In this experiment, the genotypes of the two respective stains were172 MATa ade2 his3 leu2 trp1 ura3 can1 and 196 MATɑ ade2 his4 leu2 trp1 ura3 can1. Each strain has a mutation at the his locus which means that the yeast is unable to grow unless it is in …show more content…
From the data, it was determined that there were 5-6 complementation groups. There were four absolute complementation groups: (A, B, N, Q, 1, 16, 17, 18, 22, 23, 24, 26, 27), (C, H, O, R, U, V, Y, Z, BB, 3, 6, 7, 9, 11, 12, 13, 15, 19, 21, 25), (D, F, I, 2, 8, 14, 20), and (J, K, L, 4, 5, 10, 28). E, G, M, P, W, X, and AA were determined to be possible revertants because they did not complement with any mutants from the opposite strain. S and T were determined to be 172a mutants that complemented all of the 196ɑ mutants. S and T could not be crossed since they were the same mating type. These 172a mutants could potentially be one or two groups. Each complementation group represents a gene located before the AIR intermediate. From the results of the experiment, it was determined that there were five to six genes that