Level 5 — MasteryPopulation Genetics & Speciation

Population Genetics & Speciation

90 minutes60 marksprintable — key stays hidden on paper

Level 5 — Mastery (Cross-domain: Biology + Mathematics + Computation) Time limit: 90 minutes Total marks: 60

Instructions: Answer all THREE questions. Show all derivations. Where code is requested, write clean pseudocode or Python. Use ...... notation for algebra.


Question 1 — Deriving and Perturbing Hardy–Weinberg (22 marks)

A diploid population is examined at a single autosomal locus with two alleles, AA (frequency pp) and aa (frequency q=1pq = 1-p).

(a) Starting from random mating, derive the Hardy–Weinberg genotype frequencies (p2,2pq,q2)(p^2, 2pq, q^2) and prove that these frequencies are stationary — i.e. that allele and genotype frequencies do not change from one generation to the next under the H–W assumptions. State all assumptions used. (6)

(b) In a sample of 400 individuals for a recessive condition, 36 are affected (genotype aaaa). Assuming H–W equilibrium, calculate qq, pp, and the expected number of heterozygous carriers. Then perform a chi-square goodness-of-fit test given observed genotype counts AA=202AA = 202, Aa=162Aa = 162, aa=36aa = 36 (state degrees of freedom and conclusion at α=0.05\alpha = 0.05; critical value χ0.05,12=3.84\chi^2_{0.05,1}=3.84). (8)

(c) Selection now acts against the recessive homozygote with selection coefficient ss (relative fitnesses AA:Aa:aa=1:1:1sAA:Aa:aa = 1:1:1-s). Derive the recurrence for the change in allele frequency Δq\Delta q per generation, and show that for small qq the rate of removal of aa becomes very slow. Explain the biological significance for eliminating deleterious recessives. (8)


Question 2 — Genetic Drift, Bottlenecks & a Simulation (20 marks)

(a) Define genetic drift and distinguish the bottleneck effect from the founder effect, giving one real biological example of each. (6)

(b) For a Wright–Fisher population of constant size NN diploid individuals, the expected heterozygosity HH decays as Ht=H0(112N)t.H_t = H_0\left(1 - \frac{1}{2N}\right)^t. Calculate the number of generations required for heterozygosity to fall to half its initial value when N=50N = 50. Show your working using logarithms. (6)

(c) Write pseudocode (or Python) for a Wright–Fisher simulation that, given initial allele frequency p0p_0, population size NN, and number of generations TT, returns the trajectory of allele AA's frequency. Then explain how you would estimate, from many replicate runs, the probability of fixation of allele AA, and state the theoretical expectation for that probability. (8)


Question 3 — Speciation, Isolation & Phylogenetics (18 marks)

(a) State the biological species concept and give one limitation that makes it inapplicable to certain organisms. (3)

(b) Compare allopatric and sympatric speciation, and classify each of the following isolation mechanisms as prezygotic or postzygotic: (i) temporal isolation, (ii) hybrid sterility, (iii) gametic incompatibility, (iv) hybrid inviability. (7)

(c) Four taxa W, X, Y, Z were scored for five binary morphological characters (0 = ancestral, 1 = derived):

Taxon C1 C2 C3 C4 C5
W 1 1 0 0 0
X 1 1 1 0 0
Y 1 1 1 1 1
Z 0 0 0 0 0

Using cladistic parsimony (Z as outgroup), construct the most parsimonious cladogram, identify the shared derived characters (synapomorphies) defining each clade, and explain the difference between gradualism and punctuated equilibrium in the context of the branch lengths such a tree might show. (8)

Answer keyMark scheme & solutions

Question 1

(a) Derivation & proof (6)

  • Allele frequencies in gamete pool: A=pA = p, a=qa = q. (1)
  • Random mating = random union of gametes → genotype probabilities from expansion of (p+q)2(p+q)^2: AA=p2AA = p^2, Aa=2pqAa = 2pq, aa=q2aa = q^2. (2)
  • Proof of stationarity: allele frequency of AA in offspring p=p2+12(2pq)=p2+pq=p(p+q)=p.p' = p^2 + \tfrac{1}{2}(2pq) = p^2 + pq = p(p+q) = p. (2)
  • Since p=pp'=p, genotype frequencies next generation are again p2:2pq:q2p^2:2pq:q^2 → equilibrium reached in ONE generation and stays. Assumptions: infinite population, random mating, no selection, no mutation, no migration, non-overlapping generations/diploid. (1)

(b) Calculation + chi-square (8)

  • q2=36/400=0.09q=0.3q^2 = 36/400 = 0.09 \Rightarrow q = 0.3. (1)
  • p=10.3=0.7p = 1 - 0.3 = 0.7. (1)
  • Expected carriers =2pq×400=2(0.7)(0.3)(400)=0.42×400=168= 2pq \times 400 = 2(0.7)(0.3)(400) = 0.42\times400 = 168. (1)
  • Expected counts: AA=p2400=0.49×400=196AA = p^2\cdot400 = 0.49\times400 = 196; Aa=168Aa = 168; aa=36aa = 36. (1)
  • χ2=(202196)2196+(162168)2168+(3636)236\chi^2 = \frac{(202-196)^2}{196} + \frac{(162-168)^2}{168} + \frac{(36-36)^2}{36} =36196+36168+0=0.1837+0.2143=0.398= \frac{36}{196} + \frac{36}{168} + 0 = 0.1837 + 0.2143 = 0.398. (2)
  • df = (#genotypes − 1 − #alleles estimated) = 3 − 1 − 1 = 1. (1)
  • 0.398<3.840.398 < 3.84 \Rightarrow fail to reject H0H_0; population consistent with H–W equilibrium. (1)

