Question

A normal female X H X H marries a man who has hemophilia.

What percentage of their sons will have hemophilia?

Answer 1

0%

Step-by-step explanation:

Normal female genotype =
`X^HX^H`

Genotype of a man with hemophilia =
`X^hY`

Crossing the two in marriage:

`X^HX^H` x
`X^hY`

Progeny:
`X^HX^h, X^HY, X^HX^h, and X^HY`

The two sons from the cross have the genotype
`X^HY` and are therefore free from hemophilia.

Hence, the percentage of their sons that will have hemophilia is zero (0).