At what point in cell division is a chromosome lost so that, after fertilization with a normal gamete, the result is an embryo with 45, x?

The options are missing, the options are;

I. an error in anaphase I

II. an error in anaphase II

III. an error of the first postfertilization mitosis

IV. an error in pairing

A) I or II only

B) II or IV only

C) III or IV only

D) I, II, or III only

E) I, II, III, or IV

The correct answer is A

Explanation:

Meiosis is the type of cell division undergone by reproductive cells to produce gametes that have a reduced number of chromosomes. Normally, a diploid (2n) organism like humans undergo meiosis to produce haploid gametes (n). This reduction occurs because homologous chromosomes (similar but non-identical chromosomes received from both parent) and sister chromatids are separated during the anaphase of meiosis I and II respectively. Hence each cell (gamete) at the end of meiosis now possess 23 chromosomes each as opposed to the 46 in the producing organisms.

However, an anomaly called ANEUPLOIDY can occur in which one or more chromosomes fail to separate during anaphase. This failure to separate is referred to as NON-DISJUNCTION and it results in gametes that have extra chromosomes i.e. n+1 or missing chromosomes i.e. n-1.

Note that, this error can occur in either meiosis I or II depending on whether it is homologous chromosomes or sister chromatids that fail to separate.

According to the question, a gamete with missing chromosome (n-1) is described, hence, it will undergoes fertilization to produce a zygote with 2n-1 chromosomes i.e. A diploid organism missing one chromosome (46-1) = 45 chromosomes. This particular condition is referred to as MONOSOMY.

option a and d is the correct answer

i hope this u

plz mark as the brainliest answer