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

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

Step-by-step 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%.