Answer:
50% for sons 0%for daughters
Step-by-step explanation:
the genotype of the mother is X⁻X
the genotype of the father is XY
the gametes from the mother are X⁻ andX
the gametes from the father are X and Y
to find the chance we make a Punnet square and cross each gamete
![\left[\begin{array}{ccc}X^-X&XX\\X^-Y&XY\end{array}\right]](https://img.qammunity.org/2022/formulas/biology/college/675qcx6kex1za15ssqx7y1cnfl7apxbr2t.png)
1 of the 2 males has the gene for colorblindness so there is a 50% chance that the son will be colorblind
1 of the 2 daughters has the gene for cloroblindness but only in one exemplary so the chance is 0%