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MIT 7 03 - Exam Three

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Name: ________key______________17.03 Exam Three -- 2005 KEYName: ______________________________Exam starts at 11:05 am and ends at 11:55 am.There are 8 pages including this cover page.Please write your name on each page.Only writing on the FRONT of every page will be graded.(You may use the backs, but only as scratch paper.)Question 1 17 pts________Question 2 45 pts________Question 3 20 pts________Question 4 18 pts________TOTAL out of 100_______Name: ________key______________2 1. (17 pts) You are studying the expression of the yeast gene ProA that is necessaryfor the synthesis of the amino acid proline. ProA is normally expressed only when thecell is lacking supplemental proline in the growth medium. You isolate two haploidyeast strains (ProB– and ProC–) that misregulate ProA expression.You mate a ProB– haploid strain to a wild-type haploid strain. The resultingdiploid expresses ProA properly.You mate a ProB– haploid strain to a ProA– haploid strain. The resulting diploidexpresses ProA properly.You mate a ProA– ProC– haploid strain to a ProC– haploid strain. The resultingdiploid expresses ProA when proline is present in the growth medium.You mate a ProC– haploid strain to a ProA– haploid strain. The resulting diploidexpresses ProA properly.You mate a ProB– ProC– haploid strain to a wild-type haploid strain. Theresulting diploid expresses ProA properly. You induce sporulation of this diploid, andexamine 40 tetrads. 30 (of those 40) each contain: two spores that do not express ProAwhen proline is absent from the growth medium, one spore that expresses ProA whenproline is present in the growth medium, and one spore that expresses ProA properly.(a, 6pts) Classify the ProB– mutation by its genetic properties (cis vs. trans,constitutive vs. uninducible, dominant vs. recessive).Trans, recessive, uninducibleTrans – You mate a ProB– haploid strain to a ProA– haploid strain. The resultingdiploid expresses ProA properly. This means that the A and B mutationscomplement each other and are thus in different genes. This means that B mustact in trans to A.Recessive -- You mate a ProB– haploid strain to a wild-type haploid strain. Theresulting diploid expresses ProA properly. This means B– is recessive.Uninducible – The only mutant phenotypes you see when you sporulate a diploidthat contains both the B– and C– mutations are constitutive and uninducible. C–gives constitutive, so B– must give uninducible.(b, 6pts) Classify the ProC– mutation by its genetic properties (cis vs. trans,constitutive vs. uninducible, dominant vs. recessive).Trans, recessive, constitutiveTrans – You mate a ProC– haploid strain to a ProA– haploid strain. The resultingdiploid expresses ProA properly. This means that the A and C mutationsName: ________key______________3complement each other and are thus in different genes. This means that C mustact in trans to A.Recessive -- You mate a ProC– haploid strain to a ProA– haploid strain. Theresulting diploid expresses ProA properly. This means C– is recessive.Constitutive – You mate a ProA– ProC– haploid strain to a ProC– haploid strain.The resulting diploid expresses ProA when proline is present in the growthmedium. This means that a cell that has a functional copy of A but has nofunctional copies of C expreses ProA even when it is not supposed to (i.e. whenthe cell already has proline available to it).(c, 5pts) If you a drew a linear pathway showing the regulation of ProA, which functionwould you place closer to ProA: ProB or ProC?ProBWhen you sporulate a diploid that was produced from a mating of B– C+ toB+ C–, you see mostly tetratypes. You know they are tetratypes because thetypes of spores do not come in pairs. (There are three types of spores.)Each tetratype contains: two spores that do not express ProA when proline isabsent from the growth medium (uninducible), one spore that expresses ProAwhen proline is present in the growth medium (constitutive), and one spore thatexpresses ProA properly (inducible). A tetratype resulting from this matingwould contain the spores:GENOTYPE PHENOTYPEB– C+ uninducibleB+ C– constitutiveB+ C+ inducibleB– C– NOT KNOWN PREVIOUSLYThe spore of unknown phenotype must be the double mutant, and it shows thesingle mutant phenotype of B (uninducible), so B must be more downstream inthe pathway (i.e. closer to the reporter gene).2. (45 pts) You are studying the transcriptional regulation of a mouse gene calledStringy. This gene is normally only expressed in tail cells due to the presence of a tail-specific inducer molecule in these cells. You have isolated two true-breeding mutantstrains of mice that do not spatially regulate the expression of the Stringy gene properly.The strains of mice that you have, and their corresponding phenotypes, are listed in thetable below.Genotype of mouse Phenotype of mouseWild-type Stringy expressed only in tailName: ________key______________4A– / A–Stringy not expressed anywhereB– / B–Stringy expressed in all cells in the bodyWhen you cross mice that are B– / B– to mice that are deficient in Stringy, the resultingmice only have Stringy expressed in the tail.When you cross mice that are B– / B– to mice that are A– / A–, and then cross theresulting F1 mice to each other, you get a genotypic ratio in the F2 that indicates that theA and B loci segregate independently of each other.You inject a piece of DNA containing the A– allele of the A gene into a fertilized eggproduced by the mating of two true-breeding B– mice. You then transfer this injectedfertilized egg into a pseudopregnant mouse. The mouse that is born does not expressStringy in any cells in its body.(a, 6pts) Classify the A– mutation by its genetic properties (cis vs. trans, constitutivevs. uninducible, dominant vs. recessive).Trans, dominant, uninducibleTrans, dominant, and epistatic to B -- You inject a piece of DNA containing the A–allele of the A gene into a fertilized egg produced by the mating of two true-breeding B– mice. You then transfer this injected fertilized egg into apseudopregnant mouse. The mouse that is born does not express Stringy in anycells in its body. The phenotype you are seeing in this A+/A+/A– B–/B– mouse isthe phenotype of A–, not the phenotype of B– (which is constitutive). This tellsyou three things:1) A– is dominant. This mouse has 3 copies of A and two


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MIT 7 03 - Exam Three

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