Class 11

Balbharti Yuvakbharati English 11th Digest Chapter 1.2 On To The Summit: We Reach The Top Notes, Textbook Exercise Important Questions and Answers.

Class 11 English Chapter 1.2 On To The Summit: We Reach The Top Question Answer Maharashtra Board

11th Std English Chapter 1.2 Brainstorming Question Answer

Yuvakbharati English Navneet 11th Digest PDF Free Download Maharashtra Board

Question 1.
Complete the web of different activities related to climbing.

Answer:

Question 2.
There are certain prerequisites for Mountaineering. With reference to the following points develop a short dialogue between you and your friend about mountaineering.

Answer:

• Ashish: Hey Mayank ! when is your next plan for mountaineering?
• Mayank: Yes, I am planning for some days in September – October. Are you interested in joining?
• Ashish: Yes, I am. But I’m not very confident.
• Mayank: Then we will go to Mount Kilimanjaro to start with. But before that you must begin with small practices like walking, stair climbing and may be uphill hiking a bit.
• Ashish: What are the usual equipments that we need? You must be having most of them?
• Mayank: Honestly speaking, we need a lot of equipments and yes, I have some. But let’s visit a mountaineering equipment shop one day to get an idea of the latest ones, convenient as well as inexpensive. I have the basic ones like tent, headlight, map, compass, cap, jackets, gloves, sunscreen, lip balm, etc.
• Ashish: Oh ! You have quite a lot. Let’s visit the shop tomorrow.
• Mayank: But you need to talk to a fitness expert to get the guidance because a fit climber spends less energy. He will suggest what type of fitness regime we should follow as well as can recommend a good trainer.
• Ashish: Yes, I know. Mental and physical fitness are the main ingredients for the success of a mountaineer. Tomorrow, I’ll come to your place and we’ll plan everything. We have to start as soon as possible.
• Mayank: You are most welcome. See you tomorrow.

Question 3.
Discuss various hazards and risks that a mountaineer/ trekker has to face in an expedition.
Answer:
The hazards and risks for a mountaineer/trekker may be:

1. Bad visibility caused by bad weather (mist, rain, snow) or darkness is a major hazard.
2. Snow and ice can make a simple path a very dangerous one, especially if there is a steep drop off near by.
3. Rock fall is a common mishap during heavy snow and rain.
4. Lightning can cause a major problem at some places.
5. Flash floods due to storm might be risky.
6. A mountaineer cannot ignore rain and wind. Rain makes the path towards upward climbing slippery and a strong wind can cause loss of balance of a mountaineer, making climbing impossible.
7. Hypothermia, that is, lowering of the body’s natural temperature to 32° C due to wet clothing and heat loss because of strong wind, can be a big hindrance. If it is not treated immediately, severe hypothermia may occur leading to fatalities.
8. Heat exhaustion and dehydration are very common problems because of difficult physical work.
9. Timely care should be taken by pouring water on the victim to avoid heatstroke. The patient should be urgently given cold water mixed with a small amount of salt and sugar to refresh him.

(A1)

Question 1.
Summarize, in your own words, the highly risky and dangerous journey of Tenzing and Hillary from the base to the top of Mount Everest.
Answer:
Tenzing and Hillary were well-informed about the flat side of the summit, its steep rise, rocky and snowy sides too. It was decided that Tenzing would lead the mission followed by Hillary in the beginning and this position would be alternated. They would be tied together by a rope. They started early from camp nine. Tenzing carried the flags of Britain, the United Nations, Nepal and India to be put on the top of Mount Everest.

They steadily climbed for an hour and faced the difficulties of steep rocks on the west side. The last fifty feet to the top was made easier by their togetherness and they officially announced that without making any attempts to take the credit for conquering Mt. Everest first. Four flags were put by Tenzing wearing the red scarf gifted by his friend. Both of them stayed safely at the top for fifteen minutes.

(A2)

Question (i)
Complete the web highlighting the feelings/emotions of Tenzing after reaching the summit.

Answer:

Question (ii)
Explain the qualities of Tenzing Norgay. Pick lines that show his unique qualities.
Answer:

Qualities	Lines
1. Patriotic	I told Colonel Hunt that I was carrying the Indian flag with me and I would like it to be on the top with other flags.
2. Unselfish	Our main thought was that both of us reached the top.
3. Non-controversial	We will not take part in any controversy.
4. ratitude	My first thought on reaching the top was a sense of gratitude to God.

Question (iii)
Write down the significance of the following in the context of‘On to the Summit*.
(a) Red Scarf
(b) Husiar
(c) Kerosene Flavoured Tea
(d) Ice Axe
(e) Anchor
Answer:
(a) Red Scarf: The red scarf was given to Tenzing by his good friend Lambert who was the leader of the Swiss Expedition of 1952. This scarf reminded Tenzing of their successful mission, a year ago and rejuvenated him to feel fit, energetic, and excited for the completion of another mission.

(b) Husiar: This code-mixing makes Tenzing’s knowledge about mountaineering clearer to the readers. Both Tenzing and Hillary did not take it easy while climbing down. After successful completion, they were not over-confident and were fully aware of the technique and caution needed for the downward journey.

(c) Kerosene Flavored Tea: This tea was brought by C.W.F. Noyce for Tenzing and Hillary from camp eight and he had to do a lot of climbing up and down for that. Some amount of Kerosene got mixed with the tea as it was made in a hurry. Tenzing did not make Mr. Noyce feel guilty as both the mountaineers were grateful for getting hot tea at a time when they needed it the most. The gentleman had taken so much trouble to serve them tea and they were not bothered about its flavour.

(d) Ice Axe: Tenzing used his ice-axe to tie one end of the four flags of Britain, the United Nations and Nepal and the other end was held up by him. Hillary took his pictures holding all the flags up. But he needed his ice-axe for climbing down. So he had to remove his ice-axe and buried one end of the flags in the snow on the slope down below and other end in the ice on the top. Obviously, they fell down but Tenzing had no choice but to remove his ice-axe.

(e) Anchor: An anchor is a person who is strong and reliable. He makes people feel safe and confident. In mountaineering, the man who follows while climbing has a difficult job both in climbing up and climbing down. Tenzing and Hillary decided to alternate this job of anchoring so that the person leading feel safer. This proved that their expedition was a team-work where both shared the responsibility of being the anchor in order to have a successful mission.

(A3)

Question (i)
Add suitable suffixes (- tion, ly, – ment, – ous) and prefixes (un, il, im) to the words given below. Answer:

Word	Prefix	Suffix
absolute	–	absolutely
fulfill	unfulfill	fulfillment
determine	–	determination
danger	–	dangerous
tight	untight	tightly
clear	unclear	clearly
sure	unsure	surely
legal	illegal	legally
legitimate	illegitimate	legitimately
possible	impossible	possibly

(ii) Fill appropriate letters in the blank space to get a past participle form of a word. Use the word as adjective in your own sentence.

Question (a)
s _ g n _ d
Answer:
signed – The advocate took all the signed documents to the court.

Question (b)
p _ _ z _ d
Answer:
prized – My grandmother’s diary is a prized possession for me.

Question (c)
f _ _t_ene_
Answer:
flattened – On the highway we were in great trouble due to the flattened tyre.

Question (d)
b _ i _ e _
Answer:
boiled – My sister loves to eat boiled egg.

(iii) Spot the error in the following sentences.

Question (i)
I was not afraid for die that day.
Answer:
I was not afraid to die that day.

Question (ii)
Tenzing and Hillary made on pact at the office of PM of Nepal.
Answer:
Tenzing and Hillary made a pact in the office of PM of Nepal.

Question (iii)
I has to bring my ice axe down with me.
Answer:
I had to bring my ice axe down with me.

Question (iv)
Tenzing have spent a night with Camp Eight.
Answer:
Tenzing had spent a night at Camp Eight.

Question (v)
Pact was signed from Tenzing and Hillary.
Answer:
A pact was signed between Tenzing and Hillary.

Question (vi)
I and Hillary were in no mood of talking.
Answer:
Hillarv and I were in no mood for any talking.

(A4)

Question 1.
In passive voice the doer is the objeet and the job accomplished is the subject. The verb of passive voice is formed by using the correct form of verb ‘to be’ or verb ‘to have’ + past participle of the main verb. Construct the passive voice of the following sentences accordingly.
Answer:

Active Voice	Passive Voice
(i) We have done it.	(i) It has been done by us.
(ii) We made a pact.	(ii) A pact was made by us.
(iii) I offered silent prayer in my heart.	(iii) Silent prayer in my heart was offered by me.
(iv) Colonel Hunt gave me three flags.	(iv) Three flags were given to me by Colonel Hunt.
(v) I was carrying the Indian flag.	(v) The Indian flag was being carried by me.
(vi) I took photographs of Tenzing holding aloft the flags.	(vi) Photographs of Tenzing holding aloft the flags were taken by me.
(vii) I remembered him.	(vii) He was remembered by me.
(viii) We spent the night at Camp Eight.	(viii) The night was spent at Camp Eight by us.

(A5)

Question (i)
Write in short about any adventure sport that you like and the risk involved in it.
Answer:
I like Bunjee-jumping a lot and have taken part in it at Lonavala. Bungee-jumping, I am sure, many of you, know that it is an activity where the participant jumps from a tall structure but he is obviously connected to a large elastic cord. It is thrilling and one needs a lot of courage to take part in it since a number of risks are involved in it.

Most common injuries are fracture, strains, sprains, bruising, chronic stress features. But the most dangerous risk is acute head injuries. Lots of precautions are being taken by the organizers but one needs to think whether it is worth taking such risks just for the sake of adventure.

Question (ii)
Give your opinion: whether we should or should not participate in adventure sports because –
Answer:
Every coin has two sides, so do adventure sports. Adventure sports do have some benefits.
They are as follows:
1. Boost self-confidence.
2. Stay mentally sharp.
3. Help to cope with challenges in life.
4. Spark the desire to explore.

However, the negative aspects of adventure sports can be quite disturbing.
1. They can cause serious health issues.
2. They are obviously very dangerous.
3. Regular safety precautions may not be stringently followed to prevent life threatening injuries.
It depends on the individual to assess his/her own character traits as well as the pros and cons of the sport, to come to the final decision to take part in any adventure sports with proper training.

Question (iii)
Tenzing and Hillary created history by reaching the summit of Mt. Everest. There are many mountains in Maharashtra where one can fancy his or her chances of climbing them. Imagine that you have climbed a mountain and are immensely thrilled and excited. Write a letter to your friend about it.
Answer:
XYZ,
ABC Apartment,
Flat No. 4, 29th Road,
Shivaji Nagar, Pune
11th July, 2019

Dear Manav,

How are you? I am writing to you after a long time. These days I feel so lazy to write letters, thanks to modern technologies. But today, I am very excited. I went on my first trek last week and I wanted to share with you the details of my trekking experience.

You obviously know, Maharashtra lies amidst the Western Ghats and it welcomes the rains to the beautiful ranges of the Sahayadris. Last week my close friends and I decided to go to Lonavala and from there to Lohagad Fort for trekking, which is for the beginners. We were told that Lohagad Fort trek was not very difficult. We could reach up to the fort to witness some of the most spectacular views in the Lonavala trekking region.

It took us about three to four hours to reach the summit. I must admit that the initial fear was there. But after reaching the peak and seeing the magnificent view all around, we felt that it was worth going there.

Something I must tell you here before I forget. My elder brother Deepak has joined as an executive in a good company and he was my money-bag this time. I felt so, fortunate. Coming back to my expedition, the descent was easier as we were all in a jovial mood for completing the mission (to me, it was a mission only, my first trek) successfully. We have taken beautiful photographs of the surroundings which is breathtakingly beautiful. The waterfalls, lakes, streams, rock-climbing patches, hillocks all around are real wonders of Mother Earth.

When are you planning to come to Pune? I’m eagerly waiting for showing you the photographs. How are Uncle and Aunty. Give them my regards.

Yours lovingly,
XYZ

Question (iv)
You are the college representative and your Principal has assigned you the task of writing a letter to the Divisional Officer, Satpuda Mountain Ranges, Nagpur, seeking permission for the mountaineering expedition to be organized by your college.
Answer:
Harshad Kale
College Representative
XYZ College of Arts and Commerce,
Ghatkopar (East),
Mumbai – 400 077
11th July, 2019

Divisional Officer,
Satpuda Mountain Range,
Nagpur

Subject: Permission for the mountaineering expedition to be organized by our college.

Respected Sir/Madam,

I am Harshad Kale, the College Representative of XYZ College of Arts and Commerce. On behalf of my Principal Dr. Sunil Patil I would request you to grant me the permission for organizing the mountaineering expedition.

I would like to inform you that NCC Unit of our college has planned for the mountaineering expedition from 25th to 30th July. The students who are selected for taking part are all expert trekkers and they will be accompanied by experienced mountaineers as well as our Professors who go regularly for trekking. Our Principal has instructed to arrange for all precautionary measures and is personally taking care of all arrangements.

We shall be obliged if you kindly permit us to carry on the expedition. Awaiting your positive response.

Thanking you,
Yours Sincerely Harshad Kale

Enclosed:
(a) Copy of the expedition programme.
(b) Letter of permission from Principal Dr. Sunil Patil

Question (v)
Convert the above letter into an e-mail format.
Answer:
To: abc@gmail.com
Cc: qrs@gmail.com
Bcc: (non-visible email ids)
Subject: Permission for organising mountaineering expedition

Respected Sir/Madam,

This is Harshad Kale, the representative of XYZ college of Arts and Commerce. On behalf of my teachers and Principal, I request you to grant me the permission for organising a mountaineering expedition planned by our college NCC unit from 25th to 30th July, 2019.

We shall be obliged if you permit us to organize the above mentioned expedition. I have attached the expedition programme for your kind perusal.

Awaiting your positive reply.
Thanks and regards,
Harshad Kale
XYZ College of Arts and Commerce,
Expedition programme [×]
Permission letter (Principal) [×]

Comparisons:

1. Look at the sentences given below. Find out which one is correct. If the sentence is wrong give reasons.

Question (i)
Sunita is the quieter of four sisters.
Answer:
The above sentence is wrong.
Reason: When more than two nouns are compared superlative adjective should be used, i.e., quitest

Question (ii)
Sunita is the quietest of the four sisters.
(iii) Anil’s computer is more new than mine.
Answer:
The above sentence is wrong.
Reason: ‘More new’ is the wrong form of comparative degree of the adjective ‘new’.

Question (iv)
Anil’s computer is newer than mine.
(v) I have the wonderfullest mother in the world
Answer:
The above sentence is wrong.
Reason: ‘wonderfullest’ is the wrong form of superlative degree of the adjective ‘wonderful’.

Question (vi)
I have the most wonderful mother in the world.
(vii) Aditi is more carefuller than Mary.
Answer:
The above sentence is wrong.
Reason: ‘More carefuller’ is the wrong form of comparative degree of ‘careful’.
(viii) Aditi is more careful than Mary.

Look at the following sentences and observe the changes in the three sentences. Note your responses.

Question (a)
Atul is not as bright as Milind in studies.
Answer:
Positive

Question (b)
The dining room is brighter than the kitchen.
Answer:
Comparitive

Question (c)
Anne is the brightest girl in class.
Answer:
Superlative

(A7)

Project:

Go to your college library or surf the internet for names of various mountaineers who have successfully climbed Mt. Everest. Write in your notebook about their struggles, the interesting anecdotes they have shared, their failures and the message that we get from their lives. Submit them to your teacher.

Yuvakbharati English 11th Digest Chapter 1.2 On To The Summit Additional Important Questions and Answers

Question 1.
Give reasons for the ‘rope’ being called a symbol.
Answer:
The rope that tied the two mountaineers, Hillary and Tenzing, together,was referred to as a ‘symbol’ by Tenzing because he always felt that climbing was a teamwork for them. Success, risks, hard work and failure be shared by both as two of them were tied together to fulfill an ambition which was full of hazards. They were aware of the risks involved in their mission and in all those situations, the rope would be between them as a symbol of their togetherness.

Question 2.
Discuss in pairs about any controversy in which you haive got embroiled. Also speak of your experience to the class.
Answer:
Controversies are parts and parcels of our lives and most of the times they are man-made. It happened with me also when I made an innocent comment about one of my classmates. There was a lot of hue and cry about these comments as it was reported by adding fuel to it. But I decided to talk to my classmate and the matter was resolved amicably.

Question 3.
List the ways in which Tenzing Norgay celebrates on reaching the top.
Answer:
1. Tenzing and Hillary embraced each other saying “We have done it.”
2. Tenzing showed his gratitude to God by praying silently for the fulfillment of his desire of climbing the Everest after having failed six times. His offerings were biscuits, candy and a little blue pencil which his daughter Nima had requested him to put on the top of the mountain.
3. Hillary took the photograph of Tenzing holding aloft the flags of Great Britain, Nepal, the United Nations and India.

Comprehension:

Read the extract and complete the activities given below.

Global Understanding:

Question 1.
Complete the web on climbing the summit by Tenzing and Hillary.
Answer:

Question 2.
Rewrite the given sentences in their order of occurrence.
1. I ate some biscuits and offered some to Hillary.
2. I wore the red scarf given to me by my friend, all the way up the mountain from Darjeeling.
3. I offered a silent prayer to God in my heart.
4. At the summit I felt absolutely fit.
5. My daughter Nima gave me a blue pencil, one of her prized possessions, to put at the summit as her offering to God.
Answer:
3. I offered a silent prayer to God in my heart.
5. My daughter Nima gave me a blue pencil, one of her prized possessions, to put at the summit as her offering to God.
1. I ate some biscuits and offered some to Hillary.
2. I wore the red scarf given to me by my friend, ah the way up the mountain from Darjeeling.
4. At the summit I felt absolutely fit.

Complex Factual:

Question 1.
Why didn’t it matter for Tenzing to be the first one to reach the summit?
Answer:
Both Tenzing and Hillary were not thinking of being the first to reach the summit since climbing took all their attention. Their main thought was that one could not do that job alone and only a difference of one thousand feet could decide the position which could be found by the person coming behind.

Question 2.
Why didn’t Tenzing feel tired after reaching the summit?
Answer:
After reaching the summit Tenzing felt exhilarated. He forget about the tiredness. His first thought was a sense of gratitude to god who had blessed him with the fulfillment of his desire after having failed six time.

Inference / Interpretation / Analysis

Question 1.
Explain.
“The rope was a symbol”
Answer:
The rope was a symbol of togetherness of Hillary and Tenzing as it was used to tie them, one following the other. It was also the proof that climbing the summit of Mt. Everest needed a teamwork and the two members of their team respected and considered each other important.

Question 2.
Explain.
“He gave me a big smile, showing that he understood.”
Answer:
Hillary smiled because he understood the emotion of Tenzing. The blue pencil was given to Tenzing by his youngest daughter, Nima, and it was one of her prized possessions. She parted with it with the expectation that her father would conquer the summit and the offering of the blue pencil was her way of showing gratefulness to God. Her father would fulfill her wish.

Personal Response

Question 1.
Pick up one sentence to show one quality of Tenzing that you have appreciated the most in his story. Explain your answer.
Answer:
The sentence “What does it matter whether I reached the top first or Hillary?” shows the unselfish character of Tenzing. He had the opportunity to claim that he reached the summit first as he was walking first and Hillary was behind him. But he did not take the credit as he knew it was a team work. Accomplishing the job was more important than taking credit for it.

Question 2.
Mention some of the benefits of trekking trips organized by colleges for the students.
Answer:

1. The students get more exposure and experience.
2. They get close to nature.
3. They learn to handle adverse situations.

Language Study

Question (i)
You have many more important worries. (Change the degree)
Answer:
This worry is not as important as many other worries you have.

Question (ii)
Most of the time the rope was loose. (Make negative)
Answer:
Most of the time the rope was not tight.

Question (iii)
My mind was absolutely clear. (Use past perfect tense and rewrite)
Answer:
My mind had been absolutely clear.

Question (iv)
My first thought on reaching the top was a sense of gratitude to God. (Use ‘When’ and rewrite)
Answer:
When I reached the top, my first thought was a sense of gratitude to God.

Vocabulary:

Question 1.
Add suffixes to convert the following verbs into nouns.

1. confirm
2. know
3. near
4. move

Answer:

1. confirmation
2. knowledge
3. nearness
4. movement

Question 2.
Give one word for:
1. satisfy one’s thirst
2. very happy and excited
Answer:
1. quench
2. exhilaration

Glossary:

1. anchor – support / production
2. steep – high / sharp
3. expedition – mission
4. incline – slope
5. taut – tightly
6. bulky – large
7. descent – going down / coming down
8. spared – saved
9. flushed – glowed
10. alternated – changing places
11. sloping – slanting up or down
12. summit – the highest point of a hill or mountain
13. controversy – prolonged public disagreement or heated discussion
14. pact – a formal agreement between individuals or parties
15. ridge – a long narrow piece of raised land
16. embrace – hug, hold (someone) closely in one’s arms
17. accord – agreement, be harmonious or consistent with
18. quench – satisfy one’s thirst
19. exhilarated – very happy and excited
20. gale – strong wind
21. apparently – as far as one knows or one can see.

11th Std English Questions And Answers

• Being Neighborly Class 11 Question And Answers
• On To The Summit: We Reach The Top Class 11 Question And Answers
• The Call of the Soil Class 11 Question And Answers
• Pillars of Democracy Class 11 Question And Answers
• Mrs. Adis Class 11 Question And Answers
• Tiger Hills Class 11 Question And Answers

Balbharti Yuvakbharati English 11th Digest Chapter 1.1 Being Neighborly Notes, Textbook Exercise Important Questions and Answers.

Class 11 English Chapter 1.1 Being Neighborly Question Answer Maharashtra Board

11th Std English Chapter 1.1 Brainstorming Question Answer

Yuvakbharati English Navneet 11th Digest PDF Free Download Maharashtra Board

Read the following statements and mark those that apply to you.

Question 1.
(i) I make friends easily.
(ii) I wish to be friends with someone but my friendship is rejected.
(iii) Someone has extended a hand of friendship towards me and I have not accepted it.
(iv) I have a large group of friends but no best buddy.
(v) I have a small group of close friends and have no wish to interact with anyone else.
(vi) I have cordial relationships with all but I cannot connect with anyone.
Answer:
(i) I make friends easily.
(iv) I have a large group of friends but no best buddy.

Complete the following web diagram.

Question (i)

Answer:

Question (ii)
If you see someone lonely or sad you will –
(a) try to cheer the person by talking something pleasant.
(b) try to distract the person’s attention by doing some activity together.
(c) discuss the problem if the person wants to, give a patient hearing and also try to suggest some possible solutions.

(A1)

Question (i)
Jo’s decision to make friends with the lonely boy next door proves to be a good one. Elaborate. You may begin with ‘Jo was a bold, friendly and warm person…”
Answer:
Jo was bold, friendly and warm person who observed the boy next door closely and came to the conclusion that he was hungry for friends and fun. She was sad for the boy and felt it her neighborly duty to help the boy come out of his loneliness in her own way. She took a quick decision to catch the boys attention by throwing snowballs towards the window from where he was peeping.

She took the initiative to start an immediate conversation which was well-received by the boy. He invited her home and Jo readily accepted. Jo’s friendliness made the boy feel comfortable and he enjoyed Jo’s way of talking, her humour and most importantly, her companionship. He showed her his library and together they had a gala time which he never had before.

Question (ii)
Complete the following statements. (Answers are given directly in bold)
Answer:

1. To Jo the fine house seemed like an enchanted house.
2. Jo swept a path around the garden for Beth to walk in when the sun came out.
3. Jo entered the old stone house carrying her broom.
4. In order to tidy the room, Jo had whisked things into place.

Question (iii)
Bring out the contrast between the two houses with the help of the following points:
Answer:

House of March	House of Laurence
1. Old, brown house 2. Rather bare and shabby 3. Children played all around 4. A lively household having four girls and a loving mother	(a) Stately stone mansion (b) Stately stone mansion (c) Well kept grounds (d) All quiet, curtain down at the lower windows No motherly face smiled at the windows

(A2)

Question 1.
The traits of the characters you meet in the extract are jumbled. Sort them out and write them in the appropriate columns.
(Shy, bold, gruff, friendly, withdrawn, perceptive, empathetic, playful, lonely, happy, gentlemanly, frank, mature, dull, sharp, adventurous.)

Jo	Laurie	Grandpa
adventurous	dull	gruff
bold	lonely	withdrawn
playful	gentlemanly	frank
happy	shy	sharp
friendly	mature
empathetic	sharp
perceptive	friendly

(A3)

Question (i)
Write down in your own words the way Laurie confirmed the names of the March sisters.
Answer:
The March family sometimes forgot to put the curtain down at the window and that helped Laurie, their neighbor, to observe minutely inside the March household. The sisters often call one another and lonely Laurie enjoys watching them having good time. That’s how he came to know that Beth is the one who is generally a home-bird but whenever she goes out, she carries a basket with her. Amy’s curly hair and Meg’s pretty face has also caught his attention.

Question (ii)
Give a brief account of the interaction between Grandpa and Jo.
Answer:
Grandpa and Jo had an interesting interaction as Jo had come out of her initial fear after having a closer look at him. Grandpa had overheard Jo’s comments on his portrait and Jo did not even try to deny any one of them. This pleased Grandpa immensely and he remembered Jo’s grandfather who was similarly brave and honest.

Jo frankly told Grandpa about the problem Laurie was facing because of his loneliness. She showed her concern and expressed the March sisters’ eagerness to help Laurie. They started talking informally about Hemmel family, Jo’s mother and he also invited Jo to join for tea which Jo courteously accepted. This interaction made Jo very satisfied as she could find out how good their neighbor was.

(A4)

Question (i)
Find proverbs, maxims and idioms related to ‘friendship’.
Answer:

1. Birds of a feather flock together.
2. A friend in need is a friend in deed.
3. Friendship is love with understanding.
4. To get on like a house on fire (idiom).
5. Like two peas in a pod (idiom).

Question (ii)
The Extract deals with the atmosphere of two homes. Collect the words associated with –

1. Home
2. Library
3. Garden

Answer:

1. Home: old, brown, bare, shabby, stately stone mansion, comfort, luxury, big coach house, lovely things, rich curtains, lifeless, lawn, enchanted, hidden glories, full of splendour.
2. Library: books, pictures, statues, little cabinets, coins, sleepy hollow chairs, queer tables, quaint tiles, open fireplace, bronzes.
3. Garden: large, low hedge, vines, flowers.

(A5)

Change into indirect speech.

Question (a)
“Do you like your school?” asked the boy. “Don’t go to school I’m a businessman – girl, I mean”, answered Jo.
Answer:
The boy wanted to know whether she (Jo) liked school to which Jo answered quite emphatically that she did not go to school. She further added that she was a businessman and jovially corrected the gender.

Question (b)
Jo flourished her broom as she called out… “How do you do? Are you sick?
Laurie opened the window and croaked out as hoarsely as a raven…
“Better, thank you. I’ve had a bad cold, and been shut up a week.”
Answer:
Flourishing her broom Jo asked Laufie about his well-being and enquired whether he was sick. Laurie opened the window and croaked out as hoarsely as a raven thanking Jo for her concern and informed her that he was feeling better. He further added that he had been shut up a week as he had a bad cold.

Question (c)
“The pretty one is Meg, and the curly-haired one is Amy, I believe?” – Laurie.
“How did you find that out?” – Jo.
Answer:
Laurie wanted to confirm from Jo whether the pretty one was Meg and the curly-haired was Amy. With surprise in her voice Jo enquired how he(Laurie) had found that out.

Question (d)
“I’m not afraid of anything”, returned Jo with a toss of the head.
“I don’t believe you are !” exclaimed the boy.
Answer:
With a toss of the head Jo emphatically told that she was not afraid of anything. The boy was not surprised at her claim and agreed with her completely.

(A6)

Question (i)
Narrate in 100 words an incident, that illustrates the way a friend of yours ‘made you feel happy and accepted’, at some point in your life.
Answer:
That was my first day at school. I was just five years old. When my parents left me in school and I entered the classroom. I felt so lonely that I was about to cry. I was looking around, desperately trying to find out a known or a friendly face to talk to. Suddenly, there was a pat on my back and I saw a girl standing behind me.

She held my hand and took me to the bench where she was sitting. We became friends instantly. Till today we are the best friends. I shall be very grateful to her for her acceptance of me on the very first day at school.

Question (ii)
Give reasons, for us being reluctant to make friends with some strangers, but being comfortable with some, even after meeting them for the first time.
Answer:
Strangers are always mysteries for us. But some people have the inherent simplicity which instantly attract us towards them and we long to be friends with them. We feel comfortable in their company. But there is another category of people who have the attention-catching technique of blowing their own trumpets. It is difficult to carry on normal conversation with them as they are obsessed with their ownselves.

Question (iii)
Are friends different from neighbors? Are you friends with your neighbors? Give examples and write.
Answer:
It is not necessary to have one’s friend as one’s neighbor always. If it happens that way, then one is lucky. Friendship does not depend on the residence of a person and it can be different from the friendly relation one can have with one’s neighbor.

I am very fortunate to have a very friendly family as our neighbor. We are always there for one another at the hour of need as well as sharing happiness. For example, the owner of the apartment is a doctor and he is helping us with useful advices whenever anyone in our family falls sick. My mother is a teacher and she guides the children of our neighbor with their difficulties in studies. A good neighbor is always an asset.

Question (iv)
Make a note about how people amused themselves in earlier times without TV, internet or social media for entertainment.
Answer:
In earlier times when TV, internet and social media did not make people slaves of these sorts of entertainments, people used to socialize a lot. They used to meet their friends and relatives, talk to them over a telephone, make enquirers about each other’s well-beings and exchange ideas. The human connections were more and people used to share their joys and sorrows. Gone are those days of personal relationships which have been taken over by the modern technology enslaving people.

(A7)

Question 1.
Use your imagination and extend the story in about 100 to 130 words.
Answer:
Jo had a nice time with Laurie and his grandpa having tea and snacks which she enjoyed thoroughly. Both of them were very interesting characters, nice to talk to and Jo had an entertaining evening. She was excited to be acquainted with a friendly neighbor which she had always longed for. She was happy to go back home with so much of positive feelings about their neighbor who had been a mystery for her and her family.

Her entire family always felt sad for the lonely boy Laurie but nobody could approach him for helping him. She was extremely delighted to know Laurie and his grandpa who were courteous enough to invite her for tea. She was in a hurry to share her excitement with her family. “Ah! what a pleasant day it was !” she whispered.

(A8) Project:

If you are social, like to meet new people, can emphatise and connect with peole easily, make a list of careers available to you and write in brief about them. For example: Human Resource Development or HRD.

Yuvakbharati English 11th Digest Chapter 1.1 Being Neighborly Additional Important Questions and Answers

Question 1.
Jo doesn’t want to be a pussy-cat because –
Answer:
Pussy cat symbolises lethargy. Jo was always on the look out for excitements and thrills. She was an adventurous girl who does not want to idle away her time sleeping like a pussy-cat and enjoy the warmth of the fireplace on a cold winter afternoon. She would rather find out something interesting to spend her time.

Question 2.
Guess the meaning of‘hidden glories’ in the context of the mansion mentioned in the story.
Answer:
The expression has been used in the context of the mansion where Laurie lives. It has glimpses of lovely things and a look of an enchanted house, which probably hides lots of attractions inside.

Question 3.
Explain: “That boy is suffering for society and fun”.
Answer:
The young boy Laurie is lonely and longs for having fun with friends of his age-group, play with them and enjoy life the way a boy of his age does. The absence of company of friends and fun has made him dull which is affecting him like a disease.

Question 4.
Discuss ‘as dull as tombs’ and name the figure of speech.
Answer:
The figure of speech is ‘Simile’ as the dullness of the house is directly compared to the serious and dull atmosphere in a tomb. The boy meant by the expression that his house is very boring.

Question 5.
Complete the sentence: ‘a little gentleman’ means.
Answer:
The young boy is referred to as ‘a little gentleman’ here as he talks and behaves decently with others. He has a good upbringing which has taught him to welcome guests at his place by presenting himself as well as his room tidily.

Question 6.
Make a list of gifts you give/receive to/from your friends.
The gifts I usually give/receive to/from my friends are:

1. books
2. cosmetics
3. various food items
4. accessories

Question 7.
Complete the sentence in your own words : Hunger is related to food. Laurie is ‘hungry’ for –
Answer:
Laurie is a lonely young boy who is hungry for spending happy times both at home and with friends. He belongs to a rich family where he gets everything but suitable companions to have fun with. That is why he longs for food for his mind, that is, happy times with friends.

