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Molecular Genetics > Prokaryotic Transcription 2 > Flashcards

Prokaryotic Transcription 2 Flashcards

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1
Q

what are the two subunits that show specific DA recognition in binding?

A
  • alpha subunits
  • sigma subunit
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2
Q

what are the two elements most promoters have?

A

-10 and -35

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3
Q

what is the UP element?

A
  • stimulates upward transcriptional activity
  • upstream promoter element
  • found in ribosomal structural genes
  • alpha subunits makes contact with it and recognizes the sequence from the closed promoter
  • does this by having two hairpin motifs with the two hairpins contacts both sides of the groove
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4
Q

domain 1.2

A
  • in the sigma element
  • discriminator element
  • some bacteria has additional info upstream of -10, so it recognizes this info
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5
Q

domain 3

A

recognizes the extended TATA
- binds to the extended -10

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6
Q

domain 3.2

A

in the RNA exit pores and 5’ end of transcription
- contacts first nt of transcript
- region 3.2 of the sigma subunit contributes to the binding of the 5’ initiating nucleotide in the RNA polymerase active center and facilitates promoter clearance during initiation

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7
Q

domain 4.2

A
  • uses the major groove of the sequence that is found at -35
  • also called the -35 site
  • helix turn helix
  • activator contacts and alpha CTD
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8
Q

sigma 70

A

most required functions

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9
Q

sigma S

A

stationary phase/some stress responses
- cannot interact with alpha CTD, specificity involves UP motif

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10
Q

Sigma 32

A

heat shock

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11
Q

sigma E

A

periplasmic/extracellular proteins

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12
Q

sigma 54

A

nitrogen assimilation
- It is unusual in that it can bind to the promoter in the absence of the Core. It has no inhibitory domain. This makes it more like a eukaryotic transcription factor.
- no 1.1 domain

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13
Q

sigma F

A

flagellar synthesis/chemotaxis

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14
Q

sigma fecl

A

iron metabolism/transport

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15
Q

what happens when a mutation introduces more spacing in the binding domains?

A
  • decrease binding and efficiency of promoter
  • this is because the sigma is not very flexible and has set -10 and -35 domains
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16
Q

domain 2.2

A
  • makes contact with core beta prime unit
  • connection to the core, recognizing tata and melting the promoter
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17
Q

domain 2

A
  • domain 2 is the most conserved domain
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18
Q

domain 2.3

A
  • melting: aromatic and basic aa
  • aromatic acids trp433 and tyr430 make contact and promote melting
19
Q

domain 2.4

A

base specific interactions with -10
- recognizes tata
- domain 2.4 helix recognizes TA of TATAAT on non-template strand

20
Q

domain 2.1

A

most conserved-contact with core

21
Q

domain 1.1

A

initially binds in the active site in place of DNA

22
Q

what happens if you mutate away from the consensus?

A
  • decreases the rate of closed promoter complex formation (promoter recognition)
  • if you mutate at the -35, you decrease initial contact
23
Q

what happens if you mutate the -10 site?

A
  • mutations away from the consensus may decrease the rate of open and closed promoter complex formation
  • decreased promoter recognition and melting
24
Q

what is the role of rifampicin?

A

blocks bacterial RNA polymerase and is the major drug against tuberculosis. It binds close to the active site and blocks the formation of RNA beyond 2-3 nt. Blocks exit of nascent RNA. Mimics activity of sigma, but can not be dislodged.

25
Q

what is an additional role for domain 1.1?

A
  • prevents sigma from binding promoter without first binding to the RNA polymerase core
  • when sigma is floating with no attachment, domain 1.1 folds back onto the body of sigma and covers domain 4.2
  • sigma 54 lacks a 1.1 domain
26
Q

which subunit of RNA polymerase holoenzyme partially blocks the entry of DNA into the channel?

A

sigma region 1.1

27
Q

how does the holoenzyme transition from abortive cycling to elongation?

A

Sigma loses affinity for the promoter DNA and RNA polymerase, thus allowing the enzyme to move away from the promoter as elongation proceeds (70% sigma still loosely attached). Sigma then slowly disassociates from RNA polymerase. The growing RNA chain dislodges bound sigma (3.2) from the RNA exit pore.

28
Q

what are the seven differences between RNA polymerase and DNA polymerase?

A

1) RNAPs can initiate synthesis which involves promoter recognition
2) RNAPs can melt the DNA duplex
3) RNAPs initiation is primed by a single nucleotide, not an oligo as is the case for DNAPs
4) RNAPs make multiple contacts with the 2’-OH of the incoming NTP
5) DNA scrunching occurs for RNAPs allowing abortive cycling while still retaining contact with the promoter
6) For RNAPs, the transcript is peeled away from the template; not so for DNAPs where the open cleft allows the duplex to extend out of the enzyme
7) Initiation of synthesis is regulated by many proteins for RNAPs, but not for DNAPs

29
Q

what are the steps of footprinting?

A

Step 1: end-label the DNA at a specific site
Step 2: incubate DNA with Protein (RNA polymerase in this example)
Step 3: incubate DNA:Protein complex with dilute DNase I enzyme
Step 4: remove protein and run polyacrylamide gel electrophoresis (PAGE)

30
Q

What are the two major changes in RNA polymerase as it transitions from promoter recognition phase to elongation phase of transcription?

A
  • release of sigma
  • smaller footprint (changing of conformation)
31
Q

what is the process of RNA polymerase recovering from a stall?

A

Sometimes RNA polymerase pauses due to a temporary shortage of the complementary nucleotide. Thermal agitation, RNAP backtracks & 3’end out of active site. When this occurs, restarting synthesis requires the GreA and GreB proteins to release the pause. Acidic portion of the factors GreA & GreB repositions the Mg2+ ion at the active site. This makes RNAP act as a nuclease. RNP cleaves the 3’ end so that it is properly aligned within the catalytic site of the polymerase again.
- then resumes elongation

32
Q

which eukaryotic factor performs that same function as GreA and GreB?

A

TF2S

33
Q

the most heavily transcribed genes in e. coli fire once per second

A
  • true
  • ribosomal genes are the most actively transcribed of any in e. coli
34
Q

how much does the UP element increase the activity of the promoter?

A

60 fold

35
Q

core promoters

A

-10 and -35 elements

36
Q

-10 element

A

TATAAT

37
Q

-35 element

A

TTGACA

38
Q

upstream activating sequences

A

1, 2, and 3
- FIS sites that are positively acting
- stabilizes UP
- increases activity of the promoter 5 fold

39
Q

why is the distance between the -10 and -35 sites are so conserved?

A

the single protein sigma simultaneously makes contact

40
Q

what types of base changes would lead to increases or decreases in promoter activities?

A
  • anything that makes it more like the consensus and anything that makes the distance more like the consensus
  • there are not many consensus promoters
41
Q

What is a promoter to an operon regulated by sigma 54?

A
  • glutamine synthase in cyanobacteria
  • nif in e, coli
42
Q

dual promoters

A

: the vegetative promoter is read by sigma 70 and is weak. The nif-like promoter is much stronger and is read by sigma 54. This allows the gene to be active under normal conditions and when NH4 is in short supply.

43
Q

what are the limitations to sigma 54?

A
  • it is defective and unable to melt the promoter without added ATP and the NtrC protein

Molecular Genetics (28 decks)

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