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State Board · Class 12 · Biology

Lesson 4: Molecular Basis of Inheritance

7 concepts · 56 practice questions with answers · Free to read

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DNA as the genetic material

Link to Std 9–10

Std 11 covered DNA, RNA and the cell cycle; here we see how DNA was proved to be the genetic material, and how it replicates and makes proteins.

From daily life

The ‘message’ of heredity is written in DNA, as several experiments proved.

Simple definition

Nucleic acids (DNA/RNA) carry genetic information. Transformation = a bacterium changes character by taking up DNA.

DNA experimentsDNA is the genetic material: experimentsGriffith 1928Avery et al 1944Hershey–Chase 1952S + RDNase / protease / RNase³²P (DNA) / ³⁵S (protein)Griffith: heat-killed S + live R → mouse dies; ‘transforming principle’S = capsule, virulent; R = no capsule, harmless (Streptococcus)Avery, MacLeod, McCarty: only DNase stopped transformationProtease and RNase had no effect → DNA is the principleHershey–Chase: bacteriophage; ³²P DNA entered bacteria³⁵S (protein) stayed outside; blender + centrifugeDNA has P but no S; protein has S but no PMiescher 1869: ‘nuclein’ from white cells → nucleic acid

In the figure : Griffith, Avery and Hershey–Chase.

Rule 1 : Miescher, Griffith, Avery — Miescher (1869): isolated ‘nuclein’ from nuclei of white blood cells (rich in phosphorus, acidic) → nucleic acid. Griffith (1928): two types of Streptococcus pneumoniae — S (capsulated, virulent) and R (no capsule, harmless). Four experiments: R → mouse lived; S → died; heat-killed S → lived; heat-killed S + live R → died, and live S was recovered. Conclusion: a ‘transforming principle’ from dead S changed R into S. Avery, MacLeod, McCarty (1944): treated S extract with protease, RNase, DNase; only DNase stopped transformation ⇒ the principle = DNA.

  • Easy : S = smooth + severe.
  • Medium : DNase blocked transformation.
  • Exam-type (CET) : Griffith: 1928; Avery: 1944.

Rule 2 : Hershey–Chase experiment (1952) — The virus is a bacteriophage (DNA + protein coat). One batch carried ³²P (DNA has phosphorus, protein does not), another ³⁵S (protein has sulphur, DNA does not). After infecting E. coli, a blender removed the coats and the mixture was centrifuged. ³²P was found inside the bacteria; ³⁵S stayed outside (supernatant). So DNA passed from virus to bacterium, not protein ⇒ DNA is the genetic material. Earlier most thought protein was, being varied and complex.

  • Easy : ³²P = DNA; ³⁵S = protein.
  • Medium : What entered = DNA.
  • Exam-type (CET) : Phage = bacterium eater.
Remember

P for DNA (phosphate), S for protein (sulphur).

Common mistake and CET trap

‘³⁵S labels DNA’ is wrong (protein). Trap: ‘Griffith proved DNA’ is wrong (Avery did).

Shortcut

Memory clue : "P-D, S-P: phosphorus-DNA, sulphur-protein".

Spot the difference

ExperimentYearFinding
Griffith1928Transforming principle
Avery et al.1944Principle = DNA
Hershey–Chase1952DNA enters bacteria

Practice questions

Easy
  1. 1What did Miescher isolate?
  2. 2Bacterium used by Griffith?
  3. 3Feature of S strain?
Medium
  1. 4What is the transforming principle?
  2. 5Which enzyme blocked transformation?
  3. 6³²P labels?
Challenging
  1. 7³⁵S labels?
  2. 8Year of Hershey–Chase?
Show answers (try them yourself first)
  1. 1Nuclein.
  2. 2Streptococcus pneumoniae.
  3. 3Capsulated, virulent.
  4. 4DNA.
  5. 5DNase.
  6. 6DNA.
  7. 7Protein.
  8. 81952.

DNA packaging and chromatin

From daily life

A 2 m rope fits into a tiny box only if coiled.

Simple definition

A nucleosome = DNA wound around a histone octamer; the repeating unit of chromatin.

