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Which mechanism is NOT how RNA polymerase finds a promoter? Sliding along double stranded DNA molecule back and forth Transferring off and onto the same DNA molecule in a close vicinity Sliding on a single stranded noncoding DNA molecule Transferring between distant DNA segments located on the same molecule |
Sliding on a single stranded noncoding DNA molecule |
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Fill in the blank: Transcription occurs by _____________ in a _______? DNA replication; DNA loop Base pairing; bubble of unpaired DNA Reading DNA coding sequence; terminal loop Splicing; spliceosome |
Base pairing; bubble of unpaired DNA |
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What is the rate of transcription and the rate of translation? 40-50 nt/second; 15 amino acids/sec 50-60 nt/sec; 5 amino acids/sec 800 nt/sec; 45 amino acids/sec 40-50 nt/sec; 45 amino acids/sec |
40-50 nt/second; 15 amino acids/sec |
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What is the function of RNAP Clamp/Jaws? Causes DNA bend; bending of DNA helps melt the strands and flip bases in the template strand to be accessible Changes its conformation with each cycle of new nucleotide addition and keeps in contact with the growing RNA strand Blocks exit of the nascent RNA transcript and separates it from DNA template Clamp is initially out of position; after DNA has melted, it gets repositioned to keep DNA in the active site tighter |
Clamp is initially out of position; after DNA has melted, it gets repositioned to keep DNA in the active site tighter |
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Which statement is FALSE? T7 RNAP has a DNA binding region and the active site T7 RNAP recognizes 1000 phage promoters T7 RNAP uses a specificity loop to recognize the promoter T7 RNAP is a single peptide |
T7 RNAP recognizes 1000 phage promoters |
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Which subunits form the catalytic center of RNAP? Alpha and beta Alpha and sigma Beta prime and omega Beta and beta prime |
Beta and beta prime |
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What percentage of RNAP (RNA polymerase) is present in a storage form (core) and how much is actually in elongation mode? 50%; 50% |
50%; 25% |
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What statement is FALSE in regard to how RNA polymerase (RNAP) interacts with promoter DNA vs. non-promoter DNA? Core RNAP is stored bound to nonspecific DNA sites before sigma subunit binds to core Affinity of RNAP core for loose binding sites is too high to allow it to distinguish promoter sequence from non-promoter sequence Binding of sigma subunit reduces affinity of RNAP core for non-promoter sequence and increases affinity for promoter sequence Binding of sigma causes RNAP core to bind much tighter to non-promoter sequence |
Binding of sigma causes RNAP core to bind much tighter to non-promoter sequence |
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What is the most important stage of transcription for regulation? Intron splicing from pre-mRNA Elongation Initiation Termination |
Initiation |
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Which statement is FALSE? Flexible domain of T7 RNAP is part of a single protein and it binds to a major grove recognizing T7 phage promoter T7 RNAP is a minimal enzyme, which recognizes only one promoter T7 RNAP activity is stringently regulated T7 RNAP synthesizes RNA at a rate of 200 nt/sec |
T7 RNAP activity is stringently regulated |
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What is the function of the RNAP Rudder/Lid domain? Changes its conformation with each cycle of new nucleotide addition and keeps in contact with the growing RNA strand Limits the length of RNA/DNA hybrid; separates nascent RNA chain from DNA template as it exits through the exit pore Causes DNA bend; bending of DNA helps melt the strands and flip bases in the template strand to be accessible Clamp is initially out of position; after DNA has melted, it gets repositioned to keep DNA in the active site tighter |
Limits the length of RNA/DNA hybrid; separates nascent RNA chain from DNA template as it exits through the exit pore |
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Which statement is FALSE? Formation of the open promoter complex is irreversible and involves the melting of DNA by sigma The ternary complex is the least stable promoter complex The ternary complex is formed after the first phosphodiester bond is formed Formation of the closed promoter complex is reversible |
The ternary complex is the least stable promoter complex |
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What is a promoter? 3 untranslated region Control region for gene expression Signal to terminate transcription Gene template |
Control region for gene expression |
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RNAP changes conformation during the transcription cycle. Given below are stages of transcription and the amount of promoter DNA that is protected from DNase digestion: Which one is NOT correct? Ternary enzyme complex; -34/+20 Elongation complex: -75/+20 Closed promoter complex; -55/+1 Opened promoter complex; -55/+20 |
Elongation complex: -75/+20 |
