Problem SetsFall 19947.03 Problem Set 1due in class Friday, September 16All three problemswill be graded. Some parts of these problems are trickier than theymightfirstappear.1. Yeast cells have in theirmembranesa numberof differentproteinsthattransportamino acids intothe cell. (Incidentally,it isthese transportersthat allow cellswithauxotrophicmutationsin aminoacid biosyntheticpathways to take up the neededaminoacid and thereforeto grow on richmedium). The transportsystem that isresponsiblefor uptake of the amino acidarginine willalso take up the toxiccompoundcanavanine. It is easy to isolatemutantsthat are defective in the argininetransporterby selectingfor mutantsthat are resistantto canavanine. Nine canavanine resistantmutants (canr) are isolated. Four canr mutants are isolated in a haploidyeast strainofmatingtype c¢(strains1 - 4) and five carlr mutants are isolated in a haploidstrain ofmatingtype a (strains5 - 9). All possible pairwisecrosses as well as crossesto wildtype are performed as indicatedin the table below. When the resultingdiploidcan notgrowon medium withcanavanine a (-) is indicated at the intersectionof the twoparental strains. When the resultingdiploidcan grow with canavanine a (+) isindicated.strains of mating type¢z1 2 3 4 wild typei u i m i_ i ! + mstrainsofmating type a 7 + + + + +8 + -- + -- --n m i n mwild type .....(a) Which mutants are recessive and which are dominant for canavanineresistance?(b) How many differentgenes or, more precisely, complementationgroups appearto be requiredfor uptake of canavanine. Indicatewhich mutationsare in the samecomplementationgroup. Also indicateany ambiguitiesin the assignmentofcomplementationgroups.(c) Proposea simple experimentto resolveany ambiguitiesin the assignmentofstrains5 and 9 to complementationgroups.2. (a) A new mouse mutant is discovered that has a tail shorterthan normal.When a short tailed mouse iscrossedto wildtype about half of the F1 are normalandhalf have short tails. Is the the short tailed mutationdominantor recessive? When twoshort tailed F1 mice are crossedwhat fractionof the F2 wouldyou expect to have shorttails?(b) After college you get a job as a geneticsexpert in a biotechnologycompany.Your firstjob isto make a true-breedinglinewiththe short tailed mutation. A year and: many crosses later you have stillfailed to produce a linethat only gives short tailedprogeny. Afraidthat you are goingto loseyourjob you recall somethingyou learnedmany years ago doingthe first7.03 problemset. What hypothesiscan you come upwithto explain yourfailure to isolatetrue-breedingshorttailed mice, that is, micethatare homozygousfor the shorttailed mutation(assume completepenetrance of theshort tailed mutation). Accordingto this hypothesis,when two short tailed mice arecrossedwhat fractionof the progeny would be expectedto have shorttails?(C) For a cross between two shorttailed mice approximatelyhow many progenywouldyou expect to have to scorein a typicalexperiment,to distinguishthehypothesisin part b fromthe hypothesisin part a? (For this problem an exactanalyticalsolutionwould be very hard to obtainso just assume an average outcomegiventhe expectationfor part b and then use the Chi-squaretable to find theapproximatenumber of progeny thatyou would need to score to show a significantdeviationfrom the outcomeexpectedfrom part a at the p< 0.05 significance level).(d) Anothergeneticistin your companyhas independentlyisolated another shorttailed mouse. After performingsome crossesshe findsthatthis mouse has similarpropertiesto the mutantyou are workingwith. That is, the new shorttailed strainproducesa mixtureofnormalandshorttailed micewhencrossedtowildtypeandattemptstogeneratea truebreedingshorttailedstrainhavefailed. Proposeanexperimentand interpretationof thedatathatyouwoulduseto determinewhetherthetwomutationsare allelesof the samegene.(e) If youthinkaboutit,thetestyouproposedforpartd isactuallyacomplementationtest. Giventhatcomplementationtestsare onlymeaningfulwhenperformedwithrecessivemutations,explainwhythetestyouare proposingisvalid.3. Considerthefollowingsimplepedigreewherea couplehavea sonwitha veryraretrait. The coupleisgoingto haveanotherchild(sexnotyetknown)andtheyaskyouradviceaboutthechancesthattheirnextchildwillbe affectedwiththetrait.0 r] 0 -female?(a) Assumethatthetraitisautosomalrecessivewithcompletepenetrance.Whatis. the probabilitythatthe nextchildwill havethetrait? If thenextchilddoesn'texhibitthetrait,whatisthe probabilitythattheyarea carder?(b) AssumethatthetraitisX-linkedrecessivewithcompletepenetrance.Whatistheprobabilitythatthe nextchildwillhavethetrait? If the nextchildisa girlwhat istheprobabilitythatshe willbe a carder?(¢) Assumethatthetraitisautosomaldominantwithincompletepenetrance. Iftheprobabilityofexpressionofthe trait ina heterozygoteis20=/o,what isthe probabilitythatthe nextchildwillexhibitthetrait? Ifthenextchilddoesnotexhibitthetrait,whatistheprobabilitythatthey carrythe trait?7.03 Problem Set 1 Answers1. (a) All mutants are recessive except mutant 7. Mutant 7 is dominant towild type because resulting diploids are canavanine resistant (can r) whenmated to a wild type strain. All others are recessive with respect to wild typebecause the resulting spores are canavanine sensitive (canS) when mated to awild type strain. The crucial information is the result of the cross to wildtype. Remember that the terms dominant and recessive refer to phenotypesof alleles, and that they are relative terms (i. e. A mutant phenotype may berecessive to wild type, but dominant to different mutant phenotype, as longas the phenotypes can be distinguished).(b) Illustrated below is a schematic representation of the two possibleresults of a complementation test:Noncomplementors:can r _ _can rhaploid _ haploidcan rdiplonidComplementors:can r _ _ can rhaploid ,_ateS haploidcan sdiplorlidThe lower case letters represent recessive canavanine resistantmutations while the upper case letters represent the wild type canavaninesensitive alleles of the genes. In the case of the noncomplementors, theresulting diploid will contain two recessive canavanine resistant alleles in thesame gene (genotype a/a). Therefore, canavanine will not be imported intothe ceils, and the cells will live. In the case of the complementors, thediploid 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 thecells, and the cells will die.The dominant mutant 7
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