2n=6 in a primary spermatocyte which is in metapahase of first meiotic divison. What shall be the total number of chromatids in each of the secondary spermatocyte?

Options

A. 6
B. 8
C. 24
D. 32

Answer

To solve the problem step by step, we need to understand the process of meiosis and how it affects the number of chromosomes and chromatids in spermatocytes.

### Step-by-Step Solution:

1. **Identify the Diploid Number (2n)**:
- The question states that the primary spermatocyte has a diploid number of chromosomes (2n = 6). This means that there are 6 chromosomes in the primary spermatocyte.

2. **Determine the Number of Chromosomes in Secondary Spermatocyte**:
- During the first meiotic division, the primary spermatocyte divides to form two secondary spermatocytes. Each secondary spermatocyte is haploid (n). Since the primary spermatocyte has 6 chromosomes, each secondary spermatocyte will have half of that number:
\[ n = \frac{2n}{2} = \frac{6}{2} = 3 \]
- Therefore, each secondary spermatocyte has 3 chromosomes.

3. **Determine the Number of Chromatids**:
- Each chromosome at this stage (in the secondary spermatocyte) consists of 2 chromatids because the chromosomes have already replicated during the S phase before meiosis.
- Therefore, for each of the 3 chromosomes, there are 2 chromatids:
\[ \text{Total number of chromatids} = \text{Number of chromosomes} \times \text{Number of chromatids per chromosome} \]
\[ = 3 \times 2 = 6 \]

4. **Conclusion**:
- The total number of chromatids in each secondary spermatocyte is 6.

### Final Answer:
The total number of chromatids in each secondary spermatocyte is **6**.

---