Question

You should complete your work on a sheet of paper. Cross a carrier (heterozygous) woman with normal vision with a colorblind man. What are the chances the child will be a colorblind male?

Answer 1

To determine the chances of having a colorblind male child from a cross between a heterozygous carrier woman for color blindness and a colorblind man, we need to analyze the genetic situation.

1. **Genotypes**:
- **Carrier woman** (heterozygous for color blindness): X\(^C\)X\(^c\)
- X\(^C\): Normal vision
- X\(^c\): Colorblind (allele causing color blindness)
- **Colorblind man**: X\(^c\)Y
- X\(^c\): Colorblind allele
- Y: Male chromosome

2. **Punnett Square**:

We'll set up a Punnett Square to predict the offspring.

| | X\(^C\) | X\(^c\) |
|-------|-------|-------|
| X\(^c\)| X\(^C\)X\(^c\) (carrier female) | X\(^c\)X\(^c\) (colorblind female) |
| Y | X\(^C\)Y (normal male) | X\(^c\)Y (colorblind male) |

3. **Offspring Genotypes**:

From the Punnett square, the possible genotypes of the offspring are:
- X\(^C\)X\(^c\) (Carrier female)
- X\(^c\)X\(^c\) (Colorblind female)
- X\(^C\)Y (Normal male)
- X\(^c\)Y (Colorblind male)

4. **Probability of Each Outcome**:
- 1/4 X\(^C\)X\(^c\) (carrier female)
- 1/4 X\(^c\)X\(^c\) (colorblind female)
- 1/4 X\(^C\)Y (normal male)
- 1/4 X\(^c\)Y (colorblind male)

5. **Chances of Colorblind Male**:
Among the four possible outcomes, only one (X\(^c\)Y) represents a colorblind male.

Thus, the probability that the child will be a colorblind male is **1 out of 4**, or **25%**.