CBSE Class 12

NCERT Class 12 Chemistry Solutions for Chapter 12 provides answers for the questions provided in the textbook. The answers are accurate to the best of our knowledge and are provided by subject experts. The students can refer to these or sure shot success.

The students appearing for various boards and competitive exams can find these solutions helpful for practice. Most of the questions asked in UP board, MP board, Gujarat board, CBSE, etc. are asked from these. To score well in the examinations, the students should go through these solutions atleast once after finishing the entire syllabus.

NCERT Solutions for Class 12 Chemistry Chapter 12 Aldehydes, Ketones and Carboxylic Acids

This chapter explains the structure, physical and chemical properties and applications of aldehydes, ketones and carboxylic acid. It also explains a correlation between the three. It also explains the mechanism of different reactions of aldehydes, ketones and carboxylic acids. The factors affecting the acidity of carboxylic acids and their reactions help in understanding the advanced concepts related to this chapter.

NCERT Solutions for Class 12 Chapter 12 will help you revise the chapter during the examination. It also helps in clearing doubts if any.

NCERT INTEXT QUESTIONS

Question 1:
Write the structures of the following compoimds :
(i) α -Methoxypropionaldehyde
(ii) 3-Hydroxybutanal
(iii) 2-Hydroxycyclopentanecarbaldehyde
(iv) 4-Oxopentanal
(v) Di-sec butylketone
(vi) 4-Fluoroacetophenone
Answer:

Question 2.
Write the structures of products of following reactions:

Answer:

Question 3.
Arrange the following compounds in increasing order of their boiling points :
CH₃CHO,  CH₃CH₂OH,  CH₃OCH₃, CH₃CH₂CH₃
Answer:
The increasing order of boiling points of all these compounds of comparable molecular masses is :

Explanation: We know that the boiling points of liquids are directly related to the magnitude of the intermolecular forces of attraction.

1. Hydrocarbons (alkanes) are completely non-polar. The only attractive forces in their molecules are Van der Waals forces which are quite weak. That is why propane (CH₃CH₂CH₃) has the least boiling point. It is a gas at room temperature.
2. Ethers have bent structures and are also polar. However, there is no hydrogen bonding in their molecules. The only attractive forces are dipolar forces. Therefore, boiling point of dimethyl ether (CH₃OCH₃) is higher than that of propane. However, it is also a gas at room temperature.
3. Aldehydes contain polar carbonyl group and have strong dipolar interactions in their molecules. It is more than in ethers. Therefore, the boiling point of acetaldehyde (CH₃CHO) is more than that of dimethyl ether.
   
4. Out of all the families listed, alcohols have maximum intermolecular forces in the form of hydrogen bonding
   
   As a result, ethyl alcohol (C₂H₅OH) has the maximum boiling point.

Question 4.
Arrange the following compounds in increasing order of their reactivity in nucleophilic addition reactions
(i) Ehtanal, propanaL, propanone, butanone
(ii) Benzaldehyde,p-Tolualdehyde, p-Nitrobenzaldehyde, acetophenone.
Ans: (i) Butanone < Propanone < Propanal < Ethanal .This is because as the no. of alkyl groups attached to carbonyl carbon increases, +I-effect increases. As a result, e^– density

Question 5.
Predict the products of the following reactions :

Answer:

Question 6.
Give the IUPAC names of the following compounds:

Answer:

Question 7.
Show how each of the following compounds can be converted into benzoic acid?
(i) Ethylbenzene (C.B.S.E. Outside Delhi 2017)   
(ii) Acetophenone  (C.B.S.E. Outside Delhi 2017)
(iii) Bromobenzene
(iv) Phenylethene (Styrene)
Answer:

Question 8.
Which acid of each pair would you expect to be stronger?
(i) CH₃CO₂H or FCH₂CO₂H
(ii) FCH₂CO₂H or ClCH₂CO₂H
(iii) FCH₂CH2CH₂CO₂H or CH₃CH(F)CH₂CO₂H

Answer:

NCERT EXERCISE

 Question 1.
What is meant by the following terms? Give an example in each case.
(a) Cyanohydrin
(b) Semicarbazone
(c) Acetal
(d) Oxime
(e) Cyanohydrin
(f) Ketal
(g) Aldol
(h) Schiff’s base
(i) 2, 4-D.N.P.
(ii) Imine
Answer:

Question 2.
Name the following compounds according to IUPAC system of nomenclature.
(i) CH₃CH(CH₃)CH₂CH₂CHO
(ii) CH₃CH=CHCHO
(iii) CH₃CH(CH₃)CH₂C(CH₃)₂COCH₃
(iv) OHCC₆H₄CHO(-p)
(v) CH₃CH₂COCH(C₂H₅)CH₂CH₂Cl
(vi) CH₃COCH₂COCH₃
(vii) (CH₃)₃CCH₂COOH
(viii) (CH₃)₂CHCH(CH₃)COCl
Answer:
(i) 4-Methylpentanal
(ii) But-2-enal
(iii) 3 3 5-Trimethylhexan-2-one
(iv) Benzene- 1, 4-dicarbaldehyde
(v) 6-Chloro-4-ethylhexan-3-one
(vi) Pentane-2, 4-dione
(vii) 3, 3-Dimethylbutanoic acid
(viii) 2, 3-Dimethylbutanoyl chloride

Question 3.
Draw the structures of the following compounds:
(i) 3-Methylbutanal
(ii) p-Nitropropiophenone
(iii) p-Methylbenzaldehyde
(iv) 4-Methylpent-3-en-2-one
(v) 4-Chloropentan-2-one
(vi) 3-Bromo-4-phenylpentanoic acid
(vii) pp’-Dihydroxybenzophenone
(viii) Hex-2-en-4-ynoic acid
Answer:

Question 4.
Write the IUPAC names of the following aldehydes and ketones. Also give the common names wherever possible.

Answer:

Answer:

Question 7.
Which of the following will undergo Aldol condensation, which Cannizzaro’s reaction, and which neither of these? Write the structures of the expected products in each case
(i) Methanal
(ii) 2-Methylpentanal
(iii) Benzaldehyde
(iv) Benzophenone
(v) Cyclohexanone
(vi) 1-Phenylpropanone
(vii) Phenylacetaldehyde
(viii) Butan-1-ol
(ix) 2, 2-Dimethyl butanal
Answer:
(i) Methanal (HCHO): It will give Cannizzaro’s reaction since the α-hydrogen atom is absent.

(ii) 2-Methylpentanal [CH₃CH₂CH₂CH (CH₃)CHO]: It will give Aldol condensation since the α-hydrogen atom is present.

(iii) Benzaldehyde (CgH₅CHO): It will give Cannizzaro’s reaction since a-hydrogen is not present.

(iv) Benzophenone (C₆H₅COC₆H₅): It will not give any of the two reactions. Being ketone, does not take part in Cannizzaro’s reaction. Without a-hydrogen, it fails to participate in Aldol condensation.

(vi) 1-Phenylpropanone (C₆H₅COCH₂CH₃): It will undergo Aldol condensation since the α-hydrogen atom is present.

(vii) Phenylacetaldehyde (C₆H₅CH₂CHO): It will give Aldol condensation since the α-hydrogen atom is present.

(viii) Butan-1-ol: It will not give any of the reactions.

