CBSE Class 12 Biology Ch 5 Practice Paper with Solutions

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Class 12 Biology · Chapter 5 · Practice Paper + Solutions

CBSE · NCERT

BIOLOGY – CHAPTER 5 : MOLECULAR BASIS OF INHERITANCE

Chapter-wise Practice Question Paper (Board + Competitive / NEET Level)

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Class : XIISubject : BiologyRoll No. : ..................

Time : 3:00 HoursSession : 2026–27Max. Marks : 100

General Instructions

1. All questions are compulsory. This paper is based on NCERT Class XII Biology, Chapter 5, and includes NEET-style MCQs and Assertion–Reason questions.
2. Q.1 to Q.5 are Section A (40 marks): Q.1 has 20 MCQs; Q.2, Q.3, Q.4 and Q.5 have 5 questions each; every question carries 1 mark.
3. Q.6 to Q.13 carry 2 marks each; Q.14 to Q.20 carry 3 marks each; Q.21 and Q.22 are case-based questions of 4 marks each; Q.23 to Q.25 carry 5 marks each.
4. Internal choice (OR) is given in Q.14 to Q.20 and Q.23 to Q.25. Attempt only one of the alternatives.
5. Draw neat, labelled diagrams wherever required. Solutions to all questions are given after the paper.

SECTION – A (Objective Questions · 40 Marks)

Q.1 Choose the correct option.20 × 1 = 20

(i) In a double-stranded DNA, adenine forms 30 % of the bases. The percentage of guanine is –

(a) 30 %(b) 20 %(c) 40 %(d) 70 %

(ii) The N-glycosidic bond is formed between –

(a) Phosphate and sugar(b) Two nucleotides(c) Two nitrogenous bases(d) A nitrogenous base and a pentose sugar

(iii) In B-DNA the distance between two consecutive base pairs is –

(a) 0.34 nm(b) 3.4 nm(c) 34 nm(d) 0.034 nm

(iv) In the Hershey–Chase experiment, the radioactive isotope ³²P was used to label the –

(a) Protein coat of the phage(b) Lipids of the bacterium(c) DNA of the phage(d) Cell wall of the bacterium

(v) The biochemical nature of Griffith’s ‘transforming principle’ was shown to be DNA by –

(a) Hershey and Chase(b) Avery, MacLeod and McCarty(c) Meselson and Stahl(d) Watson and Crick

(vi) A nucleosome core has a histone octamer made of –

(a) Two molecules each of H1, H2A, H2B and H3(b) Four molecules of H1 only(c) Eight molecules of H3(d) Two molecules each of H2A, H2B, H3 and H4

(vii) Euchromatin is –

(a) Loosely packed, lightly stained and transcriptionally active(b) Densely packed, darkly stained and inactive(c) Present only in prokaryotes(d) Made only of RNA

(viii) In the Meselson–Stahl experiment, after two generations in ¹⁴N medium the ratio of hybrid (¹⁵N–¹⁴N) DNA to light (¹⁴N) DNA is –

(a) 1 : 0(b) 1 : 3(c) 1 : 1(d) 3 : 1

(ix) Okazaki fragments are formed on the –

(a) Leading strand(b) Lagging strand(c) Both strands(d) RNA primer only

(x) DNA polymerase adds nucleotides in the direction –

(a) 3′ → 5′(b) Both 5′ → 3′ and 3′ → 5′(c) Randomly(d) 5′ → 3′

(xi) In a transcription unit, the promoter is located –

(a) Towards the 5′ end (upstream) of the structural gene(b) Towards the 3′ end of the structural gene(c) In the middle of the structural gene(d) Within the terminator

(xii) In eukaryotes, tRNA and 5S rRNA are transcribed by –

(a) RNA polymerase I(b) RNA polymerase II(c) RNA polymerase III(d) σ factor

(xiii) Processing of hnRNA into mRNA in eukaryotes involves –

(a) Splicing only(b) Splicing, capping and tailing(c) Tailing only(d) Translation

(xiv) The initiation codon in mRNA is –

(a) UAA(b) UAG(c) UGA(d) AUG

(xv) Out of the 64 codons, the number that code for amino acids is –

(a) 61(b) 64(c) 20(d) 3

(xvi) The genetic code is said to be degenerate because –

(a) One codon codes for many amino acids(b) It is read continuously without punctuation(c) More than one codon can code for the same amino acid(d) It is the same in all organisms

