Murali

NCERT Solutions for Class 8 Science Chapter 3 Synthetic Fibres and Plastics are part of NCERT Solutions for Class 8 Science. Here we have given NCERT Solutions for Class 8 Science Chapter 3 Synthetic Fibres and Plastics.

Board	CBSE
Textbook	NCERT
Class	Class 8
Subject	Science
Chapter	Chapter 3
Chapter Name	Synthetic Fibres and Plastics
Number of Questions Solved	15
Category	NCERT Solutions

NCERT Solutions for Class 8 Science Chapter 3 Synthetic Fibres and Plastics

NCERT TEXTBOOK EXERCISES

Question 1.
Explain why some fibres are called synthetic.
Answer.
Since man-made fibres are synthesised from petrochemicals, they are called synthetic fibres.

Question 2.
Mark the correct answer.
Rayon is different from synthetic fibres because
(a) it has a silk-like appearance
(b) it is obtained from wood pulp
(c) its fibres can also be woven like those of natural fibres
Answer.
(b) It is obtained from wood pulp.

Question 3.
Fill in the blanks with appropriate words:

1. Synthetic fibres are also called…………. or……………fibres.
2. Synthetic fibres are synthesised from raw material called………..
3. Like synthetic fibres, plastic is also a …………..

Answer.

1. artificial, man-made.
2. petrochemicals.
3. polymer.

Question 4.
Give examples which indicate that nylon fibres are very strong.
Answer.
They are used to make parachutes, and ropes for rock climbing.

Question 5.
Explain why plastic containers are favoured for storing food.
Answer.
Three main advantages of using plastic containers for storing food are:

1. They do not react with food items.
2. They do not get rusted.
3. They are light, strong and durable.

Question 6.
Explain the difference between thermoplastic and thermosetting plastics.
Answer.
Thermoplastics can be softened on heating and can be bent easily whereas thermosetting plastics cannot be softened on heating and breaks when forced to bend.

Question 7.
Explain why the following are made of thermosetting plastics.
(1) Saucepan handles
(2) Electric plugs Iswitches /plugboards.

Answer.
The above articles are made up of bakelite (a thermosetting plastic) because it is—

1. the bad conductor of heat.
2. poor conductor of electricity.

Question 8.
Categorise the materials of the following products into ‘can be recycled’ and ‘cannot be recycled’:
Telephone instruments, plastic toys, cooker handles, carry bags, ballpoint pens, plastic bowls, plastic covering on electrical wires, plastic chairs, electrical switches.
Answer.

Can be recycled	Cannot be recycled
Toys carry bags, plastic bowls, electric wire covering, plastic chairs.	Telephone instruments, cooker handles, ballpoint pens, electrical switches.

Question 9.
Rana wants to buy shirts for summer. Should he buy cotton shirts or shirts made from synthetic material? Advise Rana, giving your reason.
Answer.
I would advise Rana to buy cotton shirts as cotton shirts absorb the sweat and thus endorse cooling. Besides, they also provide aeration.

Question 10.
Give examples to show that plastics are noncorrosive in nature.
Answer.

1. It does not react with the chemical or other items stored in the containers made of it.
2. It does not get rusted when exposed to moisture and air.
3. It does not decompose when left in open for a long period.

Question 11.
Should the handle and bristles of a toothbrush be made of the same material? Explain your answer.
Answer.
The handles and bristles of a toothbrush should be made of a material which has lightweight good strength and is hygiene. “But the bristles should be soft enough so that ” they do not harm gums while the handle should be quite stiff so that it may not get bent while brushing.

Question 12.
‘Avoid plastics as far as possible.’ Comment on this advice.
Answer.
Since plastic takes several years to decompose, it is not environment friendly. It causes environmental pollution. Besides, when the synthetic material is burnt it takes a long time to get completely burnt. In the process, it releases a lot of poisonous fumes into the atmosphere causing air pollution.

Question 13.
Match the terms of Column A correctly with the phrases given in Column B.

Column A	Column B
(i) Polyester	(a) Prepared by using wood pulp
(ii) Teflon	(b) Used for making parachutes and stockings
(iii) Rayon	(c) Used to make non-stick cookwares
(iv) Nylon	(d) Fabrics do not wrinkle easily

Answer.

Column A	Column B
(i) Polyester	(d) Fabrics do not wrinkle easily
(ii) Teflon	(c) Used to make non-stick cookwares
(iii) Rayon	(a) Prepared by using wood pulp
(iv) Nylon	(b) Used for making parachutes and stockings

Question 14.
‘Manufacturing synthetic fibres are actually helping conservation of forests. ’ Comment.
Answer.
The synthetic fibres are made up of petrochemicals. So, manufacturing synthetic fibres does not depend upon plants. These synthetic fibres cater to the need of people up to great extent. Thus, the forests are not destroyed to manufacture clothes and other items. So, indirectly, we can come to the conclusion that manufacturing synthetic fibres is actually helping the conservation of forests.

Question 15.
Describe an activity to show that thermoplastic is a poor conductor of electricity.
Answer.
A thermoplastic (or plastic) is a poor conductor of electricity as shown in Fig. 3.3.
If we place a piece of copper in the gap touching points A and B (in Fig. 3.3), we observe that bulb glow. If we place plastic in the gap between A and B, we observe that the bulb will not glow. It means plastic is a poor conductor.

We hope the NCERT Solutions for Class 8 Science Chapter 3 Synthetic Fibres and Plastics help you. If you have any query regarding NCERT Solutions for Class 8 Science Chapter 3 Synthetic Fibres and Plastics, drop a comment below and we will get back to you at the earliest

NCERT Solutions for Class 8 Science Chapter 2 Microorganisms: Friend and Foe are part of NCERT Solutions for Class 8 Science. Here we have given NCERT Solutions for Class 8 Science Chapter 2 Microorganisms: Friend and Foe.

Board	CBSE
Textbook	NCERT
Class	Class 8
Subject	Science
Chapter	Chapter 2
Chapter Name	Microorganisms: Friend and Foe
Number of Questions Solved	9
Category	NCERT Solutions

NCERT Solutions for Class 8 Science Chapter 2 Microorganisms: Friend and Foe

NCERT TEXTBOOK EXERCISES

Question 1.
Fill in the blanks.

1. Microorganisms can be seen with the help of a………………
2. Blue-green algae fix………….. directly from the air and enhance the fertility of the soil.
3. Alcohol is produced with the help of……………….
4. Cholera is caused by………………….

Answer.

1. microscope
2. nitrogen
3. yeast
4. bacteria.

Question 2.
Tick the correct answer.
(a) Yeast is used in the production of
(i) sugar
(ii) alcohol
(iii) hydrochloric acid
(iv) oxygen
Answer.
(ii) alcohol

(b) The following is an antibiotic
(i) Sodium bicarbonate
(ii) Streptomycin
(iii) Alcohol
(iv) Yeast
Answer.
(ii) Streptomycin

(c) Carrier of malaria-causing protozoan is
(i) female Anopheles mosquito
(ii) cockroach
(iii) housefly
(iv) butterfly
Answer.
(i) female Anopheles mosquito

(d) The most common carrier of communicable diseases is
(i) ant
(ii) housefly
(iii) dragonfly
(iv) spider
Answer.
(ii) housefly

(e) The bread or idli dough rises because of
(i) heat
(ii) grinding
(iii) growth of yeast cells
(iv) kneading
Answer.
(iii) growth of yeast cells

(f) The process of conversion of sugar into alcohol is called
(i) nitrogen fixation
(ii) moulding
(iii) fermentation
(iv) infection
Answer.
(iii) fermentation.

Question 3.
Match the organisms in Column A with their action in Column B.

Column A	Column B
(i) Bacteria (ii) Rhizobium (iii) Lactobacillus (iv)Yeast (v) A protozoan (vi) A virus	(a) Fixing nitrogen (b) Setting of curd (c) Baking of bread (d) Causing malaria (e) Causing cholera (f) Causing AIDS (g) Producing antibodies

Answer.

Column A	Column B
(i) Bacteria	(e) Causing cholera
(ii) Rhizobium	(a) Fixing nitrogen
(iii) Lactobacillus	(b) Setting of curd
(iv) Yeast	(c) Baking of bread
(v) A protozoan	(d) Causing malaria
(vi) A virus	(f) Causing AIDS

Question 4.
Can microorganisms be seen with the naked eye? If not, how can they be seen?
Answer.
No, we cannot see microorganisms with unaided eyes. They can be seen with the help of a microscope.

Question 5.
What are the major groups of microorganisms?
Answer.
Microorganisms are classified into four major groups based on their size. These are:

1. Bacteria
2. Fungi
3. Protozoa
4. Some algae.

Viruses are another type of microorganisms.

Question 6.
Name the microorganisms which can fix atmospheric nitrogen in the soil.
Answer.
They are

• Rhizobium
• Azobhcter
• blue-green algae (such as Anabaena and Nostoc), etc.

Question 7.
Write 10 lines on the usefulness of microorganisms in our lives.
Answer.
Microorganisms are useful for us in many ways e.g.,

1. the bacterium, lactobacillus converts milk into curd.
2. bacteria are also involved in the making of cheese.
3. Acetobacter acetic is used for the production of acetic acid from alcohol.
4. Yeast is used for the commercial production of alcohol and wine.
5. Antibiotics are manufactured by growing specific microorganisms.
6. Some bacteria fix atmospheric nitrogen and increase soil fertility.
7. Bacteria are used in the preparation of medicines like antibiotics and vaccines.
8. Bacteria are used in the preservation of pickles and many other food items.
9. Yeast is also used in the baking industry for making bread, pastries, and cakes.
10. They act as cleansing agents and decompose the waste products into manure.

Question 8.
Write a short paragraph on the harmful effects of microorganisms.
Answer.
The harms caused by microorganisms are as follows:

1. Many communicable diseases, such as cholera, common cold, chickenpox, tuberculosis, etc., are caused by microorganisms.
2. Malaria is caused by a microorganism called Plasmodium carried by the female Anopheles mosquito.
3. The Female Aedes mosquito acts as a carrier of the dengue virus.
4. Anthrax is a dangerous human and cattle disease caused by a bacterium called Bacillus anthracis.
5. Several microorganisms cause diseases in plants like wheat, rice, potato, sugarcane, orange, apple, etc., and reduce the yield of the crops.
6. Food poisoning is also caused by microorganisms. They make food poisonous by producing toxic substances in the food.

Question 9.
What are antibiotics? What precautions must be taken while taking antibiotics?
Answer.
The medicines, that kill or stop the growth of the disease-causing microorganisms, are called antibiotics. Streptomycin, tetracycline, erythromycin, etc. are some of the commonly known antibiotics which are made from fungi and bacteria.
The precautions to be taken while taking antibiotics are as follows :

1. These medicines should be taken only on the advice of a qualified doctor.
2. One must complete the course prescribed by the doctor.
3. If anybody takes antibiotics when not needed, his/her body may develop resistance against that antibiotic.

We hope the NCERT Solutions for Class 8 Science Chapter 2 Microorganisms: Friend and Foe help you. If you have any query regarding NCERT Solutions for Class 8 Science Chapter 2 Microorganisms: Friend and Foe, drop a comment below and we will get back to you at the earliest

NCERT Solutions for Class 8 Science Chapter 1 Crop Production and Management are part of NCERT Solutions for Class 8 Science. Here we have given NCERT Solutions for Class 8 Science Chapter 1 Crop Production and Management.

Board	CBSE
Textbook	NCERT
Class	Class 8
Subject	Science
Chapter	Chapter 1
Chapter Name	Crop Production and Management
Number of Questions Solved	11
Category	NCERT Solutions

NCERT Solutions for Class 8 Science Chapter 1 Crop Production and Management

NCERT TEXTBOOK EXERCISES

Question 1.
Select the correct word from the following list and fill in the blanks, float, water, crop, nutrients, preparation

(a) The same kind of plants grown and cultivated on a large scale at a place is called __________.
Answer:
Crop

(b) The first step before growing crops is ___________ of the soil.
Answer:
Preparation

(c) Damaged seeds would ___________ on top of water.
Answer:
Float

(d) For growing a crop, sufficient sunlight, __________ and __________ from the soil are essential.
Answer:
water
Nutrients

Question 2.
Match items in column A with those in column B.

Column A	Column B
(i) Kharif crops	(a) Food for cattle
(ii) Rabi crops	(b) Urea and super phosphate
(iii) Chemical fertilisers	(c) Animal excreta, cow dung, urine and plant waste
(iv) Organic manure	(d) Wheat, gram, pea
	(e) Paddy and maize

 

 

 

 

 

Answer.

Column A	Column B
(i) Kharif crops	(e) Paddy and maize
(ii) Rabi crops	(d) Wheat, gram, pea
(iii) Chemical fertilisers	(b) Urea and superphosphate
(iv) Organic manure	(c) Animal excreta, cow dung, urine, and plant waste

 

 

 

 

 

Question 3.
Give two examples of each:

1. Kharif crop
2. Rabi crop

Answer.

1. Paddy, maize
2. Wheat, gram.

Question 4.
Write a paragraph in your own words on each of the following:

1. Preparation of soil
2. Sowing
3. Weeding
4. Threshing

Answer.
(1) Preparation of soil: It is the first step before growing a crop. One of the most important tasks in agriculture is to turn the soil and loosen it. This allows the roots to penetrate deep in the soil. The process of loosening and turning of soil is called tilling or ploughing which is done by a plough. Ploughs are made of wood or iron. The ploughed fields may have big pieces of soil called crumbs. These crumbs are broken and the field is levelled for sowing and for irrigation.

(2) Sowing: It is the process of putting seeds in the soil. For this purpose, good quality seeds are selected which are clean, healthy, of good variety, and give a high yield. Seeds are sown with the help of a traditional funnel-shaped tool or a seed drill. An appropriate distance between the seeds is also important to avoid overcrowding.

(3) Weeding: In a crop field many other undesirable plants may grow naturally along with the crop. These are called weeds. Weeds should be removed to protect the crops. The process of removal of weeds is called weeding. Weeding is necessary because weeds compete with the cultivated plants for space, light, water, and nutrients. Some weeds are poisonous for animals and human beings. Tilling before sowing of crops helps in uprooting and killing of weeds, which may then dry up and get mixed with the soil. The weeds must be removed before they produce flowers and seeds. They are removed either manually or by using weedicides.

(4) Threshing: After harvesting, the crops are threshed for removing grain seeds from the chaff. This is carried out with the help of a thresher or a machine called ‘combine’ which is in fact a combined harvester and thresher.

Question 5.
Explain how fertilisers are different from manure.
Answer.

Fertilisers	Manure
1.  Fertilisers are chemicals which are rich in a particular nutrient like nitrogen, phosphorus, and potassium. 2. Excessive use of fertilisers destroys soil fertility.	1. Manures are decomposed organic matter obtained from plant or animal wastes. 2. The use of manures improves soil texture as well as its water-holding capacity.

Question 6.
What is irrigation? Describe two methods of irrigation which conserve water.
Answer.
The supply of water to crops at different intervals is called irrigation. The two methods of irrigation in which water is conserved are the sprinkler system and the drip system.

(1) Sprinkler system: This system is mostly used on uneven lands where water is available in smaller quantities. In this system, perpendicular pipes having rotating nozzles on the top are joined to the main pipeline at regular intervals. When water is allowed to flow through the main pipe under pressure with the help of a pump, it escapes from the rotating nozzles. It is sprinkled on the crop as if it is raining. It is very useful for sandy soil.

(2) Drip system: In this system, the water falls drop by drop just at the position of the roots. So, it is called a drip system. It is the best technique for watering fruit plants, gardens, trees, etc. This system consists of the main pipe to which lateral pipes are joined. The specially prepared nozzles are attached to these lateral pipes. The nozzles are grounded just near the roots of the plants. It provides water drop by drop to plants. Water is not wasted at all. So, it is a boon in regions where the availability of water is poor.

Question 7.
If wheat is sown in the Kharif season, what would happen? Discuss.
Answer.
The best season for the wheat crops is from November/December to March/April. If it is sown in the Kharif season its production will be decreased considerably.

Question 8.
Explain how soil gets affected by the continuous plantation of crops in a field.
Answer.
The continuous growing of crops makes the soil deficient in certain nutrients. To avoid this, the following practices should be facilitated:

1. Crop rotation
2. Manuring the soil
3. Leaving field fallow

Question 9.
What are weeds? How can we control them?
Answer.
The undesirable plants in the field that grow naturally are called weeds. Weeds must be removed, otherwise, our own crop plants may not get sufficient water, nutrients, space, and light. So, they are removed either by manual method or by using weedicides.

The manual removal includes physical removal of weeds by uprooting or cutting them close to the ground from time to time. This is done with the help of a khurpi or a harrow. By using weedicides also, we can remove weeds. These weedicides damage only weeds and do not harm crops, e.g., 2, 4-D.

Question 10.
Arrange the following boxes in proper order to make a flow chart of sugarcane crop production.

Answer.

Question 11.
Complete the following word puzzle with the help of clues below.
Down
1. Providing water to the crops.
2. Keeping crop grains for a long time under proper conditions.
5. Certain plants of the same kind grown on a large scale.
Across
3. A machine used for cutting the matured crop.
4. A rabi crop that is also one of the pulses.
6. A process of separating the grain from the chaff.
Answer.

We hope the NCERT Solutions for Class 8 Science Chapter 1 Crop Production and Management help you. If you have any query regarding NCERT Solutions for Class 8 Science Chapter 1 Crop Production and Management, drop a comment below and we will get back to you at the earliest.

In Online Education NCERT Solutions for Class 6 English Honeysuckle Poem Chapter 6 The Wonderful Words are part of NCERT Solutions for Class 6 English. Here we have given NCERT Solutions for Class 6 English Honeysuckle Poem Chapter 6 The Wonderful Words.

Board	CBSE
Textbook	NCERT
Class	Class 6
Subject	English Honeysuckle Poem
Chapter	Chapter 6
Chapter Name	The Wonderful Words
Number of Questions Solved	22
Category	NCERT Solutions

Online Education NCERT Solutions for Class 6 English Honeysuckle Poem Chapter 6 The Wonderful Words

Stanzas For Comprehension

Read the following extracts and answer the questions that follow choosing the best option from among the given ones.
1.
Never let a thought shrivel and die
For want of a way to say it
For English is a wonderful game
And all of you can play it
All that you do is match the words
To the brightest thoughts in your head

Paraphrase :
Do not let a thought die unexpressed for want of words. Expression is a funny game in English. Everyone can play it. All that you have to do is to find proper words to match your best thoughts.

