Scott_Handout7-corrected - 3/15/2010 MCB 104; Handout #7...

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3/15/2010 MCB 104; Handout #7 GUIDE TO EPISTASIS ANALYSIS : Intro : Often, genes function together in a biological pathway. Epistasis is used to determine the order of genes in that pathway. Epistasis requires making a double mutant for two genes with distinct mutant phenotypes . If the mutant phenotypes of the two genes are identical, you will rarely learn anything from making the double mutant. When you make a double mutant there are two possibilities: 1. The double mutant has both mutant phenotypes. In this first case, the genes probably don’t function in the same pathway, so stop here; there’s nothing more to do. 2. The double mutant has only one of the mutant phenotypes. The gene with the observed phenotype is epistatic to the gene with the other phenotype. Think of epistasis as a battle between genes; there can be only one gene that wins; the gene that wins is epsitatic to the gene that loses. Step I. Ask whether the pathway is a biosynthetic pathway or a signaling pathway. A. If it’s biosynthetic, go to Step II. B. If it’s a signaling pathway, go to Step III. Step II . Biosynthetic pathways: The epistatic gene is UPSTREAM from the other gene . Think of biosynthetic pathways as a conveyor belt or an assembly line. A mutation in a biosynthetic enzyme will block the conveyor belt. The earliest place that you block the assembly line will be the place where the process stops and intermediates accumulate. A . For biosynthetic pathways, the mutations are almost always loss-of-function (lof) mutations in enzymes in the pathway. B . When you mutate an enzyme in a biosynthetic pathway, the intermediate that accumulates is the intermediate immediately upstream to the mutated enzyme . C . The pathway can be rescued by adding either the defective enzyme or an intermediate downstream of the defective enzyme(s). Adding downstream enzymes will not rescue the pathway because these enzymes were not mutated in the first place. Adding intermediates upstream of the mutated enzyme will not rescue because there is still a block at the mutant enzyme. D . Example: Intermediate 1 –(enzyme A) Intermediate 2 –(enzyme B) Product i. A lof mutation in enzyme A is epistatic to a lof mutation in enzyme B. ii. In an A - ; B - double mutant, Intermediate 1 will accumulate. iii. An A - single mutant can be rescued by adding Intermediate 2 or functioning enzyme A. Adding Intermediate 1 or enzyme B will not rescue an A - mutant. iv. Often you will see a problem where there are mutants in each enzyme in the pathway, and intermediates are added back. Go through these problems enzyme-by- enzyme. The enzyme that functions earliest will be rescued by the most intermediates because there’s a lot of stuff after it. The enzyme that functions latest will be rescued by the fewest intermediates because there’s not much stuff after it. In the above example, you may see a table like this: Enzyme A Enzyme B Intermediate 1 - - Intermediate 2 + - Product + +
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Scott_Handout7-corrected - 3/15/2010 MCB 104; Handout #7...

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