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Problem 1: C. elegans epistasis You are interested in studying vulval development in C. elegans! Because you really truly enjoy examining the bag-of-worms phenotype that results when eggs hatch inside their mother, and the creepiness of multi-vulval worms delights you! Yup, pretty cool, right? You have the genotypes listed on the following table. gf = over-expression (you have a line with a transgene containing multiple copies of each gene) lf= complete loss of function) Single mutants Multivulval lin-15(lf) lin-3(gf) let-60(gf) lin-1(lf) lin-3(lf) let-23(lf) let-60(lf) lin-45(lf) Vulvaless lin-3(lf) let-23(lf) let-60(lf) lin-45(lf) lin-15(lf) Multivulval Vulvaless Vulvaless Vulvaless lin-3(gf) Multivulval vulvaless No data No data let-60(gf) Multivulval Multivulval Multivulval Vulvaless lin-1(lf) Multivulval Multivulval Multivulval Multivulval A) Which genes in this pathway are positive regulators of vulval formation? Which genes are negative regulators of vulval formation? Positive regulators: lin-3, let-23, let-60, and lin-45 Negative regulators: lin-15, lin-1 B) How would you order these genes in a pathway? lin-3 --| lin-15 --| let-23 let-60 lin-45 --| lin-1 --| vulval formation C) To complicate matters further – another lab has recently identified another mutant, glp-1 which lacks a germline altogether. glp-1 mutants also don’t have a normal vulva. Can you order glp-1 in your pathway? Purely from an epistatis pathway, yes. glp-1 since it ablates the germline altogether is technically epistatic to all the lin- genes. However, in reality because glp-1 specifies cells that are precursors to normal germline development, glp-1; lin-15 double mutants don’t really tell you much about the lin-15 pathway. This is similar to the original example Jeff used when introducing epistatis (ie in a hairless mouse, talking about which gene specifies coat color is not particularly informative) Problem 2: You decide to rotate with Cynthia Kenyon your second quarter, and of course, she has you searching for the fountain of youth. She wants you to do a screen for new genes that make worms live longer. Explain how you start your screen. Mutagenize WT hermaphrodites. Allow them to self. What do you expect the phenotype of your F1s to be? All hermaphrodites, all WT phenotype. Wow, you get one worm in your F1 progeny that just won’t die, you call it liv-1. What do you think is up with this worm? How do you verify this? It has a dominant mutation. Cross it to a WT male to check. If dominant, half of the progeny should live long and half are wild type. If recessive, none should live long. liv/+ X +/+ à 1:1 liv/+ : +/+ Ok, now that you’ve got liv-1 all sorted out, you want to find more mutants. What do you do? Move your F1 progeny onto their own plates. Allow for self-mating. Examine the phenotypes of the F2s. Great, you’ve found a bunch of other mutants that live more than 2 times the average worm life span! Cynthia’s tells you to make sure that all the phenotypes are each caused by a single mutation. What do you do and why is this important? Back-cross to WT males. You will get 50% WT males, and 50% WT hermaphrodites. Self the hermaphrodites and pick mutants. Repeat. This is important to eliminate random background mutations and make sure that you have isolated a single mutation. Not outcrossing enough has actually caused big mistakes in the worm field! The sir2 gene in worm was originally reported to have a 40% lifespan extension, but later it was discovered that actually the reported lifespan extension was caused by a background mutation because they didn’t outcross their strains enough. Some people still believe that sir2 causes a 10% lifespan extension, but it’s a huge controversy in the field. Moral of the story- backcross your strains – at least 5x. Because this is 2012, we have lots of tools available to us that make things faster and easier- like deep sequencing. To figure out what genes are causing lifespan extension we might be able to deep sequence. Would you still want to outcross first? Why? If you were going to identify the mutation via deep sequencing, would you still need to map at all? You would still want to outcross to isolate a single mutation. Deep sequencing the entire genome would still be really expensive. You would want to map to a region of the chromosome, but you wouldn’t necessarily need to map to a small region of the genome. Problem 3: You need to map a newly identified mutation (m) by the end of your rotation in the Ashrafi lab. You decided to use a mapping strain homozygous for all three known genetic markers a, b and c. a is on Choromome I b is on Chromosome II c is on Chromosome III You know that: 1) m is a recessive mutation 2) a, b and c have a BamHI restriction site Before BamHI digestion After BamHI digestion A 300bp 300bp a 300 230 and 70 B 100 100 b 100 50 and 50 C 200 200 c 200 150 and 50 What steps do you go through before before you get to the mapping portion? Will you use F1s or F2s? Why? 1) After mating m/m strain with a test strain, you isolated F1 hermaphrodites. 2) You took F1 and put them on separate plates for self-fertilization. 3) You isolated m/m F2 progeny to figure out which of the three markers is linked to your mutation. 4) You performed PCR to amplify all three marker loci. 5) You cut the PCR products with BamHI and ran the fragments on a gel. You isolate F2s because F1s are hets and thus uninformative. What fraction of the time would you expect your mutation to segregate with your marker if it is unlinked to the chromosome that the marker is on? 75% of the time What fraction of the time would you expect your mutation to segregate with your marker if it is linked to the chromosome that the marker is on? <75% of the time – dependent on how many map units apart you are from your mutation of interest You got the following result: On which chromosome is the mutant allele found? (To make it easier to pick out, what is the genotype of each worm?) Chromosome I. Since it rarely segregates with a. What is the map position of the mutation? Remember that each lane on your gel represents two chromosomes. Map units (cM) = 100*(# of recombinant chromosomes/total # chromosomes) 2/24*100 = 8 cM