Evolution & Natural Selection

Question 1

(a) (4)

• Mutations accumulate in DNA/protein sequences over time; the more time since two lineages diverged, the more independent substitutions accumulate, so difference count ∝ divergence time. (2)
• Assumptions (any two, 1 each): (i) mutations accumulate at a roughly constant rate over time/lineages; (ii) most substitutions are neutral (not strongly selected); (iii) sites are independent / back-mutation negligible. (2)

(b) (3)

• $d_{H,T} = 21$; $d = 2\mu t \Rightarrow t = \dfrac{d}{2\mu} = \dfrac{21}{2(0.30)}$ (1 setup)
• $t = \dfrac{21}{0.60} = 35$ million years. (2)

(c) (6) UPGMA:

• Smallest distance = 12 (H, Ho). Cluster {H,Ho}; node height = 12/2 = 6. (2)
• New distances (average): to T = (21+19)/2 = 20; to Y = (45+45)/2 = 45. (1)
• Next smallest = 20 → cluster {H,Ho} with T; node height = 20/2 = 10. (2)
• Finally join Y; distance = (47 + 45)... average of {H,Ho} and T members to Y = (45+45+47)/3 ≈ 45.67; node height ≈ 22.8. (1)
• Tree: (((H,Ho):6, T):10, Y):~22.8 — H & Ho most closely related, Y is outgroup. (reasoning credited)

(d) (4)

• Claim is incorrect. (1)
• Cytochrome-c in yeast and humans is homologous: descended from a common ancestral gene (same origin, similar sequence/function). (2)
• Analogy at molecular level would require similar function arising from independent origins (e.g. convergent enzyme active sites) — not the case here since the whole protein is orthologous. (1)

(e) (3)

def count_diff(seq1, seq2):
    return sum(1 for a, b in zip(seq1, seq2) if a != b)

• Correct zip/comparison (2); returns integer count (1).

Question 2

(a) (3) — Variation: dark vs light alleles exist in the population (1). Heritability/overproduction: colour is genetically inherited and moths overproduce offspring (1). Differential survival: on dark (polluted) bark, light moths are seen by predators and die more → dark survive/reproduce more (1).

(b) (6)

• HW genotype freqs: $AA=p^2=0.01$, $Aa=2pq=0.18$, $aa=q^2=0.81$. (1)
• Mean fitness $\bar w = p^2 w_{AA} + 2pq\,w_{Aa} + q^2 w_{aa}$ $= 0.01(1)+0.18(1)+0.81(0.70) = 0.01+0.18+0.567 = 0.757$. (2)
• New $p' = \dfrac{p^2 w_{AA} + pq\,w_{Aa}}{\bar w} = \dfrac{0.01 + 0.09}{0.757} = \dfrac{0.10}{0.757}$. (2)
• $p_1 \approx 0.1321$. (1)

(c) (5)

• With $w_{AA}=w_{Aa}=1$, $w_{aa}=1-s$ (here $s=0.30$): $\bar w = 1 - s q^2$. (1)
• $p' = \dfrac{p^2 + pq}{1 - s q^2} = \dfrac{p(p+q)}{1-sq^2} = \dfrac{p}{1 - s q^2}$ (since $p+q=1$). (2)
• $\Delta p = p' - p = \dfrac{p}{1-sq^2} - p = \dfrac{p - p(1-sq^2)}{1-sq^2} = \dfrac{s p q^2}{1-sq^2} \geq 0$. (1)
• Since $s,p,q^2 \geq 0$ and denominator $>0$, $\Delta p \geq 0$ always → $p$ monotonically increases toward 1. (1)

(d) (3) — Directional selection (1): selection consistently favours one extreme phenotype (dark) (1); the population mean phenotype shifts toward dark, reducing the light phenotype over generations (1).

(e) (5)

s = 0.30
p = 0.10
gen = 0
while p <= 0.95:
    q = 1 - p
    p = p / (1 - s*q**2)        # recurrence from (c)
    gen += 1
print(gen)

• Correct recurrence line (2); correct loop/stopping condition while p <= 0.95 (2); increments generation counter (1).

Question 3

(a) (5) — Dolphin & shark streamlined form = analogous (analogous structures) (1); produced by convergent evolution — unrelated lineages evolve similar traits under similar selective pressures (aquatic predation/locomotion) (2). Finch beaks arise from divergent evolution / adaptive radiation — one ancestral species diversifies into many forms exploiting different niches (homologous beaks) (2).

(b) (6)

• Moths with longer proboscis reach nectar → survive/reproduce better → mean $P$ increases (directional selection on moth) (2).
• Plants with longer spur force the moth's body against anthers → better pollen transfer → those plants leave more offspring → mean $L$ increases (directional selection on plant) (2).
• Reciprocal selection: each partner is the selective pressure on the other → coevolutionary arms race; escalation continues, driving both traits upward (2).

(c) (4)

• $R = h^2 S = 0.5 \times 1.2 = 0.6$ cm. (2)
• New mean $= 10.0 + 0.6 = 10.6$ cm. (1)
• Natural selection: the selective agent (reaching nectar) is environmental/ecological, not a human breeder — no deliberate human choice of parents (1).

(d) (3) — Vestigial hindlimb bones are reduced, non-functional remnants (1); their presence shows whales descended from four-limbed land mammal ancestors that had functional hindlimbs (1); such structures only make sense as evolutionary "leftovers" from common ancestry, providing anatomical evidence for evolution (1).

[
  {"claim":"Human-Tuna divergence time = 35 My", "code":"d=21; mu=0.30; t=d/(2*mu); result=(t==35)"},
  {"claim":"p1 after one generation approx 0.1321", "code":"p=Rational(10,100); q=1-p; wbar=p**2+2*p*q+q**2*Rational(70,100); p1=(p**2+p*q)/wbar; result=(abs(float(p1)-0.1321)<0.001)"},
  {"claim":"Delta p simplifies to s*p*q^2/(1-s*q^2)", "code":"p,q,s=symbols('p q s'); pprime=(p**2+p*q)/(1-s*q**2); dp=simplify(pprime.subs(q,1-p)-p); target=simplify(s*p*(1-p)**2/(1-s*(1-p)**2)); result=(simplify(dp-target)==0)"},
  {"claim":"Breeder response R=0.6 and new mean 10.6", "code":"h2=0.5; S=1.2; R=h2*S; newmean=10.0+R; result=(R==0.6 and newmean==10.6)"}
]