Suppose a scientist measures the amount of DNA per cell of a particular diploid species at various stages of meiosis. She finds that the meiotic cells contain 3.7 pg, 7.3 pg, or 14.6 pg of DNA. Match each stage of the cell cycle to the corresponding amount of DNA contained within a cell at that stage. 1. metaphase II2. prophase I3. G14. after cytokinasis of meiosis II5. telophase I before cytokinesis6. G2

1. metaphase II - 7.3 pg.

2. prophase I - 14.6 pg.

3. G1 - 7.3 pg.

4. after cytokinesis of meiosis II - 3.7 pg.

5. telophase I before cytokinesis - 14.6 pg.

6. G2 - 14.6 pg.

Explanation:

The cell cycle begins at G1, where all cells in humans (except for the gametes) have 23 pairs of chromosomes, 46 in total, and without any part of the DNA being duplicated. This is the stage in which the cell will have 7.3 pg of DNA.

After G1 comes phase S, which is the phase in which the DNA is duplicated to prepare the cell for the oncoming division; so we can infer that the phase that comes after S, G2, will have twice the DNA the cell had initially: 14.6 pg.

With this amount of DNA, the cell enters the first part of meiosis, in which the homologous chromosomes will be separated into two cells that will each carry on with the second part of meiosis. In prophase I, the cell is just beginning the process of meiosis, so it has 14.6 pg of DNA. In telophase I, the cell still has that amount of DNA because cytokinesis hasn't happened yet, and that is the defining step that ends up actually dividing the cell into two.

Meiosis II begins after cytokinesis I, therefore, the homologous chromosomes have already been separated and each cell has now 7.3 pg. In metaphase II the cell has 7.3 pg, but when cytokinesis of meiosis II ends, the cell will be half the size that was just before, because the sister chromatids are separated - this is where the cell will only have 3.7 pg of DNA.

The cells produced by meiotic cells are gametes, having only 23 chromosomes and the lowest amount possible of DNA.

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