Answer:
50% or 1/2. The result remains unchanged if the husband were to have G6PD.
Explanation:
For X-linked recessive inheritance, a female (XX) needs two recessive alleles to be affected while a male needs only one (XY). It is hypothetically assumed that the Y chromosome does not carry any trait.
Assuming the allele for the disease is represented by g, a woman whose father suffered from G6PD is a carrier for the disease with genotype [tex]X^GX^g[/tex]. A normal man will have the genotype [tex]X^GY[/tex]. When the 2 marries:
[tex]X^GX^g[/tex]  x  [tex]X^GY[/tex] = [tex]X^GX^G (normal female), X^GY (normal male), X^GX^g (carrier female), X^gY (affected male)[/tex]It thus means that 50% or 1/2 of their sons will be expected to have G6PD.
Now, assuming the husband has G6PD, the mating becomes:
[tex]X^GX^g[/tex]  x  [tex]X^gY[/tex] = [tex]X^GX^g (carrier female), X^GY (normal male), X^gX^g (affected female), X^gY (affected male)[/tex]50% or 1/2 of their sons is still expected to have G6PD. The ratio remains unchanged.
