1 - RNA Pol II CTD and Co-transcriptional RNA Processing - Wilson

Question 1

Draw a simple diagram showing gene expression - from DNA to protein. state why it is important that each step is coupled to the next

Answer 1

• allows high fidelity and efficiency of mRNA production and translation in the cytoplasm

Question 2

draw a diagram showing how the 5’ end of every mRNA is capped

Answer 2

325 - 1 word

Question 3

why is mRNA capped?

Answer 3

• cap allows efficient export and translation of the mRNA in the cytoplasm
• protects against nucleases eg XrnII that will degrade the mRNA in the nucleus

Question 4

when does mRNA capping occur?

Answer 4

co-trancriptionally, within the first 20-30 nucleotides being synthesised

Question 5

draw a diagram and explain each stage of the splicing mechanism

Answer 5

325 - 1 word

Question 6

draw a diagram highlighting the overall stages for the processing of the 3’ end. include the structures that direct this processing

Answer 6

325 - 1 word

Question 7

what are the functions of the polyA tail for mRNA?

Answer 7

protects the mRNA from 3’-5’ exonucleases (eg exosome)

Question 8

how long is the pA sequence roughly?

Answer 8

around 50 nucleotides (this is the -AAUAAA sequence upstream of CA cleavage site

Question 9

draw a diagram and explain it to show the protein complexes that aid in the 3’ end processing and addition of polyA tail

Answer 9

CLEAVAGE;
- CPSF (cleavage and polyadenylation stimulatory factor) recognises pA sequence and recruits CstF (cleavage stimulatory factor). this stimulates cleaveage of the RNA
- the symplekin scaffold protein brings all of the proteins together for this cleavage reaction eg polII, CPSF, CstF
POLYADENYLATION;
- PAP (polyA polymerase) recruited and adds of As to the end
- the As recognised by PABP (polyA binding protein) this aids in mRNA translation once reached cytoplasm

Question 10

which form of mRNA (pre/normal) does the 3’ end processing occur?

Answer 10

pre-mRNA 3’ end processing

Question 11

why is it necessary to define where exons are?

how is this achieved?

Answer 11

• necessary because exons around 200 nucleotides long whereas introns can range from 10-10^5 nucleotides long
• needle in a haystack
• interplay between proteins that bind exotic sequences and proteins that bind introns required for correct splicing

Question 12

what pre-mRNA processing events are co-transcriptional? give the general reason why this happens

Answer 12

all of them;

• capping (5’)
• splicing
• cleavage and polyadenylation (3’)
• because the CTD of RNAp II is differentially phosphorylated and proteins required for these processing events bind to the CTD of polII. P of the CTD governs its interactions with various proteins eg Ser5 -> spliceosome assembly

Question 13

describe the pol II CTD in humans

Answer 13

• CTD unique to pol II and is part of the Rpb1 subunit

- humans, consists of 52 repeats of the heptad sequence YSPTSPS

Question 14

how does the CTD of pol II in humans differ to that of budding yeast?

Answer 14

• in yeast contains 26 repeat sequences not 52.

- number repeats roughly correlates to the complexity of the organism

Question 15

what does the experiment shown in Q15 325 - 1 word show?

Answer 15

• shows that the CTD is essential for splicing from DNA templates
• once we remove the CTD then we do not get any pre-mRNA splicing (no band at the bottom of the gel)

Question 16

what does the experiment shown in Q156325 - 1 word show?

Answer 16

• CTD is essential for 3’ end processing, more specifically cleavage

Question 17

what does the experiment shown in Q17 325 - 1 word show? DRB = kinase inhibitor

Answer 17

• kinases required for the P of various residues on the CTD of pol II
• as DRB conc increases, less and less processing events
• P of the CTD therefore is required for BOTH pre-mRNA splicing and 3’ end processing and cleavage
• overal; inhibits co-transcriptional processes

Question 18

name the CTD modifications of the individual residues within the repeats and state their functions. also state the kinases that P these residues (the ones that are known)

Answer 18

Y - can be P - kinase unknown
S - P during escape from initiation and the overall elongation (Cdk12)
P - Cis/trans conformational change
T - histone mRNA processing
S - initiation, RNA capping (Cdk 7) and splicing
P -Cis/trans conformational change
S - widely P, unclear function (Cdk7)

Question 19

draw a graph and include a few examples of the CTD differential Phosphorylation at different timepoints

Answer 19

325 - 1 word

Question 20

which kinase is recruited and which residue of the CTD is phosphorylated for capping?

Answer 20

ser5 is phosphorylated

by Cdk7

Question 21

draw a diagram of the capping process and the CTD

Answer 21

325 - 1 word

Question 22

describe CTD assembly of the spliceosome

Answer 22

phosphorylation of CTD recruits the U2AF

• U1snRNP bound to pre-mRNA
• U2AF recruits the PRP19C complex allowing assembly of the spliceosome
• subsequent splicing processes

Question 23

what kinases and CTD residue is invovled in 3’ end processing ?

Answer 23

Cdk12 phosphorylates Ser2 of the CTD

Question 24

draw a diagram of 3’ end processing and the CTD and explain it

Answer 24

• upstream of pA, Ser2P present at low levels
• CPSF accompanies Pol II
• transcription of pA site induces pol II pausing aided by CPSF
• triggers Ser2P by Cdk12
• CstF recruited and Cleavage and polyA take place efficiently
• the slowing down of the pol II allows efficent processing events eg cleavage

Question 25

give 2 examples of mRNA processing events that affect transcription

Answer 25

1) in yeast, transcriptional checkpoint is satisfied by spicing. ser5P of pol II = checkpoint. allows spliceosome assembly because it slows down the pol II and gives time for recruitment of spliceosome factors and assembly
2) when we have circular genome, removal of the pA site means that we get continuous transcription of the whole genome. transcription rate is not slowed down at any point. the addition of pA site -> transcription slows down, recruitment of the cleavage factors and cleavage occurs and the pol II stops transcription