Unit 7: Prokaryotic Transcription Flashcards
1
Q
TRUE OR FALSE: DNA is the genetic material of most organisms other than some viruses or viroids that have RNA
A
TRUE
2
Q
How do viruses that RNA as genetic material complete transcription?
A
their genetic material must first be transcribed to DNA first before it can be transcribed
3
Q
TRUE OR FALSE: RNA is translated into protein and this is unidirectional
A
TRUE: central dogma theory
4
Q
What does it mean that translation and transcription is coupled in prokaryotes?
A
This means that as the mRNA is being transcribed, ribosomes can bind to the newly synthesized messenger RNA and start translation of the mRNA into protein even before the Messenger RNA has been fully transcribed.
5
Q
Why does coupled translation and transcription not occur in eukaryotes?
A
In prokaryotes, there is not distinct nucleus hence they can have coupled transcription and translation. In eukaryotes transcription occurs in the nucleus and the mRNA needs to come out into the cytoplasm where the ribosomes can bind and initiate translation.
6
Q
What is meant by a polyribosome or polisome?
A
This happens when several ribosomes bind simultaneously to a single emerging RNA molecule and translate it into a protein.
7
Q
Transcription occurs in what direction? It reads off a template strand that runs in what direction?
A
Transcription is always in the 5’-3’ direction (non-template strand) and the template it is reading off should be the 3’-5’ strand called the template strand.
8
Q
The non-template strand is also called what? Why?
A
the coding strand b/c it will have the same sequence as the mRNA and caries the code for protein sequence. It runs in the 5’-3’
9
Q
What is an RNA polymerase? What is the function?
A
an enzyme that synthesizes the RNA using a DNA template (formally described as DNA-dependent RNA polymerase).
Function: to copy one strand of duplex DNA into RNA.
10
Q
The RNA sequence is complementary to which DNA strand? and identical to which DNA strand?
A
complementary to template strand
identical to coding strand
11
Q
Promoter
A
a region of DNA where RNA polymerase binds to initiate transcription.
12
Q
Start Point
A
the position on DNA corresponding to the first base incorporated into RNA
13
Q
Terminator
A
a sequence of DNA that causes RNA polymerase to terminate transcription
14
Q
Transcription Unit
A
A sequence of DNA transcribed into a single RNA, starting at the promoter and ending at the terminator. it may include more than one gene or multiple genes
15
Q
When is transcription initiated?
A
when the RNA polymerase binds to the promoter region.
16
Q
Sequences prior to the start point of transcription is described as what? Where does numbering of the start point begin?
A
upstream of transcription (sequences that are in the opposite direction from expression)
numbering begins with the start point -1 and increases as your go further upstream
17
Q
Sequences after the start point (w/in the transcribed sequence) are described as what? Where does numbering of start point begin?
A
Downstream - sequences proceeded farther in the direction of expression
numbering begins at +1
18
Q
What is a primary transcript?
A
the initial transcript that comes off the ribosome (original unmodified RNA produce corresponding to a transcription unit)
19
Q
What is the maturation process?
A
a process in which sequences at the ends of primary transcripts are cleaved off (processed) by endonucleases
-Ribosomal RNA and transfer RNA in both prokaryotes and eukaryotes undergo this process which render them stable.
20
Q
TRUE OR FALSE: Transcription occurs by base pairing
A
TRUE: just like replication and translation, transcription also occurs by base pairing in a “bubble” of unpaired DNA
21
Q
How is a transcription bubble made and In the transcription bubble, which stand is used to direct synthesis of RNA?
A
RNA polymerase binds to promoter and separates the two strands of DNA to make the transcription bubble.
A template strand is used to direct synthesis of the complementary sequence of RNA
22
Q
The transcription bubble is about 12-14 basepairs, how long is the RNA-DNA hybrid w/in the bubble?
A
8 to 9 base pairs
23
Q
What are the three stages of the transcription reaction?
