Cystic fibrosis is a rare disease caused by homozygosity for a recessive allele, and characterized by...

Cystic fibrosis is a rare disease caused by homozygosity for a recessive allele, and characterized by a build-up of mucus in the lungs and difficulty breathing. A woman whose maternal uncle had the disease is trying to determine the probability she and her husband could have an affected child. Her husband's sister died of the disease. Assuming the uncle and the sister are the only family members who had the disease, what is the chance that the couple's first child will be affected? I'm getting 1/18, but chegg says 1/64? Please verify it is 1/64 for the answer.

Solutions

Expert Solution

The answer will be 1/18.

Explanation: Let, A = dominant normal allele, a = allele for cystic fibrosis

Now, according to the question, the pedigree is drawn below. III-1 is the woman, III-2 is her husband, II-3 is her maternal uncle & III-3 is her sister-in-law. As no information is given regarding frequency of the allele in the population, we assume that woman's father is homozygous dominant, i.e., AA genotype.

As only maternal uncle of woman & sister of her husband develop the disease, I-1, I-2, II-4 & II-5 will have A/a genotype. Now in order to the woman become carrier of the disease, II-2 has to be carrier. The probability that II-2 has A/a genotype is 2/3 (We would ignore a/a genotype as she doesn't develop the disease). Now, the probability that woman will get the disease allele (Allele a) from II-2 is 1/2.

So, probability that woman has genotype A/a = Probability that woman will get the disease allele (Allele a) from II-2 x Probability that II-2 has A/a genotype = 1/2 x 2/3 = 1/3

Probability that woman's husband (III-2) has A/a genotype = 2/3 (Similar to II-2)

Now, probability that she & her husband will have affected child = 1/4 (As both are carrier for the disease)

So, overall probability that they have an affected child = Probability that woman has genotype A/a x Probability that woman's husband (III-2) has A/a genotype x Probability that she & her husband will have affected child = 1/3 x 2/3 x 1/4 = 1/18

gladiator answered 1 year ago

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