Question

A disease-associated allele shows X-linked recessive inheritance. If the mother is a carrier of the disease, the father is genetically normal, and the child born is a son, then the probability that he is born with the disease is . (rounded off to one decimal place)

Hint:

A male has only one X chromosome, so any allele on it is fully expressed, there is no second copy to mask it. Work out only what the son can get from his mother.

Answer 1

Correct Answer: 0.5

This question checks whether you know that X-linked recessive inheritance behaves differently in sons than in daughters. Set up the full cross between the parents, then read off only the son outcomes.

Take $X^A$ as the healthy allele and $X^a$ as the disease allele. The mother is a carrier, so her genotype is $X^AX^a$. The father is unaffected, so his genotype is $X^AY$.

Lay out the cross as a table of the four possible offspring, splitting the mother's two eggs against the father's two types of sperm:

	Egg carries $X^A$	Egg carries $X^a$
Sperm carries $X^A$	$X^AX^A$ daughter, unaffected	$X^AX^a$ daughter, carrier, unaffected
Sperm carries $Y$	$X^AY$ son, unaffected	$X^aY$ son, affected

1. $X^AY$ son: gets the normal allele from his mother, so he is healthy.
2. $X^aY$ son: gets the disease allele from his mother, and since he has no second X chromosome to cover for it, the allele acts as if it were dominant in him, so he is affected.

The two daughter outcomes can be dropped, since the question already tells us the child is a son. Out of the two equally likely son outcomes in the bottom row, exactly one is affected, so the chance is

\[ P = \frac{1}{2} = 0.5 \]

Let's summarize:

• A son's single X chromosome always comes from the mother, never the father.
• A carrier mother passes the disease allele to half of her sons, on average.
• A son has only one X, so a single copy of the disease allele is enough to make him affected, there is no masking copy like in a daughter.

So the probability that the son is born with the disease is $0.5$.