In the progeny of a colour blind woman and a normal man

Options

A. all the daughters are normal
B. all the sons are colour blind
C. all the sons and all the daughters are colour blind
D. all the daughters are colour blind

Answer

To solve the question regarding the progeny of a colorblind woman and a normal man, we need to understand the genetics of colorblindness, which is a sex-linked recessive trait. Here’s a step-by-step solution:

### Step 1: Understand the Genotypes
- **Colorblind Woman**: Since colorblindness is a recessive trait linked to the X chromosome, a colorblind woman must have two mutated X chromosomes. Thus, her genotype is **XcXc** (where Xc represents the mutated X chromosome).
- **Normal Man**: A normal man has one normal X chromosome and one Y chromosome. His genotype is **XY**.

### Step 2: Set Up the Punnett Square
To determine the possible genotypes of their offspring, we can set up a Punnett square. The mother can only contribute Xc (the mutated X chromosome), and the father can contribute either X or Y.

| | X (from father) | Y (from father) |
|----------|------------------|------------------|
| Xc (from mother) | XcX (Daughter, carrier) | XcY (Son, colorblind) |
| Xc (from mother) | XcX (Daughter, carrier) | XcY (Son, colorblind) |

### Step 3: Analyze the Results
From the Punnett square, we can see:
- **Daughters**: All daughters will have the genotype XcX, making them carriers of the colorblind trait (but not colorblind themselves).
- **Sons**: All sons will have the genotype XcY, making them colorblind.

### Step 4: Conclusion
Based on the analysis, we conclude that:
- All daughters will be carriers (XcX).
- All sons will be colorblind (XcY).

Thus, the correct answer to the question is:
- **Option 2: All the sons are colorblind.**