Molecular Basis of Inheritance Class 12: Concepts, Diagrams and Board Answers

What Molecular Basis of Inheritance is about

Weightage: This chapter is part of Unit VII, Genetics and Evolution, which carries 20 of the 70 theory marks together with Principles of Inheritance and Variation and Evolution.Source: CBSE Biology (044) syllabus 2026-27, course structure, Unit VII Genetics and Evolution

In real board papers: asked in 8 of the 8 board papers we hold (2023 to 2026), on average 8 marks per paper. See the questions

Key concepts

Structure of DNA and the double helix

NCERT section 5.1 The DNA (5.1.1 Structure of Polynucleotide Chain)

Nucleotide = nitrogenous base + pentose sugar (deoxyribose in DNA, ribose in RNA) + phosphate.

• Purines: adenine (A), guanine (G). Pyrimidines: cytosine (C), thymine (T, only DNA), uracil (U, only RNA).
• Base + sugar (joined at C1′ by an N-glycosidic linkage) = nucleoside (e.g. adenosine, deoxyadenosine).
• Nucleoside + phosphate (at C5′ by a phosphoester linkage) = nucleotide.
• Nucleotides join by 3′–5′ phosphodiester linkages. The chain has a free phosphate at the 5′ end and a free 3′-OH at the 3′ end. Sugar and phosphate form the backbone; bases project from it.
• RNA has an extra 2′-OH on every ribose, and uracil in place of thymine (thymine = 5-methyl uracil).

Length of DNA is given in nucleotides or base pairs (bp): φ×174 phage 5386 nucleotides; phage lambda 48,502 bp; E. coli 4.6 × 10⁶ bp; human haploid 3.3 × 10⁹ bp.

History: Friedrich Meischer (1869) identified DNA as an acidic substance in the nucleus, 'nuclein'. Watson and Crick (1953) proposed the double helix using X-ray diffraction data of Maurice Wilkins and Rosalind Franklin and Erwin Chargaff's finding that A/T and G/C ratios are constant and equal to one.

Salient features of the double helix:
1. Two polynucleotide chains; sugar-phosphate backbone outside, bases inside.
2. Antiparallel: one chain 5′→3′, the other 3′→5′.
3. Base pairing by hydrogen bonds: A=T (two H-bonds), G≡C (three H-bonds). A purine always faces a pyrimidine, so the distance between strands stays nearly uniform.
4. Right-handed coil; pitch 3.4 nm, about 10 bp per turn, so about 0.34 nm between adjacent base pairs.
5. Base pairs stack on one another; stacking plus H-bonds stabilise the helix.

Complementarity: knowing one strand's sequence gives the other's; each strand can template a copy, making two identical daughter molecules.

Chargaff's rule: in double-stranded DNA, A = T and G = C, so A + G = C + T (purines = pyrimidines).

Central dogma (Crick): DNA → RNA → protein. In some viruses information flows from RNA to DNA (reverse transcription).

Worked example

A double-stranded DNA has 18% guanine. Calculate the percentage of adenine. How many hydrogen bonds hold a 10-bp segment containing 4 G–C pairs? (2 marks)

Try it first, then show the solutionSolution

1. Method 1: G = C = 18%, so G + C = 36%; A + T = 64%; A = T = 32%. Method 2: A = (100 − 2 × 18) ÷ 2 = 32%. (1)
2. 4 G–C pairs × 3 = 12 H-bonds; 6 A–T pairs × 2 = 12 H-bonds; total 24. (1)

Answer: Adenine 32%; 24 hydrogen bonds.

Packaging of DNA

NCERT section 5.1.2 Packaging of DNA Helix

The problem: a typical mammalian cell has 6.6 × 10⁹ bp. Length = 6.6 × 10⁹ × 0.34 × 10⁻⁹ m ≈ 2.2 m, yet the nucleus is only about 10⁻⁶ m across.

Prokaryotes (E. coli): no defined nucleus, but DNA is not scattered. The negatively charged DNA is held by positively charged proteins in a region called the nucleoid, organised into large loops.

Eukaryotes:
1. Histones: positively charged, basic proteins rich in lysine and arginine (amino acids with positive side chains).
2. Eight histones form a histone octamer.
3. Negatively charged DNA wraps around the octamer = nucleosome; a typical nucleosome has about 200 bp of DNA.
4. Nucleosomes are the repeating units of chromatin, which looks like 'beads-on-string' under the electron microscope.
5. The beads-on-string fibre coils into chromatin fibres, which condense further at metaphase into chromosomes. Higher-level packing needs non-histone chromosomal (NHC) proteins.

Euchromatin versus heterochromatin:

• Euchromatin: loosely packed, stains light, transcriptionally active.
• Heterochromatin: densely packed, stains dark, inactive.

Flow chart: DNA double helix → nucleosome (DNA + histone octamer) → beads-on-string → chromatin fibre → condensed metaphase chromosome.

Worked example

The DNA of E. coli is 1.36 mm long. Calculate the number of base pairs. How many nucleosomes would a mammalian cell with 6.6 × 10⁹ bp have, assuming 200 bp per nucleosome? (2 marks)

Try it first, then show the solutionSolution

1. Number of bp = length ÷ distance per bp = 1.36 × 10⁻³ m ÷ 0.34 × 10⁻⁹ m = 4 × 10⁶ bp. Check: 4 × 10⁶ × 0.34 × 10⁻⁹ = 1.36 × 10⁻³ m. (1)
2. Nucleosomes = 6.6 × 10⁹ ÷ 200 = 3.3 × 10⁷. Check: 3.3 × 10⁷ × 200 = 6.6 × 10⁹. (1)

Answer: 4 × 10⁶ bp; 3.3 × 10⁷ nucleosomes.

