The first step in solving this problem requires an examination of the possible genotypes of the parents. As stated color-blindness is a recessive sex linked trait. So the genotypes of the parents should be:
mother's genotype (normal vision but whose father was color blind) XNXn
She is a carrier because of her heterozygous condition. For the recessive gene to be expressed one must be homozygous for it.
father's genotype (color blind) XnY
From the Punnett square it is obvious that half of the children will be color blind and the other half will be normal. More specifically there will be 1 normal boy to 1 color-blind boy to 1 normal girl to 1 color- blind girl.
XNXn one normal (carrier) girl
XNY one normal boy
XnXn one color blind girl
XnY one color blind boy
- Probability of having a color blind daughter is (1/2 x 1/2) or 1/4 or 25% (0.25) This is because both sex and colorblindness are involved in the solution.
- The probability that their first son is color blind is 50% (0.50). In this question only males are part of the solution.