Question 8.
Laurie has
(i) _________
(ii) _________
(iii) _________
He doesn’t have
(i) __________
(ii) _________
(iii) _________
Answer:
Laurie
(i) a rich house filled with loneliness,
(ii) a kind but indifferent grandpa,
(iii) half a dozen servants and a tutor Mr. Brooke,

He doesn’t have
(i) his mother.
(ii) friends and companions,
(iii) any one to go out with.

Question 9.
Describe the effect of Laurie’s words on Jo.
Answer:
Jo started talking with Laurie frankly. Her words had lots of positive effects on Laurie as he was longing for exactly those things which Jo mentioned. For their first meeting, Jo was a bit blunt but Laurie liked her bluntness since he could recognize Jo’s sincerity and kindness hidden in those words. He started feeling comfortable in Jo’s company and enjoyed every bit of humorous description of Aunt March.

Question 10.
Find out what ‘good breeding’ means.
Answer:
‘Good breeding’ means that a person is well-behaved, polite, cultured and refined, which are the results of his upbringing, training as well as family atmosphere.

Question 11.
List some of the things that you need in order to be happy.
Answer:
Things that I need in order to be happy are –
(i) a supportive family and dependable friends.
(ii) hobbies to occupy myself during my free time.
(iii) a healthy life for me as well as my family members.
(iv) sufficient money earned from a satisfying career.

Question 12.
“A fellow can’t live on books” – Explain.
Answer:
A fellow, of course, cannot live on books though books are his emotional suppdrt and in many ways, his best friend. But he also needs someone, a companion, with whom he can share his feelings, fulfill his curiosities, have fun, etc. Human contact is a necessity in a person’s life since a few words, an exchange of ideas collected from the books can work wonders giving immense pleasure.

Question 13.
List the things that Jo notices in the portrait.
Answer:

1. The gentleman in the portrait is not as handsome as her own grandfather.
2. Though the gentleman is having a grim face, his kind eyes assure that there is nothing to be afraid of him.
3. From his looks, it appears that he has tremendous will-power.

Question 14.
Find out the reason for Jo’s dismay.
Answer:
Jo loudly expressed her opinions on Laurie’s grandfather, while looking at his portrait. When she came to know that the gentleman had heard all her comments, she felt embarrassed. She felt uncomfortable to face the old gentleman and felt like running away.

Question 15.
Complete the sentence.
Answer:
In spite of Jo’s apprehensions, Grandpa is –

1. having kinder eyes than what the painting shows.
2. having a shy twinkle in his eyes which could lessen Jo’s fear.
3. quite a friendly gentleman.

Question 16.
Discuss what Jo meant by –
(i) “only trying to be neighborly, Sir.”
(ii) seems a little lonely.
(iii) splendid Christmas present.
Answer:
(i) By saying, only trying to be neighborly, Sir”. Jo means she just wanted to be friendly with Laurie as he was her neighbor. She strongly felt that neighbors should know each other well.
(ii) Jo had observed Laurie now and again and she felt Laurie badly needed company since he always eagerly looked at his neighbors as if he was missing the fun they were having. To her, he appeared to be a lonely boy longing for enjoyment with friends.
(iii) Jo remembered the beautiful Christmas present that was sent to the March family by their neighbor Mr. Laurence and she felt it was a nice gesture by their neighbor.

Question 17.
Guess the meaning of the phrase “go on being neighborly” in the context.
Answer:
The phrase “go on being neighborly” in the context of the story means being friendly and helpful to the people living in one’s neighborhood.

Question 18.
Bring out the contrast in the lives of Jo and Laurie in a few lines.
Answer:
Jo belonged to a happy family who according to Laurie, had always good times together. Laurie was hungry to have company of friends and was a lonely boy. Jo had a loving mother who used to take care of her children but, Laurie was a motherless child who badly missed his mother. Jo was frank and innocent as any child of her age, whereas, Laurie’s loneliness was sickening for him.

Question 19.
I’m happy as a cricket here. (Name and explain the figure of speech)
Answer:
Simile. The happiness of Jo is directly compared to the happiness of the insect cricket.

Question 20.
Guess the meaning of the word ‘affair’ in the context.
Answer:
The word ‘affair’ in the context means ‘responsibility/matter’.

Comprehension:

Read the extract and complete the activities given below.

Global Understanding:

Question 1.
Complete the table.
The traits of Jo and Laurie are jumbled. Sort them out and write them in appropriate columns.
(companionless, adventurous, empathetic, unenergetic) (Answers are given directly)
Answer:

Jo	Laurie
adventurous	companionless
empathetic	unenergetic

Question 2.
Pick up the statements which confirm the theme of this passage.
(a) This passage is about Jo’s family not putting down the curtain.
(b) This passage is about Jo’s confirmation about Laurie’s loneliness.
(c) This passage is about Laurie’s habit of peeping at Jo’s family.
(d) This passage brings out the contrast in the lives of Jo and Laurie.
Answer:
(b) This passage is about Jo’s confirmation about Laurie’s loneliness.
(d) This passage brings out the contrast in the lives of Jo and Laurie.

Question 3.
Complete the sentences in column ‘A’ by matching them with the clues in column ‘B’

Column ‘A’	Column ‘B’
1. Laurie inspite of being inquisitive asked no questions as ____________ .	(a) Laurie seldom laughed aloud
2. Maid was surprised as ____________ .	(b) That indicated his good breeding
3. Jo was elated ____________ .	(c) As she was successful in making Laurie laugh
4. Jo found happiness in ____________ .	(d) Reading books

Answer:

Column ‘A’	Column ‘B’
1. Laurie inspite of being inquisitive asked no questions as ____________ .	(a) That indicated his good breeding
2. Maid was surprised as ____________ .	(b) Laurie seldom laughed aloud
3. Jo was elated ____________ .	(c) Reading books
4. Jo found happiness in ____________ .	(d) As she was successful in making Laurie laugh

Question 4.
Complete the following statement with four correct information from the extract.
Jo felt Laurie needs cheering up because:

1. ________
2. ________
3. ________
4. ________

Answer:

1. He seemed lonely
2. she was being neighborly
3. She was social and empathetic
4. Laurie looked expectantly at the sisters as they bad good time.

Complex Factual:

Question 1.
What were Jo’s queries to Laurie when they had talked for the first time?
Answer:
Jo wanted to know whether Laurie was sick, how he amused himself, his liking for books and if he had any visitor or not.

Question 2.
Mention any two outcomes of Jo’s visit to Laurie’s place.
Answer:
Jo’s visit made Laurie excited in the expectation of getting a companion which he never had. It also helped him to come out of his shyness and converse with Jo freely.

Question 3.
What was Jo’s suggestion to do away with Laurie’s loneliness.
Answer:
Jo assured Laurie that the curtain at her place would never be drawn so that Laurie can spend time looking at their activities. She also suggested that Laurie could come to their home and get himself entertained by. all the members of her family.

Question 4.
Why was Laurie’s grandfather impressed with Jo?
Answer:
Laurie’s grandfather was impressed by Jo’s spirited answers like her grandfather. He also appreciated that she was brave and honest as her grandfather was.

Question 5.
Mention any four changes that occurred in grandfather after meeting Jo.
Answer:

1. Grandfather shed his strict countenance.
2. He invited Jo to come over for tea.
3. He promised to come over to meet Jo’s mother.
4. He offered Jo his arm with old fashioned courtesy (indicating his gratitude for her neighborly arrival)

Inference / Interpretation / Analysis:

Question 1.
Complete the following statement.
Jo wanted to help Laurie because –
Answer:
Jo wanted to help Laurie because Laurie was unwell and he felt lonely as he had no one of his age at home. He deserved to have fun.
OR
Give reasons
“The big eyes brightened and the mouth began to smile”.
Answer:
The brightness in the eyes of Laurie and his smiling face are proofs of his happiness of having a possible friendship with Jo. When Jo threw snowballs at Laurie, he could feel Jo’s eagerness to talk to him. His loneliness has always made him unhappy and this gesture of Jo is a welcome change for him.

Question 2.
Complete the following sentence Mr. Laurie was a Tittle gentleman’ as …
Answer:
Mr. Laurie was a Tittle gentlemen’ because he was known for offering due respect to the guest who was coming to his place. He prepared himself decently by brushing his pate, pulling on a fresh set of clothing and making an attempt to clean his room. He followed the same routine for Jo’s arrival to his house also.

Question 3.
Point out the reason for Laurie’s minute observation of Jo’s family.
Answer:
Laurie spent his lonely time looking at the fun Jo’s family was having. He enjoyed watching each member of the family eagerly as he missed all those good times at his own home. He did not have his mother and he loved these girls in the company of their mother.

Question 4.
Mention the impact of Jo’s narratives on Laurie –
1. …………….
2. ……………..
Answer:
1. Laurie enjoyed her narrative immensely and he laughed out aloud.
2. He found a sudden merriment in otherwise dull mood owing to his illness.

Question 5.
Jo wasn’t scared of Laurie’s grandfather. Give evidence from the passage quoted to you.
Answer:
Jo was a bold girl who spoke her mind when it was needed. As she looked at Mr. Laurence’s portrait she found his eyes to be kind and grew fond of him instantly. She found him to be compassionate as she spoke to him and was confident that there was nothing to be scared of that gentleman.

Question 6.
Complete the boxes with information.

Answer:

Personal Response:

Question 1.
“Girls are quiet and like to play nurse”. Do you Agree or Disagree? Justify your answer.
Answer:
I do not agree to the statement. There is no hard and fast rule about this impression about girls. There may be many boys who are very quiet and also good at nursing. It depends on the nature of a person whether he/ she is quiet, or noisy or talkative. The ability to nurse somebody is also dependant on the ability of a person irrespective of any gender.

Question 2.
Enlist the gifts that you receive from or give to your friends,
Answer:

1. Books
2. Wind Chimes
3. Coffee mugs
4. Photo frames

Question 3.
Give your suggestions in two sentences. How you will cheer up one of your lonely classmates.
Answer:
I can cheer up my lonely classmate by giving him/her company and involving in some activities together. We can sit together in the classroom, share our tiffins and invite him/her at my place on holidays.

Question 4.
What are you afraid of? Explain your answer.
Answer:
Generally I am not afraid of anything and a carefree type of person. But sometimes I worry about the loss of my near and dear ones as I am very attached to my family and friends.

Question 5.
What do you fear the most? why?
Answer:
As a student I fear examination especially the public exams as they decide the future course of action. There is always an element of uncertainty which brings in fear for exams among students.

Question 6.
How do you help your neighbor?
Answer:
I help my neighbor by making myself available when they need me. I also extend courtesy calls when I meet them.

Language Study:

Question (i)
The idea amused Jo who liked to do daring things
Answer:
Jo liked to do daring things and the idea amused her.

Question (ii)
The boy is suffering for society.
Answer:
The boy has been suffering for society.

Question (iii)
What a cozy room this is! (Rewrite as a statement)
Answer:
This room is indeed very cozy.

Question (iv)
Laurie forgot his bashfulness and grew sociable. (Remove ‘and’ to make it a simple sentence)
Answer:
Forgetting his bashfulness Laurie grew sociable.

Question (v)
Her face was very friendly and her sharp voice unusually gentle.
(Use ‘not only but also’ and rewrite)
Answer:
Her face was not only very friendly but her sharp voice was unusually gentle also.

Question (vi)
She had been so simply taught that there was no nonsense in her head. (Use ‘too’)
Answer:
She had been too simply taught to have any nonsense in her head.

Question (vii)
Laurie enjoyed that immensely. (Use ‘enjoyment’and rewrite)
Answer:
Laurie’s enjoyment at that was immense.

Question (vii)
Jo liked his good breeding. (Frame a Wh-question to get the underlined part as an answer)
Answer:
What did Jo like in him?

Question (viii)
For a minute a wild desire to run away possessed her. (Change the voice)
Answer:
For a minute she was possessed by a wild desire to run away.

Question (ix)
He isn’t as handsome as my grandfather, but I like him. (Use ‘Though’)
Answer:
Though he isn’t as handsome as my grandfather, I like him.

Question (x)
He seems a little lonely. (Frame a question to get the underlined part as answer)
Answer:
How does he seem?

Question (xi)
I shall come and see your mother. (Use a modal auxiliary showing ‘obligation’)
Answer:
I must come and see your mother.

Vocabulary:

Question 1.
Match the words in Column ‘A’ with their meanings in Column ‘B’.
Answer:

Column ‘A’	Column ‘B’
1. queer	(a) bold
2. dismal	(b) frail
3. daring	(c) unusual
4. weak	(d) dull

Answer:

Column ‘A’	Column ‘B’
1. queer	(c) unusual
2. dismal	(d) dull
3. daring	(a) bold
4. weak	(b) frail

Question 2.
Mention any 4 adjectives that describe the traits of Jo’s character in the extract.
Answer:
Character
Jo

Adjectives

1. Cheerful
2. Empathetic
3. Sociable
4. Kind

Question 3.
Give antonyms of the following words,

1. rude
2. splendid
3. funny
4. frank

Answer:

1. polite
2. ordinary
3. serious

Question 4.
Give antonyms of the following using prefix.
Answer:
1. interesting × uninteresting
2. afraid × unafraid

Question 5.
Give one word for:
Answer:

1. of a voice: low and rough – Gruff
2. Lacking courage – Cowardly
3. unpleasant – Dreadful
4. very large or great – Tremendous

Glossary:

1. shovel – spade
2. mischievous – naughty
3. to doze – to sleep lightly
4. groves – group of trees
5. hedge – row of bushes
6. shabby – broken down/dilapidated
7. stately – grand
8. mansion – big house
9. betokening – a sign of something
10. glimpses – brief/faint looks
11. frolicked – played fun games
12. enchanted – attractive/ fascinating
13. splendors – richness / luxury
14. behold – look/see
15. scandalizing – shocking /disgusting
16. queer – odd/unusual/funny
17. dismal – dull row – noise
18. flutter – tremendous/full of
19. pate – head
20. parlor – sitting room
21. briskly – quickly
22. comforting – soothing
23. sociable – friendly
24. cozy – comfortable
25. hearth – floor of fireplace
26. whisked – removed
27. beckoned – called
28. twitching – shivering
29. splendid – grand/superb
30. bother- trouble/nuisance
31. acquainted – be familiar
32. blunt – frank/straightforward
33. fidgety – restless
34. poodle – a bread of dog
35. immensely – vastly/very much
36. tweaked – pulled
37. elated – delighted
38. trifle – little
39. quaint – old-fashioned /unusual/attractive
40. velour – woven fabric
41. grim – ill-tempered/stern
42. gruff – rough
43. cowardly – fearful
44. twinkle – shining
45. dreadful – terrible
46. courtesy – politeness
47. colored up – embarrassed
48. to wait on – act as an attendant to
49. pranced – walked in an energetic way
50. wicked – playfully mischievous
51. affair – matter/responsibility
52. good breeding – being raised well/ the result of good upbringing and training for good manners.

11th Std English Questions And Answers

• Being Neighborly Class 11 Question And Answers
• On To The Summit: We Reach The Top Class 11 Question And Answers
• The Call of the Soil Class 11 Question And Answers
• Pillars of Democracy Class 11 Question And Answers
• Mrs. Adis Class 11 Question And Answers
• Tiger Hills Class 11 Question And Answers

Balbharti Maharashtra State Board 11th Biology Textbook Solutions Chapter 3 Kingdom Plantae Textbook Exercise Questions and Answers.

Kingdom Plantae Class 11 Exercise Question Answers Solutions Maharashtra Board

Class 11 Biology Chapter 3 Exercise Solutions Maharashtra Board

Biology Class 11 Chapter 3 Exercise Solutions

1. Choose the correct option.

Question (A)
Which is the dominant phase in Pteridophytes?
(a) Capsule
(b) Gametophyte
(c) Sporophyte
(d) Embryo
Answer:
(c) Sporophyte

Question (B)
The tallest living gymnosperm among the following is
(a) Sequoia sempervirens
(b) Taxodium mucronatum
(c) Zamia pygmaea
(d) Ginkgo biloba
Answer:
(a) Sequoia sempervirens

Question (C)
In Bryophytes
(a) sporophyte and gametophyte generation are independent
(b) sporophyte is partially dependent upon gametophyte
(c) gametophyte is dependent upon sporophyte
(d) inconspicuous gametophyte
Answer:
(b) sporophyte is partially dependent upon gametophyte

Question (D)
A characteristic of Angiosperm is
(a) Collateral vascular bundles
(b) Radial vascular bundles
(c) Seed formation
(d) Double fertilization
Answer:
(d) Double fertilization

Question (E)
Angiosperms differ from gymnosperms in having
(a) Vessels in wood
(b) Mode of nutrition
(c) Siphonogamy
(d) Enclosed seed
Answer:
Both (a) Monocotyledons and (d) Enclosed seed

Question 2.
How you place the pea, jowar and fern at its proper systematic position? Draw a flow chart.
Answer:

Question 3.
Complete the following table.

Groups of algae	Chlorophyceae	Phaeophyccac	Rhodophyceae
1. Stored food	Starch
2. Cell wall	—	Cellulose and algin
3. Major pigments	—		Chl-a, d and phycoerythrin

Answer:

Groups of algae	Chlorophyceae	Phaeophyccac	Rhodophyceae
1. Stored food	Starch	Mannitol, laminarin	Floridean starch
2. Cell wall	Cellulose	Cellulose and algin	Cellulose, pectin
3. Major pigments	Chl-a, b	Chl-a, c, fucoxanthin	Chl-a, d and phycoerythrin

Question 4.
Differentiate between Dicotyledonae and Monocotyledonae based on the following characters:
a. Type of roots
b. Venation in the leaves
c. Symmetry of flower
Answer:

Characters	Dicotyledonae	Monocotyledonae
1. Type of roots	Taproots	Fibrous roots
2. Venation in the leaves	Reticulate venation	Parallel venation
3. Symmetry of flower	Tetramerous or Pentamerous symmetry	Trimerous symmetry

Characters Dicotyledonae Monocotyledonae
1. Type of roots Tap roots Fibrous roots
2. V enation in the leaves Reticulate venation Parallel venation
3. Symmetry of flower Tetramerous or Pentamerous symmetry Trimerous symmetry

5. Answer the following questions.

Question (A)
We observe that land becomes barren soon after monsoon. But in the next monsoon it flourishes again with varieties we observed in season earlier. How you think it takes place?
Answer:

1. After monsoon, plants like mosses (bryophytes), ferns (pteridophytes), small herbaceous plants, etc become dry, due to which land becomes barren.
2. However, spores of bryophytes, pteridophytes and seeds of herbaceous plants, grass remain in barren land.
3. During next monsoon, these spores and seeds germinate due to availability of water and other favourable conditions.
4. Bryophytes and pteridophytes require water for reproduction. Hence they flourish during monsoon season.
5. Along with bryophytes and pteridophytes varieties of higher plants like grasses, some seasonal herbs or shrubs grow on barren land during monsoon due to favourable conditions.

Question (B)
Fern is a vascular plant. Yet it is not considered a Phanerogams. Why?
Answer:

1. Fern belongs to sub-kingdom Cryptogamae.
2. Cryptogams produce spores but do not produce seeds.
3. Also, in cryptogams the sex organs are concealed.
4. Phanerogams are seed producing plants and their sex organs are visible.
5. Hence, fern is a vascular plant. Yet it is not considered a Phanerogams.

Question (C)
Chlamydomonas is microscopic whereas Sargassum is macroscopic; both are algae. Which characters of these plants includes them in one group?
Answer:

1.  Both Chlamydomonas and Sargassum belong to division Thallophyta.
2. Members of Thallophyta range from unicellular (e.g. Chlamydomonas) to multicellular (e.g. Sargassum).
3. Both are aquatic plants containing photosynthetic pigments.
4. In both Chlamydomonas and Sargassum plant body is not differentiated into root, stem and leaves.
5. The stored food is mainly in the form of starch and its other forms.
6. Cell wall is made up of cellulose and other components. Due to these characters, both Chlamydomonas and Sargassum are included in one group i.e. Thallophyta.

Question 6.
Girth of a maize plant does not increase over a period of time. Justify.
Answer:

1. Maize plant belongs to class monocotyledonae.
2. In monocotyledonous plants, vascular bundles are closed type.
3. Thus, cambium is absent between xylem and phloem, due to which secondary growth does not occur in these plants.
4. Increase in girth of a stem occurs by secondary growth. Thus, girth of a maize plant does not increase over a period of time.

Question 7.
Radha observed a plant in rainy season on the compound wall of her school. The plant did not have true roots but root like structures were present. Vascular tissue was absent. To which group the plant may belong?
Answer:
The plant observed by Radha belongs may belong to division Bryophyta, as it shows root like structures i.e. rhizoids and absence of vascular tissue.

8. Draw neat labelled diagrams

Question 1.
Draw neat and labelled diagram of:
(A) Spirogyra
(B) Chlamydomonas
Answer:

Question (C)
Draw neat and labelled diagram of Funaria.
Answer:

Question (D)
Draw neat and labelled diagram of Nephrolepis.
Answer:

[Note: Frond: Fern leaf, originating from rhizome. It consists of blade and petiole, Blade: Main part of the frond which is rich in chlorophyll]

Question (E)
Draw neat and labelled diagram of Haplontic and Haplo-diplontic life cycle.
Answer:

Question 9.
Identify the plant groups on the basis of following features:
A. Seed producing plants
B. Spore producing plants
C. Plant body undifferentiated into root, stem and leaves
D. Plant needs water for fertilization
E. First vascular plants
Answer:
1. Phanerogams (Angiospermae and Gymnospermae)
2. Cryptogams (Thallophyta, Bryophyta and Pteridophyta)
3. Thallophyta, Bryophyta
4. Thallophyta, Bryophyta, Pteridophyta
5. Pteridophytes

Practical/Project:

Question 1.
Study the Nephrolepis plant in detail.
Answer:

1. Nephrolepis belongs to division pteridophyta.
2. They grow abundantly in cool, shady, moist places.
3. Roots are adventitious (fibrous) growing from the underground stem.
4. Leaves are well developed on the stem (Rhizome).
5. They show presence of well-developed conducting system for transportation of water and food.
6. They reproduce asexually by spores produced within sporangia, which are present in sori. These sori are located along the posterior surface of leaflets.
7. These plants have neither fruits nor flowers.
8. Some ferms are used as food, medicine or as ornamental plants.

Question 2.
Study the coralloid roots, scale leaf and megasporophyll of Cycas in detail.
Answer:
1. Coralloid roots of Cycas:
Coralloid roots of Cycas show association with blue green algae for nitrogen fixation.
Coralloid roots are coral-like, dichotomously branched and fleshy. They grow upward toward the surface of the soil. These roots arise from the lateral branches of normal roots.
2. Scale leaf of Cycas:
In Cycas leaves are dimorphic i.e. foliage leaves and scale leaves. Scale leaves are minute, membranous and brown. These are non- photosynthetic and provide protection to the stem apex.
3. Megasporophyll of Cycas:
Megasporophylls are usually arranged in compact structures called female cones or female strobili. Megasporophyll contains megasporangia (ovule) which produce megaspores.
[Students are expected to collect more information about coralloid roots, scale leaf and megasporophyll of Cycas.]

Question 10.
Observe the following diagram. Correct it and write the information in your words.

Answer:

1. The given figure indicates alternation of generation.
2. The life cycle of a plant includes two generations, sporophytic (diploid = 2n) and gametophytic (haploid = n)
3. Some special diploid cells of sporophyte divide by meiosis to produce haploid cells.
4. These haploid cells divide mitotically to produce gametophyte.
5. On maturation, gametophyte produces male and female gametes which fuse during fertilization and produce diploid zygote.
6. Diploid zygote divides by mitosis and forms diploid sporophyte.

11th Biology Digest Chapter 3 Kingdom Plantae Intext Questions and Answers

Can you recall? (Textbook Page No. 19)

Why do we call plants as producers on land?
Answer:
Plants can prepare their own food by the process of photosynthesis. Hence, they are called as producers on land.

Can you recall? (Textbook Page No. 19)

What are differences between sub-kingdoms cryptogamae and Phanerogamae?
Answer:

Cryptogamae	Phanerogamae
1. Plants belonging to this sub-kingdom are non­flowering.	Plants belonging to this sub-kingdom are flowering.
2. Sex organs are concealed.	Sex organs are visible.
3. These plants do not produce seeds.	These plants produce fruits and seeds.
4. An ovule is not formed.	An ovule is formed.
5. It is further divided into three divisions, viz.	It is further divided into two divisions, viz.
6. Thallophyta, Bryophyta and Pteridophyta.	Gymnospermae and Angiospermae.

Observe and Discuss (Textbook Page No. 19)

Collect different water samples of fresh water. Mount them on a glass slide and observe under a compound microscope. Try to identify the organisms which are visible under it.
Answer:
Micro-organisms like Paramoecium, Amoeba, blue-green algae, unicellular algae, filamentous algae can be observed under compound microscope.
[Students are expected to observe different water samples of fresh water under compound microscope and identify the organisms.]

Can you tell? (Textbook Page No. 21)

Give salient features of algae.
Answer:
Algae belongs to division Thallophyta.
Salient features of algae:
1. Habitat: Algae are mostly aquatic, few grow on other plants as epiphytes and some grow symbiotically. Some algae are epizoic i.e. growing or living non-parasitically on the exterior of living organisms.
Aquatic algae grow in marine or fresh water. Most of them are free-living while some are symbiotic.

2. Structure: Plant body is thalloid i.e. undifferentiated into root, stem and leaves. They may be small, unicellular, microscopic like Cblorella (non-motile), Chlamydomonas (motile). They can be multicellular, unbranched, filamentous like Spirogyra or branched and filamentous like Chara. Sargassum is a huge macroscopic sea weed which measures more than 60 meters in length.

3. Cell wall: The algal cell wall contains either polysaccharides like cellulose / glucose or a variety of proteins or both.
Reserve food material: Reserve food is in the form of starch and its other forms.

4. Photosynthetic pigments: Photosynthetic pigments like chlorophyll – a, chlorophyll – b, chlorophyll – c, chlorophyll – d, carotenes, xanthophylls, phycobilins are found in algae.

5. Reproduction: Reproduction takes place by vegetative, asexual and sexual method.

6. Life cycle: The life cycle shows phenomenon of alternation of generation, dominant haploid and reduced diploid phases.

Internet my friend (Textbook Page No. 20)

Write different pigments found in algae.
Answer:
Various types of photosynthetic pigments are found in algae.
1. Chlorophyll-a (Essential photosynthetic pigment) is present in all groups of algae.
2. The accessory pigments are chlorophyll-b, chlorophyll-c, chlorophyll-d, carotenes, xanthophylls and phycobilins. Phycobilins are of two types, i.e. phycocyanin and phycoerythrin.
[Students are expected to collect more information about pigments found in algae from internet.]

Can you tell? (Textbook Page No. 21)

Name the accessory pigments of algae.
Answer:
The accessory pigments are chlorophyll-b, chlorophyll-c, chlorophyll-d, carotenes, xanthophylls and phycobilins. Phycobilins are of two types, i.e. phycocyanin and phycoerythrin.
[Students are expected to collect more information about pigments found in algae from internet.]

Can you tell? (Textbook Page No. 21)

Differentiate between Chlorophyceae and Phaeophyceae.
Answer:

Chlorophyceae (Green algae)	Phaeophyceae (Brown algae)
1. Photosynthetic pigments are chlorophyll-a, chlorophyll-b.	Photosynthetic pigments are chlorophyll-a, chlorophyll-c and fucoxanthin.
2. Reserve food is in the form of starch.	Reserve food is mannitol and laminarin.
3. e.g. Chlorella, Chlamydomonas, Spirogyra, Chara, I Volvox, Ulothrix	Ectocarpus, Sargassum, Fucus, Laminaria, etc.

Can you tell? (Textbook Page No.21)

Enlist examples of Chlorophyceae and Rhodophvceae.
Answer:
1. Examples of Chlorophyceae:
Chlorella, Chlamydomonas, Spirogyra, Char a, Volvox, Ulothrix, etc.
2. Examples of Rhodophyceae:
Chondrus, Batrachospermum, Porphyra, Gelidium, Gracillaria, Polysiphonia, etc.

Internet my friend (Textbook Page No. 21)

Different forms of green, red, brown and blue green algae.
Answer:
1. Forms of green algae:
Unicellular motile: e.g. Chlamydomonas Unicellular non-motile: E.g. Chlorella Colonial forms: e.g. Volvox Filamentous branched: e.g. Cladophora, Chara Filamentous unbranched: e.g. Ulothrix, Spirogyra

2. Forms of red algae:
The red thalli of most of the red algae are multicellular, macroscopic, e.g. Gracilaria, Gelidium, Porphyra, Polysiphonia, etc. .

3. Forms of brown algae:
Simple, branched and filamentous: Sargassum, Fucus, Ectocarpus Profusely branched: Laminaria, Dictyota, Kelps (Seaweed)

4. Forms of blue-green algae:
Unicellular, colonial or filamentous, freshwater or marine water or terrestrial algae.
[Note: Blue-green algae are cyanobacteria which are photosynthetic autotrophs.]
[Students are expected to collect more information from internet.]

Internet my friend (Textbook Page No. 20)
Enlist the forms of filamentous algae.
Answer: The forms of filamentous algae:
1. Filamentous branched: e.g. Cladophora, Chara, Ectocarpus, Dictyota, etc.
2. Filamentous unbranched: e.g. Ulothrix, Spirogyra, etc.

Internet my friend (Textbook Page No. 21)

Economic importance of algae.
Answer:
(a) Many species of algae are used as food. For e.g. Chlorella (rich in cell proteins hence used as food supplement, even by space travelers), Sargassum, Laminaria, Porphyra, etc.
(b) Alginic acid is produced commercially from Kelps.
(c) Hydrocolloids like algin and carrageen are obtained from brown algae and red algae respectively.
(d) ‘Agar’ which is used as solidifying agent in tissue culture is obtained from red algae like Gelidium and Gracilaria.
(e) Brown algae like sea weeds are used a fodder for sheep, goat, etc.
[Students are expected to collect more information about the economic importance of algae.]
(f) Role of algae in environment.
Answer:
(a) Being photosynthetic, algae help in increasing the level of dissolved oxygen in their immediate environment.
(b) Algae are primary producers of energy rich compounds which forms the basis of food cycles in aquatic animals.
[Students are expected to find out more information about the role of algae in environment on internet.]

Can you recall? (Textbook Page No. 19)

Differentiate between Thallophytes and Bryophytes.
Answer:

Thallophytes	Bryophytes
1. Mostly aquatic in habitat.	Mostly terrestrial, occurs on moist and shady places.
2. Thallus may be unicellular or multicellular.	Thallus is multicellular.
3. Motile and non-motile forms are present.	Non-motile forms present, except male gametes.
4. Rhizoids are absent.	Rhizoids are present.

Can you tell? (Textbook Page No. 23)

Why Bryophyta are called amphibians of Plant Kingdom?
Answer:
Members of Bryophyta are mostly terrestrial plants which depend on water for fertilization and completion of their life cycle. Hence, they are called ‘amphibians of Plant Kingdom’.

Observe and Discuss (Textbook Page No. 21)

You may have seen Funaria plant in rainy season. Why is it called amphibious plant?
Answer:
Funaria belongs to division Bryophyta.
It is a terrestrial plant but requires water for fertilization and completion of its life cycle. Hence, it is called as an amphibious plant.

Observe and Discuss (Textbook Page No. 23)

You may have seen the various plants which do not bear flowers, fruits and seeds but they have well developed root, stem and leaves. Discuss.
Answer:
1. The plants which do not bear flowers, fruits and seeds, but have true roots, stem and leaves belong to division Pteridophyta.
2. These plants are cryptogams as they do not produce seeds and flowers.
3. They have primitive conducting system.

Can you tell? (Textbook Page No. 23)

Pteridophytes are also known as vascular Cryptogams – Justify.
Answer:
1. The reproductive organs of pteridophytes are hidden.
2. Pteridophytes do not produce flowers, fruits and seeds. They reproduce asexually by forming spores and sexually by forming gametes, hence they belong to Cryptogamae.
3. These plants possess a primitive conducting system. Thus, conduction of water and food occurs through vascular tissue.
Hence, Pteridophytes are also known as vascular Cryptogams.

