DNA Structure & Replication

Q1. (3 marks)

• Rule 1: The amount of adenine equals thymine (A = T), and guanine equals cytosine (G = C). (2) — because A pairs with T and G pairs with C by complementary base pairing.
• Rule 2: The total amount of purines equals the total amount of pyrimidines, i.e. (A + G) = (T + C). (1)

Q2. (4 marks)

• By Chargaff, A = T, so T = 20%. (1)
• A + T + G + C = 100%, so G + C = 100 − (20 + 20) = 60%. (1)
• Since G = C, each = 60/2 = 30%. (1)
• Answer: G = 30%, C = 30%, T = 20%. (1)

Q3. (4 marks)

• Antiparallel: the two strands run in opposite directions — one strand is oriented 5′→3′ while its partner runs 3′→5′. (2)
• 5′ end: the end of a DNA strand where the carbon-5 of the terminal sugar carries a free phosphate group (not linked to another nucleotide). (2)

Q4. (5 marks)

• Griffith worked with two strains of Streptococcus pneumoniae: smooth (S, virulent) and rough (R, harmless). (1)
• Living S killed mice; living R did not; heat-killed S did not kill mice. (1)
• When heat-killed S was mixed with living R and injected, the mice died, and living S bacteria were recovered. (2)
• Conclusion: a "transforming principle" passed from the dead S cells to the R cells, genetically transforming them — evidence that a chemical substance carries hereditary information. (1)

Q5. (5 marks — 1 each)

• (a) DNA helicase: unwinds/separates the two strands by breaking hydrogen bonds between bases. (1)
• (b) Primase: synthesises a short RNA primer to provide a free 3′-OH for DNA polymerase to start. (1)
• (c) DNA polymerase: adds nucleotides to the 3′ end of the growing strand, synthesising in the 5′→3′ direction. (1)
• (d) DNA ligase: joins Okazaki fragments by sealing the sugar-phosphate backbone (phosphodiester bonds). (1)
• (e) Telomerase: extends the telomeres at chromosome ends to prevent loss of DNA during replication. (1)

Q6. (4 marks)

• Semi-conservative replication: each new DNA molecule consists of one original (parental) strand and one newly synthesised strand. (3) — the original strand is conserved and serves as template.
• Scientists: Meselson and Stahl. (1)

Q7. (5 marks)

• Leading strand: synthesised continuously in the 5′→3′ direction, moving toward the replication fork. (2)
• Lagging strand: synthesised discontinuously, away from the fork, in short pieces. (2)
• These short pieces are Okazaki fragments, later joined by DNA ligase. (1)

Q8. (4 marks)

• Total bases = 3000 bp × 2 = 6000 bases. (1)
• C = 900, and G = C = 900. (1)
• A + T = 6000 − (900 + 900) = 4200; A = T so A = 4200/2 = 2100. (1)
• Answer: A = 2100. (1)

Q9. (2 marks — 1 each, any two)

• Two polynucleotide strands wound into a double helix. (1)
• Strands are antiparallel / held together by hydrogen bonds between complementary base pairs (A–T, G–C). (1)
• (Also acceptable: sugar-phosphate backbone on outside, bases inside; right-handed helix.)

Q10. (4 marks)

• Importance: it reduces the error rate of replication, maintaining the accuracy/fidelity of the genetic code and preventing harmful mutations. (2)
• Correction: when DNA polymerase inserts an incorrect (mismatched) nucleotide, it detects the mismatch, uses its exonuclease activity to remove the wrong base from the 3′ end, then inserts the correct nucleotide and continues. (2)

[
  {"claim":"Q2: If A=20%, then G=C=30% and T=20%","code":"A=20; T=A; GC=100-(A+T); G=GC/2; C=GC/2; result=(T==20 and G==30 and C==30)"},
  {"claim":"Q2: percentages sum to 100","code":"A=20; T=20; G=30; C=30; result=(A+T+G+C==100)"},
  {"claim":"Q8: 3000 bp with 900 C gives A=2100","code":"total=3000*2; C=900; G=C; A=(total-(C+G))/2; result=(A==2100)"},
  {"claim":"Q8: A=T and G=C consistency","code":"total=6000; C=900; G=900; A=2100; T=2100; result=(A==T and G==C and A+T+G+C==total)"}
]