NCERT Solutions for Class 12 Biology Chapter 5: Molecular Basis of Inheritance

These Class 12 Biology Chapter 5 solutions cover Molecular Basis of Inheritance for the NCERT 2026–27 session. The chapter explains how DNA functions as the genetic material — its double-helix structure, the experiments that proved DNA is the hereditary molecule, replication, transcription, the genetic code, translation, regulation of gene expression (the lac operon), the Human Genome Project and DNA fingerprinting. Every NCERT exercise question is reproduced verbatim and answered in clear, exam-ready CBSE style below.

Class: 12 Subject: Biology Chapter: 5 Name: Molecular Basis of Inheritance Type: Genetics & Molecular Biology Session: 2026–27
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Class 12 Biology Chapter 5 Solutions – Overview

Chapter 5, Molecular Basis of Inheritance, traces the journey from Mendel’s abstract ‘factors’ to the discovery that DNA is the genetic material. It begins with the structure of the polynucleotide chain and the Watson–Crick double helix (built on Chargaff’s rule and Franklin’s X-ray data), then describes how DNA is packaged into nucleosomes and chromatin. The classic experiments of Griffith (transforming principle), Avery–MacLeod–McCarty and Hershey–Chase establish DNA as the hereditary molecule, while the Meselson–Stahl experiment proves semiconservative replication. The chapter then explains transcription, the triplet genetic code, translation (with tRNA as the adapter and the ribosome as a ribozyme), the regulation of gene expression through the lac operon, and finally two landmark applications — the Human Genome Project and DNA fingerprinting. It is a high-weightage, scoring chapter in the Class 12 Biology exam.

Key Concepts & Definitions

DNA: a long polymer of deoxyribonucleotides; the genetic material in most organisms. Length is measured in base pairs (bp).

Chargaff’s rule: in double-stranded DNA the ratios A=T and G=C are constant, so A+G (purines) = T+C (pyrimidines).

Double helix (Watson & Crick, 1953): two antiparallel polynucleotide chains coiled right-handed; A pairs with T (2 H-bonds), G pairs with C (3 H-bonds); pitch 3.4 nm, ~10 bp per turn, 0.34 nm between adjacent bp.

Nucleosome: negatively charged DNA (~200 bp) wrapped around a positively charged histone octamer; the repeating unit of chromatin (seen as “beads-on-string”).

Central dogma: genetic information flows DNA → RNA → Protein (reverse transcription RNA → DNA in some viruses).

Semiconservative replication: each daughter DNA has one parental and one newly synthesised strand (proved by Meselson & Stahl, 1958).

Transcription unit: a promoter + structural gene + terminator; the 3’→5′ template strand is copied, the 5’→3′ strand is the coding strand.

Genetic code: triplet, degenerate, non-overlapping, commaless, nearly universal; AUG codes Methionine and is the start codon; UAA, UAG, UGA are stop codons.

lac operon: regulatory i gene + three structural genes (z, y, a) under one promoter; lactose acts as inducer that inactivates the repressor (negative regulation).

DNA fingerprinting: identification of individuals from polymorphism in repetitive (satellite) DNA, especially VNTRs (developed by Alec Jeffreys).

NCERT Exercises — Solutions

1. Group the following as nitrogenous bases and nucleosides: Adenine, Cytidine, Thymine, Guanosine, Uracil and Cytosine.

ANSWER A nitrogenous base linked to a pentose sugar (through an N-glycosidic bond) forms a nucleoside. Names ending in -osine / -idine are nucleosides. Nitrogenous bases: Adenine, Thymine, Uracil, Cytosine. Nucleosides: Cytidine (cytosine + ribose), Guanosine (guanine + ribose).

2. If a double stranded DNA has 20 per cent of cytosine, calculate the per cent of adenine in the DNA.

ANSWER By Chargaff’s rule, in double-stranded DNA, %C = %G and %A = %T. Given %C = 20%, therefore %G = 20%. So %C + %G = 20 + 20 = 40%. The remaining = 100% − 40% = 60% is shared equally by A and T. Therefore %A = 60 ÷ 2 = 30% (and %T = 30%).

3. If the sequence of one strand of DNA is written as follows:5′-ATGCATGCATGCATGCATGCATGCATGC-3′Write down the sequence of complementary strand in 5’→3′ direction.