Multiple Choice Questions.
1.
The thoughts die when

(a) they are bad
(b) they are good
(c) they are beautiful
(d) they are not expressed

2.
To give expression to the thoughts one has to

(a) work hard
(b) think well
(c) find proper words
(d) study a lot

3.
The poet feels that English has enough words

(a) to express every idea
(b) for us to learn
(c) to confuse us
(d) to write any book

4.
English is a game which is

(a) like cricket
(b) like playing cards
(c) for all persons
(d) for those who love to express themselves

5.
The adverb form of ‘wonderful’ is

(a) wonder
(b) wonderfully
(c) wondered
(d) wondering

Answers :

1. (d) they are not expressed
2. (c) find proper words
3. (a) to express every idea
4. (d) for those who love to express themselves
5. (b) wonderfully

2.
So that they come out clear and true
And handsomely groomed and fed-
For many of the loveliest things
Have never yet been said.
Word-Notes :
True-exact, सही-सहीHandsomely-beautifully, सुंदरता से। Groomed-decorated, सजा हुआ।

हिन्दी अनुवाद :
ताकि वे (विचार) स्पष्ट और सही तरह अभिव्यक्त हों और उन्हें सुंदरता से सजाया और विकसित किया जा सके। क्योंकि बहुत से सुन्दरतम विचार अभी तक अभिव्यक्ति नहीं पा सके हैं।

Paraphrase :
(Work to express your best thoughts) so that they are expressed beautifully decorated and developed. The truth is that many of the most beautiful things are still waiting for an expression.

Multiple Choice Questions

1.
The passage is taken from

(a) The Wonderful Words
(b) Beauty
(c) A House, A Home
(d) The Kite

2.
The author of the poem is

(a) L.M. Halli
(b) Mary O’ Neill
(c) Peter Dixon
(d) Shure

3.
The poem is about

(a) a groom
(b) a girl
(c) words
(d) beauty

4.
The ‘loveliest things’ are

(a) money
(b) fame
(c) beauty
(d) great ideas

5.
The noun form of ‘fed’ is

(a) feed
(b) feeding
(c) food
(d) feeling

Answers :

1. (a) The wonderful words
2. (b) Mary O’ Neill
3. (c) words
4. (d) great ideas
5. (c) food

3.
Words are the food and dress of thought
They give it its body and swing
And everyone’s longing today to hear
Some fresh and beautiful thing ;
But only words can free a thought
From its prison behind your eyes
May be your mind is holding now
A marvellous new surprise !
Word-Notes :
Swing-rhythm/progression/impetus/freedom, लय/ गति/ शक्ति/ स्वतंत्रा Longing-desiring, इछुक। Holding-keeping, रखे हुए। Marvellous wonderful, शानदार।

हिन्दी अनुवाद :
शब्द विचारों के शरीर और वस्त्र हैं। वे उन्हें उनका रूप, शक्ति और स्वतंत्रता प्रदान करते हैं। और आज हर व्यक्ति कोई ताजी और सुन्दर बात सुनने की कामना करता है, पर केवल शब्द ही तुम्हारी आँखों के पीछे छिपे कारागार से उन्हें मुक्ति दिला सकते हैं, हो सकता है इस समय भी तुम्हारे मस्तिष्क में कोई शानदार नया विचार सुरक्षित है।

Paraphrase :
Thoughts survive on words. They are their food and clothing. They give them the body and force. Everybody wants to hear some new and beautiful things today. Only the words can free a thought imprisoned in the mind. May be your mind is even now holding back a new and beautiful thought.

Multiple Choice Questions.
1.
If the words are the body, the thought is its

(a) dress
(b) food
(c) soul
(d) swing

2.
According to the poet, everyone wants to hear

(a) good music
(b) fine words
(c) a new poem
(d) new and noble thoughts

3.
The words can free a thought which is

(a) in a prison
(b) in the mind
(c) in the eyes
(d) nowhere

4.
the real beauty is in

(a) new and beautiful ideas
(b) beautiful words
(c) beautiful expression
(d) surprising words

5.
The word longing is a

(a) gerund
(b) verb
(c) noun
(d) adjective

Answers :

1. (c) soul
2. (d) new and noble thoughts
3. (b) in the mind
4. (a) new and beautiful ideas
5. (c) noun

Textual Questions

Question 1.
With your partner, complete the following sentences in your own words using the ideas in the poem.

1. Do not let a thought shrivel and die because ________ .
2. English is a ________ with words that everyone can play.
3. One has to match ________ .
4. Words are the ________ of thought.

Answer :

1. it may be a ‘marvellous new surprise’ which everyone is longing to hear.
2. wonderful game of matching thoughts.
3. the words to the brightest thought in the mind.
4. food and dress of thought.

Question 2.
In groups of four discuss the following lines and their meanings.

1. All that you do is match the words To the brightest thoughts in your head
2. For many of the loveliest things Haye never yet been said
3. And everyone’s longing today to hear Some fresh and beautiful thing
4. But only words can free a thought From its prison behind your eyes

Answer :

1. You have only to find the nearest possible words which seem to trans¬late your best thoughts.
2. Many of the most beautiful ideas are still waiting for proper expres¬sion.
3. Today everyone is very eager to hear something new and beautiful.
4. Words are imprisoned in the mind. They will remain so until and un¬less you find proper words to give them expression. This alone will give them freedom from the prison of mind.

We hope the NCERT Solutions for Class 6 English Honeysuckle Poem Chapter 6 The Wonderful Words help you. If you have any query regarding NCERT Solutions for Class 6 English Honeysuckle Poem Chapter 6 The Wonderful Words, drop a comment below and we will get back to you at the earliest

NCERT Solutions for Class 6 Maths Chapter 12 Ratio and Proportion Ex 12.1 are part of NCERT Solutions for Class 6 Maths. Here we have given NCERT Solutions for Class 6 Maths Chapter 12 Ratio and Proportion Ex 12.1.

• Ratio and Proportion Class 6 Ex 12.2
• Ratio and Proportion Class 6 Ex 12.3

Board	CBSE
Textbook	NCERT
Class	Class 6
Subject	Maths
Chapter	Chapter 12
Chapter Name	Ratio and Proportion
Exercise	Ex 12.1
Number of Questions Solved	16
Category	NCERT Solutions

NCERT Solutions for Class 6 Maths Chapter 12 Ratio and Proportion Ex 12.1

Question 1.
There are 20 girls and 15 boys in a class.
(a) What is the ratio of a number of girls to the number of boys?
(b) What is the ratio of a number of girls to the total number of students in the class?
Solution :
(a) Ratio of number of girls to the number of boys

(b) Total number of students in the class = 20 + 15 = 35
∴ Ratio of number of girls to the total number of students in the class

Question 2.
Out of 30 students in a class, 6 like football, 12 like cricket and remaining like tennis. Find the ratio of:
(a) Number of students liking football to a number of students liking tennis.
(b) Number of students liking cricket to the total number of students.
Solution :
(a) Number of students liking tennis = 30-(6+ 12) = 30 – 18 = 12
∴ Ratio of number of students liking football to number of students liking tennis

(b) Number of students liking cricket to total number of students

Question 3.
See the figure and find the ratio of:
(a) Number of triangles to the number of circles inside the rectangle.

(b) Number of squares to all the figures inside the rectangle.
(c) Number of circles to all the figures inside the rectangle.
Solution :
(a) Number of triangle inside the rectangle = 3 . , Number of circles inside the rectangle = 2 Ratio of number of triangles to the number of circles inside the rectangle = \(\frac{ 3 }{ 2 }\) =3:2.
(b) Number of squares inside the rectangle = 2 Number of all the figures inside the rectangle = 7
∴ Ratio of number of squares to all the figures 2
inside the rectangle = \(\frac{ 2 }{ 7 }\) =2:7.
(c) Ratio of number of circles to all the figures inside the rectangle = \(\frac{ 2 }{ 7 }\) =2:1.

Question 4.
Distance travelled by Hamid and Akhtar in an hour are 9 km and 12 km. Find the ratio of speed of Hamid to the speed of Akhtar.
Solution :
Ratio of speed of Hamid to the speed of

Question 5.
Fill in the following blanks:

Solution :

Question 6.
Find the ratio of the following :
(a) 81 to 108
(b) 98 to 63
(c) 33 km to 121 km
(d) 30 minutes to 45 minutes.
Solution :

Question 7.
Find the ratio of the following :
(a) 30 minutes to 1.5 hours
(b) 40 cm to 1.5 m
(c) 55 paise to ₹ 1
(d) 500 ml to 2 litres.
Solution :
(a) 1.5 hours = 1.5 × 60 minutes = 90 minutes

Question 8.
In a year, Seema earns- ₹ 1,50.000 and saves ₹ 50,000. Find the ratio of:
(a) Money that Seema earns to the money she saves.
(b) Money that she saves to the money that she spends.
Solution :
(a) Ratio of money that Seema earns to the money she saves

(b) Money that she spends
= ₹ 1,50,000 – ₹ 50,000 = ₹ 1,00,000
∴ Ratio of money she saves to the money she spends

Question 9.
There are 102 teachers in a school of 3300 students. Find the ratio of the number of teachers to the number of students.
Solution :
Ratio of the number of teachers to the number of students

Question 10.
In a college, out of4320 students, 2300 are girls. Find the ratio of:
(a) Number of girls to the total number of students.
(b) Number of boys to the number of girls.
(c) Number of boys to the total number of students.
Solution :
(a) Ratio of number of girls to the total number of students

(b) Number of boys = 4320 – 2300 = 2020
∴ Ratio of number of boys to the number of girl

(c) Ratio of number of boys to the total number of students

Question 11.
Out of 1800 students in a school, 750 opted basketball, 800 opted cricket and remaining opted table tennis. If a student can opt only one game, find the ratio of:
(a) Number of students who opted basketball to the number of students who opted table tennis.
(b) Number of students who opted cricket to the number of students opting basketball.
(c) Number of students who opted basketball to the total number of students.
Solution :
(a) Number of students opting table tennis = 1800 – (750 + 800) = 1800 -1550 = 250
∴ Ratio of number of students opting basketball to number of students opting table tennis

(b) Ratio of number of students opting cricket to the number of students opting basketball

(c) Ratio of number of students opting basketball to the total number of students

Question 12.
Cost of a dozen pens is ₹ 180 and cost of 8 ball pens is ₹ 56. Find the ratio of cost of a pen to the cost of a ball pen.
Solution :
1 dozen =12 items
∴ Cost of 12 pens = ₹ 180
∴ Cost of 1 pen = ₹ \(\frac{ 180 }{ 12 }\) = ₹ 15
∴ Cost of 8 ball pens = ₹ 56
∴ Cost of 1 ball pen = ₹ \(\frac{ 56 }{ 8 }\) = ₹ 7
∴ Ratio of cost of a pen to the cost of a ball 15 pen = \(\frac{ 15 }{ 7 }\) =15:7.

Question 13.
Consider the statement: Ratio of breadth and length of a hall 2: 5. Complete the following table that shows some possible breadths and lengths of the hall.

Solution :

Hence, the completed table is as follows:

Question 14.
Divide 20 pens between Sheela and Sangeeta in the ratio 3: 2.
Solution :
Total number of pens = 20 Ratio = 3:2
Sum of the parts = 3 + 2 = 5
∴ Sheel’s shae = \(\frac{ 3 }{ 5 }\) × 20 = 12 and, Samgeeta’s share = \(\frac{ 2 }{ 5 }\) × 20 = 8

Question 15.
Mother wants to divide ₹ 36 between her daughters Shreya and Bhoomika in the ratio of their ages. If the age of Shreya is 15 years and age of Bhoomika is 12 years, find how much Shreya and Bhoomika will get?
Solution :
Total money = ₹ 36
The ratio of the ages of Shreya and Bhoomika



Hence, Shreya will get ₹ 20 and Bhoomika will get ₹ 16.

Question 16.
Present age of the father is 42 years and that of his son is 14 years. Find the ratio of :
(a) Present age of father to the present age of the son.
(b) Age of the father to the age of son, when the son was 12 years old.
(c) Age of father after 10 years to the age of son after 10 years.
(d) Age of father to the age of son when father was 30 years old.
Solution :
(a) Ratio of the present age of father to the present age of the son

(b) Son was 12 years old 14 – 12 = 2 years before
Age of the father 2 years before = 42 – 2 = 40 years
∴ Ratio of the age of the father to the age of the son. when son was 12 years old

(c) Age of father after 10 years = 42 + 10 = 52 years
Age of son after 10 years = 14 + 10 = 24 years
∴ Ratio of age of father after 10 years to the age of son after 10 years

(d) Father was 30 years old 42 – 30 = 12 years before
Age of son 12 years before = 14 – 12 = 2 years
∴ Ratio of the age of father to the age of son when father was 30 years old

 

We hope the NCERT Solutions for Class 6 Maths Chapter 12 Ratio and Proportion Ex 12.1 help you. If you have any query regarding NCERT Solutions for Class 6 Maths Chapter 12 Ratio and Proportion Ex 12.1, drop a comment below and we will get back to you at the earliest.

NCERT Exemplar Solutions for Class 12 Biology chapter 13 Organisms and Populations

These Solutions are part of NCERT Exemplar Solutions for Class 12 Biology. Here we have given NCERT Exemplar Solutions for Class 12 Biology chapter 13 Organisms and Populations

Multiple Choice Questions

Question 1.
Autecology is the
(a) relation of heterogenous population to its environment
(b) relation of an individual to its environ-ment
(c) relation of a community to its environ-ment
(d) relation of a biome to its environment.
Answer:
(a, b) : Ecology is the study of interactions among organisms and between the organisms and their biotic and abiotic environment. The study of reciprocal relationships between every stage of development of a population/ species/individual and its environment is called autecology.

Question 2.
Ecotone is
(a) a polluted area
(b) the bottom of a lake
(c) a zone of transition between two communi¬ties
(d) a zone of developing community.
Answer:
(c) : The adjacent biotic communities do not always have sharp lines of demarcation between them. There are usually transition zones, the ecotones, between them. An ecotone often has some populations from each adjacent community and some characteristic of itself. The total number of species in ecotone is generally greater than in the adjoining communities, a phenomenon called edge effect.

Question 3.
Biosphere is
(a) a component in the ecosystem.
(b) composed of the plants present in the soil
(c) life in the outer space
(d) composed of all living organisms present on earth which interact with the physical environment.
Answer:
(d) : Biosphere or ecosphere is self-sufficient system. It includes the earth’s atmosphere (air, land, water) that sustains life. In biosphere living organisms interact with their physical environment.

Question 4.
Ecological niche is
(a) the surface area of the ocean
(b) an ecologically adapted zone
(c) the physical position and functional role of a species within the community
(d) formed of all plants and animals living at the bottom of a lake.
Answer:
(c) : Ecological niche is specific part of habitat occupied by individuals of a species which is circumscribed by its range of tolerance, range of movement, microclimate, type of food and its availability, shelter, type of predator and timing of activity. Tadpole and adult frog occupy different ecological niches as the former is herbivorous aquatic while the latter is carnivorous amphibian.

Question 5.
According to Allen’s Rule, the mammals from colder climates have
(a) shorter ears and longer limbs
(b) longer ears and shorter limbs
(c) longer ears and longer limbs
(d) shorter ears and shorter limbs.
Answer:
(d) : According to Allen’s Rule, animals in colder climates generally have smaller extremities like shorter ears and limbs. It is a mechanism to conserve heat by checking heat loss from the body. Heat loss can be minimised by decreasing the surface area to volume ratio of body. In the polar regions, small animals are rarely found because of their high surface area and less volume.

Question 6.
Salt concentration (salinity) of the sea measured in parts per thousand is
(a) 10-5
(b) 30-70
(c) 0-5
(d) 30-35
Answer:
(d) : Salinity of water bodies is generally measured in parts per thousand. It determines what kind of organisms can live in it. Salinity of the sea is 30-35 parts per thousand, while for inland waters and some hypersaline lagoons it is less than 5 and more than 100 per thousand parts, respectively. Fresh water animals generally cannot. live for long in sea water and vice versa because of osmotic problems.

Question 7.
Formation of tropical forests needs mean annual temperature and mean annual precipitation as
(a) 18-25°C and 150-400 cm
(b) 5-15°C and 50-100 cm
(c) 30-50°C and 100-150 cm
(d) 5-15°C and 100-200 cm.
Answer:
(a) : Temperature and precipitation are generally most important climatic abiotic factors that influence the geographical distribution of plants. Average temperature decreases from equator to the poles. Precipitation includes rainfall, snow, dew, etc. Tropical forests, which are very productive, need 18 to 25°C mean annual temperature and 150-400 cm mean annual rainfall.

Question 8.
Which of the following forest plants controls the light conditions at the ground?
(a) Lianas and climbers
(b) Shrubs
(c) Tall trees
(d) Herbs
Answer:
(c) : In a forest, plants get arranged in various strata according to their shade tolerance. Tall trees form the canopy of the forest i.e., roof of the forest, thus, controlling the amount of light reaching the ground.

Question 9.
What will happen to a well growing herbaceous plant in the forest if it is transplanted outside the forest in a park?
(a) It will grow normally.
(b) It will grow well because it is planted in the same locality.
(c) It may not survive because of change in its microclimate.
(d) It grows very well because the plant gets more sunlight.
Answer:
(c) : Herbaceous plants are of small height. Since, this herbaceous plant is growing in forest, it must have been adapted to the light intensity reaching it, moisture in surrounding air and soil, soil characteristics, etc. This constitutes the microclimate of this plant. If this plant is transplanted to a park outside forest, this microclimate might get lost, partially or completely, because of which plant might not be able to survive.