The search for the genetic material

NCERT section 5.2 The Search for Genetic Material (5.2.1 The Genetic Material is DNA)

1. Griffith (1928): transformation in Streptococcus pneumoniae.

• S strain: smooth shiny colonies, polysaccharide (mucous) coat, virulent (mice die of pneumonia).
• R strain: rough colonies, no coat, harmless.

Injected into mice	Result
Live S	Mice die
Live R	Mice live
Heat-killed S	Mice live
Heat-killed S + live R	Mice die; live S bacteria recovered

Conclusion: a 'transforming principle' from heat-killed S changed R into S (it could now make the coat). This must be genetic material, but its chemical nature was unknown.

2. Avery, MacLeod and McCarty (1933–44): until then the genetic material was thought to be protein. They purified proteins, DNA and RNA from heat-killed S cells. DNA alone transformed R cells. Proteases and RNases did not stop transformation; DNase did. So DNA is the transforming principle, but not all biologists were convinced.

3. Hershey and Chase (1952): unequivocal proof with bacteriophages.
1. Phages grown with radioactive phosphorus (³²P) → radioactive DNA (DNA has phosphorus, protein does not).
2. Phages grown with radioactive sulphur (³⁵S) → radioactive protein (protein has sulphur, DNA does not).
3. Each batch infects E. coli; blending removes viral coats; centrifuging separates the lighter virus particles from the heavier bacteria.
4. Bacteria infected by ³²P phages were radioactive; those infected by ³⁵S phages were not.
Conclusion: DNA, not protein, enters the bacterium and is the genetic material.

Exceptions: in some viruses RNA is the genetic material (e.g. tobacco mosaic virus, QB bacteriophage).

Worked example

In the Hershey–Chase experiment, why were ³²P and ³⁵S chosen? What would have been observed if protein had been the genetic material? (3 marks)

Try it first, then show the solutionSolution

1. Phosphorus is present in DNA but not in protein; sulphur is present in protein but not in DNA, so each label marks only one molecule. (1)
2. Actual result: bacteria infected with ³²P phages became radioactive, those infected with ³⁵S phages did not, so DNA entered the cells. (1)
3. If protein were the genetic material, the ³⁵S label would have entered the bacteria (bacteria radioactive with ³⁵S) and the ³²P label would have stayed in the removed coats. (1)

Answer: Each isotope labels only DNA (³²P) or protein (³⁵S); with protein as genetic material, ³⁵S would have entered the bacteria instead of ³²P.

Properties of genetic material and the RNA world

NCERT section 5.2.2 Properties of Genetic Material (DNA versus RNA); 5.3 RNA World

A genetic material must:
1. Replicate itself.
2. Be chemically and structurally stable.
3. Allow slow changes (mutation) needed for evolution.
4. Express itself as 'Mendelian characters'.

DNA versus RNA against these criteria:

• Replication: both can direct their own copying by base pairing; proteins cannot, so they fail the first test.
• Stability: DNA is better. RNA's 2′-OH on every nucleotide is reactive, making RNA labile and easily degraded; RNA is also catalytic, hence reactive. Thymine in place of uracil gives DNA extra stability. Griffith's experiment showed this: heat killed the bacteria but did not destroy the genetic material (separated DNA strands can re-pair).
• Mutation: both mutate; RNA mutates faster, so RNA viruses with short life spans evolve faster.
• Expression: RNA can directly code for proteins; DNA depends on RNA. The protein-making machinery evolved around RNA.

Conclusion: DNA is preferred for storing genetic information; RNA is better for transmitting it.

RNA world: RNA was the first genetic material. Essential processes (metabolism, translation, splicing) evolved around RNA, which acted as both genetic material and catalyst (some reactions are still catalysed by RNA enzymes). Being reactive, RNA was unstable, so DNA evolved from RNA with chemical changes that make it stable; its double strand and complementary copy also allow repair.

Worked example

Give two chemical reasons why DNA is a more stable genetic material than RNA, and one reason why RNA is better for expressing characters. (2 marks)

Try it first, then show the solutionSolution

1. RNA has a reactive 2′-OH group on every ribose, which makes it labile and easily degraded; DNA lacks it. DNA has thymine instead of uracil, which adds stability (and DNA is double stranded). (1)
2. RNA can directly code for protein synthesis (the machinery evolved around RNA), while DNA needs RNA to express itself. (1)

Answer: No 2′-OH and thymine (not uracil) make DNA stable; RNA directly codes for proteins.

DNA replication

NCERT section 5.4 Replication (5.4.1 The Experimental Proof; 5.4.2 The Machinery and the Enzymes)

Semiconservative scheme (Watson and Crick): the two strands separate and each acts as a template; each daughter molecule has one parental and one new strand.

Meselson and Stahl (1958) proved it in E. coli:
1. Grew E. coli for many generations in ¹⁵NH₄Cl (¹⁵N is a heavy, non-radioactive isotope), so all DNA became heavy (¹⁵N/¹⁵N).
2. Transferred cells to normal ¹⁴NH₄Cl and took samples as they divided (E. coli divides every 20 minutes).
3. Separated DNA by density in a caesium chloride (CsCl) density gradient by centrifugation.

Time after transfer	DNA bands
0 (start)	all heavy
20 min (generation I)	all hybrid (intermediate)
40 min (generation II)	½ hybrid : ½ light
80 min (generation IV)	⅛ hybrid : ⅞ light

Rule: after n generations, 2 of the 2ⁿ molecules are hybrid; the rest are light. Taylor and colleagues (1958) showed the same in Vicia faba chromosomes using radioactive thymidine.

