Warfarin resistance: a Hardy–Weinberg result that hides evolution
A population of 400 rodents is sampled from a farm before a warfarin baiting programme begins. Genotyping at the VKORC1 locus gives RR 4, RS 92, SS 304, where R is the resistance allele, S the susceptible allele, and R is dominant for survival on bait. Three years of continuous baiting follow. A second sample of 400 adults, taken from the same area after the breeding season, gives RR 36, RS 176, SS 188. A chi-square test for Hardy–Weinberg equilibrium uses df = 1 because the allele frequency is estimated from the data; the 5% critical value is 3.84. The student who collected the data concludes: 'The second sample fits Hardy–Weinberg, so the population is not evolving and the resistance allele has reached a stable frequency of 0.31.' Laboratory colonies show that, in the absence of bait, RR animals produce 30% fewer surviving offspring than SS animals, while RS animals are intermediate.
1. Calculate the allele frequencies and chi-square values for both samples. State whether each sample is consistent with Hardy–Weinberg equilibrium, and evaluate the student's conclusion.
Read the worked answer
Before baiting
- p(R) = (2×4 + 92)/800 = 100/800 = 0.125; q(S) = 0.875.
- Expected: RR = 0.125² × 400 = 6.25; RS = 2(0.125)(0.875) × 400 = 87.5; SS = 0.875² × 400 = 306.25.
- χ² = (4−6.25)²/6.25 + (92−87.5)²/87.5 + (304−306.25)²/306.25 = 0.81 + 0.23 + 0.02 = 1.06. 1.06 < 3.84, so the sample is consistent with HWE.
After three years of baiting
- p(R) = (2×36 + 176)/800 = 248/800 = 0.31; q = 0.69.
- Expected: RR = 0.0961 × 400 = 38.44; RS = 2(0.31)(0.69) × 400 = 171.12; SS = 0.4761 × 400 = 190.44.
- χ² = (36−38.44)²/38.44 + (176−171.12)²/171.12 + (188−190.44)²/190.44 = 0.16 + 0.14 + 0.03 = 0.33. Again consistent with HWE.
Evaluating the conclusion The statistics are right and the inference is wrong. Allele frequency rose from 0.125 to 0.31 - a 2.5-fold change in three years. That is evolution, by definition (change in allele frequency between generations). A Hardy–Weinberg fit says only that genotype proportions within a cohort are what random mating with those allele frequencies would produce; it says nothing about whether the allele frequencies themselves are stable. The student has confused a statement about genotype ratios with a statement about allele frequencies. The claim of a 'stable' frequency of 0.31 also needs more than one post-treatment sample to support.
2. Explain how a population can be at Hardy–Weinberg equilibrium at a locus and still be evolving, and explain the selection that produced the observed shift.
Read the worked answer
Why HWE does not mean 'not evolving'
- HWE describes the genotype frequencies expected when a population mates at random, given its allele frequencies. Those proportions are re-established after a single generation of random mating, whatever process set the allele frequencies - selection, drift, migration or a founder event.
- HWE also assumes no selection, no mutation, no migration, no drift and a large population. A chi-square that fails to reject does not prove these assumptions hold; it means the sample is too small, or selection too weak, to detect a deviation in genotype proportions.
- Evolution is measured by comparing allele frequencies between generations or between samples over time, which is exactly what the two samples allow.
The selection involved
- Warfarin imposes very strong directional selection: SS animals either die or fail to reproduce, so S alleles are removed each generation while R alleles survive in RR and RS animals.
- R is dominant for survival, which is why the increase is fast: even RS heterozygotes are protected, so most R alleles sit in heterozygotes and are shielded from selection when rare.
- The cost of resistance - RR animals leave 30% fewer surviving offspring without bait - is the counterweight. If the cost applies only to RR, and RS is intermediate, then with bait present the advantage of R greatly exceeds the cost and p rises; a balance is possible only if the cost of RR is paid even in baited populations.
- Two samples cannot distinguish 'new equilibrium' from 'still rising'. Distinguishing them requires repeated sampling over several more years.
3. Predict and justify what would happen to the resistance allele frequency over ten years if baiting stopped completely.
Read the worked answer
Prediction: p(R) would decline from 0.31, but slowly, and R would not disappear within ten years.
Reasoning
- With bait removed, the relative fitness values invert. SS animals now have the highest fitness; RR animals leave 30% fewer surviving offspring; RS are intermediate. Selection therefore acts against R.
- The decline is slow at first for a specific genetic reason. At p = 0.31, most R alleles are carried in RS heterozygotes (the frequency of R alleles in heterozygotes is pq = 0.31 × 0.69 ≈ 0.21 of all alleles, versus p² ≈ 0.096 in RR homozygotes). Selection can only remove R efficiently when it is exposed in homozygotes, so a recessive deleterious allele hidden in heterozygotes persists for many generations. The same shielding that made the allele spread fast under bait makes it linger afterwards.
- Rate also depends on generation time and population size. Rodents breed several times a year, so ten years is many generations and the decline will be substantial; in a long-lived species it would barely register.
- Complications to flag: immigration of resistant animals from untreated neighbouring farms would slow or reverse the decline; if the population crashes to a small size, genetic drift could fix or lose the allele regardless of fitness.
Practical implication: reversion is not a management strategy. Resistance should be assumed to persist, so baiting should be rotated between active ingredients rather than relied upon to 'wear off'.