Molecular Basis of Inheritance ISC Class 12 Biology: Notes

What Molecular Basis of Inheritance is about

Weightage: Molecular Basis of Inheritance is part of the unit Genetics and Evolution, which carries 15 of the 70 marks in the ISC Biology Paper 1 (Theory), shared with 2 other chapters. Questions on it can appear in any section of the paper.Source: CISCE ISC Biology syllabus, examination year 2027, distribution of marks

Key concepts

Search for the genetic material

Ideal genetic material: replicates itself; chemically and structurally stable; can mutate slowly (for evolution); expresses itself as Mendelian characters. DNA fits better than RNA: the 2'-OH of RNA makes it reactive and less stable, and thymine (instead of uracil) adds stability; RNA is better suited for expression and catalysis. Griffith (1928): Streptococcus (Diplococcus) pneumoniae has a virulent S strain (smooth, capsulated) and a non-virulent R strain (rough, no capsule). Live S injected into mice → mice die; live R → mice live; heat-killed S → mice live; heat-killed S + live R → mice die, and live S bacteria were recovered. Conclusion: some transforming principle from dead S cells changed R into S (transformation). Avery, MacLeod and McCarty (1944): purified biochemicals from heat-killed S cells; proteases and RNases did not stop transformation, but DNase did: the transforming principle is DNA. Hershey and Chase (1952): bacteriophages grown with ³²P (labels DNA) or ³⁵S (labels protein) infected E. coli; after blending and centrifugation, radioactive ³²P was found in the bacterial pellet while ³⁵S stayed in the supernatant with the phage coats. So DNA enters the bacterium and directs phage production.

Worked example

Avery, MacLeod and McCarty treated an extract of heat-killed S cells separately with protease, RNase and DNase before mixing each with live R cells. Predict the result in each case and the conclusion.

Try it first, then show the solutionSolution

1. Protease destroys proteins and RNase destroys RNA, but the DNA survives: R cells are still transformed into S cells.
2. DNase destroys the DNA: no transformation occurs.
3. So the transforming principle is DNA.

Answer: Transformation after protease and RNase, none after DNase: DNA is the genetic material

Structure of DNA and RNA

Contributors: Miescher (1869) isolated 'nuclein' (DNA) from pus cells. Chargaff's rules: in double-stranded DNA, A = T and G = C, so purines = pyrimidines (A + G = T + C); the ratio (A + T)/(G + C) varies between species. Wilkins and Franklin: X-ray diffraction images showing a helix. Watson and Crick (1953) built the double helix model: two polynucleotide chains, sugar-phosphate backbone outside, bases inside; antiparallel (one 5'→3', the other 3'→5'); bases pair by hydrogen bonds: A=T (two), G≡C (three); a purine always pairs with a pyrimidine so the diameter is uniform (2 nm); right-handed coil; pitch 3.4 nm, about 10 bp per turn, 0.34 nm between bp. Nucleotides are linked by 3'-5' phosphodiester bonds; base + sugar is joined by an N-glycosidic bond. GC-rich DNA is more stable (three hydrogen bonds per pair and stronger stacking), so it needs more heat to separate. RNA: ribose, uracil instead of thymine, usually single-stranded. Types: mRNA (carries the code), tRNA (adaptor, cloverleaf shape with anticodon; brings amino acids), rRNA (structural and catalytic in ribosomes), snRNA (small nuclear RNA, in splicing), hnRNA (heterogeneous nuclear RNA, the primary transcript in eukaryotes).

Worked example

A DNA sample has 18% thymine. Find the percentages of adenine, guanine and cytosine. Why can't this be done for RNA?

Try it first, then show the solutionSolution

1. A = T = 18%, so A + T = 36% and G + C = 64%; G = C = 32%.
2. RNA is single-stranded and has uracil, so bases are not paired throughout and Chargaff's equalities do not apply.

