis jalapa
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 on (YY) was dominant over green cotyledon (yy). What are the expected phenotypes in the F₂ generation of the cross RRYY x rryy?
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 and wrinkled seeds with yellow cotyledons
Answer:
d) Round seeds with yellow cotyledons and wrinkled seeds with yellow cotyledons

Question 6.
Test cross involves
a) Crossing between two genotypes with a 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 Fx hybrid with a double recessive genotype

Question 7.
In pea plants, yellow seeds are dominant to green. If a heterozygous yellow seed plant is crossed with a green seeded plant, what ratio of yellow and green seeded plants would you expect in FI generation?
a) 9:1
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 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 Fx generation resembles both the parents
a) Incomplete dominance
b) Law of dominance
c) Inheritance of one gene
d) Codominance
Answer:
d) Codominance

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 dihybrid 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 allele 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 FI 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 same chromosome
d) Both of the characters are controlled by more than one gene
Answer:
c) The two genes are linked and present on the same chromosome

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

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

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:
a) Epistatic

Question 19.
Pure tall plants are crossed with the pure dwarf plants. In the FI generation, all plants were tall. These tall plants of the F1 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 hybridiza tion 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) Trichomalgalandular 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:

• True breeding lines (pure breeding strains) means it has undergone continuous self-pollination having specific phenotype trait inheritance from parent to offspring.
• Mating within pure breeding lines produces offsprings having, specific parental traits that are the same in inheritance and expression for many generations.
• Parents are homozygous for every trait.

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 off Fi offsprings with either one of the parental genotypes.
• The recessive back cross helps to identify the heterozygosity of the hybrid.
• It involves the cross between the fi offspring with either of the parents dominant.

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:

• Alleles are alternative form of gene and they are responsible for differences in the phenotypic expression of a given trait. A gene for which atleast two alleles exist is to be polymorphic, so a particular gene may exist in three or more allelic forms known as multiple alleles
• eg) ABO of human blood is controlled by three alleles

Question 29.
What are the reasons for Mendels successes in his breeding experiment? Pisum sativum a wise choice, because
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 a 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
In cross of parents that are pure for contrasting traits only one form of the trait will appear in the next generation. They have hybrid or dominant trait in the phenotype.
eg) Monohybrid cross

Regarding F₁ generation the recessive allele is not lost, and it remain hidden or masked. But it reappears in the next generation.

Question 31.
Differentiate incomplete dominance and co-dominance.
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 a 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 a universal genetic material.
• Genes located in nuclear chromosomes follow Mendelian inheritance.
• Certain traits are governed by the chloroplast (or) mitochondrial genes which is known as extranuclear inheritance.
• It is a kind of Non – Mendelian inheritance.
• The cytoplasmic organelles chloroplast and mitochondrion act as inheritance vectors so-called cytoplasmic inheritance.
• It is based on self – replicating extrachromosomal unit called plasminogen in the cytoplasmic Organelles, Chloroplast, and mitochondria.

Question 33.
Describe dominant epistasis with an example.
Answer:
Epistasis can be defined as a gene interaction whereby one gene interferes with the phenotypic expression of another non-allelic gene. The gene or locus which suppresses or masks the action of a gene at another locus is called the epistatic gene. The gene or locus where expressions are suppressed by an epistatic gene is called gene hypostatic.
Dominant epistasis A dominant epistasis suppresses the expression of a non-allelic gene, (dominant (or) recessive)
The F₂ phenotypic ratio is 12:3:1

Question 34.
Explain polygenic inheritance with an example
Answer:

• Polygenic inheritance, also known as quantitative inheritance, refers to a single inherited phenotypic trait that is combined bv two or more different: genes.
  (or)
• Several genes combine to affect a single trait. A group of genes that together determine (or) contribute a characteristic of an organism is called polygenic Inheritance
  (or)
• Polyinheritance occur when one characteristic is controlled bv two or more genes.
  Eg. Human skin colour & eye colour and weight.
• H.Nilsson -Ehle (1909), a Swedish geneticist discovered a polygenic inheritance in wheat (kernel colour). Kernel colour is controlled by two genes each with two alleles, one with red kernel colour was dominant to white. He crossed the pure breeding wheat varieties dark red and a white.
• Dark red genotypes R₁R₁R₂R₂ crosed unit r₁ r₁ r₂ r₂. In the F₁ generation medium red were obtained with genotype R₂ r₁ R, r₂. So 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 is obtained.
  

The data produces a bell shaped curve which demonstrate continuous variation in wheat kernel from dark red to white in F₂ when the number F₁ were self crossed five different phenotypic classes appeared in F₂ in into ratio of 1:4:6:4:1
The phenotype ratio is Dark red :1 Medium dark red :4 Medium red : 6
light red : 4 white : 1
Hence the total ratio is 63 red : 1 white in F2 generation
1:6:15 :20:15:6:1 in generation

He found that In F₂ generation plants have Kernel’s with range of colour variation. This is due to the fact that the genes are segregating and recombination takes place.

Question 35.
Differentiate continuous variation with discontinuous variation .
Answer:
Variation is the difference between individual with in a species. This can be caused by inherited or environmental factors. It can be continuous and discontinuous. Height, and weight of the human being are best examples of continuous variation. Human blood group, gender identity and eye colour are best example of discontinuous variation

Question 36.
Explain with an example how single genes affect multiple traits and alleles the phenotype of an organism.
Answer:

• There are several patterns responsible for the inheritance of traits, gene causes one trait. But in some cases one gene is responsible for multiple traits. Sometimes two or more gene are required to produce one trait.
• It is otherwise called pleiotropy. It means, where one gene will code and control the phenotype or expression of several different and unrelated traits.
• Eg. Phenylketenuria disease.
• A gene that produces multiple or effect is called a Pleitropic gene. Multiple effects of a single gene is know as pleiotropy. A Pleitropic gene is a single gene that controls more that one trait.
• Eg. Human genetic disorder are often pleitropic ie, unusual tall height, thin finger and toes, dislocation of the lens of the eye, heart in the aorta (heart function)
• Eg : Pisum sativum plant with purple brown seeds and dark spot on the axis of the leaves were crossed with a variety of a peas having white flowers light coloured seed and no spot on the axils of the leaves, the three traits for peas colour, seed colour and a leaf axil spot all were inherited together as a single exist. This is due to the pattern of inheritances controlled by a single gene with dominant and recessive alleles,
• eg .Sickle cell anemia
• eg .Marfan syndrome
• A human genetic disorder called marfan syndrome is caused by a mutation in one gene, yet it affects many aspects of growth and development inducing height, vision and heart function. This is an example of pleiotropy or one gene affecting multiple characteristics.

• Gene also interact in pattern such a partial dominance or co-dominance, the trait is expressed a mix between two gene , Those are possibilities for one gene. Most trait are influenced by many genes. There are many different way for these gene to influence how trait is expressed.

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

• It is found in 4 O’clock plant (Mirabilis jalapa)
• There are 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) the pollen of pale green leaved plant is transferred to the stigma of dark green leaved plant, the F₁ generation of both the crosses is identical as per mendelian inheritance.
• In the reciprocal cross the F₁ plant differs from each other.
• The F₁ plant reveals the character of the plant.
• The inheritance is due to the chloroplast gene found in the ovum of the female plant which contributes the cytoplasm during fertilization.
• The male gamete contribute only the nucleus.

12th Bio Botany Guide Classical Genetics Additional Important Questions and Answers

I. Match the following

Question 1.

Answer:
(a) Tall – (iv) dwarf
(b) Purple – (i) white
(c) arial – (iii) terminal
(d) Round – (ii) wrinkled

Question 2.

Answer:
(a) Dominant epistasis – (ii) 12:3:1
(b) Duplicate genes – (iii) 15 :1
(c) Recessive epistasis – (iv) 9:3:4
(d) Complementary gene – (i) 9:7

Question 3.

Answer:
(a) Genetics – (ii) W. Batson
(b) Mendel – (iii) Father of Geetics
(c) lethal gene – (i) E. Baeur
(d) H. Nillsson Ehle – (iv) Kernel colour

Question 4.

Answer:
(a) Polygenic inherence – (iv) wheat kernel colour
(b) 4 O’clock pea plant – (iii) Mirabilis jalapa
(c) Garden pea plant – (i) Pisum sativm
(d) H. NillssanEhle – (ii) genetic material

II. Choose the correct statement

Question 5.
a. HbA and Hbs alleles of normal and single-cell hemoglobin are multiple alleles
b. HbA and Hbs alleles of normal and single-cell hemoglobin are dominant recessive allele
c. HbA and HbA alleles of normal and single cell heamoglobin are codominant allele
d. HbA and Hb & alleles of normal and single-cell hemoglobin are recessive alleles
Answer:
c) HbA and HbA alleles of normal and single cell heamoglobin are codominant allele

Question 6.
a. When alleles of the two contrasting characters are present together, one of the character ex-press and the other remains hidden. There is the law of purity of gametes.
b. When alleles of the contrasting characters are present together, one of the character express and the other remains hidden . There is a law of dominance.
c. When alleles of the contrasting characters are present together with one of the character express and the other remain hidden This is law of segregation
d. When allele of two contrasting character are present together, one of the character express and remain hidden. This is law of independent assortment.
Answer:
b) When alleles of the contrasting characters are present together, one of the character express and the other remains hidden. This is the law of dominance

Question 7.
a. Monohybrid ratio is 9:3:3:1
b. The crossing of FI to any one of the parent is called test cross
c. The phenotypic ratio of a monohybrid cross is 1:2:1
d. A cross in which parents differ in a single pair of contrasting character is called a dihybrid cross
Answer:
c) The phenotypic ratio of a monohybrid cross is 1:2:1

Question 8.
a. The hybrid progeny in the first generation is called F2
b. The major reasons for the success of Mendelian experiment was the true-breeding of Garden Pea plant
c. X and Y are examples of alleles.
d. A pedigree chart shows the genotypes of any parent.
Answer:
b) The major reason for the success of mendelian experiment was true-breeding of Garden Pea plant

III. Choose the correct pair

Question 9.
a. Discontinuous variation – qualitative inheritance
b. Continuous variation – qualitative inheritance
c. Duplicate gene – 13: 3
d. Recessive epilate – 9:7
Answer:
a) Discontinuous variation – qualitative inheritance

Question 10.
a. Monohybrid – 9:3:3:1
b. Dihybrid – 1: 2: 1
c. recessive epistasis — 9: 3 : 4
d. extra chromosomal
inheritance — Mendelian inheritance
Answer:
c) recessive epistasis – 9 : 3 : 4

Question 11.
a. Emasculation – removal of anther
b. Tt – homozygous
c. genetic constitution – phenotype
d. mono hybrid cross – law of independent
assortment
Answer:
a) Emasculation – removal of anther

Question 12.
a. polygenic trait – Traits that are controlled by multiple gene
b. Multiple alleles – A gene that is controlled by one allele
c. Pleiotropy – one gene cannot affects multiple characters
d. Phenotype – genetic makeup of an organism.
Answer:
a) polygenic trait – Traits that are controlled by multiple gene

IV. Choose the incorrect statement

Question 13.
a. A pedigree charts are shown which genes are co-dominant
b. A true-breeding is a kind of breeding where the parents would produce offspring that would carry the same phenotype
c. In polygenic inheritance, traits are determined by interaction of single gene
d. The interactions between separate genes, in which one masks the effect of another is called epistasis.
Answer:
c) In polygenic inheritance traits are determined by interaction of single gene

Question 14.
a. The outward appearance resulting from an individual’s genotype for a particular characteristic is called phenotype
b. The recessive allele of the same gene represented by lower case letter.
c. Blood group is a human characteristic that shown discrete variation
d. The name given to different form of the same gene is gametes
Answer:
d) The name given to different form of the same gene is gametes

Question 15.
a. An allele is a viable DNA, coding that occupies a given locus on a chromosome
b. An allele is an alternative form of gene
c. An organism which has two different alleles of the gene is called homozygous
d. A person with one ‘A’ blood type and one ‘B’ blood type allele would have a blood type of “AB” ”
Answer:
c) An organism which has two different alleles of the gene is called homozygous

Question 16.
a. A pleiotropic gene is a single gene that more than one trait
b. A single gene affects multiple traits and alter the phenotype of the organism called as pleiotropy
c. Marfans syndrome is an example of pleiotropy.
d. one (or) single gene that cannot affect multiple traits are called pleiotropy.
Answer:
d) one (or) single gene that cannot affect multiple traits are called pleiotropy.

V. Choose the Incorrect Pair

Question 17.
a. Genotype – Genetic makeup of organism
b. recessive – A trait that is hidden
c. probability – The chance that an event will take place
d. Independent assortment – Mendel’s first law
Answer:
d) Independent assortment – Mendel’s first law

Question 18.
a. Dominant Allele – RR
b. Recessive allele – rr
c. Heterozygous – Tt
d. Homozygous recessive – TT
Answer:
d) Homozygous recessive – TT

Question 19.
a. Intra-locus interaction – allelic interactions
b. Inter-locus interaction – non-allelic interactions
c. Epistatic – allelic interactions
d. Polygenic interaction – non-allelic interaction
Ans:
c) Epistatic – allelic interactions

Question 20.
a. Complementary gene – 9:7
b. Co -dominance -1:2:1
c. Dominant epistatics – 9:3:4
d. Inhibitor gene -13:3
Answer:
c) Dominant epistatics – 9:3:4

VI. Choose the Odd one out

Question 21.
a) Mirabilis jalapa
b) Snapdragon
c) ABO Blood system
d) Epistasis
Explanation: a,b,c are F₂ phenotypic ratio is 1:2:1
Answer:
d) Epistasis

Question 22.
a. DNA
b. mitochondrial inheritance
c. Chloroplast inheritance
d. Atavism
Explanation : a,b,c are used as genetic material.
Answer:
d) Atavism

Question 23.
a. Monohybrid cross
b. checkerboard
c. genotype
d. phenotype
Answer:
b) checkerboard

Question 24.
a) co-dominance
b) Duplicate gene
c) inhibitor gene
d) supplementary gene
Explanation : b,c and d are intergenic or non¬allele interaction
Answer:
b) Duplicate gene

VII. Assertion and Reason

Question 25.
A : Polygenic inheritance
R : Several genes combine to affect a single trait
(a) A is correct (b) R is false
(c) R is the correct explanation of A
(d) R only correct
Answer:
c) R is the correct explanation of A

Question 26.
A : Atavism is a modification of biological structure whereby an ancestral trait reappears after having been lost through evolutionary changes in the previous generation
R : Reemergence of sexual reproduction in the flowering plant Hieracium pilosella is the best example for Atavism in plants
(a) A is correct R is the correct explanation of A
(b) A only true
(c) R only True (d) A false & R is true
Answer:
(a) A is correct R is the correct explana-tion of A

Question 27.
A: The physical expression of an individual gene called phenotype
R: Phenotype is physical observable charactertics of an organism
a) A & R True
b) A & R False
c) A is correct
d) R is correct
Answer:
(a) A & R True

Question 28.
A : Interaction between two alleles of the same loci is the effect of epistasis
R : The epistasis is the kind of intergenic and allelic interaction.
(a) A is correct R is false
(b) R alone correct
(c) R & A are true
(d) R is the correct explanation of A
Answer:
(a) A is correct R is false

VIII. Choose the best answer

Question 29.
If you do dihybrid cross in Pisum sativum on the traits of pod shape and plant height, Will you get 9:3:3:1 ratio in F₂ ?
a. Yes, because they are independently assorting genes.
b. No, they are linked genes.
c. Yes, because thev are situated on different chromosomes
d. No, we can not do experiments on these two traits.
Answer:
a) Yes, because they are independently assorting genes.

Question 30.
A single characteristic is controlled by a number of genes is called
a. Inheritance
b. Epistasis
c. Polygenic inheritance
d. Co-dominance
Answer:
c) Polygenic inheritance

Question 31.
An allele is
a. a homozygous genotype
b. a heterozygous genotype
c. another word for gene
d. several possible form of gene
Answer:
c) another word for gene

Question 32.
Continuous variation is due to
a. effect of polygenes
b. effect of environment
c. effect of polygenes and environment
d. effect of one or two genes.
Answer:
c) effect of polygenes and environment

Question 33.
A variation in a characteristic in which individuals show two or a few traits with large differences between them.
a. dominant
b. continuous variation
c. discontinuous variation
d. recessive
Answer:
c)discontinuous variation

Question 34.
A trait that masks the expression of another trait when both versions of the gene are present in an individual
a. variation
b. recessive
c. co-dominance
d. dominant
Answer:
d) dominant

Question 35.
Which one of the following is not a correct pair regarding genes of pea plant,
a. Seed shape – Chromosome number 6
b. Pod colour – Chromosome number 5
c. Flower position – Chromosome number 4
d. Seed colour – Chromosome number 1
Answer:
a) Seed shape – Chromosome number 6

Question 36.
The study of heredity behaviour of several genes by Gregor Mendel.
a. Molecular genetics
b. Population genetics
c. Quantitative genetics
d. Transmission genetics
Answer:
d) Transmission genetics

Question 37.
Transmission of characters from parents to offsprings
a. variation
b. dominance
c. heredity
d. growth
Answer:
c) heredity

Question 38.
Species that shows a difference in the characteristics of the same natural population is called
a. heredity
b. variation
c. recessive
d. co -dominace
Answer:
b) variation

Question 39.
Qualitative inheritance is otherwise called
a. co – dominance
b. continuous variation
c. discontinuous variation
d. heredity
Answer:
c) discontinuous variation

Question 40.
“Experiments on plant hybrids” is a
a. book
b. research paper
c. journal
d. Magazine
Answer:
b) research paper

Question 41.
Mendels theory of inheritance is based on
a. Particulate theory
b. mass
c. hybridization
d. variation theory
Answer:
a) Particulate theory

Question 42.
Removal of the anther is called
a. Atavism
b. Epistasis
c. Hybridization
d. Emasculation
Answer:
d) Emasculation

Question 43.
Botanical name of garden pea is
a. Solanum tuberosum
Question b. Coccus nucitera
c. Pisum sativum
d. pea
Answer:
c) Pisum sativum

44.
Mendel’s experiments were rediscovered by
a. Hugo de vries & carl correns
b. E. Baur
c. H. Nilsson
d. T.H.Morgan
Answer:
a) Hugo de vries & carl correns

Question 45.
If a homozygous red flowered plant is crossed with a homozygous white flower plant then the off-spring will be_
a. All red flowered
b. Half white flowered
c. Half red flowered
d. All white flowered
Answer:
c) Half red flowered

Question 46.
…………….. is he best example for chloroplast inheritance
a. Mirabilis jalapa
b. Sorgum vulgare
c. Triticum vulgare
d. Musa paradisiaca
Answer:
a) Mirabilis jalapa

Question 47.
Among the pea plant cell which one has the ability to convert a precursor molecule into an active inform
a. Le:le
b. GA₁
c. Le
d. le
Answer:
b) GA₁

Question 48.
Gene interaction concept was introduced and explained by
a. W. Bateson
b. Morgan
c. E. Baur
d. Nilsson
Answer:
a) W. Bateson

Question 49.
An allele which has the potential to cause the death of an organism is called
a. Genetic interaction
b. lethal alleles/lethal gene
c. Atavism
d. Autism
Answer:
b) lethal alleles/lethal genes

Question 50.
The gene whose expression is interfered by non- alletic gene and prevents from exhibiting its character is known as
a. hypostatic
b. epistatic
c. metastatic
d. hipostatic
Answer:
a) hypostatic

Question 51.
Height and skin colour in human are controlled by
a. two pair of genes
b. three pair of genes
c. five pair of genes
d. a pair of genes
Answer:
b) three pair of genes

Question 52.
The genotypic ratio of monohybrid cross is
a. 3:1
b. 1:2:1
c. 3:1:1
d. 9:3:3:1
Answer:
b) 1:2:1

Question 53.
Which of the following statements are true regarding law of segregation
a. alleles separate with each other during gametogenesis
b. The segregation of factors is due to the segregation of chromosomes during meiosis
c. Law of segregation is called as law of purity of gametes
d. all of the above
Answer:
d) all of the above

Question 54.
The crossing of Fj to anyone of the parents is called
a. test cross
b. back cross
c. FI cross
d. all of these
Answer:
b) back cross

Question 55.
The character that is express in to the F₂ is called
a. recessive character
b. co-dominant character
c. dominant character
d. none of these
Answer:
c) dominant character

Question 56.
The recessive character will express in
a. F₁
b. F₂
c. both a & b
d. F₃ only
Answer:
b) F₂

Question 57.
Which of the following pair is not correct
a. KK=dominant
b. hybrid = heterogeneous
c. heterozygous = Kk
d. homozygous = Rr
Answer:
a) KK=dominat

Question 58.
What is the phenotype of wheat kernal colour for the genotype: R₁ R₁ r₂ r₂ ?
a. Dark red
b. Medium dark red
c. Medium red
d. Light red
Answer:
c) Medium red

Question 59.
Mendel worked at the rules of inheritance and arrived at the correct mechanism. But
a. without any knowledge of cellular mechanism
b. knowledge of cellular mechanism
c. heredity mechanism
d. growth mechanism
Answer:
a) without any knowledge of cellular mechanism

Question 60.
is crossing an individual of unknown one pair of a genes is called genetic genotype with a homogeneous recessive.
a. back cross
b. test cross
c. monohybrid cross
d. dihybrid cross
Answer:
b) test cross

Question 61.
……………….is the expression of a single character by the interaction of more than interaction or interaction of genes.
a. factor hypothesis/ Bateson factor hypo¬thesis
b. alternative hypothesis
c. nell hypothesis
d. All of the above
Answer:
d) All of the above

IX. One Mark Question

1. The genetic constitution of the individual is called
Answer:
Genotype

2. The observable characteristics of an organism are called
Answer:
Phenotype

3. Who is father of genetics?
Answer:
Gregor Johann Mendel

4. Name the Mendel’s published work.
Answer:
Experiments on plant Hybrids.

5. Name the publication of Mendel research work
Answer:
1899

6. What is the year of published work Mendel’s Research paper?
Answer:
The proceedings of the Brunn Society & Natural History.

