lecture 6 - transcription (general info and specifically in prokaryotes)

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44 Terms

1
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in prokaryotes, where does transcription begin? end?

  • begins at promoter region of the gene

  • ends at termination signal

2
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is transcription specific? what DNA participates?

  • yes specific

  • only genes participate

3
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what DNA strand is read by RNA pol? what does the mRNA resemble?

template/non-coding strand and makes a strand complementary to it (identical to the coding strand/non-template strand)

4
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name the required components for transcription

  • Enzyme - RNA pol

  • ATP, GTP, CTP, UTP

  • Mg++

  • DNA template

5
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explain the experiment that described how RNA pol works. what were the conclusions?

Experiment:

  • Used DNA with alternating AT
    sequence and following the
    incorporation of different dNTPs

  • Isolated new RNA molecule and
    analyzed for content

conclusions

  • RNA pol copies DNA

  • the product RNA molecule only had As paired with Us

  • RNA pol used complementary base pair rule

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full name for RNA pol

DNA dependent RNA polymerase

7
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describe the chemical mechanism of RNA synthesis

  • Requires Mg++

  • Adds nucleotides in 5’ to 3’ direction one by one

  • only one strand on DNA serves as a template in 3’ to 5’ direction but it could be either strand

8
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describe the structure of bacterial RNA pol

polymerase core

  • 2 copies of α subunit, one of each β, β’ and ω

  • α2ββ’ω (MW = 390,000 Da)

sigma factor

  • One subunit of σ (many different σ subunits, σ70 is most common)

  • σ70 (MW = 70,000 Da)

holoenzyme = core + sigma factor

9
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function of sigma factor

decides which promoter to bring to RNA pol core to begin transcription

10
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how does eukaryotic RNA pol differ from bacterial?

  • eukaryotic DNA pol has no sigma

  • uses other transcription factors

11
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compare the 5 subunits of prokaryotic RNA pol

Prokaryote) RNA polymerases have five types of
subunits: α, β, βʹ, and ω have rather constant sizes in different bacterial species, but σ varies more widely

  • σ gives specificity

12
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how are promoter sequences recognized in prokaryotes vs eukaryotes?

  • Recognized by sigma factor of bacterial RNA polymerase

  • Recognized by initiation factors eukaryotic RNA polymerase

13
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describe the promoter sequences sections

  • -35 is RNA pol recognition site

  • -10 is RNA pol binding site (TATA box aka Pribnow box)

  • transcription start: at +1 site also -10 from TATA box

<ul><li><p>-35 is RNA pol recognition site</p></li><li><p>-10 is RNA pol binding site (TATA box aka Pribnow box)</p></li><li><p>transcription start: at +1 site also -10 from TATA box</p></li></ul>
14
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define transcription unit

a sequence of DNA transcribed into a single RNA starting at the promoter and ending at the terminator (the genes transcribed)

15
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explain how the RNA pol binds to DNA (role of the core enzyme and sigma factor)

  • Core enzyme binds indiscriminately to any
    DNA.

  • Sigma factor reduces the affinity for
    sequence-independent binding and confers specificity for certain promoters

  • i.e., core enzyme binds any DNA, sigma factor will reduce stability of non-specific/random binding and increase affinity of holoenzyme to specific promoters

<ul><li><p><span style="font-family: sans-serif">Core enzyme binds indiscriminately to any</span><span><br></span><span style="font-family: sans-serif">DNA. </span></p></li><li><p><span style="font-family: sans-serif">Sigma factor reduces the affinity for</span><span><br></span><span style="font-family: sans-serif">sequence-independent binding and confers specificity for certain promoters </span></p></li><li><p><span style="font-family: sans-serif">i.e., core enzyme binds any DNA, sigma factor will reduce stability of non-specific/random binding and increase affinity of holoenzyme to specific promoters</span></p></li></ul>
16
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explain how the general elongation complex is formed

  • RNA polymerase initially contacts the region from −55 to +20.

  • When sigma dissociates, the core enzyme contracts to −30; when the enzyme moves a
    few base pairs, it becomes more compactly organized into the general elongation complex

<ul><li><p><span style="font-family: sans-serif">RNA polymerase initially contacts the region from −55 to +20. </span></p></li><li><p><span style="font-family: sans-serif">When sigma dissociates, the core enzyme contracts to −30; when the enzyme moves a</span><span><br></span><span style="font-family: sans-serif">few base pairs, it becomes more compactly organized into the general elongation complex </span></p></li></ul>
17
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describe sigma cycle

  • binding of RNA pol holoenzyme to form closed complex (RNA pol binds to ds helix)

  • formation of open complex (open helix - form replication bubble)

  • sigma factor is released now that initiation is over because it is only needed to give specificity

  • RNA pol core enzyme transcribes

<ul><li><p>binding of RNA pol holoenzyme to form closed complex (RNA pol binds to ds helix)</p></li><li><p>formation of open complex (open helix - form replication bubble)</p></li><li><p>sigma factor is released now that initiation is over because it is only needed to give specificity </p></li><li><p>RNA pol core enzyme transcribes</p></li></ul>
18
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elongation

  • generally describe the process

  • describe how RNA pol moves during elongation

  • A continuous process until termination

  • Basically, elongation is the stage when the RNA strand gets longer, thanks to the addition of new nucleotides.

