Section 14 — Solved Examples
Welcome to the dedicated problem set for Chapter 5 (Molecular Basis of Inheritance). The 32 worked examples below are arranged in three tiers — concept checks (Chargaff sums, complementary strands, experiment logic), application & scenarios (codon reading, Meselson–Stahl band predictions, lac operon switches), and analytical & multi-concept (frameshift effects, transcription-translation chains, integrated genome/fingerprinting problems).
Use this section as a last-mile revision tool. Most of these are deliberately not plain definitions — they ask you to calculate, predict and reason, because that is where Board and NEET marks are won in this chapter.
How to use this section
- Concept Checks (Q1–Q10): If you can't solve 9/10 quickly, revisit Sections 2–6.
- Application & Scenarios (Q11–Q22): The bread-and-butter of 2- and 3-mark questions — codon work, replication band logic, lac operon situations.
- Analytical & Multi-Concept (Q23–Q32): Multi-step reasoning that strings the central dogma together end to end.
Total target time: ~60 minutes for a full revision sweep.
Note: This section is for practice and revision only — there is no quiz at the end. Pair it with Section 15 (Board Exam-Pattern Questions) and Section 16 (NEET-Pattern Practice Questions).
Memory Capsule — Top Patterns to Master
Before you dive in, lock these 10 high-yield patterns in your head:
| # | Pattern / Fact | Where it's tested |
|---|---|---|
| 1 | Chargaff: A = T, G = C, so A + G = T + C = 50% | Base-percentage sums |
| 2 | A=T 2 H-bonds, G≡C 3 H-bonds → high G+C = harder to denature | Stability problems |
| 3 | Complementary strand: pair A-T, G-C and write it antiparallel | Sequence questions |
| 4 | Transcription: read the template (3'→5'); mRNA = coding strand with U for T | Strand problems |
| 5 | Meselson–Stahl: 1 gen → all hybrid; 2 gen → ½ hybrid + ½ light; n gen → 2 hybrid out of | Band predictions |
| 6 | DNA polymerase works only 5'→3' → leading continuous, lagging in Okazaki fragments + ligase | Replication logic |
| 7 | 64 codons = 61 coding + 3 stop (UAA, UAG, UGA); AUG = start + Met | Codon reading |
| 8 | Insert/delete bases not a multiple of 3 → frameshift scrambles everything downstream | Mutation effects |
| 9 | lac operon: no lactose → repressor on operator → OFF; lactose → repressor off → ON (negative regulation) | Operon scenarios |
| 10 | Fingerprinting rests on VNTR polymorphism in non-coding DNA; identical twins share the pattern | HGP / forensics |
Pro tip: Most problems here are built from one or two of these patterns. Name the pattern first, then solve.
Concept Checks (Q1–Q10)
Q1. A double-stranded DNA contains 22% adenine. Work out the percentage of each of the other three bases.
Answer: By Chargaff, A = T = 22%, so A + T = 44%, leaving 56% for G + C. Since G = C, each is 28%. So A 22%, T 22%, G 28%, C 28%.
Q2. One strand of a DNA segment reads 5'-ATGCCGTA-3'. Write its complementary strand with correct polarity.
Answer: Pair A-T, T-A, G-C, C-G and keep the strands antiparallel: 3'-TACGGCAT-5' (i.e. 5'-TACGGCAT-3' read the other way). Each base sits opposite its Watson–Crick partner.
Q3. Two DNA samples have the same length. Sample X is 60% G+C and sample Y is 30% G+C. Which melts (denatures) at a higher temperature, and why?
Answer: Sample X. Each G≡C pair has 3 hydrogen bonds versus 2 for A=T, so more G+C means more total bonds to break — a higher melting temperature.
Q4. A nucleosome contains about how many base pairs of DNA, and around what core? Roughly how long is the human nuclear DNA that has to be packed?
Answer: A typical nucleosome holds about 200 bp of DNA wrapped around a histone octamer. The full diploid human DNA is roughly 2.2 metres long, packed into a microscopic nucleus by this beads-on-string arrangement.
Q5. In Griffith's experiment, why did injecting heat-killed S strain plus live R strain kill the mice, even though neither component alone was lethal?
