Take a breath. This chapter has a reputation for being heavy, and I understand why: it arrives loaded with names, dates, enzymes and three-letter words that all look alike. But here is the honest truth — Molecular Basis of Inheritance is one long, tidy story, and once you can tell the story, the details hang on it by themselves. So we are going to walk through it together, slowly, from absolute zero.
Our whole tour uses one picture. Imagine a factory. Deep inside it there is a locked blueprint vault. There is a photocopier that duplicates the blueprint before the factory expands. There is a work-order desk that copies out just one page for today’s job. There is a dictionary on the wall that turns the work order into instructions the workers understand. There is an assembly line that builds the product. And there is a foreman who switches a line on only when there is raw material worth processing. That is it. That is the entire chapter. Every section below is one more room on the same tour.
One more thing worth knowing before we start: Chapter 5 sits inside Unit VII – Genetics and Evolution, which carries 20 marks — the highest weighting of any unit in the CBSE Class 12 Biology paper. There is no chapter in this book where careful study pays back faster. If genetics still feels shaky at the level of dominant and recessive, spend twenty minutes first with the Class 10 Heredity chapter; it is the on-ramp to everything here.
These are molecular basis of inheritance class 12 notes with solved examples: every idea is explained plainly, then immediately worked out on paper, including the numericals and the codon-decoding drills that students most often skip.
What You’ll Learn
Jump straight to any room in the factory:
- The Search For The Genetic Material
- Structure Of DNA And RNA
- DNA Packaging: Nucleosomes, Chromatin And Heterochromatin
- DNA Replication
- The Central Dogma
- Transcription In Prokaryotes And Eukaryotes
- The Genetic Code
- Translation
- Regulation Of Gene Expression: The Lac Operon
- Genome, Human And Rice Genome Projects
- DNA Fingerprinting
- Exam Strategy For This Chapter
Your Game Plan
- Read the factory story once, end to end, without stopping to memorise. You only need the shape of it.
- Come back and do the structure and replication sections properly — they carry every numerical in the chapter.
- Learn the genetic code as rules (triplet, degenerate, non-overlapping, comma-less, universal, unambiguous), not as 64 memorised lines.
- Practise decoding one mRNA into a peptide every single day for a week. Ten minutes. It becomes automatic.
- Draw the lac operon and the replication fork from memory until the labels come out in the right order.
- Finish with the worksheet at the bottom, and only then look at the answers.
Study Notes
The Search For The Genetic Material
Before anyone could study the blueprint, someone had to work out which molecule the blueprint was written on. For a long time the smart money was on protein. Proteins are varied and complicated; DNA looked like a boring polymer of four repeating units. The four experiments below overturned that, one careful step at a time.
Griffith, 1928 — the accident that started it. Frederick Griffith worked with Streptococcus pneumoniae, which comes in two forms. The S strain grows smooth colonies because it wears a polysaccharide coat; it kills mice. The R strain grows rough colonies, has no coat, and is harmless. Griffith found that heat-killed S bacteria alone did nothing, live R bacteria alone did nothing — but heat-killed S mixed with live R killed the mice, and live S bacteria could be recovered from their bodies. Something had passed out of the dead cells and permanently changed the living ones. He called it the transforming principle and, crucially, did not claim to know what it was.
Avery, MacLeod and McCarty, 1944 — naming the culprit. They purified the biochemicals from heat-killed S cells and then attacked each one with an enzyme. Adding protease (which destroys protein) did not stop transformation. Adding RNase (which destroys RNA) did not stop it either. Adding DNase (which destroys DNA) stopped transformation completely. The logic is beautiful: destroy the suspect and the crime stops happening.
Hershey and Chase, 1952 — the clincher. They used bacteriophage T2, a virus made of only DNA wrapped in protein. They grew one batch of phage with radioactive ³²P (phosphorus is in DNA, not in protein) and another with radioactive ³⁵S (sulphur is in protein, not in DNA). Each batch infected fresh bacteria; the empty phage coats were shaken off in a blender and spun down in a centrifuge. Radioactive ³²P turned up inside the bacterial pellet. Radioactive ³⁵S stayed outside, in the supernatant. Only DNA had gone in — and yet complete new phages came out.
Why DNA and not RNA? RNA carries an extra –OH group at the 2′ position of its sugar, which makes it chemically reactive and easily broken. DNA also uses thymine in place of uracil, and that swap gives DNA an extra layer of repairability. RNA mutates faster, so it evolves faster — wonderful for a virus, risky for a rose bush or a human being.
Reasoning: DNase is present, so the transforming principle (DNA) is destroyed. Transformation fails. No S colonies appear. Now the follow-up that carries the real marks: this result alone would not identify DNA, because three suspects were removed together. The experiment only works as proof because each enzyme was also used separately, and only DNase abolished transformation.
³²P batch (labelled DNA): radioactivity in the pellet — the pellet is the heavy bacterial cells, so the DNA went inside.
³⁵S batch (labelled protein): radioactivity in the supernatant — the light, empty protein coats stayed outside and were washed off.
Memory hook: P for Pellet, S for Supernatant. The first letter of the isotope tells you the fraction.
Structure Of DNA And RNA
This is the blueprint vault. Get the architecture right and half the chapter’s numericals solve themselves.
Start small. A nitrogenous base joined to a sugar is a nucleoside. Add a phosphate group and it becomes a nucleotide. String nucleotides together through 3′–5′ phosphodiester bonds and you have a polynucleotide chain, with a sugar–phosphate backbone on the outside and bases hanging inward like rungs. The phosphate groups are what make nucleic acids acidic. (If you want the pure chemistry of these units, the nucleic-acid section of the Class 12 Chemistry Biomolecules chapter covers the same molecules from the other side of the syllabus.)
