Transcription

Question 1

What polymerase is used in prokaryotic transcription?

Answer 1

RNAP holoenzyme 449 kDa, 6 subunits total
Core: α2ββ′ω
The σ subunit dissociates when RNA synthesis initiates, which carries out the polymerisation process
It doesn’t require a primer
Mg2+ and Zn2+ metal ions

Open complex - crab claw, with two pincers of the β and β′ subunits (27 Å)
Closed complex - 10 Å, formed when RNAP holoenzyme is in complex with dsDNA containing a promotor

Inner surface - positive
Outer surface - negative

Question 2

What are the roles of the subunits of RNA polymerase?

Answer 2

α2 - N-terminus domain (NTD) - assembly of RNA pol
C-terminus domain (CTD) - non specific interact with promotor

β′ - active site for RNA synthesis
β - involved in RNA synthesis

ω - assembly of RNA pol and stabilises RNA pol

σ factor - allows the specific binding of RNA pol with the promotor

Question 3

What portion of DNA is transcribed within prokaryotic DNA?

Answer 3

The antisense or non-coding strand is transcribed as the complementary transcribed sequence of RNA will be same as the coding/sense strand

Protein coding genes in prokaryotes - operons
Operons are multiple genes arranged in tandem along a single DNA strand so they are transcribed together

Sigma factors control gene expression:
Early/middle/late genes

Question 4

Describe initiation P1 of prokaryotic RNA transcription?

Answer 4

Closed complex:
RNAP binds to a promotor region in dsDNA form
Promotor - 40 bp, upstream of the transcription site (- value), it commonly interacts with -10 and -35

The dsDNA starts to unwind/melt forming a transciption bubble

Question 5

Describe a potential initiation P2 of prokaryotic RNA transcription?

Answer 5

Abortive initiation
In the open complex RNAP starts adding rNTPs (no primer required)
RNA pol starts to form short RNA stretches 9-11 nt, which are continually released

This occurs because the sigma factor blocks the RNA exit channel on the RNA pol
RNAP fails to escape the promotor, relieving the conformational tension by releasing RNA and the transcriptional bubble can relax

Question 6

Describe elongation of prokaryotic RNA transcription?

Answer 6

The sigma factor dissociates allowing the core enzyme to tightly bind (with high processivity)
RNA synthesis grows in the 5’ to 3’ direction with RNAP moving along the template strand (non-coding strand)

DNA is continuously unwound, breaking H bonds ahead of the enzyme (positively supercoiled)
DNA is rewound behind the core enzyme as H bonds are reformed (negatively supercoiled)
The DNA-RNA bp aren’t very stable but RNAP acts as a linker whilst in elongation

Question 7

Describe one type of termination of prokaryotic RNA transcription?

Answer 7

Two types: Rho independent and Rho dependent

Rho Independent - No external proteins required
2 elements are utilised:
1. Inverted repeats
2. 8 A:T bp (AT rich sequence)
When the repeats are transcribed - RNA forms aq stem loop hairpin, halting the RNA pol
The following A:U bp transcribed are the weakest bp
Therefore RNA is released ending transcription

Question 8

Describe the other type of termination of prokaryotic RNA transcription?

Answer 8

Rho dependent termination:
Rho protein = hexameric ATP dependent helicase, 419 residues

It binds ssRNA rich in C and poor in G (called rho utilisation site = Rut site)
Rho sensitive pause site = 100 nt away from Rut, where RNAP is paused
Rho uses ATP to unwind RNA-DNA double helices by translocating along ssRNA in the 5’ to 3’ direction - overtaking RNAP
During translocation it moves one nucleotide along the RNA for every ATP is hydrolyses (due to 6 subunits successively undergoing hydrolysis)

Question 9

What is significant about eukaryotic RNA transcription?

Answer 9

Eukaryotic transcription has multiple RNAPs and a more complicated control sequence and machinary

The polymerases require general transcriptional factors to recruit and position the polymerase at the promotor

Question 10

What are the eukaryotic RNA polymerases?

Answer 10

RNA pol I - synthesises the precursors for rRNAs

RNA pol II - synthesises the precursors for mRNA
It binds 2Mg2+ at the active site

RNA pol III - synthesises the precursors for tRNA and 5S rRNA

All contain 2 nonidentical large subunits and 12 different small subunits
Many are homologs to the subunits in prokaryotes and are highly conserved

Question 11

Describe the promotor region in eukaryotes?

Answer 11

It is very long

Structural genes expressed in all tissues (housekeeping genes) - GC-rich DNA
Structural genes that are selectively expressed - lack GC-rich sequences, they contain a core promotor 40-50 nt
e.g. TATA box (AT-rich) located 25 to 31 bp upstream of the transcription start site

Question 12

Describe initiation P1 of eukaryotic RNA transcription?

Answer 12

TFIID binds to the TATA box promotor region, TFIID contains a TBP - TATA binding protein
TBP bends the DNA by 45° in between the first 2 and last 2 bp of the TATA box
TBP uses H bonds and van der Waals interactions
The kink is stabilised by two Phe side chains, flanking each kink from their minor groove sides

The bent conformation creates a stage for assembly of other proteins - to form the Pre-initiation complex

TFIIA - binds and helps stabilise the binding of TFIID with the promotor

Question 13

Describe initiation P2 of eukaryotic RNA transcription?

Answer 13

TFIIB - binds and interacts with TBP & the promotor region (downstream to the TATA sequence) - helping the recruitment of RNA Pol II
TFIIF - binds to RNAP II and helps escort/recruit it to the promotor region, binding to TFIIB
TFIIE - binds to the transcription complex, helping with the binding of TFIIH
TFIIH - binds

= Pre-initiation complex (PIC)

Question 14

Describe TFIIH?