7. What is an allele?
Answer:
It is another word for a Gene.

8. Individuals show a range of traits with small difference between them.
Answer:
Continuous variation

9. When an individual show two or a few traits with large differences between them. This type of variation is called.
Answer:
discontinuous variation

10. Human height is the good example of ………….. variation.
Answer:
Continuous variation

11. Human skin colour is the good example of …………….. variation.
Answer:
Continuous variation

12. Mention any two examples of continuous variation.
Answer:
a. Human height
b. Human skin colour

13. Mention any two examples of discontinuous variation.
Answer:
Style length of Primula & Height of the garden pea.

14. A trait that makes the expression of another trait when both version of the gene are present in the individual called
Answer:
Dominant.

15. What is F₁?
Answer:
It is the first filial generation in a cross; the offspring of the parental generation.

16. The letter ‘P’ denoted in genetics is
Answer:
The parental generation in a cross

17. A variation in an inherited characteristics is
Answer:
Trait

18. One pair genes can completely makes the expression of another pair of genes known as
Answer:
Epistasis

19. Who discovered incomplete dominance?
Answer:
Correns. (Germany)

20. Crosses between F1 offsprings with either of the two parents (hybrids) are known as
Answer:
Back cross

21. Diploid organisms that have two different allele at a specific gene locus are said to be
Answer:
Heterozygous

22. TT referred as…………….
Answer:
Homogenous dominant variety.

23. ‘tt’ referred as ……………
Answer:
Homozygous recessive character.

24. ‘Tt’ denotes for …………….
Answer:
Heterogeneous hybrid variety.

25. The superiority of hybrid over either of its parents in one or more traits known as
Answer:
Hybrid vigour or Heterosis

26. The site or position of a particular gene on a chromosome is
Answer:
locus

27. An allele which has the potential to cause the death of an organism is called ……………….
Answer:
Lethal genes

28. A single gene affects multiple traits are called ……………..
Answer:
Pleiotropy

29. A single gene affects multiple traits and alter the phenotype of the organism is
Answer:
Pleiotropy

30. Several genes combine to affect a single trait of an organism.
This kind of inheritance is ……………
Answer:
Polygenic inheritance.

31. Who demonstrated first experiment on polygenic inheritance.
Answer:
Swedish Geneticist H. Nilsson – Ehle (1909)

32. Which plant to use to identify the polygenic inheritance?
Answer:
Wheat – Kernel colour (dark red & white variety)

33. List any two intragenic or allele interaction.
Answer:

1. Incomplete Dominance
2. Co-dominance

34. List any two intergenic or non-allele interaction
Answer:

1. Dominant Epistasis
2. Recessive Epistasis

35. Corren has used plant for studied incomplete dominance.
Answer:
Mirabilis jalapa (4′ O clock plant)

36. Mention the botanical name of 4′ O clock plant.
Answer:
Mirabilis jalapa.

37. Duplicate genes with cumulative effect of non-alleleic interaction is derived in
Answer:
Fruit shape in Summer squash.

38. What is the FI phenotypic ratio of inhibitor genes in the intergenic interaction?
Answer:
13:3

39. When the heterozygote exhibits a mixture of phenotypic character of both homozygous called as
Answer:
Co-Dominance.

40. Name the two gene interaction.
Answer:

1. Intralocus interaction (allelic interaction)
2. Interlocus interaction (non-allelic interaction)

41. A chart shows which genes are co-dominant. This is known as
Answer:
A pedigree charts.

42. Each character is controlled by distinct units called factor, which occur in pairs. If the pairs are heterozygous, one wiil always dominant other. This is known as
Answer:
First law of inheritance or Law of Dominance.

43. The second law of inheritance otherwise called as
Answer:
Law of Segregation.

44. Give the name of the scientists who re-discovered Mendelism
Answer:

1. Hugo Devries
2. Carl Correns
3. Erich Von Tschermak.

45. is the prerequisite for Hybridization technique.
Answer:
Emasculation.

46. Transmission of genes that occur outside the nucleus is called………………
Answer:
Cytoplasmic Inheritance or Extra Nuclear

47. Cytoplasmic inheritance are found in
Answer:
Mitochondria & Chloroplast

48. The interaction between separate gene in which one makes the effect of another
Answer:
Epistasis

49. The acquisition of traits or conditions controlled by self replicating substances within the cytoplasm. This is a type of
Answer:
Cytoplasmic Inheritance.

50. The hybrid progeny in the first generation is called as
Answer:
F₁

51. The innate tendency of offspring to resemble their parents is called
Answer:
Heredity

52. The tendency of offspring to differ from parents is called
Answer:
Variation

53. Multiple allelic inheritances is otherwise called as
Answer:
Co – dominance

54. What is the use of pedigree analysis in genetics?
Answer:
It helps in genetic counselling.

55. Who proposed the genetic theory of inheritance?
Answer:
T.H.Morgan

56. Give one good example for Atavism in plants.
Answer:
Reemergence of sexual reproduction in Hieracium pilosella.

57. In pea plant, yellow seeds are dominant to green. If a heterozygous yellow seeded plant is crossed with a green seeded plant. What ratio of yellow and green seeded plants would you expect in FI generation?
Answer:
50:50 (or) 1:1

58. Some genes have allele that prevents survival when homozygous or heterozygous. What is the kind of allele?
Answer:
Lethal alleles

59. Recessive alleles of two different genes may give the same phenotype; This kind of genes also called
Answer:
Complementary gene.

60. A gene is a functional unit of DNA which codes for a
Answer:
Polypetide chain

61. Allele are the alternative form of the
Answer:
gene

62. discovered incomplete dominance.
Answer:
Correns

63. Human blood group is an example of variation.
Answer:
Discontinuous

X. Two marks

Question 1.
Write short note on Genotype.
Answer:
It is the genetic makeup of an organism responsible for a particular trait.

Question 2.
What is phenotype?
Answer:
It is the outward appearance or observable physical attributes of that trait.

Question 3.
Briefly explain monohybrid inheritance.
Answer:
Monohybrid inheritance looks at the inheritance of a single trait (a characteristics such as eye color, round or wrinkled seed type) coded by a single gene locus on a chromosome

Question 4.
Define Mendel’s first law.
Answer:
Mendel’s first law is ‘The law of segregation’. Segregation means separation. The two alleles are separated from each other during meiosis, so each gamete produced is haploids that is contain one allele of each gene.

Question 5.
What is epistasis?
Answer:
It is a term which describes how genes interact to affect a phenotype whereby an allele at one locus prevents an allele at another locus from manifesting its effect.
(or)
One gene is effectively interfering with or masking the effects of another gene.

Question 6.
What is epistatic?
The gene that suppresses or masks the . phenotypic expression of a gene at another locus is known as epistatic.

Question 7.
Write a note on hypostatic?
Answer:
In epistatsis, the gene whose expression is interfered by non- allelic genes and prevents from exhibiting its character is known as hypostatic.

Question 8.
Describe one of the reason that made the garden pea an excellent choice of Mendel system for studying inheritance.
Answer:
It is easily available self pollinated crop.

Question 9.
What is continuous variation with examples?
Answer:
A variation in a characteristics in which individuals show a range of traits with small difference between them. Eg: Human height and skin colour.

Question 10.
Write a note on discontinuous variation with suitable examples.
Answer:
Discontinuous is a variation in characteristic in which individuals show two or a few traits with large differences between them. (Eg) Height or Length of a plant.

Question 11.
What is hybridization?
Answer:
The process of mating two individuals that differ, with the goal of achieving a certain characteristics in their offspring.

Question 12.
Briefly explain ‘F₂‘.
Answer:
The second filial generation produced when Fi individuals are self-crossed or fertilized with each other.

Question 13.
Write a short note on Punnett square or checkerboard?
Answer:
A sort of cross multiplication matrix used in the prediction of the outcome of a genetic cross, in which male and female gametes and their frequencies are arranged along the edges.

Question 14.
List out the ‘R’ gene on responsible for polygenic inheritance in wheat (kernel colour)
Answer:
Four R genes are produced dark red kernel color. Three R genes are produced medium dark red kernel colour. Two R genes are produced medium red kernel colour. One R gene is produced medium red kernel colour and absence of R genes in results in white kernel colour.

Question 15.
Explain the role of genes in the formation of purple colour in the flowers of pisum sativum.
Answer:

• It was called Pea Gene A which encodes a protein that functions as a transcription factor which is responsible for the production of anthocyanin pigment.
• So the flowers are purple. Pea plants with white flowers do not have anthocyanin, even though they have the gene that encodes the enzyme involved in anthocyanin synthesis.

Question 16.
Write a note on Mendel’s Law of Dominance.
Answer:
It states that a dominant allele expresses itself in a monohybrid cross and suppresses the expression of recessive allele. However this recessive allele for a character is not lost and remain that hidden or masked in the progenies of F:l generation and reappear in the next generation.

Question 17.
What are multiple alleles?
Answer:
Alleles are alternative form of a gene. A gene for which at least two alleles exist is said to be polymorphic. Instances in which a particular gene may exist in three or more allelic forms are known as multiple allele conditions.

Question 18.
Briefly explain Mendelian Genetics.
Answer:
The set of theories prepared by Gregor Mendel, which attempt to explain the inheritance pattern of genetic characteristics based on simple breeding experiments involving single gene on chromosome pairs.

Question 19.
Write a note on 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 20.
Explain the three kinds of plants that have recersive lethal gene in Antirrhinum sp.
Answer:

1. Green plants with chlorophyll (CC)
2. Yellowish green plants with carotenoids are referred to as pale green, golden or a urea plants (Cc)
3. White plants without any chlorophyll, (cc)
4. The genotype of the homozygous green plants is CC. The genotype of the homozy¬gous white plant is cc.

Question 21.
Write a note on incomplete dominance.
Answer:
It refers to genetic situation in which one allele does not completely dominate another allele, and therefore results in a new phenotype.
(or)
It is a form of intermediate inheritance in which one allele for a specific trait is not completely expressed over its paired allele. This results in third phenotype in which the expressed physical traits is a combination of the phenotypes of both alleles.

Question 22.
A diagram that shows the possible outcomes of breeding between two individuals.
Answer:
Punnett Square or Checkerboard

Question 23.
Write a note on Punnet Square.
Answer:
It is a square type of a diagram that shows the possible outcomes of breeding between two individuals.

Question 24.
What do you mean by genetics ?
Answer:
Genetics is the study of how living things receive common traits from previous generation.

Question 25.
What are genes ?
Answer:
Genes are functional unit of inheritance. It is the basic unit of heredity (biological information) which transmits biochemical, anatomical and behavioural traits from parents to off springs.

Question 26.
What is population Genetics ?
Answer:
It deals with heredity in groups of individuals for trait which is determined by a few genes.
(or)
Population genetics is the study of genetic variation with in population, and the examination and moddling of changes in the frequencies of gene and allele in populations over space and time.

Question 27.
Define Molecular genetics .
Answer:
It is the field of that biology that studies the structure and function of genes at a molecular level
(or)
Study of structure and function of genes at molecular level
(or)
A branch of genetics that deals with structure and function of genes at molecular leve

Question 28.
Define Mutation.
Answer:
A permanent, heritable change in the nucleotide sequence in a genes or a chromosome , the process in which such a change occurs in a gene or in chromosome.

Question 29.
What do you mean by genetic transmission ?
Answer:
Genetic transmission is the transfer of genetic information (From parent to offspring), almost synonymous with heredity, or from one location in a cell to another .

Question 30.
Define Transmission Genetics :
Answer:
The study of the mechanisms involved in the passage of gene from one generation to the next.

Question 31.
What are polygenes ?
Answer:
A gene where individual effect on a phenotype is too small to be observed but which can act together with others to produce observable variation.
(or)
Characters are determined by two or more gene pairs, and they have additive or cumulative effect. Such genes are called polygenes or multiple factors or cumulative gene. Eg. Human skin colour.

Question 32.
Define Polygene ?
Answer:
Inheritance of phenotype is determined by the combined effects of many genes with environmental factor . These gene are called as polygene

Question 33.
Mendel was successful, why?
Answer:

• He applied mathematical method of law of probability to his breeding experiments
• He used pairs of contrasting characters in their experiment.

Question 34.
Write a note on self fertization.
Answer:
> (2 Marks) Fertilisation in a plant or animal by the fusion of male and female gametes produced by the same individual
(or)
> (3 Marks) Fertilisation that occurs when male and female gamete produced by the organism unite self fertilisation occur in many protozoans and invertebrate animal. It result from self pollination in plants. Seeds fertilization allows an isolated individual organism to reproduce but restricts the genetic diversity of a community.

Question 35.
What is cross fertilisation ?
Answer:
The fertilisation of an organism by the fusion of an egg from one individual with a sperm or male gamete from a different individual’s is opposite to the self.
(or)
Cross fertilisation is a term used in the field of biological reproduction describing the fertilisation of an occurs from one individual with spermatozoa of another. It is also called allogamy.
(or)
The fusion of male and female gamete (sex cells) from different individual of the same species.
It is mostly occur in dieocious plant and in animal species which they are separate male and female individual.

Question 36.
Does pure breeding means homozygous?
Answer:
If they are pure breeding that mean they are homozygous . So A group of identical individual that always produce offspring and same phenotype when intercrossed

Question 37.
What is the relationship between pure breeding and true breeding ?
Answer:
True breeding means that the parents with also pass down a specific phenotypic trait to their offspring. True breeding organism will have a pure genotype (genetic expression of a trait) and they will produce a certain phenotype. True breed is sometime also called pure breed.

Question 38.
Write a short note on Anthocyanin pigment.
Answer:

• Anthocyanin are naturally occurring pigment of red, purple and blue.
• Anthocyanin pigments are more stable at low PH (Acidic condition) which gives a red pigment. Measurable higher the PH value of anthocyanin will provide of colour fading of the colour blue or purple.

Question 39.
What is the mean ‘progeny’?
Answer:
The word progeny is the progeny of the Latin verb “progignere” meaning “to beget” . In biology, offspring are the young born of living organism, produced either by a single organism or in the case of sexual reproduction, true organism. Collective offspring may be known as a brood or progeny. It is also called as offspring of animals or plants or the children and other descendants.

Question 40.
Point out the mechanism of Trihybird cross.
Answer:
A cross between homozygous parent that differ in three gene pairs is called to trihybrid cross. A self fertilising trihybird plants forms 8 different gemeter and 64 different zygote. So these combination of three pair crosses operating together.

Phenotypic ratio -27:9:9:9:3:3:3:1 F₂
Mendel laws of segregation and independent assortment are also applicable to three pairs of contrasting traits ie. Trihybrid cross

Question 41.
What is back cross ?
Answer:
The cross between the F₁ offspring with either of the two parents. The parent may be dominant or recessive
(or)
When F₁ individuals are crossed with one of the true parenst from which they were derived, then such cross is called back cross
Explanation

• When TT is crossed with tt we get Tt as F₁ generation
• TT x tt = Tt
• when Tt (F₁ ) is crossed with either TT or tt (parent) it is called a back cross .

Question 42.
What are the classification of gene interactions?
Answer:
Interactions take place between the alleles o the same gene.
alleles at the same locus is called intragenic or intralocus gene interactions.

• Incomplete dominance
• co dominance
• multiple alleles
• pleiotropic genes.

Question 43.
Inheritance of chloroplast and mitochondria characters are non-mendelian inheritance
pattern why?
Answer:
The chloroplast arid mitochondrial genes show special pattern of inheritance known as Extra chromosomal inheritance.

Chromosomal inheritance:
The other aspects are

• They have vegetative segregation involving cytoplasmic plasmagenes. .
• It has uniparental inheritance (only from female parent)
• Both have reduced rate of recombinations.

Question 44.
What is hybrids?
Answer:
Mendels non-true breeding plants ae heterozygous called as hybrids.

Question 45.
What is Dihybrid cross?
Answer:

• It is a genetic cross which involves individuals differing in two characters.
• Dihvbrid inheritance is the inheritance of two separate genes each with two alleles.

XI. Three marks

Question 1.
Explain Bateson’s factor hypothesis ?
Answer:
Mendelian experiments prove that a single gene controls one character. But in the post mendelion findings, various exception have been noticed, in which different types of interaction are possible between the genes. FTence the expression of a single character by the interaction of more one pair of genes is called genic interaction or interaction of genes. According to this hypothesis some character are produced by the interaction of two or more pairs factor (gene).

Question 2.
What is the human ABO phenotype blood type based on?
Answer:
It is the major human blood group system. The ABO type of a person depends on they presence of absence of two gene, A and B. These gene determine the configuration of the red blood cell surface. A person who has two A gene or an A and O gene has bloodcells of type A. There are four main group of blood A,B,AB and O. The phenotype ratio is given below.
Blood group inheritance phenotype only

Question 3.
Explain the Genetic inheritance of pattern of human blood system ?
Answer:
An individually ABO type results from the inheritance of 193 alleles is A,B,0 from each parent . The possible out comes are given below

Both A and B alleles are dominant over O. As a results individual who have an AO gene type will- have an A phenotype. People who are type O have OO genotype. In other words, they inherited a recessive ‘O’ allele from both parents . The A and B alleles are co-dominant. Therefore, if an A is inherited from one parent and a B from the other the phenotype will be AB.

Question 4.
In blood type co-dominance or incomplete dominance ?
Answer:
It is closely related to incomplete dominance is co-dominance is which both alleles are simultaneously expressed in the heterozygote. In both co-dominonce and incomplete dominance both alleles for a trait are dominate in co-dominance a hetrozygous individual express both simultaneously with out any blending. People who are to type O have OO genotype. In other words they inherited a recessive O allele from both parents. The A and B alleles are co-dominant. Therefc )re is an A is inherited from one parent and a B from other the phenotype will be AB

Question 5.
In sickle cell co-dominant or incomplete dominance ?
Answer:
sickle cell anemia is a disease, in which the haemoglobin protein is produced incorrectly and the red bloodcells have a sickle shape. A person that is homozygous recessive for the sickle cells traits wills have red blood cells that all have the incorrect haemoglobin.

Question 6.
Write a note on co-dominance ?
Answer:
Co-dominance occurs when the phenotype of both parents are simultaneously expressed in the same offspring . An example of co¬dominance occurs in the human ABO blood group

Question 7.
Across between Bbcc and Bbcc. What is the probability of Bbcc?

Answer:
Solution
Probability of Bbcc = (Probability Bb) . (Probability Cc)

Question 8.
Write a note on Homologous chromosome or homologous.
Answer:
Morphologically, physiologically and genetically similar chromosome present is a diploid cell are called homologous or homologous chromosomes. In each pair of homologous chromosomes, one chromosome maternal and the other is paternal.

Question 9.
Write a note Emasculation.
Answer:
Removal of stamen well before another is called emasculation . It is done in bud condition to prevent self -pollination.

Question 10.
What is Punnett square or checker board?
Answer:
Punnett square is a graphical representation to calculate the probability of all possible genotypes of offsprings in a genetic cross. It was developed by Reginald C.Punnett.

Question 11.
Distinguish between homozygous and heterozygous
Answer:
homozygous :

1. Organism having identical alleles for a character are homozygous.
2. It is pure or true breeding
3. They form only one type of gametes
4. (eg) Tall (TT) dwarf (tt)

heterozygous :

1. Organism having dismillar alleles for a character are heterozygous.
2. It is hybrid
3. They form more then one type of gametes.
4. es (Tt)

Question 12.
Differentiate dominant from recessive character.
Answer:

Question 13.
Differentiate between Phenotype and Genotype
Answer:
Phenotype

1. It is the physical appearance of and organism
2. It can be directly seen
3. phenotype can be determined from genotype, (eg) Tt =Tall

Genotype

1. It is the genetic constitution of an organism
2. It is determined by inheritance pattern
3. Genotype can not be determined from phenotype (eg) Tall can either Tt (or) TT

Question 14.
Listant/Point out/Enlist the several traits in pea selected by mendel
Answer:

Question 15.
Draw the flow chart for heterozygous tall X homozygous dwarf pisum sativum plants If heterozygous tall test cross
Answer:

Question 16.
Distinguish between monohybrid cross and dihybrid cross
Answer:
Monohybrid cross :

1. The cross between to pure parents differing in a single pair of contrasting character is called Monohybrid cross
2. Phenotypic ratio is 3:1
3. Genotypic ratio is 1:2:1
4.  The law of segregation is explained by this method

Dihybrid cross :

1. The cross between two pure parents differing in two pairs of contrasting character is called dihybrid cross
2. phenotype ratio is 9:3:3:1
3. genotype ratio is 1:2:2:4:1:2:1:2:1
4. The law a independent is explained by this cross.

Question 17.
Distinguish between Test cross and Back cross
Answer:
Test Cross

1. The cross between F1 hybrid and its recessive parent is called test cross
2. A test cross is always a back cross
3. Test cross determines the genetic constitution of an organism
4. Test cross produces both dominant and recessive character is equal proportion

Back Cross :

1. The cross between F1 hybrid and any one of its parents (either dominant or recessive) is called back cross.
2. A back cross is not always a test cross
3. Back cross helps in improving and obtaining desirable character
4. Back cross helps in improving and obtaining desirable character

Question 18.
What is genetic testing?
Answer:
Genetic testing is analysing an individuals genetic material to determine predisposition to a particular health condition or to confirm a diagnosis of genetic disease

Question 19.
What are genetic disorder ?
Answer:
Genetic disorders are nothing but malfunctioning of genes due to some changes in their arrangement brought by mutation. Often these disorders characterized by absence or inactive protein products.

Question 20.
Write a short note on ‘Mutation’?
Answer:
Sudden heritable change in DNA or chromosome is called mutation. There are agents which cause mutation called Mutagens. Due to mutations many abnormalities will appear in new generations which may be useful or harmful.

