Category: Class 11

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

Maharashtra State Board Class 11 English Yuvakbharati Solutions Chapter 1.1 Being Neighborly

11th English Digest Chapter 1.1 Being Neighborly Textbook Questions and Answers

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.

Balbharti Maharashtra State Board 11th Biology Textbook Solutions Chapter 16 Skeleton and Movement Textbook Exercise Questions and Answers.

Maharashtra State Board 11th Biology Solutions Chapter 16 Skeleton and Movement

1. Choose the correct option

Question (A).
The functional unit of striated muscle is …………..
a. cross-bridges
b. myofibril
c. sarcomere
d. z-band
Answer:
c. sarcomere

Question (B).
A person slips from the staircase and breaks his ankle bone. Which bones are involved?
a. Carpals
b. Tarsal
c. Metacarpals
d. Metatarsals
Answer:
b. Tarsal

Question (C).
Muscle fatigue is due to the accumulation of ……..
a. pyruvic acid
b. lactic acid
c. malic acid
d. succinic acid
Answer:
b. lactic acid

Question (D).
Which one of the following is NOT antagonistic muscle pair?
a. Flexo-extensor
b. Adductor-abductor
c. Levator-depressor
d. Sphinetro-suprinater
Answer:
d. Sphinetro-suprinater

Question (E).
Swelling of sprained foot is reduced by soaking in hot water containing a large amount of common salt,
a. due to osmosis
b. due to plasmolysis
c. due to electrolysis
d. due to photolysis
Answer:
a. due to osmosis

Question (F).
Role of calcium in muscle contraction is ……….
a. to break the cross bridges as a cofactor in the hydrolysis of ATP
b. to bind with troponin, changing its shape so that the actin filament is exposed
c. to transmit the action potential across the neuromuscular junction.
d. to re-establish the polarisation of the plasma membrane following an action potential
Answer:
b. to bind with troponin, changing its shape so that the actin filament is exposed

Question (G).
Hyper-secretion of parathormone can cause which of the following disorders?
a. Gout
b. Rheumatoid arthritis
c. Osteoporosis
d. Gull’s disease
Answer:
c. Osteoporosis

Question (H).
Select correct option between two nasal bones

Answer:
(c)

Question 2.
Answer the following questions

Question (A).
What kind of contraction occurs in your neck muscles while you are reading your class assignment?
Answer:

1. Isometric contractions occur in the neck muscles while reading class assignment.
2. These contractions are important for supporting objects in a fixed position.

Question (B).
Observe the diagram and enlist importance of ‘A’, ‘B’ and ‘C’.

Answer:

1. A – Posterior portion of vertebral foramen of atlas vertebrae; Importance – The spinal cord runs through this portion of vertebral foramen
2. B – Anterior portion of vertebral foramen of axis vertebrae; Importance – In this portion, the odontoid process of axis vertebrae forms ‘NO’ joint.
3. C – Inferior articular facet; Importance – It articulates with superior articular facet of axis and permits rotatory movement of head.

Question (C).
Raju intends to train biceps; while exercising using dumbbells, which joints should remain stationary and which should move?
Answer:
While performing exercise of biceps using dumbbells, the joint which should remain stationary are wrist joint or radiocarpal joint, ball and socket joint of shoulder. The only joint which should move is hinge joint of elbow.

Question (D).
In a road accident, Moses fractured his leg. One of the passers by, tied a wodden plank to the fractured leg while Moses
was rushed to the hospital Was this essential? Why?
Answer:

1. Fracture is a significant and traumatic injury which requires medical attention however, getting timely first aid is important.
2. If any bone is fractured, it is essential that the fractured part be immobilized to prevent further injury. It can be done with the help of any available wooden plank or batons or rulers. Thus, a wooden plank was tied to Moses’s fractured leg as a first aid for fracture.
3. A fractured bone is immobilized to prevent the sharp edges of the fractured bone from moving and cutting tissue, muscle, blood vessels and nerves. Immobilization can also help reduce pain or control shock.

Question (E).
Sprain is more painful than fracture. Why?
Answer:

1. A sprain is an injury that involves the ligaments (tissues that connect bones at joints), whereas a fracture is an injury that involves bones.
2. Sprains can be of three degree: 1^st degree: Mild with micro-tears, 2^nd degree: Partial with visible tear in ligament, 3^rd degree: Completely torn ligament.
3. If a sprain is 3^rd degree, it will be more painful than a fracture. It usually requires a surgery to fix this injury, while breaking a bone, most of the time does not require surgery.
4. Breaks or Fractures also vary greatly. Minor fractures (like stress/ hairline fractures) are much less painful than compound/ complex fractures in which the bone may be cracked into half.
5. Blood supply is essential for growth and regeneration. Bones are highly vascularized whereas, ligaments are not. This causes the bones to heal comparatively faster than severe sprains. Thus, the duration of enduring pain until the injury heals also differs.
6. Also, ligaments have a rich supply of sensory nerves, which may also be responsible for an elevated sense of pain during severe sprains.

[Note: 1^st and 2^nd degree sprains are not very serious and may be lesser painful than a fracture. Depending on the severity of the injury, intensity of pain will vary.]

Question (F).
Why a red muscle can work for a prolonged period whereas white muscle fibre suffers from fatigue after a shorter work? (Refer to chapter animal tissues.)
Answer:

1. Red muscle fibres contain large amount of myoglobin and mitochondria (site of aerobic respiration), whereas white muscles fibres contain lesser amount of myoglobin and mitochondria.
2. Myoglobin is an iron-containing pigment that carries oxygen molecules to muscle tissues. Abundance of these pigments in red muscle fibres supports higher rate of aerobic respiration, whereas white muscle fibres have less mitochondria and depend upon anaerobic respiration.
3. Anaerobic respiration in muscle white fibres leads to the production of lactic acid and accumulation of higher of levels lactic acid can result in fatigue in white muscle fibres.

Thus, red muscle fibres can perform prolonged work and show less fatigue due to accumulation of negligible amount loss or of lactic acid, whereas white muscle fibres suffer from fatigue after a shorter work due to accumulation of higher amount of lactic acid.

3. Answer the following questions in detail

Question (A).
How is the structure of sarcomere suitable for the contractility of the muscle? Explain its function according to sliding
filament theory. (Refer to chapter animal tissues.)
Answer:
i. Sarcomere is the functional unit of myofibril. It has specific arrangement of actin and myosin filaments. The components of sarcomere are organized into variety of bands and zones. Actin and myosin are referred as contractile proteins. Actin is called as thin filament whereas myosin in called as thick filament. The structure of sarcomere:

ii. ‘A’ band – dark bands present at the centre of sarcomere and contain myosin as well as actin.
‘H’ zone or Hensen’s zone – light area present at the centre of ‘A’ band
‘M’ line – present at the centre of ‘H’ zone
‘I’ band – light bands present on the either side of ‘A’ band containing only actin
Z’ line – adjacent ‘I’ bands are separated by ‘Z’ line.

iii. Sliding filament theory: It was put forth by H.E Huxley and A.F Huxley. It is also known as ‘Walk along theory’ or Ratchet theory.

• According to the sliding filament theory, the interaction between actin and myosin filaments is the basic cause of muscle contraction. The actin filaments are interdigitated with myosin filaments.
• The head of the myosin is joined to the actin backbone by a cross bridge forming a hinge joint. From this joint, myosin head cannot tilt forward or backward. This movement is an active process as it utilizes ATP.
• Myosin head contains ATPase activity. It can derive energy by the breakdown of ATP molecule. This energy can be used for the movement of myosin head.
• During contraction, the myosin head gets attached to the active site of actin filaments and pull them inwardly so that the actin filaments slide over the myosin filaments. This results in the contraction of muscle fibre.
  

Question (B).
Ragini, a 50 year old office goer, suffered hair-line cracks in her right and left foot in short intervals of time. She was worried about minor jerks leading to hair line cracks in bones. Doctor explained to her why it must be happening and prescribed medicines.

What must be the cause of Ragini’s problem? Why has it occurred? What precautions she should have taken earlier? What care she should take in future?
Answer:

1. Considering Ragini’s age, she may be undergoing menopause. After menopause, oestrogen level declines resulting in lower bone density.
2. Osteoporosis:
   • In this disorder, bones become porous and hence brittle. It is primarily age related disease and is more common in women than men.
   • Osteoporosis may be caused due to decreasing estrogen secretion after menopause, deficiency of vitamin D, low calcium diet, decreased secretion of sex hormones and thyrocalcitonin.
3. As age advances, bone resorption outpaces bone formation. Hence, the bones lose mass and become brittle. More calcium is lost in urine, sweat, etc., than it is gained through diet. Thus, prevention of disease is better than treatment by consuming adequate amount of calcium and exercise at young age.
4. A person with previous hairline fractures is more susceptible to reoccurrence of fractures. Hence, Ragini needs to take her medications and supplements properly, avoid jerky movements and maintain body weight.

Question (C).
How does structure of actin and myosin help muscle contraction?
Answer:
i. Myosin filament:

1. Each myosin filament is a polymerized protein.
   Many meromyosins (monomeric proteins) constitute one thick filament.
2. Myosin molecule consists of two heavy chains (heavy meromyosin / HMM) coiled around each other forming a double helix. One end of each of these chains is projected outwardly is known as cross bridge. This end folds to form a globular protein mass called myosin head.
3. Two light chains are associated with each head forming 4 light chains/light meromyosin / LMM.
4. Myosin head has a special ATPase activity. It can split ATP to produce energy.
5. Myosin contributes 55% of muscle proteins.
6. In sarcomere, myosin tails are arranged to point towards the centre of the sarcomere and the heads point to the sides of the myofilament band.
   

ii. Actin filament: It is a complex type of contractile protein. It is made up of three components:

1. F actin: It forms the backbone of actin filament. F actin is made up of two helical strands. Each strand is composed of polymerized G actin molecules. One ADP molecule is attached to G actin molecule.
2. Tropomyosin: The actin filament contains two additional protein strands that are polymers of tropomyosin molecules. Each strand is loosely attached to an F actin. In the resting stage, tropomyosin physically covers the active myosin-binding site of the actin strand.
3. Troponin: It is a complex of three globular proteins, is attached approx. 2/3^rd distance along each tropomyosin molecule. It has affinity for actin, tropomyosin and calcium ions. The troponin complex is believed to attach the tropomyosin to the actin. The strong affinity of troponin for calcium ions is believed to initiate the contraction process.
   

Question (D).
Justify the structure of atlas and axis vertebrae with respect to their position and function.
Answer:
i. Atlas vertebrae:

1. Atlas is the ring-like, 1^st cervical vertebrae. It has anterior, posterior arches and large lateral masses.
2. It lacks centrum and spinous process. The superior surfaces of the lateral masses are concave and are known as superior articular facets.
3. These facets articulate with the occipital condyles of the occipital bone thereby forming atlanto-occipital joints. This articulation permits ‘YES movement’ or nodding movement.
4. The inferior surfaces of the lateral masses known as inferior articular facets articulate with axis vertebrae,

ii. Axis vertebrae:

1. It is the 2^nd cervical vertebrae.
2. A peg-like process called odontoid process projects superiorly through the anterior portion of the vertebral foramen of the atlas.
3. The odontoid process forms a pivot on which the atlas and head rotate. This arrangement allows ‘NO movement’ or side to side movement of the head.
4. The articulation formed between the anterior arch of atlas, the odontoid process of the axis and between their articular facets is called as atlanto-axial joint.

Question (E).
Observe the blood report given below and diagnose the possible disorder.

Answer:
On observing Report D, it is clear that the level of uric acid is more than normal, thus the patient must be suffering from gouty arthritis.

Also, the elevated blood urea nitrogen (BUN) indicates dysfunctional liver and/ or kidneys. It generally occurs due to decrease in GFR, caused by renal disease or obstruction of urinary tract.

Question 4.
Write short notes on following points

Question (A).
Actin filament
Answer:
Actin filament: It is a complex type of contractile protein. It is made up of three components:

1. F actin: It forms the backbone of actin filament. F actin is made up of two helical strands. Each strand is composed of polymerized G actin molecules. One ADP molecule is attached to G actin molecule.
2. Tropomyosin: The actin filament contains two additional protein strands that are polymers of tropomyosin molecules. Each strand is loosely attached to an F actin. In the resting stage, tropomyosin physically covers the active myosin-binding site of the actin strand.
3. Troponin: It is a complex of three globular proteins, is attached approx. 2/3rd distance along each tropomyosin molecule. It has affinity for actin, tropomyosin and calcium ions. The troponin complex is believed to attach the tropomyosin to the actin. The strong affinity of troponin for calcium ions is believed to initiate the contraction process.
   

Question (B).
Myosin filament
Answer:
i. Myosin filament:

1. Each myosin filament is a polymerized protein.
   Many meromyosins (monomeric proteins) constitute one thick filament.
2. Myosin molecule consists of two heavy chains (heavy meromyosin / HMM) coiled around each other forming a double helix. One end of each of these chains is projected outwardly is known as cross bridge. This end folds to form a globular protein mass called myosin head.
3. Two light chains are associated with each head forming 4 light chains/light meromyosin / LMM.
4. Myosin head has a special ATPase activity. It can split ATP to produce energy.
5. Myosin contributes 55% of muscle proteins.
6. In sarcomere, myosin tails are arranged to point towards the centre of the sarcomere and the heads point to the sides of the myofilament band.
   

Question (C).
Role of calcium ions in contraction and relaxation of muscles.
Answer:
Calcium ions play a major role in contraction and relaxation of muscles.

1. Calcium ions are released from the sarcoplasm during muscle contraction and stored in sarcoplasmic reticulum during muscle relaxation.
2. When a skeletal muscle is excited and an action potential travels along the T tubule, the concentration of calcium ions increases.
3. These calcium ions bind to troponin which in turn undergoes a conformational change that causes tropomyosin to move away from the myosin-binding sites on actin. Once these binding sites are free, myosin heads bind to them to form cross-bridges and the muscle fiber contracts.
4. The decrease in calcium ion concentration in the sarcoplasmic reticulum causes tropomyosin to slide back and block the myosin binding sites on actin. This causes the muscle to relax.

Question 5.
Draw labelled diagrams

Question (A).
Synovial joint.
Answer:
i. Synovial joints / freely movable joints / diarthroses:

1. It is characterized by presence of a space called synovial cavity between articulating bones that renders free movement at the joint.
2. The articulating surfaces of bones at a synovial joint are covered by a layer of hyaline cartilage. It reduces friction during movement and helps to absorb shock.
3. Synovial cavity is lined by synovial membrane that forms synovial capsule. Synovial membrane secretes synovial fluid.
4. Synovial fluid is a clear, viscous, straw coloured fluid similar to lymph. It is viscous due to hyaluronic acid. The synovial fluid also contains nutrients, mucous and phagocytic cells to remove microbes.
   Synovial fluid lubricates the joint, absorbs shocks, nourishes the hyaline cartilage and removes waste materials from hyaline cartilage cells (as cartilage is avascular). Phagocytic cells destroy microbes and cellular debris formed by wear and tear of the joint.
5. If the joint is immobile for a while, the synovial fluid becomes viscous and as joint movement starts, it becomes less viscous.
6. The joint is provided with capsular ligament and numerous accessory ligaments. The fibrous capsule is attached to periosteum of articulating bones. The ligament helps in avoiding dislocation of joint.
7. The types of synovial joints are on follows:

1. Pivot joint: In this type of joint, the rounded or pointed surface of one bone articulates with a ring formed partly by another bone and partly by the ligament. Rotation only around its own longitudinal axis is possible. e.g. in joint between atlas and axis vertebrae, head turns side ways to form ‘NO’ joint.

2. Ball and socket joint: The ball like surface of one bone fits into cup like depression of another bone forming a movable joint. Multi-axial movements are possible. This type of joint allows movements along all three axes and in all directions. e.g. Shoulder and hip joint.

3. Hinge joint: In a hinge joint, convex surface of one bone fits into concave surface of another bone. In most hinge joints one bone remains stationary and other moves. The angular opening and closing motion (like hinge) is possible. In this joint only mono-axial movement takes place like flexion and extension. e.g. Elbow and knee joint.

4. Condyloid joint: It is an ellipsoid joint. The convex oval shaped projection of one bone fits into oval shaped depression in another bone. it is a biaxial joint because it permits movement along two axes viz, flexion, extension, abduction, adduction and circumduction is possible. e.g. Metacarpophalangeal joint.

5. Gliding joint: It is a planar joint, where the articulating surfaces of bones are flat or slightly curved. These joints are non-axial because the motion they allow does not occur along an axis or a plane. e.g. Intercarpal and intertarsal joints.

6. Saddle joint: This joint is a characteristic of Homo sapiens. Here the articular surface of one bone is saddle-shaped and that of other bone fits into saddle (each bone forming this joint have both concave and convex areas). It is a modified condyloid joint in which movement is somewhat more free. It is a biaxial joint that allows flexion, extension, abduction, adduction and circumduction.
e.g. Carpometacarpellar joint between carpal (trapezium) and metacarpal of thumb.

Question (B).
Different cartilagenous joints.
Answer:
Cartilaginous / slightly movable joints / amphiarthroses:
These joints are neither fixed nor freely movable. Articulating bones are held together by hyaline or fibrocartilages. They are further classified as

a. Synchondroses: The two bones are held together by hyaline cartilage. They are meant for growth. On completion of growth, the joint gets ossified, e.g. Epiphyseal plate found between epiphysis and diaphysis of a long bone, Rib – Sternum junction.

b. Symphysis: In this type of joint, broad flat disc of fibrocartilage connects two bones. It occurs in mid-line of the body. e.g. Intervertebral discs, manubrium and sternum, pubic symphysis.

Practical / Project :

Identify the following diagrams and demonstrate the concepts in classroom.

Answer:
The diagrams A, B and C represent Class I, Class II and Class III lever respectively.
For description:

1. Class I lever: The joint between the first vertebra and occipital condyle of skull is an example of Class I lever. The force is directed towards the joints (fulcrum); contraction of back muscle provides force while the part of head that is raised acts as
   
2. Class II lever: Human body raised on toes is an example of Class II lever. Toe acts as fulcrum, contracting calf muscles provide the force while raised body acts as resistance.
   
3. Class III lever: Flexion of forearm at elbow exhibit lever of class III. Elbow joint acts as fulcrum and radius, ulna provides resistance. Contracting bicep muscles provides force for the movement.
   

[Students are expected to perform the given activity on their own]

12th Biology Digest Chapter 16 Skeleton and Movement Intext Questions and Answers

Movements And Locomotion (Textbook Page No. 193)

Question 1.
Streaming of protoplasm, peristalsis, walking, running, etc. Which of the above-mentioned movements are internal? Which are external? Can you add few more examples?
Answer:

1. Streaming of protoplasm, peristalsis are internal movements. Walking and running are external movements.
2. Examples of internal movement: Contraction and relaxation heart, inspiration and expiration, contraction of blood vessels, etc.
3. Examples of external movement: Swimming; movement tongue, jaws, snout, tentacles, movement of ear pinna, etc.

Can you recall? (Textbook Page No. 193)

Question 1.
Which are different types of muscular tissues?
Answer:

1. Smooth / non-striated / visceral / involuntary muscles
2. Cardiac muscles
3. Skeletal / straited / voluntary muscles.

Question 2.
Name the type of muscles which bring about running and speaking.
Answer:
Skeletal muscles (Voluntary muscles)

Question 3.
Name the muscles which do not contract as per our will.
Answer:
Involuntary muscles (smooth muscles and cardiac muscles)

Question 4.
Which type of muscles show rhythmic contractions?
Answer:
Cardiac muscles

Question 5.
Which type of muscle is present in the diaphragm of the respiratory system?
Answer:
Skeletal muscle

Question 6.
State the functions of:

1. Smooth muscles
2. Cardiac muscles
3. Striated muscles

Answer:

1. Smooth muscles: They bring about involuntary movements like peristalsis in the alimentary canal, constriction and dilation of blood vessels.
2. Cardiac muscles: They bring about contraction and relaxation of the heart.
3. Striated muscles: They control voluntary movements of limbs, head, trunk, eyes, etc.

Can you recall? (Textbook Page No. 193)

Question 1.
Name the part of human skeleton situated along the vertical axis.
Answer:
Axial skeleton

Question 2.
Give an account of bones of human skull.
Answer:
Skull is made up of 22 bones. It is located at the superior end of vertebral column. The bones of skull are
joined by fixed or immovable joints except for jaw.

Skull consists of cranium or brain box and facial bones.

i. Cranium: It is made up of four median bones and two paired bones.

1. Frontal bone: It is median bone (unpaired) forming forehead, roof of orbit (eye socket) and the most anterior part of cranium. It is connected to two parietals, sphenoid and ethmoid bone.
2. Parietal bones: These paired bones form the roof of cranium and greater portion of sides of the cranium.
3. Temporal bones: These paired bones are situated laterally just above the ear on either side. Each temporal bone gives out zygomatic process that joins zygomatic bone to form zygomatic arch. Just at the base of zygomatic process is mandibular fossa, a depression for mandibles (lower jaw bone) that forms the only movable joint of the skull. This bone harbors the ear canal that directs sound waves into the ear. The processes of temporal bones provide points for attachment for various muscles of neck and tongue.
4. Occipital bone: It is a single bone present at the back of the head. It forms the posterior part and most of the base of cranium. The inferior part of this bone shows foramen magnum, the opening through which medulla oblongata connects with spinal cord. On the either sides of foramen magnum are two prominent protuberances called occipital condyles. These fit into the corresponding depressions present in 1^st vertebra.
5. Sphenoid bone: Median bone present at the base of the skull that articulates with all other cranial bones and holds them together. This butterfly shaped bone has a saddle shaped region called sella turcica. In this hypophyseal fossa, the pituitary gland is lodged.
6. Ethmoid bone: This median bone is spongy in appearance. It is located anterior to sphenoid and posterior to nasal bones. It contributes to formation of nasal septum and is major supporting structure of nasal cavity.

ii. Facial Bones: Fourteen facial bones give a characteristic shape to the face. The growth of face stops of the
age of 16.
Following bones comprise the facial bones:

1. Nasals: These are paired bones that form the bridge of nose.
2. Maxillae: These form the upper jaw bones. They are paired bones that join with all facial bones except mandible. Upper row of teeth are lodged maxillae.
3. Palatines: These are paired bones forming the roof of buccal cavity or floor of the nasal cavity.
   
4. Zygomatic bones: They are commonly called as cheek bones.
5. Lacrimal bones: These are the smallest amongst the facial bones.
   These bones form the medial wall of each orbit. They have lacrimal fossa that houses lacrimal sacs. These sacs gather tears and send them to nasal cavity.
6. Inferior nasal conchae: They form the part of lateral wall of nasal cavity. They help to swirl and filter air before it passes to lungs.
7. Vomer: The median, roughly triangular bone that forms the inferior portion of nasal septum.
8. Mandible: This median bone forms the lower jaw. It is the largest and strongest facial bone. It is the only movable bone of skull. It has curved horizontal body and two perpendicular branches i.e. rami. These help in attachment of muscles. It has lower row of teeth lodged in it.

Think about it. (Textbook Page No. 193)

Question 1.
Did you ever feel tickling in muscles?
Answer:
Yes, the tickling sensation in muscles can be felt and sometimes it is also accompanied by itching sensation.

Question 2.
What is locomotion?
Answer:
The change in locus of whole body of living organism from one place to another place is called locomotion.

Question 3.
State the four basic types of locomotory movements seen in animals.
Answer:
The four basic types locomotory movements seen in animals are:

1. Amoeboid movement: It is performed by pseudopodia, e.g. leucocytes.
2. Ciliary movement: It is performed by cilia, e.g. ciliated epithelium. In Paramoecium, cilia help in passage of food through cytopharynx.
3. Whirling movement: It is performed by flagella, e.g. sperms.
4. Muscular movement: It is performed by muscles, with the help of bones and joints.

Question 4.

Question 1.
Why do muscles show spasm after rigorous contraction?
Answer:

1. Rigorous contraction of muscles occurs during strenuous activities (swimming. running, cycling, aerobics. etc.)
2. Muscle contraction requires energy. Glucose in muscle cells breakdown during anerobic respiration resulting in accumulation of lactic acid.
3. This lactic acid buildup triggers muscle spasm around muscle cells.

Question 2.
Why do we shiver during winter?
Answer:

1. Humans are homeotherms as the can regulate their body temperature with respect to the surrounding temperature. During winter, when temperature falls, the thermoreceptors detect the change in temperature and send signals to the brain.
2. Shivering reflex i.e. rapid contraction of muscles is triggered by the brain to generate heat and raise the body temperature.

Can you tell? (Textbook Page No. 194)

Question 1.
Why are movement and locomotion necessary among animals?
Answer:

• Movement is one of the important characteristics of all the living organisms. Animals exhibit wide range of
  movements like rhythmic beating of heart, movement of diaphragm during respiration, ingestion of food,
  movement of eyeballs, etc.
• Locomotion results into change in place or location of an organism. Animals locomote in search of food, mate, shelter, breeding ground. while escaping from the enemy, etc.
• Thus, locomotion and movement are necessary to support the living of animals.

Question 2.
All locomotions are movements but all movements are not locomotion. Justify
Answer:
Locomotion occurs when body changes its position, however all movements may not result in locomotion. Thus, all locomotions are movements but all movements are not locomotion.

Question 3.
Kriti was diagnosed with knee tendon injury. She asked the doctor whether she will be able to walk due to the injury? If not then state the reason.
Answer:
Knee tendon injury affects the ability to walk. Kriti may not be able to walk freely as the tendons attached to the bones help in the movement of the parts of skeleton.

Question 4.
What are antagonistic muscles? Explain with example.
Answer:

1. The muscles that work in pairs and produce opposite action are known as antagonistic muscles, e.g. biceps and triceps of upper arm.
2. The biceps (flexors) bring flexion (folding) and triceps (extensors) bring extension of the elbow joint.
3. One member from a pair is capable of bending the joint by pulling of bones the other member is capable of straightening the same joint by pulling.
4. In antagonistic pair of muscles, one member is stronger than the other, e.g. The biceps are stronger than the triceps.

Question 5.
Describe the antagonistic muscles in detail.
Answer:
Following are the important antagonistic muscles:

1. Flexor and extensor: Flexor muscle on contraction results into bending or flexion of joint, e.g. Biceps. Extensor muscle on contraction results in straightening or extension of a joint, e.g. Triceps.
2. Abductor and adductor: Abductor muscle moves a body part away from the body axis. e.g. Deltoid muscle of shoulder moves the arm away from the body. Adductor muscle moves a body part towards the body axis, e.g. Latissimus dorsi of shoulder moves the arm near the body.
3. Pronator and supinator: Pronator turns the palm downwards and supinator turns the palm upward.
4. Levator and depressor: Levator raises a body part and the depressors lower the body part.
5. Protractor and retractor: Protractor moves forward, whereas the retractor moves backward.
6. Sphincters: Circular muscles present in the inner walls of anus, stomach, etc., for closure and opening.

Question 6.
Differentiate between:
i. Flexor and extensor muscles
Answer:

	Flexor Muscles		Extensor Muscles
a.	Flexor muscles contract and bring about the bending or flexion of joint.	a.	Extensor muscles contract and bring about the straightening or extension of joint.
b.	These muscles decrease the angle between the bones on two sides of a joint.	b.	These muscles increase the angle between the components of limb.
e.g.	Biceps	e.g.	Triceps

ii. Pronator and supinator: Pronator turns the palm downwards and supinator turns the palm upward.

Can you recall? (Textbook Page No. 194)

Question 1.
Comment on contraction of skeletal muscles.
Answer:
Skeletal muscles show quick and strong voluntary contractions. They bring about voluntary movements of the body. For mechanism of muscle contraction:

When the muscles are relaxed, the active sites remain covered with tropomyosin and troponin complex. Due to this, myosin cannot interact with active site of actin and thus contraction cannot occur.

1. When an impulse (action potential) comes to muscle through motor end plate, it spreads throughout the sarcolemma of the myofibril.
2. The transverse tubules of sarcoplasmic reticulum release large number of Ca⁺⁺ ions into sarcoplasm. These calcium ions interact with the troponin molecules and the interaction inactivates troponin-tropomyosin complex. This causes change in the structure of tropomyosin.
3. As a result, tropomyosin gets detached from the active site of actin (F actin) filament, exposing the active site for actin.
4. The myosin head cleaves the ATP to derive energy and gets attached to the uncovered active site of actin. This results into the formation of acto-myosin complex.
5. The myosin heads are now tilted backwards and pull the attached actin filament inwardly. This results in contraction of the muscle fibres.

Do You Know How (Textbook Page No. 195)

Question 1.
Do skeletal muscles contract and bring about movement and locomotion?
Answer:
When the muscles are relaxed, the active sites remain covered with tropomyosin and troponin complex. Due to this, myosin cannot interact with active site of actin and thus contraction cannot occur.

1. When an impulse (action potential) comes to muscle through motor end plate, it spreads throughout the sarcolemma of the myofibril.
2. The transverse tubules of sarcoplasmic reticulum release large number of Ca⁺⁺ ions into sarcoplasm. These calcium ions interact with the troponin molecules and the interaction inactivates troponin-tropomyosin complex. This causes change in the structure of tropomyosin.
3. As a result, tropomyosin gets detached from the active site of actin (F actin) filament, exposing the active site for actin.
4. The myosin head cleaves the ATP to derive energy and gets attached to the uncovered active site of actin. This results into the formation of acto-myosin complex.
5. The myosin heads are now tilted backwards and pull the attached actin filament inwardly. This results in contraction of the muscle fibres.