Machinery:

• DNA-dependent DNA polymerase: the main enzyme; uses a DNA template to join deoxynucleotides. It is fast (E. coli's 4.6 × 10⁶ bp are copied in about 18 minutes, about 2000 bp per second) and accurate (errors cause mutations).
• Deoxyribonucleoside triphosphates act as both substrates and energy source (their two terminal phosphates are high-energy, like ATP).
• Replication fork: the long DNA cannot be unwound along its whole length (too much energy), so replication happens in a small opening.
• DNA polymerase works only 5′→3′. So on the template with polarity 3′→5′ synthesis is continuous (leading strand), and on the template with polarity 5′→3′ it is discontinuous (lagging strand, short fragments). The fragments are joined by DNA ligase.
• Polymerases cannot start replication on their own; it begins at a definite origin of replication. This is why a DNA piece to be propagated in recombinant DNA work needs a vector, which supplies the origin.
• In eukaryotes, replication occurs in the S phase; if cell division fails after replication, polyploidy results.

Worked example

In the Meselson–Stahl experiment, what proportions of hybrid and light DNA would be found 80 minutes after transfer to the ¹⁴N medium? Show your working. (2 marks)

Try it first, then show the solutionSolution

1. 80 min ÷ 20 min = 4 generations. One heavy molecule gives 2⁴ = 16 molecules; the two original ¹⁵N strands stay in 2 of them (hybrid); 14 are light. (1)
2. Hybrid = 2/16 = 1/8 (12.5%); light = 14/16 = 7/8 (87.5%). Check by doubling: 1 hybrid pair → gen I 2 hybrid; gen II 2 hybrid + 2 light; gen III 2 + 6; gen IV 2 + 14. (1)

Answer: Hybrid 1/8 (12.5%), light 7/8 (87.5%), ratio 1 : 7.

Transcription

NCERT section 5.5 Transcription (5.5.1 Transcription Unit; 5.5.2 Transcription Unit and the Gene; 5.5.3 Types of RNA and the process of Transcription)

Transcription copies the information of one strand of a segment of DNA into RNA, following complementarity, except that A pairs with U.

Why only one strand? If both strands were copied, (1) one DNA segment would code for two different proteins, confusing the system; and (2) the two RNAs would be complementary and pair into double-stranded RNA, which cannot be translated.

Transcription unit = promoter + structural gene + terminator.

• Template strand: polarity 3′→5′; RNA polymerase reads it (it makes RNA only 5′→3′).
• Coding strand: polarity 5′→3′; same sequence as the RNA (T in place of U); it does not code for anything but all references use it.
• Promoter: towards the 5′ end (upstream) of the coding strand; binding site for RNA polymerase; it decides which strand is template. Terminator: towards the 3′ end (downstream); marks the end. Extra regulatory sequences may lie further up- or downstream.

Gene terms: a cistron is a DNA segment coding for one polypeptide. Structural genes are monocistronic (mostly eukaryotes) or polycistronic (mostly prokaryotes). Eukaryotic genes are split: exons (coding; appear in mature RNA) are interrupted by introns (intervening; removed). Regulatory sequences are loosely called regulatory genes.

Bacteria: three RNAs: mRNA (template), tRNA (brings amino acids, reads code), rRNA (structural and catalytic). One RNA polymerase makes all. It binds the promoter (initiation) with the σ (sigma) factor, extends RNA (elongation, its own catalytic ability), and stops at the terminator with the ρ (rho) factor (termination). mRNA needs no processing, and since there is no nucleus, transcription and translation can be coupled.

Eukaryotes: two extra complexities.
1. Three nuclear RNA polymerases: RNA pol I → rRNAs (28S, 18S, 5.8S); RNA pol II → hnRNA (precursor of mRNA); RNA pol III → tRNA, 5S rRNA and snRNAs.
2. The primary transcript (hnRNA) has exons and introns and is non-functional. Processing:

• Splicing: introns removed, exons joined in order.
• Capping: methyl guanosine triphosphate added at the 5′ end.
• Tailing: 200–300 adenylate residues added at the 3′ end, without a template.

The processed mRNA leaves the nucleus for translation. Split genes and splicing are thought to be ancient, reflecting the RNA world.

Worked example

The coding strand of a transcription unit reads 5′-ATGTTCCAGTGA-3′. Write (a) the template strand in the 3′→5′ direction and (b) the mRNA, with polarity. (2 marks)

Try it first, then show the solutionSolution

1. (a) Template is complementary and antiparallel: 3′-TACAAGGTCACT-5′. (1)
2. (b) mRNA has the coding-strand sequence with U for T: 5′-AUGUUCCAGUGA-3′. Check: mRNA is complementary to the template (A–U, T–A, G–C, C–G). (1)

Answer: (a) 3′-TACAAGGTCACT-5′ (b) 5′-AUGUUCCAGUGA-3′

The genetic code, mutations and tRNA

NCERT section 5.6 Genetic Code (5.6.1 Mutations and Genetic Code; 5.6.2 tRNA – the Adapter Molecule)

Deciphering the code: George Gamow (a physicist) argued that 4 bases must combine to code 20 amino acids; a triplet gives 4³ = 64 codons, more than enough. Har Gobind Khorana's chemical method made RNAs of defined sequence (homopolymers, copolymers); Marshall Nirenberg's cell-free system for protein synthesis decoded them; Severo Ochoa's enzyme (polynucleotide phosphorylase) made RNA without a template. A checkerboard of codons resulted.

