Home/Biology/Chapter 5 — Molecular Basis of Inheritance
Class 12 Biology

Molecular Basis of Inheritance

DNA structure, Griffith to Hershey–Chase, replication, transcription, genetic code, translation, lac operon, Human Genome Project and DNA fingerprinting — with SVG diagrams, a codon table, a mini transcription–translation tool, MCQs, A-R and PYQs.

10NCERT Sections
12MCQs
6Assertion–Reason
5PYQs
Section 01

The DNA

5.1Structure of Polynucleotide Chain

  • DNA length is expressed in nucleotides or base pairs (bp): φX174 = 5386 nucleotides; bacteriophage λ = 48,502 bp; E. coli = 4.6 × 10⁶ bp; haploid human = 3.3 × 10⁹ bp.
  • Nucleotide = nitrogenous base + pentose sugar + phosphate. Sugar: deoxyribose (DNA), ribose (RNA).
  • Purines: adenine, guanine. Pyrimidines: cytosine, thymine (DNA), uracil (RNA).
  • Base + sugar (1′C, N-glycosidic bond) = nucleoside; + phosphate at 5′C (phosphoester bond) = nucleotide. Nucleotides join by 3′–5′ phosphodiester bonds.
  • Free phosphate at one end = 5′ end; free 3′-OH at other = 3′ end. Backbone = sugar–phosphate; bases project from it.
  • RNA differs: extra 2′-OH on ribose and uracil (5-methyl uracil = thymine) in place of thymine.

5.1.1Double Helix Model (Watson & Crick, 1953)

DNA was first isolated as 'nuclein' by Friedrich Miescher (1869). Watson and Crick used X-ray diffraction data of Maurice Wilkins and Rosalind Franklin and Chargaff's rule (A : T and G : C ratios are 1 for double-stranded DNA).

V1 · DNA double helix — base pairing and dimensions Base pairing & antiparallel strands 5′ 3′ 3′ 5′ A T G C T A C G A T 2 H-bonds 3 H-bonds Pitch 3.4 nm ≈10 bp / turn Right-handed · 0.34 nm between base pairs
Purine always pairs with pyrimidine → uniform distance between strands.
Salient features

(i) Two polynucleotide chains, sugar-phosphate backbone outside, bases inside. (ii) Antiparallel polarity (5′→3′ and 3′→5′). (iii) A = T (2 H-bonds), G ≡ C (3 H-bonds). (iv) Right-handed; pitch 3.4 nm, ~10 bp per turn, 0.34 nm per bp. (v) Stacking of base pairs plus H-bonds stabilises the helix.

Central dogma (Crick): DNA → RNA → Protein. In some viruses the flow is reverse (RNA → DNA) — reverse transcription (Temin & Baltimore).

5.1.2Packaging of DNA Helix

  • Mammalian cell DNA: 6.6 × 10⁹ bp × 0.34 × 10⁻⁹ m ≈ 2.2 m, but a nucleus is ~10⁻⁶ m.
  • Prokaryotes: negatively charged DNA held by positively charged proteins in nucleoid, organised in large loops.
  • Eukaryotes: histones (rich in lysine and arginine, positively charged) form a histone octamer; DNA wraps around it → nucleosome (~200 bp). Nucleosomes = repeating unit of chromatin ('beads-on-string' in EM).
  • Chromatin → chromatin fibres → coiled/condensed at metaphase → chromosomes (needs NHC proteins).
  • Euchromatin: loosely packed, light-stained, transcriptionally active. Heterochromatin: densely packed, dark-stained, inactive.
V2 · DNA packaging DNA double helix Nucleosomes (200 bp each) histone octamer core Chromatin fibre Chromatin fibres + NHC proteins Chromosome metaphase
Section 03

Replication

5.4Semiconservative Replication

Watson and Crick proposed that the two strands separate and each acts as a template for a new complementary strand; each daughter DNA has one parental and one new strand — semiconservative replication.

Meselson and Stahl (1958)

Grew E. coli in ¹⁵NH₄Cl (heavy N) for many generations, then shifted to ¹⁴NH₄Cl; DNA separated by CsCl density gradient centrifugation. After 20 min (1 generation): hybrid (intermediate) DNA. After 40 min (2nd generation): equal hybrid and light DNA. (Taylor et al., 1958, confirmed it in Vicia faba using radioactive thymidine.)