Internet my friend. (Textbook Page No. 196)

Question 1.
Collect information about‘T’ tubules of sarcoplasmic reticulum.
Answer:

1. T tubules or the transverse tubules are invaginations of the sarcolemma penetrating into the myocyte interior, forming a highly branched and interconnected network that makes junctions with the sarcoplasmic reticulum.
2. These tubules are selectively enriched with specific ion channels and proteins crucial in the development of calcium transients necessary in excitation-contraction coupling, thereby facilitating a fast-synchronous contraction of the entire cell volume.
3. They are unique to straited muscle cells.
   [Source: https://www. uniprot. org/locations]

Can you tell? (Textbook Page No. 197)

Question 1.
Explain the chemical changes taking place in muscle contraction.
Answer:
The muscle undergoes various chemical changes during contraction, they are as follows:

1. A nerve impulse arrives at the motor nerve. The neurotransmitter – acetylcholine is released at the neuromuscular junction (N-M junction or motor endplate) enters into the sarcomere.
2. This leads to inflow of Na+ inside the sarcomere and generates an action potential in the muscle fibre.
3. The action potential passes down the T tubules and activates calcium channels in the T tubular membrane. Activation of calcium channel allows calcium ions to pass into the sarcoplasm. These Ca⁺⁺ ions binds to the specific sites on troponin of actin filaments and a conformational change occurs in the troponin – tropomyosin complex, thereby removing the masking of active sites for myosin on the actin filament.
4. In the myosin head, the enzyme ATPase gets activated in the presence of Ca⁺⁺ and converts ATP into ADP and inorganic phosphate.
5. This energy from ATP hydrolysis is utilized by myosin bridges or myosin heads to bind with active sites of actin and form actomyosin complex pulling the actin filaments towards the centre of sarcomere. The myosin heads are now tilted backwards and pull the attached actin filament inwardly towards them. The actin filament slides over mysosin and contraction occurs.
6. Also, the ADP needs to be converted back to ATP immediately as they required for muscular contraction, This is achieved in the muscles by the presence of another high energy compound, creatine phosphate.
7. ADP combines with creatine phosphate to produce ATP and creatinine due to which the supply of ATP for muscle contraction is restored but the level of creatine phosphate keeps decreasing and the level of creatinine keeps on increasing.
8. The creatinine formed needs to be reconverted to creatine phosphate. This is done by ATP produced during oxidation of glycogen through glycolysis.

Question 2.
Why are muscle rich in creatine phosphate?
Answer:

1. Creatine phosphate or phosphocreatine is formed from ATP, when the muscle is in relaxed state. It is a phosphorylated form of creatine.
2. Muscle cells contain creatine phosphate which acts as energy reserve as this high energy compound acts as a phosphate donor for ATP formation.
3. ATP acts as an immediate source of energy for contraction

Question 3.
Explain the mechanism of muscle contraction and relaxation.
Answer:
Mechanism of muscle contraction:

When the muscles are relaxed, the active sites remain covered with tropomyosin and troponin complex. Due to this, myosin cannot interact with active site of actin and thus contraction cannot occur.

1. When an impulse (action potential) comes to muscle through motor end plate, it spreads throughout the sarcolemma of the myofibril.
2. The transverse tubules of sarcoplasmic reticulum release large number of Ca⁺⁺ ions into sarcoplasm. These calcium ions interact with the troponin molecules and the interaction inactivates troponin-tropomyosin complex. This causes change in the structure of tropomyosin.
3. As a result, tropomyosin gets detached from the active site of actin (F actin) filament, exposing the active site for actin.
4. The myosin head cleaves the ATP to derive energy and gets attached to the uncovered active site of actin. This results into the formation of actomyosin complex.
5. The myosin heads are now tilted backwards and pull the attached actin filament inwardly. This results in contraction of the muscle fibres.

Muscle relaxation: During relaxation, all the events occur in reverse direction as that of muscle contraction.

1. When the stimulation is terminated, the actomyosin complex is broken down and myosin head gets detached from actin filaments. This process utilizes ATP.
2. Also, the Ca⁺⁺ ions are pumped back into the sarcoplasmic reticulum. This process too is an energy dependent process and utilizes ATP.
   
3. As a result, the troponin-tropomyosin complex is restored again which covers the active sites of act in filament, due to disappearance of the Ca⁺⁺ ions. The interaction between actin and myosin ceases and the actin filaments return back to their original position.
4. This results in muscle relaxation.

Question 4.
What do you understand by muscle twitch?
Answer:
Single muscle twitch:

Single muscle twitch: It is a muscle contraction initiated by a single brief-stimulation. It occurs in 3 stages: a latent period of no contraction, a contraction period and a relaxation period.

1. The involuntary contraction of muscle fibers is known as muscle twitch.
2. Muscle twitch is also known as fasciculation.
3. It is caused due accumulation of lactic acid in muscles.

Do You Know How (Textbook Page No. 198)

Question 1.
Exoskeletal components change from lower to higher group of animals. These include chitinous structures, nails, horns, hooves, scales, hair, shell, plates, fur, muscular foot, tube feet, etc.

Question 1.
Do you know any of these exoskeletal structures help in movement and locomotion?
Answer:
Nails, hooves, scales, plates, muscular foot and tube feet help in movement and locomotion.

Question 2.
How do scales and plates help in movement and locomotion?
Answer:
Scales and plates in reptiles like snakes provide grip to move on rough edgy surfaces.

Question 3.
Are scales of a fish and that of snake similar?
Answer:
Fishes have dermal scales (bony scales), whereas reptiles like snakes have epidermal scales or scutes (horny, tough extensions of outer layer of skin i.e., stratum corneum).

Question 4.
Find out more information about exoskeletal structures and their role in movement and locomotion.
Answer:
Exoskeletal structures: Exoskeleton provide support, help in movement and also provides protection from predators. The exoskeletal structures vary from organism to organism. Echinoderms have tube feet for locomotion whereas molluscs (e.g. Chiton) have muscular foot for movement and locomotion

[Students are expected to find out more information about exoskeletal structures on their own.]

Question 2.
Name the tissues that form the structural framework of the body.
Answer:
Cartilage and bone

Do you remember? (Textbook Page No. 198)

Question 1.
What are the components of skeletal system?
Answer:
The components of skeletal system are bones, tendons, ligaments and joint.

Question 2.
What type of bones are present in our body?
Answer:
Long bones, short bones, flat bones, irregular bones and sesamoid bones.

Question 3.
How do bones help in various ways?
Answer:

1. Bones form the framework of our body and thus provide shape to the body.
2. They protect vital organs thus help in the smooth functioning of body.
3. The joints between the bones help in movement and locomotion.
4. They provide firm surface for attachment of muscles.
5. They are reservoirs of calcium and form important site for hemopoiesis.

Use your brain power. (Textbook Page No. 198)

Question 1.
Can you compare bone, muscle and joint which help in locomotion with any simple machines you have studied earlier?
Answer:
Bone, muscle and joint can be compared to the simple machines called levers. Joints act as fulcrum, respective muscle generates the force required to move the bone associated with joint.

Question 2.
Explain the three types of lever found in human body.
Answer:
The three types of lever are as follows:

1. Class I lever: The joint between the first vertebra and occipital condyle of skull is an example of Class I lever. The force is directed towards the joints (fulcrum); contraction of back muscle provides force while the part of head that is raised acts as resistance.
   
2. Class II lever: Human body raised on toes is an example of Class II lever. Toe acts as fulcrum, contracting calf muscles provide the force while raised body acts as resistance.
   
3. Class III lever: Flexion of forearm at elbow exhibit lever of class III. Elbow joint acts as fulcrum and radius, ulna provides resistance. Contracting bicep muscles provides force for the movement.
   

Use your brain power. (Textbook Page No. 199)

Question 1.
Why are long bones slightly bent and not straight?
Answer:

1. Long bones include tibia, fibula, femur, humerus, radius, ulna, etc.
2. They have greater length than width. They consist of a shaft and variable number of epiphysis.
3. They are slightly bent or curved to absorb the stress of the body’s weight and evenly distribute the body weight at several different points.
4. If long bones were straight, the weight of the body would be unevenly distributed and the bone would fracture more easily.

Identify and label. (Textbook Page No. 199)

Question 1.
Identify the different bones.
Answer:

Identify and label. (Textbook Page No. 200)

Question 1.
Name A, B, C and D from the given figure and discuss in group.

Answer:
A – Coronal suture,
B – Sagittal suture,
C – Lambdoidal suture,
D – Lateral / Squamous suture

Skull has many sutures (type of immovable joints) present, out of which four prominent ones are:

1. Coronal suture: Joins frontal bone with parietals.
2. Sagittal suture: Joins two parietal bones.
3. Lambdiodal suture: Joins two parietal bones with occipital bone.
4. Lateral/squamous sutures: Joins parietal and temporal bones on lateral side.

Can you tell? (Textbook Page No. 201)

Question 1.
Give schematic plan of human skeleton.
Answer:

[Note: Numbers in the bracket indicate the number of bones.]

Can you tell? (Textbook Page No. 201)

Question 1.
Enlist the bones of cranium.
Answer:
Cranium: It is made up of four median bones and two paired bones.

1. Frontal bone: It is median bone (unpaired) forming forehead, roof of orbit (eye socket) and the most anterior part of cranium. It is connected to two parietals, sphenoid and ethmoid bone.
2. Parietal bones: These paired bones form the roof of cranium and greater portion of sides of the cranium.
3. Temporal bones: These paired bones are situated laterally just above the ear on either side. Each temporal bone gives out zygomatic process that joins zygomatic bone to form zygomatic arch. Just at the base of zygomatic process is mandibular fossa, a depression for mandibles (lower jaw bone) that forms the only movable joint of the skull. This bone harbors the ear canal that directs sound waves into the ear. The processes of temporal bones provide points for attachment for various muscles of neck and tongue.
4. Occipital bone: It is a single bone present at the back of the head. It forms the posterior part and most of the base of cranium. The inferior part of this bone shows foramen magnum, the opening through which medulla oblongata connects with spinal cord. On the either sides of foramen magnum are two prominent protuberances called occipital condyles. These fit into the corresponding depressions present in 1st vertebra.
5. Sphenoid bone: Median bone present at the base of the skull that articulates with all other cranial bones and holds them together. This butterfly shaped bone has a saddle shaped region called sella turcica. In this hypophyseal fossa, the pituitary gland is lodged.
6. Ethmoid bone: This median bone is spongy in appearance. It is located anterior to sphenoid and posterior to nasal bones. It contributes to formation of nasal septum and is major supporting structure of nasal cavity.

Can you tell? (Textbook Page No. 201)

Question 1.
Write a note on structure and function of skull.
Answer:
i. Structure of skull:
Skull is made up of 22 bones. It is located at the superior end of vertebral column. The bones of skull are joined by fixed or immovable joints except for jaw.

Skull consists of cranium or brain box and facial bones.

i. Cranium: It is made up of four median bones and two paired bones.

1. Frontal bone: It is median bone (unpaired) forming forehead, roof of orbit (eye socket) and the most anterior part of cranium. It is connected to two parietals, sphenoid and ethmoid bone.
2. Parietal bones: These paired bones form the roof of cranium and greater portion of sides of the cranium.
3. Temporal bones: These paired bones are situated laterally just above the ear on either side. Each temporal bone gives out zygomatic process that joins zygomatic bone to form zygomatic arch. Just at the base of zygomatic process is mandibular fossa, a depression for mandibles (lower jaw bone) that forms the only movable joint of the skull. This bone harbors the ear canal that directs sound waves into the ear. The processes of temporal bones provide points for attachment for various muscles of neck and tongue.
4. Occipital bone: It is a single bone present at the back of the head. It forms the posterior part and most of the base of cranium. The inferior part of this bone shows foramen magnum, the opening through which medulla oblongata connects with spinal cord. On the either sides of foramen magnum are two prominent protuberances called occipital condyles. These fit into the corresponding depressions present in 1^st vertebra.
5. Sphenoid bone: Median bone present at the base of the skull that articulates with all other cranial bones and holds them together. This butterfly shaped bone has a saddle shaped region called sella turcica. In this hypophyseal fossa, the pituitary gland is lodged.
6. Ethmoid bone: This median bone is spongy in appearance. It is located anterior to sphenoid and posterior to nasal bones. It contributes to formation of nasal septum and is major supporting structure of nasal cavity.

ii. Facial Bones: Fourteen facial bones give a characteristic shape to the face. The growth of face stops of the age of 16.
Following bones comprise the facial bones:

1. Nasals: These are paired bones that form the bridge of nose.
2. Maxillae: These form the upper jaw bones. They are paired bones that join with all facial bones except mandible. Upper row of teeth are lodged maxillae.
   
3. Palatines: These are paired bones forming the roof of buccal cavity or floor of the nasal cavity.
4. Zygomatic bones: They are commonly called as cheek bones.
5. Lacrimal bones: These are the smallest amongst the facial bones. These bones form the medial wall of each orbit. They have lacrimal fossa that houses lacrimal sacs. These sacs gather tears and send them to nasal cavity.
6. Inferior nasal conchae: They form the part of lateral wall of nasal cavity. They help to swirl and filter air before it passes to lungs.
7. Vomer: The median, roughly triangular bone that forms the inferior portion of nasal septum.
8. Mandible: This median bone forms the lower jaw. It is the largest and strongest facial bone. It is the only movable bone of skull. It has curved horizontal body and two perpendicular branches i.e. rami. These help in attachment of muscles. It has lower row of teeth lodged in it.

ii. Functions of skull:

1. It protects the brain.
2. It provides sockets for ear, nasal chamber and eyes.
3. Mandible bone of the skull helps in opening and closing of the mouth.

Internet my friend. (Textbook Page No. 201)

Question 1.
Cleft palate and cleft lip
Answer:

1. Cleft palate and cleft lip are the birth defects that occur when a baby’s lip or mouth does develop properly.
2. Cleft palate happens when the tissue that forms the roof of the mouth does not join together completely during pregnancy.
3. Cleft lip happens when the tissue that makes up the lip does not join completely before birth. This leads to formation of an opening in the upper lip.

[Students are expected to find more information about Cleft Palate and lip on internet.]

Can you tell? (Textbook Page No. 202)

Question 1.
Why skull is important for us? Enlist few reasons.
Answer:
Functions of skull:

• It protects the brain.
• It provides sockets for ear, nasal chamber and eyes.
• Mandible bone of the skull helps in opening and closing of the mouth.

Internet my friend. (Textbook Page No. 202)

Question 1.
Find out information about sinuses present in skull, functions of skull and disorder ‘sinusitis’.
Answer:
Sinuses are the hollow cavities present in the skull. They humidify the air we breathe.

i. The four types of sinuses present in the skull:

• Frontal sinuses: They are located above each eye. There are right and left frontal sinuses.
• Maxillary sinuses: They are the largest among all sinuses, located just behind the cheekbones near to upper jaws.
• Sphenoid sinuses: These are present just behind the nose.
• Ethmoid sinuses: These are present between the eyes.

ii. Functions of skull:

Functions of skull:

• It protects the brain.
• It provides sockets for ear, nasal chamber and eyes.
• Mandible bone of the skull helps in opening and closing of the mouth.

iii. Sinusitis: It is the inflammation of tissue lining the sinuses. Healthy sinuses when get blocked with mucus and germs causing infection which may lead to sinusitis.

[Students are expected to find more information about sinusitis, using the internet.]

Something interesting. (Textbook Page No. 202)

Question 1.
If police suspect strangulation, they carefully inspect hyoid bone and cartilage of larynx. These get fractured during strangulation. V arious such investigations are done in case of suspicious death of an individual where ossification of sutures in skull, width of pelvic girdle, etc. are examined to find out approximate age of victim or gender of victim, etc. You may find out information about forensic science.
Answer:
Forensic science is an application of science which is used in the matter of criminal determination and civil law. It is generally used in investigation of crimes. Forensic scientists collect, preserve and analyze the evidence during the course of investigation.
[Students are expected to find more information about forensic science on internet.]

Try this. (Textbook Page No. 202)

Question 1.
Feel your spine (vertebral column). Is it straight or curved?
Answer:
Our spinc shows four slight curves which are visible when viewed from the sides.

Question 2.
Find information about slipped disc. (Textbook page no.202)
Answer:

1. The bones of vertebral column are supported by the intervertebral discs.
2. These intervertebral discs act as shock absorbers due to which they are constantly compressed.
3. The disc consists of two parts – soft gelatinous inner part (nucleus pulposus) and tough outer ring.

If the ligaments of the intervertebral discs become injured, the pressure developed in the nucleus pulposus protrudes posteriorly or into one of the adjacent vertebrae. This is known as slipped disc.

[Students are expected to find more information using the internet.]

Can you tell? (Textbook Page No. 204)

Question 1.
Write a note on curvatures of vertebral column and mention their importance.
Answer:

1. The four curvatures in human spine are cervical, lumbar, thoracic and sacral curvatures.
2. The cervical and lumbar curvatures are secondary and convex whereas the thoracic and sacral curvatures are
   primary and concave.
3. Importance: Curvatures help in maintaining balance in upright position. They absorb shocks while walking
   and also protect the vertebrae from fracture.

Question 2.
Explain the structure of typical vertebra.
Answer:

1. Each vertebra has prominent central body called centrum.
2. The centra of human vertebrae are flat in anterio-posterior aspect. Thus, human vertebrae are amphiplatyan.
3. From the either side of the centrum are two thick short processes which unite to form an arch like structure called neural arch, posterior to centrum.
4. Neural arch forms vertebral foramen which surrounds the spinal cord.
5. Vertebral foramina of all vertebrae form a continuous ‘neural canal’. Spinal cord along with blood vessels and protective fatty covering passes through neural canal.
6. The point where two processes of centrum meet, the neural arch is drawn into a spinous process called neural spine.
   
7. From the base of neural arch, two articulating processes called zygapophyses are given out on either side. The anterior is called superior zygapophyses and posterior called inferior zygapophyses.
8. In a stack of vertebrae, inferior zygapophyses of one vertebra articulates with superior zygapophyses of next vertebra. This allows slight movement of vertebrae without allowing them to fall.
9. At the junction of zygapophyses, a small opening is formed on either side of vertebra called intervertebral foramen that allows passage of spinal nerve.
10. From the base of neural arch, lateral processes are given out called transverse processes. Neural arch, neural spine and transverse processes are meant for attachment of muscles.

Question 2.
How will you identify a thoracic vertebra?
Answer:
Thoracic vertebrae can be identified on the basis of centrum, as the centrum of the thoracic vertebrae is heart shaped.

Can you recall? (Textbook Page No. 206)

Question 1.
How does humerus form ball and socket joint? Where is it located?
Answer:
The head of humerus fits into the glenoid cavity of scapula and forms ball and socket joint. it is located in shoulder and hips.

Can you tell? (Textbook Page No. 208)

Question 1.
Differentiate between the skeleton of palm and foot.
Answer:

	Skeleton of palm	Skeleton of foot
a.	It consists of metacarpals and phalanges	It consists of metatarsals and phalanges
b.	Saddle joints and condyloid joints are in the palm.	Condyloid or saddle joints are not present the foot.

Question 2.
Explain the longest bone in human body.
Answer:
Femur: The thigh bone is the longest bone in the body. The head is joined to shaft at an angle by a short neck. It forms ball and socket joint with acetabulum cavity of coxal bone. The lower one third region of shaft is triangular flattened area called popliteal surface. Distal end has two condyles that articulate with tibia and fibula.

Internet my friend. (Textbook Page No. 212)

Question 1.
Find out information about types of fractures and how they heal.
Answer:

1. Fractures are classified based on their severity, shape or position of the fracture line or the physician who first described them.
2. Types of fractures:
   1. Open fractures: The broken ends of the bone protrude through skin.
   2. Comminuted fractures: The bone is splintered, crushed or broken into pieces at the site of impact and smaller bone fragments lie between the two main fragments.
   3. Greenstick fractures: A partial fracture in which one side of the bone is broken and the other side bends.
   4. Impacted fractures: One end of the fractured bone is forcefully driven into the inferior of the other.
   5. Pott fractures: Fracture of the distal end lateral leg bone with serious injury of the distal tibial articulation.
   6. Codes fractures: Fracture of the distal end of the lateral forearm in which the distal fragment is displaced posteriorly.
3. A fractured bone heals in four phases viz, reactive phase, fibrocartilaginous formation phase, bony callus formation phase and bone remodeling phase.

[Source: Tortora, G., Derrickson, B. Principles of Anatomy and Physiology. 15^th Edition]

[Students are expected to find out more information about healing of fractures using the internet.]

Do you remember? (Textbook Page No. 208)

Question 1.
What are joints? What are the types?
Answer:
i. A point where two or more bones get articulated is called joint or articulation or athrosis.
They are classified based on degree of flexibility or movement they permit into lastly synovial or freely movable or diarthroses type of joints.

ii. Synarthroses / fibrous joints / movable joints:
In this joint, the articulating bones are held together by means of fibrous connective tissue. Bones do not exhibit movement. Hence, it is immovable or fixed type of joint. Synarthroses are further classified into sutures, syndesmoses and gomphoses.

• Sutures: It is composed of thin layer of a dense fibrous connective tissue. Sutures are places of growth. They remain open till growth is complete. On completion of growth, they tend to ossify. Sutures may permit some moulding during childhood. Sutures are further classified into butt joint, scarf joint, lap joint and serrate joint.
• Syndesmoses: It is present where there is greater distance between articulating bones. At such locations, fibrous connective tissue is arranged as a sheet or bundle, e.g. Distal tibiofibular joint, interosseous membrane between tibia and fibula and that between radius and ulna.
• Gomphoses: In this type of joint, a cone shaped bone fits into a socket provided by other bone,
  e. g. Tooth and jaw bones.
  

iii. Cartilaginous / slightly movable joints / amphiarthroses:
These joints are neither fixed nor freely movable. Articulating bones are held together by hyaline or
fibrocartilages. They are further classified as

• Synchondroses: The two bones are held together by hyaline cartilage. They are meant for growth.
  On completion of growth, the joint gets ossified, e.g. Epiphyseal plate found between epiphysis and diaphysis of a long bone, Rib – Sternum junction.
• Symphysis: In this type of joint, broad flat disc of fibrocartilage connects two bones. It occurs in mid-line of the body. e.g. Intervertebral discs, manubrium and sternum, pubic symphysis.
  

iv. Synovial joints / freely movable joints / diarthroses:

1. It is characterized by presence of a space called synovial cavity between articulating bones that renders free movement at the joint.
2. The articulating surfaces of bones at a synovial joint are covered by a layer of hyaline cartilage. It reduces friction during movement and helps to absorb shock.
   
3. Synovial cavity is lined by synovial membrane that forms synovial capsule. Synovial membrane secretes synovial fluid.
4. Synovial fluid is a clear, viscous, straw coloured fluid similar to lymph. It is viscous due to hyaluronic acid. The synovial fluid also contains nutrients, mucous and phagocytic cells to remove microbes. Synovial fluid lubricates the joint, absorbs shocks, nourishes the hyaline cartilage and removes waste materials from hyaline cartilage cells (as cartilage is avascular). Phagocytic cells destroy microbes and cellular debris formed by wear and tear of the joint.
5. If the joint is immobile for a while, the synovial fluid becomes viscous and as joint movement starts, it becomes less viscous.
6. The joint is provided with capsular ligament and numerous accessory ligaments. The fibrous capsule is attached to periosteum of articulating bones. The ligament helps in avoiding dislocation of joint.

g. The types of synovial joints are on follows:

1. Pivot joint: In this type of joint, the rounded or pointed surface of one bone articulates with a ring formed partly by another bone and partly by the ligament. Rotation only around its own longitudinal axis is possible. e.g. in joint between atlas and axis vertebrae, head turns side ways to form ‘NO’ joint.

2. BaIl and socket joint: The ball like surface of one bone fits into cup like depression of another bone forming a movable joint. Multi-axial movements are possible. This type of joint allows movements along all three axes and in all directions. e.g. Shoulder and hip joint.

3. Hinge joint: In a hinge joint, convex surface of one bone fits into concave surface of another bone. In most hinge joints one bone remains stationary and other moves. The angular opening and closing motion (like hinge) is possible. In this joint only mono-axial movement takes place like flexion and extension. e.g. Elbow and knee joint.

4. Condyloid joint: It is an ellipsoid joint. The convex oval shaped projection of one bone fits into oval shaped depression in another bone. It is a biaxial joint because it permits movement along two axes viz, flexion, extension, abduction, adduction and circumduction is possible. e.g. Metacarpophalangeal joint.

5. Gliding joint: It is a planar joint, where the articulating surfaces of bones are flat or slightly curved. These joints are non-axial because the motion they allow does not occur along an axis or a plane. e.g. Intercarpal and intertarsal joints.

Saddle joint: This joint is a characteristic of Homo sapiens. Here the articular surface of one bone is saddle-shaped and that of other bone fits into saddle (each bone forming this joint have both concave and convex areas). It is a modified condyloid joint in which movement is somewhat more free. It is a biaxial joint that allows flexion, extension, abduction, adduction and circumduction. e.g. Carpometacarpellar joint between carpal (trapezium) and metacarpal of thumb.

Imagine. (Textbook Page No. 208)

Question 1.
If your elbow joint would be a fixed type of joint and joint between teeth and gum would be freely movable.
Answer:

1. If the elbow joint would be fixed the flexion and extension of the forearm won’t be possible. Also, rotation of the forearm and wrist would not be not possible.
2. Gomphoses is the type of joint that holds the teeth in the jaw bone. If this joint would be freely movable, we would not be able to chew and all our teeth would fall out.

Use your brain power. (Textbook Page No. 210)

Question 1.
Why are warming up rounds essential before regular exercise?
Answer:

1. Warming up before exercise stimulates the production and secretion of synovial fluid which reduces the stress on joints during exercise.
2. Also, if a joint is immobile for a while, the synovial fluid becomes viscous and as joint movement starts, it becomes less viscous.
3. Warming up increases the blood circulation, loosening the joints and increasing the blood flow. It also prepares the muscles for physical activity and prevents injuries.

Can you tell? (Textbook Page No. 211)

Question 1.
Classify various types of joints found in human body. Present the information in the form of chart. Give example of each type.
Answer:

i. A point where two or more bones get articulated is called joint or articulation or athrosis.
They are classified based on degree of flexibility or movement they permit into lastly synovial or freely movable or diarthroses type of joints.

ii. Synarthroses / fibrous joints / movable joints:
In this joint, the articulating bones are held together by means of fibrous connective tissue. Bones do not exhibit movement. Hence, it is immovable or fixed type of joint. Synarthroses are further classified into sutures, syndesmoses and gomphoses.

1. Sutures: It is composed of thin layer of a dense fibrous connective tissue. Sutures are places of growth. They remain open till growth is complete. On completion of growth, they tend to ossify. Sutures may permit some moulding during childhood. Sutures are further classified into butt joint, scarf joint, lap joint and serrate joint.
2. Syndesmoses: It is present where there is greater distance between articulating bones. At such locations, fibrous connective tissue is arranged as a sheet or bundle, e.g. Distal tibiofibular joint, interosseous membrane between tibia and fibula and that between radius and ulna.
3. Gomphoses: In this type of joint, a cone shaped bone fits into a socket provided by other bone,
   e. g. Tooth and jaw bones.
   

iii. Cartilaginous / slightly movable joints / amphiarthroses:
These joints are neither fixed nor freely movable. Articulating bones are held together by hyaline or
fibrocartilages. They are further classified as

• Synchondroses: The two bones are held together by hyaline cartilage. They are meant for growth.
  On completion of growth, the joint gets ossified, e.g. Epiphyseal plate found between epiphysis and diaphysis of a long bone, Rib – Sternum junction.
• Symphysis: In this type of joint, broad flat disc of fibrocartilage connects two bones. It occurs in mid-line of the body. e.g. Intervertebral discs, manubrium and sternum, pubic symphysis.
  

iv. Synovial joints / freely movable joints / diarthroses:

1. It is characterized by presence of a space called synovial cavity between articulating bones that renders free movement at the joint.
2. The articulating surfaces of bones at a synovial joint are covered by a layer of hyaline cartilage. It reduces friction during movement and helps to absorb shock.
   
3. Synovial cavity is lined by synovial membrane that forms synovial capsule. Synovial membrane secretes synovial fluid.
4. Synovial fluid is a clear, viscous, straw coloured fluid similar to lymph. It is viscous due to hyaluronic acid. The synovial fluid also contains nutrients, mucous and phagocytic cells to remove microbes.
   Synovial fluid lubricates the joint, absorbs shocks, nourishes the hyaline cartilage and removes waste materials from hyaline cartilage cells (as cartilage is avascular). Phagocytic cells destroy microbes and cellular debris formed by wear and tear of the joint.
5. If the joint is immobile for a while, the synovial fluid becomes viscous and as joint movement starts, it becomes less viscous.
6. The joint is provided with capsular ligament and numerous accessory ligaments. The fibrous capsule is attached to periosteum of articulating bones. The ligament helps in avoiding dislocation of joint.

g. The types of synovial joints are on follows:

1. Pivot joint: In this type of joint, the rounded or pointed surface of one bone articulates with a ring formed partly by another bone and partly by the ligament. Rotation only around its own longitudinal axis is possible. e.g. in joint between atlas and axis vertebrae, head turns side ways to form ‘NO’ joint.

2. BaIl and socket joint: The ball like surface of one bone fits into cup like depression of another bone forming a movable joint. Multi-axial movements are possible. This type of joint allows movements along all three axes and in all directions. e.g. Shoulder and hip joint.

3. Hinge joint: In a hinge joint, convex surface of one bone fits into concave surface of another bone. In most hinge joints one bone remains stationary and other moves. The angular opening and closing motion (like hinge) is possible. In this joint only mono-axial movement takes place like flexion and extension. e.g. Elbow and knee joint.

