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 chances that this child would be colour blind ?

Options

A. `25%`
B. `50%`
C. `100%`
D. Zero precent

Answer

To solve the problem of determining the chances of the daughter being color blind, we will follow these steps:

### Step 1: Understand the Genetic Background
Color blindness is a sex-linked recessive trait, which means it is carried on the X chromosome. A normal visioned man has the genotype X^N Y (where X^N represents the normal vision allele), while a color blind man has the genotype X^c Y (where X^c represents the color blind allele).

### Step 2: Determine the Genotype of the Parents
- The father of the normal visioned man (X^N Y) was color blind, so he must have the genotype X^c Y.
- The woman, whose father was also color blind, must have received the X^c allele from her father. Therefore, she can either be X^N X^c (carrier) or X^c X^c (affected). However, since she is not stated to be color blind, we assume she is a carrier: X^N X^c.

### Step 3: Set Up the Punnett Square
Now we can set up a Punnett square to determine the possible genotypes of their daughter. The father contributes either X^N or Y, and the mother contributes either X^N or X^c.

| | X^N (from father) | X^c (from mother) |
|----------|--------------------|--------------------|
| **X^N** | X^N X^N (normal) | X^N X^c (carrier) |
| **Y** | X^N Y (normal male) | X^c Y (color blind male) |

### Step 4: Analyze the Offspring
From the Punnett square, we can see the possible genotypes for their offspring:
1. X^N X^N (normal vision)
2. X^N X^c (carrier for color blindness)
3. X^N Y (normal male)
4. X^c Y (color blind male)

### Step 5: Determine the Chance of the Daughter Being Color Blind
Since the first child is a daughter, we only consider the female offspring:
- The possible female offspring are X^N X^N (normal vision) and X^N X^c (carrier).
- There are no combinations that result in a color blind daughter (X^c X^c).

Thus, the chances of their daughter being color blind is 0%.

### Final Answer
The chances that the daughter would be color blind is **0%**.

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