MECHANISM OF TRANSCRIPTION IN BACTERIA P2 Flashcards
1
Q
(Rationale of FRET Assay for σ movement relative to DNA)
Trailing edge FRET
- A fluorescence donor (D, green) is attached to the single ________
- A fluorescence acceptor (A, red) is attached to the _________
- FRET efficiency is _________ in the open promoter complex
- FRET efficiency ________ regardless of whether σ dissociates from the core, because the two probes grow farther apart in either case
A
cysteine residue ; 5’-end of the DNA ; high (solid purple line) ; decreases (dashed purple line)
2
Q
(Rationale of FRET Assay for σ movement relative to DNA)
Leading edge FRET
- The fluorescence acceptor is attached to the _________.
- FRET efficiency is ________ in the open promoter complex because the two probes are far apart.
- If sigma dissociates from core, FRET should __________
- if sigma remains bound to core, the two probes will grow closer together as the polymerase moves downstream, and FRET efficiency will _________
A
- 3’-end of the DNA
- low (dashed purple line)
- decrease (dashed purple line)
- increase (solid purple line)
3
Q
(Local DNA Melting at the Promoter)
Studies on RNA polymerase-promoter interactions showed that such complexes were much more stable at elevated temperature.
A
Chamberlin
4
Q
(Local DNA Melting at the Promoter)
This suggested that local melting of DNA occurs on tight binding to ______
High temperature would tend to ______ melted DNA
DNA melting is essential because it exposes bases of the template strand so they can base pair with bases on __________
A
polymerase ; stabilize ; incoming nucleotides
5
Q
(Local DNA Melting at the Promoter)
Provided more direct evidence for local DNA melting
Knowing the number of RNA polymerase holoenzymes bound to their DNA, _______ calculated that each polymerase caused a separation of about 10 bp
A
1978 Tao-shih Hsieh and James Wang
6
Q
Identified the base pairs that RNA polymerase melted in a T7 phage early promoter
A
1979 Ulrich Siebenlist
7
Q
On binding promoter, ______ causes melting that has been estimated at 10-17 bp in the vicinity of the transcription start site.
A
RNA polymerase
8
Q
This transcription bubble moves with the polymerase, exposing the ___________ so it can be transcribed.
A
template strand
9
Q
(Promoter Clearance)
- RNA polymerases cannot work if they do not recognize _______, so they have evolved to recognize and bind strongly to them.
- Several hypotheses have been proposed, including the idea that the energy released by forming a _______ (up to 10 nt long) is stored in a distorted polymerase or DNA, and the release of that energy in turn allows promoter clearance.
A
promoters ; short transcript
10
Q
(Promoter Clearance)
- The polymerase cannot move enough downstream to make a 10-nt transcript without doing one of three things:
- Moving briefly downstream and then snapping back to the starting position (___________)
- Stretching itself by leaving its trailing edge in place while moving its leading edge downstream (________)
- Compressing the DNA without moving itself (_________)
A
transient excursion ; inchworming ; scrunching
11
Q
(Promoter Clearance)
The E. coli RNA polymerase achieves abortive transcription by _______: drawing downstream DNA into the polymerase without actually moving and losing its ____ on promoter DNA
A
scrunching ; grip
12
Q
(Promoter Clearance)
The scrunched DNA could store enough energy to allow the polymerase to break its bonds to the promoter and begin productive ________
A
transcription
13
Q
(Structure and Function of sigma)
FOUR REGIONS:
A
Region 1, 2, 3, 4
14
Q
(Structure and Function of sigma)
A REGION THAT:
- Found only in the primary sigma’s (σ70 and σ43)
- Prevent s from binding by itself to DNA
A
Region 1
15
Q
(Structure and Function of sigma)
A REGION THAT:
- Found in all σ-factors and is the most highly conserved σ region.
- It can be subdivided into four parts, 2.1–2.4
- 2.4 is responsible for a crucial σ activity, recognition of the promoter’s -10 box.
A
REGION 2
16
Q
(Structure and Function of sigma)(Structure and Function of sigma)
A REGION THAT:
Involved in both core and DNA binding.
A
REGION 3
17
Q
(Structure and Function of sigma)
A REGION THAT:
Subdivided into 2 subregions
A
REGION 4
18
Q
Comparison of the sequences of different sigma genes reveals ______ of similarity among a wide variety of σ-factors.
A
four regions
19
Q
_______ and ______ are involved in promoter -10 box and -35 box recognition, respectively.
A
Subregions 2.4 and 4.2
20
Q
The σ-factor by itself cannot bind to DNA, but interaction with _____ unmasks a DNA-binding region of σ.