Answer: A transforming principle passed from the dead S cells to the live R cells, converting some R into virulent S bacteria. Live S were then recovered from the dead mice — showing heredity had been transferred.
Q6. In the Hershey–Chase experiment, which isotope ended up inside the bacteria, and what did this prove?
Answer: (which labels DNA) entered the bacteria, while (which labels protein) stayed outside with the empty phage coats. So the genetic material that goes in to direct new phages is DNA, not protein.
Q7. Why is RNA thought to have been the first genetic material, and why did DNA take over later?
Answer: RNA can both store information and act as a catalyst (ribozyme), so an RNA world needed no separate enzymes. But RNA is chemically reactive and unstable (the 2'-OH and uracil), so the more stable DNA evolved later as the better long-term store.
Q8. In transcription, if the template strand reads 3'-TACGGT-5', what is the mRNA sequence?
Answer: RNA is built complementary and antiparallel to the template, with U replacing T: template 3'-TACGGT-5' gives mRNA 5'-AUGCCA-3'. Note AUG appears as the first codon (start/methionine).
Q9. How many codons are there in total, how many code for amino acids, and which three are stop codons?
Answer: There are 64 codons in all; 61 code for amino acids and 3 are stop codons — UAA, UAG, UGA. AUG codes for methionine and also signals the start of translation.
Q10. In a messenger RNA, the translated stretch from the start codon up to and including the stop codon is 900 nucleotides long. How many amino acids will the resulting polypeptide contain?
Answer: Each codon is 3 nucleotides, so 900/3 = 300 codons. One of these is the stop codon (no amino acid), so the polypeptide has about 299 amino acids (the initiator Met may also be removed later).
Application & Scenarios (Q11–Q22)
Q11. A stretch of mRNA reads 5'-AUG-GCU-UUU-UAA-3'. Translate it: how many amino acids are added, and what ends the chain?
Answer: AUG = start/methionine, GCU = alanine, UUU = phenylalanine, then UAA = stop. So 3 amino acids are joined (Met–Ala–Phe); the chain ends at UAA because no tRNA reads a stop codon and a release factor terminates translation.
Q12. E. coli with fully heavy () DNA is shifted to medium. Predict the band pattern after one and after two generations.
Answer: After one generation, all DNA is a single hybrid (intermediate) band — one old heavy strand + one new light strand. After two generations, half the molecules are hybrid and half are fully light, giving a hybrid band plus a light band. This is the signature of semiconservative replication.
Q13. Continuing the Meselson–Stahl experiment, what fraction of the DNA molecules is still hybrid after three generations in ?
Answer: The two original heavy strands are always conserved, so exactly 2 molecules are hybrid out of , i.e. 1/4 (25%). The other 6/8 are fully light. The number of hybrid molecules stays at 2 while the light ones keep doubling.
Q14. A DNA polymerase can only add nucleotides in the 5'→3' direction. Explain why this forces one new strand to be made discontinuously.
Answer: The two template strands are antiparallel, but the enzyme reads only 3'→5' (building 5'→3'). On the strand running the "wrong" way the enzyme must work away from the fork in short pieces — Okazaki fragments (the lagging strand), later joined by DNA ligase; the other strand (leading) is made continuously. Hence replication is semi-discontinuous.
Q15. DNA polymerase in E. coli adds about 2000 base pairs per second. Show that this rate is consistent with the genome (~4.6 × 10⁶ bp) being replicated in about 38 minutes, and state the twin demands this places on the enzyme.
Answer: 4.6 × 10⁶ bp ÷ 2000 bp per second ≈ 2300 seconds ≈ 38 minutes, matching the observed replication time. The enzyme must therefore be both very fast and highly accurate — any uncorrected error becomes a mutation. Speed plus accuracy is the hallmark of DNA-dependent DNA polymerase.
Q16. For the transcription unit, the coding strand reads 5'-ATGGCATTC-3'. Identify the template strand and write the mRNA.
Answer: The template is the complement, antiparallel: 3'-TACCGTAAG-5'. The mRNA has the same sequence as the coding strand but with U for T: 5'-AUGGCAUUC-3'. This is why the non-template strand is called the "coding" strand.