The bases come in two families. Purines — adenine and guanine — have a double ring. Pyrimidines — cytosine, thymine (DNA) and uracil (RNA) — have a single ring.
The Watson–Crick double helix (1953). Working from the X-ray diffraction data produced by Maurice Wilkins and Rosalind Franklin, Watson and Crick proposed a model with these features:
- Two polynucleotide chains, backbones outside, bases inside.
- The chains are antiparallel: one runs 5′→3′, the other 3′→5′.
- The bases pair through hydrogen bonds: A=T with two H-bonds, G≡C with three H-bonds. Three bonds are stronger than two, so GC-rich DNA is harder to melt apart.
- The two chains coil into a right-handed helix.
- The pitch (one full turn) is 3.4 nm and contains roughly 10 base pairs, so each base pair rises 0.34 nm. The helix is about 2 nm wide.
| Feature | DNA | RNA |
|---|---|---|
| Sugar | Deoxyribose (no –OH at 2′) | Ribose (–OH present at 2′) |
| Bases used | A, G, C, T | A, G, C, U |
| Usual strandedness | Double-stranded helix | Usually single-stranded; may fold and pair internally |
| Chemical stability | High — the missing 2′-OH makes it far less reactive | Lower — the 2′-OH makes it easily hydrolysed |
| Catalytic ability | Essentially none | Yes — some RNAs are ribozymes |
| Rate of mutation | Slow | Faster |
| Main job | Long-term storage of information | Carrying, decoding and helping build (mRNA, tRNA, rRNA) |
| Repair | Thymine allows damaged cytosine to be recognised and repaired | No equivalent safeguard |
Step 1. A = T, so T = 30%.
Step 2. A + T = 60%, so G + C = 100 − 60 = 40%.
Step 3. G = C, so G = C = 40 ÷ 2 = 20% each.
Step 4. (A + T)/(G + C) = 60/40 = 1.5.
Check: purines A + G = 30 + 20 = 50%; pyrimidines T + C = 30 + 20 = 50%. They match, as Chargaff requires.
Step 1. G = C = 17%, so G + C = 34%.
Step 2. A + T = 100 − 34 = 66%.
Step 3. A = T = 66 ÷ 2 = 33%.
One line of working, full marks. The trap is doubling when you should halve — always write “G = C = 17, so G + C = 34” explicitly.
(a) Length = 1×10⁶ × 0.34 nm = 3.4×10⁵ nm = 3.4×10⁻⁴ m = 0.34 mm.
(b) Turns = 1×10⁶ ÷ 10 = 1×10⁵ turns.
(c) Nucleotides: each base pair is two nucleotides, one on each strand, so 2 × 1×10⁶ = 2×10⁶ nucleotides.
Sanity check on (a): length ÷ pitch should also give the turns — 3.4×10⁵ nm ÷ 3.4 nm = 1×10⁵. It matches.
DNA Packaging: Nucleosomes, Chromatin And Heterochromatin
Now the storage problem. Human DNA in a single nucleus is about 2.2 metres long. The nucleus is roughly 10 micrometres across. That is two metres of thread inside a container the width of a dust speck. The vault has to fold the blueprint, and fold it in a way that lets a page be pulled out on demand.
The trick is electrical. DNA is negatively charged because of its phosphate backbone. Histones are proteins unusually rich in the basic amino acids lysine and arginine, which carry positive charges. Opposite charges attract, and the DNA wraps itself around the histones without needing any covalent bond.
- Eight histone molecules — two each of H2A, H2B, H3 and H4 — form a histone octamer.
- About 200 base pairs of DNA wind around one octamer. The whole unit is a nucleosome.
- Histone H1 is not part of the octamer; it sits on the linker DNA and clamps the nucleosomes together.
- A chain of nucleosomes gives the classic “beads on a string” appearance of chromatin.
- Chromatin coils further into a thicker fibre, then loops and condenses — helped by non-histone chromosomal (NHC) proteins — into a chromosome.
Prokaryotes have no nucleus and no histones, but they still cannot leave their DNA loose. Bacterial DNA is held by a set of positively charged proteins in a region called the nucleoid, organised into large loops held in place by those proteins.
Length of DNA = 200 × 0.34 nm = 68 nm.
Compaction = 68 nm ÷ 11 nm ≈ 6-fold.
So the very first level of packing already shortens the thread about six times — and that is before the fibre coils, loops and condenses.
Step 1. Convert: 10 µm = 10 × 10⁻⁶ m = 1×10⁻⁵ m.
Step 2. Packing ratio = 2.2 ÷ 1×10⁻⁵ = 2.2×10⁵, i.e. roughly 220,000 times.
Where does the 2.2 m come from? A diploid human nucleus holds about 6.6×10⁹ bp. Length = 6.6×10⁹ × 0.34 nm = 2.244×10⁹ nm = 2.24 m. Round it to 2.2 m. (The haploid figure, 3.3×10⁹ bp, gives about 1.12 m — exactly half, as it must.)
DNA Replication
This is the photocopier room. Before a cell divides it must hand each daughter a complete blueprint — not a summary, not a sample, the whole thing. The question Watson and Crick were immediately asked was: how? Their answer, tucked into a famous one-sentence remark at the end of their 1953 paper, was that the base pairing itself suggests a copying mechanism. Separate the two strands, and each one is already a template for rebuilding its partner.
That mechanism is called semi-conservative replication: every new double helix keeps one old strand and gains one newly made strand. Half conserved, half fresh.
Proving it: Meselson and Stahl, 1958. They grew E. coli for many generations in a medium whose only nitrogen source was ¹⁵N, the heavy isotope. Every base in that DNA became heavy. Then they moved the bacteria to ordinary ¹⁴N medium and took samples after each 20-minute generation, spinning each sample in a caesium chloride density gradient, where DNA settles at the depth matching its density.
