|
In the uvr excision repair system in E. coli, which enzyme unwinds damaged DNA?
UvrC
UvrD
UvrB
UvrA
|
UvrD
|
|
In eukaryotic BER, long patch repair replaces _______ nucleotides, and a short patch repair replaces ______ nucleotides?
700-900; 24
2-10; 1
500-700; 5
1500-9000; 12
|
2-10; 1
|
|
Which statement is FALSE in regard to eukaryotic base excision repair BER?
Base removal triggers the removal and replacement of a stretch of polynucleotide, using either long patch or short path repair.
BER starts by directly removing a damaged base from DNA.
In BER, polymerase delta and epsilon replaces long stretch of nucleotides, which is 1500-9000 bases.
The nature of base removal in BER determines which polymerase is recruited to the DNA by AP1 protein, either delta epsilon or beta.
|
In BER, polymerase delta and epsilon replaces long stretch of nucleotides, which is 1500-9000 bases.
|
|
What step is NOT part of excision repair?
Dam methylation
Incision by endonuclease
Excision by exonuclease
Damage recognition
|
Dam methylation
|
|
In the uvr excision repair system in E. coli, which steps DO NOT require hydrolysis of ATP?
UvrA release from damaged DNA
UvrA recognition of damaged nucleotide
UvrD helicase DNA unwinding
UvrBC nicking on either side of damaged nucleotide
|
UvrA recognition of damaged nucleotide
|
|
What protein is uniquely linked to transcription and DNA repair in E. coli?
Mfd
Dam methylase
DNA polymerase I
Reverse transcriptase
|
Mfd
|
|
Which statement is TRUE in regard to eukaryotic transcription-linked nucleotide excision repair NER?
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.
Global genome repair pathway does not share any common steps with transcription-linked nucleotide excision repair.
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.
|
|
What is the enzyme in bacteria that directly photoreactivates pyrimidine dimers?
MutS, L
AP1 endonuclease
FEN1 endonuclease
Photolyase phr
|
Photolyase phr
|
|
What protein is uniquely linked to transcription and DNA repair in eukaryotes?
XPC recognition protein
TFIIH helicases XPB and XPD
XPG
XPF/ERCC1
|
TFIIH helicases XPB and XPD
|
|
In base excision repair, the uracil glycosylase enzyme corrects DNA damage by biased replacement of ________ to ______?
Methyl-G; unmethylated G
U-G; C-G
T-T dimer; A-A
Depurinated nucleotide; insertion of nucleotide
|
U-G; C-G
|
|
In the uvr excision repair system of E. coli, long patch repair replaces _______ nucleotides, and a short patch repair replaces ______ nucleotides?
2-10; 1
1500-9000; 12
700-900; 24
500-700; 5
|
1500-9000; 12
|
|
Which of the following is NOT the activity of Mfd protein?
It degrades RNA polymerase
It displaces RNA polymerase.
It recruits the uvrAB proteins and directs repair to the damaged template strand.
It binds to stalled RNA polymerase.
|
It degrades RNA polymerase
|
|
In the uvr excision repair system in E. coli, which enzyme routinely synthesizes DNA to replace the excised strand?
DNA polymerase I
DNA polymerase III
RNA polymerase II
DNA polymerase V
|
DNA polymerase I
|
|
Which type of DNA damage does NOT result in stopped replication or stopped transcription?
Thymine dimers
Base alkylation; DNA nicking
Depurination
Mismatch; deamination
|
Mismatch; deamination
|
|
In the uvr excision repair system in E. coli, which enzyme recruits uvrC?
DNA pol I
UvrD
UvrA
UvrB
|
UvrB
|
|
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
RecBC and RecA; restart stalled replication forks
RecF, RecO and RecR; to repair gaps due to T-T dimers left in after replication in daughter strand
RecA and SSB; bind to double stranded DNA
|
RecA and SSB; bind to double stranded DNA
|
|
Mutations in genes that encode __________ represent the Mutator phenotype.
proteins responsible of ligating DNA
proteins that direct transcription
proteins that participate in recombination and chiasmata formation
repair system proteins, or fidelity of replication proteins
|
repair system proteins, or fidelity of replication proteins
|
|
Which SOS repair proteins are motivated by RecA to self-cleave, which protein activates them?
