Transcription_probs

Transcription_probs - MCB121, C. S. Gasser MCB121 Problem...

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Unformatted text preview: MCB121, C. S. Gasser MCB121 Problem Set: Transcription Machinery 1. For each of the following statements referring to aspects of eukaryotic gene expression circle which types of genes (type I, II, or III) to which the statement applies. Circle ALL correct answers for full credit. I II III Transcription complex includes TATA-binding protein (TBP). I II III Transcription usually requires TFIID. I II III RNA product can be translated into protein. I II III RNA product can be a component of a ribosome. 2. The diagram below illustrates the results of a gel shift experiment. Two DNA fragments, DNA 1 and DNA 2, are suspected of including binding sites for two newly discovered general transcription factors (factor X and factor Y). The fragments were radioactively labeled an used in the gel shift assay with purified factors X and Y. The information above each lane indicates which components were added to the reaction run in that lane. Migration is from top to bottom. Results for the last lane (at right) are not given. Factor X X X X X Factor Y Y Y Y Y DNA 1 1 1 1 1 DNA 2 2 2 2 2 X Y 1 2 * A. For each of the following circle "T" if the statement is true, "F" if the statement in false, and "N" if the truth of the statement cannot be determined from the information given. T F N Factor X can bind to DNA 1 by itself. T F N Factors X and Y can form a heterodimer in the absence of DNA. T F N Factor X only associates with DNA 1 in the presence of Factor Y. T F N The arrow at right indicates the band corresponding to factor Y bound to DNA 1. T F N The two bands indicated by the star at right are identical in their composition. T F N Free DNA 1 migrates to the same position as free DNA 2. MCB121, C. S. Gasser B. (5) In the last lane on the gel above, draw the bands you would expect to see from a reaction containing all four components (factors X and Y, and DNAs 1 and 2). 3. The C-terminal domain (CTD) of the large subunit of RNA polymerase II consists of multiple repeats of the consensus sequence TyrSerProThrSerProSer. This region appears to have important roles in polymerase function. This region can be phosphorylated by CTD-kinase (CTD-K) activity and can be dephosphorylated by CTD phosphatase (CTDP). You have recently discovered specific inhibitors of CTD-K and CTD-P which you call IK, and IP, respectively. In vitro transcription reactions are carried out on a standard DNA template in reactions including all normally required transcription factors (with initially unphosphorylated Pol II), and including CTD-K and CTD-P. Four reactions are performed, one with no inhibitor, one with IK, one with IP, and one with both inhibitors present (inhibitors are added before adding the DNA template in each case). A. (12) What would the relative levels of transcript be in the four reactions (which reaction would produce the most, least etc. transcript)? Explain briefly. Control (no inhibitor) IK IP IK and IP B. If DNA template is not limiting, where would you expect the majority of the RNA polymerase II to be at the end of the experiment in which IK is included in the reaction? (circle one) and explain briefly. soluble in solution bound in the PIC stalled during transcription bound to the site of transcription termination 4. Transcription of the major late promoter of adenovirus-2 is dependent on a gene specific upstream stimulatory factor (USF), the general transcription factors TFIID, TFIIB and TFIIE and RNA polymerase II. (Since the publication of this work, specific factor fractions have been fractionated into multiple components.) USF binds to the upstream element, UE, whereas TFIID binds to the TATA element. The sequence of interaction is thought to be as follows; USF and TFIID first interact with the DNA to form a preinitiation complex. The subsequent binding of RNA polymerase II, TFIIB and TFIIE completes the formation of a preinitiation complex. This is followed by an energy dependent step (ATP is required) that is separate from the nucleoside triphosphate requirement for initiation and elongation. Following the hydrolysis of ATP the transcript is initiated. Footprint analysis of transcription complexes formed on the major late promoter of adenovirus-2 in the presence of various combinations of factors, a-amanitin and nucleotides MCB121, C. S. Gasser is shown below. The components present in each reaction is shown at the top of each lane. G and A are control sequencing marker lanes. A. What conclusion(s) can be drawn from these studies. Explain clearly how the results support your conclusion(s) and how this relates to the mechanism of transcription. B. Which steps in transcription are the most likely to require energy provided by hydrolysis of ATP? In other words, what is ATP used for separate from the nucleotides that are incorporated into the nascent RNA chain? Explain your answer. MCB121, C. S. Gasser Solutions 1. For each of the following statements referring to aspects of eukaryotic gene expression circle which types of genes (type I, II, or III) to which the statement applies. Circle ALL correct answers for full credit. I II III Transcription complex includes TATA-binding protein (TBP). I II III Transcription usually requires TFIID. I II III RNA product can be translated into protein. I II III RNA product can be a component of a ribosome. 2. The diagram below illustrates the results of a gel shift experiment. Two DNA fragments, DNA 1 and DNA 2, are suspected of including binding sites for two newly discovered general transcription factors (factor X and factor Y). The fragments were radioactively labeled an used in the gel shift assay with purified factors X and Y. The information above each lane indicates which components were added to the reaction run in that lane. Migration is from top to bottom. Results for the last lane (at right) are not given. Factor X X X X X Factor Y Y Y Y Y DNA 1 1 1 1 1 DNA 2 2 2 2 2 X Y 1 2 ? * A. For each of the following circle "T" if the statement is true, "F" if the statement in false, and "N" if the truth of the statement cannot be determined from the information given. T F N Factor X can bind to DNA 1 by itself. T F N Factors X and Y can form a heterodimer in the absence of DNA. T F N Factor X only associates with DNA 1 in the presence of Factor Y. T F N The arrow at right indicates the band corresponding to factor Y bound to DNA 1. T F N The two bands indicated by the star at right are identical in their composition. MCB121, C. S. Gasser T F N Free DNA 1 migrates to the same position as free DNA 2. B. (5) In the last lane on the gel above, draw the bands you would expect to see from a reaction containing all four components (factors X and Y, and DNAs 1 and 2). This lane will have all bands visible in the other lanes. In addition, because factor X can bind DNA2 and factor Y can bind DNA 1, and the two factors appear to bind each other (at least when the DNAs are present), there may possibly be a fifth band consisting of all four molecules in a complex. 