Can you tell? (Textbook Page No. 23)

Give one example of aquatic and xerophytic Pteridophytes.
Answer:
Habitat: Pteridophytes grow in moist and shady places, e.g. Ferns, Horsetail. Some are aquatic (Azolla, Marsilea), xerophytic (Equisetum) and epiphytic (Lycopodium).

Can you recall? (Textbook Page No. 19)

Give any two examples of Pteridophyta.
Answer:
Nephrolepis, Selaginella, Azolla, Marsilea, Equisetum, Lycopodium, Psilotum, Dryopteris, Pteris, Adiantum.

Can you tell? (Textbook Page No. 25)

Give general characters of Gymnosperms and Angiosperms.
Answer:
1. General characters of Gymnosperms:
(a) Types: Most of the gymnosperms are evergreen, shrubs or woody trees.
(b) Vascular tissues: They are vascular plants having xylem with tracheids and phloem with sieve cells.
(c) Flower: These are primitive group of flowering plants producing naked seeds.
(d) Body: The plant body is sporophyte. It is differentiated into root, stem and leaves.
(e) Roots: The root system is tap root type. In some gymnosperms, the roots form symbiotic association with other life forms. Coralloid roots of Cycas show association with blue green algae and roots of Pinus show association with endophytic fungi called mycorrhizae.
(f) Stem: In gymnosperms, stem is mostly erect, aerial, solid and cylindrical. Secondary growth is seen in Gymnosperms due to the presence of cambium. In Cycas it is usually unbranched, while in conifers it is branched, (e.g. Pinus, Cedrus).
(g) Leaves: The leaves are dimorphic. The foliage leaves are green, simple needle like or pinnately compound, whereas scale leaves are small, membranous and brown.
(h) Spores: Spores are produced by microsporophyll (Male) and megasporophyll (Female).

(ii) General characters of angiosperms:
(a) Habitat: Angiosperms is a group of highly evolved plants, primarily adapted to terrestrial habitat.
(b) Alternation of generations: Angiosperms show heteromorphic alternation of generation in which the sporophyte is diploid, dominant, autotrophic and independent. The gametophytes (male or female) are haploid, reduced and concealed.
(c) Spores and Sporophylls: Angiosperms are heterosporous. Microspores (commonly called pollens) are formed in microsporangia (or anthers). They develop in highly specialized microsporophyll or stamens while megaspores are formed in megasporangia (or ovules) borne on highly specialized megasporophyll called carpel.
(d) Flower: Besides the essential whorls of microsporophylls (androecium) and megasporophylls (gynoecium), there are accessory whorls namely, calyx (sepals) and corolla (petals) arranged together to form flowers.

Observe and Discuss (Textbook Page No. 23)

Observe all garden plants like Cycas, Thuja, Pinus, Sunflower, Canna and compare them. Note similarities and dissimilarities among them.
Answer:
1. When we observe garden plants like Cycas, Thuja, Pinus, Sunflower, Canna, following similarities can be observed:
Plant body is divided into root, stem and leaves.
2. When we observe garden plants like Cycas, Thuja, Pinus, Sunflower, Canna, following dissimilarities can be observed:
(a) In Cycas, Thuja and Pinus seeds are not enclosed within a fruit, whereas in Sunflower and Canna seeds are enclosed within a fruit.
(b) Plants like Cycas, Thuja, Pinus show cones bearing microsporophylls and megasporophylls, whereas sunflower and Canna plant bear flowers.
(c) In Cycas, Thuja and Pinus green, simple needle like or pinnately compound foliage leaves and brown, membranous scaly leaves can be observed, whereas in Sunflower, Canna green foliage leaves can be observed.

Can you recall? (Textbook Page No. 24)

What are the salient features of angiosperms?
Answer:
(ii) General characters of angiosperms:
(a) Habitat: Angiosperms is a group of highly evolved plants, primarily adapted to terrestrial habitat.
(b) Alternation of generations: Angiosperms show heteromorphic alternation of generation in which the sporophyte is diploid, dominant, autotrophic and independent. The gametophytes (male or female) are haploid, reduced and concealed.
(c) Spores and Sporophylls: Angiosperms are heterosporous. Microspores (commonly called pollens) are formed in microsporangia (or anthers). They develop in highly specialized microsporophyll or stamens while megaspores are formed in megasporangia (or ovules) borne on highly specialized megasporophyll called carpel.
(d) Flower: Besides the essential whorls of microsporophylls (androecium) and megasporophylls (gynoecium), there are accessory whorls namely, calyx (sepals) and corolla (petals) arranged together to form flowers.

Can you recall? (Textbook Page No. 24)

What is double fertilization?
Answer:
(a) Double fertilization is a characteristic feature of angiosperms.
(b) In this process one male gamete fuses with egg cell and another male gamete fuses with secondary nucleus, to form an embryo and endosperm respectively.

Can you recall? (Textbook Page No. 24)

Explain in brief the two classes of Angiosperms? Draw and label one example of each class.
Answer:
Two classes of Angiosperms are Dicotyledonae and Monocotyledonae.
а. Dicotyledonae:

1. These plants have two cotyledons in their embryo.
2. They have a tap root system and the stem is branched.
3. Leaves show reticulate venation.
4. Flowers show tetramerous or pentamerous symmetry.
5. Vascular bundles are conjoint, collateral and open type.
6. Cambium is present between xylem and phloem for secondary growth.
7. In dicots, secondary growth is commonly found. e. g. Helianthus annuus (Sunflower)

b. Monocotyledonae:

1. These plants have single cotyledon in their embryo.
2. They have adventitious root system and stem is rarely branched.
3. Leaves generally have sheathing leaf base and parallel venation.
4. Flowers show trimerous symmetry.
5. The vascular bundles are conjoint, collateral and closed type.
6. Cambium is absent between xylem and phloem.
7. In Monocots, except few plants secondary growth is absent, e.g. Zea mays (Maize)

Try This (Textbook Page No. 24)

Study the leaves of Hibiscus, Peepal, Canna, Grass and Tulsi. Classify them as Monocot and Dicot.
Answer:

Monocot leaves	Dicot leaves
Canna. Grass (Parallel venation)	Hibiscus, Peepal, Tulsi (Reticulate venation)

Can you tell? (Textbook Page No. 25)

(i) Distinguish between Dicotyledonae and Monocotyledonae.
Answer:
Spores and Sporophylls: Angiosperms are heterosporous. Microspores (commonly called pollens) are formed in microsporangia (or anthers). They develop in highly specialized microsporophyll or stamens while megaspores are formed in megasporangia (or ovules) borne on highly specialized megasporophyll called carpel.

(ii) Why do Dicots show secondary growth while Monocots don’t?
Answer:
(a) In dicots, vascular bundles are conjoint, collateral and open type. Cambium is present between xylem and phloem for secondary growth.
(b) Whereas in monocots, vascular bundles are conjoint, collateral and closed type. Thus, due to absence of cambium, secondary growth does not occur in majority of monocots.

Observe and Discuss (Textbook Page No. 23)

Which differences did you notice between Gymnosperms and Angiosperms?
Answer:

Gymnosperms	Angiosperms
1. In gymnosperms, the seeds arc naked.	In angiosperms, the seeds are enclosed within the fruit.
2. Plants are evergreen, shrubs or woody trees.	Plants are annual, biennial or perennial herbs, shrubs or trees, either woody or herbaceous.
3. Xylem is made up of tracheids only.	Xylem is made up of vessels and tracheids.
4. Phloem is with sieve cells only.	Phloem is with sieve tubes and companion cells.
5. Usually two types of leaves are present, i.e. green foliage leaves and scale leaves.	Leaves are of usually one type only, such as green foliage leaves.
6. Double fertilization absent.	Double fertilization occurs.

Can you tell? (Textbook Page No. 26)

What is alternation of generations?
Answer:
The sporophytic and gametophytic generations generally occur alternately in the life cycle of a plant. This phenomenon is called alternation of generations.

Can you tell? (Textbook Page No. 26)

Which phase is dominant in the life cycle of Bryophyta and Pteridophyta?
Answer:
In the life cycle of Bryophyta, gametophyte is the dominant phase whereas in the life cycle of Pteridophyta, sporophyte is the dominant phase.

11th Std Biology Questions And Answers:

• Living World Class 11 Biology Questions And Answers
• Systematics of Living Organisms Class 11 Biology Questions And Answers
• Kingdom Plantae Class 11 Biology Questions And Answers
• Kingdom Animalia Class 11 Biology Questions And Answers
• Cell Structure and Organization Class 11 Biology Questions And Answers
• Biomolecules Class 11 Biology Questions And Answers
• Cell Division Class 11 Biology Questions And Answers
• Plant Tissues and Anatomy Class 11 Biology Questions And Answers

Balbharti Maharashtra State Board 11th Chemistry Textbook Solutions Chapter 14 Basic Principles of Organic Chemistry Textbook Exercise Questions and Answers.

Basic Principles of Organic Chemistry Class 11 Exercise Question Answers Solutions Maharashtra Board

Class 11 Chemistry Chapter 14 Exercise Solutions Maharashtra Board

Chemistry Class 11 Chapter 14 Exercise Solutions

1. Answer the following :

Question A.
Write condensed formulae and bond line formulae for the following structures.

Answer:

Question B.
Write dash formulae for the following bond line formulae.

Answer:

Question C.
Write bond line formulae and condensed formulae for the following compounds
a. 3-methyloctane
b. hept-2-ene
c. 2, 2, 4, 4- tetramethylpentane
d. octa-1,4-diene
e. methoxyethane
Answer:

Question D.
Write the structural formulae for the following names and also write correct IUPAC names for them.
a. 5-ethyl-3-methylheptane
b. 2,4,5-trimethylthexane
c. 2,2,3-trimethylpentan-4-01
Answer:

Question E.
Identify more favourable resonance structure from the following. Justify.

Answer:
a.

Structure (I) will be more favourable resonance structure as structure (II) involves separation of opposite charges and the electronegative oxygen atom has a positive charge.

Both structures (I) and (II) involves separation of opposite charges, but structure (I) has a positive charge on the more electropositive ‘C’ and a negative charge on more electronegative ‘O’. Thus, structure (I) will be more favourable resonance structure.

Question F.
Find out all the functional groups present in the following polyfunctional compounds.
a. Dopamine a neurotransmitter that is deficient in Parkinson’s disease.

b. Thyroxine the principal thyroid hormone.

c. Penicillin G, a naturally occurring antibiotic

Answer:
i. Functional groups: Phenolic -OH group (Ar-OH) and primary amine (-NH₂) group are present in dopamine.
ii. Functional groups: Phenolic -OH group (Ar-OH), halide (-I), ether (Ar-O-Ar), primary amine (-NH₂) carboxylic acid (-COOH) groups are present in thyroxine.
iii. Functional groups: Secondary amide
,
carboxylic acid (-COOH), tertiary amide
,
thioether (R-S-R) groups are present in penicillin G.

Question G.
Find out the most stable species from the following. Justify.

Answer:
a. The most stable species from the given species is \(\left(\mathrm{H}_{3} \mathrm{C}\right)_{3} \dot{\mathrm{C}}\) i.e., tert-butyl radical.
This is because it has greater number of alkyl groups attached to the C-atom having unpaired electron. More the number of the alkyl groups, the greater will be +1 inductive (electron releasing) effect, and thereby greater will be the stability of the free radical.

b. The most stable species from the given species is \(\mathrm{CBr}_{3}^{-}\).
This is because it contains 3 -Br atoms, which exhibits electron withdrawing inductive effect. Carbanions are stabilized by -I inductive (electron withdrawing) effect. Larger the number of -I groups attached to the negatively charged carbon atom, lower will be the electron density on the carbon atom and higher will be its stability.

c. The most stable species from the given species is \(\stackrel{+}{\mathbf{C}} \mathbf{H}_{3}\).
This because it does not contain Cl atom, which exhibits electron withdrawing inductive effect. Carbocations are destabilized by -I inductive (electron withdrawing) effect. When more number of-I groups are attached to the positively charged carbon atom, the positive charge on the carbon atom increases further, thus destabilizing the species. Hence, the species with no -I groups will be most stable.

Question H.
Identify the α-carbons in the following species and give the total number of α-hydrogen in each.

Answer:
a.

In structure (i), C-2 and C-4 are α-carbon atoms.
Hydrogen atoms(s) attached to α-C atoms is a α-H atom. Thus, structure (i) contains 4 α-H atoms.
b.

In structure (ii), carbon atoms adjacent to C-2 are α-carbon atoms (as shown in the structure).
Thus, structure (ii) contains 6 α-H atoms.

c.

C-3 carbon atom, that is, C-atom next to (H₂C=CH-) is a α-C atom.
Thus, structure (iii) contains 2 α-H atoms.

Question I.
Identify primary, secondary, tertiary and quaternary carbon in the following compounds.

Answer:

2. Match the pairs

	Column ‘A’		Column ‘B’
i.	Inductive effect	a.	Delocalization of π  electrons
ii.	Hyperconjugation	b.	Displacement of π electrons
iii.	Resonance effect	c.	Delocalization of σ electrons
		d.	Displacement of σ electrons

Answer:
i – d,
ii – c,
iii – a

3. What is meant by homologous series ? Write the first four members of homologous series that begins with
A. CH₃CHO
B. H-C≡C-H
Also write down their general molecular formula.
Answer:
Homologous series: A series of compounds of the same family in which each member has the same type of carbon skeleton and functional group, and differs from the next member by a constant difference of one methylene group (-CH₂-) in its molecular and structural formula is called as homologous series.
A. CH₃CHO :

Comparing these molecular formulae and assigning the number of carbon atoms as ‘n’, the following general formula is deduced: C_nH_2nO/C_nH_2n-1CHO (where n = 1, 2, 3, …).

B. H-C≡C-H :

Comparing these molecular formulae and assigning the number of carbon atoms as ‘n’, the following general formula is deduced: C_nH_2n-2 (where n = 2, 3,4,….).

4. Write IUPAC names of the following

Answer:

5. Find out the type of isomerism exhibited by the following pairs.


Answer:
A. Metamerism
B. Functional group isomerism
C. Tautomerism
D. Tautomerism

6. Draw resonance srtuctures of the following :

A. Phenol
B. Benzaldehyde
C. Buta-1,3-diene
D. Acetate ion
Answer:
A. Resonance structures for phenol:

B. Resonance structures of benzaldehyde:

C. Resonance structures of Buta-1,3-diene:

D. Resonance structures of acetate ion:

7. Distinguish :

Question A.
Inductive effect and resonance effect
Answer:
Inductive effect:

1. Presence of polar covalent bond is required.
2. The polarity is induced in adjacent carbon- carbon single (covalent) bond due to a presence of influencing group (more electronegative atom than carbon).
3. Depending on the nature of influencing group it is differentiated as +I effect and -I effect.
4. The direction of the arrow head denotes the direction of the permanent electron displacement.

Resonance effect:

1. Presence of conjugated n electron system or species having an atom carrying p orbital attached to a multiple bond is required.
2. The polarity is produced in the molecule by the interaction of conjugated π bonds (or that between π bond and p orbital on the adjacent atom).
3. Depending on the nature of influencing group it is differentiated as +R and -R effect.
4. The delocalisation of n electrons is denoted by using curved arrows.

Question B.
Electrophile and nucleophile
Answer:
Electrophile:

1. Electrophile is an electron deficient species.
2. It is attracted towards negative charge (electron seeking).
3. It attacks a nucleophilic centre in the substrate and brings about an electrophilic reaction
4. It is an electron pair acceptor. (Lewis acid)
5. It can be a positively charged ion or a neutral species having a vacant orbital.
   e.g. H⁺, Br , \(\mathrm{NO}_{2}^{+}\), BF₃, AlCl₃, etc.

Nucleophile:

• Nucleophile is an electron rich species.
• It is attracted towards positive charge (nucleus seeking).
• It attacks the electrophilic centre in the substrate and brings about a nucleophilic reaction.
• It is an electron pair donor. (Lewis base)
• It can be negatively charged ion or neutral species having at least one lone pair of electrons.
  

C. Carbocation and carbanion
Answer:
Carbocation:

• It is a species in which carbon carries a positive charge.
• Positively charged carbon is sp² hybridized.
• It is electron-deficient.
• e.g. tert-Butyl carbocation, (CH₃)₃C⁺

Carbanion:

• It is a species in which carbon carries a negative charge.
• Negatively charged carbon is sp³/sp² hybridized.
• It is electron-rich.
• e.g.Methyl carbanion,
  

D. Homolysis and heterolysis
Answer:

8. Write true or false. Correct the false stament
A. Homolytic fission involves unsymmetrical breaking of a covalent bond.
B. Heterolytic fission results in the formation of free radicals.
C. Free radicals are negatively charged species
D. Aniline is heterocyclic compound.
Answer:
A. False
Homolytic fission involves symmetrical breaking of a covalent bond.
B. False
Heterolytic fission results in the formation of charged ions like cation and anion.
C. False
Free radicals are electrically neutral/uncharged species.
D. False
Aniline is a homocyclic aromatic compound.

9. Phytane is naturally occuring alkane produced by the alga spirogyra and is a constituent of petroleum. The IUPAC name for phytane is 2, 6, 10, 14-tetramethyl hexadecane. Write zig-zag formula for phytane. How many primary, secondary, tertiary and quaternary carbons are present in this molecule.
Answer:
Zig-zag formula of phytane (2,6,10,14-tetramethyl hexadecane) is as follows:

Dash formula to represent types of C-atom:

In phytane, six 1° C-atoms, ten 2° C-atoms, four 3° C-atoms are present. Phytane does not contain any quaternary carbon atom in its structure.

10. Observe the following structures and answer the questions given below.

a. What is the relation between (i) and (ii) ?
b. Write IUPAC name of (ii).
c. Draw the functional group isomer of (i).
Answer:
a. (a) and (b) are chain isomers of each other.
b. IUPAC name of structure (b) is 2-methylpropanal.
c. Functional group isomer of (a) is butanone.

11. Observe the following and answer the questions given below

a. Name the reactive intermediae produced
b. Indicate the movement of electrons by suitable arrow to produce this intermediate
c. Comment on stability of this intermediate produced.
Answer:
i. The reactive intermediates produced are methyl free radicals:

ii. Stability order of alkyl free radicals is: \(\dot{\mathrm{C}} \mathrm{H}_{3}\) < 1° <2° <3°
Hence, \(\dot{\mathrm{C}} \mathrm{H}_{3}\) produced in the above reaction is least stable and highly reactive.

12. An electronic displacement in a covalent bond is represented by following notation.

A. Identify the effect
B. Is the displacement of electrons in a covalent bond temporary or permanent.
Answer:
A. The electronic displacement represented above is inductive effect (-I effect).
B. Inductive effect is a permanent electronic effect as it depends on the electronegativity of the atoms. In the given example, the displacement of electrons is permanent as Cl is more electronegative than C.

13. Draw all the no-bond resonance structures of isopropyl carbocation.
Answer:

14. A covalent bond in tert-butyl bromide breaks in a suitable polat solvent to give ions.
A. Name the anion produced by this breaking of a covalent bond.
B. Indicate the type of bond breaking in this case.
C. Comment on geometry of the cation formed by such bond cleavage.
Answer:
A. The anion produced by breaking of the covalent C – Br bond is bromide

B. Heterolytic cleavage/fission takes place as charged ions are produced.
C. tert-Butyl carbocation formed in the given cleavage has trigonal planar geometry.

15. Choose correct options

A. Which of the following statements are true with respect to electronic displacement in covalent bond ?
a. Inductive effect operates through π bond
b. Resonance effect operates through σ bond
c. Inductive effect operates through σ bond
d. Resonance effect operates through π bond
i. a. and b
ii. a and c
iii. c and d
iv. b and c
Answer:
iii. c and d

B. Hyperconjugation involves overlap of …………. orbitals
a. σ – σ
b. σ – p
c. p – p
d. π – π
Answer:
b. σ – p

C. Which type of isomerism is possible in CH₃CHCHCH₃?
a. Position
b. Chain
c. Geometrical
d. Tautomerism
Answer:
a. Position

D. The correct IUPAC name of the compound

is ……………
a. hept-3-ene
b. 2-ethylpent-2-ene
c. hex-3-ene
d. 3-methylhex-3-ene
Answer:
d. 3-methylhex-3-ene

E. The geometry of a carbocation is …………
a. linear
b. planar
c. tetrahedral
d. octahedral
Answer:
b. planar

F. The homologous series of alcohols has general molecular formula ………..
a. C_nH_2n+1OH
b. C_nH_2n+2OH
c. C_nH_2n-2OH
d. C_nH_2nOH
Answer:
a. C_nH_2n+1OH

G. The delocaalization of electrons due to overlap between p-orbital and sigma bond is called …………….
a. Inductive effect
b. Electronic effect
c. Hyperconjugation
d. Resonance
Answer:
c. Hyperconjugation

11th Chemistry Digest Chapter 14 Basic Principles of Organic Chemistry Intext Questions and Answers

Can you recall? (Textbook Page No. 204)

Question i.
Which is the essential element in all organic compounds?
Answer:
Carbon is the essential element in all organic compounds.

Question ii.
What is the unique property of carbon that makes organic chemistry a separate branch of chemistry?
Answer:

• All organic compounds contain carbon.
• Carbon atoms show catenation property in which carbon atoms combine with other carbon atoms to form long chains and rings.
• Carbon atom can also form multiple bonds with other carbon atoms and with atoms of other elements.
• Due to this property of self-linking of carbon, a large number of organic compounds like proteins, DNA, sugar, oils, etc., are formed.

Thus, the unique property of catenation of carbon makes organic chemistry a separate branch of chemistry.

Question iii.
Which classes of organic compounds are often used in our daily diet?
Answer:
Carbohydrates (sugars), proteins (pulses), fats (edible plant and animal oil) and vitamins are the major classes of organic compounds often used in our daily diet.

Try this. (Textbook Page No. 204)

Question 1.
Find out the structures of glucose, vanillin, camphor and paracetamol using internet. Mark the carbon atoms present in them. Assign the hybridization state to each of the carbon and oxygen atom. Identify sigma (σ) and pi (π) bonds in these molecules.
Answer:
i. Structure of glucose:

a. Hybridization of carbon: In glucose, only carbon at position C-1 is sp² hybridized. On the other hand, carbons at C-2, C-3, C-4, C-5 and C-6 positions are sp³ hybridized.
b. Hybridization of oxygen: Oxygen atom attached to C-1 is sp² hybridized, rest oxygen atoms attached to carbon at C-2, C-3, C-4, C-5 and C-6 are sp³ hybridized.
[Note: Here, the open chain structure of glucose is used to answer the given questions.]

ii. Structure of vanillin:

a. Hybridization of carbon: In vanillin, carbon atoms C-1 to C-7 are sp² hybridized. Only C-8 carbon is sp³ hybridized.
b. Hybridization of oxygen: Oxygen atom bonded to C-7 sp² hybridized whereas oxygen atom bonded to C-4 and C-8 are sp³ hybridized.

iii. Structure of camphor:

a. Hybridization of carbon: In camphor, all the carbons are sp³ hybridized except the carbonyl carbon which is sp² hybridized.
b. Hybridization of oxygen: The carbonyl oxygen is sp² hybridized.

iv. Structure of paracetamol:

a. Hybridization of carbon: In paracetamol, carbons present in the ring and carbon at C-7 position are sp² hybridized. Only C-8 carbon is sp³ hybridized.
b. Hybridization of oxygen: Oxygen atom attached to carbon at ,C-1 position is sp³ hybridized. Oxygen atom attached to carbon at C-7 position is sp² hybridized.

Question 2.
i. Draw the structural formula of ethane.
ii. Draw electron-dot structure of propane.
Ans:
i. Structural formula of ethane (C₂H₆) can be drawn as follows:

ii. Electron-dot structure of propane is given as,

Where ‘•’ represents valence electrons of carbon and hydrogen.

Try this (Textbook Page No. 205)

Complete the table:

Answer:

Try this. (Textbook Page No. 206)

Question 1.
Draw two Newman projection formulae and two Sawhorse formulae for the propane molecule.
Answer:
Structural formula of propane is:

Structural formula of propane:
i. Newman projection formulae for propane molecule can be given as:

ii. Sawhorse formula for propane molecule can be given as:

Can you tell? (Textbook Page No. 208)

Question 1.
Consider the following reaction:
2CH₃ – CH₂ – CH₂ – OH + 2Na → 2CH₃ – CH₂ – CH₂ – ONa + H₂
Compare the structure of the substrate propanol with that of the product sodium propoxide. Which part of the substrate, the carbon skeleton or the OH group has undergone a change during the reaction?
Answer:
In above reaction, the -OH group of the substrate molecule has undergone a change. The H-atom of hydroxyl group (-OH) is replaced by sodium forming the product.

Activity: (Textbook Page No. 219)

Observe the structural formulae (a) and (b).
i. Find out their molecular formulae.
ii. What is the difference between them?
iii. What is the relation between the two compounds represented by these structural formulae?
Answer:
i. Molecular formula of both (a) and (b) are same i.e., C₃H₆O.
ii. Compound (a), has a ketone (-CO-) functional group (i.e., acetone) and compound (b) has an aldehyde (-CHO) functional group (i.e., propionaldehyde). Both the compounds have different functional groups.
iii. Compound (a) and (b) are isomers of each other.
[Note: Aldehydes and ketones are the functional group isomers of each other.]

Can you tell? (Textbook Page No. 222)

Question 1.
Some bond fissions are described in the following table. For each of them, show the movement of electron/s using curved arrow notation. Classify them as homolysis or heterolysis and identify the intermediate species produced as carbocation, carbanion or free radical.

Answer:

Can you recall? (Textbook Page No. 223)

i. What is meant by ‘reagent’?
ii. Identify the ‘reagent’, ‘substrate’, ‘product’ and ‘byproduct’ in the following reaction.
CH₃COCl + NH₃ → CH₃CONH₂ + HCl
Answer:
i. The reactant which reacts with a substrate to form corresponding products is known reagent.
ii.

Can you recall? (Textbook Page No. 224)

i. How is covalent bond formed between two atoms?
ii. Consider two covalently bonded atoms Q and R where R is more electronegative than Q. Will these atoms share the electron pair equally between them?
iii. Represent the above polar covalent bond between Q and R using fractional charges δ⁺ and δ^–.
Answer:
i. A covalent bond is formed between two atoms by mutual sharing of electrons so as to complete their octets or duplets (in case of elements having only one shell).

ii. A covalent bond is formed between Q and R having different electronegativities, that is, R is more electronegative than Q. In such a case, the atom R with a higher value of electronegativity pulls the shared pair of electrons to a greater extent towards itself as compared to the atom Q with lower value of electronegativity. As a result of this, the shared pair of electrons will get shifted towards atom R. Thus, both the atoms Q and R will not share the electron pair equally between them.

iii. Polar covalent bond between Q and R can be represented as:
\(\mathrm{Q}^{\delta+}-\mathrm{R}^{\delta-}\)

Try this (Textbook Page No. 225)

i. Draw a bond line structure of benzene (C₆H₆).
ii. How many C – C and C = C bonds are there in this structure?
iii. Write down the expected values of the bond lengths of the carbon-carbon bonds in benzene (Refer chapter 5).
Answer:
i. Bond line structure of benzene:

ii. In benzene, there are three alternating C – C single bonds and C = C double bonds.
[Note: In benzene, there are six C – C sigma bonds and three C – C pi bonds.]
iii. The expected values of carbon-carbon bond lengths in benzene are:

Bond	Bond length
C – C	154 pm
C = C	133 pm

Can you recall? (Textbook Page No. 225)

i. Write down two Lewis structures for ozone. (Refer chapter 5)
ii. How are these two Lewis structures related to each other?
iii. What are these two Lewis structures called?
Answer:
i. Lewis structures of ozone can be shown as follows:

ii. In these two Lewis structures, the position of the atoms is same but the position of pair of electrons (or formal charge) is different. These two Lewis structures are considered equivalent to each other.
iii. These two Lewis structures are called as resonating or contributing or canonical structures.

Internet my friend (Textbook Page No. 229)

i. Basic principles of organic chemistry:
https://authors.library.caltech.edu/25034
ii. Collect information about isomerism.
Answer:
i. Students are expected to refer the book provided in the above link to collect additional information on the basic principles of organic chemistry.

ii. https://www.compoundchem.com/2014/05/22/typesofisomerism/
chemdictionary.org/structural-isomers/
https://en.wikipedia.org/wiki/Structural_isomer
[Note: Students can use the above links as reference and collect additonal information about isomerism on their own.]

11th Std Chemistry Questions And Answers:

• Elements of Group 13, 14 and 15 Class 11 Chemistry Questions And Answers
• States of Matter Class 11 Chemistry Questions And Answers
• Adsorption and Colloids Class 11 Chemistry Questions And Answers
• Chemical Equilibrium Class 11 Chemistry Questions And Answers
• Nuclear Chemistry and Radioactivity Class 11 Chemistry Questions And Answers
• Basic Principles of Organic Chemistry Class 11 Chemistry Questions And Answers
• Hydrocarbons Class 11 Chemistry Questions And Answers
• Chemistry in Everyday Life Class 11 Chemistry Questions And Answers

Balbharti Maharashtra State Board 11th Biology Textbook Solutions Chapter 6 Biomolecules Textbook Exercise Questions and Answers.

Biomolecules Class 11 Exercise Question Answers Solutions Maharashtra Board

Class 11 Biology Chapter 6 Exercise Solutions Maharashtra Board

Biology Class 11 Chapter 6 Exercise Solutions

1. Choose correct option

Question (A)
Sugar, amino acids and nucleotides unite to their respective subunits to form ________
(a) bioelements
(b) micromolecules
(c) macromolecules
(d) all of these
Answer:
(c) macromolecules

Question (B)
Glycosidic bond is found in __________ .
(a) Disaccharide
(b) Nucleosides
(c) Polysaccharide
(d) all of these
Answer:
(d) all of these

Question (C)
Amino acids in a polypeptide are joined by _______ bond.
(a) Disulphide
(b) glycosidic
(c) hydrogen bond
(d) none of these
Answer:
(d) none of these

Question (D)
Lipids associated with cell membrane are _________ .
(a) Sphingomyelin
(b) Isoprenoids
(c) Phospholipids
(d) Cholesterol
Answer:
(c) Phospholipids

Question (E)
Linoleic, Linolenic and ________ acids are referred as essential fatty acids since they cannot be synthesized by the body and hence must be included in daily diet.
(a) Arachidonic
(b) Oleic
(c) Steric
(d) Palmitic
Answer:
(a) Arachidonic

Question (F)
Hemoglobin is a type of ________ protein, which plays indispensable part in respiration.
(a) simple
(b) derived
(c) conjugated
(d) complex
Answer:
(c) conjugated

Question (G)
When inorganic ions or metallo-organic molecules bind to apoenzyme, they together form
(a) isoenzyme
(b) holoenzyme
(c) denatured enzyme
(d) none of these
Answer:
(b) holoenzyme

Question (H)
In enzyme kinetics, Km = Vmax/2. If Km value is lower, it indicates _______
(a) Enzyme has less affinity for substrate
(b) Enzyme has higher affinity towards substrate
(c) There will be no product formation
(d) All active sites of enzyme are saturated
Answer:
(b) Enzyme has higher affinity towards substrate

2. Solve the following questions

Question (A)
Observe the following figures and write the differences between them.

Answer:

Saturated fats	Unsaturated fats
1. They contain single chain of carbon atoms with single bonds.	They contain chain of carbon atoms with one or more double bonds.
2. They are solid at room temperature.	They are liquid at room temperature.
3. They increase blood cholesterol level by depositing it in the inner wall of arteries.	They lower the blood cholesterol level and have many health benefits.
4. They do not get spoiled.	They get spoiled easily.
5. Saturated fats are obtained from animal fats, palm oil, etc.	Unsaturated fatty acids are obtained from plant and vegetable oil, etc.

3. Answer the following questions

Question (A)
What are building blocks of life?
Answer:
Life is composed of four main building blocks: Carbohydrates, proteins, lipids and nucleic acids.

Question (B)
Explain the peptide bond.
Answer:
1. The covalent bond that links the two amino acids is called a peptide bond.
2. Peptide bond is formed by condensation reaction.