LexA and UmuD2C
RecA and uvrD
LexA and DNA polymerase I
UvrA, uvrB, and uvrC
|
LexA and UmuD2C
|
|
Which statement is FALSE in regard to recombination repair of the replication errors?
A stalled fork may reverse by pairing between the two daughter strands.
A replication fork may stall when it encounters a mismatch.
A stalled replication fork may restart after repairing the damage and use helicase to move the fork forward.
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 replication fork may stall when it encounters a mismatch.
|
|
What activity is encoded by MutY and what does it do?
DNA polymerase I; repairs DNA
Dam methylase; methylates GATC site
Reverse transcriptase; synthesizes complementary DNA strand using RNA as a template
Adenosine glycosylase; creates apurinic site
|
Adenosine glycosylase; creates apurinic site
|
|
What statement is FALSE?
Roles of MutS and MutL are completely different between bacterial and MSH eukaryotic proteins.
MSH repair system in yeast is homologous to the E. coli MutS/L system.
Msh2/Msh3 complex binds DNA loops resulting from replication slippage.
Msh2/Msh6 complex binds single base mismatches, while other proteins do the repairing.
|
Roles of MutS and MutL are completely different between bacterial and MSH eukaryotic proteins.
|
|
Which gene is NOT an example of a Mutator gene?
FEN1 endonuclease
MutH, S, L
DnaQ-Polymerase III epsilon subunit
uvrD
|
FEN1 endonuclease
|
|
Which function is NOT an activity of RecA?
Act as a nuclease, which directly cleaves LexA repressor
Single strand-exchange of recombination-repair
Normal recombination
Induces latent protease activity in its target proteins
|
Act as a nuclease, which directly cleaves LexA repressor
|
|
In dam methylation mismatch repair, which protein recognizes the mismatch?
MutH
MutS
MutL
DNA polymerase III
|
MutS
|
|
Which step is NOT a part of the SOS repair?
All targets of RecA are cleaved at the dipeptide Ala-Gly
Constitutive promoter of uvrB is repressed by LexA under normal conditions
Inducible promoter of uvrB repair gene is repressed by LexA under normal conditions
RecA activates LexA to self-cleave and this inhibits repressor activity of LexA
|
Constitutive promoter of uvrB is repressed by LexA under normal conditions
|
|
In dam methylation mismatch repair, which protein acts as the nuclease and nicks the unmethylated strand?
MutL
MutS
MutY
MutH
|
MutH
|
|
Which step is NOT a part of the SOS repair?
Repair uvrB gene is regulated by dual promoter, one constitutive and one inducible
UV exposure activates RecA
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
|
RecA cuts LexA thus inhibiting the inhibitors activity
|
|
In dam methylation mismatch repair, what is the signal that causes MutH to nick the unmethylated strand?
UvrD helicase DNA unwinding
DNA excision by exonuclease VII or I
Recognition of the mismatched bases by MutL
Contact with MutL
|
Contact with MutL
|
|
Which of the following best describes the SOS repair system?
It is a system in eukaryotes, which uses photoreactivation to directly repair thymine dimers.
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.
By-pass or tolerance system that allows DNA replication across damage areas at the cost of fidelity.
|
By-pass or tolerance system that allows DNA replication across damage areas at the cost of fidelity.
|
|
When two bases are mismatched, how does the cell know which base to repair?
Dam methylation system marks the GATC sequence in the original strand and unmethylated daughter strand is repaired.
UvrA and B system recognizes the mismatch and knows which base to cut out.
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.
|
|
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.
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.
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.
|
|
In yeast mismatch repair system, which proteins ______recognize mismatches, and which are specificity factors_______?
