A colour blind man marries a woam with normal sight who has no history of colour blindness in her family. What is the probability of their grandson being colour blind?

Options

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

Answer

To solve the problem of determining the probability of a grandson being colorblind from a colorblind man and a woman with normal sight, we need to follow these steps:

### Step 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. Males have one X and one Y chromosome (XY), while females have two X chromosomes (XX). A male only needs one copy of the colorblind allele (X^c) to be colorblind, while a female needs two copies (X^cX^c).

### Step 2: Determine the Genotypes of the Parents
- The colorblind man has the genotype X^cY (where X^c represents the colorblind allele).
- The woman with normal sight has the genotype XX, as she has no history of colorblindness in her family.

### Step 3: Create a Punnett Square
We will set up a Punnett square to determine the possible genotypes of their children. The mother can contribute either of her X chromosomes, while the father can contribute either his X^c or Y chromosome.

| | 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 children:
1. X^cX (carrier female)
2. X^cX (carrier female)
3. XY (normal male)
4. XY (normal male)

### Step 5: Determine the Next Generation (Grandchildren)
Now, we need to consider the potential offspring of these children. The carrier females (X^cX) can have children with a normal male (XY). The normal males (XY) can have children with normal females (XX).

#### Possible combinations from a carrier female (X^cX) and a normal male (XY):
| | X | Y |
|---------|-------|-------|
| **X^c** | X^cX | X^cY |
| **X** | XX | XY |

From this cross, we have:
1. X^cX (carrier female)
2. X^cY (colorblind male)
3. XX (normal female)
4. XY (normal male)

### Step 6: Calculate the Probability of Colorblindness in Grandson
From the combinations above, we see that:
- 1 out of 4 combinations results in a colorblind male (X^cY).
- Therefore, the probability of the grandson being colorblind is 1 out of 4, or 25%.

### Final Answer
The probability of their grandson being colorblind is **25%** or **0.25**.

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