Question 21.
Co-dominance is an example of intragenic gene interaction. How?
Answer:

• The phenomenon in which two alleles are both expressed in the heterzygous individual is known as codominance
  Example:
• Red and white flowers of camellia, inheritance of sickle cell haemoglobin.
• ABO blood group system in human beings.
• In humanbeings, IA and IB alleles of I gene are codominant which follows mendels law of segregation.
• The co-dominance was demonstrated in plants with the help of electrophoresis or chromatography for protein or flavonoid substance.

Question 23.
What is the different between sex linked and sex influenced diseases ?
Answer:
In sex linked diseases the defected gene are present on the sex chromosomes attached to them whereas in sex influenced diseases defective gene are present on the other chromosome but affects the sex chromosomes.

Question 24.
What is Genome ?
Answer:
A complete set off gene is an organism is called genome .

Question 25.
What are lethal gene or lethal allele ?
Answer:
Lethal allele are alleles that cause the death of the organism that carries them. They are unusually a result of mutation is gene that are essential to growth or development. Lethal allele may be recessive, dominant or conditional depending on the genes or genes involved.

Question 26.
What do you mean by inheritance of sickle cell anemia in man.
Answer:
The diseases sickle cell anemia is causes by a gene (Hbs) which is lethal in homozygous condition. But has a slight denotable effect is the heterozygous conditions, producing sickle cell trait. The homozygous for this gene (Hbs/HbS) generally die of fatal anemia. The hetrozygotes or carriers for Hbs. (ie) HbA/HbS) show signs of mild anemia as their RBC become sickle – shaped in oxygen deficiency. A marriage between two carriers, therefore results in carrier and normal offspring in the ratio 2:1

Question 27.
What is cytoplasmic male sterility ?
Answer:
Plants that fail to produce functional pollengrains are said to be male-sterile. If the traits conditioning the sterility is not inherited according to mendelion rules, but is instead maternally transmitted, it is referred to as cytoplasmic male sterility(cms). So in this male-sterility is inherited maternally.
The gene for cytoplasmic male sterility is found in the mitochondrial DNA
(or)
When plants fails to produce functional pollengrain, they are called male sterile mole. Male sterility may be conditioned by either nuclear or Cytoplasmic genes. If the sterility trait is inherited is a non -Mandelian fashion, it is designated as cytoplasmic male sterility (CMS). Cytoplasmic gene are most often maternally transmitted in plants.

Question 28.
Briefly explain ‘Atavism’ with suitable examples.
Answer:
Atavism derives via French from Latin atavius, meaning “ancestor”. Avus in Latin means ‘grand father’; and its is believed that the ‘at’ is related to atta a word for “Daddy”. Atavism is a term rooted in evolutionary study referring to instances when an organism possesses trait closer to a more remote ancestor, rather than its own parents. It is modification of a biological structure whereby an ancestral traits re appears after having been lost through evolutionary changes is the previous generations.

(eg) Re-emergence of sexual reproduction in the flowering plant Hieracium pilosella is the best example for Atavism in plants

Question 29.
How to do test for homozygosity of a trait in plant.
Answer:

• To identify whether an organism exhibiting a dominant trait is homozygous or hesterozy- gous for a specific allele a scientist can perform a test cross.
• The organism in question can be crossed with an organism that is homozygous for the recessive trait – and the offspring’s of the test cross are examined.

XII. Five Marks

Question 1.

Difference between Pleiotropy and polygenic inheritance with suitable examples.
Answer:
Pleiotropy is when one gene affect multiple characters eg. Marfan syndrome and polygene inheritance is when one traits is controlled by multiple genes (eg), skin colour (or) skin pigmentation

Question 2.
Co-dominance and incomplete dominance are not the same? why?
Answer:

• In co-dominance neither allele is dominant over the other, so both will be expressed equally in the heterozygote.
• In incomplete dominance, there is an intermediate heterozygote. Such as pink flower when the parent phenotypes are red. and white.

Question 3.
Difference between Monohybrid cross and Reciprocal cross.
Answer:
Monohybrid

1.  It is one sided or both sided
2.  It is used to study the inheritance of single pair of alleles.
3. It cannot distinguish between nuclear and Cytoplasmic (or) sex linked and autosomal traits

Reciprocal cross

1. It is both sided cross in which female of one type is crossed with male of the second type and vice versa
2. It may study inheritance of one, two or more traits
3. It can distinguish between nuclear and cytoplasmic inheritance as well as sea linked l j and autosomal inheritance.

Question 4.
Difference between Monohybrid and Dihybrid cross
Answer:
Monohybrid :

1. Mono refer to single and hybrid means mixed breed
2. It is used to study the inheritance of single pair of alleles.
3. Genotype ratio is 1:2:1
4. Phenotypic ratio is 3:1
5. One pair of contrasting character are involved

Dihybrid

1. Di refers to two or double and hybrid means breed.
2. It is used to study the inheritance of two different alleles.
3. Genotype is ratio is 1:2:1:2:4:2:1:2:1
4. Phenotypic ratio is 9:3:3:1
5. Two pair of contrasting character are involved.

Question 5.
Explain Monohybird cross.
Answer:
A monohybird is a genetic cross which occurs between two individuals, focusing on the inheritance of one trait at one time. Monohybrid cross is also known as single trait cross. Two homozygous parent are selected for this cross.

Each homozygous parent in the P generation produces only one kind of gamete.
The heterozygous F] offspring produces two kinds of gamete
The heterozygous Fi offspring produces two kinds of gamete

Question 6.
Explain Dihybrid cross.
Answer:
A dihybrid cross is a genetic cross that occurs between two individuals, focusing on the inheritance of two independent traits at one time. It is also known as two trait cross.
Two parents considered for this cross have two independent traits (eg: pea colour and pea shapes of plants). Thus a dihybrid cross involves two pairs of genes. The following figure explains the process of dihybrid crossing.

F₁ phenotype: All round yellow cotyledon
Fi genotype : All RrYy
RrYy x RrYy (Fj generation selfied)
Ry Ry rY rY x Ry Ry ry ry (Haploid gametes)

How to do a Dihybrid Cross

• Analyze the data!
• Make a tally of all possible phenotypes.

In a dihybrid cross, traits are considered as not linked, and they have an equal probability of sharing up in offspring. Each pair of alleles segregates independently of the gametes. Offspring is predicted and assessed for two trait inheritance. The phenotypic ratio of the offspring generation is 9:3:3:1 in a dihybrid cross.

Question 7.
Briefly explain Trihybrid cross.
Answer:
A trihybrid cross is between two individuals that are homozygous for three different traits. (Eg: Pea shape, colour and pea shape)
(or)
A cross between homozygous parents that differ in three gene pairs, (ie: producing trihybrid) is called trihybrid cross. A seed fertilizing trihybrid plant forms 8 different gametes and 64 different zygotes. So a combination of three single pair crosses operating together. The three contrasting characters of a trihybrid crosses are

• F₂ Phenotypic ratio – 27:9:9:9:3:3:3:1
• 27 – round, green, smooth pod
• 9 – round, green, constructed pod
• 9 – round, yellow, smooth pod
• 9 – wrinkled, green, smooth pod
• 3 – round, yellow, constructed pod
• 3 – wrinkled, green, constructed pod
• 3 – wrinkled, yellow, smooth pod
• 1- wrinkled, yellow, constructed pod

Question 8.
What traits are determined by multiple alleles?
Answer:
A trait controlled by one gene but multiple allele is blood type. There are four phenotypes A, B, AB, O. Type A and B are co-dominant and ‘O’ is recessive to A and B. None are dominant. Some traits are controlled by a single gene with two alleles. Mendelian heredity had only two alternative expression or alleles. However many genes can change in several different ways or changes. Those changes give rise to several alternative states which are called multiple alleles.
(or)
Blood type is an example of a common multiple allele trait. There are three different alleles for blood type A, B & O. A is dominant to O, B is also dominant to O. A and B are both co-dominant.

Question 9.
What is a gene?
Answer:
A gene is a segment of DNA that spells out the gentic code for a particular trait. A trait is a physical characteristics.

Question 10.
What is Incomplete dominance with example.
Answer:

• Carl Correns’s (1905) experimented in 40′ clock plant, Mirabilis jalapn.
• When the pure breeding homozygous red (R¹R¹) parent is crossed with homozygous white (R¹R¹)
• The phenotype of the F₁ hybrid is heterozygous pink (R²R²)
• The F1 heterozygous phenotype differs from both the parental homozygous phenotype.
• This cross did not exhibit the character of the dominant parent but has an intermediate colour pink.
• The phenotypic and genotypic ratios were found to be same as 1:2:1 (1 red : 2 pink : 1 white). Genotypic ratio is 1 R¹R¹: 2 R¹R²: 1 R²R² in F₂ interbreed.
• In the F2 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 and R² allele codes for defective enzyme. R¹ and R² genotypes produce only enough red pigments to make the flower pink.
• Mendel’s particulate inheritance takes place in this cross which is confirmed by the reappearance of original phenoty in F₂
  
  

Question 11.
Briefly explain about lethal gene.
Answer:
Allele that cause an organism to die are called lethal alleles or lethal genes. Lethal genes are usually a result of mutations in genes that are essential to the growth or development. Lethal gene can cause death of an organism prenatally or anytime after birth. Lethal genes are first discovered by Lucien cuenot in the study of coat colour in mice.

Question 12.
Explain epistatsis and its two types.
Answer:
Epistasis is a type of polygenic interaction where one gene controls the phenotype of another gene for a trait. Both genes have an influence on the physical appearance of the traits, but the one that shows epistasis masks the effect of the other. Eg: albinism.

Dominant epistatsis: It happens when the dominant allele of one gene masks the expression of all allele of another gene.

Recessive Epistasis:
Recessive epistasis is when the recessive allele of one gene in a homozygous slate masks the phenotypic expression of the dominant allele of another gene.
(eg) Mice,
In Mice, body Colour is determined by a gene A. A is hypostatic to an allele C of another gene, which mean that C marks the expression of A. C is the presence of a gives cinnamon mice, While C in the prsence of A gives agouti mice.

Recessive Epistasis :

In dominant epistasis, the majority of the individuals are affected. There is a 12:3:1 ration.

Genes that show recessive epistasis can only mask a phenotype if two alleles are present The ratio is 9:3:4

Question 13.
Briefly explain duplicate recessive gene in Intergenic interaction (or) complementary gene interaction
Answer:
If both gene loci have homozygous recessive alleles and both of them produce identical phenotype the F2 ratio 9:3:3:1 would be 9:7. The genotype aaBB, aaBb, AAbb, Aabb and aabb produce same phenotype. Both dominant alleles when are present together only than they can complement each other. This is known as complementary gene.
Complementary Genes (9:7)
Ex: In Lathyrusodoratus ,Bateson and punnet crossed two varieties(CCpp x ccPP),each with white flowers.

• Eg: complete dominance at both gene pairs, but either recessive homozygote is epistatic to the effect of the other gene.
• In sweet pea flower colour.
• Gene pair A = purple dominant over white
• Gene pair B = colour dominant over white
• Interaction = Homozygous recessive of either gene A or B produce white

Question 14.
Explain duplicate gene with cumulative effect (9:6:1)
Answer:

• Certain phenotype traits despond on the dominant alleles of two gene loci. When dominant is present it will share its phenotype. The ratio will be 9:6;1 Eg: Fruit shape in summer squash.
• Complete dominance at both gene pair, interaction between, both dominance to give new phenotype.
• Gene pair ‘A’ sphere shape dominant over long.
• Gene pair ‘B’ sphere shape dominant over long.
• So interaction at ‘AB’ when present together form disc shaped fruit.
• Finally disc shaped fruit 9/16 Sphere shaped fruit 6/16
  

Duplicate genes with cumulative effect (9:6:1) :

Question 15.
What are Duplicate dominant gene (15:1) or duplicate gene?
Answer:

• If a dominant allele of both gene low produces the same phenotype without cumulative effect i.e., independently the ratio will be 15:1
• Eg : seed capsule of shephered’s purse complete dominance at both gene pair, but either gene when dominant, epistatic to the other.
• Gene pair ‘A’=Triangular shape dominant over ovoid
  Gene pair ‘B’=Triangular shape dominant over ovoid (double recessive)

Duplicate dominant genes (15:1):
15/16 = Triangular
1/16= Ovoid (top shaped)
AABB x aabb
Triangular ovoid
AaBb x AaBb
Triangular 

Question 16.
Explain dominant and recessive interaction (or) inhibitor gene (13:3)
Answer:
Sometimes the dominant alleles of one gene locus (A) in homozygous and heterozygous (AA, Aa) condition and homozygous recessive alleles bb of another locus (B) produces the same phenotype. The F2 ratio will become 13:3. The genotype AABB, AaBB, AAbb, Aabb and aabb produce one type of phenotype and genotype aaBb, aaBB, will produce another type of phenotype.

• Eg: Feather colour of Fowl
• Complete dominance at both gene pair, but are gene when dominant epistatic to the other and the second gene when homozygous recessive epistatic to the first.
• Gene ‘A’ colour inhibition is dominant to colour appearance.
• Gene ‘B’ colour in dominant to white.

Interaction:

• Dominant colour inhibitors prevents colour even when colour is present, colour gene, when homozygous recessive prevents colour when dominant inhibitor is present.

Dominant and recessive interaction (13:3):

13/16 = white
3/16 = coloured

Question 17.
Male sterility found in pearl maize (sorgum Vulgare) is the best example for mitochondria cytoplasmic inheritance.
Answer: Male sterility found in pearl maize (sorgum Vulgare) is the best example for mitochondria 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.

In this plant there are two types, one with normal cytoplasm (N) which is male fertile and the other one with aberrant cytoplasm (s) which is male sterile.
These types also exhibit reciprocal differences as found in Mirabilis jalapa

Recently it has been discovered that cytoplasmic genetic male sterility is common in many plant species.
This sterility maintained by the influence of both nuclear and cytoplasmic genes.
There are commonly two types of cytoplasm N (Normal) and S (Sterile)
The genes for these are found in mitochondrian.
There are also restores of fertility (Rf) genes. Even though these genes are nuclear genes, they are distinct from genetic male sterility genes of other plants.
Because the Rf genes do not have any expression of their own, unless the sterile cytoplasm is present.
Rf genes are required to restore fertility in S cytoplasm which is responsible for sterility.
So the combination of N cytoplasm with rfrf and s cytoplasm with RfRf products plants with fertile pollens, while S cytoplasm with rfrf produces only male sterile plants.

Tamilnadu State Board New Syllabus Samacheer Kalvi 9th Science Guide Pdf Chapter 18 Organization of Tissues Text Book Back Questions and Answers, Notes.

Samacheer Kalvi 9th Science Solutions Chapter 18 Organization of Tissues

9th Science Guide Organization of Tissues Text Book Back Questions and Answers

I. Choose the correct answer:

Question 1.
The tissue composed of living thin walled polyhedral cell is
(a) parenchyma
(b) pollenchyma
(c) pclerenchyma
(d) none of above
Answer:
(d) none of above

Question 2.
The fibres consists of
(a) parenchyma
(b) sclerenchyma
(c) collenchyma
(d) none of above
Answer:
(b) sclerenchyma

Question 3.
Companion cells are closely associated with
(a) sieve elements
(b) vessel elements
(c) trichomes
(d) guard cells
Answer:
(a) sieve elements

Question 4.
Which of the following is a complex tissue?
(a) Parenchyma
(b) Collenchyma
(c) Xylem
(d) Sclerenchyma
Answer:
(c) Xylem

Question 5.
Aerenchyma is found in
(a) epiphytes
(b) hydrophytes
(c) halophytes
(d) xerophytes
Answer:
(b) hydrophytes

Question 6.
Smooth muscles occur in
(a) uterus
(b) artery
(c) vein
(d) all of the above
Answer:
(d) all of the above

Question 7.
Nerve cell does not contains
(a) axon
(b) nerve endings
(e) tendons
(d) dendrités
Answer:
(c) tendons

II. Match the following.

Question 1.

Answer:

III Fill in the blanks :

1. ……………. tissues provide mechanical support to organs.
Answer:
Compound epithelium

2. Parenchyma, Collenchyma, Sclerenchyma are ……………. type of tissue.
Answer:
simple

3. ……………. and ……………… are complex issues.
Answer:
Xylem, phloem

4. Epithelial cells with cilia are fóund in ……………….. of our body
Answer:
trachea or windpipe

5. Lining of the small intestine is made up of …………………..
Answer:
columnar epithelium

IV. State whether true or false. If false, correct the statement :

1. Epithelial tissue is protective tissue in an animal body.
Answer:
True.

2. Bone and cartilage are two types of areolar connective tissue.
Answer:
False.
Correct statement: Bone and cartilage are two types of supportive connective tissue.

3. Parenchyma is a simple tissue.
Answer:
True.

4. Phloem is made up of tracheids.
Answer:
False.
Correct statement: Phloem is made up of sieve tubes.

5. Vessels are found in collenchyma.
Answer:
False.
Correct statement: Vessels are found in the xylem.

V. Answer briefly :

Question 1.
What are intercalary meristems? How do they differ from other meristems?
Answer:
Intercalary meristem lies between the region of permanent tissues and is part of primary meristem which is detached due to formation of intermittent permanent tissues. It is found either at the base of leaf e.g. Pinus or at the base of intemodes e.g. grasses.

Question 2.
What is complex tissue? Name the various kinds of complex tissues.
Answer:

• Complex tissues are made of more than one type of cells that work together as a unit.
• Complex tissues consist of parenchyma and sclerenchyma cells. However, collenchymatous cells are not present in such tissues.
• Common examples are xylem and phloem.

Question 3,.
Mention the most abundant muscular tissue found in our body. State its function.
Answer:
Connective tissue is the most abundant and widely distributed tissue. It provides a structural framework and gives support to different tissues forming organs.

Question 4.
What is skeletal connective tissue? How is it helpful in the functioning of our body?
Answer:
The supporting or skeletal connective tissues forms the endoskeleton of the vertebrate body. They protect various organs and help in locomotion. The supportive tissues include cartilage and bone.

Question 5.
Why should gametes be produced by meiosis during sexual reproduction?
Answer:
Meiosis is important as it produces gametes i.e., male or female germ cells. During meiosis, a germ cell or gamete divides to make four new sex cells. As a result of fertilization two gamates join together to form an egg or zygote. Therefore only if gametes are produced, fertilization can take place.

Question 6.
In which stage of mitosis the chromosomes align in an equatorial plate? How?
Answer:
Metaphase (meta – after) The duplicated chromosomes arrange on the equatorial plane and form the metaphase plate. Each chromosome gets attached to a spindle fibre by its centromere. The centromere of each chromosome divides into two each being associated with a chromatid.

VI. Answer in detail:

Question 1.
What are the permanent tissues? Describe the different types of simple permanent tissues.
Answer:
Permanent tissues:
Permanent tissues are those in which, growth has stopped either completely or for the time being. At times, they become meristematic partially or wholly.

Different types of simple permanent tissue :
Simple tissue: Simple tissue is homogeneous-composed of structurally and functionally similar cells. Eg : Parenchyma, collenchyma, and sclerenchyma.

Parenchyma:

• Parenchyma are simple permanent tissue composed of living cells.
• Parenchyma cells are thin-walled, oval, rounded, or polygonal in shape with well-developed spaces among them.
• In aquatic plants, parenchyma possesses intercellular air spaces and is named as aerenchyma.
• When exposed to light, parenchyma cells may develop chloroplasts and are known as chlorenchyma.

Functions:

• Parenchyma may store water in many succulent and xerophytic plants.
• It also serves the functions of storage of food reserves, absorption, buoyancy, secretion, etc

Collenchyma:

• Collenchyma is a living tissue found beneath the epidermis.
• Cells are elongated with unevenly thickened non-lignified walls. Cells have rectangular oblique or tapering ends and persistent protoplast.
• They possess thick primary non-lignified walls.
  Functions: They provide mechanical support for growing organs.

Sclerenchymà:

• Scierenchyma consists of thick-walled cells which are often lignified.
• Scierenchyma cells do not possess living protoplasts at maturity. Scierenchyma cells are grouped into

(i) fibers and (ii) sclereids.

Fibres : Elongated scierenchymatous cells, usually with pointed ends. Their walls are lignified. Fibres are abundantly found in many plants. Eg. Jute.
Sclereids:

• Sclereids are widely distributed in plant body. They are usually broad, may occur in single or in groups.
• Sclereids are isodiametric, with lignified walls. Pits are prominent and seen along the walls.
• Lumen is filled with wall materials. Sclereids are also common in fruits and seeds.

Question 2.
Write about the elements of Xylem.
Answer:
Xylem is a conducting tissue which conducts water, mineral nutrients upward from root to leaves. Xylem is also meant for mechanical support to the plant body. Xylem is composed of different kinds of elements. They are

1. xylem tracheids
2. xylem fibres
3. xylem vessels and
4. xylem parenchyma.

(i) Xylem tracheids: They are elongated or tube-like dead cells with hard, thick and lignified walls. Their ends are tapering, blunt or chisel-like. These cells are devoid of protoplast. They have large lumen without any content. Their function is conduction of water and providing mechanical support to the plant.

(ii) Xylem fibers: These cells are elongated, lignified and pointed at both the ends. Xylem fibres help in the conduction of water and nutrients from root to the leaf and also provide mechanical support to the plant.

(iii) Xylem vessels: They are long cylindrical, tube-like structures with lignified walls and wide central lumen. These cells are dead as these do not have protoplast. They are arranged in longitudinal series in which the partitioned walls (transverse walls) are perforated, and so the entire structure looks-like a water pipe. Their main function is the transport of water and minerals from root to leaf, and also to provide mechanical strength.

(iv) Xylem parenchyma: Its cells are living and thin-walled. The main function of xylem parenchyma is to store starch and fatty substances.

Question 3.
List out the differences< between mitosis and meipsis.
Answer:

VII. Higher Order Thinking Skills :

Question 1.
What is the consequence that occurs if all blood platelets are removed from the blood?
Answer:
Blood platelets play a major role in the clotting of blood whenever there is a wound/injury. If blood platelets are removed from the blood, clotting of blood will not occur. In case of any injury/surgery etc., blood will be lost from the body in excess and may even prove to be fatal.

