Answer:
Mother: XBXb
Father XBY
Explanation:
A sex linked trait is carried on the X chromosome. Lets denote the colorblindness allele as b and the normal allele as B.
For the father to be unaffected, he must not carry the trait, as he only has one X chromosome, so if he carries it, then he would be affected. That means the father's genotype must be XBY
If the sons are unaffected, that means they lack the trait on their only X chromosome. The X chromosome in boys must come from their mother (as the Y has to come from their father). Therefore, the son's genotype is XBY too.
The daughters are carriers (XBXb). This means their mother must carry the trait, as we know the father does not. However, the mother is also unaffected, so she must have only one colorblindness allele (heterozygous), therefore her genotype is XBXb.