V5 · Meselson–Stahl experiment ¹⁵N medium Heavy DNA After 20 min All hybrid (¹⁵N/¹⁴N) After 40 min Hybrid + Light (1 : 1) Shifted to ¹⁴N medium → Proportion of DNA molecules Generation 1 (20 min): 2 hybrid Generation 2 (40 min): 2 hybrid : 2 light Generation 3 (60 min): 2 hybrid : 6 light Generation 4 (80 min): 2 hybrid : 14 light Hybrid always = 2 molecules; light = 2ⁿ − 2

5.4.2Machinery and Enzymes

  • DNA-dependent DNA polymerase adds nucleotides only in the 5′→3′ direction. E. coli replicates 4.6 × 10⁶ bp in ~18 min (~2000 bp/s) and with high accuracy.
  • Deoxynucleoside triphosphates are both substrates and energy source.
  • Replication occurs at a replication fork (small opening of the helix). One strand (template 3′→5′) replicates continuously — leading strand; the other (template 5′→3′) discontinuously in short Okazaki fragments — lagging strand, joined by DNA ligase.
  • Replication starts at a specific origin of replication (ori); a vector supplies ori in recombinant DNA work.
  • In eukaryotes replication occurs in S-phase; failure of cell division after replication gives polyploidy.
V6 · Replication fork — leading and lagging strands Parental DNA ← Replication fork Leading strand (continuous, 5′→3′) Lagging strand — Okazaki fragments (discontinuous), joined by DNA ligase Fork movement → DNA polymerase works 5′→3′ only
Section 04

Transcription

5.5Transcription Unit

Definition

Copying of genetic information from one strand of DNA into RNA (only a segment of DNA and only one strand). A pairs with U.

Why not both strands? They would code for two different RNAs/proteins from one segment, and the two RNAs would be complementary, forming double-stranded RNA that cannot be translated.

  • Transcription unit = promoter + structural gene + terminator.
  • Template strand: 3′→5′ (read by RNA polymerase). Coding strand: 5′→3′ — same sequence as RNA (T in place of U).
  • Promoter — 5′ (upstream) of the gene; binding site for RNA polymerase; defines the template strand. Terminator — 3′ (downstream); ends transcription.
V7 · Transcription unit and eukaryotic RNA processing Transcription unit (bacterial) 3′ 5′ 5′ 3′ Promoter Structural gene Terminator Template strand 3′→5′ Coding strand 5′→3′ upstream (5′ of coding strand) downstream RNA polymerase moves 5′→3′ (initiation σ · elongation · termination ρ) Eukaryotic gene — processing of hnRNA Exon 1 Intron Exon 2 Intron Exon 3 hnRNA (primary transcript) — RNA polymerase II Splicing (introns removed) + capping + tailing 5′ cap Poly-A tail (200–300 A) Mature mRNA (exons joined)

5.5.2Gene, Cistron & Types of RNA

  • Gene = functional unit of inheritance. Cistron = DNA segment coding for a polypeptide. Structural gene is monocistronic (mostly eukaryotes) or polycistronic (mostly bacteria).
  • Eukaryotic genes are split: exons (appear in mature RNA) and introns (intervening, absent from mature RNA).
  • Bacteria have 3 RNAs: mRNA (template), tRNA (brings amino acids, reads code), rRNA (structural + catalytic). A single RNA polymerase makes all three.
  • RNA polymerase alone does elongation; σ (sigma) factor helps initiation and ρ (rho) factor helps termination.
  • In bacteria, transcription and translation are coupled (no nucleus, no processing).
Eukaryotic RNA polymerase Transcribes
RNA polymerase I rRNAs (28S, 18S, 5.8S)
RNA polymerase II Precursor of mRNA — hnRNA
RNA polymerase III tRNA, 5S rRNA, snRNAs
hnRNA processing: Splicing (introns out, exons joined) → capping (methyl guanosine triphosphate at 5′ end) → tailing (200–300 adenylate residues at 3′ end, template-independent) → mature mRNA leaves nucleus.
Section 05

Genetic Code

5.6Deciphering the Code

  • George Gamow: 4 bases for 20 amino acids → a triplet code (4³ = 64 codons).
  • H. G. Khorana — chemical synthesis of RNA with defined sequences; Marshall Nirenberg — cell-free protein synthesis; Severo Ochoa — polynucleotide phosphorylase enzyme.
Salient features

(i) Triplet: 61 codons code amino acids, 3 are stop codons (UAA, UAG, UGA). (ii) Degenerate: some amino acids have more than one codon. (iii) Read contiguously, no punctuation. (iv) Nearly universal (UUU = Phe from bacteria to humans; exceptions in mitochondria and some protozoa). (v) AUG codes Met and is the initiator. (vi) Unambiguous — one codon specifies only one amino acid.

V8 · The codons (mRNA, 5′→3′) — checkerboard
Try it · template strand → mRNA → protein

Enter the DNA template strand (3′→5′). mRNA is its complement (A→U, T→A, G→C, C→G), translated from the first AUG to a stop codon.