(xvii) In bacteria, peptide bond formation during translation is catalysed by –

(a) A protein enzyme(b) 23S rRNA (a ribozyme)(c) tRNA(d) mRNA

(xviii) In the lac operon, the inducer is –

(a) Glucose(b) Galactose(c) Repressor(d) Lactose (allolactose)

(xix) In the lac operon, the i gene codes for –

(a) The repressor protein(b) β-galactosidase(c) Permease(d) Transacetylase

(xx) DNA fingerprinting is based on the polymorphism in –

(a) mRNA(b) tRNA(c) VNTRs (variable number of tandem repeats)(d) Exons

Q.2 Fill in the blanks.5 × 1 = 5

(i) A nitrogenous base linked to a pentose sugar is called a ______.

(ii) The experiment using bacteriophages to prove that DNA is the genetic material was performed by ______ and ______.

(iii) The short DNA fragments formed on the lagging strand are called ______ fragments.

(iv) tRNA is called the ______ molecule because it links the codon to the amino acid.

(v) Non-coding sequences that are removed during splicing are called ______.

Q.3 Write True or False.5 × 1 = 5

(i) In DNA, adenine pairs with guanine through two hydrogen bonds.

(ii) DNA replication is semiconservative.

(iii) The template strand of DNA has 5′ → 3′ polarity.

(iv) UAA, UAG and UGA are stop codons.

(v) In the lac operon, lactose acts as an inducer.

Q.4 Match Column ‘A’ with Column ‘B’.5 × 1 = 5

Column ‘A’	Column ‘B’
(i) Griffith	(a) Semiconservative replication
(ii) Meselson and Stahl	(b) DNA fingerprinting
(iii) Hershey and Chase	(c) Transformation in Streptococcus
(iv) Jacob and Monod	(d) lac operon
(v) Alec Jeffreys	(e) Bacteriophage experiment with 32P and 35S

Q.5 Assertion (A) – Reason (R) type questions. Choose: (a) Both A and R true, R explains A; (b) Both true, R does not explain A; (c) A true, R false; (d) A false, R true.5 × 1 = 5

(i) Assertion (A): DNA is a better genetic material than RNA.
Reason (R): DNA is chemically less reactive and structurally more stable than RNA.

(ii) Assertion (A): tRNA is called the adapter molecule.
Reason (R): Eukaryotic mRNA has a 5′ cap and a 3′ poly-A tail.

(iii) Assertion (A): The genetic code is degenerate.
Reason (R): One codon codes for more than one amino acid.

(iv) Assertion (A): hnRNA is the functional mRNA of eukaryotes.
Reason (R): hnRNA undergoes splicing, capping and tailing to form mature mRNA.

(v) Assertion (A): The lac operon is switched on when lactose is present in the medium.
Reason (R): Lactose (allolactose) binds the repressor and inactivates it.

SECTION – B (Very Short Answer · 2 Marks each)

6. Differentiate between a nucleoside and a nucleotide with one example of each.(2)

7. State Chargaff’s rule. If a DNA has 28 % thymine, find the percentage of guanine.(2)

8. What are histones? What is a nucleosome?(2)

9. A DNA template strand reads 3′-TACGGATTC-5′. Write the mRNA sequence and the amino acids it codes for (AUG = Met, CCU = Pro, AAG = Lys).(2)

10. What is the significance of the 5′ cap and poly-A tail of eukaryotic mRNA?(2)

11. Why is DNA replication called semiconservative? Name the scientists who proved it.(2)

12. State any two salient features of the genetic code.(2)

13. What is DNA fingerprinting? Name the DNA sequences that make it possible.(2)

SECTION – C (Short Answer · 3 Marks each)

14. Describe any six salient features of the double-helix model of DNA given by Watson and Crick.(3)

OR

Write any six differences between DNA and RNA.

15. Describe Griffith’s transformation experiment.(3)

OR

Describe the Hershey–Chase experiment and its conclusion.

16. Explain how DNA is packaged in the eukaryotic nucleus.(3)

OR

Differentiate between euchromatin and heterochromatin.

