AP EAPCET (Agriculture & Pharmacy) · Biology (Botany & Zoology)
Molecular Basis of Inheritance
DNA structure and replication, transcription, genetic code, translation, gene regulation and the human genome project.
Eight concepts that follow the information: what DNA is, how it is proved to be the gene, how it copies, how it is read into RNA and protein, how a bacterial switch turns genes on, and what the human sequence catalogue actually found.
- AP EAPCET (Agriculture & Pharmacy)
- Hard level
- 8 concepts
- 5 practice questions
1The double helix and Chargaff's equalities
DNA is a polymer of deoxyribonucleotides. Watson and Crick's double helix has two sugar-phosphate backbones with the bases projecting inward; the strands run antiparallel — one 5' \to 3', the other 3' \to 5' — and coil right-handed with about 10 base pairs per turn and 0.34 nm between neighbouring pairs.
Adenine pairs with thymine by two hydrogen bonds and guanine with cytosine by three, so a purine always faces a pyrimidine and the helix stays roughly uniform in width. That pairing is exactly what Chargaff had already measured: in double-stranded DNA, A = T and G = C. Because the strands are complementary, the sequence of one predicts the other, and genetic information is said to flow DNA \to RNA \to protein — the central dogma — with reverse transcription as the viral exception.
Figure. Not a helical coil — just the pairing rule the coil depends on. Left strand 5'\to 3' downward, right strand 3'\to 5' downward, so the partners face each other.
How the helix is built
- Backbones outsideSugar-phosphate chains form the two rails; the nitrogenous bases project into the middle.
- AntiparallelIf one strand reads 5' \to 3' left to right, its partner reads 3' \to 5' the same way.
- Base pairsA forms two H-bonds with T; G forms three with C. Purine always opposite pyrimidine.
- ChargaffIn any double-stranded sample those pairings force A = T and G = C, so purines equal pyrimidines.
Adenine from a cytosine percentage
A double-stranded DNA sample is 22% cytosine. What percentage is adenine?
- Chargaff: C = GG = 22\%
- G + C44\%
- A + T = 100\% - 44\%56\%
- A = T, so adenine28\%
Pro tip. Chargaff's equalities hold only for double-stranded DNA. A single strand can have any base composition, so the same arithmetic on an ssDNA figure is meaningless.
A researcher measures 22% cytosine in a purified single-stranded viral DNA and concludes adenine must be 28%. The mistake is that
- Cytosine never equals guanine, even in double-stranded DNA
- Chargaff's A = T and G = C apply to double-stranded DNA, not to a single strand
- Adenine should equal cytosine, not complement it
Chargaff's rules are consequences of base pairing across two strands. A lone strand has no partner forcing A = T, so a cytosine percentage tells you nothing about adenine. The first option denies the dsDNA rule itself; the third confuses pairing partners with like-with-like.
2Packaging metres of DNA into a nucleus
Take the diploid mammalian content as 6.6 \times 10^9 base pairs and the spacing as 0.34 nm per pair, and the double helix is about 2.2 metres long — far longer than a nucleus roughly 10^{-6} m across. The length has to be folded, not shortened.
In eukaryotes the fold starts with histones. Eight histone proteins form a positively charged octamer; negatively charged DNA wraps around it to make a nucleosome of about 200 bp. Nucleosomes in a row look like beads on a string under the electron microscope, then coil into thicker fibres and finally into metaphase chromosomes with help from non-histone proteins. E. coli has no nucleus: its 4.6 \times 10^6 bp genome sits in a nucleoid as large loops held by proteins.
Figure. Packaging hierarchy that fits metres of DNA into a micron-scale nucleus. Levels are named compressions, not a histone-coil drawing.
How the length is stored
- Measure the polymerMultiply base-pair count by 0.34 nm per pair to get contour length — about 2.2 m in a typical mammalian cell.
- Wrap on histonesDNA winds on a histone octamer to form a nucleosome (~200 bp).
- Beads on a stringLinked nucleosomes are the chromatin fibre seen in EM images.
- Higher coilsFurther coiling plus non-histone proteins pack the fibre into the metaphase chromosome.
| System | Organisation | Rough scale |
|---|---|---|
| E. coli | Nucleoid loops held by proteins | 4.6 \times 10^6 bp |
| Eukaryote, first fold | Nucleosome (DNA on histone octamer) | ~200 bp per bead |
| Eukaryote, next folds | Beads-on-string → fibres → chromosome | Up to metaphase condensation |
| Mammalian contour | Fully stretched double helix | ~2.2 m from 6.6 \times 10^9 bp |
Contour length of mammalian DNA
A typical mammalian cell has 6.6 \times 10^9 bp of DNA. Taking 0.34 nm between consecutive base pairs, how long is the double helix if stretched end to end?