4. Condyloid joint: It is an ellipsoid joint. The convex oval shaped projection of one bone fits into oval shaped depression in another bone. It is a biaxial joint because it permits movement along two axes viz, flexion, extension, abduction, adduction and circumduction is possible. e.g. Metacarpophalangeal joint.

5. Gliding joint: It is a planar joint, where the articulating surfaces of bones are flat or slightly curved. These joints are non-axial because the motion they allow does not occur along an axis or a plane. e.g. Intercarpal and intertarsal joints.

Saddle joint: This joint is a characteristic of Homo sapiens. Here the articular surface of one bone is saddle-shaped and that of other bone fits into saddle (each bone forming this joint have both concave and convex areas). It is a modified condyloid joint in which movement is somewhat more free. It is a biaxial joint that allows flexion, extension, abduction, adduction and circumduction. e.g. Carpometacarpellar joint between carpal (trapezium) and metacarpal of thumb.

[Students are expected to prepare a chart on their own.]

Can you tell? (Textbook Page No. 211)

Question 1.
Human beings can hold an object in a better manner than monkeys. Why?
Answer:

1. Humans and monkeys both have five fingers including thumb, however humans can hold an object in better manner than monkeys because humans have highly developed opposable thumbs. The opposable thumb allows better grip.
2. The saddle joint in thumb allows free and independent movement to the thumb the carpometacarpellar joint between carpal (trapezium) and metacarpal of thumb makes the thumb opposable. It allows biaxial movements, i.e. flexion – extension and adduction – abduction but not rotation.

[Note: Gorillas, chimpanzees, orangutans and some other variants of apes have opposable thumb.]

Internet my friend. (Textbook Page No. 211)

Question 92.
Now a days we hear from many elderly people that they are undergoing knee replacement surgery. Find out why one has to undergo knee replacement; how it is carried out and how it can be prevented.
Answer:
Knee replacement is done in following cases:

1. Osteoarthritis: The cartilage in the knee undergoes degradation. It is caused by many factors such as muscle weakness, aging, obesity, etc.
2. Rheumatoid arthritis: It is characterised by inflammation of the synovial membrane, where it starts secreting excess of synovial fluid in the joint. This fluid exerts extensive pressure on the joint and causes severe pain.
3. Post-traumatic arthritis: This is caused due to breakage of ligament or cartilage. The breakage can be due to severe injury or accident. It causes severe pain and requires knee replacement.
4. Procedure:
   The procedure involves removal of the damaged cartilage or ligament and replaces it with artificial implant made up of either metal, plastic or both. Metal or plastic knee caps are used to cover the knees. The implant is connected to the bone and an artificial knee joint is made between them.
5. Prevention: Maintaining body weight, exercising regularly, consuming appropriate medications and supplements, etc.

[Students are expected to find out more information about knee replacement on internet]

Find out. (Textbook Page No. 212)

Question 1.
You must have heard of Sachin Tendulkar suffering from ‘tennis elbow’, a cricketer suffering from a disorder named after another game. Can common people too suffer from this disorder? Find out more information about this disorder.
Answer:

1. Tennis elbow is caused due to inflammation of tendon which joins muscles of forearm to the bone of upper arm (humerus). It is known as lateral epicondylitis.
2. It causes severe pain in the elbow. It occurs due to extensive repetitive movement of hand. This damages the tendon and increases the tenderness of the elbow joint.
3. This disorder develops not only in athletes but also in other common people whose job involves extensive movement of hand such as carpenter, painter, plumber, etc.

[Students are expected to find more information about tennis elbow on their own.]

Balbharti Maharashtra State Board 11th Biology Textbook Solutions Chapter 15 Excretion and Osmoregulation Textbook Exercise Questions and Answers.

Maharashtra State Board 11th Biology Solutions Chapter 15 Excretion and Osmoregulation

1. Choose the correct option

Question (A).
Which one of the following organisms would spend maximum energy in the production of nitrogenous waste?
a. Polar bear
b. Flamingo
c. Frog
d. Shark
Answer:
b. Flamingo

Question (B).
In human beings, uric acid is formed due to metabolism of __________.
a. amino acids
b. fatty acids
c. creatinine
d. nucleic acids
Answer:
d. nucleic acids

Question (C).
Visceral layer : Podocytes :: PCT : _______
a. Cilliated cells
b. Squamous cells
c. Columnar cells
d. Cells with brush border
Answer:
d. Cells with brush border

Question (D).
Deproteinised plasma is found in __________.
a. Bowman’s capsule
b. Descending limb
c. Glomerular capillaries
d. Ascending limb
Answer:
a. Bowman’s capsule, b. Descending limb, d. Ascending limb

Question (E).
Specific gravity of urine would _______ if level of ADH increases.
a. remain unaffected
b. increases
c. decreases
d. stabilise
Answer:
b. increases

Question (F).
What is micturition?
a. Urination
b. Urine formation
c. Uremia
d. Urolithiasis
Answer:
a. Urination

Question (G).
Which one of the following organisms excrete waste through nephridia?
a. Cockroach
b. Earthworm
c. Crab
d. Liver Fluke
Answer:
c. Crab

Question (H).
Person suffering from kidney stone is advised not to have tomatoes as it has _______.
a. seeds
b. lycopene
c. oxalic acid
d. sour taste
Answer:
c. oxalic acid

Question (I).
Tubular secretion does not take place in ________.
a. DCT
b. PCT
c. collecting duct
d. Henle’s loop
Answer:
b. PCT

Question (J).
The minor calyx ____________.
a. collects urine
b. connects pelvis to ureter
c. is present in the cortex
d. receives column of Bertini
Answer:
a. collects urine

Question (K).
Which one of the followings is not a part of human kidney?
a. Malpighian body
b. Malpighian tubule
c. Glomerulus
d. Loop of Henle
Answer:
b. Malpighian tubule

Question (L).
The yellow colour of the urine is due to presence of ___________
a. uric acid
b. cholesterol
c. urochrome
d. urea
Answer:
c. urochrome

Question (M).
Hypotonic filtrate is formed in _______
a. PCT
b. DCT
c. LoH
d. CT
Answer:
a. PCT

Question (N).
In reptiles, uric acid is stored in _____
a. cloaca
b. fat bodies
c. liver
d. anus
Answer:
a. cloaca

Question (O).
The part of nephron which absorbs glucose and amino acid is______
a. collecting tubule
b. proximal tubule
c. Henle’s loop
d. DCT
Answer:
b. proximal tubule

Question (P).
Bowman’s capsule is located in kidney in the ________
a. cortex
b. medulla
c. pelvis
d. pyramids
Answer:
a. cortex

Question (Q).
The snakes living in desert are mainly__________
a. aminotelic
b. ureotelic
c. ammonotelic
d. uricotelic
Answer:
d. uricotelic

Question (R).
Urea is a product of breakdown of ___________
a. fatty acids
b. amino acids
c. glucose
d. fats
Answer:
b. amino acids

Question (S).
Volume of the urine is regulated by__________
a. aldosterone
b. ADH
c. both a and b
d. none
Answer:

Question 2.
Answer the following questions

Question (A).
Doctors say Mr. Shaikh is suffering from urolithiasis. How it could be explained in simple words?
Answer:
Urolithiasis is the condition of having calculi in the urinary tract (which also includes the kidneys), which may pass into urinary bladder.

Question (B).
Anitaji needs to micturate several times and feels very thirsty. This is an indication of change in permeability of certain part of nephron. Which is this part?
Answer:

1. Need to micturate several times (polyuria) and feeling very thirsty (polydipsia) is a symptom of diabetes insipidus (imbalance of fluids in the body).
2. ADH prevents diuresis and due to absence of ADH, large amount of dilute urine is excreted.
3. ADH stimulates reabsorption of water from last part of DCT and entire collecting duct by increasing the permeability of cells.
4. If the permeability of these cells changes, it will result in increase in urine volume (frequent micturition) and increase in the osmolarity of blood. An imbalance in volume and osmolarity of body fluids increases thirst.

[Note: Water is reabsorbed by osmosis in PCT, DCT and descending limb of loop of Henle)

Question (C).
Effective filtration pressure was calculated to be 20 mm Hg; where glomerular hydrostatic pressure was 70 mm of Hg. Which other pressure is affecting the filtration process? How much is it?
Answer:
The other pressure affecting the filtration process is osmotic pressure of blood and filtrate hydrostatic pressure. Commonly effective filtration pressure (EFP) is represented as;
EFP = Glomerular Hydrostatic pressure in glomerulus – (Osmotic pressure of blood + Filtrate Hydrostatic pressure)
If EFP = 20 mmHg and Glomerular Hydrostatic pressure = 70 mmHg
20 = 70 – (Osmotic pressure of blood + Filtrate hydrostatic pressure)
∴ (Osmotic pressure of blood + Filtrate hydrostatic pressure) = 70-20
Then (Osmotic pressure of blood + Filtrate Hydrostatic pressure) = 50 mmHg .

[Note: Given values are insufficient to calculate the exact osmotic pressure of blood and filtrate hydrostatic pressure. The sum of the two values can be calculated to be 50 mmHg ]

Question (D).
Name any one guanotelic organism.
Answer:
Spiders, scorpions and penguins are guanotelic organisms as they excrete guanine.

Question (E).
Why are kidneys called ‘retroperitoneal’?
Answer:
Kidneys are located in abdomen. Kidneys are not surrounded by peritoneum instead they are located posterior to it. Thus, kidneys are called retroperitoneal.

Question (F).
State role of liver in urea production.
Answer:

1. Ammonia formed during the breakdown of amino acids is converted into urea in the liver of ureotelic animals.
2. This conversion takes place by the help of the ornithine / urea cycle.
3. 3 ATP molecules are used to produce one molecule of urea using the ornithine/ urea cycle. Since, the liver contains carrier molecules and enzymes necessary for urea cycle, it plays a major role in urea production.

Question (G).
Why do we get bad breath after eating garlic or raw onion?
Answer:

1. Raw onion and garlic contain volatile sulphur-containing compounds.
2. Sulphur-containing compounds have a distinctive odour which remain in the mouth after consumption of onion and garlic.
3. Also, volatile compounds (like certain sulphur containing compounds) in foodstuffs are generally excreted through the lungs and may result in bad breath.

3. Answer the following questions

Question (A).
John has two options as treatment for his renal problem : Dialysis or kidney transplants. Which option should he choose? Why?
Answer:

1. If John has two options of dialysis and kidney transplant, readily available he must opt for kidney transplant.
2. A kidney transplant, if successful, can improve the quality of life of a patient and reduce the risk of death.
3. The patient would not have to endure frequent dialysis procedures. Repeated visits for dialysis takes time and may not allow the patient to perform normal activities or go to office regularly.
4. Dialysis is regarded as a holding measure until kidney transplant can be performed or a supportive measure in those for whom a transplant would be inappropriate. However, dialysis cannot replace all the functions of a normal kidney such as production of hormones like erythropoietin, calcitriol and renin. Hence, if John has an option of kidney transplant, he must opt for it.

Question (B).
Amphibian tadpole can afford to be ammonotelic. Justify.
Answer:

1. Tadpole (larval stage of life cycle of amphibian) is aquatic. They are ammonotelic as they excrete nitrogenous waste in the form of ammonia.
2. Ammonia is very toxic and requires large amount of water for its elimination.
3. It is readily soluble in water and diffuses across the body surface and into the surrounding water.
4. Also, the water lost during excretion can be made up through the surrounding water in ammonotelic organisms.

Hence, amphibian tadpole can afford to be ammonotelic.

Question (C).
Birds are uricotelic in nature. Give reason.
Answer:

1. Birds are capable of converting ammonia into uric acid by ‘inosinic acid pathway’ in their liver.
2. Uric acid is least toxic and hence, it can be retained in the body for some time.
3. It is least soluble water hence, negligible amount of water is required for its elimination.
4. This mode of excretion can also help reduce body weight (an adaptation for flight) and those animals which
   need to conserve more water follow uricotelism.

Hence, in order to conserve water as an adaptation for flight, birds are uricotelic in nature.

Question 4.
Write the explanation in your word

Question (A).
Nitya has been admitted to hospital after heavy blood loss. Till proper treatment could be given; how did Nitya’s body must have tackled the situation?
Answer:

1. Heavy blood loss is called haemorrhage. In case of haemorrhage or severe dehydration, the osmoreceptors stimulate Antidiuretic hormone (ADH) secretion.
2. ADH is important in regulating water balance through the kidneys. For detailed mechanism of reabsorption by ADH:

Regulating water reabsorption through ADH:
Hypothalamus in the midbrain has special receptors called osmoreceptors which can detect change in osmolarity (measure of total number of dissolved particles per liter of solution) of blood.

If osmolarity of blood increases due to water loss from the body (after eating namkeen or due to sweating), osmoreceptors trigger release of Antidiuretic hormone (ADH) from neurohypophysis (posterior pituitary). ADH stimulates reabsorption of water from last part of DCT and entire collecting duct by increasing the permeability of cells.

This leads to reduction in urine volume and decrease in osmolarity of blood.

Once the osmolarity of blood comes to normal, activity of osmoreceptor cells decreases leading to decrease in ADH secretion. This is called negative feedback.
In case of hemorrhage or severe dehydration too, osmoreceptors stimulate ADH secretion. ADH is important in regulating water balance through kidneys.

In absence of ADH, diuresis (dilution of urine) takes place and person tends to excrete large amount of dilute urine. This condition called as diabetes insipidus.

[Note: Hypothalamus is a part of forebrain]

Another regulatory mechanism that must have been activated is RAAS. For detailed mechanism of electrolyte reabsorption:

Electrolyte reabsorption through RAAS:
Another regulatory mechanism is RAAS (Renin Angiotensin Aldosterone System) by Juxta Glomerular Apparatus (JGA).

Whenever blood supply (due to change in blood pressure or blood volume) to afferent arteriole decreases (e.g. low BP/dehydration), JGA cells release Renin. Renin converts angiotensinogen secreted by hepatocytes in liver to Angiotensin I. ‘Angiotensin-converting enzyme’ further modifies Angiotensin I to Angiotensin II, the active form of hormone. It stimulates adrenal cortex to release another hormone called aldosterone that stimulates DCT and collecting ducts to reabsorb more Na⁺ and water, thereby increasing blood volume and pressure.

Question 5.
Complete the diagram / chart with correct labels / information. Write the conceptual details regarding it

Question (A).

Answer:

The composition of urine depends upon food and fluid consumed by an individual. There are two ways in which it the composition is regulated. They are as follows:

i. Regulating water reabsorption through ADH
ii. Electrolyte reabsorption though RAAS
iii. Atrial Natriuretic Peptide

i. Regulating water reabsorption through ADH:
Hypothalamus in the midbrain has special receptors called osmoreceptors which can detect change in osmolarity (measure of total number of dissolved particles per liter of solution) of blood.

If osmolarity of blood increases due to water loss from the body (after eating namkeen or due to sweating), osmoreceptors trigger release of Antidiuretic hormone (ADH) from neurohypophysis (posterior pituitary). ADH stimulates reabsorption of water from last part of DCT and entire collecting duct by increasing the permeability of cells.
This leads to reduction in urine volume and decrease in osmolarity of blood.

Once the osmolarity of blood comes to normal, activity of osmoreceptor cells decreases leading to decrease in ADH secretion. This is called negative feedback.
In case of hemorrhage or severe dehydration too, osmoreceptors stimulate ADH secretion. ADH is important in regulating water balance through kidneys.

In absence of ADH, diuresis (dilution of urine) takes place and person tends to excrete large amount of dilute urine. This condition called as diabetes insipidus.

[Note: Hypothalamus is a part of forebrain]

ii. Electrolyte reabsorption through RAAS:
Another regulatory mechanism is RAAS (Renin Angiotensin Aldosterone System) by Juxta Glomerular Apparatus (JGA).

Whenever blood supply (due to change in blood pressure or blood volume) to afferent arteriole decreases (e.g. low BP/dehydration), JGA cells release Renin. Renin converts angiotensinogen secreted by hepatocytes in liver to Angiotensin I. ‘Angiotensin converting enzyme’ further modifies Angiotensin I to Angiotensin II, the active form of hormone. It stimulates adrenal cortex to release another hormone called aldosterone that stimulates DCT and collecting ducts to reabsorb more Na⁺ and water, thereby increasing blood volume and pressure.

iii. Atrial natriuretic peptide (ANP): A large increase in blood volume and pressure stimulates atrial wall to produce atrial natriuretic peptide (ANP). ANP inhibits Na⁺ and Cl^– reabsorption from collecting ducts inhibits release of renin, reduces aldosterone and ADH release too. This leads to a condition called Natriuresis (increased excretion of Na⁺ in urine) and diuresis.

Question (B).

Answer:

1. Nephrons are structural and functional units of kidney.
2. Each nephron consists of a 4 – 6 cm long, thin-walled tube called the renal tubule and a bunch of capillaries known as the glomerulus.
3. The wall of the renal tubule is made up of a single layer of epithelial cells.
4. Its proximal end is wide, blind, cup-like and is called as Bowman’s capsule, whereas the distal end is open.
5. The nephron is divisible into Ilowman’s capsule, neck, proximal convoluted tubule (PCT), Loop of Henle (LoH), distal convoluted tubule (DCT) and collecting tubule (CT).
6. The glomerulus is present in the cup-like cavity of Bowman’s capsule and both are collectively known as renal corpuscle or Malpighian body.

Question (C)

Answer:
Nephron is the structural and functional unit of kidney.
Structure of nephron:
A nephron (uriniferous tubule) is a thin walled, coiled duct, lined by a single layer of epithelial cells. Each nephron is divided into two main parts:

i. Malpighian body
ii. Renal tubule

i. Malpighian body: Each Malpighian body is about 200pm in diameter and consists of a Bowman’s capsule and glomerulus.

a. Glomerulus:
Glomerulus is a bunch of fine blood capillaries located in the cavity of Bowman’s capsule.
A small terminal branch of the renal artery, called as afferent arteriole enters the cup cavity (Bowman capsule) and undergoes extensive fine branching to form network of several capillaries. This bunch is called as glomerulus.
The capillary wall is fenestrated (perforated).

All capillaries reunite and form an efferent arteriole that leaves the cup cavity.
The diameter of the afferent arteriole is greater than the efferent arteriole. This creates a high hydrostatic pressure essential for ultrafiltration, in the glomerulus.

b. Bowman’s capsule:
It is a cup-like structure having double walls composed of squamous epithelium.
The outer wall is called as parietal wall and the inner wall is called as visceral wall.
The parietal wall is thin consisting of simple squamous epithelium.
There is a space called as capsular space / urinary space in between two walls.
Visceral wall consists of special type of squamous cells called podocytes having a foot-like pedicel. These podocytes are in close contact with the walls of capillaries of glomerulus.
There are small slits called as filtration slits in between adjacent podocytes.

ii. Renal tubule:

a. Neck:
The Bowman’s capsule continues into the neck. The wall of neck is made up of ciliated epithelium. The lumen of the neck is called the urinary pole. The neck leads to proximal convoluted tubule.

b. Proximal Convoluted Tubule :
This is highly coiled part of nephron which is lined by cuboidal cells with brush border (microvilli) and surrounded by peritubular capillaries. Selective reabsorption occurs in PCT. Due to convolutions (coiling), filtrate flows slowly and remains in the PCT for longer duration, ensuring that maximum amount of useful molecules are reabsorbed.

c. Loop of Henle :
This is ‘U’ shaped tube consisting of descending and ascending limb.
The descending limb is thin walled and permeable to water and lined with simple squamous epithelium.
The ascending limb is thick walled and impermeable to water and is lined with simple cuboidal epithelium.
The LoH is surrounded by capillaries called vasa recta.
Its function is to operate counter current system – a mechanism for osmoregulation.
The ascending limb of Henle’s loop leads to DCT.

d. Distal convoluted tubule:
This is another coiled part of the nephron.
Its wall consists of simple cuboidal epithelium.
DCT performs tubular secretion / augmentation / active secretion in which, wastes are taken up from surrounding capillaries and secreted into passing urine.
DCT helps in water reabsorption and regulation of pH of body fluids.

e. Collecting tubule:
This is a short, straight part of the DCT which reabsorbs water and secretes protons.
The collecting tubule opens into the collecting duct.

Question (D).

Answer:

The composition of urine depends upon food and fluid consumed by an individual. There are two ways in which it the composition is regulated. They are as follows:

i. Regulating water reabsorption through ADH
ii. Electrolyte reabsorption though RAAS
iii. Atrial Natriuretic Peptide

i. Regulating water reabsorption through ADH:
Hypothalamus in the midbrain has special receptors called osmoreceptors which can detect change in osmolarity (measure of total number of dissolved particles per liter of solution) of blood.

If osmolarity of blood increases due to water loss from the body (after eating namkeen or due to sweating), osmoreceptors trigger release of Antidiuretic hormone (ADH) from neurohypophysis (posterior pituitary). ADH stimulates reabsorption of water from last part of DCT and entire collecting duct by increasing the permeability of cells.
This leads to reduction in urine volume and decrease in osmolarity of blood.

Once the osmolarity of blood comes to normal, activity of osmoreceptor cells decreases leading to decrease in ADH secretion. This is called negative feedback.
In case of hemorrhage or severe dehydration too, osmoreceptors stimulate ADH secretion. ADH is important in regulating water balance through kidneys.

In absence of ADH, diuresis (dilution of urine) takes place and person tends to excrete large amount of dilute urine. This condition called as diabetes insipidus.

[Note: Hypothalamus is a part of forebrain]

ii. Electrolyte reabsorption through RAAS:
Another regulatory mechanism is RAAS (Renin Angiotensin Aldosterone System) by Juxta Glomerular Apparatus (JGA).
Whenever blood supply (due to change in blood pressure or blood volume) to afferent arteriole decreases (e.g. low BP/dehydration), JGA cells release Renin. Renin converts angiotensinogen secreted by hepatocytes in liver to Angiotensin I. ‘Angiotensin-converting enzyme’ further modifies Angiotensin I to Angiotensin II, the active form of hormone. It stimulates adrenal cortex to release another hormone called aldosterone that stimulates DCT and collecting ducts to reabsorb more Na⁺ and water, thereby increasing blood volume and pressure.

iii. Atrial natriuretic peptide (ANP): A large increase in blood volume and pressure stimulates atrial wall to produce atrial natriuretic peptide (ANP). ANP inhibits Na⁺ and Cl^– reabsorption from collecting ducts inhibits release of renin, reduces aldosterone and ADH release too. This leads to a condition called Natriuresis (increased excretion of Na⁺ in urine) and diuresis.

Question (E).

Answer:

1. When renal function of a person falls below 5 – 7 %, accumulation of harmful substances in blood begins. In such a condition the person has to go for artificial means of filtration of blood i.e. haemodialysis.
2. In haemodialysis, a dialysis machine is used to filter blood. The blood is filtered outside the body using a dialysis unit.
3. In this procedure, the patients’ blood is removed; generally from the radial artery and passed through a cellophane tube that acts as a semipermeable membrane.
4. The tube is immersed in a fluid called dialysate which is isosmotic to normal blood plasma. Hence, only excess salts if present in plasma pass through the cellophane tube into the dialysate.
5. Waste substances being absent in the dialysate, move from blood into the dialyzing fluid.
6. Filtered blood is returned to vein.
7. In this process it is essential that anticoagulant like heparin is added to the blood while it passing through the tube and before resending it into the circulation, adequate amount of anti-heparin is mixed.
8. Also, the blood has to move slowly through the tube and hence the process is slow.

Question 6.
Prove that mammalian urine contains urea.
Answer:

1. Urea is a nitrogenous waste formed by breakdown of protein (deamination of amino acids).
2. During this process, amino groups are removed from the amino acids present in the proteins and converted to highly toxic ammonia. The ammonia is finally converted to area through ornithine cycle. Thus, the urea formed is passed to kidneys and excreted out of the body through urine.
3. Reabsorption of urea (proximal tubule, collecting ducts) and active secretion of urea (Henle loop) leads to a urea circulation (urea recycling) between the lumen of the nephron and renal medulla, which is an important element of the renal urine concentration.
4. About 54 g of urea is filtered per day in the glomerular capsule, of which approximately 30 g is excreted in the urine and 24 g is reabsorbed into blood (assuming GFR is 180 litres/day).
5. Urinalysis can help detect the amount of urea in urine (Urine urea nitrogen test, urease test, etc.).

Practical / Project :

Visit to a nearby hospital or pathological laboratory and collect detailed information about different blood and urine tests.
Answer:
Testing the urine is known as urinalysis. It generally has three parts:

1. Visual examination: Check sample colour and clearness.
2. Dipstick examination: Checks for abnormal amounts of glucose, protein, etc.
3. Microscopic examination: Check for presence of RBCs. WBCs, bacteria, crystals, etc.
4. Apart from routine urine examination, specific tests may also be done. They are as follows:
   • BUN (Blood Urea Nitrogen) Test: It measures the amount of nitrogen in blood and evaluates kidney function.
   • Urease Test/ Urea Nitrogen Test: It is done to check the amount of urea in urine sample.
   • Urine albumin to creatine ratio (UACR) test: Estimates the amount of albumin in urine.

[Students are expected to collect more information and perform the given activity on their own]

12th Biology Digest Chapter 15 Excretion and Osmoregulation Intext Questions and Answers

Can you recall? (Textbook Page No. 174)

Question 1.
Why are various waste products produced in the body of an organism?
Answer:
Metabolism produces a variety of by-products, some of which need to be eliminated. Such by-produçts are called metabolic waste products.

Question 2.
How are these waste eliminated?
Answer:
Depending on the type of waste product, they are eliminated through various organs of the body:

The various excretory products produced by the human body are as follows:

1. Fluids such as water; gaseous wastes like CO₂ nitrogenous wastes like ammonia, urea and uric acid, creatinine; minerals; salts of sodium, potassium. calcium, etc. if present in body in excess are excreted through urine, faeces and sweat.
2. Pigments formed due to breakdown of haemoglobin like bilirubin (excreted through faeces) and urochrome (eliminated through urine).
3. The pigments present in consumed foodstuffs like beet root or excess of vitamins, hormones and drugs.
4. Volatile substances present in spices (eliminated through lungs).

Have you ever observed? (Textbook Page No. 174)

Question 1.
When does urine appear deeply coloured?
Answer:
Urine can appear deeply coloured due to various reasons:

• Severe dehydration resulting in production of concentrated urine.
• Consumption of foodstuff like beet root, which contain coloured pigments.
• Some medications can also cause the urine to appear deeply coloured.

Think about it. (Textbook Page No. 174)

Question 1.
Do organisms differ in type of metabolic wastes they produce?
Answer:
Yes, organisms differ in the type of metabolic wastes they produce. Some organisms excrete ammonia while some excrete urea or uric acid as metabolic wastes.

Question 2.
Do environment or evolution have any effect on type of waste produced by an organism?
Answer:

• The theory of evolution proposes that life started in an aquatic environment.
• Aquatic organisms are generally ammonotelic. It is believed that the urea cycle evolved to adapt to a changing environment when terrestrial life forms evolved.
• Arid conditions probably led to the evolution of the uric acid pathway as a means of conserving water.
• However, the correlation between evolution and type of waste production is uncertain.

Question 3.
How do thermoregulation and food habits affect saste production?
Answer:

1. To generate heat. endotherms convert the food that they eat into energy through a process called metabolism. Hence, they consume more tì.od in order to meet their energy requirements.
2. Also, carnivorous diet contains more proteins than herbivores.
3. Consumption of high protein or more food containing proteins can result in production of large amount of nitrogenous waste
4. These animals would also require more energy to eliminate the high levels oF nitrogenous wastes which build up when animal protein is digested.

Use your brain power. (Textbookpage No. 175)

Question 1.
Why ammonia is highly toxic?
Answer:

1. Ammonia is basic in nature and its retention in the body would disturb the pH of the body.
2. An increase in pH would disturb all enzyme catalysed reactions in the body and also make the plasma membrane unstable.

Hence, ammonia is highly toxic to the body.

Find out. (Textbook page No. 175)

You will study about a type of arthritis called gouty arthritis caused due to accumulation of uric acid in joints. Where does uric acid come from in case of ureotelic human beings?
Answer:

1. Uric acid produced as a waste product during the normal breakdown of nucleic acids (purines) and certain naturally occurring substances found in foods such as mushrooms. Mackerel, dried beans. etc.
2. This uric acid is generally excreted out along with urine.
3. If uric acid is produced in excess or not excreted, it accumulates in joints causing gouty arthritis.

Think about it. (Textbook Page No.175)

Endotherms consume more food in order to meet energy requirements. Also, carnivorous diet contains more proteins than herbivorous. Does it affect excretion of nitrogenous waste?
Answer:

1. To generate heat, endotherms convert the food that they eat into energy through a process called metabolism. Hence, they consume more food in order to meet their energy requirements.
2. Also, carnivorous diet contains more proteins than herbivores.
3. Consumption of high protein or more food containing proteins can result in production of large amount of nitrogenous waste.
4. These animals would also require more energy to eliminate the high levels of nitrogenous wastes which build up when animal protein is digested.

Observe and Discuss (Textbook Page No. 176)

Question 1.
These are blood reports of patients undergoing investigations for kidney function. What is creatinine? What is your observation and opinion about the findings? Why is it used as an index of kidney function?