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Which statement is FALSE? Core binds to random DNA through nonspecific, mostly electrostatic interactions, with a half-life of circa 60 min. The strongest bacterial promoter is that for Lac repressor and it reinitiates 1 time per second. The variation in the affinities of holoenzyme for promoter DNA may vary by 1,000,000-fold. Addition of sigma to the core lowers core affinity for random DNA 10,000-fold, and increases core affinity for promoter DNA circa 1,000-fold. |
The strongest bacterial promoter is that for Lac repressor and it reinitiates 1 time per second. |
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Which is the ratio of sigma to core, and how much RNAP is actually elongating? 3 sigma: 1 core; 25 % |
1 sigma: 3 core; 25% |
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In the transition from abortive cycling to elongation of transcription, which is TRUE? Sigma looses affinity for the promoter DNA and RNAP core, and pressure of growing RNA chain dislodges sigma from the RNA exit pore. DNA scrunching contributes to a stronger binding between RNAP core and sigma 3.2 subdomains. The wall separates RNA transcript from DNA template, thus facilitating abortive cycling . Sigma 1.1 subdomain pretends to be a promoter and confuses RNAP, which looses its mind, does not know what to do, and aborts transcription. |
Sigma looses affinity for the promoter DNA and RNAP core, and pressure of growing RNA chain dislodges sigma from the RNA exit pore. |
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How does RNAP holoenzyme recognize different gene promoters? By changing the conformation of it active site/DNA binding site By using hairpin structure of the sigma 3.2 subdomain, which touches the first nucleotide in the active site By binding alternative specialized sigma subunits, which recognize different cis-element sequences and various configurations of the promoter By the removal of the sigma 1.1 subdomain from its active site, thus making specific space for DNA binding available |
By binding alternative specialized sigma subunits, which recognize different cis-element sequences and various configurations of the promoter |
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Which is common for both DNA and RNA polymerases? Can recognize promoter sequences Can slide along DNA Can melt the DNA duplex Can scrunch DNA |
Can slide along DNA |
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Which statement is FALSE? If domain 1.1 is deleted, sigma will bind promoter sequence. Domain 1.1 is blocking the region where DNA is located when the promoter is melted. Domain 1.1 prevents sigma from binding promoter without first binding the RNAP core. Domain 1.1 helps melt the promoter sequence and create an open promoter complex. |
Domain 1.1 helps melt the promoter sequence and create an open promoter complex. |
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In the GST pull-down experiment, the entire bacterial core and distal promoter was used as a radioactive probe pre-bound with isolated GST-alpha CTD subunit. The outcome was a very sharp decline in the pulled-down radioactivity. What competitor was used to obtain this effect? Core promoter sequence UP element Upstream activating sequence I, II and III Discriminator |
UP element |
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Which is NOT the function of sigma 3.2? Pulls DNA template into active center (scrunching) Positions the first nucleotide in nascent RNA Facilitates promoter clearance by RNAP (displacement by the nascent RNA) Blocks exit channel for the RNA |
Pulls DNA template into active center (scrunching) |
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What is the function of heparin in EMSA (gel retardation) and DNase I footprinting assays? Heparin changes the conformation of enzymes/proteins involved in these DNA:protein binding techniques. Heparin acts like sigma 3.2 subdomain and blocks the RNA exit pore within enzyme active site. Heparin stabilizes binding of RNAP to the double stranded DNA. Heparin is negatively charged; acts as a competitor to knock off any RNAP that is not in a stable open complex formation |
Heparin is negatively charged; acts as a competitor to knock off any RNAP that is not in a stable open complex formation |
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Listed below are sigma subdomains followed by its function. Which of the following pairings has an incorrect function assigned? 3.0; contacts upstream sequence in the extended TATA box 1.2; contacts discriminator element 2.1/2.2; recognizing conserved region in RNAP core 2.3; contacts alpha CTD |
2.3; contacts alpha CTD |
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What is the function of unlabeled competitor DNA in various footprint assays? Stops double stranded DNA from melting, thus making protein binding to DNA more feasible. Substitutes for labeled probe originally bound to the protein of interest (outcompeting it) and allows for analysis of kinetics (off-constants) and/or specificity/strength of binding. Protects first nucleotides as they form first phosphodiester bond in DNA footprinting assays. Helps stabilize double stranded DNA structure and makes protein:DNA binding stronger. |
