Answer:
The correct answer would be 50%
.
Red-green colorblindness is an X-linked recessive disorder that is, the mutated gene responsible for this trait is present on the X chromosome.
Let the "R" and "r" be the alleles of the gene responsible for this trait.
R is the dominant trait which leads to the normal vision in a person. Thus, heterozygotes will have a normal vision however, they will act as the carrier of the disease.
r is the recessive allele which causes red-green colorblindness in a person when present in homozygous condition.
Now, the genotype of the mother with red-green colorblindness would be [tex]X^{r}X^{r}[/tex] and the genotype of unaffected father would be [tex]X^{R}Y[/tex].
The cross will result in the production of two daughters each heterozygous and two sons each with the colorblindness.
The probability of having a son = 1/2.
The probability of son being colorblind = 1
Thus, the probability of having a son with colorblindness = 1/2 x 1 = 1/2 or 50%