Answer:
i. Creatinine:

• Plasma creatinine is produced from catabolism of creatinine phosphate during skeletal muscle contraction.
• It provides a ready source of high energy phosphate.

ii. Observations and Opinion:
Report A indicates a value of creatinine which is higher than the normal range. This would indicate impaired kidney function.
Report B indicates high fasting blood sugar and detection of sugar in the blood is known as glucosuria. High fasting blood sugar (>126 mg/dL) is usually indicative of diabetes.

iii. Creatinine used as an index of kidney function:

• Normally blood creatinine levels remain steady because the rate of production matches his excretion in urine.
• Hence, plasma creatinine is used as an index of kidney function and its level above normal is an indication of poor renal function.

Think about it. (Textbook Page No. 176)

Question 1.
During summer, we tend to produce less urine, is it so?
Answer:

1. Generally, excess water containing wastes is lost from the body in the time of urine. sweat and faeces.
2. During summer when the surrounding temperature is high. we also lose a considerable amount of water in the form of sweat.

Thus, the kidneys retain water for maintaining the concentration of body fluids and reduce the amount of water lost through urine.

Question 2.
Marine birds like Albatross spend their life on the sea. That means the water the, drink is salty. how do they manage osnioregulation then?
Answer:

1. Marine birds like Albatross have special glands called salt glands near their nostrils.
2. These glands are capable of secreting salts by active transport and help to manage osmotic balance,

[Note: The salt glands in Albatross are located in or on the skull in the area of eyes.]

Question 3.
like ectothermic and endothermic animals, do organisms differ in the way they maintain salt balance?
Answer:
Yes, organisms differ in the way they maintain salt balance.

1. Animals can be either isosmotic to the surrounding (osmoconformers or control the internal environment independent of external environment (osmoregulators).
2. Marine organisms are mostly osmoconfòrmers because their body fluids and external environment are isosmotic in nature while fresh water forms and terrestrial organisms are osmoregulators,
3. Generally, most organisms can tolerate only a narrow range of salt concentrations. Such organism are known as stenohaline organisms.
4. Organisms which are capable of handling a wide change in salinity are called euryhaline organisms.e.g. Barnacles, clams etc.

Find out. (Textbook Page No 176)

Question 1.
How do freshwater fishes and marine fishes carry out osmoregulation?
Answer:
Osmoregulation is the process of maintaining an internal concentration of salt and water in the body of fishes.

i. Freshwater fishes:
The salt concentration inside the body of freshwater fishes is higher than their surrounding water. Due to this, water enters the body due to osmosis.
If the flow of water into the body is not regulated. fishes would swell and get bigger.
To compensate this, the kidneys produce a large amount of urine,
Excretion of large amounts of urine regulates the level of water in the body hut results in the loss of salts.
Thus, in order to maintain a sufficient salt level, special cells in the gills (chloride cells) take tip ions from
the water, which are then directly transported into the blood.

ii. Marine fishes:
Since the salt content in blood of marine fishes is much lower than that of seawater, they constantly tend to lose water and build up salt.
To replace the water loss, they continually need to drink seawater.
Since their small kidney can only excrete a relatively small amount of urine, salt is additionally excreted through gills, where chloride cells work in reverse as in freshwater fishes.

Make a table. (Textbook Page No. 178)

Question 1.
The details of modes of excretion of nitrogenous wastes.
Answer:
The three main modes of excretion in animals are as follows:

i. Ammonotelism
ii. Ureotelism
iii. Uricotelism

i. Ammonotelism:

1. Elimination of nitrogenous wastes in the form of ammonia is called as ammonotelism.
2. Ammonia is basic in nature and hence it can disturb the pH of the body, if not eliminated immediately.
3. Any change in pH would disturb all enzyme catalyzed reactions in the body and would also make the plasma membrane unstable.
4. Ammonia is readily soluble in water and needs large quantity of water to dilute and reduce its toxicity.
5. This is however an energy saving mechanism of excretion and hence all animals that have plenty of water available for dilution of ammonia, excrete nitrogenous wastes in the form of ammonia.
6. Animals that follow this mode of excretion are known as ammonotelic animals.
7. 1 gm ammonia needs about 300 – 500 ml of water for elimination.
8. Ammonotelic animals excrete ammonia through general body surface (skin), gills and kidneys.
   e.g. Ammonotelism is found in aquatic invertebrates, bony fishes, and aquatic / larval amphibians. Animals without excretory system (Protozoa) are also ammonotelic.

ii. Ureotelism:

1. Elimination of nitrogenous wastes in the form of urea is called as ureotelism.
2. Urea is comparatively less toxic and less water-soluble than ammonia. Hence, it can be concentrated to some extent in body.
3. The body requires less water for elimination.
4. Since it is less toxic and less water soluble, ureotelism is suitable for animals that need to conserve water to some extent. Hence, ureotelism is common in terrestrial animals, as they have to conserve water.
5. It takes about 50 ml H₂O for removal of 1 gm NH₂ in form of urea.
6. Ureotelic animals generally convert ammonia to urea in the liver by operating ornithine / urea cycle in which 3 ATP molecules are used to produce one molecule of urea.
   e.g. Mammals, cartilaginous fishes (sharks and rays), many aquatic reptiles, most of the adult amphibians, etc. are ureotelic.

iii. Uricotelism:

1. Elimination of nitrogenous wastes in the form of uric acid is called as uricotelism.
2. Uric acid is least toxic and hence, it can be retained in the body for some time in concentrated form.
3. It is least soluble in water. Hence there is minimum (about 5 – 10 ml for 1 gm) or no need of water for its elimination.
4. Those animals which need to conserve more water follow uricotelism. However, these animals need to spend more energy.
5. Ammonia is converted into uric acid by ‘inosinic acid pathway’ in the liver of birds, e.g. Birds, some insects, many reptiles, land snails, are uricotelic.

No.	Ureotelism	Uricotelism
i.	It is the elimination of nitrogenous waste in the form of urea.	It is the elimination of nitrogenous waste in the form of uric acid.
ii.	Excretion of urea requires less (moderate ) amount of water.	Excretion of uric acid requires negligible amount of water.
iii.	Removal of 1 gm of urea requires 50 ml of water.	Removal of 1 gm of uric acid requires 5 – 10 ml of
iv.	rea is less toxic.	Uric acid is least toxic.
e.g.	It is generally seen in terrestrial animals. Mammals, cartilaginous fishes (sharks and rays), many aquatic reptiles, most adult amphibians, etc.	It is seen in birds, some insects, many reptiles, land snails, etc.

No.	Ammonotelism	Uricotelism
i.	It is the elimination of nitrogenous waste in the form of ammonia.	It is the elimination of nitrogenous waste in the form of uric acid.
ii.	Excretion of ammonia requires plenty of water.	Excretion of uric acid requires negligible amount of water.
iii.	Removal of 1 gm of ammonia requires 300 – 500 ml of water.	Removal of 1 gm of uric acid requires 10ml of water.
iv.	Ammonia is very toxic.	Uric acid is less toxic.
e.g.	It is found in aquatic invertebrates, bony fishes and aquatic/ larval amphibians, etc.	It is seen in birds, some insects, many reptiles, land snails, etc.

[Students can Refer these and make a chart on their own.]

Use your brain power. (Textbook Page No. 178)

Question 1.
Creatinine is considered as index of kidney function. Give reason.
Answer:

1. Plasma creatinine is produced from catabolism of creatinine phosphate during skeletal muscle contraction.
2. It provides a ready source of high energy phosphate.
3. Normally blood creatinine levels remain steady because the rate of production matches its excretion in urine.
4. Hence, plasma creatinine is used as an index of kidney function and its level above normal is an indication of poor renal function.

[Note: Plasma creatinine is a waste product produced by muscles from the breakdown of a compound called ‘creatine phosphate ’.]

Make a table. (Textbook Page No. 178)

Question 1.
The excretory organs found in various animal phyla.
Answer:

Sr. No.	Animal Phyla	Excretory organs
i.	Porifera	Lack excretory organ instead rely on water transport system/ Canal system
ii.	Coelenterata	Lack specialised excretory organs. Excretion takes place through simple diffusion or through the mouth.
iii.	Ctenophora	Lack specialised excretory organs
iv.	Platyhelminthes	Protonephridia or Flame cells
v.	Aschelminthes	Excretory tube and pore
vi.	Annelida	Nephridia
vii.	Arthropoda	Malpighian tubules
viii.	Mollusca	Organ of Bojanus
ix.	Echinodermata	Lack specialized excretory organs, waste materials directly diffuse into water or are excreted through tube feet
x.	Hemichordata	Proboscis gland
xi.	Chordata	Kidney

Observe and complete. (Textbook Page No. 178)

Question 1.
Label the diagram and complete following paragraphs.

i. Kidney: A pair of ____ shaped kidneys are present on either side of the ____ from 12^th thoracic to 3^rd lumbar vertebra. Kidneys are present behind ___. Hence are called retroperitoneal. Dimensions of
each kidney are 10 × ____ × ____ cms. Average weight is ___ g in males and 135 g in ____. Outer surface is ___ and inner is concave. Notch on the inner concave surface is called ___. Renal artery enters and renal vein as well as ureter leave the kidney through hilus. Each kidney has almost 1 million functional units called ___.

ii. Ureters: A pair of ureters arise from ___ of each kidney. Each ureter is a long muscular tube 25 – 30 cm in length. Ureters open into ___ by separate openings, which are not guarded by valves. They pass obliquely through the wall of urinary bladder. This helps in prevention of ___ of urine due to compression of ureters while bladder is filled.

iii. Urinary bladder: It is a median ___ sac. A hollow muscular organ, the bladder is situated in pelvic cavity posterior to pubic symphysis. At the base of the ___ there is a small inverted triangular area called trigone. At the apex of this triangle is opening of urethra. At the two points of the base of the triangle are openings of ureters. Urinary bladder is covered externally by peritoneum. Inner to peritoneum is muscular layer. It is formed by detrusor muscles which consist of three layers of smooth muscles. Longitudinal – circular – longitudinal respectively. Innermost layer is made up of transitional ___. It helps bladder to stretch.

iv. Urethra: It is a ___ structure arising from urinary bladder and opening to the exterior of the body.
There are ___ urethral sphincters between urinary bladder and urethra.
a. Internal sphincter: Made up of ___ muscles, involuntary in nature.
b. External sphincter: Made up of ___ muscles, voluntary in nature.
Answer:

i. Kidney: A pair of bean shaped kidneys are present on either side of the backbone from 12^th thoracic to 3^rd lumbar vertebra. Kidneys are present behind peritoneum. Hence are called retroperitoneal. Dimensions of each kidney are 10 × 5 × 4 cms. Average weight is 150 g in males and 135 g in females. Outer surface is convex and inner is concave. Notch on the inner concave surface is called hilum. Renal artery enters and renal vein as well as ureter leave the kidney through hilus. Each kidney has almost 1 million functional units called nephron.

ii. Ureters: A pair of ureters arise from hilum of each kidney. Each ureter is a long muscular tube 25 – 30 cm in length. Ureters open into urinary bladder by separate openings, which are not guarded by valves. They pass obliquely through the wall of urinary bladder. This helps in prevention of backward flow of urine due to compression of ureters while bladder is filled.

iii. Urinary bladder: It is a median pear-shaped sac. A hollow muscular organ, the bladder is situated in pelvic cavity posterior to pubic symphysis. At the base of the urinary bladder there is a small inverted triangular area called trigone. At the apex of this triangle is opening of urethra. At the two points of the base of the triangle are openings of ureters. Urinary bladder is covered externally by peritoneum. Inner to peritoneum is muscular layer. It is formed by detrusor muscles which consist of three layers of smooth muscles. Longitudinal – circular – longitudinal respectively. Innermost layer is made up of transitional epithelial tissue. It helps bladder to stretch.

iv. Urethra: It is a fibromuscular tube-like structure arising from urinary bladder and opening to the exterior of the body. There are two urethral sphincters between urinary bladder and urethra.
a. Internal sphincter: Made up of detrusor muscles, involuntary in nature.
b. External sphincter: Made up of striated muscles, voluntary in nature.
If this valve is not functioning properly during inflammation of bladder, it can lead to kidney infection.

Internet is my friend. (Textbook Page No. 179)

Question 1.
Find out what is floating kidney.
Answer:

1. Floating kidney or nephroptosis, is an inferior displacement or dropping of the kidney.
2. This condition occurs when the kidney slips from its normal position because it is not held securely in place by the adjacent organs or its fat covering.
3. It generally develops in extremely thin people whose adipose capsule or renal fascia is deficient.
4. It may result in twisting of the ureter and cause blockage of urine flow. The resulting backup of urine would put pressure on the kidney and damage the tissues.
5. Twisting of the ureter may also cause pain and discomfort.
6. This condition is more common in females than males and happens commonly among one in four people.
7. Weakening of the fibrous bands that hold the kidney in place can predispose to floating kidney.

Can you recall? (Textbook Page No. 179)

Question 1.
Observe the figure carefully and label various regions of L.S. of kidney.

Answer:

Can you tell? (Textbook Page No.182)

Question 1.
Why are kidneys called ‘retroperitoneal’?
Answer:
Kidneys are located in abdomen. Kidneys are not surrounded by peritoneum instead they are located posterior to it. Thus, kidneys are called retroperitoneal.

Question 2.
Why urinary tract infections are more common in females than males?
Answer:

• The urethra in women (4 cm) is much shorter than that of males (20 cm).
• This allows easy passage of bacteria into the urinary bladder.

Hence, urinary tract infections are more common in females than males.

Question 3.
What is nephron? Which are its main parts? Why are they important?
Answer:
Nephron is the structural and functional unit of kidney.
Structure of nephron:
A nephron (uriniferous tubule) is a thin walled, coiled duct, lined by a single layer of epithelial cells. Each nephron is divided into two main parts:

i. Malpighian body
ii. Renal tubule

i. Malpighian body: Each Malpighian body is about 200pm in diameter and consists of a Bowman’s capsule and glomerulus.

a. Glomerulus:
Glomerulus is a bunch of fine blood capillaries located in the cavity of Bowman’s capsule.
A small terminal branch of the renal artery, called as afferent arteriole enters the cup cavity (Bowman capsule) and undergoes extensive fine branching to form network of several capillaries. This bunch is called as glomerulus.
The capillary wall is fenestrated (perforated).

All capillaries reunite and form an efferent arteriole that leaves the cup cavity.
The diameter of the afferent arteriole is greater than the efferent arteriole. This creates a high hydrostatic pressure essential for ultrafiltration, in the glomerulus.

b. Bowman’s capsule:
It is a cup-like structure having double walls composed of squamous epithelium.
The outer wall is called as parietal wall and the inner wall is called as visceral wall.
The parietal wall is thin consisting of simple squamous epithelium.
There is a space called as capsular space / urinary space in between two walls.
Visceral wall consists of special type of squamous cells called podocytes having a foot-like pedicel. These podocytes are in close contact with the walls of capillaries of glomerulus.
There are small slits called as filtration slits in between adjacent podocytes.

ii. Renal tubule:

a. Neck:
The Bowman’s capsule continues into the neck. The wall of neck is made up of ciliated epithelium. The lumen of the neck is called the urinary pole. The neck leads to proximal convoluted tubule.

b. Proximal Convoluted Tubule :
This is highly coiled part of nephron which is lined by cuboidal cells with brush border (microvilli) and surrounded by peritubular capillaries. Selective reabsorption occurs in PCT. Due to convolutions (coiling), filtrate flows slowly and remains in the PCT for longer duration, ensuring that maximum amount of useful molecules are reabsorbed.

c. Loop of Henle :
This is ‘U’ shaped tube consisting of descending and ascending limb.
The descending limb is thin walled and permeable to water and lined with simple squamous epithelium.
The ascending limb is thick walled and impermeable to water and is lined with simple cuboidal epithelium.
The LoH is surrounded by capillaries called vasa recta.
Its function is to operate counter current system – a mechanism for osmoregulation.
The ascending limb of Henle’s loop leads to DCT.

d. Distal convoluted tubule:
This is another coiled part of the nephron.
Its wall consists of simple cuboidal epithelium.
DCT performs tubular secretion / augmentation / active secretion in which, wastes are taken up from surrounding capillaries and secreted into passing urine.
DCT helps in water reabsorption and regulation of pH of body fluids.

e. Collecting tubule:
This is a short, straight part of the DCT which reabsorbs water and secretes protons.
The collecting tubule opens into the collecting duct.

Think about ¡t. (Textbook Page No. 182)

Question 1.
How much blood ¡s supplied to kidney?
Answer:
Around 600 ml of blood passes through each kidney per minute.

Do this. (Textbook Page No. 183)

Question 1.
Check blood reports of patients and comment about possibility of glucosuria.
Answer:
Glucosuria is the presence of glucose sugar in urine. High glucose in urine is usually indicative of diabetes mellitus.

Condition	Glucose range in urine
Normal	0 to 15 mg/dL (0 to 0.8 mmol/L)
Prediabetes	100 to 125 mg/dL (5.6 to 6.9 mmol/L)
Diabetes	126 mg/dL (7 mmol/L)

[Students can get access to sample reports on the internet and refer the above table to comment on blood reports of patients on their own.]

Use your brain power. (Textbook Page No. 185)

Question 1.
In which regions of nephron the filtrate will he isotonic to blood?
Answer:
Filtrate leasing the proximal convoluted tubule (PCT) is isotonic to the blood plasma.

Can you tell? (Textbook Page No. 185)

Question 1.
Explain the process of urine formation in details.
Answer:
Process of urine formation is completed in three steps, namely;

i. Ultrafiltration/ Glomerular filtration,
ii. Selective reabsorption,
iii. Tubular secretion / Augmentation

i. Ultrafiltration / Glomerular filtration :
Diameter of afferent arteriole is greater than the efferent arteriole. The diameter of capillaries is still smaller than both arterioles. Due to the difference in diameter, blood flows with greater pressure through the glomerulus. This is called as glomerular hydrostatic pressure (GHP) and normally, it is about 55 mmHg. GIIP is opposed by osmotic pressure of blood (normally, about 30 mm Hg) and capsular pressure (normally, about 15 mm Hg).

Hence net / effective filtration pressure (EFP) is 10 mm Hg.
EFP = Hydrostatic pressure in glomerulus – (Osmotic pressure of blood + Filtrate Hydrostatic pressure)
= 55 – (30 + 15)
= 10 mm Hg

Under the effect of high pressure, the thin walls of the capillary become permeable to major components of blood (except blood cells and macromolecules like protein).
Thus, plasma except proteins oozes out through wall of capillaries.
About 600 ml blood passes through each kidney per minute.

The blood (plasma) flowing through kidney (glomeruli) is filtered as glomerular filtrate, at a rate of 125 ml / min. (180 L/d).
Glomerular filtrate / deproteinized plasma / primary urine is alkaline, contains urea, amino acids, glucose, pigments, and inorganic ions.
Glomerular filtrate passes through filtration slits into capsular space and then reaches the proximal convoluted tubule.

ii. Selective reabsorption :
Selective reabsorption occurs in proximal convoluted tubule (PCT). It is highly coiled so that glomerular filtrate passes through it very slowly. Columnar cells of PCT are provided with microvilli due to which absorptive area increases enormously.
This makes the process of reabsorption very effective.
These cells perform active (ATP mediated) and passive (simple diffusion) reabsorption.

Substances with considerable importance (high threshold) like – glucose, amino acids, vitamin C, Ca⁺⁺, K⁺, Na⁺, Cl^– are absorbed actively, against the concentration gradient. Low threshold substances like water, sulphates, nitrates, etc., are absorbed passively.
In this way, about 99% of glomerular filtrate is reabsorbed in PCT and DCT.

iii. Tubular secretion / Augmentation :
Finally filtrate reaches the distal convoluted tubule via loop of Henle. Peritubular capillaries surround DCT. Cells of distal convoluted tubule and collecting tubule actively absorb the wastes like creatinine and ions like K⁺, H⁺ from peritubular capillaries and secrete them into the lumen of DCT and CT, thereby augmenting the concentration of urine and changing its pH from alkaline to acidic.
Secretion of H⁺ ions in DCT and CT is an important homeostatic mechanism for pH regulation of blood. Tubular secretion is the only process of excretion in marine bony fishes and desert amphibians.

Question 2.
How does counter current mechanism help concentration of urine?
Answer:
Under the conditions like low water intake or high water loss due to sweating, humans can produce concentrated urine. This urine can be concentrated around four times i.e. 1200 mOsm/L than the blood (300 mOsm/L). Hence, a mechanism called countercurrent mechanism is operated in the human kidneys. The countercurrent mechanism operating in the Limbs of Henle’s loop of juxtamedullary nephrons and vasa recta is as follows:

1. It involves the passage of fluid from descending to ascending limb of Henle’s loop.
2. This mechanism is called countercurrent mechanism, since the flow of tubular fluid is in opposite direction through both limbs.
3. In case of the vasa recta, blood flows from ascending to descending parts of itself.
4. Wall of descending limb is thin and permeable to water, hence, water diffuses from tubular fluid into tissue fluid due to which, tubular fluid becomes concentrated.
5. The ascending limb is thick and impermeable to water. Its cells can reabsorb Na⁺ and Cl^– from tubular fluid and release into tissue fluid.
6. Due to this, tissue fluid around descending limb becomes concentrated. This makes more water to move out from descending limb into tissue fluid by osmosis.
7. Thus, as tubular fluid passes down through descending limb, its osmolarity (concentration) increases gradually due to water loss and on the other hand, progressively decreases due to Na⁺ and Cl^– secretion as it flows up through ascending limb.
8. Whenever retention of water is necessary, the pituitary secretes ADH. ADH makes the cells in the wall of collecting ducts permeable to water.
9. Due to this, water moves from tubular fluid into tissue fluid, making the urine concentrated.
10. Cells in the wall of deep medullar part of collecting ducts are permeable to urea. As concentrated urine flows through it, urea diffuses from urine into tissue fluid and from tissue fluid into the tubular fluid flowing through thin ascending limb of Henle’s loop.
11. This urea cannot pass out from tubular fluid while flowing through thick segment of ascending limb, DCT and cortical portion of collecting duct due to impermeability for it in these regions.
12. However, while flowing through collecting duct, water reabsorption is operated under the influence of ADII. Due to this, urea concentration increases in the tubular fluid and same urea again diffuses into tissue fluid in deep medullar region.
13.  Thus, same urea is transferred between segments of renal tubule and tissue fluid of inner medulla. This is called urea recycling; operated for more and more water reabsorption from tubular fluid and thereby excreting small volumes of concentrated urine.
14. Osmotic gradient is essential in the renal medulla for water reabsorption by counter current multiplier system.
15. This osmotic gradient is maintained by vasa recta by operating counter current exchange system.
16. Vasa recta also have descending and ascending limbs. Blood that enters the descending limb of the vasa recta has normal osmolarity of about 300 mOsm/L.
17.  As it flows down in the region of renal medulla where tissue fluid becomes increasingly concentrated, Na⁺, Cl^– and urea molecules diffuse from tissue fluid into blood and water diffuse from blood into tissue fluid.
18. Due to this, blood becomes more concentrated which now flows through ascending part of vasa recta. This part runs through such region of medulla where tissue fluid is less concentrated.
19. Due to this, Na⁺, Cl^– and urea molecules diffuse from blood to tissue fluid and water from tissue fluid to blood. This mechanism helps to maintain the osmotic gradient.
    

Try this. (Textbook Page No. 185)

Question 1.
Read the given urine report and prepare a note on composition of normal urine.

Answer:
The composition of normal urine is as follows:

1. A volume of 1 – 2 litres of urine in 24 hours is normal. This volume can however vary considerably as it depends on fluid intake, physical activity, temperature, etc.
2. The colour of normal urine is generally pale yellow due to urochrome (pigment produced by breakdown of bile). The colour of urine may vary slightly due to urochrome concentration and diet.
3. The appearance of urine is generally clear and transparent.
4. Any form of deposits (sediments/ crystals) is generally absent in normal urine.
5. The pH of normal urine is acidic and is generally around 6.0 (Range: 4.6 to 8.0). The pH varies considerably with the diet of a person.
6. The specific gravity of urine is an average of 1.02 ( Range : 1.001 to 1.035).
7. Albumin, sugar, bile salts bile pigments, ketone bodies and casts are absent in normal urine.
8. Occult blood is generally not seen in normal urine.

Think (Textbook Page No. 185)

Question 1.
What would happen if ADH secretion decreases due to any reason?
Answer:
In absence of ADH, diuresis (dilution of urine) takes place and person tends to excrete large amount of dilute urine. This condition called as diabetes insipidus. Frequent excretion of large amount of dilute urine may cause a person to feel thirsty.

Think and appreciate. (Textbook Page No. 185)

Question 1.
How do kidneys bring about homeostasis? Is there any role of neuro endocrine system in it?
Answer:
The composition of urine depends upon food and fluid consumed by an individual. There are two ways in which it the composition is regulated. They are as follows:

i. Regulating water reabsorption through ADH
ii. Electrolyte reabsorption though RAAS
iii. Atrial Natriuretic Peptide

i. Regulating water reabsorption through ADH:
Hypothalamus in the midbrain has special receptors called osmoreceptors which can detect change in osmolarity (measure of total number of dissolved particles per liter of solution) of blood.
If osmolarity of blood increases due to water loss from the body (after eating namkeen or due to sweating), osmoreceptors trigger release of Antidiuretic hormone (ADH) from neurohypophysis (posterior pituitary). ADH stimulates reabsorption of water from last part of DCT and entire collecting duct by increasing the permeability of cells.

This leads to reduction in urine volume and decrease in osmolarity of blood.
Once the osmolarity of blood comes to normal, activity of osmoreceptor cells decreases leading to decrease in ADH secretion. This is called negative feedback.
In case of hemorrhage or severe dehydration too, osmoreceptors stimulate ADH secretion. ADH is important in regulating water balance through kidneys.

In absence of ADH, diuresis (dilution of urine) takes place and person tends to excrete large amount of dilute urine. This condition called as diabetes insipidus.
[Note: Hypothalamus is a part of forebrain]

ii. Electrolyte reabsorption through RAAS:
Another regulatory mechanism is RAAS (Renin Angiotensin Aldosterone System) by Juxta Glomerular Apparatus (JGA).
Whenever blood supply (due to change in blood pressure or blood volume) to afferent arteriole decreases (e.g. low BP/dehydration), JGA cells release Renin. Renin converts angiotensinogen secreted by hepatocytes in liver to Angiotensin I. ‘Angiotensin converting enzyme’ further modifies Angiotensin I to Angiotensin II, the active form of hormone. It stimulates adrenal cortex to release another hormone called aldosterone that stimulates DCT and collecting ducts to reabsorb more Na and water, thereby increasing blood volume and pressure.

iii. Atrial natriuretic peptide (ANP): A large increase in blood volume and pressure stimulates atrial wall to produce atrial natriuretic peptide (ANP). ANP inhibits Na⁺ and Cl^– reabsorption from collecting ducts inhibits release of renin, reduces aldosterone and ADH release too. This leads to a condition called Natriuresis (increased excretion of Na⁺ in urine) and diuresis.

No. Both ADH and RAAS are essential for homeostasis.

1. Only ADH can lower blood Na⁺ concentration by way of water reabsorption in DCI and collecting duct. whereas RAAS stimulates Na⁺ reabsorption and maintains osmolarity of body fluid.
2. Action of ADH and RAAS leads to increase in blood volume and osmolarity.
3. For mechanism of Atrial natriuretic peptide:

Atrial natriuretic peptide (ANP): A large increase in blood volume and pressure stimulates atrial wall to produce atrial natriuretic peptide (ANP). ANP inhibits Na⁺ and Cl^– reabsorption from collecting ducts inhibits release of renin, reduces aldosterone and ADH release too. This leads to a condition called Natriuresis (increased excretion of Na⁺ in urine) and diuresis.

ADH is produced by the hypothalamus and is stored and released by the posterior pituitary or the neurohypophysis in response to appropriate trigger. Hence, there is a role of the neuroendocrine system in homeostasis.

Use your brain power. (Textbook Page No. 186)

Question 1.
Can we use this knowledge in treatment of high blood pressure? Why high BP medicines are many a times diuretics?
Answer:

1. Yes, the knowledge of homoeostasis is used in the treatment of high blood pressure.
2. Some commonly used theories for treatment of high blood pressure are as follows:
   • Angiotensin II receptor blockers (ARBs) are used as medications to treat high blood pressure. These medications block the action of angiotensin II by preventing angiotensin II from binding to angiotensin II receptors on the muscles surrounding blood vessels. As a result, blood vessels enlarge (dilate), and blood pressure is reduced.
   • Another method is the use of ‘Angiotensin converting enzyme’ ACE blockers. These inhibitors inhibit activity of ACE and therefore decrease the production of angiotensin II. As a result, these medications cause the blood vessels to enlarge or dilate, and this reduces blood pressure.
3. Vasodilation reduces arterial pressure. Reduced angiotensin II leads to natriuresis (increased excretion of Na⁺ in urine) and diuresis, thereby reducing blood pressure.
4. Too much salt can cause extra fluid to build up in the blood vessels, raising blood pressure. Diuretics are substances that slow renal absorption of water and thereby cause diuresis (elevated urine flow rate) which in turn reduces blood volume and blood pressure by flushing out salt and extra fluid. Hence, high BP medicines are many a times diuretics.

Can you tell? (Textbook page no. 186)
How do skin and lungs help in excretion?
OR
Can you tell? (Textbook page no. 187)
Explain role of lungs and skin in excretion.
Answer:
Yes, various organs other than the kidney participate in excretion. They are as follows:

i. Skin:

Skin acts as an accessory excretory organ. The skin of many organisms is thin and permeable. It helps in diffusion of waste products like ammonia.
Human skin however is thick and impermeable. It shows presence of two types of glands namely, sweat glands and sebaceous glands.