Substitutes for labeled probe originally bound to the protein of interest (outcompeting it) and allows for analysis of kinetics (off-constants) and/or specificity/strength of binding. |
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What statement is FALSE? Sigma 54 is very unusual since it can bind promoter in the absence of RNAP core. Sigma 54 activates the constitutive promoter of nitrogen starvation gene glnA. Sigma 54 does not have region 1.1. Some NIF genes have dual promoters, which can be turned on either by sigma 70 or sigma 54. |
Sigma 54 activates the constitutive promoter of nitrogen starvation gene glnA. |
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How can you explain the effect of temperature on the dissociation of RNAP holoenzyme from promoter sequence? Both, closed and open promoter complexes co-exist at 37 deg and this contributes to their long half life. Open promoter complex is not formed at 15 deg; closed promoter complex is the most stable overall and lasts up to hundreds of hours as shown by the bottom curve. Transition from closed to open promoter complex happens readily at higher temperatures (DNA melting); open promoter complex is most stable at 37 deg and the least stable at 15 deg. Open promoter complex does not occur at 37 deg (no DNA melting), therefore, the half-life of the closed complex is the longest at higher temperatures. |
Transition from closed to open promoter complex happens readily at higher temperatures (DNA melting); open promoter complex is most stable at 37 deg and the least stable at 15 deg. |
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What technique can be used to evaluate protein: DNA contacts on the single-stranded DNA? Filter binding assay |
DMS footprinting |
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What is the source of energy that allows an intrinsic terminator to function? kinetic energy of the moving RNA polymerase thermal energy due to the temperature ATP provided during transcription NADP provided by metabolism |
thermal energy due to the temperature |
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What is the role of the hairpin in an intrinsic terminator? It alters the conformation of the active site of RNA polymerase causing the synthesis reaction to run backwards. The hairpin produces DNA scrunching of the non-template strand resulting in a misalignment of the RNA:DNA hybrid portion of the transcription bubble. It causes the RNA polymerase to pause so that the stretch of U:As are positioned in the RNA:DNA region of the transcription bubble. It causes the polymerase to stall, which requires either processing of the 3-end or termination to occur. |
It causes the RNA polymerase to pause so that the stretch of U:As are positioned in the RNA:DNA region of the transcription bubble. |
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Polarity: what condition must occur in order for a spontaneous STOP mutation to activate a Rho terminator embedded within a coding region? The STOP mutation must be located between the rut site and the hairpin within the same coding region that contains the rho terminator. The STOP mutation must be located upstream of the rut site within the same coding region that contains the rho terminator. The STOP mutation must occur in the coding region immediately upstream of the coding region that contains the rho terminator. The STOP mutation must be located after the hairpin of the rho terminator. |
The STOP mutation must be located upstream of the rut site within the same coding region that contains the rho terminator. |
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What is required for a Rho-dependent terminator? an inverted repeat rich in Cs downstream of a rut site an inverted repeat downstream of a very C-rich stretch of mRNA an inverted repeat upstream of a stretch of Us on the mRNA an RNA hairpin upstream of a rut site an inverted repeat in mRNA downstream of a site rich in Gs. |
an inverted repeat downstream of a very C-rich stretch of mRNA |
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True/False: Normal polycistronic RNA’s can be terminated by either intrinsic or factor-dependent terminators located downstream of the last coding region (in the 3′-untranslated region; 3′-UTR). |
True |
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Which of the following statements are TRUE regarding "€œPolarity" in terms of bacterial gene expression? All of the coding regions have the same orientation to form a polycistronic mRNA. Polarity occurs when all coding regions of a polycistronic transcript are experiencing heavy ribosome traffic. It is when termination in a coding region located near the 5-end of a polycistronic mRNA causes the loss of both transcriptional and translational expression of all genes that follow it. It occurs when a rho-dependent terminator is located in the coding region of the last gene in a polycistronic mRNA. |
It is when termination in a coding region located near the 5-end of a polycistronic mRNA causes the loss of both transcriptional and translational expression of all genes that follow it. |
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Which protein or domain within bacterial RNA polymerase contacts DNA using minor groove contact? Sigma 70 domain 2.3 Sigma 70 domain 4.2 The CTD of the alpha subunit The NTD of alpha |
The CTD of the alpha subunit |