Answer: A 18%, G 32%, C 32%

DNA packaging: nucleosomes and chromosomes

Human diploid DNA has about 6.6 × 10⁹ bp; length = 6.6 × 10⁹ × 0.34 × 10⁻⁹ m ≈ 2.2 m, packed into a nucleus about 10⁻⁶ m across. Prokaryotes: DNA (negatively charged) is held with positively charged proteins in a region called the nucleoid. Eukaryotes: positively charged basic proteins, histones (rich in lysine and arginine), form an octamer (two each of H2A, H2B, H3, H4). Negatively charged DNA wraps round it (about 200 bp per nucleosome including linker) to form a nucleosome; H1 binds the linker DNA. Nucleosomes look like 'beads on a string' under the electron microscope. The chromatin fibre coils further into solenoids (30 nm fibre), loops and finally condensed metaphase chromosomes with the help of non-histone chromosomal (NHC) proteins. Euchromatin: loosely packed, lightly staining, transcriptionally active. Heterochromatin: densely packed, darkly staining, inactive.

Worked example

Estimate the number of nucleosomes in a human diploid cell (6.6 × 10⁹ bp) if each nucleosome with its linker contains 200 bp.

Try it first, then show the solutionSolution

1. Nucleosomes = 6.6 × 10⁹ / 200
2. = 3.3 × 10⁷

Answer: About 3.3 × 10⁷ nucleosomes

Central dogma and DNA replication

Central dogma (Crick): DNA → (transcription) → mRNA → (translation) → protein; DNA → DNA (replication). Reverse transcription (basic idea): in retroviruses (e.g. HIV), RNA → DNA by reverse transcriptase. Replication is semi-conservative: each daughter molecule has one parental and one new strand. Meselson and Stahl (1958): E. coli grown for many generations in ¹⁵NH₄Cl (heavy DNA), then moved to ¹⁴N; DNA separated by CsCl density-gradient centrifugation. After one generation (20 min): all hybrid (¹⁵N/¹⁴N, intermediate density); after two: half hybrid, half light. Taylor (1958): Vicia faba root tips labelled with ³H-thymidine, then grown in unlabelled medium; autoradiography showed chromosomes replicating semi-conservatively. Mechanism: starts at an origin of replication; helicase unwinds and single-strand binding proteins (SSBP) keep strands apart; topoisomerase relieves supercoiling; primase lays an RNA primer; DNA polymerase (DNA-dependent) adds deoxynucleotides only in the 5'→3' direction, using deoxyribonucleoside triphosphates (energy and substrates). At the replication fork, the leading strand (template 3'→5') is made continuously; the lagging strand (template 5'→3') is made discontinuously as Okazaki fragments, joined by DNA ligase. Replication occurs in the S phase; it is very fast and accurate (E. coli: 4.6 × 10⁶ bp in about 18 minutes).

Worked example

E. coli with fully ¹⁵N DNA is transferred to ¹⁴N medium. What fraction of DNA molecules will be hybrid and light after three generations?

Try it first, then show the solutionSolution

1. 1 molecule → 2 after gen 1 (both hybrid) → 4 after gen 2 (2 hybrid, 2 light) → 8 after gen 3.
2. The two original ¹⁵N strands are always in two hybrid molecules: 2/8 hybrid, 6/8 light.

Answer: 25% hybrid, 75% light

Transcription and post-transcriptional processing

Transcription: copying a segment of DNA into RNA. Only one strand is transcribed. The template strand (3'→5') is read; the other, the coding strand (5'→3'), has the same sequence as the RNA (with T instead of U). If both strands were copied, two complementary RNAs would pair into double-stranded RNA and no protein would form; also two different proteins would be coded. Transcription unit: promoter (upstream, binds RNA polymerase), structural gene and terminator (downstream). Prokaryotes: one RNA polymerase makes all RNAs; the σ (sigma) factor helps initiation, the ρ (rho) factor termination. Eukaryotes: RNA polymerase I (rRNAs), II (hnRNA, the mRNA precursor), III (tRNA, 5S rRNA, snRNAs). Eukaryotic genes are split genes: coding exons interrupted by non-coding introns. Processing of hnRNA: splicing (introns removed, exons joined by snRNPs), capping (methyl guanosine triphosphate added at the 5' end), tailing (poly-A tail of 200-300 adenylates added at the 3' end); the mature mRNA is exported to the cytoplasm. Definitions: cistron (segment of DNA coding a polypeptide; the functional unit), recon (smallest unit of recombination), muton (smallest unit of mutation), monocistronic (one structural gene per transcription unit, mostly eukaryotes), polycistronic (several genes, mostly prokaryotes, e.g. lac operon).

