Lecture 6. Post-Transcriptional Regulation of Gene Expression 1

Question 1

What are the eukaryotic Pol II post-transcriptional events?

Answer 1

Possible attenuation (RNA transcript aborts)
5’ capping
Splicing
3’ end processing
Possible RNA editing
Nuclear export

Question 2

What do all eukaryotic Pol II post-transcriptional events require?

Answer 2

Factors that bind the phosphorylated C-terminal domain (CTD) of RNA PolII
The CTD of RNA PolII comprises multiple repeats of a 7 amino acid sequence YSPTSPS Phosphorylation of seryls 2 and 5 marks the transition from transcriptional initiation to
elongation

Question 3

Why is the 5’ cap put on backwards?

Answer 3

The 5; end of the transcript now masquerades as the 3’ end with a 3’ hydroxyl (no longer at risk of degreadtion)

Question 4

How is the 5’ cap added?

Answer 4

Nascent RNA transcript has phosphate removed from 5’ end by RNA triphosphatase
Guanylyl transferase cleaves diphosphate from GTP and adds the G and P to the RNA transcript
Methylate (add CH₃) G to generate 7-methylguanosine and first ribose also methylated

Question 5

What is the importance of the 5’ cap?

Answer 5

Distinguishes Pol II transcripts from other RNA molecules
Stabilises the RNA - there is no 5’ phosphate, so it is resistant to 5’ exonucleases
Aids in further processing (exon definition hypothesis) and export to the cytosol
And is required for efficient translation of mRNAs

Question 6

What is the key principle of the 5’ cap?

Answer 6

5’ capping is co-ordinated with transcription

Question 7

How does decapping of the 5’ cap take place?

Answer 7

A multi-subunit cytosolic ATP-dependent decapping enzyme complex removes the 5’ cap
It restores a 5’ phosphate on the mRNA. The mRNA can now no longer be bound by ribosomes and so can no longer be translated
The mRNA is now degraded by a 5’-3’ RNAse

Question 8

What is ivs (intervening sequence)?

Answer 8

Intron found in the 26R rRNA gene
The intron is excised from the primary transcript and the RNA can do it by itself

Question 9

How could the ivs intron be removed from the Tetrahymena nuclei?

Answer 9

α-amanitin fungal toxin (Pol II inhibitor) so no mRNAs could be made but rRNA genes were still transcribed
A nuclease inhibitor stabilises the RNA
And with ribonucleotides ATP, GTP, CTP and radioactive ³²P-UTP

Question 10

How did Cech discover the ribozyme?

Answer 10

Deproteinised the rRNA by phenol (denatures proteins) extraction and/or boiled the rRNA for five minutes. The intron was still excised and still accumulated
Cloned the rDNA and made rRNA in vitro using bacterial RNA polymerase. The intron was still excised and still accumulated
Cech defined the RNA itself was catalytic, with the new principle that RNA could catalyse functions that had previously been thought of as enzyme functions (ivs = ribozyme)

Question 11

How does the mechanism of introns function in the Tetrahymena?

Answer 11

1. The intron folds. A co-factor s held in a pocket: guanosine, GMP, GDP, GTP. The 3’-OH of the co-factor is a nucleophile that attacks the phosphate at the 5’ splice site
2. The 3’-OH of the upstream exon attacks the phosphate at the 3’ splice site
3. The exons are fused, and the intron is ultimately degraded
   Two sequential transesterifications fuse the exons and release the intron

Question 12

How does the mechanism of introns function in humans?

Answer 12

1. The intron folds and the 2’-OH of the ‘branch site’ adenosine attacks the phosphate at the 5’ splice site
2. This adenosine now has three phosphodiester bonds: one is an unusual 2’, 5’ phosphodiester bond. The 3’-OH of the upstream exon attacks the phosphate at the 3’ splice site
3. The exons are fused and the intron is released as a lariat
   Two sequential transesterifications fuse the exons and release the intron

Question 13

What controls the intron mechanism?

Answer 13

The spliceosome: an RNA and protein complex

Question 14

What is the spliceosome and what makes up the spliceosome?

Answer 14

The spliceosome: an RNA and protein complex
5 small uridine-rich snRNPs (small nuclear ribonucleoproteins) U1, U2, U4, U5, U6
Each snRNP is a small nuclear RNA (snRNA) complexed with at least seven protein subunits. There are extra proteins ~200 in total

Question 15

What allows the spliceosome to splice precisely?

Answer 15

A combination of RNA base-pairing (specificity) and protein binding (stabilisation) allows precise splicing because the splice sites and the branch site are brought close together

Question 16

Where does the spliceosome recognise specific sequences at?

Answer 16

The 5’ splice site
The branch site and
The poly pyrimidine tract (poly Y)

Question 17

How does the spliceosome function?

Answer 17

1. U1 snRNP binds the 5’ splice site: base pairing provides specificity
2. Branch binding protein (BBP) and the protein U2 auxiliary factor (U2AF) bind the branch site and associated pyrimidine tract (see later) respectively
3. U2 is recruited and displaces BBP
4. A pre-formed trimer of U4, U5 and U6 associates
5. A molecular rearrangement (loss of U1 and U4) activates the complex
6. The 2’-OH of the branch adenine attacks the 5’ splice site phosphate
7. The lariat is formed and U6 is activated, guiding the 3’-OH of the upstream exon that attacks the 3’ splice site phosphate
8. The exons fuse, and the intron is excised as a lariat. The snRNPs are recycled and the intron is degraded.

Question 18

What are the key principles of how the spliceosome works?