Salient features:
1. Triplet: 61 codons code for amino acids; 3 are stop codons (UAA, UAG, UGA).
2. Degenerate: some amino acids have more than one codon.
3. Read contiguously, with no punctuation (comma-less).
4. Nearly universal: UUU codes phenylalanine from bacteria to humans; exceptions in mitochondria and some protozoans.
5. AUG has dual roles: codes methionine and is the initiator codon.
6. Unambiguous and specific: one codon codes only one amino acid.

Point mutation: in sickle-cell anaemia a single base change in the β-globin gene replaces glutamate with valine.

Frameshift mutations (NCERT's sentence 'RAM HAS RED CAP'):

• Inserting or deleting one or two bases shifts the reading frame from that point, changing every later codon.
• Inserting or deleting three (or a multiple of three) bases adds or removes whole codons; the frame after that point stays the same.

tRNA, the adapter: Crick proposed an adapter that reads the code and carries an amino acid. tRNA (earlier called sRNA, soluble RNA) has an anticodon loop complementary to the codon and an amino-acid acceptor end (3′ end). tRNAs are specific for each amino acid; there is a special initiator tRNA; there are no tRNAs for stop codons. Secondary structure is a clover-leaf; the real 3-D shape is an inverted L.

Worked example

An mRNA reads 5′-AUG UUU UUC UUC UUU UUU UUC-3′. Write the polypeptide. If you are given only Met-Phe-Phe, can you write a unique mRNA? Which two features of the code does this show? (3 marks)

Try it first, then show the solutionSolution

1. AUG = Met; UUU and UUC = Phe. Polypeptide: Met-Phe-Phe-Phe-Phe-Phe-Phe. (1)
2. Reverse: Phe has two codons (UUU, UUC), so Met-Phe-Phe could be AUG UUU UUU, AUG UUU UUC, AUG UUC UUU or AUG UUC UUC: no unique mRNA. (1)
3. Features: the code is degenerate (several codons per amino acid) yet unambiguous (each codon gives one amino acid); AUG also serves as the start codon. (1)

Answer: Met-(Phe)₆; the reverse is not unique because the code is degenerate (but unambiguous).

Translation

NCERT section 5.7 Translation

Translation is the polymerisation of amino acids into a polypeptide, in the order set by the mRNA. Amino acids are joined by peptide bonds, which need energy.

Steps:
1. Charging (aminoacylation) of tRNA: amino acids are activated with ATP and linked to their own (cognate) tRNA. Two charged tRNAs close together favour peptide bond formation.
2. Ribosome: made of structural RNAs and about 80 different proteins; exists as a small and a large subunit when inactive. Translation begins when the small subunit meets an mRNA. The large subunit has two sites where successive aminoacyl-tRNAs sit close enough to bond. The ribosome is also the catalyst: in bacteria, 23S rRNA is the ribozyme that forms the peptide bond.
3. Translational unit: the mRNA stretch from the start codon (AUG) to a stop codon; mRNA also has untranslated regions (UTRs) at the 5′ end (before the start) and the 3′ end (after the stop), needed for efficient translation.
4. Initiation: the ribosome binds the mRNA at AUG, which only the initiator tRNA recognises.
5. Elongation: charged tRNAs bind their codons by anticodon–codon pairing; the ribosome moves codon by codon; amino acids are added one at a time.
6. Termination: a release factor binds the stop codon; the finished polypeptide is released.

Two roles of the ribosome: it binds the mRNA and provides sites for charged tRNAs (a platform), and its rRNA catalyses peptide bonds (ribozyme).

Counting rule: number of amino acids = (nucleotides from AUG to the end of the stop codon ÷ 3) − 1.

Worked example

A translational unit, from the first base of AUG to the last base of the stop codon, has 903 nucleotides. How many amino acids will the polypeptide have? What happens at the stop codon? (2 marks)

Try it first, then show the solutionSolution

1. Codons = 903 ÷ 3 = 301; the last is a stop codon with no tRNA, so amino acids = 301 − 1 = 300. Check: 300 amino acid codons × 3 = 900 + 3 (stop) = 903. (1)
2. At the stop codon a release factor binds, translation ends and the polypeptide is released from the ribosome. (1)

Answer: 300 amino acids; a release factor terminates translation.

Regulation of gene expression: the lac operon

NCERT section 5.8 Regulation of Gene Expression (5.8.1 The Lac operon)

Levels of regulation in eukaryotes: (i) transcription (primary transcript), (ii) processing (splicing), (iii) transport of mRNA from nucleus to cytoplasm, (iv) translation.

Genes are expressed when needed: E. coli makes β-galactosidase (which splits lactose into galactose and glucose) only when lactose is present. So metabolic, physiological or environmental conditions regulate gene expression; development of an embryo is also coordinated gene regulation.

In prokaryotes, control is mainly at the rate of transcription initiation. Accessory proteins act positively (activators) or negatively (repressors). Repressors bind the operator, next to the promoter; each operon has its own operator and repressor.

Operon: a polycistronic structural gene regulated by a common promoter and regulatory genes (e.g. lac, trp, ara, his, val operons). Jacob and Monod were the first to explain one (lac).

Parts of the lac operon:

• i gene (regulatory; 'i' from inhibitor): makes the repressor, all the time (constitutively).
• p: promoter; o: operator.
• z gene → β-galactosidase (hydrolyses lactose to galactose and glucose).
• y gene → permease (lets β-galactosides into the cell).
• a gene → transacetylase.