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Which element of the bacterial promoter binds the CTD of the alpha subunit? -35 site |
UP |
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At the promoter for the Ara pBAD operon, how does the DNA loop cause repression? by creating torsional strain in the DNA at the promoter which prevents the transition from closed to open promoter complex. by blocking the binding of AraC protein by blocking access to the core promoter by RNA polymerase by causing RNA polymerase to bind to the "dead-end" promoter |
by blocking access to the core promoter by RNA polymerase |
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Which of the following is FALSE regarding CAP? Most CAP-dependent promoters have a good match to the -35 site consensus. Class II CAP-dependent promoters always have a single binding site for the CAP dimer located at position -41.5. In the Lac promoter, CAP binding increases formation of the closed promoter complex. |
Most CAP-dependent promoters have a good match to the -35 site consensus. |
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In Class II CAP-dependent promoters, it is possible to have CAP located at both the -41.5 and -61 positions. True |
false |
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In the Cytr promoter (purine metabolism), the Cytr regulatory protein binds in between the two CAP dimers. This is an example of what type of regulation? (best answer) Attenuation |
Anti-activation |
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A strain of E. coli has two copies of the lac operon: one in the chromosome and the other on a plasmid. You observe that when you grow up one of the colonies that it produces half as much of all of the proteins encoded by the lac operon when glucose is absent and lactose is present (the operon should be ON). Which of the following statements is the most likely explanation of these results? A mutation occurred in an Operator site (O1) for one of the operons. A mutation occurred in the core promoter (-10 or -35 sites) of one of the two operons. A mutation occurred that deleted the gene encoding CAP. A mutation occurred in the gene encoding the Lac repressor. |
A mutation occurred in the core promoter (-10 or -35 sites) of one of the two operons. |
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The Lac repressor bound to IPTG prefers Operator DNA 10,000 times more than non-Operator DNA. True |
true |
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In base excision repair, the uracil glycosylase enzyme corrects DNA damage by biased replacement of ________ to ______? Methyl-G; unmethylated G Depurinated nucleotide; insertion of nucleotide T-T dimer; A-A U-G; C-G |
U-G; C-G |
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Which statement is FALSE in regard to eukaryotic base excision repair BER? In BER, polymerase delta and epsilon replaces long stretch of nucleotides, which is 1500-9000 bases. BER starts by directly removing a damaged base from DNA. The nature of base removal in BER determines which polymerase is recruited to the DNA by AP1 protein, either delta epsilon or beta. Base removal triggers the removal and replacement of a stretch of polynucleotide, using either long patch or short path repair. |
In BER, polymerase delta and epsilon replaces long stretch of nucleotides, which is 1500-9000 bases. |
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Which of the following is NOT the activity of Mfd protein? It recruits the uvrAB proteins and directs repair to the damaged template strand. It binds to stalled RNA polymerase. It degrades RNA polymerase It displaces RNA polymerase. |
It binds to stalled RNA polymerase. |
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In the uvr excision repair system in E. coli, which enzyme routinely synthesizes DNA to replace the excised strand? |
DNA polymerase I |
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In eukaryotic BER, long patch repair replaces _______ nucleotides, and a short patch repair replaces ______ nucleotides? |
2-10; 1 |
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What step is NOT part of excision repair? Dam methylation Incision by endonuclease Damage recognition Excision by exonuclease |
Dam methylation |
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What protein is uniquely linked to transcription and DNA repair in E. coli? |
Mfd |
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Which statement is TRUE in regard to eukaryotic transcription-linked nucleotide excision repair NER? Global genome repair pathway does not share any common steps with transcription-linked nucleotide excision repair. Yeast Rad4 recognizes damaged DNA and this process involves flipping out thymine dimers. In eukaryotic nucleotide excision repair DNA polymerase V synthesized DNA after damage is removed. The large subunit of RNA polymerase is degraded; TFIIH remains and recruits XP proteins to repair DNA damage. |
The large subunit of RNA polymerase is degraded; TFIIH remains and recruits XP proteins to repair DNA damage. |
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Which step is NOT a part of the SOS repair? LexA repressor binds to the SOS boxes in repair genes promoters and represses their activity under normal conditions RecA cuts LexA thus inhibiting the inhibitors activity Repair uvrB gene is regulated by dual promoter, one constitutive and one inducible UV exposure activates RecA |
