Slide 1Slide 2Slide 3Slide 4Slide 5Slide 6Slide 7Slide 8Slide 9Slide 10Slide 11MCB502 Midterm ExamName:___KEY______________________Each problem is worth 10 points.Problem 1. You are investigating the regulation of the GAT14 gene. In your studies you find that the Spit6 transcription factor can induce the expression of GAT14 when the cells are exposed to sun light. In addition, you find that the homolog to the histone-like protein H-NS influences transcription of GAT14. To determine the mechanism of Spit6 action and how GAT14 is regulated you perform the experiments shown below.In addition to the shown data, you utilize the chromatin immunoprecipitation assay to discover that H-NS has 2 binding sites in the GAT14 locus at positions centering +2200 and -600 whereas Spit6 only occupies the GAT14 locus at a site centered +1400.Question (10 points) How does H-NS and Spit6 each contribute to GAT14 regulation? Include in your answer explanations for why the mobility of GAT14 DNA changes when exposed to Spit6, H-NS, and Spit6 + H-NS.EMSAIn vitro transcriptionSpit6 - - - -H-NS - - - - GAT14 DNA + + + + + + + + + +FreeProbe**Free probe is the entire GAT14 DNA locus.H-NS and Spit6 stabilize looped/bent forms of the GAT14 locusH-NS closes the GAT14 locus to repress transcriptionSpit6 opens the locus to promote transcriptionSpit6 can outcompete H-NS for effects on DNA, which leads to transcriptional activation1. Gene regulatory mechanisms are generally conserved between prokaryotes and eukaryotes. True or False (circle one) 2. Signaling pathways involving a kinase and a transcription factor are called a two-component system.3. Histone-like bind to curved DNA in prokaryotes.4. Pioneering factors are sequence-specific DNA binding proteins used to invade heterochromatin.5. Partial antagonist are referred to as such since these molecules only mildly reduce a transcriptional response. True or False (circle one) Problem 2 (2 pts each question)Question A (5 points) What assay could you use to determine the mechanism(s) for why the gene expression patterns remain the same or change under all your experimental conditions? In your answer include why the assay you choose would be useful for differentiating both the changes in gene expression profiles and cell morphology phenotypes.Problem 3You find 3 lentiviral stocks (lentiviruses infect all mammalian cell types found in your laboratory and are used to express individual genes) in your old lab mates refrigerator labeled A, B and C. When you infect embryonic stem cells (ESC) with either stock A or B the ESCs differentiate into long, multinucleated cells whereas stock C causes the ESCs to enlarge in size and accumulate a high level of lipids. To your surprise infection of fibroblast (skin) cells with stock B also causes the cells to differentiate into long, multinucleated cells. However, stocks A and C have no effect on the fibroblasts. When you check global gene expression you find that the patterns are the same in the ESCs differentiated with either stock A or B or when the fibroblast cells are treated with stock B. However, stock C induces an unique gene expression profile in the ESCs and the gene profiles of the fibroblasts are comparable whether the cells are untreated or exposed to stocks A or C.Question B (5 points) Why might the protein expressed using lentiviral stock B impact fibroblast cells whereas stocks A and C have no effect?RNA Microarray, RNA-Seq, DNase-Seq or MNase-Seq as these assays will identify either the global gene expression patterns or global areas of open chromatin that correlate with gene expression patternsGene expression patterns correlate with cell typesStock B expresses a pioneering transcription factor whereas stocks A and C express non-pioneering transcription factorsProblem 4 (2 points each)1. Which subunits are part of the core eukaryotic RNA polymerase II (circle all that apply)? A. TAF250B. Rpb1C. aD. TFIIBE. All of the above2. FRAP or FLIP is used to measure the dynamics of protein fusions in live cells. 3. Consensus sequence is the most favored DNA sequence recognized by a transcription factor.4. Histone chaperones are unable to slide nucleosomes but are able to remove histone octamers from DNA.5. Circle the name(s) of the assay(s) used to measure transcription rates.A. Electrophoretic-mobility shift assayB. Northern blotC. Run-off assayD. Quantitative RT-PCRE. Run-on assayProblem 5A new strain of virus (DV1) has risen. The original version (CCV1) causes symptoms of a common cold. Unfortunately, the new strain causes mortality in ~30% of the very young (<6 years of age) and ~20% of the old (>85 years of age) infected individuals compared to middle aged people. A single nucleotide polymorphism (SNP) array revealed only 1 nucleotide difference between the two viral genomes. The mutation was located 112 bp upstream of an ORF that influences virulence. In parallel with the SNP analysis, it was discovered that glucocorticoid treatment reduced the mortality rate to ~10 and 15%, respectively. In light of both results, you carry out an electrophoretic mobility shift assay (shown below) in an attempt to understand why the new strain is so dangerous and why glucocorticoids are beneficial. As a probe you use a DNA fragment that is only 9 base pairs in length with the SNP site in the middle since you found using DNA footprinting that the central 5 bases of this site are the only nucleotides contacted under all conditions.Question A (5 points)What is a plausible mechanism for the shift to virulence in the DV1 strain?Cortisol - - - + - + - + - + - + - + SNP DNA - + - - + + - - + + - - + +WT DNA + - + + - - + + - - + + - -FreeProbeYoung Middle Old- -Question B (5 points)What is the molecular basis for the difference in virulence between the young and old patients and why is there a difference in the efficacy of glucocorticoid treatment?Mutation converts a response element into a tethering elementA tethering partner is only present in young and old individualsThe tethering protein has a different affinity for the DNA-bound transcription factor (e.g., tethering partner is a different protein in young and old)Hormone addition has a lower efficacy in old individualsCircle each of the following statements as true (T) or false (F)1. T or F Prokaryotic transcription typically involves multiple
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