Question

What mutants are in the same complementation group as M4? Select as many as apply.

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

if the hybrid of two mutants shows wildtype phenotype then two mutants complement each other, that is the mutants are alleles of different genes, so the hybrid is heterozygous at both loci,

If the hybrid fails to complement each other, the mutants are due to mutations in the same gene, so the hybrid has two different mutant alleles of the same gene.

m4 fails to complement with m1. and it complements with all others, so m4 and m1 are alleles of the same gene.

so m1 and m4 belong to the same complementation group.

m6 fails to complement with all others, so it is a dominant mutation, so we cannot determine whether it is the allele of which gene.

so the answer is A) M1