A marriage between normal visioned man and colour blind woman will produce offspring

Options

A. Colour blind sons and `50%` carrier daughters
B. `50%` colourblind sons and `50%` carrier daughters
C. Normal males and carrier daughters
D. Colour blind sons and carrier daughters

Answer

To solve the question regarding the offspring produced from a marriage between a normal visioned man and a colorblind woman, we can follow these steps:

### Step 1: Understand the Genetics of Color Blindness
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), and a colorblind woman has the genotype X^c X^c (where X^c represents the colorblind allele).

### Step 2: Determine the Genotypes of the Parents
- The normal visioned man: X^N Y
- The colorblind woman: X^c X^c

### Step 3: Set Up a Punnett Square
To find the possible genotypes of the offspring, we can set up a Punnett square. The man can contribute either X^N or Y, while the woman can only contribute X^c.

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

### Step 4: Analyze the Offspring
From the Punnett square, we can see the following possible offspring:
- 2 daughters: X^N X^c (carrier for color blindness)
- 2 sons: X^c Y (colorblind)

### Step 5: Calculate the Ratios
From the analysis:
- 50% of the offspring are daughters who are carriers for color blindness.
- 50% of the offspring are sons who are colorblind.

### Conclusion
Therefore, the marriage between a normal visioned man and a colorblind woman will produce:
- 2 colorblind sons (100% of sons)
- 2 carrier daughters (100% of daughters)

Thus, the correct answer is that there will be colorblind sons and carrier daughters.

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