• Sweat glands are distributed all over the skin. They are abundant in the palm and facial regions.
  These simple, unbranched, coiled, tubular glands open on the surface of the skin through an opening called sweat pore. Sweat is primarily produced for thermoregulation but it also excretes substances like water, NaCl, lactic acid and urea.
• Sebaceous glands are present at the neck of hair follicles. They secrete oily substance called sebum.
  It forms a lubricating layer on skin making it softer. It protects skin from infection and injury.

ii. Lungs:

Lungs are the accessory excretory organs. They help in excretion of volatile substances like CO₂ and water vapour produced during cellular respiration. Along with CO₂, lungs also remove excess of H₂O in the form of vapours during expiration. They also excrete volatile substances present in spices and other food stuff.

Can you tell? (Textbook Page No. 187)

Question 1.
When does kidney produce renin? Where is it produced in kidney?
Answer:
Kidney produces renin whenever blood supply (due to change in blood pressure or blood volume) to afferent arteriole decreases (e.g. low BP/dehydration).
The juxtaglomerular Apparatus (JGA) cells secrete renin.

Question 2.
Explain how electrolyte balance of blood plasma maintained.
Answer:
The composition of urine depends upon food and fluid consumed by an individual. There are two ways in which it the composition is regulated. They are as follows:

i. Regulating water reabsorption through ADH
ii. Electrolyte reabsorption though RAAS
iii. Atrial Natriuretic Peptide

i. Regulating water reabsorption through ADH:
Hypothalamus in the midbrain has special receptors called osmoreceptors which can detect change in osmolarity (measure of total number of dissolved particles per liter of solution) of blood.

If osmolarity of blood increases due to water loss from the body (after eating namkeen or due to sweating), osmoreceptors trigger release of Antidiuretic hormone (ADH) from neurohypophysis (posterior pituitary). ADH stimulates reabsorption of water from last part of DCT and entire collecting duct by increasing the permeability of cells.

This leads to reduction in urine volume and decrease in osmolarity of blood.
Once the osmolarity of blood comes to normal, activity of osmoreceptor cells decreases leading to decrease in ADH secretion. This is called negative feedback.
In case of hemorrhage or severe dehydration too, osmoreceptors stimulate ADH secretion. ADH is important in regulating water balance through kidneys.

In absence of ADH, diuresis (dilution of urine) takes place and person tends to excrete large amount of dilute urine. This condition called as diabetes insipidus.
[Note: Hypothalamus is a part of forebrain]

ii. Electrolyte reabsorption through RAAS:
Another regulatory mechanism is RAAS (Renin Angiotensin Aldosterone System) by Juxta Glomerular Apparatus (JGA).

Whenever blood supply (due to change in blood pressure or blood volume) to afferent arteriole decreases (e.g. low BP/dehydration), JGA cells release Renin. Renin converts angiotensinogen secreted by hepatocytes in liver to Angiotensin I. ‘Angiotensin converting enzyme’ further modifies Angiotensin I to Angiotensin II, the active form of hormone. It stimulates adrenal cortex to release another hormone called aldosterone that stimulates DCT and collecting ducts to reabsorb more Na and water, thereby increasing blood volume and pressure.

iii. Atrial natriuretic peptide (ANP): A large increase in blood volume and pressure stimulates atrial wall to produce atrial natriuretic peptide (ANP). ANP inhibits Na⁺ and Cl^– reabsorption from collecting ducts inhibits release of renin, reduces aldosterone and ADH release too. This leads to a condition called Natriuresis (increased excretion of Na⁺ in urine) and diuresis.

No. Both ADH and RAAS are essential for homeostasis.

1. Only ADH can lower blood Na⁺ concentration by way of water reabsorption in DCI and collecting duct. whereas RAAS stimulates Na⁺ reabsorption and maintains osmolarity of body fluid.
2. Action of ADH and RAAS leads to increase in blood volume and osmolarity.
3. For mechanism of Atrial natriuretic peptide:

Atrial natriuretic peptide (ANP): A large increase in blood volume and pressure stimulates atrial wall to produce atrial natriuretic peptide (ANP). ANP inhibits Na⁺ and Cl^– reabsorption from collecting ducts inhibits release of renin, reduces aldosterone and ADH release too. This leads to a condition called Natriuresis (increased excretion of Na⁺ in urine) and diuresis.

Can you tell? (Textbook Page No. 187)

Question 1.
What is the composition of sweat?
Answer:
Sweat is composed of water, NaCl, lactic acid and urea.

Internet my friend. (Textbook Page No. 189)

Question 1.
Treatments other than surgical removal of kidney stone like Lithotripsy. (Breaking down of kidney stones using shock waves).
Answer:
a. Cystoscopy and ureteroscopy:
During cystoscopy, the doctor uses a cystoscope to look inside the urethra and bladder to find a stone in the urethra or bladder.
During ureteroscopy, the doctor uses a ureteroscope, which is longer and thinner than a cystoscope, to see detailed images of the lining of the ureters and kidneys.

The doctor inserts the cystoscope or ureteroscope through the urethra to see the rest of the urinary tract. Once the stone is found, the doctor can remove it or break it into smaller pieces.
The doctor performs these procedures in the hospital with anesthesia.

b. Percutaneous nephrolithotomy:
The doctor uses a thin viewing tool, called a nephroscope, to locate and remove the kidney stone.
The doctor inserts the tool directly into your kidney through a small cut made in your back.
For larger kidney stones, the doctor also may use a laser to break the kidney stones into small pieces. The doctor performs percutaneous nephrolithotomy in a hospital with anesthesia.

c. Generally for smaller stones doctors recommend drinking lots of water, consuming pain relievers and consuming medicines like alpha blocker to relax the ureter muscles, and help pass the kidney stones more quickly and with less pain

[Students are expected to find more information using the internet.]

Question 2.
Dietary restrictions suggested for kidney patients.
Dietary restrictions for kidney patients include the following:

1. Drinking large amounts of water.
2. Reduce consumption of oxalate rich food like rhubarb, beets, okra, spinach, Swiss chard, sweet potatoes, nuts, tea, chocolate and soy products.
3. Follow a diet low in salt and animal protein.
4. Reduce consumption of calcium supplements (if any) but consume appropriate amount of calcium in food.

[Students are expected to find more information using the internet.]

Balbharti Maharashtra State Board 11th Biology Textbook Solutions Chapter 14 Human Nutrition Textbook Exercise Questions and Answers.

Maharashtra State Board 11th Biology Solutions Chapter 14 Human Nutrition

1. Choose the correct option

Question A.
Acinar cells are present in ……………..
a. liver
b. pancreas
c. gastric glands
d. intestinal glands
Answer:
b. pancreas

Question B.
Which type of teeth is maximum in number in the human buccal cavity?
a. Incisors
b. Canines
c. Premolars
d. Molars
Answer:
d. Molars

Question C.
Select odd one out on the basis of digestive functions of tongue.
a. Taste
b. Swallowing
c. Talking
d. Mixing of saliva in food
Answer:
c. Talking

Question D.
Complete the analogy:
Ptyalin: Amylase : : Pepsin : …………….. .
a. Lipase
b. Galactose
c. Proenzyme
d. Protease
Answer:
d. Protease

2. Answer the following questions

Question A.
For the school athletic meet, Shriya was advised to consume either Glucon-D or fruit juice but no sugarcane juice. Why it must be so?
Answer:
Sugarcane juice contain disaccharides. Disaccharides take time to digest i.e. breaking into monosaccharides, Glucon — D and fruit juices contain monosaccharide. Therefore, for instant supply of energy during athletic meet Glucon – D or fruit juices are preferred and not sugarcane.

Question B.
Alcoholic people may suffer from liver disorder. Do you agree? Explain your answer.
Answer:

1. Liver disorder in alcoholic people may occur after years of heavy drinking.
2. Most of the alcohol in the body is broken down in the liver by an enzyme called alcohol dehydrogenase, which transforms ethanol into a toxic compound called acetaldehyde (CH₃CHO).
3. ver consumption of alcohol leads to cirrhosis (distorted or scarred liver) and eventually to liver failure.
   Therefore, alcoholic people may suffer from liver disorder.

Question C.
Digestive action of pepsin comes to a stop when food reaches small intestine. Justify.
Answer:
Pepsin acts in acidic medium thus it is active in stomach. There is alkaline condition in the small intestine. pH of small intestine is very high for pepsin to work. Therefore, pepsin gets denatured in the small intestine.

Question D.
Small intestine is very long and coiled. Even if we jump and run, why it does not get twisted? What can happen if it gets twisted?
Answer:

1. Mesentery is a tissue that is located in the abdomen. It attaches the small intestine to the wall of the abdomen and keeps it in place and therefore it does not get twisted while running and jumping.
2. If small intestine gets twisted, the affected spot may block the food, liquid passing through it. It may sometimes cut off the blood flow if the twist is very severe. If this happens the surrounding tissue may die and can cause serious problems.

3. Write down the explanation

Question A.
Digestive enzymes are secreted at appropriate time in our body. How does it happen?
Answer:

1. The digestive enzymes and juices are produced in sequential manner and at a proper time.
2. These secretions are under neurohormonal control.
3. Sight, smell and even thought of food trigger saliva secretion.
4. Tenth cranial nerve stimulates secretion of gastric juice in stomach.
5. Even the hormone gastrin brings about the same effect.

B. Explain the structure of tooth. Explain why human dentition is considered as thecodont, diphydont and heterodont.
Answer:

1. Structure of tooth:
   • A tooth consists of the portion that projects above the gum called crown and the root that is made up of two or three projections which are embedded in gum.
   • A short neck connects the crown with the root.
   • The crown is covered by the hardest substance of the body called enamel which is made up of calcium phosphate and calcium carbonate.
   • Basic shape of tooth is derived from dentin which is a calcified connective tissue.
   • The dentin encloses the pulp cavity. It is filled with connective tissue pulp. It contains blood vessels and nerves.
   • Pulp cavity has extension in the root of the tooth called root canal.
   • The dentin of the root of tooth is covered by cementurn which is a bone like substance that attaches the root to the surrounding socket in the gum.
2. Human dentition is described as thecodont, diphyodont and heterodont.
3. It is called the codont type because each tooth is fixed in a separate socket present in the jaw bones by gomphosis type of joint.
4. It is called diphyodont type because we get only two sets of teeth, milk teeth and permanent teeth.
5. It is called heterodont type because humans have four different type of teeth like incisors, canines, premolars and molars.
   

Question C.
Explain heterocrine nature of pancreas with the help of histological structure.
Answer:
Pancreas:

1. Pancreas is a leaf shaped heterocrine gland present in the gap formed by bend of duodenum under the stomach.
2. Exocrine part of pancreas is made up of acini, the acinar cells secrete alkaline pancreatic juice that contains various digestive enzymes.
3. Pancreatic juice is collected and carried to duodenum by pancreatic duct.
4. The common bile duct joins pancreatic duct to form hepato-pancreatic duct. It opens into duodenum.
5. Opening of hepato-pancreatic duct is guarded by sphincter of Oddi.
6. Endocrine part of pancreas is made up of islets of Langerhans situated between the acini.
7. It contains three types of cells a-cells which secrete glucagon, P-cells which secretes insulin and 5 cells secrete somatostatin hormone.
8. Glucagon and insulin together control the blood-sugar level.
9. Somatostatin hormone inhibits glucagon and insulin secretion.

4. Write short note on

Question A.
Position and function of salivary glands.
Answer:
Salivary Glands:

• There are three pairs of salivary glands which open in buccal cavity.
• Parotid glands are present in front of the ear.
• The submandibular glands are present below the lower jaw.
• The glands present below the tongue are called sublingual.
• Salivary glands are made up of two types of cells.
• Serous cells secrete a fluid containing digestive enzyme called salivary amylase.
• Mucous cells produce mucus that lubricates food and helps swallowing.

Question B.
Jaundice
Answer:

1. Jaundice is a disorder characterized by yellowness of conjunctiva of eyes and skin and whitish stool.
2. It is a sign of abnormal bilirubin metabolism and excretion.
3. Jaundice develops if excessive break down of red blood cells takes place along with increased bilirubin level than the liver can handle or there is obstruction in the flow of bile from liver to duodenum.
4. Bilirubin produced from breakdown of haemoglobin is either water soluble or fat soluble.
5. Fat soluble bilirubin is toxic to brain cells.
6. There is no specific treatment to jaundice.
7. Supportive care, proper rest are the treatments given to the patient.
   [Note: Treatment ofjaundice will depend on the underlying cause of it. For example, hepatitis-induced jaundice would require treatment which includes antiviral or steroid medications ]

Question 5.
Observe the diagram. This is histological structure of stomach. Identify and comment on significance of the layer marked by arrow.

Answer:
The layer marked in the diagram represents glandular epithelium of mucosa.
Significance of the glandular epiihelium of mucosa:
Goblet cells of the epithelial layer of a mucous membrane secrete mucus which lubricates the lumen of the alimentary canal. This helps in movement of food through the gastrointestinal tract.

Question 6.
Find out pH maxima for salivary amylase, trypsin, nucleotidase and pepsin and place on the given pH scale

Answer:
Salivary amylase = 6.8
Trypsin = 8
Nucleotidase = 7.5
Pepsin = 2

Question 7.
Write the name of a protein deficiency disorder and write symptoms of it.
Answer:

1. Kwashiorkor is a protein deficiency disorder.
2. This protein deficiency disorder is found generally in children between one to three years of age.
3. Children suffering from Kwashiorkor are underweight and show stunted growth, poor brain development, loss of appetite, anaemia, protruding belly, slender legs, bulging eye, oedema of lower legs and face, change in skin and hair colour.

Question 8.
Observe the diagram given below label the A, B, C, D, E and write the function of A, C in detail.

Answer:
A- Bile duct, B- Stomach, C- Common hepatic duct, D- Pancreas, E- Gall Riadder

Functions: Bile duct: It carries hile from the gall bladder and empties it into the tipper part of the small intestine. Common hepatic duct: It drains bile from the liver. It helps in transportation of waste from liver and helps in digestion by releasing bile.
[Note: Labels (A) and (O) have been modified for the better understanding of the students]

Practical / Project : Here are the events in the process of digestion. Fill in the blanks and complete the flow chart.

Answer:

11th Biology Digest Chapter 14 Human Nutrition Intext Questions and Answers

Can you recall? (Textbook Page No. 161)

Question 1.
What is nutrition?
Answer:

1. Nutrition is the sum of the processes by which an organism consumes and utilizes food substances,
2. WHO defines nutrition as the intake of food, considered in relation to the body’s dietary needs.
3. The term nutrition includes the process like ingestion, digestion, absorption, assimilation and egestion.

Question 2.
Enlist life processes that provide us energy to perform different activities.
Answer:
The life processes which are essential and provide us energy are nutrition and respiration.

Think about it (Textbook Page No. 161)

Question 1.
Our diet includes all necessary nutrients. Still we need to digest it. Why is it so?
Answer:

1. Digestion is a very important process of converting complex, noil-diffusible and non-absorbable food substances into simple, diffusible and assimilable substances.
2. Our diet includes all necessary nutrients, which are in the form of complex substances like carbohydrates, proteins, fats and vitamins.
3. These complex substances are converted into simple, diffusible and assimilable substances through the process of digestion.
   Hence, there is a need for digestion of food.

Human Digestive System (Textbook Page No. 161)

Question 1.
Label the diagram
Answer:

Do you know? (Textbook Page No. 162)

Question 1.
Who controls the deglutition?
Answer:
The process of swallowing is called deglutition. Medulla oblongata controls the deglutition.

Question 2.
Is deglutition voluntary or involuntary?
Answer:

• Deglutition consists of three phases: oral phase, pharyngeal phase and oesophagal phase.
• The oral phase is voluntary whereas the pharyngeal and oesophagal phases are involuntary.
  [Source: Goya!, R. K., & Mashimo, H. (2006,.). Physio!o’ of oral, pharyngeal, and esophageal motility. GI Motility online.]

Use your brain power (Textbook Page No. 165)

Question 1.
Draw a neat labelled diagram of human alimentary canal and associated glands in situ.
Answer:

Question 2.
Write a note on human dentition.
Answer:

1. Human dentition is described as thecodont, diphyodont and heterodont.
2. It is called thecodont type because each tooth is fixed in a separate socket present in the jaw bones by gomphosis type of joint.
3. It is called diphyodont type because we get only two sets of teeth, milk teeth and permanent teeth.
4. It is called heterodont type because humans have four different type of teeth like incisors, canines, premolars and molars.

Question 3.
Muscularis layer in stomach is thicker than that in intestine. Why is it so?
Answer:
Muscularis layer in stomach is thicker than that of intestine because food is churned and gastric juices are mixed in the stomach whereas in intestine only absorption takes place.

Question 4.
Liver is a vital organ. Justify.
Answer:

1. Kupffer cells of liver destroy toxic substances, dead and worn-out blood cells and microorganisms.
2. Bile juice secreted by liver emulsifies fats and makes food alkaline.’
3. Liver stores excess of glucose in the form of glycogen.
4. Deamination of excess amino acids to ammonia and its further conversion to urea takes place in liver.
5. Synthesis of vitamins A, D, K and BI2 takes place in liver.
6. It also produces blood proteins like prothrombin and fibrinogen.
7. During early development, it acts as haemopoietic organ.
   Therefore, liver is a vital organ.

Internet my friend: (Textbook Page No. 171)

Question 1.
Collect the different videos of functioning of digestive system,
Answer:
[Note: Students can scan the adjacent Q.R code to get conceptual clarity with the aid of a relevant video.]

Find out (Textbook Page No. 162)

Question 1.
What will be the dental formula of a three years old child?
Answer:
The dental formula of a three-year-old child will be: I \(\frac{2}{2}\), C \(\frac{1}{1}\), M \(\frac{2}{2}\) = \(\frac{2,1,2}{2,1,2}\)
i. e. 5 × 2 = 10 teeth in each jaw = 20 teeth.
As a child has 20 teeth by the age of three.

Question 2.
What is dental caries and dental plaque? How can one avoid it?
Answer:

• Dental caries are tooth decay or cavities caused by acids secreted by bacteria. Dental caries may be yellow or black in color.
• Dental plaques also known as tooth plaque is a soft, sticky film which forms on the teeth regularly. It is colourless to pale yellow in colour.
• Tooth decay and dental plaque can be prevented by brushing teeth twice a day with a fluoride containing tooth paste.
• Rinsing mouth thoroughly with a mouth wash and use of dental floss or interdental cleaners to clean teeth daily can help to avoid dental caries and dental plaque.

Internet my friend (Textbook Page No. 162)

Question 1.
Find out the role of orthodontist and dental technician.
Answer:
a. Orthodontics is a specialization in dental profession. Orthodontist straightens the crooked teeth, locates problem in patients’ teeth and their overall oral development. They might use X-rays, plaster molds or dental appliances like retainers and space maintainers to correct the problems,

b. Dental technicians are the ones which improves patients’ appearance, ability to chew and speech. They make dentures, crowns, bridges and dental braces.

Question 2.
What is a root canal treatment?
Answer:

• Root canal treatment is also known as endodontic treatment.
• It is a dental treatment of removing infection from inside of a tooth.
• Root canal is hollow section of tooth which contains the nerve tissue, blood vessels and other cells, this is also known as pulps.
• Crown and root are a part of tooth. Crown is present above the gum while root is embedded in the gum.
  e. Pulp which is present inside the root canal nourishes the tooth and provides moisture to the surrounding material.
• The nerves present inside the pulp sense hot cold temperatures as pain.
• First step of a root canal treatment is removal of dead pulp tissues by making a hole on the surface of tooth.
• In second step, the dentist cleans and decontaminates the area and fills the hollow area with adhesive cement in order to seal the canal completely.
• The tooth is dead after the therapy and the patient no longer feel any pain but the tooth becomes more fragile than ever.
• The last step of root canal is adding a crown or filling. Until the crown or filling is complete, patient is not supposed to chew or bite using that tooth. After the crown or filling patient can use that tooth as before.

Find out (Textbook Page No. 163)

Question 1.
You must have heard about appendicitis. It is inflammation of appendix. Find more information about this disorder.
Answer:

1.  Appendicitis is a condition where there is inflammation of appendix.
2. Appendix is a vestigial organ. It is a linger shaped pouch that projects from colon on the lower right side of the abdomen.
3. Appendicitis pain is very severe. It initially starts from the navel and then moves.
4. It occurs in the people of age group between 10 to 30.
5. Surgical removal is the standard treatment for appendicitis.
6. Symptoms: Nausea and vomiting, loss of appetite, low grade fever, constipation, abdominal bloating, severe pain in the right side of the abdomen.
7. Appendicitis is caused when there is blockage in the lining of the appendix that results in infection. The bacteria multiply rapidly and causes inflammation and it is then filled with pus.
8. If not treated properly appendix can rupture which can lead to further complications.
   [Students can use above answer for reference and find more information about appendicitis.]

Question 2.
What is heartburn? Why do we take antacids to control it?
Answer:
Heart burn is a problem created when stomach contents (acid) are forced back up to oesophagus. It causes a burning pain in lower chest.

Antacids are bases and help to treat heartburn by neutralizing the stomach acid. The key ingredients of antacids are calcium carbonate, magnesium hydroxide, aluminium hydroxide or sodium bicarbonate.

Activity (Textbook Page No. 163)

Make a model of human digestive system in a group.
Answer:
[Students are expected to perform this activity on their own.]

Always Remember (Textbook Page No. 166)

Question 1.
Food remains for a very short time in mouth but action of salivary amylase continues for further IS to 30 minutes till gastric juice mixes with food in the stomach. Why do you think it stops after the food gets mixed with gastric juice?
Answer:

1. The gastric juices are mixed with food in the stomach.
2. The pH of the stomach is 1.0-2.0 which is very acidic. Such high level of acidity leads to denaturation of salivary amylase’s protein structure.
3. On the other hand, pH 6.8 is required for salivary amylase to carry out the activity which is not found in stomach. Thus, activity of salivary amylase is stopped when food is mixed with gastric juice.

Internet my friend (Textbook Page No. 167)

Question 1.
How are bile pigments formed?
Answer:

1. When old and worn out red blood cells are destroyed by macrophages in liver, the globin portion of hemoglobin is split off and heme is converted to biliverdin.
2. Most of this biliverdin is converted to bilirubin, which gives bile its major pigmentation.
   [Source http://www.biologydiscussion.com/human-physiology/digestive-system/bile-pigments/bile-pigments-origin-and-formation-digestive-juice-human-biology/81803]

Think about it (Textbook Page No. 167)

Question 1.
How can I keep my pancreas healthy? Can a person live without pancreas?
Answer:

1. Pancreas can be kept healthy by:
   • Eating proper balanced and low-fat diet, with plenty of whole grains, fruits and vegetables.
   • Regular exercise and maintaining a healthy weight.
   • Limiting alcohol consumption and avoid smoking.
   • Adequate intake of water.
   • Regular checkups.
2. The pancreas is a gland that secretes digestive enzymes and insulin which is needed for a person to survive.
3. Without pancreas the person will develop diabetes and will have to take insulin for the rest of the life.
4. Without pancreas the body’s ability to absorb nutrients also decreases.
   Hence, though a person can survive without pancreas he may have to remain dependent on the medicines for survival.

Do it yourself? (Textbook Page No. 167)

Question 1.
You have studied the representation of enzymatic actions in the form of reactions.
Write the reactions of pancreatic enzymes.
Answer:

Do it yourself (Textbook Page No. 168)

Question 1.
Observe the following reactions and explain in words.

Answer:

1. Maltase acts on maltose to form glucose.
2. Sucrase acts on sucrose to form glucose and fructose.
3. Lactase acts on lactose to form glucose and galactose.
4. Dipeptidase acts on dipeptides to form amino acids.
5. Emulsified fats are converted into fatty acids and glycerol by lipase.

Use your brain power (Textbook Page No. 168)

Question 1.
Make a flow chart for digestion of carbohydrate.
Answer:

Question 2.
What is a proenzyme? Enlist various proenzymes involved in process of digestion and state their function.
Answer:
Proenzymes are synthesized in cells as an inactive precursor that undergo some modification before becoming catalytically active.
The various proenzymes involved in process of digestion are as follows:

• Pepsinogen: Pepsinogen when converted into its active form pepsin acts on proteins to form peptones and proteoses.
• Trypsinogen: Trypsinogen when converted to it active form trypsin converts proteins, proteoses and peptones to polypeptides.
• Chymotrypsinogen: Chymotrypsinogen when converted to active form chymotrypsin it converts polypeptides to dipeptides.

Question 3.
Differentiate between Chyme and Chyle.
Answer:

No.	Chyme	Chyle
a.	Chyme is a semi-fluid acidic mass of partially digested food.	Chyle is an alkaline slurry which contains various nutrients ready for absorption.
b.	Chyme leaves stomach and enters the small intestine.	Chyle leaves small intestine and enters large intestine.

Question 4.
Digestion of fats take place only after the food reaches small intestine. Give reason.
Answer:
Digestion of fats takes place in small intestine because the presence of fats in small intestine stimulates the release of pancreatic lipase from pancreas and bile from liver. Pancreatic lipases hydrolyze fat molecules into fatty acids and monoglycerides and bile brings about emulsification of fats. Therefore, digestion of fats occur when food reaches small intestine.

Observe and Discuss (Textbook Page No. 169)

Question 1.
Action of digestive juice in your group.
Answer:

Digestive juices	Action
Saliva	Saliva contains salivary amylase which breaks down starch into maltose.
Gastric juice	HC1 breaks converts inactive pepsinogen into its active form pepsin. Pepsin then breakdown proteins into peptones and proteoses.
Pancreatic juice	Pancreatic amylase acts on glycogen and starch and converts those into disaccharides. Enterokinase converts trypsinogen into trypsin (active form). Trypsin converts proteins, proteoses, peptones to polypeptides. Chymotrypsin converts polypeptides to dipeptides. Nucleases digest nucleic acids to pentose sugar.
Intestinal enzymes	Maltase converts maltose to glucose. Sucrase converts sucrose to glucose and fructose. Lactase converts lactose to glucose and galactose. Dipeptidases converts dipeptides to amino acids. Lipase converts emulsified fats into fatty acids and monoglycerides.
Bile juice	It brings about emulsification of fats.

Can you recall? (Textbook Page no. 170)

Question 1.
What is balanced diet?
Answer:
Balanced diet is a diet which contains proper amount of carbohydrates, fats, vitamins, proteins and minerals to maintain a good health.

Question 2.
Explain the terms undernourished, over-nourished and malnourished in details.
Answer:

• Undernourished: When supply of nutrients is less than the minimum amount of nutrients or food required for good health is called undernourished.
• Over-nourished: The intake of nutrients is excessive. In over-nourished the amount of nutrients exceeds the amount required for normal growth.
• Malnourished: Malnourished is a condition where a person’s diet does not contain right amount of nutrients.

Do you know? (Textbook Page No. 170)

Question 1.
What is gross calorific value?
Answer:
The amount of heat liberated by complete combustion of lg food in a bomb calorimeter is termed as gross calorific (gross energy) value.

Question 2.
What is physiological value?
Answer:
The actual energy produced by 1 g food is its physiological value.

Question 3.
Name the following
Energy content of food in animals is expressed in terms of?
Answer:
Heat Energy

Question 4.
Complete the following table representing Gross calorific value and physiological value of food component.

Food Component	Gross calorific value (Kcal/g)	Physiological value (Kcal/g)
Fats	(A)	9.0
(B)	5.65	4.0
Carbohydrates	(C)	(D)

Answer:

Food Component	Gross calorific value (Kcal/g)	Physiological value (Kcal/g)
Fats	9.45	9.0
Proteins	5.65	4.0
Carbohydrates	4.1	4.0

Find out (Textbook Page No. 171)

Question 1.
Find out the status of nialnutrition among children in Maharashtra and efforts taken by the government to overcome the situation. Search for various NGOs working in this field.
Answer:
93,783 children have been diagnosed with severe acute malnutrition and 5.7 lakh with moderate acute malnutrition in Maharashtra.
Steps taken by government to overcome malnutrition:

1. Promotion of infant and young child feeding practices.
2. Management of malnutrition at community and facility level by trained service providers.
3. Treatment of children with severe acute malnutrition at special units called the Nutrition Rehabilitation Centres (NRCs), set up at public health facilities.
4. A special program to combat micronutrient deficiencies of Vitamin A, Iron and Folic acid.
5. The initiatives like Mother and Child protection card, village health and nutrition days, are taken by the government for addressing the nutrition concerns in children, pregnant women and lactating mothers.

Various NCOs working in this field:

1. Akshay Patra
2. Fight Hunger Foundation,
3. Feeding India,
4. No Hungry child
   [Source: http://pib.nic.in/newsite/PrintRelease.aspx?relid=l 13725; https://yourstory.com/2016/10/world- food-day-ngosj
   [Note: Students can use above answer as reference and find more information from the internet.]

Question 2.
Are jaundice and hepatitis same disorders?
Answer:
Jaundice and Hepatitis are two different disorders.

Jaundice: Jaundice occurs when the rate of bilirubin production exceeds the rate of its elimination. It causes yellowing of skin and eyes.

Hepatitis: It is a disease where there is inflammation of liver. It may be caused because of infection, over alcohol consumption, immune system disorder etc.

Do you know (Textbook Page No. 171)

Question 1.
Alcoholism causes different disorders of liver like steatosis (fatty liver), alcoholic hepatitis, fibrosis and cirrhosis. Collect more information on these disorders and try to increase awareness against alcoholism in society. Collect information about NGOs working against alcoholism.
Answer:
Steatosis (fatty liver): Steatosis is accumulation of fat in the liver. Treatment can help but it cannot be cured. Major risk factors are obesity and Diabetes type II, it is also associated with excessive alcohol consumption. Fatigue, weight loss and abdominal pain are some symptoms. It is a benign condition but in very smaller number of patients it can lead to liver failure. Treatment involves diet and exercise to reduce obesity.

