Population Genetics & Speciation

Question 1 (10 marks)

(a) Let $p$ = frequency of allele $A$, $q$ = frequency of allele $a$. Since these are the only two alleles at the locus, every allele copy is either $A$ or $a$, so $p + q = 1$. (1 mark definitions, 1 mark constraint)

(b) Under random mating, gametes unite independently. Probability an offspring gets $A$ from each parent independently:

• $AA$: $p \times p = p^2$
• $Aa$: $pq + qp = 2pq$ (A from father & a from mother, OR reverse)
• $aa$: $q \times q = q^2$

(1 mark for random-union principle; 1 mark each for $p^2$, $2pq$, $q^2$ — max 4)

(c) Sum $= p^2 + 2pq + q^2 = (p+q)^2 = 1^2 = 1$ since $p+q=1$. (1 mark expansion, 1 mark conclusion)

(d) Any four (½ mark each): No mutation (allele freqs not changed); No migration/gene flow (no new alleles in/out); No selection (all genotypes equally fit); Random mating (genotype freqs = product of allele freqs); Infinitely large population (no drift). (2 marks total)

Question 2 (10 marks)

(a) $q^2 = 45/500 = 0.09$, so $q = \sqrt{0.09} = 0.3$. (1 mark $q^2$, 1 mark $q$)

(b) $p = 1 - q = 1 - 0.3 = 0.7$. (1 mark)

(c)

• $AA = p^2 \times 500 = 0.49 \times 500 = 245$ (2 marks)
• $Aa = 2pq \times 500 = 2(0.7)(0.3)\times 500 = 0.42 \times 500 = 210$ (2 marks)

(Check: 245 + 210 + 45 = 500 ✓)

(d) Dominant phenotype = $AA + Aa = 245 + 210 = 455$. Fraction heterozygous $= 210/455 = 0.4615 \approx 46.2\%$. (1 mark numerator/denom, 1 mark division, 1 mark answer)

Question 3 (12 marks)

(a) (4 marks: 1 per definition, 1 per example)

• Bottleneck: a sharp reduction in population size (e.g. disaster, hunting) so surviving alleles are a random subset — reduced diversity. Example: northern elephant seals / cheetahs.
• Founder effect: a new population started by a few individuals carrying only a fraction of the parent population's alleles. Example: Amish/Afrikaner populations, or island colonisation.

(b) (3 marks) In a small sample, chance sampling error is large — a single death or non-breeding event changes allele frequencies substantially. In large populations random gains and losses average out (law of large numbers), so frequencies stay near expected values. Drift ∝ $1/(2N)$.

(c) (3 marks) Migrants carry alleles between populations; interbreeding mixes allele pools, making populations genetically more similar (homogenising). Gene flow opposes local adaptation when migrants from a differently-adapted population introduce alleles ill-suited to local conditions (e.g. mainland moths repeatedly diluting a locally camouflaged island population).

(d) (2 marks) Mutation is the only process that creates new alleles/variation, so all variation ultimately originates from it (ultimate source). But mutation rates are very low (~$10^{-5}$–$10^{-8}$ per locus/gen), so per-generation frequency change is tiny; selection and drift act on the variation mutation supplies.

Question 4 (10 marks)

(a) (3 marks) A species is a group of actually or potentially interbreeding natural populations that are reproductively isolated from other such groups and produce fertile offspring. (2 marks) Limitation (1 mark): cannot apply to asexual organisms / fossils / ring species / hybridising species.

(b) (4 marks)

• Allopatric: speciation with a geographic barrier separating populations (physical isolation → independent divergence). (2)
• Sympatric: speciation without geographic separation — divergence within the same area. Mechanism: polyploidy (esp. plants), disruptive selection, or habitat/host-plant differentiation. (2)

(c) (1 mark each)

• (i) Prezygotic — temporal (seasonal) isolation.
• (ii) Postzygotic — hybrid sterility (reduced hybrid fertility).
• (iii) Prezygotic — gametic isolation (gamete/pollen incompatibility).

Question 5 (10 marks)

(a) (2 marks) The outgroup is a taxon known to have diverged before the ingroup. Its character states are assumed ancestral (state 0), letting us polarise characters — determine which states are derived (1) vs ancestral (0) rather than assuming.

(b) (4 marks) The characters form a perfectly nested pattern. Most parsimonious tree:

O ─────────────────────
        │ C1
        W ──────────────
            │ C2
            X ──────────
                │ C3
                Y ──────
                    │ C4
                    Z

Branch order: (O,(W,(X,(Y,Z)))). Each synapomorphy arises once (C1 unites W+X+Y+Z; C2 unites X+Y+Z; C3 unites Y+Z; C4 defines Z). (2 marks correct nesting, 2 marks characters mapped to correct nodes)

(c) (1 mark each)

• Monophyletic clade: an ancestor and ALL its descendants — e.g. {X, Y, Z} defined by shared C2.
• Synapomorphy: a shared derived character uniting a clade — e.g. C3 (present in Y and Z).
• Plesiomorphy: an ancestral character state — e.g. state 0 for C4 in O, W, X, Y (absence is ancestral).

(d) (1 mark) Most parsimonious = the tree requiring the fewest evolutionary changes (character-state transitions).

Question 6 (8 marks)

(a) (4 marks)

• Gradualism: slow, steady, continuous morphological change over long time; predicts many intermediate/transitional fossils forming smooth series. (2)
• Punctuated equilibrium: long periods of stasis (little change) interrupted by rapid bursts of change, often at speciation events; predicts few transitionals and abrupt appearances in the fossil record. (2)

(b) (4 marks)

• Oparin–Haldane: early Earth's reducing atmosphere + energy (lightning, UV) allowed abiotic synthesis of organic molecules that accumulated in a "primordial soup," forming precursors to life. (2)
• Miller–Urey: sparked a simulated early atmosphere ($CH_4, NH_3, H_2, H_2O$) and produced amino acids — showing organic building blocks can form abiotically. (1) Limitation: the atmosphere may not have been so strongly reducing; it did not produce life itself, only monomers. (1)

[
  {"claim":"q = sqrt(45/500) = 0.3","code":"q=sqrt(Rational(45,500)); result = (q==Rational(3,10))"},
  {"claim":"p = 0.7 when q=0.3","code":"q=Rational(3,10); p=1-q; result = (p==Rational(7,10))"},
  {"claim":"AA=245 and Aa=210 in N=500","code":"p=Rational(7,10); q=Rational(3,10); AA=p**2*500; Aa=2*p*q*500; result = (AA==245 and Aa==210)"},
  {"claim":"heterozygote fraction of dominant phenotype = 210/455","code":"AA=245; Aa=210; frac=Rational(Aa,AA+Aa); result = (frac==Rational(210,455) and abs(float(frac)-0.4615)<0.001)"},
  {"claim":"HW genotype frequencies sum to 1","code":"p,q=symbols('p q'); result = simplify((p**2+2*p*q+q**2).subs(q,1-p))==1"}
]