ANSWER The two strands are antiparallel and complementary (A–T, G–C). First pair each base of the given strand: Given (5’→3′): A T G C A T G C …  →  complement (3’→5′): T A C G T A C G … Now write the complementary strand in the required 5’→3′ direction (i.e. read the antiparallel partner end-to-end): 5′-GCATGCATGCATGCATGCATGCATGCAT-3′

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.

ANSWER The mRNA has the same sequence as the coding strand (and same 5’→3′ polarity), except that thymine (T) is replaced by uracil (U). Coding strand: 5′-ATGCATGCATGCATGCATGCATGCATGC-3′ mRNA: 5′-AUGCAUGCAUGCAUGCAUGCAUGCAUGC-3′

5. Which property of DNA double helix led Watson and Crick to hypothesise semi-conservative mode of DNA replication? Explain.

ANSWER The key property was the complementary base pairing between the two antiparallel strands (A pairs with T, G pairs with C). Because the strands are complementary, the base sequence of one strand exactly fixes the sequence of the other. Watson and Crick reasoned that the two strands could separate, and each could then act as a template for the synthesis of a new complementary strand. As a result, each of the two daughter DNA molecules would contain one old (parental) strand and one newly synthesised strand — this is the semiconservative mode. Their own words were that “the specific pairing we have postulated immediately suggests a possible copying mechanism for the genetic material.”

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.

ANSWER Polymerases are named by their template and their product:
EnzymeTemplateProductProcess
DNA-dependent DNA polymeraseDNADNAReplication
DNA-dependent RNA polymeraseDNARNATranscription
RNA-dependent RNA polymerase (RNA replicase)RNARNARNA replication (some RNA viruses)
RNA-dependent DNA polymerase (reverse transcriptase)RNADNAReverse transcription (retroviruses)

7. How did Hershey and Chase differentiate between DNA and protein in their experiment while proving that DNA is the genetic material?

ANSWER Hershey and Chase (1952) used the fact that DNA contains phosphorus but no sulphur, while protein contains sulphur but no phosphorus. They grew bacteriophages in two separate media: one with radioactive phosphorus (32P), which labelled only the viral DNA, and another with radioactive sulphur (35S), which labelled only the viral protein coat. The labelled phages were allowed to infect E. coli. The empty viral coats were then detached by agitating in a blender, and the bacteria were separated by centrifugation. Bacteria infected with 32P phages became radioactive (DNA entered the cell), while bacteria infected with 35S phages were not radioactive (protein stayed outside). This proved that only DNA, not protein, passed from the virus into the bacterium — so DNA is the genetic material.

8. Differentiate between the followings:(a) Repetitive DNA and Satellite DNA(b) mRNA and tRNA(c) Template strand and Coding strand

ANSWER (a) Repetitive DNA vs Satellite DNA
Repetitive DNASatellite DNA
Short DNA sequences repeated many times in the genome.A subset of repetitive DNA that separates as small peaks (satellites) from bulk DNA during density-gradient centrifugation.
Includes both interspersed and tandem repeats.Classified by base composition and repeat number into micro-satellites, mini-satellites, etc.; shows high polymorphism and is the basis of DNA fingerprinting.
(b) mRNA vs tRNA
mRNA (messenger RNA)tRNA (transfer RNA)
Carries the genetic message from DNA; acts as the template for protein synthesis.Acts as an adapter; brings the correct amino acid and reads the codon.
Linear; has codons, a start and stop codon and untranslated regions (UTRs).Clover-leaf secondary structure (inverted-L in 3-D); has an anticodon loop and an amino-acid acceptor end; specific for each amino acid.
(c) Template strand vs Coding strand
Template strandCoding strand
Has 3’→5′ polarity; is actually copied by RNA polymerase.Has 5’→3′ polarity; is displaced and not copied.
Its sequence is complementary to the mRNA.Its sequence is the same as the mRNA (except T in place of U); all reference points are defined on it.

9. List two essential roles of ribosome during translation.

ANSWER (i) The ribosome provides the platform and binding sites (in the large subunit) that hold the charged tRNAs and the growing peptide close together, and it moves along the mRNA codon by codon so that amino acids are added in the correct order. (ii) The ribosome acts as a catalyst (ribozyme) for peptide-bond formation — in bacteria the 23S rRNA of the large subunit catalyses the joining of amino acids.

