DNA Structure & Replication

Question 1 (10 marks)

(a) Rule: In double-stranded DNA, $\%A = \%T$ and $\%G = \%C$ (1). This holds because A pairs specifically with T (2 H-bonds) and G with C (3 H-bonds) (1); each base on one strand is matched by its complement on the other, so their amounts must be equal (1).

(b) Since $\%A = \%T$: $\%T = 28\%$ (1). $\%A + \%T = 56\%$, so $\%G + \%C = 100 - 56 = 44\%$ (1). Since $\%G = \%C$: $\%G = \%C = 44/2 = 22\%$ each (2). Answer: T = 28%, G = 22%, C = 22%.

(c) The claim is generally false for single-stranded DNA (1). Chargaff's rule arises from complementary base pairing between the two strands (1); a single strand has no complementary partner within it, so A and T need not be equal (1).

Question 2 (12 marks)

(a) After one round: (i) Semi-conservative: each daughter molecule = one old (heavy) + one new (light) strand → all DNA is hybrid (intermediate/HL density) (2). (ii) Conservative: the parental duplex stays intact and a wholly new light duplex is made → a mix of heavy (HH) and light (LL), no hybrid (2). Since the actual result after one round was a single intermediate band, conservative model is ruled out and semi-conservative supported (2).

(b) Track strand composition. Start: 1 HH molecule (2 heavy strands).

• Heavy strands never increase — there remain only 2 heavy strands total throughout.
• After 3 rounds: number of molecules = $2^3 = 8$ (1).
• The 2 heavy strands each pair with a new light strand → 2 hybrid (HL) molecules (1).
• Remaining $8 - 2 = 6$ molecules are LL (1).
• Ratio HL : LL = $2 : 6 = 1 : 3$ (1).

(c) Fraction hybrid after $n$ rounds $= \dfrac{2}{2^n} = \dfrac{1}{2^{n-1}}$ (2).

Question 3 (14 marks)

(a) (award up to 10, ~1.5 each component correctly placed and explained)

1. Helicase unwinds and separates the two parental strands by breaking H-bonds, forming the replication fork.
2. Primase synthesises a short RNA primer providing a free 3′-OH to start synthesis.
3. DNA polymerase extends from the primer, adding nucleotides 5′→3′ and pairing them complementarily to the template.
4. Leading strand is synthesised continuously toward the fork (template read 3′→5′).
5. Lagging strand is synthesised discontinuously, away from the fork, in short pieces.
6. Okazaki fragments are these short lagging-strand pieces, each needing its own primer.
7. DNA ligase seals the nicks between fragments after primers are replaced, joining the sugar-phosphate backbone. Marks: correct order + correct role for each = up to 10.

(b) DNA polymerase can only add to a free 3′-OH, i.e. it synthesises strictly 5′→3′ (1). The two template strands are antiparallel (1). On the strand whose template runs 5′→3′ toward the fork, continuous synthesis toward the fork is impossible; polymerase must move away from the fork (1). So as the fork opens, synthesis is repeatedly re-primed and made in short backward fragments — i.e. discontinuously (1).

Question 4 (8 marks)

(a) Template: $5'-\text{TACGGATCG}-3'$ Complement (antiparallel): $3'-\text{ATGCCTAGC}-5'$ (3) Correct antiparallel labelling of 5′/3′ ends (1). (Equivalently written 5′→3′: $5'-\text{CGATCCGTA}-3'$.)

(b) DNA polymerase adds nucleotides to the 3′ end of the growing strand (1); it therefore builds the new strand 5′→3′ (1). To pair antiparallel bases it must read the template in the 3′→5′ direction (1). These are consistent because the new strand and template are antiparallel, so moving 3′→5′ along the template corresponds to extending 5′→3′ on the new strand (1).

Question 5 (10 marks)

(a) On the lagging strand, each Okazaki fragment needs an RNA primer (1). When the terminal primer at the chromosome end is removed, there is no upstream 3′-OH for polymerase to extend into that gap (1), so the very end cannot be fully replicated (1). Chromosomes therefore shorten a little each division (1).

(b) Telomerase carries its own RNA template and extends the 3′ end of the parental strand with repeated telomere sequences (1), providing room for primer placement so the end is replicated without losing coding DNA (1). Cells lacking telomerase progressively shorten telomeres, leading to replicative senescence / cell ageing (or apoptosis) (1).

(c) Proofreading: performed by DNA polymerase itself during synthesis, using its 3′→5′ exonuclease to remove a just-added mismatched base (1). Mismatch repair: a separate system acting after replication, detecting and excising mismatches that escaped proofreading (1). Difference: timing/agent — proofreading is co-synthetic by the polymerase; mismatch repair is post-replicative by dedicated enzymes (1).

Question 6 (6 marks)

(a) $^{32}P$ labels DNA (phosphate backbone; protein has no phosphate) (1); $^{35}S$ labels protein (sulfur in cysteine/methionine; DNA has no sulfur) (1). Because each isotope tags only one molecule type, the two can be tracked independently (1).

(b) $^{32}P$ (DNA) entered the bacteria while $^{35}S$ (protein) stayed outside in the phage coat (1). Since only DNA entered and directed production of new phage (1), DNA — not protein — is the genetic material (1).

[
  {"claim":"If A=28% then T=28, G=C=22% and totals 100%","code":"A=28; T=A; GC=100-A-T; G=GC/2; C=G; result=(T==28 and G==22 and C==22 and A+T+G+C==100)"},
  {"claim":"After 3 rounds from HH in light medium: 2 hybrid, 6 light, molecules=8, ratio HL:LL=1:3","code":"n=3; total=2**n; hybrid=2; light=total-hybrid; result=(total==8 and hybrid==2 and light==6 and Rational(hybrid,light)==Rational(1,3))"},
  {"claim":"Fraction hybrid after n rounds = 1/2**(n-1); check n=1,2,3 give 1,1/2,1/4","code":"f=lambda n: Rational(2,2**n); result=(f(1)==1 and f(2)==Rational(1,2) and f(3)==Rational(1,4) and f(3)==Rational(1,2**(3-1)))"},
  {"claim":"Complement of 5'-TACGGATCG-3' read 5'->3' is CGATCCGTA","code":"comp={'A':'T','T':'A','G':'C','C':'G'}; t='TACGGATCG'; c=''.join(comp[b] for b in t)[::-1]; result=(c=='CGATCCGTA')"}
]