Understanding X-Linked Inheritance

Fabry disease is an X-linked disorder. This means the gene responsible is located on the X chromosome.

Parental Genotypes

We are given:

• A phenotypically normal father. Since males have one X and one Y chromosome (XY), his genotype is X^AY, where X^A represents the normal allele.
• A carrier mother. Since females have two X chromosomes (XX), and she is a carrier, she has one normal allele and one affected allele. Her genotype is X^AX^a, where X^a represents the allele causing Fabry disease.

Inheritance Pattern for Sons

Sons inherit their Y chromosome from their father and their X chromosome from their mother.

Therefore, a son's phenotype is determined solely by the X chromosome he inherits from his mother.

Possible Outcomes for Sons

The mother (X^AX^a) can pass on either X^A or X^a to her children.

• If she passes on X^A, the son will be X^AY (phenotypically normal).
• If she passes on X^a, the son will be X^aY (will have Fabry disease).

Each outcome has an equal probability.

Calculating the Probability

The probability of the mother passing the affected allele (X^a) to her son is:

$ P(\text{Son inherits } X^a) = \frac{1}{2} $

This corresponds to 50%.

Alternatively, using a Punnett square for the cross X^AY (father) x X^AX^a (mother):

Out of the four possible genotypes, one (X^aY) results in a son with Fabry disease. This represents 1 out of 4, or 25% of the total offspring.

Final Probability

The probability for a phenotypically normal father and a carrier mother to have a son with Fabry disease is 25%.