Malaria control and the sickle-cell allele
In a Kenyan district in 2005, before widespread bed-net distribution, 45% of children aged 2-10 years had malaria parasitemia, and the HbS allele frequency was 0.12. Children with HbSS usually died before age 5 (fitness about 0.10). HbAS children had about 80% protection against severe malaria (fitness 1.0), while HbAA children had fitness about 0.85 because of malaria mortality. By 2018, bed nets and indoor residual spraying had reduced parasitemia to 8%. HbSS mortality remained high, but the HbAS advantage was estimated at only 10% (HbAS fitness 1.0, HbAA fitness 0.98). A 2019 sample of 1,500 adults found an HbS frequency of 0.105. A reporter concludes that malaria control is removing the sickle-cell allele from the population and predicts near-zero frequency within 50 years.
1. Using the 2005 fitness values, explain why the HbS allele remained near 0.12 despite strong selection against HbSS. Estimate the equilibrium frequency under heterozygote advantage.
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Let p be HbS and q HbA. Fitnesses are HbSS 0.10, HbAS 1.0, and HbAA 0.85. Heterozygote advantage maintains a polymorphism when the heterozygote is fittest. At equilibrium, Δp=0, giving p_eq = (w_AS - w_AA)/[(w_AS - w_AA) + (w_AS - w_SS)] = (1.0 - 0.85)/[(1.0 - 0.85) + (1.0 - 0.10)] = 0.15/1.05 ≈ 0.143. The observed 0.12 is close, and the difference is plausible from drift, gene flow, or imperfect fitness estimates. Selection against HbSS removes alleles, but HbAS children survive malaria better than HbAA and pass on the allele, so the loss is balanced. This is not simple directional selection; it is balancing selection. The calculation also shows why the allele does not disappear quickly even though HbSS is lethal: most HbS alleles in the population are in heterozygotes.
2. After malaria control, fitnesses are approximately HbSS 0.10, HbAS 1.00, and HbAA 0.98. Starting from p=0.105, calculate the expected change per generation and predict the allele frequency after two generations. Does this support near-zero frequency in 50 years?
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Use p' = p(p w_SS + q w_AS)/wbar, with q=0.895. wbar = q²(0.98) + 2pq(1.0) + p²(0.10) = 0.8010(0.98) + 0.18795 + 0.011025(0.10) = 0.7850 + 0.18795 + 0.00110 ≈ 0.9741. p' = 0.105[(0.105)(0.10) + (0.895)(1.0)]/0.9741 = 0.105(0.0105 + 0.895)/0.9741 = 0.105(0.9055)/0.9741 ≈ 0.0976. Δp ≈ -0.0074, about 0.7 percentage points per generation. After two generations p ≈ 0.083 if selection stays constant. Fifty years is only about two human generations at 25 years, so the prediction of near-zero is unsupported. The allele declines slowly because HbSS homozygotes are rare when p is low; most selection acts on a small fraction of the population. Medical treatment of HbSS would reduce selection further and slow decline, while drift and migration could raise or lower p.
3. Critique the reporter's conclusion and identify what additional data would make the prediction more reliable.
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The reporter equates relaxed malaria selection with elimination, but selection is only one force. The observed drop from 0.12 to 0.105 over about 14 years is small and could reflect sampling error: with n=1500 and p=0.105, the standard error of p is sqrt(pq/2n) = sqrt(0.105 × 0.895/3000) ≈ 0.0056, so 0.12 is about 2.7 SE away; not decisive but suggestive. Gene flow from nearby populations with higher malaria, nonrandom mating, and genetic drift in a finite population can matter. Improved treatment of HbSS changes fitness and thus the equilibrium. To predict reliably, we need multi-generation allele frequency data, actual fitness estimates by genotype in the current environment, migration rates, population size, and cause-specific mortality. A model with these parameters could forecast, and the forecast should have uncertainty. The claim that malaria control is removing the allele is also misleading because the allele is not being removed directly; its advantage is shrinking.