Question 10.
If a population of 50 Paramecium present in pool increases to 150 after an hour, what would be the growth rate of population?
(a) 50 per hour
(b) 200 per hour
(c) 5 per hour
(d) 100 per hour
Answer:
(d). Population of Paramecium at time t = 50
Population of Paramecium 1 hour after t= 150
Growth rate of population = 150 – 50 per
hour = 100 per hour

Question 11.
What would be the percent growth or birth rate per individual per hour for the same population mentioned in the previous question (Question 10)?
(a) 100
(b) 200
(c) 50
(d) 150
Answer:
(b) : Initial number of Paramecium = 50
Number of Paramecium after 1 hour = 150 Birth rate

Question 12.
A population has more young individuals compared to the older individuals. What would be the status of the population after some years?
(a) It will decline.
(b) It will stabilise.
(c) It will increase.
(d) It will first decline and then stabilise.
Answer:
(c) : If in a population more young individuals are present as compared to older individuals, population will increase after some years. This is because of the reason that number of individuals in pre-reproductive age is high and thus, more number of individuals will enter the reproductive age in coming years. Such a population can be represented by an age structure forming an upright pyramid.

Question 13.
What parameters are used for tiger census in our country’s national parks and sanctuaries?
(a) Pug marks only
(b) Pug marks and faecal pellets
(c) Faecal pellets only
(d) Actual head counts
Answer:
(b) : In our country, parameters for tiger census in national parks and sanctuaries include both pug marks and faecal pellets. They can be used to determine the number as well as distribution of tiger population. New techniques of excreta DNA analysis and camera trappings are considered more accurate for tiger census.

Question 14.
Which of the following would necessarily decrease the density of a population in a given habitat?
(a) Natality > mortality
(b) Immigration > emigration
(c) Mortality and emigration
(d) Natality and immigration
Answer:
(c) : Population density in a given habitat is influenced by four processes :

• Natality : It is the number of births in a population during a given period of time.
• Mortality : It is the number of deaths in a population during a given period of time.
• Immigration : Number of individuals coming from other habitats in a given period of time
• Emigration : Number of individuals exiting from a given pop-.Tation in a given period of time.
• Mortality and emigration will, therefore decrease the population density while natality and immigration will increase it.
  

Question 15.
A protozoan reproduces by binary fission. What will be the number of protozoans in its population after six generations?
(a) 128
(b) 24
(c) 64
(d) 32
Answer:
(c) : By binary fission, an individual protozoan will divide in two. In this way after six generations their number will be 64.

Question 16.
In 2005, for each of the 14 million people present in a country, 0.028 were born and 0. 008 died during the year. Using exponential equation, the number of people present in 2015 is predicted as
(a) 25 millions
(b) 17 millions
(c) 20 millions
(d) 18 millions.
Answer:

Question 17.
Amensalism is an association between two species where
(a) one species is harmed and other is benefited
(b) one species is harmed and other is unaffected
(c) one species is benefited and other is , unaffected
(d) both the species are harmed.
Answer:
(b) : Amensalism is the interaction between two populations in which oneis adversely affected, whereas other is apparently, neither harmed nor benefited. For example, Penicilliitm does not allow the growth of Staphylococcus bacterium by secreting certain chemicals.

Question 18.
Lichens are the associations of
(a) bacteria and fungus
(b) algae and bacterium
(c) fungus and algae
(d) fungus and virus.
Answer:
(c) : Lichens are association (mutually beneficial) between fungus and alga. The fungal partner is mycobiont and algal partner is phycobiont. Lichens can grow in extremely inhospitable conditions. In many ecosystems they are the pioneer species. The role of mycobiont is to provide body structure and anchorage and absorption of minerals and water. The role of phycobiont is to manufacture food through photosynthesis for itself and also for fungus.

Question 19.
Which of the following is a partial root parasite?
(a) Sandal wood
(b) Mistletoe
(c) Orobanche
(d) Ganoderma
Answer:
(a) : Parasites can be divided into : holo- parasitesandhemiparasites. Holoparasitesare those which are dependent on their host for all of their requirements, while hemiparasites are those, which receive only a part of their nourishment from host. Holoparasites and hemiparasites are also known as complete and partial parasites, respectively. Sandal wood is partial root parasite, which synthesises its own food but is dependent on host’s roots for water and inorganic nutrients. Mistletoe is partial stem parasite. Orobanche (Broomrope) is complete root parasite. Ganoderma, a fungus, is parasitic on hardwood.

Question 20.
Which one of the following organisms reproduces sexually only once in its lifetime?
(a) Banana plant
(b) Mango
(c) Tomato
(d) Eucalyptus
Answer:
(c) : Since tomato is an annual plant, so it reproduces sexually only once in its life time.

Very Short Answer Type Questions

Question 1.
Species that can tolerate narrow range of temperature are called …………
Answer:
stenothermal organisms

Question 2.
What are eurythermic species?
Answer:
Eurythermal organisms are those which can tolerate wide range of temperature variations, e.g., most mammals and birds.

Question 3.
Species that can tolerate wide range of salinity are called
Answer:
eurvhaline species

Question 4.
Define stenohaline species.
Answer:
Species which can tolerate only a narrow range of salinity are called stenohaline species e.g., sharks.

Question 5.
What is the interaction between two species called?
Answer:
The interaction between two species is called interspecific interaction, it can be positive, negative or neutral.

Question 6.
What is commensalism?
Answer:
Commensalism isanassociation between two organisms, in which one is benefited and second is neither harmed nor benefited. E.g., interaction between sucker fish and shark.

Question 7.
Name the association in which one species produces poisonous substance or a change in environmental conditions that is harmful to another species.
Answer:
In amensalism one organism inhibits the growth of the other. This inhibition is done by secreting chemicals called allochemics.

Question 8.
What is mycorrhiza?
Answer:
Symbiotic association between fungus and the roots of higher plants is called mycorrhiza. The fungus helps the plant in absorption of essential nutrients from the soil, while the plant in turn provides the fungus with energy yielding carbohydrates.

Question 9.
Emergent land plants that can tolerate the salinities of the sea are called .
Answer:
halophytes

Question 10.
Why do high altitude areas have brighter sunlight and lower temperatures as compared to the plains?
Answer:
Brighter sunlight at high altitudes is due to thinner air and less pollution. High altitudes have lower temperature because as we go higher, air pressure decreases that causes the temperature to be colder at high altitudes.

Question 11.
What is homeostasis?
Answer:
Homeostasis is the ability to maintain constant body temperature and osmotic concentration irrespective of the environmental conditions.

Question 12.
Define aestivation.
Answer:
The period of dormancy or suspended metabolism during summer months to escape from heat and unfavourable conditions is called aestivation or summer sleep, c.g., snails undergo aestivation to avoid heat and dessication.

Question 13.
What is diapause and its significance?
Answer:
Diapause is the stage of suspended development. During stress period, many zooplanktons and larvae of insects suspend their development and enter a stage of dormancy. They resume their growth during favourable conditions. This helps organisms to escape harsh, extreme conditions.

Question 14.
What would be the growth rate pattern, when the resources are unlimited?
Answer:
When resources like food, space, etc., are unlimited for a population, it grows in an exponential or geometric ratio resulting in a J-shaped growth curve, c.g., growth of algal bloom.

Question 15.
What are the organisms that feed on plant sap and other plant parts called?
Answer:
Organisms feeding on plant sap and other plant parts are called phytophagous.

Question 16.
What is high altitude sickness? Write its symptoms.
Answer:
High altitude sickness is experienced at high altitudes (> 3500 m) like Leh, Rohtang Pass, etc. It is due to low atmospheric pressure at high altitude due to which the body does not get enough oxygen. Its symptoms include nausea, fatigue, heart palpitation.

Question 17.
Give a suitable example for commensalism.
Answer:
The cattle egret and grazing cattle show an example of commensalism. The egrets always forage close to where cattle are grazing, because the cattle, as they move, stir up and flush out insects from the vegetation which are otherwise difficult for egrets to find and catch.

Question 18.
Define ectoparasite and endoparasite and give suitable examples.
Answer:
Ectoparasites live on the surface of the host. They suck blood (in animals), or juice (in plants), c.g., lice on humans and ticks on dogs.

Endoparasites live inside the body of host Most of them spend a part of their life cycle in another host. Their life cycle is more complex because of their extreme specialisation, c.g., Plasmodium in RBCs and Taenia solium in intestine.

Question 19.
What is brood parasitism? Explain with the help of an example.
Answer:
Brood parasitism is an example of parasitism in which the parasitic bird lays its eggs in the nest of its host and the host incubates them. During the course of evolution, the eggs of the parasitic bird have evolved to resemble the host’s eggs in size and colour to reduce the chances of the host bird detecting the foreign eggs and ejecting them from the nest. Most common example is koel whose female lays its eggs in the crow’s nest.

Short Answer Type Questions

Question 1.
Why are coral reefs not found in the regions from West Bengal to Andhra Pradesh but are found in Tamil Nadu and on the east coast of India?
Answer:
Coral reefs are found in zone with high salt concentration (salinity), optimal temperature and with a less siltation condition which fairly facilitate the corals to colonise. In case of high siltation and very high freshwater inflow, coral reefs do not colonise.

Question 2.
If a freshwater fish is placed in an aquarium containing sea water, will the fish be able to survive? Explain giving reasons.
Answer:
freshwater fish is placed in aquarium having sea water, then it would not be able to survive due to osmoregulatory problems. Outer hypertonic environment having high salt concentration would cause exosmosis and fish would die.

Question 3.
Why do all the freshwater organisms have contractile vacuoles whereas majority of marine organisms lack them?
Answer:
Osmotic concentration of freshwater organisms is hypertonic to the surrounding water. The water enters their body by endosmosis. Excess water is expelled out of the body through contractile vacuole. Osmotic concentration of most of the marine organisms is almost equal to surrounding water, i.e., isotonic to surroundings, therefore, they need pot expel extra water out of their body and hence lack contractile vacuole.

Question 4.
Define heliophytes and sciophytes. Name a plant from your locality that is either heliophyte or sciophyte.
Answer:
Plants adapted to bright light are known as heliophytes and are commonly called sun plants. E.g., sunflower.
Plants growing in partial shade or low intensity of light are called sciophytes or shade plants. E.g., money plant.

Question 5.
Why do submerged plants receive weaker illumination than exposed floating plants in a lake?
Answer:
There is a light zonation in deep lakes and oceans. Floating plants are found in upper part of lake and receive more light, however less light penetrates into deep lake and thus submerged plants receive weaker illumination of light.

Question 6.
In a sea shore, the benthic animals live in sandy, muddy and rocky substrata and accordingly developed the following adaptations.
(a) Burrowing
(b) Building cubes
(c) Holdfasts/peduncle
Find the suitable substratum against each adaptation.
Answer:
ln a sea shore, water current restrict distribution of organisms. In streamed areas of ocean, animals are strong swimmers or possess attaching organs such as pedunde, or live under stone, in burrows etc. Burrowing animals like tubeworm, Nereis are strong swimmers. Burrowing, building cubes and hold fast or penduncle are found in sandy muddy and rocky substratum respectively.

Question 7.
Categorise the following plants into hydrophytes, halophytes, mesophytes and xerophytes. Give reasons for your answer.
(a) Salvinia
(b) Opuntia
(c) Rhizophora
(d) Mangifera
Answer:
(a) Salvinia – Hydrophyte – Free floating aquatic plant
(b) Opuntia – Xerophyte – Grows in hot, dry conditions with less rainfall.
(c) Rhizophora – Halophyte – Grows in saline soil.
(d) Mangifera – Mesophyte – Grows in mesic (moderate) conditions with high moisture.

Question 8.
In a pond, we see plants which are free-floating; rooted-submerged; rooted emergent; rooted with floating leaves. Write the type of plants against each of them.”

Answer:

Question 9.
The density of a population in a habitat per unit area is measured in different units. Write the unit of measurement against the following:
(a) Bacteria …………
(b) Banyan …………
(c) Deer ………….
(d) Fish …………..
Answer:
(a) Bacteria – Number per unit volume
(b) Banyan tree – Biomass per unit area
(c) Deer – Number per unit area
(d) Fish – Weight per unit area

Question 10.

(a) Label the three tiers 1, 2, 3 given in the above age pyramid.
(b) What type of population growth is represented by the above age pyramid?
Answer:
(a)
1 – Pre-reproductive individuals
2 – Reproductive individuals
3 – Post-reproductive individuals

(b) Given age pyramid has broad base or triangular structure which indicates a rapidly expanding population with a high percentage of young individuals and only few old individuals.

Question 11.
In an association of two animal species, one is a termite which feeds on wood and the other is a protozoan Trichonympha present in the gut of the termite. What type of association they establish?
Answer:
Termites and Trichonympha show mutualistic relationship. Termites feed on wood though they do not possess enzymes for digesting the same. Termites harbour cellulose digesting flagellates (e.g., Trichonympha campanula) for this purpose. Flagellates are unable to live independently. Termites would die of starvation in the absence of flagellates.

Question 12.
Lianas are vascular plants rooted in the ground and maintain erectness of their stem by making use of other trees for support. They do not maintain direct relation with those trees. Discuss the type of association the lianas have with the trees.
Answer:
Lianas are woody vascular climbers. They maintain their erectness of stem by using the support of the tree without any direct relation with it. In this relation, lianas are benefited, but the tree neither gets benefit nor is harmed. It represents commensalism, in which one is benefited and other remains unaffected.

Question 13.
Give the scientific names of any two microorganisms inhabiting the human intestine.
Answer:
Escherichia coli and Lactobacillus are two microorganisms which inhabit human intestine.

Question 14.
What is a tree line?
Answer:
The tree line is the edge of the habitat till which trees are capable of growing. Beyond the tree line, trees cannot tolerate the environmental conditions, like, temperature, humidity. Beyond this limit trees are not found.

Question 15.
Define ‘zero population growth rate’. Draw a age pyramid for the same.
Answer:
When birth rate and death rate become equal and there is no increase in population, it is called zero population growth. It is obtained when the number of pre-reproductive and reproductive individuals is almost equal and post-reproductive individuals are comparatively

Question 16.
List any four characters that are employed in human population census
Answer:
Census is an official counting of population and preparing data considering various population characteristics, such as :

1. Birth rate
2. Sex ratio
3. Death rate
4. Age distribution

Question 17.
Give one example for each of the following types.
(a) Migratory animal
(b) Camouflaged animal
(c) Predator animal
(d) Biological control agent
(e) Phytophagous animal
(f) Chemical defence agent
Answer:
(a) Migratory animal – Arctic tern
(b) Camouflaged animal – Leaf insect
(c) Predator animal – Lion
(d) Biological control agent – Gambusia
(e) Phytophagous – Butterfly
(f) Chemical defense agent – Cardiac glycosides produced by Calotropis

18.
Fill in the blanks:

Species  A	Species B	Type of Interaction	Example
+	–
+	+
+		Commensalism

Answer:

Species A	Species B	Type of Interaction	Example
+	–	Predation	Tiger and Deer
+	+	Mutualism	Lichen
+	O	Commensalism	Sucker fish and shark

Question 19.
Observe the set of 4 figures A, B, C and D and, answer the following questions.

1. Which one of the figures shows mutualism?
2. What kind of association is shown in D?
3. Name the organisms and the association in
4. What role is the insect performing in B?
   

Answer:

1. Fig. A shows mutualism, in plant- animal relationship. Plants need the help of animals for pollinating their flowers. Animals get rewards in the form of pollen and nectar.
2. Fig. D shows predation where, tiger is the predator and deer is a prey.
3. Fig. C shows commensalism between the cattle egret and grazing cattle. The cattle egret and grazing cattle show an example of commensalism. The egrets always forage close to where cattle are grazing, because the cattle, as they move, stir up and flush out insects from the vegetation which are otherwise difficult for egrets to find and catch.
4. In fig. B, insect is scavenging, i.c., it is ‘ feeding on dead and decaying matter.

Long Answer Type Questions

Question 1.
Comment on the following figures 1,2 and 3. A, B, C, D, G, P, Q, R and S are species.
Answer:
In fig. 1, all the individuals are of same species and show intraspecific competition. Here, all the members of species have similar requirements of food, light, water, space, shelter and mate.
In Fig. 2, there are individuals of three different species (A, B and C) and have interspecific competition. Here two or more populations usually belonging to same trophic level or feeding habit compete with one another for natural resources. E.g., tigers and leopards may compete for same prey in forest, trees, shrubs, herbs and vines compete for water, sunlight, nutrients, pollinators, etc.
In Fig. 3, three different biotic communities interact with one another and with their physical environment and exchange energy and material with surroundings. It represents a biome.

Question 2.
An individual and a population has certain characteristics. Name these attributes with definitions.
Answer:
An individual or an organism is the smallest unit of ecological hierarchy.Characteristics of an individual are :

1. It performs all body functions indepen-dently, but cannot live in isolation.
2.  It depends on both biotic and abiotic environment.
3. It reproduces to give rise to young ones.
4. Organism is well adapted to the environment and has a definite life span.

Population is the total number of individuals of a species in a specific geogra¬phical area.
Characteristics of population are:

1. Natality : It refers to the number of births during a given period in the population that are added to the initial density.
2. Mortality : It is the number of deaths in the population during a given period.
3. Immigration : It is the number of individuals of the same species that have come into the habitat from elsewhere during the time period.
4. Emigration : It is the number of individuals of the population who left the habitat and gone elsewhere during the time period.
5. Sex ratio : Number of females produced per 1000 males.

Question 3.
The following diagrams are the age pyramids of different populations. Comment on the status of these populations.

Answer:
Fig. A shows a triangular age pyramid. It indicates that number of individuals in pre-reproductive age group are much more than those in reproductive age group, which in turn are more than the individuals in post reproductive age group. This type of population shows rapid growth and is known as expanding population. Fig. B is a bell shaped age pyramid. Here, individuals in pre-reproductive age group and reproductive age group are almost same. Such population show the zero growth rate because death rate and birth rate are equal. The population is termed as stable or stationary population.
Fig. C shows an urn shaped age pyramid.
Here, . number of individuals in pre- reproductive age group are less than number of individuals in reproductive age group. Number of post-reproductive individuals is also sizeable. Death rate is more than the birth rate and this pyramid shows declining population.

Question 4.
Comment on the growth curve given below.

Answer:
The given growth model shows logistic growth, where the resources are limited and a given habitat can support certain number of individuals. The limit beyond which population cannot grow further is known as carrying capacity. Population shows sigmoid growth curve, (S-shaped curve) and has following phases – lag phase, log phase, exponential phase and stationary phase.
This population growth is called Verhulst – Pearl logistic growth, it has following equation :

where,
N = Population density at a time
K = Carrying capacity
r = Intrinsic rate of natural increase.

