Question

in humans, hemophilia is a recessive sex - linked trait.
cross a male with hemophilia with a female that is homozygous normal.
this recessive disorder is on the x chromosome.
males only have to inherit 1 recessive allele to have hemophilia, while females have to have 2 recessive alleles to have hemophilia (ex. x^hy and x^hx^h)
place the male’s gamete alleles on top and the female’s gamete alleles on the side of the punnett square.
%kids expressing hemophilia=
%kids carrying hemophilia=
there are some allele and percentage options at the bottom: x^h, x^h, y, x^hx^h, x^hx^h, x^hy, 0%, 25%, 50%

Explanation:

Step1: Identify parental genotypes

Father (hemophiliac male): $X^hY$; Mother (homozygous normal female): $X^HX^H$

Step2: List parental gametes

Father's gametes: $X^h$, $Y$; Mother's gametes: $X^H$, $X^H$

Step3: Fill Punnett Square

	$X^h$	$Y$
$X^H$	$X^HX^h$	$X^HY$

Step4: Calculate affected kids

Only males with $X^hY$ express hemophilia; no offspring have this genotype.

Step5: Calculate carrier kids

Females $X^HX^h$ are carriers; 2 out of 4 offspring are carriers.

Answer:

Punnett Square filled:
Father's gametes (top): $X^h$, $Y$
Mother's gametes (side): $X^H$, $X^H$
Square cells (top-left to bottom-right): $X^HX^h$, $X^HY$, $X^HX^h$, $X^HY$

%Kids expressing hemophilia= 0%
%Kids carrying hemophilia= 50%