  • During elongation, RNA polymerase "walks" along one strand of DNA, known as the template strand, in the 3' to 5' direction. For each nucleotide in the template, RNA polymerase adds a matching (complementary) RNA nucleotide to the 3' end of the RNA strand.

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how does RNA pol ensure the transcript is correct?

  • kinetic proofreading

  • nucleotide proofreading

both done by RNA pol

20
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describe kinetic proofreading

  • when nucleotides are mis-matched they tend to dissociate more rapidly then when correctly matched

  • Nucleotide is removed by pyrophosphorolysis

  • then another nucleotide is added again (try again)

<ul><li><p><span style="font-family: sans-serif">when <mark data-color="yellow">nucleotides are mis-matched</mark> they tend to<mark data-color="yellow"> dissociate more rapidly</mark> then when correctly matched</span></p></li><li><p><span style="font-family: sans-serif">Nucleotide is removed by pyrophosphorolysis</span></p></li><li><p><span style="font-family: sans-serif">then another nucleotide is added again (try again)</span></p></li></ul>
21
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describe nucleotide proofreading

  • RNA polymerase reverses direction to remove mismatched bases (usually create a bulge)

  • RNA pol goes forward again once done proofreading

<ul><li><p><span style="font-family: sans-serif">RNA polymerase reverses direction to remove mismatched bases (usually create a bulge)</span></p></li><li><p><span style="font-family: sans-serif">RNA pol goes forward again once done proofreading</span></p></li></ul>
22
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describe generally how termination occurs

  • what are the two main types in bacteria?

  • The DNA sequences required for
    termination are located upstream of
    the terminator sequence. Formation
    of a hairpin (ds) in the RNA may be
    necessary.

  • hairpin tells RNA pol to stop/disassemble

  • 2 main types in bacteria:

    • Rho-independent

    • rho-dependent

23
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what are the two features of the Rho-independent termination pathway?

  • indirect repeats containing lot’s of G-C pairs which forms stem+loop (hairpin)

  • sequence of As in DNA to cause a string of Us in the RNA

these cause the hairpin configuration which causes RNA to dissociate

24
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describe the Rho-independent termination pathway

  • once RNA synthesis is done

  • hairpin forms

  • regions of As and Ts in DNA cause RNA to be released when RNA pol reaches this segment

<ul><li><p>once RNA synthesis is done</p></li><li><p>hairpin forms</p></li><li><p>regions of As and Ts in DNA cause RNA to be released when RNA pol reaches this segment</p></li></ul>
25
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describe the hairpin structure

  • has double stranded G-C rich region in the stem which allows for strong binding

  • has a single stranded U-run which will have weaker binding with DNA’s As so is good for releasing the RNA

  • the hairpin does not need to be perfectly complementary

<ul><li><p>has double stranded G-C rich region in the stem which allows for strong binding</p></li><li><p>has a single stranded U-run which will have weaker binding with DNA’s As so is good for releasing the RNA</p></li><li><p>the hairpin does not need to be perfectly complementary</p></li></ul>
26
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what are intrinsic terminators?

palindromic regions that form hairpins
(stem and loop) varying in length from 7 to 20 bp

27
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describe Rho’s structure

  • Rho monomer has 2 domains

    • Rho has an N-terminal, RNA- binding domain

    • C-terminal ATPase domain

  • A hexamer in the form of a gapped ring can form and bind RNA along the exterior of the N-terminal domains

<ul><li><p><span style="font-family: sans-serif">Rho monomer has 2 domains</span></p><ul><li><p><span style="font-family: sans-serif">Rho has an N-terminal, RNA- binding domain </span></p></li><li><p><span style="font-family: sans-serif">C-terminal ATPase domain</span></p></li></ul></li><li><p><span style="font-family: sans-serif">A hexamer in the form of a gapped ring can form and bind RNA along the exterior of the N-terminal domains</span></p></li></ul>
28
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describe the Rho-dependent termination pathway

  • Rho factor attaches to RNA’s rut site (rho utilization site)

  • rho translocates along RNA until it reaches
    the RNA–DNA hybrid in RNA polymerase,
    where it releases the RNA from the DNA.

  • rho which has ATPase activity causes release of RNA polymerase (rho helicase)

29
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what makes the RNA more likely to undergo Rho-dependent termination?

  • no GC rich regions

  • lack of sequence repeated A residues (therefore no run of Us added to RNA)

30
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how did they find promoter sequences?