Msh3; Msh5 and Msh4
Msh5; Msh3 and Msh4
Msh2; Msh3 and Msh6
Msh6; Msh2 and Msh5
|
Msh2; Msh3 and Msh6
|
|
What step is NOT a part of the SOS repair?
After damage, RecA is continuously activated and, therefore, SOS response is irreversible.
RecA gene is induced approximately 50-fold, which results in protein induction of 1200 x50=60,000 molecules/cell upon UV.
LexA repressor protein normally keeps levels of LexA, RecA, and excision repair enzymes low.
LexA gene itself is de-repressed when LexA is self-cleaved.
|
After damage, RecA is continuously activated and, therefore, SOS response is irreversible.
|
|
Mismatch repair: Which arrow designates MutS?
(
1 (yellow on T)
2 (purple on mCTAG and GATC)
3 (Blue blob)
4 (yellow on G)
|
4 (yellow on G)
|
|
What statement is FALSE?
Replication in uvr-/recA- double mutant results in production of DNA fragments as long as distance between individual T-T dimers.
In uvr (excision repair) mutants, additional RecA mutation eliminates all remaining repair capabilities.
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.
|
|
Which induction conditions do NOT trigger the SOS response?
Crosslinking and alkylating chemical agents
UV irradiation
Mismatch mutations
Thymine shortage
|
Mismatch mutations
|
|
Mismatch repair: Which arrow designates UvrD?
MutSLH mismatch repair-Exam fig.png
1 (yellow on T)
2 (purple on mCTAG and GATC)
3 (Blue blob)
4 (yellow on G)
|
3 (Blue blob)
|
|
Mismatch repair: Which arrow designates the protein that determines which strand will be replaced?
(Identifies the new strand.)
1 (yellow on T)
2 (purple on mCTAG and GATC)
3 (Blue blob)
4 (yellow on G)
|
…
|
|
Mismatch repair: Which arrow designates MutH?
1 (yellow on T)
2 (purple on mCTAG and GATC)
3 (Blue blob)
4 (yellow on G)
|
2 (purple on mCTAG and GATC)
|
|
What is the function of RNAP Bridge?
Protects first nucleotides as they form first phosphodiester bond
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
Dynamically changes its conformation with each cycle of new nucleotide addition; keeps in contact with the growing RNA strand as enzyme moves forward
|
Dynamically changes its conformation with each cycle of new nucleotide addition; keeps in contact with the growing RNA strand as enzyme moves forward
|
|
What percentage of RNAP (RNA polymerase) is present in a storage form (core) and how much is actually in elongation mode?
40%; 60%
50%; 50%
25%; 75%
50%; 25%
|
50%; 25%
|
|
What statement is FALSE in regard to how RNA polymerase (RNAP) interacts with promoter DNA vs. non-promoter DNA?
When sigma is released from the RNAP, the core and RNA transcript are bound extremely tight onto DNA until termination
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
By reducing the stability of non-promoter complexes sigma allows RNAP core to find promoter sequence much faster
|
By increasing the stability of non-promoter complexes sigma allows RNAP core to slide along DNA much faster, and find promoter sequence easier
|
|
Which statement is FALSE?
T7 RNAP activity is stringently regulated
T7 RNAP synthesizes RNA at a rate of 200 nt/sec
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
|
|
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
Clamp is initially out of position; after DNA has melted, it gets repositioned to keep DNA in the active site tighter
Causes DNA bend; bending of DNA helps melt the strands and flip bases in the template strand to be accessible
|
Limits the length of RNA/DNA hybrid; separates nascent RNA chain from DNA template as it exits through the exit pore
|
|
Which statement is FALSE?
The ternary complex is the least stable promoter complex
Formation of the open promoter complex is irreversible and involves the melting of DNA by sigma
Formation of the closed promoter complex is reversible
The ternary complex is formed after the first phosphodiester bond is formed
|
The ternary complex is the least stable promoter complex
|
|
How many different types of subunits are there in bacterial RNAP holoenzyme and what are their names?
