If a diploid cell is arrested in Anaphase I and Anaphase II than what is correct for the stages respectively :-

Options

A. 2n,4c and 2n,2c
B. 2n,4c and n,2c
C. 2n,2c and n,2c
D. 4n,4c and 2n,2c

Answer

To solve the question regarding a diploid cell arrested in Anaphase I and Anaphase II, we need to analyze the stages of meiosis and the chromosomal content at each stage.

### Step-by-Step Solution:

1. **Understanding Diploid Cells**:
- A diploid cell (2n) contains two sets of chromosomes. For example, in humans, this is 46 chromosomes (2n = 46).

2. **Meiosis Overview**:
- Meiosis consists of two main stages: Meiosis I and Meiosis II.
- Meiosis I includes Prophase I, Metaphase I, Anaphase I, and Telophase I.
- Meiosis II includes Prophase II, Metaphase II, Anaphase II, and Telophase II.

3. **Chromosome Number in Meiosis I**:
- At the beginning of Meiosis I, the diploid cell is 2n (2 sets of chromosomes).
- During Anaphase I, homologous chromosomes are separated.
- Therefore, the number of chromosomes in Anaphase I is still 2n, but the chromatid number doubles because each chromosome consists of two sister chromatids.
- Hence, the chromosomal content at Anaphase I is 4c (where c represents the amount of DNA in a single chromatid).

4. **Chromosome Number in Meiosis II**:
- After Meiosis I, the cell divides into two haploid cells (n), each containing half the number of chromosomes (but still with sister chromatids).
- In Anaphase II, the sister chromatids are separated, resulting in each daughter cell having n chromosomes.
- Since each chromosome in Anaphase II consists of one chromatid, the chromosomal content is n (where n is half of 2n).

5. **Conclusion**:
- Therefore, if a diploid cell is arrested in Anaphase I, it has 4c (chromatids), and if it is arrested in Anaphase II, it has n (chromosomes).

### Final Answer:
- In Anaphase I: 4c (chromatids)
- In Anaphase II: n (chromosomes)