- Generation 0: one heavy band (¹⁵N/¹⁵N).
- Generation 1: a single band of intermediate density — hybrid ¹⁵N/¹⁴N. This one result already kills the conservative model, which predicted two separate bands.
- Generation 2: two bands in equal amounts — half hybrid, half light. This kills the dispersive model, which predicted a single band getting steadily lighter.
A parallel experiment by Taylor and colleagues (1958) on the root tips of Vicia faba (faba bean), using radioactive thymidine, showed the same semi-conservative pattern in the chromosomes of a higher organism.
Because the two parent strands are antiparallel, the fork opens in one direction but the two templates point opposite ways:
- On the template that runs 3′→5′ towards the fork, the new strand can be built continuously, chasing the fork. This is the leading strand.
- On the other template, building 5′→3′ means moving away from the fork. So the polymerase waits for a stretch to open, makes a short piece, drops off, and starts again further back. These short pieces are Okazaki fragments, and together they form the lagging strand.
- DNA ligase then seals the nicks between the fragments into one continuous strand.
The other players: helicase unwinds the duplex at the fork; topoisomerase relieves the twisting strain ahead of it; primase lays down the short RNA primer that gives polymerase its first free 3′ end; and the whole thing starts at a defined sequence called the origin of replication (ori). The building blocks are deoxyribonucleoside triphosphates, which do double duty — they are both the substrate and the energy source, since breaking off the two extra phosphates releases the energy for polymerisation.
In eukaryotes replication happens in the S phase of interphase, and because the genome is so large it starts at many origins at once rather than one.
The logic: the two original heavy strands are never destroyed. They simply end up in two different molecules, each paired with a new light strand. So there are always exactly 2 hybrid molecules, while the total number of molecules doubles each generation (2ⁿ).
Gen 1: 2 molecules, 2 hybrid → 100% hybrid, 0% light.
Gen 2: 4 molecules, 2 hybrid → 50% hybrid, 50% light.
Gen 3: 8 molecules, 2 hybrid → 25% hybrid, 75% light.
General formula: fraction hybrid = 2/2ⁿ = 1/2ⁿ⁻¹. And note: no fully heavy molecule ever reappears after generation 1.
Step 1. 18 minutes = 18 × 60 = 1080 s.
Step 2. Total rate = 4.6×10⁶ ÷ 1080 ≈ 4260 bp per second.
Step 3 — the part most students miss. The circular chromosome has one origin, and replication proceeds bidirectionally, so two forks are working at once. Rate per fork = 4260 ÷ 2 ≈ 2130 bp per second, which is where the commonly quoted figure of “about 2000 bp/second” comes from.
If a question asks for “the rate of polymerisation”, give the per-fork figure and say so. If it asks how fast the genome is copied, give the total.
Step 1. If the upper strand is 5′ on the left, it is 3′ on the right; the lower strand must be the reverse — 3′ on the left, 5′ on the right.
Step 2. New strands grow 5′→3′. For growth to point rightwards (towards the fork), the new strand needs its 3′ end on the right, so its template must have its 3′ end on the left. That is the lower strand.
Step 3. Therefore the lower strand is the leading-strand template, and the upper strand is the lagging-strand template. Each Okazaki fragment on it has its 5′ end nearer the fork and its 3′ end pointing away.
Shortcut for the exam: find the template whose 3′ end faces the oncoming fork — that one is leading. Everything else follows.
The Central Dogma
Francis Crick proposed the central dogma as a simple statement about the flow of information in a cell: DNA → RNA → protein. The blueprint stays in the vault; a work order is copied out; the work order is read on the assembly line. Add the photocopier — DNA making more DNA — and you have the whole map.
The one important exception is reverse transcription. Some viruses — retroviruses, such as HIV — carry their genome as RNA and use an enzyme called reverse transcriptase to make DNA from that RNA. Howard Temin’s work on such viruses is why the dogma is stated as a general rule rather than a law.
DNA → DNA: replication • DNA-dependent DNA polymerase • nucleus (S phase).
DNA → RNA: transcription • DNA-dependent RNA polymerase • nucleus.
RNA → protein: translation • the ribosome, whose 23S rRNA acts as the peptidyl transferase • cytoplasm.
RNA → DNA: reverse transcription • reverse transcriptase • only in retrovirus-infected cells.
Notice the naming habit: an enzyme called DNA-dependent RNA polymerase tells you its template first and its product second. Read the name and you already know the reaction.
Transcription In Prokaryotes And Eukaryotes
The work-order desk. Nobody carries the master blueprint out onto the factory floor — it would be damaged, and most of it is irrelevant to today’s job. Instead a clerk copies out the one page that is needed. That copy is RNA, and making it is transcription.
A transcription unit has three parts, and their order matters:
- A promoter, upstream, where RNA polymerase binds. It defines which strand is the template and therefore which way transcription runs.
- The structural gene itself — the part actually copied.
- A terminator, downstream, which tells the polymerase to stop.
The mechanics are the same everywhere. RNA polymerase binds the promoter, the duplex opens locally, and ribonucleotides are added 5′→3′ against the template. In bacteria, an accessory protein called the sigma (σ) factor helps with initiation, and another called the rho (ρ) factor assists termination; the core enzyme handles elongation.
Where prokaryotes and eukaryotes part company is in the tidying up afterwards.
A bacterium has no nuclear envelope, so the ribosome can start translating an mRNA while the far end of it is still being transcribed. There is no time and no need for processing. A eukaryote transcribes in the nucleus and translates in the cytoplasm, which buys time for three processing steps — and the primary transcript, hnRNA, needs all three:
- Splicing: eukaryotic genes are split genes. The coding stretches are exons (expressed) and the non-coding stretches in between are introns (intervening). Introns are cut out and exons joined.
