pp 11 Flashcards

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1
Q

What is transcription?

A

DNA-templated synthesis of RNA, with immediate product from the DNA template. Double stranded DNA is unwound and becomes single stranded DNA.

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2
Q

What is primary transcript?

A

Pre-mRNA processed to become mRNA and they may be cleaved into tRNA or rRNA.

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3
Q

What is the enzyme mechanism of RNA synthesis?

A

Ribonucleotides are added from an rNTP to the growing transcript through formation of a phosphodiester bond. The direction of the synthesis is 5’ to 3’ and is catalysed by RNA polymerase.

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4
Q

What are the different subunits of RNA polymerase in E.coli?

A
  1. core enzyme: B subunit, B’ subunit, two a subunits and w subunit
  2. core enzyme and sigma form a holoenzyme, o70
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5
Q

Why are sigma factors important?

A

The presence of sigma subunit allows recognition of authentic RNA polymerase binding sites.

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6
Q

Explain the filter binding assay experiment (nitrocellulose)

A
  1. mix radioactive T7 phage DNA and RNA polymerase
  2. Add excess non-radioactive T7 phage DNA
  3. If it is bound tight, radioactive DNA will stay bound to RNA and stay on filter, if not it will go through.
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7
Q

What is the conclusion of the assay protein-DNA binding?

A

Holoenzyme binds the DNA more than the core, when temperature is increased, it promotes tigher DNA binding, because DNA melting promotes strong binding.

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8
Q

What are the RNA polymerase binding sites?

A

Promoters

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9
Q

How does the polymerase bind the promoter?

A

holoenzyme “scans” DNA until it finds promoter. It then loosely binds it, creating closed promoter complex. Holoenzyme will then melt DNA at the promoter site, forming open promoter complex, binding polymerase TIGHT

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10
Q

What are the promoter sequences?

A

sequences recognized by RNA polymerase holoenzyme core

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11
Q

How are the promoters recognized?

A

core promoter elements are recognized by the sigma subunit

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12
Q

What are the core promoter elements?

A
  • -10 region upstream transcription site
  • -35 region upstream
  • UP element bounded by a subunit
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13
Q

What are the consensus sequences in E.coli?

A

10 box: TAtAaT
35 box: TTGACa

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14
Q

What mutations weaken promoter binding in RNA polymerase?

A
  • down mutations: increase deviation from consensus sequence
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15
Q

What mutations strengthen promoter binding?

A

Up mutations: decrease deviation from the consensus sequence

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16
Q

How do we do a direct analysis of protein-DNA interactions

A

Using DNAse footprinting, we can identify where/what sequences sigma factor binds

17
Q

How is DNAse footprinting used?

A

when protein binds to DNA, there is protection that protects the site from digestion by DNAse.
1. take DNA strand and amplify
2. label one end with pieces of DNA and treat with DNAse
3. it will randomly cleave DNA, wont touch part where protein is
4. gel electrophoresis, when theres a gap, theres a protein

18
Q

What has DNAse footprinting proven?

A
  1. Different sigma factors recognize different consensus sequences in the core promoter elements
  2. Different sigma factors dominate in the cell under different conditions
19
Q

Why is UP element used?

A

gene expression is needed at high levels, the consensus sequences for -10, -35 elements is not strong enough, therefore the UP element can stimulate transcription by a factor of 30

20
Q

How does RNA polymerase bind UP unit?

A

RNA polymerase binds to core with sigma factor, it then binds to the promoters in -35 and -10 region, however, the a subunits will interact with the UP element. This produces a high level of transcription

21
Q

What are the stages of transcription?

A
  1. Transcription initiation
  2. Transcription elongation
  3. Transcription termination
22
Q

What is the role of the sigma factor in transcription initiation?

A

Sigma factor is involved in specificity of gene transcription. It causes tight binding between RNA polymerase and promoters. This is because the binding depends on the melting of DNA that opens promoter complex. This causes sigma factor to dissociate from the core after enabling polymerase-promoter binding

23
Q

What briefly happens during initiation?

A
  1. Closed promoter complex formed
  2. DNA melts: closed to open complex
  3. Abortive transcription
  4. RNA polymerase comes loose from promoter – promoter clearance. The transcript will then become long enough to form a stable hybrid with template
24
Q

What is abortive transcription?

A

Polymerization of early nucleotides (polymerase is stuck at promoter) draws downstream DNA into the polymerase without actually moving or losing its grip on promoter.

25
Q

How is sigma factor reusable?

A

It is recycled through sigma cycle. When sigma factor is released from holoenzyme, it can associate with a core enzyme.

26
Q

What briefly happens during elongation?

A
  1. Core enzyme continues elongation of RNA
  2. Sigma factor dissociates, reused in initiation
  3. nucleotides are added one after another in 5’ to 3’ direction
  4. RNA polymerase is proccessive and wont be released until transcription is finished
27
Q

What is the function of the core polymerase?

A

It has the RNA synthesizing machinery

28
Q

How are the transcription bubble formed?

A

17 bp of DNA is unwound and forms the bubble. Positive supercoils form due to unwinding and movement of polymerase ahead of polymerase. Negative supercoils form behind the polymerase (are removed by topoisomerases)

29
Q

Why are transcription bubbles formed?

A

This enables RNA polymerase to access the template strand

30
Q

What is the topology of elongation?

A

polymerase will maintain a short, melted region of template DNA, and is unwounded ahead of the polymerase, the strain is relaxed by topoisomerase

31
Q

What are the different terminators?

A
  1. intrinsic terminators: function with the RNA polymerase by itself without help of other proteins
  2. Depends on auxilary factor called RHo (p): they are called p-dependent terminators
32
Q

What does Rho-independent termination depend on?

A

-Inverted repeat followed immediately by
-T-rich region in nontemplate strand of gene

33
Q

What is the purpose of inverted repeats and T-rich region?

A

Predisposes a transcript to form a hairpin structure. The string of U’s downstream of hairpin causes RNA polymerase to pause

34
Q

In Rho-dependent termination what is the mechanism of Rho?

A

There has to be a rut site where rho (p) can bind transcript. Rho can then feed itself the RNA (ATP-hydrolysis driven). It goes from 5’ to 3’ along mRNA to polymerase and will then cause the separation of mRNA from polymerase.