Question

A prospective groom, who is healthy, has a sister with cystic fibrosis (CF), an autosomal recessive disease. Their parents are normal. The brother plans to marry a woman who has no history of CF in her family. What is the probability that they will produce a CF child? 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 1

Answer:

Explanation:

From the given information:

Since the parent are normal; Let assume that their trait is Pp

The affected sister  will be recessive in nature, so the trait will be pp

Now, the husband who is the groom will be a carrier with a probability of 2/3 since her sister is affected.

Also, the wife will be carrying a heterozygous trait since she is normal as well.

However, using Hardy-Weinberg assumption:

$q^2 = \dfrac{1}{2500}= 0.0004$

$q = \sqrt{0.0004}$

q = 0.02

Recall that:

$p+q =1 \\ p = 1 - q \\p = 1 - 0.02 \\p = 0.98$

Thus, $p^2 = (0.98)^2$

$p^2 = 0.9604$

$2pq = 2 \times 0.98 \times 0.022$

2pq = 0.0392

Finally, the required probability that there will be an affected offspring amongst the family is:

$= \dfrac{1}{4} \times \dfrac{2}{3} \times 0.0392$

$= \dfrac{1}{153}$