- Capping: a methylated guanosine triphosphate cap is added to the 5′ end.
- Tailing: a poly-A tail of roughly 200–300 adenylate residues is added to the 3′ end.
Only after capping, tailing and splicing does the transcript become mature mRNA and leave the nucleus.
| Feature | Prokaryotic transcription | Eukaryotic transcription |
|---|---|---|
| Where | Cytoplasm (nucleoid region) | Nucleus |
| RNA polymerase | One enzyme makes all RNA types | Three separate polymerases (I, II, III) |
| Accessory factors | σ factor for initiation, ρ factor for termination | A large set of transcription factors |
| Transcript type | Often polycistronic — several genes on one mRNA | Monocistronic — one gene per mRNA |
| Introns | Generally absent | Present; removed by splicing |
| Processing | Essentially none needed | Capping, tailing and splicing all required |
| Coupling | Transcription and translation can occur together | Separated in space and time |
| RNA polymerase | What it transcribes in eukaryotes | Why it matters |
|---|---|---|
| RNA polymerase I | rRNAs — 28S, 18S and 5.8S | Builds the bulk of the ribosome, the assembly line itself |
| RNA polymerase II | hnRNA, the precursor of mRNA | Makes the work orders; this is the one gene-expression questions are about |
| RNA polymerase III | tRNA, 5S rRNA and snRNAs | Supplies the adaptors and the small helper RNAs |
Step 1 — mRNA. Pair each template base with its RNA partner (A→U, T→A, G→C, C→G), keeping the same left-to-right order, because the mRNA runs antiparallel to the template:
mRNA = 5′–AUGGCAUUCCGGUAA–3′
Step 2 — coding strand. It is the DNA complement of the template: 5′–ATGGCATTCCGGTAA–3′.
Step 3 — the check. Replace every T in the coding strand with U: ATGGCATTCCGGTAA → AUGGCAUUCCGGUAA. It matches the mRNA exactly. If it does not match, you have made an error somewhere — go back.
Always write the 5′ and 3′ labels. An unlabelled sequence is ambiguous and examiners deduct for it.
Step 1. Total intron length = 900 + 750 + 600 + 450 = 2700 nucleotides.
Step 2. Exonic length = 5400 − 2700 = 2700 nucleotides.
Step 3. In a split gene the exons and introns alternate and the transcript begins and ends with an exon, so with 4 introns there are 5 exons.
Worth noticing: half of this primary transcript was thrown away. That is completely normal in eukaryotes, and it is one reason the human genome contains far more DNA than protein-coding information.
The Genetic Code
The dictionary on the factory wall. The work order is written in a four-letter alphabet (A, U, G, C). The product is built from twenty different amino acids. Something has to translate one into the other, and that something is the genetic code.
Start with the arithmetic, because the arithmetic is the argument. If one base coded for one amino acid, you could only specify 4 amino acids. If two bases made a code word, 4² = 16 — still not enough for 20. Three bases give 4³ = 64 code words, comfortably more than enough. So the code must be at least a triplet code, and experiment confirmed it is exactly a triplet code.
The people who cracked it: George Gamow proposed the mathematical case for a triplet code; Marshall Nirenberg used a cell-free protein-synthesising system to read out codons; Har Gobind Khorana chemically synthesised RNA molecules with known repeating sequences, which made the readings unambiguous; and Severo Ochoa’s enzyme, polynucleotide phosphorylase, made those defined RNAs possible.
Of the 64 codons, 61 are sense codons that specify amino acids and 3 are stop codons — UAA, UAG and UGA — which specify no amino acid at all and act as full stops. AUG is special: it codes for methionine and serves as the start codon. Only two amino acids have a single codon each: methionine (AUG) and tryptophan (UGG).
Here is the complete standard code. You are not expected to memorise it — a codon table is normally supplied when it is needed — but you are expected to read one quickly and without errors.
| 1st+2nd base | 3rd = U | 3rd = C | 3rd = A | 3rd = G |
|---|---|---|---|---|
| UU_ | UUU → Phe | UUC → Phe | UUA → Leu | UUG → Leu |
| UC_ | UCU → Ser | UCC → Ser | UCA → Ser | UCG → Ser |
| UA_ | UAU → Tyr | UAC → Tyr | UAA → STOP | UAG → STOP |
| UG_ | UGU → Cys | UGC → Cys | UGA → STOP | UGG → Trp |
| CU_ | CUU → Leu | CUC → Leu | CUA → Leu | CUG → Leu |
| CC_ | CCU → Pro | CCC → Pro | CCA → Pro | CCG → Pro |
| CA_ | CAU → His | CAC → His | CAA → Gln | CAG → Gln |
| CG_ | CGU → Arg | CGC → Arg | CGA → Arg | CGG → Arg |
| AU_ | AUU → Ile | AUC → Ile | AUA → Ile | AUG → Met (START) |
| AC_ | ACU → Thr | ACC → Thr | ACA → Thr | ACG → Thr |
| AA_ | AAU → Asn | AAC → Asn | AAA → Lys | AAG → Lys |
| AG_ | AGU → Ser | AGC → Ser | AGA → Arg | AGG → Arg |
| GU_ | GUU → Val | GUC → Val | GUA → Val | GUG → Val |
| GC_ | GCU → Ala | GCC → Ala | GCA → Ala | GCG → Ala |
| GA_ | GAU → Asp | GAC → Asp | GAA → Glu | GAG → Glu |
| GG_ | GGU → Gly | GGC → Gly | GGA → Gly | GGG → Gly |
The adaptor: tRNA. Crick predicted, before anyone had seen one, that there must be a molecule with two faces — one that recognises the codon and one that carries the amino acid. That is transfer RNA. It has an anticodon loop complementary to the codon, and an amino acid acceptor end at the other extremity. Drawn flat it looks like a clover leaf; in reality it is folded into a compact L. There is a special initiator tRNA for AUG, and — this is the examinable point — there are no tRNAs for the stop codons. That absence is exactly why translation stops.
