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

Step 1: Set up the genotypes.
The disease allele shows X-linked recessive inheritance. Let \(X^A\) stand for the normal allele and \(X^a\) stand for the disease allele.
The mother is a carrier, so her genotype is \(X^A X^a\). She carries one normal allele and one disease allele, and the normal allele masks the disease allele in her, so she is not affected.
The father is genetically normal, so his genotype is \(X^A Y\).

Step 2: Work out what a son inherits.
A son always gets his Y chromosome from his father and his single X chromosome from his mother.
The father's X chromosome goes only to daughters, so the father's genotype has no effect on whether a son is affected.
So the son's fate depends only on which of the mother's two X alleles he receives.

Step 3: Apply the segregation probability.
During meiosis, the mother's two X chromosomes, one carrying \(X^A\) and one carrying \(X^a\), segregate with equal chance.
So a son gets \(X^A\) with probability \(0.5\) and \(X^a\) with probability \(0.5\).
A male is hemizygous for the X chromosome, meaning he has only one copy, so there is no second allele to mask a recessive allele.
If he gets \(X^a\), he is affected, even though the allele is called recessive, because there is no dominant partner allele to hide it in a male.

Step 4: Final Answer.
The probability that the son is born with the disease equals the probability that he receives the \(X^a\) allele from his mother.
\[ P(\text{affected son}) = 0.5 \]
\[ \boxed{0.5} \]