Red-Green colorblindness is a X linked recessive disorder. A woman who is colorblind marries a man with normal vision. a. how would you write the woman’s genotype? b. How would you write the man’s genotype? c. Conduct a Punnett square to determine the chances that the couple has a son who is colorblind. Write the probability in ¼, 2/4 ,3/4 or 4/4 d. What is the chance that they have a colorblind daughter. Punnett square Write the probability in ¼, 2/4 ,3/4 or 4/4
The Correct Answer and Explanation is:
Correct Answer:
a. The woman’s genotype: XᶜXᶜ
b. The man’s genotype: XY
c. Chance of a colorblind son: 2/4
d. Chance of a colorblind daughter: 0/4
Explanation
Red-green colorblindness is caused by a gene located on the X chromosome. It is a recessive disorder, meaning a person must inherit two copies of the defective gene (if female) or one copy (if male) to be colorblind.
a. Since the woman is colorblind and this is an X-linked recessive trait, she must carry two defective X chromosomes. Therefore, her genotype is XᶜXᶜ (where Xᶜ represents the X chromosome with the colorblind gene).
b. The man has normal vision and is male. Males only have one X chromosome and one Y chromosome. Since he is not colorblind, his genotype is XY (with a normal X chromosome).
Now, we use a Punnett square to determine the possible genotypes of their children.
Punnett Square:
| X (male) | Y (male) | |
|---|---|---|
| Xᶜ (female) | XᶜX | XᶜY |
| Xᶜ (female) | XᶜX | XᶜY |
From the Punnett square:
- Sons (XY combinations):
- XᶜY (colorblind)
- XᶜY (colorblind)
So, 2 out of 2 sons are colorblind → 2/4 total probability.
- Daughters (XX combinations):
- XᶜX (carrier, not colorblind)
- XᶜX (carrier, not colorblind)
So, 0 out of 2 daughters are colorblind → 0/4 total probability.
Conclusion:
This couple will not have any daughters who are colorblind, but all daughters will be carriers. There is a 2/4 (or 50%) chance that a child will be a colorblind son. The key factor is that the mother passes on an Xᶜ to every child, while the father determines the sex of the baby through the contribution of either an X or a Y.
