A couple wants to start a family, but they are concerned that their child might have cystic fibrosis. After taking a
family history, you determine that while neither of them have the disease, the woman had a sister with cystic fibrosis
(with unaffected parents), and the man's father also had cystic fibrosis. What do you tell them?
What will be an ideal response?
ANSWER: To produce a child affected by cystic fibrosis, each parent must pass on a recessive allele.
The man is heterozygous for the cystic fibrosis trait, since he must have inherited a recessive
allele from his father, but is not sick. Since the woman's sister had the disease but her parents
did not, both parents must be heterozygous. The probability that she is also heterozygous is
1/4 + 1/4 = 1/2. If both she and her partner are heterozygous, the probability of having a child
with cystic fibrosis is equal to the chance of each of them passing on a recessive allele: 1/2 x
1/2 = 1/4. However, there is only a 1/2 chance that the woman has the mutant allele;
therefore, the probability that these individuals could have a child affected by cystic fibrosis
is 1/4 x 1/2 = 1/8.
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What will be an ideal response?
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