Question 8.
How will you convert ethanal to the following compounds?
(i) Butane-1, 3-diol
(ii) But-2-enal
(iii) But-2-enoic acid.
Answer:
(i) Ethanal to butane -1, 3-diol

(ii) Ethanal to but-2-enal

(iii) Ethanal into but-2-enoic acid

Question 9.
Write the structural formulae and names of four possible aldol condensation products from propanal and butanal. In each case, indicate which aldehyde serves as a nucleophile and which as an electrophile.
Answer:
Both propanal and butanal have a-hydrogen atoms present. These can undergo self aldol condensation as well as cross aldol condensation to give four compounds as follows:
(i) Condensation involving propanal: It is a case of a self aldol condensation.

(ii) Condensation involving butanal: It is self aldol condensation.

(iii) Condensation involving butanal (electrophile) and propanal (nucleophile):  It is cross-aldol condensation.

(iv) Condensation involving propanal (electrophile) and butanal (nucleophile): It is cross-aldol condensation.

Question 10.
An organic compound with molecular formula CgHjoO forms 2, 4-DNP derivative, reduces Tollen’s reagent, and undergoes Cannizzaro’s reaction. On vigorous oxidation, it gives 1, 2 benzene dicarboxylic acid. Identify the compound.      (C.B.S.E. Outside Delhi 2012; Haryana Board 2013)
Answer:
Since the compound forms 2, 4-DNP derivative on reacting with 2, 4-DNP, it is a carbonyl compound. As the compound reduces Tollen’s reagent and undergoes Cannizzaro’s reaction, it is an aldehyde and not a ketone. The data further reveals that the compound on vigorous oxidation gives 1, 2-benzene dicarboxylic acid. This clearly shows that in the compound which is of aromatic nature, CHO group is present at position-1 and C₂H₅ side chain at position-2. The given compound is 2-ethyl benzaldehyde. The reactions involved are given below :

Question 11.
An organic compound [A] with molecular formula C₈H₁₆O₂ was hydrolysed with dilute sulphuric acid to give a carboxylic acid [B] and alcohol [C]. Oxidation of [C] with chromic acid produced [B]. The alcohol [C] on dehydration gave but-1-ene. Write equations for the reactions involved. (C.B.S.E. 2008 Supp., C.B.S.E. 2009)
Answer:
(i) The available data shows that the compound [A] upon hydrolysis gave a carboxylic acid [B] and alcohol [C]. It must be an ester.
(ii) Since the alcohol [C] upon oxidation with chromic acid gave back the carboxylic acid [B], both the acid and alcohol must have the same number of carbon atoms (four each).
(iii) The alcohol [C] upon dehydration gave an alkene.
The equations for the reactions are given:

Question 12.
Arrange the following compounds in increasing order of their property as indicated:
(i) Acetaldehyde, Acetone, Di-tert butyl ketone, Methyl tert-butyl ketone (reactivity towards HCN)
(ii) CH₃CH₂CH(Br)COOH, CH₃CH(Br)CH₂COOH, (CH₃)₂CH COOH, CH₃CH₂CH₂COOH (acid strength).
(iii) Benzoic acid, 4-Nitrobenzoic acid, 3,4-Dinitrobenzoic add, 4-Methoxybenzok acid (acid strength).
Answer:
(i) The reactivity of aldehydes and ketones towards HCN addition decreases as the +1 – effect of the alkyl groups increases. Secondly, it decreases with increase in steric hindrance to the nucleophilic attack by CN^– at the carbonyl carbon. Thus the decreasing order of reactivity towards HCN is,

(ii) We know that the + I-effect decreases while -I-effect increases the acidic strength of carboxylic acids. Since + I-effect of isopropyl group is more than that of propyl group, therefore, (CH₃)₂CHCOOH is a weaker acid than CH₃CH₂CH₂COOH. Further since -I-effect decreases with distance, therefore CH₃CH₂CHBrCOOH is a stronger acid than CH₃CHBrCH₂COOH. Thus, the overall acid strength increases in the order:

(iii) Since electron-donating groups decrease the acidic strength, therefore, 4-methoxy benzoic acid is a weaker acid than benzoic acid. Further, since electron-withdrawing groups increase the acidic strength, therefore, both 4-nitrobenzoic acid and 3,4-dinitrobenzoic acid are stronger acids than benzoic acid. Further due to the presence of an additional -NO₂ group at /w-position with respect to -COOH group, 3,4-dinitrobenzoic acid is a stronger acid than 4-nitrobenzoic acid. Thus, the overall acidic strength increases in the order:4-methoxy benzoic acid < benzoic acid < 4-nitrobenzoic acid < 3,4-dinitrobenzoic acid.

Question 13.
Give chemical tests to distinguish between the following pairs of compounds :
(i) Propanal and propanone  (C.B.S.E. Delhi 2011, 2012)
(ii) Phenol and benzoic acid
(iii) Acetophenone and benzophenone
(iv) Benzoic acid and ethyl benzoate  (C.B.S.E. Outside Delhi 2009, 2011)
(v) Pentan-2-one and pentane-3-one
(vi) Benzaldehyde and acetophenone (C.B.S.E. Outside Delhi 2015)
(vii) Ethanal and propanal (C.B.S.E. Outside Delhi 2009, 2011, 2012)
Answer:
(i) Propanal and propanone:  Propanal will give a silver mirror upon heating with Tollen’s reagent but propanone will not respond.
(ii) Phenol and benzoic acid: Benzoic acid will give brisk effervescence with sodium hydrogen carbonate (NaHC0₃) but phenol will not respond.
(iii) Acetophenone and benzophenone: Acetophenone is a methyl ketone. It will give a yellow precipitate upon heating with I₂ and NaOH. Benzophenone will not respond.
(iv) Benzoic acid and ethyl benzoate: Benzoic acid will give brisk effervescence with sodium hydrogen carbonate (NaHC0₃) but ethyl benzoate (ester) will not respond.
(v) Pentan-2-one and pentan-3-one: Pentan-3-one is a methyl ketone and will give a yellow precipitate upon heating with I₂ and NaOH. Pentan-3-one will not respond.
(vi) Benzaldehyde and acetophenone: The distinction can also be made by iodoform test. Acetophenone will give yellow precipitate while benzaldehyde will not react.
(vii) Ethanal and propanal: Ethanal will respond to iodoform test and give yellow precipitate. Propanal will not react.

