ANSWERS TO Problem set questions from Exam 3 Eukaryotic Gene Regulation Genome Modifications in Eukaryotes Population Genetics Characterizing novel pathways that control the expression of yeast genes 1 You are studying regulation of the yeast enzyme glutamine synthetase GS which is encoded by the GLN1 gene a they are both activators because recessive trans uninducible b 3 2 1 2 3 1 2 1 3 c glu 2 1 gln 3 d never express GS no matter what conditions e 300 250 is the glu response element is a UAS may bind GLN2 200 100 is the gln response element is a UAS may bind GLN3 50 1 is the promoter f with neither amino acid added 50 with only glu added 50 with only gln added 0 g with neither amino acid added 0 with only glu added 0 with only gln added 0 2 Consider a eukaryotic gene regulatory pathway where a small molecule X activates the expression of a reporter gene a X A B reporter gene OR X B A reporter gene b A always activates B regardless of whether X is present upper model OR A always activates the reporter gene regardless of whether B is activated or not lower model c if the upper model is correct you would get a 1 PD 1 NPD 4 TT ratio PD NPD TT Constitutive uninducible regulated Constitutive uninducible uninducible Uninducible regulated constitutive Uninducible regulated uninducible if the lower model is correct you would get a 1 PD 1 NPD 4 TT ratio PD NPD TT Constitutive constitutive regulated Constitutive constitutive constitutive Uninducible regulated constitutive Uninducible regulated uninducible d the F1s would all be constitutive regardless of the model for the F2s if the upper model is correct you would get a ratio of 9 constitutive 3 uninducible 3 regulated 1 uninducible if the lower model is correct you would get a ratio of 9 constitutive 3 constitutive 3 regulated 1 uninducible 3 You are studying the regulation of an enzyme in yeast a both trans b Type 1 regulated 1 2 regulated 1 2 uninducible 1 2 uninducible 1 2 Type 2 regulated 1 2 constitutive 1 2 uninducible 1 2 uninducible 1 2 c no d uninducible e inducer 1 2 reporter gene Type3 constitutive constitutive uninducible uninducible 1 2 1 2 1 2 1 2 4 You have discovered a gene in yeast that is involved in repairing damaged DNA a repressor because constitutive trans recessive b UV Reg1 Rad66 c UV Reg1 Reg2 Rad66 OR UV Reg2 Reg1 Rad66 d constitutive e UV Reg2 Reg1 Rad66 f region 2 is the Reg1 binding site and region 6 is the promoter g The DNA sequence of regions 3 5 are not necessary for proper Rad66 regulation and the spacing between elements 2 and 6 is not critical 5 In the examples of gene expression that we have covered in class the genes were regulated in some way or another a yes at 27 cM b zero c 1 reporter gene 2 d Con 1 binds region 4 and Con2 binds region 2 Altering the genomes of mice Transgenics and Gene targeting 1 You hypothesize that a loss of function of the Pindrop gene is the cause of the recessive phenotype of deafness in a strain of mice called the Ard strain a i gene targeting ii the Pindrop gene with a antibiotic resistance gene disrupting that gene iii ES cells iv wild type ES cells v at the Pindrop gene locus vi yes it would make a chimera vii you would have to breed the chimeric heterozygote to wild type to generate a non chimeric heterozygote You would then have to mate two non chimeric heterozygotes together and 1 4 of their offspring would be the animal you are looking for viii if the mouse is deaf then Pindrop is required If the mouse can hear then Pindrop is not required b breed the mice from above to the Ard strain If the resulting animals are deaf then Pindrop was mutated in the Ard strain If the resulting animals can hear then a different gene was mutated in the Ard strain c i transgenic ii the wild type human Pindrop gene iii egg iv homozygous Pindrop mutant egg v randomly vi no vii none viii if the mouse is deaf then human and mouse are not interchangeable If the mouse can hear then the human and mouse genes are interchangeable 2 In mammals including humans and mice growth hormone a protein is speculated to play a prominent role in determining adult size a i transgenic ii the wild type mouse GH gene iii egg iv wild type egg v randomly vi no vii none viii if the mouse is bigger then more copies of GH do yield larger mice If the mouse is not bigger then more copies of GH is ineffective b i gene targeting ii the GH gene with a antibiotic resistance gene disrupting that gene iii ES cells iv wild type ES cells v at the GH gene locus vi yes it would make a chimera vii you would have to breed the chimeric heterozygote to wild type to generate a non chimeric heterozygote You would then have to mate two non chimeric heterozygotes together and 1 4 of their offspring would be the animal you are looking for viii If the mouse is smaller then one copy of GH is not sufficient to control size normally If the mouse is normal size then one copy of GH is enough for sufficient GH function c breed two mice from part b and 1 4 of their offspring will have zero copies of GH d breed two mice from part a and 1 4 of their offspring will have four copies of GH e i transgenic ii the wild type mouse GH gene iii egg iv an egg created by the breeding of two mice from part d v randomly vi no vii none viii if the mouse is bigger then more copies of GH do yield larger mice If the mouse is not bigger then more copies of GH is ineffective f breed two mice from part e and 1 4 of their offspring will have six copies of GH 3 As we will study later in the semester there are genes in the human and mouse genomes that control cell proliferation a yes b yes c a transgenic mouse because the process is quicker and requires less mating d add a mutant copy of the Ras gene e the mutant copy of the Ras gene would insert randomly f wild type fertilized egg g no further steps h no i yes j you don t have a choice k disrupt both copies of the Rb gene l the Rb gene that is disrupted by a gene encoding antibiotic resistance which would insert at the endogenous Rb locus m wild type ES cells n you would have to breed the chimeric heterozygote to wild type to generate a nonchimeric heterozygote You would then have to mate two non chimeric heterozygotes together and 1 4 of their offspring would be the animal …
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