RecA cuts LexA thus inhibiting the inhibitors activity |
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In dam methylation mismatch repair, which protein recognizes the mismatch? |
MutS |
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What statement is FALSE? In uvr (excision repair) mutants, additional RecA mutation eliminates all remaining repair capabilities. Replication in uvr-/recA- double mutant results in production of DNA fragments as long as distance between individual T-T dimers. The uvr-/recA- double mutants can tolerate up to 50 thymine dimers. In retrieval repair, uvr (excision repair) and rec (recombination) pathways are interconnected. |
The uvr-/recA- double mutants can tolerate up to 50 thymine dimers. |
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In dam methylation mismatch repair, which protein acts as the nuclease and nicks the unmethylated strand? |
MutH |
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When two bases are mismatched, how does the cell know which base to repair? UvrA and B system recognizes the mismatch and knows which base to cut out. Dam methylation system marks the GATC sequence in the original strand and unmethylated daughter strand is repaired. Cell is not able to distinguish new and old DNA strands, hence it does not repair a mismatch. RecBC recognizes mismatch in the daughter strand and corrects it by replacing DNA fragment using the Okazaki fragment. |
Dam methylation system marks the GATC sequence in the original strand and unmethylated daughter strand is repaired. |
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In dam methylation mismatch repair, what is the signal that causes MutH to join the Mut complex and to nick the unmethylated strand? Recruitment of MutL Recognition of the GATC site by MutS DNA excision by exonuclease VII or I UvrD helicase DNA unwinding |
Recognition of the GATC site by MutS |
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In dam methylation mismatch repair, which protein translocates to the GATC site and is able to bind to two sites simultaneously, thus making a DNA loop? |
MutS |
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Which SOS repair proteins are motivated by RecA to self-cleave, which protein activates them? LexA and UmuD2C LexA and DNA polymerase I UvrA, uvrB, and uvrC RecA and uvrD |
LexA and UmuD2C |
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Which induction conditions do NOT trigger the SOS response? Thymine shortage Crosslinking and alkylating chemical agents UV irradiation Mismatch mutations |
Mismatch mutations |
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Which function is NOT an activity of RecA? Normal recombination Single strand-exchange of recombination-repair Act as a nuclease, which directly cleaves LexA repressor Induces latent protease activity in its target proteins |
Act as a nuclease, which directly cleaves LexA repressor |
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Which step is NOT a part of the SOS repair? RecA activates LexA to self-cleave and this inhibits repressor activity of LexA Constitutive promoter of uvrB is repressed by LexA under normal conditions All targets of RecA are cleaved at the dipeptide Ala-Gly Inducible promoter of uvrB repair gene is repressed by LexA under normal conditions |
Constitutive promoter of uvrB is repressed by LexA under normal conditions |
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What statement is FALSE? Msh2/Msh6 complex binds single base mismatches, while other proteins do the repairing. MSH repair system in yeast is homologous to the E. coli MutS/L system. Roles of MutS and MutL are completely different between bacterial and MSH eukaryotic proteins. Msh2/Msh3 complex binds DNA loops resulting from replication slippage. |
Roles of MutS and MutL are completely different between bacterial and MSH eukaryotic proteins. |
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What step is NOT a part of the SOS repair? LexA gene itself is de-repressed when LexA is self-cleaved. RecA gene is induced approximately 50-fold, which results in protein induction of 1200 x50=60,000 molecules/cell upon UV. After damage, RecA is continuously activated and, therefore, SOS response is irreversible. LexA repressor protein normally keeps levels of LexA, RecA, and excision repair enzymes low. |
After damage, RecA is continuously activated and, therefore, SOS response is irreversible. |
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Retrieval or recombination-repair systems in E. coli do NOT use these proteins (______) to perform these (_______) functions? RecBC and RecF; help associate RecA with single stranded DNA RecA and SSB; bind to double stranded DNA RecF, RecO and RecR; to repair gaps due to T-T dimers left in after replication in daughter strand RecBC and RecA; restart stalled replication forks |
RecA and SSB; bind to double stranded DNA |
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Mutations in genes that encode __________ represent the Mutator phenotype. proteins that direct transcription proteins responsible of ligating DNA repair system proteins, or fidelity of replication proteins proteins that participate in recombination and chiasmata formation |
