Solutions For An Introduction to Genetic Analysis Chapter 2 Problem 65P

b. The probability of having the first child to be color-blind boy is probability of being color-blind X probability of being a boy.

Probability, for the first child being color blind is 1/2 (the mother should be heterozygous X^C/C^c and the father is X^c/Y)

Probability the first child being a boy is 1/2

So, the probability of the first child being color blind boy is 1/2 ×1/2 = 1/4

c. The cross is X^C/X^c × X^c/Y -- X^C/X^c, X^C/Y, X^c/X^c, X^c/Y

The girls will be one normal X^C/X^c and one color blind X^c/X^c

d. In the cross XC/Xc × Xc/Y

X ^C /Y is a normal boy and Xc/Y is color blind boy.

X ^C /X ^c is a normal girl and X^c/X^c is a color blind girl.

In both sexes, the ratio is 1: 1, for normal to color blind.