Both husband and wife have normal vision though their father were colour blind and mother did not have any gene for colour blindness .The probability of their daughter becoming colour blind is :

Options

A. `0%`
B. `25%`
C. `50%`
D. `75%`

Explanation

To solve the problem of determining the probability of the daughter becoming colorblind when both parents have normal vision but their fathers were colorblind, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Genetics of Color Blindness**:
- Color blindness is an X-linked recessive trait. This means that the gene responsible for color blindness is located on the X chromosome, and a female must have two copies of the colorblind allele (XcXc) to express the trait, while a male only needs one (XcY).

2. **Determine the Genotypes of the Parents**:
- The husband has normal vision, but his father was colorblind. This means he must have inherited one normal X chromosome from his mother and one Y chromosome from his father. Therefore, his genotype is XcY (where Xc represents the normal vision allele).
- The wife also has normal vision, and her father was colorblind. Similarly, she must have inherited one normal X chromosome from her mother and one X chromosome from her father. Thus, her genotype is also XcX (where Xc represents the normal vision allele).

3. **Set Up the Punnett Square**:
- To find the possible genotypes of their children, we can set up a Punnett square using the parents' genotypes:
- Husband (XcY) x Wife (XcX)

| | Xc (from wife) | X (from wife) |
|--------|----------------|----------------|
| Xc (from husband) | XcXc (carrier daughter) | XX (normal daughter) |
| Y (from husband) | XcY (normal son) | XY (normal son) |

4. **Analyze the Results**:
- From the Punnett square, we can see that:
- Daughters: XcX (carrier) and XX (normal)
- Sons: XcY (normal) and XY (normal)
- None of the daughters can be colorblind because they either have one normal vision allele or are carriers.

5. **Calculate the Probability**:
- Since both daughters produced (XcX and XX) do not have the genotype for color blindness (XcXc), the probability of their daughter becoming colorblind is 0%.

### Final Answer:
The probability of their daughter becoming colorblind is **0%**.

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