Alcoholic hepatitis: Alcoholic Hepatitis is liver inflammation caused by excessive consumption of alcohol. It occurs in people who drink heavily for many years. Symptoms like yellowing of skin and eye, accumulation of fluid in stomach which leads to increase in stomach size. Treatments like completely stopping of alcohol consumption, hydration and nutrition care are carried out. Administration of steroid drugs reduces liver inflammation.

Fibrosis: There is significant scarring of liver tissue in this condition. Fibrosis itself does not cause any symptoms. Diagnosis includes doctor’s evaluation, blood tests and imaging tests, liver biopsy. Treatments include stopping the consumption of alcohol. There are no such effective drugs for curing of fibrosis.

Cirrhosis: It is a chronic liver damage caused due to various reasons which leads to irreversible scarring of liver and liver failure. Causes of cirrhosis are chronic alcohol abuse and hepatitis. Patients may experience fatigue, weakness and weight loss. In later stages, patients may develop jaundice, abdominal swelling and gastrointestinal bleeding. In advanced stage, a liver transplant is required.

NGOs working against alcoholism:

1. Muktangan Rehabilitation Centre
2. Anmol Jeevan Foundation
3. Sankalp Rehabilitation Trust
4. Kripa Foundation
5. Harmony Foundation
6. Hands for you Rehab Centre

Balbharti Maharashtra State Board 11th Biology Textbook Solutions Chapter 13 Respiration and Energy Transfer Textbook Exercise Questions and Answers.

Maharashtra State Board 11th Biology Solutions Chapter 13 Respiration and Energy Transfer

1. Choose the Correct option.

Question (A)
The reactions of the TCA cycle occur in
(A) ribosomes
(B) grana
(C) mitochondria
(D) endoplasmic reticulum
Answer:
(C) mitochondria

Question (B)
In eukaryotes the complete oxidation of a molecule of glucose results in the net gain of
(A) 2 molecules of ATP
(B) 36 molecules of ATP
(C) 4 molecules of ATP
(D) 38 molecules of ATP
Answer:
(D) 38 molecules of ATP

Question (C)
Which step of Krebs cycle operates substrate-level phosphorylation?
(A) ∝-ketoglutarate → succinyl CoA.
(B) Succinyl CoA → succinate
(C) Succinate → fumarate
(D) Fumarate → malate
Answer:
(B) Succinyl CoA → succinate

2. Fill in the blanks with suitable words.

Question 1.
A. Acetyl CoA is formed from __________ and co-enzyme A.
B. In the prokaryotes ________ molecules of ATP are formed per molecule of glucose oxidised.
C. Glycolysis takes place in ________ .
D. F₁ – F₀ particles participate in the synthesis of _________ .
E. During glycolysis _________ molecules of NADH⁺H⁺ are formed.
Answer:
A. pyruvic acid
B. 2/38
C. cytoplasm
D. ATP
E. 2
[Note: ii. In prokaryotes, during anaerobic respiration 2 ATPs are formed per glucose and 38 ATPs are formed during aerobic respiration.]

3. Answer the following questions

Question (A)
When and where does anaerobic respiration occur in man and yeast?
Answer:
1. In absence of oxygen, anaerobic respiration takes place in skeletal muscles of man during vigorous exercise.
2. Anaerobic respiration occurs in the cytoplasm of the yeast cell.

Question (B)
Why is less energy produced during anaerobic respiration than in aerobic respiration?
Answer:
Anaerobic respiration produces less energy because:

1. Incomplete breakdown of respiratory substrate takes place.
2. Some of the products of anaerobic respiration can be oxidised further to release energy which shows that anaerobic respiration does not liberate the whole energy contained in the respiratory substrate.
3. NADH₂ does not produce ATP, as electron transport is absent.
4. Only 2 ATP molecules are generated from one molecule of glucose during anaerobic respiration.

Question (C)
Which is the site for ETS in mitochondrial respiration?
Answer:
The inner mitochondrial membrane is the site for ETS in mitochondrial respiration.

Question (D)
Which compound is the terminal electron acceptor in aerobic respiration?
Answer:
Molecular oxygen is the terminal electron acceptor in aerobic respiration.

Question (E)
What is RQ.? What is its value for fats?
Answer:
1. Respiratory quotient (R.Q.) or respiratory ratio is the ratio of volume of CO₂ released to the volume of O₂ consumed in respiration.
2. R.Q. = Volume of CO₂ released / Volume of O₂ consumed

Question (F)
What are respiratory substrates? Name the most common respiratory substrate.
Answer:
Respiratory substrates are the molecules that are oxidized during respiration to release energy which can be used for ATP synthesis. Carbohydrates, fats and proteins are the common respiratory substrate. Glucose is the most common respiratory substrate.

Question (G)
Write explanatory notes on:

Question (i)
Glycolysis
Answer:
Glycolysis is a process where glucose is broken down into two molecules of pyruvic acid, hence called glycolysis (glucose-breaking). It is common to both aerobic and anaerobic respiration. It occurs in the cytoplasm of the cell. It involves ten steps.
Glycolysis consists of two major phases:
1. Preparatory phase (1-5 steps).
2. Payoff phase (6-10 steps).
1. Preparatory phase:
a. In this phase, glucose is phosphorylated twice by using two ATP molecules and a molecule of fructose 1,6-bisphosphate is formed.
b. It is then cleaved into two molecules of glyceraldehyde-3-phosphate and dihydroxy acetone phosphate. These two molecules are 3-carbon carbohydrates (trioses) and are isomers of each other.
c. Dihydroxy acetone phosphate is isomerised to second molecule of glyceraldehyde-3-phosphate.
d. Therefore, two molecules of glyceraldehyde-3- phosphate are formed.
e. Preparatory phase of glycolysis ends.

2. Payoff phase:
a. In this phase, both molecules of glyceraldehyde-3-phosphate are converted to two molecules of 1,3- bisphoglycerate by oxidation and phosphorylation. Here, the phosphorylation is brought about by inorganic phosphate instead of ATP.
b. Both molecules of 1, 3-bisphosphoglycerate are converted into two molecules of pyruvic acid through series of reactions accompanied with release of energy. This released energy is used to produce ATP (4 molecules) by substrate-level phosphorylation.

Question (ii)
Write explanatory notes on: Fermentation by yeast
Answer:
Alcoholic fermentation is a type of anaerobic respiration where the pyruvate is decarboxylated to acetaldehyde. The acetaldehyde is then reduced by NADH+H⁺ to ethanol and Carbon dioxide. Since ethanol is produced during the process, it is termed alcoholic fermentation.

Question (iii)
Write explanatory notes on: Electron transport chain
Answer:

1. NADH₂ and FADH₂ produced during glycolysis, connecting link reaction and Krebs cycle are oxidized with the help of various electron carriers and enzymes.
2. These carriers and enzymes are arranged on inner mitochondrial membrane in the form of various complexes as complex I, II, III, VI and V.
3. NADH+H⁺ is oxidised by NADH dehydrogenase (complex I) and it’s electrons are transferred to ubiquinone (coenzyme Q-CoQ) present on inner membrane of mitochondria. Reduced ubiquinone is called as ubiqunol.
4. FADH₂ is oxidised by complex II (Succinate dehydrogenase) and these electrons are also transferred to CoQ.
5. During oxidation of NADH+H⁺ and FADH₂ , electrons and protons are released but only electrons are canned forward whereas protons are released into outer chamber of mitochondria (intermembrane space).
6. Ubiquinol is oxidised by complex-III (Cytochrome bcl complex) and it’s electrons are transferred to cytochrome C. Cytochrome C is a small, iron-containing protein, loosely associated with inner membrane. It acts as a mobile electron carrier, transferring the electrons between complex III and IV.
7. Cytochrome C is oxidised by complex IV or cytochrome C oxidase consisting of cytochrome a and a₃. Electrons are transferred by this complex to the molecular oxygen. This is terminal oxidation.
8. Reduced molecular oxygen reacts with protons to form water molecule called as metabolic water.
9. Protons necessary for this are channelled from outer chamber of mitochondria into inner chamber by F₀ part of oxysome (complex V) present in inner mitochondrial membrane.
10. This proton channelling by F₀ is coupled to catalytic site of F₁ which catalyses the synthesis of ATP from ADP and inorganic phosphate. This is oxidative phosphorylation.
11. As transfer of protons is accompanied with synthesis of ATP, this process is named as ‘Chemiosmosis’ by Peter Mitchell.

Significance of ETS:

1. Major amount of energy is generated through ETS or terminal oxidation in the form of ATP molecules.
2. Per glucose molecule 38 ATP molecules are formed, out of which 34 ATP molecules are produced through ETS.
3. Oxidized coenzymes such as NAD and FAD are regenerated from their reduced forms (NADH+H⁺ and FADH₂) for recycling.
4. In this process, energy is released in a controlled and stepwise manner to prevent any damage to the cell.
5. ETS produces water molecules.

Question (H)
How are glycolysis, TCA cycle and electron transport chain-linked? Explain.
Answer:
Glycolysis, TCA cycle and electron transport chain are linked in the following manner:

1. The coenzymes are initially present in the form of NAD⁺ and FAD⁺ which latter get reduced to NADH+H⁺ and FADH+H⁺ by accepting the hydrogen from organic substrate during glycolysis, link reaction and Krebs cycle.
2. During glycolysis, glucose is oxidised to two molecules of pyruvic acid with net gain 2 molecules of NADH+H⁺.
3. This pyruvic acid undergoes link reaction to form two molecules of acetyl CoA and two molecules of NADH+H⁺.
4. Acetyl CoA, thus formed enters into the Krebs cycle and it gets completely oxidised to C0₂ and H₂0; with a net gain of 6 NADH+H+ and 2 FADH+H⁺ are formed.
5. During ETS, reduced coenzymes are reoxidized to NAD⁺ and FAD⁺ with a net gain of 34 ATPs. The ATPs thus formed are used during glycolysis.
6. The oxidized NAD⁺ and FAD⁺ will again accept the hydrogen from organic substrate. Thus, reduced coenzymes are converted back to their oxidized forms by dehydrogenation to keep the process going.

Question (I)
How would you demonstrate that yeast can respire both aerobically and anaerobically?
Answer:
Respiration in yeast can be demonstrated with the help of an experiment.
Anaerobic respiration in yeast:

1. A pinch of dry baker’s yeast suspended in water containing 10ml of 10% glucose in a test tube (test tube A).
2. The surface of the liquid is covered with oil to prevent entry of air and the test tube is closed tightly with rubber stopper to prevent leakage.
3. One end of a short-bent glass tube is inserted through it to reach the air inside the tube.
4. Other end of the glass tube is connected by a polyethylene or rubber tubing to another bent glass tube fitted into a stopper.
5. The open end of the glass tube (delivery tube) is dipped into lime water containing in a test tube
   (Tube B).
6. Stoppers of both the tubes are fitted tightly to prevent leakage of gases. First test tube is placed in warm water (37° C-38° C) in a beaker.
7. Lime water gradually turns milky, indicating the evolution of carbon dioxide from the yeast preparation.
8. Level of the lime water in the delivery tube does not rise, showing that there is no decline in volume of gas in test tube A and consequently no utilization of oxygen by yeast. Preparation is stored for a day or two.
9. When we open the stopper of tube A we will notice a smell of alcohol indicating the formation of ethanol.
10. From this activity it may be inferred that yeast respires anaerobically to ferment glucose to ethanol and carbon dioxide.

Aerobic respiration in yeast: Experiment explained can be carried out for demonstrating aerobic respiration in yeast.

1. If the level of the lime water in the test tube B rises, indicating intake of oxygen, hence the level of volume of gas rises.
2. The preparation tube is stored for a day or two, if no smell of alcohol is noticed it indicates that the yeast respires aerobically.

Question (J)
What is the advantage of step wise energy release in respiration?
Answer:
In ETS energy is released in step wise manner to prevent damage of cells.

1. A stepwise release of the chemical bond energy facilitates the utilization of a relatively higher proportion of that energy in ATP synthesis.
2. Activities of enzymes for the different steps may be enhanced or inhibited by specific compounds. This provides a means of controlling the rate of the pathway and the energy output according to need of the cell.
3. The same pathway may be utilized for forming intermediates used in the synthesis of other biomolecules like amino acids.

Question (K)
Explain ETS.
Answer:

1. NADH₂ and FADH₂ produced during glycolysis, connecting link reaction and Krebs cycle are oxidized with the help of various electron carriers and enzymes.
2. These carriers and enzymes are arranged on inner mitochondrial membrane in the form of various complexes as complex I, II, III, VI and V.
3. NADH+H⁺ is oxidised by NADH dehydrogenase (complex I) and it’s electrons are transferred to ubiquinone (coenzyme Q-CoQ) present on inner membrane of mitochondria. Reduced ubiquinone is called as ubiqunol.
4. FADH₂ is oxidised by complex II (Succinate dehydrogenase) and these electrons are also transferred to CoQ.
5. During oxidation of NADH+H⁺ and FADH₂ , electrons and protons are released but only electrons are canned forward whereas protons are released into outer chamber of mitochondria (intermembrane space).
6. Ubiquinol is oxidised by complex-III (Cytochrome bcl complex) and it’s electrons are transferred to cytochrome C. Cytochrome C is a small, iron-containing protein, loosely associated with inner membrane. It acts as a mobile electron carrier, transferring the electrons between complex III and IV.
7. Cytochrome C is oxidised by complex IV or cytochrome C oxidase consisting of cytochrome a and a₃. Electrons are transferred by this complex to the molecular oxygen. This is terminal oxidation.
8. Reduced molecular oxygen reacts with protons to form water molecule called as metabolic water.
9. Protons necessary for this are channelled from outer chamber of mitochondria into inner chamber by F₀ part of oxysome (complex V) present in inner mitochondrial membrane.
10. This proton channelling by F₀ is coupled to catalytic site of F₁ which catalyses the synthesis of ATP from ADP and inorganic phosphate. This is oxidative phosphorylation.
11. As transfer of protons is accompanied with synthesis of ATP, this process is named as ‘Chemiosmosis’ by Peter Mitchell.

Question (L)
Discuss “The respiratory pathway is an amphibolic pathway”.
OR
Question (M)
Why is Krebs cycle referred as amphibolic pathway?
Answer:

1. Respiration is considered as a catabolic process; however, it is not entirely correct in case of Krebs cycle.
2. Many reactions of Krebs cycle involve oxidation of acetyl CoA to release energy and C02.
3. However, the breakdown of respiratory substrates provides intermediates like a-ketoglutarate, oxaloacetate are used as precursors for synthesis of fatty acids, glutamic acid and aspartic acid respectively.
4. Thus, as the same respiratory process acts as catabolic as well as anabolic pathway for synthesis of various intermediate metabolic products, it is called amphibolic pathway.

Question (N)
The common pathway for both aerobic and anaerobic respiration is
(A) Krebs cycle
(B) Glycolysis
(C) ETS
(D) Terminal oxidation
Answer:
(B) Glycolysis

4. Compare

Question (A)
Photosynthesis and respiration.
Answer:

Photosynthesis	Respiration
(a) It takes place in the cells containing chlomplasts.	It takes place in all living cells of higher organisms.
(b) It occurs in chloroplast.	It occurs in cytoplasm and mitochondria.
(c) It is an energc trapping process.	It is an energy releasing process.
(d) It is an anabolic process.	It is a catabolic process.
(e) This process requires C02 and FLO.	This process requires sugar and 02.
(f) Light is necessary for photosynthesis.	Light is not necessary for aerobic respiration.
(g) End products are carbohydrates and oxygen.	End products can be C02 and H20 or ethanol or lactic acid and energy.

Question (B)
Aerobic respiration and Anaerobic respiration
Answer:

Aerobic respiration	Anaerobic respiration
(a) It takes place in higher organisms.	It takes place in lower organisms.
(b) It takes place in cytoplasm and mitochondria.	It takes place in cytoplasm.
(c) It involves the participation of free molecular oxygen.	It does not involve participation of free molecular oxygen.
(d) Oxidation of food is complete.	Oxidation of food is incomplete.
(e) It produces C02 and H20.	It produces C02 and C2H5OH.
(f) It releases more energy, i.e. 38 ATP.	It releases less energy, i.e. 2 ATP.
(g) Overall equation: C6H1206 + 602 → 6C02 + 6H20 + Energy	Overall equation: C6H1206 → 2C2H5 OH + 2C02 + Energy

5. Differentiate between

Question (A)
Respiration and combustion.
Answer:

Respiration	Combustion
(a) It is a biochemical and stepwise process.	It is physiochemical and spontaneous process.
(b) It occurs inside the cells.	It is a non-cellular process.
(c) Energy is released in steps.	Large amount of energy is released at a time.
(d) No light is produced in respiration.	Light may be produced in combustion.
(e) It is controlled by enzymes.	It is not controlled by enzymes.
(f) A number of intermediates are produced.	No intermediates are produced.

Question (B)
Distinguish between Glycolysis and Krebs cycle.
Answer:

Glycolysis/EMP pathway	Krebs cycle/TCA cycle/ Citric acid cycle
1. Glycolysis is common in both aerobic and anaerobic respiration.	Krebs cycle occurs only in aerobic respiration.
2. It takes place in the cytoplasm.	It takes place in the mitochondria.
3. C02 is not released.	C02 is released.
4. Total amount of energy produced = 8 ATP.	Total amount of energy produced = 24 ATP.
5. It is linear pathway.	It is cyclic pathway.
6. Pyruvic acid is the end product.	C02 and H2Q are the end products.

Question (C)
Aerobic respiration and fermentation.
Answer:

Aerobic respiration	Fermentation
1. It takes place in higher organisms.	It takes place in both higher and lower organisms.
2. It takes place in cytoplasm and mitochondria	It takes place in cytoplasm.
3. It involves the participation of free molecular oxygen.	It does not involve participation of free molecular oxygen.
4. It involves many steps – glycolysis, link reaction, Krebs cycle and ETS.	It involves only glycolysis, decarboxylation and reduction, (alcoholic fermentation)
5. Oxidation of food is complete.	Oxidation of food is incomplete.
6. It produces C02 and H20.	It produces either ethanol or lactic acid and C02 depending upon the type of fermentation.
7. It releases more energy, i.e. 38 ATP.	It releases less energy, i.e. 2 ATP.

Question 6.
Identify the cycle given below. Correct it and fill in the blanks and write description of it in your own
Answer:

1. Krebs cycle or citric acid cycle is the second phase of aerobic respiration which takes place in the matrix of the mitochondria.
2. The acetyl CoA formed during the link reaction undergoes aerobic oxidation.
3. This cycle serves a common oxidative pathway for carbohydrates, fats and proteins.
4. In mitochondria pyruvic acid is decarboxylated and the remaining 2-carbon fragment is combined with a molecule of coenzyme A to form acetyl-CoA.
5. This reaction is an oxidative decarboxylation process and produces H⁺ ions and electrons along with carbon dioxide. During the process NAD⁺ is reduced to NADH+H⁺.
6. P-oxidation of fatty acids also produces acetyl-CoA as the end product.
7. Acetyl-CoA from both sources is condensed with oxaloacetic acid to form citric acid. Citric acid is oxidized step-wise by mitochondrial enzymes, releasing carbon dioxide.
8. Regeneration of oxaloacetic acid occurs to complete the cycle.
9. There are four steps of oxidation in this cycle, catalyzed by dehydrogenases (oxidoreductases) using NAD⁺ or FAD⁺ as the coenzyme.
10. The coenzymes are consequently reduced to NADH+H⁺ and FADH₂ respectively. These transfer their electrons to the mitochondrial respiratory chain to get reoxidised.
11. One molecule of GTP (ATP) is also generated for every molecule of citric acid oxidized.

Practical / Project:

Question 1.
Make Powerpoint Presentation on Glycolysis, Krebs Cycle and Conduct the group discussion on it in classroom.
[Note: Students are expected to perform above activity on their own.]

11th Biology Digest Chapter 13 Respiration and Energy Transfer Intext Questions and Answers

Can you recall? (Textbook Page No. 151)

(i) Which nutrients are used for energy production?
Answer:
Nutrients like carbohydrates, fats and proteins are used for energy production.

(ii) Why do organisms take up oxygen and release carbon dioxide?
Answer:
a. At cellular level, organisms require energy to carry out different metabolic activities.
b. The energy is made available by oxidizing/breaking the food.
Therefore, oxygen is required by aerobic organisms for breaking the food and carbon dioxide is released as a byproduct of oxidation.

Use your brainpower (Textbook Page No. 152)

Why is glycolysis considered as biochemical proof of evolution?
Answer:

1. Glycolysis does not require oxygen. Hence it might have been used by earlier organisms for energy production, as there was no free oxygen in atmosphere of primitive earth.
2. Glycolysis is the first metabolic pathway, an ancient pathway which is common to both aerobic and anaerobic organisms.
3. All cells have glycolysis in their metabolic pathway.
4. Upto pyruvate the pathway is similar to all aerobic and anaerobic organisms. Later, the fate of pyruvic acid can be either C0₂ or ethanol or lactic acid depending upon the type of organism.
5. Hence it is considered as a biochemical proof of evolution.

Use your brainpower (Textbook Page No. 152)

(i) What is role of Mg⁺⁺, Zn⁺⁺ in various steps of glycolysis?
Answer:
a. Mg⁺⁺ and Zn⁺⁺ are the cofactors that are tightly bound to enzymes and helps the enzymes to perform their functions.
b. They regulate the activity of the most important enzymes like Hexokinase, Phosphoffuctokinase, Triose phosphate dehydrogenase, Phosphoglycerate kinase, Enolase, Pyruvate kinase.

(ii) Why some reactions of glycolysis are reversible and some irreversible?
Answer:
Irreversible chemical reactions:
Some chemical reactions can occur in only one direction i.e. these reactions are irreversible reactions. The reactants can change to the products, but the products cannot change back to the reactants.

Reversible chemical reactions:

1. Some chemical reactions can occur in both directions i.e. these reactions are reversible reactions. In this case the reactants change to the products and the products can change back to the reactants, atleast under specific conditions.
2. Out of ten, four are irreversible reactions which involves the enzyme kinase that is required for phosphorylation reactions, these reactions involve large negative energy AG, hence the reactions are irreversible.
3. Other reversible reactions do not involve high negative energy hence are reversible.

Use your brainpower (Textbook Page No. 152)

Why do athletes like sprinters have higher proportion of white muscle fibers?
Answer:
1. The white muscle fibres produce energy in a very short period of time that is required for fast and severe work. Thus, the energy becomes immediately available to the athletes.
2. On the other hand, the red muscle produce energy over a prolonged period of time, hence athletes have higher proportion of white muscle fibers.

Can you recall? (Textbook Page No. 151)

Which steps are involved in aerobic respiration?
Answer:
It involves glycolysis, acetyl CoA formation (connecting link reaction), Krebs cycle, electron transfer chain reaction and terminal oxidation.

Can you recall? (Textbook Page No. 151)

What is aerobic and anaerobic respiration?
Answer:
For anaerobic respiration: Anaerobic respiration is the cellular respiration that does not involve the atmospheric oxygen. It is also called as fermentation. It involves glycolysis where the product of glycolysis i.e. pyruvate is converted to either lactic acid or ethanol and for aerobic respiration.
1. Aerobic respiration occurs in the presence of free molecular oxygen during oxidation of glucose.
2. In this type of respiration, the glucose is completely oxidized to C0₂ and H₂0 with release of large amount of energy. It involves glycolysis, acetyl CoA formation (connecting link reaction), Krebs cycle, electron transfer chain reaction and terminal oxidation.

Use your brainpower (Textbook Page No. 157)

Do the plants breath like animals? If yes, how and why?
Answer:

1. Yes, plants breath like animals because they also require energy to carry out different metabolic activities. Hence, plants take up oxygen required for respiration and they also give out C0₂.
2. Plants take care of their gas exchange needs. The stomata and lenticels are important for this purpose.
3. Leaves are well adapted for gaseous exchange during photosynthesis.
4. Large amount of gases is exchanged. In plants, each living cell is located quite close to the surface of the plants.

Internet my friend (Textbook Page No. 155)

What is effect of carbon monoxide poisoning on cytochromes?
Answer:

1. At sub-cellular level, carbon monoxide is toxic for mitochondria.
2. It alters the mitochondrial respiratory chain at the cytochrome c oxidase level (complex IV of the mitochondrial respiratory chain) and causes inhibition of ETS.
3. This inhibition leads to the development of symptoms observed in acute CO poisoning, and in some diseases related to smoking.
4. These symptoms include headache, nausea, vomiting, dizziness, weakness, difficulty in concentration or confusion, visual changes, syncope, seizures, abdominal pain and muscle cramping.

Can you recall? (Textbook Page No. 151)

Which is most preferred nutrient among carbohydrate, protein and fat for energy production? Why?
Answer:

1. The preferred nutrient is carbohydrate because it quickly supplies energy as compared to other nutrients.
2. Carbohydrates are easy to digest as compared to fats.
3. The RQ of carbohydrate is 1. Hence allows complete oxidation of food. Thus, the preferred nutrient is carbohydrate.

Internet my friend (Textbook Page No. 158)

Calculate the RQ for different respiratory substrates using appropriate formula.
Answer:
The RQ for different respiratory substrates are:
1. Carbohydrates (R.Q. is 1)
When carbohydrates are used as substrate, equal volumes of C0₂ and 0₂ are released and consumed respectively, thus its R.Q. is 1.
C₆ H₁₂ O₆ + 6O₂ → 6 C0₂ + 6H₂0
R.Q. = 6C0₂ / 60₂ = 1.0

2. Fats (R.Q. is less than 1)
Substrates like fats are poorer in oxygen than carbohydrates. Thus, more oxygen is utilized for its complete oxidation.
2(C₅₁ H₉₈ O₆) + 145O₂ → 102CO₂ + 98H₂O + Energy
R.Q. = C0₂ / 0₂ = 102 / 145 = 0.7

3. Protein respiration (R.Q. is less than 1)

1. When proteins serve as respiratory substrate, they are first degraded to amino acids.
2. Then, amino acids are converted into various intermediates of carbohydrates.
3. However, amino acids have low proportion of O₂ as compared to carbohydrates.
4. Thus, they require more O₂ during their complete oxidation and value of R.Q. becomes less than 1.
5. In case of proteins, the R.Q. is approximately 0.9.

Balbharti Maharashtra State Board 11th Biology Textbook Solutions Chapter 2 Systematics of Living Organisms Textbook Exercise Questions and Answers.

Maharashtra State Board 11th Biology Solutions Chapter 2 Systematics of Living Organisms

1. Choose the correct option.

Question (A)
Which of the following shows single-stranded RNA and lacks protein coat?
(a) Bacteriophage
(b) Plant virus
(c) Viroid
(d) Animal virus
Answer:
(c) Viroid

Question (B)
Causative agent of red tide is ________ .
(a) Dinoflagellate
(b) Euglenoid
(c) Chrysophyte
(d) Lichen
Answer:
(A) Dinoflagellate

Question (C)
Select the odd one out for Heterotrophic bacteria.
(a) Nitrogen-fixing bacteria
(b) Lactobacilli
(c) Methanogens
(d) Cyanobacteria
Answer:
(c) Methanogens or (d) Cyanobacteria

Question (D)
Paramoecium: Ciliated Protist :: Plasmodium: _______ .
(a) Amoeboid protozoan
(b) Ciliophora
(c) Flagellate protozoan
(d) Sporozoan
Answer:
(d) Sporozoan

2. Answer the following

Question (A)
What are the salient features of Monera?
Answer:
Salient features of Kingdom Monera:

1. Size: The organisms included in this kingdom are microscopic, unicellular and prokaryotic.
2. Occurrence: These are omnipresent. They are found in all types of environment which are not generally inhabited by other living beings.
3. Nucleus: These organisms do not have well-defined nucleus. DNA exists as a simple double-stranded circular single chromosome called as nucleoid. Apart from the nucleoid they often show presence of extrachromosomal DNA which is small circular called plasmids.
4. Cell wall: Cell wall is made up of peptidoglycan (also called murein) which is a polymer of sugars and amino acids.
5. Membrane-bound cell organelles: Membrane-bound cell organelles like mitochondria, chloroplast, endoplasmic reticulum are absent. Ribosomes are present, which are smaller in size (70S) than in eukaryotic cells.
6. Nutrition: Majority are heterotrophic, parasitic or saprophytic in nutrition. Few are autotrophic that can be either photoautotrophs or chemoautotrophs.
7. Reproduction: The mode of reproduction is asexual or with the help of binary fission or budding. Very rarely, sexual reproduction occurs by conjugation method.
8. Examples:
   Archaebacteria: e.g. Methanobacillus, Thiobacillus, etc.
   Eubacteria: e.g. Chlorobium, Chromatium, and Cyanobacteria e.g. Nostoc, Azotobacter, etc.

Question (B)
What will be the shape of a bacillus and coccus type of bacteria?
Answer:
The shape of bacillus type of bacteria is rod-shaped and coccus is spherical.

Question (C)
Why is binomial nomenclature important?
Answer:
Binomial nomenclature is important because:

1. The binomials are simple, meaningful and precise.
2. They are standard since they do not change from place to place.
3. These names avoid confusion and uncertainty created by local or vernacular names. The organisms are known by the same name throughout the world.
4. The binomials are easy to understand and remember.
5. It indicates phylogeny (evolutionary history) of organisms.
6. It helps to understand inter-relationship between organisms.

