RNA Processing Flashcards

1
Q

What is a Selex ribozyme?

A

It is a systematic selection of catalytic activity in DNA that is evolved as RNA using PCR.

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2
Q

How are ribozymes like proteins?

A

They coordinate metals (Mg) and have catalytic activity. However, ribozymes can replicate themselves, unlike proteins!

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3
Q

Other than cells, where are ribozymes found?

A

Many viruses.

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4
Q

Why are group I and II introns considered ribozymes?

A

Because they perform self-splicing.

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5
Q

Where in the RNA sequence do riboswitches form?

A

In the 5’ UTR, preventing RNA pol from continuing transcription in some cases, such as during guanine binding.

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6
Q

What are snoRNAs?

A

Small nucleolar RNAs that help process and chemically modify rRNAs

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7
Q

What is the order of mRNA processing?

A
  1. capping
  2. splicing
  3. 3’-end formation and polyadenylation
  4. transport
  5. proofreading
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8
Q

How can prokaryotic DNA be transcribed and translated at the same time?

A

There is no nucleus separating the process, and transcripts have no introns that need spliced.

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9
Q

Describe the mRNA cap structure.

A
  • 5’-5’ link to guanine to stabilize the transcript (triphosphate linkage)
  • signals for splicing and export
  • binds eIF4E and promotes translation
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10
Q

Describe the process of mRNA capping.

A
  • phosphohydrolase removes one phosphate from 5’ end nucleotide and GTP is added by the phosphate groups
  • guanylyl transferase removes two phosphates from the GTP, leaving a total of 3 phosphates
  • guanine 7-methyl transferase adds methyl to cap from SAM donor
  • 2’-O-methyl transfersae adds methyl to cap from SAM in the last step (2 methyls total)
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11
Q

Where are mRNA capping factors found?

A

Bound to RNA pol II CTD. (as do 3’ end polyadenylation factors and splicing factors)

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12
Q

What is the result of phosphorylation of CTD by the capping process?

A
  • increases rate of elongation
  • promotes CTD association with splicing factors
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13
Q

Describe the process of 3’ polyadenylation of mRNA.

A

After capping and splicing:

  • cleavage and polyadenylation specificity factor (CPSF) recognizes AAUAAA signal
  • cleavage stimulation factor (Cstf) endonuclease cleaves poly(A) site 10-30nt downstream of cleavage recognition signal
  • poly(A) polymerase and PAP with ATP add A’s (~200) using no template. First 12 A’s are slow, the rest are rapid.
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14
Q

When does mRNA splicing finish?

A

After polyadenylation.

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15
Q

What is the role of poly A binding proteins (PABPII)?

A

They bind the polyA tail and stimulate PAP activity.

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16
Q

How do group I and II self splicing factors differ? Relate?

A

Group I uses guanosine co-factor, while group II use adenine within the mRNA sequence.

They are both found in mtDNA.

Introns encode endonuclease.

Folding of structure causes catalysis

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17
Q

Where is spliceosome-catalyzed splicing of pre-mRNA found?

A

In the nucleus of eukaryotes.

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18
Q

Very basically, describe the reaction that is used by splicing machinery.

A

2’ OH of 3’ adenine attacks upstream phosphate, forming a lariat and breaking off 5’ end. That 5’ end OH then attacks phosphate of free 3’ end to close the gap and release the lariat for degradation. These are called transesterification reactions.

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19
Q

What do we call the adenine involved in splicing?

A

The branch point.

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20
Q

Why is the accuracy of splicing critical?

A

Even a 1nt mistake will destroy the protein open reading frame.

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21
Q

Of the two splice sites, which is the splice donor and which is the splice acceptor?

A

3’OH of the 5’ splice site is the donor (because convention is 5’->3’), and 5’PO4 of 3’ splice site is the acceptor.

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22
Q

How many snRNA/protein complexes (snRPS) are found in the core spliceosome?

A
  1. U1, U2, U3, U4, U5, U6.
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23
Q

Describe the intron/exon boundaries and how they allow for these regions to be recognized by splicing machinery.