Lactose absent (operon OFF): the repressor binds the operator and blocks RNA polymerase, so z, y, a are not transcribed.

Lactose present, no glucose (operon ON): a little lactose enters through the small amount of permease always present. Lactose (or allolactose), the inducer, binds and inactivates the repressor; RNA polymerase reaches the promoter and transcribes z, y, a.

Lactose used up: repressor is active again and the operon switches off. So this is regulation of enzyme synthesis by its substrate.

Notes: glucose and galactose cannot induce the lac operon. Control by the repressor is negative regulation; the lac operon also has positive regulation, which is beyond this level.

Worked example

An E. coli mutant has lost its i gene. Predict whether z, y and a are transcribed (a) when lactose is absent and (b) when lactose is present, and explain why the mutant wastes energy. (2 marks)

Try it first, then show the solutionSolution

1. Without the i gene no repressor is made, so nothing binds the operator and RNA polymerase can always transcribe z, y and a: the operon is on with or without lactose (constitutive expression). (1)
2. When lactose is absent, the cell still makes β-galactosidase, permease and transacetylase that it cannot use, wasting energy and materials; normal cells avoid this because the repressor switches the operon off. (1)

Answer: (a) Transcribed (b) transcribed: the operon is always on; enzymes are made even when there is no lactose, wasting energy.

Human Genome Project

NCERT section 5.9 Human Genome Project (5.9.1 Salient Features; 5.9.2 Applications and Future Challenges)

Launched in 1990, completed in 2003 (13 years), coordinated by the U.S. Department of Energy and the National Institutes of Health, with the Wellcome Trust (U.K.) as a major partner and contributions from Japan, France, Germany, China and others. It was a mega project:

• About 3 × 10⁹ bp; at the initial estimate of 3 US dollars per bp, the cost would be about 9 billion US dollars.
• Printed at 1000 letters per page and 1000 pages per book, one cell's sequence would fill about 3300 books.
• The data needed high-speed computers, giving rise to bioinformatics.

Goals: identify all the approximately 20,000–25,000 genes; sequence the 3 billion base pairs; store the data in databases; improve analysis tools; transfer technologies to other sectors such as industry; address ethical, legal and social issues (ELSI).

Methods:

• Expressed Sequence Tags (ESTs): identify all genes expressed as RNA.
• Sequence annotation: sequence the whole genome (coding and non-coding) blindly, then assign functions.
• DNA is broken into random fragments, cloned in hosts (bacteria, yeast) using BAC (bacterial artificial chromosome) and YAC (yeast artificial chromosome) vectors, then sequenced by automated sequencers based on Frederick Sanger's method. Overlapping fragments were aligned by computer programs. Chromosome 1 was the last of the 24 to be completed (May 2006). Genetic and physical maps used restriction-site polymorphism and microsatellites.

Other genomes sequenced: bacteria, yeast, Caenorhabditis elegans, Drosophila, and plants (rice and Arabidopsis).

Salient features:
1. 3164.7 million bp.
2. Average gene 3000 bases; largest known, dystrophin, 2.4 million bases.
3. About 30,000 genes (far fewer than earlier estimates of 80,000–1,40,000); 99.9% of bases identical in all humans.
4. Functions unknown for over 50% of discovered genes.
5. Less than 2% of the genome codes for proteins.
6. Repeated sequences form a very large part; they have no direct coding role but tell us about chromosome structure, dynamics and evolution.
7. Chromosome 1 has the most genes (2968); Y the fewest (231).
8. About 1.4 million SNPs (single nucleotide polymorphisms, 'snips'): useful for locating disease-related sequences and tracing human history.

Applications: new ways to diagnose, treat and prevent disorders; studying all genes or transcripts of a tissue at once; solving problems in health care, agriculture, energy and environmental remediation using genomes of other organisms.

Worked example

99.9% of base sequences are the same in all humans. Taking the human genome as 3 × 10⁹ bp, at about how many positions would two people differ? Why is DNA fingerprinting used rather than full sequencing to compare people? (2 marks)

Try it first, then show the solutionSolution

1. 0.1% of 3 × 10⁹ = 3 × 10⁹ × 0.001 = 3 × 10⁶ bp. Check: 3 × 10⁹ ÷ 1000 = 3 × 10⁶. (1)
2. Sequencing whole genomes every time is slow and expensive; fingerprinting quickly compares only highly polymorphic repetitive regions. (1)

Answer: About 3 × 10⁶ positions; fingerprinting is a fast, cheap comparison of polymorphic regions.

DNA fingerprinting

NCERT section 5.10 DNA Fingerprinting

Principle: compare repetitive DNA, where a short stretch is repeated many times. Such sequences show high polymorphism between individuals.

Satellite DNA: in density gradient centrifugation, bulk DNA forms a major peak and repetitive DNA forms small peaks, the satellite DNA. It is classified (by A:T or G:C richness, length of segment and number of repeats) into micro-satellites, mini-satellites, etc. It usually codes for no protein but forms a large part of the genome.

DNA polymorphism: inheritable variation in DNA, arising from mutation. A variant (allele) at a locus present in the population at a frequency above 0.01 is a polymorphism. Polymorphisms are more likely in non-coding DNA, because mutations there do not harm reproduction and so accumulate over generations.

Why it works for identification: DNA from every tissue of a person (blood, hair follicle, skin, bone, saliva, sperm) shows the same polymorphism, and polymorphisms are inherited from parents.

