What is the likelihood that children of a woman heterozygous for colorblindness and man with normal color vision will be colorblind?
To determine the likelihood of children being colorblind based on the given genetic information, let's consider the following:
The woman is heterozygous for colorblindness, which means she carries one normal allele (N) and one colorblind allele (C). The man has normal color vision, so he has two normal alleles (N).
Colorblindness is an X-linked recessive trait, meaning that it is carried on the X chromosome. Males have one X chromosome and one Y chromosome, while females have two X chromosomes.
To determine the possible genotypes of their children, we can create a Punnett square:
```
| N | C
-------
N | NN | NC
C | CN | CC
```
- NN: Normal vision (female)
- NC: Carrier of colorblindness (female)
- CN: Normal vision (male)
- CC: Colorblind (male)
From the Punnett square, we can see that there are four possible outcomes:
- 50% chance of daughters being carriers (NC): These daughters will not be colorblind themselves, but they will carry the colorblind allele and can pass it on to their children.
- 25% chance of sons being colorblind (CC): These sons will inherit the colorblind allele from their mother and will have colorblindness.
- 25% chance of sons having normal vision (CN): These sons will inherit the normal allele from their mother and will have normal color vision.
Therefore, the likelihood that children of this couple will be colorblind is 25% for sons and 0% for daughters.