3; alpha, beta, sigma
4; alpha, beta, beta prime, omega
7; alpha, beta, beta prime, omega, sigma, and delta
6; alpha, beta, beta prime, omega, and sigma
|
6; alpha, beta, beta prime, omega, and sigma
|
|
What is a promoter?
Gene template
Control region for gene expression
Signal to terminate transcription
3 untranslated region
|
Control region for gene expression
|
|
Which statement is FALSE?
T7 RNAP has a DNA binding region and the active site
T7 RNAP recognizes 1000 phage promoters
T7 RNAP is a single peptide
T7 RNAP uses a specificity loop to recognize the promoter
|
T7 RNAP recognizes 1000 phage promoters
|
|
What is the size of the transcription bubble and RNA/DNA hybrid in it?
12-14 nt; 8-9 nt
10-12 nt; 5-6 nt
8-9 nt; 6-7 nt
16-18 nt; 10-12 nt
|
12-14 nt; 8-9 nt
|
|
What is the function of RNAP Wall?
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
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
|
|
Fill in the blank: Transcription occurs by _____________ in a _______?
Base pairing; bubble of unpaired DNA
Reading DNA coding sequence; terminal loop
Splicing; spliceosome
DNA replication; DNA loop
|
Base pairing; bubble of unpaired DNA
|
|
What is the function of RNAP Clamp/Jaws?
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
Changes its conformation with each cycle of new nucleotide addition and keeps in contact with the growing RNA strand
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
|
|
What is the most important stage of transcription for regulation?
Initiation
Intron splicing from pre-mRNA
Elongation
Termination
|
Initiation
|
|
Sigma reduces affinity of RNAP core for non-promoter sequences _______fold and increases affinity for specific promoter DNA _______fold?
15,000; 800
10,000; 1,000
1,000; 10,000
5,000; 500
|
10,000; 1,000
|
|
What statement is FALSE in regard to how RNA polymerase (RNAP) interacts with promoter DNA vs. non-promoter DNA?
Binding of sigma causes RNAP core to bind much tighter to non-promoter sequence
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
Core RNAP is stored bound to nonspecific DNA sites before sigma subunit binds to core
|
Binding of sigma causes RNAP core to bind much tighter to non-promoter sequence
|
|
What two enzymes help RNAP to eliminate supercoiling generated by the mechanism of transcription?
Reverse transcriptase; topoisomerase
Helicase; DNA polymerase I
Gyrase; topoisomerase
DNA polymerase I; gyrase
|
Gyrase; topoisomerase
|
|
What is DNA scrunching?
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, when sigma 1.1 mimics DNA in the DNA channel, and the actual DNA molecule tries to enter the channel
Occurs during transcription when double stranded DNA sinks into the channel and hits the RNAP active site wall
Occurs during transcription when length of the nascent transcript is limited by RNAP rudder to 6-9 nts
|
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
|
|
Which statement is FALSE?
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.
When the open promoter complex is formed, sigma 1.1 is displaced from DNA channel and replaced by 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.
|
|
What is abortive cycling?
When RNAP pools template strand into its active site
When RNA polymerase hops on and off the DNA molecule searching for promoter sequence
When RNAP transcribes 2-9 nt, then restarts again and does not leave the promoter
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
|
|
Which subunit of bacterial RNAP is required for promoter specificity?
Sigma
Alpha
Beta
Omega
|
Sigma
|
|
How many Mg2+ ions are present in the RNAP active site during transcription?
2
0
1
5
|
2
|
|
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
Transferring between distant DNA segments located on the same molecule
Sliding on a single stranded noncoding DNA molecule
|
Sliding on a single stranded noncoding DNA molecule
|
|
What is the rate of transcription and the rate of translation?
40-50 nt/second; 15 amino acids/sec
800 nt/sec; 45 amino acids/sec
40-50 nt/sec; 45 amino acids/sec
50-60 nt/sec; 5 amino acids/sec
|
40-50 nt/second; 15 amino acids/sec
|
|
What is a promoter?
Signal to terminate transcription
Gene template
3 untranslated region
Control region for gene expression
|
Control region for gene expression
|
|
Which subunits form the catalytic center of RNAP?