Question 5.
A population of Paramecium caudatum was grown in a culture medium. After 5 days the culture medium became overcrowded with Paramecium and had depleted nutrients. What will happen to the population and what type of growth curve will the population attain? Draw the growth curve.
Answer:
Paramecium caudatum when grown in a culture medium will show logistic or sigmoid growth curve. Such type of growth occurs when a population is growing in a habitat with ‘limited resources. The population of Paramecium initially shows a lag phase followed by phase of exponential growth as the nutrients and space will be abundant in nutrient medium. When the resources get depleted, the population density starts decreasing and finally the population density reaches carrying capacity beyond which no further growth is possible.

Question 6.
Discuss the various types of positive interactions between species.
Answer:
In positive population interactions, both members involved are either benefited or one is benefited and other remains unaffected. Types of positive interactions are as follows :
(1) Mutualism – An interaction between two organisms of different species where both the partners are benefited. Examples of symbiosis or mutualism are lichens, symbiotic nitrogen fixation, mycorrhizae etc.
Lichen is a composite entity which is formed jointly by an alga (phycobiont) and a fungus (mycobiont). The main body of the lichen is formed of fungus. The fungus also provides, water, minerals and shelter to the alga. The alga manufactures food not only for itself but also for the fungus. This interaction or relationship allows the lichen to grow in highly hostile environment like bare rock,

(2) Commensalism – It is the relationship between two living individuals of different species in which one is benefited while the other is neither harmed nor benefited except to a negligible extent. E.g., sucker fish attaches itself to the under-surface of shark with the help of its dorsal fin which is modified into holdfast. Sucker fish gets a free ride. It is widely dispersed and remains protected from its predators. Sucker fish detaches itself from shark w’hen the latter is feeding to obtain smaller pieces of food.

Question 7.
In an aquarium two herbivorous species of ‘fish are living together and feeding on phytoplanktons. As per the Gause’s principle, one of the species is to be eliminated in due course of time, but both are surviving well in the aquarium. Give possible reasons.
Answer:
According to Gause’s ‘Competitive Exclusion Principle’ two closely related species competing for same resources cannot co-exist indefinitely and the competitively inferior one will be eliminated eventually. If two herbivorous species of fish are co-existing in same ecological niche, then it might be possible that the two herbivorous species of fish may have developed different specialisations, i.e., they may have evolved mechanisms to encourage co-existence, e.g., resource partitioning.

Question 8.
While living in and on the host species, the animal parasite has evolved certain adaptations. Describe these adaptations with examples.
Answer:
Parasites living inside or outside the bodies of host species evolve certain special adaptations to have better survival. The few parasitic adaptations are as follow’s:

1. Parasites living in intestine usually develop a covering called cuticle to resist digestive enzvmes of the host, e.g., Ascnris.
2. Endoparasites undergo anaerobic respi-ration.
3. Some develop adhesive structures to avoid expulsion, e.g., hooks and suckers in tapeworm.
4. Loss of locomotorv organs and absence of joined appendages, e.g., Sacculimi.
5. High rate of reproduction to compensate loss during transfer from one host to another.
6. Simplified sense organs and nervous system.

Question 9.
Do you agree that regional and local variations exist within each biome? Substantiate your answer with suitable example.
Answer:
Abiome is a major ecological communitv or complex of communities characterised by a major vegetation type and distinct landscape
characters. Different types of flora and fauna exist in a biome. Climate is the main factor that determines the type of soil which in turn determines the type of vegetation. The type of vegetation and climate together determine the type of animal population inhabiting the area. The organisms of a biome are adapted to the climatic conditions associated with the region. There are no distinct boundaries between adjacent biomes which merge gradually with each other.

World major biomes include tropical forests, savanna, deserts, polar regions, chaparral, temperate grasslands, temperate deciduous forests, coniferous forests and tundra. Each biome has regional and local variations so that there is wide variety of habitats and ecological niches. For example, the temperate broad leaf and mixed forests biome include a number of distinct ecoregions with characteristic species. In the south, the middle Atlantic coastal forest occupies the flat Atlantic coastal plain, from the eastern shore of Maryland and Delaware to just south of Georgia. The river swamp or bottom land forests in this ecoregion are dominated by majestic bald cypress and swamp tupelo. Farther north, the north eastern forest type is typified by white and red oak.

Question 10.
Which element is responsible for causing soil salinity? At what concentration does the soil become saline?
Answer:
The most important element responsible for causing soil salinity is Na present as NaCl in the soil. Chlorides, nitrates, sulphates and carbonates of potassium, magnesium and ^sodium; chlorides and nitrates of calcium are important elements to make soil more saline.

Question 11.
Does light factor affect the distribution of organisms? Write a brief note giving suitable examples of either plants or animals.
Answer:
Light is an important abiotic factor responsible for growth, development and distribution of plants. There are three parameters of light.

1. Light intensity,
2. Light quality
3. Light duration

(1) Decreasing light intensity at various depths in ocean determines distribution of plants.
Green algae – found along the shoreline Brown algae – deeper level Red algae – Very deep in water

(2) Terrestrial communities are regulated by light. The dominant species occupy highest canopy level to utilise maximum light. The quantity and quality of light result in different types of plants growing in different areas in various states.

(3) Day length varies, when we move from equator to poles, it plays important role in plant distribution. According to duration of light available to the plants, they can be long day plants, short day plants or day neutral plants. The flowering period varies depending on the availability and duration of light.

Question 12.
Give one example for each of the following:

1. Eurythermal plant species
2. A hot water spring organism
3. An organism seen in deep ocean trenches
4. An organism seen in compost pit
5. A parasitic angiosperm
6. A stenothermal plant species
7. Soil organism
8. A benthic animal
9. Antifreeze compound seen in antarctic fish
10. An organism which can conform

Answer:

1. Eurvthermal plant species – Artemcsin.
2. A hot water spring organism – Thermits aquatints
3. An organism seen in deep ocean trenches – Euplectclla
4. An organism seen in compost pit – Pseudomonas
5. A parasitic angiosperm – Cuscuta
6. A stenothermal plant species – Coconut
7. Soil organism – Earthworm
8. A benthic animal – Sponge
9. Antifreeze compound seen in antarctic fish – Glycoprotein
10. An organism which can conform – Asterias

We hope the NCERT Exemplar Solutions for Class 12 Biology chapter 13 Organisms and Populations help you. If you have any query regarding .NCERT Exemplar Solutions for Class 12 Biology chapter 13 Organisms and Populations, drop a comment below and we will get back to you at the earliest.

NCERT Exemplar Solutions for Class 12 Biology chapter 12 Biotechnology and Its Applications

These Solutions are part of NCERT Exemplar Solutions for Class 12 Biology. Here we have given NCERT Exemplar Solutions for Class 12 Biology chapter 12 Biotechnology and Its Applications

Multiple Choice Questions

Question 1.
Bt cotton is not
(a) a GM plant
(b) insect resistant
(c) a bacterial gene expressing system
(d) resistant to all pesticides.
Answer:
(d) : Several transgenic plants have been developed. One such plant is Bt cotton in which the Bt toxin from the bacterium Bacillus thuringiensis has been used as a biological insecticide. The choice of genes depends upon the crop and targeted pest. Bt cotton is resistant to insects but is not pesticide resistant.

Question 2.
C-peptide of human insulin is
(a) a part of mature insulin molecule
(b) responsible for formation of disulphide bridges
(c) removed during maturation of pro-insulin to insulin
(d) responsible for its biological activity.
Answer:
(c) : In mammals, including humans, insulin is synthesized as a pro-hormone which needs to be processed before it becomes a fully mature and functional hormone. It contains an extra, stretch called the C peptide which is removed during maturation into insulin.

Question 3.
GEAC stands for
(a) Genome Engineering Action Committee
(b) Ground Environment Action Committee
(c) Genetic Engineering Approval Committee
(d) Genetic and Environment Approval Committee.
Answer:
(c) : Genetic modification of organisms can have unpredictable results when such organisms are introduced into the ecosystem. Therefore, the Indian Government has set up organizations such as GEAC (Genetic Engineering Approval Committee), which makes decisions regarding the validity of GM research and the safety of introducing GM- organisms for public services.

Question 4.
a-1 antitrypsin is
(a) an antacid
(b) an enzyme
(c) used to treat arthritis
(d) used to treat emphysema.
Answer:
(d) : Disorders of a-l-antitrypsin protein include a-l-antitrypsin deficiency, an autosomal, codominant hereditary disorder in which deficiency of a-l-antitrypsin leads to a chronic uninhibited tissue breakdown. This causes the degradation of lung tissue, and eventually leads to characteristic manifestations of pulmonary emphysema.

Question 5.
A probe which is a molecule used to locate specific sequences in a mixture of DNA or RNA molecules could be
(a) a single stranded RNA
(b) a single stranded DNA
(c) either RNA or DNA
(d) can be ss DNA but not ss RNA.
Answer:
(a, b) : The molecular probes are usually single stranded pieces of DNAs (sometimes RNAs) labelled with radio isotopes such phosphorus-32. Molecular probes are available for many genetic disorders such as Duchenne muscular dystrophy, cystic fibrosis, Tay-Sachs disease.

Question 6.
Choose the correct option regarding retrovirus.
(a) An RNA virus that can synthesise DNA during infection.
(b) A DNA virus that can synthesise RNA during infection.
(c) A ssDNA virus.
(d) A dsRNA virus.
Answer:
(a) : A retrovirus is a ssRNA virus that stores its nucleic acid in the form of an mRNA genome and targets a host cell as an obligate parasite. Once inside the host cell cytoplasm, the virus uses its own reverse transcriptase enzyme to produce DNA from its RNA genome (the reverse of usual pattern, thus retro).

Question 7.
The site of production of ADA in the body is
(a) erythrocytes
(b) lymphocytes
(c) blood plasma
(d) osteocytes.
Answer:
(b) : Lymphocytes are a kind of white blood cells present in bone marrow. ADA (adenosine deaminase) is an enzyme that is present in lymphocytes and is very important for the immune system to function.

Question 8.
A protoxin is
(a) a primitive toxin
(b) a denatured toxin
(c) toxin produced by protozoa
(d) inactive toxin.
Answer:
(d)

Question 9.
Pathophysiology is the
(a) study of physiology of pathogen
(b) study of normal physiology of host
(c) study of altered physiology of host
(d) none of the above.
Answer:
(c) : Pathophysiology means the functional changes in the affected person associated with or resulting from a disease or injury or a syndrome.

Question 10.
The trigger for activation of toxin of Bacillus thuringiensis is
(a) acidic pH of stomach
(b) high temperature
(c) alkaline pH of gut
(d) mechanical action in the insect gut.
Answer:
(c) : The Bt toxin proteins exist as inactive protoxins but once an insect ingests the inactive toxin it is converted into an active form of toxin due to the alkaline pH of the
alimentary canal that solubilises the crystals. The activated toxin binds to the surface of midgut epithelial cells and create pores which causes cell swelling and lysis and finally cause death of the insect.

Question 11.
Golden rice is
(a) a variety of rice grown along the yellow river in China
(b) long stored rice having yellow colour tint
(c) a transgenic rice having gene for (3-carotene)
(d) wild variety of rice with yellow coloured grains.
Answer:
(c) : Golden rice is a transgenic variety of rice (Oryza sativa) which contains good quantities of (3-carotene (provitamin A- inactive state of vitamin A). (3-carotene is a principle source of vitamin A. Since the grains (seeds) of the rice are yellow in colour due to (3-carotene, the rice is commonly called golden rice.

Question 12.
In RNAi, genes are silenced using
(a) ssDNA
(b) dsDNA
(c) dsRNA
(d) ssRNA.
Answer:
(c) : RNAi takes place in all eukaryotic organisms as a method of cellular defense. This method involves silencing of a specific mRNA due to a complementary dsRNA molecule that binds to and prevents translation of the mRN A (silencing). The source of this complementary RNA could be from an infection by viruses having RNA genomes or mobile genetic elements (transposons) that replicate via an RNA intermediate.

Question 13.
The first clinical gene therapy was done for the treatment of
(a) AIDS
(b) cancer
(c) cystic fibrosis
(d) SCID (Severe Combined Immuno Defici-ency resulting form deficiency of ADA).
Answer:
(d) : The first clinical gene therapy was given in 1990 to a 4-year old girl with adenosine deaminase (ADA) deficiency. This enzyme is very important for the immune system to function. ADA deficiency can lead to Severe Combined Immuno Deficiency (SCID).

Question 14.
ADA is an enzyme which is deficient in a genetic disorder SCID. What is the full form of ADA?
(a) Adenosine deoxyaminase
(b) Adenosine deaminase
(c) Aspartate deaminase
(d) Arginine deaminase
Answer:
(b) Refer answer 13.

Question 15.
Silencing of a gene could be achieved through the use of
(a) RNAi only
(b) antisense RNA only
(c) both RNAi and antisense RNA
(d) none of the above.
Answer:
(c) : For RNAi refer answer 12.
Antisense RNA is an RNA molecule whose base sequence is complementary to that of the sense RNA. It can undergo base pairing with its comlementary mRNA and block the gene expression either by preventing access for ribosomes to translate the mRNA or by triggering degradation of the dsRNA by ribonucleases. Flavr savr variety of tomato is produced using this antisense technology in which an artificial gene inserted for antisense RNA prevented expression of genes that cause ripening. It has been found that RNA interference is more effective than antisense RNA.

Very Short Answer Type Questions

Question 1.
In view of the current food crisis, it is said, that we need another Green Revolution. Highlight the major limitations of the earlier Green Revolution.
Answer:
BID Limitations of Green Revolution are:

1. Increased crop yield is not sufficient to feed the increasing human population.
2. Fertilisers and pesticides which are used to increase the food production are not good for environment.
3. Green revolution was agrochemical based, and as agrochemicals are expensive so, small farmers cannot afford them.

Question 2.
Expand GMO. How is it different from a hybrid?
Answer:
GMO is Genetically Modified Organism.
Differences between GMO and hybrid are as follows:

GMO	Hybrid
(1) GMO is obtained by inserting foreign gene into an organism.	Hybrid is produced by crossing two superior individuals
(2) Change in genotype is precisely controlled.	Change in genotype depends on chance segregation or aggregation during fertilisation.
(3) New traits are introduced	Existing traits of parents are reshuffled.

Question 3.
Differentiate between diagnostics and therapeutics. Give one example and for each category.
Answer:
Diagnosis refers to the detection of disease and understanding its pathophysiology (symptoms, causes etc.) so that it can be cured. It involves ELISA, PCR, blood tests etc. Whereas, therapeutics is treatment of disease by providing proper medication or replacing defective genes by normal ones. E.g., Vaccines, production of humulin, antibiotics etc.

Question 4.
Give the full form of ELISA. Which disease can be detected using it? Discuss the principle underlying the test.
Answer:
HELISA is Enzyme Linked Immuno- sorbant Assay. ELISA is used for detecting HIV, Hepatitis-B virus etc. diseases. ELISA f is based on principle of antigen-antibody interaction. An antibody (Ab) reacts with the concerned antigen (Ag) in a highly specific manner and forms Ag-Ab complex. A second antibody conjugated with enzyme specific to a second site on the test protein is added. The enzyme causes colour change and this colour change is analysed to identify the condition.

Question 5.
Can a disease be detected before its symptoms appear? Explain the principle involved.
Answer:
Yes, a disease can be detected when symptoms are not yet visible due to very low
coqnt of pathogens, by the technique called Polymerase Chain Reaction (PCR). PCR multiplies pathogen nucleic acids and by analysing them, disease can be diagnosed. By PCR, even very low amounts of DNA can be detected and amplified. It is used to detect HIV and many other genetic disorders etc.

Question 6.
Write a short note on biopiracy highlighting the exploitation of developing countries by the developed countries.
Answer:
Biopiracy is exploitation of bioresources of a country by organisations and multi-nationals for commercial exploitation with or without patent but without any Access and Benefit Sharing Agreement (ABA). Bioresources or biological resources are all those organisms which can provide commercial benefits. They are abundant in developing countries which are poor in technology though the countries have traditional knowledge related to bioresources. On the other hand, developed countries are poor in bioresources but are rich in technology. Traditional knowledge helps in saving time, effort and expenditure in developing refined product for commercialisation. Therefore, based on traditional knowledge, institutions and companies of industrialised nations are collecting and exploiting bioresources of other nations by getting them patented.

Question 7.
Many proteins are secreted in their inactive form. This is also true of many toxic proteins produced by microorganisms. Explain how the mechanism is useful for the organism producing the toxin?
Answer:
Bacillus thuringiensis is one such organism which produces a toxic protein in its inactivated form (protoxin). The benefit of doing so is that the bacterium itself remains unaffected from the toxic effects but once an insect ingests it, the alkaline pH of insect’s gut changes it into active form. The activated toxin binds to the surface of midgut epithelial cells and creates pores which cause cell swelling and lysis and death of an insect, but the bacterium remains unaffected.

Question 8.
While creating genetically modified organisms, genetic barriers are not respected. How can this be dangerous in the long run?
Answer:
Genetically modified organisms can be dangerous in long run due to following reasons :

1. It is possible that insects might become resistant to Bt or other crops that have been genetically modified to produce their own pesticides.
2. Genetically engineered herbicide crop plants and weeds may cross breed, resulting in transfer of herbicide resistance genes from crops into weeds. Therefore, weeds would also be herbicide tolerant.
3. The enzyme produced by antibiotic resistance gene can cause allergies, because it is a foreign protein.
4. Genetically engineered microbes may be uncontrollable if get released in ecosystems. They may also transfer virulence to other microbial populations.
5. GMOs can be dangerous to other organisms in the ecosystem. Such as honeybees show toxicity to pollens of insect resistant crops.

Question 9.
Why has the Indian Parliament cleared the second amendment of the country’s patents bill?
Answer:
The Indian Parliament has recently also cleared second amendment of the Indian Patents Bill because there has been growing realisation of injustice, inadequate compensation and benefit sharing between developed and developing countries. Such patent bills prevent unauthorised exploitation of bioresources and traditional knowledge of a country and consider emergency provisions and development initiative.