Question (C)
How many types of polysaccharides you know?
Answer:
There are two types of polysaccharides:
1. Homopolysaccharides: It contains same type of monosaccharides. E.g. Starch, glycogen, cellulose.
2. Heteropolysaccharides: It contains two or more different monosaccharides. E.g. Hyaluronic acid, heparin, hemicellulose.

Question (D)
Enlist the significance of carbohydrates.
Answer:
Significances of carbohydrates are as follows:

1. Carbohydrates provide energy for metabolism.
2. Glucose is the main substrate for ATP synthesis.
3. Lactose, a disaccharide present in the milk provides energy to babies.
4. Polysaccharide serves as a structural component of cell membrane, cell wall and reserved food as starch and glycogen.

Question (E)
What is reducing sugar?
Answer:
1. A sugar that serves as a reducing agent due to presence of free aldehyde or ketone group is called a reducing sugar.
2. These sugars reduce the Benedict’s reagent (Cu²+ to Cu⁺) since they are capable of transferring hydrogens (electrons) to other compounds, a process called reduction.
3. All monosaccharides are reducing sugars.

Question (F)
Enlist the examples of simple proteins and their significance.
Answer:
Examples of simple proteins are: E.g.: Albumins and histones.
Significance:
1. Albumin:
a. % It is the main protein in the blood.
b. It maintains the pressure in the blood vessels.
c. It helps in transportation of substances like hormone and drugs in the body.
2. Histones:
a. It is the chief protein of chromatin.
b. They are involved in packaging of DNA into structural units called nucleosomes.

Question (G)
Describe the secondary structure of protein with examples.
Answer:

1. There are two types of secondary structure of protein: a-helix and P-pleated sheets.
2. The polypeptide chain is arranged in a spiral helix. These spiral helices are of two types: a-helix (right handed) and P-helix (left handed).
3. This spiral configuration is held together by hydrogen bonds.
4. The sequence of amino acids in the polypeptide chain determines the location of its bend or fold and the position of formation of hydrogen bonds between different portions of the chain or between different chains. Thus, peptide chains form an a-helix structure.
5. Example of a-helix structure is keratin.
6. In some proteins two or more peptide chains are linked together by intermolecular hydrogen bonds. Such structures are called P-pleated sheets.
7. Example of P-pleated sheet is silk fibres.
8. Due to formation of hydrogen bonds peptide chains assume a secondary structure.

Question (H)
Explain the induced fit model for mode of enzyme action.
Answer:
1. The induced fit model shows that enzymes are flexible structures in which the active site continually reshapes by its interactions with the substrate until the time the substrate is completely bound to it. It is also the point at which the final form and shape of the enzyme is determined.
2. Three-Dimensional conformation:
a. All enzymes have specific 3-dimensional conformation.
b. They have one or more active sites to which substrate (reactant) combines.
c. The points of active site where the substrate joins with the enzyme is called substrate binding site.

Question (I)
What is RNA? Enlist types of RNA.
Answer:
1. RNA stands for Ribonucleic Acid. It is a long single stranded polynucleotide chain which helps in protein synthesis, functions as a messenger and translates messages coded in DNA into protein.
2. There are three types of RNA:
mRNA (messenger RNA), rRNA (ribosomal RNA) and tRNA (transfer RNA)

Question (J)
Describe the concept of metabolic pool.
Answer:
1. Metabolic pool is the reservoir of biomolecules in the cell on which enzymes can act to produce useful products as per the need of the cell.
2. The concept of metabolic pool is significant in cell biology because it allows one type of molecule to change into another type E.g. Carbohydrates can be converted to fats and vice-versa.

Question (K)
How do secondary metabolites useful for mankind?
Answer:
1. Drugs developed from secondary metabolites have been used to treat infectious diseases, cancer, hypertension and inflammation.
2. Morphine, the first alkaloid isolated from Papaver somniferum is used as pain reliver and cough suppressant.
3. Secondary metabolites like alkaloids, nicotine, cocaine and the terpenes, cannabinol are widely used for recreation and stimulation.
4. Flavours of secondary metabolites improve our food preferences.
5. Tannins are added to wines and chocolate for improving astringency.
6. Since most secondary metabolites have antibiotic property, they are also used as food preservatives.
7. Glucosinolates is a secondary metabolite which is naturally present in cabbage imparts a characteristic flavour and aroma because of nitrogen and sulphur-containing chemicals. It also offers protection to these plants from many pests.

4. Solve the following questions

Question (A)
Complete the following chart.

Protein	Physiological role
Collagen	(i)
(ii)	Responsible for muscle contraction
Immunoglobulin	(iii)
(iv)	Significant in Respiration
Fibrinogen	(v)

Answer:

Protein	Physiological role
1. Collagen	Provides strength and plays structural role
2. Myosin & Actin	Responsible for muscle contraction
3. Immunoglobulin	Protects the body from infection
4. Haemoglobin	Significant in Respiration
5. Fibrinogen	Responsible for normal clotting of blood.

Question (B)
Answer the following with reference

i. Name the type of bond formed between two polypeptides.
ii. Which amino acid is involved in the formation of such bond?
iii. Amongst I, II, III and IV structural level of protein, which level of structure includes such bond? Answer:
i. Disulfide bond.
ii. Cysteine
iii. Tertiary structure.
[Note: Quaternary structure of protein also have disulfide bond, for stabilization of protein structure.!

Question (C)
Match the following items given in column I and II.

Column I	Column 11
1. RNA	(a) Induced fit model
2. Yam plant	(b) Flax seeds
3. Koshland	(c) Hydrolase
4. Omega – 3 – fatty acid	(d) Uracil
5. Sucrase	(e) Anti-fertility pills

Answer:

Column I	Column 11
1. RNA	(d) Uracil
2. Yam plant	(e) Anti-fertility pills
3. Koshland	(a) Induced fit model
4. Omega – 3 – fatty acid	(b) Flax seeds
5. Sucrase	(c) Hydrolase

5. Long answer questions

Question (A)
What are biomolecules? Explain the building blocks of life.
Answer:
Biomolecules are essential substances produced by our body which are necessary for life.
The building blocks of life are carbohydrates, lipids, proteins and nucleic acids.
1. Carbohydrates:
a. Carbohydrates are biomolecules made from carbon, hydrogen and oxygen.
b. The general formula of carbohydrates is (CH20) n.
c. They contain hydrogen and oxygen in the same ratio as in water (2:1).
d. Carbohydrates can be broken down to release energy.
e. Based on sugar units, carbohydrates are classified into three types: Monosaccharides, disaccharides and polysaccharides.

2. Lipids:
a. These are group of substances with greasy consistency with long hydrocarbon chain containing carbon, hydrogen and oxygen.
b. In lipids hydrogen to oxygen ration is greater than 2:1.
c. Lipid is a broader term used for fatty acids and their derivatives.
d. They are soluble in organic solvents (non-polar solvents).
e. Fatty acids are organic acids which are composed of hydrocarbon chain ending in carboxyl group (COOH) ….
f. These are divided into: Saturated fatty acids and unsaturated fatty acids.
g. Fatty acids are basic molecules which form different kinds of lipids.
h. Lipids are classified into three types:
Simple lipids, Compound lipids, Derived lipids.

3. Proteins:
a. Proteins are large molecules containing amino acid units ranging from 100 to 3000.
b. They have higher molecular weight.
c. In proteins, amino acids are linked together by peptide bonds which join the carboxyl group of one amino acid residue to the amino group of another residue.
d. A protein molecule consists of one or more polypeptide chains.
e. Proteins contain any or all twenty naturally occurring amino acid types.
f. Proteins have different structures like primary structure, secondary structure, tertiary structure and quaternary structure.
g. Proteins are classified into three types:
Simple proteins: Simple proteins on hydrolysis yield only amino acids. E.g. Histones and albumins. Conjugated proteins: It consists of a simple protein united with some non-protein substance. E.g. Haemoglobin.
Derived proteins: These proteins are not found in nature as such but are derived from native protein molecules on hydrolysis. E.g. Metaproteins, peptones.

4. Nucleic Acids:
a. Nucleic acids are macromolecules composed of many small units or monomers called nucleotides.
b. Each nucleotide is formed of three components i.e. pentose sugar, a nitrogen base and a phosphate (phosphoric acid).
c. When sugar combine with nitrogenous base it forms nucleoside. Nucleotides can be called as nucleoside phosphate.
d. There are two types of nucleic acids, i.e. DNA and RNA.
DNA (Deoxyribonucleic acid) is a genetic material of a cell. It is double stranded helix. Each strand of helix is made up of deoxyribose nucleotides.
RNA (Ribonucleic Acid) is a single stranded structure having fewer nucleotides as compared to DNA. The strands may be straight or variously folded upon itself. It is made up of nucleotides.

Question (B)
Explain the classes of carbohydrates with examples.
Answer:
Based on number of sugar units, carbohydrates are classified into three types namely, monosaccharides, disaccharides and polysaccharides.
1. Monosaccharides:
a. Monosaccharides are the simplest sugars having crystalline structure, sweet taste and soluble in water.
b. They cannot be further hydrolyzed into smaller molecules.
c. They are the building blocks or monomers of complex carbohydrates.
d. They have the general molecular formula (CH20)n, where n can be 3, 4, 5, 6 and 7.
e. They can be classified as triose, tetrose, pentose, etc.
f. Monosaccharides containing the aldehyde (-CHO) group are classified as aldoses e.g. glucose, xylose, and those with a ketone(-C=0) group are classified as ketoses. E.g. ribulose, fructose.

2. Disaccharides:
a. Disaccharide is formed when two monosaccharide react by condensation reaction releasing a water molecule. This process requires energy.
b. A glycosidic bond forms and holds the two monosaccharide units together.
c. Sucrose, lactose and maltose are examples of disaccharides.
d. Sucrose is a nonreducing sugar since it lacks free aldehyde or ketone group.
e. Lactose and maltose are reducing sugars.
f. Lactose also exists in beta form, which is made from P-galactose and p-glucose.
g. Disaccharides are soluble in water, but they are too big to pass through the cell membrane by diffusion.

3. Polysaccharides:
a. Monosaccharides can undergo a series of condensation reactions, adding one unit after the other to the chain till a very large molecule (polysaccharide) is formed. This is called polymerization.
b. Polysaccharides are broken down by hydrolysis into monosaccharides.
c. The properties of a polysaccharide molecule depends on its length, branching, folding and coiling.
d. Examples: Starch, glycogen, cellulose.

Question (C)
Describe the types of lipids and mention their biological significance.
Answer:
Lipids are classified into three main types:
1. Simple lipids:
a. These are esters of fatty acids with various alcohols. Fats and waxes are simple lipids.
b. Fats are esters of fatty acids with glycerol (CH₂OH-CHOH-CH₂OH).
c. Triglycerides are three molecules of fatty acids and one molecule of glycerol.
d. Unsaturated fats are liquid at room temperature and are called oils. Unsaturated fatty acids are hydrogenated to produce fats e.g. Vanaspati ghee.

Biological significance:
a. Fats are a nutritional source with high calorific value and they act as reserved food materials.
b. In plants, fat is stored in seeds to nourish embryo during germination.
c. In animals, fat is stored in the adipocytes of the adipose tissue.
d. Fats deposited in subcutaneous tissue act as an insulator and minimize loss of body heat.
e. Fats deposited around the internal organs act as cushions to absorb mechanical shocks.
f. Wax is another example of simple lipid. They are esters of long chain fatty acids with long chain alcohols.
g. They are found in the blood, gonads and sebaceous glands of the skin.
h. Waxes are not as readily hydrolyzed as fats.
i. They are solid at ordinary temperature.
j. Waxes form water insoluble coating on hair and skin in animals, waxes form an outer coating on stems, leaves and fruits.

2. Compound lipids:
a. These are ester of fatty acids containing other groups like phosphate (Phospholipids), sugar (glycolipids), etc.
b. They contain a molecule of glycerol, two molecules of fatty acids and a phosphate group or simple sugar.
c. Some phospholipids such as lecithin also have a nitrogenous compound attached to the phosphate group.
d. Phospholipids have both hydrophilic polar groups (phosphate and nitrogenous group) and hydrophobic non-polar groups (hydrocarbon chains of fatty acids).
e. Glycolipids contain glycerol, fatty acids, simple sugars such as galactose. They are also called cerebrosides.
Biological significance:
a. Phospholipids contribute in the formation of cell membrane.
b. Large amounts of glycolipids are found in the brain white matter and myelin sheath.

3. Derived Lipids:
a. They are composed of fused hydrocarbon rings (steroid nucleus) and a long hydrocarbon side chain.
b. One of the most common sterols is cholesterol.
Biological significance:
a. It is widely distributed in all cells of the animal body, but particularly in nervous tissue.
b. Cholesterol exists either free or as cholesterol ester.
c. Adrenocorticoids, sex hormones (progesterone, testosterone) and vitamin D are synthesized from cholesterol.
d. Cholesterol is not found in plants.
e. Sterols exist as phytosterols in plants.
f. Yam Plant (Dioscorea) produces a steroid compound called diosgenin. It is used in the manufacture of antifertility pills, i.e. birth control pills.

Question (D)
Explain the chemical nature, structure and role of phospholipids in biological membrane.
Answer:
Chemical nature: Phospholipids are amphiphilic in nature. As they have hydrophilic head and hydrophobic tail.
Structure: It contains an alcohol, two fatty acid chains and a phosphate group.
Role: Phospholipids forms the membranes around the cells and cellular organelles. They form a lipid bilayer membrane. The phospholipids are arranged tail to tail. It serves as a barrier against movement of any ions or polar compounds into and out of the cell.

Question (E)
Describe classes of proteins with their importance.
Answer:
On the basis of structure, proteins are classified into three categories:
1. Simple proteins:
a. Simple proteins on hydrolysis yield only amino acids.
b. These are soluble in one or more solvents.
c. Simple proteins may be soluble in water.
d. Histones of nucleoproteins are soluble in water.
e. Globular molecules of histones are not coagulated by heat.
f. Albumins are also soluble in water but they get coagulated on heating.
g. Albumins are widely distributed e.g. egg albumin, serum albumin and legumelin of pulses are albumins.
Importance: They are involved in structural components; they also act as a storage kind of protein.
Some are associated with nucleic acids in nucleoproteins of cell.

2. Conjugated proteins:
a. Conjugated proteins consist of a simple protein united with some non-protein substance.
b. The non-protein group is called prosthetic group e.g. haemoglobin.
c. Globin is the protein and the iron containing pigment haem is the prosthetic group.
d. Similarly, nucleoproteins have nucleic acids.
e. Proteins are classified as glycoproteins and mucoproteins.
f. Mucoproteins are carbohydrate-protein complexes e.g. mucin of saliva and heparin of blood.
g. Lipoproteins are lipid-protein complexes e.g. conjugate protein found in brain, plasma membrane, milk etc. Importance: They are involved in structural components of cell membranes and organelles.
They also act as a transporter.
Some conjugated proteins are important in electron transport chain in respiration.

3. Derived proteins:
a. These proteins are not found in nature as such.
b. These proteins are derived from native protein molecules on hydrolysis.
c. Metaproteins, peptones are derived proteins.
Importance: They act as a precursor for many molecules which are essential for life.

Question (F)
What are enzymes? How are they classified? Mention example of each class.
Answer:
1. Enzymes are biological macromolecules which act as a catalyst and accelerates the reaction in the body.
2. Enzymes are classified into six classes:
a. Oxidoreductases: These enzymes catalyze oxidation and reduction reactions by the transfer of hydrogen and/or oxygen, e.g. alcohol dehydrogenase

b. Transferases: These enzymes catalyse the transfer of certain groups between two molecules, e.g. glucokinase

c. Hydrolases: These enzymes catalyse hydrolytic reactions. This class includes amylases, proteases, lipases etc. e.g. Sucrase

d. Lyases: These enzymes are involved in elimination reactions resulting in the removal of a group of atoms from substrate molecule to leave a double bond. It includes aldolases, decarboxylases, and dehydratases, e.g. fumarate hydratase.

e. Isomerases: These enzymes catalyze structural rearrangements within a molecule. Their nomenclature is based on the type of isomerism. Thus, these enzymes are identified as racemases, epimerases, isomerases, mutases, e.g. xylose isomerase.

f. Ligases or Synthetases: These are the enzymes which catalyze the covalent linkage of the molecules utilizing the energy obtained from hydrolysis of an energy-rich compound like ATP, GTP e.g. glutathione synthetase, Pyruvate carboxylase.

Question (G)
Explain the properties of enzyme? Describe the models for enzyme actions.
Answer:
1. Proteinaceous Nature:
All enzymes are basically made up of protein.

2. Three-Dimensional conformation:
a. All enzymes have specific 3-dimensional conformation.
b. They have one or more active sites to which substrate (reactant) combines.
c. The points of active site where the substrate joins with the enzyme is called substrate binding site.

3. Catalytic property:
a. Enzymes are like inorganic catalysts and influence the speed of biochemical reactions but themselves remain unchanged.
b. After completion of the reaction and release of the product they remain active to catalyze again.
c. A small quantity of enzymes can catalyze the transformation of a very large quantity of the substrate
into an end product.
d. For example, sucrase can hydrolyze 100000 times of sucrose as compared with its own weight.

4. Specificity of action:
a. The ability of an enzyme to catalyze one specific reaction and essentially no other is perhaps its most significant property. Each enzyme acts upon a specific substrate or a specific group of substrates.
b. Enzymes are very sensitive to temperature and pH.
c. Each enzyme exhibits its highest activity at a specific pH i.e. optimum pH.
d. Any increase or decrease in pH causes decline in enzyme activity e.g. enzyme pepsin (secreted in stomach)shows highest activity at an optimum pH of 2 (acidic)

5. Temperature:
a. Enzymes are destroyed at higher temperature of 60-70°C or below, they are not destroyed but become inactive.
b. This inactive state is temporary and the enzyme can become active at suitable temperature.
c. Most of the enzymes work at an optimum temperature between 20°C and 35°C.

There are two types of models:
1. Lock and Key model:
a. Lock and Key model was first postulated in 1894 by Emil Fischer.
b. This model explains the specific action of an enzyme with a single substrate.
c. In this model, lock is the enzyme and key is the substrate.
d. The correctly sized key (substrate) fits into the key hole (active site) of the lock (enzyme).

2. Induced Fit model (Flexible Model):
a. Induced Fit model was first proposed in 1959 by Koshland.
b. This model states that approach of a substrate induces a conformational change in the enzyme.
c. It is the more accepted model to understand mode of action of enzyme.
d. The induced fit model shows that enzymes are rather flexible structures in which the active site continually reshapes by its interactions with the substrate until the time the substrate is completely bound to it.
e. It is also the point at which the final form and shape of the enzyme is determined.
[Note: Temperature is a factor affecting enzyme activity and not a property of enzyme.]

Question (H)
Describe the factors affecting enzyme action.
Answer:
The factors affecting the enzyme activity are as follows:
1. Concentration of substrate:
a. Increase in the substrate concentration gradually increases the velocity of enzyme activity within the limited range of substrate levels.
b. A rectangular hyperbola is obtained when velocity is plotted against the substrate concentration.
c. Three distinct phases (A, B and C) of the reaction are observed in the graph.
Where V = Measured velocity, V_max = Maximum velocity, S = Substrate concentration,
K_m = Michaelis-Menten constant.
d. K_m or the Michaelis-Menten constant is defined as the substrate concentration (expressed in moles/lit) to produce half of maximum velocity in an enzyme catalyzed reaction.
e. It indicates that half of the enzyme molecules (i.e. 50%) are bound with the substrate molecules when the substrate concentration equals the Km value.
f. K_m value is a constant and a characteristic feature of a given enzyme.
g. It is a representative for measuring the strength of ES complex.
h. A low K_m value indicates a strong affinity between enzyme and substrate, whereas a high Km value reflects a weak affinity between them.
1. For majority of enzymes, the K_m values are in the range of 10_-5 to 10_-2 moles.

2. Enzyme Concentration:

a. The rate of an enzymatic reaction is directly proportional to the concentration of the substrate.
b. The rate of reaction is also directly proportional to the square root of the concentration of enzymes.
c. It means that the rate of reaction also increases with the increasing concentration of enzyme and the rate of reaction can also decrease by decreasing the concentration of enzyme.

3. Temperature:

a. The temperature at which the enzymes show maximum activity is called Optimum temperature.
b. The rate of chemical reaction is increased by a rise in temperature but this is true only over a limited range of temperature.
c. Enzymes rapidly denature at temperature above 40°C.
d. The activity of enzymes is reduced at low temperature.
e. The enzymatic reaction occurs best at or around 37°C which is the average normal body temperature in homeotherms.

4. Effect of pH:

a. The pH at which an enzyme catalyzes the reaction at the maximum rate is known as optimum pH.
b. The enzyme cannot perform its function beyond the range of its pH value.

5. Other substances:
a. The enzyme action is also increased or decreased in the presence of some other substances such as co-enzymes, activators and inhibitors.
b. Most of the enzymes are combination of a co-enzyme and an apo-enzyme.
c. Activators are the inorganic substances which increase the enzyme activity.
d. Inhibitor is the substance which reduces the enzyme activity.

Question (I)
What are the types of RNA? Mention the role of each class of RNA.
Answer:
There are three types of cellular RNAs:
1. messenger RNA (mRNA),
2. ribosomal RNA (rRNA),
3. transfer RNA (tRNA). ‘

1. Messenger RNA (mRNA):
a. It is a linear polynucleotide.
b. It accounts 3% of cellular RNA.
c. Its molecular weight is several million. , d. mRNA molecule carrying information to form a complete polypeptide chain is called cistron.
e. Size of mRNA is related to the size of message it contains.
f. Synthesis of mRNA begins at 5’ end of DNA strand and terminates at 3’ end.

Role of messenger RNA:
It carries genetic information from DNA to ribosomes, which are the sites of protein synthesis.

2. Ribosomal RNA (rRNA):
a. rRNA was discovered by Kurland in 1960.
b. It forms 50-60% part of ribosomes.
c. It accounts 80-90% of the cellular RNA.
d. It is synthesized in nucleus.
e. It gets coiled at various places due to intrachain complementary base pairing.
Role of ribosomal RNA: It provides proper binding site for m-RNA during protein synthesis.

3. Transfer RNA (tRNA):
a. These molecules are much smaller consisting of 70-80 nucleotides.
b. Due to presence of complementary base pairing at various places, it is shaped like clover-leaf.
c. Each tRNA can pick up particular amino acid.
d. Following four parts can be recognized on tRNA
1. DHU arm (Dihydroxyuracil loop/ amino acid recognition site
2. Amino acid binding site
3. Anticodon loop / codon recognition site
4. Ribosome recognition site.
e. In the anticodon loop of tRNA, three unpaired nucleotides are present called as anticodon which pair with codon present on mRNA.
f. The specific amino acids are attached at the 3′ end in acceptor stem of clover leaf of tRNA.
Role of transfer RNA: It helps in elongation of polypeptide chain during the process called translation.

Question (J)
What is metabolism? How metabolic pool is formed in the cell.
Answer:

1. Metabolism is the sum of the chemical reactions that take place within each cell of a living organism and provide energy for vital processes and for synthesizing new’ organic material.
2. Metabolic pool in the cell is formed due to glycolysis and Krebs cycle.
3. The catabolic chemical reaction of glycolysis and Krebs cycle provides ATP and biomolecules. These biomolecules form the metabolic pool of the cell.
4. These biomolecules can be utilized for synthesis of many important cellular components.
5. The metabolites can be added or withdrawn from the pool according to the need of the cell.

Question 6.
If double stranded DNA has 14% C (cytosine) what percent A (adenine), T (thymine) and G (guanine) would you expect?
Answer:
A purine always pairs with pyrimidine.
Adenine pairs with thymine and cytosine pairs with guanine.
Therefore, as per the given data If cytosine = 14% then guanine = 14%.
According to Chargaff s rule,
(C+G) = 14+ 14 = 28%
Therefore, (A+T) = 72%
So, A= 36%, T= 36%, G = 14%.

Question 7.
Name
1. The reagent used for testing for reducing sugar.
2. The form in which carbohydrate is transported in a plant.
3. The term that describes all the chemical reactions taking place in an organism.
Answer:
1. Benedict’s reagent
2. Sucrose
3. Metabolism

Practical / Project:

Question 1.
Perform an experiment to study starch granules isolated from potato.
Answer:
Isolation of starch granules from potato:

1. Peal the potato with a clean knife.
2. Grind the potato till the homogenous mixture is formed.
3. Then strain the mixture through a cheese cloth into a beaker.
4. Keep it standing for some time.
5. Throw the supernatant and fill the beaker containing starch with water.
6. Stir it well and again allow the starch to settle.
7. After sometime, again through the supernatant.
8. Repeat this for 2-3 times.
9. Collect the white starch in the watch glass and keep it in the oven for drying.

To study the isolated starch granules:
1. Examination under microscope:
Examine starch granules under microscope by using a mixture of equal volumes of glycerol and distilled water.
Result: The potato starch granules appears transparent granules. They are irregularly shaped.
2. Using Iodine solution:
Boil a little amount of starch with water. Cool it. Add iodine solution to it.
Result: The solution changes colour to blue. This indicates the presence of starch.

Question 2.
Study the action of enzyme urease on urea.
Answer:
Urease is an enzyme which exists in a dimer form. It has two active sites which are highly specific and only bind to urea or hydroxy urea. The active sites of urease contain nickel atoms. Urease catalyzes the hydrolysis of substrate urea into carbon dioxide and ammonia. It attacks the nitrogen and carbon bond in amide compounds and forms alkaline product like ammonia.

11th Biology Digest Chapter 6 Biomolecules Intext Questions and Answers

Can you recall? (Textbook Page No. 59)

(i) Which are different cell components?
Answer:
a. The three main components of any cell are: Cell membrane, Cytoplasm, Nucleus.
b. The components present in both plant and animal cells are: Endoplasmic reticulum, ribosomes, golgi apparatus, lysosomes, mitochondria, vacuoles.
c. The components present in plant cell and not in animal cell: Cell wall and plastids.
d. The components present in animal cell and not in plant cell: Cilia and flagella.

(ii) What is the role of each component of cell?
Answer:
The role of each component of a cell is as follows:
a. Cell membrane: Cell membrane separates the cytoplasmic contents from external environment.
b. Cytoplasm: Site for metabolic activities and organelles.
c. Nucleus: It is the control center of the cell. Genetic material is present in the nucleus.
d. Endoplasmic reticulum: It produces, processes and transports proteins and lipids.
e. Ribosomes: Ribosome is the site for protein synthesis.
f. Golgi apparatus: It is involved in modifying, sorting and packing of proteins for secretion. It also transports lipids around the cell.
g. Lysosomes: It is involved in digestion of worn out organelles and waste removal.
h. Mitochondria: It is responsible for production of energy.
i. Vacuoles: It has various functions like storage, waste disposal, protection and growth.
j. Cell wall: It provides strength and support to the cell.
k. Plastids: They are responsible for production and storage of food. It also contains photosynthetic pigments (Chloroplasts).
l. Cilia and flagella: Help in motility.

Can you tell? (Textbook Page No. 62)

What are carbohydrates?
Answer:

1. The word carbohydrates mean ‘hydrates of carbon’.
2. They are also called saccharides.
3. They are biomolecules made from just three elements: carbon, hydrogen and oxygen with the general formula C_x(H₂0)y.
4. They contain hydrogen and oxygen in the same ratio as in water (2:1).
5. Carbohydrates can be broken down (oxidized) to release energy.

Can you tell? (Textbook Page No. 62)

(i) Enlist the natural sources, structural units and functions of the following polysaccharides.
a. Starch
b. Cellulose
c. Glycogen
Answer:
a. Starch:
1. Natural Sources: Cereals (wheat, maize, rice), root vegetables (potato, cassava etc.)
2. Structural units: Starch consist of two types of molecules – Amylose and amylopectin.
3. Functions: It acts as a reserve food and supply energy.

b. Cellulose:
1. Natural sources: Plant fibers (cotton, flax, hemp, jute, etc.), wood.
2. Structural units: It is made from p glucose molecules.
3. Functions: It in a major component of cell wall. It provides structural support.

c. Glycogen:
1. Natural sources: Fruits, starchy vegetables, whole grain foods.
2. Structural units: It consists of linear chains of glucose residues. The glucose is linked linearly by a (1 → 4) glycosidic bonds and branches are linked to the linear chain by a (1 → 6) glycosidic bonds.
3. Functions: It is stored in liver and muscles and it readily provides energy when the blood glucose level decreases.

(ii) The exoskeleton of insects is made up of chitin. This is a ________.
(A) mucoprotein
(B) lipid
(C) lipoprotein
(D) polysaccharide
Answer:
polysaccharide

(iii) List names of structural polysaccharides.
Answer:
Arabinoxylans, cellulose, chitin, pectin.

(iv) Write a note on oligosaccharide and glycosidic bond.
Answer:
Oligosaccharides:
a. A carbohydrate polymer comprising of two to six monosaccharide molecules is called oligosaccharide.
b. They are linked together by glycosidic bond.
c. They are classified on the basis of monosaccharide units:
Disaccharides: These are the sugars containing two monosaccharide units and can be further hydrolysed into smaller components. E.g.: Sucrose, maltose, lactose, etc.
Trisaccharides: These contain three monomers. E.g. Raffmose.
Tetrasaccharides: These contain four monomers. E.g.: Stachyose.

Glycosidic bond:
a. Glycosidic bond is a covalent bond that forms a linkage between two monosaccharides by a dehydration reaction.
b. It is formed when a hydroxyl group of one sugar reacts with the anomeric carbon of the other.
c. Glycosidic bonds are readily hydrolyzed by acid but resist cleavage by base.
d. There are two types of glycosidic bonds: a-glycosidic bond and P-glycosidic bond.

Can you tell? (Textbook Page No. 63)

What are lipids? Classify them and give at least one example of each.
Answer:
Lipids:
Lipids are a group of heterogeneous compounds like fats, oils, steroids, waxes, etc.
They are macro-biomolecules.
These are group of substances with greasy consistency with long hydrocarbon chain containing carbon, hydrogen and oxygen.

Lipids are classified into:
1. Saturated fatty acids: They contain single chain of carbon atoms with single bonds.
E.g. Palmitic acid, stearic acid
2. Unsaturated fatty acids: They contain one or more double bonds between the carbon atoms of the hydrocarbon chain.
a. Simple lipids: These are esters of fatty acids with various alcohols.
E.g. Fats, wax.
b. Compound lipids: These are ester of fatty acids containing other groups like phosphate (Phospholipids), sugar (glycolipids), etc.
E.g. Lecithin
c. Sterols: They are derived lipids. They are composed of fused hydrocarbon rings (steroid nucleus) and a long hydrocarbon side chain.
E.g. Cholesterol, phytosterols.

Find out (Textbook Page No. 63)

(i) Why do high cholesterol level in the blood cause heart diseases?
Answer:
a. When there is high level of cholesterol in the blood, the cholesterol builds up on the walls of arteries causing a condition called atherosclerosis (a form of heart disease).
b. Because of this the arteries are narrowed and the blood flow to the heart is slowed down.
c. The blood carries oxygen to the heart, but because of this condition enough blood and oxygen does not reach to the heart and causes heart diseases.
d. If the condition increases, the supply of oxygen and blood is completely cut off to the heart and this can lead to heart attack.

(ii) Polyunsaturated fatty acids are believed to decrease blood cholesterol level. How?
Answer:
a. The liver converts polyunsaturated fatty acids into ketones instead of cholesterol.
b. Therefore, polyunsaturated fatty acids are transported directly to tissues for oxidation without leaving behind any lipoprotein in the form of cholesterol as it is seen in the case of saturated fatty acids.
c. Thus, polyunsaturated fatty acids are believed to decrease blood cholesterol level.

Can you tell? (Textbook Page No. 64)

Which of the following is a simple protein?
(A) nucleoprotein
(B) mucoprotein
(C) chromoprotein
(D) globulin
Answer:
Globulin

Can you tell? Textbook Page No. 64)

What are conjugated proteins? How do they differ from simple ones? Give one example of each.
Answer:
1. Conjugated proteins consist of a simple protein attached with some non-protein substance. The non-protein group is called prosthetic group.
2. The conjugated protein functions in interaction with other chemical group whereas simple proteins contain only amino acids and no other chemical group attached to it.
3. Example of conjugated protein is haemoglobin. Globin is the protein and iron containing pigment and haem is the prosthetic group.