2. Which are not true cells in the blood? Why?
Answer:
Red blood cells or erythrocytes cannot be considered as true cells since they have a nucleus only in the early stages. A mature RBC lacks a nucleus which is the controlling centre of all living cells.

Intext Activities

ACTIVITY – 1
Rinse your mouth with water. Using a toothpick or ice-cream stick, scrap superficial cells from the inner side of the cheek and spread it on a clean glass slide. Dry the glass slide with the scrap cells taken from the inner side of the cheek. Add two drops of methylene blue stain. Identify the cells under low and high power of the microscope.
Solution:
1. Large irregularly shaped cells with cell walls.
2. Dark blue nucleus at the central part of each cell.
3. Lightly stained cytoplasm colour in each cell.

9th Science Guide Organization of Tissues Additional Important Questions and Answers

I. Choose the correct answer :

Question 1.
A meristematic tissue consists of
(a) immature cells that are capable of undergoing cell division.
(b) mature cells
(c) non-living cells
(d) sclerenchyma cells
Answer:
(a) immature cells that are capable of undergoing cell division

Question 2.
Two long bones of the hand are dislocated in a person who met with an accident. Which among the following may be the possible reason?
(a) Tendon injury
(b) Break of skeletal muscle
‘(c) Ligament tear
(d) Rupture of Areolar tissue
Answer:
(c) Ligament tear

Question 3.
Non-striated muscles are found in
(a) blood vessels
(b) gastric glands
(c) urinary bladder
(d) all of these
Answer:
(d) all of these

Question 4.
Which of the following is not found in a neuron?
(a) Sarcolemma
(b) Dendrite
(c) Myelin sheath
(d) Axon
Answer:
(a) Sarcolemma

Question 5.
Cylindrical, unbranched multinucleated cells are
(a) striated muscle cells
(b) smooth muscles
(c) cardiac muscles
(d) none of the above.
Answer:
(a) striated muscle cells

Question 6.
The matrix of the bone is rich in
(a) elastin
(b) reticular fibres
(c) collagen
(d) myosin
Answer:
(c) collagen

Question 7.
Which muscles act involuntarily?
(i) Striated muscles
(ii) Smooth muscles
(iii) Cardiac muscles
(iv) Skeletal muscles
(a) (i) and (ii)
(b) (ii) and (iii)
(c) (iii) and (iv)
(d)(i) and (iv)
Answer:
(b) (ii) and (iii)

Question 8.
Tendon connects
(a) cartilage with muscles
(b) bone with skeletal muscles
(c) ligament with muscles
(d) bone with bone
Answer:
(b) bone with skeletal muscles

Question 9.
In a certain type of cell division, the diploid number of chromosomes is reduced to half. This kind of division occurs in
(a) testis
(b) ovary
(c) both ovary and testis
(d) all body cells
Answer:
(c) both ovary and testis

Question 10.
……………… is derived from the ground meristem.
(a) Cortex
(b) Epidermis
(c) Xylem
(d) Cambium
Answer:
(a) Cortex

Question 11.
The function of phloem fibres is …………….to the plant body
(a) passage of food
(b) Store food
(c) mechanical strength
(d) preparation of food
Answer:
(c) mechanical strength

Question 12.
The …………….. epithelium is also known as pavement membrane.
(a) Ciliated
(b) Squamous
(c) Cuboidal
(d) Glandular
Answer:
(b) Squamous

Question 13.
Elastic structures that connect bones to bones are called ………………..
(a) muscles
(b) tendons
(c) ligaments
(d) areolar tissue
Answer:
(c) ligaments

Question 14.
………………… is seen in unicellular animals.
(a) Mitosis
(b) meiosis
(e) Amitosis
(d) none of the above
Answer:
(c) Amitosis

Question 15.
The disappearance of spindle fibres is seen in …………….
(a) metaphase
(b) prophase
(c) anaphase
(d) telophase
Answer:
(d) telophase

Question 16.
The ………………… is a single, long fiber like process that develops from the cyton.
(a) dendron
(b) axon
(c) dendrite
(d) neurilemma
Answer:
(b) axon

Question 17.
Bouquet stage refers to ……………..
(a) diakinesis
(b) leptotene
(c) zygotene
(d) pachytene
Answer:
(b) leptotene

Fill in the blanks :

1. The ……………. tissues are made up of more than one type of cells and these woks together as a unit.
Answer:
complex

2. The two types of skeletal connective tissues are ……………. and …………….
Answer:
bone, cartilage

3. Humans have 46 chromosomes. Their sperms and eggs will have ……………. chromosomes each.
Answer:
23

4. During pairing of chromosomes in meiosis, the …………….. chromosomes come to lie side by side.
Answer:
homologous

5. The word meristem is derived from a Greek word ………………
Answer:
Meristos

6. Cork cambium is an example of ……………… meristem.
Answer:
secondary

7. The meristem found at the base of intemodes is called ……………….
Answer:
intercalary meristem

8. In apple, paranchyma stores ………………
Answer:
sugar

9. Fibres are extensively longer ranging from 20 mm to 550 mm ……………….
Answer:
corchorus capsularis (jute)

10. During mieosis in pachytene, stages the paired chromosomes are called ………………….
Answer:
bivalents

11. Mitosis was discovered by …………………..
Answer:
Flemming

12. Both smooth and cardiac muscles are …………….. in nature.
Answer:
involuntary

13. ………………. is a non-flexible skeletal connective tissue.
Answer:
Bone

14. ……………. acts as a fat reservoir.
Answer:
Adipose tissue

15. …………….. epithelium is seen in sweat glands.
Answer:
Cuboidal

16. Genetic variations occur in meiosis because of …………….
Answer:
crossing over

III. State whether True or false. If false, write the correct statement:

1. Epithelial layer does not allow regulation of materials between the body and the external environment.
Answer:
False.
Correct statement: Epithelium is involved in the absorption and elimination of waste.

2. Striated and non-striated tissues are types of epithelial tissues.
Answer:
False.
Correct statement: They are types of muscular tissues.

3. Spindle formation occur in amitosis.
Answer:
False.
Correct statement: Spindle formation occur in mitosis.

4. Movement of food in the alimentary canal is because of cardiac muscles.
Answer:
False.
Correct statement: Movement of food is alimentary canal by the rhythmic con t no-tom and relaxation of the muscular nails of the alimentary canal.
Correct statement: Movement of food in the alimentary canal.

5. A mature RBC lacks a nucleus.
Answer:
True

6. Excessive pulling of bones causes a sprain.
Answer:
False.
Correct statement: Sprain is caused by excessive pulling of ligaments.

7. Glandular epithelium gives a stratified appearance.
Answer:
False.
Correct statement: Compound epithelium given a stratified appearance.

8. Sieve cells have no companion cells.
Answer:
True.

9. Conduction can be bidirectional in phloem tissue.
Answer:
True.

10. White blood corpuscles contain respiratory pigment hemoglobin.
Answer:
False.
Correct statement: Red blood corpuscles contain respiratory pigment hemoglobin.

IV. Assertion and Reason type:

Direction: In each of the following questions, a statement of Assertion is given and a corresponding statement of Reason is given just below it. Of statements, given below, mark the correct answer as
(a) Both Assertion and Reason are true and Reason is the correct explanation of Assertion.
(b) Both Assertion and Reason are true that Reason is not the correct explanation of Assertion.
(c) Assertion is true but Reason is false.
(d) Both Assertion and Reason are false.

Question 1.
Assertion: Non-striated muscles are said to be voluntary in nature.
Reason: Non-striated muscles are under the control of our will.
Answer:
(d) Both Assertion and Reason are false

Question 2.
Assertion: Materials are exchanged between epithelial and connective tissues by diffusion.
Reason: Blood vessels are absent in epithelial tissue.
Answer:
(a) Assertion and Reason are true and Reason is the correct explanation of Assertion

V. Answer in one or two sentences :

Question 1.
Name the two types of sclerenchyma cells.
Answer:
Sclerenchyma cells are grouped into

1. fibres and
2. sclereids.

Question 2.
Name the components of xylem and phloem.
Answer:
Xylem is composed of :

• Xylem tracheids
• Xylem fibres
• Xylem vessels
• Xylem parenchyma

Phloem is composed of:

• Sieve elements
• Companion cells
• Phloem fibres
• Phloem parenchyma

Question 3.
Name the tissue that connects muscle to bone in humans.
Answer:
Tendons join skeletal muscles to bones in our body.

Question 4.
Name the tissue that stores excess fat in our body.
Answer:
Adipose tissue.

Question 5.
Name the connective tissue with a fluid matrix.
Answer:
Blood and lymph

Question 6.
Name the tissue present in the brain.
Answer:
Nervous tissue.

Question 7.
What is plate meristem?
Answer:
These cells divide into two planes resulting in an increase in the area of an organ.
Eg: Leaf formation.

Question 8.
Differentiate collenchyma and sclerenchyma.
Answer:

Question 9.
Mention the type of epithelium seen in alveoli of lungs.,
Answer:
Squamous epithelium.

Question 10.
Name the supportive connective tissues.
Answer:
Cartilage and bone

Question 11.
Name the cartilage cells present in the matrix.
Answer:
Chondrocytes.

Question 12.
What is the role of RBC?
Answer:
RBC contains a respiratory pigment called hemoglobin which is involved in the transport of oxygen to tissues.

Question 13.
Mention the stages of meiotic Prophase -I.
Answer:
Leptotene, Zygotene, Pachytene, Diplotene, Diakinesis.

Question 14.
What is the significance of Meiosis?
Answer:
The constant number of chromosomes in a given species is maintained by meiotic division.

Question 15.
Draw a shoot apex and label the meristem’s parts.
Answer:

VI. Short Answer Type :

Question 1.
How would you differentiate meristematic and permanent tissue?
Answer:

Question 2.
Differentiate fibres from sclereids.
Answer:

Question 3.
Which tissue is the main component of tendons and ligaments? How do they differ in function?
Answer:
Dense Connective Tissue is a fibrous connective tissue densely packed with fibres and fibroblasts. It is the principal component of tendons and ligaments.

a. Tendons: Cord-like, strong, structures that join skeletal muscles to bones. Tendons have great strength and limited flexibility. They consist of parallel bundles of collagen fibres, between which are present rows of fibroblasts.

b. Ligaments: They are highly elastic structures and have great strength which connects bones to bones. They contain very little matrix. They strengthen the joints and allow normal movement.

Question 4.
What are the fibres present in the connective tissue proper?
Answer:
Connective tissue proper: Connective tissue proper consists of collagen fibres, elastin fibres and fibroblast cells.

Areolar tissue: It has cells and fibres loosely arranged in a semi-fluid ground substance called matrix. It takes the form of fine threads crossing each other in every direction leaving small spaces called areolae. It joins skin to muscles, fills space inside organs and is found around muscles, blood vessels and nerves. It helps in repair of tissues after injury and fixes skin to underlying muscles.

Adipose Tissue: Adipose tissue is the aggregation of fat cells or adipocytes spherical or oval in shape. It serves as fat reservoir. The matrix consists of collagen fibres, elastin fibres and fibroblast cells.

Question 5.
How are collagen fibres organized in dense connective tissues?
Answer:

• Dense connective tissue is a fibrous connective tissue densely packed with fibres and fibroblasts. It is the principal component of tendon and ligaments.
• Tendons consist of parallel bundles of collagen fibres, between which are present rows of fibroblasts.
• Ligaments are highly elastic structures and contain very little matrix.

Question 6.
Write one point of difference between
a) Bone and cartilage.
b) Simple and compound epithelial tissue.
Answer:
a)

b)

Question 7.
Why is blood considered to be a fluid connective tissue?
Answer:

• The blood and the lymph are the fluid connective tissues which link different parts of the body. The cells of the connective tissue are loosely spaced and are embedded in an intercellular matrix.
• Blood contains corpuscles which are red blood cells (erythrocytes), white blood cells (leucocytes), and platelets. In this fluid connective tissue, blood cells move in a fluid matrix called plasma. The plasma contains inorganic salts and organic substances. It is a main circulating fluid that helps in the transport of nutrient substances.

Question 8.
Give the sequence of the events occurring during prophase of mitosis.
Answer:

• During this stage, chromosomes become short and thick and are clearly visible inside the nucleus.
• Centrosome splits into two daughter centrioles and occupies opposite poles of the cell.
• Each centriole is surrounded by aster rays. Spindle fibres appear between the two centrioles.
• The nuclear membrane and nucleolus disappear gradually.

Question 9.
Why is meiosis called reductional division and mitosis as equational division?
Answer:

• In mitosis one parent cell divides into two identical daughter cells, each with a nucleus having the same amount of DNA, same number of chromosomes and genes as the parent cells. It is therefore called an equational division.
• Meiosis is called reduction division because the chromosome number is reduced to haploid (n) from diploid (2n) in the daughter cells.

Question 10.
What is terminalization?
Answer:
In the stage of diplotene of meiotic prophase I, chiasmata begin to move along the length of the chromosome from the centromere towards the end resulting in terminalization.

Question 11.
What is a tetrad?
Answer:
The chromosomes are visible as long paired twisted threads. The pairs so formed are called bivalents. Each bivalent now contains four chromatids (tetrad stage) in pachytene of mieotic prophase I. The condition of bivalent containing four chromatids are called tetrad stage.

Question 12.
What is crossing over?
During pachytene of mieotic prophase I, the chromatids break and the broken segments are interchanged between homologous chromosomes. The points of exchange are the chiasmata. This is called .crossing over.

Question 13.
What is bouquet stage?
Answer:
During leptotene of mieotic prophase I, the chromosomes become uncoiled and assume long thread-like structures and take up a specific orientation inside the nucleus. They form a bouquet stage.

Question 14.
What is zygotene?
Answer:
It is one of the stages of mieoticphophase I. Two homologous chromosomes approach each other and begin to pair. The pairing of homologous chromosomes is called synapsis.

Question 15.
Explain amitosis.
Answer:
It is the simplest model of cell division and occurs in unicellular animals, aging cells and in foetal membranes. During amitosis, the nucleus elongates first, and a constriction appears in it which deepens and divides the nucleus into two, followed by this cytoplasm divides resulting in the formation of two daughter cells.

Question 16.
Write the salient features of the compound epithelium.
Answer:

• It consists of more than one layer of cells and gives a stratified appearance. Hence, they are also known as stratified epithelial cells.
• The main function of this epithelium is to give protection to the underlying tissues against mechanical and chemical stress.
• They also cover the dry surface of the skin, the moist surface of the buccal cavity, and the pharynx.

Question 17.
Write a note the significance of mitosis.
Answer:

• This equational division results in the production of diploid daughter cells (2n) with equal distribution of genetic material (DNA).
• In multicellular organisms growth, organ development and an increase in body size are accomplished through the process of mitosis.
• Mitosis helps in the repair of damaged and wounded tissues by the renewal of the lost cells.

Question 18.
Draw a neuron and label the parts.
Answer:

VII. Answer in detail:

Question 1.
What are meristems? Describe the distribution and functions of various types of meristems.
Answer:
Meristematic tissues are group of immature cells that are capable of undergoing cell division. In plants, the meristem is found in zones where growth can take place. Example: apex of stem, root, leaf primordia, vascular cambium, cork cambium, etc.,

Types of Meristems based on position:
On the basis of their position in the plant, meristems are of three types: Apical meristem, Intercalary meristem, and Lateral meristem.

Apical meristem: These are found at the apices or growing points of root and shoot and bring about an increase in length.

Intercalary meristem: It lies between the region of permanent tissues and is part of the primary meristem. It is found either at the base of leaf e.g. pinus or at the base of internodes e.g. grasses.

Lateral Meristem: These are arranged parallel and causes the thickness of the plant part.

Functions: Meristems are actively dividing tissues of the plant, that are responsible for primary (elongation) and secondary (thickness) growth of the plant.

Question 2.
Give one reason for the following:
a. Blood is fluid connective tissue.
b. Skeletal muscles are voluntary muscles.
c. Heart muscles are involuntary in nature.
Answer:
The blood and the lymph are the fluid connective tissues which link different parts of the body. The cells of the connective tissue are loosely spaced and are embedded in an intercellular matrix.

(a) Blood: Blood contains corpuscles which are red blood cells (erythrocytes), white blood cells (leucocytes), and platelets. In this fluid connective tissue, blood cells move in a fluid matrix called plasma. The plasma contains inorganic salts and
organic substances. It is a main circulating fluid that helps in the transport of nutrient substances.

(b) They work under our control and are also known as voluntary muscles. They are not under the control of our will and so are called involuntary muscles.

(c) Cardiac muscle: It is a special contractile tissue present in the heart. The muscle fibers are cylindrical, branched, and uninucleate The branches join to form a network called as itercalated disc which are unique distinguishing features of the
cardiac muscles. The contraction of cardiac muscle is involuntary and rhythmic.

Question 3.
Explain simple epithelium and its types.
Answer:
Simple Epithelium :
1. It is formed of a single layer of cells. It forms a lining for the body cavities and ducts.
2. Simple epithelium is further divided into the following types.

• Squamous epithelium
• Cuboidal epithelium
• Columnar epithelium
• Ciliated epithelium
• Glandular epithelium

(i) Squamous epithelium :

• It is made up of thin, flat cells with prominent nuclei. These cells have irregular boundaries and bind with neighbouring cells.
• The squamous epithelium is also known as pavement membrane, which forms the delicate lining of the buccal cavity, alveoli of lungs, proximal tubule of kidneys, blood vessels etc.
• It protects the body from mechanical injury, drying and invasion of germs.
  

(ii) Cuboidal epithelium:

• It is composed of single layer of cubical cells.
• The nucleus is round and lies in the centre.
• This tissue is present in the thyroid vesicles, salivary glands, sweat glands, exocrine pancreas.
• It is also found in the intestine and tubular part of the nephron (kidney tubules) as microvilli that increase the absorptive surface area.
• Their main function is secretion and absorption.
  

(iii) Columnar epithelium:

• It is composed of a single layer of slender, elongated and pillar like cells.
• Their nuclei are located at the base.
• It is found lining the stomach, gall bladder, bile duct, small intestine, colon, oviducts and forms a mucous membrane. ‘
• They are mainly involved in secretion and absorption.
  

(iv) Ciliated epithelium :

• Certain columnar cells bear numerous delicate hair like out growths called ilia and are called ciliated epithelium.
• Their function is to move particles or mucus in a specific direction over the epithelium.
• It is seen in the trachea of wind-pipe, bronchioles of respiratory tract, kidney tubules and fallopian tubes of oviducts.
  

(v) Glandular epithelium :

• Epithelial cells are often modified to form specialized gland cells which secrete chemical substances at the epithelial surface.
• This lines the gastric glands, pancreatic tubules and intestinal glands.
  

Question 4.
Explain the components of phloem tissue.
Answer:
Phloem is a complex tissue and consists of the following elements :
(i) Sieve elements
(ii) Companion cells
(iii) Phloem fibres
(iv) Phloem parenchyma

(i) Sieve elements :

• The conducting elements of phloem are collectively called as Sieve elements.
• Sieve tubes are elongated, tube-like slender cells placed end to end. The transverse walls at the ends are perforated and are known as sieve plates.
• The main function of sieve tubes is translocation of food, from leaves to the storage organs of the plants.

(ii) Companion cells : These are elongated cells attached to the lateral wall of the sieve tubes. A companion cell may be equal in length to the accompanying sieve tube element or the mother cell may be divided transversely forming a series of companion cells.

(iii) Phloem parenchyma : The phloem parenchyma are living cells which have cytoplasm and nucleus. Their function is to store food materials.

(iv) Phloem fibers : Sclerenchymatous cells associated with primary and secondary phloem are commonly called phloem fibers. These cells are elongated, lignified and provide mechanical strength to the plant body.

Question 5.
Write a note on blood and its components.
Answer:
Blood is a fluid connective tissue.
Blood contains corpuscles which are red blood cells (erythrocytes), white blood cells (leucocytes) and platelets. In this fluid connective tissue, the blood cells move in a fluid matrix called. The plasma contains inorganic salts and organic substances. It is a main circulating fluid that helps in the transport of nutrient substances.

Red blood corpuscles (Erythrocytes):

• The red blood corpuscles are oval shaped, circular, biconcave disc-like and lack nucleus when mature (mammalian RBC).
• They contain a respiratory pigment called haemoglobin which is involved in the transport of oxygen to tissues.

White blood corpuscles (Leucocytes): They are larger in size, contain distinct nucleus and are colourless. They are capable of amoeboid movement and play an important ‘ role in body’s defense mechanism. WBC’s are of two types :
(i) Granulocytes.
(ii) Agranulocytes.

(i) Granulocytes have irregular shaped nuclei and cytoplasmic granules. They include the neutrophils, basophils and eosinophils. Agranulocytes lack cytoplasmic granules and include the lymphocytes and monocytes.

Blood platelets : They are minute, anucleated, fragile fragments of giant bone marrow called mega karyocytes They play an important role in blood clotting mechanism.

VIII. Higher Order Thinking Skills :

Question 1.
Identify the figure given below

(a) Label the parts A, B and C.
(b) What is the chemical composition of the tissue?
Answer:
(a) T.S. of Bone
(A) Lamellae
(B) Lacunae
(C) Central (Haversian canal)
(b) The matrix of the bone is rich in calcium salts and collagen fibres which gives the bone its strength.
(c) C – Haversian canal

Question 2.
Identify figures A and B.

(a) …………… epithelium forms the outer lining of the buccal cavity.
(b) ………………. epithelium consist of ceils that are tall and pillar-like.
(c) Which one allows diffusion of substances?
(d) Which is called pavement epithelium?
(e) Which epithelium lines the gastrointestinal tract and epiglottis?
Answer:
Figure A – Squamous Epithelium
Figure B. – Glandular epithelium
(a) Squamous
(b) Columnar
(c) Columnar epithelium
(d) Squamous epithelium
(e) Columnar epithelium

Question 3.
If cell (A) has undergone one mitotic division and another cell (B) has completed its meiotic division. The number of cells produced in A and B would be
Cell A: Cell B :
Answer:
Cell A : 2 daughter cells.
Cell B : 4 daughter cells.