Answer 14

TFIIH (large 9 subunits) - 2 subunits have ATPase activity - acting like a helicase, melting the promotor to induce the open complex formation
7 subunits have kinase activity to phosphorylate the CTD of RNA pol II in order to escape the promotor and start elongation

Question 15

Describe elongation of eukaryotic RNA transcription?

Answer 15

RNA pol II is phosphorylated this stimulates elongation
TFEb (kinase protein) - phosphorylates serine residues in the CTD of RNA pol II
Serine 5 -> allows release of PIC
Serine 2 -> conformational change in RNA pol II to elongate

TFIIS - helps polymerase overcome transcription blocks and helps increase the rate of transcription where slow
mRNA is produced in the 5’ to 5’ direction and 5’ capping takes place

Question 16

What are the protective measures that take place on mRNA?

Answer 16

5’ cap protects mRNAs from 5’ degradation and facilitates translation (bound elongation factor)

The polyA tail protects mRNA from 3’ degradation - helps export mRNA from the nucleus and facilitates translations

Splicing provides opportunities for protein diversity in tissue, development etc…

Question 17

Describe termination of eukaryotic RNA transcription?

Answer 17

There is no defined termination sites for RNA polymerase II
It stops at various distances downstream of the protein-coding gene

At the end the protective measures take place

Question 18

How is the 5’ cap formed?

Answer 18

Takes place whilst RNA is being transcribed (soon after initiation), around 30 nt long
At the 5’ end of mRNAs:
1. RNA triphosphatase removes a gamma phosphate
2. Guanylyation by a guanylyltransferase adds a guanidine cap (uses GTP), forming 5’-5’ triphosphate and PPi
3. Guanine is methylated by guanine-7-methyltransferase, using SAM to supply the methyl group

Question 19

How is the polyA tail formed?

Answer 19

CstF and CPSF are recruited to the CTD of RNA pol II
CPSF = Cleavage and Polyadenylation Specificity Factor - binds AAUAAA on RNA
CstF = Cleavage Stimulation Factor - binds a GU-rich sequence downstream of AAUAAA

CstF cleaves the mRNA and the dissociates
CPSF then recruits poly A polymerase assing 200 adenine residues at the 3’ end (using ATP) = poly A tail
Poly A binding protein binds to the polyA tail and CPSF is releasesd
PolyA binding protein prevents degredation of the polyA tail

Question 20

What mature mRNAs lack a polyA tail?

Answer 20

Histone mRNAs

They have a protective hairpin to stop the degradation of the mRNA

Question 21

What does splicing do?

Answer 21

Removes introns from eukaryotic genes and splices together the exons from pre-mRNA or heterogenous nuclear RNAs (hnRNAs)
pre-mRNA contains 8 introns on average and are 4-10x largers than exons (average 1800 nt)

Question 22

How does splicing occur?

Answer 22

The splicing occurs via two transesterification reactions
1. A 2′,5′-phosphodiester bond forms between an intron adenosine residue and the intron’s 5′-terminal phosphate group
2. The 5′ exon is released and the intron forms a lariat intermediate (loop in the intron) - 1 exon detached
3. The exon’s free 3′-OH group displaces the 3′ end of the intron, forming a phosphodiester bond with the 5′-terminal phosphate of the other exon
= spliced product

Question 23

How much energy does splicing involve?

Answer 23

Splicing process proceeds without free energy input
The transesterification reactions preserve the free energy of each cleaved phosphodiester bond through the formation of a new one

Question 24

What is splicing mediated by?

Answer 24

Small nuclear RNAs (snRNAs) which form small nuclear ribonucleoproteins (snRNPs - snurps) in the spliceosome

The spliceosome carries out the two transesterification reactions
U1-U6 snRNAs are at the heart of spliceosome action - enhancing the nucleophilicity of the attacking OH group and stabilising the leaving group
There is a snRNP core protein, consisting of 7 Sm proteins - they bind to a conserved AAUUUGUGG sequence called Sm RNA motif
The Sm proteins form the heteroheptameric ring = 70 Å diameter

Question 25

What is alternative mRNA splicing?

Answer 25

Alternative combinations of exons - yeilding multiple proteins from a single gene
= isoforms
Isoforms of genes lead to protein diversity

Exons can be retained or skipped, introns may be excised or retained, and the positions of 5′and 3′splice sites can be shifted to make exons longer or shorter

Question 26

How does the spliceosome recognise small exons within the introns?

Answer 26

Regulatory proteins recognise these sequences and the splice site - to control which exons are spliced together

Exonic splicing enhancers (ESEs)
Exonic splicing silencers (ESSs)
Intronic splicing enhancers (ISEs)
Intronic splicing silencers (ISSs)

Question 27

What can affect splicing?

Answer 27

15% of human genetic diseases are caused by point mutations - that result in pre-mRNA splicing defects

Activating preexisiting ‘cryptic splice sites’ or generating incorrect new ones

Question 28

What happens to the mRNA after 5 cap, polyA tail and splicing?

Answer 28

mRNA can be edited - additional PTM using guide RNAs (gRNAs)

Mature eukaryotic mRNAs are actively transported from the nucleus to the cytoplasm
Mature mRNA is identified by m7G caps, polyA tails and the exon junction complex (EJC) that attach during splicing
Nucleoporins allow the diffusion of molecules but mRNA requires active transport to pass through a nuclear pore complex (NPC)