Can you tell? (Textbook Page No. 64)

All Proteins are made up of the same amino acids; then how proteins found in human beings and animals may be different from those of other?
Answer:
The proteins found in human beings and animals may be different from those of others because the ratio of amino acids present in the protein differs.

Can you tell? (Textbook Page No. 67)

What is a nucleotide? How is it formed? Mention the names of all nucleotides.
Answer:
1. Nucleotide is a unit which consists of a sugar, phosphate and a base. Nucleotides are basic units of nucleic acids.
2. The nitrogen base and a sugar form a nucleoside. In a nucleoside, nitrogenous base is attached to the first carbon atom (C-1) of the sugar and when a phosphate group gets attached with that of the carbon (C-5) atom of the sugar molecule a nucleotide molecule is formed.
3. The names of all nucleotides are:

Base	Nucleotides of RNA	Nucleotides of DNA
Adenine	Adenylate	Deoxydenylate
Guanine	Guanylate	Deoxyguanylate
Cytosine	Cytidylate	Deoxy cytidylate
Thymine	–	Deoxythymidylate
Uracil	Uridylate	–

Can you tell? (Textbook Page No. 67)

Describe the structure of DNA molecule as proposed by Watson and Crick.
Answer:

1. According to Watson and Crick, DNA molecule consists of two strands twisted around each other in the form of a double helix.
2. The two strands i.e. polynucleotide chains are supposed to be in opposite direction so end of one chain having 3′ lies beside the 5′ end of the other.
3. One turn of the double helix of the DNA measures about 34A.
4. It consists paired nucleotides and the distance between two neighboring pair nucleotides is 3.4A.
5. The diameter of the DNA molecule has been found be 20A.

Can you tell (Textbook Page No. 70)

Name the chemical found in the living cell which has necessary message for the production of all enzymes required by it.
Answer:
DNA found in the nucleus of a living cell has necessary message for the production of all enzymes required by it. DNA forms mRNA through the process of transcription. This mRNA through the process of translation forms proteins.

Can you tell? (Textbook Page No. 67)

Difference between DNA and RNA is because of
(A) sugar and base
(B) sugar and phosphate
(C) phosphate and base
(D) sugar only
Answer:
Sugar and base

Can you tell? (Textbook Page No. 67)

Differentiate between DNA and RNA.
Answer:

DNA	RNA
1. It is a genetic material of majority of the organisms.	It is a genetic material only of some viruses.
2. It is double stranded.	It is single stranded.
3. Deoxyribose sugar is present.	Ribose sugar is present.
4. Nitrogen bases like Adenine, Guanine, Cytosine, Thymine are present.	Nitrogen bases like Adenine, Guanine, Cytosine, Uracil are present.
5. Specific base pairing is observed.	Nitrogen bases do not form pair.
6. Total number of purines is equal to total number of pyrimidine. Thus, purine to pyrimidine ratio is 1:1.	Amount of purine and pyrimidine may or may not be equal.
7. It is present in nucleus.	It is present in nucleus and cytoplasm.
8 It is responsible for determining hereditary characters and for formation of RNA.	It takes part in protein synthesis.

Can you tell? (Textbook Page No. 70)

Co-enzyme is ________
(A) often a metal
(B) often a vitamin
(C) always as organic molecule
(D) always an inorganic molecule
Answer:
Always as organic molecule

Can you tell? (Textbook Page No. 70)

(i) Which enzyme is needed to digest food reserve in castor seed?
(A) Amylase
(B) Diastase
(C) Lipase
(D) Protease
Answer:
Lipase

(ii) List the important properties of enzymes.
Answer:
a. Proteinaceous Nature
b. Three-Dimensional conformation
c. Catalytic property
d. Specificity of action
e. Temperature

Try this: (Textbook Page No. 70)

To demonstrate the effect of heat on the activities of inorganic catalysts and enzymes.
Answer:
1. Using MnO₂ and Enzymes without any heat treatment:
Mn0₂ and cellular enzymes (catalase/peroxidase) causes breakdown of H₂0₂ and evolution of oxygen.
2. Using Mn0₂ and Enzymes after heat treatment:
Oxygen evolves in the H₂0₂ solution containing boiled and cooled Mn0₂. But oxygen does not evolve in the tube containing the enzyme.
3. This confirms that heat affects the enzyme and inactivates it whereas heat does not have any effect on inorganic catalyst.

11th Std Biology Questions And Answers:

• Living World Class 11 Biology Questions And Answers
• Systematics of Living Organisms Class 11 Biology Questions And Answers
• Kingdom Plantae Class 11 Biology Questions And Answers
• Kingdom Animalia Class 11 Biology Questions And Answers
• Cell Structure and Organization Class 11 Biology Questions And Answers
• Biomolecules Class 11 Biology Questions And Answers
• Cell Division Class 11 Biology Questions And Answers
• Plant Tissues and Anatomy Class 11 Biology Questions And Answers

Balbharti Maharashtra State Board 11th Biology Textbook Solutions Chapter 12 Photosynthesis Textbook Exercise Questions and Answers.

Photosynthesis Class 11 Exercise Question Answers Solutions Maharashtra Board

Class 11 Biology Chapter 12 Exercise Solutions Maharashtra Board

Biology Class 11 Chapter 12 Exercise Solutions

1. Choose correct option

Question (A)
A cell that lacks chloroplast does not
(a) evolve carbon dioxide
(b) liberate oxygen
(c) require water
(d) utilize carbohydrates
Answer:
(b) liberate oxygen

Question (B)
Energy is transferred from the light reaction step to the dark reaction step by
(a) chlorophyll
(b) ADP
(c) ATP
(d) RuBP
Answer:
(c) ATP

Question (C)
Which one is wrong in photorespiration?
(a) It occurs in chloroplasts
(b) It occurs in day time only
(c) It is characteristic of C₄-plants
(d) It is characteristic of C₃-plants
Answer:
(c) It is characteristic of C₄-plants

Question (D)
Non-cyclic phosphorylation differs from cyclic photophosphorylation in that former
(a) involves only PS
(b) Include evolution of 02
(c) involves formation of assimilatory power
(d) both (b) and (c)
Answer:
(d) both (b) and (c)

Question (E)
For fixation of 6 molecules of C0₂ and formation of one molecule of glucose in Calvin cycle, requires
(a) 3 ATP and 2 MADPPE
(b) 18 ATP and 12 NADPH₂
(c) 30 ATP and 18 NADPH₂
(d) 6 ATP and 6 NADPIT₂
Answer:
(b) 18 ATP and 12 NADPH₂

Question (F)
In maize and wheat, the first stable products formed in bundle sheath cells respectively are
(a) OAA and PEPA
(b) OAA and OAA
(c) OAA and 3PGA
(d) 3PGA and OAA
Answer:
(c) OAA and 3PGA

Question (G)
The head and tail of chlorophyll are made up of
(a) porphyrin and phytin respectively
(b) pyrrole and tetrapyrrole respectively
(c) porphyrin and phytol respectively
(d) tetrapyrole and pyrrole respectively
Answer:
(c) porphyrin and phytol respectively

Question (H)
The net result of photo-oxidation of water is release of ……………. .
(a) electron and proton
(b) proton and oxygen
(c) proton, electron and oxygen
(d) electron and oxygen
Answer:
(c) proton, electron and oxygen

Question (I)
For fixing one molecule of C0₂ in Calvin cycle are required.
(a) 3ATP + 1NADPFE
(b) 3ATP + 2NADPH₂
(c) 2ATP + 3NADPH₂
(d) 3ATP + 3NADPFE
Answer:
(b) 3ATP + 2NADPH₂

Question (J)
In presence of high concentration of oxygen, RuBP carboxylase converts RuBP to …………… .
(a) Malic acid and PEP
(b) PGA and PEP
(c) PGA and malic acid
(d) PGA and phosphoglycolate
Answer:
(d) PGA and phosphoglycolate

Question (K)
The sequential order in electron transport from PSII to PSI of photosynthesis is
(a) FeS, PQ, PC and Cytochrome
(b) FeS, PQ, Cytochrome and PC
(c) PQ, Cytochrome, PC and FeS
(d) PC, Cytochrome, FeS, PQ
Answer:
(c) PQ, Cytochrome, PC and FeS

2. Answer the following questions

Question (A)
Describe the light-dependent steps of photosynthesis. How are they linked to dark reactions?
Answer:
The light dependent steps of photosynthesis include cyclic and non-cyclic photophosphorylation,
1. Cyclic photophosphorylation:
a. Illumination of photosystem-I causes electrons to move continuously out of the reaction center of photosystem-I and back to it.
b. The cyclic electron-flow is accompanied by the photophosphorylation of ADP to yield ATP. This is termed as Cyclic photophosphorylation.
c. Since this process involves only pigment system I, photolysis of water and consequent evolution of oxygen does not take place.

2. Non-cyclic photophosphorylation:
a. It involves both photosystems- PS-I and PS-II.
b. In this case, electron transport chain starts with the release of electrons from PS-II.
c. In this chain high energy electrons released from PS-II do not return to PS-II but, after passing through an electron transport chain, reach PS-I, which in turn donates it to reduce NADP to NADPH.
d. The reduced NADP⁺ (NADPH) is utilized for the reduction of CO₂ in the dark reaction.
e. Electron-deficient PS-II brings about oxidation of water-molecule. Due to this, protons, electrons and oxygen atom are released.
f. Electrons are taken up by PS-II itself to return to reduced state, protons are accepted by NADP⁺ whereas oxygen is released.
g. As in this process, high energy electrons released from PS-II do not return to PS-II and it is accompanied with ATP formation, this is called Non-cyclic photophosphorylation.

3. Link between light-dependent and dark reactions:

1. The light reaction gives rise to two important products, a reducing agent NADPH₂ and an energy rich compound ATP. Both these are utilized in the dark phase of photosynthesis.
2. ATP and NADPH₂ molecules function as vehicles for transfer of energy of sunlight into dark reaction leaving to carbon fixation. In this reaction C0₂ is reduced to carbohydrate.
3. During dark reaction, ATP and NADPH₂ are transformed into ADP, iP and NADP which are transferred to the grana in which light reaction takes place.

(B)

Question (a)
Distinguish between Respiration and Photorespiration
Answer:

Respiration	Photorespiration
1. Occurs in all aerobic and anaerobic organisms.	Occurs in C3 plants under high temperature, bright light, high oxygen and low C02 concentration.
2. A light independent process, occurs in both presence and absence of light.	A light dependent process, occurs in presence of Jight.
3. Produce energy rich molecules like ATP, GTP, FADH2, NADH2	Do not produce energy rich molecules such as ATP.
4. Respiration is an energy-producing process.	Photorespiration is an energy wastage process.

Question (b)
Distinguish between Cyclic photophosphorylation and Non-cyclic photophosphorylation
Answer:

Cyclic photophosphorylation	Non – cyclic photophosphorvlution
1. Electrons emitted by chlorophyll return back to the same chlorophyll.	The electrons emitted by chlorophyll do not return back to the same chlorophyll.
2. First electron acceptor is FRS.	First electron acceptor is CO – Q
3. It forms only ATP.	NADPH2 and ATP are formed.
4. Does not involve photolysis of H20.	Involves photolysis of H20.
5. No evolution of 02.	There is evolution of 02.
6. Only Photosystem-I (P700) is involved in this cycle.	Both Photosystem PS-I (P700) as well as PS-II (P680) are involved.

Question (C)
Answer the following questions.
1. What are the steps that are common to C₃ and C₄ photosynthesis?
2. Differentiate between C₃ and C₄ plants.
Answer:
Steps that are common to C₃ and C₄ photosynthesis are Carboxylation, Reduction, Glucose synthesis, Regeneration.
[Note: Students are expected to refer the given Q.R code for detail understanding the common steps between C₁ and C₄ plants.]

Question (D)
Are the enzymes that catalyze the dark reactions of carbon fixation located inside the thylakoids or outside the thylakoids?
Answer:
Carbon fixation occurs in the stroma by series of enzyme catalyzed steps. The enzymes that catalyze the dark reactions of carbon fixation are located outside the thylakoids.

Question (E)
Calvin cycle consists of three phases, what are they? Explain the significance of each of them.
Answer:
The entire process of dark reaction was traced by Dr. Melvin Calvin along with his co-worker, Dr. Benson. Hence, the process is called as Calvin cycle or Calvin- Benson cycle. Since the first stable product formed is a 3-carbon compound, it is also called as C₃ pathway and the plants are called C₁₄ plants.
Calvin carried out experiments on unicellular green algae (Chlorella), using radioactive isotope of carbon, C₁₄ as a tracer.
It is also called synthesis phase or second phase of photosynthesis.

The cycle is divided into the following phases:
1. Carboxylation phase:
a. Carbon dioxide reduction starts with a five-carbon sugar ribulose-l,5-bisphosphate (RuBP). It is a 5- carbon sugar with two phosphate groups attached to it.
b. RuBP reacts with CO₂ to produce an unstable 6 carbon intermediate in the presence of Rubisco.
c. It immediately splits into 3 carbon compounds called 3-phosphoglyceric acid.
d. RuBisCO is a large protein molecule and comprises 16% of the chloroplast proteins.

2. Glycolytic reversal:
a. 3-phosphoglyceric acid form 1,3-diphosphoglyceric acid by utilizing ATP molecule.
b. These are then reduced to glyceraldehyde-3-phosphate (3-PGA) by NADPH supplied by the light reactions of photosynthesis.
c. In order to keep Calvin cycle continuously running there must be sufficient number of RuBP and regular supply of ATP and NADPH.
d. Out of 12 molecules of 3-phosphoglyceraldehyde, two molecules are used for synthesis of one glucose molecule.

3. Regeneration of RuBP:
a. 10 molecules of 3-phosphoglyceraldehyde are used for the regeneration of 6 molecules of RuBP at the cost of 6 ATP.
b. Therefore, six turns of Calvin cycle are needed to get one molecule of glucose.
Significance:
1. Carboxylation: RuBisCO is the most abundant enzyme in the world. It is responsible for fixing carbon in the form of C0₂ into sugar. As a result of Carboxylation, the first stable product of carbon fixation i.e. 3- PGA is synthesized.
2. Reduction/Glycolytic reversal: NADPH2 donates electrons to 1, 3-Bisphoshoglycerate to form 3- phosphoglyceraldehyde molecules. During this process ADP and NADP are generated which are used in light reaction.
3. Regeneration of RuBP: Some 3-phosphoglyceraldehyde molecules are involved in production of glucose while others are recycled to regenerate the 5-carbon compound RuBP which used to accept new carbon molecules. Thus, regeneration of RuBP is required for Calvin cycle to run continuously.

Question (F)
Why are plants that consume more than the usual 18 ATP to produce 1 molecule of glucose favoured in tropical regions?
Answer:

1. C₄ plants are favoured in tropical regions as they require 30 ATP to produce 1 molecule of glucose.
2. High temperature in tropical regions leads to closure of stomata to reduce rate of transpiration. Due to this availability of C0₂ decreases.
3. PEP carboxylase present in mesophyll cells can fix C0₂ even at low concentration. This helps the plant in efficient assimilation of atmospheric carbon dioxide.
4. C₄ plants contain a special leaf anatomy called Kranz anatomy which minimizes the losses due to photorespiration.
5. It helps C₄ plants to survive in conditions of high daytime temperatures, intense sunlight and low moisture.

Question (G)
What is the advantage of having more than one pigment molecule in a photocenter?
Answer:
Advantages of having more pigment molecules in a photocenter are as follows:
1. Having more than one pigment molecule in photocenter means more sunlight being captured and thus facilitating more effective light reaction.
2. It will provide protection to chlorophyll molecule against photo-oxidation.
3. More pigments will capture more energy to start the initial reactions, which is not possible by single pigment.

Question (H)
Why does chlorophyll appear green in reflected light and red transmitted light? Explain the significance of these phenomena in terms of photosynthesis.
Answer:
1. Chlorophyll is a light absorbing pigment. It absorbs light in red and blue regions of the visible light spectrum. Absorption spectrum of chlorophyll for red light is maximum so chlorophyll appears red in transmitted light. Green light is not absorbed but reflected so chlorophyll appear green in reflected light.
2. Chlorophyll predominantly absorbs red and violet-blue light and it allows plants to use this light as a form of energy for photosynthesis process.
3. It is most effective wavelength of light in photosynthesis as it has exactly right amount of energy to excite electrons of chlorophyll and boost them out of their orbits to higher energy level.

Question (I)
Explain why photosynthesis is considered the most important process in the biosphere.
Answer:
Photosynthesis is considered to be the most important process in the biosphere due to following reasons:
1. Photosynthesis is the biochemical process through which all plants (primary producers) produce food.
2. It is responsible for release of oxygen in the atmosphere.
3. Heterotrophs are directly or indirectly dependent on autotrophs for energy and other related resources. Therefore, photosynthesis is considered the most important process in the biosphere.

Question (J)
Why is photolysis of water accompanied with non-cyclic photophosphorylation?
Answer:
1. Photolysis of water provides new electrons to Photosystem – II.
2. The water molecule is lysed into three components:
a. Protons (H⁺) which are used as part of reactions that makes NADPH.
b. Second component formed is electrons which replaces the electrons lost by PS-II.
c. The third component is oxygen (0₂) which is released into the atmosphere.
3. Photosystem I sends electrons to reduce NADP+.
4. Then, Photosystem II sends replacement electrons to Photosystem I.
5. Finally, photolysis of water replaces the electrons lost by Photosystem II.
6. Water is the ultimate source of electrons for photosynthesis.
Therefore, photolysis of water is accompanied with non – cyclic photophosphorylation.

Question (K)
In C-4 plants, why is C-3 pathway operated in bundle sheath cells only?
Answer:

1. Decarboxylation of malic acid occurs in bundle sheath cells of C₄ plants. Due to which concentration of C0₂ increases in bundle sheath cells.
2. The enzymes required for Calvin cycle i.e. RuBisCO is present in bundle sheath cells.
3. In presence of high concentration of C0₂, RuBisCO acts as carboxylase and bring about carboxylation of RuBP.
4. Hence, in C-4 plants, C-3 pathway is operated in bundle sheath cells only.

Question (L)
What would have happened if C-4 plants did not have Kranz anatomy?
Answer:
Photorespiration would occur if C₄ plants did not have Kranz anatomy.

Question (M)
Why does RuBisCO carry out preferentially carboxylation than oxygenation in C₄ plants?
Answer:

1. In C₄ plants, C0₂ taken from the atmosphere is accepted by a 3-carbon compound, phosphoenolpyruvic acid in the chloroplasts of mesophyll cells.
2. This leads to the formation of 4-carbon compound oxaloacetic acid with the help of enzyme phosphoenolpyruvate carboxylase.
3. It is converted to another 4-carbon compound called malate.
4. Malate is transported to chloroplasts of bundle sheath cells where malate is converted to pyruvate and releases C0₄ in the cytoplasm thus increasing the concentration of C0₂ in the bundle sheath cells.
5. Chloroplasts of bundle sheath cells contains enzymes of Calvin cycle.
6. Thus, due to high concentration of C0₂, RuBisCO participates in carboxylation and not in oxygenation.

Question (N)
What would have happened if plants did not have accessory pigments?
Answer:

1. Accessory pigments are light absorbing molecules which are found in photosynthetic organisms.
2. They transfer the absorbed light to chlorophyll-a and thus increasing the photosynthetic rate.
3. In absence of accessory pigments less amount of light will be absorbed and also there would be no protection provided to chlorophyll molecule from photo-oxidation.

Question (O)
How can you identify whether the plant is C₃ or C₄? Explain / Justify.
Answer:

1. By observing the cross section of a leaf we can identify whether the plant is a C₃ plant or a C₃ plant.
2. C₄ plants possess a special anatomy of leaves called Kranz anatomy. In Kranz anatomy two types of chloroplasts are present, agranal in bundle sheath cells and granal in mesophyll cells.
3. In C₃ plants Kranz anatomy is absent.

Question (P)
In C₄ plants, bundle sheath cells carrying out Calvin cycle are very few in number. Then also, C₄ plants are highly productive. Explain.
Answer:

1. C₄ plants have special type of leaf anatomy called Kranz anatomy.
2. In C₄ plants, C0₂ fixation occurs twice.
3. In these plants, chloroplasts of mesophyll cells contain enzyme PEP carboxylase which fixes atmospheric C0₂.
4. Thus, first C0₂ fixation occurs in mesophyll cells.
5. Decarboxylation of malic acid in bundle sheath cells results in increase in C0₂ concentration.
6. Thus, RuBisCO acts as carboxylase and brings about carboxylation of RuBP.
7. Due to this oxygenation of RuBP and photorespiration is prevented.
8. Thus, despite of having less number of bundle sheath cells carrying out Calvin cycle, C₄ plants are highly productive.

Question (Q)
What is functional significance of Kranz anatomy?
Answer:

1. Leaves of C₄ plants show some structural peculiarities called Kranz anatomy.
2. The chloroplast of mesophyll cells contain enzyme PEP Carboxylase, which can fix C0₂ at low concentration.
3. Thus, light reaction and evolution of 0₂ occurs in mesophyll cells.
4. Decarboxylation of malate occurs in bundle sheath cells, which results in release of C0₂, due to which concentration of C0₂ in bundle sheath cells increases.
5. Enzyme RuBisCO present in bundle sheath cells acts as carboxylase in presence of high C0₂ concentraion and catalyses carboxylation of RuBP.
6. Thus, possibility of oxygenation of RuBP is avoided and photorespiration does not take place.

3. Correct the pathway and name it.

Question 1.
Correct the pathway and name it.
Answer:

1. The pathway shown is C₄ pathway.
2. M. D. Hatch and C. R. Slack while working on sugarcane found four carbon compounds (dicarboxylic acid) as the first stable product of photosynthesis.
3. It occurs in tropical and sub-tropical grasses and some dicotyledons.
4. The first product of this cycle is a 4-carbon compound oxaloacetic acid. Hence it is also called as C₄ pathway and plants are called C₄ plants.

Mechanism:

1. C0₂ taken from atmosphere is accepted by a 3-carbon compound, phosphoenolpyruvic acid in the chloroplasts of mesophyll cells, leading to the formation of 4-C compound, oxaloacetic acid with the help of enzyme phosphoenolpyruvate carboxylase.
2. It is converted to another 4-C compound, malic acid.
3. It is transported to the chloroplasts of bundle sheath cells.
4. Malic acid (4-C) is converted to pyruvic acid (3-C) with the release of C0₂ in the cytoplasm.
5. Thus, concentration of C0₂ increases in the bundle sheath cells.
6. Chloroplasts of these cells contain enzymes of Calvin cycle.
7. Because of high concentration of C0₂, RuBP carboxylase participates in Calvin cycle and not photorespiration.
8. Sugar formed in Calvin cycle is transported into the phloem.
9. Pyruvic acid generated in the bundle sheath cells re-enter mesophyll cells and regenerates
   phosphoenolpyruvic acid by consuming one ATP.
10. Since this conversion results in the formation of AMP (not ADP), two ATP are required to regenerate ATP from AMP.
11. Thus, C₄ pathway needs 12 additional ATP.
12. The C₃ pathway requires 18 ATP for the synthesis of one glucose molecule, whereas C4 pathway requires 30 ATP.
13. Thus, C₄ plants are better photosynthesizers as compared to C₃ plants as there is no photorespiration in these plants.

4. Is there something wrong in following schematic presentation? If yes, correct it so that photosynthesis will be operated.

Question 1.
Is there something wrong in following schematic presentation? If yes, correct it so that photosynthesis will be operated.
Answer:

Non-cyclic photophosphorylation:
a. It involves both photosystems- PS-I and PS-II.
b. In this case, electron transport chain starts with the release of electrons from PS-II.
c. In this chain high energy electrons released from PS-II do not return to PS-II but, after passing through an electron transport chain, reach PS-I, which in turn donates it to reduce NADP to NADPH.
d. The reduced NADP⁺ (NADPH) is utilized for the reduction of CO₂ in the dark reaction.
e. Electron-deficient PS-II brings about oxidation of water-molecule. Due to this, protons, electrons and oxygen atom are released.
f. Electrons are taken up by PS-II itself to return to reduced state, protons are accepted by NADP⁺ whereas oxygen is released.
g. As in this process, high energy electrons released from PS-II do not return to PS-II and it is accompanied with ATP formation, this is called Non-cyclic photophosphorylation.

Practical/ Project:

Question 1.
Draw schematic presentation of different processes/ cycles/ reactions related to photosynthesis.
Answer:
Cyclic photophosphorylation:
a. Illumination of photosystem-I causes electrons to move continuously out of the reaction center of photosystem-I and back to it.
b. The cyclic electron-flow is accompanied by the photophosphorylation of ADP to yield ATP. This is termed as Cyclic photophosphorylation.
c. Since this process involves only pigment system I, photolysis of water and consequent evolution of oxygen does not take place.

Non-cyclic photophosphorylation::
a. It involves both photosystems- PS-I and PS-II.
b. In this case, electron transport chain starts with the release of electrons from PS-II.
c. In this chain high energy electrons released from PS-II do not return to PS-II but, after passing through an electron transport chain, reach PS-I, which in turn donates it to reduce NADP to NADPH.
d. The reduced NADP⁺ (NADPH) is utilized for the reduction of CO₂ in the dark reaction.
e. Electron-deficient PS-II brings about oxidation of water-molecule. Due to this, protons, electrons and oxygen atom are released.
f. Electrons are taken up by PS-II itself to return to reduced state, protons are accepted by NADP⁺ whereas oxygen is released.
g. As in this process, high energy electrons released from PS-II do not return to PS-II and it is accompanied with ATP formation, this is called Non-cyclic photophosphorylation.

Interdependence of light and dark reactions:

1. The light reaction gives rise to two important products, a reducing agent NADPH₂ and an energy rich compound ATP. Both these are utilized in the dark phase of photosynthesis.
2. ATP and NADPH₂ molecules function as vehicles for transfer of energy of sunlight into dark reaction leaving to carbon fixation. In this reaction C0₂ is reduced to carbohydrate.
3. During dark reaction, ATP and NADPH₂ are transformed into ADP, iP and NADP which are transferred to the grana in which light reaction takes place.

Calvin cycle: The entire process of dark reaction was traced by Dr. Melvin Calvin along with his co-worker, Dr. Benson. Hence, the process is called as Calvin cycle or Calvin- Benson cycle. Since the first stable product formed is a 3-carbon compound, it is also called as C₃ pathway and the plants are called C₁₄ plants.
Calvin carried out experiments on unicellular green algae (Chlorella), using radioactive isotope of carbon, C₁₄ as a tracer.
It is also called synthesis phase or second phase of photosynthesis.
The cycle is divided into the following phases:
1. Carboxylation phase:
a. Carbon dioxide reduction starts with a five-carbon sugar ribulose-l,5-bisphosphate (RuBP). It is a 5- carbon sugar with two phosphate groups attached to it.
b. RuBP reacts with CO₂ to produce an unstable 6 carbon intermediate in the presence of Rubisco.
c. It immediately splits into 3 carbon compounds called 3-phosphoglyceric acid.
d. RuBisCO is a large protein molecule and comprises 16% of the chloroplast proteins.

2. Glycolytic reversal:
a. 3-phosphoglyceric acid form 1,3-diphosphoglyceric acid by utilizing ATP molecule.
b. These are then reduced to glyceraldehyde-3-phosphate (3-PGA) by NADPH supplied by the light reactions of photosynthesis.
c. In order to keep Calvin cycle continuously running there must be sufficient number of RuBP and regular supply of ATP and NADPH.
d. Out of 12 molecules of 3-phosphoglyceraldehyde, two molecules are used for synthesis of one glucose molecule.

3. Regeneration of RuBP:
a. 10 molecules of 3-phosphoglyceraldehyde are used for the regeneration of 6 molecules of RuBP at the cost of 6 ATP.
b. Therefore, six turns of Calvin cycle are needed to get one molecule of glucose.

Photorespiration: Mechanism:
1. Photorespiration involves three organelles chloroplast, peroxisomes and mitochondria and occurs in a series of cyclic reactions which is also called PCO cycle. (Photosynthetic Carbon Cycle)
2. Enzyme Rubisco acts as oxygenase at higher concentration of O₂ and photorespiration begins.
3. When RuBP reacts with 0₂ rather than C0₂ to form a 3-carbon compound (PGA) and 2-carbon compound phosphoglycolate.
4. Phosphoglycolate is then converted to glycolate which is shuttled out of the chloroplast into the peroxisomes.
5. In Peroxisomes, glycolate is converted into glyoxylate by enzyme glycolate oxidase.
6. Glyoxylate is further converted into amino acid glycine by transamination.
7. In mitochondria, two molecules of glycine are converted into serine (amino acid) and C0₂ is given out.
8. Thus, it loses 25% of photosynthetically fixed carbon.
9. Serine is transported back to peroxisomes and converted into glycerate.
10. It is shuttled back to chloroplast to undergo phosphorylation and utilized in formation of 3-PGA, which get utilized in C₃ pathway.
Hatch-Slack pathway: M. D. Hatch and C. R. Slack while working on sugarcane found four carbon compounds (dicarboxylic acid) as the first stable product of photosynthesis.
It occurs in tropical and sub-tropical grasses and some dicotyledons.
The first product of this cycle is a 4-carbon compound oxaloacetic acid. Hence it is also called as C₄ pathway and plants are called C₄ plants.
Mechanism:

1. C0₂ taken from atmosphere is accepted by a 3-carbon compound, phosphoenolpyruvic acid in the chloroplasts of mesophyll cells, leading to the formation of 4-C compound, oxaloacetic acid with the help of enzyme phosphoenolpyruvate carboxylase.
2. It is converted to another 4-C compound, malic acid.
3. It is transported to the chloroplasts of bundle sheath cells.
4. Malic acid (4-C) is converted to pyruvic acid (3-C) with the release of C0₂ in the cytoplasm.
5. Thus, concentration of C0₂ increases in the bundle sheath cells.
6. Chloroplasts of these cells contain enzymes of Calvin cycle.
7. Because of high concentration of C0₂, RuBP carboxylase participates in Calvin cycle and not photorespiration.
8. Sugar formed in Calvin cycle is transported into the phloem.
9. Pyruvic acid generated in the bundle sheath cells re-enter mesophyll cells and regenerates
   phosphoenolpyruvic acid by consuming one ATP.
10. Since this conversion results in the formation of AMP (not ADP), two ATP are required to regenerate ATP from AMP.
11. xi. Thus, C₄ pathway needs 12 additional ATP.
12. The C₃ pathway requires 18 ATP for the synthesis of one glucose molecule, whereas C4 pathway requires 30 ATP. Thus, C₄ plants are better photosynthesizers as compared to C₃ plants as there is no photorespiration in these plants.
13. CAM Pathway: In CAM plants, malic acid accumulates during night, which is formed from Oxaloacetic acid in presence of the enzyme malate dehydrogenase.

Question 2.
Check the effects of different factors on photosynthesis under the guidance of teacher.
Answer:
External factors which affect photosynthesis are as follows:
1. Light:
a. It is an essential factor as it supplies the energy necessary for photosynthesis.
b. Quality and intensity of light affects the photosynthesis.
c. Highest rate of photosynthesis takes place in red light followed by blue light.
d. The rate of photosynthesis considerably decreases in plants which are growing under a forest canopy.
e. In most of the plants, photosynthesis is maximum in bright diffused sunlight.
f. Uninterrupted and continuous photosynthesis for a very long period of time may be sustained without any visible damage to the plant.

2. Carbon dioxide:
The main source of C0₂ in land plants is the atmosphere, which contains only 0.3% of the gas.
b. Under normal conditions of temperature and light, carbon dioxide acts as a limiting factor in photosynthesis.
c. Increase in concentration of CO₂ increases the photosynthesis.
d. Increase in C0₂ to about 1% is advantageous to most of the plants.
e. Higher concentration of the gas has an inhibitory effect on photosynthesis.

3. Temperature:
a. Like all other physiological processes, photosynthesis also needs a suitable temperature.
b. The optimum temperature at which the photosynthesis is maximum is 25-30 °C. Except in plants like Opuntia, photosynthesis takes place at as high as 55 °C.
c. This is the maximum temperature. Minimum temperature is temperature at which photosynthesis process just starts.
d. In the presence of sufficient light and CO₂, photosynthesis increases with the rise of temperature till it becomes maximum. After that there is a decrease or fall in the rate of the process.

