Problem set questions from Exam 3 – Eukaryotic Gene Regulation, Genome Modificationsin Eukaryotes, Population GeneticsCharacterizing novel pathways that control the expression of yeast genes1. You are studying regulation of the yeast enzyme glutamine synthetase (GS), whichis encoded by the GLN1 gene. You have isolated two mutants, designated gln2– andgln3–, that give decreased GS activity. Mating of either gln2– or gln3– haploids to wild-type haploids produce heterozygous diploids that show normal amounts of GSexpression. When you cross either a gln2– or gln3– haploid strain to a gln1– haploidstrain, the resulting diploids show normal expression of GS.(a) From these experiments, classify the gln2– and gln3– mutations in terms of theirbasic genetic properties (dominant vs recessive, cis vs trans, uninducible vsconstitutive), and explain the rationale behind your conclusions. Based on theseproperties, make a proposal for the nature of the wild-type regulatory functions of theGLN2 and GLN3 genes.(b) Diagram two different linear models and one parallel model that could illustrate howthe GLN1 gene is regulated by the wild-type GLN2 and GLN3 genes.The GLN1 gene shows a rather complex regulation in response to two different aminoacids. When either glutamate (glu) or glutamine (gln) is added to the medium, theamount of GS expression diminishes; when both glutamate and glutamine are added tothe medium, GS expression is shut off completely. The effects of different mutants onthe response to glu and gln are shown below. Units of GS activity – + glu + gln +glu& glnwild type 100 50 50 0gln1– 0 0 0 0gln2– 50 50 0 0gln3– 50 0 50 0(c) Which of the models from part (b) best fits these experimental results? Diagram acomplete model for the regulation of GLN1 that includes the effects of glu, gln, wild-typeGLN2, and wild-type GLN3.(d) Based on your model for part (c), how would you expect a gln2– gln3– doublemutant to behave?Next, you decide to evaluate the cis regulatory DNA sequences found in front of theGLN1 open reading frame. To do this, you first fuse these regulatory DNA sequencesto the LacZ coding sequence, and then place this hybrid gene on an appropriate yeastplasmid. You find that cells carrying this reporter gene express LacZ activity under thesame conditions that GS is expressed in wild-type cells. This tells you that the cisregulatory region you have selected contains all of the necessary cis-acting sequencesfor normal regulation. The figure below shows the effect of six different 50-basepair-long deletions in the cis regulatory region on the amount of ß-galactosidase activityexpressed by the reporter gene.-300 -250 -200 -150 -100 -50 +1 Units of ß-galactosidase | | | | | | | – + glu + gln wt _________________________________LacZ 100 50 501 ___ _________________________ LacZ 50 50 02 ________ ____________________ LacZ 100 50 503 _____________ _______________LacZ 50 0 504 __________________ __________LacZ 50 0 505 _______________________ ____LacZ 100 50 506 ____________________________ _LacZ 0 0 0(e) Describe the cis-acting elements in the GLN1 cis regulatory region that are evidentfrom these experiments, giving both their positions and as much of their wild-typefunction as you can deduce.(f) How many units of ß-galactiosidase would you expect to be expressed in gln2–mutant yeast from a reporter gene construct carrying deletion #1…with neither amino acid added?with only glu added?with only gln added?(g) How many units of ß-galactiosidase would you expect to be expressed in gln2–mutant yeast from a reporter gene construct carrying deletion #4…with neither amino acid added?with only glu added?with only gln added?2. Consider a eukaryotic gene regulatory pathway where a small molecule X activatesthe expression of a reporter gene. You have isolated loss-of-function mutations in twodifferent genes, A and B, both of which give uninducible expression of the reporter.Genes A and B are not linked to each other and neither gene is linked to the reporter.(a) Assuming that the regulatory factors encoded by A and B act in series, there are twopossible orders in which these two regulatory factors can act. Draw out these twomodels showing the relationships between the wild-type regulatory functions of A andB, and the reporter. Also be sure to indicate where and how the inducer X acts.(b) In order to distinguish between the two models from part (a), an epistasis test wouldbe useful. Because the mutations that have been isolated in the A and B genes havethe same phenotype (uninducible), it is not possible to perform an epistasis test usingthese alleles. Fortunately, you are able to isolate an allele of gene A that givesconstitutive expression of the reporter. This allele, called As, causes a dominantphenotype of constitutive reporter expression. Describe in molecular terms how theallele As affect the normal regulatory function of A, given each of your models frompart (a).(c) Assume that you are studying this regulatory pathway in yeast and you wish toperform an epistasis test by constructing the As B– double mutant. To do this, youcross a MATa As B+ haploid strain to a MATα A+ B– haploid strain, and inducesporulation of the resulting diploid. You examine the resulting tetrads. For each of thetwo models from part (a), give the types of tetrads that you would expect and theirrelative frequencies. The tetrad types should be described by the phenotypes(constitutive, uninducible, or regulated) of the four spores in each tetrad.(d) Now assume that you are studying this regulatory pathway in Drosophila. Toperform the epistasis test, you cross a As/As B+/B+ male to a A+/A+ B–/B– female.For each of the two models from part (a), give the expected phenotype of the F1 fliesfrom this cross. Now you cross the F1 flies among themselves to produce F2 flies. Foreach of the two models, give the expected ratio of constitutive, uninducible, or regulatedphenotypes among the F2 flies.3. You are studying the regulation of an enzyme in yeast. To begin your analysis ofthis regulation, you first fuse the cis regulatory
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