Alpha and sigma
Alpha and beta
Beta prime and omega
Beta and beta prime
|
Beta and beta prime
|
|
How is the bacterial core promoter recognized by RNAP?
Discriminator element binds subdomain 3.2 and this helps melt the promoter
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
Strong contacts between sigma and alpha CTDs contribute to promoter recognition
|
Through contacts of sigma subdomains 2.4/2.3 and 4.2 and cis-acting promoter elements
|
|
How can you explain the effect of temperature on the dissociation of RNAP holoenzyme from promoter sequence?
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.
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.
Both, closed and open promoter complexes co-exist at 37 deg and this contributes to their long half life.
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.
|
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.
|
|
Listed below are sigma subdomains followed by its function. Which of the following pairings has an incorrect function assigned?
4.2; melting of the -35 element
1.1; blocking DNA binding site within RNAP holoenzyme until finding a real promoter
2.3; aromatic amino acids contribute to promoter melting of the -10 element
2.4; base-specific interactions with the -10 element
|
4.2; melting of the -35 element
|
|
What is the function of heparin in EMSA (gel retardation) and DNase I footprinting assays?
Heparin stabilizes binding of RNAP to the double stranded DNA.
Heparin acts like sigma 3.2 subdomain and blocks the RNA exit pore within enzyme active site.
Heparin is negatively charged; acts as a competitor to knock off any RNAP that is not in a stable open complex formation.
Heparin changes the conformation of enzymes/proteins involved in these DNA:protein binding techniques.
|
Heparin is negatively charged; acts as a competitor to knock off any RNAP that is not in a stable open complex formation.
|
|
Listed below are sigma subdomains followed by its function. Which of the following pairings has an incorrect function assigned?
Hairpin 3.2; keeps contact with the first nucleotide of the nascent RNA until abortive cycling or elongation occurs
Extended helix 2.3/2.4; recognizes and melts -10 element
H-T-H 4.2; recognizes -35 element
H-T-H 1.1; blocks the exit pore of RNA
|
H-T-H 1.1; blocks the exit pore of RNA
|
|
How do sigma factors recognize promoter sequence?
By recognition of specific DNA cis-elements at position -10 and the equivalent of -35, as well as promoter configuration (distance between cis-elements)
By sliding, hopping, or transfer between very AT-rich sequences located along DNA helix
By recognizing RNA polymerase core bound to the promoter elements and recruiting sigma factor
By counting 10 and 35 nucleotides upstream from the start of transcription on one side of the helix
|
By recognition of specific DNA cis-elements at position -10 and the equivalent of -35, as well as promoter configuration (distance between cis-elements)
|
|
Which statement is FALSE?
Domain 1.1 is blocking the region where DNA is located when the promoter is melted.
Domain 1.1 helps melt the promoter sequence and create an open promoter complex.
Domain 1.1 prevents sigma from binding promoter without first binding the RNAP core.
If domain 1.1 is deleted, sigma will bind promoter sequence.
|
Domain 1.1 helps melt the promoter sequence and create an open promoter complex.
|
|
Sigma 54 has a safety check that prevents it from continuously expressing the glutamine synthase gene. What is the basis for this control?
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.
GreA and GreB proteins are necessary to activate sigma 54.
DNA looping and ATP hydrolysis inhibit sigma 54 from activating NIF genes under control conditions.
Negative factor NtrC inhibits transcription regulated by sigma 54 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.
|
|
What is the function of unlabeled competitor DNA in various footprint assays?
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.
Stops double stranded DNA from melting, thus making protein binding to DNA more feasible.
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.
|
|
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
Discriminator
Upstream activating sequence I, II and III
|
UP element
|
|
Which is NOT the function of sigma 3.2?