Question 14.
How will you prepare the following compounds from benzene? You may use any inorganic reagent and any organic reagent having not more than one carbon atom.
(i) Methylbenzoate
(ii) m-Nitrobenzoic acid
(iii) p-Nitrobenzoic acid
(iv) Phenylacetic acid
(v) p-nitrobenzaldehyde
Answer:
(i) Benzene to methylbenzoate

(ii) Benzene to m-nitrobenzoic acid


(iii) Benzene to p-nitrobenzoic acid


(vi) Benzene to phenylacetic acid


(v) Benzene to p-nitrobenzaldehyde

Question 15.
How will you bring about the following conversions in not more than two steps?
(i) Propanone to propene   (C.B.S.E. Delhi 2009, Uttarakhand Board 2009)
(ii) Propanal to butanone
(iii) Ethanol to 3-hydroxybutanal (C.B.S.E. Outside Delhi 2012)
(iv) Benzaldehyde to benzophenone (C.B.S.E. Outside Delhi 2012)
(v) Benzaldehyde to 3-PhenyIpropan-1-ol
(vi) Benzaldehyde to a-Hydroxyphenylacetic acid
(vii) Benzoic acid to benzaldehyde (C.B.S.E. Delhi 2009, Outside Delhi 2017)
(viii) Benzene to m-nitroacetophenone
(ix) Benzoic acid to /n-nitrobenzyl alcohol.      (C.B.S.E. Delhi 2012)
Answer:
(i) Propanone to propene

(ii) Propanal to butanone

(iii) Ethanol to 3-hydroxybutanal

(iv) Benzaldehyde to benzophenone

(v) Benzaldehyde to 3-phenylpropan-1-ol

(vi) Benzaldehyde to a-Hydroxyphenylacetic acid

(vii) Benzoic acid to benzaldehyde

(viii) Benzene to m-nitroacetophenone

(ix) Benzoic acid to m-nitrobenzyl alcohol

Question 16:
Describe the following:
(i) Acylation
(ii) Cross-aldol condensation
(iii) Cannizzaro’s reaction
(iv) Decarboxylation.
Answer:

Question 17:
Complete each synthesis by giving missing starting material, reagent, or products.


Answer:

Question 18:
Give a plausible explanation for each of the following:
(i) Cyclohexanone forms cyanohydrin in good yield but 2, 4, 6-trimethylcyclohexanone does not.
(ii) There are two – NH₂ groups in semicarbazide. However, only one is involved in the formation of semicarbazone.
(iii) During the preparation of esters from carboxylic acid and alcohol in the presence of an acid catalyst, the water or the ester should be removed as fast as it is formed.
Answer:

In cyclohexanone, the attack of CN“ ion (nucleophile) can easily take place at the carbonyl carbon atom. However, in 2, 4, 6-trimethylcyclohexanone, the three CH₃ groups being electron releasing in nature (+ I effect) will considerably increase the electron density on the carbonyl carbon atom and the nucleophile attack does not seem to be feasible. Moreover, the two —CH₃ substituents at the ortho positions will also hinder the attack of nucleophile CN^– ion on the carbonyl group.

(ii) The structural formula of semi-carbazide is NH₂NHCONH₂. Although both the amino groups have lone electron pairs, one of these is in conjugation with the electron-withdrawing carbonyl group and acquires a positive charge. Therefore, it is not in a position to act as the nucleophile, and only one -NH₂ group is involved in the formation of semicarbazone.

(iii) The esterification carried in the presence of acid is of reversible nature and the reverse reaction is called ester hydrolysis.

In order that the reaction may proceed in the forward direction, ester or water formed in the reaction must be removed. Sulphuric acid added in esterification helps in removing molecules of H₂0 as it is a dehydrating agent.

Question 19:
An organic compound contains 69-77% carbon, 11-63% hydrogen and the rest is oxygen. The molecular mass of the compound is 86. It does not reduce Tollen’s reagent but forms an addition compound with sodium hydrogen sulphite and gives a positive iodoform test. On vigorous oxidation, it gives ethanoic acid and propanoic acid. Write the possible structure of the compound. (C.B.S.E. Delhi 2008, 2009, Uttarakhand Board 2015)
Answer:
Step I: Calculation of molecular formula of the compound
Percentage of oxygen = 100 – (% C + % H) = 100 – (69.77 + 11.63) = 18.6%

Step II. Predicting the structure of the compound

• Since the compound forms an addition compound with NaHS0₃, it must be a carbonyl compound.
• As the compound does not reduce Tollen’s reagent but gives a positive iodoform test, it must contain in it a methyl
  

Keeping in view these characteristics, the compound is CH₃CH₂CH₂COCH₃ (Pentan-2-one).
All the reactions in which pentan-2-one participates, are given for the benefit of the students.

Question 20.
Although phenoxide ion has more number of resonating structures than carboxylate ion, carboxylic acid is a stronger acid than phenol. Why?
Answer:

We hope the NCERT Solutions for Class 12 Chemistry Chapter 12 Aldehydes, Ketones, and Carboxylic Acids help you. If you have any query regarding NCERT Solutions for Class 12 Chemistry Chapter 12 Aldehydes, Ketones, and Carboxylic Acids, drop a comment below and we will get back to you at the earliest.

Here we are providing NCERT Solutions for Class 12 English Flamingo Chapter 2 Lost Spring. Students can get Class 12 English Lost Spring NCERT Solutions, Questions and Answers designed by subject expert teachers.

Lost Spring NCERT Solutions for Class 12 English Flamingo Chapter 2

Lost Spring NCERT Text Book Questions and Answers

Lost Spring Think as you read 

Question 1.
What is Saheb looking for in the garbage dumps? Where is he and where has he come from?
Answer:
Unlike his parents who sifted through the garbage dumps for their survival, Saheb took it to be a treasure trove, wondrous and magical. He sometimes chanced upon a coin and hoped of finding more. He lived in Seemapuri. His family had arrived from Bangladesh in 1971.

Question 2.
What explanations does the author offer for the children not wearing footwear?
Answer:
The author disagreed with the usual explanation that is offered for the children going barefoot as a part of tradition in India. She felt it was only an excuse for the lack of money. They could ill-afford shoes as they lived in “a perpetual state of poverty”.

Question 3.
Is Saheb happy working at the tea-stall? Explain.
Answer:
Saheb was secure working at a tea-stall where he received daily wages and was given regular meals. However, it can be guessed that he was unhappy as he does not answer the writer when asked if he was happy. The writer also noticed that his face no longer carried the carefree look. He looked burdened with responsibilities.

Question 4.
What makes the city of Firozabad famous?
Answer:
Firozabad was famous for its bangles. Most families in Firozabad were engaged in making bangles. It engaged most of the families in its central industry. They worked around furnaces, welding glass and making bangles.

Question 5.
Mention the hazards of working in the glass bangles industry?
Answer:
Bangles were manufactured in glass furnaces with high temperatures, in dingy cells without air or light. As a consequence, the children, who slogged away in cloistered rooms close to the hot furnaces, often lost the brightness of their eyes, even their vision.

Question 6.
How is Mukesh’s attitude to his situation different from that of his family?
Answer:
Mukesh’s father worked as a tailor and as a bangle maker. He passed on his bangle-making skill to his family.
However, Mukesh dreamt of becoming a car mechanic, and wanted to break away from the occupation his family had been involved in for generations.

Lost Spring Understanding the text

Question 1.
What could be some of the reasons for the migration of people from villages to cities?
Answer:
Although it is difficult for the people to relocate from villages to cities, migration of such a nature continues unabated. People migrate for various reasons. These could be:

• de-fragmentation of land holdings
• lack of job opportunities
• lack of physical infrastructure no factories or other forms of employment, no medical support, no educational institutions, etc.
• lack of public health amenities such as sewage, drainage, etc.
• inability to deal with environmental hazards such as rains, storms, etc.
• glamour of the city life lures youngsters
• limited opportunities for progress
• aspirations for a better lifestyle

Question 2.
Would you agree that promises made to the poor children are rarely kept? Why do you think this happens in the incidents narrated in the text?
Answer:
Promises made to the poor children are rarely kept. Promises are made both at the national and international levels to provide healthy lives, quality education, protection against abuse, exploitation and violence, and combating HIV or AIDS. Yet, it is estimated that 246 million children are engaged in child labour. Of those, almost 171 million work in hazardous conditions, such as in mines, with chemicals and pesticides in agriculture or with dangerous machinery.

