Taylor is heterozygous for two autosomal gene pair ( CcDd ). He is colour - blind . What is the probability of gametes having c, d and colour - blind genes ?

Options

A. `1/2`
B. `1/4`
C. `1/8`
D. `1/16`

Answer

To solve the problem, we need to determine the probability of Taylor's gametes carrying the alleles for c (colorblind), d (recessive trait), and the colorblind gene, which is sex-linked.

**Step-by-step Solution:**

1. **Identify Taylor's Genotype:**
- Taylor is heterozygous for two autosomal gene pairs, represented as CcDd.
- He is colorblind, which indicates that he has the recessive allele for colorblindness on his X chromosome. Therefore, his genotype for the sex-linked trait would be X^cY (where X^c represents the X chromosome with the colorblind allele).

2. **Determine Gamete Formation:**
- Taylor can produce gametes that include combinations of the alleles from his genotype.
- For the autosomal genes (CcDd), he can produce gametes with the following combinations:
- CD
- Cd
- cD
- cd
- Since he is colorblind (X^cY), his gametes will also include either X^c or Y.

3. **Calculate the Probability of Each Allele:**
- For the alleles C and c:
- Probability of C = 1/2
- Probability of c = 1/2
- For the alleles D and d:
- Probability of D = 1/2
- Probability of d = 1/2
- For the sex-linked colorblind gene:
- Probability of X^c (colorblind) = 1/2
- Probability of Y (not colorblind) = 1/2

4. **Combine Probabilities:**
- We need to find the probability of getting c (1/2), d (1/2), and X^c (1/2) together in one gamete.
- The combined probability is calculated by multiplying the individual probabilities:
\[ P(c) \times P(d) \times P(X^c) = \left(\frac{1}{2}\right) \times \left(\frac{1}{2}\right) \times \left(\frac{1}{2}\right) = \frac{1}{8} \]

5. **Conclusion:**
- The probability of Taylor's gamete carrying the alleles for c, d, and the colorblind gene is \(\frac{1}{8}\).

**Final Answer:**
The probability of gametes having c, d, and the colorblind gene is \(\frac{1}{8}\).

---