5.6.1Mutations and Genetic Code

  • Point mutation: change of a single base pair. Classic example: sickle-cell anaemia — β-globin glutamate → valine.
  • Frameshift mutation: insertion or deletion of 1 or 2 bases alters the reading frame from that point (e.g. RAM HAS BRE DCA P).
  • Insertion/deletion of three bases (or multiples) adds/removes whole codon(s); the reading frame stays unaltered (RAM HAS BIG RED CAP).

5.6.2tRNA — the Adapter Molecule

  • Crick postulated an adapter. tRNA (once called sRNA) has an anticodon loop (bases complementary to the codon) and an amino acid acceptor end.
  • tRNAs are specific for each amino acid; there is an initiator tRNA but no tRNA for stop codons.
  • Secondary structure: clover-leaf; actual 3-D shape: inverted L.
Section 06

Translation

5.7Protein Synthesis

  • Polymerisation of amino acids into a polypeptide by peptide bonds, in the order set by mRNA.
  • Charging (aminoacylation) of tRNA: amino acids are activated with ATP and linked to their cognate tRNA.
  • Ribosome = structural RNAs + ~80 proteins; large and small subunits. Large subunit has two sites for successive amino acids. 23S rRNA in bacteria is a ribozyme catalysing peptide bond formation.
  • Initiation: small subunit binds mRNA; initiator tRNA recognises AUG. Elongation: charged tRNAs bind codons through anticodons; ribosome moves codon to codon. Termination: a release factor binds the stop codon and releases the polypeptide.
  • UTRs (untranslated regions) lie at 5′ end (before AUG) and 3′ end (after stop) of the translational unit and aid efficient translation.
V9 · Translation on the ribosome 5′ 3′ AUG GCU UUC UAA 5′ UTR 3′ UTR Large subunit Small subunit P site A site Met Ala growing polypeptide UAC CGA mRNA is read 5′→3′, one codon at a time ① Initiation: AUG + initiator tRNA ② Elongation: peptide bond, ribosome moves ③ Termination: release factor at stop codon Charged tRNA anticodon pairs with mRNA codon · ribosome (23S rRNA) forms peptide bonds
Section 07

Regulation of Gene Expression — Lac Operon

5.8Levels of Regulation

In eukaryotes: transcriptional level, processing level (splicing), transport of mRNA to cytoplasm, translational level. In prokaryotes the main control is at transcription initiation, via activators and repressors binding operators adjacent to promoters.

5.8.1The Lac Operon (Jacob & Monod)

Gene Codes for
i (regulatory) Repressor (made constitutively). i from 'inhibitor'.
z β-galactosidase — hydrolyses lactose → galactose + glucose
y Permease — increases permeability to β-galactosides
a Transacetylase
  • No lactose: repressor binds the operator → RNA polymerase blocked → operon OFF.
  • Lactose (inducer; also allolactose) binds the repressor and inactivates it → RNA polymerase transcribes z, y, a → operon ON.
  • This is negative regulation (by repressor). Glucose or galactose cannot act as inducers. A very low basal expression must always exist so lactose can enter the cell.
V10 · lac operon — OFF and ON Lactose ABSENT — operon OFF i gene Promoter Operator z (β-gal) y (permease) a (transacetylase) Repressor bound RNA pol ✗ No transcription → no enzymes Lactose PRESENT — operon ON i gene Promoter Operator z y a Inactive Inducer (lactose) RNA pol ✓ polycistronic mRNA → β-gal, permease, transacetylase
Section 08

Human Genome Project & DNA Fingerprinting

5.9Human Genome Project (HGP)

  • Launched 1990, a 13-year project, completed 2003; coordinated by the US Department of Energy and NIH, with Wellcome Trust (UK) and Japan, France, Germany, China. Associated with the rise of bioinformatics.
  • Scale: ~3 × 10⁹ bp at US $3/bp ≈ $9 billion; printed (1000 letters/page, 1000 pages/book) it would need 3300 books.
  • Goals: identify all ~20,000–25,000 genes; sequence the 3 billion bp; store in databases; improve data-analysis tools; transfer technology to industry; address ethical, legal and social issues (ELSI).
  • Methods: (1) ESTs — identify genes expressed as RNA; (2) Sequence annotation — sequence the whole genome (coding + non-coding) and assign functions later. DNA was cut into fragments, cloned in bacteria/yeast using BAC and YAC vectors, sequenced by automated sequencers based on the Frederick Sanger method, aligned by computer programs.
  • Chromosome 1 was the last sequenced (May 2006). Model organisms also sequenced: bacteria, yeast, C. elegans, Drosophila, rice, Arabidopsis.
Salient features of human genome Value
Total size 3164.7 million bp
Average gene ~3000 bases; largest = dystrophin (2.4 million bases)
Number of genes ~30,000 (earlier estimates 80,000–1,40,000)
Same in all humans 99.9% of bases
Coding portion Less than 2%; functions unknown for over 50% of genes
Repeated sequences Very large portion; no direct coding function; inform chromosome structure and evolution
Chromosomes Chr 1 has most genes (2968); Y has fewest (231)
SNPs ~1.4 million single-nucleotide polymorphisms ('snips')

