If the number of bivalents is 10 in prophase - l what will be the number of the chromosomes during late anaphase - II in each cell ?

Options

A. 10 with two chromatids each
B. 10 with one chromatid each
C. 20 with two chromatids each
D. 20 with one chromatid each

Answer

To solve the problem, we need to understand the relationship between bivalents, chromatids, and chromosomes during the stages of meiosis.

### Step-by-Step Solution:

1. **Understanding Bivalents in Prophase I**:
- In prophase I of meiosis, homologous chromosomes pair up to form bivalents. Each bivalent consists of two homologous chromosomes, and each chromosome has two sister chromatids.
- Given that there are 10 bivalents, this means there are 10 pairs of homologous chromosomes.

2. **Calculating Total Chromatids**:
- Since each bivalent is made up of two chromosomes, and each chromosome has two sister chromatids, the total number of chromatids can be calculated as follows:
- Total chromatids = Number of bivalents × 2 (chromosomes per bivalent) × 2 (chromatids per chromosome)
- Total chromatids = 10 × 2 × 2 = 40 chromatids.

3. **Transition to Anaphase II**:
- During meiosis, after the first meiotic division (which separates homologous chromosomes), the second meiotic division occurs. In anaphase II, the sister chromatids of each chromosome are separated.
- Therefore, each of the 20 chromosomes (since 40 chromatids were present, and they are separated into 20 chromosomes) will be pulled apart into individual chromatids.

4. **Counting Chromosomes in Late Anaphase II**:
- At the end of anaphase II, each cell will have half the number of chromatids that were present at the start of prophase I.
- Since we calculated there were 20 chromosomes (after the first meiotic division), each cell will have 20 individual chromatids.
- Therefore, the number of chromosomes in each cell during late anaphase II will be 20.

### Final Answer:
The number of chromosomes during late anaphase II in each cell will be **20**.