A colourblind man marries a woman with normal sight who was has no history of colour blindness in her family. What is the probability of their sons being colourblind?

Options

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

Answer

To solve the problem of determining the probability of the sons of a colorblind man and a woman with normal sight (who has no history of colorblindness in her family) being colorblind, we can follow these steps:

### Step 1: Understand the Genetics of Colorblindness
Colorblindness is an X-linked recessive disorder. This means that the gene responsible for colorblindness is located on the X chromosome. Males have one X and one Y chromosome (XY), while females have two X chromosomes (XX).

### Step 2: Determine the Genotype of the Parents
- The colorblind man has the genotype **X^cY** (where X^c represents the colorblind allele).
- The woman with normal sight, who has no history of colorblindness in her family, is assumed to have the genotype **XX** (both X chromosomes are normal).

### Step 3: Set Up the Punnett Square
To find the probability of their sons being colorblind, we can set up a Punnett square. The possible gametes from each parent are:
- Father (X^cY): can contribute X^c or Y
- Mother (XX): can contribute X or X

The Punnett square will look like this:

| | X | X |
|-------|-----|-----|
| **X^c** | X^cX | X^cX |
| **Y** | XY | XY |

### Step 4: Analyze the Offspring
From the Punnett square, we can see the possible genotypes of the offspring:
- **X^cX**: Female, carrier of colorblindness (not colorblind)
- **XY**: Male, normal vision

### Step 5: Determine the Probability for Sons
Since we are only interested in the sons:
- The only genotype for sons is **XY**, which means they will have normal vision.
- There are no sons with the genotype **X^cY** (colorblind) in this scenario.

### Conclusion
The probability of their sons being colorblind is **0** or **nil**.

### Final Answer
The probability of their sons being colorblind is **0**.

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