In some sectors their presence is kept under wraps where they toil away as domestic servants in homes, and labouring behind the walls of workshops. Millions of girls work as domestic servants and as unpaid household help and are especially vulnerable to exploitation and abuse. As the world looks ahead to prosperity, many children barely have a future, trapped in the conditions of poverty, conflict, and a degraded environment.

Just like Saheb and Savita, their future continues to be bleak. The children become disillusioned and often turn cynical. This happens because of utter poverty and the failure of the government to provide social security to the people.

Question 3.
What forces conspire to keep the workers in the bangle industry of Firozabad in poverty?
Answer:
Despite a government ban on child labour, 20,000 children in Firozabad work in horrific conditions to support their poor families. The workers are exposed to hazards such as blindness, tuberculosis, bronchitis, etc. In spite of working in such hazardous conditions the children are underpaid.They are forced to lead this life of poverty, as bangle-making continues to be their only means of livelihood. They can barely afford two square meals a day. They lack money and enterprise to do anything except carry on the business of

making bangles. The young men follow in the path of their elders as the profession is carried from one generation to the next. Years of mind-numbing toil kills their drive and their ability to dream. They lack the education and awareness to organise themselves into a cooperative and escape the vicious circle created by middlemen. The fear of the police and lack of leadership keep them back They remain caught in a web of poverty, burdened by the stigma of caste. The bureaucrats and the politicians exploit them further.

Lost Spring Talking about the text

Question 1.
How, in your opinion, can Mukesh realise his dream?
Answer:
Mukesh, one of the many children in Firozabad, aspired to be a motor mechanic. His dreams were unlike those of his peers, who worked in bangle manufacturing units amidst appalling conditions. Most of the people there, caught in the vicious circle, were bom and died in the same miserable plight as their forefathers. Mukesh, however, dared to dream.

He was determined to go to a garage and learn how to become a garage mechanic. He realized that the garage was a long way from his home, yet he was resolute and decided to walk all his way there. He dreamt of driving cars that he saw hurtling down the streets of his town. His passion and perseverance would help him achieve his goal. Mukesh was able to dream of breaking away from tradition, and that was the first step towards the realisation of his dreams.

Question 2.
Mention the hazards of working in the glass bangle industry.
Answer:
Child labour gives rise to a situation where the children are forced to work in dangerous and unhealthy conditions that scar them physically, emotionally, and mentally for the rest of their lives. The glass and glassware industry in India is concentrated in Firozabad. These factories produce a number of glass items, such as bangles, chandeliers, wine glasses, beads, crockery, bulbs, and cut glass items. The industry employs about 8,000 to 50,000 children, some as young as eight years old. The factory floor is like an inferno, due to the intense heat, poor ventilation, broken glass, dangling electric wires, and lack of protective equipment.

Often, glass splinters injure the workers, and pieces of glass cut into the bare feet of children. The children bump into each other and may scorch their bodies. Children are seen walking barefoot over glass littered floors, some with scarred eyes and burnt scalps. Child workers in the glass factories in Firozabad suffer from mental regression, asthma, bronchitis, eye problems, liver ailments, skin bums, chronic anaemia, and tuberculosis. Studies conducted at the Maulana Azad Medical College, in New Delhi, show genetic damage in the body cells of the labourers who have worked close to furnace heat for three years or more.Children, working in factories, often suffer from emotional, mental, and psychological scars.

Question 3.
Why should child labour be eliminated and how?
Answer:
There are various harmful effects of child labour. These include:
(a) Economic exploitation: Children may only receive one quarter of adult wages.
(b) Long working hours: Some children are expected to work for excessive hours, often up to 12-16 hours per day.
(c) Loss of educational opportunities: Children, who work, either give up their school education or find that their educational performance declines because of their work.
(d) Physical harm: Working children experience physical harm in a number of ways in terms of:require concerted efforts from all sections of the society to make a dent in the problem. It can be eliminated by:

• Increased risk of accidents
• Risk of physical violence from people in authority
• Theft is a risk faced by children who work as street vendors
• Risk of illness from poor hygiene and exposure to bad weather
• Harmful effects of chemicals

(e) Abuse and exploitation: It is essentially a socio-economic problem inextricably linked to poverty and illiteracy.
It will

• Legislative action plan
• Focusing general developmental programmes for benefitting child labour
• Subsidising education
• Providing basic necessities

Lost Spring Thinking about language

Question.
Carefully read the following phrases and sentences taken from the text and name the figures of speech used.
Answer:

• Saheb-e-Alam which means the lord of the universe is directly in contrast to what Saheb was in reality. The figure of speech used – irony
• “Drowned in an air of desolation”. The figure of speech used – hyperbole
• “Seemapuri, a place on the periphery of Delhi yet miles away from it, metaphorically.” The figure of speech
  used –  metaphor/irony
• “For the children it is wrapped in wonder; for the elders it is a means of survival.” The figure of speech
  used – contrast
• “As her hands move mechanically like the tongs of a machine, I wonder if she knows the sanctity of the bangles she helps make.” The figure of speech used – simile
• “She still has bangles on her wrist, but no light in her eyes.” The figure of speech used – paradox/contrast
• “Few airplanes fly over Firozabad.” The figure of speech used – metaphor/sarcasm/contrast
• “Web of poverty”, the figure of speech used – metaphor
• “Scrounging for gold”, the figure of speech used – metonymy/hyperbole
• “And survival in Seemapuri means rag-picking. Through the years, it has acquired the proportions of a fine art.” The figure of speech used – hyperbole/sarcasm
• “The steel canister seems heavier than the plastic bag he would carry so lightly over his shoulders.” The figure of speech used – metaphor

Lost Spring Extra Questions and Answers

Lost Spring Short Answer Questions

Question 1.
Who was Saheb? Where was he and where had he come from? What did he look for in the garbage dumps?
Answer:
Saheb was a child who had been forced by circumstances to become a ragpicker. His family had migrated from the green fields of Dhaka, Bangladesh in 1971. They had been forced to move out because storms had swept away all they had. They had shifted to Delhi to make a living. They lived in the slums of Seemapuri. Each day the child went looking for money in garbage heaps.

Question 2.
Saheb was a victim of circumstances. Justify.
Answer:
Saheb had once lived in the green fields of Dhaka but the storms swept away their fields and homes. Consequently, he ended up in Delhi as a ragpicker. There was an inherent desire in him to attend school and study. This could not be fulfilled because of poverty. When Anees suggested that he go to school, he was excited and a few days later asked her if her school was ready.

Question 3.
Bring out the irony in Saheb’s name.
Answer:
Saheb’s name was “Saheb-e-Alam” Ironically, it meant, lord of the universe. But that was something he would never know. Even if he did, he would have found it hard to believe. He roamed the streets barefoot scrounging the garbage heaps, but hardly managed to get one full meal.

Question 4.
Explain: “Seemapuri, a place on the periphery of Delhi, yet miles away from it, metaphorically”.
Answer:
Geographically, Seemapuri is a place on the outskirts of Delhi. It housed migrants from Bangladesh, who earned their living as ragpickers. A run-down place that lacked amenities of sewage, drainage, or running water, it was unlike the life of glitter and glamour in Delhi. People in Delhi lived a luxurious life in contrast to the poverty prevailing in Seemapuri.

