Question

G6PD deficiency in humans is an X-linked recessive trait. If a homozygous dominant female (XGXG) has children with an affected male (XgY), then what is the probability of the children being carriers?

Answer 1

Answer:

Only the daughters will be carriers and the probability is 1 or 100%

Explanation:

Homozygous dominant female = $X^{G} X^{G}$

Affected male = $X^{G}Y$

$X^{G} X^{G}$   x   $X^{G}Y$

Offspring = $X^{G}X^{g}$ (female), $X^{G}X^{g}$ (females), $X^{G}Y$ (male), $X^{G}Y$ (male)

For X-linked traits, male gender can either be affected or not affected, they cannot be carriers because the Y chromosome is hypothesized to not carry any gene. Since the two X chromosomes in female carry genes, they can be carriers for X-linked traits.

Considering the female children only, the probability is 1 or 100%.