A normal-visioned man whose father was colour blind, marries a woman whose father was also colour blind. They have their first child as a daughter. What are the chances that this child would be colour blind?

Options

A. 1
B. 0
C. 0.25
D. 0.5

Explanation

To solve the problem regarding the chances of the daughter being color blind, we will follow these steps:

### Step 1: Understand the Genetics of Color Blindness
Color blindness is an X-linked recessive trait. This means that the gene responsible for color blindness is located on the X chromosome. Males (XY) have one X chromosome, while females (XX) have two X chromosomes.

### Step 2: Determine the Genotypes of the Parents
- The father is normal-visioned but has a color-blind father. Therefore, he must have the genotype X^N Y (where X^N represents the normal vision allele).
- The mother has a color-blind father, which means she must be a carrier. Therefore, her genotype is X^C X^N (where X^C represents the color-blind allele).

### Step 3: Set Up the Punnett Square
To find the possible genotypes of their offspring, we can set up a Punnett square using the gametes from each parent.

- The father's gametes: X^N (normal vision) and Y (male)
- The mother's gametes: X^C (color blind) and X^N (normal vision)

### Step 4: Fill in the Punnett Square
The Punnett square will look like this:

| | X^C (Mother) | X^N (Mother) |
|----------|---------------|---------------|
| X^N (Father) | X^N X^C (Carrier Female) | X^N X^N (Normal Female) |
| Y (Father) | X^C Y (Color Blind Male) | X^N Y (Normal Male) |

### Step 5: Analyze the Results
From the Punnett square, we can see the possible offspring genotypes:
1. X^N X^C (Carrier Female)
2. X^N X^N (Normal Female)
3. X^C Y (Color Blind Male)
4. X^N Y (Normal Male)

### Step 6: Determine the Chances of a Color Blind Daughter
In this case, the daughter can either be a carrier (X^N X^C) or normal (X^N X^N). There are no combinations that result in a color blind daughter (X^C X^C) because the father can only provide one X chromosome (X^N).

### Conclusion
The chances that their daughter would be color blind is 0%.