17. Describe the Meselson and Stahl experiment and state its conclusion.(3)

OR

Explain the process of DNA replication, mentioning the leading and lagging strands.

18. Explain a transcription unit. What are the template and coding strands?(3)

OR

Explain the post-transcriptional processing of hnRNA in eukaryotes.

19. Write any six features of the genetic code.(3)

OR

Describe the structure and role of tRNA.

20. Explain the lac operon as an example of regulation of gene expression in bacteria.(3)

OR

State six salient features of the Human Genome Project (HGP).

SECTION – D (Case-Based Questions · 4 Marks each)

21. Read the passage and answer the questions that follow.(4)

In an experiment, E. coli was grown for many generations in a medium containing ¹⁵NH₄Cl so that all its DNA was heavy. The cells were then transferred to a medium containing ¹⁴NH₄Cl. After 20 minutes (one generation) and after 40 minutes (two generations), DNA was extracted and centrifuged in a caesium chloride (CsCl) density gradient.

(i) What type of DNA is found after 20 minutes?(1)

(ii) What is the composition of DNA after 40 minutes?(1)

(iii) Name the scientists and state the conclusion. What would be observed after one generation if replication were conservative?(2)

22. Read the passage and answer the questions that follow.(4)

A segment of the template strand of a gene is 3′-TAC TTT GGG ACT-5′. It is transcribed and translated (AUG = Met, AAA = Lys, CCC = Pro, UGA = stop; UAA = stop).

(i) Write the sequence of the mRNA formed.(1)

(ii) Write the amino acid sequence of the polypeptide formed.(1)

(iii) If a mutation changes the second codon AAA into UAA, what will be the effect on the polypeptide?(2)

SECTION – E (Long Answer · 5 Marks each)

23. Describe the structure of a polynucleotide chain and the double-helix model of DNA. How is DNA packaged in eukaryotes?(5)

OR

Describe the experiments that proved that DNA is the genetic material (Griffith, Avery–MacLeod–McCarty and Hershey–Chase).

24. Describe the process of DNA replication. Explain how Meselson and Stahl proved that it is semiconservative.(5)

OR

Describe the process of transcription in bacteria and the processing of the transcript in eukaryotes.

25. Describe the genetic code and the process of translation (protein synthesis).(5)

OR

Explain DNA fingerprinting (principle, steps and applications) and the salient features of the Human Genome Project.

********

CBSE · NCERT

BIOLOGY – CHAPTER 5 : COMPLETE SOLUTIONS

Molecular Basis of Inheritance · Answer Key with Step-by-Step Explanations

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Class : XIISubject : BiologyMax. Marks : 100

Q.1 Quick Answer Key (MCQ)20 × 1 = 20

i	(b)	ii	(d)	iii	(a)	iv	(c)	v	(b)
vi	(d)	vii	(a)	viii	(c)	ix	(b)	x	(d)
xi	(a)	xii	(c)	xiii	(b)	xiv	(d)	xv	(a)
xvi	(c)	xvii	(b)	xviii	(d)	xix	(a)	xx	(c)

Q.1 Solutions with Explanation

(i) Ans. (b) 20 %

Chargaff: A = T = 30 % so A + T = 60 %; G + C = 40 %, therefore G = C = 20 %.

(ii) Ans. (d) A nitrogenous base and a pentose sugar

Base + sugar (N-glycosidic linkage) = nucleoside; phosphate joins the sugar by a phosphoester bond; nucleotides are joined by 3′–5′ phosphodiester bonds.

(iii) Ans. (a) 0.34 nm

Pitch is 3.4 nm with 10 base pairs per turn, so each base pair is 0.34 nm from the next.

(iv) Ans. (c) DNA of the phage

DNA contains phosphorus (labelled by ³²P), while protein contains sulphur (labelled by ³⁵S). Only ³²P entered the bacteria, proving DNA is the genetic material.

(v) Ans. (b) Avery, MacLeod and McCarty

Avery, MacLeod and McCarty purified biochemicals from heat-killed S cells; proteases and RNase did not stop transformation but DNase did.

(vi) Ans. (d) Two molecules each of H2A, H2B, H3 and H4

The histone octamer = (H2A, H2B, H3, H4) × 2; about 200 bp of negatively charged DNA wraps around it.