Question 4.
Identify the stage of mitosis from the below picture. List the chromosomal events in this stage.

Answer:
Mitotic anaphase
(i) The centromeres attaching the two chromatids divide and the two daughter chromatids of each chromosome separate and migrate towards the two opposite poles.
(ii) The migration of the daughter chromosomes is achieved by the contraction of spindle fibres.

Question 5.
Identify the following relationship
Cuboidal : Epithelial
Cardiac : ………..
Granulocytes : …………
Osteocytes : ………….
Answer:
Cardiac : Muscular
Granulocytes : Blood cells
Osteocytes : Bone cells

Question 6.
Umbilical cord blood is collected at the time of child birth and stored in stem cell banks? Reason out.
Answer:

• Umbilical cord blood consists of stem cells, they are undifferentiated cells which undergo unlimited divisions and give rise to one or more different types of cells. – Embryonic stem cells differentiate into different tissues and organs.
• Stem cells can be used in the treatment of certain degenerative diseases in future.

Question 7.
How do WBC help in defence?
Answer:
They are capable of amoeboid movement and play an important role. They engulf or destroy foreign bodies.

Tamilnadu State Board New Syllabus Samacheer Kalvi 11th English Guide Pdf Prose Chapter 4 Tight Corners Text Book Back Questions and Answers, Summary, Notes.

Tamilnadu Samacheer Kalvi 11th English Solutions Prose Chapter 4 Tight Corners

11th English Guide Tight Corners Text Book Back Questions and Answers

I. Choose the most appropriate answer for the following questions:

Question a.
‘Tight Corner’ means a _______.
i. difficult situation
ii. crowded corner
iii. tragic incident
iv. fierce fight
Answer:
i. difficult situation

Question b.
Barbizon refers to a _______.
i. kind of paint
ii. type of architecture
iii. region in Britain
iv. French school of painters
Answer:
i. kind of paint

Question c.
The narrator visited the sale-room as he _______.
i. wished to see an auction
ii. had a painting to sell
iii. was persuaded by his friend
iv. wanted to buy a painting
Answer:
iii. was persuaded by his friend

Question d.
The narrator had been a safe contributor at the auction, as _______.
i. there were bidders quoting higher prices
ii. he had a sound financial background
iii. his friend had lent him money
iv. he did not make any bidding
Answer:
i. there were bidders quoting higher prices

Question e.
“And I got it.” Here ‘it’ refers to the _______.
i. picture he wanted to buy
ii. money he asked for
iii. card to participate in the auction
iv. amount he had to pay
Answer:
ii. money he asked for

2. Answer the following questions:

Question a.
What is a tight corner? What happens when one finds oneself in a tight corner?
Answer:
Tight corner refers to the difficult or critical situation that one faces in his life. The person who finds himself in a tight corner becomes stressful both physically and mentally.

Question b.
What is the difference between a physical and mental tight comer?
Answer:
A physical tight corner is something which is visualized in person on spot. One can overcome this if he has extreme courageousness. Mental tight corner affects the whole system of a man as his mind is filled with stress till he comes out of it. In fact, it is more dangerous than physically tight corner.

Question c.
Why did the narrator visit Christie’s?
Answer:
The narrator visited Christie’s as his friend persuaded him to see the auction inside.

Question d.
The narrator heard his own voice saying,” and fifty”.What does this suggest?
Answer:
The narrator without his knowledge and any understanding of the situation said, ‘and forty’.

Question e.
What was the narrator’s financial condition?
Answer:
The narrator had exactly sixty-three pounds in the bank and he did not have securities even for five hundred pounds.

Question f.
The narrator could not pretend to have made a mistake in bidding. Why?
Answer:
The narrator could not pretend to have made a mistake in bidding because already he made many biddings earlier which made others think of him as a bloatocrat. Moreover, a genuine mistake of such a kind would have been rectified at once.

Question g.
What could have been the best way for the narrator, to get himself out of the tight comer?
Answer:
The best way for the narrator to get himself out of the tight corner was to confess his poverty to one of Christie’s staff and having the picture put up again.

Question h.
Why did the narrator feel he could have welcomed a firing party?
Answer:
It was his thought of bidding for fun which made him get caught in a tight corner. If he welcomed a firing party that would bring his death and he need not be humiliated in front of others.

Question i.
What was the bidder’s offer to the narrator?
Answer:
The bidder’s offer was to give fifty guineas to the narrator.

Question j.
How did the narrator take advantage of the situation?
Answer:
The narrator took advantage of the situation by asking a hundred guineas from the bidder who offered four thousand guineas for big Daubigny.

Text Inside Questions:

Question a.
Describe the activity that was going on in the saleroom at king street.
Answer:
The place was full. They were selling Barbizon pictures and getting tremendous sums even for little bits of things.

Question b.
What can you say about the author’s attitude when he high – handedly participated in the auction?
Answer:
An author is a nonchalant person who tries to have some fun in his life. At the same time, he knows his limitations.

Question c.
Why was the author sure he would not be caught?
Answer:
The author decided to bid safely by just raising the stake a little bit and leave it for real millionaires to go ahead. Thus he was sure that he would not be caught.

Question d.
What made the author ignore his friend’s warning?
Answer:
The author ignored his friend’s warning just because he liked to have some fun and was sure that he was not going to run any risks.

Question e.
How had the author managed the auction without getting involved in the deal?
Answer:
Although many bids ended up in four figures, they were started with a modest price of fifty to a hundred guineas only. He ventured till the figures reached only upto three digits. Thus he managed the auction without getting involved in the deal.

Question f.
What came as a shock to the author?
Answer:
There was bidding for four thousand guineas and as usual, he added fifty guineas to it. But to his surprise, none of them bid more than that.

Question g.
What did the falling of the hammer indicate?
Answer:
The falling of the hammer indicated “closure of the bid” and it mandated the highest bidder to pay and collect his purchase.

Question h.
What made the friend laugh heartily?
Answer:
The narrator had to pay four thousand and fifty guineas for his bidding. In reality, he had only sixty-three pounds. This made his friend laugh heartily.

Question i.
What kind of excuses did the narrator think he could make?
Answer:
The author speculated on the possibility of confessing his poverty to one of Christie’s staff and request to put up the picture for sale once again.

Question j.
Why did the friend desert the narrator, a second time?
Answer:
The narrator was standing on the outskirts of the little knot of buyers to pay the amount and get the picture. His friend who joined the narrator had a look at his face and could not control his laughter. Thus he deserted the narrator, a second time.

Question k.
How does the narrator describe the man who approached him?
Answer:
The man was a messenger of the high gods who wore a green baize apron and spoke in husky cockney tones.

Question l.
How does the Narrator show the presence of mind in the sudden turn of events?
Answer:
The man who bid for the picture first was ready to pay fifty guineas to the narrator. At that moment the narrator asked for a hundred guineas which shows his presence of mind.

Question m.
The narrator would not forget two things about his friend What are they?
Answer:
The author’s friend only persuaded him to go to Christie’s auction. Secondly, he was the only witness to the author ’s mental agony in trying to get out of the crisis.

3. Form a meaningful summary of the lesson by rewriting the numbers in the correct sequence:

1. The narrator had only 63 pounds with him and did not know how to manage the situation
2. The narrator thought of all his relations from whom he could borrow
3. Unfortunately, he had made the highest bid.
4. The narrator entered Christie’s as his friend persuade him to visit the sale-room.
5. Every time someone else made a higher bid and the narrator was not caught.
6. The narrator on a sudden impulse added 50 more guineas, to the amount offered.
7. His friend joined him then but left immediately unable to control his laughter.
8. He even thought of borrowing from moneylenders and confessing the truth to the staff at Christie’s.
9. The picture was declared sold to the narrator.
10. After some time a picture was put up and a bid for 4000 guineas was a raise
11. A sudden stroke of luck befell the narrator when he heard that the agent who had made the bid of 4000 guineas and buy the picture.
12. The narrator kept bidding just for fun.
13. The picture was given away to the other bidder and the narrator was saved from humiliation.
14. His friend had left the place roaring with laughter at the narrator’s predicament.
15. The narrator was quite happy at the offer but demanded 100 guineas instead of the 50. Now there was no need for him to make any payment.

Answers:

1. 8
2. 10
3. 6
4. 1
5. 5
6. 4
7. 12
8. 11
9. 7
10. 3
11. 13
12. 2
13. 15
14. 9
15. 14

4. Answer the following questions in a paragraph of about 100-150 words:

Question a.
Narrate the circumstances that led to the narrator getting into a tight corner, by his own folly.
Answer:
Lucas learned that an auction was in progress. His friend suggested that they peeped in, to watch the fun. Despite the caution from his friend, he started bidding at moderate rates. He had only 63 pounds in his account. A bidder was supposed to have a minimum of 500 pounds to take part in the bid. As bidding for most of the paintings were started with two or three digits in guineas, the author sailed through raising the stakes of many paintings and staying behind watching millionaires bid with higher prices. But one painting viz big Daubigny was launched at an offer price of 4000 Guineas.

Only one bidder showed interest. The rest were in silence. The author heard himself say “and fifty”. After seconds of uncomfortable silence, the dealer banged the hammer indicating the acceptance of the narrator’s offer of 4050 guineas for the painting. It was only then the narrator realized that he was in a tight comer. He wished a firing squad would be welcomed to eliminate him and put an end to his mental agony. He had no friend or relative or even money lenders who could extend him a loan to raise the money. He had got into a mess of his own choice.

“Auction houses run a rigged game. They know exactly how many people will be bidding on work and exactly who they are. In a gallery, works of art just need to pay. ”

Question b.
Trace the thoughts that went on in the mind of the narrator, when picture after picture was put up and sold at the auction.
Answer:
The narrator started bidding for fun and got into a difficult position of paying four thousand and fifty guineas for a picture which was useless for him. He had only sixty-three pounds with him and didn’t know how to pay for it.

He handed over his card to the clerk and without seeing the picture put for sale he, was thinking about the names of uncles and other persons from whom he might borrow money. He wondered of the money lenders who would as promissory notes.

He also thought of confessing his poverty to one of Christie’s staff and make them put up the picture again. All his thoughts ended in vain as the staff of Christie’s seemed unsympathetic and he was sure that they wouldn’t believe it to be a mistake.

Question c.
Explain how the narrator got out of the tight corner that he was in.
Answer:
When the author was perplexed beyond measure and was even ready to welcome a firing squad to bail him out of the current crisis, a divine chance presented itself to the narrator. The narrator had stupidly given an open bid to buy “big Daubigny” for 4050 guineas when he had only 63 pounds in his bank account.

However hard he tried, he could not recall the name of an “uncle” or a friend who could extend him a loan to cover the price of the painting. To delay disgrace, he was standing at the end of the queue of the successful bidders. Like a providential intervention, a mediator from the starting bidder who was ready to take the same painting for 4000 guineas enquired the narrator in a husky cockney tone if he was the gentleman who had bought, “big Daubigny”.

The narrator admitted it. To the narrator’s great relief, the mediator said the first bidder wanted to know if he would take 50 guineas for his interest. The author should have embraced him and wept for joy for bailing him out of a potential disgrace. But he made the best use of the opportunity exhibiting his guile, by asking him if that was the most he could offer. The mediator said that there was no harm in asking for some more. The narrator said he would take a hundred guineas. When the man left to find out the possibility both the author and his friend laughed.

But when the author saw the cheque for a hundred guineas, he became serious. He said with joy and shock, “of all the luck! well, I’m hanged”. Thus the narrator had a narrow escape from a tight comer. One could even say that the narrator escaped by the skin of his teeth.

“Call it a narrow escape, maybe it’s your lucky day. ”

Vocabulary:

Auction House Puzzler: (Text Book Page No. 111)

You have come across many terms associated with an auction, in the lesson. Now solve the crossword puzzle with words from the lesson. Make use of the clues given:

Across:

1. conducts auction
2. a protective garment
3. strip with numbers
4. offer

Down

1. painter
2. school of painting
3. auction house
4. painting

Answer:

Reading:

Read the following passage and answer the questions that follow:

The Stationmaster’s Supreme Sacrifice by Sanchari Pal (Adapted)

1. Thirty-three years ago, on the night of December 2, 1984, Bhopal was hit by a catastrophe that had no parallel in the world’s industrial history. An accident at the Union Carbide pesticide plant in Bhopal had released almost 30 tons of a highly toxic gas called methyl isocyanate, turning the city into a vast gas chamber.

The result was a nightmare; more than 600,000 people were exposed to the deadly gas cloud that left thousands dead and much more breathless, blind, and in agonizing pain. Few people know that during the Bhopal gas tragedy a heroic stationmaster risked his own life to save others.

2. On the evening of December 3, 1984, Ghulam Dastagir was settling down in his office to complete some pending paperwork. This work kept him in his office till lam in the night, when he emerged to check the arrival of the Gorakhpur Mumbai Express.

As he stepped on to the platform, the deputy stationmaster felt his eyes burn and a queer itching sensation in his throat. He did not know that poisonous fumes leaking from Union Carbide’s pesticide factory were stealthily enveloping the railway station.

3. Beginning to choke, Dastagir did not know then that twenty-three of his railway colleagues, including his boss, station superintendent Harish Dhurve, had already died. It was later reported that Dhurve had heard about the deadly gas and had immediately tried stopping the movement of trains passing through Bhopal before collapsing in his office chamber.

His suddenly worsening health and years of experience told Dastagirthat something was very wrong. Though he did not fully comprehend what was happening, he decided to act immediately when he did not get any response from the station master. He alerted the senior staff at nearby stations, like Vidisha and Itarsi, to suspend all train traffic to Bhopal.

4. However, the jam-packed Gorakhpur Kanpur Express was already standing at the platform and its departure time was 20 minutes away. Listening to his gut instinct, Dastagir summoned his staff and told them to immediately clear the train for departure. When they asked if they should wait until the order to do so came from the head office, Dastagir replied that he would take complete responsibility for the train’s early departure.

He wanted to ensure that the train left immediately, without any delay. His colleagues later recalled that Dastagir could barely stand and breathe as he spoke to them. Breaking all rules and without taking permission from anyone, he and his brave staff personally flagged off the train.

5. But Dastagir’s work was not done. The railway station was filling up with people, desperate to flee the fumes. Some were gasping, others were vomiting, and most were weeping. Dastagir chose to remain on duty, running from one platform to another, attending, helping, and consoling victims.

He also sent an SOS to all the nearby railway offices, asking for immediate medical help. As a result, four ambulances with paramedics and railway doctors arrived at the station. It was winter and the gas was staying low to the ground, a thick haze poisoning everything in its path.

Besieged by hordes of suffering people, the station soon resembled the emergency room of a large hospital. Dastagir stayed at the station, steadfastly doing his duty, knowing that his family was out there in the ill-fated city. That day all he had for his protection was a wet handkerchief on his mouth.

6. Ghulam Dastagir’s devotion to duty saved the lives of hundreds of people. However, the catastrophe didn’t leave him unscathed. One of his sons died on the night of the tragedy and another developed a lifelong skin infection.

Dastagir himself spent his last 19 years shuttling in and out of hospitals; he developed a painful growth in the throat due to prolonged exposure to toxic fumes. When he passed away in 2003, his death certificate mentioned that he was suffering from diseases caused as a direct result of exposure to MIC (Methyl Isocyanate) gas.

A memorial has been built at platform No.1 to pay tribute to those who sacrificed their lives in the line of duty on the fateful night of December 3, 1984. However, Ghulam Dastagir, who died later, is not one of them. A forgotten hero whose sense of duty and commitment saved countless lives, Dastagir’s story deserves to be recognized and remembered by our fellow countrymen.

Answer the following questions:

Question (i)
Why was the accident at union carbide unparalleled in the word’s industrial history?
Answer:
The accident was unparalleled in the world s Industrial history because it affected more than 600,000 people.

Question (ii)
How was Dastagir affected by the poisonous gas?
Answer:
He developed a painful growth in the throat due to prolonged exposure to toxic fumes.

Question (iii)
What was the action taken by the station superintendent?
Answer:
As soon as he heard about the deadly gas, he tried stopping the movement of trains through Bhopal.

ஆசிரியரைப் பற்றி:

எ.வே.லூக்காஸ் (1868-1938) ஒரு ஆங்கில நகைச்சுவை எழுத்தாளர், கட்டுரையாளர். நாடக ஆசிரியர், வாழ்க்கை வரலாற்று எழுத்தாளர், புத்தக வெளியீட்டாளர், கவிஞர், நாவலாசிரியர், சிறுகதை எழுத்தாளர், மற்றும் பத்திரிக்கை ஆசிரியர். லண்டன் புறநகர் பகுதியில் பிறந்தவர். தன் 16 வயதில் புத்தக விற்பனையாளரின் உதவியாளராக பணியாற்றியவர்.

பின்னர் பத்திரிக்கை துறையில் ஆர்வம் கொண்டு பிரிட்டனில் உள்ளூர் பத்திரிக்கையிலும், லண்டன் மாத இதழ் பத்திரிக்கையிலும் பணியாற்றினார். பெர்னார்ட் பார்ட்டன் என்ற குவாக்கர் கவிஞரின் வாழ்க்கை வரலாற்றை எழுத அவர் பணிக்கப்பட்டார்.

அதை சிறப்பாக எழுதியதால் சார்லஸ் லேம்பின் புத்தகங்களை மதிப்பிடும் வாய்ப்புக் கிடைத்தது. பின்பு இவர் 1904ல் பஞ்ச் என்ற இதழில் வாழ்நாள் முழுவதும் பணியாற்றினார். தன் சிறு கட்டுரைக்காக பிரபலமானார். பல பாடல்களையும், நாடகங்களையும் எழுதியுள்ளார்.

பாடச் சுருக்கம்:

இப்பாட பகுதியில் கதாசிரியர் தான் ஒரு இக்கட்டான சூழ்நிலையில் மாட்டிக் கொண்டு பின்னர் அதிலிருந்து எவ்வாறு தன் சாமர்த்தியத்தியத்தால் தப்பித்துக் கொள்கிறார் என்பதை தெளிவாக சொல்லியிருக்கிறார். தன் நண்பருடன் ஓவியங்களை ஏலம்விடும் இடத்திற்கு செல்கிறார்.

எல்லோரும் ஒவ்வொன்றாய் ஏலம்விட்டுக்கொண்டும், படங்களை வாங்கிக்கொண்டும் இருக்கையில் ஆசிரியர் விளையாட்டாக ஒரு ஓவியத்தை ஏலம் கேட்கிறார். ஆனால் இவரின் வங்கிக் கணக்கில் 63 பவுண்டுகள் மட்டுமே உள்ளது. ஆனால் விளையாட்டாக ஏலம் கேட்டிருக்கும் தொகையோ 4050 இனியாக்கள்.

விளையாட்டாக ஏலம் கேட்டு மாட்டிக் கொள்கிறார். இந்த சிக்கலாள தருனத்திலிருந்து தன் சமயோதித புத்திக்கூர்மையால் எப்படி இவர் இந்த இக்கெட்டான சூழ்நிலையில் இருந்து தன்னை காத்து கொள்கிறார் என்று இக்கட்டுரையில் விரிவாகக் காண்போம்.

Tight Corners Summary in Tamil

எங்கள் பேச்சி சிக்கலான நிலையில் மாட்டிக்கொள்ளும் நிகழ்வுகளைப்பற்றி ஓடிக்கொண்டிருந்தது, அதில் வாழத் தெரிந்தவர்கள் சாகசம் நிறைந்தவர்களாகவும் மற்றும் சமரசம் செய்ய தெரிந்தவர்களாகவும் இருப்பார்கள்.

ஒரு மனிதன் கடலோரப் பகுதியில் வடகிழக்கு France யில் பேரலையில் மாட்டிக்கொண்தாகவும், பின்பு சாமர்த்தியமாக தன் வலிமையினால் தப்பிவிட்டதாகவும் சொன்னார். மற்றொருவர் காயப்பட்ட புலியால் nதாக்கப்பட்டபோது யானையின் மீது இருந்தாக கூறினார்.

மூன்றாமவர் அவர் எரியும் வீட்டின் மூன்றாவது மாடியில் இருந்தார் எனவும் கூறினார். நான்காமவர் போரில் ஏவுகனையால் தாக்கப்பட்டார் எனவும் கூறினார்.

அவர்களில் ஒருவர் ஆனால் நீங்கள் எல்லோரும் உடல் ரீதியாக மாட்டிக்கொண்டவர்களை பற்றியே பேசுகிறீர்கள்”. கண்டிப்பாக அவர்கள் மனநிலையை விட உடல்நிலை இருக்கம் கொண்டவர்கள். நான் மிக மோசமான சிக்கலில் christies ல் இருந்தபோது சிக்கிக்கொண்டேன்”.

“Christie’s?” (க்ரைஸ்டீஸ்)

“ஆம். லண்டனில் பெரிதாக வணிகம் நடைபெறும் தெருவில் (St. Jame’s street) உள்ள எனது பழைய வெளிநாட்டு நண்பருடன் மதிய உணவு சாப்பிட்டேன், பின்னர் king streetயை கடந்தபோது, அவர் விற்பனை அறையை பார்வையிட என்னை வற்புறுத்தினார். அந்த இடம் மக்களால் நிறைந்து இருந்தது.

அவர்கள் Barbizon படங்களை விற்றனர், ஒவ்வொரு சிறு சிறு பொருட்களையும், படங்களையும் இரண்டாயிரம், மூவாயிரம் என நல்ல விலைக்கு விற்றனர். அவற்றில், காட்டுப்படங்கள், மாலை நேரத்து குளங்கள், மெய்ப்பர் ஆடுமேய்க்கும் சிறுவன் மற்றும் எப்போதும் போல உள்ள சாதாரன தலைப்பிலான படங்கள் இருந்தன.