Question 5.
What does garbage mean to adults and the children in the slum?
Answer:
Garbage meant different things to the adults and to the children in the slum. To the adults in Seemapuri, rag¬picking meant survival. It had assumed proportions of fine art. On the other hand, to the children garbage was like a mysterious package. They scrounged through it to discover unknown valuables.

Question 6.
Saheb is resigned to his fate and does not covet for what he considers is beyond his means. Justify.
Answer:
Saheb, a poor ragpicker, had resigned himself to his fate. He knew the areas that were out of bounds for him. He used to stand by the fenced gate of the club and was content watching others play tennis. He ventured into the club, to swing when no one was around. He had accepted his place in the society where he had to subsist on the items discarded by the privileged—tennis shoes, shirt and shorts. He gladly accepted work at a tea stall although it robbed him of his freedom.

Question 7.
How was Mukesh different from Saheb?
Answer:
Saheb was more resigned to his fate and had given up the freedom he enjoyed as a ragpicker for a salaried job at a tea stall. On the other hand, Mukesh insisted on being his own master. He was determined to be a motor mechanic. He was not prepared to compromise his dreams and give in like Saheb. He had even chalked out a path to achieve his dreams.

Question 8.
What did most slum dwellers do for a living in Firozabad?
Answer:
Firozabad is known for its bangles; it is the centre of India’s glass-blowing industry. Many people are employed in this industry. Families have spent generations working around furnaces, welding glass and making bangles. Since child labour is cheap, this place has around twenty thousand children who work in the hot furnaces. These children often lose vision before they become adults because of the environment they work in.

Question 9.
Describe the scene in Mukesh’s house as viewed by the narrator.
Answer:
Mukesh was another young child who had been forced by poverty into child labour. He lived in a dilapidated shanty with garbage strewn around. His house was a half-built shack, thatched with dead grass and a wobbly iron door. When the narrator visited the place, she noticed a firewood stove with a large vessel of spinach leaves. She also noticed a frail very young woman cooking. Later she realized she was wife of Mukesh’s elder brother. The narrator also saw Mukesh’s grandmother and his father, who were weak having spent their . lives making bangles.

Question 10.
How do you know that everyone in Mukesh’s family had resigned to their fate?
Answer:
Like most people in Firozabad, Mukesh was bom in the caste of bangle makers. Mukesh’s father was a poor bangle maker who had worked hard, first as a tailor. But despite slogging all his life, he had not been able to renovate a house or educate his two sons. Hence he had no option but to pass on the art of making bangles to his sons. Mukesh’s grandmother was an old woman who had watched her husband go blind but she did not complain. She accepted it as her fate of her “god-given lineage”. It was only Mukesh who dreamt of breaking out and being a mechanic.

Question 11.
The future of the slum dwellers in Firozabad is as bleak as their present. Why does the writer feel so?
Answer:
The future of the slum dwellers in Firozabad was as bleak as their present. The families were trapped in poverty, a curse that continued for generations. They also faced the stigma of having been bom in a lower caste and were victimized by the inhuman sahukars, the middlemen, the policemen, the keepers of law, the bureaucrats and the politicians. With all forces working against them, they were unable to defy the norms.

Question 12.
Why didn’t the people in Firozabad organize themselves into a cooperative to fight the system?
Answer:
Despite being exploited, the people in Firozabad were unable to organize themselves into a cooperative to escape from being manipulated and fight the system. Had they organized themselves into cooperatives, they ran the risk of running into trouble with the authorities. Moreover, there was no leader among them who could lead them. They were trapped in the vicious cycle of poverty, indifference, greed and injustice.

Question 13.
Mukesh dreams of a different future. What does he dream of? How does he want to accomplish his dreams?
Answer:
Mukesh was bom in a family of poor slum dwellers. He has been taught to work in the glass factory but Mukesh did not want to follow the traditional profession. He dreamt of being a motor mechanic. He wanted to learn about cars and was determined to leam the skill in order to ensure for himself a better life.

Question 14.
Explain: Few airplanes fly over Firozabad.
Answer:
Mukesh wanted to be a motor mechanic and he was prepared to walk to the garage to leam. He never dreamt of flying a plane as to the slum dwellers in Firozabad, planes were a far-fetched reality. Because of the limited exposure in the slums of Firozabad, Mukesh dreamt within his means.

Lost Spring Long Answer Questions

Question 1.
Describe the miserable plight of the people in Seemapuri.
Answer:
Seemapuri was a locality on the outskirts of Delhi which housed unlawful residents who came from Bangladesh back in 1971. This area was a place where thousands of ragpickers lived. The people lived in structures of mud, with roofs of tin and tarpaulin. There was no sewage system, no drainage or running water. The migrant poor lived there for more than thirty years without an identity, without permits but with ration . cards that got their names on voters’ lists. The children in these slums grew up to become partners in survival as ragpickers. To them, garbage heaps were like gold mine that would ensure their daily bread and a roof over their heads.

Question 2.
“But promises like mine abound in every comer of his bleak world.” What does the writer mean?
Answer:
The narrator, Anees, met Saheb every morning looking for money in the garbage. Saheb confessed to the narrator that he scrounged the rubbish heaps as he had nothing better to do. He longed to go to school but there was not one in the neighbourhood. When Anees asked him half joking that if she started a school, would he join, he consented very gladly. In fact he was so enamoured with the idea that a few days later he asked her if her school was ready. Anees was embarrassed at having made a promise that she was not serious about. She felt that most people made promises to children like him that were never fulfilled.

Question 3.
What do people have to say about people walking barefoot? What is the writer’s opinion?
Answer:
The writer narrates an experience when she asked a child why they were barefoot. One replied that his mother did not bring them down from the shelf, while the other boy felt that he would throw them off anyway. Yet another boy expressed his desire for shoes. The writer recalled having seen children walking barefoot. She had been informed that people walked barefoot not because of lack of money but due to the tradition in India. She wondered if this was an excuse to explain away the perpetual poverty. She had also heard of a boy from Udipi, who prayed every morning for a pair of shoes.

Question 4.
Explain: The steel canister seems heavier than the plastic bag.
Answer:
The writer, one morning, saw Saheb on his way to the milk booth carrying a steel canister. He had relinquished his job as a ragpicker and had taken up employment in a tea stall for eight hundred rupees.His face had lost the carefree look when he roamed the streets like a vagabond. As a ragpicker he seemed carefree, constantly looking for things—“wrapped in wonder”.

At the tea stall he was now burdened with the responsibility of a job. He was literally and metaphorically fraught with the weight of the steel canister. The canister was heavier than the plastic bag that he carried “lightly over his shoulder”. The plastic bag was light because the bag gave him the freedom to lead a life that was not governed by adults. He was no longer the master of his own free will after he was burdened by the job at the tea stall.

Question 5.
Describe the poverty of living conditions of the people in Firozabad.
Answer:
Firozabad is a city known for bangles and glass industry. However the people working in the industry led a pitiable life. They lived in dilapidated, dingy houses in cloistered lanes that were foul smelling and clogged with garbage. Their homes were hovels with crumbling walls, shaky doors, no windows, and crowded with families of humans and animals living together.

