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.

Please explain answer in terms of gene conversion.

Explanation / Answer

The haploid meiotic products of 2:2 ratio are :- 2 cells having prototropic wild type leu+ gene, one cell having mutated (auxotrophic) leu-gene, and one cell having not having the gene at all and is unable to grow on leucine.

The haploid meiotic products of 3:1 ratio are :- 3 cells having prototropic wild type leu+ gene. On sequencing these 3 cells, it was found that all three had wild type gene.

One cell had leu-. On sequencing it was found that a GA mutation is there at +243 position in leu- cell. This means that a transition at 243 position upstream of leu+ gene leads to gene conversion, converting leu+ gene into leu-. The leu- gene of the mutant has replaced by its homologous leu+ gene at the time of replication.

Transition mutation can occur either due to error in replication or is induced. This is a point mutation.

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