எந்த ஏலமும் மூன்று இலக்க எண்களுக்கு மேல் ஏலத்திற்கு போகவில்லை. நான் வேடிக்கையாக ஏலத்தை கேட்டேன். என்னிடம் அறுபத்தி மூன்று பவுண்ட் மட்டுமே வங்கியில் இருந்தது. ஐந்நூறு பவுண்ட கடன் பெற நான் ஒரு பெரிய பணக்காரர் (bloatocrat) போல ஏல விற்பணையாளருடன் தலையை ஆட்டினேன்.

நீ கண்டிப்பாக மாட்டிக் கொள்வாய், என் நண்பன் என்னிடம் கூறினான் “இல்லை, நான் மாட்டிக்கொள்ள மாட்டேன்,” என்று கூறினேன். நான் எந்த விதமான இடர்களுக்கும் உள்ளாகமாட்டேன்” என்று நான் சொன்னேன்.

“நீண்ட நேரம் நான் சிக்கல் ஒன்றிலும் மாட்டிக்கொள்ளவில்லை ஒன்றும் செய்யவில்லை. பின்னர் ஒரு ஓவியத்தை ஏலத்திற்கு கொண்டுவந்தது வைத்தார்கள். சிவப்பு முகத்தை கொண்ட தொப்பியை அணிந்திருந்த ஒரு புதிய மனிதர் ஒரு படத்தை ஏலத்திற்காக முன்னே வைத்தார். யாரும் கேட்க இயலாத விலையை ஓவியத்திற்கு கேட்டு அனைவரையும் வியப்பில் ஆழ்த்தினர்.

முந்தைய ஏலம் (lots) நான்கு இலக்கங்களில் விற்கப்பட்டிருந்தாலம் ஐம்பது அல்லது நூறு Guineas என்று தொடங்கி, நான்கிலக்க எண்ணைத் தொட்டது. சிறப்பான முடிவை (crescendo) நோக்கி நான் அடிக்கடி பாதுகாப்பாக பங்களித்தேன். ஆனால் சிறிது நேரத்தில் புதியப்படம் வைக்கப்பட்டவுடன் வியாபாரி பரப்பரப்பாக ஆரம்பத் தொகையாக “நான்கு ஆயிரம் Guineas” என்று கூறினார்.

மிகுந்த சலசலப்பான உற்சாக ஒலி எழுந்தது. முடிவில் என் குரல் கூறியது, ”ஐம்பது!”. ஆழ்ந்த அமைதி நிலவியது. அப்போது ஏல அறிவிப்பாளர் ஏலத்தொகையை கேட்டவரையும் பின்பு எல்லோரையும் பார்த்தார்.

வியப்புடனும், அதிர்ச்சியுடனும் சிவப்பு முகம் வியாபாரி உயிரற்றவர் போல் தோன்றினார். அவர் தன்னுடைய பலத்தை பயன்படத்தியிருக்கிறார் என்று இப்போது நான் உணர்ந்தேன்.

“நான்கு ஆயிரத்து ஐம்பது Guineas வழங்கப்படும்”, திரும்பவும் அறையை நோட்டமிட்டுக்கொண்டு ஏலம் விடுபவர் கூறினார்.

எனது இதயதுடிப்பு நின்றது. இரத்தம் உறைந்தது (congealed). எந்த சத்தமும் இன்றி என் நண்பனின் கட்டுப்படுத்தப்பட்ட (smothered) சத்தம் மட்டுமே கேட்டது.

“நான்கு ஆயிரத்து ஐம்பது Guineas” நான்காயிரத்து ஐம்பதிற்கு ஏதாவது கேள்வி உண்டா ? பிறகு சுத்தியல் அடிக்கப்பட்டு ஏலம் முடிக்கப்பட்டது.

ஏல அறையின் உள்ளே இருக்க எனக்கு நெருக்கடியாக இருந்தது!, அறுபத்து மூன்று பவுன்டிற்கு நானூறு பவுன்ட் விலை மதிப்பு பெறாத எனக்கு பிடிக்காத அந்தப்படத்தை வேண்டாத, இந்நாளில் உயர்ந்த விலையான நான்காயிரத்து ஐம்பது guineas ற்கு வாங்கினேன்.

கனிவான ஆறுதல் பெற என் நண்பனை நோக்கி திரும்பினேன். நான் பார்த்தபோது என்னை தனியே விட்டு சென்றிருந்தான்; அந்த நிமிடம் பயந்தேன், ஆனால், என் நண்பன், தனிமையான இடத்தில் நின்று என் நிலையை பார்த்து சிரித்தான்.

நான் இதைக் கண்டு அதிர்ச்சியடைந்தேன். பணம் சேகரிப்பவரிடம் என் கார்டை ஒப்படைக்க நான் அங்கு இயல்பாக (nonchalantly) இருந்தேன். அடுத்து வரும் பிரச்சனையை சமாளிக்க நான் யோசித்தேன்- படங்கள் மேல் படங்கள் வந்து விற்பனை ஆகிறது .நான் எதையும் பார்க்கவில்லை.

நான் ஓடி சென்று கடன் வாங்க சாத்தியமாக இருக்குமென்று மாமாவின் (Uncles) பெயர் மனத்திரையுல் வருகிறதா எனப் பார்த்தேன். ஆனால் வரவில்லை. மறுபடியும் இப்படத்தை ஏலம் விட சாத்தியக்கூறு உள்ளதா? என வினவினேன். எனது ஏழ்மையை நான் Christies உள்ள ஒரு பணியாளரிடம் எடுத்துக்கூறினேன்.

அந்த படத்தை திரும்பவும் ஏலம் விட சொன்னேன். இதுதான் சிறந்த வழி – அனைத்து முயற்சியும் செய்தபிறகு நான் இந்த ஏலத்தை எவ்வாறு செய்ய போகிறேன், அந்த பணியாளர் வளமாக காட்சியளித்தாலும் இரக்கமற்றவர், இது ஒரு தவறு என்று யாரும் நம்பவில்லை. எத்தகைய தவறும் ஒரு நேரத்தில் சரி செய்யப்பட வேண்டும்.

சரியான நேரம் விற்பனை முடிவுக்கு வந்தது. நான் வியாபாரிகள் இருக்கும் வெளி இடத்திற்கு சென்றேன். அவர்கள் காசோலை எழுதிக்கொண்டும் வழிமுறைகள் சொன்னார்கள். எப்போதும் போல் நான்தான் கடைசி. அந்த நேரம் என் நண்பனுடன் சேர முயன்றேன், என்னை பார்த்தவுடன் அவன் கைக்குட்டையால் தன்முகத்தை மூடிக் கொண்டான்.

விதியின் படி நான் தனியாக விடப்பட்டேன் என் வாழ்நாளில் அதைபோல் முட்டாள்தனமாக நான் உணர்ந்ததில்லை. இரக்கமில்லாத மனிதரை பார்த்ததுமில்லை.

நேர்மையுள்ள வாழ்க்கையில் சில நேரங்களில் நல்லொழுக்கத்திற்கு அப்பால் வெகுமதிகள் பெற்றுள்ளது என்பதை உணர்ந்தேன். எனது காதில் ஒரு குரல் திடீரென சொன்னது, “மன்னித்துக் கொள்ளுங்கள், சார், நீங்கள் பெரிய மனிதர், பெரிய Daubigny படத்தை வாங்கிய சீமான் நீங்கள் தானே?”

“நான் தான் என்று ஒப்புக்கொண்டேன்”.

நன்று, நான்கு ஆயிரம் Guineas உங்களுக்கு அளித்தபெரிய மனிதர் உங்கள் ஏலத்தை ஐம்பது Guineasற்கு பெற்றுக் கொள்வீர்களா? என தெரிந்துகொள்ள நினைக்கிறார்.”

உயர்ந்த கடவுளின் தூதர் ஒரு கரடுமுரடான பச்சை கம்பளி மேலங்கி அணிந்து கொழகொழவென்ற Cockney குரலில் பேசினார். அவரை கட்டி தழுவி சந்தோசத்தில் நனைந்தேன். ஐம்பது Guineas எடுத்துகொண்டிருப்பேன். ஏன் நான் குறைந்த காசை (farthings எடுத்து) கொள்ள வேண்டும்.

”இது தான் கேட்கக் கூடிய அதிகபட்ச விலையா?” என்று கேட்டேன். “அவனிடம் சொல்லுங்கள் நான் நூறு எடுத்து கொள்கிறேன்”, நான் சொல்லி பிறகு பெற்றுக்கொண்டேன்.

என் நண்பனை நான் காணும் போது நானும் சிரித்து கொண்டிருந்தேன், ஆனால் அவன் அந்த காசோலை ‘ பார்த்தவுடன் மயங்கினான்”. “எல்லாம் அதிர்ஸ்டம். நல்லது நான் தொடங்குகிறேன்” என்றான்.

நான் அழைக்கவில்லையென்றால் நீ christies வந்திருக்க முடியாது என்பதை மறந்துவிடாதே” என்று நண்பன் சொன்னான், “நான் அதை மறக்க மாட்டேன்” என்று கூறினேன். “இது அழிக்கமுடியாத (indelibly) நெருப்பாக என் நெஞ்சில் இருக்கும். எனது முடி வெள்ளையாக மாறாது, பாருங்கள்?”

Tamilnadu State Board New Syllabus Samacheer Kalvi 11th Maths Guide Pdf Chapter 7 Matrices and Determinants Ex 7.4 Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 11th Maths Solutions Chapter 7 Matrices and Determinants Ex 7.4

Question 1.
Find the area of the triangle whose vertices are (0, 0), (1, 2) and (4, 3)
Answer:
The given points are (0, 0), (1, 2) and (4, 3)
Area of the triangle with vertices
(x₁, y₁), (x₂, y₂) and (x₃, y₃) is

∴ The area of the triangle with vertices
(0, 0), (1, 2) and (4, 3) is

Area cannot be negative. Taking positive value, we have
Required area Δ = \(\frac{5}{2}\) sq.units.

Question 2.
If (k, 2), (2, 4) and (3, 2) are vertices of the triangle of area 4 square units then determine the value of k.
Answer:
Given Area of the triangle with vertices (k, 2), (2, 4) and (3, 2) is 4 square units.
The area of the triangle with vertices
(x₁, y₁) , (x₂, y₂) and (x₃, y₃) is

Given Δ = 4, (x₁, y₁) = (k , 2), (x₂, y₂) = (2 , 4) and (x₃, y₃) = (3 , 2)

± 4 = k(4 – 2) – 2 (2 – 3) + 1(4 – 12)
± 4 = k × 2 – 2 × – 1 – 8
± 4 = 2k + 2 – 8
± 4 = 2k – 6
2k – 6 = 4 or 2k – 6 = -4
2k = 4 + 6 or 2k = – 4 + 6
2k = 10 or 2k = 2
k = 5 or k = 1
Required values of k are 1, 5.

Question 3.
Identify the singular and non – singular matrices.
(i) \(\left[ \begin{matrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{matrix} \right] \)
(ii) \(\left[ \begin{matrix} 2 & -3 & 5 \\ 6 & 0 & 4 \\ 1 & 5 & -7 \end{matrix} \right] \)
(iii) \(\left[ \begin{matrix} 0 & a\quad -\quad b & k \\ b-\quad a & 0 & 5 \\ -k & -5 & 0 \end{matrix} \right] \)
Answer:
(i) \(\left[ \begin{matrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{matrix} \right] \)

|A| = 1 (45 – 48) – 2(36 – 42) + 3(32 – 35)
|Al = – 3 – 2 × – 6 + 3 × – 3
|A| = – 3 + 12 – 9
|A| = – 12 + 12 = 0
∴ A is a singular matrix.

(ii) \(\left[ \begin{matrix} 2 & -3 & 5 \\ 6 & 0 & 4 \\ 1 & 5 & -7 \end{matrix} \right] \)
Answer:

|B| = 2(0 – 20) + 3 (- 42 – 4) + 5(30 – 0)
|B| = -40 + 3 × – 46 + 150
|B| = -40 – 138 + 150
|B| = -178 + 150 ≠ 0
∴ B is non singular.

(iii) \(\left[ \begin{matrix} 0 & a\quad -\quad b & k \\ b-\quad a & 0 & 5 \\ -k & -5 & 0 \end{matrix} \right] \)

|C| = 0 – (a – b) (0 + 5k) + k(-5 (b – a) – 0)
|C| = -5k (a – b) – 5k (b – a)
|C| = -5k (a – b) + 5k(a – b)
|C| = o
∴ C is a singular matrix.

Question 4.
Determine the values of a and b so that the following matrices are singular:
(i) A = \(\left[ \begin{matrix} 7 & 3 \\ -2 & a \end{matrix} \right] \)
(ii) B = \(\left[ \begin{matrix} b\quad -\quad 1 & 2 & 3 \\ 3 & 1 & 2 \\ 1 & -2 & 4 \end{matrix} \right] \)
Answer:
(i) A = \(\left[ \begin{matrix} 7 & 3 \\ -2 & a \end{matrix} \right] \)
|A| = \(\left[ \begin{matrix} 7 & 3 \\ -2 & a \end{matrix} \right] \)
|A| = 7a + 6
Given that A is singular
∴ |A| = 0
7a + 6 = 0 ⇒ a = \(\frac{-6}{7}\)

(ii) B = \(\left[ \begin{matrix} b\quad -\quad 1 & 2 & 3 \\ 3 & 1 & 2 \\ 1 & -2 & 4 \end{matrix} \right] \)
|B| = \(\left[ \begin{matrix} b\quad -\quad 1 & 2 & 3 \\ 3 & 1 & 2 \\ 1 & -2 & 4 \end{matrix} \right] \)
= (b – 1 )(4 + 4) – 2(12 – 2) + 3(- 6 – 1)
= 8 (b – 1) – 20 – 21
= 8b – 8 – 41
|B| = 8b -49
Given that B is singular
∴ |B| = 0
8b – 49 = 0 ⇒ b = \(\frac{49}{8}\)

Question 5.
If cos 2θ = 0, determine

Answer:
Given cos 2θ = 0

Question 6.
Find the value of the product

Answer:

Tamilnadu State Board New Syllabus Samacheer Kalvi 11th Maths Guide Pdf Chapter 7 Matrices and Determinants Ex 7.3 Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 11th Maths Solutions Chapter 7 Matrices and Determinants Ex 7.3

Solve the following problems by using Factor Theorem:

Question 1.
Show that \(\left| \begin{matrix} x & a & a \\ a & x & a \\ a & a & x \end{matrix} \right| \) = (x – a)² (x + 2a)
Answer:

By putting x = a , we have three rows of |A| are identical. Therefore (x – a)² is a factor of |A|
Put x = – 2a in |A|

∴ x + 2a is a factor of |A|. The degree of the product of the factors (x – a)² (x + 2a) is 3.
The degree of tfie product of the leading diagonal elements x . x . x is 3.
∴ The other factor is the contant factor k.

a³ [ – 1 (1 – 1) – 1 ( – 1 – 1) + 1 (1 + 1)] = k . 4a³
a³ [o + 2 + 2 ] = 4 ka³
4a³ = 4 ka³
k = 1

Question 2.
Show that \(\left| \begin{matrix} b\quad +\quad c & a\quad -\quad c & a\quad -\quad b \\ b\quad -\quad c & c\quad +\quad a & b\quad -\quad a \\ c\quad -\quad b & c\quad -\quad a & a\quad +\quad b \end{matrix} \right| \) = 8 abc
Answer:


since two columns identical
= bc × 0 = 0
∴ a – 0 is a factor. That is, a is a factor.
Put b = 0 in |A|

since two columns identical
= ca × 0 = 0
∴ b – 0 is a factor. That is, a is a factor.
Put c = 0 in |A|

since two columns identical
= ab × 0 = 0
∴ c – 0 is a factor. That is, c is a factor.
The degree of the product of the factors abc is 3.
The degree of the product of leading diagonal elements (b + c) (c + a) (a + b) is 3.
∴ The other factor is the constant factor k.

Question 3.
Solve that \(\left| \begin{matrix} x\quad +\quad a & b & c \\ a & x\quad +\quad b & c \\ a & b & x\quad +\quad c \end{matrix} \right| \) = 0
Answer:

Put x = 0

x = 0 satisfies the given equation. x = 0 is a root of the given equation, since three rows are identical. x = 0 is a root of multiplicity 2. Since the degree of the product of the leading diagonal elements (x + a) (x + b) (x + c) is 3. There is one more root for the given equation.
Put x = – (a + b + c)

∴ x = – (a + b + c) satisfies the given equation.
Hence, the required roots of the given equation are x = 0, 0 , – (a + b + c)

Question 4.
Show that \(\left| \begin{matrix} b\quad +\quad c & a & { a }^{ 2 } \\ c\quad +\quad a & b & { b }^{ 2 } \\ a\quad +\quad b & c & { c }^{ 2 } \end{matrix} \right| \) = (a + b + c) (a – b) (b – c) (c – a)
Answer:

Since two rows are idenctical
|A| = 0
since two rows are idenctical
|A| = 0
∴ a – b is a factor of | A |. The given determinant is in cyclic symmetric form in a , b and c. Therefore, b – c and c – a are also factors. The degree of the product of the factors (a – b) (b – c) (c – a) is 3 and the degree of the product of the leading diagonal elements (b + c) . b . c² is 4.
Therefore, the other factor is k (a + b + c).

5(18 – 12) – 1(36 – 12) + 1(12 – 6) = 12k
5 × 6 – 24 + 6 = 12k
30 – 24 + 6 = 12k
12 = 12 ⇒ k = 1

Question 5.
Solve \(\left| \begin{matrix} 4\quad -\quad x & 4\quad +\quad x & 4\quad +\quad x \\ 4\quad +\quad x & 4\quad -\quad x & 4\quad +\quad x \\ 4\quad +\quad x & 4\quad +\quad x & 4\quad -\quad x \end{matrix} \right| \) = 0
Answer:

Put x = 0

∴ x = 0 satisfies the given equation. Hence x = 0 is a root of the given equation. since three rows are identical, x = 0 is a root of multiplicity 2.

Since the degree of the product of the leading diagonal elements (4 – x) (4 – x) (4 – x) is 3. There is one more root for the given equation.


∴ x = – 12 is a root of the given equation.
Hence, the required roots are x = 0 , 0 , – 12

Question 6.
Show that \(\left| \begin{matrix} 1 & 1 & 1 \\ x & y & z \\ { x }^{ 2 } & { y }^{ 2 } & { z }^{ 2 } \end{matrix} \right| \) = (x – y) (y – z) (z – x)
Answer:

|A| = 0 since two columns identical
∴ x – y is a factor of A. The given determinant is in the cyclic symmetric form in x, y, and z. Therefore, y – z and z – x are also factors of |A|.

The degree of the product of the factors (x – y) (y – z) (z – x) is 3 and the degree of the product of the leading diagonal elements 1, y, z² is 3. Therefore, the other factor is the constant factor k.

Put x = 0, y = 1, z = -1 we get

Expanding along the first column
1 (1 + 1) = 2k
2 = 2k ⇒ k = 1

Tamilnadu State Board New Syllabus Samacheer Kalvi 11th Maths Guide Pdf Chapter 7 Matrices and Determinants Ex 7.2 Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 11th Maths Solutions Chapter 7 Matrices and Determinants Ex 7.2

Question 1.
without expanding the determinant,

Answer:

= s (a² + b² + c²) × 0
since two columns are equal.
= 0

Question 2.

Answer:

Question 3.
Prove that

Answer:



= 2abc [0 – b(0 – ac) + c(ab – 0)]
= 2 abc [ abc + abc ]
= 2 abc × 2abc
Δ = 4 a²b²c²

Question 4.
Prove that

Answer:

= a [b(1 + c) + c (1)] – 0 – c [0 – b]
= a[b + bc + c] + bc
= ab + abc + ac + bc
= abc + ab + bc + ac
= abc

Question 5.

Answer:

Question 6.
show that \(\left| \begin{matrix} x\quad +\quad 2a & y\quad +\quad 2b & z\quad +\quad 2c \\ x & y & z \\ a & b & c \end{matrix} \right| \) = 0
Answer:
Let Δ = \(\left| \begin{matrix} x\quad +\quad 2a & y\quad +\quad 2b & z\quad +\quad 2c \\ x & y & z \\ a & b & c \end{matrix} \right| \)

Question 7.
Write the general form of a 3 × 3 skew- symmetric matrix and prove that its determinant is 0.
Answer:
A square matrix A = [ a_ij ]3 × 3 is a skew – symmetric matrix if a_ij = – a_ij for all i,j and the elements on the main diagonal of a skew – symmetric matrix are zero.


= 0 – a₁₂ (0 + a₁₃ a₂₃) + a₁₃ (a₁₂ a₂₃ – 0)
= – a₁₂ a₁₃ a₂₃ + a₁₃ a₁₂ a₂₃
= 0
Hence the determinant of a skew – symmetric matrix is 0.

Question 8.

Prove that a, b, c are in G. P or α is a root of ax² + 2bx + c = 0.
Answer:

Question 9.

Answer:

Question 10.
If a, b, c are p^th, q^th and r^th terms of an A.P, find the value of \(\left| \begin{matrix} a & b & c \\ p & q & r \\ 1 & 1 & 1 \end{matrix} \right| \)
Answer:
Given a, b, c are p^th, q^th and r^th terms of an A.P.
t_p = a = A + (p – 1)D,
t_q = b = A + (q – 1)D,
t_r = c = A + (r – 1) D
where A – first term , D – Common difference of the AP.

Question 11.

Answer:

Question 12.
If a , b , c , are all positive, and are p^th, q^th and r^th terms of a G.P., show that \(\left| \begin{matrix} log\quad a & p & 1 \\ log\quad b & q & 1 \\ log\quad c & r & 1 \end{matrix} \right| \) – 0
Answer:
Given a, b, c are the p^th, q^th and r^th terms of a G.P.
∴ a = AR^p-1, b = AR^q-1, c = AR^r-1
where A is the first term , R – common ratio.

Question 13.
Find the value of

Answer:

Question 14.