Worked example

The coding strand of a gene segment is 5'-ATGCTTAGC-3'. Write the template strand and the mRNA.

Try it first, then show the solutionSolution

1. Template strand is complementary and antiparallel: 3'-TACGAATCG-5'.
2. mRNA has the coding-strand sequence with U for T: 5'-AUGCUUAGC-3'.

Answer: Template 3'-TACGAATCG-5'; mRNA 5'-AUGCUUAGC-3'

Genetic code

Discovery: George Gamow argued that 20 amino acids need a triplet code (4³ = 64 combinations); Nirenberg and Matthaei with cell-free synthesis (poly-U → polyphenylalanine), Har Gobind Khorana (synthetic RNAs with defined repeats) and Severo Ochoa (polynucleotide phosphorylase) deciphered it. Codon: a sequence of three bases on mRNA coding for one amino acid. Essential features: (1) triplet: 64 codons, 61 code for amino acids; (2) three stop codons (UAA, UAG, UGA) code for none; (3) AUG is the start codon and also codes for methionine; (4) unambiguous and specific: one codon codes for only one amino acid; (5) degenerate: some amino acids have more than one codon (valine: GUU, GUC, GUA, GUG); (6) read continuously in a fixed frame, without punctuation (commaless); (7) nearly universal (some exceptions in mitochondria and some protozoans); (8) non-overlapping. tRNA reads codons with its anticodon. Mutations: a frame-shift (insertion/deletion of one or two bases) changes all later codons; three bases inserted or deleted add or remove one amino acid without a frame shift.

Worked example

A point mutation changes the codon UAU (tyrosine) in the middle of an mRNA to UAA. What happens to the protein, and what is this kind of mutation called?

Try it first, then show the solutionSolution

1. UAA is a stop codon, so translation ends at this point.
2. A shortened (truncated) polypeptide is released, usually non-functional: a nonsense mutation, caused here by a single base substitution.

Answer: Translation stops early, giving a truncated protein (nonsense mutation)

Translation in prokaryotes

Protein synthesis on ribosomes (70S in prokaryotes: 50S + 30S subunits). Activation (charging of tRNA): each amino acid is activated by ATP and attached to its tRNA by a specific aminoacyl-tRNA synthetase, forming aminoacyl-tRNA. Initiation: the small subunit binds mRNA at the start codon AUG; the initiator tRNA carrying formyl-methionine (fMet) pairs with AUG; the large subunit joins. The ribosome has A (aminoacyl), P (peptidyl) and E sites. Elongation: the next charged tRNA enters the A site (codon-anticodon pairing); a peptide bond forms between amino acids, catalysed by the 23S rRNA (a ribozyme, peptidyl transferase); the ribosome moves one codon along mRNA (translocation, 5'→3'). Termination: a stop codon (UAA, UAG, UGA) reaches the A site; a release factor binds; the polypeptide is released and the ribosome subunits separate. mRNA has untranslated regions (UTRs) at both ends that help efficient translation. Several ribosomes may translate one mRNA at once (polysome).

Worked example

An mRNA from start to stop codon has 903 nucleotides (including the stop codon). How many amino acids will the polypeptide have (ignore removal of fMet)?