Mukesh’s house was no better. It was a half-built shack. One part of the house was thatched with dead grass and it had a wobbly iron door. Most of the houses were similar dark huts. The children worked under flickering .oil lamps with their parents, welding pieces of coloured glass into circles of bangles. Their eyes were more adjusted to the dark than to the light outside. Hence, they often ended up losing their eyesight before they became adults. The people were exploited by sahukars, the middlemen, the policemen, the keepers of law, the bureaucrats and the politicians.

Question 6.
Describe the writer’s visit to Mukesh’s place.
Answer:
Anees visited Mukesh’s house in Firozabad, a place known for its glass industry. Mukesh was bom in the caste of bangle makers. Anees noticed a weak young woman, the wife of Mukesh’s elder brother, cooking . the evening meal for the whole family. She was very young but as the daughter-in-law of the house, was in charge of three men—her husband, Mukesh and their father.

Mukesh’s father was an impoverished bangle maker. He had worked hard, first as a tailor, then a bangle maker. Despite years of relentless labour, he had neither been able to renovate a house, nor send his two sons to school.Mukesh’s grandmother had watched her husband go blind with the dust from polishing the glass of bangles. But she had accepted it as her fate. She felt that “god-given lineage” could never be broken.

Question 7.
Mukesh is not like the others. His “dream(s) loom like a mirage amidst the dust of streets that fill his town Firozabad.” Justify.
Answer:
Mukesh was born in a family riddled with poverty. He and his family were leading a difficult life that was not uncommon to the people of that socio-economic strata. But that had not deterred his desire to lead a different life. Mukesh was determined to be his own master. He had decided to become a motor mechanic and was . determined to leam to drive a car.

When Anees heard of that, she felt that Mukesh’s dreams were like a mirage—unattainable because it was difficult for him to break out of the generations of bangle-making tradition. She was convinced it would be difficult for Mukesh to achieve his unconventional dream.

Question 8.
In your opinion, can Mukesh realize his dream?
Answer:
Yes, Mukesh could certainly achieve his dream as he dared to dream in the first place. He was unlike most people who spent their lives doing what their families had done for generations. Mukesh had a tangible plan in action to realize his dreams. He was determined to go to a garage and leam more about cars. Despite the fact that the garage was at a distance, he was resolute. “I will walk,” he said. His passion and perseverance would certainly help him achieve his goals.
OR
No, Mukesh will not be able to realize his dream because there were thousands of families trapped in poverty ‘ who face the stigma of caste system. To cap it all, they live with insensitive people who exploit the situation. The inhuman sahukars, the middlemen, policemen, the keepers of law, the bureaucrats and the politicians— all work against them. These poor people are unable to come out of their misery because they lack education or leadership. Mukesh’s dream will too die a death like many others of his station.

Lost Spring Value Based Question

Question 1.
What societal evils are depicted in the “Stories of Stolen Childhood”?
Answer:
Anees Jung voices her concern over the exploitation of children in hazardous jobs such as bangle making and rag-picking. Grinding poverty and thoughtless traditions result in the loss of childhood innocence and education. They are denied a life of dignity, having been born into and conditioned by a life of poverty.

The miserable plight of Saheb-e-Alam and Mukesh brings out the grinding poverty and traditions which condemn children to a life of exploitation. It also spells out the callousness of the society towards the underprivileged.

Saheb, a ragpicker, was a young boy who had been denied education and was engaged in ragpicking as a profession. Mukesh was a bangle maker, born into the bangle-making legacy of his poor family. He, however, nurtures dreams of becoming a motor mechanic someday. Through examples of these slum children, constricted by the narrow bounds of poverty and child labour, the author voices the relevant concerns of societal evils.

NCERT Solutions for Class 12 Biology Chapter 6 Molecular Basis of Inheritance

These Solutions are part of NCERT Solutions for Class 12 Biology. Here we have given NCERT Solutions for Class 12 Biology Chapter 6 Molecular Basis of Inheritance

Question 1.
Group the following as nitrogenous bases and nucleosides:
Adenine, cytidine, thymine, guanosine, uracil, and cytosine.
Solution:
Adenine, Guanosine, Thymine, Uracil, and Cytosine are nitrogenous bases. (Adenine and Guanosine → Purine, Thymine, Uracil and Cytosine → Pyrimidine) Cytidine is a nucleoside.

Question 2.
If a double-stranded DNA has 20 percent of cytosine, calculate the percent of adenine in the DNA.
Solution:
According to Chargaff’s rule, in a double-stranded DNA, the total number of cytosine molecules will be equal to the number of guanine molecules and the number of adenine molecules will be equal to the number of thymine molecules. Therefore, if a double-stranded DNA has 20 percent of cytosine then the guanine will also be 20 per cent. The remaining 60% will consist of adenine and thymine in equal amount. Thus adenine will be 30%.

Question 3.
If the sequence of one strand of DNA is written as follows:
5′-ATGCATGCATGCATGCATGCA
TGCATGC-3′
Write down the sequence of complementary strand in 5′ -> 3′ direction.
Solution:
5′-GCATGCATGCATGCATGCAT G C ATG CAT-3′.

Question 4.
If the sequence of the coding strand in a transcription unit is written as follows: 5′-ATGCATGCATGCATGCATGCATGCATGC-3′ Write down the sequence of mRNA.
Solution:
If the sequence of coding strand is :
5′ – ATGCATGCATGCATGCATGCATGCATGC – 3′
Then template strand is :
3′ – TACGTACGTACGTACGTACGTACGTACG – 5′
The mRNA will be formed on the template strand in 5′ —> 3’ direction. Thus mRNA sequence will be:
5′-AUGCAUGCAUGCAUGCAUGCAUGCAUGC-3′
Thymine in DNA is substituted by uracil in RNA.

Question 5.
Which property of DNA double helix led Watson and Crick to hypothesise a semi-conservative mode of DNA replication? Explain.
Solution:
The two strands of DNA show complementary base pairing. This property of DNA led Watson and Crick to suggest a semi-conservative mechanism of DNA replication in which one strand of a parent is conserved while the other complementary strand formed is new.

Question 6.
Depending upon the chemical nature of the template (DNA or RNA) and the nature of nucleic acids synthesised from it (DNA or RNA), list the types of nucleic acid polymerases
Solution:
DNA dependent DNA polymerases and DNA dependent RNA polymerases.

Question 7.
How did Hershey and Chase differentiate between DNA and protein in their experiment while proving that DNA is the genetic
material?
Solution:
They raised 2 types of bacteriophages

• On radioactive phosphorous (³²P)
• On radioactive sulphur (³⁵S).

³⁵S gets into protein and ³²P into DNA When both bacteriophages infected bacteria differently and by shaking them, the viral protein coat was separated

After raising these bacteria it was found that those infected with ³²P bacteriophage → radioactivity were found. But with ³⁵S → no radioactivity was found.

Question 8.
Differentiate between the following:

1. Repetitive DNA and Satellite DNA
2. Template strand and Coding strand
3. mRNA and tRNA

Solution:

1. Differences between repetitive DNA and satellite DNA are as follows:
   
   
2. Differences between template strand and coding strand are as follows:
   
   
3. Differences between mRNA and tRNA are as follows:
   
   
   
   

Question 9.
List two essential roles of ribosome during translation
Solution:
Two essential roles of the ribosome during translation are:

1. One of the RNA acts as a peptidyl transferase ribozyme for the formation of peptide bonds.
2. The ribosome provides sites for attachment of mRNA and charged tRNA for polypeptide synthesis.