Mutations and the reading frame. Because the code is comma-less, the ribosome keeps the frame only by counting in threes from the start codon. Insert or delete a single base and every codon after that point is redrawn — a frameshift mutation. Substitute one base for another and only one codon changes — a point mutation. The classic example is sickle-cell anaemia: in the β-globin gene the sixth codon changes from GAG to GUG, so glutamic acid becomes valine, and that single swap reshapes the whole haemoglobin molecule.
Singlet: 4¹ = 4 code words — far fewer than 20. Rejected.
Doublet: 4² = 16 code words — still fewer than 20. Rejected.
Triplet: 4³ = 64 code words — more than enough. Accepted.
Of those 64: 61 sense codons and 3 stop codons (UAA, UAG, UGA). Since 61 codons share only 20 amino acids between them, the surplus shows up as degeneracy — on average about three codons per amino acid.
Follow-up often asked: a quadruplet code would give 4⁴ = 256 codons. It would work, but it would waste bases and make every frameshift even more damaging — nature took the smallest sufficient option.
(a) Original. AUG = Met, CAU = His, UGG = Trp, AAG = Lys, CCC = Pro, UAA = STOP.
→ Met–His–Trp–Lys–Pro (5 amino acids).
(b) After deleting the C. The message becomes AUGAUUGGAAGCCCUAA. Re-read it in threes from the start: AUG = Met, AUU = Ile, GGA = Gly, AGC = Ser, CCU = Pro, and then only AA is left — an incomplete codon, and the stop signal has vanished.
→ Met–Ile–Gly–Ser–Pro, and the ribosome would read on into whatever follows.
Comment: a single deleted base changed four of the five amino acids and destroyed the stop codon. That is the whole point of a frameshift. Compare it with a point mutation such as GAG→GUG in sickle-cell anaemia, where exactly one amino acid changes and the reading frame survives.
Translation
The assembly line. The work order has arrived in the cytoplasm; now it has to be turned into a product. Translation is the process of joining amino acids in the order dictated by the mRNA, and the machine that does it is the ribosome.
A ribosome has two subunits that stay apart until work begins. The large subunit contains the catalytic site that forms the peptide bond — and remarkably, the catalyst is not a protein at all. The 23S rRNA of the large subunit acts as the peptidyl transferase: it is a ribozyme, an RNA enzyme.
The steps, in order:
- Charging (aminoacylation) of tRNA. Each tRNA is loaded with its correct amino acid using energy from ATP. An uncharged tRNA is useless.
- Initiation. The mRNA binds the small subunit, the start codon AUG is located, the initiator tRNA settles onto it, and the large subunit joins.
- Elongation. Charged tRNAs arrive one by one, each anticodon pairing with the next codon. Peptidyl transferase forms the peptide bond, the ribosome moves on by exactly three bases, and the used tRNA leaves.
- Termination. A stop codon arrives. No tRNA can read it, so a release factor binds instead, the finished polypeptide is let go, and the subunits separate.
Step 1 — find the start. The sequence begins with AUG, so the reading frame starts at base 1.
Step 2 — cut into threes: AUG • GCC • UCU • UGC • AAG • UAA.
Step 3 — read the dictionary: AUG = Met, GCC = Ala, UCU = Ser, UGC = Cys, AAG = Lys, UAA = STOP.
Answer: Met–Ala–Ser–Cys–Lys — 5 amino acids.
The check that saves marks: 18 bases ÷ 3 = 6 codons, but one of them is a stop, so 6 − 1 = 5 amino acids. If your peptide has 6 residues you have wrongly translated the stop codon.
AUG • UCA • AGC • UGG • ACA • UAA → Met • Ser • Ser • Trp • Thr • STOP.
Answer: Met–Ser–Ser–Trp–Thr, 5 amino acids.
The trap: UCA and AGC are different codons but both give serine — that is degeneracy in action, and students often assume they must have made a mistake and “correct” one of them. They are right the first time. Note also UGG (tryptophan), one of only two amino acids with a single codon.
Step 1. Each amino acid needs one codon: 498 × 3 = 1494 nucleotides.
Step 2. Add the stop codon, which takes 3 more nucleotides but adds no amino acid: 1494 + 3 = 1497 nucleotides.
Step 3. The gene is double-stranded, so the coding region is 1497 base pairs. Its length = 1497 × 0.34 nm ≈ 509 nm.
Why “minimum”? Because a real gene also carries UTRs and, in a eukaryote, introns. The question asks for the coding minimum, so say so in one line and you protect the mark.
Regulation Of Gene Expression: The Lac Operon
Meet the foreman. A bacterium floating in your gut does not need to make lactose-digesting enzymes when there is no lactose about — that would be like running a production line all night with no raw material coming in. So the cell keeps that line switched off, and switches it on only when lactose actually arrives. The switch is the lac operon, worked out by François Jacob and Jacques Monod.
An operon is a cluster of genes controlled as a single unit. The lac operon has six parts, and the order is examinable:
- i — the regulator gene. It is transcribed all the time and makes the repressor protein. (The i stands for inhibitor, which is a helpful accident.)
- p — the promoter. Where RNA polymerase docks.
- o — the operator. The actual switch: a short sequence the repressor can sit on.
- z — codes β-galactosidase, which splits lactose into glucose and galactose.
- y — codes permease, which lets lactose into the cell.
- a — codes transacetylase.
Lactose absent — the line is off. The i gene makes repressor protein continuously. The repressor binds tightly to the operator. RNA polymerase can dock at the promoter but cannot move past the blocked operator, so z, y and a are not transcribed and no enzymes are made.
