A normal man and a heterozygous color blind "carrier" woman have a baby. If they have a girl, what is the probability that the girl will be colorblind. If they have a boy, what is the probability that the boy will be colorblind?

First of all, carriers do not show the effects in their phenotype, so she cannot be colorblind.

You can do a Punnett square to find your answer.

However, to save you some time, the probability is 50% in both cases.

To determine the probability of the baby being colorblind, we need to understand the inheritance pattern of color blindness. Color blindness is a recessive trait that is usually associated with the X chromosome.

In this scenario:

1. The man is "normal," which means he has two functional copies of the color vision genes.
2. The woman is a "carrier" of color blindness, which means she has one functional copy of the gene and one non-functional copy.

With these considerations, let's calculate the probabilities for each gender:

1. Probability of a girl being colorblind:
- Since the girl's father is normal, he can only contribute a functional copy of the gene.
- The mother, being a carrier, has a 50% chance of passing the non-functional copy of the gene to her daughter.
- Therefore, the probability of a girl being colorblind is 50%.

2. Probability of a boy being colorblind:
- Since a boy inherits one X chromosome from his mother, the boy has a 50% chance of getting the non-functional copy of the color vision gene from his carrier mother.
- However, if the boy inherits the non-functional gene copy, he will be colorblind since he does not have another X chromosome with a functional gene to compensate.
- Therefore, the probability of a boy being colorblind is also 50%.

To summarize:
- If they have a girl, the probability of the girl being colorblind is 50%.
- If they have a boy, the probability of the boy being colorblind is also 50%.