Calculating X-linked Recessive Disease Probability in Son

This question concerns the probability of inheriting an X-linked recessive disease. We are given that the mother is a carrier and the father is normal. We need to find the probability that their son is born with the disease.

Understanding Inheritance Patterns

• X-linked recessive inheritance means the disease-causing allele is located on the X chromosome, and it only causes the disease when an individual has two copies of the allele (in females) or one copy (in males, as they only have one X chromosome).
• Carrier Mother: A carrier female has one normal allele (let's denote it as X^R) and one disease allele (X^r). Her genotype is X^RX^r.
• Normal Father: A normal male has one normal X chromosome and one Y chromosome. His genotype is X^RY.
• Son's Inheritance: A son inherits his Y chromosome from his father and his X chromosome from his mother.

Probability Calculation

To determine if the son has the disease, we look at the X chromosome he inherits from his mother. The mother has two possible alleles to pass on: X^R or X^r.

• The probability of the mother passing the normal allele (X^R) is 1/2. If the son inherits this, his genotype is X^RY, and he is normal.
• The probability of the mother passing the disease allele (X^r) is 1/2. If the son inherits this, his genotype is X^rY. Since the disease is X-linked recessive, this genotype results in the son having the disease.

Therefore, the probability that the son is born with the disease is the probability of inheriting the X^r allele from the mother, which is:

$ P(\text{Son is affected}) = P(\text{Mother passes } X^r) = \frac{1}{2} = 0.5 $

Final Answer Check

The calculated probability is 0.5. Rounded off to one decimal place, this remains 0.5.