A
core
21
Q
(The Role of the a-Subunit in UP Element Recognition)
- RNA polymerase itself can recognize an ___________ (UP element)
- ________ recognizes the core promoter element
- a-subunit of the core polymerase is responsible for recognizing the _______.
A
upstream promoter element ; Sigma-factor ; UP element
22
Q
(The Role of the a-Subunit in UP Element Recognition)
__________________
- Made E. coli strains with mutations in the a-subunit
Purified subunits:
1. __________ with a normal a- subunit;
2. _____ - a polymerase whose a-subunit was missing 94 amino acids from its C-terminus;
3. ______ - a polymerase whose a-subunit contained a cysteine (C) in place of the normal arginine (R) at position.
A
Richard Gourse and colleagues ;
- wild-type polymerase
- a-235
- R265C
23
Q
(The Role of the a-Subunit in UP Element Recognition)
The size of the band on the gel tells which promoter, and the intensity depends on how much transcription occurred with that promoter.
Different promoters can distinguish by their bands.
A
YES
24
Q
(The Role of the a-Subunit in UP Element Recognition)
Used limited proteolysis analysis to show that the a-subunit N-terminal and C-terminal domains fold independently to form two domains that are tethered together by a flexible linker
A
Richard Gourse, Richard Ebright, and their colleagues
25
The ___________ has an independently folded C-terminal domain that can recognize and bind to a promoter’s UP element.
RNA polymerase a-subunit
26
This allows very tight binding between _________ and __________
polymerase and promoter
27
After initiation of transcription is accomplished, the core continues to _______the RNA, adding one nucleotide after another to the growing RNA chain.
elongate (elongation)
28
____________- contains the RNA synthesizing machinery , so the ____ is the central player in elongation.
Core polymerase ; core
29
(Core Polymerase Functions in Elongation)
- involved in phosphodiester bond formation
- these subunits also participate in DNA binding
β- and β’- subunits
30
(Core Polymerase Functions in Elongation)
has several activities, including assembly of the core polymerase.
a-subunit
31
(The Role of β in Phosphodiester Bond Formation)
First to investigate the individual core subunit
1970 Walter Zillig
32
(The Role of β in Phosphodiester Bond Formation)
______________ electrophoresed the core enzyme on cellulose acetate in the presence of urea
- Urea is a denaturing agent that can separate the individual polypeptides in a complex protein.
- Urea is a mild denaturant that is relatively easy to remove.
- It is easier to renature a urea-denatured polypeptide than an SDS-denatured one
Alfred Heil and Zillig
33
(The Role of β in Phosphodiester Bond Formation)
provided more evidence for the notion that β plays a role in elongation, using a technique called ____________
This technique is to label an enzyme with a derivative of a normal substrate that can be cross-linked to protein
1987 M. A. Grachev and colleagues ; affinity labeling
34
(The Role of β in Phosphodiester Bond Formation)
The core subunit β lies near the _________ of the RNA polymerase where phosphodiester bonds are formed.
active site
35
(The Role of β in Phosphodiester Bond Formation)
The σ-factor may also be near the ___________, at least during the initiation phase..
nucleotide-binding site
36
(Structure of the Elongation Complex: The RNA–DNA Hybrid)
________________
applied a transcript walking technique, together with RNA–DNA cross-linking, to prove that an RNA–DNA hybrid really does occur within the elongation complex and that this hybrid is 8–9 bp long.
Nudler and Goldfarb and their colleagues
37
(Structure of the Elongation Complex: The RNA–DNA Hybrid)
The RNA–DNA hybrid within the E. coli elongation complex extends from position ________ to ________ with respect to the 3'-end of the nascent RNA
The T7 hybrid appears to be _____ long.
