Proportion of colour blind children when normal man marries a carrier woman is

Options

A. 0.25
B. 0.5
C. 0.75
D. 1

Answer

To solve the question regarding the proportion of colorblind children when a normal man marries a carrier woman, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Genetics of Colorblindness**:
- Colorblindness is an X-linked recessive trait. This means that the gene responsible for colorblindness is located on the X chromosome, and a male (XY) will express colorblindness if he inherits the affected X chromosome (Xc).

2. **Identify the Genotypes of the Parents**:
- The normal man has the genotype XY (normal vision).
- The carrier woman has the genotype XXc (one normal vision allele and one colorblind allele).

3. **Set Up a Punnett Square**:
- Create a Punnett square to visualize the possible combinations of alleles from the parents. The male can contribute either X (normal) or Y (male chromosome), and the female can contribute either X (normal) or Xc (colorblind).

| | X (from father) | Y (from father) |
|----------|------------------|------------------|
| X (from mother) | XX (normal daughter) | XY (normal son) |
| Xc (from mother)| XcX (carrier daughter) | XcY (colorblind son) |

4. **Analyze the Offspring Genotypes**:
- From the Punnett square, we can see the possible offspring:
- XX: Normal daughter
- XY: Normal son
- XcX: Carrier daughter (normal vision)
- XcY: Colorblind son

5. **Count the Proportion of Colorblind Children**:
- Out of the four possible offspring, only one (XcY) is colorblind.
- Therefore, the proportion of colorblind children is 1 out of 4, which can be expressed as a fraction: 1/4.

6. **Convert the Fraction to Decimal**:
- The fraction 1/4 is equivalent to 0.25.

7. **Final Conclusion**:
- The proportion of colorblind children when a normal man marries a carrier woman is 0.25 or 25%.