Problem Sets Fall 1994 7 03 Problem Set 1 due in class Friday September 16 All three problems will be graded Some parts of these problems are trickier than they might firstappear 1 Yeast cells have in their membranes a number of different proteinsthat transportamino acids intothe cell Incidentally it is these transportersthat allow cells withauxotrophicmutationsin amino acid biosyntheticpathways to take up the needed amino acid and therefore to grow on rich medium The transport system that is responsible for uptake of the amino acid arginine will also take up the toxic compound canavanine It is easy to isolatemutants that are defective in the arginine transporter by selectingfor mutantsthat are resistantto canavanine Nine canavanine resistant mutants canr are isolated Four canr mutants are isolated in a haploid yeast strain of mating type c strains 1 4 and five carlr mutants are isolated in a haploid strain of mating type a strains5 9 All possible pairwisecrosses as well as crosses to wild type are performed as indicated in the table below When the resultingdiploidcan not grow on medium with canavanine a is indicated at the intersectionof the two parental strains When the resultingdiploidcan grow with canavanine a is indicated strains of mating type z 1 strainsof mating type a 2 3 4 wild type i u i m i i m 7 8 n m i n m wild type a Which mutants are recessive and which are dominant for canavanine resistance b How many differentgenes or more precisely complementationgroups appear to be requiredfor uptake of canavanine Indicate which mutationsare in the same complementationgroup Also indicateany ambiguities in the assignment of complementation groups c Propose a simple experimentto resolve any ambiguities in the assignment of strains5 and 9 to complementationgroups 2 a A new mouse mutant is discovered that has a tail shorterthan normal When a short tailed mouse is crossed to wild type about half of the F1 are normal and half have short tails Is the the short tailed mutationdominantor recessive When two short tailed F1 mice are crossed what fractionof the F2 would you expect to have short tails b After college you get a job as a genetics expert in a biotechnologycompany Your firstjob is to make a true breeding line with the short tailed mutation A year and many crosses later you have still failed to produce a line that only gives short tailed progeny Afraid that you are going to lose your job you recall something you learned many years ago doing the first 7 03 problem set What hypothesiscan you come up with to explain your failure to isolatetrue breeding shorttailed mice that is mice that are homozygousfor the short tailed mutation assume complete penetrance of the short tailed mutation Accordingto this hypothesis when two short tailed mice are crossedwhat fraction of the progeny would be expected to have short tails C For a cross between two short tailed mice approximately how many progeny would you expect to have to score in a typical experiment to distinguishthe hypothesisin part b from the hypothesisin part a For this problem an exact analytical solutionwould be very hard to obtain so just assume an average outcome given the expectationfor part b and then use the Chi square table to find the approximate number of progeny that you would need to score to show a significant deviationfrom the outcome expected from part a at the p 0 05 significance level d Another geneticist in your company has independently isolated another short tailed mouse After performing some crosses she finds that this mouse has similar propertiesto the mutantyou are workingwith That is the new shorttailed strain producesa mixtureof normaland shorttailed mice whencrossedto wild typeand attemptsto generatea true breedingshorttailedstrainhavefailed Proposean experimentand interpretation of the datathat youwoulduseto determinewhetherthe two mutationsare allelesof the samegene e If youthinkaboutit the testyouproposedfor partd isactuallya complementation test Giventhatcomplementation testsare onlymeaningfulwhen performedwithrecessivemutations explainwhythe testyouare proposingis valid 3 Considerthe followingsimplepedigreewherea couplehave a sonwith a very raretrait The coupleisgoingto have anotherchild sexnotyetknown andtheyask youradviceaboutthe chancesthattheirnext childwillbe affectedwiththe trait 0 r 0 female a Assumethatthe traitisautosomalrecessivewithcompletepenetrance Whatis the probabilitythatthe nextchildwill havethe trait If the nextchilddoesn texhibitthe trait what isthe probabilitythattheyare a carder b Assumethatthe traitisX linkedrecessivewithcompletepenetrance Whatis the probabilitythatthe nextchild willhave thetrait If the nextchildisa girlwhat isthe probabilitythatshe willbe a carder Assumethatthe traitis autosomaldominantwith incompletepenetrance If the probabilityof expressionof the trait in a heterozygoteis 20 o what isthe probability thatthe nextchildwill exhibitthe trait If the next childdoesnotexhibitthe trait whatis the probabilitythat they carrythe trait 7 03 Problem Set 1 Answers 1 a All mutants are recessive except mutant 7 Mutant 7 is dominant to wild type because resulting diploids are canavanine resistant can r when mated to a wild type strain All others are recessive with respect to wild type because the resulting spores are canavanine sensitive canS when mated to a wild type strain The crucial information is the result of the cross to wild type Remember that the terms dominant and recessive refer to phenotypes of alleles and that they are relative terms i e A mutant phenotype may be recessive to wild type but dominant to different mutant phenotype as long as the phenotypes can be distinguished b Illustrated below is a schematic representation results of a complementation test of the two possible Noncomplementors can r can haploid r haploid can r diplonid Complementors can r haploid can r ate S haploid can s diplorlid The lower case letters represent recessive canavanine resistant mutations while the upper case letters represent the wild type canavanine sensitive alleles of the genes In the case of the noncomplementors the resulting diploid will contain two recessive canavanine resistant alleles in the same gene genotype a a Therefore canavanine will not be imported into the ceils and the cells will live In the case of the complementors the diploid will contain a functional wild type allele of each of two genes genotype A a B b In this case canavanine will still be imported into
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