-1 to position -8 or -9 ; 8 bp
38
(Structure of the Elongation Complex: Structure of the Core Polymerase)
____________ would give the best resolution for the clearest picture of the structure of the elongation complex, but it requires three-dimensional crystals
X-ray crystallography
39
(Structure of the Elongation Complex: Structure of the Core Polymerase)
__________________
- Crystallized the core polymerase from another bacterium, ____________
- Obtained a crystal structure to a resolution of 3.3 Å Crystal structure of the T. aquaticus polymerase is the best window right now on the structure of a bacterial polymerase
1999 Seth Darst and colleagues ; Thermus aquaticus
40
(Structure of the Elongation Complex: Structure of the Core Polymerase)
X-ray crystallography on the Thermus aquaticus RNA polymerase core has revealed an enzyme shaped like a ______ designed to grasp DNA
A channel through the enzyme includes the ____________ and ____________
crab claw ; catalytic center and the rifampicin-binding site
41
(Structure of the Holoenzyme–DNA Complex)
_______________
Bound the T. aquaticus holoenzyme to the “fork-junction” DNA
Darst and Colleagues
42
(Structure of the Holoenzyme–DNA Complex)
- mostly double-stranded, including the -35 box
- has a single-stranded projection on the nontemplate strand in the -10 box region, beginning at position -11
“Fork-junction” DNA
43
(Structure of the Holoenzyme–DNA Complex)
The crystal structure of a Thermus aquaticus holoenzyme–DNA complex mimicking an open promoter complex reveals several things:
1. DNA is bound mainly to the σ-subunit, which makes all the important interactions with the promoter DNA
2. The predicted interactions between amino acids in region 2.4 of σ and the -10 box of the promoter are really possible.
3. Three highly conserved aromatic amino acids are predicted to participate in promoter melting.
4. Two invariant basic amino acids in s are predicted to participate in DNA binding.
YES
44
(Structure of the Elongation Complex)
________________
- Presented the x-ray crystal structure of the Thermus thermophilus RNA polymerase elongation complex at 2.5Å resolution
This complex contained:
- _____ of downstream double-stranded DNA that had yet to be melted by the polymerase
- _____ of RNA–DNA hybrid
- _____ of RNA product in the RNA exit channel
2007- Dmitry Vassylyev and Colleagues ; 14 bp ; 9 bp ; 7 nt
45
(Structure of the Elongation Complex)
______ residue in the β subunit inserts into the minor groove of the downstream DNA
Two important consequences:
1. It could prevent the DNA from _____ backward or forward in the enzyme
2. It could induce the _________ of the DNA through the enzyme
Valine ; slipping ; screw-like motion
46
(Structure of the Elongation Complex)
Only one base-pair of DNA (at position +1) is melted and available for base-pairing with an incoming nucleotide, so only one nucleotide at a time can bind specifically to the complex.
Several forces limit the length of the RNA–DNA hybrid:
1. The length of the cavity in the enzyme that accommodates the ____
2. A hydrophobic pocket in the enzyme at the end of the cavity that traps the first ________ displaced from the hybrid.
hybrid ; RNA base
47
(Structure of the Elongation Complex)
The RNA product in the exit channel assumes the ______ of one-half of a double-stranded RNA.
shape
48
(Structure of the Elongation Complex)
It can readily form a _____ to cause pausing or even termination of transcription.
hairpin
49
(Structure of the Elongation Complex)
Structural studies of the enzyme with an inactive substrate analog and the antibiotic streptolydigin have identified a _________ for the substrate that is catalytically inactive but could provide for checking that the substrate is the correct one.
preinsertion state
50
(Topology of Elongation)
Does the core, moving along the DNA template, maintain the local melted region created during initiation?
- Common sense tells us that it does because this would help the RNA polymerase “______” the bases of the template strand and therefore insert the correct bases into the transcript
- Experimental evidence also demonstrates that this is so
read
51
(Topology of Elongation)
A class of enzymes that can introduce transient breaks into DNA strands and relax this kind of strain.
Topoisomerases
52
(Topology of Elongation)
Strain due to twisting a double-helical DNA causes the helix to tangle up like a twisted rubber band.
Supercoiling
53
(Topology of Elongation)
The supercoiled DNA
Supercoil or Superhelix
54
(Topology of Elongation)
Build up in front of the advancing polymerase
If the mutant cannot relax positive supercoils, these build up in the DNA that is being transcribed.
Positive supercoils
55
(Topology of Elongation)
Form behind the polymerase.
Negative supercoils accumulate during transcription in topoisomerase mutants that cannot relax that kind of superhelix
Negative supercoils
56
(Topology of Elongation)
_____________ involves the polymerization of nucleotides as the RNA polymerase travels along the template DNA.
Elongation of transcription
57
(Topology of Elongation)
As it moves, the _________ maintains a short melted region of template DNA.
polymerase
58
(Pausing and Proofreading)
___________ frequently pauses, or even backtracks, during elongation
RNA polymerase
59
(Pausing and Proofreading)
______ allows ribosomes to keep pace with the RNA polymerase, and it is also the first step in termination.
Pausing
60
(Pausing and Proofreading)
_____________ aids proofreading by extruding the 3'-end of the RNA out of the polymerase, where misincorporated nucleotides can be removed by an inherent nuclease activity of the polymerase, stimulated by auxiliary factors.
