Lecture 20 (RR8): RNA Processing I Flashcards
What are the similarities between the 3 RNA polymerases.
All three eukaryotic RNA polymerases share common features :
➲ all of them are multimeric protein complexes
➲Some subunits show significant homology with bacterial RNA polymerase
➲all of these subunits are more or less essential
Why is the CTD of large subunit of RNA Pol II is unique among the RNA Polymerases?
It is very critical, when this CTD is in its phosphorylated form, it is associated with highly transcribed genes.
- When stretched out, it forms a long scaffold
- CTD is made up of many repeats (heptapeptide repeats)
- YSPTSPS in humans – occurs 52 times (repeated)
- YSPTSPS in yeast – occurs 26 times (if you get rid of this sequence the yeast die) - The repeat becomes phosphorylated on S5 and then S2
- TFIIH and its associated protein kinase activity will phosphorylate Serine 5 of this heptapeptide
- This is associated with getting into an elongation phase
Transcription steps detailed
1) Initiation
* Mediator, RNA pol II and all the transcription factors have come together. NTPs are there, lots of ATP in the cell and TFHII will then el the DNA at the same time.
* The CDK7 subunit of TFIIH will phosphorylate serine 5 on the hepta peptide sequence.
* RNA Pol II will then squeak away from the promoter and generate an RNA molecule but it does not switch to full blown elongation, it moves slowly and inefficiently.
* The tail of the carboxy terminal domain of a large subunit of RNA Pol II is phosphorylated.
* The phosphorylated carboxy terminal domain on serine 5 acts as an important platform to recruit enzymes that are required to modify the five prime end of that emerging pre mRNA.
2) Pausing and then elongation
* Post-initiated RNA pol complex comes to the first nucleosome and stops, with all S5 phosphorylated by TFIIH
* RNA Polymerase II pausing after initiation is an important regulated step.
* RNA Polymerase II pausing allows for a change in factors from those that actively block elongation (NELF) to a new set of factors that enhance elongation (DSIF, SPT6 and PAF).
* It waits for S2 to be phosphorylated by P-TEFb (or CK9)
* S2 is recognized by splicing factors
* RNA Pol can elongate only after Serine 2 has been phosphorylated –> Takes place when RNA Pol II gets held up in a nucleosome
* This change is dependent on CDK9/P-TEFb mediated phosphorylation of the CTD.
* Not only does it change the RNA Pol to complex and switch it from a promoter clearance mode to an elongation mode but this new phosphorylation of the CTD on all those heptapeptide repeats also changes the CTD so that a number of enzymes and proteins required for a number of RNA processing events associate with the CTD: splicing factors, polyadenylation factors and export factors. They are all bound to the CTD in its serine 2 phosphorylated mode.
* The CTD is largely disordered and is critical for the formation/inclusion of RNA Pol II into the condensates. If you get rid of the phosphorylation of CTD you die.
Modification to the pre-mRNA upon the initiation of transcription at the 5’ end
Maturation of nascent transcripts must occur to form functional mRNAs
* Precursor mRNAs (pre-mRNA) are modified at their 5’ end
* Once initiated and start to form pre-mRNA, as it emerges from the transcription bubble, it needs to be capped
* Capping enzyme will recognize 5’ end of pre-mRNA. A Cap is put on by an enzyme that recognizes the phosphorylation of S5 of CTD. The capping enzyme recognizes the serine 5 phosphorylation that were catalyzed by TFIIH during the initiation reaction. The phosphorylated CTD recruits the capping enzyme and once its capping interacts with the serine 5 phosphorylation on the CTD it becomes activated. Everything is in close proximity, the capping enzyme is close to the 5’ end of the emerging pre mRNA and this is one reason why the phosphorylation event at initiation is so important.
* It’s modified so that enzymes that degrade RNAs won’t chew it up
It will remove a γ phosphate
* A 7’ methylguanylate CAP is added to the 5’ terminal nucleotide through an unusual 5’-5’ linkage
* RNAase can’t degrade pre-mRNA from 5’ end – it’s protected
- In animal cells and in higher plants, the 2’ hydroxyl of the ribose group of the first base is methylated
- In vertebrates the second base is also methylated
- All mRNAs are capped (distinguishes them). Addition of the CAP protects the pre-mRNA from enzymatic degradation, while facilitating nuclear export and recognition by translation initiation factors
- All of these modifications are to ensure that the work that has been put in to generate the RNA is not wasted the moment that it emerges into the nuclear environment, it is to protect what you have started by adding on these chemical modifications that enhance its stability.
