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3130 > Chapter 6 > Flashcards

Chapter 6 Flashcards

1
Q

What are the components of an E. coli polymerase holoenzyme?

A

β’, β, σ, α (2x) , ω

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2
Q

T/F: The ω component is necessary for holoenzyme function.

A

False

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3
Q

T/F: SDS-PAGE gels denature.

A

True

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4
Q

Core refers to:

A

The holoenzyme without the σ subunit.

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5
Q

T/F: Transcription is less efficient when σ is missing from core.

A

True

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6
Q

What would you expect to see transcription-wise if you had nicked eukaryotic DNA with holoenzyme?

A

Normal transcription

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7
Q

What would you expect to see transcription-wise if you had nicked eukaryotic DNA with only core?

A

Medium transcription

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8
Q

What would you expect to see transcription-wise if you had intact bacteriophage DNA with holoenzyme?

A

Normal transcription

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9
Q

What would you expect to see transcription-wise if you had intact bacteriophage DNA with only core?

A

Only basal transcription

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10
Q

What is necessary for RNA polymerase to have specificity?

A

σ is required, otherwise both strands will be transcribed

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11
Q

What is specificity?

A

Action where only the template strand is transcribed and the coding strand is not

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12
Q

Describe the experiment used to show specificity in transcription.

A

DNA was incubated with core and holoenzymes. RNase was added to each. RNase only works on single-stranded RNA, so if transcription was non-specific, the RNase would not work to degrade the molecules. ~30% was found to be resistant to RNase.

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13
Q

Define promoter

A

Polymerase binding site.

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14
Q

What can a filter binding assay determine?

A

Amount of protein-DNA binding. Temperature at which there is more binding affinity. How binding affinity is affected by salt concentration.

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15
Q

Describe the process and theory of a filter binding assay.

A

Double-stranded DNA will not bind to a nitrocellulose filter, while single-stranded DNA and proteins will do so. dsDNA can be combined with proteins and then run through the filter to determine the strength of binding because the dsDNA-protein complex will bind. Measurement of the quantity of DNA on the filter is then measured (often radio-labeled DNA which is then measured with a scintillator).

Specifically, the two can be mixed together and aliquots taken at intervals. From this a curve can be obtained to show binding over time.

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16
Q

What are the weights of the components of the holoenzyme?

A

β - 160 kD
β’ - 150 kD
σ - 70 kD
α - 40kD
ω - 10 kD

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17
Q

Which of the following are true about σ?
a. σ directs RNA polymerase to start at specific promoters.
b. σ is not necessary for transcription to occur in a specific manner.
c. σ can be recycled.
d. Two of the above.
e. None of the above

A

d (a and c)

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18
Q

What are the steps in performing a filter binding assay?

A
  1. Purify enzymes.
  2. Incubate with 3H labeled DNA whose promoters are recognized by the polymerase being used.
  3. Add excess unlabelled DNA.
  4. Over time, remove aliquots and pass them through a nitrocellulose filter.
  5. Measure radioactivity on the filter with a scintillation counter.
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19
Q

In the filter binding assay, the level of holoenzyme bound to DNA remained fairly constant over time, while the amount of core bound to DNA diminished. Interpret these results.

A

Holoenzyme binds more tightly to DNA than core. Becuase the level stayed fairly constant, it was not dissociating from the DNA and then being outcompeted by the excess DNA. With core, the amounts quickly decreased because as core fell off, it was being outcompeted by non-labelled DNA.

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20
Q

START AT PAGE 3 OF NOTES

A
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21
Q
A
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22
Q

FOLLOWING IS LECTURE 2, IN-CLASS

A
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23
Q

Does σ always dissociate from core after initiation?

A

No, it may be random. Alternatively, experimental methods may have caused σ dissociation, so it may not happen at all in vivo.

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24
Q

What does FRET stand for?

A

Fluorescence Resonance Energy Transfer

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25
Q

Is trailing edge FRET a good method for determining whether σ is released from core during transcription? Why or why not?

A

No, because in trailing edge FRET the acceptor is attached to the 5’ end of DNA and the donor is attached to σ. The distance is measured between the two. If σ detaches, the distance will increase. But if it doesn’t detach, the distance will also increase. Therefore, the data will not tell you which scenario occurred.

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26
Q

What is the theory behind FRET?

A

Fluorescent molecules can be used to measure distance without appreciably changing the behaviour of the molecule. A fluorescence donor will transfer energy to a fluorescence acceptor and the amount of energy transferred can be measured to determine the distance between the two molecules.

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27
Q

What experiment would be used to determine whether σ dissociates from core during transcription?

A

Leading edge FRET.

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28
Q

What is the difference between Leading Edge FRET and Trailing Edge FRET?

A

In leading edge, the acceptor molecule is on the 3’ end of the DNA strand, while in trailing edge it is on the 5’ end.

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29
Q

When and why will polymerase be unable to dissociate from DNA?

A

In low salt (low Mg2+) because the polymerase slows and eventually stops.

