You have a wild type strain of yeast that grows well at any temperature between

Question

You have a wild type strain of yeast that grows well at any temperature between 10C and 37C. You mutagenize wild-type haploid (mat a) yeast and immediately plate about 200 cells onto each of 1000 plates containing complete medium and grow them at 25C until colonies form. After colonies grow up on these plates, you make a replica of each plate onto two more plates; you incubate the first copy at 10C and the second at 25C, and look for mutants that grow on the 25C copy but not on the 10C copy. In this way you obtain 2 independent mutants that are cold sensitive (die at 10C but live at 25C).

(a)Was your mutant hunt a selection or a screen?

B)What do you conclude about each mutant?

Explanation / Answer

(a) Was your mutant hunt a selection or a screen?

Ans: It is a screen

Conclusions:

(b) You mate cold-sensitive mutant #1 to wild type haploid (mate) yeast. The resulting diploids had a wild type phenotype (grow at 10C). What do you conclude?   (The result for mutant #2 was the same).

Recessive       cld       = wt phenotype

                        +

(c) For mutant 1, you take the diploids from (b) and sporulate them (induce meiosis) to analyze tetrads. All four spores of each tetrad were grown at 25C and then tested for growth at 10C. In every tetrad, there were 2 wt and 2 cold-sensitive spores. (The result for mutant #2 was the same).

Mutant phenotype is caused by one mutation (as opposed to two independently assorting mutations)

          cld       diploid    ----------> gives [cld cld + +] tetrads

            +

(d) From the spores obtained in (c), you cross a cold-sensitive mat a spore from mutant #1 to a cold-sensitive mat spore from mutant #2 and grow at 25C to obtain a diploid. You test for growth at 10C and find that the diploid has a wild-type phenotype. What do you conclude?

Mating types don’t matter here except to see that the cross is possible

You know that both mutations are recessive

Therefore: Complementation (two genes)        cld1     +       

                                                                            +     cld2

(e) You sporulate the diploid from (d) to analyze tetrads again. For each type of tetrad below (PD, NPD, and TT), state the genotypes of the 4 spores (disregarding mating type gene), and predict the phenotype for growth at 10C (CS or WT):

           PD                                       NPD                              TT

cld1     +          (CS)                 cld1 cld2 (CS)             cld1     +        (CS)

cld1     +          (CS)                 cld1 cld2 (CS)             cld1    cld2     (CS)

+        cld2       (CS)                 +        +   (WT)              +        +        (WT)

+        cld2       (CS)                 +        +   (WT)              +       cld2     (CS)

Note I assumed that the double mutant is (CS). More elaborate interactions are theoretically possible (eg suppression) but it is not valid to assume them if no reason is given to support the hypothesis. The answer (CS?) is also acceptable.

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