  • did DNA footprint analysis → looks to see if there are parts of the DNA protected from enzymes/DNAase (which covered by sigma)

  • sequences regions protected by RNA pol

<ul><li><p>did DNA <strong>footprint </strong>analysis → looks to see if there are parts of the DNA protected from enzymes/DNAase (which covered by sigma)</p></li><li><p>sequences regions protected by RNA pol</p></li></ul>
31
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how does the location of the promoter mutation effect transcription?

  • mutations that effect transcription occur in -35 region and -10 region

  • mutations in between the -35 and -10 region do not effect transcription

32
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can mutations increase transcription/enzyme activity

yes if they make the sequence more similar to the consensus sequence (mutations that make the sequence less like the consensus sequence have the opposite effect)

33
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describe abortive initiation

  • RNA pol falls off promoter and tries again

  • happens if RNA pol binds to the wrong promoter, binds wrong or not enough RNA has been made yet to stabilize binding

  • Happens in the first 8- 10 nucleotides, after
    that the association becomes more stable

<ul><li><p>RNA pol falls off promoter and tries again</p></li><li><p>happens if RNA pol binds to the wrong promoter, binds wrong or not enough RNA has been made yet to stabilize binding </p></li><li><p><span style="font-family: sans-serif">Happens in the first 8- 10 nucleotides, after</span><span><br></span><span style="font-family: sans-serif">that the association becomes more stable</span></p></li></ul>
34
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describe sigma 70

  • function of the genes it activated

  • consensus sequence

  • “house-keeping” genes expressed in all growing cells

  • -35: TTGACA

  • spacer is 16-18 bp

  • -10: TATAAT (perfect -10 sequence)

35
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what sigma factor is responsible for transcribing osmotic shock gene -osmY

  • how was this discovered?

  • sigma 38 is responsible

  • able to figure it out because with increasing osmotic shock, more sigma 38 genes were transcribed and less sigma70 genes were transcribed

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describe sigma factor’s mechanism (for all sigma’s except sigma 54)

  • sigma of holoenzyme binds to DNA

  • sigma opens up DNA strand

  • abortive initiation may occur

  • initiation begins (for real this time)

  • promoter is cleared

  • elongation occurs

37
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describe sigma factor 54’s mechanism

  • responsible for nitrogen take up

  • requires an IHF protein (DNA binding and bending protein)

  • requires AAA+ protein that hydrolyses a ATP

STEPS

  • AAA+ binds upstream DNA and then sigma 54

  • IHF binds the DNA and bends it

  • AAA+ hydrolyzes an ATP on the DNA

<ul><li><p>responsible for nitrogen take up</p></li><li><p>requires an IHF protein (DNA binding and bending protein)</p></li><li><p>requires AAA+ protein that hydrolyses a ATP</p></li></ul><p>STEPS</p><ul><li><p>AAA+ binds upstream DNA and then sigma 54</p></li><li><p>IHF binds the DNA and bends it </p></li><li><p>AAA+ hydrolyzes an ATP on the DNA</p></li></ul>
38
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explain what happens when early promoters are transcribed by bacteriophage T4 (cascade effect)

  • early transcription is performed by host RNA pol, promoter are recognized by host sigma factor (sigma 20)

  • several genes are transcribed, one makes a phage specific sigma (comes from phage)

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explain what happens when middle promoters are transcribed by bacteriophage T4 (cascade effect)

  • Middle transcription is performed by host RNA polymerase with phage sigma factor made during the early transcription phase

  • genes in this region code for another new sigma factor

40
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explain what happens when late promoters are transcribed by bacteriophage T4 (cascade effect)

  • Late transcription if performed by host RNA
    polymerase and phage sigma factor made
    during the middle transcription phase

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explain what happens when late promoters are transcribed by bacteriophage T4 (cascade effect)

  • Each new factor affinity for new promoters and therefore transcription from those genes starts, the order of transcription cannot be changed as each step is dependent on what happened at the previous step

42
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explain cascade transcription in sporulation

Transcription of phage SPO1 genes is controlled
by two successive substitutions of the sigma
factor that change the initiation specificity.

<p><span style="font-family: sans-serif">Transcription of phage SPO1 genes is controlled</span><span><br></span><span style="font-family: sans-serif">by two successive substitutions of the sigma</span><span><br></span><span style="font-family: sans-serif">factor that change the initiation specificity.</span></p>
43
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what are antiterminators?

regions that contain a terminator that no longer
read as terminators usually because there has been a modification to RNA polymerase so it no longer recognizes the terminator sequence

  • control transcription

<p><span style="font-family: sans-serif">regions that contain a terminator that no longer</span><span><br></span><span style="font-family: sans-serif">read as terminators usually because there has been a modification to RNA polymerase so it no longer recognizes the terminator sequence</span></p><ul><li><p>control transcription</p></li></ul>
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how do antiterminators affect RNA pol?

An anti-termination protein can act on RNA polymerase to enable it to read through a specific terminator (helps it ignore the terminator)

  • keeps RNA pol bound to DNA

  • allows termination to occur at the 2nd termination signal (bypasses the first)