A human male is heterozygous for two autosomal genes A & B and homozygous dominant for `3^(rd)` autosomal gene C. The individual is having sry gene (S) on its Y– chromosome. What will be the probability of sperms to carry abcS condition?

Options

A. `(1)/(4)`
B. `(1)/(8)`
C. 0
D. `(1)/(16)`

Answer

To solve the problem of determining the probability of sperm carrying the genotype ABCS, we will break it down step-by-step.

### Step 1: Identify the Genotype of the Male
The male is heterozygous for two autosomal genes A and B, and homozygous dominant for the third autosomal gene C. His genotype can be represented as:
- For gene A: Aa (heterozygous)
- For gene B: Bb (heterozygous)
- For gene C: CC (homozygous dominant)
- For the SRY gene on the Y chromosome: S (hemizygous)

### Step 2: Determine the Possible Gametes
To find the possible gametes produced by this male, we need to consider the combinations of alleles from genes A, B, and C, along with the SRY gene.

- For gene A (Aa), the possible alleles are A and a.
- For gene B (Bb), the possible alleles are B and b.
- For gene C (CC), the only allele is C (since it is homozygous).
- For the SRY gene (S), the only allele is S (since it is on the Y chromosome).

### Step 3: Combine the Alleles to Form Gametes
The combinations of alleles from A, B, and C, along with the SRY gene, can be represented as follows:
1. From A: A
2. From B: B
3. From C: C
4. From SRY: S

Thus, the combinations of gametes can be:
1. ABCS (from A, B, C, S)
2. AbCS (from A, b, C, S)
3. aBCS (from a, B, C, S)
4. abCS (from a, b, C, S)

### Step 4: Count the Total Gametes
Now we can count the total number of unique gametes that can be formed:
- From gene A: 2 options (A or a)
- From gene B: 2 options (B or b)
- From gene C: 1 option (C)
- From the SRY gene: 1 option (S)

The total number of gametes is calculated by multiplying the number of options:
Total gametes = 2 (from A) × 2 (from B) × 1 (from C) × 1 (from S) = 4 gametes.

### Step 5: Identify the Probability of the Desired Gamete
Out of the 4 possible gametes, only one of them is ABCS. Therefore, the probability of obtaining a sperm that carries the ABCS condition is:
Probability = Number of favorable outcomes / Total outcomes = 1 / 4.

### Final Answer
The probability of the sperm carrying the ABCS condition is **1/4** or **25%**.

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