Lactose present — the line switches on. A little lactose leaks into the cell and is converted to allolactose, which is the true inducer. Allolactose binds the repressor and changes its shape so that it can no longer grip the operator. The repressor falls off, the operator is free, RNA polymerase runs through, and a single polycistronic mRNA is made carrying all three structural genes at once. Enzymes appear, lactose is digested, and when the lactose is gone the inducer disappears, the repressor recovers its shape, and the line shuts down again.
Reasoning: with no working repressor, nothing ever binds the operator. The operator is permanently free, so RNA polymerase transcribes z, y and a whether lactose is there or not.
Answer: β-galactosidase is produced constitutively — all the time, in both conditions. The cell wastes energy making an enzyme it may not need.
The teaching point: this mutant is the proof that the repressor is the thing doing the switching off. Remove the off-switch and the line never stops.
1. Before lactose: repressor bound to o; polymerase stalled at p; no mRNA; no enzymes. Operon OFF.
2. Just after lactose arrives: permease and β-galactosidase already exist in trace amounts, so a little lactose gets in and is converted to allolactose. Allolactose binds the repressor; the repressor changes shape and releases the operator.
3. During growth on lactose: polycistronic mRNA is transcribed; β-galactosidase, permease and transacetylase accumulate; lactose is digested. Operon ON.
4. After lactose is removed: allolactose is used up; the repressor returns to its active shape, re-binds the operator, transcription stops. Existing enzymes are gradually diluted and degraded. Operon OFF again.
Notice the elegance: the system is self-limiting. The signal that switches it on is the very molecule the enzymes destroy.
Genome, Human And Rice Genome Projects
A genome is the complete set of genetic material in an organism — the entire blueprint library, not one page of it. Once sequencing became fast enough, biologists asked the obvious next question: could we simply read the whole thing?
The Human Genome Project (HGP) ran from 1990 to 2003, coordinated in the United States by the Department of Energy and the National Institutes of Health, with the Wellcome Trust in Britain and laboratories in several other countries taking part. It was a genuinely enormous undertaking — NCERT quotes an estimated cost of around US $9 billion — and it is usually called the first great megaproject of biology.
Its stated goals were to identify all the genes in human DNA, determine the sequence of the roughly three billion base pairs, store the information in databases, improve the tools for analysing it, transfer the technologies to industry, and deal with the ethical, legal and social issues (ELSI) that such data would raise.
Two broad approaches were used. Expressed Sequence Tags (ESTs) concentrated only on the parts of the genome that are actually transcribed into RNA — fast, but it skips everything else. Sequence Annotation sequenced the whole genome, coding and non-coding alike, and worked out functions afterwards. In practice DNA was broken into fragments, cloned in vectors such as BACs and YACs, sequenced by the Sanger method on automated sequencers, and the overlapping pieces were reassembled by computer — which is why bioinformatics grew up alongside the project.
The findings that examiners ask about:
- The human genome contains about 3164.7 million base pairs — roughly 3.2 billion, and usually rounded to 3.3×10⁹ bp in calculations.
- An average gene is about 3000 bases long; the largest known human gene, dystrophin, runs to about 2.4 million bases.
- The estimated number of genes came out around 30,000 — startlingly few, given that earlier guesses ran to 80,000 or more.
- Less than 2% of the genome codes for protein. Repetitive sequences make up a very large share of the rest.
- 99.9% of the nucleotide bases are identical in all human beings. Everything that distinguishes us sits in the remaining fraction.
- The function of over 50% of the discovered genes was unknown at the time of reporting.
- Chromosome 1 carries the most genes (about 2968); the Y chromosome the fewest (about 231).
- About 1.4 million locations of single-base differences, called SNPs, were identified — the raw material for tracing disease genes.
The rice genome. Rice matters because it feeds more people than any other crop, and because its genome is small enough to be tractable and similar enough to other cereals to be useful as a reference. The International Rice Genome Sequencing Project, with substantial Indian participation, sequenced both cultivated subspecies, indica and japonica. The rice genome is roughly 430 million base pairs — about one-eighth the size of the human genome — yet it is generally credited with a larger number of genes than ours. Knowing the sequence lets breeders find genes for drought tolerance, pest resistance and yield directly, instead of waiting for them to show up by chance in a field trial. The same logic connects back to how crop plants are bred and improved in the Sexual Reproduction in Flowering Plants chapter.
E. coli: 4.6×10⁶ × 0.34 nm = 1.564×10⁶ nm = 1.56 mm.
Rice (4.3×10⁴ kb = 4.3×10⁷ bp): 4.3×10⁷ × 0.34 nm = 1.462×10⁷ nm = 1.46 cm.
Human (haploid): 3.3×10⁹ × 0.34 nm = 1.122×10⁹ nm = 1.12 m.
Ratios: human ÷ rice ≈ 77; human ÷ E. coli ≈ 717. So a millimetre and a half of DNA runs a bacterium; a metre of it runs you.
A note on a number you may meet: some books quote the E. coli chromosome as 1.36 mm long. Work backwards — 1.36 mm ÷ 0.34 nm gives exactly 4.0×10⁶ bp, not 4.6×10⁶. Whichever figure your textbook uses, show your working with 0.34 nm per base pair; a clearly worked answer is always credited.
DNA Fingerprinting
The security badge at the factory gate. If 99.9% of human DNA is the same in everyone, then identifying an individual by sequencing the whole genome would be absurdly wasteful — you would read three billion letters to find a handful of differences. DNA fingerprinting is the clever shortcut: instead of reading everything, read only the stretches that vary the most.