4. Water:
a. Water is necessary for photosynthetic process.
b. An increase in water content of the leaf results in the corresponding increase in the rate of photosynthesis.
c. Thus, the limiting effect of water is not direct but indirect.
d. It is mainly due to the fact that it helps in maintaining the turgidity of the assimilatory cells and the proper hydration of their protoplasm.
[Students can refer the given information and perform this activity on their own]

11th Biology Digest Chapter 12 Photosynthesis Intext Questions and Answers

Can you recall? (Textbook Page No. 138)

(i) Why energy is essential in different life processes?
Answer:
a. Energy is the basic requirement of life.
b. Without energy no work can be done.
c. All living organisms need energy for reproduction and survival.
d. Sun is the main source of energy, and that energy should be transformed into the usable forms for living organisms to carry out life processes.
Therefore, energy is essential in different life processes.

(ii) How do we get energy?
Answer:
a. Sun is the main source of energy.
b. Plants utilize sunlight, carbon dioxide and water for the process called photosynthesis to produce sugars.
c. Animals make use of these sugars provided by the plants in their own cellular energy factories called mitochondria. Thus, energy is obtained.

Use your brainpower (Textbook Page No. 138)

Justify: All life on earth is ‘bottled solar energy’.
Answer:

1. Life on earth is dependent on solar energy directly or indirectly.
2. Plants by carrying out photosynthesis converts solar energy into chemical energy by producing carbohydrates.
3. Humans and animals depend on plants for food. Basically, life on earth depends totally on photosynthesis for food and energy.
4. Therefore, all life on earth is bottled solar energy.

Can you tell? (Textbook Page No. 140)

Draw well labeled diagram of ultrastructure of chloroplast.
Answer:

1. The chloroplasts are discoid and lens shaped in higher plants. Chloroplast is bounded by a double membrane.
   System of chlorophyll bearing a double-membrane sac is present inside the stroma.
2. These are stacked one above the other to form grana.
3. Individual sacs in each granum is are known as thylakoid.
4. All the pigments chlorophylls, carotenes and xanthophylls are located in thylakoid membranes.
5. These pigments are fat soluble and are present in lipid part of membrane also they absorb light of specific spectrum in the visible regions.

Use your brainpower (Textbook Page No. 140)

The photosynthetic lamellae taken out from a chloroplast and suspended in a nutrient medium in the presence of C0₂ and light. Will they synthesize sugar or not?
Answer:
Photosynthetic ladmellae will not synthesize sugar because sugar synthesis occurs only in stroma, as all the enzymes required for sugar synthesis are present there. In photosynthetic lamellae only light reactions occur. Thus, lamellae cannot synthesize sugar even when C0₂, light and other nutrients are provided.

Internet my friend (Textbook Page No. 139)

Collect information: Why does chlorophyll appear red in reflected light and green in transmitted light?
Answer:
Chlorophyll is a light absorbing pigment. It absorbs light in red and blue regions of the visible light spectrum. Absorption spectrum of chlorophyll for red light is maximum so chlorophyll appears red in transmitted light. Green light is not absorbed but reflected so chlorophyll appear green in reflected light. [Note: Chlorophyll appear red in transmitted light and green in reflected light.[

Activity 1 (Textbook Page No. 139)

Grind the spinach leaves in small quantity of acetone / nail paint remover. Mix the contents properly and filter with filter paper in test tube. Test tube contains green filtrate. Take the test tube in dark-room and put a flash of torch on it. Now, solution appears red. Why does this occur? Which phenomenon is this? Discuss this with your physics, chemistry and biology teachers.
Answer:
Chlorophyll is the green pigment present in chloroplast. It absorbs light in red and blue region of visible spectrum. It does not absorb green light and thus the green light is reflected which is why it appears green. In this experiment, the chlorophyll in test tube appears red when a flash torch is put on it in the dark room.

This is because when the electrons of the chlorophyll molecule are excited in dark in the absence of electron transport chain the electrons release their energy in the form of red light as they return to ground state. This phenomenon observed here is transmission of light.

Activity 2 (Textbook Page No. 139)

To separate the chloroplast pigments by paper chromatography. Concentrate the extracted chlorophyll solution by evaporation. Apply a drop of it at one end, 2cm away from edge of a strip of chromatography paper and allow it to dry thoroughly. Take a mixture of petroleum ether and acetone in the ratio of 9:1 at temperature of 40 to 60°C. Hang the strip in the jar with its loaded end dipping in the solvent. Close the jar tightly and keep it for an hour. The pigments separate into distinct green and yellow bands of chlorophyll and carotenoid respectively.
Answer:
Pigments are the molecules which reflects only certain wavelengths of visible light. Chromatography is the technique used to separate the chloroplast pigments. Carotenes form yellow-orange band, chlorophyll forms blue-green band, chlorophyll b forms yellow-green bands.

Can you tell? (Textbook Page No. 139)

Tomatoes, carrots and chillies are red in colour due to the presence of pigments. Name the pigment. Answer:
Red colour pigment present in tomatoes, carrots and chillies is lycopene.

Think about it (Textbook Page No. 140)

Large number of gas bubbles are evolved during day time in a pond of water.
Answer:
Photosynthesis occurs in the presence of sunlight. During photosynthesis, plants give out oxygen and take in carbon dioxide. The plants present underwater carryout photosynthesis and release oxygen. Hence, large number of gas bubbles are evolved during day time in a pond.

Think about it (Textbook Page No. 141)

Does moonlight support photosynthesis?
Answer:
The reactions of photosynthesis take place in the presence of sunlight. The intensity of moonlight is several thousand times less than that of direct sunlight which is insufficient for the light dependent phase of photosynthesis. As the sun sets, rate of photosynthesis also decreases. Therefore, moonlight does not support photosynthesis.

Can you tell? (Textbook Page No. 145)

How chlorophyll – a is excited? Show it with a diagram.
Answer:

1. Chlorophyll-a is an essential photosynthetic pigment as it converts light energy into chemical energy and acts as a reaction centre.
2. Initially, it lies at ground state or singlet state but when it absorbs or receives photons (solar energy), it gets activated and goes in excited state or excited second singlet state.
3. In the excited state, chlorophyll-a emits an electron. The emitted electron is energy rich, i.e. has extra amount of energy.
4. Due to the loss of electron (e_–), chlorophyll-a becomes positively charged. This is the ionized state.
5. Chlorophyll-a molecule cannot remain in the ionized state for more than 10‘9 seconds. Hence the photo-chemical reaction or electron transfer occurs very fast.
6. The energy rich electron is then transferred through various electron acceptors and donors (carriers).
7. During the transfer, the electron emits energy which is utilized for the synthesis of ATP. This shows that light energy is converted into chemical energy in the form of ATP.

Can you tell? (Textbook Page No. 140)

What made Hill to perform his experiment?
Answer:
Robert Hill proved that the source of oxygen evolved during photosynthesis is water and not carbon dioxide. Hence, it is called Hill’s Reaction.

1. In this experiment, Hill cultured isolated chloroplasts in a medium containing C0₂ free water, haemoglobin and ferric compound.
2. Ferric salts and haemoglobin were added in the medium as hydrogen and oxygen acceptors respectively.
3. When the suspension was illuminated, he observed that haemoglobin turned into oxyhaemoglobin (red colour).
4. This confirmed that water must have oxidized releasing 0₂, that reacted with haemoglobin. Reduction of ferric compound was also indicated by change in colour.
5. The H₂O molecule oxidized to evolve 0₂ as a by-product. Thus, Hill proved that the source of evolving 0₂ is H₂0 and not C0₂.
6. This process of splitting up of water molecules under the influence of light in the presence of chlorophyll is called Photolysis of water or Hill Reaction.
7. Hill’s reaction can be represented as follows:

Can you tell? (Textbook Page No. 145)

Draw a flowchart of non-cyclic photophosphorylation.
Answer:
Non-cyclic photophosphorylation:
a. It involves both photosystems- PS-I and PS-II.
b. In this case, electron transport chain starts with the release of electrons from PS-II.
c. In this chain high energy electrons released from PS-II do not return to PS-II but, after passing through an electron transport chain, reach PS-I, which in turn donates it to reduce NADP to NADPH.
d. The reduced NADP⁺ (NADPH) is utilized for the reduction of CO₂ in the dark reaction.
e. Electron-deficient PS-II brings about oxidation of water-molecule. Due to this, protons, electrons and oxygen atom are released.
f. Electrons are taken up by PS-II itself to return to reduced state, protons are accepted by NADP⁺ whereas oxygen is released.
g. As in this process, high energy electrons released from PS-II do not return to PS-II and it is accompanied with ATP formation, this is called Non-cyclic photophosphorylation.

Can you tell? (Textbook Page No. 145)

Describe Calvin’s cycle.
Answer:
The entire process of dark reaction was traced by Dr. Melvin Calvin along with his co-worker, Dr. Benson. Hence, the process is called as Calvin cycle or Calvin- Benson cycle. Since the first stable product formed is a 3-carbon compound, it is also called as C₃ pathway and the plants are called C₁₄ plants.
Calvin carried out experiments on unicellular green algae (Chlorella), using radioactive isotope of carbon, C₁₄ as a tracer.
It is also called synthesis phase or second phase of photosynthesis.
The cycle is divided into the following phases:
1. Carboxylation phase:
a. Carbon dioxide reduction starts with a five-carbon sugar ribulose-l,5-bisphosphate (RuBP). It is a 5- carbon sugar with two phosphate groups attached to it.
b. RuBP reacts with CO₂ to produce an unstable 6 carbon intermediate in the presence of Rubisco.
c. It immediately splits into 3 carbon compounds called 3-phosphoglyceric acid.
d. RuBisCO is a large protein molecule and comprises 16% of the chloroplast proteins.

2. Glycolytic reversal:
a. 3-phosphoglyceric acid form 1,3-diphosphoglyceric acid by utilizing ATP molecule.
b. These are then reduced to glyceraldehyde-3-phosphate (3-PGA) by NADPH supplied by the light reactions of photosynthesis.
c. In order to keep Calvin cycle continuously running there must be sufficient number of RuBP and regular supply of ATP and NADPH.
d. Out of 12 molecules of 3-phosphoglyceraldehyde, two molecules are used for synthesis of one glucose molecule.

3. Regeneration of RuBP:
a. 10 molecules of 3-phosphoglyceraldehyde are used for the regeneration of 6 molecules of RuBP at the cost of 6 ATP.
b. Therefore, six turns of Calvin cycle are needed to get one molecule of glucose.

Can you tell? (Textbook Page No. 147)

Summarise the photosynthetic reaction.
Answer:
6C0₂ + 12H₂0 → C₂H₁₂0₆ + 60₂ + 6H₂0
1. Photosynthesis is a two step process.
The light dependent reactions convert the light energy from the sun into chemical energy.
The light independent reactions convert the chemical energy to synthesize carbohydrates.
2. Light dependent reactions: Light is absorbed by chlorophyll which results in the production of ATP. Photolysis of water produce oxygen and hydrogen. The hydrogen and ATP are used in the light independent reactions and the oxygen is released from stomata.
3. Light independent reactions: ATP and hydrogen are transferred to the site of light independent reactions. The hydrogen is combined with carbon dioxide to form complex organic compounds.
The ATP provides the required energy to power these anabolic reactions and fix the carbon molecules.

Can you tell? (Textbook Page No. 147)

Summarise the photosynthetic reaction.
Answer:
6C0₂ + 12H₂0 → C₂H₁₂0₆ + 60₂ + 6H₂0
1. Photosynthesis is a two step process.
The light dependent reactions convert the light energy from the sun into chemical energy.
The light independent reactions convert the chemical energy to synthesize carbohydrates.
2. Light dependent reactions: Light is absorbed by chlorophyll which results in the production of ATP. Photolysis of water produce oxygen and hydrogen. The hydrogen and ATP are used in the light independent reactions and the oxygen is released from stomata.
3. Light independent reactions: ATP and hydrogen are transferred to the site of light independent reactions. The hydrogen is combined with carbon dioxide to form complex organic compounds.
The ATP provides the required energy to power these anabolic reactions and fix the carbon molecules.

Can you tell? (Textbook Page No. 147)

C₄ plants are more productive. Why?
Answer:

1. Photorespiration is considered to be a wasteful process in plants. It is an energy consuming process in plants which ultimately leads to reduction in final yield of plants.
2. During C₃ photosynthesis, 25% of the carbon dioxide fixed has to pass through photorespiratory process.
3. This decreases the photosynthetic productivity.
4. In C₄ plants, photorespiration is absent and hence they have better productivity.

Can you tell? (Textbook Page No. 147)

Xerophytic plants survive in high temperature. How?
Answer:

1. Xerophytic plants are those that have adapted to dry environments.
2. They have adapted to arid conditions by storing water in stems.
3. Stomata of these plants remain closed during day time to reduce the rate of transpiration to bare minimum.
4. Leaves are modified into spines or are reduced in size to check the loss of water due to transpiration.
5. The waxy surfaces of xerophytic plants prevent the loss of moisture.
6. Thus, they are able to survive in high temperature.

Can you tell? (Textbook Page No. 147)

Compare C₄ plants and CAM plants.
Answer:

C4 Plants	CAM Plants
1. These are mostly tropical and subtropical plants.	These are mostly xerophytic plants.
2. Leaves show Kranz anatomy.	Leaves does not show Kranz anatomy.
3. Stomata is open during day time.	Stomata is open during night time.
4. Photorespiration is not easily detectable.	Photorespiration is detectable in afternoon.
5. Carbon fixation takes place in mesophyll cells and Calvin Cycle takes place in bundle sheath cells.	Photosynthesis takes place in the mesophyll cells but carbon fixation takes place at night and Calvin cycle happens during day.
e.g. Sugarcane, maize, jowar, Amaranthus, etc.	e.g. Kalanchoe, Opuntia, Aloe, etc.

11th Std Biology Questions And Answers:

• Morphology of Flowering Plants Class 9 Biology Questions And Answers
• Animal Tissue Class 9 Biology Questions And Answers
• Study of Animal Type : Cockroach Class 9 Biology Questions And Answers
• Photosynthesis Class 9 Biology Questions And Answers
• Respiration and Energy Transfer Class 9 Biology Questions And Answers
• Human Nutrition Class 9 Biology Questions And Answers
• Excretion and Osmoregulation Class 9 Biology Questions And Answers
• CSkeleton and Movement Class 9 Biology Questions And Answers

Balbharti Maharashtra State Board 11th Biology Textbook Solutions Chapter 8 Plant Tissues and Anatomy Textbook Exercise Questions and Answers.

Plant Tissues and Anatomy Class 11 Exercise Question Answers Solutions Maharashtra Board

Class 11 Biology Chapter 8 Exercise Solutions Maharashtra Board

Biology Class 11 Chapter 8 Exercise Solutions

1. Choose the correct option.

Question (A)
Location or position of meristematic regions is divided into _______ types.
(A) one
(B) two
(C) three
(D) none of the above
Answer:
(C) three

Question (B)
Cambium is also called
(A) apical meristem
(B) intercalary meristem
(C) lateral meristem
(D) none of the above
Answer:
(C) lateral meristem

Question (C)
Collenchyma is a type of ________ tissue.
(A) living
(B) dead
(C) living and dead
(D) none of the above
Answer:
(A) living

Question (D)
_______ is a complex permanent tissue.
(A) Parenchyma
(B) Sclerenchyma
(C) Chlorenchyma
(D) Xylem
Answer:
(D) Xylem

Question (E)
Mesophyll tissue is present in ________ .
(A) root
(B) stem
(C) leaf
(D) flower
Answer:
(C) leaf

2. Answer the following questions

Question (A)
A fresh section was taken by a student but he was very disappointed because there were only few green and most colourless cells. Teacher provided a pink colour solution. The section was immersed in this solution and when observed it was much clearer. What is the magic?
Answer:
1. The pink coloured solution given by teacher must be a saffanin stain.
2. Saffanin is used to stain plant tissues, especially lignified tissues such as cell wall and xylem.

Question (B)
While observing a section, many scattered vascular bundles could be seen. Teacher said, in spite of this large number the stem cannot grow in girth. Why?
Answer:

1. Students must have observed monocot stems.
2. It is because, monocot stem shows scattered vascular bundles.
3. In monocot stem, vascular bundles are closed i.e. without cambium.
4. Thus, secondary growth does not occur which is required for increase in girth. Hence, in spite of having large number of scattered vascular bundles, monocot stems do not grow in girth.

Question (C)
A section of the stem had vascular bundles, where one tissue was wrapped around the other. How will you technically describe it?
Answer:
Concentric vascular bundle:
a. When one vascular tissue is completely encircling the other, it is called as concentric vascular bundle.
b. When phloem is encircled by xylem, it is called as leptocentric vascular bundle, whereas when xylem is encircled by phloem, it is called as hadrocentric vascular bundle.
c. When xylem is encircled by phloem on both faces, it is called as amphicribral vascular bundle. When phloem is encircled by xylem on both faces it is called as amphivasal vascular bundle.

Question (D)
There were two cut logs of wood lying in the campus. One had growth rings and other didn’t. Teacher said it is due to differences in their pattern of grow th which is dependent on season. How?
Answer:
1. It is possible that one of the cut logs was of a tropical tree, whereas the other was of a temperate tree. Since tropical trees grow in a similar manner all year, growth rings are not apparent. Another explanation for this could be that the log which had growth rings must be of an old tree which has experience many seasons, whereas the log without growth rings must be of younger tree, that has not been subjected to seasonal changes and hence not developed prominent growth rings.

2. Growth rings are formed due cambial activity during favourable and non-favourable climatic conditions.

3. During favourable conditions, spring wood (early wood) is formed which has broader xylem bands, lighter colour, tracheids with thin wall and wide lumen, fibres are less in number, low density. Whereas, during unfavourable conditions, autumn wood (late wood) is formed which has narrow xylem band, darker in colour, lumen is narrow and walls are thick with abundant fibres, high density.

4. Spring wood and autumn wood that appear as alternate light and dark concentric rings, constitute an annual ring or growth ring.

5. These growth rings can be used to estimate the age of the tree. These are found more in older trees as compare to younger tree.

Question (E)
While on the trip to Kashmir, Pintoo observed that cut portions of large trees show distinct rings, which he never found in Maharashtra. Why is so?
Answer:
1. Cut portions of large tress show distinct rings which are annual rings formed due to activity of cambium during favourable and non-favourable climatic conditions.
2. Kashmir falls under temperate region where the climatic conditions are not uniform through the year. In the spring season, conditions are favourable due to which cambium is active, whereas in autumn season, conditions are unfavourable due to which cambium is less active. This leads to formation of spring wood and autumn wood that appear as alternate light and dark concentric rings, constitute an annual ring or growth ring.
3. Maharashtra falls under tropical region where climatic conditions are favourable throughout the year. In tropical areas, continuous growth of secondary xylem occurs. Thus, trees growing in tropical regions show less or no annual rings as compared to trees in temperate region.

Question (F)
A student was observing a slide with no label under microscope. The section had some vascular bundles scattered in the ground tissue. It is section of a monocot stem! He exclaimed. No! it is section of fern rachis, said the teacher. Teacher told to observe vascular bundle again. Student agreed, Why?
Answer:

1. In fern rachis, the number of vascular bundles is less as compared to number of vascular bundles in monocot stem. In monocot stem, vascular bundles are numerous.
2. In fern rachis, xylem consists of only tracheids whereas in monocot stem, xylem consists of vessels (protoxylem and metaxylem) as well as tracheids. Monocot stem shows presence of lysigenous cavity just below protoxylem.
3. In fern rachis, phloem consists of only sieve cells whereas in monocot stem, phloem consists of sieve tubes and companion cells. Thus, a student must hav e observed these differences in the given section and agreed to teacher’s statement that the given section is of fern rachis and not of monocot stem.

Question (G)
Student found a wooden stopper in lab. He was told by an old lab attendant that it is there for many years. He kept thinking how it did not rot?
Answer:
1. Wooden stopper or cork is obtained from the phellem (cork) part of a bark.
2. Phellem (cork) is impervious in nature and does not allow entry of water due to suberized walls.
3. Due to this it does not rot and remains as it is for many years.

Question (H)
Student while observing a slide of leaf section observed many stomata on the upper surface. He thought he has placed slide upside down. Teacher confirmed it is rightly placed. Explain.
Answer:
1. In a dicot leaf, stomata are generally absent on upper epidermis but are present on lower epidermis. Thus, the student must have thought that he has placed slide upside down.
2. According to teacher, the section was placed rightly, thus the given section must be of monocot leaf.
3. It is because, in monocot leaf stomata are present on both upper and lower epidermis.

3. Write short notes on the following points.

Question (A)
Structure of stomata.
Answer:

1. Small gateways in the epidermal cells are called as stomata.
2. Stoma is controlled or guarded by specially modified cells called guard cells.
3. These guard cells may be kidney-shaped (dicot) or dumbbell-shaped (monocot), collectively called as stomata.
4. Guard cells have chloroplasts to carry out photosynthesis.
5. Change in turgor pressure of guard cells causes opening and closing of stomata, which enables exchange of gases and water vapour.
6. Stomata are further covered by subsidiary cells.
7. Stoma, guard cells and subsidiary cells form a unit called stomatal apparatus.

Question (B)
Write a short note on secondary growth.
OR
With the help of neat and labelled diagram explain the secondary growth in dicot stem.
Answer:
Secondary growth:

1. Dicotyledonous plants and gymnosperms exhibit increase in girth of root and stem.
2. In dicot stem, secondary growth begins with the formation of a continuous cambium ring.
3. The cambium present between the primary xylem and primary phloem of a vascular bundle is called intrafascicular cambium.
4. The cells of medullary rays adjoining these intrafascicular cambium strips become meristematic (regain the capacity to divide) and form the interfascicular cambium.
5. Thus, a complete and continuous ring of vascular cambium is formed.
6. The cambium ring cuts off new cells, towards both inner and outer sides.
7. The cells that are cut-off towards pith (inner side) mature into secondary xylem and cells that are cut-off towards periphery mature into secondary phloem.
8. Generally, amount of secondary xylem is more than the secondary phloem.

Question (C)
Write a short note on peculiarity of a sclerenchyma cell wall.
Answer:
Peculiarity of a sclerenchyma cell wall:
1. Cell wall of sclerenchyma is evenly thickened due to uniform deposition of lignin.
2. Cell wall of sclereids is extremely thick and strongly lignified.

4. Differentiate

Question (A)
Differentiate between vascular bundles of monocot and dicot.
Answer:

1. Vascular bundle of monocot and dicot root.
2. Vascular bundle of monocot and dicot stem.
3. Vascular bundle of monocot and dicot leaf.

Question (B)
Differentiate between xylem and phloem.
Answer:

Xylem	Phloem
1. It is a dead complex tissue.	It is a living complex tissue.
2. It is composed of xylem, tracheids, vessels, xylem fibres and xylem parenchyma.	It is composed of sieve tubes, sieve cells, companion cells, phloem parenchyma and phloem fibres.
3. It is also known as wood.	It is also known as bast.
4. The cell walls are thick due to lignin.	The cell walls are thin.
5. Xylem conducts water and minerals from roots to the stem and leaves. It also provides mechanical strength to the plant parts.	It is the chief food conducting tissue of vascular plants responsible for translocation of food from leaves to other plant parts.

5. Draw neat labelled diagrams

Question (A)
T.S. of dicot leaf.
Answer:
1. Structure of dorsiventral leaf: The mesophyll tissue is differentiated into palisade and spongy parenchyma in a dorsiventral leaf. This type is very common in dicot leaf. The different parts of this leaf are as follows:
2. Upper epidermis: It consists of a single layer of tightly packed rectangular, barrel shaped, parenchymatous cells which are devoid of chloroplast. A distinct layer of cuticle lies on the outside of the epidermis. Stomata are generally absent.
3. Mesophyll: Between upper and lower epidermis, there is chloroplast-containing photosynthetic tissue called mesophyll It is differentiated into Palisade parenchyma and Spongy parenchyma.
a. Palisade parenchyma:
Palisade parenchyma is present below upper epidermis and consists of closely packed elongated cells. The cells contain abundant chloroplasts and help in photosynthesis.
b. Spongy parenchyma:
Spongy parenchyma is present below palisade tissue and consists of loosely arranged irregularly shaped cells with intercellular spaces. The spongy parenchyma cells contain chloroplast and are in contact with the atmosphere through stomata.

4. Vascular system: It is made up of a number of vascular bundles of varying size depending upon the venation. Each one is surrounded by a thin layer of parenchymatous cells called bundle sheath. Vascular bundles are closed. Xylem lies towards upper epidermis and phloem towards lower epidermis. Cambium is absent, hence there is no secondary growth in the leaf.
5. Lower epidermis: It consists of a single layer of compactly arranged rectangular, parenchymatous cells. A thin layer of cuticle is also present. The lower epidermis contains a large number of microscopic pores called stomata. There is an air-space called substomatal chamber at each stoma.

Question (B)
T.S. of Monocot root.
Answer:

Question (C)
Draw neat labelled diagrams of T.S. of dicot stem.
Answer:

Question 6.
Write the information related to diagram given below.

Answer:


[Note: The labelled part can be considered as the ‘region of maturation ’ of root apical however, the region of maturation does not contain meristematic tissue ]
Classification of meristematic tissue based on its position:
1. Apical meristem:
a. It is produced from promeristem and forms growing point of apices of root, shoot and their lateral branches.
b. It brings about increase in length of plant body and is called as apical initials.
c. Shoot apical meristem is terminal in position whereas in root it is subterminal i.e. located behind the root cap.

2. Intercalary meristem:
a. Intercalary meristematic tissue is present in the top or base area of node.
b. Their activity is mainly seen in monocots.
c. These are short lived.

3. Lateral meristem:
a. It is present along the sides of central axis of organs.
b. It takes part in increasing girth of stem or root, e.g. Intrafascicular cambium.
c. It is found in vascular bundles of gymnosperms and dicot angiosperms.

Question 7.
Identify the following diagrams, label it and prepare a chart of characteristics.
Answer:
1. Figure ‘c’

Question 8.
Distinguish between dicot and monocot leaf on the basis of following characters.
Answer:

Characters	Dicot leaf	Monocot leaf
Stomata	Stomata are restricted to lower epidermis. Guard cells of stoma are kidney shaped.	Stomata occur on both epidermis. Guard cells of stoma are dumbbell shaped.
Intercellular space	More intercellular spaces due to presence of spongy parenchyma.	Less intercellular spaces as mesophyll is not differentiated into spongy and palisade tissue.
Venation	Reticulate venation	Parallel venation
Vascular bundle	Vascular bundles of varying size. The size of the vascular bundles is dependent on the size of the veins which vary in thickness in dicot leaf.	Vascular bundles are nearly of similar size (Except in main veins).
Mesophyll cells	Mesophyll tissue is differentiated into palisade parenchyma and spongy parenchyma.	Mesophyll tissue is not differentiated into palisade parenchyma and spongy parenchyma.

Practical/ Project:

Question 1.
Prepare detail anatomical charts with diagrammatic representation of dicot and monocot plants.
Answer:
Anatomy of dicot root: The transverse section of a typical dicotyledonous root shows following anatomical features:
1. Epiblema: It is the outermost single layer of cells without cuticle. Some epidermal cells prolong to form unicellular root hairs.
2. Cortex: It is made up of many layers of thin walled parenchyma cells. Cortical cells store food and water.
3. Exodermis: After the death of epiblema, outer layer of cortex become cutinized and is called Exodermis.

4. Endodermis:
The innermost layer of cortex is called Endodermis.
The cells are barrel-shaped and their radial walls bear Casparian strip or Casparian bands composed of suberin. Near the protoxylem, there are unthickened passage cells.

5. Stele: It consists of pericycle, vascular bundles and pith.
a. Pericycle: Next to the endodermis, there is a single layer of thin walled parenchyma cells called pericycle. It forms outermost layer of stele or vascular cylinder.
b. Vascular bundle: Vascular bundles are radial. Xylem and Phloem occur in separate patches arranged on alternate radii. Xylem is exarch in root that means protoxylem vessels are towards periphery and metaxylem elements are towards centre. Xylem bundles vary from two to six number, i.e. they may be diarch, triarch, tetrarch, etc.
Connective tissue: A parenchymatous tissue is present in between xylem and phloem.
c. Pith: The central part of stele is called pith. It is narrow and made up of parenchymatous cells, with or without intercellular spaces.
6. At a later stage cambium ring develops between the xylem and phloem causing secondary growth.

Anatomy of monocot stem: A transverse section of maize (monocot) stem shows the following structures:

1. Epidermis: It is single-layered and without trichomes.
2. Hypodermis: It is sclerenchymatous.
3. Ground tissue: It consists of thin walled parenchyma cells. It extends from hypodermis to the centre. It is not differentiated into cortex, endodermis, pericycle and pith.
4. Vascular bundles: Vascular bundles are numerous and are scattered in ground tissue. Each vascular bundle is surrounded by a sclerenchymatous bundle sheath. Vascular bundles are conjoint, collateral and closed (without cambium). Xylem is endarch and shows lysigenous cavity.
5. Pith: Pith is absent.

Anatomy of dicot leaf:

1. Structure of dorsiventral leaf: The mesophyll tissue is differentiated into palisade and spongy parenchyma in a dorsiventral leaf. This type is very common in dicot leaf. The different parts of this leaf are as follows:
2. Upper epidermis: It consists of a single layer of tightly packed rectangular, barrel shaped, parenchymatous cells which are devoid of chloroplast. A distinct layer of cuticle lies on the outside of the epidermis. Stomata are generally absent.
3. Mesophyll: Between upper and lower epidermis, there is chloroplast-containing photosynthetic tissue called mesophyll It is differentiated into Palisade parenchyma and Spongy parenchyma.
a. Palisade parenchyma:
Palisade parenchyma is present below upper epidermis and consists of closely packed elongated cells. The cells contain abundant chloroplasts and help in photosynthesis.
b. Spongy parenchyma:
Spongy parenchyma is present below palisade tissue and consists of loosely arranged irregularly shaped cells with intercellular spaces. The spongy parenchyma cells contain chloroplast and are in contact with the atmosphere through stomata.
4. Vascular system: It is made up of a number of vascular bundles of varying size depending upon the venation. Each one is surrounded by a thin layer of parenchymatous cells called bundle sheath. Vascular bundles are closed. Xylem lies towards upper epidermis and phloem towards lower epidermis. Cambium is absent, hence there is no secondary growth in the leaf.
5. Lower epidermis: It consists of a single layer of compactly arranged rectangular, parenchymatous cells. A thin layer of cuticle is also present. The lower epidermis contains a large number of microscopic pores called stomata. There is an air-space called substomatal chamber at each stoma.

Anatomy of monocot leaf:

1.It is single layered, present on both sides of the leaf.
It consists of compactly arranged rectangular transparent parenchymatous cells.
Both the surfaces contain stomata.
Both the surfaces have a distinct layer of cuticle.
2. Mesophyll:
Mesophyll is not differentiated into palisade and spongy tissue.
3. Vascular bundle:
These are conjoint, collateral and closed.

Question 2.
Observe different slides related to anatomy of flowering plants under the guidance of teacher.
[Students are expected to perform this practical own their own.]

11th Biology Digest Chapter 8 Plant Tissues and Anatomy Intext Questions and Answers

Can you recall? (Textbook Page No. 85)

(i) Which component brings about important processes in the living organisms?
Answer:
Cell is the component that brings about important processes in the living organisms.

(ii) What is tissue?
Answer:
A group of cells having essentially a common function and origin is called as tissue.

(iii) Explain simple and complex tissue.
Answer:
a. Simple tissue:
1. They are made up of only one type of cells.
2. They are found in all the plant parts.
3. They perform many functions.
4. Simple tissues in plants are Parenchyma, Collenchyma, Sclerenchyma.

b. Complex tissue:
1. They are made up of many types of cells.
2. They are found only in the vascular regions of the plant.
3. They mainly perform the function of conduction of food and water.
4. Complex tissues in plants are Xylem and Phloem.

(iv) Complete the flow chart.
Organisms → Organs → Cells
Answer:
Organism → Organ system → Organs → Tissue system → Tissue → Cells

Can you tell? (Textbook Page No. 86)

Enlist the characteristics of meristematic tissue.
Answer:
Characteristics of meristematic tissue:

1. It is a group of young, immature cells.
2. These are living cells with ability to divide in the regions where they are present.
3. These are polyhedral or isodiametric in shape without intercellular spaces.
4. Cell wall is thin, elastic and mainly composed of cellulose.
5. Protoplasm is dense with distinct nucleus at the centre and vacuoles if present, are very small.
6. Cells show high rate of metabolism.

