Color blindness, a sex-linked recessive trait, is caused by the recessive allele \( X^c \) on the X chromosome. Normal vision is determined by the dominant allele \( X^C \). Males, possessing one X and one Y chromosome, display a phenotype dictated by their single X chromosome.
- The female is color-blind, indicating a genotype of \( X^c X^c \) (homozygous recessive).
- The male has normal vision, with a genotype of \( X^C Y \) (hemizygous for the dominant allele).
To ascertain the probability of their son being color-blind, a Punnett square is employed to analyze potential offspring genotypes.
The female (\( X^c X^c \)) can only contribute an \( X^c \) gamete. The male (\( X^C Y \)) can contribute either an \( X^C \) or a \( Y \) gamete. The resultant offspring genotypes are:
\[
\begin{array}{c|cc} & X^c & X^c
\hline X^C & X^C X^c & X^C X^c
Y & X^c Y & X^c Y
\end{array}
\]
- Daughters (\( X^C X^c \)): They inherit \( X^C \) from their father and \( X^c \) from their mother. They are carriers with normal vision, as the dominant \( X^C \) allele masks the recessive \( X^c \).
- Sons (\( X^c Y \)): They inherit \( X^c \) from their mother and \( Y \) from their father. Due to the presence of the recessive \( X^c \) allele and the absence of a second X chromosome to counteract it, they are color-blind.
Given that all sons will possess the genotype \( X^c Y \), they will all be color-blind.
Consequently, the probability of their son being color-blind is \( 100\% \).