(vii) Ans. (a) Loosely packed, lightly stained and transcriptionally active

Euchromatin is the loosely packed, light-stained, transcriptionally active chromatin; heterochromatin is dense, dark-stained and inactive.

(viii) Ans. (c) 1 : 1

After 1st generation: all hybrid. After 2nd: 4 DNA molecules – 2 hybrid and 2 light = 1 : 1.

(ix) Ans. (b) Lagging strand

On the lagging strand (template with 5′→3′ polarity) DNA is made discontinuously as short Okazaki fragments, later joined by DNA ligase.

(x) Ans. (d) 5′ → 3′

DNA polymerase polymerises only in the 5′ → 3′ direction (adds to the 3′-OH of the growing chain).

(xi) Ans. (a) Towards the 5′ end (upstream) of the structural gene

The promoter lies upstream (5′ side) of the structural gene (on the coding strand) and is the binding site of RNA polymerase.

(xii) Ans. (c) RNA polymerase III

RNA pol I – 28S, 18S, 5.8S rRNA; RNA pol II – hnRNA (precursor of mRNA); RNA pol III – tRNA, 5S rRNA and snRNAs.

(xiii) Ans. (b) Splicing, capping and tailing

Introns are removed and exons are joined (splicing); a methyl guanosine triphosphate cap is added at the 5′ end and a poly-A tail (200–300 adenylate residues) at the 3′ end.

(xiv) Ans. (d) AUG

AUG codes for methionine and acts as the start codon; UAA, UAG, UGA are stop codons.

(xv) Ans. (a) 61

61 codons code for the 20 amino acids; 3 are stop codons.

(xvi) Ans. (c) More than one codon can code for the same amino acid

Some amino acids are coded by more than one codon (e.g. 61 codons for 20 amino acids).

(xvii) Ans. (b) 23S rRNA (a ribozyme)

The 23S rRNA of the large subunit acts as a ribozyme (peptidyl transferase); this supports the RNA-world idea.

(xviii) Ans. (d) Lactose (allolactose)

Lactose (allolactose) binds the repressor and inactivates it, so the operator is free and RNA polymerase can transcribe z, y and a genes.

(xix) Ans. (a) The repressor protein

The i gene (regulatory gene) codes for the repressor; z → β-galactosidase, y → permease, a → transacetylase.

(xx) Ans. (c) VNTRs (variable number of tandem repeats)

VNTRs are repetitive satellite DNA sequences whose number of repeats differs from person to person; Alec Jeffreys used them as probes.

Q.2 Fill in the blanks5 × 1 = 5

(i) Nucleoside

(ii) Hershey and Chase (1952)

(iii) Okazaki

(iv) Adapter

(v) Introns

Q.3 True / False5 × 1 = 5

(i) False.

Adenine pairs with thymine (2 H-bonds); guanine pairs with cytosine (3 H-bonds).

(ii) True.

Each daughter DNA has one parental strand and one newly synthesised strand (Meselson and Stahl).

(iii) False.

The template strand has 3′ → 5′ polarity; RNA is synthesised in the 5′ → 3′ direction.

(iv) True.

They do not code for any amino acid and terminate translation.

(v) True.

Lactose (allolactose) inactivates the repressor and switches the operon on.

Q.4 Matching5 × 1 = 5

(i) – (c)  ·  (ii) – (a)  ·  (iii) – (e)  ·  (iv) – (d)  ·  (v) – (b)

Griffith – transformation; Meselson and Stahl – semiconservative replication; Hershey and Chase – bacteriophage experiment; Jacob and Monod – lac operon; Alec Jeffreys – DNA fingerprinting.

Q.5 Assertion – Reason5 × 1 = 5

(i) Ans. (a)

Both true; RNA has a reactive 2′-OH group and is labile, whereas DNA (with thymine and no 2′-OH) is more stable, so it is better for storing genetic information.

(ii) Ans. (b)

Both are true but unrelated; R does not explain A (tRNA reads the codon and carries the amino acid).

(iii) Ans. (c)

A is true (more than one codon for an amino acid). R is false: a codon is unambiguous and specific, coding for only one amino acid.

(iv) Ans. (d)

A is false: hnRNA is the primary transcript (precursor); after processing it becomes functional mRNA. R is true.