A
1) Initiation - RNA poly binds to promoter site on the DNA as a closed complex (closed means DNA stays double stranded). The RNA poly then opens up the DNA helix, in the promoter region, which also includes the start site to form the transcription bubble (which is an open complex and begins synthesis of complementary RNA strand)
2) Elongation - polymerase synthesizes RNA: Involves the RNA strand by the addition of new nucleotides to the 3’ end of the growing chain. As a result there is a RNA-DNA hybrid inside the bubble. Behind the bubble the DNA reforms the double strand helix.
3) Termination - when transcription stops and RNA poly and RNA are released. The polymerase recognizes certain sequences called terminator sequences that signal transcription to stop further addition of nucleotides to the RNA chain. No more phosphodiester bonds formed and the bubble collapses.
24
Q
Why is the initiation stage sometimes extended for a short period of time before the elongation stage?
A
because there may be several abortive attempts where short RNA transcripts of about 10 nucleotides are synthesized as the polymerase stays bound to the promoter. The short nucleotides are released and synthesis is restarted until the polymerase can finally extend the chain and clear the promoter.
25
TRUE OR FALSE: In bacterial there are multiple RNA polymerase that transcribe messenger RNA, ribosomal RNA, and transfer RNA
FALSE: In bacteria there is a single RNA polymerase that transcribes Messenger RNA, ribosomal RNA and transfer RNA. unlike eukaryotes that have three different polymerases; Pol I, Pol II, and Pol III.
26
What is a holoenzyme?
the RNA polymerase form that is competent to initiate transcription. It consists of five subunits of the core enzyme and a sigma factor.
The core enzyme is made up of two alpha subunits, beta, beta-prime, and omega subunits
27
What is the C-terminal domain (CTD)?
the domain of RNA polymerase that is involved in stimulating transcription by contact w/regulatory proteins.
-the two alpha subunits of the holoenzyme recognize the promoter via the c-terminal domains.
28
The core enzyme is made up of five subunits, what are the roles of these subunits?
1 and 2) Two alpha subunits - form a dimer and serve as a scaffold for the assembly of the core enzyme. Also recognize the promoter via its c-terminal domain and serve as binding sites for many factors that regulate transcription.
3 and 4) beta and beta prime - form the bulk of the core enzyme and is the catalytic center. mutations in the genes encoding for these two subunits affect all stages of transcriptions. BB' form a crablike claw in which the DNA lies
5) Omega subunit - involved ini enzyme assembly and regulatory function
29
What is the role of the sigma factor?
the sigma factor is primarily responsible for promoter recognition and ensures that the RNA poly initiates transcription from specific sites but also reduces binding of RNA poly to non-specific sites.
-also important in the melting reaction of DNA
30
The core enzyme of the RNA polymerase binds generally to DNA through what kind of association?
through electrostatic interactions as protein is basic and binds to DNA which is acidic.
31
TRUE OR FALSE: The core enzyme can recognize promoters
FALSE: the core enzyme cannot recognize promoter an needs sigma factor, which recognizes promoters
32
Why does the sigma factor change the DNA binding properties of RNA polymerase?
so that its affinity for general DNA is reduced and its affinity for promoter is increased.
-The dissociation constant for the core enzyme to DNA is reduced by a factor of 10,000 and at the same time the Association constant of the holoenzyme to the promoter is increased thousand fold. it also increases the half life of the complex, thus contributing to its stability as well
33
When is the sigma factor usually released from the core enzyme?
When the RNA chain reaches less than ~10 nucleotides in length, leaving the core enzyme responsible for elongation
34
How does a polymerase find promoter sequences among the four million base pairs in the e.coli genome?