What are the 3 modifications made to the pre-mRNA.
CTD is critical for ensuring that the 5’ end is stabilized by adding the 7’ methylguanylate CAP, there is appropriate elongation that takes place, during elongation there are splicing factors required to make a mature mRNA in close proximity, and after elongation has been terminated, there is poly - adenylation.
1) 5’ capping
2) splicing
3) Poly A tail
5’ Capping
Eukaryotic pre-mRNA primary transcripts must undergo addition processing after transcription. The resulting mature mRNA will be translated into protein.
The 5’ end of the mRNA is capped with a 7-methylguanosine.
* This processing occurs co-transcriptionally (i.e when the gene is still actively being transcribed)
* TFIIH phosphorylates Serine 5 on the RNAPII CTD heptapeptide repeat after initiation
* This phosphylation allows the 5’ capping enzyme to bind and activate
* Focuses the 5’ capping activity to pre-mRNAs transcribed from RNAPII ONLY (not RNAPI or RNAPIII transcripts)
5” capping is important because:
1. Stabilizes the pre-mRNA: protects it from being digested by RNA-digesting enzymes
2. Allows elongation to proceed at a faster rate
- Serine 2 on the CTD is phophorylated and Negative Elongation Factor (NELF) is released from RNAPII
From pre-mRNA to mRNA: splicing
- Unlike bacterial genes, eukaryotic genes can (and most of them do) have introns
- Exons are those regions of a transcript that are present in the mature mRNA, whereas introns get spliced out from the primary RNA transcript and are not part of the mature mRNA
Introns
- Introns are not junk – they can encode regulatory information
- Introns are more common in higher eukaryotes than in lower ones (introns correlate with the complexity of the organism).
- Introns were discovered, because a discrepancy between mRNA size and gene size was observed for some genes
- If you lined up a cDNA sequence and the genomic sequence, there were large chunks missing in the cDNA (the mRNA was missing sequences, not the same as the DNA).
- They were even visualized by hybridization experiments, in which the mRNA of a gene was annealed to its corresponding coding DNA strand in its genomic context
- The mRNA of the adenovirus hexon gene hybridized to the DNA fragment containing the hexon gene forms a RNA-DNA hybrid, however the intron sequences in the DNA loop out since they have no hybridization partner sequence in the mRNA.
Intron Borders
Intron borders are highly conserved
* Comparison of a large number of introns indicated that their border sequences are conserved. Where the sequence no longer matches, you know that must be an intron boundary.
* These features are useful for the in silico prediction of mRNA sequences based on genomic sequencing data
* These are invariant, you need these in introns: GU is at the 5’ end, AG is at the 3’ end highly conserved boundaries (GU AG rule for splicing).
* There is a pyrimidine rich region somewhere in the middle, closer to the 3’ end (aprox 15 bases from 3’ end of intron).
* Just upstream of that region, there is an Adenine nucleotide highly conserved (Branch point nucleotide). This is also invariant, you need this A to be present and near the pyrimidine rich region. Need it for splicing to go forward
The two trans-esterification reactions
- The way that the bulging out occurs presents the 2’ hydroxyl in a way that it will interact with the phosphorus group in the 5’ nucleotide of an intron segment that is going to be removed. The hydroxyl group of the residue at the branch point attacks the 5’ phosphate group of the first intron residue (first trans-esterification), leading to a free 3’ end of the preceding exon and a cyclic RNA structure referred to as a “lariat”. All mediated by the branch point nucleotide.
- First trans-esterification reaction gives rise to an exon with a free hydroxyl group at the 3’ end and an intron that has a lariat at the 5” end.
- The free 3’ end of the preceding exon then attacks the 5’ phosphate group of the first residue in the following exon (second trans-esterification), resulting in the joining of the two exons and the release of the intro lariat. All directed by some of those snRNPs that are interaction with the intron.