Question 10.
Give any two reasons why the patent on basmati should not have gone to an American Company.
Answer:
Patents on basmati should not have gone to an American Company because:

1. Presence of rice goes back thousands of years in Asia’s agricultural history. There are over 2,00,000 varieties of rice in India alone.
2. The “new” variety of basmati having the US patent had actually been derived from Indian farmer’s varieties by crossing Indian basmati with semi-dwarf varieties.

Question 11.
How was insulin obtained before the advent of rDNA technology? What were the problems encountered?
Answer:
Before the advent of rDNA technology, insulin was extracted from pancreas of slaughtered pigs and cattle, as insulin is secreted by (3-cells of islets of Langerhans of pancreas. Problem encountered was that this insulin was different from human insulin and caused some undiserable side effects such as allergy.

Question 12.
With respect to understanding diseases, discuss the importance of transgenic animal models.
Answer:
Animals having foreign gene inserted in their genome by rDNA technology are called transgenic animals. Many transgenic animals have been developed to increase our understanding of how genes contribute to the development of disease so that investigation of new treatments for diseases is made possible. Now transgenic models exist for many human diseases such as cancer, cystic fibrosis, rheumatoid arthritis, Alzheimer’s disease, haemophilia, thalessaemia, etc. Such transgenic animals have been created which code for particular products of therapeutic value such as human protein (a-l-antitrypsin) which is used to treat emphysema (sheep), tissue plasmogen activator (goat), blood clotting factors VIII and IX (sheep) and lactoferrin (cow).

Question 13.
Name the first transgenic cow. Which gene was introduced in this cow?
Answer:
First transgenic cow was ‘Rosie’. The gene introduced was the gene for human a lactalbumin. It resulted in production of a lactalbumin (human protein) rich milk which is for more nutritious than normal milk.

Question 14.
PCR is a useful tool for early diagnosis of an infectious disease. Elaborate.
Answer:
Early detection of disease refers to diagnosis before the symptoms are seen. In this case, pathogen (bacteria or virus) are present in very less number, i.e., not capable of producing visible symptoms. Here, PCR is used to multiply their genetic material to get enough quantities to be analysed. E.g., it is used to detect HIV virus in AIDS patient, gene mutations in suspected cancer patients, etc.

Question 15.
What is GEAC and what are its objectives?
Answer:
GEAC is Genetic Engineering Approval Committee, which has been set up by Indian government.
Objectives of GEAC are as follows:

1. To ensure safety of Genetically Modified Organisms (GMO) for public services, i. c., safely using GM organisms for food, medicine etc.
2. To take decisions regarding validity of GM research.

Question 16.
For which variety of Indian rice, the patent was filed by USA company?
Answer:
USA company got patent rights on basmati rice of India. They claimed to develop “new” variety by crossing basmati rice with semi-dwarf varieties.

Question 17.
Discuss the advantages of GMO.
Answer:
USAdvantages of Genetically Modified Organisms (GMO) are:

1. GM crops are more tolerant to abiotic stresses.
2. They are disease resistant.
3. GM plants have increased nutritional value.
4. Transgenic mice are being used for testing vaccine safety.
5. Transgenic animals have been used + as model for studying many human diseases.

Short Answer Type Questions

Question 1.
Gene expression can be controlled with the help of RNA. Explain the method with an example.
Answer:
Silencing of gene expression using a dsRNA is called RNA interference (RNAi). It involves silencing of specific mRNA due to complementary dsRNA molecule that binds to and prevent translation of n;RNA. Using Agrobacterium vectors, nematode specific genes have been introduced into the host plant (tobacco plant). The introduction of DNA was such that it produced both sense and anti¬sense RNA in the host cells. These two RNAs being complementary to each other formed a dsRNA (double stranded RNA) that initiated RNAi.
Different steps involved in making tobacco plant resistant to nematode are briefly described below:

1. RNase enzyme called ‘dicer’ cuts all dsRNA molecules into small interfering RNAs (siRNAs) (21-23 nucleotides long).
2. Each siRN A complexes with ribonucleases (distinct from dicer) to form an RNA- induced silencing complex (RISC).
3. The siRNA unwinds and RISC is activated.
4. The activated RISC targets complementary mRNA molecules. The siRNA strands act as guides where the RISCs cut the transcripts in an area where the siRNA binds to the wRNA. This destroys the 77! RNA.
5. When 77/RNA of the parasite is destroyed no parasite proteins are synthesised. It results in the death of the parasite (nematode) in the transgenic host. Thus the transgenic plant gets itself protected from the parasite.

Question 2.
Ignoring our traditional knowledge can prove costly in the area of biological patenting. Justify.
Answer:
Ignoring traditional knowledge is harmful for developing countries as developed nations take advantage of this. Developing nations are rich in traditional knowledge and biodiversity but are poor in finances and thus are not able to exploit modern applications for commercialisation of bioresources. E.g., basmati rice of India was patented by US company which was later challenged by Indian government.

Question 3.
Highlight any four areas where genetic modification of plants has been useful.
Answer:
Four areas where genetic modification of plants has been successful are :

1. Tolerance of abiotic stress like drought and salinity.
2. Better nutritional value.
3. Increased efficiency of mineral utilisation.
4.  Reduced post harvest losses.

Question 4.
What is a recombinant DNA vaccine? Give two examples.
Answer:
Recombinant DNA vaccines are vaccines that are produced by genetic modification technique. For production of such vaccines, microbes are programmed to produce desired antigenic fragment. E.g., vaccine against hepatitis B virus consists of a portion of the viral protein coat produced by genetically modified yeast.

Question 5.
Why is it that the line of treatment for a genetic disease is different from infectious diseases?
Answer:
Infectious diseases are due to infection with a pathogen. They are treated by killing or inhibiting the growth of pathogen by taking various drugs or by strengthening our immune system. Genetic disorders are due to defect in genes, and subsequent defective protein and enzyme formation and subsequent can be cured by gene therapy, which involves replacement of faulty genes by healthy genes.

Question 6.
Discuss briefly how a probe is used in molecular diagnostics.
Answer:
The molecular probes are usually pieces of ssDNA (or RNA) labelled with radio isotopes such as 32P. These are used for molecular diagnosis of various diseases such as Duchenne muscular dystrophy, cystic fibrosis, Tay-Sachs disease etc.

In molecular diagnosis, a single stranded DNA or RNA joined with a radioactive molecule (probe) is allowed to hybridise with its complementary strand followed by detection using autoradiography.

Question 7.
Who was the first patient who was given gene therapy? Why was the given treatment recurrent in nature?
Answer:
The first clinical gene therapy was given in 1990 to a 4 – year old girl named Ashanti de Silva with adenosine deaminase (ADA) deficiency. This enzyme is very important for the immune system to function as its deficiency causes SCID. In some children, ADA deficiency can be cured by bone marrow transplantation. However, in others it can be treated by enzyme replacement therapy, in which functional ADA is given to the patient by injection. But in both approaches the patients are not completely cured.

As a first step towards gene therapy, lymphocytes, a kind of white blood cells, are extracted from the bone marrow of the patient and are grown in a culture outside the body. A functional ADA cDNA (using a retroviral vector) is then introduced into these lymphocytes, which are reinjected to the patient’s bone marrow. But as these cells do not always remain alive, the patient requires periodic infusion of such genetically engineered lymphocytes. However, if the isolated gene from bone marrow cells producing ADA is introduced into cells at early embryonic stages, it can be a permanent cure.

Question 8.
Taking examples under each category, discuss upstream and downstream processing.
Answer:
A biotechnological manufacturing process can be separated into upstream processes and downstream processes. Upstream processing refers to the entire process of selecting and isolating cells to be used, their cultivation, cell banking and culture expansion of the cells until final harvest (termination of the culture and collection of the live cell batch) is achieved. For example, in production of humulin, upstream processing includes cell line preparation by obtaining human insulin gene, and inserting it into E. coli cells and media and equipment preparation.

Downstream processing is the part where the final harvest is purified and quality checked to make it suitable for commercial applications. Suitable preservatives may also be added. For example, in production of humulin, downstream processing includes lysis of the bacterial cells, separation of cell components from the products, synthesis of active insulin by joining A and B chains to produce mature insulin and its purification to obtain highly purified insulin, fit for medicinal use.

Question 9.
Answer:
Antigens (Ag) are substances which when introduced into the body, stimulate the production of antibodies. Antibodies (Ab) are immunoglobulins which are produced in the body in response to antigens or foreign bodies.
Two diagnostic kits based on Ag and Ab are :

1. ELISA which is used for detection of diseases like HIV.
2. Pregnancy test kit.

Question 10.
ELISA technique is based on the principles of antigen-antibody interaction. Can this technique be used in the molecular diagnosis of a genetic disorder, such as phenylketonuria?
Answer:
Yes, ELISA can be used in diagnosis of a genetic disorder, such as phenylketonuria. In this process antibody against the enzyme which is responsible for the metabolism of phenylalanine i.e. phenylalanine hydroxylase, is used to develop the ELISA based diagnostic technique. The patient has deficiency of this enzyme, thus will test negative while a normal individual which has this enzyme in its body, will test positive.

Question 11.
How is a mature, functional insulin hormone different from its prohormone form?
Answer:
The mature insulin consists of two polypeptide chains – chain A and chain B. The pro-hormone insulin contains an extra C-peptide which is removed during maturation and is not present in mature insulin hormone.

Question 12.
Gene therapy is an attempt to correct a genetic defect by providing a normal gene into the individual. By this, the normal function can be restored. An alternate method would be to provide the gene product (protein/enzyme) known as enzyme replacement therapy, which would also restore the function. Which in your opinion is a better option? Give reason for your answer.
Answer:
Gene therapy is better to cure genetic defect than enzyme replacement therapy because in enzyme replacement therapy a functional ADA is given to the patient by injection and patients do not have functional T-lymphocytes to provide immune response against pathogens and thus patient is not completely cured. In gene therapy, the isolated gene from bone marrow cells producing ADA can be introduced into cells at early embryonic stages and therefore can provide permanent cure.

Question 13.
Transgenic animals are the animals in which a foreign gene is expressed. Such animals can be used to study the fundamental biological process, phenomenon as well as for producing products useful for mankind. Give one example for each type.
Answer:
Transgenic animals used to study fundamental biological processes include transgenic mice in which some specific genes are deleted or replaced with nonfunctional genes thought to be associated with fundamental processes like ageing. Then, the effect of this gene’s absence is studied.

A transgenic animal used to produce products useful for mankind may be exemplified by rosie, a transgenic cow that has gene for human lactalbumin inserted in its genome which is expressed in its milk making it more nutritious.

Question 14.
When a foreign DNA is introduced into an organism, how is it maintained in the host and how is it transferred to the progeny of the organism?
Answer:
The foreign genes are inserted into the genome of an organism using recombinant DNA technology to produce a transgenic organism. This foreign gene is maintained in the host organism because as its cells undergo division, rDNA also gets replicated alongwith the cell chromosomes and daughter cells receive copies of rDNA. Similarly, during reproduction, reproductive cells get the rDNA along ,with the other chromosomes and as a result, the transgenic trait is transferred to the progeny. Chances are 100%, if the organism reproduces asexually.

Question 15.
Bt cotton is resistant to pests, such as iepidopteran, dipterans and coleopterans. is Bt cotton also resistant to other pests as well?
Answer:
Bt cotton is made resistant against specific pests only. Soil bacterium Bacillus thuringiensis produces proteins that kills certain insects like lepidopterans (tobacco budworm), coleopterans (beetles) and dipterans (flies) etc. It does not have gene which is effective against all type of insect pests. Therefore, other pests may attack cotton plant.

Long Answer Type Questions

Question 1.
A patient is suffering from ADA deficiency. Can he be cured? How?
Answer:
Gene therapy is the technique of genetic engineering used to replace a faulty gene by a normal healthy functional gene. The first clinical gene therapy was given in 1990 to a 4 years old girl with adenosine deaminase deficiency (ADA deficiency). This enzyme is very important for the immune system to function. Severe combined immunodeficiency (SCID) is caused due to defect in the gene for the enzyme adenosine deaminase. SCID patient lacks functional T-lymphocytes and, therefore, fails to fight the infecting pathogens.

To perform gene therapy, lymphocytes are extracted from the patient’s bone marrow and a normal functional copy of human gene coding for ADA is introduced into these lymphocytes with the help of retroviral vector. The cells so treated are reintroduced into the patient’s bone marrow. The lymphocytes produced by these cells contain functional ADA gene which reactivates the victim’s immune system.

But, as these lymphocytes do not divide and are short lived, so periodic infusion of genetically engineered lymphocytes is required. This problem can be overcome, if stem cells are modified at early embryonic stage.

Question 2.
Define transgenic animals. Explain in detail any four areas where they can be utilised.
Answer:
The organisms which have their DNA manipulated to possess and express a foreign or extra gene are known as transgenic animals. Various areas where transgenic animals can be used are as follows:

1. Transgenic animals produce useful products, such as human protein (a-1- antitrypsin) used to treat emphysema. Attempts are being made for treatment of phenylketonuria and cystic fibrosis.
2. Transgenic mice are being used for testing the safety of vaccines before they are used for human beings. E.g., they are being used to test the safety of polio vaccine.
3. Transgenic animals act as models for human diseases like cancer, cystic fibrosis, rheumatoid arthritis, Alzheimer’s disease etc., and their possible new methods of their treatment.
4. Transgenic animals are specifically developed to study, how genes are regulated and how they affect normal functioning of the body and its development. E.g. study of complex factors involved in growth, such as insulin like growth-factor (IGF).

Question 3.
You have identified a useful gene in bacteria. Make a flow chart of the steps that you would follow to transfer this gene to a plant.
Answer:

Question 4.
Highlight five areas where biotechnology has influenced our lives.
Answer:
Various areas where biotechnology has influenced our lives are :

1. Gene therapy – A collection of methods that allows correction of a gene defect diagnosed in a person, child or an embryo.
2. Molecular diagnosis – Early detection and treatment of diseases before their symptoms start appearing.
3. Production of proteins using rDNA technique – Several proteins have been produced in abundance for curing certain diseases. They include insulin, growth hormone, interferons, vaccines etc.
4. Agricultural applications – rDNA technology has developed transgenic plants which can tolerate drought, various diseases and have increased productivity.
5. Industrial applications – Enzymes are synthesised and used to produce sugar, cheese and detergents etc.

Question 5.
What are the various advantages of using genetically modified plants to increase the overall yield of the crop?
Answer:
Advantages of genetically modified plants are as follows :

1. Tolerance – GM plants are resistant to abiotic stresses (drought, cold, salinity, heat).
2. Pest resistance – GM plants are pest resistant and can reduce the utilisation of chemical insecticides or pesticides e.g. Bt cotton.
3. Disease resistance – Genetically modified plants are resistant to various diseases caused by bacteria, virus, fungus etc.
4. Reduced post harvest losses – GM plants have helped to reduce post harvest losses, e.g. flavr savr transgenic tomatoes.
5. Increased efficiency of mineral usage – GM plants can more efficiently utilise soil minerals and thus prevent early exhaustion of fertility of soil.
6. Increased nutritional value of food – GM plants have enhanced nutritional value of food, e.g. golden rice is rich in vitamin A.

Question 6.
Explain with the help of one example how genetically modified plants can
(a) reduce usage of chemical pesticides.
(b) enhance nutritional value of food crops.
Answer:
(a) GM crops are pest resistant crops. E.g., a nematode Meloidogyne incognita infects the roots of tobacco plant and causes lots of damage by reducing the yield. Using Agrobacterium vectors nematode specific genes are introduced into the host plant, as a result they produce both sense and antisense RNA. This initiates RNAi i.e. a specific segment of RNA is made silent and is unable to produce the protein required by the nematode. The nematode dies in such a transgenic host. By this way, using biotechnological technique the transgenic plant gets protected by itself without using chemical pesticides,

(b) GM plants have increased nutritional vdlue of food. E.g., golden rice is a transgenic variety of rice, which contains good quantities of (3 carotene (provitamin A). Since the contents of vitamin A are very low in rice, so genetically engineered rice have been produced by introducing three genes associated with synthesis of carotene. The grains of transgenic rice are rich in provitamin and reduce the occurrence of vitamin A deficiency diseases.

Question 7.
List the disadvantages of insulin obtained from the pancreas of slaughtered cows and pigs.
Answer:
Disadvantages of insulin obtained from slaughtered cows and pigs are as follows :

1. As insulin from slaughtered animals is produced in vary small amounts, so it requires killing of large number of animals.
2. Insulin produced by animals is slightly different from human insulin, therefore it sometimes does not respond well and causes allergy.
3. It is unethical to kill so many animals to obtain a drug.
4. Due to infection by microorganisms, some animals might produce contaminated insulin.
5. Its production is a more time taking process and the supply is also limited.

Question 8.
List the advantages of recombinant insulin.
Answer:
Advantages of recombinant insulin are as follows:

1. Recombinant insulin exactly resembles human insulin in structure and is commonly called humulin.
2. It is available in pure form, therefore, chances of contamination are very little.
3. Its production does not involve killing of animals.
4. There is no immune response or allergy or any other side effects.
5. There is no shortage of supply.

Question 9.
What is meant by the term biopesticide? Name and explain the mode of action of a popular biopesticide.
Answer:
A biopesticide is a living organism and not a chemical substance. Its product or gene can kill the pest, thus it is used for pest control and is thus named so. It is highly specific and safe for the environment. It does not cause any pollution in the environment. Bt toxin is a biopesticide produced from Bacillus thuringiensis.

Bt toxins are proteins that kill certain insects like lepidopterans, coleopterans and dipterans. This protein exists in inactive form in bacteria and does not cause any harm to it. Once an insect ingests the inactive toxin, it gets converted into active form of toxin due to alkaline pH of alimentary canal of the insect, that solubilises the crystals.

The activated toxin binds to the surface of the midgut epi thelial cells of the insect and create pores which cause swelling and lysis and finally the death of insect. Bt toxin crystals are used as pesticide in agriculture. The protein crystals are available commercially as liquids which are sprayed on the leaves. When a insect pest ingests them, the toxins get inactivated inside the insect and kill it. With advances in the field of biotechnology, genetic engineering techniques have been used to incorporate pest resistance in crop plants. The Bt toxin genes are isolated from Bacillus thuringiensis and incorporated into several crop plants like cotton. Bt cotton farming has shown good results in various parts of the country.

