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Class 12BiologyBiologyChapter 5

Molecular Basis of Inheritance

≈ 65 min read · 16 Prelims and 1 Mains questions from this chapter
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Start here: why this chapter matters

In simple wordsThe previous chapter treated genes like cards in a hand without asking what the cards are made of. This chapter opens the pack. A gene is a stretch of a long chemical thread, DNA, written in a four-letter alphabet. The cell photocopies the thread before dividing (replication), makes working copies of single pages in RNA (transcription), and uses a dictionary of three-letter words to turn each page into a protein (translation).

This is the fifth chapter of NCERT's Class 12 Biology textbook (pages 79–109 of the 2026-27 reprint), the second of Unit VII, Genetics and Evolution. It has ten sections — the DNA (5.1), the search for genetic material (5.2), the RNA world (5.3), replication (5.4), transcription (5.5), the genetic code (5.6), translation (5.7), regulation of gene expression (5.8), the Human Genome Project (5.9) and DNA fingerprinting (5.10) — with Figures 5.1–5.16, an unnumbered central-dogma sketch, Table 5.1 (the codon chart), a summary and 14 exercise questions. Every DNA and RNA sequence on this page is generated and checked in the source file, so you can trust the pairings.

  • Neighbouring chapters: the previous chapter, Principles of Inheritance and Variation, gave the rules (Mendel, linkage, sickle-cell as a point mutation); this one gives the molecule. The next, Evolution, uses the RNA world, mutation and DNA polymorphism; the later biotechnology chapters use restriction enzymes, vectors, PCR and the origin of replication met here. Class 11 Biomolecules introduced nucleotides.
  • Prelims — science: UPSC has asked about tRNA as the adapter (1985), base pairing (1985), genes acting through protein synthesis (1986), complementarity as DNA's key feature (2001), the operon concept (1998), genetics pioneers including Khorana and Kornberg (2001), James Watson (2008), DNA fingerprinting (1991, 1997, 2000), genome sequencing (1997, 2017, 2026), the transcriptome (2016), microsatellite DNA (2023) and viruses that infect bacteria (2025).
  • GS3 Science and Technology: genome sequencing, genomic medicine, forensic DNA and bioinformatics all start here. GS2: DNA testing and privacy (Mains 2024) — the science of a paternity test is in Section 10.
WordWhat it means here
Nucleoside / nucleotideBase + sugar (e.g. adenosine) / nucleoside + phosphate (the monomer of DNA or RNA)
Purine / pyrimidineTwo-ring bases A and G / one-ring bases C, T (DNA) and U (RNA)
Base pair (bp)A paired with T, or G with C, across the two strands; DNA length is counted in bp
5′ and 3′ endsThe free phosphate end and the free 3′-OH end of a strand; strands are written 5′→3′
Nucleosome / chromatinDNA wound round eight histones / the whole DNA–protein thread in the nucleus
Replication / transcription / translationDNA → DNA / DNA → RNA / RNA → protein
Template / coding strandThe strand read by RNA polymerase (3′→5′) / the other strand, whose sequence the RNA copies
Exon / intronPart of a gene kept in mature RNA / part removed by splicing
Codon / anticodonThree-base word in mRNA / its complementary three bases on tRNA
OperonA set of bacterial genes under one promoter and operator, switched together
Polymorphism / VNTRA DNA variant common in a population (frequency above 0.01) / a variable number of tandem repeats
Reading NCERT's chapter opening (page 79)The opening page carries a small cover image titled ‘Watson–Crick model for semiconservative DNA replication’, a QR code, the list of ten sections, and an introduction: Mendel's ‘factors’ were unexplained in his time; over the next hundred years DNA was shown to be the genetic material for most organisms, while RNA acts as genetic material in some viruses and otherwise as messenger, adapter, structural and sometimes catalytic molecule. The point: the chapter moves from structure (5.1) to proof (5.2–5.3) to processes (5.4–5.8) to genomes (5.9–5.10).

At a glance — the six ideas to carry away

  • DNA: polymer of deoxyribonucleotides (base + deoxyribose + phosphate); double helix (Watson and Crick, 1953, from Wilkins and Franklin's X-ray data): antiparallel strands, A=T (2 H-bonds), G≡C (3), purine opposite pyrimidine, right-handed, pitch 3.4 nm, ~10 bp per turn, 0.34 nm per bp. Chargaff: A/T = G/C = 1.
  • Packaging: 6.6 × 10⁹ bp ≈ 2.2 m per diploid cell; DNA (−) wraps a histone octamer (+; lysine, arginine) → nucleosome (~200 bp) → beads-on-string → chromatin fibre (NHC proteins) → chromosome. Euchromatin active, heterochromatin inactive. Central dogma: DNA → RNA → protein.
  • Genetic material: Griffith 1928 (transforming principle, Streptococcus pneumoniae); Avery–MacLeod–McCarty 1944 (only DNase stops transformation); Hershey–Chase 1952 (³²P DNA enters, ³⁵S protein does not). RNA is the genetic material in TMV, Qβ; RNA world: RNA first, DNA more stable.
  • Replication: semiconservative (Meselson–Stahl 1958, ¹⁵N/¹⁴N, CsCl; Taylor, Vicia faba). DNA polymerase works 5′→3′: continuous on one template, discontinuous on the other (joined by ligase); needs an origin. Transcription: template strand 3′→5′; promoter, structural gene, terminator; σ starts, ρ stops.
  • Eukaryotes: RNA pol I (rRNA 28S, 18S, 5.8S), II (hnRNA), III (tRNA, 5S rRNA, snRNA); split genes — splicing, capping, tailing. Code: triplet, 64 = 61 + 3 stop (UAA, UAG, UGA), degenerate, no punctuation, nearly universal, AUG = Met/start (Gamow, Khorana, Nirenberg, Ochoa). tRNA = adapter; ribosome's rRNA is a ribozyme.
  • lac operon (Jacob and Monod): i (repressor), z (β-galactosidase), y (permease), a (transacetylase); lactose = inducer; negative regulation. HGP 1990–2003: 3164.7 Mbp, ~30,000 genes (NCERT), <2% codes protein, 99.9% identical, 1.4 million SNPs. DNA fingerprinting (Alec Jeffreys): VNTR mini-satellites, Southern blot; forensics, paternity.

1. DNA: the polynucleotide chain and the double helix

In simple wordsPicture a ladder twisted into a spiral staircase. The two side rails are made of sugar and phosphate; each rung is a pair of bases. The rungs only fit as A with T and G with C, so if you know one rail's letters you can write the other's — the way a key tells you the shape of its lock.

Size. DNA length is counted in nucleotides or base pairs and is characteristic of an organism: bacteriophage φX174 has 5386 nucleotides, bacteriophage lambda 48,502 bp, Escherichia coli 4.6 × 10⁶ bp, and the haploid human genome 3.3 × 10⁹ bp.

The chain. A nucleotide has three parts: a nitrogenous base, a pentose sugar (ribose in RNA, deoxyribose in DNA) and a phosphate. Bases are purines (adenine, guanine) or pyrimidines (cytosine, thymine, uracil); cytosine is in both nucleic acids, thymine only in DNA, uracil in its place in RNA. A base joined to the 1′ carbon of the sugar by an N-glycosidic bond makes a nucleoside (adenosine, guanosine, cytidine, uridine; deoxy- forms in DNA); a phosphate on the 5′ carbon makes a nucleotide. Nucleotides join by 3′–5′ phosphodiester bonds into a chain whose backbone is sugar–phosphate, with bases projecting from it; one end has a free 5′ phosphate (the 5′ end), the other a free 3′-OH (the 3′ end). RNA has an extra –OH at the 2′ carbon of every ribose, and uracil instead of thymine (5-methyl uracil).

Reading NCERT's Figure 5.1 (page 80)A single strand drawn left to right: phosphate circles, CH₂ links and pentagon-shaped sugars alternate, with bases A, T, G and C hanging below; ‘5′ phosphate’ is labelled at the left end and ‘3′ hydroxyl’ at the right. The point: a strand has direction — it starts with a phosphate on carbon 5′ and ends with an –OH on carbon 3′.

The discovery. Friedrich Miescher (NCERT spells it ‘Meischer’) first identified DNA as an acidic substance in the nucleus in 1869 and called it ‘nuclein’. Its structure stayed unknown until 1953, when James Watson and Francis Crick proposed the double helix from X-ray diffraction data of Maurice Wilkins and Rosalind Franklin. A hallmark was base pairing, supported by Erwin Chargaff's observation that in double-stranded DNA the ratios A : T and G : C are constant and equal to one. Because the strands are complementary, one strand's sequence predicts the other's, and each could serve as a template for an identical daughter molecule.

  • Two polynucleotide chains; sugar–phosphate backbone outside, bases inside.
  • The chains are antiparallel: one runs 5′→3′, the other 3′→5′.
  • Bases pair by hydrogen bonds: A with T by two, G with C by three. A purine always faces a pyrimidine, which keeps the distance between strands almost uniform.
  • The chains coil right-handedly; the pitch is 3.4 nm with roughly 10 bp per turn, so about 0.34 nm between base pairs.
  • Base pairs stack on each other, which, with the H-bonds, stabilises the helix.
5′3′strand read 5′ → 3′ATPPTAPPGCPPCGPPTAPPATPPGCPPCG3′-OH5′partner strand also 5′ → 3′, in the opposite direction (anti-parallel)A (purine, 2 rings)G (purine)T (pyrimidine, 1 ring)C (pyrimidine)deoxyribose sugarPphosphate (backbone)dashes = H-bonds: A=T two, G≡C three · this piece: A 4 = T 4, G 4 = C 4
A stretch of double-stranded DNA drawn as a ladder — our schematic for NCERT Figures 5.2 and 5.3 (page 82); the bottom strand is generated as the complement of the top one and Chargaff's ratios are asserted in the source file.
Reading NCERT's Figures 5.2 and 5.3 (page 82)Figure 5.2 draws two strands of four nucleotides each, the upper running 5′ to 3′ and the lower 3′ to 5′, with A–T joined by two dashed lines and G–C by three, labelled ‘hydrogen bonds’. Figure 5.3 is a coloured ribbon model of the right-handed helix with ‘Base pairs’ and ‘Sugar phosphate backbone’ labelled and a key pairing adenine with thymine and guanine with cytosine. The point: antiparallel strands, 2 versus 3 H-bonds, and a helix whose rungs are all the same width.