- Spacing0.34 \times 10^{-9} m/bp
- 6.6 \times 10^9 \times 0.34 \times 10^{-9}6.6 \times 0.34
- 6.6 \times 0.342.244 m \approx 2.2 m
Pro tip. NCERT's 2.2 m uses the diploid content 6.6 \times 10^9 bp. The haploid figure quoted elsewhere in the same chapter is 3.3 \times 10^9 bp — half the length if you stretch only one genome copy.
A nucleus is about a micrometre across, yet mammalian DNA is about two metres long. The cell solves this mainly by
- Deleting most of the DNA so the polymer is shorter
- Folding the same polymer onto histones and higher-order coils
- Replacing DNA with the much shorter RNA genome
Packaging keeps the base-pair count and shortens the occupied space. Deleting DNA would lose genes; RNA is not the nuclear genome of a mammal. The nucleosome is the first folding step, not a shortening of the sequence.
3Proving the gene is DNA: Hershey and Chase
Griffith had already shown that a heat-killed smooth pneumococcus could transform a live rough strain into a virulent one, so some chemical 'transforming principle' carried heredity. Avery's group pointed at DNA, but the cleanest proof that the gene entering a cell is DNA, not protein, came from Hershey and Chase in 1952.
They labelled bacteriophages two ways: radioactive phosphorus in a medium that marks DNA, and radioactive sulfur in a medium that marks protein. After infection they sheared empty phage coats off the bacteria and asked which radioactivity went inside. Only ^{32}P — the DNA label — entered the cells and directed new phage production. Protein coats stayed outside.
Figure. Follow only the phosphorus tube: DNA label goes in with the injection; empty protein coats stay out. The sulfur tube (not drawn) leaves radioactivity outside.
How the labels decide
- Mark DNAGrow phages with ^{32}P: DNA contains phosphorus, protein does not, so only DNA is radioactive.
- Mark proteinGrow another batch with ^{35}S: protein contains sulfur, DNA does not.
- Infect and separateAllow attachment, then agitate to remove empty coats from the bacterial surface.
- Read the pelletBacterial cells carry ^{32}P inside; ^{35}S stays with the discarded coats. New phages follow the DNA.
In a Hershey–Chase run, the bacterial pellet after blending is radioactive, and the phage coats spun off are not. The phages must have been grown with
- ^{35}S, because protein is what enters the cell
- ^{32}P, because DNA is what enters the cell
- Both labels together, because both macromolecules enter equally
Radioactivity in the pellet means the labelled molecule went inside. Only DNA is labelled by ^{32}P, and only DNA enters. ^{35}S would leave radioactivity with the coats; dual labelling is not how the decisive tubes were set up.
4Semiconservative replication and the fork
Watson and Crick saw at once that complementary strands suggest a copy mechanism: each strand templates a new partner. Meselson and Stahl proved the copy is semiconservative. They grew E. coli for many generations in heavy ^{15}N so all DNA was dense, then switched to light ^{14}N. After one generation every duplex banded at intermediate density — one old heavy strand paired with one new light strand. After two generations, half the duplexes were hybrid and half were fully light. Conserved whole-molecule or fully dispersive patterns were ruled out.
At the fork, DNA polymerases add nucleotides only 5' \to 3'. The template read 3' \to 5' therefore supports continuous leading-strand synthesis, while the template oriented 5' \to 3' forces discontinuous synthesis of short fragments later joined by DNA ligase. Deoxyribonucleoside triphosphates are both substrates and energy sources. In E. coli the origin of replication starts the process; polymerases alone cannot begin a chain without a primer.
Figure. Polymerase polarity, not preference, creates the two behaviours: one new strand can follow the fork; the other must be made as fragments and sealed.
What the fork must do
- Open a forkOnly a short stretch of helix is unwound at a time; both strands template new DNA there.
- Leading strandOn the 3' \to 5' template, polymerase runs continuously 5' \to 3' toward the fork.
- Lagging strandOn the 5' \to 3' template, synthesis is discontinuous: short fragments later sealed by ligase.
- Origin requiredReplication starts at a defined origin — which is why a cloning vector must carry one.
