Carrier Screening and Bayes: Two Positive Tests
A regional screening programme in Kagawa offers a carrier test for a recessive metabolic disorder. In the local population, 1 person in 25 is a carrier. The test is not diagnostic: it gives a positive result for 90% of carriers and for 5% of non-carriers. A couple, both from the region and with no family history, take the test. Both receive a positive result on this test. A counsellor tells them that because the test is '90% accurate', there is a 90% chance each is a carrier, and therefore a 0.9 × 0.9 × 0.25 = 20% chance their child will be affected. The couple asks for a second, independent test with 99% sensitivity and 99% specificity. Assume the disorder is autosomal recessive, the population is in Hardy–Weinberg equilibrium for this locus, and the two tests are independent conditional on carrier status.
1. Calculate the correct posterior probability that one parent is a carrier after the first positive test, and explain the counsellor's error.
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Use Bayes. Prior P(C) = 1/25 = 0.040; P(not C) = 0.960. P(+|C) = 0.90; P(+|not C) = 0.05.
P(C|+) = 0.040 × 0.90 / (0.040 × 0.90 + 0.960 × 0.05) = 0.036 / (0.036 + 0.048) = 0.036/0.084 = 0.429.
So the probability is about 43%, not 90%. The counsellor confused the test's sensitivity P(+|C) with the predictive value P(C|+), and ignored the low base rate: most positives come from the large non-carrier group, because 5% of 0.96 is larger than 90% of 0.04.
2. Using the correct carrier probabilities, what is the risk that their child is affected after only the first test? Compare this with the counsellor's figure and with the risk before any testing.
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Given both parents test positive and the tests are independent conditional on status, each has carrier probability 0.429. For an autosomal recessive disorder, an affected child requires both parents to be carriers and both to pass the variant: P = 0.429² × 0.25 = 0.184 × 0.25 = 0.046, about 4.6%.
The counsellor's 20% used 0.9 instead of 0.429 for each parent, overstating the risk roughly four-fold (0.9²/0.429² ≈ 4.4).
Baseline before testing: 0.04² × 0.25 = 0.0004, about 0.04%. The first test raises the risk about 100-fold, but it is still not 20%.
3. Both parents now test positive on the second, independent test. Find the updated carrier probability and the final child risk. What should the counsellor communicate?
Read the worked answer
Second test: prior 0.429, sensitivity 0.99, false-positive 0.01.
P(C|++) = 0.429 × 0.99 / (0.429 × 0.99 + 0.571 × 0.01) = 0.424 / (0.424 + 0.0057) = 0.987.
Child risk = 0.987² × 0.25 = 0.243, about 24%. Even two positives leave about a 1.3% chance each parent is not a carrier, and the child risk is about one in four, not near certainty or 20%.
The counsellor should quote the posterior and the child risk separately, and state the assumptions: population frequency, independent tests, and no consanguinity.