Question

Question 17. Saccharomyces cerevisiae (also called budding yeast or baker's yeast) normally makes LARGE colonies on the Petri dish. The fungus can exist as either a haploid or as a diploid. As a geneticist, you decide to isolate haploid mutants with altered size of colonies. You collect 6 such haploid mutants, which you arbitrarily name M1, M2, M3, M4, M5, and M6. To determine how many genes are represented by your mutants, you do a series of crosses resulting in diploids, i.e. you mate two haploid strains to produce a diploid strain. The complementation data are shown below. "+" indicates wild type (large) colony size and" - " indicates smaller (mutant) colony size. WT = wild type. I WT| m1 | m2 | M3 | MA | MS | 5 M2 M3 M4 M5 WT + M1 + M2 + + M3 + + + T M4 + + M5 + + M6

Answer 1

Aus :- p Since M3 is the belongs to the complementation group of M₂ M₂ mutant also show the phenotype of m2 mutant so Mi, M3 double haploid mutant will be Tiny. Please rate my Auswer