So remember that Hemophilia is a recessive, x linked trait.
For a woman to have hemophilia, she must have the trait linked to both x chromosomes. For a man to have hemophilia, he must have the trait linked to his singular x chromosome.
For this, I will be making a Punnett Square to determine the Possibilities:
[tex]\left[\begin{array}{cccc}&&X_h&Y\\&&-&-\\X_h&|&X_hX_h&X_hY\\X_h&|&X_hX_h&X_hY\end{array}\right][/tex]
A.
Since there is a 1/2 chance for their offspring to be a boy, and of that 1/2 both would be hemophiliacs, there is a 1/2 chance for the child to be a hemophiliac male.
B.
As previously mentioned, for a woman to have hemophilia, they would need to have that trait attached to both x chromosomes. Since there is a 1/2 chance for their offspring to be a girl, and of that 1/2 both would be hemophiliacs (since they both carry the trait on both chromosomes), there is a 1/2 chance that they will have a hemophiliac female.
C.
So a carrier is someone who carries a recessive trait, but it isn't displayed due to the dominant trait masking it. With x-linked traits, only women can be carriers since they carry more than one x chromosome. What this asks is the probability of an offspring having the trait attached to only 1 of the x chromosomes. Looking at the girls, since both carry the traits on both x chromosomes, there is 0 chance of a carrier.
D.
So symptom free is as it seems, without the hemophilia trait. Looking at the table, since all the offspring contain the hemophilia trait, there is 0 chance that any of their offspring will go symptom free.
