In humans, red-green color-blindness is controlled by a gene on the X chromosome. For females, normal color vision is dominant to color-blindness. A convenient way to represent the gene is to use an X (for the X chromosome) with a superscript for the particular allele. Thus, if we use the letter “B” for the normal gene, the normal allele would be X^B and the color-blindness allele would be X^e. Genotypes are listed below. What are the phenotypes with respect to sex and color vision of each? X^B X^B – female with normal color vision X^B X^b – color-blind man Based on this, draw your own Punnett square on your own paper to determine their possible offspring. Submit a picture of your work with this assignment. What is the probability (out of 4) that they will have a color-blind daughter? What is the probability (out of 4) that they will have a color-blind son?
The Correct Answer and Explanation is:
Correct Answer:
Genotypes and Phenotypes:
- Xᴮ Xᴮ – Female with normal color vision
- Xᵇ Y – Male with color blindness
Punnett Square Cross:
Parent Genotypes:
- Mother: Xᴮ Xᴮ
- Father: Xᵇ Y
Now let’s set up the Punnett Square:
| Xᴮ (mother) | Xᴮ (mother) | |
|---|---|---|
| Xᵇ (father) | Xᴮ Xᵇ | Xᴮ Xᵇ |
| Y (father) | Xᴮ Y | Xᴮ Y |
Offspring Genotypes and Phenotypes:
- Xᴮ Xᵇ – Female, normal vision (carrier, not color-blind)
- Xᴮ Xᵇ – Female, normal vision (carrier)
- Xᴮ Y – Male, normal vision
- Xᴮ Y – Male, normal vision
Final Probabilities:
- Probability of a color-blind daughter = 0 out of 4 = 0%
- Probability of a color-blind son = 0 out of 4 = 0%
Explanation
In humans, red-green color blindness is a sex-linked recessive trait, meaning the gene responsible is found on the X chromosome and not on the Y. Males (XY) only need one copy of the color-blind gene (Xᵇ) to express the trait, while females (XX) need two copies (Xᵇ Xᵇ) to be color blind. If a female has only one Xᵇ, she will be a carrier but will still have normal vision.
In this problem, the mother is Xᴮ Xᴮ, meaning she has two normal alleles and is not a carrier. The father is Xᵇ Y, meaning he is color-blind.
When we cross these genotypes in a Punnett square, the mother can only pass on the Xᴮ allele, and the father can pass on either Xᵇ or Y. The resulting children will be:
- All daughters will be Xᴮ Xᵇ: not color-blind, but carriers.
- All sons will be Xᴮ Y: they inherit the Xᴮ from their mother and Y from their father, so they will have normal vision.
Since none of the daughters are Xᵇ Xᵇ, no daughter will be color-blind. And none of the sons inherit the Xᵇ allele from the mother (because she doesn’t have it), so no son will be color-blind either.
Thus, the chance of having a color-blind daughter or son is 0 out of 4, or 0%.