The central dogma. Soon after, Francis Crick proposed that genetic information flows DNA → RNA → protein. In some viruses information flows from RNA to DNA — reverse transcription (NCERT's in-text question).

DNAmRNAProteintranscriptiontranslationreplicationreverse transcription — RNA → DNA in some viruses (e.g. retroviruses)Crick’scentral dogma:information flowsDNA → RNA → protein
The central dogma with its exception — our schematic for NCERT's unlabelled sketch on page 82.
Reading NCERT's central-dogma sketch (page 82)An unnumbered diagram: a circular arrow labelled ‘replication’ around DNA, an arrow ‘transcription’ to mRNA and an arrow ‘translation’ to protein, captioned ‘Central dogma’. The point: three processes, three arrows — and the text adds the reverse arrow for some viruses.
Common confusions
  • Nucleoside = base + sugar; nucleotide = base + sugar + phosphate. Adenine, thymine, uracil and cytosine are bases; cytidine and guanosine are nucleosides (exercise 1).
  • Purines: A, G (two rings). Pyrimidines: C, T, U (one ring). Mnemonic: ‘pure As Gold’.
  • Chargaff's rule holds for double-stranded DNA across both strands: A = T, G = C, purines = pyrimidines. It does not say A + T = G + C, and it does not hold within one strand or in single-stranded genomes.
  • Pitch 3.4 nm per turn; 0.34 nm per base pair — do not swap them.
  • RNA differs from DNA chemically in two ways: ribose with a 2′-OH, and uracil instead of thymine.
  • DNA is right-handed in the Watson–Crick model.
Beyond NCERT
  • Watson, Crick and Wilkins shared the 1962 Nobel Prize in Physiology or Medicine; Rosalind Franklin had died in 1958, and Nobel Prizes are not awarded posthumously.
  • A G≡C pair, with three hydrogen bonds, holds more strongly than A=T, so DNA richer in G and C separates (‘melts’) at a higher temperature.
  • Reverse transcription is carried out by the enzyme reverse transcriptase, found in retroviruses such as HIV (discovered by Temin and Baltimore, 1970).

2. Packaging: from two metres of DNA to a chromosome

In simple wordsEach of your cells holds about two metres of DNA in a nucleus a few thousandths of a millimetre across — like stuffing kilometres of thread into a tennis ball. The cell solves it the way you store a long rope: wind it on spools (histones), coil the spools, and coil the coils.

Multiplying the number of base pairs in a typical mammalian cell by the spacing between them gives 6.6 × 10⁹ bp × 0.34 × 10⁻⁹ m/bp ≈ 2.2 metres — far longer than the nucleus (about 10⁻⁶ m). In prokaryotes such as E. coli, which lack a defined nucleus, the negatively charged DNA is held by positively charged proteins in a region called the nucleoid, organised in large loops.

In eukaryotes, histones — basic proteins rich in the positively charged amino acids lysine and arginine — form a unit of eight, the histone octamer. Negatively charged DNA wraps around it to form a nucleosome, which typically contains 200 bp. Nucleosomes are the repeating units of chromatin, seen under the electron microscope as ‘beads-on-string’. Chromatin is packed into fibres that coil and condense at metaphase into chromosomes; this higher-level packing needs non-histone chromosomal (NHC) proteins. Loosely packed, lightly staining euchromatin is transcriptionally active; densely packed, dark-staining heterochromatin is inactive.

From the double helix to a metaphase chromosome (schematic, not to scale)DNA double helix(negatively chargedphosphates)H1Nucleosome: DNA (~200 bp)around a histone octamer (8)histones rich in Lys, Arg (+)‘Beads-on-string’chromatin (seenunder the EM)Coiled chromatin fibre(higher packing needsnon-histone (NHC) proteins)Metaphase chromosome(most condensed)In an interphase nucleus:euchromatin: loose,stains light, activeheterochromatin: dense,stains dark, inactive6.6 × 10⁹ bp × 0.34 nm ≈ 2.2 m of DNA≈ 3.3 × 10⁷ nucleosomes per cell(our calculation)
Levels of DNA packing — our schematic for NCERT Figure 5.4 (page 83); the 2.2 m and nucleosome count are computed in the source file.
Reading NCERT's Figures 5.4a and 5.4b (page 83)Figure 5.4a draws a green sphere labelled ‘Histone octamer’ (‘Core of histone molecules’) with a pink DNA tube wrapped around it and a rod labelled ‘H1 histone’ across the DNA. Figure 5.4b is an electron micrograph of chromatin showing dark beads strung along a thread. The point: the nucleosome is the basic unit of packing, and it is visible under the EM.
NCERT's in-text calculationWorkingAnswer
Length of DNA in a mammalian cell6.6 × 10⁹ bp × 0.34 × 10⁻⁹ m≈ 2.2 m
Base pairs in E. coli DNA 1.36 mm long1.36 × 10⁻³ m ÷ 0.34 × 10⁻⁹ m4.0 × 10⁶ bp
Nucleosomes in a mammalian cell6.6 × 10⁹ bp ÷ 200 bp≈ 3.3 × 10⁷
Common confusions
  • Histones are basic and positive (lysine, arginine); DNA is acidic and negative (phosphates). Opposite charges hold the nucleosome together.
  • Octamer = 8 histone molecules; H1 sits outside it. A nucleosome holds about 200 bp.
  • Euchromatin = loose, light, active; heterochromatin = dense, dark, inactive.
  • Prokaryotes have a nucleoid, not a nucleus, and no nucleosomes of this kind.
  • NCERT inconsistency: 1.36 mm of E. coli DNA gives 4.0 × 10⁶ bp, but page 80 quotes 4.6 × 10⁶ bp (which would be about 1.56 mm). For the in-text question show the working and write 4 × 10⁶ bp; for ‘genome size of E. coli’ write 4.6 × 10⁶ bp.

4. Replication: semiconservative copying

In simple wordsUnzip a zip-fastener and use each half as a guide to build a new matching half. You end with two complete zips, each made of one old half and one new half. That is semiconservative replication — and Meselson and Stahl proved it by making the old halves heavier than the new ones.

Watson and Crick saw at once that specific pairing ‘suggests a possible copying mechanism’ (1953): the strands separate and each acts as a template for a new complementary strand, so each daughter molecule has one parental and one new strand — semiconservative replication.

Reading NCERT's Figure 5.6 (page 88)A double helix whose base pairs are lettered (GC, AT, TA, CG…) opens from the bottom into two daughter helices; in each, one ribbon is the original strand and the other a newly made one in a different colour, with 5′ and 3′ marked. The point: each daughter keeps one old strand, and the new strand's bases are fixed by pairing.

The experimental proof (1958). Matthew Meselson and Franklin Stahl grew E. coli for many generations on ¹⁵NH₄Cl, so all DNA contained the heavy, non-radioactive isotope ¹⁵N, and could be told from normal DNA by caesium chloride (CsCl) density-gradient centrifugation. They moved the cells to ¹⁴NH₄Cl and sampled DNA as the cells divided (every 20 minutes). After one generation all DNA was of hybrid (intermediate) density; after two, half was hybrid and half light. Similar experiments with radioactive thymidine on Vicia faba (faba bean) by Taylor and colleagues (1958) showed that chromosomes too replicate semiconservatively.

Meselson and Stahl (1958): E. coli moved from ¹⁵N to ¹⁴N medium; DNA spun in a CsCl density gradientlight ¹⁴N/¹⁴Nhybrid ¹⁵N/¹⁴Nheavy ¹⁵N/¹⁵N0 min (in ¹⁵N)all heavy20 min · generation Iall hybrid40 min · generation II1/2 hybrid : 1/2 light80 min · generation IV1/8 hybrid : 7/8 lightto ¹⁴NBars: red = ¹⁵N (old) strand, pale = ¹⁴N (new) strand · band thickness ∝ share of DNA · the 80-min column answers NCERT’s question (our simulation)
Meselson and Stahl's results, generated by a strand-by-strand simulation in the source file — our schematic for NCERT Figure 5.7 (page 89), extended to 80 minutes.
Reading NCERT's Figure 5.7 (page 89)A red ¹⁵N-DNA helix (‘Heavy’) passes by an arrow marked ‘20 min’ to two helices each of one red ¹⁵N and one green ¹⁴N strand (‘Generation I’, ‘Hybrid’), and by ‘40 min’ to four helices, two hybrid and two all-green (‘Generation II’, ‘Light’ and ‘Hybrid’). Under each stage a centrifuge tube shows the band position, with an arrow for ‘Gravitational force’. The point: the hybrid band after one generation rules out conservative replication; the half-light pattern after two confirms semiconservative.
⤵ Meselson and Stahl's logic
Grow in ¹⁵Nmany generations; all DNA heavy (¹⁵N/¹⁵N)
Switch to ¹⁴Nnew strands are now light
Generation I (20 min)one band, hybrid ¹⁵N/¹⁴N — not one heavy plus one light band, so not conservative
Generation II (40 min)hybrid and light in equal amounts — each molecule keeps one old strand
80 min (4 generations)2 of 16 molecules hybrid, 14 light: 1/8 hybrid, 7/8 light (in-text question)

Machinery and enzymes. The main enzyme is DNA-dependent DNA polymerase, which uses a DNA template to join deoxynucleotides. It must be fast and accurate: E. coli's 4.6 × 10⁶ bp are copied in about 18 minutes (compare the human diploid 6.6 × 10⁹ bp), and mistakes become mutations. Replication is energetically expensive: deoxyribonucleoside triphosphates act both as substrates and as the energy source (their two terminal phosphates are high-energy, as in ATP).

