Wild-type yeast are able to biosynthesize their own leucine. A haploid yeast had

Question

Wild-type yeast are able to biosynthesize their own leucine. A haploid yeast had been mutagenized to look for leu– mutants (those that cannot make their own leucine but must have it supplied) and those that were unable to grow on leucine were selected and crossed to a wild-type haploid strain. You analyze the meiotic products from the heterozygous diploid Saccharomyces strain.

After separating the 4 haploid meiotic products, which are haploid yeast and can grow, you test to see if the for their ability to grow on leucine again. You get most of the meiotic products resulting in a 2 leu+ : 2 leu– but there is 1 mutant that you separated where you saw 3 leu+ : 1 leu–. You sequence the leu-6 gene in this mutant and find that for each of the 3+ meiotic products the leu-6 gene is WT but for the one that is mutant there is a G->A mutation at position +243. Explain why you got 3:1 and not 2:2 for your meiotic products.  

Explanation / Answer

3 : 1 ratio was obtained because in one of the meiotic products there was a mutation (at position 243) which reverted the mutatnt leu- back into wild type leu+. Due to this phenomena the leu-6 gene was able to now synthesize wild type protein which could make its own leucine. Thus, because of this point mutation the 2 : 2 ratio was changed to 3 : 1.

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