5.10DNA Fingerprinting

Quick method to compare DNA of two individuals (since 99.9% of sequence is common). It uses repetitive DNA (satellite DNA — separate as small peaks in density-gradient centrifugation; micro- and mini-satellites), which does not code for protein but shows high polymorphism.

  • Polymorphism arises by mutation; an inheritable variant at a locus with frequency > 0.01 in the population = DNA polymorphism. It is higher in non-coding DNA.
  • Technique (developed by Alec Jeffreys, uses VNTR probes): isolate DNA → cut with restriction endonucleases → separate fragments by gel electrophoresis → transfer to nylon membrane (Southern blotting) → hybridise with labelled VNTR probe → autoradiography shows band pattern.
  • Uses: forensics (same pattern from blood, hair follicle, skin, saliva, semen), paternity testing (polymorphism is inherited), genetic diversity studies.
Practice

📝 Solved Examples with Step Marking

📋 2 Marks

Q1. The length of E. coli DNA is 1.36 mm. Calculate the number of base pairs.

Formula 1 Mark

No. of bp = total length ÷ distance between consecutive bp (0.34 nm).

Calculation 1 Mark

1.36 mm = 1.36 × 10⁻³ m; bp = (1.36 × 10⁻³) ÷ (0.34 × 10⁻⁹) = 4 × 10⁶ bp.

4 × 10⁶ base pairs
📋 2 Marks

Q2. Coding strand is 5′-TACGTACG-3′. Write the template strand and mRNA.

Template strand 1 Mark

Complementary and antiparallel: 3′-ATGCATGC-5′.

mRNA 1 Mark

Same as coding strand with U for T: 5′-UACGUACG-3′.

📋 3 Marks

Q3. In Meselson–Stahl experiment, what proportions of hybrid and light DNA are present after 80 minutes (E. coli divides every 20 min)?

Generations 1 Mark

80 ÷ 20 = 4 generations → 2⁴ = 16 DNA molecules.

Hybrid 1 Mark

Each of the 2 original ¹⁵N strands stays in one molecule → 2 hybrid.

Light 1 Mark

16 − 2 = 14 light.

Hybrid : Light = 2 : 14 = 1 : 7
Test Yourself

✍️ MCQ Practice

Score: 0 / 12
New Pattern

🎯 Assertion–Reason

(A) Both true, R explains A. (B) Both true, R does NOT explain A. (C) A true, R false. (D) A false, R true.

Score: 0 / 6
Exam Focus

📖 Important Board Questions

NCERT

Q. Describe the Hershey–Chase experiment. What did it prove?

Answer 3 Marks

Phages were grown on ³²P (labels DNA) or ³⁵S (labels protein). After infecting E. coli, blending and centrifugation, only ³²P-labelled DNA was found inside bacteria; ³⁵S protein stayed outside. So DNA is the genetic material.

NCERT

Q. Why is DNA a better genetic material than RNA?

Answer 2 Marks

RNA has a reactive 2′-OH so it is labile and degrades easily; DNA lacks it and has thymine, is double-stranded with repair, hence chemically less reactive and structurally more stable.

NCERT

Q. Describe the structure of the lac operon and how it is regulated by lactose.

Answer 3 Marks

Regulatory gene i (repressor) + promoter, operator + structural genes z, y, a. Without lactose, repressor binds operator and blocks RNA polymerase. Lactose (inducer) binds and inactivates the repressor, so transcription of z, y, a occurs.

NCERT

Q. Write the salient features of the genetic code.

Answer 3 Marks

Triplet; 61 sense + 3 stop codons; degenerate; contiguous reading without punctuation; nearly universal; AUG is start codon (Met); unambiguous.

NCERT

Q. What is DNA fingerprinting? Mention its applications.

Answer 3 Marks

Technique to identify individuals using polymorphism in repetitive (VNTR) DNA. Steps: DNA isolation, restriction digestion, gel electrophoresis, Southern blotting, hybridisation with labelled probe, autoradiography. Uses: forensics, paternity testing, population diversity studies.