repair system proteins, or fidelity of replication proteins |
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Which statement is FALSE in regard to recombination repair of the replication errors? A replication fork may stall when it encounters a mismatch. A stalled fork may reverse by pairing between the two daughter strands. The structure of the stalled fork is the same as Holliday junction and may be converted to a duplex and ds breaks by resolvases. A stalled replication fork may restart after repairing the damage and use helicase to move the fork forward. |
A replication fork may stall when it encounters a mismatch. |
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Which gene is NOT an example of a Mutator gene? MutH, S, L DnaQ-Polymerase III epsilon subunit FEN1 endonuclease uvrD |
FEN1 endonuclease |
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Default repair systems show bias in error correction. Which statement is FALSE? MutS/L removes T from GT and CT mismatch pairs, and is not dependent on GATC methylation. MutM removes oxidated dGTP that is paired with C, but is not able to hydrolyze it as a free nucleotide . MutS/L removes T from GT and CT mismatch pairs, and this depends on GATC methylation. MutY removes A from CA and GA mismatches, and does not use MutS/L system. |
MutS/L removes T from GT and CT mismatch pairs, and this depends on GATC methylation. |
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In yeast mismatch repair system, which proteins ______recognize mismatches, and which are specificity factors_______? Msh2; Msh3 and Msh6 Msh6; Msh2 and Msh5 Msh5; Msh3 and Msh4 Msh3; Msh5 and Msh4 |
Msh2; Msh3 and Msh6 |
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In dam methylation mismatch repair, what is the signal that causes MutH to nick the unmethylated strand? Contact with MutL UvrD helicase DNA unwinding Recognition of the mismatched bases by MutL DNA excision by exonuclease VII or I |
Contact with MutL |
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Which of the following best describes the SOS repair system? By-pass or tolerance system that allows DNA replication across damage areas at the cost of fidelity. Repair system based on the variation of the nonhomologous end joining mechanism. Repair system that recognizes, which strand is the parental strand, and which is daughter, and fixes mutations in the new strand. It is a system in eukaryotes, which uses photoreactivation to directly repair thymine dimers. |
By-pass or tolerance system that allows DNA replication across damage areas at the cost of fidelity. |
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What activity is encoded by MutY and what does it do? Adenosine glycosylase; creates apurinic site Reverse transcriptase; synthesizes complementary DNA strand using RNA as a template DNA polymerase I; repairs DNA Dam methylase; methylates GATC site |
Adenosine glycosylase; creates apurinic site |
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What is the function of RNAP Wall? Blocks exit of the nascent RNA transcript Causes DNA bend; bending of DNA helps melt the strands and flip bases in the template strand to be accessible Protects first nucleotides as they form first phosphodiester bond Helps stabilize double stranded DNA structure |
Causes DNA bend; bending of DNA helps melt the strands and flip bases in the template strand to be accessible |
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What is abortive cycling? When RNAP transcribes 2-9 nt, then restarts again and does not leave the promoter When RNAP pools template strand into its active site When RNA polymerase hops on and off the DNA molecule searching for promoter sequence When free RNAP core looses sigma subunit and clears the promoter |
When RNAP transcribes 2-9 nt, then restarts again and does not leave the promoter |
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What is the function of RNAP Bridge? Dynamically changes its conformation with each cycle of new nucleotide addition; keeps in contact with the growing RNA strand as enzyme moves forward Causes DNA bend; bending of DNA helps melt the strands and flip bases in the template strand to be accessible Blocks exit of the nascent RNA transcript Protects first nucleotides as they form first phosphodiester bond |
Dynamically changes its conformation with each cycle of new nucleotide addition; keeps in contact with the growing RNA strand as enzyme moves forward |
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What statement is FALSE in regard to how RNA polymerase (RNAP) interacts with promoter DNA vs. non-promoter DNA? By reducing the stability of non-promoter complexes sigma allows RNAP core to find promoter sequence much faster When sigma is released from the RNAP, the core and RNA transcript form a ternary complex with DNA By increasing the stability of non-promoter complexes sigma allows RNAP core to slide along DNA much faster, and find promoter sequence easier When sigma is released from the RNAP, the core and RNA transcript are bound extremely tight onto DNA until termination |
By increasing the stability of non-promoter complexes sigma allows RNAP core to slide along DNA much faster, and find promoter sequence easier |