(c) Selection recurrence (8)

  • Mean fitness wˉ=p2(1)+2pq(1)+q2(1s)=1sq2\bar w = p^2(1) + 2pq(1) + q^2(1-s) = 1 - sq^2. (2)
  • Frequency of aa next generation: q=pq(1)+q2(1s)1sq2=q(1sq)1sq2.q' = \frac{pq(1) + q^2(1-s)}{1 - sq^2} = \frac{q(1-sq)}{1-sq^2}. (2)
  • Δq=qq=q(1sq)q(1sq2)1sq2=sq2(1q)1sq2.\Delta q = q' - q = \dfrac{q(1-sq) - q(1-sq^2)}{1-sq^2} = \dfrac{-s q^2(1-q)}{1-sq^2}. (2)
  • For small qq, Δqsq20\Delta q \approx -sq^2 \to 0 as q0q\to0: removal slows dramatically because deleterious recessive alleles are "hidden" in heterozygotes and not exposed to selection. Hence recessive alleles are never fully eliminated by selection alone. (2)

Question 2

(a) Definitions (6)

  • Genetic drift = random change in allele frequencies between generations due to sampling error, strongest in small populations. (2)
  • Bottleneck = drastic reduction in population size (e.g. by disaster) followed by recovery, reducing genetic diversity — example: northern elephant seal / cheetah. (2)
  • Founder effect = new population started by a small number of colonising individuals carrying a non-representative allele sample — example: Amish (polydactyly) / Galápagos colonisers. (2)

(b) Half-life of heterozygosity (6)

  • 12=(11/(250))t=(0.99)t\tfrac12 = (1 - 1/(2\cdot50))^t = (0.99)^t. (2)
  • ln(0.5)=tln(0.99)\ln(0.5) = t\ln(0.99). (1)
  • t=ln0.5ln0.99=0.69310.0100568.97t = \dfrac{\ln 0.5}{\ln 0.99} = \dfrac{-0.6931}{-0.01005} \approx 68.97. (2)
  • 69 generations. (1)

(c) Simulation pseudocode + fixation (8)

import numpy as np
def wright_fisher(p0, N, T):
    p = p0
    traj = [p]
    for _ in range(T):
        # number of A alleles drawn ~ Binomial(2N, p)
        k = np.random.binomial(2*N, p)
        p = k / (2*N)
        traj.append(p)
    return traj
  • Correct binomial sampling of 2N2N alleles each generation. (3)
  • Estimating fixation probability: run RR replicate simulations to a large TT (or until p=0p=0 or p=1p=1); fixation probability \approx (number of runs ending at p=1p=1)/RR. (3)
  • Theoretical expectation: for a neutral allele the probability of fixation equals its initial frequency, Pfix=p0P_{fix} = p_0. (2)

Question 3

(a) Species concept (3)

  • A species = a group of actually or potentially interbreeding natural populations that are reproductively isolated from other such groups (produce fertile offspring). (2)
  • Limitation: cannot apply to asexual/clonal organisms, fossils, or ring species / cases with ongoing hybridisation. (1)

(b) Speciation & isolation (7)

  • Allopatric: population split by a geographic barrier, gene flow stopped, divergence in isolation. (2)
  • Sympatric: speciation without geographic separation (same area) — e.g. via polyploidy, disruptive selection, host shift. (2)
  • Classification: (i) temporal = prezygotic; (ii) hybrid sterility = postzygotic; (iii) gametic incompatibility = prezygotic; (iv) hybrid inviability = postzygotic. (3, ½ each rounded)

(c) Cladistics + tectonics of change (8)

  • Using Z (all 0) as outgroup, character polarity: C1,C2 derived in W,X,Y; C3 derived in X,Y; C4,C5 derived in Y only. (2)
  • Nested synapomorphies:
    • C1+C2 → clade (W, X, Y) [ingroup].
    • C3 → clade (X, Y).
    • C4+C5 → autapomorphies of Y.
  • Most parsimonious cladogram: (Z,(W,(X,Y))) — W branches first within ingroup, then X and Y are sister taxa. (3)
  • Gradualism = evolutionary change accumulates slowly and steadily → tree would show roughly proportional, even branch lengths. Punctuated equilibrium = long stasis punctuated by rapid change at speciation events → most morphological change concentrated at nodes with short intense bursts and long unchanging branches. (3)

[
  {"claim":"q=0.3, p=0.7, carriers=168 under HWE for aa=36/400","code":"q=sqrt(Rational(36,400)); p=1-q; carriers=2*p*q*400; result=(q==Rational(3,10) and p==Rational(7,10) and carriers==168)"},
  {"claim":"chi-square statistic approx 0.398","code":"chi=Rational((202-196)**2,196)+Rational((162-168)**2,168)+Rational(0,36); result=abs(float(chi)-0.398)<0.01"},
  {"claim":"heterozygosity half-life at N=50 is about 69 generations","code":"import sympy; t=log(Rational(1,2))/log(Rational(99,100)); result=abs(float(t)-68.97)<0.1"},
  {"claim":"delta q simplifies to -s*q**2*(1-q)/(1-s*q**2)","code":"s,q=symbols('s q'); p=1-q; dq=simplify((q*(1-s*q))/(1-s*q**2)-q); target=simplify(-s*q**2*(1-q)/(1-s*q**2)); result=simplify(dq-target)==0"}
]