3. Write short notes

Question (A)
Write a note on useful and harmful bacteria.
Answer:
(i) Useful bacteria:
Most of the bacteria act as a decomposer. They breakdown large molecules in simple molecules or minerals. Examples of some useful bacteria:
Lactobacillus’. It helps in curdling of milk.
Azotobacter. It helps to fix nitrogen for plants.
Streptomyces: It is used in antibiotic production such as streptomycin.
Methanogens: These are used for production of methane (biogas) gas from dung.
Pseudomonas spp. and Alcanovorax borkumensis: These bacteria have the ability to destroy the pyridines and other chemicals. Hence, used to clear the oil spills.

(ii)Harmful bacteria:
This includes disease causing bacteria. They cause various diseases like typhoid, cholera, tuberculosis, tetanus, etc. Examples of some harmful bacteria:
Salmonella typhi: It is a causative organism of typhoid.
Vibrio cholerae: It causes cholera.
Mycobacterium tuberculosis’. It causes tuberculosis.
Clostridium tetani: It causes tetanus.
Clostridium spp.: It causes food poisoning.
Many forms of mycoplasma are pathogenic.
Agrobacterium , Erwinia, etc are the pathogenic bacteria causing plant diseases.
Animals and pets also suffer from bacterial infections caused by Brucella, Pastrurella, etc.

Question (B)
Write short note on five kingdom system.
Answer:
Five kingdom system of classification was proposed by R.H. Whittaker in 1969. This system shows the phylogenetic relationship between the organisms.
The five kingdoms are:

1. Kingdom Monera
2. Kingdom Protista
3. Kingdom Plantae
4. Kingdom Fungi
5. Kingdom Animalia

Question (C)
Write short note on useful fungi.
Answer:
Economic importances of fungi are as follows:
1. Role of fungi in medicine:
(a) Antibiotic penicillin is obtained from Penicillium.
(b) Drugs like cyclosporine, immunosuppressant drugs, precursors of steroid hormones, etc are isolated from fungi.

2. Role of fungi in industries:
(a) Yeast is used in bread making. It causes dough to rise and make the bread light and spongy. It is also used in breweries or wine making industries. Sugars present in grapes are fermented by using yeast. This results in production of alcohol which is used for making wine.
(b) Lichen is a symbiotic association of algae and fungi are used in preparation of litmus paper which is used as acid-base indicator.

3. Role of fungi in food:
(a) Fungi like mushrooms are consumed as a food. These are rich source of protein.
(b) Fungi genus Penicillium helps in ripening of cheese.

4. Role of fungi as biocontrol agents:
(a) Fungi help to control growth of weeds.
(b) Pathogenic fungi like Fusarium sp., Phytophthorapalmivora, Alternaria crassa, etc act as mycoherbicides.

Question 4.
Complete tree diagram in detail.

Answer:

5. Draw neat labelled diagrams

Question (A)
Draw neat and labelled diagram of Paramoecium.
Answer:
Characteristics:

1. It belongs to kingdom Protista. It is further classified as animal like protist.
2. It lacks cell wall.
3. It shows heterotrophic and holozoic nutrition.
4. It is a ciliated protozoan where locomotion is due to cilia.
5. It has gullet (a cavity) which opens on the cell surface.

Quesiton (B)
Draw neat and labelled diagram of Euglena.
Answer:
Characteristics:
It belongs to kingdom Protista. It is further classified into euglenoids.

1. Dinoflagellates:

1. They are aquatic (mostly marine) and autotrophic (photosynthetic).
2. They have wide range of photosynthetic pigments which can be yellow, green, brown, blue and red.
3. The cell wall is made up of cellulosic stiff plates.
4. A pair of flagella is present, hence they are motile.
5. They are responsible for famous ‘red tide’. E.g. Gonyaulax. It makes sea appear red.

2. Euglenoids:

1. They lack cell wall but have a tough covering of proteinaceous pellicle.
2. Pellicle covering provides flexibility and contractibility to Euglena.
3. They possess two flagella, one short and other long.
4. They behave as heterotrophs in absence of light but possess pigments, similar to that of higher plants, for photosynthesis.

Question (C)
Draw a neat labelled diagram of TMV.
Answer:

Question 6.
Complete chart and explain in your word.

Answer:

Depending upon the host, viruses are classified into three types as:
1. Plant virus
2. Animal virus
3. Bacterial virus (Bacteriophage)

1. Plant virus:
(a) Generally, they are rod shaped or cylindrical with helical symmetry.
(b) Majority of plant viruses have RNA as their genetic material. (Exception: Cauliflower Mosaic Virus has double stranded DNA as genetic material)
(c) Plant viruses cause disease in plants, e.g. Tobacco Mosaic Virus (TMV).

2. Animal virus:
(a) Generally, they are polyhedral in shape with radial symmetry.
(b) They have either DNA or RNA as genetic material.
(c) It causes disease to majority of animals including human beings, e.g. Influenza virus.

3. Bacteriophage:
(a) They have tadpole-like shape.
(b) They infect bacteria and hence are called as bacteriophage.
(c) Bacteriophages were discovered by Twort.
(d) Bacteriophages have double stranded DNA as the genetic material.
(e) Its body consists of head, collar and tail.

Question 7.
Identify the following diagram, label it and write detail information in your words.



Answer:
The given figure represents Bacteriophage.

A.

B.

c.

D.

E.

F.

Question 8.
The scientific name of sunflower is given below. Identify the correctly written name.
(A) Helianthus annus
(B) Helianthus Annus
(C) Helianthus annuus L.
(D) Helianthus annuus l.
Answer:
The correctly written scientific name of sunflower is Helianthus annuus L.

Question 9.
Match the following.

Kingdom	Examples
1. Monera	a. Riccia
2. Protista	b. Cyanobacteria
3. Plantae	c. Rhizopus
4. Fungi	d. Diatoms

Answer:

Kingdom	Examples
1. Monera	b. Cyanobacteria
2. Protista	d. Diatoms
3. Plantae	a. Riccia
4. Fungi	c. Rhizopus

Question 10.
Complete the following.
1. Plant-like Protista – [ ]
2. [ ] – Entamoeba

Practical /Project:

Question 1.
Make a group of students. Observe living organisms in your school/college campus and try to write their characters with respect to habit, habitat, mode of nutrition, growth- determinate or indeterminate, type of reproduction – vegetative reproduction, asexual reproduction, sexual reproduction. With the help of similarity and dissimilarity, try to classify organisms into different categories. Similar work should implement for animal group.
Answer:
The common living organisms observed near school/college are:
1. Plants
Habit: Herb, shrub, tree, etc.
Habitat: Terrestrial or aquatic
Mode of nutrition: Autotrophic
Growth: Indeterminate
Types of reproduction: Vegetative, asexual and sexual reproduction.

2. Animal e.g. dog, cats, cow, etc.
Habitat: Terrestrial
Mode of nutrition: Heterotrophs
Growth: Determinate
Types of reproduction: Only sexual reproduction

3. Birds e.g. Crow, sparrow, etc.
Habitat: Aviary (shows diverse habitat)
Mode of nutrition: Heterotrophs Growth: Determinate
Types of reproduction: Only sexual reproduction
[Note: Students are expected to collect more information about characteristics of living organisms and classify them into different categories]

Question 2.
Find out types of lichens and its economic importance.
Answer:
Types of lichens are:
1. Based on fungal components:
(a) Ascolichens:
In this category, the fungal partner belongs to Ascomycetes group of fungi.
(b) Basidiolichens:
Here, the fungal partner belongs to Basidiomycetes group of fungi.
(c) Deuterolichens:
In this category, the fungal partner belongs to Deuteromycetes group of fungi.

2. Based on their forms:
(a) Crustose lichen:
These lichens show crust-like growth.
These lichens grow on rocks and bark of the trees,
e. g. Graphis, Lecanora, Haematomma, etc.
(b) Foliose lichen:
These lichens grow on trees in the hilly regions.
The thallus is like a dry forked leaf,
e. g. Parmelia, Collema, Peltigera
(c) Fruticose lichen:
These lichens are seen on the branches of trees hanging down.
They are cylindrical, well branched and pendulous, with hair-like outgrowths,
e. g. Usnea, Cladonia, Alectoria, etc.

3. Economic importance of lichens:

(a) Lichen as food and fodder:
Many species of lichens are used as food by animals including man. Lichens contain a substance lichenin which is similar to carbohydrate making them edible. Parmelia is used in curry powder in India. Lichens like Cladonia, Citraria, Evernia, Parmelia are used as fodder as they form a favourite food for reindeers and cattles.

(b) Lichens in medicine:
Lichens contain usnic acid due to which they are used in medicines. Usnea and Cladonia species are used as an antibiotic against Gram positive bacteria.
Species like Lobaria, Citraria are useful in respiratory disease like T.B., Peltigera is useful in hydrophobia, Parmelia is used in treatment of epilepsy, whereas Usnea is used in urinary disease. Some lichens are also used in medicine due to their anticarcinogenic property.

(c) Industrial use of lichens:
1. Lichens are used in various dyes for colouring fabrics.
2. Species like Rocella and Lasallia are used in preparation of litmus paper which is acid-base indicator.
3. In Sweden and Russia, lichens are used for production of alcohol.
4. Orcein is a biological stain obtained from Orchrolechia androgyna and O. tortaria.
5. Some lichens are also used in tanning process in leather industry.
6. Evernia and Ramalina are the sources of essential oils which are used in preparation of soaps and other cosmetics.

(d) Other uses of lichens:
1. Lichens are used in cosmetics.
2. Some lichens like Everniaprunastri also known as oakmoss is used in making perfumes.
3. Lichen is also used as a preservative for beer.

11th Biology Digest Chapter 2 Systematics of Living Organisms Intext Questions and Answers

Can you tell? (Textbook Page No. 7)

Enlist uses of taxonomy?
Answer:
Uses of taxonomy are as follows:

1. It is used to assign each organism an appropriate place in a systematic framework of classification.
2. It is used to group animals and plants by their characteristics and relationships.
3. It is used to classify organisms based upon their similarities and differences.
4. It is used for nomenclature of an organism. Assigning a name to an organism is essential for its identification without confusion throughout the scientific world.
5. It is used to serve as an instrument for identification of an organisms. A newly isolated organism can be placed to its nearest relative or can be identified as a new organism with unknown characteristics.
6. It becomes easier to understand the evolutionary trends in different groups of organisms.

Can you tell? (Textbook Page No. 7)

Which characters of organisms are visible characters?
Answer:
The visible characters of organisms include habit, colour, mode of respiration, growth, reproduction, etc.

Can you tell? (Textbook Page No. 7)

What is evolution?
Answer:

1. It is believed that the life originated on earth in its very simple form.
2. Constant struggle of the early living beings gave rise to more and more perfect forms of life.
3. This struggle and progress are evolution which led to formation of diverse life forms.

Can you tell? (Textbook Page No. 7)

What is DNA barcoding?
Answer:
DNA barcoding is a new method for identification of any species based on its DNA sequence, which is obtained from a tiny tissue sample of the organism under study.

Can you tell? (Textbook Page No. 7)

Name the recent approaches in taxonomy.
Answer:
The recent approaches in taxonomy includes:

1. Morphological Approach
2. Embryological Approach
3. Ecological Approach
4. Behavioral Approach / Ethological Approach
5. Genetical Approach / Cytological Approach
6. Biochemical Approch
7. Numerical Taxonomy

Can you tell? (Textbook Page No. 9)

Make a flow chart showing taxonomic hierarchy.
Answer:
Kingdom → Sub-kingdom → Division / Phylum → Class → Cohort / Order → Family Genus → Species

Do Yourself (Textbook Page No. 16)

Complete the table (given on textbook Page No.16) through collecting information about sunflower, tiger with characteristic features.
(i) Sunflower:

Category	Taxon	Characteristics
Kingdom	Plantae	Autotrophic, photosynthetic, cell wall present.
Sub-kingdom	Phanerogamae	Seed producing plants, reproductive structures are visible.
Division	Angiospermae	Seeds are enclosed within the fruit.
Class	Dicotyledonae	Two cotyledons, tap root system, reticulate venation, pentamerous symmetry of flower, vascular bundle open.
Order	Asterales	Capitulum inflorescence, showing ray florets and disc florets.
Family	Asteraceae	Aster family
Genus	Helianthus	–
Species	annuus	–

(ii) Tiger:

Category	Taxon	Characteristics
Kingdom	Animalia	Multicellular eukaryotes, cell wall absent, heterotrophic nutrition.
Phylum	Chordata	Notochord present
Class	Mammalia	Presence of mammary gland
Order	Carnivora	Carnivorous in nature
Family	Felidae	Cat-like mammals
Genus	Panthera	Large cats
Species	tigris	–

Can you tell? (Textbook Page No. 9)

Why horse and ass are considered to be two different species or animals?
Answer:
1. Species is a group of organisms that can interbreed under natural conditions to produce fertile offsprings.
2. Horse and ass (donkey) are considered to be two different species or animals, because they cannot interbreed under natural condition to produce fertile offspring.

Internet my friend: (Textbook Page No. 9)

(i) Collect the information about most recent system of classification of living organisms and Kingdom System of Classification, e.g. Search for APG system of classification for Plants.
Answer:
[Note: Students are expected to collect more information about most recent system of classification of living organisms and Kingdom System of Classification from internet on their own.]

(ii) Collect the information about classification systems for all types of organisms.
Answer:
[Note: Students are expected to collect more information about classification systems for all types of organisms from internet on their own.]

Can you recall? (Textbook Page No. 6)

What is Five Kingdom system of classification?
Answer:
Five kingdom system of classification was proposed by R.H. Whittaker in 1969. This system shows the phylogenetic relationship between the organisms.
The five kingdoms are:

1. Kingdom Monera
2. Kingdom Protista
3. Kingdom Plantae
4. Kingdom Fungi
5. Kingdom Animalia

Can you tell (Textbook Page No. 14)

Classify fungi into their types.
Answer:
Fungi are classified into four types on the basis of their structure, mode of spore formation and fruiting bodies as follows:
1. Phycomycetes:
Members of this class are commonly called as algal fungi.
These are consisting of aseptate coenocytic hyphae.
They grow well in moist and damp places on decaying organic matter as well as in aquatic habitats or as parasites on plants.
e.g. Mucor, Rhizopus (bread mold), Albugo (parasitic fungus on mustard).

2. Ascomycetes:
These are commonly called as sac fungi.
These are multicellular. Rarely they are unicellular (e.g. Yeast).
Hyphae are branched and septate.
They can be decomposers, parasites or coprophilous (grow on dung).
Some varieties of this class are consumed as delicacies such as morels and truffles.
Neurospora is useful in genetic and biochemical assays.
e.g. Aspergillus, Penicillium, Neurospora, Claviceps, Saccharomyces (unicellular ascomycetes).

3. Basidiomycetes:
These are commonly called as club fungi.
They have branched septate hyphae.
e.g. Agaricus (mushrooms), Ganoderma (bracket fungi), Ustilago (smuts), Puccinia (rusts), etc.

4. Deuteromycetes:
It is a group of fungi which are known to reproduce only asexually.
They are commonly called imperfect fungi.
They are mainly decomposers, while few are parasitic, e.g. Alternaria.

Can you tell? (Textbook Page No. 14)

Write a note on economic importance of fungi.
Answer:
Economic importances of fungi are as follows:
1. Role of fungi in medicine:
(a) Antibiotic penicillin is obtained from Penicillium.
(b) Drugs like cyclosporine, immunosuppressant drugs, precursors of steroid hormones, etc are isolated from fungi.

2. Role of fungi in industries:
(a) Yeast is used in bread making. It causes dough to rise and make the bread light and spongy. It is also used in breweries or wine making industries. Sugars present in grapes are fermented by using yeast. This results in production of alcohol which is used for making wine.
(b) Lichen is a symbiotic association of algae and fungi are used in preparation of litmus paper which is used as acid-base indicator.

3. Role of fungi in food:
(a) Fungi like mushrooms are consumed as a food. These are rich source of protein.
(b) Fungi genus Penicillium helps in ripening of cheese.

4. Role of fungi as biocontrol agents:
(a) Fungi help to control growth of weeds.
(b) Pathogenic fungi like Fusarium sp., Phytophthorapalmivora, Alternaria crassa, etc act as mycoherbicides.

Can you tell? (Textbook Page No. 14)

Why are fungi considered as heterotrophic organisms?
Answer:
In fungi, chloroplast is absent, thus they cannot synthesize their own food by photosynthesis. Fungi decompose the organic matter by breaking down with the help of enzymes from which they absorb nutrients. Thus, exhibiting heterotrophic mode of nutrition.

Can you tell? (Textbook Page No. 14)

What are coenocytic hyphae?
Answer:
1. In filamentous fungi, body consists of mycelium which is formed by a network of hyphae.
2. When these hyphae are non-septate, multinucleated, they are known as coenocytic hyphae.

Can you tell? (Textbook Page No. 14)

(i) How are fungi different from plants?
Answer:
Fungi are different from plants because:
(a) They lack chloroplast hence, do not perform photosynthesis and are heterotrophic in nutrition. Whereas plants are autotrophic and prepare their own food by photosynthesis.
(b) They are separated from Plantae based on their saprophytic mode of nutrition.
(c) Fungi are decomposers of ecosystem whereas plants are producers of ecosystem.
(d) In fungi, cell wall is made up of fungal cellulose or chitin. Whereas in plants, cell wall is made up of cellulose and pectic compounds.

(ii) Have you seen any diseased plant in your farm?
Answer:
Yes, I have seen some diseased plants in our farm.
There are different pathogens like fungi, bacteria, viruses that cause diseases in plants.
The common plant diseases are:
(a) Leaf rust disease: It is caused by fungus Puccinia triticina. It is the most common rust disease of wheat.
(b) Blight disease in rice: It is caused by harmful bacteria Xanthomonas oryzae. It causes wilting of seedlings and yellowing and drying of leaves.
(c) Early blight of potato: It is caused by fungi Alternaria solani. It causes ‘bulls eye’ patterned leaf spots and tuber blight on potato.
(d) Crown gall disease: It is caused by Agrobacterium tumefaciens. This pathogen infects the plant and forms rough surfaced galls on stem and roots.
[Students are expected to write their observations about diseased plants found informs]

Can you tell? (Textbook Page No. 14)

Complete the following table:
Answer:

Plantae	Animalia
1. Autotrophic mode of nutrition.	Heterotrophic mode of nutrition.
2. They do not show locomotion.	They show locomotion.
3. Cell wall is present.	Cell wall is absent.
4. Chloroplast present.	Chloroplast absent.
5. They do not possess nervous system.	They possess well developed nervous system, i
6. Reproduction can be both sexual and asexual.	Mainly shows sexual reproduction.

Can you tell? (Textbook Page No. 15)

Why are viruses called infectious nucleoproteins?
Answer:
1. Viruses are acellular, highly infectious and ultramicroscopic.
2. Viruses possess their own genetic material in the form of either DNA or RNA, but never both. The genetic material in viruses is covered by a protein coat (capsid), hence called nucleoprotein.
3. They do not show any activity outside the body of host but once they enter their specific host cell, they start multiplying within the living host cells.
4. Viruses lack their own metabolic machinery, they make use of the cellular machinery of the host i.e. ribosome for the synthesis of protein during their reproduction and therefore, they cause severe infection. Thus, they are called infectious nucleoproteins.

Can you tell? (Textbook Page No. 15)

Describe genetic material in plant and animal viruses as well as in bacteriophages.
Answer:
The genetic material in different viruses is as given below:
1. Plant virus: (b) Majority of plant viruses have RNA as their genetic material. (Exception: Cauliflower Mosaic Virus has double-stranded DNA as genetic material)
2. Animal virus: (b) They have either DNA or RNA as genetic material.
3. Bacteriophage: (d) Bacteriophages have double-stranded DNA as the genetic material.

Can you tell? (Textbook Page No. 15)

Differentiate between viruses and viroids.
Answer:

Viruses	Viroids
1. They have high molecular weight.	They have low molecular weight.
2. They are larger in size.	They are smaller in size.
3. They can infect plant, animals and bacteria.	They mainly infect plants.
4. The genetic material can be ss-RNA, ds-RNA or DNA.	The genetic material is single stranded circular RNA.
5. Protein coat is present.	Protein coat is absent.
6. mosaic disease is a plant disease caused by viruses.	Tomato chloric dwarf is a plant disease caused by viroids.

Internet my friend. (Textbook Page No. 15)

In modern medicine, certain infectious neurological diseases were found to be transmitted by abnormally folded proteins. These proteins are called prions. The word prion comes from ‘proteinaceous infectious particle’, e.g. mad cow disease in cattle, Jacob’s disease in human.
Find more information about prions.
Answer:
Prions:
1. A prion is a misfolded form of a protein generally present in brain cells.
2. When the prion gets into a cell containing the normal form of the protein, the prion somehow converts normal protein molecules to the misfolded prion versions.
3. Several prions then aggregate into a complex that can convert other normal proteins to prions.
4. Prions can be transmitted through blood, surgical instruments and contaminated food.
5. Diseases caused by prions are Bovine Spongiform Encephalopathy in cattles, Kuru and Creutzfeldt – Jakob disease in humans.
[Note: Students are expected to search for more information about Prions on internet]

Balbharti Maharashtra State Board 11th Biology Textbook Solutions Chapter 1 Living world Textbook Exercise Questions and Answers.

Maharashtra State Board 11th Biology Solutions Chapter 1 Living world

1. Choose the correct option.

Question (A)
Which is not a property of living beings?
(a) Metabolism
(b) Decay
(c) Growth
(d) Reproduction
Answer:
(b) Decay

Question (B)
A particular plant is strictly seasonal plant. Which one of the following is best suited if it is to be studied in the laboratory?
(a) Herbarium
(b) Museum
(c) Botanical garden
(d) Flower exhibition
Answer:
(a) Herbarium

Question (C)
A group of students found two cockroaches in the classroom. They had a debate whether they are alive or dead. Which life property will help them to do so?
(a) Metabolism
(b) Growth
(c) Irritability
(d) Reproduction
Answer:
(c) Irritability

Question 2.
Distinguish between botanical gardens, zoological parks and biodiversity parks with reference to characteristics.
Answer:

Botanical Gardens	Zoological Parks	Biodiversity Parks
1. Plants of different varieties collected from different parts of the world are grown in vivo in a scientific and systematic manner in a botanical garden.	Zoological parks are places where wild animals are kept in captivity.	It is an assemblage of species that      form    self-sustaining communities on degraded/ barren landscape.
2. It is a type of ex situ conservation.	It is a type of ex situ conservation.	It is a type of in situ conservation.
3. It is related to conservation of various flora.	It is related to conservation of various fauna.	It is related to conservation of all biodiversity.

3. Answer the following questions

Question (A)
Jijamata Udyan, the famous zoo in Mumbai has acclimatised Humboldt penguins. Why should penguins be acclimatised when kept at a place away from their natural habitat?
Answer:

1. Zoological park (zoo) is a type of ex-situ conservation in which wild animals are kept in captivity.
2. Humboldt penguins are native to South America and the surrounding environment differs significantly at Jijamata Udyan (zoo) in Mumbai.
3. In order to ensure that these penguins survive longer and are healthy they need to be acclimatised (adjust) to their new environment slowly.
4. If they are not acclimatised or the facilities in the zoo where the penguins are kept are not optimal/ suitable, they may develop abnormal stress and exhibit unusual behaviours due to it.
5. These penguins may also be more prone to contracting certain diseases, since they are suited to living in a particular climatic condition.
6. The enclosure of these penguins consists of water pool, air handling units and a chiller system to maintain temperatures between 12 – 14°C, where the penguins were kept for around 8 to 10 days to get acclimatised to their new environment before allowing any visitors inside the zoo.

Hence, Humboldt penguins need to be acclimatised to their new surroundings, when kept at a place away from their natural habitat.

Question (B)
Riya found a peculiar plant on her visit to Himachal Pradesh. What are the ways she can show it to her biology teacher and get information about it?
Answer:

1. Riya can press and mount the plant specimen on a herbarium sheet and preserve the dried plant material, until she returns back from her visit.
2. She can also write any available information regarding the collected specimen on the herbarium sheet, which can be useful for further studies with her biology teacher.
3. Various taxonomical aids can be useful to get information about this peculiar plant.

[Note: In order to conserve the local flora, Riya can collect photographs ofplant and describe it’s structure to her teacher.]

Question (c)
At Andaman, authorities do not allow tourists to collect shells from beaches. Why must it be so?
Answer:

1. Seashells are an important part of the coastal ecosystem and are crucial for the survival of various marine creatures.
2. They provide material for building nests of birds and also act as a substratum for attachment of algae, sea grass, sponges and various microbes.
3. Fishes use shells for hiding from predators, whereas hermit crabs use shells as temporary shelters.
4. Removal of seashells from seashores may also indirectly affect the rate of shoreline erosion.

Hence, in an attempt to protect the ecosystem, authorities in Andaman do not allow tourists to collect shells from beaches.

Question (D)
Why do we have greenhouse in botanical gardens?
Answer:

1. Greenhouse is a structure with suitable walls and a roof in which plants are grown under regulated climatic conditions.
2. Most botanical gardens exhibit ornamental plants which require stringent/ optimum climatic conditions for their growth and/or flowering.
3. The greenhouse associated with botanical gardens are also used to grow and propagate those plants that may not survive seasonal changes.

Question (E)
What do you understand from terms like in situ and ex situ conservation?
Answer:
1. In situ conservation: It includes conservation of species in their natural habitats. Grazing, cultivation and collection of products from the forests is banned in such areas. Legally protected areas include national parks, wildlife sanctuaries and biosphere reserves.
2. Ex situ conservation: It includes conservation of species outside their natural habitats. Species are conserved in botanical gardens, culture collections and zoological parks.

4. Write short notes

Question (A)
Write a short note on role of human being in biodiversity conservation.
Answer:

1. Due to rapid increase in human population and industrialization, humans have over-utilized natural resources; leading to degradation of the environment and hence only humans can help conserve the ecosystem.
2. Humans are capable of conserving and improving the quality of nature and thus, can play a major role in biodiversity conservation.
3. In order to conserve biodiversity and its environmental resources, humans must use the resources rationally and avoid excessive degradation of environment.
4. Human beings are stakeholders of the environment and need to come together to overcome pollution and improve the environment quality in order to conserve biodiversity. E.g. Ban or limit on use of harmful products (plastic, chemicals, etc.) that are toxic to various birds, animals, etc.
5. Human beings also play a role in conservation of biodiversity by establishment of various sites for in situ (national parks, wildlife sanctuaries and biosphere reserves) and ex situ (botanical gardens, culture collections and zoological parks) conservation.

Question (B)
Importance of botanical garden
Answer:
The importance of botanical gardens is as follows:

1. It is a place where there is an assemblage of living plants maintained for botanical teaching and research purpose.
2. Botanical gardens are important for their records of local flora.
3. Botanical gardens provide facilities for the collection of living plant materials for botanical studies.
4. Botanical gardens also supply seeds and material for botanical investigations.
5. The development of botanical gardens in any country is associated with its history of civilization, culture, heritage, science, art, literature and various other social and religious expressions.
6. Botanical gardens besides possessing an outdoor garden may contain herbaria, research laboratory, greenhouses and library.
7. Botanical gardens are not only important for botanical studies, but also to develop tourism in the country.

Question 5.
How can you, as an individual, prevent the loss of biodiversity?
Answer:
As individuals, we can prevent loss of biodiversity in the following ways:

1. Increasing awareness about environmental issues. Making posters that provide more information about biodiversity conservation, to raise public awareness.
2. Increased support and/ or active participation in government policies and actions laid down for conservation of biodiversity.
3. Protect various plant and animal species in our surrounding.
4. Set up bird and bat houses wherever possible.
5. Prevent felling of trees especially native plants or trees in a particular area.
6. Reduce, recycle and reuse resources. Especially, reduce pollution and use of plastic bags and other materials that are potential threats for the environment.
7. Use environment friendly products, segregate and dispose garbage correctly.
8. Convince people about the importance of trees and the need to participate in tree plantation campaign.
9. Obey the rules that fall under Biodiversity Act.

[Students can use the given points as reference and mention additional preventive measures on their own.]

Practical / Project:

Question 1.
Make a herbarium under the guidance of your teacher.
[Students are expected to perform the given activity by themselves under the guidance of their teacher.]

Question 2.
Find out information about any one sacred grove (Devrai) in Maharashtra.
Answer:
Sacred groves in Maharashtra are located in districts like Ahmednagar, Bhandara, Chandrapur, Jalgaon, Kolhapur, Nashik, Pune, Raigad, Ratnagiri, Sangli, Satara, Sindhudurg, Thane, Yavatmal.
[Source: Data as per C.P.R. Environment Education Centre, Chennai.]
e. g. Sacred grove of Parinche valley, Pune district of Maharashtra:

The Parinche valley region is comprised of the inaccessible rear part of the Purandhar fort and its surrounding valley region and is situated about 63 km to the southeast of Pune city and 18 km from Saswad town. The total area of the valley region is about 132 sq. km. Parinche is the biggest village and a nodal place in the valley. The majority (12) of the documented groves are located in the Kaldari and Pangare zones. The size of the sacred groves has however reduced due to various human related activities that have taken place in recent years.

The biggest sacred grove in the Parinche valley belongs to Buvasaheb of Tonapewadi and spreads over an area of 4.80 hectares. The forest types are unique to the groves. Presence of key species in the sacred groves varies from region to region. Two key tree species, i.e. Terminalia bellerica and Ficus spp., are present in these sacred groves which have almost disappeared from the surrounding areas.

Large buttressed trees are another important feature of well-preserved sacred groves. The presence of these tree species indicates the vegetation of the past and also the type of potential vegetation that can be regenerated in these regions.
[Source: Waghchaure, C. K., Tetali, P., Gunale, V. R., Antia, N. H., & Birdi, T. J. (2006). Sacred Groves of Parinche Valley of Pune District of Maharashtra, India and their Importance. Anthropology & Medicine, 13(1), 55-761
[Students can refer the given answer and search for more information about other sacred groves on their own.]