Bands after the ^{15}N \to ^{14}N switch
Cells with fully heavy (^{15}N/^{15}N) DNA are moved into ^{14}N medium. What density classes exist after one generation and after two, if replication is semiconservative?
- Startall duplexes ^{15}N/^{15}N (heavy)
- After 1 generation: each old strand pairs with new ^{14}Nall duplexes hybrid (^{15}N/^{14}N)
- After 2 generations: hybrid templates again½ hybrid + ½ light (^{14}N/^{14}N)
Pro tip. One generation with only a hybrid band kills the conservative model (which would still show a heavy parental band). Two generations with hybrid still present kills a fully dispersive smear into one ever-lighter band.
DNA polymerase can polymerise only 5' \to 3'. At a replication fork, discontinuous synthesis therefore occurs on the template that runs
- 3' \to 5' in the direction of fork movement, so the new strand can be continuous
- 5' \to 3' in the direction of fork movement, forcing short back-filled fragments
- Either way at random, because ligase chooses the strand
Continuous synthesis needs a template read 3' \to 5' as the fork advances. The opposite template forces the polymerase to work in short stretches away from the fork opening — the discontinuous fragments ligase later joins. Ligase seals nicks; it does not pick which strand is discontinuous.
5Transcription and eukaryotic RNA processing
Transcription copies a DNA template into RNA. RNA polymerase reads the template 3' \to 5' and builds RNA 5' \to 3', so the RNA sequence matches the coding (non-template) strand except that uracil replaces thymine. A transcription unit has a promoter where polymerase binds, the structural gene, and a terminator.
Bacteria use one RNA polymerase for all RNA types; a sigma factor helps initiation and a rho factor helps termination, while the core enzyme alone can elongate. Because there is no nucleus, translation can start on an mRNA before transcription finishes. Eukaryotes split the labour: RNA polymerase II makes hnRNA (the mRNA precursor). Introns are spliced out and exons joined; a methyl-guanosine cap is added at the 5' end and 200–300 adenylates are tailed at the 3' end before the mature mRNA leaves the nucleus.
Figure. Bacterial mRNA skips this flowchart and can be translated while still being transcribed. The eukaryotic path is what creates a nucleus–cytosol boundary for gene expression.
From DNA to exportable mRNA
- Bind and openPolymerase binds the promoter and opens the helix over the start site.
- ElongateNTPs are polymerised complementary to the template; a short RNA–DNA hybrid stays in the enzyme.
- TerminateAt the terminator, RNA and polymerase release (rho assists in many bacterial cases).
- Process (eukaryotes)hnRNA is capped, spliced and tailed; only then does mRNA exit for translation.
Writing the mRNA from a template
A DNA template strand reads 3'-TAC GGA ATC-5'. Write the mRNA transcribed from it.
- Template 3'-TAC GGA ATC-5'read 3' \to 5'
- Complement with U for T5'-AUG CCU UAG-3'
- Read the codonsAUG (start), CCU (Pro), UAG (stop)
Pro tip. The mRNA looks like the coding strand with T replaced by U. If your answer matches the template instead of its complement, you transcribed the wrong strand.
In a eukaryotic nucleus, which molecule is ready to be exported for translation?
- hnRNA still containing introns, because splicing happens on the ribosome
- mRNA after capping, splicing and tailing
- The DNA coding strand itself, copied into the cytosol
Export follows processing: introns removed, 5' cap and 3' poly-A tail added. Unspliced hnRNA is not the cytosolic message, and DNA does not leave as the template for cytosolic ribosomes.
6The genetic code: triplet, degenerate, nearly universal
Translation needs a mapping from nucleotide polymer to amino-acid polymer. With four bases and twenty amino acids, George Gamow argued the codon must be at least three letters: two-letter words give only 4^2 = 16 combinations, too few, while three-letter words give 4^3 = 64. Khorana's synthetic RNAs and Nirenberg's cell-free system filled the 64-codon table.
Sixty-one codons specify amino acids and three — UAA, UAG, UGA — stop translation. The code is degenerate (many amino acids have synonyms), read continuously without commas, and nearly universal from bacteria to humans, with known exceptions in mitochondria and some protozoa. AUG both codes for methionine and starts most open reading frames.
Figure. Counting argument for a triplet code: one- and two-letter words cannot name twenty amino acids; three-letter words give sixty-four.
Features worth stating in one breath
- TripletEach codon is three nucleotides; 61 sense + 3 stop = 64.
- DegenerateMost amino acids are encoded by more than one codon.
- No commasCodons are read in a contiguous frame from the start.