DNA packagingDNA packagingDNANucleosomeSolenoid 30 nmChromosomeHuman DNA ≈ 2.2 m; nucleus ≈ 10⁻⁶ m → tight coilingProkaryote: nucleoid; 40–50 loops; HU proteins, gyraseHistone octamer = H2A, H2B, H3, H4 (two each)Histones: lysine + arginine = positive; DNA negativeNucleosome: ~200 bp; 146 bp wrapped; H1 seals‘Beads-on-string’ → 30 nm solenoid → loops → chromosomeHeterochromatin: dense, inactive; euchromatin: open, activeNHC proteins for higher-order folding

In the figure : DNA → nucleosome → chromosome.

Rule 1 : Packaging in prokaryotes — Human DNA is ~2.2 m long but the nucleus ~10⁻⁶ m, so coiling is essential. E. coli: cell 2–3 µm; nucleoid (circular naked DNA, ~4.6 million bp, ~1.1 mm). Negatively charged DNA is circular → 350 µm; 40–50 loops (RNA connectors) → 30 µm; each loop supercoiled → 2 µm. Helpers: positively charged HU proteins (histone-like), DNA gyrase, topoisomerase I.

  • Easy : 40–50 loops.
  • Medium : HU = histone-like.
  • Exam-type (CET) : Nucleoid = no membrane.

Rule 2 : Packaging in eukaryotes — Histones are rich in lysine and arginine (basic), positively charged, binding negative DNA. Octamer = two each of H2A, H2B, H3, H4. Nucleosome: ~200 bp; about 146 bp wound 1¾ turns round the octamer; the rest is linker DNA; H1 binds where DNA enters and leaves. Under the electron microscope ‘beads-on-string’ (10 nm) → 6 nucleosomes make a solenoid (30 nm) → loops → metaphase chromosome; higher levels need NHC proteins (Kornberg 1974). Heterochromatin (dense, dark, inactive; near centromere/telomere) versus euchromatin (open, light, active).

  • Easy : Octamer = 8 histones.
  • Medium : 146 bp wrapped.
  • Exam-type (CET) : Euchromatin = active.
Remember

H2A, H2B, H3, H4: two of each = 8.

Common mistake and CET trap

‘H1 is part of the octamer’ is wrong. Trap: ‘histones are negative’ is wrong.

Shortcut

Memory clue : "Hetero = hard and dark; eu = easy and open".

Spot the difference

PointHeterochromatinEuchromatin
StateCondensedOpen
ActivityInactiveActive
StainDarkLight

Practice questions

Easy
  1. 1Length of human DNA?
  2. 2Loops in prokaryotes?
  3. 3Histones in octamer?
Medium
  1. 4Why are histones positive?
  2. 5bp in a nucleosome?
  3. 6Role of H1?
Challenging
  1. 7Solenoid diameter?
  2. 8What is NHC?
Show answers (try them yourself first)
  1. 1~2.2 m.
  2. 240–50.
  3. 3H2A, H2B, H3, H4 (×2).
  4. 4Lysine + arginine.
  5. 5~200 (146 wound).
  6. 6Binds DNA at entry/exit.
  7. 730 nm.
  8. 8Non-histone chromosomal proteins.

DNA replication

From daily life

A photocopy keeps the original; in DNA every copy keeps one old strand.

Simple definition

Replication = DNA making an identical copy of itself (S phase). Semiconservative = each new DNA has one old and one new strand.

DNA replicationDNA replicationLeading: continuous 5′→3′Lagging: Okazaki fragmentsReplication fork (Y) — S phaseSemiconservative: each new DNA = 1 old + 1 new strandMeselson–Stahl 1958: ¹⁵N → ¹⁴N, CsCl gradient1st gen: all hybrid ¹⁴N¹⁵N; 2nd: hybrid + lightHelicase unwinds; SSBP prevents rejoiningPrimase: RNA primer; DNA polymerase adds 5′→3′Ligase joins Okazaki fragments; topoisomerase relieves strainAutocatalytic = DNA copies itself; heterocatalytic = RNA/proteinOrigin ‘O’: endonuclease nick; bidirectional

In the figure : Replication fork and Okazaki fragments.

