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A couple has three children, one of whom was diagnosed with cystic fibrosis (aut

ID: 64424 • Letter: A

Question

A couple has three children, one of whom was diagnosed with cystic fibrosis (autosomal recessive trait). The mother's father had a brother who died of the disease and the father's father had two sisters who died of the disease. a) Draw a pedigree that represents the generations. b) What is the probability that the mother Is a carrier? c) The couple plan on having a fourth child, what is the probability that the child will have cystic fibrosis? d) What is the probability that the fourth child will be heterozygote? . e) What Is the probability that the father's father Is heterozygote? f) What is the probability that the mother's father is homozygote?

Explanation / Answer

b). Since both the mother's father and father's father had a history of disease, and they have a child with cystic fibrosis, both are 100% carriers.

Cross between two heterozygous carries will have the offspring with the following genotypes.

Cc* Cc = CC, Cc, Cc, cc. Means they are having 25% of chances to have an affected child at each birth.

c). The probability of having an affected child is 25% or 1/4th.

d). Since, 50% of the offspring are carriers at each birth, the probability of the fourth child to be heterozygous is 1/2.

e). The probability that the father’s father is heterozygous is 50%. This is because, he had a brother who died of the disease, means both of his parents are carriers. And no information is given that he had an affected child but had a carrier daughter (she might have inherited from her mother!). But we can surely say that his parents are heterozygous. As the probability for a heterozygous couple to have a heterozygous child is 1/2, the answer is 50% or 1/2.

f). As the probability for a heterozygous couple to have a normal child is 1/4, the probability that the mother's father is homozygous is 25% or 1/4.

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