Question

John B and Sarah are seriously dating, and they are dreaming of the children they may have together. John becomes worried because he remembers that Sarah is colorblind. He is concerned because, even though he is not colorblind, he doesn’t want his children to have an eyesight deficiency that could prevent them from being able to see gold. (colorblindness is x-linked recessive)

What are the chances that their child, regardless of gender, will be colorblind? %

Is it possible for these prospective parents to have a colorblind daughter?

Is it possible for these prospective parents to have a colorblind son?

Answer 1

To analyze the chances of John and Sarah having a child who is colorblind, we need to understand the inheritance pattern of colorblindness, which is X-linked recessive.

1. **Understanding Parent Genotypes**:
- **John (not colorblind, XY)**: Since John is not colorblind, his genotype for the colorblind allele is \(X^C Y\), where \(X^C\) is the normal vision X chromosome.
- **Sarah (colorblind, XX)**: As a colorblind female, her genotype is \(X^c X^c\), where \(X^c\) represents the X chromosome carrying the colorblind allele.

2. **Possible Gametes**:
- John can contribute either \(X^C\) (normal vision) or \(Y\) (male determining).
- Sarah can only contribute \(X^c\) (colorblind alleles).

3. **Punnett Square Analysis**:
We can set up a Punnett square to determine the possible genotypes of their children.

| | \(X^C\) (from John) | \(Y\) (from John) |
|------|----------------------|---------------------|
| \(X^c\) (from Sarah) | \(X^C X^c\) (normal vision with colorblind allele) | \(X^c Y\) (colorblind son) |
| \(X^c\) (from Sarah) | \(X^C X^c\) (normal vision with colorblind allele) | \(X^c Y\) (colorblind son) |

4. **Offspring Genotypes**:
- Daughters (XX): \(X^C X^c\) (normal vision, carriers of the colorblind allele)
- Sons (XY): \(X^c Y\) (colorblind)

### Probability Calculations:

- **Daughter outcomes**:
- All daughters will be \(X^C X^c\) (normal vision but carriers)

- **Son outcomes**:
- All sons will be \(X^c Y\) (colorblind)

### Summary of Chances:
- The chance that their child, regardless of gender, will be colorblind:
- Since only sons can be colorblind (and daughters are carriers, not affected), we need to consider the sex of the child.
- If we consider one of each gender, 50% of children could be sons. Therefore, the probability of having a colorblind child overall is 50% (the chance of having a son) × 100% (the chance he is colorblind, if he is a son) = **50%** chance of having a colorblind child.

### Specific Questions:
- **Is it possible for these prospective parents to have a colorblind daughter?**
- **No**, because daughters will always be carriers (normal vision), and they cannot express the colorblind trait since they would have one normal allele from their father.

- **Is it possible for these prospective parents to have a colorblind son?**
- **Yes**, there is a 100% chance that any sons they have will be colorblind (as shown in the Punnett square).

### Final Answers:
1. Probability of any child being colorblind: **50%**
2. Can they have a colorblind daughter? **No**
3. Can they have a colorblind son? **Yes**