Live figure

DNA replication

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Rule 1 : Semiconservative mode: Meselson–Stahl — Watson and Crick suggested it from the antiparallel, complementary nature of the strands. Meselson and Stahl (1958): E. coli grown for generations in ¹⁵N (heavy) then moved to ¹⁴N. With 6 M CsCl density-gradient centrifugation: generation 1 (20 min) = all hybrid (¹⁴N¹⁵N); generation 2 (40 min) = half hybrid + half light (¹⁴N). This ruled out the conservative (separate heavy and light) and dispersive models. Autocatalytic function = DNA making DNA; heterocatalytic = making RNA/protein.

  • Easy : Generation 1: 100% hybrid.
  • Medium : Generation 2: 50% hybrid + 50% light.
  • Exam-type (CET) : Once, in S phase.

Rule 2 : Process and enzymes — 1 Nucleotide activation: ATP makes dNTPs (dATP, dGTP, dCTP, dTTP). 2 Origin ‘O’: endonuclease nicks one strand. 3 Unwinding: helicase breaks hydrogen bonds → Y-shaped replication fork (bidirectional); SSBPs keep strands apart; topoisomerase/gyrase relieve strain. 4 Primase lays an RNA primer; DNA polymerase adds only 5′→3′, so the leading strand is continuous and the lagging strand is made in pieces (Okazaki fragments). 5 Primers are replaced and gaps filled; DNA ligase joins fragments. Eukaryotes have many origins.

  • Easy : Polymerase works 5′→3′ only.
  • Medium : Okazaki fragments = lagging strand.
  • Exam-type (CET) : Ligase = joiner.
Remember

H-S-P-P-L: helicase, SSBP, primase, polymerase, ligase.

Common mistake and CET trap

‘Replication is conservative’ is wrong. Trap: ‘Okazaki fragments occur on the leading strand’ is wrong.

Shortcut

Memory clue : "Leading = one long piece; lagging = lots of pieces".

Spot the difference

EnzymeRole
HelicaseUnwinds strands
PrimaseMakes RNA primer
DNA polymeraseBuilds new strand 5′→3′
LigaseJoins fragments

Practice questions

Easy
  1. 1When does replication occur?
  2. 2Semiconservative means?
  3. 3Isotopes in Meselson–Stahl?
Medium
  1. 4DNA after 1st generation?
  2. 5Role of helicase?
  3. 6Where are Okazaki fragments?
Challenging
  1. 7Which enzyme joins them?
  2. 8Role of SSBP?
Show answers (try them yourself first)
  1. 1S phase.
  2. 2One old + one new strand.
  3. 3¹⁵N and ¹⁴N.
  4. 4All hybrid.
  5. 5Breaks hydrogen bonds.
  6. 6Lagging strand.
  7. 7DNA ligase.
  8. 8Keeps strands apart.

Transcription and RNA processing

From daily life

A page of the master book (DNA) is photocopied (mRNA) and taken to the workshop (ribosome).

Simple definition

Transcription = making RNA from one strand of DNA. Central dogma: DNA → RNA → protein (Crick, 1958).

TranscriptionCentral dogma and transcriptionDNAmRNAProteinTranscriptionTranslationReversetranscription(Temin, Baltimore)Central dogma: Crick 1958; DNA → RNA → proteinTranscription unit: promoter – structural gene – terminatorTemplate (antisense) 3′→5′; coding (sense) = like RNA (T→U)RNA polymerase: 5′→3′; starts at promoter, ends at terminatorEukaryote: hnRNA (exons + introns) → splicingCapping (5′, methyl-G) + tailing (3′, poly-A)Introns removed, exons joined = mature mRNAThree RNAs: mRNA, tRNA, rRNA; Pol I, II, III

In the figure : Central dogma and transcription unit.

Rule 1 : Transcription unit and process — A transcription unit = promoter (RNA polymerase binds), structural gene, terminator (3′ end). The template (antisense) strand runs 3′→5′; the coding (sense) strand has the same sequence as RNA (U for T). RNA polymerase adds complementary RNA 5′→3′ (A–U, G–C). Prokaryotes have one polymerase; eukaryotes Pol I (rRNA), Pol II (mRNA), Pol III (tRNA). Three RNAs: mRNA, tRNA, rRNA. Retroviruses use reverse transcription: RNA → DNA (Temin, Baltimore, 1970).

  • Easy : Promoter = start; terminator = end.
  • Medium : Sense = same as RNA.
  • Exam-type (CET) : T → U.