Three proposed mechanisms:
1) the enzyme may move in a one-dimensional random walk along the DNA (SLIDING)
2) given the intricately folded nature of the chromosome in bacterial nucleoid, having bound to one sequence on the chromosome the enzyme is now closer to other sites, reducing the time needed for dissociation and rebinding to another site (intersegment transfer or hopping)
3) while bound nonspecifically to one site, the enzyme may exchange DNA sites until a promoter is found (direct transfer/intrasegment transfer )
35
Initially, the initiation complex w/the sigma factor cover how many base pairs? What happens when sigma dissociates
Initially: -75 base pairs (region -55 to +20)
when sigma dissociates: the core enzyme contracts to -30 and as the enzyme moves forward a few base pairs, the core enzyme compacts further forming the elongation complex
36
Initial elongation complex forms at ____ bases, may lose sigma, and loses contacts from ____ to _____.
forms at 10 bases
loses contact from -35 to -55
37
General elongation complex forms at ____ to _____ bases and covers ____ to _____ base pairs.
forms at 15 to 20 bases and covers 30-40 base pairs
38
What is a ternary complex?
the complex in initiation of transcription that consists of RNA poly , DNA, sigma factor, and core enzyme. as well as dinucleotide that represents the first to bases in the RNA product.
39
What is meant by abortive initiations? When does this occur in transcription?
After the DNA template is positioned properly in the enzyme complex, the next step is the incorporation of the first two nucleotides and formation of the phosphodiester bond to form the ternary complex.
-After each base is added, there is a certain probability that the enzyme will release the RNA chain resulting in abortive initiation products
After release of the abortive product, the enzyme begins synthesizing RNA at position +1. This is repeated until the polymerase succeeds in escaping the promoter
40
What is a conserved sequence?
sequences in which particular nucleic acid or protein are compared and the same individual bases or amino acids are always found at particular locations
From the slides: if one was to compare sequences from different examples of the protein or a particular nucleic acid and you would find that the same individual bases or amino acids are always found at the same location, one would call this a conserved sequence.
41
TRUE OR FALSE: A promoter is defined by the presence of short conserved sequences at specific locations
FALSE:
A promoter is defined by the presence of short CONSENSUS sequences at specific locations
42
What are the two six base pair elements that are key to the recognition of the promoter region?
1) the TATAAT sequence that is centered on -10 base pairs upstream of the start site. This is called -10 element, TATA box, or pribnow box
2) the TTGACA sequence which is -35 base pairs upstream the start point. called the -35element.
both of these factors bind to the sigma factor
43
What is the distance between the -10 element and -35 element ? Why is this an important feature ?
The distance is about 16 to 19 base pairs and is conserved in about 90% of promoters.
The distance is critical b/c given the helical nature of the DNA it determines the appropriate separation of the two interacting regions in RNA polymerase and also the geometrical orientation of the two sites w/respect to one another
44
What is the UP element?
a sequence in bacteria adjacent to the promoter, 20 base pair upstream of the -35 element, that enhances transcription
-it is believed to interact w/the c-terminal domains of the two alpha subunits and ultimately enhance transcription
45
What is the discriminant element?
The sequences on either side of the -10 element label extended ten element which is upstream of the TATA box and the other element downstream of the TATA box called the discriminant element also interacts with RNA polymerase and contributes to promoter efficiency.
46
What are the additional elements that can affect promoter efficiency?
- UP element
- discriminant element
- down mutations
- up mutations
47
What is a down mutation?
decrease promoter efficiency and results in decreased transcription
48
What are up mutations?
mutations that result in increased transcription and increase the similarity of the -10 and -35 elements to the consensus (bringing these elements closer to 17 base pairs increase promoter efficiency)
49
What is affected when there are down mutations in the -35 and -10 sequence ?
- Mutations in the -35 sequence can affect initial binding of RNA polymerase
- Mutations in the -10 sequence can affect binding or the melting reaction that converts a closed to an open complex
50
Why might the sigma factor be required for the melting of DNA>
b/c the Sigma 70 contacts the -10 element, the extended -10 element, and the discriminator element and it remains in contact with these bases after the DNA has unwound in this region suggesting that the Sigma factor is required for the melting of the DNA
51
What is the difference between free sigma and sigma that is part of the core enzyme?