Technique (Alec Jeffreys): used a satellite DNA probe called VNTR (Variable Number of Tandem Repeats), a mini-satellite. Southern blot hybridisation steps:
1. Isolate DNA.
2. Digest with restriction endonucleases.
3. Separate fragments by gel electrophoresis.
4. Blot (transfer) to a synthetic membrane (nitrocellulose or nylon).
5. Hybridise with a labelled VNTR probe.
6. Detect hybridised fragments by autoradiography.

VNTR copy number varies between chromosomes and people, so VNTR size ranges from 0.1 to 20 kb, giving a band pattern unique to each person except identical (monozygotic) twins. PCR has made it so sensitive that DNA from a single cell is enough.

Uses: forensic identification, paternity testing, and studying population and genetic diversity (and evolutionary biology).

Worked example

In a paternity dispute, a child's DNA fingerprint has 8 bands. 4 match the mother. Of the other 4, all match man A and only 1 matches man B. Who is likely to be the father? Explain the reasoning. (2 marks)

Try it first, then show the solutionSolution

1. Each band (VNTR allele) of a child comes from either the mother or the father, since polymorphisms are inherited. (1)
2. The 4 non-maternal bands must come from the father; all 4 match man A, so A is likely the father; B shares only 1 (by chance), so B is excluded. (1)

Answer: Man A, because he accounts for all the bands not inherited from the mother.

Molecular Basis of Inheritance in real life

Each concept, met in an everyday situation. Read the situation, answer the question in your own words, then compare with the model answer. The one-line hook is what to remember.

Topics in this chapter

• DNA Replication
• Transcription
• Genetic Code and Translation

Molecular Basis of Inheritance: key facts

Chargaff's rule

$A=T;G=C;A+G=C+T;\%A=\frac{100-2(\%C)}{2}$

For double-stranded DNA only.

Helix dimensions

$\text{pitch}=3.4\text{nm};10\text{bp per turn};0.34\text{nm per bp}$

Length of DNA = number of bp × 0.34 × 10⁻⁹ m.

Hydrogen bonds

$A=T:2;G\equiv C:3;\text{total}=2(\text{AT pairs})+3(\text{GC pairs})$

More G–C pairs make DNA harder to separate.

Meselson–Stahl after n generations

$\text{hybrid}=\frac{2}{2^n};\text{light}=\frac{2^n-2}{2^n}$

E. coli divides every 20 minutes.

Codons and amino acids

$4^3=64\text{codons}=61\text{sense}+3\text{stop};\text{amino acids}=\frac{\text{nucleotides (AUG to stop)}}{3}-1$

Stop codons: UAA, UAG, UGA. Start: AUG (Met).

Central dogma

$\text{DNA}\stackrel{\text{replication}}{\to }\text{DNA};\text{DNA}\stackrel{\text{transcription}}{\to }\text{RNA}\stackrel{\text{translation}}{\to }\text{protein}$

Reverse transcription (RNA → DNA) occurs in some viruses.

Eukaryotic RNA polymerases

$\text{Pol I}\to 28S,18S,5.8S\text{rRNA};\text{Pol II}\to \text{hnRNA};\text{Pol III}\to \text{tRNA},5S\text{rRNA},\text{snRNA}$

Bacteria have a single RNA polymerase (with σ and ρ factors).

Lac operon genes

$i\to \text{repressor};z\to \beta \text{-galactosidase};y\to \text{permease};a\to \text{transacetylase}$

Inducer: lactose or allolactose (not glucose or galactose).

Molecular Basis of Inheritance: diagrams to practise

Practise each one on paper: draw it from the checklist, then tick off every label.

How the board asks Molecular Basis of Inheritance

The question patterns to expect, and the method that works for each.

Tricks to remember Molecular Basis of Inheritance

Short memory aids. Say each one aloud once, then use it on a question the same day.

• Pure As Gold: Purines are A and G

  PUre As Gold = PUrines Adenine, Guanine. CUT the PYe = C, U, T are PYrimidines.

  Use it when: Purine versus pyrimidine questions.

• AT 2, GC 3

  A–T has 2 hydrogen bonds (both letters straight-lined); G–C has 3 (curvy letters hold tighter).

  Use it when: Counting hydrogen bonds.

• Hershey–Chase: P for Polynucleotide, S for Shell

  ³²P labels DNA (the polynucleotide inside); ³⁵S labels the protein shell (coat).

  Use it when: Which isotope labelled which molecule.

• Pol I, II, III: 'rRNA, mRNA, tRNA = 1-2-3 Ribosomes Make Tools'

  I → rRNA (28S, 18S, 5.8S); II → hnRNA (mRNA); III → tRNA, 5S rRNA, snRNA.

  Use it when: Eukaryotic RNA polymerases.

• Cap the head, tail the end

  Capping at 5′ (methyl guanosine triphosphate); tailing at 3′ (200–300 A's); splicing removes introns.

  Use it when: Processing of hnRNA.

• Lac operon 'i z y a': 'I Zip Your Apron'

  i = inhibitor (repressor); z = β-galactosidase; y = permease; a = transacetylase.

  Use it when: Genes and products of the lac operon.

• Stop codons: 'U Are Away, U Are Gone, U Go Away'

  UAA, UAG, UGA.

  Use it when: Genetic code questions.

Molecular Basis of Inheritance: board-level solved questions

Ordered from board level to HOTS. Cover the solution, write your own, then compare step by step.

Case-study practice (Section D style)

Read the situation, then answer every part. In the board paper this is a 4-mark question split into 1 + 2 + 1 marks.

Where the marks are

How a typical answer is split into step marks. Show every step: a right final answer alone rarely earns full marks.

Where students lose marks in Molecular Basis of Inheritance

The slips examiners see most often, and what to write instead.