Those stretches are repetitive DNA. When genomic DNA is separated by density, the bulk forms one large peak and the repetitive sequences form small side peaks — hence the name satellite DNA. Satellite DNA does not code for protein, but it is highly polymorphic: the number of times a short sequence is repeated at a given site differs enormously between individuals. Such a site is a VNTR — a Variable Number of Tandem Repeats, and it is the basis of the whole technique. The method was developed by Alec Jeffreys in the mid-1980s; in India it was established by Lalji Singh at the Centre for Cellular and Molecular Biology, Hyderabad.
The classical procedure, in order:
- Isolation of DNA from the sample — blood, hair root, saliva, bone, semen.
- Digestion with restriction endonucleases, which cut the DNA into fragments at specific sequences.
- Separation of the fragments by size using gel electrophoresis — DNA is negatively charged, so it migrates towards the anode, and smaller fragments travel further.
- Blotting — transferring the separated fragments from the fragile gel onto a nitrocellulose or nylon membrane (Southern blotting).
- Hybridisation with a labelled VNTR probe, which sticks only to the repeat sequences it matches.
- Detection by autoradiography, giving the familiar ladder of bands — the fingerprint.
Applications: forensic identification in criminal investigations, parentage disputes, identification of victims in disasters, settling immigration claims, and studies of genetic diversity in populations and in wildlife conservation.
Step 1 — assign the maternal bands. The child’s bands 2 and 4 are present in the mother, so they are accounted for.
Step 2 — what must come from the father? Bands 5 and 8.
Step 3 — test each man. Man A has 5 and 8. ✓ Man B has 5 but not 8. ✗
Answer: Man A cannot be excluded; Man B is excluded.
Say it carefully. DNA fingerprinting excludes with certainty and includes with very high probability. Write “cannot be excluded as the father” rather than “is definitely the father” — it is both better science and the phrasing that earns the mark.
Exam Strategy For This Chapter
Chapter 5 sits in Unit VII – Genetics and Evolution, worth 20 marks, the heaviest unit on the paper. In practice the marks in this chapter cluster into four very predictable kinds of question.
- Experiment questions (Griffith, Avery, Hershey–Chase, Meselson–Stahl). Answer with the structure: what was done → what was observed → what it proves. The third part is where the marks are, and it is the part students leave out.
- Diagram questions (replication fork, lac operon, transcription unit, tRNA). Draw large, label everything, and put the polarity marks on. A neat labelled diagram often carries more marks than the paragraph beside it.
- Numericals (Chargaff, length, base-pair counts, amino-acid counts). Write the formula, substitute with units, then compute. Never present a bare number.
- Compare-and-contrast (DNA vs RNA, template vs coding strand, euchromatin vs heterochromatin, prokaryotic vs eukaryotic transcription). Always answer these in a table, never in prose — it is faster to write and easier to mark.
Once your notes are solid, move to whole papers under a clock. Working through the CBSE Class 12 Biology sample paper will show you very quickly which of the four question types you are still slow at.
Molecular Basis Of Inheritance Class 12 Important Questions — Rapid Revision
Use this as a spoken quiz the night before. Cover the answers, say each one out loud, and mark the ones that stumble.
- Who proved DNA is the genetic material, and how? Avery, MacLeod and McCarty by enzyme elimination; Hershey and Chase by isotope labelling in bacteriophage T2.
- Why is DNA a better genetic material than RNA? No 2′-OH, so it is chemically stable; thymine instead of uracil allows repair; it mutates more slowly.
- State Chargaff’s rules. A = T, G = C, therefore purines = pyrimidines in double-stranded DNA.
- What holds a nucleosome together? Negatively charged DNA wrapped on a positively charged histone octamer (two each of H2A, H2B, H3, H4), about 200 bp per nucleosome.
- What did the second generation of Meselson–Stahl show? Equal hybrid and light bands — ruling out the dispersive model.
- Why is the lagging strand discontinuous? Because polymerase can only build 5′→3′, so on one template it must work away from the fork in short Okazaki fragments.
- Which strand is the template? The one running 3′→5′; the promoter decides which strand that is.
- Name the three eukaryotic RNA polymerases and their products. I → rRNA; II → hnRNA (mRNA precursor); III → tRNA, 5S rRNA and snRNAs.
- What are the stop codons and the start codon? UAA, UAG, UGA; AUG, which also codes methionine.
- What is the peptidyl transferase? The 23S rRNA of the large ribosomal subunit — a ribozyme.
- Give the lac operon gene order and the inducer. i–p–o–z–y–a; the inducer is allolactose.
- What fraction of the human genome codes for protein? Less than 2%.
- What does a VNTR probe detect? Differences in the number of tandem repeats at a satellite-DNA locus — not the base sequence.
DNA Replication Class 12 Numericals: The Only Four Formulas You Need
Every numerical in this chapter — and there are more of them each year — comes out of these four lines. Copy them onto the inside cover of your notebook.
| You are asked for | Formula | Worked one-liner |
|---|---|---|
| Length of a DNA molecule | length = number of bp × 0.34 nm | 10⁶ bp → 3.4×10⁵ nm = 0.34 mm |
| Number of base pairs from a length | bp = length ÷ 0.34 nm | 3.4×10⁵ nm → 10⁶ bp |
| Number of helical turns | turns = bp ÷ 10, or length ÷ 3.4 nm | 10⁶ bp → 10⁵ turns |
| Number of nucleotides | nucleotides = 2 × bp (double-stranded) | 10⁶ bp → 2×10⁶ nucleotides |
| Base percentages | A = T, G = C, and A + T + G + C = 100 | A = 30% → T = 30%, G = C = 20% |
| Amino acids from a coding sequence | amino acids = (coding nucleotides ÷ 3) − 1 for the stop | 1497 nt → 498 amino acids |
Practice Worksheet
Ten original questions, in the mark pattern CBSE actually uses. Do them on paper with the codon table above beside you, then open the answers. Writing the answer badly and checking it is worth ten times more than reading a model answer you never attempted.