Q17. A single base is inserted near the start of a gene's coding sequence. Predict the effect on the protein, and contrast it with inserting three bases.
Answer: One inserted base causes a frameshift — every codon downstream is misread, so the protein is almost entirely wrong (often truncated at a new stop codon). Inserting three bases (a multiple of 3) only adds one extra amino acid and keeps the rest of the reading frame intact.
Q18. In the lac operon, predict what happens to enzyme production (a) when lactose is absent and (b) when lactose is added.
Answer: (a) No lactose → the repressor (from gene i) binds the operator and blocks RNA polymerase → operon OFF, no enzymes. (b) Lactose → it binds and inactivates the repressor, which leaves the operator → RNA polymerase transcribes z, y, a → β-galactosidase, permease and transacetylase are made.
Q19. A mutant E. coli has a defective i gene that produces a non-functional repressor. Predict the lac operon's behaviour with and without lactose.
Answer: A non-functional repressor can never bind the operator, so RNA polymerase is never blocked. The operon is constitutively ON — the enzymes are made whether or not lactose is present (constitutive expression). This shows the repressor is the OFF switch in negative regulation.
Q20. State the products of the lac operon's z, y and a genes, and explain why it makes sense to control them as one unit.
Answer: z → β-galactosidase (splits lactose into galactose + glucose), y → permease (lets lactose into the cell), a → transacetylase. All three handle lactose, so grouping them under one promoter lets the cell switch the whole lactose-metabolising kit on or off together, only when lactose is available.
Q21. Two suspects' DNA fingerprints are compared with DNA from a crime scene. Suspect 1's bands match the crime-scene sample exactly; Suspect 2's do not. What does this indicate, and on what molecular variation does it rest?
Answer: The matching bands point to Suspect 1 as the source of the crime-scene DNA. The technique rests on VNTR polymorphism — the highly variable copy number of tandem repeats in non-coding satellite DNA, which gives each person an (almost) unique banding pattern.
Q22. Arrange the steps of DNA fingerprinting in order, and name the technique used to transfer fragments to the membrane.
Answer: Order: isolation of DNA → digestion with restriction endonucleases → separation by gel electrophoresis → blotting onto a nylon/nitrocellulose membrane → hybridisation with a labelled VNTR probe → detection by autoradiography. The transfer step is Southern blotting.
Analytical & Multi-Concept (Q23–Q32)
Q23. A DNA molecule has 1500 base pairs. (a) How many hydrogen bonds hold it together if 40% of the base pairs are G–C? (b) Which would be tougher to separate — this molecule or one with 20% G–C?
Answer: (a) G–C pairs = 40% of 1500 = 600 pairs × 3 bonds = 1800; A–T pairs = 900 × 2 = 1800; total = 3600 H-bonds. (b) The molecule here (40% G–C) is tougher to separate, because more G≡C pairs mean more 3-bond pairs overall.
Q24. A template strand reads 3'-TAC-CGG-AAT-ACT-5'. Transcribe it to mRNA and then translate it (use AUG = Met, the rest read in order).
Answer: mRNA (complementary, antiparallel, U for T) = 5'-AUG-GCC-UUA-UGA-3'. Translating: AUG = methionine (start), GCC = alanine, UUA = leucine, then UGA = stop. So the peptide is Met–Ala–Leu before termination.
Q25. Three statements about replication:
- DNA polymerase can initiate a new chain on its own without a primer-like start.
- The leading strand is synthesised continuously towards the fork.
- DNA ligase seals the gaps between Okazaki fragments on the lagging strand.
Which are correct?
Answer: (2) and (3) are correct; (1) is wrong. (1) ✗ DNA polymerase cannot start on its own — synthesis must begin at a defined origin. (2) ✓ the leading strand grows continuously toward the fork. (3) ✓ ligase joins the discontinuous Okazaki fragments.
Q26. After the shift to in a Meselson–Stahl-type experiment, a student samples DNA after four generations. Give (a) the number of hybrid molecules out of the total, and (b) the percentage of light DNA.
