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 solve the problem of determining the chances of a child being colorblind when a colorblind man marries a normal vision woman whose father was colorblind, we will follow these steps:

### Step-by-Step Solution:

1. **Identify the Genotypes:**
- The colorblind man has the genotype **X^cY** (where X^c represents the colorblind allele).
- The normal vision woman has the genotype **XX**. Since her father was colorblind, she must have inherited a normal vision allele from her mother, making her genotype **XX**.

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

3. **Set Up a Punnett Square:**
- Create a Punnett square to visualize the combinations of the gametes from both parents.
- The rows will represent the father's gametes (X^c and Y), and the columns will represent the mother's gametes (X).

| | X | X |
|-------|-----|-----|
| **X^c** | X^cX | X^cX |
| **Y** | XY | XY |

4. **Analyze the Offspring:**
- From the Punnett square, we can see the possible genotypes of the offspring:
- **X^cX** (Daughter, colorblind carrier)
- **X^cX** (Daughter, colorblind carrier)
- **XY** (Son, normal vision)
- **XY** (Son, normal vision)

5. **Determine the Chances of Colorblindness:**
- Since the question specifies that the first child is a son, we only consider the male offspring.
- Both sons have the genotype **XY**, which means they do not have colorblindness.

6. **Conclusion:**
- The chances that a son from this couple will be colorblind is **0%**.

### Final Answer:
The chances that a child would be colorblind is **0%**.