Long DNA cannot be unwound along its whole length, so copying happens in a small opening, the replication fork. Polymerases add nucleotides only in the 5′→3′ direction. So on the template with polarity 3′→5′ synthesis is continuous; on the template with polarity 5′→3′ it is discontinuous, and the fragments are joined by DNA ligase. Polymerases cannot start on their own, and replication begins only at a definite origin of replication — which is why a piece of DNA propagated in recombinant-DNA work needs a vector that supplies an origin. In eukaryotes replication takes place in the S phase; if the cell fails to divide after replication, polyploidy results.

5′3′3′5′Parental (template) strandsunwound at the fork5′Continuous synthesis(template 3′→5′;the ‘leading’ strand)5′Discontinuous synthesis(template 5′→3′): shortpieces, each made 5′→3′away from the forkgaps sealedby DNA ligaseDNA-dependent DNA polymerase adds nucleotides only 5′ → 3′ · fork opens upward (schematic)
A replication fork — our schematic for NCERT Figure 5.8 (page 91).
Reading NCERT's Figure 5.8 (page 91)Two blue parental strands, marked 5′ and 3′ at the top, split into a Y labelled ‘Template DNA (parental strands)’. Along the left arm a single red arrow runs toward the fork (‘Continuous synthesis’); along the right arm short red arrows point away from it (‘Discontinuous synthesis’); the new strands are labelled ‘Newly synthesised strands’ with their own 5′ and 3′ ends. The point: the 5′→3′ rule forces one strand to be made in pieces.
Number (NCERT)Check (our calculation)
E. coli: 4.6 × 10⁶ bp in 18 min ‘≈ 2000 bp per second’4.6 × 10⁶ ÷ 1080 s ≈ 4259 bp/s in all; with two forks moving out from one origin, ≈ 2130 bp/s per fork — so NCERT's figure is a per-fork rate
Meselson–Stahl at 80 minutes80 ÷ 20 = 4 generations → 2 hybrid : 14 light = 1/8 : 7/8
Common confusions
  • Semiconservative = each daughter has one old and one new strand. Conservative (both old together) and dispersive models were ruled out by Meselson–Stahl.
  • ¹⁵N is a heavy, stable isotope separated by density (CsCl), not detected by radioactivity.
  • DNA polymerase synthesises only 5′→3′. Continuous synthesis on the 3′→5′ template; discontinuous on the 5′→3′ template; ligase joins the pieces.
  • dNTPs are both substrate and energy source.
  • NCERT number to handle: ‘≈ 2000 bp per second’ is about right per replication fork, not for the whole chromosome (about 4300 bp/s from the numbers given).
  • The two original ¹⁵N strands survive in exactly two hybrid molecules in every later generation.
Beyond NCERT
  • E. coli has a single origin (oriC); replication runs in both directions from it, giving two forks.
  • The continuously made strand is called the leading strand, the other the lagging strand, and its pieces Okazaki fragments. Helicase unwinds the helix, topoisomerase relieves twisting, and primase lays down a short RNA primer because DNA polymerase cannot start a new chain.
  • Arthur Kornberg isolated the first DNA polymerase, from E. coli (1956), and shared the 1959 Nobel Prize with Severo Ochoa.

5. Transcription: DNA to RNA

In simple wordsYou do not photocopy a whole library to use one recipe; you copy one page. Transcription copies one gene, from one strand, into RNA. The promoter is the ‘start reading here’ sign, the terminator the ‘stop’ sign. In eukaryotes the copy is also edited — the irrelevant passages (introns) are cut out before it leaves the nucleus.

Transcription copies genetic information from one strand of DNA into RNA, again by complementarity, except that adenine now pairs with uracil. Unlike replication, only a segment of DNA and only one strand are copied. Why not both? The two RNAs would differ in sequence (complementary is not identical), so one DNA segment would code for two different proteins; and two complementary RNAs made together would pair into double-stranded RNA that could not be translated.

The transcription unit has three regions: a promoter, the structural gene and a terminator. Since DNA-dependent RNA polymerase also works only 5′→3′, the strand with polarity 3′→5′ is the template strand. The other strand, 5′→3′, has the same sequence as the RNA (with T for U) and is displaced; though it codes for nothing, it is called the coding strand, and all reference points are given relative to it. The promoter lies upstream (towards the 5′ end of the coding strand) and binds RNA polymerase — its position defines which strand is template; swapping promoter and terminator would swap the definitions. The terminator lies downstream (3′) and usually ends transcription. Further regulatory sequences may lie upstream or downstream.

A transcription unit (polarity named by the coding strand)3′5′5′3′PromoterTerminatorstart sitetemplate strand (3′→5′) — read by RNA polymerasecoding strand (5′→3′) — same sequence as the RNA, T for Ustructural geneupstream (5′ side)downstream (3′ side)coding strand5′-ATGGCCAAATTTGGCTAA-3′template strand3′-TACCGGTTTAAACCGATT-5′mRNA5′-AUGGCCAAAUUUGGCUAA-3′mRNA = template’s complement (A→U, T→A, G→C, C→G) = coding strand with U in place of T · codons AUG GCC AAA UUU GGC UAA
A transcription unit and one gene's three sequences, all generated from the coding strand in the source file — our schematic for NCERT Figure 5.9 (page 92).
Reading NCERT's hypothetical sequence and Figure 5.9 (page 92)NCERT prints a 24-base pair: template 3′-ATGCATGC…-5′ over coding 5′-TACGTACG…-3′, and asks for the RNA (answer: 5′-UACGUACGUACGUACGUACGUACG-3′ — the coding strand with U for T; checked in the source file). Figure 5.9 draws two lines, the upper ‘Template strand’ arrowed 3′←5′ and the lower ‘Coding strand’ 5′→3′, with a red ‘Promoter’ box at the left, a bent arrow at the ‘Transcription start site’, ‘Structural gene’ in the middle and a blue ‘Terminator’ box at the right. The point: read the template, copy the coding strand.

Transcription unit and the gene. A gene is the functional unit of inheritance, but hard to define as a DNA stretch: DNA coding for tRNA or rRNA is also a gene. A cistron is a DNA segment coding for one polypeptide; structural genes are monocistronic (mostly eukaryotes) or polycistronic (mostly bacteria). Eukaryotic genes are split: coding or expressed sequences, exons, appear in mature RNA; intervening sequences, introns, do not. Promoters and regulatory sequences also affect inheritance, so they are loosely called regulatory genes even though they code for no RNA or protein.

In bacteria three major RNAs — mRNA (template), tRNA (brings amino acids and reads the code) and rRNA (structural and catalytic) — are all made by a single DNA-dependent RNA polymerase. It binds the promoter (initiation), uses nucleoside triphosphates to build RNA by complementarity while opening the helix (elongation; only a short stretch of RNA stays bound), and at the terminator the RNA and polymerase fall off (termination). The polymerase itself can only elongate: it associates briefly with the sigma (σ) factor to initiate and the rho (ρ) factor to terminate. Bacterial mRNA needs no processing and there is no nuclear membrane, so translation can begin before transcription ends — the two are coupled.

Reading NCERT's Figure 5.10 (page 93)Three rows of a red-and-blue DNA helix. ‘Initiation’: an orange RNA polymerase carrying a σ (sigma factor) sits at a yellow promoter. ‘Elongation’: the polymerase opens a bubble and a yellow RNA strand emerges, σ having left; a green terminator lies ahead. ‘Termination’: the RNA is released and the polymerase, now with ρ (‘Rho factor’), leaves the DNA. The point: one enzyme, two helpers — σ to start, ρ to stop.

In eukaryotes there are two extra complexities. (i) At least three RNA polymerases in the nucleus (besides those of organelles) divide the work, as in the table. (ii) The primary transcript contains exons and introns and is non-functional; splicing removes introns and joins exons in order. The hnRNA is also capped — an unusual nucleotide, methyl guanosine triphosphate, is added to the 5′ end — and tailed — 200–300 adenylate residues are added to the 3′ end without a template. The fully processed RNA, now mRNA, leaves the nucleus. Split genes are probably an ancient feature, and splicing reflects the dominance of the RNA world.

Eukaryotic nuclear enzymeTranscribes
RNA polymerase IrRNAs — 28S, 18S and 5.8S
RNA polymerase IIthe precursor of mRNA, heterogeneous nuclear RNA (hnRNA)
RNA polymerase IIItRNA, 5S rRNA and snRNAs (small nuclear RNAs)
Gene (DNA)promoterexon 1intronexon 2intronexon 3terminatortranscription by RNA polymerase IIhnRNA(primary transcript)capping (5′) and tailing (3′)Capped + tailedcapAAAA…Apoly-A 200–300splicing: introns cut out, exons joined in ordermRNA(to cytoplasm)exon 1exon 2exon 3AAAA…Aintrons removed (degraded)
From split gene to mature mRNA — our schematic for NCERT Figure 5.11 (page 94); exon lengths are summed in the source file.
Reading NCERT's Figure 5.11 (page 94)A helix being transcribed by an orange polymerase releases a yellow-and-orange RNA. Below, the RNA gains a ‘Cap’ (mGppp) at its 5′ end (‘Capping’), is labelled with ‘Exon’ and ‘Intron’ segments, undergoes ‘RNA splicing’ (introns looped out as small circles), gets a green ‘Poly A tail’ (‘Polyadenylation’), and ends as ‘Messenger RNA’ of exons only. The point: cap, tail and splice turn hnRNA into mRNA.
⤵ Making a eukaryotic mRNA
TranscriptionRNA polymerase II copies the whole split gene into hnRNA (exons + introns)
Cappingmethyl guanosine triphosphate added at the 5′ end
Tailing200–300 adenylates added at the 3′ end, without a template
Splicingintrons removed; exons joined in a defined order
Exportmature mRNA leaves the nucleus for translation in the cytoplasm
Common confusions
  • Template strand (3′→5′) is read; coding strand (5′→3′) has the RNA's sequence — the RNA is the coding strand with U for T, and the complement of the template (exercise 4: coding 5′-ATGC…-3′ → mRNA 5′-AUGC…-3′).
  • Promoter = upstream, binds RNA polymerase; terminator = downstream. Directions are named on the coding strand.
  • σ initiates, ρ terminates; the core polymerase only elongates.
  • Pol I → big rRNAs (28S, 18S, 5.8S); pol II → hnRNA/mRNA; pol III → tRNA, 5S rRNA, snRNA. 5S is the odd one out — made by pol III.
  • Exons stay in mature RNA, introns are removed — even an exon can be non-coding at its ends. Bacteria: no introns, no capping, coupled transcription–translation.
  • Monocistronic = eukaryotes mostly; polycistronic = bacteria (e.g. the lac operon).
Beyond NCERT
  • The cap is usually written 7-methylguanosine (m⁷G); it protects mRNA and helps ribosomes bind. Splicing is carried out by the spliceosome, built of snRNAs and proteins.
  • Alternative splicing lets one gene give several proteins by joining exons in different combinations — one reason humans have far fewer genes than once expected.
  • Split genes were discovered in 1977 (Phillip Sharp and Richard Roberts; Nobel Prize 1993).

