The ATL buffer mimicked the inside of an animal cell to keep the DNA stable, while the Pro K stripped the DNA of proteins. Lysis buffer (AL) was added to lyse the cells and membranes. Ethanol was also added because it prevented the DNA from dissolving in the solution and caused into precipitate. The sample was spun in a microcentrifuge after adding two different wash buffers and elution buffer. As groups, we set up a PCR master mix on ice for 7 reactions that included the reagents: H2O, 10X Buffer, 25mM MgCl2, Forward Primer, Reverse Primer, 10mM dNTP’s, and taq was added in after. This helped in reducing the likelihood of errors in pipetting each reagent for each reaction and reducing the time it would take to do so. Moreover, we were only able to make 7 reactions worth when we should have had about 10. The first time we made the master mix we did not pipette gently and too many bubbles formed, so we had to use whatever was left of the reagents to make the second batch. As a result, our group only had one negative control and by the time we had used the master mix for our individual samples, not much was left for the negative control. We then ran our PCR reactions on 2% agarose gel because the DNA fragments we wanted to observe were 100 bp and 400 bp. We mized the appropriate reagents and amounts to make the gel. The agarose …show more content…
We had bands at less than 100 bp due to primer molecules sticking together, suggesting no DNA was present for the primers to attach to. Our PCR reactions failed and Dr. Monsen- Collar provided us with successful gels to analyze and use for our calculations. The gel provided the genoytpes of 21 students, 13 students were homozygous for the insert, 5 students were heterozygous for the insert, and 3 students were homozygous for no insert. These were my observed absolute allele frequencies and I was able to calculate the observed relative frequencies by solving for p and q. The frequency of p was 2 times the 13 p alleles plus the 5 alleles from the heterozygotes divided by a total of 42 alleles in the class. The calculated p frequency is .738. The frequency of q was 2 times the 3 q alleles plus the 5 alleles from the heterozygotes divided by the total 42 alleles. The calculated q frequency is .262. The two frequencies sum to 1. The expected relative frequencies then are as follows, p2=.545, q2=.0686, and 2pq=.387. To convert to expected absolute frequencies I multiplied each by 21 to get p2=11.45, q2=1.44, and 2pq=8.13. The chi square test was performed where the difference in the observed and expected was squared, divided by the expected and all added together. My chi square value is 3.105 and there are 2 degrees of freedom. The corresponding p value was about .10, indicating that the