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?

ANSWER Lactose works as the inducer: it binds to the repressor protein (made constitutively by the i gene) and inactivates it, freeing the operator so that RNA polymerase can transcribe the structural genes (z, y, a). The enzyme β-galactosidase produced then hydrolyses the lactose into glucose and galactose, so the lactose is gradually used up. As the lactose level falls, there is no longer enough inducer to keep the repressor inactive. The repressor again becomes active, binds to the operator and blocks RNA polymerase. Hence the lac operon switches off some time after the lactose is added (it stays on only as long as lactose is present).

11. Explain (in one or two lines) the function of the followings:(a) Promoter(b) tRNA(c) Exons

ANSWER (a) Promoter: a DNA sequence located upstream (towards the 5′-end of the coding strand) that provides the binding site for RNA polymerase and defines the template and coding strands, so transcription can begin. (b) tRNA: the adapter molecule that reads the codon on mRNA through its anticodon and carries the corresponding amino acid to the ribosome during translation. (c) Exons: the coding (expressed) sequences of a gene that are retained in the mature mRNA after splicing and are translated into protein.

12. Why is the Human Genome project called a mega project?

ANSWER The HGP was called a mega project because of its enormous scale, cost and duration: It aimed to sequence all ~3 × 109 base pairs of the human genome and to identify all ~20,000–25,000 genes. At an estimated US $3 per base pair, the total cost was about 9 billion US dollars. If printed (1000 letters per page, 1000 pages per book), the sequence from a single cell would fill about 3300 books. It was a 13-year international project (1990–2003) coordinated by the U.S. Dept. of Energy and NIH, with partners from the U.K., Japan, France, Germany, China and others, and it needed high-speed computing (giving rise to bioinformatics).

13. What is DNA fingerprinting? Mention its application.

ANSWER DNA fingerprinting is a quick technique to identify and compare individuals by detecting differences (polymorphism) in specific repetitive DNA sequences — especially VNTRs (Variable Number of Tandem Repeats), a type of mini-satellite. Since 99.9% of human DNA is identical, these highly polymorphic regions give a unique banding pattern for each person (except identical twins). It was developed by Alec Jeffreys. Applications: (i) Forensic science — identifying criminals/victims from blood, hair, skin, semen, etc.; (ii) Paternity and maternity testing in disputes; (iii) Determining population and genetic diversity and studying evolution.

14. Briefly describe the following:(a) Transcription(b) Polymorphism(c) Translation(d) Bioinformatics

ANSWER (a) Transcription: the process of copying genetic information from the template strand of DNA into RNA. Following base complementarity (with adenine pairing uracil instead of thymine), only one strand of a segment of DNA is copied. RNA polymerase binds the promoter (initiation), polymerises ribonucleotides in the 5’→3′ direction (elongation), and stops at the terminator (termination). (b) Polymorphism: variation at the genetic (DNA) level. An inheritable mutation that occurs in a population at a frequency greater than 0.01 is called a DNA polymorphism. Such variations (more common in non-coding DNA) accumulate over generations and form the basis of genetic mapping and DNA fingerprinting. (c) Translation: the process of polymerising amino acids into a polypeptide. Amino acids are activated and charged onto tRNAs; the ribosome binds the mRNA at the start codon (AUG) recognised by the initiator tRNA, then moves codon by codon adding amino acids (joined by peptide bonds) until a release factor binds a stop codon and the polypeptide is released. (d) Bioinformatics: a branch of biology that uses computational tools to store, retrieve, organise and analyse the huge amount of biological (especially sequence) data generated by projects like the HGP.

Extra Practice Questions

Short Answer Type Questions

Q1. State Chargaff’s rule and give one of its uses.

ANSWERIn double-stranded DNA the amount of adenine equals thymine and guanine equals cytosine (A=T, G=C), so purines = pyrimidines. It supported Watson and Crick’s base-pairing scheme and lets us calculate the percentage of one base if another is known.

Q2. Why are the two strands of DNA said to be antiparallel?

ANSWERThe two strands run in opposite directions: if one strand has 5’→3′ polarity, the partner strand has 3’→5′ polarity. This opposite orientation is described as antiparallel.

