Cystic fibrosis carrier risk in a couple
Priya and Tom are planning a family. Neither has cystic fibrosis (CF), an autosomal recessive condition. Priya's brother has CF, but Priya's parents do not. Tom has no known family history of CF. The couple are referred to a genetic counsellor. Priya is offered a carrier test; Tom is told that, in his population, about 1 in 25 people is a carrier. Priya asks, 'If my brother is affected, why is my chance of being a carrier only two in three rather than one in two?' Tom asks whether he should be tested first, because 'it is more likely to be positive'. The counsellor explains that carrier testing looks for common CFTR variants but does not detect every possible variant. The couple must decide whether to have testing before trying to conceive.
1. Using genetic reasoning, explain why Priya's chance of being a carrier is 2/3.
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Let A be the normal allele and a the CF allele. Priya's brother is aa, so each parent must carry a. Both parents are unaffected, so both are Aa. A cross Aa × Aa gives 1 AA : 2 Aa : 1 aa. Priya is unaffected, so she cannot be aa. Of the three equally likely unaffected outcomes, two are Aa. Therefore P(Priya is a carrier) = 2/3, not 1/2. The 1/2 figure would be the chance that an unaffected child of a carrier and a non-carrier is a carrier, which is not this family's situation. This assumes simple Mendelian inheritance and no new mutation.
2. Calculate the probability that their first child would have CF if Tom is assumed to have the population carrier risk. Show each step.
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Assume Tom's carrier probability is 1/25 and is independent of Priya's. P(child affected) = P(Priya carrier) × P(Tom carrier) × P(child aa | both carriers). = 2/3 × 1/25 × 1/4 = 2/300 = 1/150 ≈ 0.0067, or about 0.67% (roughly 1 in 150). This is a conditional estimate: if Tom is actually a carrier, the chance is 1/4; if he is not, it is effectively zero, ignoring new mutations. The population figure is an average and may not fit Tom's ancestry, so a genetic counsellor would use the most appropriate carrier frequency.
3. Evaluate Tom's statement that his test is 'more likely to be positive', and explain how the counsellor's note about undetected variants affects the interpretation of a negative test.
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Tom's statement is mistaken: his prior chance of being a carrier, about 4%, is much lower than Priya's, about 67%, so a test on Priya is more likely to reveal a carrier. Testing Tom can still be useful because a negative result would lower the couple's risk, but the counsellor's warning matters: a standard test detects common CFTR variants, not all. A negative test therefore reduces but does not eliminate Tom's chance of being a carrier; the residual risk comes from rare or undetected variants. The couple should interpret any result with the counsellor, and testing is a personal decision with implications for family members. A negative result should not be described as 'no risk'.