ANSWERS TO Problem set questions from Exam 3 – Eukaryotic Gene Regulation,Genome Modifications in 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.(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 = 50with only glu added = 50with only gln added = 0(g) with neither amino acid added = 0with only glu added = 0with only gln added = 02. Consider a eukaryotic gene regulatory pathway where a small molecule X activatesthe expression of a reporter gene.(a) X A B reporter geneOR X B A reporter gene(b) A always activates B, regardless of whether X is present (upper model)ORA always activates the reporter gene, regardless of whether B is activated or not (lowermodel)(c) if the upper model is correct, you would get a 1 PD: 1 NPD: 4 TT ratioPD NPD TTConstitutive uninducible regulatedConstitutive uninducible uninducibleUninducible regulated constitutiveUninducible regulated uninducibleif the lower model is correct, you would get a 1 PD: 1 NPD: 4 TT ratioPD NPD TTConstitutive constitutive regulatedConstitutive constitutive constitutiveUninducible regulated constitutiveUninducible regulated uninducible(d) the F1s would all be constitutive, regardless of the modelfor the F2s: if the upper model is correct, you would get a ratio of9 constitutive: 3 uninducible: 3 regulated: 1 uninducibleif the lower model is correct, you would get a ratio of9 constitutive: 3 constitutive: 3 regulated: 1 uninducible3. You are studying the regulation of an enzyme in yeast.(a) both trans(b)Type 1 Type 2 Type3regulated (1+ 2+) regulated (1+ 2+) constitutive (1– 2+)regulated (1+ 2+) constitutive (1– 2+) constitutive (1– 2+)uninducible (1– 2–) uninducible (1– 2–) uninducible (1+ 2–)uninducible (1– 2–) uninducible (1+ 2–) uninducible (1+ 2–)(c) no(d) uninducible(e) inducer --] 1 --] 2 reporter gene4. 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 Rad66ORUV 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 wereregulated 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 2Altering the genomes of mice -- Transgenics and Gene targeting1. You hypothesize that a loss of function of the Pindrop gene is the cause of therecessive 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 anon-chimeric heterozygote. You would then have to mate two non-chimericheterozygotes together, and 1/4 of their offspring would be the animal you are lookingfor. viii) if the mouse is deaf, then Pindrop is required. If the mouse can hear, thenPindrop is not required.(b) breed the mice from above to the Ard strain. If the resulting animals are deaf, thenPindrop was mutated in the Ard strain. If the resulting animals can hear, then adifferent 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 themouse can hear, then the human and mouse genes are interchangeable.2. In mammals, including humans and mice, growth hormone (a protein) is speculatedto 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 themouse 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 anon-chimeric heterozygote. You would then have to mate two non-chimericheterozygotes together, and 1/4 of their offspring would be the animal you are lookingfor. viii) If the mouse is smaller, then one copy of GH is not sufficient to control sizenormally. If the mouse is normal size, then one copy of GH is enough for sufficient GHfunction.(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 themouse 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 GH3. As we will study later in the semester, there are genes in the human and mousegenomes that control cell proliferation.(a) yes(b) yes(c) a transgenic mouse, because the
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