Rule 2 : RNA processing in eukaryotes — Eukaryotic genes are split: exons (expressed) and introns (intervening). The primary transcript is hnRNA. Processing: splicing (remove introns, join exons), capping (methyl-guanosine at 5′) and tailing (poly-A at 3′). The mature mRNA moves from nucleus to cytoplasm. Prokaryotes have no such processing; their mRNA can be polycistronic.

  • Easy : Exon = expressed.
  • Medium : 5′ cap + 3′ tail.
  • Exam-type (CET) : hnRNA → mRNA.
Remember

Exon = expressed; intron = intervening (removed).

Common mistake and CET trap

‘Introns are translated’ is wrong. Trap: ‘the coding strand is the template’ is wrong.

Shortcut

Memory clue : "Cap at the head (5′), tail at the tail (3′)".

Spot the difference

RNA PolMakes
Pol IrRNA
Pol IImRNA
Pol IIItRNA

Practice questions

Easy
  1. 1Who proposed the central dogma?
  2. 2Role of promoter?
  3. 3Polarity of template?
Medium
  1. 4What replaces T in RNA?
  2. 5What is hnRNA?
  3. 6What is splicing?
Challenging
  1. 7Where is the cap?
  2. 8Who found reverse transcription?
Show answers (try them yourself first)
  1. 1Crick (1958).
  2. 2Binds RNA polymerase.
  3. 33′→5′.
  4. 4U (uracil).
  5. 5Primary transcript.
  6. 6Removing introns.
  7. 75′ end.
  8. 8Temin and Baltimore.

Genetic code and mutation

From daily life

In Morse code symbols make a letter; here three bases make one amino acid.

Simple definition

A codon = three mRNA bases specifying one amino acid. A mutation = sudden change in DNA sequence.

Genetic codeGenetic code and mutationAUGStart / MetUAAUAGUGAStop codons64 = 61 + 3Triplet code; 64 codons; 61 for amino acids, 3 stopDegenerate: many codons = one amino acid (wobble)Commaless, non-overlapping, non-ambiguous, universalCodon = mRNA 5′AUG3′; anticodon = tRNA 3′UAC5′Mutation: sudden change in DNA sequence; raw material of variationPoint mutation: one base pair (sickle-cell)Frame-shift: 1–2 bases added/lost → reading frame shifts3 (or multiples) bases: only amino acids lost/gained

In the figure : Start/stop codons and mutation.

Live figure

Genetic code and mutation

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Rule 1 : Properties of the genetic code — Sets of 3 of the 4 bases = 4³ = 64 codons; 61 code for amino acids and 3 are stop (UAA, UAG, UGA). AUG = start (methionine). Properties: triplet; commaless (no gaps); non-overlapping; degenerate (many codons for one amino acid — wobble: the third base varies); non-ambiguous (one codon = one amino acid); universal (bacteria to humans). Codon on mRNA 5′AUG3′; anticodon on tRNA 3′UAC5′.

  • Easy : 64 = 61 + 3.
  • Medium : AUG = start; UAA/UAG/UGA = stop.
  • Exam-type (CET) : Degenerate ≠ ambiguous.

Rule 2 : Types of mutation — Mutation, with recombination, supplies variation and the raw material of evolution. Point mutation: change in a single base pair (e.g. sickle-cell: GAG → GUG, Glu → Val). Frame-shift mutation: adding/losing 1 or 2 bases shifts the reading frame and changes every later amino acid. Adding/losing 3 (or multiples of 3) bases only adds/removes a few amino acids; the frame stays. Deletion / insertion / duplication can alter the chromosome.

  • Easy : 1–2 bases = frame-shift.
  • Medium : 3 bases = frame retained.
  • Exam-type (CET) : Point = single base.
Remember

UAA, UAG, UGA = ‘U Are Away / U Are Gone / U Go Away’ (stop).

Common mistake and CET trap

‘The code overlaps’ is wrong. Trap: ‘AUG is a stop codon’ is wrong (start).

Shortcut

Memory clue : "Start = AUG; three stops = UAA, UAG, UGA".