Try it first, then show the solutionSolution

1. Codons = 903/3 = 301, including one stop codon.
2. Amino acids = 301 − 1 = 300.

Answer: 300 amino acids

Lac operon, Human Genome Project and DNA fingerprinting

Regulation of gene expression: in prokaryotes mainly at transcription initiation. Lac operon (Jacob and Monod) in E. coli: a polycistronic unit with a regulator gene i (makes the repressor), promoter p, operator o, and three structural genes: z (β-galactosidase, hydrolyses lactose into glucose and galactose), y (permease, lets lactose in) and a (transacetylase). No lactose: the repressor binds the operator and blocks RNA polymerase: genes off. Lactose present (inducer, as allolactose): it binds the repressor, changing its shape so it cannot bind the operator; RNA polymerase transcribes z, y, a: inducible operon, negative regulation. Human Genome Project (1990-2003): goals: identify all genes (about 20 000-25 000), determine the sequence of the 3 × 10⁹ bp, store information in databases, improve tools for analysis, transfer technology to industry, address ethical, legal and social issues (ELSI). Methodologies: (1) Expressed Sequence Tags (ESTs): identify genes expressed as RNA; (2) Sequence annotation: sequence the whole genome (coding and non-coding) and later assign functions; DNA was cut into fragments, cloned in BAC and YAC vectors, sequenced by automated sequencers and assembled by computer. Salient features: 3164.7 million bp; average gene 3000 bases, largest dystrophin (2.4 million bases); less than 2% codes for proteins; large repetitive sequences; chromosome 1 has most genes (2968), Y fewest (231); about 1.4 million SNPs. Applications: understanding diseases, diagnosis and treatment, human evolution. DNA fingerprinting (Alec Jeffreys): based on VNTRs (variable number tandem repeats, minisatellites), which differ between individuals. Technique: isolate DNA; cut with restriction enzymes; separate fragments by gel electrophoresis; transfer to nylon membrane (Southern blotting); hybridise with a labelled VNTR probe; detect by autoradiography: a unique band pattern. Applications: forensic identification, paternity disputes, population and evolutionary studies. Ethical issues (briefly): privacy of genetic data, consent, misuse for discrimination, accuracy of sample handling.

Worked example

In the lac operon, what happens if the regulator gene i is mutated so that the repressor cannot bind the operator?

Try it first, then show the solutionSolution

1. The operator is always free, so RNA polymerase can always transcribe z, y and a.
2. The enzymes are made continuously, even without lactose (constitutive expression).

Answer: Genes expressed all the time (constitutive)

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

• Search for the Genetic Material
• Structure of DNA and Chargaff's Rules
• DNA Replication

Molecular Basis of Inheritance: key facts

Chargaff's rules

$A=T,G=C,A+G=T+C$

Double-stranded DNA only.

DNA length

$\text{length}=\text{bp}\times 0.34nm$

Human diploid ≈ 2.2 m; E. coli ≈ 1.56 mm.

Helix dimensions

$\text{pitch}=3.4nm,10\text{bp per turn},\text{diameter}=2nm$

Right-handed B-DNA.

Meselson-Stahl after n generations

$\text{hybrid fraction}=\frac{2}{2^n}$

Rest is light (¹⁴N/¹⁴N) for n ≥ 2.

Number of codons

$4^3=64(61\text{sense}+3\text{stop})$

AUG = start (Met).

Amino acids from mRNA

$\text{amino acids}=\frac{\text{coding nucleotides}}{3}-1\text{(stop)}$

Count from AUG to the stop codon.

Central dogma

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

Reverse transcription: RNA → DNA in retroviruses.

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: Adenine, Guanine. Pyrimidines: C, T, U ('CUT the Py').

  Use it when: Base types.

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

  UAA, UAG, UGA.

  Use it when: Termination codons.

• Leading leads, lagging lags

  Leading strand continuous towards the fork; lagging in Okazaki pieces, sealed by ligase.

  Use it when: Replication.

• Lac: i-p-o-z-y-a

  'I Prefer Our Zesty Yellow Apples'.

  Use it when: Order of the lac operon.

• Griffith Gave, Avery Analysed, Hershey Hammered it

  Transformation shown → DNA identified → phage proof.

  Use it when: Order of experiments.

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 5-mark question split into 1 + 1 + 1 + 2 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 T in RNA codons

  CISCE examiners' comments, 2024: asked for the fourth valine codon after GUU, GUC and GUA, candidates wrote GUT, using thymine in an RNA codon.

