One in Five Hundred, or One in Four?
In a block of villages, 6 children with the same rare autosomal recessive disorder are born among 3,000 births over five years. The medical officer uses Hardy–Weinberg to estimate the carrier frequency and gets q ≈ 0.045 and 2pq ≈ 8.6%. A couple in the block, who are first cousins, have one affected child; the woman is 10 weeks pregnant and asks about the risk for this baby. The health worker tells them not to worry because the disorder affects 'only one in five hundred births'. The couple's grandparents were siblings. No other affected children have been born in the couple's own families. A student volunteer argues that since 8.6% of people are carriers, the chance of the new baby being affected is 0.086 × 0.086 × 0.25.
1. Is the 8.6% carrier estimate valid for this block? State the Hardy–Weinberg assumptions, identify the most likely violation, and say whether the figure over- or under-states the risk that two related people will both be carriers.
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Assumptions: a large population; random mating (panmixia); no mutation; no migration; no selection; and, in practice, no appreciable genetic drift.
Calculation: q² = 6/3,000 = 0.002, so q ≈ 0.045, p ≈ 0.955, and 2pq ≈ 0.086.
Violation: the block is small and marriages are consanguineous - these parents are first cousins whose grandparents were siblings. Mating is therefore not random, and a small isolated population also invites founder effect and drift, so the allele frequency itself may not be typical.
Direction of error: as a description of the gene pool the figure may be roughly right, but as a statement about these two people it understates the risk. Relatives can inherit the same allele identical by descent from a shared ancestor, so the probability that both are carriers is much greater than 2pq × 2pq. In an inbred union the affected frequency is better described by q² + Fpq, where F is the coefficient of inbreeding; a positive F adds cases that pure random mating would not produce. The health worker's reassurance is therefore misleading for this couple.
2. What is the risk that this couple's next child is affected? Show your reasoning and state every assumption you rely on.
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Reasoning step by step.
- The disorder is autosomal recessive, so an affected child is aa.
- Each parent is phenotypically unaffected but has already produced an aa child. An unaffected parent of an affected child must carry one a allele, so both are obligate heterozygotes, Aa.
- Cross Aa × Aa gives 1/4 AA, 1/2 Aa and 1/4 aa.
- Each pregnancy is an independent event; the previous affected child does not 'use up' the risk.
Risk: 1 in 4 (25%) for every pregnancy, and the same for a boy or a girl because the gene is autosomal.
Assumptions: full penetrance; no new mutation; no misattributed parentage; both parents truly heterozygous (already established by the affected child); no prenatal screening result yet that would change the conditional probability.
Useful add-ons: the recurrence risk for an affected child's unaffected sibling is 2/3, not 1/4, because that sibling is known not to be aa. Candidates often confuse the two questions - the risk for the next pregnancy (1/4) versus the carrier probability of an existing healthy sibling (2/3).
3. The volunteer computed 0.086 × 0.086 × 0.25 ≈ 0.0018, about 1 in 540. Identify the error and explain the correct logic.
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The error is using population priors for people whose genotypes are already established.
2pq = 0.086 is the probability that a randomly chosen person from the block is a carrier. That number is appropriate only when nothing is known about the person's genotype. Here, both parents have an affected child, so the probability that each is a carrier is effectively 1, not 0.086. This is conditional probability: P(carrier | affected child) ≈ 1, whereas P(carrier) before that information was 2pq. Multiplying 0.086 twice is double-counting the population information while ignoring the pedigree.
Correct calculation: 1 × 1 × 1/4 = 1/4.
Why the volunteer's answer looks plausible: 1 in 540 is close to the population incidence of 1 in 500, which is exactly the figure that applies to unrelated couples drawn at random. It is not a coincidence - it is the same calculation. The trap is that a consanguineous couple is no longer a random draw: the two carrier states are correlated through common descent, so the joint probability of both being carriers is far higher than the product of two independent population frequencies. This correlation is the whole reason the couple's risk is elevated and why genetic counselling for consanguineous marriages cannot be done with a population carrier rate alone.