Spot the difference

TermMeaning
DegenerateMany codons: one amino acid
Non-ambiguousOne codon: one amino acid
CommalessNo gaps
UniversalSame in all organisms

Practice questions

Easy
  1. 1What is a codon?
  2. 2Total codons?
  3. 3Stop codons?
Medium
  1. 4Start codon?
  2. 5What is degeneracy?
  3. 6Wobble involves which base?
Challenging
  1. 7What causes frame-shift?
  2. 8Example of point mutation?
Show answers (try them yourself first)
  1. 1A triplet of bases.
  2. 264.
  3. 3UAA, UAG, UGA.
  4. 4AUG.
  5. 5Many codons for one amino acid.
  6. 6The third base.
  7. 7Adding/losing 1–2 bases.
  8. 8Sickle-cell anaemia.

Translation: protein synthesis

From daily life

In a kitchen (ribosome) a helper (tRNA) reads the recipe (mRNA) and adds ingredients.

Simple definition

Translation = making a polypeptide on a ribosome by joining amino acids as dictated by mRNA codons.

TranslationTranslation: protein synthesisE P AmRNARibosome: P, A, E sitesInitiator tRNA: Met (UAC)tRNA: cloverleaf; 3′ CCANeeds: amino acids, mRNA, tRNA, ribosome, ATP, Mg⁺⁺, enzymes1 Initiation: small subunit at 5′; AUG + Met-tRNA in P-site2 Elongation: tRNA to A-site → peptide bond → translocationFirst tRNA leaves from E-site3 Termination: stop codon at A-site; release factorProkaryote: formyl-methionine; eukaryote: methionineUTR = untranslated regions; mRNA is short-livedtRNA = adapter (anticodon + amino-acid end)

In the figure : Ribosome, tRNA and translation.

Rule 1 : tRNA and ribosome — tRNA = adapter molecule: cloverleaf (2D), inverted L (3D); the anticodon loop is complementary to the codon; the amino acid attaches at the 3′ CCA end; each amino acid has its own tRNA; no tRNA for stop codons. Ribosome: small + large subunits (joined by Mg⁺⁺); one site for mRNA and 3 for tRNA: P (peptidyl), A (aminoacyl), E (exit). Needs: amino acids (~20), mRNA, tRNA, ribosome, ATP, Mg⁺⁺, enzymes, elongation/translocation/release factors.

  • Easy : tRNA: 3′ CCA.
  • Medium : P, A, E sites.
  • Exam-type (CET) : No tRNA for stop.

Rule 2 : Three steps of translation — 1 Initiation: amino acids activated by ATP; small subunit binds mRNA's 5′ end; the initiator tRNA (anticodon UAC; Met — formyl-Met in prokaryotes) binds AUG; the large subunit joins; initiator tRNA sits in the P-site. 2 Elongation: the next tRNA enters the A-site; a peptide bond forms; the ribosome moves one codon (translocation); the first tRNA leaves from the E-site; repeated (a peptide bond takes < 0.1 s). 3 Termination: a stop codon at the A-site; release factor binds; the polypeptide is freed; subunits separate. UTRs (5′ and 3′) on mRNA are not translated; mRNA is short-lived.

  • Easy : I-E-T: initiation, elongation, termination.
  • Medium : First tRNA goes to P-site.
  • Exam-type (CET) : Release factor = stop.
Remember

PAE: peptidyl → aminoacyl → exit (the tRNA's path).

Common mistake and CET trap

‘Translation occurs in the nucleus’ is wrong (cytoplasm / ribosome). Trap: ‘a tRNA reads the stop codon’ is wrong.

Shortcut

Memory clue : "Add at A, pass to P, exit at E".

Spot the difference

SiteRole
PPeptidyl-tRNA
AIncoming aminoacyl-tRNA
EtRNA exit

Practice questions

Easy
  1. 1Where does translation occur?
  2. 2Why is tRNA an adapter?
  3. 3Initiator codon?
Medium
  1. 4Anticodon of initiator tRNA?
  2. 5tRNA sites of ribosome?
  3. 6Where does the peptide bond form?
Challenging
  1. 7What causes termination?
  2. 8What is UTR?
Show answers (try them yourself first)
  1. 1On ribosomes.
  2. 2Reads codon and carries amino acid.
  3. 3AUG.
  4. 4UAC.
  5. 5P, A, E.
  6. 6Between amino acids at P and A.
  7. 7Stop codon + release factor.
  8. 8Untranslated region.

Gene regulation, genomics and DNA fingerprinting

From daily life

A lamp uses power only when switched on; genes too switch on only when needed.