Can you tell? (Textbook Page No. 86)

Classify meristematic tissue on the basis of origin.
Answer:
Classification of meristematic tissue on the basis of origin:
1. Promeristem / Primordial meristem:
a. It is also called as embryonic meristem.
b. It usually occupies very minute area at the tip of root and shoot.

2. Primary meristem:
a. It originates from the primordial meristem and occurs in the plant body from the beginning, at the root and shoot apices.
b. Cells are always in active state of division and give rise to permanent tissues.

3. Secondary meristem:
a. These tissues develop from living permanent tissues during later stages of plant growth hence are called as secondary meristems.
b. This tissue occurs in the mature regions of root and shoot of many plants.
c. Secondary meristem is always lateral (to the central axis) in position e.g. Fascicular cambium, inter fascicular cambium, cork cambium.

Can you tell? (Textbook Page No. 89)

Write a note on parenchyma.
Answer:
Parenchyma:

1. It is a type of simple permanent tissue.
2. Cells in this tissue are thin walled, isodiametric, round, oval to polygonal or elongated in shape.
3. Cell wall is composed of cellulose.
4. Cells are living with prominent nucleus and cytoplasm with large vacuole.
5. Parenchyma has distinct intercellular spaces. Sometimes, cells may show compact arrangement.
6. The cytoplasm of adjacent cells is interconnected through plasmodesmata and thus forms a continuous tissue.
7. This is less specialized permanent tissue.
8. Occurrence:
   These cells are distributed in all the parts of a plant body viz. epidermis, cortex, pericycle, pith, mesophyll cells, endosperm, xylem and phloem.
9. Functions:
   These cells store food, water, help in gaseous exchange, increase buoyancy, perform photosynthesis and different functions in plant body.
10. Dedifferentiation in parenchyma cells develops vascular cambium and cork cambium at the time of
    secondary growth.

Can you tell? (Textbook Page No. 89)

Describe sclerenchyma fibres.
Answer:
Sclerenchyma fibres:
1. Fibres are thread-like, elongated and narrow structures with tapering and interlocking end walls.
2. Fibres are mostly in bundles. Pits are narrow, unbranched and oblique.
They provide mechanical strength.

Can you tell? (Textbook Page No. 89)

Sketch and label T.S. of phloem tissue.
Answer:
T.S. of phloem tissue: Structure of phloem:
1. Phloem is a living tissue. It is also called as bast.
2. It is responsible for conduction of organic food material from source (generally leaf) to a sink (other plant parts).
3. On the basis of origin, it can be protophloem (first formed) and metaphloem (latterly formed).
4. It is composed of sieve elements (sieve cells and sieve tubes), companion cells, phloem parenchyma and phloem fibres.

2. Sieve elements:
a. Sieve tubes are long tubular conducting channel of phloem.
b. These are placed end to end with bulging at end walls.
c. The sieve tube has sieve plate formed by septa with small pores.
d. The sieve plates connect protoplast of adjacent sieve tube cells.
e. The sieve tube cell is a living cell with a thin layer of cytoplasm, but loses its nucleus at maturity.
f. The sieve tube cell is connected to companion cell through phloem parenchyma by plasmodesmata.
g. Sieve cells are found in lower plants like pteridophytes and gymnosperms and sieve tubes are found in angiosperms.
h. The cells are narrow, elongated with tapering ends and sieve area located laterally.

3. Companion cells:
a. These are narrow elongated and living.
b. Companion cells are laterally associated with sieve tube elements.
c. Companion cells have dense cytoplasm and prominent nucleus.
d. Nucleus of companion cell regulates functions of sieve tube cells through simple pits.
e. From origin point of view, sieve tube cells and companion cell are derived from same cell. Death of the one result in death of the other type.

4. Phloem parenchyma:
a. Cells of phloem parenchyma are living, elongated found associated with sieve tube and companion cells.
b. Their chief function is to store food, latex, resins, mucilage, etc.
c. The cells carry out lateral conduction of food material.
d. These cells are absent in most of the monocots.

5. Phloem fibres (Bast fibres):
a. Phloem fibres are the only dead tissue among this unit.
b. They are sclerenchymatous.
c. They are generally absent in primary phloem, but present in secondary phloem.
d. These cells have with lignified walls and provide mechanical support.
e. They are used in making ropes and rough clothes.

Can you tell? (Textbook Page No. 92)

Concentric vascular bundles are always closed. Describe.
Answer:

1. When one vascular tissue is completely encircling the other, it is called as concentric vascular bundle.
2. When cambium is not present between xylem and phloem, it is known as closed vascular bundle.
3. Due to absence of cambium between xylem and phloem, concentric vascular bundles are always closed.

Can you tell? (Textbook Page No. 92)

How is the structure of vascular bundles of the root?
Answer:

1. Vascular bundles of the root are radial.
2. In radial vascular bundles, complex tissues are situated separately on separate radius as separate bundle.
3. The xylem and phloem bundles are arranged alternating with each other.

Can you tell? (Textbook Page No. 92)

Why vascular bundles of dicot stem are described as conjoint collateral and open?
Answer:
Vascular bundles of dicot stem are described as conjoint collateral and open because;
1. In dicot stem, the complex tissue is collectively present as neighbours of each other on the same radius in the form of xylem inside and phloem outside. Such type of vascular bundles are called as conjoint and collateral.
2. In dicot stem, a strip of cambium is present between xylem and phloem. Hence, it is called as open vascular bundle.

Can you tell? (Textbook Page No. 92)

How is the arrangement of vascular bundles in dicot and monocot stem?
Answer:
1. Vascular bundle in dicot stem: Vascular bundles are conjoint, collateral, open, and are arranged in a ring. Each one is composed of xylem, phloem and cambium. Xylem is endarch. A strip of cambium is present between xylem and phloem.
2. Vascular bundle in monocot stem: Vascular bundles are numerous and are scattered in ground tissue. Each vascular bundle is surrounded by a sclerenchymatous bundle sheath. Vascular bundles are conjoint, collateral and cloused (without cambium). Xylem is endarch and shows lysigenous cavity.

11th Std Biology Questions And Answers:

• Living World Class 11 Biology Questions And Answers
• Systematics of Living Organisms Class 11 Biology Questions And Answers
• Kingdom Plantae Class 11 Biology Questions And Answers
• Kingdom Animalia Class 11 Biology Questions And Answers
• Cell Structure and Organization Class 11 Biology Questions And Answers
• Biomolecules Class 11 Biology Questions And Answers
• Cell Division Class 11 Biology Questions And Answers
• Plant Tissues and Anatomy Class 11 Biology Questions And Answers

Balbharti Maharashtra State Board 11th Biology Textbook Solutions Chapter 10 Animal Tissue Textbook Exercise Questions and Answers.

Animal Tissue Class 11 Exercise Question Answers Solutions Maharashtra Board

Class 11 Biology Chapter 10 Exercise Solutions Maharashtra Board

Biology Class 11 Chapter 10 Exercise Solutions

1. Choose correct option

Question (A)
The study of structure and arrangement of tissue is called as _______ .
(a) anatomy
(b) histology
(c) microbiology
(d) morphology
Answer:
(b) histology

Question (B)
_______ is a gland which is both exocrine and endocrine.
(a) Sebaceous
(b) Mammary
(c) Pancreas
(d) Pituitary
Answer:
(c) Pancreas

Question (C)
_______ cell junction is mediated by integrin.
(a) Gap
(b) Hemidesmosomes
(c) Desmosomes
(d) Adherens
Answer:
(b) Hemidesmosomes

Question (D)
The protein found in cartilage is _______ .
(a) ossein
(b) haemoglobin
(c) chondrin
(d) renin
Answer:
(c) chondrin

Question (E)
Find the odd one out.
(a) Thyroid gland
(b) Pituitary gland
(c) Adrenal gland
(d) Salivary gland
Answer:
(d) Salivary gland

2. Answer the following questions

Question (A)
Identify and name the type of tissues in the following:

1. Inner lining of the intestine
2. Heart wall
3. Skin
4. Nerve cord
5. Inner lining of the buccal cavity

Answer:

1. Epithelial tissue (Columnar epithelium)
2. Cardiac muscles (Muscular tissue)
3. Epithelial tissue (Stratified epithelium)
4. Nervous tissue
5. Epithelial tissue (Ciliated epithelium)

Question (B)
Why do animals in cold regions have a layer of fat below their skin?
Answer:
1. In adipose tissues, fats are stored in the form of droplets.
2. The adipose tissue acts as good insulator and helps retain heat in the body. This helps in survival of animals in the colder regions. Hence, animals in cold regions have a layer of fat below their skin.

Question (C)
What enables the ear pinna to be folded and twisted while the nose tip can’t be twisted?
Answer:
1. The ear pinna (outer ear) is made up of a thin plate of elastic cartilage and is connected to the surrounding.
2. The nose tip is made up of elastic cartilage. However, several bones and cartilage make up the bony- cartilaginous framework of the nose.
Hence, even though the tip of the nose is made up of elastic cartilage, it cannot be twisted like the ear pinna due to presence of bony-cartilaginous framework.

Question (D)
Sharad touched a hot plate by mistake and took away his hand quickly. Can you recognize the tissue and its type responsible for it?
Answer:
1. Nervous and muscular tissues are responsible for this action
2. Nervous tissue recognizes the stimuli whereas muscular tissue allows responding to the stimuli.

Question (E)
Priya got injured in an accident and hurt her long bone and later on she was also diagnosed with anaemia. What could be the probable reason?
Answer:
1. The centre of long bones (diaphysis) contains bone marrow, which is a site of production of blood cells (red blood cells).
2. Any severe injury to the bone marrow can affect rate of haematopoiesis (formation of blood cells).
3. A low count of erythrocytes (red blood cells) is characterised as anaemia. Hence, an injury to Priya’s long bone might have resulted in anaemia.

Question (F)
Supriya stepped out into the bright street from a cinema theatre. In response, her eye pupil shrunk. Identify the muscle responsible for the same.
Answer:
Smooth muscles (Involuntary muscles) are responsible for shrinking of eye pupil.

3. Answer the following questions

Question (A)
What is cell junction? Describe different types of cell junctions.
Answer:
1. Cell junctions: The epithelial cells are connected to each other laterally as well as to the basement
membrane by junctional complexes called cell junctions.
2. The different types of cell junctions are as follows:
a. Gap Junctions (GJs): These are intercellular connections that allow the passage of ions and small molecules between cells as well as exchange of chemical messages between cells.
b. Adherens Junctions (AJs): They are involved in various signalling pathways and transcriptional regulations.
c. Desmosomes (Ds): They provide mechanical strength to epithelial tissue, cardiac muscles and meninges.
d. Hemidesmosomes (HDs): They allow the cells to strongly adhere to the underlying basement membrane. These junctions help maintain tissue homeostasis by signalling.
e. Tight junctions (TJs): These junctions maintain cell polarity, prevent lateral diffusion of proteins and ions.

Question (B)
Describe in brief about areolar connective tissue with the help of suitable diagram.
Answer:
Areolar tissue is a loose connective tissue found under the skin, between muscles, bones, around organs, blood vessels and peritoneum. It is composed of fibres and cells.
The matrix of areolar tissues contains two types of fibres i.e. white fibres and yellow fibres.
a. White fibres: They are made up of collagen and give tensile strength to the tissue.
b. Yellow fibres: They are made up of elastin and are elastic in nature.
The four different types of cells present in this tissue are as follows:
a. Fibroblast: Large flat cells having branching processes. They produce fibres as well as polysaccharides that form the ground substance or matrix of the tissue.
b. Mast cells: Oval cells that secrete heparin and histamine.
c. Macrophages: Amoeboid, phagocytic cells.
d. Adipocytes (Fat cells): These cells store fat and have eccentric nucleus.

Question (C)
Describe the structure of multipolar neuron.
Answer:
A neuron is the structural and functional unit of the nervous tissue. A neuron is made up of cyton or cell body and cytoplasmic extensions or processes.
1. Cyton:
The cyton or cell body contains granular cytoplasm called neuroplasm and a centrally placed nucleus. The neuroplasm contains mitochondria, Golgi apparatus, RER and Nissl’s granules.
2. Cytoplasmic extensions or processes:
(a) Dendron: They are short, unbranched processes.
The fine branches of a dendron are called dendrites.
Dendrites carry an impulse towards the cyton.

(b) Axon: It is a single, elongated and cylindrical process.

1. The axon is bound by the axolemma.
2. The protoplasm or axoplasm contains large number of mitochondria and neurofibrils.
3. The axon is enclosed in a fatty sheath called the myelin sheath and the outer covering of the myelin sheath is the neurilemma. Both the myelin sheath and the neurilemma are parts of the Schwann cell.
4. The myelin sheath is absent at intervals along the axon at the Node of Ranvier.
5. The fine branching structure at the end of the axon (terminal arborization) is called telodendron.

Question (D)
How to differentiate the skeletal and the smooth muscles based on their nucleus?
Answer:
Skeletal muscles contain nucleus arranged at periphery. Striated or smooth muscles are with centrally placed single large oval nucleus therefore, skeletal and smooth muscle fibres can be identified.

Question 4.
Complete the following table.
Answer:

Cell / Tissue / Muscles	Functions
1. Cardiac muscles	Cardiac muscles bring about contraction and relaxation of heart
2. Tendons	Connect skeletal muscles to bones
3. Chondroblast cells	Produce and maintain cartilage matrix
4. Mast cells	Secrete heparin and histamine

Question 5.
Match the following.

‘A’ Group	B’ Group
1. Muscle	(a) Perichondrium
2. Bone	(b) Sarcolemma
3. Nerve cell	(c) Periosteum
4. Cartilage	(d) Neurilemma

Answer:

‘A’ Group	B’ Group
1. Muscle	(c) Periosteum
2. Bone	(a) Perichondrium
3. Nerve cell	(b) Sarcolemma
4. Cartilage	(d) Neurilemma

Practical / Project:

Question 1.
To study the different tissues with the help of permanent slides in your college laboratory.
Answer:
Students may observe permanent slides of different tissues like epithelial tissue, connective tissue, muscular tissue and nervous tissue slides in laboratory.
[Students are expected to perform this activity on their own.]

Question 2.
Collect the information about the exercise to keep muscles healthy and strong.
Answer:

1. Muscles become stronger when we are physically active.
2. Physical activities like walking, jogging, lifting weights, playing tennis, climbing stairs, jumping, and dancing are good ways to exercise our muscles.
3. Apart from this, swimming and biking can also be considered as good workouts for muscles.
4. Different kinds of activities, work different muscles. Hence, it is essential to perform various types of physical activities.
5. Also, activities that increase our breath rate, help in exercising our heart muscle as well.
   [Students are expected to collect more information on their own.]

11th Biology Digest Chapter 10 Animal Tissue Intext Questions and Answers

Can you recall? (Textbook Page No. 116)

What is tissue?
Answer:
A group of cells having the same origin, same structure and same function is called ‘tissue’.

Do you know? (Textbook Page No. 116)

Number of cells in human body.
Answer:
There are about 100 trillion of 200 different types of cells in the human body.

Can you tell? (Textbook Page No. 119)

Explain basic structure of epithelial tissue and mention its types.
Answer:
The characteristics of epithelial tissues are as follows:

1. Epithelial tissue forms a covering on inner and outer surface of body and organs.
2. The cells of this tissue are compactly arranged with little intercellular matrix.
3. The cells rest on a non-cellular basement membrane.
4. The epithelial cells are polygonal, cuboidal or columnar in shape.
5. A single nucleus is present at the centre or at the base of the cell.
6. The tissue is avascular and has a good regeneration capacity.
7. The major function of the epithelial tissue is protection. It also helps in absorption, transport, filtration and secretion.

The different types of epithelial tissues are as follows:
1. Simple epithelium: Epithelial tissue made up of single layer of cells is known as simple epithelium. Simple epithelium is further classified into:
a. Squamous Epithelium
b. Cuboidal Epithelium
c. Columnar Epithelium
d. Ciliated Epithelium
e. Glandular Epithelium
f. Sensory epithelium
g- Germinal epithelium

2. Compound epithelium: Epithelium composed of several layers is called compound epithelium. Compound epithelium is further classified into:
a. Stratified epithelium
b. Transitional epithelium

Epithelial tissue has good capacity of regeneration. Give reason.
Answer:
Epithelial tissue rests on a basement membrane which acts as a scaffolding on which epithelium can grow and regenerate after injuries.

Can you recall? (Textbook Page No. 116)

Where is squamous epithelium located?
Answer:
Location: It is present in blood vessels, alveoli, coelom, etc.

Can you tell? (Textbook Page No. 119)

Write a note on glandular epithelium.
Answer:
Structure:
1. The cells of the glandular epithelium can be columnar, cuboidal or pyramidal in shape.
2. The nucleus of these cells is large and situated towards the base.
3. Secretory granules are present in the cell cytoplasm.
4. Glands consist of glandular epithelium. The glands may be either unicellular (goblet cells of intestine) or multicellular (salivary gland), depending on the number of cells.
5. Types: Depending on the mode of secretion, multicellular glands can be further classified as duct bearing glands (exocrine glands) ad ductless glands (endocrine glands).
a. Exocrine glands: These glands pour their secretions at a specific site. e.g. salivary gland, sweat gland, etc.
b. Endocrine glands: These glands release their secretions directly into the blood stream, e.g. thyroid gland, pituitary gland, etc.
6. Function: Glandular epithelium secretes mucus to trap the dust particles, lubricate the inner surface of respiratory and digestive tracts, secrete enzymes and hormones, etc.
Heterocrine glands
1. Heterocrine glands or composite glands have both exocrine and endocrine function.
2. Pancreas is called a heterocrine gland because it secretes the hormone insulin into blood which is an endocrine function and enzymes into digestive tract which is an exocrine function.

Use your brain power? (Textbook Page No. 118)

When do the transitional cells change their shape?
Answer:
Transitional cells change their shape depending on the degree of distention (stretch) needed. As the tissue stretches, the transitional cells start changing shape from round and globular to thin and flat.

Can you tell? (Textbook Page No. 119)

How do cell junctions help in functioning of epithelial tissue?
Answer:
1. Cell junctions: The epithelial cells are connected to each other laterally as well as to the basement
membrane by junctional complexes called cell junctions.
2. The different types of cell junctions are as follows:
a. Gap Junctions (GJs): These are intercellular connections that allow the passage of ions and small molecules between cells as well as exchange of chemical messages between cells.
b. Adherens Junctions (AJs): They are involved in various signalling pathways and transcriptional regulations.
c. Desmosomes (Ds): They provide mechanical strength to epithelial tissue, cardiac muscles and meninges.
d. Hemidesmosomes (HDs): They allow the cells to strongly adhere to the underlying basement membrane. These junctions help maintain tissue homeostasis by signalling.
e. Tight junctions (TJs): These junctions maintain cell polarity, prevent lateral diffusion of proteins and junctions.

Can you tell? (Textbook Page No. 122)

Give reason.
As we grow old, cartilage becomes rigid.
Answer:
Calcified cartilage is a type of cartilage that becomes rigid due to deposition of salts in the matrix. This reduces the flexibility of joints in old age and cartilage becomes rigid.

Can you recall? (Textbook Page No. 116)

Enlist functions of bone.
Answer:
Bones support and protect different organs and help in movement.

Can you tell? (Textbook Page No. 122)

(i) Give reason. Bone is stronger than cartilage.
Answer:
a. Bone is rigid, non-pliable, dense connective tissue characterised by the hard matrix called ossein (made up of calcium salt hydroxyapatite). An outer tough membrane called periosteum encloses the matrix. The matrix is arranged in the form of concentric layers called lamellae. Bones are well vascularized and possess blood vessels and nerves that pierce through the periosteum,
b. Cartilage is a pliable supportive connective tissue. On comparison with bones, cartilage is thin, avascular and flexible. In cartilage, a sheath of collagenous fibres called perichondrium encloses the matrix.
Hence, a bone is stronger than a cartilage.

(ii) Explain histological structure of mammalian bone.
Answer:
a. The bone is characterised by hard matrix called ossein which is made up of mineral salt hydroxy apatite (Ca₁₀ (P0₄)₆ (OH)₂).
b. An outer tough membrane called periosteum encloses the matrix.
c. Blood vessels and nerves pierce through the periosteum.
d. The matrix is arranged in the form of concentric layers called lamellae.
e. Each lamella contains fluid filled cavities called lacunae from which fine canals called canaliculi radiate.
f. The canaliculi of adjacent lamellae connect with each other as they traverse through the matrix.
g. Active bone cells called osteoblasts and inactive bone cells called osteocytes are present in the
lacunae.
h. The mammalian bone shows the peculiar haversian system.
i. The haversian canal encloses an artery, vein and nerves.

Can you recall? (Textbook Page No. 122)

How can exercise improve your muscular system?
Answer:
1. Exercise can improve both muscular strength and stamina endurance.
2. Exercises are commonly grouped into two types depending on the effect they have on the body:
a. Aerobic exercises: such as cycling, walking, and running. They increase muscular endurance and cardiovascular health, etc.
b. Anaerobic exercises: such as weight training or sprinting, increase muscle strength, etc.
3. Anaerobic exercies: It comprises brief periods of physical exertion and high-intensity, strength-training activities.
Anaerobic exercise is a physical exercise intense enough to cause lactate to form.
It is used by athletes to promote strength, speed and power; and by body builders to build muscle mass.

Can you recall? (Textbook Page No. 122)

How many skeletal muscles are present in human body?
Answer:
There are over 650 named skeletal muscles in the human body.

Can you tell? (Textbook Page No. 125)

Differentiate between medullated and non-medullated fibre.
Answer:

Medullated fibre	Non – Medullated fibre
1. Medullary sheath is present around the axon hence also known as Myelinated nerve fibre.	Medullary sheath is absent hence also known as Non-myelinated nerve fibre.
2. They have nodes of Ranvier at regular intervals.	They do not have nodes of Ranvier.
3. Saltatory conduction takes place in medullated nerve fibres.	Saltatory conduction is not seen in non-medullated nerve fibre.
4. These nerve fibres conduct the nerve impulse faster.	These nerve fibres conduct nerve impulse at slow rate.
5. These fibres appear white in colour due to an insulating fatty layer (myelin sheath).	These fibres appear grey in colour due to absence of fatty layer.
6. Schwann cell of this nerve fibre secrete myelin sheath.	Schwann cell of this nerve fibre does not secrete myelin sheath.
7. Cranial nerves of vertebrates are medullated.	Nerves of autonomous nervous system are non-

Internet is my friend. (Textbook Page No. 125)

Learn about transmission of impulse from one neuron to another.
Answer:

1. A nerve impulse is transmitted from one neuron to another through junctions called synapses.
2. A synapse is formed by the membranes of a pre-synaptic neuron and a post-synaptic neuron, which may or may not be separated by a gap called synaptic cleft.
3. There are two types of synapses, namely, electrical synapses and chemical synapses.
4. Electrical synapses: The membranes of pre- and post-synaptic neurons are in very close proximity.
   Thus, electrical current can flow directly from one neuron into the other across these synapses.
   Impulse transmission across an electrical synapse is faster.
5. Chemical synapse: The membranes of the pre- and post-synaptic neurons are separated by a fluid- filled space called synaptic cleft.
6. Chemicals called neurotransmitters are involved in the transmission of impulses at these synapses.
7. The axon terminals contain vesicles filled with these neurotransmitters.
8. When an impulse arrives at the axon terminal, it stimulates the movement of the synaptic vesicles towards the membrane where they fuse with the plasma membrane and release their neurotransmitters into the synaptic cleft.
9. The released neurotransmitters bind to their specific receptors, present on the post-synaptic membrane.
10. This binding opens ion channels and allows the entry of ions which can generate a new potential in the post-synaptic neuron.

[Students are expected to refer the given information and collect more information from the internet.]
[Note: Students can scan the adjacent QR code to get conceptual clarity with the aid of a relevant video ]

Observe and Discuss (Textbook Page No. 125)

Explain the structure of nerve.

Answer:

1. Each spinal nerve consists of many axons and contains layers of protective connective tissue coverings.
2. Axons are enclosed in a fatty sheath called myelin sheath.
3. Individual axons within a nerve are wrapped in an endoneurium (innermost layer).
4. Groups of axons with their endoneurium are arranged in bundles called fascicles.
5. Each fascicle is wrapped in perineurium (middle layer).
6. The outermost covering over the entire nerve is the epineurium. The epineurium extends between fascicles.
7. Many blood vessels nourish the nerve and are present within the perineurium and epineurium.
   [Source: Tortora. G, Derrickson. B. Principles of Anatomy and Physiology. 11th Edition.]

11th Std Biology Questions And Answers:

• Morphology of Flowering Plants Class 9 Biology Questions And Answers
• Animal Tissue Class 9 Biology Questions And Answers
• Study of Animal Type : Cockroach Class 9 Biology Questions And Answers
• Photosynthesis Class 9 Biology Questions And Answers
• Respiration and Energy Transfer Class 9 Biology Questions And Answers
• Human Nutrition Class 9 Biology Questions And Answers
• Excretion and Osmoregulation Class 9 Biology Questions And Answers
• CSkeleton and Movement Class 9 Biology Questions And Answers

Balbharti Maharashtra State Board 11th Chemistry Textbook Solutions Chapter 11 Adsorption and Colloids Textbook Exercise Questions and Answers.

Adsorption and Colloids Class 11 Exercise Question Answers Solutions Maharashtra Board

Class 11 Chemistry Chapter 11 Exercise Solutions Maharashtra Board

Chemistry Class 11 Chapter 11 Exercise Solutions

1. Choose the correct option.

Question A.
The size of colloidal particles lies between
a. 10^-10 m and 10^-9 m
b. 10^-9 m and 10^-6 m
c. 10^-6 m and 10^-4 m
d. 10^-5 m and 10^-2 m
Answer:
b. 10^-9 m and 10^-6 m

Question B.
Gum in water is an example of
a. true solution
b. suspension
c. lyophilic sol
d. lyophobic sol
Answer:
c. lyophilic sol

Question C.
In Haber process of production of ammonia K₂O is used as
a. catalyst
b. inhibitor
c. promotor
d. adsorbate
Answer:
c. promotor

Question D.
Fruit Jam is an example of-
a. sol
b. gel
c. emulsion
d. true solution
Answer:
b. gel

2. Answer in one sentence :

Question A.
Name type of adsorption in which van der Waals focres are present.
Answer:
Physical adsorption or physisorption.

Question B.
Name type of adsorption in which compound is formed.
Answer:
Chemical adsoiption or chemisorption.

Question C.
Write an equation for Freundlich adsorption isotherm.
Answer:
Freundlich proposed the following empirical equation for adsorption of a gas on solid.
\(\frac{x}{\mathrm{~m}}\) = k P^1/n (n > 1) ……(i)
where,
x = Mass of the gas adsorbed
m = Mass of the adsorbent
\(\frac{x}{\mathrm{~m}}\) = Mass of gas adsorbed per unit mass of adsorbent
P = Equilibrium pressure
k and n are constants which depend on the nature of adsorbate, adsorbent and temperature.

3. Answer the following questions:

Question A.
Define the terms:
a. Inhibition
b. Electrophoresis
c. Catalysis.
Answer:
a. Inhibition:
The phenomenon in which the rate of chemical reaction is reduced by an inhibitor is called inhibition.

b. Electrophoresis:
The movement of colloidal particles under an applied electric potential is called electrophoresis.

c. Catalysis:
The phenomenon of increasing the rate of a chemical reaction with the help of a catalyst is known as catalysis.

Question B.
Define adsorption. Why students can read blackboard written by chalks?
Answer:

• Adsorption is the phenomenon of accumulation of higher concentration of ‘one substance on the surface of another (in bulk) due to unbalanced/unsatisfied attractive forces on the surface.
• When we write on blackboard using chalk, the chalk particles get adsorbed on the surface of the blackboard.

Hence, students can read blackboard written by chalks.

Question C.
Write characteristics of adsorption.
Answer:
Following are the characteristics of adsorption:

• Adsorption is a surface phenomenon.
• It depends upon the surface area of the adsorbent.
• It involves physical forces (van der Waals forces) or chemical forces (chemical or covalent bonds).
• Adsorbate is always present in higher concentration on the surface of an adsorbent than in the bulk.
• Adsorption is dependent on temperature (of the surface) and pressure (of adsorbate gas).
• It takes place with the evolution of heat (with some exceptions).

Question D.
Distinguish between Lyophobic and Lyophilic sols.
Answer:
Lyophobic sols (colloids):

1. Lyophobic sols are formed only by special methods.
2. They are irreversible.
3. These are unstable and hence, require traces of stabilizers.
4. Addition of small amount of electrolytes causes precipitation or coagulation of lyophobic sols.
5. Viscosity of lyophobic sol is nearly the same as the dispersion medium.
6. Surface tension of lyophobic sol is nearly the same as the dispersion medium.

Lyophilic sols (colloids):

1. Lyophilic sols are formed easily by direct mixing.
2. They are reversible.
3. These are self-stabilized.
4. Addition of large amount of electrolytes causes precipitation or coagulation of lyophilic sols.
5. Viscosity of lyophilic sol is much higher than that of the dispersion medium.
6. Surface tension of lyophilic sol is lower than that of dispersion medium.

Question E.
Identify dispersed phase and dispersion medium in the following colloidal dispersions.
a. milk
b. blood
c. printing ink
d. fog
Answer:

Colloidal dispersion	Dispersed phase	Dispersion medium
Milk	Liquid	Liquid
Blood	Solid	Liquid
Printing ink	Solid	Liquid
Fog	Liquid	Gas

Question F.
Write notes on :
a. Tyndall effect
b. Brownian motion
c. Types of emulsion
d. Hardy-Schulze rule
Answer:
a. Tyndall effect:
i. Tyndall observed that when light passes through true solution, the path of light through it cannot be detected.
ii. However, if the light passes through a colloidal dispersion, the particles scatter some light in all directions and the path of the light through colloidal dispersion becomes visible to observer standing at right angles to its path.
iii. The phenomenon of scattering of light by colloidal particles and making path of light visible through the dispersion is referred as Tyndall effect and the bright cone of the light is called Tyndall cone.
iv. Tyndall effect is observed only when the following conditions are satisfied.

• The diameter of the dispersed particles is not much smaller than the wavelength of light used.
• The refractive indices of dispersed phase and dispersion medium differ largely.

v. Significance of Tyndall effect:

• It is useful in determining number of particles in colloidal system and their particle size.
• It is used to distinguish between colloidal dispersion and true solution.

b. Brownian motion:
i. The colloidal or microscopic particles undergo ceaseless random zig-zag motion in all directions in a fluid. This motion of dispersed phase particles is called Brownian motion.
ii. Cause of Brownian motion:

• Particles of the dispersed phase constantly collide with the fast-moving molecules of dispersion medium (fluid).
• Due to this, the dispersed phase particles acquire kinetic energy from the molecules of the dispersion medium.
• This kinetic energy brings about Brownian motion.

c. Types of emulsion:
iii. There are two types of emulsions:
a. Emulsion of oil in water (o/w type): An emulsion in which dispersed phase is oil and dispersion medium is water is called emulsion of oil in water.
e.g. 1. Milk consists of particles of fat dispersed in water.
2. Other examples include vanishing cream, paint, etc.
b. Emulsion of water in oil (w/o type): An emulsion in which dispersed phase is water and dispersion medium is oil is called emulsion of water in oil.
e.g. 1. Cod liver oil consists of particles of water dispersed in oil.
2. Some other examples of this type include butter, cream, etc.

d. Hardy-Schulze rule:
i. Generally, greater the valency of the flocculating ion added, greater is its power to cause precipitation. This is known as Hardy-Schulze rule.
ii. In the coagulation of negative sol, the flocculating power follows the following order:
Al³⁺ > Ba²⁺ > Na⁺
iii. Similarly, in the coagulation of positive sol, the flocculating power is in the following order:
[Fe (CN)₆]^4- > PO₄^3- > SO₄^2- > Cl^–

Question G.
Explain Electrophoresis in brief with the help of diagram. What are its applications ?
Answer:
i. Electrophoresis: Electrophoresis set up is shown in the diagram below.

