A prospective groom, who is not affected by cystic fibrosis (CF), has a sister with cystic fibrosis, an autosomal recessive disease. Their parents are also not affected. The brother plans to marry a woman who has no history of CF in her family. What is the probability that they will have a child with CF? They are both Caucasian and the overall frequency of CF in the Caucasian population is 1/2500 - that is, 1 affected child per 2500 (assume the population meets the Hardy-Weinberg assumptions).

Answer: In punnett square number 1, we see the cross between AA and aa, and the result is 100% Aa individuals, so there is 0% chances of haveing a child with CF. In punnett square number 2, we see the cross between Aa and aa and there is 50% chances of having a child with CF (aa genotype).

p = 0.98

q = 0.02

p² = 0.9604

q² = 0.0004

2pq = 0.0392.

Explanation:

Due to technical difficulties, the explanation has been attached as a word document.

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