Question 10.
Name the five key tools for accomplishing the tasks of recombinant DNA technology. Also mention the functions of each tool.
Answer:
Five key tools for recombinant DNA technology are :

1. Cleaving enzymes – Restriction endonucleases cleave DNA at specific points and exonucleases cut DNA at the terminal end.
2. DNA ligase – Enzyme used for sealing gaps in DNA fragment and for joining foreign gene with plasmid DNA.
3. Vectors – Vectors are DNA molecule that carry foreign DNA segment and replicate inside host cell, e.g., plasmid.
4. Competent host – Cell capable of transformation, i.e., able to take foreign DNA.
5. Eysing enzymes – To open the cell to isolate the DNA.

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

NCERT Exemplar Solutions for Class 12 Biology chapter 15 Biodiversity and Conservation

These Solutions are part of NCERT Exemplar Solutions for Class 12 Biology. Here we have given NCERT Exemplar Solutions for Class 12 Biology chapter 15 Biodiversity and Conservation

Multiple Choice Questions

Question 1.
Which of the following countries has the hiqhest biodiversity?
(a) Brazil
(b) South Africa
(c) Russia
(d) India
Answer:
(a) : The environment of Brazil is characterised by high biodiversity. Brazil’s large area comprises different ecosystems, which together sustain some of the world’s greatest biodiversity. It has the most known species of plants (55,000), freshwater fish (3,000) and mammals (700). Birds and reptiles are also found in abundance.

Question 2.
Which of the following is not a cause for loss of biodiversity?
(a) Destruction of habitat
(b) Invasion by alien species
(c) Keeping animals in zoological parks
(d) Over-exploitation of natural resources
Answer:
(c) : Keeping animals in zoological parks is not a cause for loss of biodiversity rather it is a method of conservation of biodiversity.

Question 3.
Which of the following is not an invasive alien species in the Indian context?
(a) Lantana
(b) Cynodon
(c) Parthenium
(d) Eichhornia
Answer:
(b) : Lantana, Eichhornia and Parthenium are all exotic species, which had been introduced in India. Lantana camara has  replaced many species in forests of Uttar Pradesh and Madhya Pradesh. Eichhornia (water hyacinth) has clogged water bodies including wetlands resulting in death of several aquatic plants and animals. Parthenium has pushed out several herbs and shrubs from open places in the plains.

Question 4.
Where among the following will you find pitcher plant?
(a) Rain forest of North-East India
(b) Sunderbans
(c) Thar Desert
(d) Western Ghats
Answer:
(a) : Pitcher plant is mainly found in rain forest of North-East India. This plant grows in the soil which has low nitrate. Hence, they obtain their nitrogen nutrition by trapping insects.

Question 5.
Which one of the following is not a feature of biodiversity hotspots?
(a) Large number of species
(b) Abundance of endemic species
(c) Mostly located in the polar regions
(d) Mostly located in the tropics
Answer:
(c) : Hotspots are areas of high endemism and high species richness. All over the world, some 34 such spots have been identified, including 3 in India (Western Ghats, Indo- Burma and Himalayas). The environmental conditions in polar regions do not favour large number of species or species richness.

Question 6.
Match the animals given in Column A with their location in Column B.

Choose the correct match from the following.
(a) i-A, ii-C, iii-B, iv-D
(b) i-D, ii-C, iii-A, iv-B
(c) i-C, ii-A, iii-B, iv-D
(d) i-C, ii-A, iii-D, iv-B
Answer:
(d)

Question 7.
What is common to the following plants: Nepenthes, Psilotum, Rauwolfa and Aconitum?
(a) All are ornamental plants.
(b) All are phylogenic link species.
(c) All are prone to over exploitation.
(d) All are exclusively present in the Eastern Himalayas.
Answer:
(c) : Nepenthes, Psilotum, Rauwolfia and Aconitum have either medicinal or ornamental importance. Hence, they are prone to over exploitation.

Question 8.
The one-horned rhinoceros is specific to which of the following National Parks?
(a) BhitarKanika
(b) Bandipur
(c) Kaziranga
(d) Corbett park
Answer:
(c)

Question 9.
Amongst the animal groups given below, which one has the highest percentage of endangered species?
(a) Insects
(b) Mammals
(c) Amphibians
(d) Reptiles
Answer:
(c) : The percentage number of endangered species in the list of threatened ones is 19% mammals, 17%birds, 21% reptiles, 22% amphibians.

Question 10.
Which one of the following is an endangered plant species of India?
(a) Rauwolfia serpentina
(b) Santalum album (Sandal wood)
(c) Cycas beddqmei
(d) All of the above
Answer:
(d)

Question 11.
What is common to Lantana, Eichhornia and African catfish?
(a) All are endangered species of India.
(b) All are key stone species.
(c) All are mammals found in India.
(d) All the species are neither threatened nor indigenous species of India.
Answer:
(d) : Lantana, Eichhornia and African catfish are exotic species or non-native species. Exotic species are often introduced for their economic and other uses. They often become invasive .and drive away the local species. These species are considered to be second major cause of extinction of species. Exotic species have proved harmful to both aquatic and terrestrial ecosystems.

Question 12.
The extinction of passenger pigeon was due to
(a) increased number of predatory birds
(b) over exploitation by humans
(c) non-availability of the food
(d) bird flu virus infection.
Answer:
(b) : Passenger pigeons (Ectopistes migratorious) were once one of the most abundant birds on our planet, living in North
AmeVica. But hunting on a very massive scale as well as deforestation, lead to their extinction. Many cruel ways were used for their hunting. The last individual of this species died in 1914 in Cincinnati Zoo.

Question 13.
Which of the following statements is correct?
(a) Parthenium is an endemic species of our country.
(b) African catfish is not a threat to indigenous catfishes.
(c) Steller’s sea cow is an extinct animal.
(d) Lantana is popularly known as carrot grass.
Answer:
(c) : Steller’s sea cow became extinct due to over exploitation by humans.

Question 14.
Among the ecosystem mentioned below, where can one find maximum biodiversity?
(a) Mangroves
(b) Desert
(c) Coral reefs
(d) Alpine meadows
Answer:
(c) : Coral reefs have the highest biodiversity with its macrobiota representing about 4-5% of the described global biota.

Question 15.
Which of the following forests is known as the ‘lungs of the planet Earth’?
(a) Taiga forest
(b) Tundra forest
(c) Amazon rainforest
(d) Rain forests of North East India
Answer:
(c) : Amazon rainforests are also called as the lungs of the planet earth. This rainforest in South America is the largest and most diverse tropical rain forest on our planet. It contributes around 20% of the total oxygen in our planet.

Question 16.
The active chemical drug reserpine is obtained from
(a) Datura
(b) Rauwolfia
(c) Atropa
(d) Papaver.
Answer:
(b) : Reserpine is an alkaloid that has been used as an anti hypertensive drug i.e., drug to lower blood pressure. It is obtained from the dried roots of Raiavolfia serpentina (Sarpagandha).

Question 17.
Which of the following groups exhibits more species diversity?
(a) Gymnosperms
(b) Algae
(c) Bryophytes
(d) Fungi
Answer:
(d)

Question 18.
Which of the below mentioned regions exhibit less seasonal variations?
(a) Tropics
(b) Temperates
(c) Alpines
(d) Both (a) & (b)
Answer:
(a) : There are no unfavourable seasons in tropics. Continued favourable environment has helped tropical organisms to gain more niche specialisation and increased diversity. Hence, tropics exhibit lesser seasonal variations.

Question 19.
The historic convention on Biological Diversity held in Rio de Janeiro in 1992 is known as
(a) CITES Convention
(b) The Earth Summit
(c) G-16 Summit
(d) MAB Programme.
Answer:
(b) : Earth Summit at Rio de Janeiro (1992), Brazil, promoted Convention on Biological Diversity (CBD) which was signed by 152 nations. Its recommendations came into effect on 29th December 1993. India became a party to this Convention on Biological Diversity in May, 1994.

Question 20.
What is common to the techniques (1) in vitro fertilisation, (2) Cryo preservation and (3) tissue culture?
(a) All are in situ conservation methods.
(b) All are ex situ conservation methods.
(c) All require ultra modern equipment and large space.
(d) All are methods of conservation of extinct organisms.
Answer:
(b) : Conservation of biodiversity can be broadly classified into two types : in situ (on-site) and ex situ (off site). Examples of in situ strategies include National parks, sanctuaries, Biosphere reserves etc. It is the preferred method to maintain species of wild animals in their natural habitats. This approach helps in conservation of total ecosystem. Ex situ approaches include conservation of those organisms, whose species may become extinct or decline heavily in number (due to a variety of reasons) in their natural habitat. This includes seed banks, cryopreservation, tissue culture, in vitro fertilisation etc.

 Very Short Answer Type Questions

Question 1.
What characteristics make a community stable?
Answer:
Characteristics which make community stable are :

1. Resistance to occasional disturbance (natural or man made).
2. Resistance to invasion by alien species.

Question 2.
What could have triggered mass extinctions of species in the past?
Answer:
The exact reason for mass extinctions of species in the past is not known. Scientists however assume that drastic environmental changes that followed events such as asteroid or meteorite impact or volcanic eruptions lead to mass extinctions.

These environmental changes include fall in temperature causing global cooling that trapped sea water in polar ice caps leading to lowering of sea level; or rise in temperature leading to disrupted ocean circulation patterns, green house effect and depletion of ozone layer etc.

Question 3.
What accounts for the greater ecological diversity of India?
Answer:
Greater ecological biodiversity in India is due to greater geographical and topographical diversity in the form of rainforest, deciduous forests, temperate forests, deserts, mangroves, wet lands alpine meadows etc.

Question 4.
According to David Tilman, greater the diversity, greater is the primary productivity. Can you think of a very low diversity man-made ecosystem that has high productivity?
Answer:
Man made ecosystems like agricultural cropfields of wheat, maize, sorghum, paddy etc., show high productivity but low diversity.

Question 5.
What does ‘Red’ indicate in the IUCN Red list (2004)?
Answer:
Red’ being the sign of danger, in IUCN Red list (2004) indicates threatened species. These species are under various degrees of extinction risk.

Question 6.
Explain as to how protection of biodiversity hotspots alone can reduce up to 30% of the current rate of species extinction.
Answer:
Hotspots are the areas that are rich in species, have high endemism, and are under constant threat. Although it bears only 2.4% of land area of the world, but by protecting hotspots the current rate of extinction of species can be reduced upto 30%.

Question 7.
What is the difference between endemic and exotic species?
Answer:
Endemic species refers to plant or animal species that is restricted to one or a few localities in its distribution. Endemic species are usually confined to islands and are vulnerable to extinction whereas exotic species is a species of organism that is not native to a locality having been moved there from its natural range by humans or other agents. Some alien species, such as rats, are introduced mainly by accident in cargoes or transport vessels, while others are transferred intentionally, often for their ornamental or economic value.

Question 8.
How does species diversity differ from ecological diversity?
Answer:
Species diversity is the variety in the number and richness of the species of a region. Species diversity is product of both species richness or evenness or equitability, i.e., species richness weighed by species evenness. For example, big cats, like tiger (Panthera tigris), and lion (Panthera leo) belong to the same genus Panthera but they all differ at the species level.

Ecosystem diversity explains the diverse number of niches, trophic levels and various ecological processes that sustain energy flow, food webs and recycling of nutrients. Within the geographical area, there may be variable number of ecosystems/habitats.

Question 9.
Why is genetic variation important in the plant Rauwolfia vomitoria
Answer:
Rnumnifin vomitoria grows in different ranges of Himalaya. This plant is a source of an alkaloid ‘reserpine’ which has medicinal value. Genetic variation is important for potency and concentration of active chemical reserpine present in it.

Question 10.
What is Red Data Book?
Answer:
International Union of Conservation of Nature and Natural Resources which which is now called World Conservation Union (WCU) maintains a red data book or red list which is a catalogue of taxa facing risk of extinction.
Red list has eight categories of species:

1. extinct
2. extinct in wild
3. critically endangered
4. endangered
5. vulnerable
6. lower risk
7. data deficiency
8. not evaluated.

Question 11.
Define gene pool.
Answer:
The sum total of all the genes found in an inbreeding population is known as gene pool.

Question 12.
What does the term ‘frugivorous’ mean?
Answer:
The term frugivorous is used for fruit eating animals.

Question 13.
What is the expanded form of IUCN?
Answer:
IUCN stands for ‘International Union of Conservation of Nature and Natural Resources’.

Question 14.
Define the terms (1) Bioprospecting (2) Endemism
Answer:

1. Bioprospecting refers to exploring molecular, genetic and species level products of economic importance.
2. Endemism refers to presence of some species in particular region only and nowhere else.

Question 15.
What is common to the species shown in figures A and B?

Answer:
Both are invasive weed species.

Question 16.
What is common to the species shown in figure A and B?

Answer:
Both are keystone species.

Short Answer Type Questions

Question 1.
How is the presently occurring species extinction different from the earlier mass extinctions?
Answer:
In earlier time, extinction of species occurred mainly due to natural causes or calamities like volcanic eruption, asteroid impact which lead to drastic environmental changes, etc., while at present human activities like deforestation, over-exploitation, pollution, intensive agriculture, introduction of alien species, etc., are the major causes of species extinction.

Question 2.
Of the four major causes for the loss of biodiversity alien species invasion, habitat loss and fragmentation, over-exploitation and co-extinctions which according to you is the major cause for the loss of biodiversity? Give reasons in support.
Answer:
Out of thy four causes for the loss of biodiversity, habitat loss and fragmentation seems to be the major cause of biodiversity losses. Loss of habitats result in annihilation of plants, microorganisms and forcing out of animals which in alien lands die out after some time. Fragmentation of habitats results in disruption of complex interactions amongst species, destruction of species in the cleared regions, annihilation of species restricted to deeper undisturbed parts of forests and decreased biodiversity in the habitat fragments. Animals requiring large territories (e.g., mammals, birds) are badly affected. Migrating animals would go astray and get killed.

Question 3.
Discuss one example, based on your day-to-day observations, showing how loss of one species may lead to the extinction of another.
Answer:
It has been observed that certain mutualistic relationships exist in nature. Extinction of one will automatically cause extinction of other. E.g., if the host fish becomes extinct, all the parasites exclusively found on it will also become extinct.

Question 4.
A species-area curve is drawn by plotting the number of species against the area. How is it that when a very large area is considered the slope is steeper than that for smaller areas?
Answer:
species-area relationship graph, Z is slope of line or regression coefficient which generally is 0.1-0.2 regardless of taxonomic group or region. When species area relationship is considered for a large area the slope of line becomes steeper having a value of 0.6-1.2. This is because the larger area has more food availability and other resources, so obviously more species can thrive there.

Question 5.
It is possible that productivity and diversity of a natural community remain constant over a time period of, say one hundred years?
Answer:
Although a typical community maintains itself, more or less in equilibrium with the prevailing conditions of the environment, in nature, communities are never stable; rather, they are dynamic, changing more or less regularly over time and space. They are never found permanently in complete balance with their component species, or with the physical environment. Environment is always changing over a period of time due to

1. variations in climatic and physiographic factors, and
2. The activities of the species of the communities themselves.
   These influences bring about marked changes in the dominants of the existing community, which is thus, sooner or later replaced by another community at the same place.

Question 6.
There is greater biodiversity in tropical/ subtropical regions than in temperate region. Explain.
KPlffl
Higher diversity in tropical areas is because of the following reasons:

1. Speciation is a function of time. Temperate areas have undergone frequent glaciation in the past. It killed most of the species. No such disturbance occurred in tropics where species continued to flourish and evolve undisturbed for millions of years.
2. There are no unfavourable seasons in tropics. Continued favourable environment has helped tropical organisms to gain more niche specialisation and increased diversity.
3. More solar energy is available in tropics. This promotes higher productivity and increased biodiversity.
4. Resource availability is higher in tropics.
5. There is reduced competition in tropics due to favourable environment.
6. Rate of extinction is low in tropics.

Question 7.
Why are the conventional methods not suitable for the assessment of biodiversity of bacteria?
Answer:
All forms of bacteria cannot be cultured in normal laboratory conditions which creates a problem in studying their morphological and biochemical characteristics. Biodiversity can be studied by examining morphological and biochemical characters. Such conventional methods cannot be applied for assessment of biodiversity of bacteria.

Question 8.
What criteria should one use in categorising a species as threatened?
Answer:
A threatened species is one which is unable to realise its full biotic potential and is, therefore, liable to become extinct. The inability of realising full biotic potential is due to depletion of food, habitat deterioration, over-exploitation, alien species, etc.

Question 9.
What could be the possible explanation for greater vulnerability of amphibians to extinction as compared to other animal groups?
Answer:
Amphibians are the group of animals which can live both on land and in water. They do not have scales and their skin is permeable to gases. A vast majority of amphibian species have to maintain moist skin surfaces because a significant amount of breathing occurs through skin. They easily get affected by the fluctuations in the surrounding environment.

Their eggs lack shell and hence are prone to dessication in an unprotective environment. Variations in environmental patterns can affect their physiological and reproductive features thereby affecting the survival and continuation of their race which accounts for their greater vulnerability to extinctions as compared to other animal groups.

Moreover, as amphibians need both aquatic and terrestrial habitats to survive, threats to either habitat can affect their population. Hence, amphibians are more vulnerable to habitat modification than the organisms which require only one habitat type.

Question 10.
Howido scientists extrapolate the total number of species on Earth?
Answer:
Scientists make a statistical comparison of the temperate-tropical species richness of an exhaustively studied group of insects and extrapolate this ratio to other groups of animals and plants to come up with a gross estimate of the total number of species on earth.

