A Negative Carrier Test in an Autosomal Recessive Family
A 29-year-old woman's 12-year-old brother has cystic fibrosis, an autosomal recessive condition. Both of her parents are unaffected, and she is unaffected. Her partner is 31 years old, has no family history of cystic fibrosis, and comes from a population in which the carrier frequency is 1 in 25. The couple seeks genetic counselling. A standard mutation panel detects 90% of carriers in this population and has no false positives; it does not detect every possible mutation. The woman tests negative. The partner has not been tested. The counsellor explains that a negative test does not make her carrier risk zero, and uses a Bayesian calculation rather than simply quoting the population carrier frequency. Assume the test's sensitivity applies to her, and that the partner's prior risk is the population carrier frequency. The couple wants to know the risk to their first child before deciding whether the partner should be tested.
1. Calculate the woman's posterior probability of being a carrier after her negative test.
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Before testing, the woman is an unaffected sibling of an affected child of two unaffected carrier parents. Her prior probability of being a carrier is therefore (2/3).
If she is a carrier, the panel misses her with probability (1-0.90=0.10). If she is not a carrier, a negative result has probability 1, because there are no false positives.
Bayes' theorem: [P(\text{carrier}\mid\text{negative})=\frac{(2/3)(0.10)}{(2/3)(0.10)+(1/3)(1)}.] [=\frac{0.0667}{0.0667+0.3333}=0.1667.] Her posterior carrier risk is about (1/6), or 16.7%. It is not zero and not 1/25. The negative test lowers her risk from 66.7% to 16.7%, but it cannot eliminate the risk because the panel misses 10% of carriers.
2. What is the probability that their first child is affected if the partner is not tested?
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Both parents must be carriers, and both must transmit the mutant allele. The woman's posterior carrier probability is (1/6). The partner's prior carrier probability is (1/25). If both are carriers, the chance that a child is affected is (1/4).
Multiply the independent probabilities: [P=\frac{1}{6}\times\frac{1}{25}\times\frac{1}{4}=\frac{1}{600}\approx0.00167.] This is about 0.17%, or 1 in 600. The population carrier frequency is used only for the partner, because he has no family history; the woman's risk comes from her family history and her negative test result.
3. If the partner also tests negative, compute the revised child risk and comment on what dominates the residual risk.
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For the partner, the prior carrier probability is (1/25). After a negative test, [P(\text{carrier}\mid\text{negative})=\frac{(1/25)(0.10)}{(1/25)(0.10)+(24/25)(1)}.] [=\frac{0.004}{0.004+0.96}=0.00415\approx\frac{1}{241}.] The child risk becomes [P=\frac{1}{6}\times\frac{1}{241}\times\frac{1}{4}=\frac{1}{5784}\approx0.000173.] That is about 0.017%, or 1 in 5800.
The woman's residual 1/6 carrier probability dominates the remaining risk. Partner testing lowers the risk about tenfold but cannot remove it, because both tests can miss mutations. If the brother's specific mutations were known, testing the woman directly for those familial mutations would be more informative than the panel.