When the sigma factor is not part of the core enzyme, the N terminal domain blocks off its DNA binding region. Once the Sigma factor binds to the core and enzyme it changes its conformation such that the end terminal N swings from the DNA binding domain as well as the two DNA binding domains separate out from each other. this change in effect elongates the Sigma factor so that it can extend over 2 turns of the DN, allowing its DNA binding regions to interact w/the promoter
52
What is the footprinting technique?
a technique for identifying the site of DNA bound by some protein by virtue of the protection of bonds in this region against attacks by nucleases.
endonucleases not have access to the phosphodiester bonds when there is a bound protein and thus would be protected from cleavage
53
Using the fingerprinting method, how were scientist able to identify the points of contact of the RNA polymerase w/the DNA?
one way was to modify the DNA before it bound to the RNA polymerase. If the modification prevented binding to the poly, then that particular base position was essential for contact.
If the RN poly DNA promoter complex was modified then comparing the pattern of protective bands w/that of free DNA and of the unmodified complex we can identify sites where the enzyme has protected the promoter from being modified. These wold be regions where the bands have disappeared. Other bands that have increased in intensity would suggest that these sites were where the DNA was in a confirmation that exposed it much better to the cleavage agent. Most of the contact points for RNA polymerase w/the promoter is in the -10 and -35 element. The points of contact are primarily on the case of the DNA. Promoter recognition and melting of the DNA occur concurrently w/melting beginning base flipping where the base is at 11 and 17 are flipped out into pockets in the sigma factor. This begins strand separation and unwinding begins at the right end of -11 and propagates down to just past the start site at +3
54
What is DNA scrunching?
occurs when the enzyme is attempting to escape from the promoter. The enzyme does not move down the template but rather pulls the first few nucleotides of the downstream DNA into itself.
55
What is the brownian ratchet mechanism?
the movement of the nucleic acid polymerase requires breaking and remaking the bonds of the nucleotides at fixed positions relative to the enzyme. the nucleotides in these positions change every time the polymerase moves up along the template.
the energy for binding the correct nucleotide stabilizes the active conformation and also prevents backtracing
STEPS
1_ polymerase contacts nucleic acid
2_new base is added and bonds are broken
3_ enzyme moves and remakes bonds
56
What is the role of pausing and proofreading in elongation?
Pausing is physiologically important as it allows translation to keep pace w/transcription and it can also be used as the first step in termination of transcription
57
How can a stalled RNA polymerase restart transcription?
by cleaving the RNA transcript to generate a new 3' end that is now positioned in the active site.
58
Apart from RNA polymerase having the main role in cleavage when RNA poly is stalled, what other factors (accessory factors) stimulate activity of RNA poly?
In e.coli: GreA and GreB
in eukaryotes: TF2S
these proteins insert a domain deep into the RNA polymerase close to the catalytic site allowing it to behave as an endonuclease
59
What are the two types of terminators and how do they differ?
1) intrinsic terminators - recognized solely by RNA polymerase itself w/o the requirement for any cellular factors
2) rho-dependent terminator-require cellular proteins called rho
60
What are the two main features found in intrinsic terminators?
1) recruitment of the GC rich hairpin to form in the secondary structure of the transcribing RNA
2_ there has to be a stretch of up to seven uracil residues in RNA and correspondingly 7 thymine residues in the DNA following the hairpin stem but before the actual site of termination. This means that the DNA sequences required for termination are actually located upstream of the terminator sequence.
61
Presence of secondary hairpin structure leads to ?
misalignment of the 3'-OH groups in the active site. and this can lead to destabilization of the elongation complex ultimately leading to termination.
62
How is antitermination at rRNA different than antitermination at Lambda phage? What is the role of rRNA antitermination?