Question 11.
Humans benefit from diversity of life. Give two examples.
Answer:
Humans derive numerous benefits from diversity of living organisms. These are discussed as follows:
(1) Useful products : Plant species provide a variety of useful products such as timber, crops, fruits and vegetables, gums, resins, dyes, fragrance, perfumes, waxes, lubricants, hydrocarbons, rubber, latex, tannins, paper, tea, coffee, dry-fruits etc. Animals like goat, hen etc., serve as good source of food to humans. Milk, flesh, honey, egg, etc. are all useful animal products. Similarly, animal species provide, wool, fur, skin, leather, lac, silk, waxes, lubricants, pearls, ivory, horns, antlers etc. Large number of substances with therapeutic properties are obtained from variety of plant species. For example, quinine is obtained from the bark of plant Cinchona to combat malaria; taxol from the bark of trees Taxus brevifolia, Taxus baccata for treating cancer; reserpine from Rauwolfia serpentina for treating blood pressure and schizophrenia.

(2) Indirect benefits : Oxygen which made the earth hospitable and is indispensable for survival of almost all living organisms comes from plants which release it in the earth’s atmosphere by the process of photosynthesis. Plants also help in maintaining water cycle, replenishment of water table, bringing rains, preventing soil erosion and flood etc.
Aesthetic values of various natural ecosystems provide the opportunity of ecotourism and hence means of earning to humans.

Question 12.
List any two major causes other than anthropogenic causes for the loss of biodiversity.
Answer:
Loss of biodiversity may be due to:
(1) Natural extinction – Species with small population are always in danger of extinction due to natural causes like inbreeding depression, increased number of predators, development of more competitive species and environmental fluctuations like severe drought, severe winter, harsh summer, excess rain, floods, etc.

(2) Mass extinction – Many species dis-appeared due to catastrophes like glaciation, volcanoes, meteoriteimpact etc. Mass extinction has occurred many times in geological history. E.g., disappearance of dinosaurs coupled with loss of more than 50% of the existing species at the end of cretaceous period.

Question 13.
What is an endangered species? Give an example of an endangered plant and animal species each?
Answer:
Endangered species are those species which are facing high risk of extinction in the near future due to decrease in their habitat and excessive predation or poaching. Example of endangered animal is red panda (Ailurus fulgens). Example of endangered plant is Lycopodium nutans.

Question 14.
What are sacred groves and their role in biodiversity conservation?
Answer:
Sacred groves are forest patches around places of worship which are held in high esteem by tribal communities. They are the most undisturbed forest patches (island of pristine forests) which are often surrounded by highly degraded landscapes.
Some examples of sacred groves are-

1. Khasi and Jaintia Hills in Meghalaya.
2. Aravalli Hills of Rajasthan.
3. Sarguja, Chand and Batsar areas of Madhya pradesh.
4. Western Ghats of Maharashtra and Madhya pradesh.

Role , of sacred groves in biodiversity conservation are as follows:
Sacred groves are serving as refugia for a number of rare, endangered and endemic species. Not a single branch is allowed to be cut from these forests. As a result many endemic species which are rare or have become extinct ‘ elsewhere can be seen to flourish here.

Question 15.
Suggest a place where one can go to study coral reefs, mangrove vegetation and estuaries.
Answer:
Place suggested for studying coral reefs is Andaman and Nicobar island. Mangrove vegetation can be well studied in West Bengal sunderbans and estuaries can be studied in coastal areas of Kerala.

Question 16.
1s it true that there is more solar energy available in the tropics? Explain briefly.
Answer:
Tropics lie on either side of equator between 30° south latitudes. It is limited in latitude by the tropic of cancer in the northern hemisphere and the tropic of capricorn in the southern hemisphere. The tropics include the areas on earth surrounding the equator where the sun reaches a subsolar point (a point directly overhead). These regions receive the most direct sunlight and heat energy from the sun and are the hottest.

Question 17.
What is co-extinction? Explain with a suitable example?
Answer:
When a species becomes extinct, the plant or animal species associated with it in an obligatory way also becomes extinct. This is called co-extinction. E.g., if the host fish becomes extinct, all the parasites exclusively found on it will also become extinct.

Long Answer Type Questions

Question 1.
Elaborate how invasion by an alien species reduces the species diversity of an area.
Answer:
Non-native or alien species are often introduced inadvertently for their economic and other uses. They often become invasive and drive away the local species. These species are considered to be second major cause of extinction of species. Exotic species have proved harmful to both aquatic and terrestrial ecosystems. This can be explained as follows:

Water hyacinth (Eichhornia crassipes) was introduced in Indian waters to reduce pollution. It has clogged water bodies including wetlands at many places resulting in death of several aquatic plants and animals. Nile Perch (a predator fish) was introduced in lake Victoria of South Africa. It killed and eliminated ecologically unique assemblage of over 200 native species of small cliched fish.

Question 2.
How can you, as an individual, prevent the loss of biodiversity?
Answer:
Biodiversity has great importance to mankind. But certain human activities are leading to biodiversity losses. In order to maintain biodiversity we all need to put an effort to save our ecosystem. Following measures can be undertaken to prevent biodiversity losses:

1. Educate people about the importance of wildlife and their conservation.
2. Implementing laws imposed by the government to protect biodiversity.
3. Avoiding over-exploitation of natural re-sources.
4. Promoting afforestation.
5. Avoiding the introduction of alien species.
6. Checking habitat loss and fragmentation of species.
7. Reducing pollution, etc.

Question 3.
Can you think of a scientific explanation, besides analogy used by Paul Ehrlich, for the direct relationship Answer:
Species diversity is formally measured as an index combining numbers and proportion of each species present in an ecosystem. A species diversity index reflects the number of links in a food web. The relationship between species diversity and ecosystem stability has been the most studied and debated topic since 1950.

Paul Ehrlich in his rivet popper hypothesis tried to prove the dependence of ecosystem stability on species diversity. Another scientific explanation to his theory is as follows :
If we imagine that only one food chain is operating in an ecosystem where only one species is occupying each trophic level then this type of food chain is most susceptible to destruction as loss or extinction of any species can lead to destruction of other species dependent on it for food. On the other hand, if many alternatives are available at each trophic level in the food sequence a food web will be formed decreasing the dependence of a species on a particular species for food. Then even extinction of one species will not affect any other species due to availability of alternative food. That is why, in nature food chains never operate singly rather many food chains are interconnected to form food web which gives stability to the ecosystem.

Question 4.
Though the conflict between humans and wildlife started with the evolution of man, the intensity of conflict has increased due to the activities of modern man. Justify your answer with suitable examples.
Answer:
Since the humans evolved on earth, they have been exploiting wildlife for their survival and continuity. Initially the degree of exploitation of wildlife was very less but with increasing civilisation and modernisation, the humans are becoming more demanding. Their ever increasing population is inventing new ways for better survival and existence which at times are interferring with the ways of nature leading to extinction of wildlife species. Hence, the conflict between humans and wildlife continues and has been worsened.
This can be explained as follows:
(1) Destruction of habitat and fragmentation : Over-population, urbanisation and industrialisation require additional land every year. It can come through destruction or fragmentation of natural habitats of wild animals through filling wetlands, ploughing grasslands, cutting down trees, burning a forest and clearing some area of vegetation. Loss of habitat results in annihilation of plants, microorganisms and forcing out of animals Which in alien lands die out after some time. Fragmentation of habitats results in disruption of complex interactions amongst species, destruction of species in the cleared regions, annihilation of species restricted to deeper undisturbed parts of forests and decreased biodiversity in the habitat fragments.

(2) Indiscriminate hunting : Hunting of animals for food, hide, tusks, horn, at excessive levels may lead to species extinction. Dodo bird and passenger pigeon have already become extinct. Many species of fish, molluscs, sea turtle and whales are facing the risk of extinction.

(3) Introduction of alien species:

• Non-native or alien species are often introduced inadvertantly for their economic and other uses. They often become invasive and drive away the local species. These species are considered to be second major cause of extinction of species.
• Carrot grass has replaced herbs and shrubs of open spaces.
• Water hyacinth has become dominant species in pools and ponds.
• Pollution : Excessive use of pesticides has polluted both groundwater and surface 6. water bodies. Many sensitive species have disappeared.
• Intensive agriculture : Spread of agriculture is at the cost of wetlands, grasslands and forests. Destruction of habitats results in extinction of species. Intensive agriculture is also based on a few high yielding varieties. As a result, there is reduction in the genetic diversity. It increases vulnerability of the crop plants to sudden attack by pathogens and pests.

Question 5.
What is an ecosystem service? List any four important ecosystem services provided by the natural ecosystems. Are you in favour or against levying a charge on the service provided by the ecosystem?
Answer:
A wide range of economic, environmental and aesthetic benefits provided by ecolpgical process of an ecosystem are called ecosystem services. Services of healthy forest ecosystem are –

1. Forests provide food in the form of roots, tubers, leaves fruits etc.
2. They provide timber for building purposes of houses, ships, railway sleepers, sport good, agriculture tools etc.
3. Number of useful products like camphor, essential oils, tannins dyes gums, resins, drugs are obtained from forests.
4. Trees in forests produce oxygen during photosynthesis and keep environment cool by regulating transpiration and precipitation.
5. Trees prevent soil erosion floods and provide shelter.
6. Many insect pollinators help in pollination of plants thereby bring about flower and fruits formation.
7. Microbes help in decomposition of waste products and recycling of nutrients. Ecosystem provides lots of services. Instead of putting a price tag or levying a charge against service a sincere effort should be made to protect ecosystem to continue its services.

Question 6.
Describe the consumptive use value of biodiversity as food, drugs and medicines, fuel and fibre with suitable examples.
Answer:
The consumptive use value of biodiversity as food, drugs and medicines, fuel and fibre has been described as follows :
(1) Source of food : Several thousand species of edible plants and animals are known. However, 85% of the world’s food production is met by cultivating less than 20 plant species. Three carbohydrate-rich crops namely, wheat, corn (maize) and rice alone yield nearly two-third of the food production. To meet the demands of increasing human population, man is not only exploring new varieties of plants but also animal food. Biodiversity is also used as a source material for breeding improved varieties. To improve the desired traits, commercial/domesticated. species are crossbred with their wild relatives.

In this way, disease resistant and high yielding varieties of crops (e.g.,wheat, rice, maize, sugarcane) and fruits have been developed. For example, cross breeding of wild rice species has helped ! in developing new varieties which are resistant to four main rice diseases. Similarly, potato has been made resistant to late blight disease, potato mosaic virus, five races of cyst nematodes etc., through crossbreeding experiments. Also, hybrid animals varieties have been produced to increase the production of milk, meat, eggs etc. This indicates the need for protecting biodiversity for breeding programmes in agriculture, horticulture, floriculture, animal husbandry, apiculture, sericulture, lac culture, piggery, poultry and fishery.

(2) Fibers : A variety of plant species such as cotton, flax, hemp, jut, Agave, etc., are the major sources of fibers. More and more variety of plants are being explored for obtaining superior fibers.

(3) Useful products : Plant species provide a variety of useful products such as gums, resins, dyes, fragrance, perfumes, waxes, lubricants, hydrocarbons, rubber, latex, tannins, paper, tea, coffee, dry-fruits etc. Similarly, animal species provide, wool, fur, skin, leather, honey, lac, silk, waxes, lubricants, pearls, ivory, horns, antlers etc.

(4) Drugs and medicines : Large number of substances with therapeutic properties are obtained from variety of plant species. For example, quinine is obtained from the bark of plant Cinchona to combat
malaria ; taxol from the bark of Taxus for treating cancer; morphine from Papaver somniferum for pains; reserpine from Rauwolfia serpentina for treating blood pressure and schizophrenia. Ayurvedic medicines available in the market for treating innumerous diseases in man are based on plant product.

(5) As fuel wood : 80% of forest wood (most trees) are used as fuel wood. E.g., Acacia nilotica, Albizzia sp. Mangifera indica etc.

Question 7.
Species diversity decreases as we move away from the equator towards the poles. What could be the possible reasons?
Answer:
Species diversity decreases as we move away from the equator towards the poles, because of the following reasons :

1. Temperature decreases and conditions become harsh.
2. Both the amount and intensity of solar radiation decreases.
3. Limited resource availability.
4. Higher competition due to unfavourable environment.
   Speciation is generally a function of time and environmental stability, so if conditions are too harsh, it is difficult for species to survive.

Question 8.
Explain briefly the ‘rivet popper hypothesis’ of Paul Ehrlich.
Answer:
According to this hypothesis proposed by Paul Ehrlich (1981), the relationship between species richness and ecosystem functioning is non-linear, and may follow a variety of possible trajectories. The loss of a few species (or rivets holding together an aeroplane) will create no problem in the beginning, but beyond a certain point losses will cause catastrophic effect. Loss of rivets or key species that drive major ecosystem functions is a more serious threat for the ecosystem. Besides, the rich biodiversity is not only essential for ecosystem health, but imperative for the very survival of the human beings.

Question 9.
The relation between species richness and area for a wide variety of taxa turns out to be a rectangular hyperbola. Give a brief explanation.
Answer:
German naturalist and geographer Alexander von Humboldt while exploring the wilderness of South American jungles found that within a region the species richness increased with increasing area but upto a certain limit. The relationship between species richness and area turned out to be rectangular hyperbola for a wide variety of taxa Whether they are birds, bats, freshwater fishes or flowering plants. On a logarithmic scale it is a straight line and is represented by equation: log S = log C + Z log A

Here S is species richness, Z is slope of line or regression coefficient, C is Y intercept while A is area. The relationship between area and species richness in most cases is represented by a rectangular hyperbola which indicates that if we take a larger area under consideration then the number of species increases because more individuals get included in the sample and large areas are environmentally more heterogenous than small areas. But this increase in number is neither uniform nor unlimited. Ecologists have proposed wide range of factors determining the slope and elevation of species area curve. These factors include relative balance between immigration and extinction, rate and magnitude of disturbance predator-prey dynamics, etc.

We hope the NCERT Exemplar Solutions for Class 12 Biology chapter 15 Biodiversity and Conservation help you. If you have any query regarding .NCERT Exemplar Solutions for Class 12 Biology chapter 15 Biodiversity and Conservation, drop a comment below and we will get back to you at the earliest.

NCERT Exemplar Solutions for Class 12 Biology chapter 11 Biotechnology: Principles and Processes

These Solutions are part of NCERT Exemplar Solutions for Class 12 Biology. Here we have given NCERT Exemplar Solutions for Class 12 Biology chapter 11 Biotechnology: Principles and Processes

Multiple Choice Questions

Question 1.
Rising of dough is due to
(a) multiplication of yeast
(b) production of CO2
(c) emulsification
(d) hydrolysis of wheat flour starch into sugars.
Answer:
(b)

Question 2.
An enzyme catalysing the removal of nucleotides from the ends of DNA is
(a) endonuclease
(b) exonuclease
(c) DNA ligase
(d) HindII
Answer:
(b) : Exonucleases remove nucleotides from the terminal ends (either 5′ or 3′) of DNA of one strand of duplex. Endonucleases make cut at specific position within the DNA. DNA ligases are known as joining or sealing enzymes. Ligases form phosphodiester bonds between adjacent nucleosides and covalently link two individual fragments of DNA. Hindll is restriction endonuclease, it cuts the DNA and produces blunt end.

Question 3.
The transfer of genetic material from one bacterium to another through the mediation of a vector like virus is termed as
(a) transduction
(b) conjugation
(c) transformation
(d) translation.
Answer:
(a)

Question 4.
Which of the given statements is correct in the context of observing DNA separated by agarose gel electrophoresis? 
(a) DNA can be seen in visible light.
(b) DNA can be seen without staining in visible light.
(c) Ethidium bromide stained DNA can be seen in visible light.
(d) Ethidium bromide stained DNA can be seen under exposure to UV light.
Answer:
(d) : Electrophoresis is a technique of separation of molecules such as DNA, RNA or protein, under the influence of an electric field, so that they migrate in the direction of electrode bearing the opposite charge, viz, positively charged molecules move towards cathode (-ve electrode) and negatively charged molecules travel towards anode (+ve electrode) through a medium/ matrix. Since DNA fragments are negatively charged molecules, they can be separated by allowing them to move towards the anode (+ve electrode) under an electric field through a matrix of agarose gel.

The DNA fragments separate’according to their size through the agarose gel, with smaller fragments moving farther away as compared to larger ones. The DNA fragments can be visualised by staining them with ethidium bromide followed by exposure to UV radiations. Bright orange coloured bands of DNA can be observed. The separated DNA bands are then cut out from the agarose gel and extracted from the gel piece, this step is known as elution.

Question 5.
‘Restriction’ in restriction enzyme refers to
(a) cleaving of phosphodiester bond in DNA by the enzyme.
(b) cutting of DNA at specific position only.
(c) prevention of the multiplication of bacteriophage in bacteria.
(d) all of the above.
Answer:
(c) : A restriction enzyme is an enzyme that cuts DNA by hydrolysing phosphodiester bonds at or near a specific recognition nucleotide sequences known as restriction sites. The term ‘restriction’ refers to the function of an enzyme restricting the propagation of foreign DNA of bacteriophage in the host bacterium. These enzymes are found in bacteria and provide a defense mechanism against invading virus. The host DNA is protected by a modification enzyme that modifies the prokaryotic DNA and blocks cleavage. Together, these processes form restriction modification system.

Question 6.
Which of the following is not required in the preparation of a recombinant DNA molecule?
(a) Restriction endonuclease
(b) DNA ligase
(c) DNA fragments
(d) E.coli
Answer:
(d) : To produce a recombinant DNA following procedure is followed :

• Genetic material is isolated by using the enzyme, lysozyme or cellulase or chitinase for animal, plant or fungal cell respectively.
• Purified DNA is cut at specific sites by restriction enzymes.
• DNA fragments are separated using agarose gel electrophoresis and the gene or fragment of interest is amplified using PCR.
• After the cutting of the source DNA and the vector DNA with specific restriction enzyme, the cut out ‘gene of interest’ from the source DNA and the cut vector with space are mixed and ligase enzyme is added. This results in the formation of a rDNA or hybrid DNA or chimeric DNA.
• The ultimate aim of recombinant DNA technology is to produce a desirable protein. The foreign gene gets expressed under appropriate condition, culturing methods are used to produce higher yields of the desired protein. Therefore, E. coli is not required for preparation of a recombinant DNA molecule. we have to break the cell open to release DNA and other macromolecules

Question 7.
In agarose gel electrophoresis, DNA molecules are separated on the basis of their
(a) charge only
(b) size only
(c) charge to size ratio
(d) all of the above.
Answer:
(b) : Electrophoresis is a technique of separation of molecules, such as DNA, RNA or proteins on the basis of their size, under the influence of an electric field. So they migrate towards electrode of opposite charge. DNA fragments separate according to the size through pores of agarose gel.

