A colour blind man marries the daughter of a colour blind lady and a normal man. Then in the progeny

Options

A. none of ther daughters will be colour blind
B. all her sons are colour blind
C. all her daughters are colour blind
D. half her sons are colour blind

Answer

To solve the problem regarding the inheritance of color blindness in the progeny of a color blind man and the daughter of a color blind lady and a normal man, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Genetics of Color Blindness**:
- Color blindness is an X-linked recessive disorder. This means that the gene responsible for color blindness is located on the X chromosome. Males (XY) have only one X chromosome, so if they inherit the color blindness allele (Xc), they will express the trait. Females (XX) need two copies of the color blindness allele (XcXc) to express the trait; if they have one normal allele (XXc), they will be carriers but not express color blindness.

2. **Identify the Genotypes of the Parents**:
- The color blind man has the genotype XcY (where Xc represents the color blind allele).
- The daughter of a color blind lady (XcXc) and a normal man (XY) must be a carrier, so her genotype is XcX (one color blind allele from her mother and one normal allele from her father).

3. **Set Up a Punnett Square**:
- We will cross the genotypes of the color blind man (XcY) with the carrier daughter (XcX).
- The Punnett square will look like this:

```
Xc X
----------------
Xc | XcXc | XcX |
----------------
Y | XcY | XY |
----------------
```

4. **Analyze the Offspring**:
- From the Punnett square, we can see the possible genotypes of the offspring:
- XcXc (color blind female)
- XcX (carrier female)
- XcY (color blind male)
- XY (normal male)

5. **Determine the Phenotypes**:
- The phenotypes based on the genotypes are:
- XcXc: Color blind female (1 out of 4)
- XcX: Carrier female (1 out of 4)
- XcY: Color blind male (1 out of 4)
- XY: Normal male (1 out of 4)

6. **Calculate the Ratios**:
- In the progeny:
- Daughters: 1 color blind (XcXc) and 1 carrier (XcX) → 0 daughters are color blind.
- Sons: 1 color blind (XcY) and 1 normal (XY) → 50% of sons are color blind.

7. **Conclusion**:
- The correct interpretation of the progeny is that none of the daughters will be color blind, and half of the sons will be color blind.

### Final Answer:
- **None of her daughters will be color blind, and half of her sons will be color blind.**