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Which statement is FALSE? When the open promoter complex is formed, sigma 1.1 is displaced from DNA channel and replaced by DNA. In the free holoenzyme of bacterial RNAP, the N-terminal domain of sigma-70 blocks DNA channel by mimicking interaction with DNA. As bent DNA template is in the RNAP active site, it presses on the sigma 3.2 domain, thus releasing it from the holoenzyme. As DNA enters the channel in the RNAP active site, it begins to bend 90 degrees and melt as the strands open up close to start of transcription. |
As bent DNA template is in the RNAP active site, it presses on the sigma 3.2 domain, thus releasing it from the holoenzyme. |
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What is DNA scrunching? Occurs during transcription, when sigma 1.1 mimics DNA in the DNA channel, and the actual DNA molecule tries to enter the channel Occurs during transcription when length of the nascent transcript is limited by RNAP rudder to 6-9 nts Occurs during transcription when double stranded DNA sinks into the channel and hits the RNAP active site wall Occurs during transcription and abortive cycling; 6-9 nts of DNA template are pulled into the RNAP active site where the template is bunched up |
Occurs during transcription and abortive cycling; 6-9 nts of DNA template are pulled into the RNAP active site where the template is bunched up |
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Sigma reduces affinity of RNAP core for non-promoter sequences _______fold and increases affinity for specific promoter DNA _______fold? 10,000; 1,000 15,000; 800 1,000; 10,000 5,000; 500 |
10,000; 1,000 |
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Which subunit of bacterial RNAP is required for promoter specificity? Omega Sigma Beta Alpha |
Sigma |
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What is the size of the transcription bubble and RNA/DNA hybrid in it? |
12-14 nt; 8-9 nt |
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How many Mg2+ ions are present in the RNAP active site during transcription? 1 0 5 2 |
2 |
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What two enzymes help RNAP to eliminate supercoiling generated by the mechanism of transcription? Helicase; DNA polymerase I Reverse transcriptase; topoisomerase Gyrase; topoisomerase DNA polymerase I; gyrase |
Gyrase; topoisomerase |
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What are three roles of alpha subunit? Catalytic center, enzyme assembly, promoter specificity Enzyme assembly, catalytic center, recruitment of trans-activators Structural, promoter specificity, recruitment of trans-activators Enzyme assembly, promoter recognition, interactions with transcription activators |
Enzyme assembly, promoter recognition, interactions with transcription activators |
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How many different types of subunits are there in bacterial RNAP holoenzyme and what are their names? 4; alpha, beta, beta prime, omega 3; alpha, beta, sigma 6; alpha, beta, beta prime, omega, and sigma |
6; alpha, beta, beta prime, omega, and sigma |
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Which protein(s) helps RNAP to recover from a stall caused by the temporary shortage of nucleotides, and how? RNAP core active site; by scrunching more DNA template into the catalytic site. GreA and NtrC; by pushing nascent RNA out of the exit pore, thus helping to convert scrunched RNA/DNA hybrid. NtrC; using ATP it provides the energy to continue elongation. GreA and GreB; reposition Mg2+ ions in the active site, which makes RNAP cleave trailing end off nascent RNA to align it correctly in the catalytic site. |
GreA and GreB; reposition Mg2+ ions in the active site, which makes RNAP cleave trailing end off nascent RNA to align it correctly in the catalytic site. |
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How is the bacterial core promoter recognized by RNAP? Discriminator element binds subdomain 3.2 and this helps melt the promoter Strong contacts between sigma and alpha CTDs contribute to promoter recognition Upstream promoter element recruits 2 copies of omega subunit Through contacts of sigma subdomains 2.4/2.3 and 4.2 and cis-acting promoter elements |
Through contacts of sigma subdomains 2.4/2.3 and 4.2 and cis-acting promoter elements |
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What is NOT the activity of the antibiotic rifampicin in fighting tuberculosis? Binds close to the active site and blocks the formation of RNA beyond 2-3 nts. Prevents exit of growing RNA chain from RNAP itself. It is in contact with nascent RNA past nucleotides 1-3. Can be easily dislodged by other antibiotics. |
Can be easily dislodged by other antibiotics. |
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Which sigmas can be used to transcribe genes during some stress conditions? Sigma S, sigma 32 Sigma A, sigma E Sigma fecl, sigma E Sigma 70, sigma F |
Sigma S, sigma 32 |
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Sigma 54 has a safety check that prevents it from continuously expressing the glutamine synthase gene. What is the basis for this control? DNA looping and ATP hydrolysis inhibit sigma 54 from activating NIF genes under control conditions. Sigma 54 is unable to melt the promoter without added ATP and a helper protein NtrC bound to the enhancer element at least 70 bp away. Negative factor NtrC inhibits transcription regulated by sigma 54 under control conditions. GreA and GreB proteins are necessary to activate sigma 54. |
