Carrier Risk and Screening in an Isolated Population
A public-health team is counselling a couple in a town of about 4,000 people where an autosomal recessive metabolic disorder is present. The disorder affects 1 newborn in 400, and the population is assumed to be in Hardy–Weinberg equilibrium for this locus. The woman is healthy, but she has an affected brother; both her parents are unaffected. The man is healthy and has no family history of the disorder. The couple wants to know the probability that their first child will be affected. Separately, the clinic offers a carrier-screening test with 95% sensitivity and 98% specificity. A healthy woman from the same town with no family history tests positive and asks whether she should conclude that she is a carrier. The counsellor must distinguish between the couple's actual risk and the predictive value of a positive screen, and must state the assumptions behind each calculation.
1. Calculate the probability that the couple's first child will be affected. Show each step and state any assumption you make about the man's carrier status.
Read the worked answer
Step 1: Find the carrier frequency in the population. If the disease incidence is q² = 1/400, then q = 1/20 = 0.05 and p = 0.95. The carrier frequency is 2pq = 2(0.95)(0.05) = 0.095, or 9.5%.
Step 2: Find the woman's carrier probability. Her parents are unaffected but have an affected child, so both must be carriers. Among their unaffected children, the possible genotypes are AA, Aa, and aA, giving a carrier probability of 2/3.
Step 3: Multiply the risks. The man has no family history, so use the population carrier frequency 0.095 as his prior probability. If both are carriers, the chance of an affected child is 1/4. P(affected child) = (2/3) × 0.095 × (1/4) = 0.0158, or about 1.6%.
Check: The answer is not 1/4, because the parents are not both known carriers. It is also not simply 0.095/4, because the woman's family history raises her carrier probability above the population value.
2. The healthy woman with no family history tests positive. Calculate the probability that she is actually a carrier, and explain why the result is not 95%.
Read the worked answer
Use Bayes' theorem with the carrier frequency as the prior, not the disease incidence. In this town, P(carrier) = 0.095 and P(non-carrier) = 0.905.
The test has sensitivity P(positive | carrier) = 0.95 and specificity P(negative | non-carrier) = 0.98, so P(positive | non-carrier) = 0.02.
P(carrier | positive) = (0.95 × 0.095) / [(0.95 × 0.095) + (0.02 × 0.905)] = 0.09025 / (0.09025 + 0.0181) = 0.09025 / 0.10835 ≈ 0.833.
So the probability is about 83%, not 95%. Sensitivity is the chance of a positive result given carrier status; the patient needs the reverse conditional probability, which also depends on how common carriers are and on the false-positive rate. A common error is to use the disease incidence 1/400 as the prior, which would wrongly give a much lower value near 11%.
3. State two assumptions of the Hardy–Weinberg calculation that may be doubtful in a town of 4,000 people, and explain the likely direction of error in the carrier-frequency estimate.
Read the worked answer
Two assumptions are random mating and a large, closed population with no migration, selection, or mutation.
In a town of 4,000, genetic drift can make the actual allele frequency differ from 0.05, and a founder effect or limited migration can raise or lower the carrier frequency substantially. Consanguinity or assortative mating increases homozygosity, so the recessive disease incidence could be higher than q² predicts even if the allele frequency is unchanged. Selection against affected individuals would slightly reduce q over time.
Therefore the calculated carrier frequency 0.095 should be treated as an estimate with real uncertainty; the couple's risk could be higher in a consanguineous or founder population and lower if the town is genetically diverse.