  Wrong working

  GUT

  Right working

  GUG

  ✓Do this: Codons are on mRNA: only A, U, G, C.

  Usually costs: 1 mark.

• ✗Mistake: Muddled Griffith experiment

  CISCE examiners' comments, 2024: candidates did not name the R and S strains, swapped their virulence or said R was heat-killed instead of S, did not follow the sequence of steps, missed the word 'transforming' (some named protein as the genetic material), and said mice die when both strains were heat-killed.

  Wrong working

  Heat-killed R + live S → mice die, so protein transforms.

  Right working

  Heat-killed S + live R → mice die; transforming principle (DNA). Both heat-killed → mice survive.

  ✓Do this: Learn the four injections in order as a table.

  Usually costs: 2-3 marks.

• ✗Mistake: Guessing abbreviations

  CISCE examiners' comments, 2025: for SSBP candidates wrote random expansions such as 'single synthetic bioprotein' and misspelt 'strand' as 'stand'.

  Wrong working

  Single stand binded portion.

  Right working

  Single-strand binding proteins.

  ✓Do this: Keep a list of expansions: SSBP, VNTR, EST, HGP, hnRNA, snRNA, BAC, YAC.

  Usually costs: 1 mark.

• ✗Mistake: Wrong reasons for strand choice and GC content

  CISCE examiners' comments, 2025: for the failed enzyme synthesis candidates confused replication with transcription, chose a strand without explanation or mentioned introns; for high GC content in organisms of hot places, they credited heat-resistant proteins instead of the stability of GC pairs; and they overlooked that RNA is single-stranded with uracil, so Chargaff's rules do not apply.

  Wrong working

  GC-rich because hot organisms have heat-resistant proteins.

  Right working

  G≡C has three hydrogen bonds, so GC-rich DNA is thermally more stable.

  ✓Do this: Answer about DNA chemistry when the question is about DNA.

  Usually costs: 1 mark each.

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

   Okazaki fragments are joined by:
   (a) DNA polymerase
   (b) helicase
   (c) DNA ligase
   (d) primase

   Hint

   Sealing enzyme.

   Answer

   (c)

2. Q2 · 1 mark · MCQBoard level

   Which RNA polymerase transcribes hnRNA in eukaryotes?
   (a) I
   (b) II
   (c) III
   (d) all

   Hint

   mRNA precursor.

   Answer

   (b)

3. Q3 · 1 mark · Very short answerBoard level

   Define cistron.

   Hint

   Functional unit.

   Answer

   A segment of DNA coding for one polypeptide.

4. Q4 · 1 mark · Very short answerBoard level

   Name the enzyme in retroviruses that makes DNA from RNA.

   Hint

   Reverse transcription.

   Answer

   Reverse transcriptase

5. Q5 · 2 marks · Short answerHarder board level

   A DNA molecule has 1200 bp with 30% adenine. Find the number of guanine bases and of hydrogen bonds.

   Hint

   Total bases 2400.

   Answer

   A = T = 30% of 2400 = 720 each; G = C = 20% = 480 each. Hydrogen bonds = 720 × 2 + 480 × 3 = 1440 + 1440 = 2880.

6. Q6 · 1 mark · Very short answerHarder board level

   Calculate the length of E. coli DNA (4.6 × 10⁶ bp).

   Hint

   × 0.34 nm.

   Answer

   4.6 × 10⁶ × 0.34 × 10⁻⁹ m = 1.56 × 10⁻³ m ≈ 1.56 mm.

7. Q7 · 3 marks · Short answerHarder board level

   Give four properties of an ideal genetic material and explain why DNA is preferred to RNA for storage.

   Hint

   2'-OH.

   Answer

   Replication; chemical and structural stability; mutability (slow change); expression as Mendelian characters. RNA's 2'-OH makes it reactive and easily degraded, and uracil is less stable than thymine; DNA's double strand allows repair, so it is better for storage.