-anti termination in rRNA does not contain N or Q like factors that are seen in Lambda phage bu they still prevent premature termination at hairpin sequences or at weak rho dependent terminators
63
How can antitermination control transcription?
by determining whether RNA polymerase terminates or reads through a particular terminator
64
What is the nut site?
an acronym for the N utilization site, the sequence of DNA that is recognized by the N antitermination factor
65
How does phage lambda use antiterminatino systems?
phages do not encode their own RNA polymerase or include their own sigma factor instead they use the host to carry out its transcription. IT does this using antitermination. Specifically for regulation of early and late transcription - the two systems work by different mechanisms (N and Q)
66
Following phage infection which protein is expressed? What is the role of this protein?
The anti termination protein N, which forms a complex with the host proteins called nus factors and this complex modifies the RNA polymerase such that it no longer responds to terminators.
67
In phage, where does the anti termination complex form?
on nascent RNA at the sequence called the nut sequence or N utilization site. this prevents termination at that site.
68
Aside from N, what other protein is produced by the antiterminator transcript? How does it differ from N?
the Q antiterminator
unlike N, Q is bound to DNA, and like N it travels with the RNA polymerase and somehow interferes with termination.
N- required for early genes
Q- required for later genes in fatal
69
What is a cascade of sigma factors?
a cascade is created when one sigma factor is required to transcribe the gene coding for the next sigma factor
ex: the early genes of phage SPO1 are transcribed by host RNA polymerase
70
How do sigma factors work in a cascade manner?
1) The phage SPO1, which infects bacillus subtilis, has three stages of gene expression.
The early genes are transcribed immediately upon infection. and these are transcribed by the bacteria holoenzyme.
2) The early gene 28 codes for a new Sigma factor that replaces the bacterial Sigma.
3) During the middle phase, the gene 28 core enzyme starts transcribing the phage middle genes; middle gene 33 and 34 code for proteins that replace gp28. Now the product of these middle genes replaced the gp28
4) finally the gp33 in the gp34 core enzyme complex start transcribing the late genes
71
What is the heat shock response?
a set of loci/genes that is activated in response to an increase in temperature that cause proteins to denature and other abuse to the cell.
- all organisms have this response
- the gene products usually include chaperones that act on denatured protein
72
What do heat-shock chaperones and proteases do?
they aid in either refolding denatured protein or degrading them in order to reduce the levels of unfolded proteins.
73
In e.coli what sigma recognizes promoters of the heat shock team?
sigma 32, encoded by RPOH gene recognizes promoters of the heat shock system
-Sigma e is also induced by accumulation of unfolded proteins in the periplasm and outer membrane
74
How does the heat-shock response have feedback regulation
-As the chaperones and proteases help in refolding/degrading proteins, the level of unfolded proteins decreases and the proteases now start degrading the Sigma32, thus returning sigma 32 levels to normal.
75
What is an anti-sigma factor?
a protein that binds to a sigma factor to inhibit its ability to utilize special promoters
76
How many sigma factors does e.coli have?
7, which causes RNA polymerase to initiate at a set of promoteres that are defined by very specific -35 and -10 sequences
77
What determines the set of promoters at which transcription is initiated?
the sigma factor associated w/the core enzyme determines the set of promoters at which transcription is initiated.
-E.g: holoenzyme w/sigma 70 recognizes one set of promoters. If another sigma factor is bound to the core enzyme it will cause the enzyme to recognize a different set of promoters
78
Why is supercoiling an important feature of transcription?
negative supercoiling increases the efficiency of some promoters by assisting the melting reaction.
-Transcription generates positive supercoils ahead of the enzyme and negatvie supercoils behind it and these must be removed by gyrase and topoisomerase
79
The efficiency of the promotores is helped much by negative supercoiling. why should that be?
the promoter's dependence on supercoiling is based on the free energy needed to melt the DNA , which in turn is dependent on the DNA sequence. The higher the GC content in the promoter sequence, the more dependent the promoter would be on supercoiling to melt the DNA. As transcription progresses, positive supercoils are produced ahead of the enzyme, while negative supercoils are produced behind it. positive supercoiling would mean that the Helix is more tightly wound while negative supercoiling is one in which the DNA is partially unwound. Thus, transcription is not only affected by supercoiling but it also affects the actual structure of the DNA.
