Q:

Aneuploidy in meiosis

I am quite confused about nondisjuction in anaphase II. If the double chromosome is not segregated, isn't it double. On pictures they are drawn as single but I don't understand why, and if they are double do they duplicate themselves in the S phase?

What is confusing me in picture above is cell (n+1) , why are all chromosomes single? I thought, after end of division in the third cell (labeled as n+1) there should be 3 chromosomes with single chromatid,  and 1 chromosome with 2 chromatid joined at centromeric region, as it is drawn in picture above that show nondisjunction in cell during anaphase ii . When and how do these 2 chromatids get physically separated?

A:

From Study.com >>
On the off chance that nondisjunction happens amid anaphase II of meiosis II, it implies that no less than one set of sister chromatids did not isolate. 
In this situation, two cells will have the typical haploid number of chromosomes. Additionally, one cell will have an additional chromosome (n + 1) and one will miss a chromosome (n - 1) due to the nondisjunction of the chromosomes.
Here is a picture reference taken from Google.ca >>

In the image provided in the question, the chromosomes are not indicated clear and concisely, so it is very hard to follow through the logic of segregation of the daughter cells to (n + 1) and (n - 1) chromosomes.
As you can see from the updated image reference in the answer, we can notice that the cell that underwent nondisjunction in anaphase 2 gives rise to 2 daughter cells, in which, one cell receives the non-segregated chromosome plus one copy of a sister chromatid which makes it (n + 1) and the other daughter cell receives one segregated sister chromatid ONLY which makes it (n - 1). 
On an additional note, from Biology Exams 4 U >>

Fusion of n+1 gamete with normal gamete (n) = 2n + 1 or Trisomic.
Fusion of n-1 gamete with normal gamete (n) = 2n - 1 or Monosomic.
In short, Nondisjuntion in meiosis II leads to trisomy (2n+1) or
  Monosomy (2n-1).

