If a cell has a diploid number of 8 (2n = 8) before meiosis, how many chromosomes will be in each of the four daughter cells after meiosis II if one pair of homologous chromosomes experiences nondisjunction in meiosis I? Show how you got the answer by drawing out the process, starting with the diploid cell undergoing meiosis I (where nondisjunction occurs) and then meiosis II to get the four daughter cells.

The final product is four gametes, two of them with 5 chromosomes, and the other two with 3 chromosomes each.

Explanation:

If nondisjunction occurs during meiosis 1, a pair of homologous chromosomes fail to separate, and one of the daughter cells will have the two chromosomes while the other cell will not get any chromosome from the pair.

If meiosis 1 occurs normally, but nondisjunction occurs in meiosis 2, sister chromatids fail to separate.  

The usual process of meiosis produces four daughter haploid cells (n) from a diploid germ cell (2n). Each daughter cell is haploid because they have half the number of chromosomes of the original one.  

If the diploid number of the original cell is 8 (2n=8), then under normal conditions, each haploid daughter cell should have 4 chromosomes (n = 4).  

But in the exposed example, one pair of homologous chromosomes experiences nondisjunction during meiosis I (in the attached file, you will recognize this pair as the red one). The other chromosomes separate as usual. So one of the daughter cells will have one extra chromosome than expected (five instead of four), and the other daughter cell will lack one chromosome (three instead of four). Meiosis II occurs normally. The final result is the formation of four gametes, two of them with 5 chromosomes, and the other two with 3 chromosomes each.

what in the weirdo : o

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