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Two adenine deficient plates were then divided into four sections each. The first plate was labeled with HA0, HA1, HA2, and HB1, and had each quadrant’s corresponding yeast type lightly spread onto it. The second plate was labeled with HB1xHA0, HB1xHA1, HA1xHA2, and HB1xHA2, and had each quadrant’s corresponding yeast cross from the incubated plate lightly spread onto it. For the spreading process, a clean toothpick was used to drag a small amount of the required yeast in a straight line near the edge of the plate. Then a new toothpick was used to drag a single line up from the first line of yeast cells and then spread them in a zig zag fashion. The goal of this process was to spread the yeast cells to roughly one cell thickness. The two adenine deficient agar plates were then left to incubate at 30^oC for a week.

The purpose of this study is to examine how mutations are affected when different organisms, mutated or not mutated, are mated. Once that has been determined, tetrad analysis can then be used to determine linkage as well as the genotypes of given phenotypes. Yeast are an excellent candidate to test these with due to their simple phenotypes, small size, and quick growth. 

The findings of this lab were sound and gave a solid understanding of the passing of mutations to offspring as well as how to determine genotypes using tetrad analysis. The only thing that I would change about this lab is for a more conclusive tetrad cross to be constructed. Because of the certain dead cultures, it is impossible to tell whether the genes are linked or not. Presenting a complete tetrad cross would allow students to more accurately predict whether the genes in question are linked or not. 

As the results state, HA0 is the only haploid yeast cell that can grow without adenine, while HA1, HA2, and HB1 cannot grow without adenine. This shows that HA0 can synthesize its own adenine, while HA1, HA2, and HB1 have mutations which halt adenine production by the yeast cell, though it is unknown where these mutations are in the genome. 

Two adenine deficient plates were then divided into four sections each. The first plate was labeled with HA0, HA1, HA2, and HB1, and had each quadrant’s corresponding yeast type lightly spread onto it. The second plate was labeled with HB1xHA0, HB1xHA1, HA1xHA2, and HB1xHA2, and had each quadrant’s corresponding yeast cross from the incubated plate lightly spread onto it. For the spreading process, a clean toothpick was used to drag a small amount of the required yeast in a straight line near the edge of the plate. Then a new toothpick was used to drag a single line up from the first line of yeast cells and then spread them in a zig zag fashion. The goal of this process was to spread the yeast cells to roughly one cell thickness. The two adenine deficient agar plates were then left to incubate at 30oC for a week.

A clean adenine infused agar gel plate was then divided into four sections and labeled with the crosses that would be carried out. The quadrants were labeled HB1xHA0, HB1xHA1, HA1xHA2, and HB1xHA2. Small amounts of the two strains for each quadrant were mixed together without cross contamination and spread into small squares. The plate was then incubated at 30oC for a week. At the end of the week, all crosses had grown and displayed a cream-colored phenotype.

Tetrad analysis is also a valuable tool in determining the genotypes of certain phenotypes, as well as linkage between two genes. If linkage is occurring between genes then the NPD is very rare to see compared to the PD and TT. If linkage is not occurring then the ratios of PD:NPD:TT are roughly 1:1:4. Once linkage has been determined and the ratios have been confirmed, it is easy to determine which phenotype is PD, NPD, or TT and simply match it to the correct genotype. 

For the tetrad dissection crosses, since it is now known that HB1 and HA2 complement each other, their genotypes were able to be figured out at (ade1-/ade2+) for HB1 and (ade1+/ade2-) for HA2. Knowing this information, the parental ditypes (PD) can be determined as (ade1-/ade2+) and (ade1+/ade2-). This means that the non-parental ditypes (NPD) are (ade1+/ade2+) (ade1-/ade2-) and the tetratypes (TT) are (ade1+/ade2-), (ade1-/ade2+), (ade1+/ade2+), and (ade1-/ade2-). 

Once the two adenine deficient plates were incubated for a week, the results were observed. For the first plate, it was found that HA0 could grow without adenine present, while HA1, HA2, and HB1 could not. For the second plate, it was found that HB1xHA0 and HB1xHA2 could grow without adenine present, while HB1xHA1 and HA1xHA2 could not. 

Two adenine deficient plates were then divided into four sections each. The first plate was labeled with HA0, HA1, HA2, and HB1, and had each quadrant’s corresponding yeast type lightly spread onto it. The second plate was labeled with HB1xHA0, HB1xHA1, HA1xHA2, and HB1xHA2, and had each quadrant’s corresponding yeast cross from the incubated plate lightly spread onto it. For the spreading process, a clean toothpick was used to drag a small amount of the required yeast in a straight line near the edge of the plate. Then a new toothpick was used to drag a single line up from the first line of yeast cells and then spread them in a zig zag fashion. The goal of this process was to spread the yeast cells to roughly one cell thickness. The two adenine deficient agar plates were then left to incubate at 30oC for a week.