A woman is a carrier of the X-linked recessive color blindness gene. She has children with a man with normal color vision. Which of the following is true of their offspring? a. All the males will be color blind; b. All the females will be carriers; c. Half the females will be color blind; d. Half the males will be color blind.

Verified step by step guidance

1. Understand the genetics involved: Color blindness is an X-linked recessive trait. This means the gene causing color blindness is located on the X chromosome and is recessive.
2. Analyze the mother's genotype: Since the woman is a carrier and not color blind, her genotype must be XcX, where Xc represents the X chromosome carrying the color blindness gene.
3. Analyze the father's genotype: The man has normal color vision and thus his genotype is XY, where Y represents the Y chromosome which does not carry the color blindness gene.
4. Determine the possible genotypes of the offspring: The mother can pass on either Xc or X, and the father can pass on either X or Y. This results in four possible combinations for their children: XcX, XX, XcY, XY.
5. Conclude the probabilities: From the combinations, XcY represents a color blind male (since Y does not mask the recessive Xc), and XY represents a normal male. Thus, half of the male children (XcY and XY) will be color blind, making option 'd' correct.