Q3. What is a nucleosome and what is its DNA content?

ANSWERA nucleosome is the repeating unit of chromatin formed when negatively charged DNA wraps around a positively charged histone octamer. A typical nucleosome contains about 200 bp of DNA helix and appears as a “bead” in the beads-on-string structure.

Q4. Why is DNA a more stable genetic material than RNA?

ANSWERRNA has a reactive 2′-OH group on every ribose making it labile and easily degraded, and it can be catalytic (reactive). DNA lacks this 2′-OH, is double-stranded (the two complementary strands re-anneal and allow repair), and uses thymine in place of uracil, all of which make DNA chemically less reactive and structurally more stable.

Q5. List two salient features of the genetic code.

ANSWER(i) It is a triplet code — three nucleotides specify one amino acid (61 sense codons + 3 stop codons). (ii) It is degenerate — some amino acids are coded by more than one codon. (It is also non-overlapping, commaless and nearly universal.)

Long Answer Type Questions

Q1. Describe the Meselson and Stahl experiment that proved DNA replication is semiconservative.

ANSWERMeselson and Stahl (1958) grew E. coli for many generations in a medium where the only nitrogen source was 15NH4Cl (heavy isotope 15N). This made all the DNA “heavy” (15N), which sediments lower than normal 14N DNA in a CsCl density gradient. They then transferred the cells to normal 14N medium and extracted DNA at intervals. After one generation (20 min, the division time of E. coli), all the DNA had an intermediate/hybrid density — one heavy and one light strand. After two generations (40 min), the DNA was made of equal amounts of hybrid and light DNA. This pattern is only possible if each strand acts as a template and each daughter molecule keeps one parental strand — proving semiconservative replication. A similar result with radioactive thymidine in Vicia faba by Taylor confirmed it in chromosomes too.

Q2. Explain the structure and mechanism of regulation of the lac operon in the presence and absence of lactose.

ANSWERThe lac operon (elucidated by Jacob and Monod) consists of one regulatory gene (i, which codes for the repressor) and a common promoter and operator controlling three structural genes: z (β-galactosidase, hydrolyses lactose to glucose + galactose), y (permease, increases lactose entry) and a (transacetylase). In the absence of lactose, the active repressor binds the operator and blocks RNA polymerase, so the operon is switched off. In the presence of lactose, lactose (the inducer) binds and inactivates the repressor, freeing the operator; RNA polymerase then transcribes the structural genes and the lactose-metabolising enzymes are made. Because the substrate (lactose) itself induces the enzymes, this is regulation of enzyme synthesis by substrate, and control by the repressor is called negative regulation.

Q3. Explain how a frameshift mutation arises, using a sentence of three-letter words as a model.

ANSWERThe genetic code is read in non-overlapping triplets, like a sentence of three-letter words. Take “RAM HAS RED CAP.” If one extra letter (B) is inserted, the message is misread from that point: “RAM HAS BRE DCA P”; inserting two letters (BI) gives “RAM HAS BIR EDC AP” — in both cases the reading frame shifts and all codons after the insertion are wrong. These are frameshift insertion mutations (deleting one or two bases does the same). However, inserting or deleting three letters (e.g. BIG) gives “RAM HAS BIG RED CAP” — only one codon is added and the reading frame downstream is restored, because three (or a multiple of three) bases insert or remove whole codons without shifting the frame.

MCQs & Assertion–Reason

1. In B-DNA, the distance between two adjacent base pairs is approximately:

(a) 3.4 nm    (b) 0.34 nm    (c) 20 nm    (d) 34 nm

2. Guanine pairs with cytosine through how many hydrogen bonds?

(a) one    (b) two    (c) three    (d) four

3. The experiment that gave the unequivocal proof that DNA is the genetic material was performed by:

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

4. A nucleosome is formed when DNA wraps around:

(a) a histone octamer    (b) RNA polymerase    (c) non-histone proteins    (d) a ribosome

5. The total number of codons that act as stop (terminator) codons is:

(a) 1    (b) 2    (c) 3    (d) 61

6. AUG codon is special because it:

(a) is a stop codon    (b) codes for methionine and acts as the start codon    (c) codes for tryptophan only    (d) has no function