• ✗Mistake: Writing sequences without polarity or in the wrong direction

  Wrong working

  Complementary strand of 5′-ATGC-3′ is 5′-TACG-3′.

  Right working

  Complementary strand is 3′-TACG-5′, written 5′→3′ as 5′-GCAT-3′.

  ✓Do this: Always label 5′ and 3′; when asked for 5′→3′, reverse the complementary sequence.

  Usually costs: 1 mark.

• ✗Mistake: Saying the template strand has the same sequence as mRNA

  ✓Do this: The coding strand has the mRNA sequence (with T for U); the template strand is complementary to mRNA.

  Usually costs: 1 mark.

• ✗Mistake: Calling ¹⁵N radioactive

  ✓Do this: ¹⁵N is a heavy, non-radioactive isotope; it is separated from ¹⁴N by density in a CsCl gradient. Radioactive isotopes were used by Hershey–Chase (³²P, ³⁵S) and Taylor (thymidine).

  Usually costs: ½ to 1 mark.

• ✗Mistake: Saying lactose binds the operator

  ✓Do this: The repressor binds the operator; lactose (inducer) binds the repressor and inactivates it.

  Usually costs: 1 mark.

• ✗Mistake: Counting the stop codon as an amino acid

  ✓Do this: Stop codons have no tRNA; amino acids = codons from AUG to stop minus one.

  Usually costs: 1 mark in numerical questions.

• ✗Mistake: Writing that lagging strand synthesis is 3′→5′

  ✓Do this: Both strands are made 5′→3′; the lagging strand is made discontinuously in short fragments joined by DNA ligase.

  Usually costs: 1 mark.

NCERT exercise guide: Molecular Basis of Inheritance

What each exercise in the NCERT chapter tests, how to approach it, and a question of the same kind solved step by step. Questions here are written in our own words.

Molecular Basis of Inheritance: previous year questions

• Asked in 8 of the 8 board papers we hold (2023 to 2026), on average 8 marks per paper; 27 questions in all, counting OR alternatives. In 4 of 4 sample papers.

Counted from the real papers on our previous year papers page, not estimated. Average marks are over every board paper we hold, with an OR pair counted once. Chapter tags are ours.

Solved board questions from Molecular Basis of Inheritance

Real questions from recent papers. Try each one first, then open our step-by-step solution with marks per step.

Every Molecular Basis of Inheritance question, by year

Practice questions on Molecular Basis of Inheritance

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

1. Q1 · 1 mark · MCQBoard level

   Which of these pairs is correct?
   (a) Adenine – pyrimidine (b) Cytidine – nucleoside (c) Uracil – nucleoside (d) Guanosine – nitrogenous base

   Hint

   Base + sugar = nucleoside.

   Answer

   (b) Cytidine – nucleoside

2. Q2 · 1 mark · Assertion-reasonBoard level

   Assertion (A): Replication is continuous on one strand and discontinuous on the other.
   Reason (R): DNA polymerase can polymerise only in the 5′→3′ direction.
   (a) Both A and R are true and R is the correct explanation of A.
   (b) Both A and R are true but R is not the correct explanation of A.
   (c) A is true but R is false.
   (d) A is false but R is true.

   Hint

   Antiparallel strands, one-way enzyme.

   Answer

   (a)

3. Q3 · 1 mark · MCQHarder board level

   In eukaryotes, the 5S rRNA is transcribed by:
   (a) RNA polymerase I (b) RNA polymerase II (c) RNA polymerase III (d) all three

   Hint

   Not with the other rRNAs.

   Answer

   (c) RNA polymerase III

4. Q4 · 2 marks · Very short answerBoard level

   Write the sequence of the complementary strand of 5′-AAGCTTCGAGTC-3′ in the 5′→3′ direction. (2 marks)

   Hint

   Complement, then reverse.

   Answer

   Complement (aligned): 3′-TTCGAAGCTCAG-5′ (1). Written 5′→3′: 5′-GACTCGAAGCTT-3′ (1).

5. Q5 · 2 marks · Very short answerHarder board level

   A DNA molecule has 1000 base pairs, of which 30% are adenine (as a percentage of all bases). How many hydrogen bonds hold the two strands together? (2 marks)

   Hint

   First find the number of A–T and G–C pairs.

   Answer

   2000 bases; A = T = 600 bases, so 600 A–T pairs; G = C = 400, so 400 G–C pairs (1). H-bonds = 600 × 2 + 400 × 3 = 1200 + 1200 = 2400. Check: pairs 600 + 400 = 1000 (1).

6. Q6 · 2 marks · Very short answerBoard level

   List four features of the genetic code. (2 marks)

   Hint

   Triplet, degenerate...

   Answer

   Triplet (61 sense + 3 stop codons); degenerate; read contiguously without punctuation; nearly universal; AUG codes Met and is the start codon (any four, ½ each).

7. Q7 · 3 marks · Short answerHarder board level

   Explain how Avery, MacLeod and McCarty identified the transforming principle. Why were some biologists still not convinced? (3 marks)

   Hint

   Enzymes that digest each molecule.

   Answer

   They purified proteins, DNA and RNA from heat-killed S cells and found that DNA alone transformed R cells (1). Proteases and RNases did not affect transformation, but DNase stopped it, so the agent was DNA (1). Protein was then widely thought to be the genetic material, and conclusive proof came only from Hershey and Chase (1952) (1).

8. Q8 · 3 marks · Short answerHarder board level

   Describe the processing of hnRNA into mRNA in eukaryotes, and suggest why split genes are thought to be ancient. (3 marks)

   Hint

   Three steps.