(v) Ans. (a)

Both true; the inactive repressor cannot bind the operator, so RNA polymerase transcribes the structural genes.

SECTION – B (2 Marks each)

6. Ans.

Nucleoside = nitrogenous base + pentose sugar (N-glycosidic bond), e.g. adenosine / cytidine. [1]
Nucleotide = nucleoside + phosphate group (phosphoester bond), e.g. adenylic acid (AMP) / deoxyadenylic acid. [1]

7. Ans.

Chargaff’s rule: in DNA, the amount of A equals T and G equals C (purines = pyrimidines). [1]
T = 28 % → A = 28 %; A + T = 56 %, so G + C = 44 %; G = 22 %. [1]

8. Ans.

Histones are positively charged, basic proteins rich in lysine and arginine, which bind the negatively charged DNA. [1]
A nucleosome is the repeating unit of chromatin: about 200 bp of DNA wrapped around a histone octamer (H2A, H2B, H3, H4 × 2). [1]

9. Ans.

mRNA: 5′-AUGCCUAAG-3′ (T→A, A→U, C→G, G→C). [1] Codons AUG | CCU | AAG → Met – Pro – Lys. [1]

10. Ans.

Capping (methyl guanosine triphosphate at the 5′ end) and tailing (200–300 adenylate residues at the 3′ end) are added to hnRNA after splicing. [1] They protect the mRNA from degradation, help its transport out of the nucleus and make it ready for translation. [1]

11. Ans.

In replication, the two strands separate and each acts as a template for a new strand, so each daughter DNA has one old (parental) strand and one new strand. [1] It was proved by Meselson and Stahl (1958) using E. coli and ¹⁵N, and by Taylor et al. in Vicia faba. [1]

12. Ans.

(i) The code is a triplet (64 codons; 61 code for amino acids and 3 are stop codons). (ii) It is degenerate (more than one codon for an amino acid). (iii) It is unambiguous and specific, universal, non-overlapping and comma-less; AUG is the start codon. (Any two, 1 mark each)

13. Ans.

DNA fingerprinting is a technique to identify an individual by comparing DNA band patterns, based on differences (polymorphism) in repetitive DNA sequences; developed by Alec Jeffreys. [1] It uses VNTRs (variable number of tandem repeats, a type of satellite DNA) whose repeat number varies between individuals. [1]

SECTION – C (3 Marks each)

14. Ans.

(i) DNA is made of two polynucleotide chains, backbone of sugar–phosphate, bases project inside. (ii) The two chains are antiparallel (5′→3′ and 3′→5′). (iii) Bases pair by H-bonds: A = T (2 bonds), G ≡ C (3 bonds); a purine always pairs with a pyrimidine, keeping a uniform distance. (iv) The chains are coiled in a right-handed helix. (v) Pitch 3.4 nm, 10 bp per turn, so adjacent base pairs are 0.34 nm apart. (vi) Base stacking of the planar bases adds to the stability of the helix. [½ × 6]

OR

DNA	RNA
Sugar is deoxyribose	Sugar is ribose
Bases: A, T, G, C	Bases: A, U, G, C
Double-stranded	Usually single-stranded
Chemically less reactive, more stable	2′-OH makes it reactive and less stable
Genetic material of most organisms	Genetic material of some viruses; mRNA, tRNA, rRNA in cells
Replicates itself	Made from DNA by transcription (no self-replication in cells)

[½ × 6]

15. Ans.

Frederick Griffith (1928) used Streptococcus pneumoniae: the S strain (smooth, capsule, virulent) and the R strain (rough, no capsule, avirulent). [½] (1) S-strain injected → mouse dies. (2) R-strain → mouse lives. (3) Heat-killed S-strain → mouse lives. (4) Heat-killed S + live R → mouse dies, and live S bacteria were recovered. [1½] He concluded that some ‘transforming principle’ from the heat-killed S cells transferred to R bacteria and transformed them into virulent S; its nature (DNA) was later shown by Avery, MacLeod and McCarty. [1]

OR

Alfred Hershey and Martha Chase (1952) grew bacteriophages on a medium with radioactive ³²P (labels DNA) or ³⁵S (labels protein) and allowed them to infect E. coli. [1] After infection the culture was blended (to remove phage coats from the cells) and centrifuged. [1] Bacteria with ³²P (DNA) were radioactive, while ³⁵S (protein) stayed in the supernatant. Hence DNA, not protein, is the genetic material. [1]

