Assume the cell has a diploid number of 4. If the above chromosomes are in the tetrad or bivalent stage of meiosis, which statement below is correct about the events of anaphase I?

Options

A. Two of the tetrads will go to one cell, and two will go to the other
B. Sister chromatids will break apart and each will go in one of four cells.
C. One member of each homologous pair will go into each of the two cells.
D. The tetrads will separate but each new cell will still be diploid since each chromosome has two chromatids.

Explanation

### Step-by-Step Solution:

1. **Understanding the Diploid Number**:
- The diploid number of chromosomes in the cell is given as 4. This means that there are two sets of chromosomes, each containing 2 homologous pairs.

2. **Identifying Tetrads in Meiosis**:
- During meiosis, specifically in prophase I, homologous chromosomes pair up to form structures known as tetrads or bivalents. In this case, with a diploid number of 4, there will be 2 tetrads present.

3. **Events of Anaphase I**:
- In anaphase I of meiosis, the homologous chromosomes (the tetrads) are separated and pulled to opposite poles of the cell. Each homologous chromosome consists of two sister chromatids.

4. **Distribution of Chromosomes**:
- As the homologous chromosomes are separated, one member of each homologous pair moves to one pole, while the other member moves to the opposite pole. This results in two cells, each receiving one chromosome from each tetrad.

5. **Final Chromosome Count**:
- After anaphase I, each of the two resulting cells will have 2 chromosomes (half of the original diploid number), which are now haploid.

6. **Conclusion**:
- The correct statement about the events of anaphase I is that one member of each homologous pair will go into each of the two cells.