If chromosome complement 44+XY of a gamete mother cell suffers a non-disjunction at the of first meiotic division. Which sets of gametes will be correct

Options

A.
B.
C.
D.

Answer

To solve the problem regarding the non-disjunction event during the first meiotic division of a gamete mother cell with a chromosome complement of 44 + XY, we will follow these steps:

### Step-by-Step Solution:

1. **Understanding Non-Disjunction**:
Non-disjunction is the failure of homologous chromosomes or sister chromatids to separate properly during cell division. In this case, it occurs during the first meiotic division.

2. **Chromosome Composition**:
The given chromosome complement is 44 autosomes and 2 sex chromosomes (XY). This means the total number of chromosomes is 46.

3. **Meiosis Overview**:
During meiosis, a diploid cell undergoes two rounds of division (meiosis I and meiosis II) to produce four haploid gametes. In meiosis I, homologous chromosomes are separated, and in meiosis II, sister chromatids are separated.

4. **Effect of Non-Disjunction in Meiosis I**:
If non-disjunction occurs during the first meiotic division, the homologous chromosomes (in this case, the X and Y chromosomes) do not separate. This results in one daughter cell receiving both X and Y chromosomes, while the other daughter cell receives none.

5. **Gamete Formation**:
After the first meiotic division, we will have:
- One cell with 44 autosomes + XX (both X chromosomes)
- One cell with 44 autosomes + YY (both Y chromosomes)
- Two cells with 44 autosomes + 0 sex chromosomes (neither X nor Y)

6. **Final Gamete Set**:
Therefore, the possible sets of gametes after the second meiotic division (which will separate the sister chromatids) will be:
- 44 + XX
- 44 + YY
- 44 + 0 (no sex chromosomes)

### Conclusion:
The correct sets of gametes produced from the non-disjunction event during the first meiotic division of a mother cell with a chromosome complement of 44 + XY are:
- 44 + XX
- 44 + YY
- 44 + 0