Question 8.
The most important feature in a plasmid to be used as a vector is
(a) origin of replication (or/)
(b) presence of a selectable marker
(c) presence of sites for restriction endonuclease
(d) its size.
Answer:
(a,b,c & d) : Origin of replication (On), a selectable marker, sites for restriction 1 endonuclease and its size, all are importantfeatures required to facilitate cloning into a vector. A good DNA vector should be able to replicate autonomously in the host cell, for which it needs to have an origin of replication site (Ori). This is important for the replication of inserted gene. A prokaryotic DNA has a single Ori while eukaryotic DNA may have more than one Ori. Selectable marker helps in identifying and eliminating non-transformants and selectively permitting the growth of transformants. Cloning sites (recognition sites) are the sites where the DNA is cut by a restriction endonuclease. A vector should be ideally less than 10 kb in size because large DNA molecules can break down during purification procedure.

Question 9.
While isolating DNA from bacteria, which of the following enzymes is not used?
(a) Lysozyme
(b) Ribonuclease
(c) Deoxyribonuclease
(d) Protease
Answer:
(c) : In order to cut the DNA with restriction enzymes, it needs to be in pure form, free from other macromolecules. Since the DNA is enclosed by the membranes,we haive to break the cell open to release DNA and other macromolecules like RNA, proteins, polysaccharides and lipids. It is obtained by treating the bacterial cells/ plant or animal tissue with enzymes such as lysozyme (bacteria), cellulase (plant cells) and chitinase (fungus). DNA is interwined with proteins like histones. RNA can be removed by treatment with ribonuclease while proteins can be removed by treatment with protease. Other molecules are removed by appropriate treatments and purified DNA ultimately precipitates out after the addition of chilled ethanol.

Question 10.
Which of the following has popularised the PCR (polymerase chain reaction)?
(a) Easy availability of DNA template.
(b) Availability of synthetic primers.
(c) Availability of cheap deoxyribonudeo- tides.
(d) Availability of ‘thermostable’ DNA poly-merase.
Answer:
(d) : The final step of PCR is extension, wherein Taq DNA polymerase (isolated from a thermophilic bacterium Thermits aquaticus) synthesises the DNA region between the primers, using dNTPs (deoxynucleoside triphosphates) and Mg²⁺. The primers are extended towards each other so that the DNA segment lying between the two primers is copied. The optimum temperature for this polymerisation step is 72CC. Taq polymerase remains active during high temperature induced denaturation of double stranded DNA.

Question 11.
An antibiotic resistance gene in a vector usually helps in the selection of
(a) competent cells
(b) transformed cells
(c) recombinant cells
(d) none of the above.
Answer:
(b) :The ligation of alien DNA is carried out at a restriction site present in one of the two antibiotic resistance genes. For example, you can ligate a foreign DNA at the BamHl site of tetracycline resistance gene in the vector pBR322. The recombinant plasmids will lose tetracycline resistance other macromolecules like RNA, proteins, polysaccharides and lipids. It is obtained by treating the bacterial cells/ plant or animal tissue with enzymes such as lysozyme (bacteria), cellulase (plant cells) and chitinase (fungus). DNA is interwined with proteins like histones. RNA can be removed by treatment with ribonuclease while proteins can be removed by treatment with protease. Other molecules are removed by appropriate treatments and purified DNA ultimately precipitates out after the addition of chilled ethanol.

Question 12.
Significance of ‘heat shock’ method in bacterial transformation is to facilitate
(a) binding of DNA to the cell wall
(b) uptake of DNA through membrane transport proteins
(c) uptake of DNA through transient pores in the bacterial cell wall
(d) expression of antibiotic resistance gene.
Answer:
(c) : Transformation is a process by which a cell takes up naked DNA fragment from the environment, incorporates it into its own chromosomal DNA and finally expresses the trait controlled by the incoming DNA. Since DNA is a hydrophilic molecule, it cannot pass through membranes, so the bacterial cells must be made competent to take up DNA. This is done by treating them with a specific concentration of a divalent cation, such as calcium (Ca²⁺) which increases the efficiency with which DNA enters the bacterium through pores in its cell wall. Recombinant DNA (rDNA) can then be forced into such cells by incubating the cell with recombinant DNA on ice, followed by placing them briefly at 42°C (heat shock), and then putting them back on ice. This enables the bacteria to take up the recombinant DNA.

Question 13.
The role of DNA ligase in the construction of a recombinant DNA molecule is
(a) formation of phosphodiester bond between two DNA fragments
(b) formation of hydrogen bonds between sticky ends of DNA fragments
(c) ligation of all purine and pyrimidine bases
(d) none of the above.
Answer:
(a) : DNA ligases (joining or sealing enzymes) are also called genetic gum. They join two individual fragments of double  straqded DNA by forming phosphodiester bonds between them. Thus, they help in sealing gaps in DNA fragments. Therefore, they act as a molecular glue.

Question 14.
Which of the following is not a source of restriction endonuclease?
(a) Haemophilus influenzae
(b) Escherichia coli
(c) Entamoeba coli
(d) Bacillus amyloliquifaciens
Answer:
(c)

Source	Restriction endonuclease
(1) Haemophilus influenzae	Hindll and HmdIII
(2) Escherichia coli	EcoRI and EcoRII
(3)Bacillus amyloliquefaciens	Bam HI

Question 15.
Which of the following steps are catalysed by Taq polymerase in a PCR reaction?
(a) Denaturation of template DNA.
(b) Annealing of primers to template DNA.
(c) Extension of primer end on the template DNA.
(d) All of the above.
Answer:
(c) : Refer answer 10.

Question 16.
A bacterial cell was transformed with a recombinant DNA that was generated using a human gene. However, the transformed cells did not produce the desired protein. Reasons could be
(a) human gene may have intron which bacteria cannot process
(b) amino acid codonsfor humans and bacteria are different
(c) human protein is formed but degraded by bacteria
(d) all of the above.
Answer:
(a) : Eukaryotic genes do not function properly when transferred into bacterial cell because introns are present in eukaryotic cells but are absent in prokaryotic cells. Hence, when bacterial cell is transformed with recombinant DNA which is generated using human gene, it could not process it. As a result, no desired protein will be produced.

Question 17.
Which of the following should be chosen for best yield if one were to produce a recombinant protein in large amounts?
(a) Laboratory flask of largest capacity
(b) A stirred-tank bioreactor without in-lets and out-lets
(c) A continuous culture system
(d) Any of the above
Answer:
(c) : The cells having cloned genes of interest can be grown on a small scale in the laboratory. The cultures may be used tor extracting and purifying the desired protein. The cells can also be multiplied in a continuous culture system where the used medium is passed out from one side and fresh medium is added from the other side to maintain the cells in their physiologically most active log/ exponential phase – rapid multiplication of the cells. This type of culturing method produces a larger biomass to get higher yields of desired protein.

Question 18.
Who among the following was awarded the Nobel Prize for the development of PCR technique?
(a) Herbert Boyer
(b) Hargovind Khurana
(c) KaryMullis 
(d) Arthur Kornberg
Answer:
(c)

Question 20.
Which of the following statements does not hold true for restriction enzyme?
(a) It recognises a palindromic nucleotide sequence.
(b) It is an endonuclease.
(c) It is isolated from viruses.
(d) It produces the same kind of sticky ends in , different DNA molecules.
Answer:
(c) : More than 900 restriction enzymes have been isolated from over 230 strains of bacteria each of which recognise different recognition sequences. No restriction enzyme has been isolated from viruses.

Very Short Answer Type Questions

Question 1.
How is copy number of the plasmid vector related to yield of recombinant protein?
Answer:
The copy number is the number of copies of plasmids compared with the amount of chromosomal DNA found in single bacterial cell. The copy number influences the stability of plasmid. The rDNA can multiply as many times as the copy number of vector plasmid. So, higher the copy number of plasmid, higher would be the copy number of gene and protein yield would be high.

Question 2.
Would you choose an exonuclease while producing a recombinant DNA molecule?
Answer:
For producing recombinant DNA, exonuclease is not added because it removes nucleotides from terminal ends (either 5′ or 3′) and it acts on single strand of DNA. It does not recognise any specific sequence.

Question 3.
What does ‘H’, ‘in’, ‘d’ and ‘III’ refer to in the enzyme Hindlll?
Answer:
ln naming of restriction endonuclease, the first letter of enzyme is the first letter of bacterium’s, generic name, “H”-Haemophilus, “in” represents two letters of species name- influenzae, “d” is for strain “Rd”, and “HI” indicates the order in which enzyme is synthesised.

Question 4.
Restriction enzymes should not have more than one site of action in the cloning site of a vector. Comment.
Answer:
Restriction enzymes should preferably have single recognition site because presence of more than one recognition sites within the vector will generate several fragments, which will complicate the gene cloning.

Question 5.
What does ‘competent’ refer to in competent cells used in transformation experiments?
Answer:
Competent cells are the cells that have altered cell wall so that they can take up foreign DNA easily. Competent cell is the one that is able to undergo genetic transformation.

Question 6.
What is the significance of adding proteases at the time of isolation of genetic material (DNA)?
Answer:
Genes located on the DNA are interwined with proteins such as histones, which are removed by proteases. It is important to add protease during DNA isolation because DNA has to be free from any other .macromolecule such as RNA, proteins to get cut with restriction endonuclease.

Question 7.
While doing a PCR/denaturation’step is missed. What will be its effect on the process?
Answer:
Denaturation is the first step in PCR reaction. If denaturation does not take place, annealing and extension will also not take place. Therefore, there will be no amplification.

Question 8.
Name a recombinant vaccine that is currently being used in vaccination program.
Answer:
HepaH tis-R surface antigen (HBsAg) is currently used as vaccine in vaccination programme against hepatitis virus.

Question 9.
Do biomolecules (DNA, protein) exhibit biological activity in anhydrous conditions?
Answer:
DNA and protein do not show biological activity in anhydrous conditions because in non-aqueous conditions rigidity of biomolecules increases due to weakening of hydrogen bond strength.

Question 10.
What modification is done on the Ti plasmid Agrobacterium tumefaciens to convert it into a cloning vector?
Answer:
Agrobacterium tumefaciens, a pathogen of several dicot plants is able to deliver T-DNA to transform normal cell into tumour and direct tumour cells to produce chemicals required by pathogen. The Ti (tumor inducing) plasmid has been modified (disarmed) by removing gene responsible for causing tumour and inserting gene to be used as selectable marker. Modified plasmid is non-pathogenic to plants and delivers the gene of interest.

Short Answer Type Questions

Question 1.
What is meant by gene cloning?
Answer:
Gene cloning is formation of multiple identical copies of any template DNA. For cloning, an alien DNA is linked with origin of replication of a vector, so that it can replicate and multiply itself in the host organism.

Question 2.
Both a wine maker and a molecular biologist who had developed a recombinant vaccine claim to be biotechnologists. Who in your opinion is correct?
Answer:
Biotechnology is technological employment of biological entities and processes to generate products and services useful to man. In this case a wine maker has utilised a strain of yeast, which is commonly used for making wine by fermentation. The molecular biologist has used cloned gene for an antigen and antigen is used as vaccine. Thus, both can be considered as biotechnologists.

Question 3.
A recombinant DNA molecule was created by ligating a gene to a plasmid vector. By mistake, an exonuclease was added to the tube containing the recombinant DNA. How does this affect the next step in the experiment i.e. bacterial transformation?
Answer:
Recombinant DNA is formed by ligating an alien DNA with plasmid DNA, a circular, self-replicating, extra-chromosomal, dsDNA. Therefore, rDNA is also circular DNA and exonucleases cleave DNA at terminal ends (either 5′ or 3′) in a linear DNA segment. Thus, addition of exonuclease to a tube containing rDNA would not affect bacterial transformation.

Question 4.
Restriction enzymes that are used in the construction of recombinant DNA are endonucleases which cut the DNA at ‘specific- recognition sequence’. What would be the disadvantage if they do not cut the DNA at specific recognition sequence?
Answer:
If endonucleases used in the recombinant DNA technology do not cut DNA at specific locations, they may cut the DNA at random location. The sticky ends will not be generated to ligate the DNA segments and recombinant DNA would not be formed.

Question 5.
A plasmid DNA and a linear DNA (both are of the same size) have one site for a restriction endonuclease. When cut and separated on agarose gel electrophoresis, plasmid shows one DNA band while linear DNA shows two fragments. Explain.
Answer:
A plasmid DNA is a circular DNA. When this circular DNA having one site for restriction endonuclease is cleaved with an enzyme, it would form a strand of linear DN? and would be visible as single band on agarose gel. On the other hand, when linear DNA with one restriction site is cleaved, it would produce two fragments and thus two bands would be seen.

Question 6.
How does one visualise DNA on an agarose gel?
Answer:
The separated DNA fragments on an agarose gel can be visualised only after staining the DNA with ethidium bromide followed by exposure to UV radiations as bright orange coloured bands.

Question 7.
A plasmid without a selectable marker was chosen as vector for cloning a gene. How does this affect the experiment?
Answer:
If a cloning vector does not have a selectable marker, then it would not be possible to distinguish between transformants (host bacterium having rDNA) and non-transformants. Therefore, an ideal cloning vector should have selectable markers for the selection of transformants.

Question 8.
A mixture of fragmented DNA was electrophoresed in an agarose gel. After staining the gef with ethidium bromide, no DNA bands were observed. What could be the reason?
Answer:
There could be various reasons for DNA bands not visible after electrophoresis, such as:

1. DNA might be impure, i.e., have not been isolated properly and is associated with RNA and proteins.
2. Restriction endonucleases have not been added in appropriate amount.
3. Electrodes might not have been connected correctly.
4. Stained bands have not been visualised under UV rays.
5. Electrodes were put in opposite orientation in the gel assembly, that is anodes towards the well (where DNA is loaded).
6. Ethidium bromide has not been added to stain DNA fragments.

Question 9.
Describe the role of CaCI₂ in the preparation of competent cells.
Answer:
Competent cell is one which can take up naked DNA fragment and incorporate it into its own chromosomal DNA. DNA is a hydrophilic molecule and. cannot pass through membranes, so bacterial cell can be made competent by CaCH method. In this method, bacterial cell is treated with divalent cation Ca²⁺ which increases the efficiency with which DNA enters the bacterium through pores in its cell wall.

Question 10.
What would happen when one grows a recombinant bacterium in a bioreactor but forgets to add antibiotic to the medium in which the recombinant is growing?
Answer:
Antibiotics do not allow other bacteria to grow in the medium. In the absence of antibiotics, the desirable bacteria may not be able to grow to its optimum level. The plasmids may also be lost, since maintaining a high copy number of plasmids is a metabolic burden on the bacteria.

Question 11.
Identify and explain steps A, B and C in the PCR diagram given below.

Answer:
Polymerase Chain Reaction (PCR) involves amplification of gene of interest.
A single PCR amplification cycle involves three basic steps:

1. A- Denaturation : It is separation of two strands of DNA by heating the DNA strand at a temperature of 94 – 96 °C. Each single strand of target DNA is used as template DNA.
2. B-Annealing : Hybridisation of two oligo-nucleotide primers to each of the ssDNA template. It is carried out at a temperature of 40- 60 °C.
3. C-Extension : It is the final step of PCR. In this, enzyme Taq DNA polymerase synthesises the DNA region between the primers, using dNTPs and Mg²⁺.

Question 12.
Name the regions marked A, B and C.

Answer:
A – BarnHI
B – PtI
C – ampR

Long Answer Type Questions

Question 1.
For selection of recombinants, insertional inactivation of antibiotic marker has been superceded by insertional inactivation of a marker gene coding for a chormogenic substrate. Give reasons.
Answer:
Selection of recombinants by insertional inactivation of an antibiotic has been superceded by insertional inactivation of selectable marker because selection of recombinants due to inactivation of antibiotics is a cumbersome procedure, because it requires simultaneous plating on two plates having different antibiotics (ampicillin and tetracyiine). Therefore, alternative selectable markers have been developed which differentiate recombinants from non-recombinants on the basis of their ability to produce colour in the presence of chromogenic substrate. In this, a recombinant DNA is inserted within the coding sequence of an enzyme p-galactosidase.

This results in the inactivation of the enzyme (insertional inactivation). The presence of chromogenic substrate gives blue coloured colonies if the plasmid in the bacteria does not have an insert. Presence of insert results into insertional inactivation of the (1-galactosidase and the colonies do not produce any colour, and these are identified as recombinant colonies.

Question 2.
Describe the role of Agiobacterium tumefaciens in transforming a plant cell.
Answer:
Aemhactprium tumefaciens is a bacterium that causes tumours in plants. It has ability to transform plant cells. For this reason it has become an important tool in plant improvement by genetic engineering.
The plasmid is disarmed by deletion of the tumour inducing gene. These modified bacteria can still transform plant cells. The part of Ti plasmid transferred into plant cell DNA, is called T-DNA. This T-DNA with desired DNA spliced into it, is inserted into the chromosomes of the host plant where it produces copies of itself. But it no longer produces tumours. Such plant cells are then cultured, induced to multiply and differentiate to form plantlets. Thus the ability of Agrobactcrium to transfer genes to the plants has been exploited in genetic engineering for plant improvement programme.

Question 3.
Illustrate the design of a bioreactor. Highlight the difference between a flask in your laboratory and a bioreactor which allows cells to grow in a continuous culture system.
Answer:
Bioreactors are vessels in which raw materials are biologically converted into specific products by microbes, plant and anirhal cells and their enzymes. They are allowed to synthesise the desired proteins which are finally extracted and purified from cultures.

Small volume of cultures are usually employed in laboratories for research and production of less quantities of products. Large scale production of products is carried out in bioreactors. The most commonly used bioreactors are of stirring type, that have provision for batch culture or continuous culture. In continuous culture, the culture medium is added and the product is taken out continuously.

Flask is used in laboratory for testing and provides batch type of culture. With flask no gadget can be connected. Bioreactors are used commercially and provide continuous culture. With a bioreactor, gadgets can be attached to give better functioning. A bioreactor provides optimal conditions for optimal growth (temperature pH, substrate, salts, vitamins, oxygen).

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