Question

Glucose-6-phosphate dehydrogenase deficiency (G6PD) is inherited as an X-linked recessive gene 43) in humans. A woman whose father suffered from G6PD marries a normal man. (a) What proportion of their sons is expected to be G6PD? (b) If the husband were not normal but were G6PD deficient, would you change your answer in part (a)?

Answer 1

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 $X^GX^g$. A normal man will have the genotype $X^GY$. When the 2 marries:

$X^GX^g$   x   $X^GY$ = $X^GX^G (normal female), X^GY (normal male), X^GX^g (carrier female), X^gY (affected male)$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:

$X^GX^g$   x   $X^gY$ = $X^GX^g (carrier female), X^GY (normal male), X^gX^g (affected female), X^gY (affected male)$50% or 1/2 of their sons is still expected to have G6PD. The ratio remains unchanged.