Final answer:
An error in meiosis II due to problems with microtubules during anaphase can lead to non-disjunction, resulting in two daughter cells with an extra set of Qq chromosomes and two with none, rather than the options provided, signaling an unequal distribution of genetic material.
Step-by-step explanation:
If a cell undergoing meiosis II experiences a problem with its microtubules that prevents proper separation of the paired chromosomes carrying alleles Q and q during anaphase, it results in a phenomenon known as non-disjunction. This error leads to an inheritance where both chromosomes (in this case, both alleles Q and q) go to one daughter cell while the other daughter cell receives no copy. As a result, two of the four daughter cells will receive an extra chromosome (they will be disomic containing both Q and q), while the other two daughter cells will lack this chromosome (they will be nullosomic).
The most likely effect on the daughter cells is that they will have an unequal number of chromosomes. More specifically for the question, since separation did not occur, it can result in two of the daughter cells having a genotype of Qq (the disomic cells) while the other two cells would be lacking both alleles (the nullosomic cells). This scenario does not match any of the options provided exactly because we expect the daughters to have different chromosomal content, with two having an abnormal extra set and two lacking it.