• The diagram shows U tube set up in which two platinum electrodes are dipped in a colloidal solution.
• When electric potential is applied across two electrodes, colloidal particles move towards one or other electrode.
• The movement of colloidal particles under an applied electric potential is called electrophoresis.
• Positively charged particles move towards cathode while negatively charged particles migrate towards anode and get deposited on the respective electrode.

ii. Applications of electrophoresis:

• On the basis of direction of movement of the colloidal particles under the influence of electric field, it is possible to know the sign of charge on the particles.
• It is also used to measure the rate of migration of sol particles.
• Mixture of colloidal particles can be separated by electrophoresis, since different colloidal particles in mixture migrate with different rates.

Question H.
Explain why finely divided substance is more effective as adsorbent?
Answer:

• Adsorption is a surface phenomenon and hence, the extent of adsorption depends upon surface area of the adsorbent.
• Adsorption increases with increase in surface area of the adsorbent.
• Finely divided powdered substances provide larger surface area for a given mass.

Hence, finely divided substance is more effective as adsorbent.

Question I.
What is the adsorption Isotherm?
Answer:
The relationship between the amount of a substance adsorbed per unit mass of adsorbent and the equilibrium pressure (in case of gas) or concentration (in case of solution) at a given constant temperature is called an adsorption isotherm.

Question J.
Aqueous solution of raw sugar, when passed over beds of animal charcoal, becomes colourless. Explain.
Answer:

• When aqueous solution of raw sugar is passed over beds of animal charcoal, charcoal adsorbs the coloured particles from the raw sugar.
• Thus, due to the adsorption of coloured particles, raw sugar becomes colourless when passed over beds of animal charcoal.

Question K.
What happens when a beam of light is passed through a colloidal sol?
Answer:
i. When a beam of light is passed through colloidal sol, it is observed that the colloidal particles scatter some of the incident light in all directions.
ii. Because of this scattering of light, the path of light through the colloidal dispersion becomes visible to observer standing at right angles to its path and the phenomenon is known as Tyndall effect.
iii.

Question L.
Mention factors affecting adsorption of gas on solids.
Answer:
Adsorption of gases on solids depends upon the following factors:

• Nature of adsorbate (gas)
• Nature of solid adsorbent
• Surface area of adsorbent
• Temperature of the surface
• Pressure of the gas

Question M.
Give four uses of adsorption.
Answer:
i. Catalysis (Heterogeneous catalysis):

• The solid catalysts are used in many industrial manufacturing processes.
• For example, iron is used as a catalyst in manufacturing of ammonia, platinum in manufacturing of sulphuric acid, H₂SO₄ (by contact process) while finely divided nickel is employed as a catalyst in hydrogenation of oils.

ii. Gas masks:

• It is a device which consists of activated charcoal or mixture of adsorbents.
• It is used for breathing in coal mines to avoid inhaling of the poisonous gases.

iii. Control of humidity: Silica and alumina gels are good adsorbents of moisture.
iv. Production of high vacuum:

• Lowering of temperature at a given pressure, increases the rate of adsorption of gases on charcoal powder. By using this principle, high vacuum can be attained by adsorption.
• A vessel evacuated by vacuum pump is connected to another vessel containing coconut charcoal cooled by liquid air. The charcoal adsorbs the remaining traces of air or moisture to create a high vacuum.

Question N.
Explain Bredig’s arc method.
Answer:

• Colloidal sols can be prepared by electrical disintegration using Bredig’s arc method.
• This process involves vaporization as well as condensation.
• Colloidal sols of metals such as gold, silver, platinum can be prepared by this method.
• In this method, electric arc is struck between electrodes of metal immersed in the dispersion medium.
• The intense heat produced vapourizes the metal which then condenses to form particles of colloidal sol.

Question O.
Explain the term emulsions and types of emulsions.
Answer:
i. A colloidal system in which one liquid is dispersed in another immiscible liquid is called an emulsion.
ii. There are liquid-liquid colloidal systems in which both liquids are either completely or partially immiscible.
iii. There are two types of emulsions:
a. Emulsion of oil in water (o/w type): An emulsion in which dispersed phase is oil and dispersion medium is water is called emulsion of oil in water.
e.g. 1. Milk consists of particles of fat dispersed in water.
2. Other examples include vanishing cream, paint, etc.
b. Emulsion of water in oil (w/o type): An emulsion in which dispersed phase is water and dispersion medium is oil is called emulsion of water in oil.
e.g. 1. Cod liver oil consists of particles of water dispersed in oil.
2. Some other examples of this type include butter, cream, etc.

4. Explain the following :

Question A.
A finely divided substance is more effective as adsorbent.
Answer:

• Adsorption is a surface phenomenon and hence, the extent of adsorption depends upon the surface area of the adsorbent.
• Adsorption increases with an increase in surface area of the adsorbent.
• Finely divided powdered substances provide a larger surface area for a given mass. Hence, a finely divided substance is more effective as an adsorbent.

Question B.
Freundlich adsorption isotherm, with the help of a graph.
Answer:
Graphical representation of the Freundlich adsorption isotherm:

i. Freundlich proposed the following empirical equation for adsorption of a gas on solid.
\(\frac{x}{\mathrm{~m}}\) = k P^1/n (n > 1) ………(i)
where,
x = Mass of the gas adsorbed
m = Mass of the adsorbent
\(\frac{x}{\mathrm{~m}}\) = Mass of gas adsorbed per unit mass of adsorbent
P = Equilibrium pressure
k and n are constants which depend on the nature of adsorbate, adsorbent and temperature.
ii. The graphical representation of Freundlich equation is as shown in the adjacent plot of x/m vs ‘P’.
iii. In case of solution, P in the equation (i) is replaced by the concentration (C) and thus,
\(\frac{x}{\mathrm{~m}}\) = k C^1/n ………(ii)
iv. By taking logarithm on both sides of the equation (ii),
we get
log \(\frac{x}{\mathrm{~m}}\) = log k + \(\frac{1}{n}\) log C ……..(iii)
v. On plotting a graph of log \(\frac{x}{\mathrm{~m}}\) against log C or log P, a straight line is obtained as shown in the adjacent plot. The slope of the straight line is and intercept on Y-axis is log k.
vi. The factor \(\frac{1}{n}\) ranges from 0 to 1. Equation (iii) holds good over limited range of pressures.
a. When \(\frac{1}{n}\) → 0, \(\frac{x}{\mathrm{~m}}\) → constant, the adsorption is independent of pressure.
b. When \(\frac{1}{n}\) = 1, \(\frac{x}{\mathrm{~m}}\) = k P, i.e., \(\frac{x}{\mathrm{~m}}\) ∝ P, the adsorption varies directly with pressure.
c. The experimental isotherms tend to saturate at high pressure.

5. Distinguish between the following :

Question A.
Adsorption and absorption. Give one example.
Answer:
Adsorption:

• Adsorption is a surface phenomenon as adsorbed matter is concentrated only at the surface and does not penetrate through the surface to the bulk of adsorbent.
• Concentration of the adsorbate is high only at the surface of the adsorbent.
• It is dependent on temperature and pressure.
• It is accompanied by evolution of heat known as heat of adsorption.
• It depends on surface area.
  e.g. Adsoiption of a gas or liquid like acetic acid by activated charcoal.

Absorption:

• Absorption is a bulk phenomenon as absorbed matter is uniformly distributed inside as well as at the surface of the bulk of substance.
• Concentration of the absorbate is uniform throughout the bulk of the absorbent.
• It is independent of temperature and pressure.
• It may or may not be accompanied by any evolution or absorption of heat.
• It is independent of surface area.
  e.g. Absorption of water by cotton, absorption of ink by blotting paper.

Question B.
Physisorption and chemisorption. Give one example.
Answer:
Physisorption:

1. In physisorption, the forces operating are weak van der Waals forces.
2. It is not specific in nature as all gases adsorb on all solids. For example, all gases adsorb on charcoal.
3. The heat of adsorption is low and lies in the range 20-40 kJ mol^-1.
4. It occurs at low temperature and decreases with an increase of temperature.
5. It is reversible.
6. Physisorbed layer may be multimolecular layer of adsorbed particles under high pressure.
   e.g. At low temperature N₂ gas is physically adsorbed on iron.

Chemisorption:

1. In chemisorption, the forces operating are of chemical nature (covalent or ionic bonds).
2. It is highly specific and occurs only when chemical bond formation is possible between adsorbent and adsorbate. For example, adsorption of oxygen on tungsten, hydrogen on nickel, etc.
3. The heat of adsorption is high and lies in the range 40-200 kJ mol^-1.
4. It is favoured at high temperature, however, the extent of chemical adsorption is lowered at very high temperature due to bond breaking.
5. It is irreversible.
6. Chemisorption forms monomolecular layer of adsorbed particles.
   e.g. N₂ gas chemically adsorbed on iron at high temperature forms a layer of iron nitride, which desorbs at very high temperature.

6. Adsorption is surface phenomenon. Explain.
Answer:
Consider a surface of a liquid or a solid.

• The molecular forces at the surface of a liquid are unbalanced or in unsaturation state.
• In solids, the ions or molecules at the surface of a crystal do not have their forces satisfied by the close contact with other particles.
• Because of the unsaturation, solid and liquid surfaces tend to attract gases or dissolved substances with which they come in close contact. Thus, the substance accumulates on the surface of solid or liquid i.e., the substance gets adsorbed on the surface.

Hence, adsorption is a surface phenomenon.

7. Explain how the adsorption of gas on solid varies with
a. nature of adsorbate and adsorbent
b. surface area of adsorbent
Answer:
i. a. Nature of adsorbate:
1. All solids adsorb gases to some extent. It is observed that gases having high critical temperature liquify easily and can be readily adsorbed.
2. The gases such as SO₂, Cl₂, NH₃ which are easily liquefiable are adsorbed to a larger extent as compared to gases such as N₂, O₂, H₂, etc. which are difficult to liquify.
3. Thus, the amount of gas adsorbed by a solid depends on the nature of the adsorbate gas i.e., whether it is easily liquefiable or not.

b. Nature of adsorbent: Substances which provide large surface area for a given mass are effective as adsorbents and adsorb appreciable volumes of gases.
e.g. Silica gel and charcoal are effective adsorbents due to their porous nature.

ii. Surface area of the adsorbent:

• Adsorption is a surface phenomenon. Hence, the extent of adsorption increases with increase in surface area of the adsorbent.
• Finely divided substances, rough surfaces, colloidal substances are good adsorbents as they provide larger surface area for a given mass.

Note: Critical temperature of some gases and volume adsorbed.

8. Explain two applications of adsorption.
Answer:
i. Catalysis (Heterogeneous catalysis):

• The solid catalysts are used in many industrial manufacturing processes.
• For example, iron is used as a catalyst in manufacturing of ammonia, platinum in manufacturing of sulphuric acid, H₂SO₄ (by contact process) while finely divided nickel is employed as a catalyst in hydrogenation of oils.

ii. Gas masks:

• It is a device which consists of activated charcoal or mixture of adsorbents.
• It is used for breathing in coal mines to avoid inhaling of the poisonous gases.

9. Explain micelle formation in soap solution.
Answer:

• Soap molecule has a long hydrophobic hydrocarbon chain called tail which is attached to hydrophilic ionic carboxylate group, called head.
• In water, the soap molecules arrange themselves to form spherical particles that are called micelles.
• In each micelle, the hydrophobic tails of soap molecules point to the centre and the hydrophilic heads lie on the surface of the sphere.
• As a result of this, soap dispersion in water is stable.

10. Draw labelled diagrams of the following :
a. Tyndall effect
b. Dialysis
c. Bredig’s arc method
d. Soap micelle
Answer:
a. Tyndall effect:

b. Dialysis:

c. Bredig’s arc method:

d. Soap micelle:

Activity :
Collect the information about methods to study surface chemistry.
Answer:
Following are the few methods that are employed to study surface chemistry.
i. X-ray photoelectron spectroscopy:
It is a surface-sensitive spectroscopic technique which is used to measure elemental composition of the surface, to determine elements that are present as contaminants on the surface, etc.

ii. Auger electron spectroscopy:
It is a common analytical technique which is used to study surfaces of materials.

iii. Temperature programmed desorption (TPD):
Adsorbed molecules get desorbed when the surface temperature is increased. TPD technique is used to observe these desorbed molecules and helps in providing information about binding energy between the adsorbate and adsorbent.

iv. Scanning Electron Microscopy:
In this technique, a scanning electron microscope is used to focus electron beam over the surface of the sample to be examined. The electron beam interacts with the sample and an image is obtained. This image provides information about surface structure and composition of the sample.

[Note: Students are expected to collect additional information about surface chemistry on their own.]

11th Chemistry Digest Chapter 11 Adsorption and Colloids Intext Questions and Answers

Can you tell? (Textbook Page No. 160)

Question 1.
What is adsorption?
Answer:
Adsorption is the phenomenon of accumulation of higher concentration of one substance on the surface of another (in bulk) due to unbalanced/unsatisfied attractive forces on the surface.

Try this. (Textbook Page No. 161)

Question 1.
Dip a chalk in ink. What do you observe?
Answer:
When a chalk is dipped in ink, it is observed that the ink molecules are adsorbed at the surface of chalk and the surface becomes coloured, while the solvent of the ink goes deeper into the chalk due to absorption.

Internet my friend. (Textbook Page No. 172)

Question i.
Brownian motion
Answer:
Students can search relevant videos on YouTube to visualize Brownian motion.

Question ii.
Collect information about Brownian motion.
Answer:
i. The colloidal or microscopic particles undergo ceaseless random zig-zag motion in all directions in a fluid. This motion of dispersed phase particles is called Brownian motion.
ii. Cause of Brownian motion:

• Particles of the dispersed phase constantly collide with the fast-moving molecules of dispersion medium (fluid).
• Due to this, the dispersed phase particles acquire kinetic energy from the molecules of the dispersion medium.
• This kinetic energy brings about Brownian motion.

Internet my friend. (Textbook Page No. 172)

Question 1.
Collect information about surface chemistry.
Answer:

• Surface or interface represents the boundary which separates two bulk phases.
  e.g. Boundary between water and its vapour is a liquid-gas interface.
• Certain properties of substances, particularly of solids and liquids, depend upon the nature of the surface.
• An interface usually has a thickness of a few molecules. However, its area depends on the size of the bulk phase particles.
• Commonly considered bulk phases may be pure compounds or solutions.
• A number of important phenomena, namely, dissolution, crystallization, heterogeneous catalysis, electrode processes and corrosion take place at an interface.
• Thus, study of chemistry of surfaces is critical to many applications in industry, analytical investigations and day-to-day activities such as cleaning and softening of water.
• The branch of chemistry which deals with the nature of surfaces and changes occurring on the surfaces is called surface chemistry.
• Study of surfaces requires a rigorously clean surface. An ultra-clean metal surface can be obtained under very high vacuum, of the order of 10^-8 to 10^-9 pascal.
• Adsorption, catalysis and colloids (such as emulsions and gels) are some of the important aspects of surface chemistry.

[Note: Students are expected to collect additional information about surface chemistry on their own.]

Activity. (Textbook Page No. 172)

Question 1.
Calculate surface area to volume ratio of spherical particle. See how the ratio increases with the reduction of radius of the particle. Plot the ratio against the radius.
Answer:
The graph below shows that as the radius of the spherical particle decreases, the surface to volume ratio increases steadily.

11th Std Chemistry Questions And Answers:

• Elements of Group 13, 14 and 15 Class 11 Chemistry Questions And Answers
• States of Matter Class 11 Chemistry Questions And Answers
• Adsorption and Colloids Class 11 Chemistry Questions And Answers
• Chemical Equilibrium Class 11 Chemistry Questions And Answers
• Nuclear Chemistry and Radioactivity Class 11 Chemistry Questions And Answers
• Basic Principles of Organic Chemistry Class 11 Chemistry Questions And Answers
• Hydrocarbons Class 11 Chemistry Questions And Answers
• Chemistry in Everyday Life Class 11 Chemistry Questions And Answers

Balbharti Maharashtra State Board 11th Chemistry Textbook Solutions Chapter 15 Hydrocarbons Textbook Exercise Questions and Answers.

Hydrocarbons Class 11 Exercise Question Answers Solutions Maharashtra Board

Class 11 Chemistry Chapter 15 Exercise Solutions Maharashtra Board

Chemistry Class 11 Chapter 15 Exercise Solutions

1. Choose correct options

Question A.
Which of the following compound has the highest boiling point?
a. n-pentane
b. iso-butane
c. butane
d. neopentane
Answer:
a. n-pentane

Question B.
Acidic hydrogen is present in :
a. acetylene
b. ethane
c. ethylene
d. dimethyl acetylene
Answer:
a. acetylene

Question C.
Identify ‘A’ in the following reaction:

a. KMnO₄/H⁺
b. alkaline KMnO₄
c. dil. H₂SO₄/1% HgSO₄
d. NaOH/H₂O₂
Answer:
a. KMnO₄/H⁺

Question D.
Major product of chlorination of ethyl benzene is :
a. m-chlorethyl benzene
b. p-chloroethyl benzene
c. chlorobenzene
d. o-chloroethylbenzene
Answer:
b. p-chloroethyl benzene

Question E.
1 – chloropropane on treatment with alc. KOH produces :
a. propane
b. propene
c. propyne
d. propyl alcohol
Answer:
b. propene

2. Name the following :

Question A.
The type of hydrocarbon that is used as lubricant.
Answer:
Waxes

Question B.
Alkene used in the manufacture of polythene bags.
Answer:
Ethene

Question C.
The hydrocarbon said to possess carcinogenic property.
Answer:
Benzene

Question D.
What are the main natural sources of alkane?
Answer:
Crude petroleum and natural gas.

Question E.
Arrange the three isomers of alkane with malecular formula C₅H₁₂ in increasing order of boiling points and write their IUPAC names.
Answer:
The three isomers of alkane with molecular formula C₅H₁₂ are as follows:

The increasing order of their boiling point is I > II > III.

Question F.
Write IUPAC names of the products obtained by the reaction of cold concentrated sulphuric acid followed by water with the following compounds.
a. propene
b. but-1-ene
Answer:
a. propene:

b. but-1-ene:

Question G.
Write the balanced chemical reaction for preparation of ethane from
a. Ethyl bromide
b. Ethyl magnesium iodide
Answer:
a. Preparation of ethane from ethyl bromide:

b. Preparation of ethane from ethyl magnesium iodide:

Question H.
How many monochlorination products are possible for
a. 2-methylpropane ?
b. 2-methylbutane ?
Draw their structures and write their IUPAC names.
Answer:
a. Possible monochlorination products for 2-methylpropane:

b. Possible monochlorination products for 2-methylbutane:

Question I.
Write all the possible products for pyrolysis of butane.
Answer:
Possible products for pyrolysis of butane are:

Question J.
Which of the following will exhibit geometical isomerism ?

Answer:
Compound (c) will exhibit geometrical isomerism.

Question K.
What is the action of following on ethyl iodide ?
a. alc. KOH
b. Zn, HCl
Answer:
a. Action of alc. KOH on ethyl iodide:

b. Action of Zn/HCl on ethyl iodide:

Question L.
An alkene ‘A’ an ozonolysis gives 2 moles of ethanal. Write the structure and IUPAC name of ‘A’.
Answer:
Structure of A: CH₃ – CH = CH – CH₃
IUPAC name of A: But-2-ene

Question M.
Acetone and acetaldehyde are the ozonolysis products of an alkene. Write the structural formula of an alkene and give IUPAC name of it.
Answer:
The structural formula of alkene:

IUPAC name is 2-methylbut-2-ene.

Question N.
Write the reaction to convert
a. propene to n-propyl alcohol.
b. propene to isoproyl alcohol.
Answer:
a.

b.

Question O.
What is the action of following on but-2-ene ?
a. dil alkaline KMnO₄
b. acidic KMnO₄
Answer:
a. Action of dil. alkaline KMnO₄ on but-2-ene:

b. Action of acidic KMnO₄ on but-2-ene:

Question P.
Complete the following reaction sequence :

Answer:

Question Q.
Write the balanced chemical reactions to get benzene from
a. Sodium benzoate.
b. Phenol.
Answer:
a. Sodium benzoate:
When anhydrous sodium benzoate is heated with soda lime, it undergoes decarboxylation and gives benzene.

b. Phenol:
When vapours of phenol are passed over heated zinc dust, it undergoes reduction and gives benzene.

Question R.
Predict the possible products of the following reaction.
a. chlorination of nitrobenzene,
b. sulfonation of chlorobenzene,
c. bromination of phenol,
d. nitration of toluene.
Answer:
a. Nitro group is meta directing group. So, chlorination of nitrobenzene gives m-chloronitrobenzene.

b. Chloro group is ortho and para directing group. So, sulphonation of chlorobenzene gives p-chlorobenzene sulphonic acid and o- chlorobenzene sulphonic acid.

c. Phenolic -OH group is ortho and para directing group. So, bromination of phenol gives p-bromophenol and o-bromophenol.

d. Methyl group is ortho and para directing group. So, nitration of toluene gives p-nitrotoluene and o-nitrotoluene.

3. Identify the main product of the reaction

Answer:
a.

b.

c.

d.

4. Read the following reaction and answer the questions given below.

a. Write IUPAC name of the product.
b. State the rule that governs formation of this product.
Answer:
a. IUPAC name of the product: 1 -Bromo-2-methylpropane
b. Anti-Markownikov’s rule/Kharasch effect/peroxide effect: It states that, the addition of HBr to unsymmetrical alkene in the presence of organic peroxide (R-O-O-R) takes place in the opposite orientation to that suggested by Markovnikov’s rule.

5. Identify A, B, C in the following reaction sequence :

Answer:

6. Identify giving reason whether the following compounds are aromatic or not.

Answer:
A.

Compound is non-aromatic since it has 4π electrons and hence, does not obey Huckel rule of aromaticity.

B.

Compound is non-aromatic since it has 4π electrons and hence, does not obey Huckel rule of aromaticity.

C.

Compound is aromatic since it has 6π electrons and hence, obeys Huckel rule of aromaticity.

D.

Compound is aromatic since it has 6n electrons and hence, obeys Huckel rule of aromaticity.

7. Name two reagents used for acylation of benzene.
Answer:
The two reagents used for acylation of benzene are:
i. CH₃COCl (acetyl chloride) and anhydrous AlCl₃
ii. (CH₃CO)₂O (acetic anhydride) and anhydrous AlCl₃

8. Read the following reaction and answer the questions given below.

A. Write the name of the reaction.
B. Identify the electrophile in it.
C. How is this electrophile generated?
Answer:
A. The name of the reaction is Friedel-Craft’s alkylation reaction.
B. The electrophile in the reaction is ⁺CH₃.
C. The electrophile ⁺CH₃ is generated as follows:

Activity:

Prepare chart of hydrocarbons and note down the characteristics.
Answer:

Characteristics of hydrocarbons:

• They are chemical compounds that are formed from only hydrogen and carbon atoms.
• Both ‘C’ and ‘H’ share an electron pair forming covalent bonds.
• One of the special properties of carbon is its ability to form double and triple bonds (unsaturation). Saturated hydrocarbons are alkanes and cycloalkanes while the unsaturated hydrocarbons are the aromatics, alkenes and alkynes.
• All hydrocarbons are insoluble in water, their boiling point increases as the size of alkane increases.
• All hydrocarbons can reach complete oxidation.
• Hydrocarbons are mainly used as fuel for transport and industry.

[Note: Students are expected to collect additional information on hydrocarbons on their own.]

11th Chemistry Digest Chapter 15 Hydrocarbons Intext Questions and Answers

Can you recall? (Textbook Page No. 233)

Question i.
What are hydrocarbons?
Answer:
The compounds which contain carbon and hydrogen as the only elements are called hydrocarbons.

Question ii.
Write structural formulae of the following compounds: propane, ethyne, cyclobutane, ethene, benzene.
Answer:

Do you know? (Textbook Page No. 233)

Question 1.
Why are alkanes called paraffins?
Answer:
i. Alkanes contain only carbon-carbon and carbon-hydrogen single covalent bonds.
ii. They are chemically less reactive and do not have much affinity for other chemicals.
Hence, they are called paraffins.

Internet my friend. (Textbook Page No. 233)

Question 1.
Collect information about hydrocarbon.
Answer:

• In organic chemistry, a hydrocarbon is an organic compound consisting of carbon and hydrogen as the only elements.
• They are examples of group 14 hydrides.
• Alkanes, cycloalkanes, aromatic hydrocarbons are different types of hydrocarbons.
• Most of the hydrocarbons found on earth occur naturally in crude oil.
• They mainly undergo substitution, addition or combustion reactions.
• Most hydrocarbons are flammable and toxic.
• They are the primary energy source in the form of combustible fuel source.

[Note: Students are expected to collect additional information on their own]

Use your brain power! (Textbook Page No. 234)

Question 1.
i. Write the structures of all the chain isomers of the saturated hydrocarbon containing six carbon atoms.
ii. Write IUPAC names of all the above structures.
Answer:
The structural formulae and names of all possible isomers having molecular formula C₆H₁₄ are as follows:

Note:
Alkanes and isomer number

Number of Carbon	Alkane	Number of isomers
1	Methane	No structural isomer
2	Ethane	No structural isomer
3	Propane	No structural isomer
4	Butane	Two
5	Pentane	Three
6	Hexane	Five

Can you recall? (Textbook Page No. 235)

Question i.
What is a catalyst?
Answer:
A catalyst is a substance that can be added to a reaction to increase the reaction rate without getting consumed in the process.
e.g. Ni is used as a catalyst in the catalytic hydrogenation of alkenes or alkynes.

Question ii.
What is addition reaction?
Answer:
When a compound combines with another compound to form a product that contain all the atoms in both the reactants, it is called an addition reaction.

Try this (Textbook Page No. 235)

Question 1.
Transform the following word equation into balanced chemical equation and write at least 3 changes that occur at molecular level during this chemical change.
\(\text { 2-Methylpropene + Hydrogen } \stackrel{\text { catalyst }}{\longrightarrow} \text { Isobutane }\)
Answer:

Three changes which occur at molecular level include:
Step 1: Adsorption of reactants: Reactants (alkene and hydrogen) get adsorbed on the catalytic surface.
Step 2: Formation of a product: Hydrogen atoms are added across the double bond of 2-methylpropene which results in the formation of product isobutane.
Step 3: Desorption: Product formed on the catalytic surface is readily desorbed making catalytic surface available for other molecules.

Use your brain power! (Textbook Page No. 236)

Question 1.
Why are alkanes insoluble in water and readily soluble in organic solvents like chloroform or ether?
Answer:

• The solubility of any substance is governed by the principle of like dissolves like. This means polar compounds are soluble in polar solvents while nonpolar compounds are soluble in nonpolar solvents.
• Alkanes consist of C – C and C – H nonpolar covalent bonds and thus, they are nonpolar in nature, whereas water is a polar solvent.
• The dipole-dipole forces that exist between water molecules is much stronger than the forces of attraction between alkane and water molecules.

Hence, alkanes are insoluble in water and readily soluble in organic solvents like chloroform or ether.

Can you recall? (Textbook Page No. 238)

Question 1.
What is the product which is poisonous and causes air pollution formed by incomplete combustion of alkane?
Answer:
When alkanes are subjected to incomplete combustion, it forms carbon monoxide and carbon (soot).
i. 2CH_4(g) + 3O_2(g) → 2CO_(g) + 4H₂O_(g)
ii. CH_4(g) + O_2(g) → C_(s) + 2H₂O_(l)

Can you recall? (Textbook Page No. 238)

Question i.
What are alkenes?
Answer:
Alkenes are unsaturated hydrocarbons containing at least one carbon-carbon double bond.

Question ii.
Calculate the number of sigma (σ) and pi (π) bonds in 2-methylpropene.
Answer:

Question iii.
Write the structural formula of pent-2-ene.
Answer:

Can you tell? (Textbook Page No. 241)

Question i.
Explain by writing a reaction, the main product formed on heating 2-methylbutan-2-ol with concentrated sulphuric acid.
Answer:

Question ii.
Will the main product in the above reaction show geometrical isomerism?
Answer:
No, the major product, i.e., 2-methylbut-2-ene does not show geometrical (or cis-trans) isomerism.

Can you tell? (Textbook Page No. 244)

Question 1.
Propan-1-ol and 2-methypropan-1-ol are not prepared by hydration method. Why?
Answer:
Propan-1-ol and 2-methylpropan-1-ol cannot be prepared by hydration of propene and 2-methylprop-1-ene because the addition reaction follows Markovnikov’s rule.

Use your brainpower. (Textbook Page No. 244)

Question 1.
On ozonolysis, an alkene forms the following carbonyl compounds. Draw the structure of unknown alkene from which these compounds are formed: HCHO and CH₃COCH₂CH₃
Answer:
The structure of alkene which produces a mixture of HCHO and CH₃COCH₂CH₃ on ozonolysis is

Use your brain power! (Textbook Page No. 245)

Question 1.
Write the structure of monomer from which each of the following polymers are obtained.

Answer:

	Polymer	Monomeric unit
i.	Teflon	CF2 – CF2 Tetrafluoroethene
ii.	Polypropene	H3C – CH = CH2 Propene
iii.	Polyvinyl chloride	H2C = CHCl Vinyl chloride

Can you tell? (Textbook Page No. 246)

Question i.
What are aliphatic hydrocarbons?
Answer:
Aliphatic hydrocarbons are hydrocarbons containing carbon and hydrogen joined together in straight chain or branched chain. They may be saturated (alkanes) or unsaturated (alkenes or alkynes).

Question ii.
Compare the proportion of carbon and hydrogen atoms in ethane, ethene and ethyne. Which compound is most unsaturated with hydrogen?
Answer:
Ethane
C : H = 2 : 6 = 1 : 3
Ethene
C : H = 2 : 4 = 1 : 2
Ethyne
C : H = 2 : 2 = 1 : 1
From the above proportion it is clear that ethyne with 1 : 1 ratio of C : H is most unsaturated with hydrogen (50%) as compared to ethane (25%) and ethene (33.33%).

Can you tell? (Textbook Page No. 247)

Question 1.
Why is sodamide used in dehydrohalogenation of vicinal dihalides to remove HX from alkenyl halide in place of alcoholic KOH?
Answer:

• Sodamide (NaNH₂) is a strong base and hence, helps in complete conversion of alkenyl halide formed in the first step to form alkynes.
• The base (KOH or NaOH) used in first step gives alkynes in poor yield and hence, stronger bases such as NaNH₂ on KNH₂ are used in second step.

Use your brainpower! (Textbook Page No. 247)

Question 1.
Convert: 1-Bromobutane to hex-1-yne
Answer:

Can you tell? (Textbook Page No. 248)

Question 1.
Alkanes and alkenes do not react with lithium amide. Give reason.
Answer:
i. The sp hybrid carbon atom in terminal alkynes is more electronegative than the sp² carbon in ethene or the sp³ carbon in ethane.
ii. Due to high electronegative character of carbon in terminal alkynes, hydrogen atom can be given away as proton (H⁺) to very strong base as shown in the reactions below.

iii. Further, since s-character decreases from sp to sp² to sp³ carbon atom, the relative acidity of alkanes, alkenes and alkynes is in the following order: H – C = C – H > H₂C = CH₂ > H₃C – CH₃
Hence, alkenes and alkanes do not react with lithium amide.

Use your brain power! (Textbook Page No. 248)

Question 1.
Arrange following hydrocarbons in the increasing order of acidic character: propane, propyne, propene.
Answer:
Propyne > propene > propane

Use your brain power! (Textbook Page No. 249)

Question 1.
Convert: 3-Methylbut-l-yne into 3-methylbutan-2-one
Answer:

Can you recall? (Textbook Page No. 249)

Question i.
What are aromatic hydrocarbons?
Answer:
Benzene and all compounds that have structures and chemical properties resembling benzene are called as aromatic hydrocarbons.

Question ii.
What are benzenoid and non-benzenoid aromatics?
Answer:
Benzenoid aromatics are compounds having at least one benzene ring in the structure.
e.g. Benzene, naphthalene, anthracene, phenol, etc.,
Non-benzenoid aromatics are compounds that contain an aromatic ring, other than benzene. e.g. Tropone, etc.

Can you recall? (Textbook Page No. 254)

Question 1.
What is decarboxylation?
Answer:
The reaction which involves removal of a carboxyl group (-COOH) in the form of carbon dioxide (CO₂) is known as decarboxylation reaction.
R – COOH → R – H + CO₂

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