A normal man (XY) undergoes meiosis. In which part of meiosis would nondisjuncti

Question

A normal man (XY) undergoes meiosis. In which part of meiosis would nondisjunction have to occur to produce a gamete with two X chromosomes? Is it Meiosis I? Meiosis II? Or both?

A guanine nucleotide triphosphate spontaneously tautomerizes into the enol form just prior to being incorporated into a growing DNA polymer. What nucleotide change will occur if the G returns to the keto form prior to DNA replication? For the answers below the nucleotide to the left is on top, i.e. A-T would be A on the top and T on the bottom.

Explanation / Answer

Under normal circumstances, Meiosis produces gametes (egg and sperm) which have haploid(n) chromosome. When fertilization occur, haploid cells fuse to form diploid zygote. The number of chromosome is reduced in meiosis. Meiosis produces four haploid gametes. The process take place in two division, meiosis I and II

However in certain conditions, chromosomes do not separate properly during meiosis . This is known as nondisjunction and results in gametes with either less chromosome or more chromosome as compared to normal chromosome. Non disjunction can occur either during meiosis I and II(mostly during anaphase). When nondisjunction take place during anaphase I of meiosis I, it have one pair of homologous chromosomes which did not separate. The end result is two cells may have an extra copy of one of the chromosome and two cells that are missing. In humans, n + 1 designates a cell with 23+1 chromosomes ,a total of 24 chromosomes. n - 1 designates a cell missing a in which there are only 22 chromosomes.

When nondisjunction occurs during anaphase II of meiosis II, one pair of sister chromatids did not get separated. In this scenario, two cells will have the normal haploid number of chromosomes. Additionally, one cell will have an extra chromosome (n + 1) and one will be missing a chromosome (n - 1).

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