6. The genetic code, mutations and tRNA

In simple wordsDNA has only four letters but proteins use twenty amino acids. One letter per amino acid gives 4 words; two letters give 16 — not enough. Three letters give 64, more than enough. So the code is written in three-letter words (codons), read one after another with no spaces, and several words can mean the same amino acid — like ‘colour’ and ‘color’.

Copying nucleic acid into nucleic acid is easy to picture by complementarity, but no complementarity exists between nucleotides and amino acids. Evidence that changes in genetic material changed amino acids led to the idea of a genetic code. Deciphering it needed physicists, organic chemists, biochemists and geneticists. The physicist George Gamow argued that 4 bases coding for 20 amino acids need combinations, and that a triplet code (4³ = 64 codons) would suffice. Har Gobind Khorana's chemical method made RNAs of defined base combinations (homopolymers and copolymers); Marshall Nirenberg's cell-free system for protein synthesis finally cracked the code; Severo Ochoa's enzyme, polynucleotide phosphorylase, polymerised RNA of defined sequence without a template. The result was a checkerboard of codons.

Second baseFirst baseThirdbaseUCAGUUUUPheUUCPheUUALeuUUGLeuUCUSerUCCSerUCASerUCGSerUAUTyrUACTyrUAAStopUAGStopUGUCysUGCCysUGAStopUGGTrpUCAGCCUULeuCUCLeuCUALeuCUGLeuCCUProCCCProCCAProCCGProCAUHisCACHisCAAGlnCAGGlnCGUArgCGCArgCGAArgCGGArgUCAGAAUUIleAUCIleAUAIleAUGMet (start)ACUThrACCThrACAThrACGThrAAUAsnAACAsnAAALysAAGLysAGUSerAGCSerAGAArgAGGArgUCAGGGUUValGUCValGUAValGUGValGCUAlaGCCAlaGCAAlaGCGAlaGAUAspGACAspGAAGluGAGGluGGUGlyGGCGlyGGAGlyGGGGlyUCAG64 codons = 61 for the 20 amino acids + 3 stop (UAA, UAG, UGA) · AUG = Met and start · Leu, Ser, Arg 6 codons each; Met, Trp 1
The standard genetic code, generated codon by codon in the source file (61 sense codons, 3 stops) — our layout of the information in NCERT Table 5.1 (page 96).
Reading NCERT's Table 5.1 (page 96)A 4 × 4 checkerboard headed ‘First position’ (rows U, C, A, G), ‘Second position’ (columns U, C, A, G) and ‘Third position’ (U, C, A, G down the right), each cell listing four codons with three-letter amino-acid names, e.g. UUU Phe, AUG Met, UAA Stop. The point: read a codon as row → column → line; notice that most amino acids are coded by codons differing only in the third base.
  • The codon is a triplet: 61 codons code for amino acids; 3 code for none and act as stop codons.
  • Some amino acids have more than one codon — the code is degenerate.
  • Codons are read contiguously: there is no punctuation.
  • The code is nearly universal — UUU means phenylalanine from bacteria to humans — with some exceptions in mitochondrial codons and some protozoans.
  • AUG has two jobs: it codes for methionine and is the initiator codon.
  • UAA, UAG, UGA are stop (terminator) codons.

NCERT's in-text exercise: AUG UUU UUC UUC UUU UUU UUC translates to Met-Phe-Phe-Phe-Phe-Phe-Phe. Going back from Met-Phe×6 to the mRNA is harder: phenylalanine has two codons (UUU, UUC), so there are 2⁶ = 64 possible mRNAs. Forward translation is unambiguous; reverse is not — the code is specific but degenerate (both checked in the source file).

Mutations and the code. Large deletions or rearrangements can remove or add whole genes. A point mutation changes one base pair: in the β-globin gene, one change turns glutamate into valine and causes sickle-cell anaemia (GAG → GUG in mRNA; Chapter 4). Inserting or deleting one or two bases shifts the reading frame from that point — a frame-shift mutation — whereas inserting or deleting three (or a multiple of three) adds or removes whole codons and leaves the frame intact after them. NCERT's sentence of three-letter words shows it:

Change to RAM HAS RED CAPRead in threes (computed)Frame after the change
insert BRAM HAS BRE DCA Pshifted
insert BIRAM HAS BIR EDC APshifted
insert BIGRAM HAS BIG RED CAPrestored — one extra word
delete RRAM HAS EDC APshifted
delete RERAM HAS DCA Pshifted
delete REDRAM HAS CAPrestored — one word lost
Reading NCERT's RAM HAS RED CAP example (pages 97–98)A worked text example rather than a figure: inserting B, BI and BIG after HAS, and deleting R, E and D one by one, then regrouping the letters in threes. The point: one or two bases inserted or deleted garble everything downstream; three keep the frame.

tRNA, the adapter. Francis Crick reasoned that amino acids cannot read the code themselves, so an adapter must read a codon on one side and carry an amino acid on the other. tRNA — then called soluble RNA (sRNA) — was known before the code but its adapter role came later. It has an anticodon loop with bases complementary to the codon and an amino-acid acceptor end. tRNAs are specific for each amino acid; a special initiator tRNA starts translation; there are no tRNAs for stop codons. Its secondary structure looks like a clover leaf; the real 3-D molecule is a compact inverted L.

5′3′ — amino-acid acceptor endPheAAG5′3′mRNAUUCanticodoncodonlooploopanticodon loopClover-leaf = how the strandpairs with itself (secondarystructure). In 3-D the tRNAfolds into a compactinverted L:acceptor endanticodon endCodon 5′-UUC-3′ (Phe) pairs withanticodon 3′-AAG-5′ — antiparallel
tRNA as the adapter between codon and amino acid; the anticodon is generated from the codon in the source file — our schematic for NCERT Figure 5.12 (page 98).
Reading NCERT's Figure 5.12 (page 98)Two clover-leaf tRNAs stand on a red mRNA line (5′ to 3′): one carries Ser at its 3′ end and has anticodon UCA over codon AGU; the other carries Tyr with anticodon AUG over codon UAC; ‘Anticodon’ and ‘Codon’ are labelled. The point: the anticodon pairs with the codon, and the amino acid on top is fixed by which tRNA it is (both codon–amino acid pairs agree with Table 5.1).
Common confusions
  • 64 codons = 61 sense + 3 stop; there are 20 amino acids. AUG is both Met and start; there is no ‘start-only’ codon in NCERT's scheme.
  • Degenerate = one amino acid, several codons. Unambiguous/specific = one codon, one amino acid. Do not swap them.
  • ‘Universal’ is nearly universal: mitochondria and some protozoans differ.
  • Frame-shift = insertion or deletion of 1 or 2 bases (not 3). A point substitution (sickle-cell) changes one codon only.
  • The anticodon is on tRNA, the codon on mRNA; they pair antiparallel. No tRNA reads stop codons — release factors do.
  • Khorana: chemical synthesis of defined RNAs; Nirenberg: cell-free system; Ochoa: polynucleotide phosphorylase; Gamow: argued for a triplet. UPSC 2001 keyed Khorana for ‘establishing the complete genetic code’.
Beyond NCERT
  • Khorana, Nirenberg and Robert Holley (who worked out a tRNA's structure) shared the 1968 Nobel Prize in Physiology or Medicine.
  • Nirenberg and Matthaei's first result (1961) was that poly-U RNA makes polyphenylalanine — UUU = Phe.
  • In human mitochondria, UGA codes for tryptophan instead of stop — an example of the code being only nearly universal.

7. Translation: RNA to protein

In simple wordsThe ribosome is a reading-and-assembly machine. It clamps onto the mRNA, reads one three-letter word at a time, lets the matching tRNA ‘delivery truck’ unload its amino acid, links it to the growing chain, and moves on — until it meets a ‘stop’ word.

Translation polymerises amino acids into a polypeptide in the order set by the mRNA's bases; amino acids are joined by peptide bonds, whose formation needs energy. So first the amino acids are activated with ATP and linked to their own tRNA — charging of tRNA, or aminoacylation. Two charged tRNAs brought close together favour peptide-bond formation, and a catalyst speeds it.

The ribosome is the protein factory: structural RNAs plus about 80 different proteins, existing when inactive as a large and a small subunit. Translation begins when the small subunit meets an mRNA. The large subunit has two sites where successive amino acids bind close enough to form a peptide bond, and the ribosome itself catalyses the bond — in bacteria the 23S rRNA is the enzyme, a ribozyme. A translational unit is the mRNA stretch from the start codon (AUG) to a stop codon; untranslated regions (UTRs) lie before the start (5′ UTR) and after the stop (3′ UTR) and are needed for efficient translation.

