DNA Structure & Replication

Question 1

(a) (4 marks)

• Chargaff rule 1: in dsDNA, %A = %T and %G = %C (1). Rule 2: within a single strand, purines ≈ pyrimidines (%A+%G ≈ %T+%C) approximately (1).
• Rule 1 is forced because A pairs only with T and G only with C across antiparallel complementary strands, so every A on one strand demands a T opposite it → equal totals (1).
• Rule 2 is an empirical near-equality not mandated by pairing — a single strand's own sequence is unconstrained by its complement's composition; it holds only statistically, so it is not a structural necessity (1).

(b) (5 marks)

• Total bases = 2 × 3120 = 6240 (1).
• A = T (Chargaff), so A = 1810 ⇒ T = 1810 (1).
• G + C = 6240 − (A+T) = 6240 − 3620 = 2620 (1).
• G = C = 2620/2 = 1310 each (1).
• G+C fraction = 2620/6240 = 0.4199 ≈ 0.42 (42%) (1).

(c) (9 marks) Semi-conservative: each original heavy strand persists forever; at n=0 all duplexes heavy.

• After 1 generation each heavy duplex → 2 hybrid (H:L). No fully-heavy remain because each duplex splits (1).
• Number of heavy strands stays 2 (constant); total strands = $2\cdot 2^n$; total duplexes = $2^n$ (1).
• Hybrid duplexes = number using a heavy strand = 2 (the two original heavy strands), for all n≥1 (1).

Closed forms (n ≥ 1): $H_n = 0,\quad B_n = \frac{2}{2^n}=2^{1-n},\quad L_n = 1 - 2^{1-n}$ (2)

Evaluations:

• Disproving conservative model: at n=1 the conservative model predicts 50% fully-heavy + 50% fully-light and NO hybrid; observation of 100% hybrid (single intermediate band) at n=1 disproves conservative replication (2).

(d) (6 marks)

• At generation 2 semi-conservative predicts TWO bands: hybrid (½) at intermediate density and light (½) at low density (2).
• Dispersive predicts ONE band, but at generation 2 it would sit at a density ¾ of the way toward light (each strand ¼ heavy on average), i.e. a single intermediate band — no fully-light DNA (2).
• Observation of a distinct fully-light band coexisting with hybrid at n=2 (two resolved bands) is incompatible with dispersive, which forbids any pure-light DNA and predicts one uniform band → semi-conservative confirmed (2).

Question 2

(a) (6 marks)

• Polymerase extends only at the free 3'-OH ⇒ synthesis strictly 5'→3' (1).
• Leading strand template runs 3'→5' into the fork, so its new strand is made continuously toward the moving fork (1).
• Lagging strand template runs 5'→3' into the fork, so new strand must be made away from the fork in short pieces as template is exposed → discontinuous (1).
• Helicase unwinds/separates the double helix at the fork (1). Primase lays RNA primers to give a 3'-OH start (1). Ligase seals nicks between Okazaki fragments (1).

(b) (4 marks)

• Two forks share the load; each covers half: $4.6\times10^6 / 2 = 2.3\times10^6$ bp per fork (1).
• Time = $2.3\times10^6 / 1000 = 2300$ s (1).
• = 2300/60 = 38.33 min (1) ≈ 38 min (1).

(c) (4 marks)

• Lagging template total (both forks) = $2.3\times10^6$ nt (each fork's lagging strand ≈ half chromosome) (1).
• Fragments = $2.3\times10^6 / 1200 \approx 1917$ (1).
• Ligation events ≈ fragments per strand − 1, ≈ 1917 (approximately equal; ligations ≈ fragments) (1).
• Assumption: fragments are uniformly 1200 nt and no gaps/overlaps; each fork's lagging strand ≈ half the genome (1).

(d) (6 marks)

def count_okazaki(strand, frag_len):
    import math
    fragments = math.ceil(strand / frag_len)  # fragments to cover the strand
    primers = fragments                       # one primer per fragment
    ligations = max(fragments - 1, 0)         # k fragments -> k-1 joints
    return (fragments, primers, ligations)
 
print(count_okazaki(2_300_000, 1200))

Output: (1917, 1917, 1916) (fragments and primers 1917; ligations 1916).

• Correct ceil handling (1), primers=fragments (1), ligations=fragments−1 (1), valid function (1), correct numeric output (2).

Question 3

(a) (5 marks)

• Lagging strand's terminal RNA primer, when removed, leaves a gap that polymerase cannot fill (no upstream 3'-OH) ⇒ chromosome shortens each round = end-replication problem (2).
• Telomeres are repetitive non-coding end sequences acting as expendable buffers (1). Telomerase carries an RNA template and extends the 3' overhang, allowing primer placement and restoring length (1).
• Somatic cells lack active telomerase, so telomeres shorten until critically short, triggering senescence — the replicative (Hayflick) limit (1).

(b) (6 marks)

• Start rate = $1\times10^{-5}$ (1).
• After proofreading (99% corrected → 1% remain): $1\times10^{-5}\times 0.01 = 1\times10^{-7}$ (2).
• After mismatch repair (99.9% corrected → 0.1% remain): $1\times10^{-7}\times 0.001 = 1\times10^{-10}$ (2).
• Final = $1\times10^{-10}$ = 1 error per $10^{10}$ bases, N ≈ $1\times10^{10}$ (1).

(c) (5 marks)

• Expected errors = $3.2\times10^9 \times 1\times10^{-10} = 0.32$ (3).
• ≈ 0.3 errors per genome copy (1).
• Significance: on average fewer than one mutation per genome replication, so the fidelity is high enough to preserve genetic information across cell divisions while still permitting rare mutations for evolution (1).

[
  {"claim":"Q1b: G=C=1310, GC fraction = 0.42","code":"A=1810;T=1810;total=2*3120;GC=total-(A+T);G=GC/2;frac=GC/total;result=(G==1310 and C:=GC/2==1310 and abs(frac-0.4199)<0.001)"},
  {"claim":"Q1c: B and L at n=1,2,3","code":"B=lambda n:2**(1-n);L=lambda n:1-2**(1-n);result=(B(1)==1 and L(1)==0 and B(2)==Rational(1,2) and L(2)==Rational(1,2) and B(3)==Rational(1,4) and L(3)==Rational(3,4))"},
  {"claim":"Q2b: replication time approx 38 min","code":"t=(4.6e6/2)/1000/60;result=abs(t-38.33)<0.5"},
  {"claim":"Q2c/d: fragments=1917, ligations=1916","code":"import math;f=math.ceil(2.3e6/1200);lig=f-1;result=(f==1917 and lig==1916)"},
  {"claim":"Q3b/c: final error rate 1e-10 and expected errors 0.32","code":"rate=1e-5*0.01*0.001;errs=3.2e9*rate;result=(abs(rate-1e-10)<1e-15 and abs(errs-0.32)<1e-6)"}
]