Question

What mutants are in the same complementation group as M2? 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

Answer: Option B is correct
Explanation:
Same complementation group = Allelic mutations = Mutations located at the same locus

In a cross between two recessive mutants, the two mutants are said to be allelic if the F1 progeny exhibits the mutant phenotype.
In a cross between two recessive mutants, the two mutants are said to be non-allelic if the F1 progeny exhibits the WT phenotype.

In the given case,
Complementation groups:
1. M1 = M4
2. M2 = M3
3. M5
4. M6

It should be noted that M6 is a dominant mutation.