16. Ans.

A human cell has about 2.2 metres of DNA that must fit in a nucleus. [½] Negatively charged DNA is wrapped around the histone octamer (basic proteins with lysine and arginine) – about 200 bp per nucleosome; this gives ‘beads-on-string’ chromatin. [1] Chromatin is further coiled and condensed with the help of non-histone chromosomal (NHC) proteins to form chromatin fibres and then the chromosome at metaphase. [1] (In bacteria DNA is organised as a nucleoid with positively charged proteins and RNA.) [½]

OR

Euchromatin	Heterochromatin
Loosely packed	Densely packed
Light-stained	Dark-stained
Transcriptionally active	Transcriptionally inactive

[3]

17. Ans.

E. coli was grown for many generations in a medium with ¹⁵NH₄Cl (heavy nitrogen), so that all DNA was heavy. [½] It was then transferred to a ¹⁴NH₄Cl medium and samples were taken after 20 min (1 generation) and 40 min (2 generations); DNA was separated by CsCl density gradient centrifugation. [1] Result: after 20 min – all DNA intermediate (hybrid, ¹⁵N–¹⁴N); after 40 min – equal amounts of hybrid and light DNA. [1] Conclusion: DNA replication is semiconservative. [½]

OR

Replication starts at the origin of replication; helicase unwinds the helix to form a replication fork (topoisomerase relieves strain). [1] DNA-dependent DNA polymerase adds nucleotides only in the 5′→3′ direction using the parental strand as template (deoxynucleoside triphosphates give energy). [½] On the template with 3′→5′ polarity, synthesis is continuous (leading strand); on the template with 5′→3′ polarity it is discontinuous – short Okazaki fragments (lagging strand) joined by DNA ligase. [1½]

18. Ans.

A transcription unit has three regions: promoter (5′ end, upstream; binding site of RNA polymerase), structural gene and terminator (3′ end; stops transcription). [1½] The strand with 3′→5′ polarity acts as the template strand; the other strand with 5′→3′ polarity, whose sequence is the same as RNA (with U in place of T), is the coding strand. [1½]

OR

Eukaryotic genes are split genes (exons + introns); the primary transcript is hnRNA. [½] Splicing: introns are removed and exons joined in a defined order. [1] Capping: methyl guanosine triphosphate added to the 5′ end. [½] Tailing: 200–300 adenylate residues added at the 3′ end. [½] The processed mRNA is transported out of the nucleus for translation. [½]

19. Ans.

(i) It is a triplet code (64 codons). (ii) 61 codons code for amino acids; 3 are stop codons (UAA, UAG, UGA). (iii) AUG is the start codon (also codes for methionine). (iv) The code is degenerate. (v) It is unambiguous and specific – one codon codes for one amino acid. (vi) It is universal, non-overlapping and comma-less (continuous). [½ × 6]

OR

tRNA is the adapter molecule (proposed by Francis Crick) that reads the code and brings amino acids. [½] Its secondary structure is clover-leaf with an anticodon loop (complementary to the codon), a D-loop, a T-loop and an amino-acid acceptor end (3′ end); the actual 3-D structure is L-shaped. [1½] There is one tRNA for each amino acid (and an initiator tRNA for methionine); it is charged by aminoacyl-tRNA synthetase using ATP. [1]

20. Ans.

Operon = a group of structural genes with a common promoter and regulator. The lac operon has a regulatory gene i (codes for the repressor) and structural genes z (β-galactosidase), y (permease), a (transacetylase) with a promoter and operator. [1] No lactose: the repressor binds the operator and blocks RNA polymerase – operon off. [1] Lactose present: lactose (allolactose, the inducer) binds the repressor and inactivates it; RNA polymerase transcribes z, y, a – operon on. (Jacob and Monod; regulation is negative and by substrate.) [1]