- Start and stopAUG starts (and codes Met); UAA, UAG, UGA terminate.
| Property | What it means | Exam peg |
|---|---|---|
| Triplet | Three bases per codon | 4^3 = 64 words |
| Degenerate | Synonyms for many amino acids | Not ambiguous — one codon, one meaning |
| Nearly universal | Same mapping across most life | Mitochondrial exceptions exist |
| Start | AUG = Met + initiator | Dual role |
| Stop | UAA, UAG, UGA | No amino acid |
Why the codon cannot be a doublet
There are 4 RNA bases and 20 amino acids to encode. Show that a two-base codon is too small and a three-base codon is large enough in combinatorial terms.
- Doublet words: 4^216 < 20 amino acids
- Triplet words: 4^364 \geq 20 amino acids
- Sense vs stop in the real table61 sense + 3 stop
Pro tip. Combinatorics only shows a triplet is large enough. Proof that nature actually uses triplets needed the synthetic-polymer experiments — the arithmetic alone does not pick which of the 64 words mean what.
UUU codes for phenylalanine in E. coli. In a human cytosolic ribosome the same codon almost always codes for
- A different amino acid, because each species has its own code
- Phenylalanine, because the code is nearly universal
- A stop, because UUU is reserved as a terminator in animals
Near-universality is the examinable claim: UUU is Phe from bacteria to humans in the standard code. Species-specific codes are the wrong default; UUU is not a stop (UAA/UAG/UGA are). Mitochondrial exceptions exist but are not the human cytosolic rule.
7Translation: charging, ribosome, peptide bond
Translation polymerises amino acids in the order the mRNA codons dictate. Each amino acid is first activated with ATP and attached to its cognate tRNA — aminoacylation or 'charging'. Only charged tRNAs can donate amino acids on the ribosome.
The ribosome, built from rRNA and dozens of proteins, has a small subunit that engages mRNA and a large subunit with sites that hold successive tRNAs close enough for peptide-bond formation. In bacteria the 23S rRNA is the peptidyl transferase ribozyme. An open reading frame runs from a start codon to a stop codon; untranslated sequences flank that unit. When a stop codon arrives, release factors free the polypeptide.
Figure. Energy accounting sits in the first box: the peptide bond is cheap once the tRNAs are charged.
How a polypeptide is built
- Charge tRNAsAminoacyl-tRNA synthetases link each amino acid to its tRNA using ATP.
- InitiateThe small subunit finds the start; initiator tRNA sits on AUG; the large subunit joins.
- ElongateA site accepts the next charged tRNA; a peptide bond forms; the ribosome translocates one codon.
- TerminateA stop codon recruits release factors; the chain leaves and subunits separate.
Two charged tRNAs sit in the ribosome ready for the first peptide bond. The ATP that made that bond energetically favourable was spent mainly during
- Peptide-bond formation itself on the 23S ribozyme
- Aminoacylation of the tRNAs before they arrived
- Termination, when release factors hydrolyse GTP
NCERT stresses that amino acids are activated and attached to tRNA first; bringing two charged tRNAs together then favours peptide-bond formation. The ribozyme accelerates the bond; it is not where the activation ATP was spent. Termination GTP is later and unrelated to the first bond's activation.
8The lac operon: inducible negative control
Jacob and Monod's lac operon is the textbook inducible switch in E. coli. One regulatory gene i is transcribed constitutively to make a repressor. Three structural genes share a promoter and operator: z (β-galactosidase, splits lactose), y (permease, imports β-galactosides), and a (transacetylase). Together they let the cell use lactose.
Without inducer the repressor binds the operator and blocks RNA polymerase. Lactose (or allolactose) inactivates the repressor, so polymerase can transcribe z, y and a — regulation of enzyme synthesis by the substrate itself. Glucose or galactose cannot act as inducers. Repressor control is negative regulation; NCERT notes positive control also exists but leaves it aside at this level.
Figure. The map order is examinable: i then promoter, operator, and the three structural genes. Inducer does not bind the operator — it disables the repressor protein.
Off, then on
- Always a little repressorThe i gene makes repressor continuously; a low basal lac expression still lets the first lactose molecules in via permease.
- Operator blockedRepressor on the operator prevents polymerase from reading z, y, a.
- Inducer arrivesLactose/allolactose binds repressor and pulls it off the operator.
- Polycistronic mRNAOne transcript covers all three structural genes so the pathway enzymes appear together.