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30
Q

What happens when you add core to stopped transcription due to low salt?

A

Transcription resumes because σ dissociates from the holoenzyme.

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31
Q

In the low-salt experiment, rifampicin-resistant holoenzyme was used. Transcription without rifampicin was higher than that with it. Does this mean rifampicin was involved in σ sensitivity?

A

No, it is an indicator that core was not 100% resistant to rifampicin.

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32
Q

What is an open promoter complex?

A

The complex formed by tight binding between RNA polymerase and a prokaryotic promoter. It is “open” in that at least 10 bp of the DNA duplex open up or seperate.

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33
Q

What is a closed promoter complex?

A

The complex formed by relatively loose binding between RNA polymerase and a prokaryotic promoter. It is “closed” in the sense that the DNA duplex remains intact, with no “opening up” or melting of base pairs.

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34
Q

What is the stochastic model?

A

Stochastic = random. This is the idea that σ may change how it associates with core, being more tightly bound to core during initiation and looser during elongation. This would make it available for other cores if it dissociates.

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35
Q

How many base pairs does RNA polymerase cause to melt at the transcription start site?

A

10-17

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36
Q

What does OPC stand for?

A

Open Promoter Complex

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37
Q

What is S1 and what does it do?

A

It is a nuclease and it cuts single-stranded sites.

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38
Q

Which are the purines?

A

A and G

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39
Q

Which are the pyrimidines?

A

C and T

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40
Q

What are primary σ factors?

A

σ factors that transcribe vegetative genes

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41
Q

What are alternate σ factors?

A

σ factors that transcribe specialized genes, like those for heat shock

42
Q

T/F: σ factors have barely any conserved regions.

A

False

43
Q

What does σ region 1 do and where is it found?

A

Region 1 is only found in primary σ molecules. It prevents σ from binding to DNA unless it is part of core.

44
Q

What does σ region 2 do?

A

Region 2, specifically 2.4, binds to the -10 box.

45
Q

Which region of σ is the most conserved?

A

Region 2

46
Q

How many parts is σ region 2 subdivided into?

A

4

47
Q

What is special about σ region 2?

A

It is the most highly conserved. In E. coli and B. subtilis, their 2.4 regions recognize the same -10 box sequence and are 95% similar.

48
Q

What does σ region 3 do?

A

It is involved in binding to core and DNA. We know very little about it.

49
Q

What does σ region 4 do?

A

Binds to the -35 box, specifically region 4.2 does so.

50
Q

Why do we believe σ region 4.2 binds to DNA and where?

A

Because it contains a helix-turn-helix motif which is common to DNA binding proteins. It binds the -35 box.

51
Q

What are the two most conserved σ regions?

A

2 and 4

52
Q

What binds to the -10 and -35 boxes?

A

σ 2.4 and σ 4.2, respectively.

53
Q

What is one reason we believe σ can’t bind to DNA without core?

A

β’ helps σ bind to the non-template strand in the -10 region of the promoter.

54
Q

Describe the experiment used to determine UP element recognition.

A

Polymerases with three types of α subunits were tested, one missing 94 amino acids from the C-terminal end, one with a point mutation R265C, and the wild-type. They were combined with two different DNA types, one with a UP promoter and the other without.

The 94C did not distingush between promoters with or without the subunit. R265C destroyed the ability to recognize the UP promoter. WT had strong transcription when UP was present and similar transcription to the other two when it was not.

55
Q

What is a DNase footprinting assay?

A

A method of determining where a protein binds to a DNA strand.

56
Q

How is a DNase footprinting assay performed?

A
  1. DNA is radio-labeled on one strand.
  2. Protein is bound to the DNA.
  3. The DNA-protein complex is treated with a small amount of DNase, preferably an amount that will cut each strand once.
  4. DNase is washed away.
  5. Protein is removed.
  6. DNA is melted
  7. Fragments are electrophoresed on a polyacrlamide gel, along with a control and size ladder.
  • Usually multiple concentrations of protein are run so it is shown that the protection is from the protein and not chance.
57
Q

What were the results of the footprinting experiment discussed in class?

A

Wild-type polymerase covered the core promoter and the UP element. The -94C mutant only covered the core element. Therefore, the C-terminal region is necessary to bind to the UP element.

58
Q

What is limited proteolysis?

A
59
Q

What did limited proteolysis show in regards to the α subunit of RNA polymerase?

A
60
Q

What is streptolydigin?

A

An antibiotic that blocks elongation.

61
Q

T/F: The β’ subunit of core is involved in both initiation and elongation.

A

False (it’s the β subunit that is involved in both processes)

62
Q

What are the steps of affinity labelling?

A
63
Q

How do you read the result of affinity labelling?

A

SDS-PAGE gel

64
Q

What species was used to determine the shape of core?

A

Thermus aquaticus

65
Q

What is the shape of core?

A

A clawed hand shape

66
Q

T/F: β’ and σ make up the bulk of core.