Even without these factors, the polymerase can carry out ______
Backtracking ; proofreading
61
(Pausing and Proofreading)
The mismatched nucleotide at the end of a nascent RNA plays a role in this process by contacting two key elements at the active site:
Metal II
Water molecule.
62
(Pausing and Proofreading)
Auxiliary proteins known as _____ and ______ stimulate an inherent RNase activity of the polymerase to cleave off the end of the growing RNA, removing the misincorporated nucleotide, and allowing transcription to resume.
GreA and GreB
63
(Pausing and Proofreading)
Produces only short RNA end fragments 2–3 nt long, and can prevent, but not reverse transcription arrest.
GreA
64
(Pausing and Proofreading)
Can produce RNA end fragments up to 18 nt long, and can reverse arrested transcription
GreB
65
(TERMINATION OF TRANSCRIPTION)
When the polymerase reaches a _______ at the end of a gene it falls off the template, releasing the RNA.
terminator
66
(TERMINATION OF TRANSCRIPTION)
E. coli cells contain about equal numbers of two kinds of terminators:
1. ___________ - function with the RNA polymerase by itself without help from other proteins
2. _____ - depends on an auxiliary factor
Intrinsic terminators ; Rho (ρ)
67
(TERMINATION OF TRANSCRIPTION)
___________ or ________ depends on terminators consisting of two elements: an inverted repeat followed immediately by a T-rich region in the nontemplate strand of the gene
Rho-Independent Termination ; Rho-independent, or intrinsic, termination
68
(Inverted Repeats and Hairpins)
Such a sequence is _______ around its center, indicated by the dot
it would read the same if rotated 180 degrees in the plane of the paper, and if we always read the strand that runs 5'→3' left to right.
symmetrical
69
(Inverted Repeats and Hairpins)
The transcript of this sequence is self-complementary (underlined G.)
Self-complementary bases can pair to form a hairpin as follows:
1. UACGAAGUUCGUA
2. U-A, A-U, C-G, G-C, A-U, A, U , G
70
(The Structure of an Intrinsic Terminator)
The E. coli trp operon contains a DNA sequence called an _________ that causes premature termination of transcription.
attenuator
71
(The Structure of an Intrinsic Terminator)
trp attenuator contains the two elements:
- inverted repeat
- a string of T’s in the nontemplate DNA strand
yes
72
(The Structure of an Intrinsic Terminator)
The inverted repeat in the trp attenuator is not perfect, but 8 bp are still possible, and 7 of these are strong G–C pairs, held together by three hydrogen bonds.
- Small loop occurs at the end of this hairpin because of the U–U and A–A combinations that cannot base-pair.
- Furthermore, one A on the right side of the stem has to be “looped out” to allow 8 bp instead of just 7. Still, the hairpin should form and be relatively stable.
YES
73
(A Model for Termination)
Two Important Clues
1. First, _________ are found to destabilize elongation complexes that are stalled artificially (not at strings of rU–dA pairs)
2. Second, _______ in which half of the inverted repeat is missing still stall at the strings of rU–dA pairs, even though no hairpin can form.
hairpins ; terminators
74
(A Model for Termination)
A normal __________ satisfies the first condition by causing a hairpin to form in the transcript, and the second by causing a string of U’s to be incorporated just downstream of the hairpin.
Intrinsic terminator
75
(Rho-Dependent Termination)
Discovered rho as a protein that caused an apparent depression of the ability of RNA polymerase to transcribe certain phage DNAs in vitro.
Jeffrey Roberts
76
(Rho-Dependent Termination)
_____________ a result of termination.
Whenever rho causes a termination event, the polymerase has to ________ to begin transcribing again.
And because initiation is a time-consuming event, less net transcription can occur
Depression ; Reinitiate
77
How does Rho do its job?
- Rho is able to bind to RNA at a so-called _________ OR _________
- Has __________ that can provide the energy to propel it along an RNA chain.
- rho loading site, or rho utilization (rut) site
- ATPase activity
78
Rho-dependent terminators consist of an ________, which can cause a hairpin to form in the transcript, but no string of T’s.
inverted repeat
79
Rho binds to the RNA polymerase in an _________
elongation complex.
80
When the RNA transcript has grown long enough, rho binds to it via a _________, forming an RNA loop between the polymerase and rho.
rho loading site
81
___ continues to feed the growing transcript through itself until the polymerase pauses at a terminator.
Rho
82
This pause allows rho to tighten the RNA loop and trap the elongation complex.
Rho then dissociates the RNA–DNA hybrid, terminating _________.
transcription
