When a germ cell undergoes meiosis, which event would MOST likely result in four aneuploid daughter cells?

A. misalignment of the chromosomes during metaphase I

B. misalignment of the chromosomes during anaphase I

C. misalignment of the chromosomes during metaphase II
D. misalignment of the chromosomes during anaphase II

B. misalignment of the chromosomes during anaphase I

Explanation:

Aneuploidy refers to an error in the chromosomal number of a cell. This occurs when the chromosomes fail to separate equally into individual cells, a process called MEIOTIC NONDISJUNCTION. Meiotic disjunction can occur in either the anaphase stage of meiosis I or meiosis II, however, different outcomes are produced.

When chromosomes fail to separate at meiosis I, it results in two gametes that lack that particular chromosome (n-1) and two gametes with two copies of the chromosome (n+1). This means that the four daughter cells will have an incorrect number of chromosomes, hence, the misalignment of the chromosomes during anaphase I will result in four aneuploid daughter cells.

answer; both surfaces of the membrane are nonpermeable

this is more of a statement than a question. is there answers to go along with this, or is it opinion based?