If a colourblind woman marries a normal visioned man, their sons will be -

Options

A. All normal visioned
B. One-half colourblind and one-half normal
C. Three-fourths colourblind and one-fourth normal
D. All colourblind

Answer

To solve the question about the offspring of a colorblind woman and a normal visioned man, we will analyze the genetics involved in colorblindness, which is an X-linked recessive trait.

### Step-by-Step Solution:

1. **Understanding Colorblindness**:
Colorblindness is primarily caused by a defect in genes located on the X chromosome. It is an X-linked recessive trait, meaning that males (XY) are more likely to express the trait because they have only one X chromosome. Females (XX) need to have the defective gene on both X chromosomes to express colorblindness.

2. **Genotype of the Parents**:
- The colorblind woman will have the genotype **XcXc** (both X chromosomes carry the colorblindness allele).
- The normal visioned man will have the genotype **XY** (one normal X chromosome and one Y chromosome).

3. **Gamete Formation**:
- The colorblind woman can only produce gametes with the **Xc** chromosome (since both her X chromosomes are affected).
- The normal visioned man can produce two types of gametes: **X** (normal vision) and **Y** (male).

4. **Punnett Square Analysis**:
We can set up a Punnett square to determine the possible offspring:
- The mother contributes **Xc** (colorblind) gametes.
- The father contributes either **X** (normal) or **Y** (male).

The possible combinations from the Punnett square are:
- **XcX** (daughter, carrier of colorblindness)
- **XcY** (son, colorblind)

5. **Conclusion about Offspring**:
- All daughters (XcX) will be carriers of the colorblindness trait but will not express it because they have one normal vision X chromosome.
- All sons (XcY) will be colorblind because they inherit the X chromosome carrying the colorblindness allele from their mother.

### Final Answer:
All sons will be colorblind.