Simple definition

An operon = group of genes controlled together. Genome = an organism's complete genetic information. DNA fingerprinting = identifying individuals from VNTRs.

Lac operonLac operon, genomics, DNA fingerprintingiPOzyaRegulator i → repressor; O = operator; z, y, a = structuralLac operon (Jacob–Monod 1961): inducible, negative controlNo lactose: repressor binds O → no mRNALactose (allolactose) inactivates repressor → z, y, a onz = β-galactosidase; y = permease; a = transacetylaseGenome (Winkler 1920); genomics: structural + functionalHGP: 1990–2003; ~3 billion bp; ~33,000 genes (book); HUGODNA fingerprinting: VNTR (minisatellites), Alec JeffreysUses: forensics, paternity, disease diagnosis, ancestry

In the figure : Lac operon and genomics.

Rule 1 : Lac operon — Jacob and Monod (1961). In E. coli three enzymes digest lactose: z = β-galactosidase (lactose → galactose + glucose), y = permease (lets lactose in), a = transacetylase. Parts: regulator (i) gene → repressor protein; promoter (P) (RNA polymerase); operator (O); structural genes z, y, a; inducer (not part of the operon). No lactose: the repressor binds O → transcription stops. With lactose (allolactose inducer): it binds and inactivates the repressor → O is free → one polycistronic mRNA for z-y-a. So it is inducible, negative control; it switches off when the inducer is used up.

  • Easy : i–P–O–z–y–a.
  • Medium : Inducer = lactose (allolactose).
  • Exam-type (CET) : Repressor binds O.

Rule 2 : Genomics, HGP, DNA fingerprinting — Genome (Winkler, 1920) = an organism's complete genetic information (one haploid set). Genomics: structural (mapping, sequencing) and functional (roles); uses: agriculture (transgenics), medicine, gene therapy, forensics. Human Genome Project: 1990 to 2003; HUGO; US + UK, France, Germany, Japan, India, China; about 3 billion bp; ~33,000 genes per the book; spurred bioinformatics. DNA fingerprinting: identifying people from differences in VNTR (minisatellite) repeats (Alec Jeffreys); uses: forensics, paternity, disease diagnosis, ancestry.

  • Easy : HGP: 1990–2003.
  • Medium : 3 billion bp.
  • Exam-type (CET) : VNTR differs between people.
Remember

Lac: i-P-O-z-y-a = ‘I Pay Only for z-y-a’ (to learn the order).

Common mistake and CET trap

‘The inducer is part of the operon’ is wrong. Trap: ‘operon is on without lactose’ is wrong.

Shortcut

Memory clue : "Repressor = red light (stop); inducer = switches it on".

Spot the difference

ConditionRepressorOperon
No lactoseBinds OOff
Lactose presentInactivatedOn

Practice questions

Easy
  1. 1Who proposed the lac operon?
  2. 2What does z code for?
  3. 3What does y code for?
Medium
  1. 4Inducer?
  2. 5Repressor binds where?
  3. 6Duration of HGP?
Challenging
  1. 7Basis of DNA fingerprinting?
  2. 8Who coined ‘genome’?
Show answers (try them yourself first)
  1. 1Jacob and Monod.
  2. 2β-galactosidase.
  3. 3Permease.
  4. 4Lactose (allolactose).
  5. 5Operator.
  6. 61990–2003.
  7. 7VNTR.
  8. 8Winkler (1920).

Quick revision of the whole lesson

ConceptKey fact
ExperimentsGriffith → Avery → Hershey–Chase
PackagingOctamer H2A, H2B, H3, H4; 146 bp
ReplicationSemiconservative; Okazaki
TranscriptionPromoter–gene–terminator; splicing
Code64 = 61 + 3; AUG start
TranslationP, A, E; initiation–elongation–termination
RegulationLac operon; HGP 1990–2003

🎯 Most important points for the exam

  • Griffith, Avery, Hershey–Chase.
  • Nucleosome and chromatin.
  • Meselson–Stahl, Okazaki fragments.
  • Transcription, splicing.
  • Codon, mutation.
  • Steps of translation.
  • Lac operon, HGP, fingerprinting.
Study mantra

Understand the concept → draw the figure → memorise the key points → solve questions yourself.