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A man whose brother has cystic fibrosis wants to know his risk of having an affected child. The prevalence of cystic fibrosis is 1 in 1600 individuals. Which of the following is the risk in this case?
A.1/8
B.1/16
C.1/60
D.1/120
E.1/256

no clue

1/16?

The answer is D 1/120

The answer is: D

According to the Hardy-Weinberg equilibrium, the frequency of heterozygotes (2pq) is twice the square root of the rare homozygote frequency (q2). The man in the question has a 2/3 chance of being a carrier and a 1/20 chance that his wife is a carrier. His risk for an affected child is 2/3 × 1/20 × 1/4 = 1/120.