5′3′ACUCAAUGGCCAAAUUUGGCUAAGCAUUMetAlaLysPheGlystop5′ UTR3′ UTRstart (AUG)stop (UAA)large subunitsmall subunittRNAUUUsite 1 (P)LysAlaMettRNAAAAsite 2 (A)PheCharged tRNAs (amino acid attachedusing ATP) pair their anticodonswith the codonsPeptide bond (arrow) formed,catalysed by rRNA — 23S rRNAin bacteria, a ribozymeribosome moves one codon at a time towards 3′Our gene: Met-Ala-Lys-Phe-Gly — a release factor at UAA ends the chain · UTRs are not translated
Our example mRNA being translated; codons, anticodons and the peptide are all computed in the source file — our schematic for NCERT Figure 5.13 (page 99).
Reading NCERT's Figure 5.13 (page 99)A green ribosome sits on a yellow mRNA (5′ CUCUUGGGUCCGCAGUUAAUUUCUAUC 3′). Two tRNAs inside it have anticodons CGU and CAA paired with codons GCA (Ala) and GUU (Val); a growing chain Gly-Leu-Tyr-Ser-Leu-Gly-Ser-Ala rises from the left tRNA; a used tRNA leaves on the left and a new one carrying Asn arrives on the right. The point: two neighbouring codons are read at a time and the chain is passed from one tRNA to the next (the pairings in the figure check out against the code).
⤵ Translation in four stages
Chargingeach amino acid is activated by ATP and joined to its own tRNA (aminoacylation)
Initiationthe small subunit binds the mRNA; the initiator tRNA reads the start codon AUG; the large subunit joins
Elongationa charged tRNA pairs its anticodon with the next codon; rRNA links the amino acids by a peptide bond; the ribosome moves one codon towards the 3′ end
Terminationa release factor binds the stop codon; the finished polypeptide is released
Common confusions
  • Transcription (DNA → RNA, nucleus in eukaryotes, RNA polymerase) vs translation (RNA → protein, cytoplasm, ribosome). Exam questions swap them.
  • Charging (aminoacylation) uses ATP and happens before the tRNA reaches the ribosome.
  • The peptide bond is made by rRNA — a ribozyme — not by a ribosomal protein.
  • UTRs are part of the mRNA but are not translated; they lie outside AUG…stop.
  • Two roles of the ribosome (exercise 9): it binds mRNA and provides the sites where charged tRNAs meet codons, and it catalyses peptide-bond formation.
  • NCERT's ‘about 80 proteins’ fits a eukaryotic ribosome; bacterial ribosomes have fewer (Beyond NCERT).
Beyond NCERT
  • Ribosomes are named by sedimentation: bacterial 70S (50S + 30S subunits) and eukaryotic cytoplasmic 80S (60S + 40S); S values do not add. The two sites NCERT mentions are usually called the P (peptidyl) and A (aminoacyl) sites, with an E (exit) site as well.
  • Many antibiotics (e.g. streptomycin, tetracycline) work by blocking bacterial 70S ribosomes, which differ from ours.
  • Several ribosomes can read one mRNA at once — a polysome.

8. Regulation of gene expression: the lac operon

In simple wordsA bacterium does not keep its lactose-digesting kit running when there is no lactose — that would waste energy. A guard protein (the repressor) sits on the switch and keeps the genes off. When lactose arrives, it pulls the guard off the switch, and the genes turn on. When the lactose is used up, the guard returns.

Gene expression, ending in a polypeptide, can be regulated at several levels. In eukaryotes: (i) transcription (forming the primary transcript), (ii) processing (regulating splicing), (iii) transport of mRNA from nucleus to cytoplasm, and (iv) translation. Metabolic, physiological or environmental conditions decide which genes are expressed; embryonic development and differentiation also rest on coordinated gene regulation. If E. coli has no lactose to use, it need not make β-galactosidase, the enzyme that splits lactose into galactose and glucose.

In prokaryotes the main control point is the rate of transcriptional initiation. Accessory proteins act on RNA polymerase at the promoter — positively (activators) or negatively (repressors). Often a repressor binds an operator next to the promoter; each operon has its own operator and repressor (the lac operator binds only the lac repressor). In an operon, a polycistronic structural gene is controlled by a common promoter and regulatory genes — lac, trp, ara, his, val operons are examples.

The lac operon was worked out by the geneticist François Jacob and the biochemist Jacques Monod — the first transcriptionally regulated system explained. It has one regulatory gene, i (from ‘inhibitor’, not ‘inducer’), coding for the repressor, and three structural genes: z (β-galactosidase), y (permease, which lets β-galactosides into the cell) and a (transacetylase) — all needed for lactose metabolism. The repressor is made all the time (constitutively) and binds the operator, blocking RNA polymerase. Lactose, the substrate of β-galactosidase, is the inducer: when glucose, the preferred carbon source, is absent and lactose is supplied, lactose enters through permease (a very low level of expression is always present, or it could not enter), and lactose or allolactose binds and inactivates the repressor, so RNA polymerase transcribes the operon. This is regulation of enzyme synthesis by its substrate, and control by a repressor is negative regulation; the lac operon is also under positive regulation, beyond NCERT's scope.

Lactose absent: operon OFFpipozyarepressor mRNArepressor (made all the time)repressor on the operatorRNA pol✕RNA polymerase cannot pass:z, y, a are not transcribedLactose present: operon ONpipozyainducer (lactose or allolactose) bound:inactive repressor cannot bind oone lac mRNA (polycistronic)β-galactosidasesplits lactosepermeaselets lactose intransacetylasep = promoter · i = regulator gene (i from ‘inhibitor’) · o = operator · z, y, a = structural genes · negative regulation
The lac operon off and on — our schematic for NCERT Figure 5.14 (page 101).
Reading NCERT's Figure 5.14 (page 101)Two rows of the gene strip p, i, p, o, z, y, a. ‘In absence of inducer’: an arrow from i gives a blue ‘Repressor mRNA’ and an orange ‘Repressor’, which binds the operator (o), with the note that it prevents RNA polymerase from transcribing the operon. ‘In presence of inducer’: the inducer joins the repressor (‘Inactive repressor’), and the structural genes give a green ‘lac mRNA’ that is translated into β-galactosidase, permease and transacetylase. The point: the same repressor is always made; the inducer decides whether it can sit on the operator.
⤵ The lac switch
i genemakes the repressor all the time
No lactoserepressor binds the operator
Operon offRNA polymerase blocked; z, y, a not transcribed
Lactose presentinducer (lactose/allolactose) binds and inactivates the repressor
Operon onone mRNA for z, y, a → β-galactosidase, permease, transacetylase
Lactose used upno inducer left; repressor binds the operator again — operon shuts down (exercise 10)
Common confusions
  • i stands for inhibitor; its product, the repressor, is made constitutively.
  • z = β-galactosidase, y = permease, a = transacetylase — in that order.
  • The repressor binds the operator, not the promoter; RNA polymerase binds the promoter.
  • The inducer is lactose (or allolactose); glucose and galactose cannot induce the operon.
  • Lac regulation by the repressor is negative regulation; the operon shuts down again once lactose is exhausted.
  • Operon concept = Jacob and Monod (UPSC 1998); one gene–one enzyme = Beadle and Tatum; mutation theory = de Vries.
Beyond NCERT
  • Allolactose, a by-product of β-galactosidase acting on lactose, is the natural inducer; lab work often uses IPTG, which induces but is not broken down.
  • The positive control NCERT leaves out is catabolite repression: when glucose is low, cAMP rises and the CAP–cAMP complex helps RNA polymerase bind — so the operon is fully on only with lactose present and glucose absent.
  • Jacob, Monod and André Lwoff shared the 1965 Nobel Prize. The trp operon is the classic repressible operon, switched off by its end-product, tryptophan.

9. The Human Genome Project

In simple wordsImagine typing out a book of three thousand million letters that nobody can read in one go. You shred many copies into overlapping scraps, read each scrap, and let a computer piece the scraps back together by their overlaps. That, in essence, is how the human genome was read.

If genetic make-up lies in DNA sequence, then individuals who differ must differ in sequence somewhere. With cloning and fast sequencing available, the Human Genome Project (HGP) was launched in 1990 — a ‘mega project’. At about 3 × 10⁹ bp and an early estimate of US $3 per base pair, it would cost about US $9 billion; typed in books of 1000 pages with 1000 letters each, one cell's sequence would fill about 3300 books. Storing, retrieving and analysing such data drove the growth of bioinformatics.

Goals of the HGP (NCERT)
1Identify all the approximately 20,000–25,000 genes in human DNA
2Determine the sequence of the 3 billion chemical base pairs
3Store the information in databases
4Improve tools for data analysis
5Transfer related technologies to other sectors such as industry
6Address the ethical, legal and social issues (ELSI) the project may raise

It was a 13-year project coordinated by the US Department of Energy and the National Institutes of Health, with the Wellcome Trust (UK) a major partner and contributions from Japan, France, Germany, China and others; it was completed in 2003. Knowing how DNA varies between people could transform diagnosis, treatment and prevention of disorders. Genomes of non-human model organisms — bacteria, yeast, Caenorhabditis elegans (a free-living nematode), Drosophila, rice and Arabidopsis — have also been sequenced, with uses in health care, agriculture, energy and environmental remediation.

Methods. One approach identified all genes expressed as RNA — Expressed Sequence Tags (ESTs). The other sequenced the whole genome, coding and non-coding, and assigned functions later — sequence annotation. Total DNA was broken into random small fragments, cloned in hosts (bacteria, yeast) using vectors called BAC (bacterial artificial chromosome) and YAC (yeast artificial chromosome) to amplify them, and sequenced on automated sequencers based on Frederick Sanger's method (Sanger also developed protein sequencing). Overlapping fragments were aligned by computer, annotated and assigned to chromosomes; the last, chromosome 1, was completed in May 2006 (of 24: 22 autosomes, X and Y). Genetic and physical maps used polymorphism in restriction-enzyme sites and repetitive DNA called microsatellites.