80
How do topoisomerase I and gyrase assist w/supercoiling?
The topoisomerase I removes/relaxes the negative supercoils and the gyrase introduces positive supercoils in DNA and these help to relieve the stress that builds up at the transcription bubbles
As RNA moves along the DNA it generates positive supercoils ahead of it and leaves negative supercoils behind. The enzymes gyrase and topoisomerase I are required to introduce and remove negative supercoils, respectively, to restore normal coiling of DNA.
81
What is polarity?
the effect of a mutation in one gene in influencing the expression (at transcription or translation) of subsequent genes in the same transcription unit.
82
What is the rho factor?
1) a protein that binds to nascent RNA at the rut site and translocates along the RNA until it catches up w/the RNA polymerase.
rho binds to nascent RNA at the rut site (rho utilization site, which is a sequence of RNA that is recognized by the rho termination factor.
83
What is a common feature of the rut site?
a HIGH percentage of C residues and LOW percentage of G residues and no secondary structure
84
Translocation of rho along the RNA is driven by what?
ATP hydrolysis
85
Describe the rho dependent termination mechanism
1) rho binds to transcript at rut site and follows the RNA polymerase using ATP hydrolysis
2) Hairpin structure forms and this causes the polymerase to pause allowing rho to catch up to the polymerase. (THIS PAUSING IS IMPORTANT b/c otherwise rho may not have enough time to get to the polymerase
3) Rho has helicase activity that it uses to unwind the RNA-DNA hybrid transcript and interact w/the poly to release the RNA (NEED TO LEARN THIS PART)
- no strings of U's are needed just hairpin secondary structure formation (NEED TO LEARN THIS PART)
86
what happens when the rho factor, during translocation, encounters a ribosome that is carrying out translation?
- in the wild-type form the rho dependent termination sites within the transcription unit are hidden by the translating ribosome and hence the rho factor cannot act on the downstream RNA polymerases. Rho attaches but the ribosome impedes its movement and transcription continues.
- In nonsense mutations, the ribosome will fall off/dissociate, allowing the rho factor to access the RNA polymerase. Transcription terminates but it does so prematurely and thus prevents distal genes from being expressed.
87
Describe the antiterminator complex and rho
An antiterminator complex seen in ribosomal RNA are formed by proteins that inhibit the action of rho, allowing the polymerase to transcribe through certain Terminator sites
88
What is a readthrough? How does it differ from antitermination?
a readthrough occurs at transcription or translation when RNA polymerase or the ribosome ignores a termination signal b/c of a mutation of the template or the behavior of an accessory factor. Whereas antitermination is a mechanisims of transcription control, in which termination is prevented at a SPECIFIC TERMINATOR SITE, allowing the rna polymerase to read into the genes beyond it.
89
Which parts of the sigma factor interact directly with bases in the -10 and -35 elements? Which parts make contact with the promoter - downstream and upstream?
Region 2 (2.3 and 2.4) and region 4 (4.2) directly interact with bases in the -10 and -35 elements
DOWNSTREAM:region 1.2 makes contact with the promoter just downstream of the -10 element
UPSTREAM:region 3 contacts bases just upstream of the same element
90
The sigma factor 70 in e.coli recognizes which elements? What does this suggest about the sigma factor
contacts with bases principally on the non-template strand of the - 1 0 element, the extended - 10 element, and the discriminator region, and it continues to hold these contacts after the DNA has been unwound in this region.
suggestion that the sigma factor is required for the melting of the DNA
91
Q) what is the structure of free sigma (sigma factor that is not bound to the core enzyme)
the N terminal domain blocks off its DNA binding region.