Answer:

Question 15.
Without expanding, evaluate the following determinants:
(i) \(\left| \begin{matrix} 2 & 3 & 4 \\ 5 & 6 & 8 \\ 6x & 9x & 12x \end{matrix} \right| \)
(ii) \(\left| \begin{matrix} x\quad +\quad y & y\quad +\quad z & z\quad +\quad x \\ z & x & y \\ 1 & 1 & 1 \end{matrix} \right| \)
Answer:
(i) \(\left| \begin{matrix} 2 & 3 & 4 \\ 5 & 6 & 8 \\ 6x & 9x & 12x \end{matrix} \right| \)

(ii) \(\left| \begin{matrix} x\quad +\quad y & y\quad +\quad z & z\quad +\quad x \\ z & x & y \\ 1 & 1 & 1 \end{matrix} \right| [

Question 16.
If A is a Square, matrix, and |A| = 2, find the value of |A A^T|.
Answer:
|A| = 2 (Given) |A^T| = 2
Now |AA^T| = |A| |A^T| = 2 × 2 = 4.

Question 17.
If A and B are square matrices of order 3 such that |A| = -1 and |B| = 3, find the value of |3 AB|.
Answer:
Given |A| = -1 : |B| = 3
Given A and B are square matrices of order 3.
∴ |kAB| = k³ |AB|
Here k = 3 ∴ |3AB| = 3³ |AB|
= 27 |AB|
= 27 (-1) (3)
= -81

Question 18.
If λ = – 2, determine the value of

Answer:

Expanding along the first row
Δ = 0 + 4 [4 × 0 – (- 1 ) ( 13)] + [4 × -13 – 0 × – 1]
= 4 [0 + 13] + 1 [- 52 + 0]
= 52 – 52 = 0

Question 19.
Determine the roots of the equation
[latex]\left| \begin{matrix} 1 & 4 & 20 \\ 1 & -2 & 5 \\ 1 & 2x & { 5x }^{ 2 } \end{matrix} \right| \) = 0
Answer:
\(\left| \begin{matrix} 1 & 4 & 20 \\ 1 & -2 & 5 \\ 1 & 2x & { 5x }^{ 2 } \end{matrix} \right| \) = 0 ………… (1)
Put x = -1 then (1) ⇒

∴ x = – 1 satisfies equation (1)
Hence x = – 1 is a root of equation (1)
Put x = 2 then ……….. (1)

[Property 4: If two rows (columns) of a determinant are identical then its determinant value is zero.]

∴ x = 2 satisfies equation (1)
Hence x = 2 is a root of equation (1)
Hence the required roots are x = -1 , 2

Question 20.
Verify that det (AB) = (det A) (det B) for

Answer:

= 4 [-10 (0 – 9 × 19) – 5 (0 + 17 × 19) + 1 (32 × 9 + 17 × 26)]
= 4 [1710 – 5 × 323 + 288 + 442]
= 4 [1710 – 1615 + 730]
= 4 [2440 – 1615]
= 4 × 825
det (AB) = 3300 …….. (1)

= 4(0 – 21) – 3 (- 5 – 14) – 2 (3 – 0)
= -84 – 3 × – 19 – 6
= -84 + 57 – 6
= -90 + 57
det A = -33 ………… (2)

= 1 (20 – 0) – 3 (- 10 – 0) + 3 (-14 – 36)
= 20 + 30 + 3 × – 50
= 50 – 150
det A = – 100 ……….. (3)

From equations (2) and (3)
(det A) (det B) = – 33 × – 100
(detA) (det B) = 3300 ………… (4)
From equations (1) and (4), we have
det (AB) = (det A) (det B)

Question 21.
Using cofactors of elements of second row, evaluate |A|, where A = \(\left[ \begin{matrix} 5 & 3 & 8 \\ 2 & 0 & 1 \\ 1 & 2 & 3 \end{matrix} \right] \)
Answer:

|A| = a₂₁ A₂₁ + a₂₂ A₂₂ + a₂₃ A₂₃
= 2 × 7 + 0 × 7 + 1 × – 7
= 14 – 7
|A| = 7

Tamilnadu State Board New Syllabus Samacheer Kalvi 11th Maths Guide Pdf Chapter 7 Matrices and Determinants Ex 7.1 Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 11th Maths Solutions Chapter 7 Matrices and Determinants Ex 7.1

Question 1.
Construct an m × n matrix A = [a_ij], where a_ij is given by
(i) a_ij = \(\frac{(\mathbf{i}-2 \mathbf{j})^{2}}{2}\) with m = 2 , n = 3
(ii) a_ij = \(\frac{|3 \mathbf{i}-4 \mathbf{j}|}{4}\) with m = 3 , n = 4
Answer:
(i) a_ij = \(\frac{(\mathbf{i}-2 \mathbf{j})^{2}}{2}\) with m = 2 , n = 3
To construct 2 × 3 matrices.

(ii) a_ij = \(\frac{|3 \mathbf{i}-4 \mathbf{j}|}{4}\) with m = 3 , n = 4
To construct a 3 × 4 matrices.

Question 2.
Find the value of p, q, r and s if

Answer:

Equating the corresponding entries
⇒ p² – 1 = 1
⇒ p² = 1 + 1 = 2
p = ± \(\sqrt{2}\)
-31 – q³ = -4
-q³ = -4 + 31 = 27
q³ = -27 = (-3)³
⇒ q = -3
r + 1 = \(\frac{3}{2}\)
⇒ r = \(\frac{3}{2}\) – 1 = \(\frac{3-2}{2}\) = \(\frac{1}{2}\)
s – 1 = π
⇒ s = – π + 1 (i.e.,) s = 1 – π
So, p = ± \(\sqrt{2}\), q = -3, r = 1/2 and s = 1 – π

Question 3.
Determine the value of x + y if
\(\left[ \begin{matrix} 2x\quad +\quad y & 4x \\ 5x\quad -\quad 7 & 4x \end{matrix} \right] \) = \(\left[ \begin{matrix} 7 & 7y\quad -\quad 13 \\ y & x\quad +\quad 6 \end{matrix} \right]\)
Answer:
\(\left[\begin{array}{cc}{2 x+y} & {4 x} \\ {5 x-7} & {4 x}\end{array}\right]=\left[\begin{array}{cc}{7} & {7 y-13} \\ {y} & {x+6}\end{array}\right]\)
⇒ 2x + y = 7 ………….. (1)
4x = 7y – 13 ………….. (2)
5x – 7 = y …………… (3)
4x = x + 6 ……………. (4)
from (4) 4x – x = 6
3x = 6 ⇒ x = \(\frac{6}{3}\) = 2
Substituting x = 2 in (1), we get
2(2) + y = 7 ⇒ 4 + y = 7 ⇒ y = 7 – 4 = 3
So x = 2 and y = 3
∴ x + y = 2 + 3 = 5

Question 4.
Determine the matrices A and B if they satisfy 2A – B + \(\left[ \begin{matrix} 6 & -6 & 0 \\ -4 & 2 & 1 \end{matrix} \right] \) = 0 and A – 2B = \(\left[ \begin{matrix} 3 & 2 & 8 \\ -2 & 1 & -7 \end{matrix} \right] \)
Answer:

Question 5.
If A = \(\left[ \begin{matrix} 1 & a \\ 0 & 1 \end{matrix} \right] \), then compute A⁴
Answer:

Question 6.
Consider the matrix A_α = \(\left[ \begin{matrix} cos\quad α & -\quad sin\quad α \\ sin\quad α & cos\quad α \end{matrix} \right] \)
(i) Show that A_αA_β = A_(α+β)
Answer:

(ii) Find all possible real values α satisfying the condition A_α + A_α^T = I
Answer:

Question 7.
If A = \(\left[ \begin{matrix} 4 & 2 \\ -1 & x \end{matrix} \right] \) and such that (A – 2I) (A – 3I) = 0, find the value of x.
Answer:

Question 8.
If A = \(\left[ \begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ a & b & -1 \end{matrix} \right] \), show that A² is a unit matrix.
Answer:

Question 9.
If A = \(\left[ \begin{matrix} 1 & 0 & 2 \\ 0 & 2 & 1 \\ 2 & 0 & 3 \end{matrix} \right]\) and A³ – 6A² + 7A + kI = 0, find the value of k.
Answer:

Equating the corresponding entries – 2 + k = 0 ⇒ k = 2
∴ The required value of k is k = 2

Question 10.
Give your own examples of matrices satisfying the following conditions in each case:
(i) A and B such that AB ≠ BA
(ii) A and B such that AB = 0 = BA, A ≠ 0 and B ≠ 0.
(iii) A and B such that AB = 0 and BA ≠ 0
Answer:
(i) A and B such that AB ≠ BA

(ii) A and B such that AB = 0 = BA, A ≠ 0 and B ≠ 0.

(iii) A and B such that AB = 0 and BA ≠ 0

Question 11.
Show that f(x) f(y) = f(x + y) , where f(x) = \(\left[ \begin{matrix} cos\quad x & -\quad sin\quad x & 0 \\ sin\quad x & cos\quad x & 0 \\ 0 & 0 & 1 \end{matrix} \right] \)
Answer:

Question 12.
If A is a square matrix such that A² = A, find the value of 7A – (I + A )³
Answer:
Given A² = A
So 7A – (I + A)³ = 7A – (I + 3A + 3A² + A³]
= 7A – I – 3A – 3 A² – A³
Given A² = A
7A – I – 3A – 3A – A³ = -I + A – A³
= -I + A – (A² × A)
= -I + A – (A × A) = -I + A – A²
= -I + A – A = -I
So the value of 7A – (I + A)³ = -I.

Question 13.
Verify the property A (B + C) = AB + AC, when the matrices A, B, and C are given by

Answer:

Question 14.
Find the matrix A which satisfies the matrix relation A\(\left[ \begin{matrix} 1 & 2 & 3 \\ 4 & 5 & 6 \end{matrix} \right] \) = \(\left[ \begin{matrix} -7 & -8 & -9 \\ 2 & 4 & 6 \end{matrix} \right] \)
Answer:

Question 15.
If A^T = \(\left[ \begin{matrix} 4 & 5 \\ -1 & 0 \\ 2 & 3 \end{matrix} \right] \) and B = \(\left[ \begin{matrix} 2 & -1 & 1 \\ 7 & 5 & -2 \end{matrix} \right] \)
verify the following
(i) (A + B)^T = A^T + B^T = B^T + A^T
(ii) (A – B)^T = A^T – B^T
(iii) (B^T)^T = B
Answer:

(i) (A + B)^T = A^T + B^T = B^T + A^T

(ii) (A – B)^T = A^T – B^T

(iii) (B^T)^T = B

Question 16.
If A is a 3 × 4 matrix and B is a matrix such that both A^TB and BA^T are defined, what is the order of the matrix B?
Answer:
A is a matrix of order 3 × 4
So AT will be a matrix of order 4 × 3
AT B will be defined when B is a matrix of order 3 × n
BA^T will be defined when B is of order m × 4
from (1) and (2) we see that B should be a matrix of order 3 × 4

Question 17.
Express the following matrices as the sum of a symmetric matrix and a skew – symmetric matrix:
(i) \(\left[ \begin{matrix} 4 & -2 \\ 3 & -5 \end{matrix} \right] \)
Answer:



A = \(\frac{1}{2}\)(A + A^T) + \(\frac{1}{2}\)(A – A^T
Thus A is expressed as a sum of a symmetric and skew-symmetric matrix.

(ii) \(\left[ \begin{matrix} 3 & 3 & -1 \\ -2 & -2 & 1 \\ -4 & -5 & 2 \end{matrix} \right] \)
Answer:

Question 18.
Find the matrix A such that

Answer:

Equating like entries
2a – d = -1, 2b – e = – 8, 2c – f = -10
a = 1, b = 2, c = -5

2a – d = -1 ⇒ 2 × 1 – d = – 1
⇒ 2 + 1 = d ⇒ d = 3

2b – e = – 8 ⇒ 2 × 2 – e = – 8
⇒ 4 + 8 = e ⇒ e = 12

2c – f = -10 ⇒ 2 × – 5 – f = -10
⇒ – 10 – f = -10 ⇒ f = 0

Question 19.
If A = \(\left[ \begin{matrix} 1 & 2 & 2 \\ 2 & 1 & -\quad 2 \\ x & 2 & y \end{matrix} \right] \) is a matrix such that AA^T = 9I, find the values of x and y.
Answer:

Equating the corresponding entries
x + 4 + 2y = 0 ………… (1)
2x + 2 – 2y = o ………… (2)
x + 4 + 2y = 0 ………… (3)
2x + 2 – 2y = 0 ………… (4)
x² + 4 + y² = 9 ………… (5)

Substituting the value of y in equation (1) we have
x + 4 + 2x – 1 = 0
x + 4 – 2 = 0 ⇒ x = – 2
Substituting x = – 2 and y = – 1 in equation(5) we have
(5) ⇒ (-2)² + 4 + (- 1)² = 9
4 + 4 + 1 = 9
9 = 9
∴ The required values of x and y are
x = – 2 and y = – 1

Question 20.
(i) For what value of x, the matrix A = \(\left[ \begin{matrix} 0 & 1 & -\quad 2 \\ -\quad 1 & 0 & { x }^{ 3 } \\ 2 & -\quad 3 & 0 \end{matrix} \right] \) is skew – symmetric
Answer:

The matrix A is skew-symmetric if A = – A^T

Equating the corresponding entries
x³ – 3 = 0
x³ = 3 ⇒ x = 3^1/3

(ii) If \(\left[ \begin{matrix} 0 & p & 3 \\ 2 & { q }^{ 2 } & -\quad 1 \\ r & 1 & 0 \end{matrix} \right] \) is skew – symmetric find the values of p, q and r.
Answer:

A is skew-symmetric if A = – A^T

Equating the corresponding entries.
p = – 2 , r = – 3
q² = -q² ⇒ q² + q² = 0
⇒ 2q² = 0 ⇒ q = 0
∴ The required values are
p = – 2 , q = 0 , r = – 3

Question 21.
Construct the matrix A = [a_ij]_3×3 , where a_ij = 1 – j. State whether A is symmetric or skew – symmetric.
Answer:

a_ij = i – j
a₁₁ = 1 – 1 = 1
a₁₂ = 1 – 2 = – 1
a₁₃ = 1 – 3 = – 2
a₂₁ = 2 – 1= 1
a₂₂ = 2 – 2 = 1
a₂₃ = 2 – 3 = – 1
a₃₁ = 3 – 1= 2
a₃₂ = 3 – 2 = 1
a₃₃ = 3 – 3 = 0

Question 22.
Let A and B be two symmetric matrices. Prove that AB = BA if and only if AB is a symmetric matrix.
Answer:
Given A and B two symmetric matrices.
∴ A = A^T and B = B^T
First, let us assume AB = BA.
Let us prove AB is a symmetric matrix.

∴ AB is a symmetric matrix.
conversely let us assume that AB is a symmetric matrix.
we prove AB = BA
AB is symmetric then

Question 23.
If A and B are symmetric matrices of the same order, prove that
(i) AB + BA is a symmetric matrix
(ii) AB – BA is a skew-symmetric matrix.
Answer:
Given A and B are symmetric matrices
⇒ – A^T = A and B^T = B
(i) To prove AB + BA is a symmetric matrix.
Proof: Now (AB + BA)^T = (AB)^T + (BA)^T = B^TA^T + A^TB^T
= BA + AB = AB + BA
i.e. (AB + BA)^T = AB + BA
⇒ (AB + BA) is a symmetric matrix.

(ii) To prove AB – BA is a skew symmetric matrix.
Proof: (AB – BA)^T = (AB)^T – (BA)^T = B^TA^T – A^TB^T = BA – AB
i.e. (AB – BA)^T = – (AB – BA)
⇒ AB – BA is a skew symmetric matrix.

Question 24.
A shopkeeper in a Nuts and Spices shop makes gift packs of cashew nuts, raisins, and almonds. The pack contains 100 gm of cashew nuts, 100 gm of raisins, and 50 gm of almonds. Pack – II contains 200 gm of cashew nuts, 100 gm of raisins, and 100 gm of almonds. Pack -III contains 250 gm of cashew nuts, 250 gm of raisins, and 150 gm of almonds. The cost of 50 gm of cashew nuts is ₹ 50, 50 gm of raisins is ₹ 10, and 50 gm of almonds is ₹ 60. What is the cost of each gift pack?
Answer:

Let us consider 50 gm of cashew nuts as one unit, 50 gms of raisins as one unit 50 gm of almonds as one unit.
∴ The Gift pack matrix becomes

Also given 50 gms of Cashew nuts cost = Rs. 50
50 gms of Raisins cost = Rs. 10
50 gms of Almonds cost = Rs. 60
∴ Cost matrix is B = [50 10 60]
∴ Cost of cash gift pack = BA

∴ Cost of I gifts Pack = Rs. 180
Cost of II gift Pack = Rs.340
Cost of III gift Pack = Rs. 480

Tamilnadu State Board New Syllabus Samacheer Kalvi 11th English Guide Pdf Prose Chapter 3 Forgetting Text Book Back Questions and Answers, Summary, Notes.

Tamilnadu Samacheer Kalvi 11th English Solutions Prose Chapter 3 Forgetting

11th English Guide Forgetting Text Book Back Questions and Answers

I. Based on your understanding of the essay, answer the following questions in one or two sentences each:

Question a.
What does Lynd actually wonder at?
Answer:
Robert Lynd wonders at the efficiency of human memory. He is amazed at the ordinary man’s capacity to remember phone numbers, addresses of friends, appointments for lunch and dinner and many names of actors, actresses and leading players in popular games.

Question b.
Name a few things that a person remembers easily.
Answer:
A person remembers telephone numbers, addresses of his friends, dates of good vintages, appointments for lunch and dinner, etc.

Question c.
How do psychologists interpret forgetfulness?
Answer:
Psychologists believe that humans forget what they don’t want to remember, like taking pills.

Question d.
What is the commonest type of forgetfulness, according to Lynd?
Answer:
According to Lynd the commonest type of forgetfulness occurs in the matter of posting letters.

Question e.
What does the author mean when he says the letter in his pocket leads an unadventurous life?
Answer:
The poet forgets the letters kept in his pocket. Whenever the friend enquires about the unposted letters, it embarrasses him. Then he is forced to produce the evidence of his guilt (i.e.,) the unposted letters. This awkward humiliation is said to be unadventurous.

Question f.
What are the articles the writer forgets most often?
Answer:
The writer forgets books, umbrellas, and walking sticks most often.

Question g.
Who are the citizens of dreamland? Why?
Answer:
Boys who return from cricket and football matches tend to forget bats and balls. Their minds are filled with a vision of the playing field. Their heads are among the stars. They are said to be the citizens of dreamland.

Question h.
What is common about the ‘angler’ and the ‘Poet’?
Answer:
The angler forgets his fishing rod and the poet forgets to post a letter just because their mind is filled with glorious matter.

2. Based on your reading, answer the following questions in two to four sentences each:

Question a.
What made people wonder about the absentmindedness of their fellow beings?
Answer:
The publication of articles lost by train travellers astonished many readers. Old people did not forget much. In fact, young men have forgotten bats and balls on their return from matches.

Question b.
What are our memories filled with?
Answer:
Our memories are filled with the names of actors and actresses, cricketers, footballers, and murderers.

Question c.
When does human memory work with less than its usual capacity?
Answer:
Human memory works with less than its usual capacity in matters like taking medicine. The author explains that human memory represents the willingness to remember certain things. It forgets what it does not wish to remember. Humans are blessed with “selective amnesia”

Question d.
Why according to Lynd should taking medicines be one of the easiest actions to remember?
Answer:
Taking medicines is one of the easiest actions to remember as it should be taken before, during or after meals. The meal itself is a reminder of it.

Question e.
How do the chemists make fortunes out of the medicines people forget to take?
Answer:
The forgotten medicines tend to aggravate the illness. Like a vicious cycle, again they are forced to buy costlier medicines. Thus people who forget to take medicines contribute to the fortunes of chemists.

Question f.
The list of articles lost in trains suggests that sportsmen have worse memories than their ordinary serious-minded fellows. Why does Lynd say this?
Answer:
Sportsmen have worse memories as when they return from the game they have their imagination still filled with a vision of the playing field. They are abstracted from the world outside them and their memories prevent them from remembering small prosaic things.

Question g.
What kind of absent-mindedness is regarded as a virtue by Lynd?
Answer:
Scientists, poets, anglers, and philosophers forget prosaic things. Their minds are absorbed in lofty thoughts and glorious imaginations that they forget ordinary things. Socrates, Tagore, and Einstein had the virtue of absent-mindedness. Einstein usually forgot to change his rocks. Once he even forgot his own house address. The absent-mindedness of such great personalities is a virtue. As they make the best of life, they have no time to remember the mediocre.

Question h.
Narrate the plight of the baby on its day out.
Answer:
The baby taken out by its father was left outside a public house just as the father slipped in for a glass of beer. His wife who came shopping saw the baby and took it home deciding to teach a lesson to her husband. To her surprise, the husband came forgetting all about the baby.

3. Answer the following in a paragraph of about 100-150 words each:

Question a.
You have borrowed a branded cricket bat from your reluctant friend for an outstation match. After returning home you realize you have absent-mindedly left it in the hotel room. Write a letter of apology and regret to your friend.
Answer:
822, Old Peter Road,
Trichy.

Dear Akshay,
Hope this letter of mine would find you in the best of health. First of all, I thank you very much for lending me your branded cricket bat for my match in Chennai. Though you were reluctant at first, you were kind enough to lend it to me later. I really played well with that bat and scored the highest run rate.

Truly it is the luckiest bat. After the match, I kept it safe in the hotel room where I stayed. Because of my weariness, I had a sound sleep that day and was in a hurry to catch my train for the return journey. In that hurry, I forgot to take your bat. Only after reaching Trichy, I realized that I absent-mindedly left your bat in the hotel room itself. I truly regret for the mistake committed by me and beg your pardon.

I know pretty well that it is your precious bat. I am also aware of the fact that you won’t forgive me easily for my action. I have made arrangements to bring back the bat here which may take some time. Kindly bear the inconvenience prevailed and try to forgive me.

With lots of regrets,

Yours affectionately,
Arun.