Answer 18

In eukaryotic cells spliceosomes assemble at specific RNA sequences on the capped RNA by making specific base-pair bonds. Proteins stabilise the spliceosome. Molecular rearrangement activates the spliceosome. Two spatially and temporally regulated transesterifications fuse the exons and release the intron as a lariat.

Question 19

How does U1 find the splice site?

Answer 19

1. U1 is also recruited by the phosphorylated CTD of the large subunit of Pol II. U1 scans the growing mRNA for the 5’ splice site and binds it.
2. The nascent chain keeps growing, forming a loop, while BBP/U2AF scan for a branch site/pyrimidine tract
3. The spliceosome assembles and is released

Question 20

What regulates the assembly of a spliceosome?

Answer 20

The process depends on the strength of binding of spliceosome components, so some introns are removed co-transcriptionally, others post transcriptionally
Spliceosome assembly is regulated by RNA Pol II and by the strength of the target binding sequences

Question 21

What are the splicing signals of the spliceosome?

Answer 21

Consensus sequences: individual motifs have different affinities for spliceosome components – and so some are more efficient at stimulating splicing than others

Question 22

What makes U1, U2, U4 and U5 snRNPs?

Answer 22

RNA Pol II

Question 23

What happens in the biogenesis of U1, U2, U4 and U5 snRNPs?

Answer 23

Transcription and capping in the nucleus - shipped to cytosol
Cytosolic loading of a ring of seven Sm proteins loaded as a ring onto the snRNA by SMN (survival motor neuron) protein.
Cap hypermethylated signal to transport and return to nucleus. Maturation into U1 snRNP by association with other proteins

Question 24

What are the functions of the protein components of snRNPs?

Answer 24

The snRNA provides specificity by base pairing with target: the proteins provide stability
Splicing associated proteins ‘define’ the exons

Question 25

What is the ‘exon definition’ hypothesis?

Answer 25

Fidelity of splice site recognition
Exons are ‘defined’ by the binding of proteins during transcription: this allows U1 to find the 5’ splice site

Question 26

What does fidelity of initial splice-site recognition require?

Answer 26

Requires SR protein (rich in Ser, Arg) family members binding to one or more exonic splicing enhancers (ESEs) , together with splicing factors assembling on the flanking 5′ and 3′ splice sites.
Individual hnRNP (heterogeneous nuclear ribonucleoproteins) proteins may also bind to exon sequences, acting as splicing repressors, and stable hnRNP complexes package the large intron sequences. Terminal exon recognition involves cap–5′ splice site interactions and 3′ splice site- poly A site interactions.

Question 27

What happens when there is a strong consensus poly Y tract present?

Answer 27

3’ end of intron strongly defined: normal splicing

Question 28

What happens when there is a weak consensus poly Y tract present?

Answer 28

Sometimes 3’ end of intron still defined: normal splicing
Or 3’ end of intron not defined. Spliceosome scans for next U2AF (poly Y tract) binding site. Alternative splicing/exon skipping

Question 29

What regulates alternative splicing?

Answer 29

The sequence of the poly Y tract

Question 30

Besides from the poly Y tract, how else can alternative splicing be controlled?

Answer 30

In a cell-type specific manner

Question 31

What does alternative splicing generate and why?

Answer 31

Generates protein diversity which can be controlled developmentally because U2AF is a family of proteins whose members have different abilities to find poly Y tract

Question 32

What causes royal haemophilia?

Answer 32

A mutation in intron 3 of the F9 (blood clotting Factor IX) gene generates an alternative 3’ splice site: the resultant protein is truncated and non-functional

Question 33

What causes spinal muscular atrophy (SMA)?

Answer 33

SMN1 mutation
SMN1 protein is responsible for loading about 90% of Sm proteins: SMN2 protein about 10%
SMN deficiency caused by SMN1 mutation leads to widespread splicing deficiency, especially in spinal motor neurons

Question 34

What is SMA severity generally correlated with?

Answer 34

SMA severity generally correlates inversely with SMN2 copy number, which varies from 0 to 8 in the normal population

Question 35

How is poly (A) added to the 3’ end of the pre-mRNA?

Answer 35

G/U rich area cleaved after the CA region and is degraded in the nucleus
Polymerase adds poly A to the end of the CA region (runs for ~250 residues)

Question 36

What cleavage sites are recruited for the poly (A) addition at the 3’ end of the pre-mRNA?

Answer 36

The poly(A) site is bound by CPSF
The cleavage site is bound by CF
The G/U rich region is bound by CStF
Interactions between CStF and CPSF provide a recruitment site for the polymerase (PAP)

Question 37

What happens in the cleavage stage of the poly (A) addition at the 3’ end of the pre-mRNA?

Answer 37

RNA cleavage by CF provides a 3’-OH for PAP which adds adenine (A) residues
This is SLOW adneylation until ~12 residues are added
Poly(A) binding protein (PABP) is added between CA and PAP causes rapid adenylation
Approx ~200 nt added and then PAP disengages

Question 38

What is the nuclear export function of polyadenylation?

Answer 38

Pol II CTD also loads a nuclear export receptor following polyadenylation. Substituting the polyadenylation site with a ribosomal RNA cleavage site produces an mRNA that is cleaved but not polyadenylated, reducing mRNA transport by about 10- fold

Question 39

What is the translation function of polyadenylation?

Answer 39

Pseudocircularisation of the transcript. The poly(A) tail and PABP interact via eIF4E and eIF4E with the methylated cap at the 5’-end to promote translation

Question 40

What is the stability function of polyadenylation?

Answer 40

In the cytoplasm, mRNAs are degraded from the 3’-end first, indicating the importance of protecting the 3’-end. The addition of the poly(A) tail and subsequently the binding of poly(A)- binding protein (PABP) reduces degradation rate and increases the half-life of the mRNA