Blocks exit channel for the RNA
Positions the first nucleotide in nascent RNA
Facilitates promoter clearance by RNAP (displacement by the nascent RNA)
Pulls DNA template into active center (scrunching)
|
Pulls DNA template into active center (scrunching)
|
|
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%
1 sigma: 3 core; 75%
1 sigma: 1 core; 50 %
|
1 sigma: 3 core; 25%
|
|
Which sigmas can be used to transcribe genes during some stress conditions?
Sigma S, sigma 32
Sigma fecl, sigma E
Sigma A, sigma E
Sigma 70, sigma F
|
Sigma S, sigma 32
|
|
Which is common for both DNA and RNA polymerases?
Can melt the DNA duplex
Can recognize promoter sequences
Can slide along DNA
Can scrunch DNA
|
Can slide along DNA
|
|
In the transition from abortive cycling to elongation of transcription, which is TRUE?
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.
The wall separates RNA transcript from DNA template, thus facilitating abortive cycling .
DNA scrunching contributes to a stronger binding between RNAP core and sigma 3.2 subdomains.
Sigma looses affinity for the promoter DNA and RNAP core, and pressure of growing RNA chain dislodges sigma from the RNA exit pore.
|
Sigma looses affinity for the promoter DNA and RNAP core, and pressure of growing RNA chain dislodges sigma from the RNA exit pore.
|
|
How does RNAP holoenzyme recognize different gene promoters?
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 using hairpin structure of the sigma 3.2 subdomain, which touches the first nucleotide in the active site
By changing the conformation of it active site/DNA binding site
|
By binding alternative specialized sigma subunits, which recognize different cis-element sequences and various configurations of the promoter
|
|
Which statement is FALSE?
Core binds to random DNA through nonspecific, mostly electrostatic interactions, with a half-life of circa 60 min.
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.
|
The strongest bacterial promoter is that for Lac repressor and it reinitiates 1 time per second.
|
|
Which protein(s) helps RNAP to recover from a stall caused by the temporary shortage of nucleotides, and how?
NtrC; using ATP it provides the energy to continue elongation.
GreA and NtrC; by pushing nascent RNA out of the exit pore, thus helping to convert scrunched RNA/DNA hybrid.
RNAP core active site; by scrunching more DNA template into 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.
|
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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What technique can be used to evaluate protein: DNA contacts on the single-stranded DNA?
DMS footprinting
DNase I footprinting
Filter binding assay
GST pool down assays
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DMS footprinting
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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
2.3; contacts alpha CTD
1.2; contacts discriminator element
2.1/2.2; recognizing conserved region in RNAP core
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2.3; contacts alpha CTD
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What statement is FALSE?
Sigma 54 does not have region 1.1.
Sigma 54 activates the constitutive promoter of nitrogen starvation gene glnA.
Some NIF genes have dual promoters, which can be turned on either by sigma 70 or sigma 54.
Sigma 54 is very unusual since it can bind promoter in the absence of RNAP core.
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Sigma 54 activates the constitutive promoter of nitrogen starvation gene glnA.
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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.
Can be easily dislodged by other antibiotics.
It is in contact with nascent RNA past nucleotides 1-3.
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Can be easily dislodged by other antibiotics.
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In the diagram below showing bacterial RNA polymerase, what is indicated by the circle and arrow?
(orange blob)
coding strand
template strand
non-coding strand
newly transcribed RNA
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newly transcribed RNA
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What is required for a Rho-dependent terminator?
an inverted repeat downstream of a very C-rich stretch of mRNA
an RNA hairpin upstream of a rut site
an inverted repeat rich in Cs downstream of a rut site
an inverted repeat in mRNA downstream of a site rich in Gs.
an inverted repeat upstream of a stretch of Us on the mRNA
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an inverted repeat downstream of a very C-rich stretch of mRNA
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What is the role of the hairpin in an intrinsic terminator?
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 alters the conformation of the active site of RNA polymerase causing the synthesis reaction to run backwards.
It causes the polymerase to stall, which requires either processing of the 3-end or termination to occur.
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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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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?
The As are on the template strand.
The As should come before the hairpin, not after.
The hairpin region has too many mismatched base pairs.
The As are on the coding strand.
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The As are on the coding strand.