⤵ How the genome was sequenced
Isolate DNAtotal DNA from cells
Fragmentbreak into random, relatively small, overlapping pieces
Cloneinsert into BAC or YAC vectors; amplify in bacteria or yeast
Sequenceautomated sequencers based on Sanger's method
Assemblecomputer programs align the overlaps into long sequences
Annotate and mapassign functions and place sequences on chromosomes
1. Fragments as sequenced (random order)GATCCTAGCAGATTCAGCTAGCTACGGATCcomputerfinds overlaps2. Aligned on their 4-base overlapsGATTCAGCTAGCTACGGATCGATCCTAGCA3. Consensus = the genome stretchGATTCAGCTACGGATCCTAGCAOnly one order of the three reads fits (checked in the source file) — HGP did this for millions of fragments with software
Why overlapping fragments are needed: a toy 22-base ‘genome’ rebuilt from three shuffled reads — our example; the unique assembly is checked in the source file.
Reading NCERT's Figure 5.15 (page 103)A cartoon sequence: a woman → a cell with its nucleus → a chromosome → a DNA double helix with base pairs → a desktop computer whose screen shows letters such as TGTAC and GTACA. The point: the genome runs from person to cell to chromosome to DNA to data — sequencing ends in a computer.
Salient feature (NCERT 5.9.1)Figure
Size of the human genome3164.7 million bp
Average gene / largest known gene3,000 bases / dystrophin, 2.4 million bases
Total genes (estimate)30,000 — far below earlier estimates of 80,000–1,40,000
Bases identical in all humans99.9 per cent
Genes of unknown functionover 50 per cent
Genome coding for proteinsless than 2 per cent; repeated sequences make up a very large part
Most / fewest geneschromosome 1 (2968) / Y (231)
Single-base differences (SNPs, ‘snips’)about 1.4 million locations

Applications. Whole-genome sequences and high-throughput methods let researchers study all genes, all the transcripts of a tissue, organ or tumour, or how thousands of genes and proteins work in networks — instead of one gene at a time. Repetitive sequences have no direct coding function but shed light on chromosome structure, dynamics and evolution; SNPs help locate disease-associated sequences and trace human history.

Position on genomic data and genetic privacy (attributed)What it argues
Genome researchers and public genome projectsopen, shared genome data speed discovery of disease genes and drugs; large, diverse datasets are needed so populations such as India's are not left out
Privacy and civil-liberties advocatesgenomes are uniquely identifying and reveal relatives' information too; even ‘anonymised’ data can be re-identified, so consent and strict limits on use and sharing are essential
Patients' groups and bioethicistsgenetic results must not lead to discrimination in jobs, insurance or marriage; testing should come with counselling
Community and indigenous groupscommunities whose samples are used should give collective consent and share in the benefits, rather than see data and profits flow elsewhere
Common confusions
  • HGP: 1990–2003, 13 years, led by the US DoE and NIH; the Wellcome Trust was the main other partner. Chromosome 1 was finished last, in 2006.
  • NCERT inconsistency on gene number: the goals say 20,000–25,000 genes, the salient features ‘30,000’. Quote ‘about 30,000 (NCERT)’ if asked from the book; current estimates are about 20,000 protein-coding genes (Beyond NCERT).
  • Genome size appears as 3 × 10⁹ (goals), 3.3 × 10⁹ (haploid, page 80) and 3164.7 million bp (features) — the same quantity, rounded differently.
  • ESTs = expressed genes; annotation = assigning function after whole-genome sequencing. BAC/YAC = vectors; Sanger = sequencing method.
  • Less than 2% codes for protein; 99.9% identical between people; 1.4 million SNPs; chromosome 1 most genes, Y fewest.
Why this matters for GS3Genome sequencing underpins precision medicine, newborn and carrier screening, pathogen surveillance (as in the COVID-19 pandemic) and crop breeding for disease resistance and drought tolerance (UPSC 2017). India's own genome effort and genomic-data governance are recurring current-affairs themes; this section supplies the vocabulary — sequencing, annotation, SNPs, bioinformatics, the transcriptome (UPSC 2016).
Beyond NCERT
  • A ‘working draft’ of the human genome was announced in 2001; the remaining gaps were closed by the Telomere-to-Telomere consortium, which published a truly complete sequence in 2022.
  • Yeast (Saccharomyces cerevisiae) was the first eukaryote whose whole genome was sequenced, in 1996 (UPSC 1997).
  • GenomeIndia, funded by the Department of Biotechnology (DBT), aims to build a catalogue of genetic variation of the Indian population by sequencing about 10,000 genomes from across the country; it is a national project, not part of the international HGP (UPSC 2026).
  • The transcriptome is the full set of RNA (mRNA) molecules expressed by a cell, tissue or organism at a time; the proteome is the full set of proteins.

10. DNA fingerprinting

In simple wordsMost of our DNA is the same as everyone else's, so comparing whole genomes is wasteful. Instead, look at a few places where a short ‘word’ is repeated — say 5 times in one person and 12 in another. Measure the lengths of those stretches and you get a barcode that is almost unique to each person and half-inherited from each parent.

If 99.9 per cent of bases are the same in all humans, then in a genome of 3 × 10⁹ bp about 3 × 10⁶ bases differ (NCERT's in-text question). Sequencing everyone's DNA to find them would be slow and costly; DNA fingerprinting compares individuals quickly. It targets repetitive DNA, in which a short stretch is repeated many times. In density-gradient centrifugation bulk DNA forms a major peak and repetitive DNA forms small peaks — satellite DNA, classed by base composition (A:T- or G:C-rich), segment length and number of repeats into micro-satellites, mini-satellites and so on. These normally code for no protein, form a large part of the genome, and are highly polymorphic. DNA from any tissue of a person — blood, hair follicle, skin, bone, saliva, sperm — shows the same pattern, which makes it a forensic tool; and because polymorphisms are inherited, it is the basis of paternity testing.

Polymorphism (variation at the genetic level) arises from mutation in somatic or germ cells. A germ-line mutation that does not seriously impair reproduction can spread through a population. An allelic variant is traditionally called a DNA polymorphism if more than one variant at a locus occurs with a frequency greater than 0.01. Mutations in non-coding DNA are more likely to persist, since they may have no immediate effect on reproduction, so variation accumulates there over generations — from single-nucleotide changes to large-scale ones — and matters for evolution and speciation.

The technique was developed by Alec Jeffreys, using as probe a highly polymorphic satellite DNA called VNTR (Variable Number of Tandem Repeats), a mini-satellite: a short sequence repeated in tandem, its copy number differing between chromosomes and individuals, so the VNTR ranges in size from 0.1 to 20 kb. After hybridisation with the VNTR probe the autoradiogram shows many bands of different sizes, a pattern characteristic of an individual and different for everyone except monozygotic (identical) twins. PCR (Chapter 9) has raised sensitivity so that DNA from a single cell is enough. Besides forensics, fingerprinting is used to study population and genetic diversity; many probes are now used.

⤵ DNA fingerprinting by Southern blot (as first used)
Isolate DNAfrom blood, hair follicle, skin, bone, saliva or sperm
Digestcut with restriction endonucleases
Electrophoresisseparate the fragments by size in a gel
Blottransfer the separated fragments to a nitrocellulose or nylon membrane
Hybridisewith a radiolabelled VNTR probe
Detectautoradiography shows the band pattern — the DNA fingerprint
Forensic comparisonPaternity testCrime sceneSuspect 1Suspect 2MotherChildMan XMan Y2468101214repeatcountshorter →runs fartherCrime-scene bands= Suspect 2Child: every band isin Mother or Man Y;Man X lacks bands15 and 13Each person: 3 VNTR loci × 2 alleles = 6 bands; repeat numbers are our invented data (match computed in the source file)
A forensic match and a paternity check on one gel; the band sets, the match and the paternity test are computed from invented repeat counts in the source file — our schematic in the spirit of NCERT Figure 5.16 (page 107).
Reading NCERT's Figure 5.16 (page 107)Pairs of paternal and maternal chromosomes 7, 2 and 16 of individuals A and B are drawn as bars with different numbers of coloured repeat boxes; the same for DNA from a crime scene (C). Arrows lead to a gel with lanes C, A and B and a scale ‘Number of short tandem repeats’ from 1 to 12; ‘Amplified repeats, separated by size on a gel, give a DNA fingerprint’. The caption explains that the two alleles of a chromosome carry different copy numbers and that the crime-scene pattern matches B, not A. The point: each band is one allele, its position set by its repeat number.
UseWhat is compared
Forensics (crime scenes)bands from trace evidence (hair, saliva, semen, blood) against suspects' bands
Paternity and maternity disputesevery band of the child must come from one parent or the other
Identifying remains, disaster victimsbands of remains against relatives
Population and genetic diversity, evolutionhow often band variants occur in different populations or species (UPSC 2023: microsatellites)
Position on forensic DNA databases (attributed)What it argues
Police and forensic agenciesa national DNA database speeds identification of offenders, missing persons and unidentified bodies, and can clear the wrongly accused
Privacy and civil-liberties groupsstoring profiles of suspects, undertrials or relatives risks surveillance and misuse; the Supreme Court recognised privacy as a fundamental right in Justice K.S. Puttaswamy v. Union of India (2017)
Forensic scientists and defence lawyersa match is a statistical statement; contamination, mixed samples and lab quality need accreditation and independent review
Parliamentary and legal scrutinysafeguards — consent, deletion of profiles, oversight bodies — must be written into law before databases expand
Common confusions
  • Repetitive DNA = any repeated sequence; satellite DNA = repetitive DNA that separates as small peaks in density-gradient centrifugation (exercise 8a). VNTRs are mini-satellites; microsatellites are shorter repeats.
  • Fingerprinting uses non-coding, highly polymorphic DNA — not genes.
  • Polymorphism = variant with frequency above 0.01 in a population; it arises by mutation.
  • Identical (monozygotic) twins have the same fingerprint; fraternal twins do not.
  • Order of steps: isolate → digest → electrophoresis → blot → hybridise → autoradiograph. Southern blotting is for DNA.
  • NCERT slip: page 105 says ‘compare two sets of 3 × 10⁶ base pairs’; it should be 3 × 10⁹ (the genome) — 3 × 10⁶ is the number of differences.
  • DNA fingerprinting (an individual) ≠ DNA barcoding (identifying a species from a short standard gene region, Beyond NCERT).
Beyond NCERT
  • Alec Jeffreys developed the method at the University of Leicester in 1984. Modern forensic profiles use PCR-amplified short tandem repeats (STRs), a kind of microsatellite.
  • India's Centre for DNA Fingerprinting and Diagnostics (CDFD) is in Hyderabad (UPSC 2001, 2008).
  • Under Section 112 of the Indian Evidence Act, 1872, a child born during a valid marriage is presumed legitimate; courts have weighed DNA tests against this presumption and the right to privacy (UPSC Mains 2024).

