Discussion+on+exercise+2

After cutting out the individual chromosomes, our group measured the length of the long and short arms of each chromosome and plotted them against each other. This was to help establish a size relationship so that we could determine which chromosomes were sister chromatids.

With that information, we were able to put together the following karyotype. As you will notice, all the chromosomes, except one, have three sets. Effectively, it is a triploid cell (with that one exception). We will discuss two scenarios that could have led to such a karyotype.

1. Parent 1 would supply a diploid gamete and Parent 2 would supply a haploid gamete with a single missing chromosome (the result of a non-disjunction event during Meiosis I or II). The diploid gamete also obviously resulted from complete non-disjunction, in which one daughter cell received all the chromatids and the other received none. So, each gamete in this case should have had 7 chromosomes, making the normal 14 diploid cell. However, the diploid parent (#1) provided 14 chromosomes and the haploid parent (#2) provided only 6, making the 20 total that is observed. Thus, a triploid is produced (minus that one set).

2. In the second scenario, Parent 1 would again be a diploid gamete resulting from non-disjunction. However, 1 of the sister chromatids did separate correctly, leaving the gamete cell with 13 chromosomes. Parent 2 would then be normal haploid gamete with 7 chromosomes. This total again adds to the 20 chromosomes that are observed.