Red-green color blindness is caused by a sex-linked recessive allele. A color-blind man marries a woman with normal vision whose father was color blind. What is the probability that they will have a color-blind daughter? What is the probability that their first son will be color-blind? (Note the different wording in the two questions.)
The Correct Answer and Explanation is:
Correct Answer:
- Probability of having a color-blind daughter: 25 percent
- Probability that their first son will be color-blind: 50 percent
Explanation:
Red-green color blindness is a sex-linked recessive trait, which means the gene responsible is located on the X chromosome. Males have XY sex chromosomes, and females have XX. Since males only have one X chromosome, if they inherit the allele for color blindness (represented as Xᵇ), they will express the trait. Females need two copies of the allele (XᵇXᵇ) to be color blind; if they have only one (XᴮXᵇ), they are carriers but not affected.
Step 1: Determine genotypes
- Color-blind man: XᵇY
- Woman with normal vision but whose father was color-blind:
- Her father: XᵇY (color-blind)
- Therefore, she must be a carrier: XᴮXᵇ
Step 2: Use a Punnett Square
Cross:
Mother (XᴮXᵇ) × Father (XᵇY)
| Xᴮ (mother) | Xᵇ (mother) | |
|---|---|---|
| Xᵇ (father) | XᴮXᵇ (carrier daughter) | XᵇXᵇ (color-blind daughter) |
| Y (father) | XᴮY (normal son) | XᵇY (color-blind son) |
Step 3: Analyze results
From the Punnett square:
- Daughters:
- 50 percent chance to be carriers (XᴮXᵇ)
- 50 percent chance to be color blind (XᵇXᵇ)
So, the chance of a daughter being color blind = 1 out of 4 total outcomes = 25 percent
- Sons:
- 50 percent chance to be normal (XᴮY)
- 50 percent chance to be color blind (XᵇY)
So, the chance that the first son will be color blind = 50 percent
This difference arises because the daughter must inherit two Xᵇ alleles, while the son only needs one.
