From Birth Incidence to Carrier Frequency: What Hardy-Weinberg Can and Cannot Tell You
A recessive metabolic disorder is present in 1 in 10,000 newborns in a large, randomly mating population. Affected individuals survive past infancy but almost never reproduce; carriers are clinically normal. A public-health team wants to estimate the carrier frequency from birth incidence, predict how quickly the disease allele would decline without treatment, and design a pilot screen of 1,000 adults to find carriers. A colleague argues that because the disease is lethal, the allele should vanish within a few generations, and proposes screening just 1,000 random adults to quantify carrier frequency precisely. Assume no mutation, migration, or heterozygote advantage.
1. Estimate the allele frequency and carrier frequency under Hardy-Weinberg, and evaluate whether a sample of 1,000 adults is adequate.
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Allele frequency
- Affected frequency = q^2 = 1/10,000 = 1.0 x 10^-4
- q = sqrt(1.0 x 10^-4) = 0.010; p = 0.990
Carrier frequency
- 2pq = 2(0.990)(0.010) = 0.0198, about 2% (roughly 1 in 50)
Is 1,000 adults enough?
- Expected carriers = 1,000 x 0.0198 = about 20
- Binomial standard deviation = sqrt(1000 x 0.0198 x 0.9802) = about 4.4
- Approximate 95% interval: roughly 11 to 29 carriers, i.e. carrier frequency about 1.1% to 2.9%.
Judgment The screen is enough to confirm the carrier frequency is on the order of 2% and to detect large deviations, but it is not 'precise' - the relative uncertainty is several tens of percent, and a 1,000-person sample has little power to detect modest differences between subgroups. The colleague's expectation of precision is unrealistic; the estimate is a rough anchor only.
2. Predict the rate at which the allele frequency falls if affected individuals never reproduce, and evaluate the colleague's claim that the allele should disappear within a few generations.
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Selection against a recessive The allele is hidden in heterozygotes, so selection removes it only through homozygotes. Each generation the frequency change is approximately:
- delta q = -q^2 / (1 + q) or, to good approximation at small q, delta q = -q^2
- At q = 0.010, delta q = -(0.010)^2 = -1.0 x 10^-4 per generation
Timescale Using 1/q_t = 1/q_0 + t: to halve q from 0.010 to 0.005 takes about 100 generations (roughly 2,000-2,500 years in humans).
Verdict on the claim The colleague is wrong. Because nearly all copies of the allele sit in unaffected carriers, selection acts on only a tiny slice of the allele pool, and the decline is glacial while q is small. Selection against a recessive is strongest only when q is already large. No mutation or heterozygote advantage was assumed here; both would slow or halt the decline.
3. The pilot screen later finds that in a subgroup with frequent consanguineous marriages, birth incidence of the disorder is still 1 in 10,000. Explain how this changes the allele-frequency inference and quantify the bias if the inbreeding coefficient is F = 0.01.
Read the worked answer
Why the simple calculation breaks Hardy-Weinberg assumes random mating. With inbreeding, homozygotes are overrepresented:
- frequency(affected) = q^2 + F p q
- With F = 0.01 and observed incidence 1.0 x 10^-4: q^2 + 0.01 q (1 - q) = 1.0 x 10^-4
- 0.99 q^2 + 0.01 q - 1.0 x 10^-4 = 0
- q = [-0.01 + sqrt((0.01)^2 + 4(0.99)(1.0 x 10^-4))] / (2 x 0.99)
- q = (-0.01 + sqrt(4.96 x 10^-4)) / 1.98 = (-0.01 + 0.0223)/1.98 = about 0.0062
- Carrier frequency = 2pq = about 1.2%, roughly 1 in 80, not 1 in 50.
Interpretation The naive HWE calculation attributed all affected births to random combination of two copies, so it overestimated q by roughly 60% and carriers by a similar factor. In the inbred subgroup, most affected individuals inherit two copies identical by descent, so incidence alone cannot pin down q without knowing F.
Judgment Before publishing carrier estimates or setting screening thresholds, the team must either measure F directly, genotype a sample, or restrict the HWE calculation to the outbred portion of the population. Reporting a single carrier frequency for a structured population would mislead counseling.