Question

A man with hemophilia (a recessive, sex-linked condition) has a daughter without the condition. She marries a man who does not have hemophilia. What is the probability that their daughter will have hemophilia? Their son? If they have four sons, what is the probability that all will be affected?

Answer 1

Since the condition is sex-linked, it must be related to the X chromosome.

Let
`X^(A)` represent the dominant allele and
`X^(a)` the recessive one, the daugther must have a
`X^(a)` chromosome from the father (the only one he could pass) and a
`X^(A)` from the mother, otherwise she wouldn't be female (
`X^(a)Y`) or would have the condition (
`X^(a)X^(a)`).

If the man does not have the condition, he has the dominant allele on the X chromosome, represented as
`X^(A)Y`. You can use Punett squares to represent their cross.

\begin{center}\begin{tabular} \ & X^{a} & X^{A} \\ \ X^{A} & X^{A}X^{a} & X^{A}X^{A} \\ \ Y & X^{a}Y & X^{A}Y \\ \end{tabular}\end{center}

None of the possible daugthers could have the condition, but there is a 50% chance of their son having it (1/2).

In total, since there is a chance of 1/2 for their son to be affected, the probability of having four sons with the condition is 1/2⁴, or 6.25%.