OR

(i) The human genome has about 3 × 10⁹ bp (3164.7 million bases). (ii) About 30,000 genes were estimated (average gene 3000 bases; the largest gene, dystrophin, has 2.4 million bases). (iii) 99.9 % of nucleotide bases are the same in all people. (iv) The functions of over 50 % of discovered genes were unknown. (v) Less than 2 % of the genome codes for proteins; a large part is repeated sequences. (vi) Chromosome 1 has the most genes (2968) and the Y chromosome the fewest (231); about 1.4 million SNPs were identified. [½ × 6]

SECTION – D (Case-Based · 4 Marks each)

21. Ans.

(i) Only hybrid (intermediate density) DNA – each molecule has one ¹⁵N strand and one ¹⁴N strand – forming a single band. [1]

(ii) Equal amounts (1 : 1) of hybrid DNA and light (¹⁴N) DNA. [1]

(iii) Meselson and Stahl; DNA replication is semiconservative. [1] If it were conservative, two separate bands (one heavy, one light) and no intermediate band would appear after one generation. [1] [2]

22. Ans.

(i) 5′-AUG AAA CCC UGA-3′. [1]

(ii) AUG – AAA – CCC – UGA(stop) → Met – Lys – Pro (UGA terminates translation). [1]

(iii) UAA is a stop codon, so translation ends after methionine. [1] Only a very short peptide (Met) is formed – a truncated, non-functional protein (nonsense mutation). [1] [2]

SECTION – E (Long Answer · 5 Marks each)

23. Ans.

Nucleotide: base (purines A, G; pyrimidines C, T; U in RNA) + deoxyribose sugar + phosphate. Nucleoside = base + sugar (N-glycosidic bond); nucleotides are linked by 3′–5′ phosphodiester bonds into a polynucleotide chain with a free 5′-phosphate and a free 3′-OH end. [1½]
Double helix (Watson and Crick, 1953): two antiparallel chains; right-handed helix; A pairs with T (2 H-bonds), G with C (3 H-bonds) (Chargaff: A = T, G = C); pitch 3.4 nm; 10 bp per turn; 0.34 nm between base pairs. [1½]
Packaging: negatively charged DNA is wound round the histone octamer (2 × H2A, H2B, H3, H4) – about 200 bp – forming a nucleosome (beads-on-string chromatin); further folded with non-histone chromosomal proteins into chromatin fibres and the metaphase chromosome. Euchromatin – loose, active; heterochromatin – dense, inactive. [2]

OR

Griffith (1928): Streptococcus pneumoniae; heat-killed S + live R killed mice and live S cells were recovered – a ‘transforming principle’ transformed R to S. [1½]
Avery, MacLeod and McCarty (1933–44): purified biochemicals from heat-killed S cells; protein-digesting enzymes (protease) and RNase did not affect transformation, but DNase inhibited it – so DNA is the transforming principle. [1½]
Hershey and Chase (1952): phages labelled with ³⁵S (protein) or ³²P (DNA) infected E. coli; after blending and centrifugation only ³²P (DNA) was found inside the bacteria. This proved DNA is the genetic material. [2]

24. Ans.

Replication is the copying of DNA; it occurs in the S-phase; DNA-dependent DNA polymerase needs a primer and uses deoxynucleoside triphosphates as substrates and energy source. [½]
Steps: origin of replication → helicase unwinds the strands (replication fork). The polymerase adds nucleotides 5′→3′; on the 3′→5′ template synthesis is continuous (leading strand); on the 5′→3′ template it is discontinuous with Okazaki fragments (lagging strand) joined by ligase. Each daughter DNA = one old + one new strand. (E. coli copies its 4.6 × 10⁶ bp in about 38 minutes.) [2]
Meselson–Stahl (1958): E. coli grown in ¹⁵N, shifted to ¹⁴N; CsCl density-gradient centrifugation – after 20 min only hybrid DNA; after 40 min equal hybrid and light DNA – semiconservative. [2½]

OR

Transcription unit: promoter, structural gene, terminator; template strand 3′→5′. [1]
Initiation: RNA polymerase binds the promoter with the σ factor. Elongation: the enzyme moves along the template, uses nucleoside triphosphates (A, U, G, C) and synthesises RNA in the 5′→3′ direction by complementary pairing (no primer). Termination: at the terminator the ρ factor releases the RNA. In bacteria one RNA polymerase makes mRNA, tRNA and rRNA. [2]
Eukaryotes: three polymerases (I – rRNA, II – hnRNA, III – tRNA, 5S rRNA, snRNA); hnRNA undergoes splicing (introns removed), capping and tailing to form functional mRNA. [2]