Question 10.
In the medium where E. coli was growing, lactose was added, which induced the lac operon. Then, why does lac operon shut down some time after addition of lactose in the medium?
Solution:
The lac operon is regulated by the amount of lactose in the medium where the bacteria are grown. When the amount of lactose is exhausted in the medium, the lac operon shuts down.

Question 11.
Explain (in one or two lines) the function of the followings:

1. Promoter
2. tRNA
3. Exons

Solution:

1. Promoter: It is located at the 5′ end of the transcription unit and provides site for attachment of transcription factors (TATA Box) and RNA polymerase.
2. tRNA: It takes part in the transfer of activated amino acids from cellular pool to ribosome so that they can take part in protein formation.
3. Exons: In eukaryotes, DNA is mosaic of exons and introns. Exons are coding sequences of DNA which are both transcribed and translated.

Question 12.
Why is the Human Genome Project called a mega project?
Solution:
The human genome was a megaproject that aimed to sequence every base in the human genome. The estimated cost of the project would be a billion (1 billion = 100 crores) US dollars.

Question 13.
What is DNA fingerprinting? Mention its application.
Solution:
DNA fingerprinting is the identification of differences in specific regions of DNA sequences based on DNA polymorphism, repetitive DNA, and satellite DNA.
Application of DNA fingerprinting: Settling, paternity disputes and identity of criminal by different DNA profiles in forensic laboratories.

Question 14.
Briefly describe the following:

1. Transcription
2. Polymorphism
3. Translation
4. Bioinformatics

Solution:
1. Transcription – It is the process of copying genetic information from the anti-sense or template strand of the DNA into RNA. It is meant for taking the coded information from DNA in nucleus to the site where it is required for protein synthesis. Principle of complementarity is used even in transcription. The exception is that uracil is incorporated instead of thymine opposite adenine of template. The segment of DNA that takes part in transcription is called transcription unit. It has three components

• • a promoter,
  • the structural gene and
  • a terminator.

2. Polymorphism – It is the variation at genetic level, arisen due to mutations. Such variations are unique at particular site of
DNA. They occur approximately once in every 500 nucleotides or about 107 times per genome. These are due to deletions, insertions, and single-base substitutions. These alterations in healthy people, occur in non-coding regions of DNA and do not code for any protein but are heritable. The polymorphism in DNA sequences is the basis of genetic mapping of human genome as well as DNA fingerprinting.

3. Translation – It is the mechanism by which the triplet base sequence of mRNA guides the linking of a specific sequence of amino acids to form a polypeptide chain (protein) on ribosomes in the cell cytoplasm. All the protein that a cell needs are synthesised by the cell within itself.
The raw materials required in protein synthesis are ribosomes, amino acids, mRNA, tRNAs and amino acyl tRNA synthetase. Mechanism of protein synthesis involves following steps:

• • Activation of amino acids
  • Charging or aminoacylation of tRNA
  • Initiation
  • Elongation (Polypeptide chain formation)
  • Termination

The ribosomes move along the mRNA ‘reading’ each codon in turn. Molecules of transfer RNA (tRNA), each bearing a particular amino acid, are brought to their correct positions along the mRNA, molecule base pairing occurs between the bases of the codons and the complementary base triplets of tRNA. In this way, amino acids are assembled in the correct sequence to form the polypeptide chain.

4. Bioinformatics – Bioinformatics is the combination of biology, information technology and computer science. Basically, bioinformatics is a recently developed science which uses information technology to understand biological phenomenon. It broadly involves the computational tools and methods used to manage, analyse and manipulate volumes of biological data.

We hope the NCERT Solutions for Class 12 Biology Chapter 6 Molecular Basis of Inheritance help you. If you have any query regarding NCERT Solutions for Class 12 Biology Chapter 6 Molecular Basis of Inheritance, drop a comment below and we will get back to you at the earliest.

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

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

Question 1.
Make a chart (with diagrammatic representation) showing a restriction enzyme, the substrate DNA on which it acts, the site at which it cuts DNA and the product it produces.
Solution:
Restriction enzyme – Eco RI
Source – Escherichia coli RY13

Question 2.
Can you list 10 recombinant proteins which are used in medical practice? Find out where they are used as therapeutics.
Solution:
The proteins produced through rDNA technology and being used in the medical practice include rh-Growth Hormone, r-Human insulin, Erythropoietin, Follicle stimulating hormone, Interferon, Insulin-like growth factor, Tissue Plasminogen Activator, factor VIII, DNase, the Envelope protein of hepatitis B virus.

Question 3.
From what you have learned, can you tell whether enzymes are bigger or DNA is bigger in molecular size? How did you know?
Solution:
Enzymes are bigger than DNA as they are proteins and proteins are macromolecules made of amino acids which are bigger than nucleotides. This can also be proved by gel electrophoresis, where denatured protein would not move but denatured DNA will move to a distance. Protein synthesis is regulated by small portions of DNA, called genes.

Question 4.
What would be the molar concentration of human DNA in a human cell? Consult your teacher.
Solution:
The average molecular weight of a nucleotide in human DNA is 130.86. The molecular weight of human DNA will therefore be 6 x 10⁹ nucleotides (based on the human genome project) x 130-86 = 784-56 x 109 gm/mol. The molar concentration of DNA can be calculated accordingly.
The molarity can be calculated as
Molar Concentration = \(\frac { No. of molecules }{ Molecular Weight }\)

Question 5.
Do eukaryotic cells have restriction endonucleases? Justify your answer.
Solution:
Restriction enzymes also called ‘the molecular scissors’ are used to break DNA molecules. These enzymes are present in many bacteria where they function as a part of their defence mechanism called a restriction-modification system. The molecular basis of this system was explained first by W. Arber in 1965. The restriction-modification system consists of two components;

A restriction enzyme which identifies the introduced foreign DNA and cuts into pieces and is called restriction endonuclease,

The second component is a modification enzyme in which methylation is done. Once a base in a DNA sequence is modified by the addition of a methyl group, the restriction enzymes fail to recognise and could not cut that DNA. This is how a bacterium modifies and therefore, protects its own chromosomal DNA from cleavage by these restriction enzymes. Eukaryotic cells do not have restriction endonucleases (or restriction-modification system). The DNA molecules of eukaryotes are heavily methylated by a modification enzyme, called methylase. Eukaryotes exhibit some different mechanisms to counteract viral attacks.

Question 6.
Besides better aeration and mixing properties, what other advantages do stirred tank bioreactors to have over shake flasks?
Solution:
Shake flask is used for growing microbes and mixing the desired materials on a small scale in the laboratory. However, the large-scale production of a desired biotechnological product requires large stirred tank bioreactors.
Besides better aeration and mixing properties, bioreactors have the following advantages:

• It has an oxygen delivery system.
• It has a foam control, temperature, and pH control system.
• Small volumes of culture can be withdrawn periodically.

Question 7.
Collect 5 examples of palindromic DNA sequences by consulting your teacher. Better try to create a palindromic sequence by following base-pair rules.
Solution:
palindromes in DNA are base pair sequences that are the same when read forward (left to right) or backward (right to left) from a central axis of symmetry. For example, the following sequences read the same on the two strands in the 5′ → 3′ direction as well as 3′ → 5′ direction.

