DNA Structure & Replication

Question 1 [10 marks]

(a) In dsDNA, Chargaff: A = T and G = C. (1) A = 28% ⇒ T = 28%. (1) A + T = 56%, so G + C = 100 − 56 = 44%. (1) G = C = 22% each. (1) Answer: T = 28%, G = 22%, C = 22%.

(b) Chargaff's rules (A=T, G=C) apply only to double-stranded DNA because pairing is between the two strands. (1) On a single strand there is no requirement that A equals T, so a value of 32% A is perfectly possible. (1) The single strand's A pairs with the T of the complementary strand, not with T on the same strand — so no equality constraint holds within one strand. (1)

(c) Higher melting temperature ⇒ more hydrogen bonds/stronger stacking. (1) G–C pairs have three H-bonds vs A–T's two, so more energy is needed to separate them. (1) Therefore the second organism has a higher G–C content (more G≡C pairs). (1)

Question 2 [12 marks]

(a) Semi-conservative: each new molecule = one old + one new strand. (grid below)

• After 1 round: all molecules intermediate (¹⁵N/¹⁴N hybrid). → 100% intermediate. (2)
• After 2 rounds: 4 molecules: 2 intermediate + 2 light → 50% intermediate, 50% light. (2)
• After 3 rounds: 8 molecules: 2 intermediate + 6 light → 25% intermediate, 75% light (1:3). (2)

(b) Conservative prediction:

• After 1 round: 50% heavy + 50% light (no intermediate). (1)
• After 2 rounds: 25% heavy + 75% light. (1) The actual result after 1 round is a single intermediate band with no fully heavy or fully light DNA. (1) Conservative replication requires a heavy band to persist after 1 round; its absence, plus the single intermediate band, rules out conservative and supports semi-conservative. (1)

(c) The two original ¹⁵N strands are never destroyed — they are conserved intact within two molecules, becoming a vanishingly small fraction but never fully eliminated. (2) [So heavy strands persist, though intermediate density becomes undetectable.]

Question 3 [12 marks]

(a) Template: 5'-TACGGATCCGTA-3' Complement (antiparallel): 3'-ATGCCTAGGCAT-5', equivalently 5'-TACGGATCCGTA-3'...

Careful: complement of each base, written antiparallel: Template 5'-T A C G G A T C C G T A-3' New strand runs antiparallel: 3'-A T G C C T A G G C A T-5' (1 for correct bases, 1 for complement A↔T/G↔C, 1 for correct 5'/3' labelling). (3)

(b) The template is read 3'→5' by polymerase; synthesis is 5'→3'. The template above is written 5'→3' left-to-right. With the fork opening left→right, this template's 3' end is on the right, so polymerase moves right→left — against the fork direction ⇒ this strand is made discontinuously = lagging strand. (2 for conclusion, 2 for polarity justification.) (4)

(c) Primase lays down a short RNA primer providing a free 3'-OH, because DNA polymerase cannot initiate synthesis de novo — it can only extend an existing 3' end. (1) Leading strand needs ~1 primer (continuous). (1) Lagging strand needs many primers (one per Okazaki fragment). (1)

(d) The lagging strand product is affected. (1) Ligase seals the nicks between Okazaki fragments; without it the fragments remain unjoined, so the lagging strand stays fragmented. (1)

Question 4 [10 marks]

(a) On the lagging strand, synthesis requires an RNA primer at the extreme 3' end of the template. (1) When the terminal primer is removed, there is no upstream 3'-OH for polymerase to extend into that gap. (1) So the very end cannot be filled. (1) Each round the daughter strand is therefore shorter at the 5' end → chromosome shortens (end-replication problem). (1)

(b) Normal somatic cells have little/no telomerase, so telomeres shorten each division until a limit (senescence) after ~50 divisions (Hayflick limit). (1.5) Cancer cells reactivate telomerase, which extends telomeres and prevents critical shortening, allowing unlimited division. (1.5)

(c) Bacterial chromosomes are circular — there are no free ends. (1) Replication proceeds around the loop and the last primer's gap can be filled by extension from adjacent DNA and sealed by ligase. (1) Hence no terminal shortening occurs. (1)

Question 5 [6 marks]

(a) When polymerase inserts a mismatched base, the 3'→5' exonuclease removes the incorrect nucleotide from the newly made 3' end. (1) It must act 3'→5' because the error is at the growing 3' end, which is where the mismatch just occurred. (1) Removing it allows correct re-synthesis, greatly lowering the error rate (proofreading). (1)

(b) Sulfur (³⁵S) is present in protein (in cysteine/methionine) but essentially absent from DNA; phosphorus (³²P) is abundant in the DNA backbone but absent from protein. (1) This lets the two molecules be tracked separately. (1) The ³²P (DNA) entered the bacteria, while ³⁵S (protein) stayed outside, proving DNA is the genetic material. (1)

[
  {"claim":"Chargaff Q1: T=28, G=C=22 sum to 100 with A=28","code":"A=28; T=A; GC=100-A-T; G=GC/2; C=GC/2; result=(T==28 and G==22 and C==22 and A+T+G+C==100)"},
  {"claim":"Meselson-Stahl semi-conservative after 3 rounds: 2 intermediate, 6 light of 8","code":"gens=3; total=2**gens; inter=2; light=total-inter; result=(total==8 and inter==2 and light==6 and light/inter==3)"},
  {"claim":"After 2 rounds semi-conservative: 50% intermediate 50% light","code":"total=2**2; inter=2; light=total-inter; result=(inter/total==Rational(1,2) and light/total==Rational(1,2))"},
  {"claim":"Complement bases count: template TACGGATCCGTA has 12 bases and A-count of complement equals template T-count","code":"tem='TACGGATCCGTA'; comp={'A':'T','T':'A','G':'C','C':'G'}; new=''.join(comp[b] for b in tem); result=(len(tem)==12 and new.count('A')==tem.count('T'))"}
]