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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?
four (stop codon between rut and hairpin)
one (stop codon before rut and hairpin)
three (no stop codon)
two (stop codon before rut and hairpin)
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three (no stop codon)
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In the diagram below showing bacterial RNA polymerase, what is indicated by the circle and arrow?
(yellow blob)
coding strand
newly transcribed RNA
template strand
non-coding strand
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coding strand
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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 occurs when a rho-dependent terminator is located in the coding region of the last gene in a polycistronic mRNA.
All of the coding regions have the same orientation to form 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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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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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
False
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True
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In the diagram below, which configuration would result in NO expression of RNA encoding regions C and D?
four (stop codon between rut and hairpin)
one (stop codon before rut and hairpin)
three (no stop codon)
two (stop codon before rut and hairpin)
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one (stop codon before rut and hairpin)
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In the diagram below showing bacterial RNA polymerase, what is indicated by the circle and arrow?
(red blob)
template strand
rudder domain
newly transcribed RNA
coding strand
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template strand
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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 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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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 the source of energy that allows an intrinsic terminator to function?
thermal energy due to the temperature
kinetic energy of the moving RNA polymerase
ATP provided during transcription
NADP provided by metabolism
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thermal energy due to the temperature
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What is the energy source that provides movement to Rho?
It is derived from breaking the phosphodiester bonds during transcription.
ATP hydrolysis
temperature of the cell
sunlight
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ATP hydrolysis
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In the diagram of the Lac repressor shown below, which number corresponds to a DNA binding domain ("hinge helix") that interacts with the minor groove of the Operator site?
Number 3 (blue and red on top)
Number 2 (blue bottom)
Number 4 (red and blue mix)
Number 1 (red bottom left)
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…
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Which element of the bacterial promoter binds the CTD of the alpha subunit?
-10 site
-35 site
CAP
UP
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UP
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At the promoter for the lac operon, how does the DNA loop block transcription?
by blocking access to the core promoter by RNA polymerase
by creating torsional strain in the DNA at the promoter which prevents the transition from closed to open promoter complex.
causes excessive abortive cycling
by blocking the binding of CAP
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by creating torsional strain in the DNA at the promoter which prevents the transition from closed to open promoter complex.
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In the diagram of the Lac repressor shown below, which number corresponds to a DNA binding domain that interacts with the major groove of the Operator site?
Number 3 (blue and red on top)
Number 2 (blue bottom)
Number 4 (red and blue mix)
Number 1 (red bottom left)
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Number 1 (red bottom left)
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In the diagram of the Lac repressor shown below, which number corresponds to region of the protein containing leucine zippers and is responsible for holding the dimers together?
Number 3 (blue and red on top)
Number 2 (blue bottom)
Number 4 (red and blue mix)
Number 1 (red bottom left)
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Number 3 (blue and red on top)
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The Lac repressor bound to IPTG prefers Operator DNA 10,000 times more than non-Operator DNA.
True
False
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True
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Which statement is TRUE regarding the operon promoter shown in the figure below?
This is an example of Positive Control.
This is an example of Negative Control.
The binding of the Active Activator decreases frequency of open promoter formation.
The binding of the small molecule is an example of catabolite repression.
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This is an example of Positive Control.
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The Lac repressor is present in only 10 copies per cell in E. coli. Why is this so?
The mRNA has no 7′-methyl G cap at the 5′-end.
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).
The mRNA has an exceptionally long 5′-UTR which inhibits the ribosome in its search for the Shine Dalgarno (ribosome binding site).
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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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In the diagram of the Lac repressor shown below, which number corresponds to the domain that binds the inducer molecule (allo-lactose or IPTG)?
Number 3 (blue and red on top)
Number 2 (blue bottom)
Number 4 (red and blue mix)
Number 1 (red bottom left)
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Number 4 (red and blue mix)
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Which protein or domain within bacterial RNA polymerase contacts DNA using minor groove contact?