92
Q) How does the N terminal domain of sigma 70 change when it binds to the core and enzyme? What does this change do to the sigma factor
Once the Sigma factor binds to the core and enzyme it changes its conformation such that the N terminal end swings out from the DNA binding domain as well as the two DNA binding domains separate out from each other allowing DNA binding regions to interact with the promoters.
this change in effect elongates the sigma factor so that it can extend over 2 turns of the DNA
93
TRUE OR FALSE: RNA polymerase and DNA-protein interactions are the same for promoter recognition and DNA melting
true
94
what is footprinting?
a technique for identifying the site on DNA bound by some protein by virtue of the protection of bonds in this region against attack by nucleases. AKA it identifies DNA-binding sites for proteins because these sites are protected against nicking - endonucleases do not have access to the phosphodiester bonds here and cleavage does not occur
95
Describe the footprinting method
1) a DNA sequence bound to protein is paritally digested
2) the DNA molecule is radially labeled at one end of a strand allowing for identification of the position of which cleavage occured.
3) the partial digestion results in the generation of different size fragments which are put through gel electrophoresis, separating fragments according to size
4) if DNA was not bound to any protein than every phosphodiester bond would be cleaved in one or another molecule. DNA bound to protein will be subjected to partial digestion and in the gel this will show a region with an absence of bands - suggesting cleavage failed to occur.
96
How can footprinting be used to get the nucleotide sequence at the binding site?
by coupling this with the sequencing reaction one can get the nucleotide sequence of the binding site as well
97
most of the points of contact of the RNA pol with the DNA promoter are found where? How was footprinting used to identify these points of contact.
Most of the contact points for RNA polymerase with the promoter is in the -10 and -35 element. the points of contact are primarily on one case of the DNA.
scientists were able to identify the points of contact of the RNA polymerase with the DNA. one way was to modify the DNA before it bound to the RNA polymerase. if the modification prevented binding to the polymerase then we know that that particular base position was essential for contact. If the RNA polymerase DNA promoter complex was modified then comparing the pattern of protective bands with that of free DNA and of the unmodified complex , we can identify sites where the enzyme has protected the promoter from being modified. These would be regions where the bands have disappeared. Other bands that have increased in intensity would suggest that these sites were where the DNA was in a confirmation that exposed it much better to the cleavage agent.
98
TRUE OR FALSE: Promoter recognition and melting of DNA occur concurrently
TRUE:promoter recognition and melting of the DNA occur concurrently with melting beginning with base flipping where the base is a 11 and T7 are flipped out into pockets in the Sigma factor. this begins strand separation and unwinding begins at the right end of -11 and propagates down to just past the start site at +3.
99
TRUE OR FALSE: Sigma factor and core enzyme recycle (are released) at the same time during transcription?
False: at different points, the core enzyme is released at termination and the sigma factor is released when newly formed RNA is about 10 nucleotides long (aka it's released first)
100
TRUE OR FALSE: The RNA exit channel is occupied by a domain of the sigma factor, which allows RNA synthesis to proceed.
the RNA exit channel is occupied by a domain of the Sigma factor and must be DISPLACED so that RNA synthesis can continue
101
What is DNA scrunching and when does this occur?
when the core enzyme binds to the promoter it attempts to escape and the enzyme does not move down the template but rather pulls the first few nucleotides of the downstream DNA into itself -this process is called DNA scrunching and occurs when the enzyme is attempting to escape the promoter
102
What follows after the core enzyme escapes the abortive mode/attempts?
the enzyme can now form a true elongation complex and move into the elongation phase
103
How does DNA enter RNA polymerase? How do nucleotides enter?
The downstream clamp holds the DNA in position as it enters the RNA polymerase. Due to the wall of protein in the active site the DNA is forced to make a 90 degree turn. nucleotides enter the active site from below through another channel called the secondary channel or the pore.
