Color blindness is a sex-linked recessive trait caused by a recessive allele (\( X^c \)) on the X chromosome. Normal vision is due to the dominant allele (\( X^C \)). Since males have one X and one Y chromosome, their phenotype depends on the single X chromosome they carry.
- The woman is color-blind, so her genotype is \( X^c X^c \) (homozygous recessive, as she expresses the recessive trait).
- The man has normal vision, so his genotype is \( X^C Y \) (hemizygous for the dominant allele).
To determine the probability of their son being color-blind, we analyze the possible genotypes of their offspring using a Punnett square.
The woman (\( X^c X^c \)) can only contribute an \( X^c \) gamete. The man (\( X^C Y \)) can contribute either an \( X^C \) or a \( Y \) gamete. The possible 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 one \( X^C \) from the father and one \( X^c \) from the mother. They are carriers (normal vision, as the dominant allele \( X^C \) masks the recessive \( X^c \)).
- Sons (\( X^c Y \)): They inherit the \( X^c \) from the mother and the \( Y \) from the father. Since the \( X^c \) carries the recessive allele and there is no second X chromosome to mask it, they are color-blind.
Since all sons will have the genotype \( X^c Y \), they will all be color-blind.
Thus, the probability that their son will be color-blind is \( 100\% \).