7. In the lac operon, the z gene codes for:

(a) permease    (b) transacetylase    (c) β-galactosidase    (d) repressor

8. The fragments synthesised discontinuously on the lagging strand are joined by:

(a) DNA polymerase    (b) DNA ligase    (c) helicase    (d) primase

9. In eukaryotes, the RNA polymerase that transcribes the precursor of mRNA (hnRNA) is:

(a) RNA polymerase I    (b) RNA polymerase II    (c) RNA polymerase III    (d) reverse transcriptase

10. The technique of DNA fingerprinting was initially developed by:

(a) Frederick Sanger    (b) Alec Jeffreys    (c) Har Gobind Khorana    (d) Marshall Nirenberg

Answer key: 1-(b), 2-(c), 3-(c), 4-(a), 5-(c), 6-(b), 7-(c), 8-(b), 9-(b), 10-(b).

For each Assertion–Reason question, choose: (A) Both true and the Reason correctly explains the Assertion; (B) Both true but the Reason is not the correct explanation; (C) Assertion true, Reason false; (D) Assertion false, Reason true.

A-R 1. Assertion: In double-stranded DNA the amount of adenine equals that of thymine.

Reason: Adenine pairs with thymine through two hydrogen bonds following Chargaff’s rule.

A-R 2. Assertion: DNA replication is semiconservative.

Reason: Each daughter DNA molecule has one parental strand and one newly synthesised strand.

A-R 3. Assertion: During transcription only one of the two DNA strands is copied into RNA.

Reason: If both strands were copied they would form double-stranded RNA and code for two different proteins, complicating gene expression.

A-R 4. Assertion: The genetic code is non-degenerate.

Reason: Each amino acid is coded by one and only one codon.

A-R 5. Assertion: RNA is a better store of genetic information than DNA.

Reason: The 2′-OH group of ribose makes RNA reactive and easily degradable.

Answer key: 1-(A), 2-(A), 3-(A), 4-(D), 5-(D).

Common Mistakes to Avoid

Watch out for these

  • Confusing the complementary strand (DNA, has T) with the mRNA (RNA, has U) — and forgetting to write sequences in the asked 5’→3′ direction.
  • Mixing up the template strand (3’→5′, copied) and the coding strand (5’→3′, not copied but same sequence as mRNA).
  • Reversing the H-bond count — A–T has two H-bonds, G–C has three.
  • Saying the genetic code is “universal” without the word nearly (there are mitochondrial and protozoan exceptions).
  • Writing that lactose is the repressor — lactose is the inducer that inactivates the repressor.
  • Stating that the 35S (protein-labelled) phages made the bacteria radioactive in the Hershey–Chase experiment — it was the 32P (DNA-labelled) phages.

Exam tips for this chapter

Learn the landmark experiments as labelled flow-steps (Griffith → Avery et al. → Hershey–Chase for “DNA is genetic material”; Meselson–Stahl for semiconservative replication) — these are repeated favourites. Memorise the exact numbers: 0.34 nm between bp, 3.4 nm pitch, ~10 bp per turn, 200 bp per nucleosome, 61 + 3 codons, ~3.16 billion bp in the human genome. For sequence problems, always show your working (pair the bases, then flip to 5’→3′). Practise the lac operon “ON/OFF” explanation and the frameshift “RAM HAS RED CAP” model, as both carry full marks when explained step by step.

Frequently Asked Questions

What is Class 12 Biology Chapter 5 Molecular Basis of Inheritance about?

It explains how DNA acts as the genetic material — covering the double-helix structure, DNA packaging, the Griffith, Avery and Hershey–Chase experiments, semiconservative replication, transcription, the genetic code, translation, the lac operon, the Human Genome Project and DNA fingerprinting.

If a DNA has 20% cytosine, what is the percentage of adenine?

By Chargaff’s rule, %C = %G = 20%, so C+G = 40%. The remaining 60% is shared by A and T, giving %A = 30% (and %T = 30%).

Why did Watson and Crick propose semiconservative replication?

Because the two DNA strands are complementary, each separated strand can act as a template to build a new complementary strand, so each daughter DNA keeps one old and one new strand — the semiconservative pattern.

Are these Class 12 Biology Chapter 5 solutions free?

Yes. All ClearStudy NCERT Solutions for Class 12 Biology are free and follow the official NCERT textbook for the 2026–27 session.

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