A colourblind man marries a normal vision woman whose father was colourblind. They have a first child as a son. What are the chances that a child would be colourblind.

Options

A. `25%`
B. `50%`
C. `100%`
D. `0%`

Answer

To determine the chances of a child being colorblind in this scenario, we need to analyze the genetic inheritance of colorblindness, which is an X-linked recessive trait.

### Step-by-Step Solution:

1. **Identify the Genotypes of the Parents:**
- The colorblind man has the genotype X^cY, where X^c represents the X chromosome with the colorblind allele.
- The normal vision woman has the genotype XX, but since her father was colorblind, she must be a carrier. Therefore, her genotype is X^cX.

2. **Determine the Possible Gametes:**
- The man can produce two types of gametes: X^c (colorblind) and Y (normal male).
- The woman can produce two types of gametes: X^c (colorblind) and X (normal vision).

3. **Create a Punnett Square:**
- Set up a Punnett square to determine the possible genotypes of their offspring.
- The rows represent the father's gametes (X^c and Y), and the columns represent the mother's gametes (X^c and X).

| | X^c | X |
|-------|------|-------|
| X^c | X^cX^c (colorblind girl) | X^cX (carrier girl) |
| Y | X^cY (colorblind boy) | XY (normal vision boy) |

4. **Analyze the Offspring Genotypes:**
- From the Punnett square, we see the following possible offspring:
- X^cX^c (colorblind girl) - 1 out of 4
- X^cX (carrier girl) - 1 out of 4
- X^cY (colorblind boy) - 1 out of 4
- XY (normal vision boy) - 1 out of 4

5. **Calculate the Probability of Colorblindness:**
- There are two scenarios where the child could be colorblind:
- A colorblind girl (X^cX^c) - 1 out of 4
- A colorblind boy (X^cY) - 1 out of 4
- Therefore, the total probability of having a colorblind child (either a boy or a girl) is:
- 1/4 (girl) + 1/4 (boy) = 2/4 = 1/2 or 50%.

### Conclusion:
The chances that a child would be colorblind is **50%**.