A woman with normal vision, but whose father was colour blind, marries a colourblind man. Suppose that the fourth child of this couple was a boy. This boy :

Options

A. Must have normal colour vision
B. Will be partially colourblind since he is heterozygous for the colourblind mutant allele
C. Must be colourblind
D. May be colourblind or may be of normal vision

Answer

To solve the question regarding the inheritance of color blindness in the offspring of a woman with normal vision (but whose father was colorblind) and a colorblind man, we can follow these steps:

### Step 1: Understand the Genetic Background
Color blindness is a sex-linked trait, specifically located on the X chromosome. The gene for color blindness is recessive.

### Step 2: Determine the Genotypes of the Parents
- The woman has normal vision but her father was colorblind. This means she must be a carrier for the colorblind allele. Therefore, her genotype is X^N X^c (where X^N = normal vision allele and X^c = colorblind allele).
- The colorblind man has the genotype X^c Y (since males have one X and one Y chromosome).

### Step 3: Set Up a Punnett Square
To find the possible genotypes of their children, we can set up a Punnett square:

| | X^c (father) | Y (father) |
|---------|---------------|---------------|
| X^N (mother) | X^N X^c (carrier daughter) | X^N Y (normal son) |
| X^c (mother) | X^c X^c (colorblind daughter) | X^c Y (colorblind son) |

### Step 4: Analyze the Possible Offspring
From the Punnett square, we can see the possible genotypes of the offspring:
- Daughters: X^N X^c (carrier, normal vision) and X^c X^c (colorblind)
- Sons: X^N Y (normal vision) and X^c Y (colorblind)

### Step 5: Determine the Fourth Child's Possibilities
The question states that the fourth child is a boy. The possible genotypes for a son are:
- X^N Y (normal vision)
- X^c Y (colorblind)

Thus, the boy could either be colorblind or have normal vision.

### Conclusion
The correct answer is: "may be colorblind or may be of normal vision."

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