PS 5 answers

PS 5 answers - LS4 Problem Set 5, 2010 Mutagenesis Screens...

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LS4 Problem Set 5, 2010 Mutagenesis Screens Answers Problem 1 . A) ½ will be homozygous carriers, and ¼ will be wild-type, since the cross was between two heterozygous F2 carriers. B) There is a 2/3 chance that the cross will result in mutant progeny, since 2/3 of the non-mutant flies should be heterozygous for the mutation. C) ½ of the progeny should be mutant (the other half will be heterozygous), since the cross is between a homozygous mutant and a heterozygous carrier. Problem 2. These are two independently segregating mutations: half of the spores will be auxotrophic for adenine, half will be auxotrophic for leucine, a quarter will be auxotrophic for both, and a quarter will be auxotrophic for neither. Therefore, you expect that A) one quarter of the spores will be able to grow on minimal media, B) ½ will be able to grow on leucine-supplemented media, and C) ½ will be able to grow on adenine-supplemented media. Problem 3 . A) Mutation A is closest to the centromere since it is homozygosed most often in a gynogenetic diploid. B) Mutation A is probably not in the same gene as B or C since it seems to be closer to a centromere than the other two. Mutations B and C are about equidistant from a centromere, but we don’t know whether they’re on the same chromosome (or chromosome arm). We can determine whether they’re in the same
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PS 5 answers - LS4 Problem Set 5, 2010 Mutagenesis Screens...

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