Q1 (1 mark)
Show Answer
Q2 (2 marks)
Show Answer
Step 2. A + T = 44%, therefore G + C = 100 − 44 = 56%.
Step 3. G = C, so G = C = 28% each.
Step 4. Purines = A + G = 22 + 28 = 50%. Pyrimidines = T + C = 22 + 28 = 50%.
Ratio of purines to pyrimidines = 1 : 1, exactly as Chargaff’s rules require for double-stranded DNA.
Q3 (3 marks)
Show Answer
Converting: 8.5×10⁴ nm ÷ 10³ = 85 µm (which is 0.085 mm).
(b) Turns = bp ÷ 10 = 2.5×10⁵ ÷ 10 = 2.5×10⁴ turns.
Cross-check: length ÷ pitch = 8.5×10⁴ nm ÷ 3.4 nm = 2.5×10⁴. ✓
(c) Nucleotides = 2 × bp = 2 × 2.5×10⁵ = 5×10⁵ nucleotides (each base pair contributes one nucleotide to each of the two strands).
Q4 (3 marks)
Show Answer
Step 2 — split into triplets: AUG • AAG • CUU • CCA • GGG • UAG.
Step 3 — read the code:
AUG = Methionine (also the start signal)
AAG = Lysine
CUU = Leucine
CCA = Proline
GGG = Glycine
UAG = STOP — no amino acid; a release factor binds here.
Polypeptide: Met–Lys–Leu–Pro–Gly — 5 amino acids.
Check: 18 nucleotides ÷ 3 = 6 codons; 6 − 1 stop = 5 amino acids. ✓
Q5 (3 marks)
Show Answer
mRNA = 5′–AUGCAUUGGAAGCCCUAA–3′
(b) The template strand is the complement of the coding strand and runs antiparallel to it:
template = 3′–TACGTAACCTTCGGGATT–5′
(c) AUG • CAU • UGG • AAG • CCC • UAA → Met • His • Trp • Lys • Pro • STOP.
Peptide = Met–His–Trp–Lys–Pro, 5 amino acids.
Note: UGG is tryptophan, one of only two amino acids specified by a single codon (the other being AUG for methionine).
Q6 (5 marks)
Show Answer
Observation after generation 1: a single band of intermediate density (¹⁵N/¹⁴N hybrid). Every molecule had one heavy parent strand and one new light strand. This eliminates the conservative model, which predicted two distinct bands (one fully heavy, one fully light).
Observation after generation 2: two bands in equal amounts — 50% hybrid and 50% light. This eliminates the dispersive model, which predicted a single band of steadily decreasing density.
Prediction for generation 3. The two original heavy strands are never destroyed; they simply end up in two different molecules. So there are always exactly 2 hybrid molecules, while the total doubles each generation.
Total molecules = 2³ = 8. Hybrid = 2. Light = 6.
→ 25% hybrid and 75% light (a ratio of 1 : 3).
Conclusion. DNA replication is semi-conservative. The same conclusion was reached independently by Taylor and co-workers using radioactive thymidine in the root tips of Vicia faba.
Q7 (3 marks)
Show Answer
i — regulator gene; makes the repressor protein continuously.
p — promoter; the binding site for RNA polymerase.
o — operator; the switch, where the repressor binds.
z — codes β-galactosidase, which hydrolyses lactose into glucose and galactose.
y — codes permease, which increases lactose entry into the cell.
a — codes transacetylase.
If the operator were deleted: the repressor would have nowhere to bind, so it could never block transcription. RNA polymerase would read straight through from the promoter into z, y and a regardless of whether lactose is present. The three enzymes would be made constitutively — the same outcome as an i⁻ mutant, but arrived at from the other end of the switch.
Q8 (2 marks)
Show Answer
2. Transcript type: prokaryotic mRNA is often polycistronic (several genes on one transcript); eukaryotic mRNA is monocistronic.
3. Processing: prokaryotic transcripts need essentially no processing and translation can begin before transcription finishes; eukaryotic hnRNA must be capped at the 5′ end, tailed with 200–300 adenylate residues at the 3′ end, and spliced to remove introns before it leaves the nucleus.
Q9 (5 marks)
Show Answer
Steps, in order:
1. Isolation of DNA from the sample (blood, hair root, saliva, bone).
2. Digestion with restriction endonucleases into fragments.
3. Separation of fragments by size using gel electrophoresis (DNA is negatively charged and moves to the anode; smaller fragments travel further).
4. Blotting — transfer of the separated fragments onto a nitrocellulose or nylon membrane (Southern blotting).
5. Hybridisation with a labelled VNTR probe.
6. Detection by autoradiography, producing the characteristic band pattern.
Why VNTRs? About 99.9% of human DNA is identical between individuals, and coding genes are among the most conserved parts of all — they would give almost the same pattern for everybody. VNTR (satellite) sequences are non-coding, so they are under little selective pressure and are free to vary; the number of tandem repeats at a given locus differs enormously between individuals. That high polymorphism is exactly what makes discrimination between people possible.
Q10 (3 marks)
Show Answer
120 × 3 = 360 nucleotides.
The stop codon adds 3 more nucleotides but contributes no amino acid:
360 + 3 = 363 nucleotides.
(b) The DNA is double-stranded, so 363 coding nucleotides correspond to 363 base pairs.
Length = 363 × 0.34 nm = 123.42 nm, i.e. about 123 nm.
Why “minimum”? A real mRNA also carries a 5′-UTR and a 3′-UTR, and a real eukaryotic gene contains introns as well — so the actual gene would be considerably longer. State that assumption in one line and the mark is safe.