1. 5′ – G – G – A – T – C – C – 3′
   3′ – C – C – T – A – G – G – 5′
2. 5′ – G – A – A – T – T – C – 3′
   5′ – C – T – T – A – A – G – 5′
3. 5′ – A – A – G – C – T – T – 3′
   3′ – T – T – C – G – A – A – 5′
4. 5′ – G – T – C – G – A – C – 3′
   3′ – C – A – G – C – T – G – 5′
5. 5′ – A – C – T – A – G – T – 3′
   3′ – T – G – A – T – C – A – 5′

Question 8.
Can you recall meiosis and indicate at what stage recombinant DNA is made?
Solution:
A recombinant DNA is made in the pachytene stage of prophase I by crossing over during meiosis cell division. Recombination nodules are visible in a synaptonemal complex in the pachytene sub-stage. Crossing over occurs in this time between chromatids than recombinant DNA is formed.

Question 9.
Can you think and answer how a reporter enzyme can be used to monitor the transformation of host cells by foreign DNA in addition to a selectable marker?
Solution:
Transformation is a process through which a piece of DNA is introduced into a host bacterial cell. Normally, the genes encoding resistance to antibiotics such as ampicillin, tetracycline, etc., are considered useful selectable markers to differentiate between transformed and non-transformed bacterial cell. In addition to these selectable markers, an alternative selectable marker has been developed to differentiate transformed and non-transformed bacterial cell on the basis of their ability to produce colour in the presence of a chromogenic substance.

A recombinant DNA is inserted in the coding sequence of an enzyme (5-galactosidase (reporter enzyme). If the plasmid in the bacterium does not have an insert, the presence of a chromogenic substance gives blue coloured colonies, presence of insert results into insertional inactivation of (3-galactosidase and, therefore, the colonies do not produce any colour, these colonies are marked as transformed colonies.

Question 10.
Describe briefly the following :
(a) Origin of replication
(b) Bioreactors
(c) Downstream processing
Solution:

(a) Origin of Replication (Ori):- It is a DNA sequence which is specialised to initiate replication Bacterial chromosomes and plasmids possess a single origin of replication Eukaryote chromosomes to have a number of origin of replication. Replication proceeds bidirectionally from the site of origin of replication. The sequence also possesses nearby replication control which determines the number of copies it would form. Therefore the selected plasmid should have an origin of replication that supports a high copy number.

(b) Bio-Reactor:
Bio reactor used in biotechnology is generally 100-1000 litre cylindrical metal. container with a curved base to facilitate mixing of contents. The culture medium containing all nutrients, salts vitamins, hormones etc. is added along with the inoculum of transformed cells with recombinant DNA. A stirrer helps in mixing and optimum availability of nutrients to culture cells. The supply of oxygen is maintained if the cells function better under aerobic conditions. Foam is kept under control. Gadgets are attached for knowing the temperature and pH of the contents. Corrections are made when required. There is a sampling port where a small volume of culture can be withdrawn to know the growth of cells and concentration of the extractable product.

(c) Downstream processing:
It is the recovery of product from fully grown genetically modified cells, its purification, and preservation. It is carried out after the sampling report indicates the completion of the biosynthetic phase and the presence of the optimum product in the cells. After leaving a part of the cellular mass of inoculum, the rest is crushed and chemically treated to separate the product. The separated product is purified and then formulated with suitable preservatives. Clinical traits are carried out to know its use and any immediate or long-term adverse effect. Every batch of the product has to pass through strict quality control testing. Of course, the procedure and vigour of downstream processing and quality control vary from product to product.

Question 11.
Explain briefly
(a) PCR
(b) Restriction enzymes and DNA
(c) Chitinase
Solution:
(a) Polymerase chain reaction (PCR) is a technique of synthesizing multiple copies of the desired gene (DNA) in vitro. This technique was developed by Kary Mullis in 1985. It is based on the principle that a DNA molecule, when subjected to high temperature, splits into two strands due to denaturation. These single-stranded molecules are then converted to double-stranded molecules by synthesising new strands in presence of enzyme DNA polymerase. Thus, multiple copies of the original DNA sequence can be generated by repeating the process several times. The basic requirements of PCR are, DNA template, two nucleotide primers usually 20 nucleotides long, and enzyme DNA polymerase which is stable at high temperature (usually Taq polymerase):
The working mechanism of PCR is as follows:

First of all, the target DNA (DNA segment to be amplified) is heated to high temperature (94 to 96° C). Heating results in the separation of two strands of DNA. Each of the two strands of the target DNA now acts as a template for the synthesis of a new DNA strand. This step is called denaturation.

Denaturation is followed by annealing. During this step, two oligonucleotide primers hybridise to each single-stranded template DNA in presence of excess synthetic oligonucleotides. Annealing is carried out at lower temperatures (40° – 60°C).

The third and final step is an extension. During this step, the enzyme DNA polymerase synthesizes the DNA segment between the primers. Usually, Taq DNA polymerase, isolated from a thermophilic bacterium Thermus aquatics, is used in most cases. The two primers extend towards each other in order to copy the DNA segment lying between the two primers. This step requires the presence of deoxynucleoside triphosphates (dNTPs) and Mg²⁺ and occurs at 72°C.
The above-mentioned three steps complete the first cycle of PCR. The second cycle begins with denaturation of the extension product of the first cycle and after completing the extension step, two cycles are completed. If these cycles are repeated many times, the DNA segment can be amplified approximately a billion times, i.e., one billion copies of desired DNA segment are made.

(b) Restriction enzymes are used to break DNA molecules. They belong to a larger class of enzymes called nucleases. Restriction enzymes are of three types – exonucleases, endonucleases, and restriction endonucleases.

Exonucleases: They remove nucleotides from the terminal ends (either 5′ or 3′) of DNA in one strand of the duplex.

Endonucleases: They make cuts at specific positions within the DNA. These enzymes do not cleave the ends and involve only one strand of the DNA duplex.

Restriction endonucleases: These were found by Arber in 1963 in bacteria. They act as “molecular scissors” or chemical scalpels. They recognise the base sequence at palindrome sites in the DNA duplex and cut its strands. Three main types of restriction endonucleases are type I, type II, and Type III. Out of the three types, only type II restriction enzymes are used in recombinant DNA technology because they can be used in vitro to recognise and cut within specific DNA sequences typically consisting of 4 to 8 nucleotides.

(c) Chitinase is a lysing enzyme that dissolves the fungal cell wall. It results in the release of DNA along with several other macromolecules.

Question 12.
Discuss with your teacher and find out how to distinguish between:
(a) Plasmid DNA and chromosomal DNA
(b) Exonuclease and endonuclease
(c) RNA and DNA
Solution:
(a) Plasmid DNA is naked double-stranded DNA that forms a circle with no free ends. It is associated with few proteins. It is smaller than the host chromosome and can be easily separated.

Chromosomal DNA is a double-stranded linear DNA molecule associated with large proteins. This DNA exists in relaxed and supercoiled forms and provides a template for replication and transcription. It has free ends.

(b) Differences between exonucleases and endonucleases are as follows :

(c) Differences between DNA and RNA are given in the following table:

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