104
How does translocation proceed?
by the brownian ratchet mechanisms
the energy for binding the correct nucleotide stabilizes the active conformation and also prevents backtracking
105
What does the movement of nucleic acid polymerase require?
The movement of the nucleic acid polymerase requires breaking and remaking the bonds to the nucleotides at fixed positions relative to the enzyme. the nucleotides in these positions change every time the polymerase moves up a base along the template.
106
During elongation, why do polymerase pause and backtrack?
pausing is physiologically important b/c it allows translation to keep pace w/transcription and it is the first step in termination
107
What happens to the 3’ end of the nascent RNA when transcription gets blocked by DNA damage?
when transcription gets blocked there is misalignment of the 3’ end of the nascent RNA with the active site. hence there has to be repositioning of that 3’-OH group back into the active site so the incoming nucleotide can attack it to form the phosphodiester bond.
108
How is misalignment of the 3’end due to blocked transcription realigned?
The manner in which realignment is accomplished is by the RNA polymerase itself cleaving the last few nucleotides and generating a 3’ end that is now positioned in the active site. Besides the RNA polymerase having the main role in the cleavage process its activity is stimulated by other factors. In E coli it has been identified as GreA and GreB, while in eukaryotes it is TF2S. These proteins insert a domain deep into the RNA polymerase close to the catalytic site.
109
What are two types of terminators?
1) Intrinsic terminators - recognized solely by RNA polymerase itself w/o the requirements of any cellular factors
2) rho dependent terminators - require cellular protien called rho
110
What must happen to the hydrogen bonds between RNA-DNA hybrid for termination to occur?
termination requires that all the hydrogen bonds between the RNA-DNA hybrid be broken after which the DNA reforms its double Helix.
111
Why has most of the information scientists have got about bacterial RNA transcription termination has been from in vitro experiments?
Most of the information scientists have got about bacterial RNA transcription termination has been from in vitro experiments. the reason being, since there are no specific 3’ end modifications as part of the post transcriptional processing as one sees in eukaryotes, it is difficult to determine if the 3’ end of the RNA, isolated from a living cell, is actually the termination site or if it has been subjected to exonucleases or even an endonuclease. one of the caveats part in vitro experiments is that since these are achieved by changes in ionic strength or temperature, it does not reflect the true conditions in a living cell. hence one cannot be sure if the termination site identified in the in vitro experiment would be the same site in the cell.
112
TRUE OR FALSE: if the in vitro experiment site matches with the sites detected in RNA isolated from living cells then there is sufficient confidence in stating that this is a termination site
TRUE
113
what are the two main features found in intrinsic terminators? what does this say about the DNA sequences required for termination?
1) first recruitment of the GC rich hairpin to form in the secondary structure of the transcribing RNA.
2) there has to be a stretch of up to seven “U” residues, or uracil residues in RNA and correspondingly 7 thymine residues in the DNA following the hairpin stem but before the actual site of termination.
this means that the DNA sequences required for termination are actually located upstream of the Terminator sequence
114
Intrinsic termination requires recognition of what? and signals for termination lie mostly where
1) intrinsic termination requires recognition of a terminator sequence ( GC rich region and single stranded U residues) in DNA that codes for a hairpin structure in the RNA product (hairpin structure is formed by palindromic sequences
2) the signals for termination lie mostly w/in sequences already transcribed by RNA polymerase and thus termination relies on scrutiny of the template and or the RNA product that the polymerase is transcribing
115
What forces contribute to the destabilization of the elongation complex leading to termination in intrinsic termination?
The RNA-DNA hybrid plays a major role thermodynamically in the stabilization of the elongation complex. so when there is a stretch of U residues, the RNA uracil within DNA adenine RNA-DNA hybrid that is produced is much weaker than the GC pairs. In addition the secondary hairpin structures that are formed by the palindromic sequence lead to the misalignment of the 3’-OH groups in the active site. both of these forces contributed the destabilization of the elongation complex leading to termination