11th Biology Digest Chapter 1 Living world Intext Questions and Answers

Can you recall? (Textbook Page No. 01)

Whether all organism are similar? Justify your answer.
Answer:
No, all organisms are not similar.

1. Organisms on the earth exhibit great diversity.
2. Organisms are grouped as microbes, plants (autotrophs), animals (heterotrophs) and decomposers.
3. Different microbes and decomposers have various shapes and sizes.
4. Plants can be further classified on their shape, size, structure, mode of reproduction, etc. Plants also differ greatly based on the locations in which they are found, e.g. Snowy, desert, forest, aquatic, etc.
5. Even animals show a high degree of variation. They are classified as unicellular, multicellular, invertebrates, vertebrates, etc. Also, based on the environment in which they live, they are classified as terrestrial, aerial, aquatic and amphibians.

Can you tell? (Textbook Page No. 01)

Whether all organisms prepare their own food?
Answer:
No, all organisms do not prepare their own food. Organisms that prepare their own food are known as autotrophs (e.g. Green plants, certain microbes). These organisms prepare their own food in the presence of sunlight, water and carbon dioxide.

Can you recall? (Textbook Page No.01)

what is the difference between living and non-living things?
Answer:

Living Things	Non-living Things
(a) Living things show growth from within.	Non-living things show growth by accumulation of materials on their surface.
(b) They reproduce asexually or sexually, except mules, sterile worker bees, infertile males.	They do not reproduce.
(c) They perform metabolism in order to obtain energy.	No metabolic changes occur in non-living things.
(d) They show irritability and respond to changes in their surroundings.	They do not show irritability.
(e) They undergo ageing and eventually die.	Non-living things do not have a finite life span.

Can we call? (Textbook Page No. 01)

Can we call reproduction as inclusive character of life?
Answer:
No, we cannot call reproduction as an inclusive character of life. Certain organisms like mules and worker bees do not reproduce and are still living. Thus, reproduction cannot be considered as an all inclusive defining characteristic of living organisms.

Can you tell? (Textbook Page No. 01)

Which feature can be considered as all-inclusive characteristic of life? Why?
Answer:
Metabolism can be considered as an all-inclusive (defining) feature of life since it is exhibited by all living organisms and does not take place in non-living things. Another all-inclusive characteristic of life is responsiveness or irritability. This is a unique property of living beings since all living beings are conscious of their surroundings.

Think about it. (Textbook Page No. 01)

(i) Can metabolic reactions demonstrated in a test tube (called ‘in vitro’ tests) be called living?
Answer:
(a) The sum total of all the chemical reactions occurring in the body is known as metabolism and no non-living object exhibits metabolism.
(b) However, metabolic reactions can be demonstrated outside the body in a test tube (cell-free medium).
(c) Thus, isolated metabolic reaction (s) outside the body of an organism, performed in a test tube is neither living nor non-living.
(d) Metabolic reactions occurring in vitro are living reactions but not living things.

(ii) Now a days patients are declared ‘brain dead’ and are on life support. They do not show any sign of self-consciousness. Are they living or non-living?
Answer:
The brain controls all life processes. Hence, when a patient is declared as ‘brain dead’, he does not carry out any of the inclusive defining characters of living things (e.g. metabolism, consciousness, etc.) and is completely dependent on machines. Since, such patients do not show any sign of self-consciousness, these patients cannot exactly be called as living.

Can you tell? (Textbook Page No. 01)

How can we study large number of organisms at a glance?
Answer:
Systematic study of organisms with the help of taxonomical aids can be used to study a large number of organisms at a glance.

Can you tell? (Textbook Page No. 03)

What are the essentials of a good herbarium?
Answer:
The essentials of a good herbarium are as follows:

1. It is essential to identify and label the collected specimen correctly.
2. Specimens should be stored in a dry place.
3. The plants are usually pressed and mounted on the sheet of paper known as herbarium sheets. Some plants are not suitable for pressing or mounting, like succulents, seeds, cones, etc. They need to be preserved in suitable liquid like formaldehyde, acetic alcohol, etc.
4. In order to preserve the specimen for longer durations, acid-free paper, special glues and inks must be used to mount the specimen so that the specimen does not deteriorate.
5. The specimens should be dried well before preparing a herbarium in order to prevent rotting of specimen.
6. It is also essential to record the date, place of collection along with detailed classification and highlighting with its ecological peculiarities, characters of the plant on a sheet.

Local names of plant specimens and name of the collector may be added. This information is given at lower right comer of sheet and is called ‘label’.

Why does the loss of biodiversity matter? (Textbook page no. 03)
Answer:

1. The loss of biodiversity is an moral and ethical issue.
2. Biodiversity helps to maintain stability in an ecosystem.
3. Humans share the environment with various other organisms and harm to these species can result in loss of biodiversity.
4. The loss of even one variety of organisms can affect the entire ecosystem. Hence, due to all these reasons, loss of biodiversity matters.

Find out. (Textbook Page No. 04)

Human being is at key position in maintaining biodiversity of earth. Find out more information about the following.

(i) Laws to protect and conserve biodiversity in India.
Answer:
a. Forest (Conservation) Act, 1980
b. Biological Diversity Act, 2002
c. Wildlife (Protection) Act, 1972
d. Environment Protection Act, 1986
[Students can find out more laws to protect and conserve Biodiversity in India]

(ii) Environmental effects of ambitious projects like connecting rivers or connecting cities by constructing roads.
Answer:
Connecting rivers or connecting cities by constructing roads have the following environmental effects:
(a) They form barriers to animals.
(b) Construction of roads requires cutting down of trees and results in large scale deforestation.
(c) They occupy large land resources resulting in loss of habitat of various species.
(d) It can alter the water flow pattern and damage many ecosystems.
(e) Increase in air, water, soil and noise pollution can disturb various animals and birds, thus affecting their behavioural pattern.

(iii) Did bauxite mining in Western Ghats affect critically endangered species like – Black panther, different Ceropegia spp., Eriocaulon spp. ?
Answer:
(a) The Western Ghats, is one of the global biodiversity hotspots and retains more than 30% of all plant, aquatic, reptile, amphibian and mammal species found in India.
(b) Recently, this ecologically sensitive region has been subjected to various developmental activities that have adversely affected the flora and fauna of the region.
(c) Bauxite mining is one such activity which has had significant negative impact on the local environment. To access bauxite ore deposits, the above-ground vegetation needs to be completely removed, causing large scale deforestation. The vegetation in the adjoining area is also affected due to dumping.
(d) The major threats of this activity include vegetation loss, forest fragmentation and biodiversity loss.
(e) Since most mines fall in Eco-Sensitive Zones (ESZ), it has seriously affected the flora and fauna of the Western Ghats.
(f) Black panthers have frequently been spotted at various locations in the Western Ghats and mining in these areas can seriously affect their health and numbers.
(g) Certain species of Ceropegia and Eriocaulon that are endemic in the area have been reported to be critically endangered.
[Source: Chandore A. (2015) Endemic and threatened flowering plants of Western Ghats with special reference to Konkan region of Maharashtra. Journal of Basic Sciences. 2 (21-25)]
Hence it is most likely that bauxite mining in Western Ghats has adversely affected the critically endangered species like – Black panther, different Ceropegia spp., Eriocaulon spp.

Internet my friend. (Textbook Page No. 02)

Collect information about Prof. Almeida, Prof. V. N. Naik, Dr. A. V. Sathe, Dr. P. G. Patwardhan with reference to their taxonomic work and biodiversity conservation.
Answer:
1. Prof. Almeida:
Prof. (Dr.) Marselin R. Almeida was a renowned Plant Taxonomist and Medicinal Plant Consultant of India. He was a curator at the Blatter Herbarium (Mumbai). He discovered four new species of pteridophytes from Bombay presidency. His work includes – Pteridophytes of Maharashtra and Flora of Mahabaleshwar. He has contributed to the Flora of Maharashtra, Sawantwadi and its adjoining areas along with Dr. S. M. Almeida.

2. Prof. V. N. Naik:
Prof. V. N. Naik is a renowned ‘Angiosperms Taxonomist’ of India. He completed the Flora of Marathwada. He has produced 15 Ph.D., 110 research articles and 6 books. His book on ‘Taxonomy of Angiosperms’ (Tata McGraw-Hill Education, 1984) is widely used throughout the world. He is currently a faculty of Dr. Babasaheb Ambedkar Marathwada University, Aurangabad.

3. Dr. A. V. Sathe:
Collection and taxonomic studies of mushrooms in Maharashtra started around 1974. Prof. A.V. Sathe and his team were amongst the first to begin these studies. They recorded 75 species distributed in 43 genera. These species were collected from Maharashtra, Karnataka and Kerala. The collection of these species was documented in the form of a Monograph on Agaricales.
[Source: Borkar P., Doshi A., Navathe D. (2015) Mushroom diversity of Konkan region of Maharashtra, India. Journal of Threatened Taxa. 7(10): 7625-7640]

4. Dr. P. G. Patwardhan:
Dr. Patwardhan and his associates at the M.A.C.S. Research Institute, Pune-renamed as Agharkar Research Institute (ARI), Pune have performed detailed studies on lichens. His school is in possession of over 600 species of crustose lichens, obtained after intensive collection programmes. These specimens have been deposited in the Ajarekar Mycological Herbarium in the Department of Mycology and Plant Pathology at the M.A.C.S. Research Institute, Pune.
[Source:
http://lib.unipune.ac.in:8080/xmlui/bitstreamfhandle/l23456789/7451/07_introduction.pdf? sequence=7&is Allowedly]
[Students are expected to find more information on their own.]

Can you tell? (Textbook Page No. 03)

Why should we visit botanical gardens, museums and zoo?
Answer:

1. Botanical gardens, museums and zoos are taxonomical aids which can be used to study biodiversity.
2. Botanical gardens have a wide range of plant species that are protected and preserved which can be observed and studied.
3. Museums help gain information about various plants and animals that are preserved and may even be extinct. They act as reference hubs for biodiversity studies.
4. Zoos provide information about various animals.

They also harbour certain endangered animals and help us understand the role of biodiversity conservation. They can also be visited to study the food habits and behaviour of animals. Hence, we should visit botanical gardens, museums and zoos.

Can you tell? (Textbook Page No. 03)

What is ‘ex-situ’and ‘in-situ’ conservation?
Answer:
1. In situ conservation: It includes conservation of species in their natural habitats. Grazing, cultivation and collection of products from the forests is banned in such areas. Legally protected areas include national parks, wildlife sanctuaries and biosphere reserves.
2. Ex situ conservation: It includes conservation of species outside their natural habitats. Species are conserved in botanical gardens, culture collections and zoological parks.

Internet my friend. (Textbook Page No. 04)

(a) Collect information about botanical gardens, zoological parks and biodiversity hotspots in India.
Answer:
a. Botanical gardens in India:

1. Acharya Jagadish Chandra Bose Indian Botanic Garden	Kolkata
2. Lloyd Botanical Garden	Darjeeling
3. National Botanical Research Institute	Lucknow
4. Botanical Garden of the Forest Research Institute	Dehradun
5. The State Botanical Garden	Odisha
6. Botanical Garden	Saharanpur
7. Government Botanical Garden	Ootacamund

b. Zoological Parks in India:

Zoological parks	Location	Type of animals
1. Rajiv Gandhi Zoological Park	Pune [Katraj]	Reptiles, mammals, birds. They have a snake park.
2. Jijamata Udyan	Mumbai	Endangered species of animals and birds.
3. Nehru Zoological Park	Hyderabad	3500 species of birds, animals and reptiles.
4. Indira Gandhi Zoological Park	Vishakhapatnam	Primates, carnivores, small mammals, reptiles and birds.
5. Padmja Naidu Himalayan Zoological Park	Darjeeling	Endangered animals like snow leopards, red pandas, gorals (mountain goat), Siberian tigers and a variety of endangered bird species.
6. Allen Forest Zoo	Kanpur	Hyena, Bear, Rhinoceros, Hippopotamus, Langoor, Musk deer. Ostrich, Emu, Crane etc.
7. Lucknow Zoo	Lucknow	Royal Bengal Tiger, White Tiger, Gibbon, Black Bear, Asiatic Elephant, Great pied, Horn Bill etc.
8. Alipore Zoological Gardens	Kolkata	Royal Bengal Tiger, African Lion, Hippopotamus, Great Indian One-homed Rhinoceros.
9. The Madras Crocodile Bank Trust	Chennai	Crocodiles and many species of turtles, snakes and lizards.
10. Parassinikkadavu Snake Park	Kannur	Spectacled Cobra, King Cobra, Russell’s Viper, Krait and Pit Viper.

c. Biodiversity hotspots in India:

Biodiversity Hotspots

1. The Eastern Himalayas (Arunachal Pradesh, Bhutan, Eastern Nepal)

2. Indo – Burma (Purvanchal Hills, Arakan Yoma, Eastern Bangladesh)

3. The Western Ghats and Srilanka

[Students are expected to use the given table as reference and collect more information on their own.]

(ii) Collect information of endemic flora and fauna of India.
Answer:
(a) Endemic flora:
Albizia sikharamensis (Mimosaceae), Argvreia arakuensis (Convolvulaceae), Arundinella setosa (Poaceae), Acacia diadenia (Mimosaceae), Citrus assamensis (Rutaceae), Magnolia bailloni (Magnoliaceae), etc.

(b) Endemic fauna:
Bare Bellied Hedgehog (Paraechinus nudiventris), Andaman Shrew (Crocidura andamanensis), Aruanchal Macaque (Macaca munzala), Car Nicobar Rat (Rattus palmarum), Peter’s Tube-nosed Bat (Harpiola grisea) etc.
[Source: http://faunaofindia.nic.in/PDFVolumes/spb/056/index.pdf]
[Students are expected to use the given sources and find more information on their own.

Balbharti Maharashtra State Board 11th Chemistry Textbook Solutions Chapter 16 Chemistry in Everyday Life Textbook Exercise Questions and Answers.

Maharashtra State Board 11th Chemistry Solutions Chapter 16 Chemistry in Everyday Life

1. Choose the correct option

Question A.
Oxidative Rancidity is …………….. reaction
a. addition
b. substitution
c. Free radical
d. combination
Answer:
c. Free radical

Question B.
Saponification is carried out by ……………..
a. oxidation
b. alkaline hydrolysis
c. polymerisation
d. Free radical formation
Answer:
b. alkaline hydrolysis

Question C.
Aspirin is chemically named as ……………..
a. Salicylic acid
b. acetyl salicylic acid
c. chloroxylenol
d. thymol
Answer:
b. acetyl salicylic acid

Question D.
Find odd one out from the following
a. dettol
b. chloroxylenol
c. paracetamol
d. trichlorophenol
Answer:
c. paracetamol

Question E.
Arsenic based antibiotic is
a. Azodye
b. prontosil
c. salvarsan
d. sulphapyridine
Answer:
c. salvarsan

Question F.
The chemical used to slow down the browning action of cut fruit is
a. SO₃
b. SO₂
c. H₂SO₄
d. Na₂CO₃
Answer:
b. SO₂

Question G.
The chemical is responsible for the rancid flavour of fats is
a. Butyric acid
b. Glycerol
c. Protein
d. Saturated fat
Answer:
a. Butyric acid

Question H.
Health benefits are obtained by consumption of
a. Saturated fats
b. trans fats
c. monounsaturated fats
d. all of these
Answer:
c. monounsaturated fats

2. Explain the following :

Question A.
Cooking makes food easy to digest.
Answer:

• During the cooking process, high polymers of carbohydrates or proteins are hydrolysed to smaller polymeric units.
• The uncooked food mixture is a heterogeneous suspension which becomes a colloidal matter on cooking.
• As a result, the constituent nutrient molecules present in cooked food are smaller in size and hence, easier to digest, than the uncooked food.

Hence, cooking makes food easy to digest.

Question B.
On cutting some fruits and vegetables turn brown.
Answer:
i. Cutting of fruits and vegetables damage the cells, resulting in release of chemicals.
ii. Depending on the pH of fruits/vegetables, polyphenols are released.
iii. Due to the action of an enzyme, these polyphenols react with oxygen present in the air and get oxidised to form quinones.

iv. Quinones further undergo reactions including polymerization, which results in the formation of brown coloured products called as tannins.
Thus, on cutting, some fruits and vegetables turn brown.

Question C.
Vitmin E is added to packed edible oil.
Answer:

• Vitamin E is a very effective natural antioxidant.
• The phenolic – OH group present in the structure of vitamin E is responsible for its antioxidant activity.
• Also, the long chain of saturated carbon atoms makes it fat soluble.

Therefore, when vitamin E is added to packed edible oil, it prevents the oxidative rancidity of the oil.

Question D.
Browning of cut apple can be prolonged by applying lemon juice.
Answer:

• Browning of cut apple is due to the oxidation of polyphenols at a particular pH to quinones, which further undergoes polymerization to form brown coloured tannins.
• This browning reaction can be prolonged or slowed down by using reducing agents or by changing the pH.
• Applying lemon juice (i.e., citric acid) on the cut apple, lowers the pH at the surface of the apple. This prevents the oxidation reaction. Thus, browning of cut apple can be prolonged by applying lemon juice.

Question E.
A diluted solution (4.8 % w/v) of 2,4,6-trichlorophenol is employed as antiseptic.
Answer:

• 2,4,6-Trichlorophenol (TCP) is more potent antiseptic than phenol.
• It has low corrosive effects as compared to phenol, if used in lower concentrations.

Hence, diluted solution (4.8% w/v) of 2,4,6-trichlorophenol is used as antiseptic.

Question F.
Turmeric powder can be used as antiseptic.
Answer:

• Turmeric powder contains an active ingredient called curcumin.
• Curcumin has antiseptic properties; thus, it is used for wound healing or applied on bruise.

Hence, turmeric powder can be used as antiseptic.

3. Identify the functional groups in the following molecule :

Answer:

4. Give two differences between the following

Question A.
Disinfectant and antiseptic
Answer:

Disinfectant	Antiseptic
1. Disinfectants are applied on non-living surfaces like floors, instruments, sanitary ware, etc. to kill wide range of microorganisms.	1. Antiseptics are applied on the surface of living tissues in order to sterilise them.
2. Disinfectants cannot be applied on wounds.	2. Antiseptics can be directly applied on wounds.
3. p-chloro-o-benzyl phenol	3. Iodine, boric acid, iodoform, dettol, etc.

Question B.
Soap and synthetic detergent
Answer:

Soap	Synthetic detergent
1. Soaps can be broadly classified into two types, i.e., toilet soaps (prepared using KOH) and laundry soaps (prepared using NaOH).	1. Synthetic detergents are of three types, i.e., anionic, cationic and nonionic detergents.
2. Soaps cannot be used in hard water.	2. Synthetic detergents can be used in soft water as well as in hard water.

Question C.
Saturated and unsaturated fats
Answer:

Saturated fats	Unsaturated fats
1. In saturated fat, long chains of tetrahedral carbon atoms in the fatty acid get closely packed together.	1. In unsaturated fats, the presence of one or more C = C bond in long chains of fatty acids, prevent molecules from packing closely together.
2. In saturated fats, the van der Waals forces between long saturated chains are strong. Hence, their melting points are higher than unsaturated fats.	2. In unsaturated fats, the van der Waals forces between long unsaturated chains are weak. Hence, their melting points are lower than saturated fats.

Question D.
Rice flour and cooked rice
Answer:

Rice flour	Cooked rice
1. Rice flour can be stored for a long period of time. It has a long shelf life.	1. Cooked rice cannot be stored for a longer period of time. It has very short shelf life.
2. Rice flour is uncooked food and hence, it is difficult to digest.	2. Cooked rice is easier to digest.

5. Match the pairs.

A group	B group
A. Paracetamol	a. Antibiotic
B. Chloramphenicol	b. Synthetic detergent
C. BHT	c. Soap
D. Sodium stearate	d. Antioxidant
	e. Analgesic

Answer:
A – e,
B – a,
C – d,
D – c

6. Name two drugs which reduce body pain.
Answer:
Aspirin and paracetamol are the two drugs that reduce body pain.

7. Explain with examples

Question A.
Antiseptics
Answer:
i. Antiseptics are used to sterilise surfaces of living tissue when the risk of infection is very high, such as during surgery or on wounds.
ii. Commonly used antiseptics include inorganics like iodine and boric acid or organics like iodoform and some phenolic compounds.

e.g.

• Tincture of iodine (2-3% solution of iodine in alcohol-water mixture) and iodoform serve as powerful antiseptics and is used to apply on wounds.
• A dilute aqueous solution of boric acid is a weak antiseptic used for eyes.
• Various phenols are used as antiseptics. A dilute aqueous solution of phenol (carbolic acid) is used as antiseptic; however, phenol is found to be corrosive in nature. Many chloro derivatives of phenols are more potent antiseptics than the phenol itself. They can be used with much lower concentrations, which reduce their corrosive effects.
• Two of the most common phenol derivatives in use are trichlorophenol (TCP) and chloroxylenol (which is an active ingredient of antiseptic dettol).
• Thymol obtained from oil of thyme (a spice plant) has excellent non-toxic antiseptic properties.

Question B.
Disinfectant
Answer:

• Disinfectants are non-selective antimicrobials.
• They kill a wide range of microorganisms including bacteria.
• They are used on non-living surfaces for example, floors, instruments, sanitary ware, etc.
• Various phenols can be used as disinfectants.
  e.g. p-Chloro-o-benzyl phenol is used as a disinfectant in all-purpose cleaners.

Question C.
Cationic detergents
Answer:
Cationic detergents: These are quaternary ammonium salts having one long chain alkyl group.
e.g. Ethyltrimethylammonium bromide: [CH₃(CH₂)₁₅ – N⁺(CH₃)₃]Br^–

Question D.
Anionic detergents
Answer:
Anionic detergents: These are sodium salts of long chain alkyl sulphonic acids or long chain alkyl substituted benzene sulphonic acids.
e.g. Sodium lauryl sulphate: CH₃(CH₂)₁₀CH₃O\(\mathrm{SO}_{3}^{-}\)Na⁺

Question E.
Non-ionic detergents
Answer:
Nonionic detergents: These are ethers of polyethylene glycol with alkyl phenol or esters of polyethylene glycol with long chain fatty acid.
e.g. a. Nonionic detergent containing ether linkage:

b. Nonionic detergent containing ester linkage: CH₃(CH₂)₁₆ – COO(CH₂CH₂O)_nCH₂CH₂OH

8. Explain : mechnism of cleansing Action of soap with flow chart.
Answer:
The following flow chart shows mechanism of cleansing action of soap:

9. What is meant by broad spectrum antibiotic and narrow spectrum antibiotics?
Answer:
Antibiotics which are effective against wide range of bacteria are known as broad spectrum antibiotics, while antibiotics which are effective against one group of bacteria are known as narrow spectrum antibiotics.

10. Answer in one senetence

Question A.
Name the painkiller obtained from acetylation of salicyclic acid.
Answer:
Aspirin is the pain killer obtained from acetylation of salicylic acid.

Question B.
Name the class of drug often called as painkiller.
Answer:
Analgesics are the class of drug often called as painkiller.

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

Question D.
Draw the structure of chloroxylenol and salvarsan.
Answer:
Structure of chloroxylenol:

Structure of salvarsan:

Question E.
Write molecular formula of Butylated hydroxy toulene.
Answer:
Molecular formula of butylated hydroxytoluene is C₁₅H₂₄O.

Question F.
What is the tincture of iodine ?
Answer:
Tincture of iodine is a 2-3% solution of iodine in alcohol-water mixture.

Question G.
Draw the structure of BHT.
Answer:

Question I.
Write a chemical equation for saponification.
Answer:

Question J.
Write the molecular formula and name of

Answer:
Molecular formula: C₉H₈O₄
Name: Aspirin

11. Answer the following

Question A.
Write two examples of the following.
a. Analgesics
c. Antiseptics
d. Antibiotics
e. Disinfectant
Answer:

No.	Drug type	Examples
i.	Analgesics	Aspirin, paracetamol
ii.	Antiseptics	Dettol, thymol
iii.	Antibiotics	Penicillin, sulphapyridine
iv.	Disinfectant	Phenol, p-Chloro-o-benzyl phenol

Question B.
What do you understand by antioxidant ?
Answer:

• An antioxidant is a substance that delays the onset of oxidant or slows down the rate of oxidation of foodstuff.
• It is used to extend the shelf life of food.
• Antioxidants react with oxygen-containing free radicals and thereby prevent oxidative rancidity.
  e.g. Vitamin E is a very effective natural antioxidant.

Activity :

Collect information about different chemical compounds as per their applications in day-to-day life.
Answer:

No.	Chemical compound	Applications
i.	Vinegar(CH3COOH)	Preservation of food, salad dressing, sauces, etc.
ii.	Magnesium hydroxide [Mg(OH)2]	Common component of antacids (used to relieve heartburn, acid indigestion and stomach upset.)
iii.	Baking soda (NaHCO3)	Cooking, antacid, toothpaste, etc.
iv.	Sodium benzoate (C6H5COONa)	Used as food preservative

[Note: Students can use the above information as reference and collect additional information on their own.]

11th Chemistry Digest Chapter 16 Chemistry in Everyday Life Intext Questions and Answers

Can you recall? (Textbook Page No. 261)

Question i.
What are the components of balanced diet?
Answer:
Carbohydrates, proteins, lipids (fats and oil), vitamins, minerals and water are the components of balanced diet.

Question ii.
Why is food cooked? What is the difference in the physical states of uncooked and cooked food?
Answer:

• Food is cooked in order to make it easy to digest.
• Also, the raw or uncooked food may contain harmful microorganisms which may cause illness. Cooking of food at high temperature kills most of these microorganisms.
• Raw/uncooked food materials like dried pulses, vegetables, meat, etc. are hard and thus, not easily chewable while cooked food is soft and tender, therefore, easily chewable.

Question iii.
What are the chemicals that we come across in everyday life?
Answer:
Detergents, shampoos, medicines, various food flavours, food colours, etc. are different types of chemicals that we come across in everyday life.

Just think (Textbook Page No. 261)

Question i.
Why is food stored for a long time?
Answer:
Food (like various cereals, pulses, pickles) is stored for a long time to make it available in all seasons.

Question ii.
What methods are used for preservation of food?
Answer:
Various physical and chemical methods are used for preservations of food.

• Physical methods like, addition of heat, removal of heat, removal of water, irradiation, etc., are used in order to preserve food.
• Chemical methods like, addition of sugar, salt, vinegar, etc. are employed for preservation of food.

Question iii.
What is meant by quality of food?
Answer:
Food quality can be described in terms of parameters such as flavour, smell, texture, colour and microbial spoilage.

Can you recall? (Textbook Page No. 263)

Question i.
How is Vanaspati ghee made?
Answer:
Vanaspati ghee is prepared by hydrogenation of oils. Hydrogen gas is passed through the oils at about 450 K in the presence of nickel catalyst to form solid edible fats like vanaspati ghee.

Question ii.
What are the physical states of peanut oil, butter, animal fat, Vanaspati ghee at room temperature?
Answer:

Example	Physical state
Peanut oil	Liquid
Butter	Semi-solid
Animal fat	Solid/semi-solid
Vanaspati ghee	Solid/semi-solid

Can you tell? (Textbook Page No. 264)

Question 1.
When is an antipyretic drug used?
Answer:
An antipyretic drug is used to reduce fever (that is, it lowers body temperature when a fever is present).

Question 2.
What type of medicine is applied to a bruise?
Answer:
Antiseptic such as tincture of iodine is applied on a bruise in order to prevent the exposed living tissue from getting infected.

Question 3.
What is meant by a broad spectrum antibiotic?
Answer:
Antibiotics which are effective against wide range of bacteria are known as broad spectrum antibiotic.

Question 4.
What is the active principle ingredient of cinnamon bark?
Answer:
Cinnamaldehyde is the principle active ingredient of cinnamon bark.

Can you tell? (Textbook Page No. 268)

Question i.
Can we use the same soap for bathing as well as cleaning utensils or washing clothes? Why?
Answer:
No, we cannot use the same soap for bathing as well as cleaning utensils or washing clothes due to the following reasons:

• Chemical composition of each type of soap or cleansing material is different.
• Nature, acidity, texture, reactivity towards water (i.e., hard water or soft water), reactivity towards microorganisms, stains are different for each type of soap.
• Depending on these qualities, soaps are classified and used accordingly.
  e.g. pH of soaps used for bathing purpose is different than that of the soap which is used for cleaning utensils.

Thus, we cannot use the same soap for bathing as well as cleaning utensils or washing clothes.

Question ii.
How will you differentiate between soaps and synthetic detergent using borewell water?
Answer:
Borewell water is hard water. Soaps and synthetic detergents react differently with hard water.

1. Soap: Soaps are insoluble in hard water. Borewell water (hard water) contains Ca²⁺ and Mg²⁺ ions. Soaps react with these ions to form insoluble magnesium and calcium salts of fatty acids. These salts precipitate out as gummy substance or form scum.
2. Synthetic detergents: Synthetic detergents can be used in hard water as well. They contain molecules (components) which form soluble calcium and magnesium salts.

Thus, soaps will form scum in borewell water but synthetic detergents will not.

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

Maharashtra State Board 11th Chemistry Solutions Chapter 15 Hydrocarbons

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

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

Maharashtra State Board 11th Chemistry Solutions Chapter 14 Basic Principles of Organic Chemistry

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. methoxy ethane
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 to 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 a reference and collect additional information about isomerism on their own.]