NCERT's in-text questions — short answers

The chapter asks many questions in the running text. Short outlines follow, in page order.

Page 82: Compare the structure of purines and pyrimidines. Can you find out why the distance between two polynucleotide chains in DNA remains almost constant?
Purines (A, G) have two rings; pyrimidines (C, T) have one. Every base pair is one purine plus one pyrimidine (A with T, G with C), so every rung has the same width and the strands stay a nearly constant distance apart (see the diagram in Section 1).
Page 83: In some viruses the flow of information is from RNA to DNA. Can you suggest a simple name to the process?
Reverse transcription (by reverse transcriptase, as in retroviruses such as HIV).
Page 83: If the length of E. coli DNA is 1.36 mm, can you calculate the number of base pairs in E. coli?
1.36 × 10⁻³ m ÷ 0.34 × 10⁻⁹ m per bp = 4.0 × 10⁶ bp. (NCERT elsewhere gives 4.6 × 10⁶ bp for E. coli; the 1.36 mm figure gives 4 × 10⁶.)
Page 83: Theoretically, how many such beads (nucleosomes) do you imagine are present in a mammalian cell?
6.6 × 10⁹ bp ÷ 200 bp per nucleosome ≈ 3.3 × 10⁷ nucleosomes.
Page 85: Can you think of any difference between DNAs and DNase?
DNA is the nucleic acid (genetic material); DNase is an enzyme — a protein — that digests DNA. Avery's team used DNase to show transformation needs DNA.
Page 87: Can you recall the two chemical differences between DNA and RNA?
(1) Sugar: deoxyribose in DNA, ribose (with a 2′-OH) in RNA. (2) Base: thymine in DNA, uracil in RNA.
Page 89: Can you recall what centrifugal force is, and think why a molecule with higher mass/density would sediment faster?
Spinning creates a centrifugal force proportional to mass; in a CsCl density gradient each DNA settles where its density equals the medium's, so denser ¹⁵N DNA moves further down than ¹⁴N DNA.
Page 90: If E. coli was allowed to grow for 80 minutes then what would be the proportions of light and hybrid densities DNA molecule?
80 min = 4 generations: of 16 molecules, 2 hybrid and 14 light — 1/8 hybrid : 7/8 light (1 : 7).
Page 92: Can you now write the sequence of RNA transcribed from the above DNA?
5′-UACGUACGUACGUACGUACGUACG-3′ — the coding strand's sequence with U in place of T.
Page 96: If following is the sequence of nucleotides in mRNA, predict the sequence of amino acid coded by it: -AUG UUU UUC UUC UUU UUU UUC-
Met-Phe-Phe-Phe-Phe-Phe-Phe.
Page 97: Now try the opposite. Met-Phe-Phe-Phe-Phe-Phe-Phe: predict the nucleotide sequence in the RNA. Do you face any difficulty? Which two properties of the genetic code have you learnt?
AUG followed by six codons each UUU or UUC — e.g. AUG UUU UUU UUU UUU UUU UUU, but there are 64 possible answers. The difficulty shows the code is degenerate (several codons per amino acid), while the forward direction shows it is unambiguous/specific (one codon, one amino acid).
Page 101: Can you think for how long the lac operon would be expressed in the presence of lactose?
Only as long as lactose (the inducer) remains; once it is used up the repressor binds the operator again and the operon switches off.
Page 105: Assuming human genome as 3 × 10⁹ bp, in how many base sequences would there be differences?
0.1 per cent of 3 × 10⁹ = 3 × 10⁶ bases.

NCERT exercise — answer outlines

Reading NCERT's Summary and Exercises (pages 108–109)The Summary restates that DNA stores and RNA mainly transfers information, DNA is the more stable genetic material though RNA evolved first, A pairs with T by two and G with C by three hydrogen bonds, replication is semiconservative, eukaryotic genes are split and spliced, tRNA reads the triplet code, rRNA is a ribozyme, the lac operon is regulated by its substrate, and the HGP and DNA fingerprinting rest on DNA polymorphism. Fourteen exercise questions follow — classification, base-pairing arithmetic, sequence writing, experiments, comparisons, the lac operon, the HGP and fingerprinting.

Try each question first, then open the outline. For sequences, write the 5′ and 3′ ends every time.

1. Group the following as nitrogenous bases and nucleosides: Adenine, Cytidine, Thymine, Guanosine, Uracil and Cytosine.
Nitrogenous bases: adenine, thymine, uracil, cytosine. Nucleosides (base + sugar): cytidine, guanosine.
2. If a double stranded DNA has 20 per cent of cytosine, calculate the per cent of adenine in the DNA.
G = C = 20%, so G + C = 40% and A + T = 60%; A = T, so A = 30%.
3. If the sequence of one strand of DNA is written as follows: 5'-ATGCATGCATGCATGCATGCATGCATGC-3'. Write down the sequence of complementary strand in 5'→3' direction.
5′-GCATGCATGCATGCATGCATGCATGCAT-3′ (complement each base, then read it from its own 5′ end).
4. If the sequence of the coding strand in a transcription unit is written as follows: 5'-ATGCATGCATGCATGCATGCATGCATGC-3'. Write down the sequence of mRNA.
5′-AUGCAUGCAUGCAUGCAUGCAUGCAUGC-3′ — the coding strand with U in place of T.
5. Which property of DNA double helix led Watson and Crick to hypothesise semi-conservative mode of DNA replication? Explain.
Complementary base pairing between antiparallel strands (A–T, G–C). Because each strand fixes the sequence of its partner, the strands can separate and each serve as a template, giving two daughters each with one old and one new strand.
6. Depending upon the chemical nature of the template (DNA or RNA) and the nature of nucleic acids synthesised from it (DNA or RNA), list the types of nucleic acid polymerases.
(i) DNA-dependent DNA polymerase (replication); (ii) DNA-dependent RNA polymerase (transcription); (iii) RNA-dependent DNA polymerase — reverse transcriptase (retroviruses); (iv) RNA-dependent RNA polymerase — replicase of RNA viruses (Beyond NCERT).
7. How did Hershey and Chase differentiate between DNA and protein in their experiment while proving that DNA is the genetic material?
They labelled them differently: ³²P goes only into DNA (protein has no phosphorus) and ³⁵S only into protein (DNA has no sulphur). After infection, blending and centrifugation, ³²P was inside the bacteria and ³⁵S in the supernatant.
8. Differentiate between the followings: (a) Repetitive DNA and Satellite DNA (b) mRNA and tRNA (c) Template strand and Coding strand
(a) Repetitive DNA: sequences repeated many times; satellite DNA: repetitive DNA that separates from bulk DNA as small peaks in density-gradient centrifugation (micro- and mini-satellites), highly polymorphic. (b) mRNA: carries the codons from DNA, linear, template for translation; tRNA: small clover-leaf adapter with an anticodon and an amino-acid acceptor end. (c) Template strand (3′→5′) is read by RNA polymerase; coding strand (5′→3′) has the same sequence as the RNA (T for U) and is not copied.
9. List two essential roles of ribosome during translation.
(1) It binds the mRNA and provides the sites where charged tRNAs pair with successive codons. (2) Its rRNA (23S in bacteria) catalyses peptide-bond formation — a ribozyme.
10. In the medium where E. coli was growing, lactose was added, which induced the lac operon. Then, why does lac operon shut down some time after addition of lactose in the medium?
β-galactosidase breaks lactose down. When the lactose (inducer) is used up, the repressor — made all the time — is no longer inactivated, binds the operator again and blocks transcription.
11. Explain (in one or two lines) the function of the followings: (a) Promoter (b) tRNA (c) Exons
(a) Promoter: DNA sequence upstream of the gene where RNA polymerase binds; it also defines which strand is the template. (b) tRNA: adapter that reads a codon with its anticodon and brings the matching amino acid. (c) Exons: coding/expressed sequences of a split gene that remain in the mature mRNA.
12. Why is the Human Genome Project called a mega project?
About 3 × 10⁹ bp to sequence at an early estimated US $3 per bp (~US $9 billion); data enough for about 3300 books of 1000 pages; needed high-speed computing and bioinformatics; 13 years (1990–2003) and international partners; goals included all genes, databases, tools, technology transfer and ELSI.
13. What is DNA fingerprinting? Mention its application.
A quick way to compare individuals' DNA by the band pattern of highly polymorphic repetitive DNA (VNTRs), obtained by restriction digestion, electrophoresis, blotting, probe hybridisation and autoradiography. Uses: forensic identification, paternity disputes, population and genetic-diversity studies, evolution.
14. Briefly describe the following: (a) Transcription (b) Polymorphism (c) Translation (d) Bioinformatics
(a) Transcription: copying one strand of a gene into RNA by RNA polymerase, from promoter to terminator; in eukaryotes followed by capping, tailing and splicing. (b) Polymorphism: inherited variation at a locus with more than one allele at frequency above 0.01, arising from mutation; basis of mapping and fingerprinting. (c) Translation: polymerising amino acids on ribosomes in the order of mRNA codons, using charged tRNAs; initiation at AUG, elongation, termination at a stop codon. (d) Bioinformatics: computer-based storage, retrieval and analysis of biological sequence data, which grew with the HGP.

