In this lab we are using Saccharomyces cerevisiae to test two genes for simple mendilan inheritance. This also models what Mendel did when he crossed two heterozygote plants, the strains we are using represent the gametes of the pea plants so we get the expected ratios of an independent assortment. Yeast has an advantage since the genotypes of the gametes are visible be looking at the haploid strains. The genes we are working with are T and R. The T mutant is deficient in making tyrosine so they will be unable to grow on minimal media. The R mutant has a mutation in the adenine pathway that causes a build up of a red pigment this one also will not grow. The typical ratio in an independently assertive genes would by 9:3:3:1 but because lacking in either of the two amino acids would show no growth the actual pattern was 9:4:3 since 2 of the phenotypes are indistinguishable. We crossed four different types of yeast HAO and HBO, which are wild type, HAR and HBR which are mutant for R, HAT and HBT mutant for T, and HART and HBRT which are mutant for both R and T. When plated on minimal media the both crosses that result in double mutants and adenine mutants will show red no growth. The color of the resulting crosses was recorded form the YED plate and growth was recorded from minimal media plates. The results showed the 9:4:3 ratio.
             The yeast Saccharomyces cerevisiae is a unicellular eukaryote that can exist stably as a haploid or diploid. Haploids are one of two mating types, 'a' or 'a'. An 'a' cell can mate with an 'a' cell to form an 'a/a' diploid. Under certain conditions of nutrient starvation, the diploid cell will sporulate, resulting in two haploid 'a' cells and two haploid 'a' cells.
             The cell cycle of yeast is similar to most other eukayotic cells and is commonly broken into 4 stages, G1, S, G2, and M. In G1 the smaller daughter cell grows so that it can go through the cell cycle, it is during this phase that the fate ...