Answer:
a: 1/4
b: 1/2
c: 1/4
d: 0
Explanation:
Since the wife had a color-blind father (X^cY), she should be a carrier for the disease. Her genotype would be "X^cX". Fathers do not transmit the X-linked disorders to their sons, so the genotype of the husband is "XY".
a. The probability that the couple's first child will be a normal son: 1/4
b. The probability that the couple's first child will be a normal daughter: 1/2
c. The probability that the couple's first child will be a color-blind son: 1/4
d. The probability that the couple's first child will be a color-blind daughter: Zero