3. The C-terminal domain (CTD) of the large subunit of RNA polymerase II consists of multiple repeats of the consensus sequence TyrSerProThrSerProSer. This region appears to have important roles in polymerase function. This region can be phosphorylated by CTD-kinase (CTD-K) activity and can be dephosphorylated by CTD phosphatase (CTDP). You have recently discovered specific inhibitors of CTD-K and CTD-P which you call IK, and IP, respectively. In vitro transcription reactions are carried out on a standard DNA template in reactions including all normally required transcription factors (with initially unphosphorylated Pol II), and including CTD-K and CTD-P. Four reactions are performed, one with no inhibitor, one with IK, one with IP, and one with both inhibitors present (inhibitors are added before adding the DNA template in each case). A. (12) What would the relative levels of transcript be in the four reactions (which reaction would produce the most, least etc. transcript)? Explain briefly. Control (no inhibitor) - Most transcript. IK - No transcript formed. IP - Some transcript formed, less than control. IK and IP - No transcript formed. Phosphorylation is required for promoter clearance and extension. Therefore, no transcript will be formed in either reaction where IK is present. One round of transcription can occur in the presence of IP, but only unphosphorylated PolII can bind the promoter so a second round of transcription will not be initated. Multiple rounds of transcription can occur when no inhibitors are present. B. If DNA template is not limiting, where would you expect the majority of the RNA polymerase II to be at the end of the experiment in which IK is included in the reaction? (circle one) and explain briefly. soluble in solution bound in the PIC phosphorylation is necessary for promoter clearance, the polymerase will form a PIC, but will not be able to dissociate from it. stalled during transcription bound to the site of transcription termination 4. Transcription of the major late promoter of adenovirus-2 is dependent on a gene specific upstream stimulatory factor (USF), the general transcription factors TFIID, TFIIB and MCB121, C. S. Gasser TFIIE and RNA polymerase II. (Since the publication of this work, specific factor fractions have been fractionated into multiple components.) USF binds to the upstream element, UE, whereas TFIID binds to the TATA element. The sequence of interaction is thought to be as follows; USF and TFIID first interact with the DNA to form a preinitiation complex. The subsequent binding of RNA polymerase II, TFIIB and TFIIE completes the formation of a preinitiation complex. This is followed by an energy dependent step (ATP is required) that is separate from the nucleoside triphosphate requirement for initiation and elongation. Following the hydrolysis of ATP the transcript is initiated. Footprint analysis of transcription complexes formed on the major late promoter of adenovirus-2 in the presence of various combinations of factors, a-amanitin and nucleotides is shown below. The components present in each reaction is shown at the top of each lane. G and A are control sequencing marker lanes. A. What conclusion(s) can be drawn from these studies. Explain clearly how the results support your conclusion(s) and how this relates to the mechanism of transcription. Conclusions from left panel: There is a clear footprint in the region of UE and TATA confirming that these regions are bound to USF and TFIID (compare lanes - and UD in both panels). The addition of BE, BII or EII do not significantly change the footprint (compare lane UD with UDBE, UDBII and UDEII). The addition of all three components to UD extends the footprint beyond the start site. Consequently, both TFIIB and TFIIE are essential for the close association of RNA polymerase II with the start site. Conclusions from right panel: MCB121, C. S. Gasser The complete reaction containing nucleotides results in a footprint virtually identical to that of the complex formed in the absence of RNA polymerase II. This indicates that RNA polymerase has cleared the promoter and that USF and TFIID remain bound. The inclusion of a-amanitin, which inhibits elongation, prevents promoter clearance (compare lane Cx+aA+NTPs with Cx). The slight change in footprint indicates that the polymerase has moved slightly. These experiments show clearly the conformational changes that occur during the course of initiation. TFIID and USF are DNA binding proteins that in the presence of TFIIB and TFIIE facilitate the binding of RNA polymerase II to a region encompassing the start site. Following initiation, RNA polymerase leaves the promoter without disrupting the interaction of USF and TFIID. B. Which steps in transcription are the most likely to require energy provided by hydrolysis of ATP? In other words, what is ATP used for separate from the nucleotides that are incorporated into the nascent RNA chain? Explain your answer. The transition from a closed to an open promoter complex requires melting of the DNA in the region of the start site and consequently may be dependent on the input of energy. This change must be associated with major conformational changes and seems to be a likely step for the utilization of ATP. ATP is also required for the phosphorylation of the C-terminal domain of RNA polymerase subunit IIa. This phosphorylation likely induces a major conformational change which may facilitate the release of RNA polymerase II from the preinitiation complex. ...
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This note was uploaded on 04/06/2011 for the course MCB 121 taught by Professor Gasser during the Winter '09 term at UC Davis.

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