An animal has a diploid chromosome number of 60. Suppose that the first meiotic division of a germ cell is normal, but a single dyad in one of the two daughter cells undergoes non-disjunction in meiosis II. How many chromosomes would be present in each of the four gametes that result from that meiosis?

Two gametes end up with 30 chromosomes, one gamete will have 31 and one will have 29.

Explanation:

During meiosis, a single diploid cell divides into four haploid cells with half the chromosome number than the parent cell.

During meiosis I, homologous chromosomes separate and the chromosome number in the two daughter cells goes down by half. During meiosis II, the "sister" chromatids separate, and the chromosome number in the respective daughter cells remains the same.

Non-disjunction of a single dyad in meiosis II causes that one of the daughter cells will have an extra chromosome, and another will be lacking one. The two other gametes will be normal.

I drew a simple example of what would happen during meiosis of a 2n=4 cell if there was non-disjunction in a single dyad during meiosis II.