Sigma 54 is unable to melt the promoter without added ATP and a helper protein NtrC bound to the enhancer element at least 70 bp away. |
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Listed below are sigma subdomains followed by its function. Which of the following pairings has an incorrect function assigned? Extended helix 2.3/2.4; recognizes and melts -10 element H-T-H 1.1; blocks the exit pore of RNA Hairpin 3.2; keeps contact with the first nucleotide of the nascent RNA until abortive cycling or elongation occurs H-T-H 4.2; recognizes -35 element |
H-T-H 1.1; blocks the exit pore of RNA |
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In the diagram below showing bacterial RNA polymerase, what is indicated by the circle and arrow? (orange in middle) non-coding strand |
newly transcribed RNA |
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In the diagram below showing bacterial RNA polymerase, what is indicated by the circle and arrow? (red) template strand |
template strand |
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Termination: The diagram below shows a GC-rich hairpin sequence of DNA followed by a stretch of A’s. Why is this NOT an intrinsic terminator? AAAA’s on 5′ top strand |
The As are on the coding strand. |
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What is the energy source that provides movement to Rho? temperature of the cell It is derived from breaking the phosphodiester bonds during transcription. sunlight ATP hydrolysis |
ATP hydrolysis |
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In the diagram below showing bacterial RNA polymerase, what is indicated by the circle and arrow? (yellow) |
coding strand |
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In the diagram below, which configuration would result in NO expression of RNA encoding regions C and D? 1: stop in B before rut 2: stop in A 3: Stop in D 4: Stop in B between rut and hairpin |
1: stop in B before rut |
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In the diagram below, which configuration would result in NO expression of coding region D at the translational level, but there would still be mRNA present? (Rut in mid B, hairpin late B) 1: stop in B before rut 2: stop in A 3: Stop in D 4: Stop in B between rut and hairpin |
3: Stop in D |
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Which of the following statements are TRUE regarding "Polarity" in terms of bacterial gene expression? Polarity occurs when all coding regions of a polycistronic transcript are experiencing heavy ribosome traffic. It is when termination in a coding region located near the 5-end of a polycistronic mRNA causes the loss of both transcriptional and translational expression of all genes that follow it. All of the coding regions have the same orientation to form a polycistronic mRNA. It occurs when a rho-dependent terminator is located in the coding region of the last gene in a polycistronic mRNA. |
It is when termination in a coding region located near the 5-end of a polycistronic mRNA causes the loss of both transcriptional and translational expression of all genes that follow it. |
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Which of the following represent the association constant of a DNA binding protein? [free repressor-DNA complex]/[free repressor] x [total cellular DNA] [free repressor] x [total cellular DNA]/ [free repressor-DNA complex] [free repressor-DNA complex] minus [total cellular DNA]/[bound repressor] [bound repressor] x [free repressor]/[total cellular DNA] |
[free repressor-DNA complex]/[free repressor] x [total cellular DNA] |
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The Lac repressor is present in only 10 copies per cell in E. coli. Why is this so? The promoter has a very poor match to the Sigma 70 consensus, and the mRNA has a very short 5′-UTR (untranslated region). The mRNA has no 7′-methyl G cap at the 5′-end. The mRNA has an exceptionally long 5′-UTR which inhibits the ribosome in its search for the Shine Dalgarno (ribosome binding site). The mRNA has a stop codon that terminates translation early, and the promoter requires CAP. |
The promoter has a very poor match to the Sigma 70 consensus, and the mRNA has a very short 5′-UTR (untranslated region). |
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Which of the following pathways are regulated by repression caused by the accumulation end-product? Lactose degradation Utilization of arabinose as a nutrient substrate Glucose catabolism Tryptophan biosynthesis |
Tryptophan biosynthesis |
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At the promoter for the lac operon, how does the DNA loop block transcription? by blocking the binding of CAP by creating torsional strain in the DNA at the promoter which prevents the transition from closed to open promoter complex. by blocking access to the core promoter by RNA polymerase causes excessive abortive cycling |
by creating torsional strain in the DNA at the promoter which prevents the transition from closed to open promoter complex. |
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