A
  • The 5’ exon:intron sequence is AG:GU
  • The 3’ intron:exon sequence is CAG:C
  • when exons come together, sequence will be AGC
  • upstream of 3’ splice site is a poly-pyrimidine track
  • 30-40nt at each end of the intron are necessary for splicing
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24
Q

How do introns and exons vary in length?

A

Introns are variable, from 10-10^5 nt long. Exons within the same gene are around the same length.

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25
Q

When do the first steps of splice site recognition take place?

A

After 5’ capping.

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26
Q

Describe splice site recognition by the spliceosome.

A
  • U1 snRNP binds the 5’ splice site (hybridization directed by snRNAs)
  • BBP binds branch point and U2AF binds pyrimidine track
  • U2 snRNP replaces BBP and U2AF on the branch point
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27
Q

What does hybridization of U2AF to the branch point of the intron accomplish?

A

It kicks out the adenine, allowing it to be activated.

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28
Q

How do snRNAs become mature?

A

They are exported to the cytoplasm for final maturtion and associate there with proteins. snRNPs are then imported back to the nucleus for splicing.

29
Q

What is the role of the U4, U6, and U5 triple snRNP?

A
  • U4 and U6 are held together by base pair interactions and are separated upon splicing, causing U4 to leave the complex
  • U6 displaces U1 at 5’ site and forms the the active site that catalyzes the first phosphoryl-transferase reaction
  • U5 holds 5’ and 3’ splice sites together
30
Q

Define hnRNA

A

heterogeneous nuclear RNA. another term for pre-mRNA

31
Q

How does the spliceosome achieve specificity?

A

Two proteins recognizing the same site:

  • U1 and U6 snRNPs bind the same area, so they must both base pair properly with the 5’ splice site.
  • BBP and U2 both must recognize the branch point properly.
  • U2 and U6, while bound to the pre-mRNA, must also bind one another to put the RNA in the correct orientation for splicing.
32
Q

Which components of the spliceosome bind cotranscriptionally and which do not?

A
  • U1, BBP, U2AF, and U2 bind cotranscriptionally
  • U4/U6, U5 do not
33
Q

What is the primary role of U5 snRNP?

A

It holds the spliceosome together.

34
Q

What is the main form of splicing variation in eukaryotes?

A

Alternative splicing.

35
Q

What is the purpose of alternative splicing?

A

To create diverse but related proteins from a single gene (e.g. human beta globin gene)

36
Q

Other than binding specificity, how does splicing maintain high fidelity and avoid translating incorrect mRNAs?

A
  • introns contain many stop codons to prevent them from forming a protein.
  • 5’ splice site machinery only sees this part of the mRNA, as it interacts upon pre-mRNA emergence from the polymerase
  • it also helps that processing factors are already associated with CTD and can act quickly on emerging mRNA
  • capping is required before subsequent events take place
37
Q

Which proteins mark the exon-exon junction?

A

exon junction complex. they play a role in proofreading of the transcript by the ribosome

38
Q

Which factor generates cross-exon recognition complexes (the exon region between U2 and U1 bound by SR protein between neighboring introns)?

A

U2AF

39
Q

Compare the lengths of the cross exon recognition complex with the spliceosome.

A

The spliceosome varies in length, but because the cross exon recognition complex contains exons it is more constant in length.

40
Q

What is the role of the SR protein?

A

It binds to exon in cross-exon recognition complex and helps in recognition of intron-exon boundaries.

41
Q

Why are cryptic splice sites important?

A

They allow for alternative splicing, and there activity as a splice site can be activated or repressed by certain factors.

42
Q

How can cryptic splice sites cause disease (e.g. misplicing of beta globin gene can cause anemia)

A
  • normal mRNA is formed from three exons
  • If U2 can’t find the 3’ splice site of exon 3 (due to mutation at this spot and activation of cryptic splice site within upstream intron), exon 3 may be extended
  • exon 2 may be spliced out if 3’ splice site is lost (mutation)
  • A fourth exon can be added if new splice site is introduced by mutation (cryptic 3’ splice site activated and new 5’ splice site from mutation)
43
Q

What percentage of human genes undergo alternative splicing?