Question b.
Kahlil Gibran states ‘Forgetfulness is a form of freedom’ Write an article for your school magazine, linking your ideas logically and giving appropriate examples.
Answer:
Forgetting is deemed by many people leading prosaic lives as a mistake or an inefficiency of mind. But in reality, forgetfulness is freedom. Osho is right in his opinion of forgetfulness. In fact, it liberates painful memories and unpleasant things. We need to “let go” painful memories of the past and be free to aspire for better things in life. Robert Frost in his poem, “Let go” talks about a mediocre person’s inability to let go of things that hurt them. The capacity to forget hurtful memories is a real blessing.

If the human mind does not have the capacity to forget, life would be miserable for every one of us. The human mind is such a wonderful machine that it retains what is most important for personal or professional growth and allows the other things to slip away from the bank of memory. But young ones should remember to remember important assignments, deadlines for submission of homework, examination time-tables, and hall tickets before leaving for examination.

To assist memory we can have a checklist before leaving for school. It is often said, “If you fail to plan, you plan to fail.” So, my dear friends, I appeal to you to love whatever work you do. The brain retains in memory whatever you do with great passion, love, and involvement. For a successful life, a strong memory is indispensable. So, cultivate a strong memory. However, I appeal to you to forget failures, betrayals, and hurts to grow into a happy and healthy person.

“Sometimes we survive by forgetting.”

Question c.
Will you sympathize or ridicule someone who is intensely forgetful? Write an essay justifying your point of view.
Answer:
It is a general fact that all human beings are absent-minded at times. I really sympathize with the person who is intensely forgetful. His extreme level of forgetfulness reveals that he is a creative person and a genius. We have heard of great Scientists who are often forgetful. A person becomes absent-minded based on two facts.

One is when his mind is completely filled with stressful thoughts. Another reason is, he may be a creative person whose mind is always thinking of creating something new and forgets the present. Whatever be the reason there is no use ridiculing them.

On the other hand, we can help or guide them to note down important information in their diaries so that they can see to it when they forget something. In Kahlil Gibran’s point of view “Forgetfulness is a form of freedom”. So it can be rightly concluded that those who enjoy that freedom are really blessed.

d) Find the antonyms of the following words in the puzzle and shade them with a pencil. The first one has been done for you:

Answer:

1. Seldom x Often
2. admitted x denied
3. methodical x disorderly
4. reality x fantasy
5. fact x fiction.
6. virtue x vice
7. vile x good
8. indignant x delighted
9. relish x hate

Now, read the following biographical extract on Sujatha Rangarajan, a Science fiction writer, and answer the questions that follow:

1. Sujatha is the allonym of the Tamil author S. Rangarajan and it is this name that is recognized at once by the Tamil Sci-Fi reading community. You might have seen the Tamil movie ‘Endhiran’ where the robot Chitti exhibits extraordinary talents in an incredible manner. The robot could excel a human being in any act, beyond one’s imagination.

Jeeno, a robotic dog which appeared in Sujatha’s science fiction novel “En Iniya lyandhira” (My Dear Robot) formed the basis of Chitti’s character. Like Chitti, Jeeno was an all-rounder who could cook, clean, and fight. High-tech computer technology terms are used in the story. Jeeno, a pet robot, plays an important role throughout the story. As the story proceeds, it behaves and starts to think on its own like a human and instructs Nila, a human being, on how to proceed further in her crises.

2. In the preface of En lniya Iyandhira the writer states the reason for his attraction to the genre: Science gives us the wonderful freedom to analyse thousands and thousands of alternative possibilities. While using it, and while playing with its new games, a writer needs to be cautious only about one thing. The story should draw some parallels or association from the emotions and desires of the present humankind.

Only then it becomes interesting. Jeeno, the robot dog, was intelligent. But the character became popular only because of the robot’s frequently displayed human tendencies’ It is no wonder that all his works echo these words and will remain etched in the minds of the readers who enjoy reading his novels to have a wonderful lifetime experience.

3. It was Sujatha, who set the trend for sci-fi stories. He had tracked the origin from Mary Shelly’s Frankenstein to his short stories. He has written 50 sci-fi short stories and these were published in various Tamil magazines. His stories have inspired many readers to extend their reading to English sci-fi writers like Isaac Asimov.

The themes were bold, even if there was a dependence on very well – established characterization of English fiction. Sujatha opened up a new world to us with his writings on holograms, computers, and works like ‘En Iniya lyanthira’ inspire many to study computer science.

4. He has been one of the greatest writers for more than four decades. He combined reasoning and science in his writings. Being a multifaceted hi-fi and sci-fi humanistic author, he expressed his views distinctively. He was the one who took Tamil novels to the next level.

As an MIT alumnus and an engineer at BHEL, he was very good at technology. He could narrate sci-fi stories impressively. His readers always enjoyed reading all his detective and sci-fi novels which featured the most famous duo ‘Ganesh’ and ‘Vasanth’.

5. Sujatha has played a crucial role as a playwright for various Tamil movies which have fascinated movie lovers. Hence, it is fathomable that the writer’s perspective of future India enthuses every reader and paves a new way to reading sd-fl stories in English.

Find words from the passage which mean the same as the following:

Question 1.
difficult to believe (para 1)
Answer:
incredible

Question 2.
a style or category of art, music or literature (para2)
Answer:
genre

Question 3.
having many sides (para 4)
Answer:
multifaceted

Question 4.
capable of being understood (para 5)
Answer:
fathomable.

ஆசிரியரைப் பற்றி:

ராபர்ட் வில்சன் லிண்ட் (1879-1949) ஒரு ஐனஷ் எழுத்தாளர். 20ம் நூற்றாண்டில் வாழ்ந்த கட்டுரையாளர்களில் மிகச்சிறந்தவர். சிறந்த பத்திரிக்கையாளராக தன் பணியைத் தொடங்கினார். தினசரி செய்திதாள்’, ‘புதிய செய்தி, நாடு போன்ற பல பத்திரிக்கைகளில் அதிகமான கட்டுரைகளை எழுதியுள்ளரர்.

தன் படைப்புகள் அனைத்தும் வாசிப்பவரின் ஆர்வத்தை தூண்டக் கூடிய நகைச்சவை, மகிழ்ச்சி, வஞ்சப்புகழ்சி, விமர்சனம் அடிப்படையில் அமைந்திருக்கும்.

1947ல் இவருக்கு குயின்ஸ் பல்கலைக்கழகத்தால் இலக்கியத்திற்கான கௌரவ முனைவர் பட்டம் வழங்கப்பட்டது. இலக்கியத்திற்காக இவருக்கு ராயல் சொனசட்டியால் வெள்ளி பதக்கமும், டைம்ஸ் நிறுவனத்தால் தங்க பதக்கமும் வழங்கப்பட்டது. என்ற இந்த கட்டுரையில் மறதியை பற்றியும், அதன் இயல்பையும் நகைச்சுவையாக எழுதியுள்ளார்.

பாடத்தைப் பற்றி:

இந்த கட்டுரையில் ராபர்ட் லிண்ட் மனிதர்களில் உள்ள மறதிக்கான அடிப்படைக் காரணங்களைப் பற்றி தெளிவாக கூறுகிறார். நாம் எதை மறந்து போகிறோம், அப்படி மறந்து போவதால் ஏற்படும் விளைவுகள், ஏன் மறந்து போகிறோம் என்று பலவிதமான வினாக்களுக்க விடையையும் தருகிறார். மறத்தலைப் பற்றி தெளிவாக இக்கட்டுரையில் காண்போம்.

Forgetting Summary in Tamil

ரயிலில் செல்லும் பயணிகள் தவரவிட்ட பொருட்களை இப்போது லண்டன் நிலையத்தில் விற்பனைக்கு உள்ளதாக அறிவித்தனர். அதை வாசித்த மக்கள் அவர்கள் மறதி மனப்பாங்கை நினைத்து திகைத்தனர். புள்ளி விவரப்படி நான் சந்தேகப்பட்டது போல் இவ்வாறு மறந்து போகுதல் பொதுவான நிகழ்வுதான்.

இவை மனித நினைவின் திறன் மற்றும் திறன் இல்லாததை சொல்லி அதிசயப்பட வைக்கிறது. நவீன மனிதன் கைபேசி எண்களைக்கூட நினைவில் வைத்திருப்பான். அவன் நண்பரின் முகவரியையும் நினைவில் வைத்திருப்பான். பழங்காலத்தில் நடந்த நல்ல நிகழ்வுகளை கூட அவன் நினைத்துப்பார்க்கிறான்.

மதிய உணவு மற்றும் இரவு சாப்பாட்டிற்கான குறிப்பை அவன் ஞாபகம் வைத்திருப்பான். அவனது நினைவுகள் நடிகர், நடிகைகள், கிரிகிகெட் வீரர்கள் மற்றும் கால்பந்து வீரர்கள் மற்றும் கொள்ளையர்கள் என நெரிசலாக இருக்கும்.

கோடை காலத்தில் அவன் நன்றாக உணவு அருந்திய உயர்ரக ஹோட்டலையும், கடந்து சென்ற ஆகஸ்ட் பருவநிலையும் அவனால் சொல்ல முடியும். அவனது சாதாரண வாழ்விலும், அவன் எதையெல்லாம் நினைவு கூற நினைக்கிறானோ அதை அனைத்தையும் நினைவுப்படுத்துவான்.

லண்டனில் உள்ள ஆண்கள் எல்லோரும் காலையில் ஆடை அணியும் போது தங்களின் ஆடைகளின் சிறு துண்டினை மறப்பதுண்டா? நூற்றில் ஒருவர் கூட இல்லை. ஏன் ஆயிரத்தில் ஒருவர் கூட இல்லை. எத்தனை பேர் வீட்டை விட்டு வெளியில் செல்லும் போது வீட்டின் முன் கதவை அடைக்காமல் செல்வோம்.

ஒரு நாள் முழுதும் அவ்வாறு போகிறோம், நாம் படுக்கைக்கு செல்லும் வரை நமது செயலை தெளிவாக செய்கிறோம். ஆனால் ஒரு சாதாரண மனிதன் மேல் மாடிக்கு செல்வதற்கு முன் விளக்குகளை அணைக்க மறக்கிறான்.

சில நேரத்தில் நாம் நமது நினைவுகள் சாதாரணமாக செயல்படுவதை விட குறைந்து செயல்படும். ஒரு முதுநிலை மனிதர் மருத்துவர் அவருக்கு பரிந்துரை செய்ததை மறவாமல் எடுத்து செல்கிறார் என நினைக்கிறேன்.

இது ஆச்சரியம் தரக்கூடிய விஷயம் தான். மருந்துகள் என்பது இயல்பாக நம் நினைவில் இருக்கக்கூடியவை. விதியின் அடிப்படையில் அவை சாப்பாட்டிற்கு முன் அல்லது சாப்பாட்டிற்கு பின்பு மற்றும் உணவு என்ன என்பது கூட நினைவில் இருக்கும்.

உண்மை என்னவென்றால் சில ஒழுக்க அரக்கர்கள் அவர்களது மருந்துகளை ஞாபகம் வைத்திருப்பார்கள்.சில உளவியலாளர்கள் நம்மிடம் கூறுவது நாம் மறக்க நினைக்கும் விஷயத்தை மறக்கிறோம், ஏனெனில் அவை மிகுந்த வெறுப்பான மருந்தாக இருக்கும்; மனிதர்கள் குறிப்பிட்ட நேரத்தில் சாப்பிட மறக்கிறார்கள்.

என்னைப்போல் மருந்துக்கு நீண்ட பக்தனாக இருப்பவர்கள் வெறுப்பாக ஆர்வமில்லாமல் (unwillingly) மறந்து விடுகிறோம். புதிய, பரவலாக விளம்பரப்படுத்தப்படும் சிகிச்சை எனக்கு மிகவும் மகிழ்ச்சியளிக்கிறது.

நான் மருந்துகளை என் பையில் வைத்திருந்தாலும், அதை மறந்து, ஒரு மணி நேரம் கழித்து அதை எடுத்து சாப்பிடுவேன். மருத்துவரின் பொக்கிஷம் (fortunes)அவரின் மருந்தை மக்கள் மறந்து சாப்பிடாமல் இருப்பது.

பொதுவாக நான் மறந்துபோவதாக நினைப்பது கடிதம் அனுப்புவதிலே. பொதுவாக என்னை பார்க்க (சந்திக்க) வருபவரிடம் தயக்கத்துடன் எனது முக்கியமான கடிதத்தை அனுப்ப சொல்வேன். கடிதத்தை கொடுக்கும் முன் என் மீது நம்பிக்கை வர வைப்பேன். என்னிடம் கடிதத்தை அனுப்ப சொல்பவர்கள் என்னைப்பற்றி முழுதும் அறியாதவர்கள்.

நானே எடுத்து சென்றாலும் ஒரு பில்லர் பெட்டியை தாண்டிய பிறகு அடுத்த பெட்டியில் போட ஞாபகம் வரும். கையில் வைத்திருப்பது பதிலாக அதை என் சட்டை பையில் வைத்து அப்படியே மறந்துவிடுவேன்.

அதன் பிறகு, இது ஒரு மகிழ்ச்சியில்லா வாழ்க்கை. சங்கிலிப்போன்ற பிரச்சனைகள், எண்ணற்ற சொல்லமுடியா கேள்விகளை கேட்பது போன்று, என்னை வற்புறுத்தி என்னுடைய குற்ற உணர்வுகளை வெளிப்படுத்த வைக்கும்.

இவை அனைத்தும் மற்றவரின் கடிதம் என்பதால் ஈடுபாடு இல்லாமல் இருக்கலாம், சில கடிதங்கள் நான் எழுத நினைத்தது கூட நான் அனுப்ப மறந்துள்ளேன்.

நான் ரயிலில், Taxi யில் பொருட்களை தவறவிட்டவர்களைப் போல மிகச்சிறந்த மறதியாளன் அல்ல. என் புத்தகத்தையும், Walking stick யும் தவிர மற்ற எல்லாவற்றையும் நினைவுபடுத்திக் கொள்வேன். Walking stick வைத்திருப்பது நடக்க கூடிய காரியம் அல்ல. பழையகால ஆசை அதன் மேல் உண்டு, அடிக்கடி நான் அதை வாங்குவேன்.

எனது நண்பன் வீட்டுக்கு அல்லது ஒரு ரயில் பயணத்திற்கு பிறகு மற்றொன்றை தொலைத்துவிடுவேன். தொலைத்து விடுவேன் என்ற பயத்தில் குடை எடுத்து செல்வதில்லை. வாழ்வில் குடையை நான் தொலைத்தது இல்லை – குள்ளமான குடையை கூட தொலைத்தது உண்டா?

நம்மில் பலர், ஞாபகம் மறதியால் பல பொருட்களை பயணங்களில் இழந்திருக்கிறோம். சாதாரண மனிதன் சேரவேண்டிய இடத்தை அடையும் போது தன் பையையும் பொருளை பத்திரமாக கொண்டு செல்கிறான். அந்த ஆண்டில் ரயிலில் பொருட்களை தவறவிட்டவர்களின் பட்டியலில் பெரும்பாலானோர் இளைஞர்களே. சாதாரண மனிதனை விட விளையாட்டு வீரனுக்கு ஞாபகமின்மை அதிகமாக உள்ளது.

கிரிக்கெட் பேட், கால்பந்து போன்ற எண்ணிலடங்கா பொருட்களே மறக்கப்பட்டுள்ளன. தெளிவாக புரிந்துகொள்ள, ஆண்கள் விளையாடி விட்டு வீடு திரும்பும் போது விளையாட்டு திடலின் நினைவே இருக்கும் – அவர்கள் தலைவர்கள் நட்சத்திரங்கள் மத்தியிலும் – அவர்கள் சிறந்த செயல் (exploit) மற்றும் குறைகளை நினைத்து பார்ப்பார்கள்.

நினைக்க கூடிய (Abstracted) வகையில் உலகம் அவர்களுக்கு வெளியே இருக்கும். நினைவுகளில் சில மந்தமான (Prosaic) செயல்கள் அவர்களுடன் எடுத்து செல்ல நேரிடும்.

மீதி நாட்களில் அவர்கள் கனவு உலகத்தின் குடியுரிமை கொண்டவர்கள். இதேபோல், சந்தேகமின்றி, மீன்பிடிப்பவர்கள் தூண்டிலை மறப்பார்கள். பொதுவாக மீன் பிடிப்பவரை சொல்வது எதன் அடிப்படையில் நியாயப்படுத்த என தெரியவில்லை.

மனிதர்களிள் அவர்கள்தான் கற்பனையாளர்கள், அம்மனிதன் புதிதாக உருவாக்கும் கற்பனையோடு அவன் வீட்டுக்கு செல்லும் போது அது அவன் குணங்களின் சிறு மறதிமனப்பாங்கு தன்மையை காட்டுகிறது.

எதார்த்ததில் மீன் பிடிப்பதை அவர் மறந்துவிட்டு பிறகு Utopia மீன்பிடிப்பை, அச்சத்தை மீறி கற்பனை செய்கிறார். விளையாட்டின் நினைவுகளை மறப்பது நன்மைதான். அவன் மீன்பிடிப்பை மறக்கலாம். ஒருகவிஞன் தனது கடிதத்தை மறக்கலாம், ஏனெனில் அவர் சிந்தனை முற்றிலும் பெருமைக்குரிய விஷயங்கள் நிறைந்திருக்கும்.

மறதிமனப்பான்மை என்னை பொறுத்தவரை சிறந்த குணம்தான். மறதிமனப்பான்மை கொண்டவனது வாழ்க்கை சிறந்ததாக இருக்கும். சாதாரண (mediocre) விஷயங்கள் நினைவுப்படுத்த அவனுக்கு நேரம் இருக்காது. Socrates அல்லது Coleridge நம்பி கடிதத்தை அனுப்ப சொல்வதற்கு சமம்? அவர்களுக்கு செயலில் ஆர்வம் உள்ளது.

கேள்வி என்னவென்றால் நல்ல நினைவுகளை தக்கவைப்பது நல்லது என்று அடிக்கடி பேசப்பட்டு வருகிறது. மனிதனின் தவறான நினைவுகளில் தான் சிறந்தவன் என தோன்றும். அனைத்தும் நினைவில் வைத்திருக்கும் மனிதன் இயந்திரம். அவன் முதல் அறிவாளி என மதிக்கப்படுவான்.

சில இடங்களில் குழந்தைகள் மற்றும் மனிதரின் சிறந்த நினைவுகளை பேச சிறந்தவன் இல்லை. சிறந்த எழுத்தாளர்கள், இசை உருவாக்குபவர்கள் மொத்தத்தில் மிகுந்த ஆற்றல் கொண்ட நினைவுகள் கொண்டவர்கள் என நான் நினைக்கிறேன். நினைவுகள் தான் அவர்கள் கலையின் பாதி சகாப்தம்.

அடுத்ததாக அரசியல் மேதைகள் முற்றிலும் மோசமான நினைவாற்றால் கொண்டவர்கள். இரண்டு அரசியல் மேதைகளை ஒரே செயலைப்பற்றி பேச செய்தால் என்ன நடக்கும். எடுத்துக்காட்டாக அமைச்சரவைக் கூட்டத்தில் ஒவ்வொருவரும் மற்றொருவர் கதையை உண்மையாக வடித்து (seive) தைரியமாக (grid) உரைப்பார்கள்.

ஒவ்வொரு அரசியல் வாதியின் சுயகுறிப்பு மற்றும் பேச்சு மொழி சவால் நிறைந்ததாக இருக்கும் , இந்த உலகம் இன்னும் சிறந்த அரசியல் வாதியை கொண்டுவரவில்லை. ஒரு சிறந்த கவிஞன் மிகுந்த நினைவாற்றல் மற்றும் புத்திகூர்மை உள்ளவனாக இருக்க வேண்டும்.

அதே நேரத்தில், சிறந்த நினைவாற்றால் கொண்ட மனிதரை மதிக்க வேண்டும். நான் ஒரு அப்பாவை அறிந்தவரை அவர் குழந்தையை (Perambulator) குழந்தைகளுக்கான வண்டியில் வைத்து அதிகாலையில் பொது இடம் ஒன்றுக்கு பீர் அருந்த சென்றார்.

சிறிது நேரம் கழித்து அவரது மனைவி அதே இடத்திற்கு பொருட்களை வாங்க வந்தார். அங்கே அவர் தூங்கிக்கொண்டிருக்கும் அவர் குழந்தையை பார்க்கிறார்.

கணவனின் செயலால் கோபம் (Indignant) கொண்டார். சரியான பாடம் கற்பிக்க நினைத்தாள். அவன் அந்த வண்டியை வீட்டிற்கு கொண்டு சென்றார். அவன் வெளியே வந்து பார்க்கும் போது வண்டி அங்கே இல்லை.

அவன் வீட்டிற்கு சென்றான், கவலையான முகத்துடனும் நடுங்கிய (shivering) உதடுகளுடனும் மனைவி முன் நின்று குழந்தையை திருடிவிட்டார்கள் எனக் கூறினான். அவளுக்கு எப்படி எரிச்சல் (vexation) இருந்திருக்கும். இருந்தும் மதிய உணவின் சில நேரத்திற்கு முன்பு சிரித்தும் சந்தோஷப்படுத்தியும் கேட்டார்.

சரி, என் அன்பே, இன்று மதிய சாப்பாடு என்ன? அனைத்து நிகழ்வுகளையும் (குழந்தை மற்றும் நடந்த நிகழ்வுகளை) மறந்து விட்டு செயல்பட்டாள். எத்தனை ஆண்கள் ஞானிகள் விட குறைந்த மறதி மனப்பான்மை பெற்றிருப்பார்கள்? என்று நினைத்து நானும் பயப்படுகிறேன்.

புத்திசாலித்தனமாக திறமையான நினைவுகளுடன் நாம் பிறந்திருக்கிறோம், அப்படி இல்லை எனில், எந்த ஒரு நவீன நகரத்திலும் குடும்பத்தின் நிறுவனம் உயிர்வாழ முடியாது.