A

False, β’ and β make up the bulk of the molecule.

67
Q

T/F: The two α subunits are found in the “hinge” of core polymerase.

A

True

68
Q

Which ions are present in RNA polymerase?

A

Two Mg2+ (one is chelated, the other shuttles back and forth) and one Zn2+ (keeps the polymerase properly aligned but does not participate in enzymatic action).

69
Q

How do we know Mg2+ is important for catalytic activity of RNA polymerase?

A

There is a conserved 7-aa string found in the β’ subunit of all prokaryotes that has three aspartic acid residues that hold the Mg2+ in place. Mutating any one of these residues out is lethal for the cell as core can form the OPC, but no catalytic activity happens.

70
Q

What will RNA polymerase do or not do if there is no Mg2+ in the active site?

A

The polymerase will be able to form the OPC, but it will be unable to form phosphodiester bonds.

71
Q

What does rifampicin do and how?

A

It prevents initiation by blocking the exit of the polymerase. This is why it does not inhibit elongation.

72
Q

Which amino acids interact with the -10 box when σ attaches to DNA?

A

HENQ

73
Q

Which amino acids interact with the phosphate backbone of DNA and why?

A

Arg and Lys (R and K) because they are positively charged.

74
Q

Which amino acids assist in base-stacking the single-stranded DNA in RNA polymerase?

A

Tyr, Phe, and Trp. They are aromatic and the aromatic ring assists with stacking and thus stabilizes the ssDNA.

75
Q

Describe the competing theories of opening and closing the DNA strand.

A

1) The polymerase could rotate around the DNA template, but this would cost a lot of energy and the RNA would be wound around the DNA.
2) The polymerase could move along the DNA and the DNA ahead would unwind and then rewind behind the polymerase. This would induce strain but could be mitigated by topoisomerases.

76
Q

What is an intrinsic terminator?

A

A type of termination that does not require any outside elements.

77
Q

What does an intrinsic terminator require?

A

An inverted repeat, followed immediately by a T-rich region in the non-template strand.

78
Q

What does an inverted repeat do in regards to termination?

A

It causes a hairpin to form.

79
Q

What does a T-rich region in the non-template strand do?

A

It creates an area of weak base pairing, which, when combined with the inverted-repeat hairpin, causes the RNA transcript to terminate.

80
Q

Why does iodo-CTP form stronger base pairs with G?

A

Iodo-CTP is more electronegative.

81
Q

What does inocene triphosphate do when substituted for GTP?

A

It weakens bonds because it only forms two hydrogen bonds instead of 3.

82
Q

What changes can be made to DNA to weaken termination?

A

Changing the sequence in the T-rich region.
Using inocenetriphosphate instead of GTP which weakens the hairpin structure.

83
Q

What method(s) can be used to strengthen termination?

A

Using Iodo-CTP instead of normal CTP, which strengthens the C-G bond and thus the hairpin.

84
Q

What is the accepted model for termination?

A

Polymerase pauses at weak dA-rU base pairs –>
Hairpin forms, further destabilizing the DNA-RNA hybrid –>
Hairpin could pull RNA out or cause the transcription bubble to collapse

85
Q

What happens if half the inverted repeat is removed from a termination sequence?

A

The polymerase pauses

86
Q

What happens if you add an oligonucleotide complementary to an inverted repeat that is missing half of its sequence?

A

The RNA will release.

87
Q

In the simplest terms, when will intrinsic termination occur?

A

When something base-pairs to the transcript upstream of something that causes the polymerase to pause.

88
Q

T/F: Rho-dependent terminators contain neither an inverted repeat, nor a T-string

A

False. They have an inverted repeat but no T-string.

89
Q

What is rho?

A

A protein that causes termination.

90
Q

What does rho accomplish?

A

It decreases the net rate of RNA synthesis. Chain elongation decreases as rho concentration increases.

91
Q

T/F: Rho binds to RNA polymerase before binding to RNA

A

True

92
Q

T/F: Rho does not contain an RNA binding domain.

A

False

93
Q

T/F: Rho will bind to any available RNA transcript

A

False, a loading site on the RNA is necessary.

94
Q

T/F: Rho is a dimer

A

False, rho is a hexamer

95
Q

T/F: Half of the rho subunits have ATPase activity

A

False, all subunits have ATPase activity

96
Q

T/F: Rho’s ATPase activity is activated upon binding to RNA

A

True (likely powers helicase activity)

97
Q

Describe rho’s mechanism of action

A

1) Rho attaches to the RNA polymerase

2a) Once long enough, rho binds to the RNA transcript,
2b) with the RNA passing through the center of the hexamer and forming a loop.

3) Rho progressively tightens the loop until the RNA-DNA hybrid dissociates

n.b. Hudak finds 2b and 3 to be unlikely, but admits it is the best of our knowledge at this point.

98
Q

T/F: Rho has helicase activity

A

True

99
Q

T/F: Rho has topoisomerase activity

A

False

100
Q
A

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