Question

cross a normal vision male with a carrier female.
the recessive disorder and on the x chromosome.
males only have to inherit 1 recessive allele to be color blind, while females have to have 2 recessive alleles to be colorblind (xc xc and xc y)
\place the male’s gamete alleles on top and the female’s gamete alleles on the side of the punnett square\
<< father
f1 phenotypes:
% kids not colorblind =
%kids colorblind =
aa mother

Explanation:

Step1: Define parental genotypes

Normal male (father): $X^CY$, carrier female (mother): $X^CX^c$

Step2: Identify parental gametes

Father's gametes: $X^C$, $Y$; Mother's gametes: $X^C$, $X^c$

Step3: Fill Punnett Square

	$X^C$	$Y$
$X^c$	$X^CX^c$	$X^cY$

Step4: Classify phenotypes

• Non-colorblind: $X^CX^C$, $X^CY$, $X^CX^c$ (3 out of 4)
• Colorblind: $X^cY$ (1 out of 4)

Step5: Calculate percentages

Non-colorblind: $\frac{3}{4} \times 100\% = 75\%$; Colorblind: $\frac{1}{4} \times 100\% = 25\%$

Answer:

Punnett Square (father's gametes top, mother's side):

	$X^C$	$Y$
$X^c$	$X^CX^c$	$X^cY$

% Kids not Colorblind = 75%
% Kids Colorblind = 25%