Biology Help Needed.

Question

For an haploid n = 3 cell, compare the end result if a single nondisjunction event happens in anaphase I with

nondisjunction that occurs in anaphase II.

Please show how to draw what is happening, and explain what is happening. Thanks!

Answer 1

When nondisjunction occurs in anaphase I, the homologous paternal and maternal chromosomes do not separate. The result is that one cell has all four sister chromatids for that chromosome, and the other cell has none. When this cell continues through meiosis II, the sister chromatids will separate. Two of the cells would have both a paternal and a maternal copy of that chromosome, for a total of four chromosomes. The other two cells would lack that chromosome, and so would only have one copy each of the unaffected chromosomes.

When nondisjunction occurs in anaphase II, the maternal and paternal chromosomes have already segragated normally. The cells at this point have three sets of sister chromatids, Sister chromatids are identical copies of unresolved chromosomes made during replication. So if nondisjunction happens at this stage one cell will have four chromosomes and another cell will have two chromosomes. The difference can be seen in the meiotic products that have the extra chromosome. If nondisjunction occurred in anaphase i, then those two chromosomes will have different alleles. If nondisjunction occurred in anaphase II, those two chromosomes will have the same alleles.

Hope this helps