g If a pair of sister chromatids fail to separate during anaphase II (meiosis II), what will be the chromosome number of the four resulting gametes with respect to the normal haploid number (n)

Question

Nyla Hogan

Biology

6 September, 21:58

g If a pair of sister chromatids fail to separate during anaphase II (meiosis II), what will be the chromosome number of the four resulting gametes with respect to the normal haploid number (n)

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Answers (1)

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Iliana Mann · Answer 1

The result will be two gametes with a normal haploid number (n), one gamete containing an extra chromosome (n + 1) and one gamete missing one chromosome (n - 1).

Explanation:

Nondisjunction is caused by the failure in the mechanism of separation of homologous chromosomes or sister chromatids during meiosis and/or mitosis. The failure in the separation of sister chromatids during meiosis II is referred to as 'secondary nondisjunction' and it produces two daughter cells with abnormal chromosome numbers.

The fusion of one gamete containing an extra chromosome (n + 1) with a normal (n) gamete will lead to trisomy (2n + 1), while the fusion of one gamete missing one chromosome (n - 1) with a normal gamete will lead to monosomy (2n-1).