   Answer

   Splicing removes introns and joins exons in order (1). Capping adds methyl guanosine triphosphate at the 5′ end; tailing adds 200–300 adenylate residues at the 3′ end without a template (1). Introns are thought to be a reminder of antiquity, and splicing reflects the dominance of the RNA world, where RNA acted as catalyst (1).

9. Q9 · 5 marks · Long answerHarder board level

   (a) Describe the methods used in the Human Genome Project. (3)
   (b) List four salient features of the human genome. (2) (5 marks)

   Hint

   ESTs, annotation, BAC/YAC, Sanger.

   Answer

   (a) Two approaches: ESTs (identifying all genes expressed as RNA) and sequence annotation (sequencing the whole genome, coding and non-coding, then assigning functions) (1). DNA was broken into random fragments and cloned in bacteria and yeast using BAC and YAC vectors (1). Fragments were sequenced on automated sequencers based on Sanger's method and overlapping sequences aligned by computer programs, then annotated and assigned to chromosomes (1). (b) 3164.7 million bp; about 30,000 genes; 99.9% identical in all humans; less than 2% codes for proteins; chromosome 1 has most genes (2968), Y the fewest (231); about 1.4 million SNPs (any four, ½ each) (2).

10. Q10 · 3 marks · Short answerHOTS / competency

    Suppose E. coli with heavy (¹⁵N/¹⁵N) DNA is grown in ¹⁴N medium for 60 minutes and then returned to ¹⁵N medium for one more generation. What DNA bands will appear and in what proportions? (3 marks)

    Hint

    Track strands, not molecules.

    Answer

    After 60 min in ¹⁴N (3 generations): 8 molecules = 2 hybrid (¹⁵N/¹⁴N) + 6 light (¹⁴N/¹⁴N); strands: 2 heavy + 14 light (1). One generation in ¹⁵N: every old strand pairs with a new heavy strand: 2 heavy old strands → 2 heavy (¹⁵N/¹⁵N) molecules; 14 light strands → 14 hybrid molecules (1). So 16 molecules: 2 heavy : 14 hybrid = 1 heavy : 7 hybrid, and no light DNA (1).

11. Q11 · 2 marks · Very short answerHOTS / competency

    A gene has three exons of 150, 210 and 93 bases, separated by two introns. The first exon begins with AUG and the last exon ends with a stop codon; there are no UTRs. How many amino acids will the protein have? (2 marks)

    Hint

    Introns are spliced out.

    Answer

    Mature mRNA coding region = 150 + 210 + 93 = 453 bases = 151 codons (1). The last is a stop codon, so 150 amino acids. Check: 150 × 3 + 3 = 453 (1).

12. Q12 · 2 marks · Very short answerHarder board level

    Explain why the lac operon is not switched on by glucose, and why a low level of permease must always be present. (2 marks)

    Hint

    Inducer specificity; entry of lactose.

    Answer

    Only lactose (or allolactose) binds and inactivates the lac repressor; glucose and galactose cannot, so they do not induce the operon (1). Lactose must first enter the cell to act as inducer, and its entry needs permease, so a very low basal level of the operon's expression is always needed (1).

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1. 1.Who first identified DNA and what did he call it?Show answer

   Friedrich Meischer (1869); he called it 'nuclein'.

2. 2.What is the pitch of the DNA helix and the number of bp per turn?Show answer

   3.4 nm; about 10 bp per turn.

3. 3.Which amino acids make histones positively charged?Show answer

   Lysine and arginine.

4. 4.How many bp of DNA does a typical nucleosome contain?Show answer

   About 200 bp.

5. 5.Name two viruses with RNA as genetic material.Show answer

   Tobacco mosaic virus and QB bacteriophage.

6. 6.What did the Avery–MacLeod–McCarty experiment show about DNase?Show answer

   DNase stopped transformation, while proteases and RNases did not, so DNA was the transforming principle.

7. 7.What is the role of σ and ρ factors?Show answer

   σ (sigma) helps RNA polymerase initiate transcription; ρ (rho) helps it terminate.

8. 8.Which rRNA acts as a ribozyme in bacteria?Show answer

   23S rRNA, which catalyses peptide bond formation.

9. 9.What are UTRs?Show answer

   Untranslated regions of mRNA at the 5′ end (before AUG) and 3′ end (after the stop codon), needed for efficient translation.

10. 10.Which chromosome has the most genes and which the fewest?Show answer

    Chromosome 1 (2968 genes) and Y (231 genes).

11. 11.What are BAC and YAC?Show answer

    Bacterial artificial chromosomes and yeast artificial chromosomes, vectors used to clone DNA fragments in HGP.

12. 12.Who developed DNA fingerprinting and which probe did he use?Show answer

    Alec Jeffreys; a VNTR (mini-satellite) probe.

Quick revision checklist

The night before the exam: can you do each of these without looking?

• ☐Can you list five features of the Watson–Crick double helix?
• ☐Can you calculate base percentages using Chargaff's rule and DNA length from bp?
• ☐Can you describe Hershey–Chase and Meselson–Stahl experiments with results?
• ☐Can you draw a replication fork showing leading and lagging strands?
• ☐Can you write the mRNA from a given coding or template strand with polarity?
• ☐Can you explain capping, tailing and splicing and the three eukaryotic RNA polymerases?
• ☐Can you list the features of the genetic code and explain frameshift mutations?
• ☐Can you draw the lac operon in the off and on states and list five salient features of the human genome?

Molecular Basis of Inheritance: frequently asked questions