25. Ans.

Genetic code: triplet; 64 codons (61 sense, 3 stop – UAA, UAG, UGA); AUG start; degenerate, unambiguous, universal, non-overlapping, comma-less (Nirenberg, Khorana, Ochoa). [1]
Charging of tRNA: amino acid + ATP with aminoacyl-tRNA synthetase → aminoacyl-tRNA. [½]
Initiation: the small ribosomal subunit binds the mRNA at the start codon AUG (after the 5′ UTR) and the initiator tRNA carrying methionine; the large subunit joins. [1]
Elongation: the ribosome has two sites for tRNAs; the aminoacyl-tRNA whose anticodon matches the next codon enters; the 23S rRNA (ribozyme, peptidyl transferase) forms the peptide bond; the ribosome moves by one codon (5′→3′). [1½]
Termination: at a stop codon a release factor binds, and the complete polypeptide is released. Several ribosomes on one mRNA form a polysome. [1]

OR

DNA fingerprinting (Alec Jeffreys) identifies people by variations in repetitive DNA (VNTR, satellite DNA) that are different in every individual except identical twins. [½]
Steps: isolate DNA → digest with restriction endonucleases → separate fragments by gel electrophoresis → transfer (Southern blotting) to nylon/nitrocellulose → hybridise with labelled VNTR probe → detect by autoradiography. [1½]
Applications: forensic science, paternity testing, studying genetic diversity of populations. [½]
HGP: launched 1990, completed 2003; ~3 × 10⁹ bp; ~30,000 genes; 99.9 % bases identical among people; < 2 % codes for proteins; SNPs identified; methods – ESTs and sequence annotation; Sanger’s sequencing; use of BAC/YAC vectors. [2½]

QUICK REVISION · HIGH-YIELD POINTS (NEET / CBSE)

1. DNA: nucleoside = base + sugar; nucleotide = + phosphate; 3′–5′ phosphodiester bond. Chargaff: A = T, G = C. B-DNA: pitch 3.4 nm, 10 bp per turn, 0.34 nm per bp; A = T (2 H-bonds), G ≡ C (3 H-bonds).
2. Length of DNA = bp × 0.34 nm: E. coli (4.6 × 10^6 bp) ≈ 1.36 mm; human haploid (6.6 × 10^9 bp) ≈ 2.2 m. Nucleosome = histone octamer (H2A, H2B, H3, H4 × 2) + ~200 bp DNA.
3. Euchromatin – loose, light, active; heterochromatin – dense, dark, inactive. Genetic material proofs: Griffith (1928), Avery–MacLeod–McCarty (DNase), Hershey–Chase 1952 (32P-DNA, 35S-protein).
4. Replication: semiconservative (Meselson–Stahl, 15N/14N, CsCl); 5′→3′ only; leading (continuous), lagging (Okazaki fragments + ligase). After n generations in 14N: 2 hybrid DNAs; the rest light.
5. Transcription unit = promoter (upstream, 5′) + structural gene + terminator; template strand 3′→5′; RNA made 5′→3′; σ and ρ factors; Pol I – rRNA, Pol II – hnRNA, Pol III – tRNA, 5S rRNA.
6. hnRNA → splicing (introns out), 5′ cap (methyl guanosine triphosphate), 3′ poly-A tail (200–300 A) → mRNA.
7. Genetic code: triplet, 64 codons, 61 sense + 3 stop (UAA, UAG, UGA), AUG start, degenerate, universal, non-overlapping, comma-less. tRNA = adapter (Crick); 23S rRNA = ribozyme for peptide bond.
8. lac operon: i (repressor), z (β-galactosidase), y (permease), a (transacetylase); lactose/allolactose = inducer; negative regulation (Jacob and Monod).
9. HGP (1990–2003): 3 × 10^9 bp, ~30,000 genes, 99.9 % identical, < 2 % coding, ESTs + annotation, SNPs; chromosome 1 largest gene count, Y fewest.
10. DNA fingerprinting (Alec Jeffreys): VNTR; restriction digestion → electrophoresis → Southern blot → probe hybridisation → autoradiography; uses: forensic, paternity, diversity.

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