Answer:
50%
Explanation:
Let's assume that the X linked allele "a" gives the disorder. Therefore, the genotype of the affected man would be X^aY while the genotype of the carrier woman would be X^aX. A cross between X^aY man and X^aX woman would produce progeny in following phenotype ratio: 1 affected daughter: 1 affected son: 1 normal but carrier daughter: 1 normal son.
Therefore, there are 50% chances that their children can express the disorder.