A

about 95%

44
Q

Which factors regulate alternative splicing?

A

Negative and positive trans-acting factors

45
Q

Describe the role of alternative splicing in male vs female development.

A
  • in males: normal splicing of Sxl (nonfunctional), normal splicing of Tra (nonfunctional), and alternative splicing of Dsx due to weak cryptic site (1st and 3rd exons expressed, represses female genes)
  • in females: cryptic promotor in Sxl activated by XX to make more Sxl protein (active), alternative splicing of Tra by Sxl repression (active Tra), and alternative splicing of Dsx by Tra activation of weak 3’ splice site on intron 1. !st and second Dsx exons expressed, represses male genes.
46
Q

Describe regulation of alternative 3’ end polyadenylation in B cells.

A
  • some introns contain weak poly(A) sites. In resting B cells with low levels of Cstf, only the strong sites in the last exon will be bound
  • in active B cells with high levels of Cstf, the strong sites are filled and leftover Ctsf will bind weak sites on some mRNA molecules, causing a short mRNA containing the intron without the last exon.
47
Q

Describe regulation of mRNA transport from the nucleus.

A
  • pre-mRNA must have CBC, SR proteins, PABP’s, and EJC protein bound (dictate exons to leave nucleus)
  • must be missing lncRNAs that were bound to introns (lncRNAs dictate introns to stay)
48
Q

Upon leaving the nucleus, which factors on mRNA are exchanged?

A

CBC is exchanged for eIF4G and eIF4E

49
Q

Describe nonsense mediated decay of mRNA in the nucleus.

A
  • normal translation takes place when stop codon is reached after the last EJC protein, indicating that all introns are spliced out.
  • in nonsense mediated decay, if an EJC protein is encountered after a stop codon, translation will stop and the mRNA degraded, because that indicates the stop codon is within an intron that was not spliced out.
50
Q

Describe nonstop decay.

A

If the ribosome translates all the way to the polyA tail without reaching a stop codon, the peptide and mRNA transcript will be degraded.

51
Q

How is natural mRNA decay regulated?

A
  • how long PABP stays on poly A tail (PABP protects from degradation)
  • translational status of mRNA: speed of translation can affect mRNA decay
  • environmental factors
  • cis-acting elements: AU rich elements that are bound by AU binding proteins protect mRNA stability
52
Q

What is the mechanism of mRNA degradation?

A
  1. polyA tail is shortened to a point where PABP cannot bind and protect it
  2. decapping followed by 5’->3’ (exonuclease) and 3’->5’ degradation (exosome)

(both directions prevent partially degraded mRNA from being translated)

53
Q

How is the polyA tail degraded?

A

polyA specific nuclease (PARN). Its activity is stimulated by CAP structure

54
Q

What is the mRNA decapping complex?

A

dcp1/dcp2

55
Q

Where does RNA degradation occur?

A

Within processing bodies, or P-bodies, within the cytoplasm where all machinery for degradation comes together in one place.

56
Q

Which polymerase transcribes 5S RNA?

A

RNA pol III

57
Q

How is rRNA incorporated into ribosomes?

A

It is expressed as one long transcript from RNA pol I and cleaved into certain lengths (18S, 5.8S, 28S - 5S made elsewhere)

58
Q

Define snoRNA

A

Small nucleolar RNAs (snoRNAs) are a class of small RNA molecules that primarily guide chemical modifications of other RNAs, mainly ribosomal RNAs, transfer RNAs and small nuclear RNAs.

59
Q

Describe tRNA processing

A
  • RNase P cleaves 5’ leader end of tRNA sequence
  • exonuclease D removes 3’ seuqnece
  • CCA added to 3’ end by nucleotidyltransferase
  • intron at anticodon loop removed by endonuclease and ligase
  • 10% of all other bases are modified post transcriptionally
60
Q

Why is C->U editing in RNA transcripts used?

A

To create new stop codons.

61
Q

Where is the signal for mRNA localization found?

A

3’ UTR

62
Q

What are RNA regulons?

A

multiple mRNAs involved with a process are coordinately regulated by an RBP/miRNAs bound to cis-acting sites.

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68
Q
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