Mr. Siddarth is suffering from hypertrichosis and phenylketonuria. His father is heterozygous for phenylketonuria. The probability of Siddarth's sperm having one recessive autosomal allele and holandric gene is

Options

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

Answer

To solve the problem, we need to analyze the genetic conditions mentioned and determine the probability of Mr. Siddarth's sperm carrying one recessive autosomal allele and a holandric gene.

### Step-by-Step Solution:

1. **Understand the Genetic Disorders**:
- **Hypertrichosis** is a condition often linked to the Y chromosome, meaning it is a holandric trait (passed from father to son).
- **Phenylketonuria (PKU)** is an autosomal recessive disorder. This means that a person must inherit two copies of the recessive allele (one from each parent) to express the disease.

2. **Identify Mr. Siddarth's Genotype**:
- Since Mr. Siddarth suffers from PKU, he must have the genotype "aa" (where "a" is the recessive allele for PKU).
- For hypertrichosis, since it is a holandric trait, Mr. Siddarth must have the Y chromosome carrying the gene for hypertrichosis. Therefore, he has the genotype "Yh" (where "h" represents the allele for hypertrichosis).

3. **Determine the Father's Genotype**:
- Mr. Siddarth's father is heterozygous for PKU, meaning his genotype is "Aa" (where "A" is the normal allele and "a" is the recessive allele for PKU).
- The father’s genotype for hypertrichosis is "Y" (assuming he does not express the trait since it is not mentioned).

4. **Gametes Produced by Mr. Siddarth**:
- Mr. Siddarth can produce sperm with the following combinations:
- For PKU: Since he is "aa", all sperm will carry "a".
- For hypertrichosis: Since he has "Yh", he can produce sperm with either "Y" (without hypertrichosis) or "Yh" (with hypertrichosis).

5. **Calculate the Probability**:
- The probability of Mr. Siddarth's sperm carrying one recessive autosomal allele (which is always "a") and the holandric gene (which can be either "Y" or "Yh") is:
- Probability of getting "a" = 1 (since all sperm will carry "a").
- Probability of getting "Yh" = 1/2 (since half of the sperm will carry the holandric gene for hypertrichosis).
- Therefore, the combined probability is:
\[ P(a \text{ and } Yh) = P(a) \times P(Yh) = 1 \times \frac{1}{2} = \frac{1}{2} \]

### Final Answer:
The probability of Mr. Siddarth's sperm having one recessive autosomal allele and a holandric gene is **1/2**.