Answer: Total molecules = . (a) Only the 2 original heavy strands survive, so 2 of 16 molecules are hybrid (1/8). (b) The remaining 14/16 are fully light = 87.5% light DNA. The hybrid never disappears but its proportion keeps falling.
Q27. A point mutation changes the mRNA codon GAG to GUG in the β-globin gene. (a) What amino-acid change results? (b) Name the disease and classify the mutation type.
Answer: (a) GAG codes for glutamic acid (Glu) and GUG codes for valine (Val), so Glu is replaced by Val at position 6. (b) This causes sickle-cell anaemia; it is a substitution / missense (point) mutation — a single base change with a single amino-acid change but severe effects.
Q28. Why is a stop codon able to end translation even though it is a perfectly normal triplet of bases? Tie your answer to tRNA.
Answer: Translation works because every codon is read by a complementary tRNA carrying an amino acid. There are no tRNAs for UAA, UAG or UGA, so when the ribosome reaches a stop codon, no tRNA can pair with it. A release factor instead binds and frees the completed polypeptide — termination.
Q29. Trace the path of a single gene's information from DNA to functional protein, naming the enzyme/molecule responsible at each step (the central dogma).
Answer: DNA → RNA → protein. DNA-dependent RNA polymerase transcribes the template strand into mRNA (template read 3'→5', A pairs with U). The mRNA leaves the nucleus; on the ribosome (whose rRNA is a ribozyme), tRNA adapters read each codon and deliver the matching amino acid, joined by peptide bonds — translation — yielding the polypeptide that folds into the functional protein.
Q30. Integrative — Human Genome Project: The human genome is ~ bp with ~30,000 genes, yet less than 2% codes for protein and any two people are 99.9% identical. (a) Where does the 0.1% difference mostly lie? (b) How is that 0.1% exploited practically?
Answer: (a) The 0.1% variation lies largely in non-coding, repetitive (satellite) DNA and as single-nucleotide differences (SNPs), since changes there rarely affect survival and accumulate freely. (b) This variation — especially VNTRs — is exploited in DNA fingerprinting for forensics and parentage testing, where copy-number differences give person-specific band patterns.
Q31. A researcher finds a DNA-rich extra peak (satellite DNA) when genomic DNA is spun in a density gradient, and notes it is highly polymorphic. (a) Why is satellite DNA so variable? (b) Why does this make it ideal for fingerprinting rather than coding genes?
Answer: (a) Satellite DNA is mostly non-coding, so mutations there rarely harm the organism and accumulate over generations — producing high polymorphism (VNTRs). (b) Coding genes are conserved (mutations are usually selected against), so they vary little between people; the freely varying satellite/VNTR regions give the person-to-person differences that fingerprinting needs.
Q32. Master integrative question. A gene's coding strand reads 5'-ATG-AAA-TTT-GGG-TAA-3'.
(a) Write the mRNA. (b) Translate it, naming start and stop. (c) Now a single base (C) is inserted right after the start codon's AUG. Qualitatively, what happens to the protein?
Answer:
(a) mRNA = coding strand with U for T = 5'-AUG-AAA-UUU-GGG-UAA-3'.
(b) AUG = Met (start) → AAA = lysine → UUU = phenylalanine → GGG = glycine → UAA = stop. Peptide: Met–Lys–Phe–Gly.
(c) Inserting one base after AUG causes a frameshift: the downstream codons are all re-grouped and misread, so the amino-acid sequence after methionine is completely altered and a premature stop often appears — yielding a non-functional protein. This shows why the genetic code's comma-less triplet reading makes single-base indels so damaging.
End of Section 14
You have now worked through 32 examples spanning DNA structure and Chargaff sums, packaging, the classic experiments, replication and Meselson–Stahl predictions, transcription, the genetic code, translation, the lac operon, the Human Genome Project and DNA fingerprinting. For more sustained practice, move on to Section 15 (Board Exam-Pattern Questions) and Section 16 (NEET-Pattern Practice Questions).
Self-assessment: Aim for 24 / 32 correct on a first attempt. If you scored below 18, revisit Sections 2 and 6 (structure, replication) and 8–11 (transcription, code, translation, lac operon) — that is where most Board and NEET marks in this chapter come from.