PSwk6key - Problem set Week 6 MCB110, Spring ‘11...

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Unformatted text preview: Problem set Week 6 MCB110, Spring ‘11 1. T/F __ F __ Sigma is part of the RNA polymerase core enzyme in E. coli __ T__ The strength of a promoter is determined by its resemblance to the consensus sequence __ F__ Like DNA polymerase, RNA polymerase also requires a primer and thus cannot start de novo synthesis __ F __ The template strand is also the coding strand __T_ The sequential hydrolysis of high ­energy phosphate bonds on incoming triphosphate nucleotides drives the equilibrium of the overall reaction toward chain elongation during transcription __ F __ There is only one version of the sigma that is used by E. coli __ T__ Termination of transcription in E. coli sometimes utilizes the protein Rho __ F__ Rho is ATP ­independent __ F __ The Lac operon is only controlled by negative transcription regulation __ F __ RNA polymerase has to be just as accurate as DNA polymerase 2. Finding promoter elements that are recognized by an RNA polymerase a. You’ve discovered a new polymerase in a species of Archaea and decided that you want to map where it binds in its own genome. What technique could you use to do this? What kind of information do you need to obtain before you perform this experiment? DNaseI footprinting assay of genomic DNA that’s broken up into smaller fragments to determine which elements the polymerase binds. You’ll need to know the approximate regions of DNA (often times the sequence) to synthesize/express and radiolabel since this is an in vitro assay that uses recombinant protein and known DNA molecule. b. Briefly outline the experiment including the negative control 1. You label the DNA on one end, usually radiolabel such as 32P 2. Digest the sample, both the negative control (DNA alone) and the experimental tube (DNA+polymerase), with low concentration of DNaseI to ensure each DNA molecule is on average only cut once. 3. Separate the DNA from protein and denature the duplex by running both samples in denaturing gel (high urea concentration e.g.) 4. Separate the fragments by size via electrophoresis, which will reveal the pieces of DNA extending from the labeled end to the site of DNaseI cleavage. 5. Run a sequencing gel side by side to get the sequence of the binding region c. What do you need to leave out of your reaction to ensure that you are mapping the polymerase’s promoter binding site and not also the regions of DNA it transcribes? You need to leave out rNTPs (aka NTPs) to lock the polymerase to the promoter ­ binding site, that way it won’t have promoter clearance 3. What would the gel pattern look like for Maxam ­Gilbert sequencing of the following nucleotide: 5’ ­CATAGCCGTAAT ­3’? Indicate which end, top or bottom of the gel, would be the 5’ G ___ ___ G+A ___ ___ ___ ___ ___ ___ C+T ___ ___ ___ ___ ___ ___ C ___ ___ ___ 5’ 4. What would you expect to be the outcomes for the following mutants? Indicate what happens to the polymerase, or its interaction with DNA, and/or transcription output a. mutant sigma subunit incapable of binding the core polymerase Core enzyme cannot interact with the promoter element, no transcript b. a mutant sigma subunit that binds the core polymerase more tightly than its wild ­ type version Sigma cannot dissociate from the core enzyme to allow elongation. So polymerase remains stuck to the promoter site, generates a few nt before stopping. c. mutant alpha subunit Polymerase cannot bind regulatory sequences (other than the core promoter elements recognized by sigma) or other transcription activators, so cannot respond to transcription activating signals. Often no transcript d. mutant omega subunit RNAP would disassemble, no transcription e. mutant beta subunit Cannot make phosphodiester bonds, no transcripts. f. mutant beta’ subunit Cannot bind DNA, no transcripts 5. What are the RNA elements required for Rho ­independent transcription termination? Both a hairpin structure formed by inverted repeats with non ­complementry sequences in between, and UUU’s immediately after the hairpin. Since A ­U bonds are not as stable at A ­T bonds, RNAP would stall at the hairpin and fall off when encountering the A ­U’s (DNA template ­mRNA). 6. Describe the mechanism by which Rho can terminate transcription. Rho loads onto a C ­rich sequence in the mRNA (rut site), and translocates 5’ ­>3’ along the mRNA to separate it from the DNA template as a helicase. It hydroylzes 1 ATP for translocating 1nt of mRNA, where each of its 6 subunits undergoes 4 catalytic states to move mRNA molecule in and out of the hexameric ring. 7. What could be one advantage of coding genes as operons, thus generating polycistronic mRNA? This way, mRNA encoding for proteins involved in the same pathway could all be synthesized after one stimulus; highly efficient 8. What would be expect to be the outcome for the following mutant? a. LacI that cannot bind Lac operon The Lac operon would always be expressed, even when no lactose is present, though the degree of its expression is still modulated by the presence of glucose through cAMP ­CAP. So it’s not full off but not fully on either. b. LacI that cannot bind lactose or IPTG Lac repressor will not dissociate from DNA, so the expression of the Lac gene will always be repressed even in the presence of lactose and little glucose. c. Mutant LacI that binds glucose but not lactose The Lac genes will be expressed when glucose is present, even though they are not needed—wasted energy. But it cannot be expressed when glucose is absent while lactose is present, such that they cannot make use of this alternate energy source d. CAP that cannot bind cAMP Even when glucose is present and only lactose is available, only very low levels of Lac transcripts are made. Thus you only get basal levels of transcription rather than huge amount of transcription of Lac operon. e. Adenylate cyclase that is constitutively (i.e. always) inactive No cAMP would be made, thus CAP cannot serve as an activator, so again as in d, expression of Lac gene would be very low. 9. What key difference between eukaryotes and prokaryotes makes the regulation of the trp operon possible in bacteria? Transcription and translation are coupled in prokaryotes due to the lack of a nuclear membrane. 10. Describe how the amino acid tryptophan can serve as an allosteric effector? Trp can bind to the tryptophan repressor which causes a conformational change in the repressor that then allows structural elements of the repressor to bind to the operator DNA. 11. The trp leader sequence affects transcription of the trp operon. Explain the outcome of transcription of the trp operon for both low levels and high levels of tryptophan. Under low levels of trp the ribosome stalls on region 1 of the leader sequence of the nascent mRNA. This allows regions 2 and 3 to form a hairpin and not regions 3&4. The 2/3 hairpin is not a terminator signal therefore RNA Polymerase continues transcribing the trp operon. Under high levels of trp the ribosome translates through the nascent mRNA and regions 3 &4 can form a stem ­loop or hairpin. When 3/4 form a stem loop this combined with the following polyU sequence serve as a termination signal for transcription and RNAP falls of the DNA. 12. What are the possible outcomes from deleting the following sequences from the trp leader sequence? a. Region 1 No translation coupled transcription regulation would occur, and region 2 will be free to pair with region 3 since they have stronger sequence complementarity, and transcription would not be terminated and only regulated by the Trp repressor, even when tryptophan is already around—waste of energy. b. Region 2 Region 3 and 4 would pair all the time even when there’s no Trp around and the ribosome stalls in region 1. Transcription would be terminated every time. c. Region 3 The Trp operon would not be regulated by attenuation, no hairpin could form (2 and 4 are not really complementary of each other). Transcription would not be terminated and it’s only regulated by trp repressor. d. Region 4 You’ll shift the UUU’s up, so when region 2 and 3 pair when ribosome stalls at region 1 due to low tryptophan, transcription is still terminated and no new tryptophan can be made. ...
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This document was uploaded on 09/12/2011.

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