8. Q8 · 2 marks · Short answerHarder board level

   Explain the roles of aminoacyl-tRNA synthetase and the 23S rRNA in translation.

   Hint

   Charging; peptide bond.

   Answer

   Aminoacyl-tRNA synthetase attaches the correct amino acid to its tRNA using ATP (charging). 23S rRNA in the large subunit catalyses peptide bond formation (peptidyl transferase, a ribozyme).

9. Q9 · 3 marks · Short answerBoard level

   State five features of the genetic code.

   Hint

   Triplet...

   Answer

   Triplet; degenerate; unambiguous; nearly universal; commaless and non-overlapping; AUG start and three stop codons (any five).

10. Q10 · 2 marks · Short answerHarder board level

    Describe two methodologies and two salient features of the Human Genome Project.

    Hint

    ESTs; annotation.

    Answer

    Methods: Expressed Sequence Tags (identifying expressed genes) and sequence annotation (sequencing the whole genome then assigning functions). Features: 3164.7 million bp; fewer than 2% code for protein; about 20 000-25 000 genes; chromosome 1 has most genes, Y fewest; about 1.4 million SNPs (any two).

11. Q11 · 5 marks · Long answerHarder board level

    (a) Describe the structure of DNA according to Watson and Crick. [3]
    (b) Name the contributions of Chargaff and Franklin. [1]
    (c) Why is DNA called a polymer with antiparallel strands? [1]

    Hint

    Pitch, pairing.

    Answer

    (a) Two polynucleotide chains coiled into a right-handed double helix; sugar-phosphate backbone outside, bases inside; A pairs with T by 2 H-bonds and G with C by 3; purine always opposite pyrimidine giving 2 nm diameter; pitch 3.4 nm, 10 bp per turn. (b) Chargaff: A = T, G = C; Franklin: X-ray diffraction showing a helix. (c) It is made of many nucleotides joined by phosphodiester bonds; one strand runs 5'→3' and the other 3'→5'.

12. Q12 · 2 marks · Short answerHOTS / competency

    After four generations of Meselson-Stahl's experiment, what percentage of DNA molecules would be hybrid? Would any be fully heavy? Explain.

    Hint

    2/2ⁿ.

    Answer

    16 molecules; only 2 contain an original ¹⁵N strand, so 2/16 = 12.5% hybrid and 87.5% light. None are heavy, because each original heavy strand always pairs with a new light strand.

13. Q13 · 2 marks · Short answerHOTS / competency

    Why is the lac operon described as negatively regulated and inducible? What would happen if glucose and lactose were both absent?

    Hint

    Repressor role.

    Answer

    Negative: a repressor protein switches it off. Inducible: the substrate (lactose) switches it on by inactivating the repressor. With neither sugar, the repressor stays on the operator and the genes remain off, saving energy.

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 isolated nuclein?Show answer

   Friedrich Miescher.

2. 2.Radioactive isotope that labels phage DNA?Show answer

   ³²P.

3. 3.Distance between base pairs in DNA?Show answer

   0.34 nm.

4. 4.What is a nucleosome?Show answer

   A histone octamer wrapped with about 200 bp of DNA.

5. 5.Function of DNA ligase?Show answer

   Joins Okazaki fragments (seals nicks).

6. 6.Start codon and its amino acid?Show answer

   AUG; methionine.

7. 7.Structural genes of lac operon?Show answer

   z (β-galactosidase), y (permease), a (transacetylase).

8. 8.Basis of DNA fingerprinting?Show answer

   Variable number tandem repeats (VNTRs).

Quick revision checklist

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

• ☐Can you describe Griffith's four injections and conclusion?
• ☐Can you explain Hershey-Chase with isotopes?
• ☐Can you give the double helix dimensions and Chargaff's rules?
• ☐Can you draw a replication fork with enzymes?
• ☐Can you explain capping, tailing and splicing?
• ☐Can you list the features of the genetic code?
• ☐Can you explain the lac operon on and off?
• ☐Can you outline DNA fingerprinting steps and HGP goals?

Molecular Basis of Inheritance: frequently asked questions