A normal visioned man whose father was colour blind, marries a woman whose father was also colour, blind. They have their first child as a daughter. what are the chance that this child would be colour blind?

Options

A. `100%`
B. zero percent
C. `25%`
D. `50 %`

Answer

To solve the problem of determining the chances of a daughter being color blind when her father and mother both have color blind fathers, we will follow these steps:

### Step 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 person needs two copies of the recessive allele (in females) or one copy (in males) to express the trait.

### Step 2: Determine the Genotypes of the Parents
- The father of the normal visioned man is color blind. Therefore, his genotype is XcY (where Xc represents the color blind allele).
- Since the normal visioned man has normal vision, his genotype must be XY (he has one normal X chromosome and one Y chromosome).

- The mother has a color blind father. Therefore, her father's genotype is XcY. Since the mother is normal visioned, she must have one normal X chromosome and one color blind allele. Thus, her genotype is XcX (she is a carrier).

### Step 3: Determine the Possible Genotypes of the Daughter
The daughter will inherit one X chromosome from each parent:
- From the father (XY), she can only inherit the normal X chromosome (X).
- From the mother (XcX), she can inherit either the normal X chromosome (X) or the color blind X chromosome (Xc).

### Step 4: Create a Punnett Square
To visualize the inheritance, we can create a Punnett square:

| | X (from father) | Xc (from mother) |
|---------|------------------|-------------------|
| **X** | XX (normal) | XXc (carrier) |

### Step 5: Analyze the Results
From the Punnett square, we see that:
- There is a 50% chance that the daughter will be XX (normal vision).
- There is a 50% chance that the daughter will be XXc (carrier, but not color blind).

### Conclusion
The daughter cannot be color blind because she must inherit one normal X chromosome from her father. Therefore, the probability that the daughter will be color blind is **0%**.