The CTD of the alpha subunit
The NTD of alpha
Sigma 70 domain 2.3
Sigma 70 domain 4.2
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The CTD of the alpha subunit
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Which of the following pathways are regulated by repression caused by the accumulation end-product?
Utilization of arabinose as a nutrient substrate
Tryptophan biosynthesis
Glucose catabolism
Lactose degradation
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Tryptophan biosynthesis
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Which statement is TRUE regarding the operon promoter shown in the figure below?
The circled molecule is a transactivator protein.
This promoter requires the circled molecule in order to express the operon.
The circled molecule is an aporepressor.
The circled molecule is an active repressor.
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The circled molecule is an active repressor.
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Which of the following represent the association constant of a DNA binding protein?
[free repressor-DNA complex] minus [total cellular DNA]/[bound repressor]
[free repressor-DNA complex]/[free repressor] x [total cellular DNA]
[free repressor] x [total cellular DNA]/ [free repressor-DNA complex]
[bound repressor] x [free repressor]/[total cellular DNA]
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[free repressor-DNA complex]/[free repressor] x [total cellular DNA]
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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 gene encoding the Lac repressor.
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.
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A mutation occurred in the core promoter (-10 or -35 sites) of one of the two operons.
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At the promoter for the Ara pBAD operon, how does the DNA loop cause repression?
by causing RNA polymerase to bind to the "dead-end" promoter
by blocking access to the core promoter by RNA polymerase
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
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by blocking access to the core promoter by RNA polymerase
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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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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)
Derepression
Anti-activation
Induction
Attenuation
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Anti-activation
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In which of the following E. coli promoters does CAP binding facilitate the formation of the Open Promoter Complex?
Galactose (Gal P)
Tryptophan (Trp)
Lactose (Lac)
Arabinose (Ara)
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Galactose (Gal P)
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In Class III CAP-dependent promoters, it is possible to have CAP located at both the -41.5 and -61 positions.
True
False
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True
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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.
In the Lac promoter, CAP binding increases formation of the closed promoter complex.
Class II CAP-dependent promoters always have a single binding site for the CAP dimer located at position -41.5.
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Most CAP-dependent promoters have a good match to the -35 site consensus.
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In the diagram below of the lac operon, what is the sugar composition of the media?
no glucose and no lactose
Glucose present and no lactose
Glucose and lactose are present
No glucose, but lactose is present
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no glucose and no lactose
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In E. coli, approximately _______ promoters are activated by CAP.
300
100
15
1,000
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100
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Which of the following statements regarding CAP is TRUE?
CAP usually binds to its’ promoter elements when glucose is present in the cell.
CAP requires cyclic AMP to bind DNA as a tetramer
CAP binds to C-rich promoter elements and causes the DNA to melt.
CAP requires cyclic AMP to bind DNA as a dimer.
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CAP requires cyclic AMP to bind DNA as a dimer.
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At which class(es) of CAP-dependent promoters is CAP always in contact with both the CTD and NTD of the alpha subunit of RNA polymerase?
Class III
Class I
Class II
Both Class I and Class II
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Class II
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In the diagram below of the ara operon, what is the sugar composition of the media?
no glucose and no arabinose
Glucose and arabinose are present
No glucose, but arabinose is present
Glucose present and no arabinose
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no glucose and no arabinose
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Anti-activation can sometimes occur by an ADP-ribosylation-dependent mechanism. What is the mechanism regarding promoters regulated by catabolite repression?
ADP-ribose moieties are added to domain 4.2 of sigma 70 which blocks its’ interaction with CAP at Class II CAP-dependent promoters.
Cyclic ADP-ribose moieties are formed which block the inhibitory effects of glucose on the adenylate cyclase enzyme.
ADP-ribose moieties are added to the CAP monomers at the AR1 and AR2 sites which inactivates CAP.
The CTD of the alpha subunit of RNA polymerase is modified by the addition of ADP-ribose moieties which block its’ interaction with CAP.
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The CTD of the alpha subunit of RNA polymerase is modified by the addition of ADP-ribose moieties which block its’ interaction with CAP.
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