Solve this sex-linked traits practice problem

We know that hemophilia is a recessive trait, and the only way to express a recessive trait is to have a homozygous mixture. So, the genotypes probably look like this:

BB = normal blood clotting
Bb = carrier for hemophilia
bb = hemophiliac

Because the mother does NOT have hemophilia, we will be using BB x Bb to find out the alleles of their children. Because this is sex-linked traits practice, our alleles will be the following.

normal = X^BY
(uppercase B represents no hemophilia)

X^BX^b

Let’s cross them and see what we get!

X^B Y

X^B | X^BX^B X^BY

X^b | X^BX^b X^bY

As you can see, 0/4 of the females (represented by XX) will have hemophilia. Again, it is only present when there are two recessive alleles and none of them satisfy that.

2/4 of the females will be carriers (X^BX^b).

As for the boys, 1/4 will have normal clotting and 1/4 will have hemophilia. None of the males will simply be carriers.

I hope I helped!
Feel free to ask me for more assistance (if needed); I’ll gladly help! :)

The answer is false