E. coli is grown with glucose as the only sugar. The lac structural genes stay off mainly because
- Glucose binds the repressor and locks it onto the operator
- Glucose is not an inducer, so the repressor keeps blocking the operator
- The i gene is repressed whenever glucose is present
Inducer means lactose/allolactose. Glucose neither inactivates the lac repressor nor, in the NCERT account you need here, turns the operon on. The i gene is constitutive, not glucose-repressed in this story. (Catabolite repression is the 'positive regulation' NCERT sets aside.)
Notes
- DNA is a double helix (Watson-Crick) with antiparallel strands and complementary base pairing A=T, G≡C; Chargaff's rule states A = T and G = C.
- DNA replication is semiconservative (proved by Meselson and Stahl); DNA polymerase synthesises 5'→3', so the leading strand is continuous and the lagging strand forms Okazaki fragments.
- Transcription: RNA polymerase reads the 3'→5' template to make mRNA; in eukaryotes the hnRNA is capped, tailed and spliced (introns removed).
- The genetic code is triplet, degenerate, universal and non-overlapping; AUG is the start codon (methionine) and UAA, UAG, UGA are stop codons.
- Translation: the ribosome reads mRNA codons while tRNA delivers amino acids; the *lac* operon of *E. coli* illustrates inducible gene regulation.
Formulas
- Chargaff's rule: A = T, G = C (so purines = pyrimidines)
- Base pairing: A=T (2 H-bonds), G≡C (3 H-bonds)
- Genetic code: 64 codons (61 sense + 3 stop); start codon AUG
- Replication is semiconservative; synthesis is 5'→3'
- Human genome ≈ 3.2 × 10⁹ base pairs, ~20,000–25,000 genes
Exam traps & shortcuts
- Central dogma: DNA → (transcription) → RNA → (translation) → protein, with reverse transcription (RNA→DNA) in retroviruses.
- The lagging strand is discontinuous (Okazaki fragments) because polymerase only works 5'→3'.
- The lac operon is switched ON by lactose, which inactivates the repressor — a form of negative regulation.
Reference tables
Name the polymerisation direction and the main polymerase before you open a mechanism question.
| Process | Template → product | Polymerisation | Key enzyme / note |
|---|---|---|---|
| Replication | DNA → DNA | 5' \to 3' only | DNA polymerases; ligase seals lagging fragments |
| Transcription | DNA → RNA | RNA 5' \to 3' | RNA polymerase (σ/ρ in bacteria) |
| Translation | mRNA → protein | N → C peptide | Ribosome; 23S rRNA is the ribozyme in bacteria |
| Reverse transcription | RNA → DNA | Retroviral exception | Central-dogma reverse flow |
These are the figures the chapter asks you to recognise, not to derive.
| Observation | NCERT figure |
|---|---|
| Human genome size | 3164.7 million bp |
| Estimated gene count | About 30,000 (far below older 80,000–140,000 guesses) |
| Protein-coding fraction | Less than 2% |
| Identity across humans | 99.9% of bases the same |
| Largest gene cited | Dystrophin, 2.4 million bases |
| Chromosome extremes | Chr 1 most genes (2968); Y fewest (231) |
| SNPs catalogued | About 1.4 million locations |
| Fingerprinting probe class | VNTR mini-satellites (Jeffreys) |
Recap
Read only this the night before.
- Chargaff
- In dsDNA, A = T and G = C. A single strand ignores those equalities.
- Length
- 6.6 \times 10^9 bp × 0.34 nm ≈ 2.2 m of mammalian DNA — folded on nucleosomes, not shortened.
- Hershey–Chase
- ^{32}P (DNA) enters the bacterium; ^{35}S (protein) stays in the coat.
- Replication
- Semiconservative; polymerase only 5' \to 3', so one strand is discontinuous and ligase joins the fragments.
- Transcription
- mRNA matches the coding strand with U for T. Eukaryotic hnRNA is capped, spliced and tailed before export.
- Code
- Triplet; 61 sense + 3 stop; degenerate; nearly universal; AUG starts and codes Met.
- lac
- Repressor from i blocks O until lactose/allolactose inactivates it — negative inducible control. Glucose is not an inducer.
- HGP / fingerprint
- ~3.16 billion bp, ~30,000 genes, <2% coding; VNTRs make individual band patterns (except identical twins).
Practise Molecular Basis of Inheritance
Reading is free and needs no account. Practice, mocks and progress live in the app.
- 5 exam-style questions on this topic, with explanations
- A 5-question practice set that ends the chapter
- Timed mocks scored with the real marking scheme
- Readiness tracked per topic, kept on your device