To solve the problem, we need to determine the probability of Taylor's gametes containing the alleles for colorblindness (represented by the recessive allele c), the recessive allele for the second gene (d), and the X chromosome carrying the colorblind gene (Xc).
### Step-by-Step Solution:
1. **Identify the Genotype of Taylor**:
- Taylor is heterozygous for two autosomal gene pairs, which means his genotype is CcDd.
- Additionally, he is colorblind, which is a sex-linked trait. For males, the genotype for colorblindness can be represented as XcY, where Xc is the X chromosome with the colorblind allele.
2. **Determine Gamete Formation**:
- When forming gametes, each allele from the heterozygous pair can segregate independently.
- For the autosomal genes:
- The possible alleles from Cc are C or c (probability of c = 1/2).
- The possible alleles from Dd are D or d (probability of d = 1/2).
- For the sex-linked gene:
- The possible alleles from XcY are Xc or Y (probability of Xc = 1/2).
3. **Calculate the Probability of Desired Alleles**:
- We are interested in the probability of obtaining the alleles c, d, and Xc in a single gamete.
- The probability of getting c from Cc = 1/2.
- The probability of getting d from Dd = 1/2.
- The probability of getting Xc from XcY = 1/2.
4. **Combine the Probabilities**:
- Since the events are independent, we multiply the probabilities:
\[
P(c, d, Xc) = P(c) \times P(d) \times P(Xc) = \left(\frac{1}{2}\right) \times \left(\frac{1}{2}\right) \times \left(\frac{1}{2}\right) = \frac{1}{8}
\]
5. **Conclusion**:
- Therefore, the probability of a gamete having the alleles c, d, and the colorblind gene (Xc) is \( \frac{1}{8} \).
### Final Answer:
The probability of gametes having c, d, and color-blind genes is \( \frac{1}{8} \).
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