- The products are a lassu that gets excised and the spliced RNA molecule.
Role of the small nuclear RNAs in splicing?
Small nuclear RNAs are essential for spliceosome functionality:
* Highly conserved regions correspond to sequential events for the excision process during splicing.
* At the 5’ end of the intron, at the intron-exon boundary, the sequence corresponds to a complementary sequence of small nuclear RNAs that direct proteins to specify where the reaction must occur
* The upstream region corresponds to the small nuclear RNA U1 snRNA
- The U1 snRNA will interact through Watson-Crick pairing with sequences present at the exon-intron boundary U2 snRNA will interact with the branch point forming Watson-Crick pairs
- Gives rise to a bulged adenine nucleotide (branch point nucleotide is always bulged out and does not contribute to the watson crick base pairing).
- Bulge makes it more exposed and is important for putting the branch point A nucleotide at proximity to what will happen at the 5’ end.
* The spliceosome consists of five snRNPs (small nuclear ribonucleoprotein particles), which in turn consist of a particular RNA molecule termed snRNA (small nuclear RNA; U1, U2, U4, U5 or U6) and six to ten proteins.
* These snRNAs are essential for splicing, contacting the intron’s 5’ border (U1 snRNA) and branch point region (U2 snRNA) and contacting each other, while being held in place by the proteins surrounding them
Remember: the adenosine will bulge out and because it does not pair. This sets up the next steps, which include 2 transesterification reactions: one immediately after the other. FInal products: excised lariat intron and weird RNA that forms a loop through a 2’ 5’ linkage. Other products in contiguous phosphate exon.
Mutations that block splicing and how to restore these.
Compensatory mutations indicate that RNA: RNA interactions are required for splicing
* Base pairing between the U1 snRNA and U2 snRNA is critical for the splicing.
* It was found that you could block or inhibit the splicing reaction in vitro simply by putting an oligonucleotide that was complementary to that sequence around the 5’ end → blocking the ability of U1 snRNA to carry out its function.
* A single base modification or single mutation in the conserved sites can eliminate splicing. Non complementary binding of U1 snRNA and pre mRNA → block spicing all together
* U1 shares limited sequence homology with the 5’ splice site
* By introducing pre-mRNAs that possess a compensatory mutation, splicing can be entirely restored
* If you mutate a U1 snRNA so it can complement and bind to the mutation introduced, you can restore splicing – it’s not the sequence that’s important, but it’s the binding of the U1 snRNA to that sequence that is critical for splicing
* RNA:RNA pairing is critical for U1 function in splicing
how can the splicing reaction be visualized?
The splicing reaction can be visualized in vitro using radiolabeled RNA substrates (probes):
* Using radiolabeled RNA substrates, each intermediate splicing product can be separated and quantified during a splicing reaction in vitro.
Spliceosome Cycle
The spliceosome cycle:
* The snRNPs sequentially assemble on the intron. U1 snRNP will recognize the 5’ end and U2 snRNP will recognize the branch point adenine nucleotide
1) The U4, U5, and U6 snRNPs recognize the U1-U2 complex and then bind to it
* After assembly of all five snRNPs, extensive rearrangements in the RNA-RNA interactions between the pre- mRNA and snRNAs occur → become the spliceosome.
2) U1 and U4 exit the complex leaving the active spliceosome.
3) First transesterification
* Following formation of the spliceosome the trans-esterification reactions can occur.
* First transesterification reaction immediately followed by second transesterification reaction. ATP hydrolysis is not required for the splicing (trans-esterification reactions) per se, but rather for the rearrangements of the spliceosome components.
4) The lariat intron gets spliced out after the second transesterification reaction and snRNPs fall off the complex and get recycled to carry out the reactions
What are the productss of the spliecesome cycle?
- The products that come out of this are the lariat intron, the lassu that got excised and all the snRNPs that leave the complex and will re being the same reaction on another downstream intron.
- The lariat intron is a problem because it has a 2’ 5’ linkage and regular ribonucleases can’t attack it.
- The designated enzyme that is devoted to catalyzing or disrupting that 2’ 5’ linkage and catalyzing the formation of a linear RNA molecule from this lariat intron.
- These intron segments can sometimes be absolutely giant.