What to remember

Numbers

Genome sizes (NCERT)φX174 5386 nt; lambda 48,502 bp; <i>E. coli</i> 4.6 × 10⁶ bp; human haploid 3.3 × 10⁹ bp (diploid 6.6 × 10⁹)
Double helixpitch 3.4 nm; ~10 bp per turn; 0.34 nm per bp; right-handed
Hydrogen bondsA=T 2; G≡C 3
DNA in a mammalian cell≈ 2.2 m; nucleus ≈ 10⁻⁶ m
Nucleosome8 histones (octamer); ~200 bp of DNA
Key datesMiescher 1869; Griffith 1928; Avery–MacLeod–McCarty 1944 (work 1933–44); Hershey–Chase 1952; Watson–Crick 1953; Meselson–Stahl 1958; Taylor (<i>Vicia faba</i>) 1958
E. colidivides every 20 min; replicates its DNA in 18 min; ≈ 2000 bp/s (per fork, our check)
Meselson–Stahlgen I all hybrid; gen II 1 : 1 hybrid : light; 80 min 1 : 7
RNA polymerases (eukaryotes)I: 28S, 18S, 5.8S rRNA; II: hnRNA; III: tRNA, 5S rRNA, snRNA
Poly-A tail200–300 adenylates
Genetic code64 codons = 61 + 3 stop (UAA, UAG, UGA); AUG = Met/start; 20 amino acids
lac operoni (repressor); z β-galactosidase; y permease; a transacetylase
HGP1990–2003 (13 years); ~US $9 billion estimate; 3300 books; chromosome 1 done May 2006
Human genome (NCERT)3164.7 million bp; average gene 3000 bases; dystrophin 2.4 million; ~30,000 genes; <2% codes protein; 99.9% identical
Genes per chromosomechromosome 1: 2968 (most); Y: 231 (fewest)
SNPsabout 1.4 million locations
Differences between two people0.1% of 3 × 10⁹ = 3 × 10⁶ bases
Polymorphism thresholdvariant frequency above 0.01
VNTR size0.1 to 20 kb

One-liners

  • Chargaff: in double-stranded DNA A = T and G = C.
  • A purine always pairs with a pyrimidine, keeping the helix width constant.
  • Histones are positively charged because they are rich in lysine and arginine.
  • Griffith found transformation; Avery's team showed it was DNA; Hershey and Chase proved it.
  • ³²P labels DNA; ³⁵S labels protein.
  • DNA is better for storing information, RNA for transmitting it.
  • Replication is semiconservative — proved with ¹⁵N and CsCl centrifugation.
  • DNA polymerase works only 5′→3′, so one new strand is discontinuous and joined by ligase.
  • The template strand is read 3′→5′; the coding strand has the RNA's sequence.
  • σ starts and ρ stops bacterial transcription.
  • Eukaryotic hnRNA is capped, tailed and spliced to give mRNA.
  • The code is triplet, degenerate, comma-less and nearly universal.
  • tRNA is the adapter; there are no tRNAs for stop codons.
  • The peptide bond is made by rRNA — a ribozyme.
  • The lac repressor binds the operator; lactose is the inducer.
  • DNA fingerprinting uses VNTRs; only identical twins share a pattern.

Traps to avoid

  • Chargaff's rule does not mean A + T = G + C, and it does not hold within one strand.
  • ³²P went into the bacteria; ³⁵S stayed outside in the supernatant.
  • ¹⁵N is not radioactive; it is separated by density.
  • The coding strand is not read; the template strand is.
  • 5S rRNA is made by RNA polymerase III, not I.
  • Frame-shift = insertion or deletion of one or two bases; three bases keep the frame.
  • AUG codes for methionine and is also the start codon.
  • The i gene's name comes from inhibitor, not inducer.
  • Glucose and galactose cannot induce the lac operon.
  • NCERT gives the human gene number as both 20,000–25,000 and 30,000.
  • ‘3 × 10⁶ base pairs’ on page 105 should be 3 × 10⁹; 3 × 10⁶ is the number of differences.
  • DNA fingerprinting identifies individuals; DNA barcoding identifies species.

How to strategise

Prelims — what UPSC asks

  • Core molecular biology: tRNA is the adapter that translates the code (1985); base pairs are A–T and C–G (1985); genes act through protein synthesis (1986); complementarity of the two strands is what suits DNA to store and pass on information (2001); Jacob and Monod gave the operon concept (1998, a match with de Vries, Darwin and Beadle–Tatum); and the genetics pioneers match — Lederberg, Morgan, Kornberg, Khorana (2001); James Watson worked in genetics (2008).
  • Genomes and fingerprints: DNA fingerprinting solves crimes (1991), establishes paternity (1997) and works on trace hair, saliva and semen (2000); yeast was the genome completed in 1996 (1997); genome sequencing helps crops — markers, faster breeding, host–pathogen studies (2017); the transcriptome is the full range of mRNA (2016); microsatellite DNA studies evolutionary relationships (2023); GenomeIndia is DBT-funded and catalogues Indian genetic diversity, but is not part of the HGP (2026); and viruses do infect bacteria and change host transcription (2025, none of the three statements correct).
  • Key notes: 1985 ‘genetic code is made up by the four letter alphabet forming the pairs’ is loosely worded — the code is read as triplet codons, not pairs; the keyed answer (d) A–T, T–A, C–G, G–C is right as the base-pairing rule. 2001 genetics match: the key (d) stands — Lederberg (conjugation, transduction), Morgan (sex-linkage), Kornberg (DNA polymerase), Khorana (genetic code); Ochoa is the distractor though he also contributed. 1997 yeast is Beyond NCERT as a date — the weakest link kept.
  • Likely next: Chargaff arithmetic, template vs coding strand, RNA polymerase I/II/III, codon properties, lac operon genes, HGP numbers, SNPs, VNTR vs STR, DNA barcoding vs fingerprinting, RNA vaccines and RNAi (built on transcription and translation), and genome-editing news.

Mains — where it appears and how to answer

  • GS2 2024 asked about the right to privacy under Article 21 and the law on DNA testing of a child in the womb to establish paternity; Section 10 supplies the science — what a DNA profile is, why every band of a child must come from a parent, and why one cell suffices with PCR. The legal half needs the polity chapters.
  • Use this chapter as the science base for GS3 answers on genome sequencing (GenomeIndia, precision medicine, pathogen surveillance), biotechnology in agriculture and health, and forensic DNA. Frame: one line of molecular biology → application → Indian programme or institution (DBT, CDFD) → attributed concerns (privacy, consent, discrimination, lab quality) → way forward.
  • A two-line base-pairing or codon example can make a technical point quickly; keep claims about laws and bills dated and attributed.

Self-check

Tap an option. NCERT's own exercise questions come first; the rest test the figures and common confusions.

1. A double-stranded DNA has 20% cytosine. What percentage of its bases is adenine?

2. In the Hershey–Chase experiment, radioactivity was found inside the bacteria when the phages had been grown with:

3. E. coli grown for many generations in ¹⁵N is moved to ¹⁴N. After 80 minutes (four generations), the DNA is:

4. The coding strand of a gene reads 5′-ATGCAT-3′. The mRNA transcribed from this gene reads:

5. Which RNA polymerase transcribes tRNA and 5S rRNA in eukaryotes?

6. Inserting two bases into the middle of a gene most likely causes:

7. In the lac operon, the repressor protein is coded by the:

8. DNA fingerprinting depends mainly on:

UPSC questions from this chapter

Open any question to attempt it in the Question bank — your attempt is saved to your progress.

Prelims

16 questions · newest first
2026Prelims GSWhich of the following statements with regard to Genome India Project is/are correct? 1. It is a part of the Human Genome Project. 2. The project is funded by the Department of Biotechnology (DBT), Government of India. 3. Its primary aim is to build a catalogue of genetic diversity of the Indian population. Select the answer using the code given below :Attempt →2025Prelims GSConsider the following statements: I. No virus can survive in ocean waters. II. No virus can infect bacteria. III. No virus can change the cellular transcriptional activity in host cells. How many of the statements given above are correct?Attempt →2023Prelims GS&#x27;Microsatellite DNA&#x27; is used in the case of which one of the following?Attempt →2017Prelims GSWith reference to agriculture in India, how can the technique of ‘genome sequencing’, often seen in the news, be used in the immediate future ? 1. Genome sequencing can be used to identify genetic markers for disease resistance and drought tolerance in various crop plants. 2. This technique helps in reducing the time required to develop new varieties of crop plants. 3. It can be used to decipher the host-pathogen relationships in crops. Select the correct answer using the code given below :Attempt →2016Prelims GSIn the context of the developments in Bioinformatics, the term ‘transcriptome’, sometimes seen in the news, refers toAttempt →2008Prelims GSNobel Prize winning scientist James D. Wastson is known for his work in which area ?Attempt →2001Prelims GSWhich of the following features of DNA makes in uniquely suited to store and transmit genetic information from generation to generation ?Attempt →2001Prelims GSMatch List I with List II and select the correct answer using the codes given below the Lists : Codes: List I (Achievement in genetics) List II (Scientists) I. Discovery of tramsdiction and Conjugation in bacteria (A) Khurana II. Establishing the sex-linked inheritance (B) Kornberg III. Isolation of DNA polymerase form E. coli (C) Lederberg IV. Establishing the complete genetic code (D) Morgan - (E) OchoaAttempt →2000Prelims GSAssertion (A) : “DNA Finger printing” has become a powerful tool to establish paternity and identity of criminals in rape and assault cases. Reason (R) : Trace evidences such as hairs, saliva and dried semen are adequate for DNA analysis.Attempt →1998Prelims GSMatch List I with List II and select the correct answer using the codes given below the lists : Codes: List I List II I. Theory of Mutation (A) Beadk and Tatum II. Theory of Evolution (B) Jacob and Monod III. One gene one enzyme hypothesis (C) Darwin IV. Operon concept (D) De VriesAttempt →1997Prelims GSWhich one of the following techniques can be used to establish the paternity of a child ?Attempt →1997Prelims GSThe sequencing of the entire genome (the totality of all genes) of an organism was completed in 1996. The organism wasAttempt →1991Prelims GSWhich one of the following techniques has helped to solve many mysteries involving murders, robberies and rapes ?Attempt →1986Prelims GSInfluence of genes in controlling the activity of living organisms is throughAttempt →1985Prelims GSThe genetic code can be translated through an intermediate, adaptor molecule byAttempt →1985Prelims GSGenetic code is made up by the four letter alphabet forming the pairsAttempt →

Mains

1 questions
2024GS2Right to Privacy is intrinsic to life and liberty and is inherently protected under Article 21 of the constitution. Explain. In this reference discuss the law relating to DNA testing of Child in the womb to establish its paternity. (Answer in 250 words)Write answer →

16 Prelims and 1 Mains UPSC questions are linked to this chapter. More from Biology.