Final answer:
The probability of having a normal daughter and a colorblind son when the father has normal vision and the mother is a carrier for colorblindness is less than 0.5 because there's a 25% chance for those events to occur together. Colorblindness is an X-linked recessive trait, making the father's contribution crucial for determining the sex of the child and the inheritance of the trait. Therefore correct option is B
Step-by-step explanation:
The question is about the probability of having a normal daughter and a colorblind son when the father has normal color vision and the mother is heterozygous for normal vision (a carrier for colorblindness). As colorblindness is an X-linked recessive trait, a mother who carries the gene has a 50% chance of passing on the recessive allele to her sons. Therefore, there is a 50% chance that any son will be colorblind. Considering daughters, since they receive one X chromosome from their father (who has a normal vision and therefore passes on only normal X allele) and one from their mother, there is a 0% chance of them being colorblind, but a 50% chance they will be carriers like their mother.
When we combine the probabilities for each event (normal daughter and colorblind son), we find that there's a 25% chance (0.5 son * 0.5 daughter) for them to occur simultaneously, since the events are independent. Thus, the correct answer to this question would be Probability less than 0.5.
Determining the sex of the offspring, it's the father who determines the sex because he provides either an X or a Y chromosome, while the mother only provides X chromosomes. In this question, we care about sex-linked inheritance, particularly an X-linked recessive condition like colorblindness, where males are more frequently affected because they have only one X chromosome.