Chapter 13 - Mechanism of Transcription

Question 1

How does RNA polymerase initiate transcription?

Answer 1

The novo - does not need a primer. Requires that DNA template be brought into the polymerase active site and held stably in a helical conformation and that the initiating ribonucleotide be brought into the active site and held steely on the template while the next NTP is presented with correct geometry for the chemistry of polymerisation to occur.

Question 2

How can multiple RNA polymerase transcribe the same gene at the time?

Answer 2

RNA product does not remain base-paired to the template DNA strand: the enzyme displaces the growing chain only a few nucleotides behind where each ribonucleotide is added. Therefore, the cell can synthesise a large number of transcripts from a single gene in a short time.

Question 3

What are the mechanical difference between transcription and replication?

Answer 3

Transcription is less accurate than replication. Transcription only produces transient copies. Transcription selectively copies only certain parts of the genome and makes anywhere from one to several hundreds or thousand copies of any given section.

Question 4

Name the subunits of RNA pol in prokaryotic cells.

Answer 4

Alpha, alpha’, beta, beta’ and omega.

Beta and Beta’ form the active site.
Omega has a nuclear function.

Question 5

What is the difference between the type of RNA polymerases?

Answer 5

Pol I and Pol III are involved in transcribing specialised, RNA-encoding genes. Pol I transcribes the large rRNA precursor genes. Pol III transcribes tRNA genes, some small nuclear RNA genes, and the 5S rRNA genes.

Question 6

Name the RNA subunits in eukaryotic cells.

Answer 6

RNAP I - RPA1, RPA2, RPC5, RPC9, RPB6.
RNAP II - RPB1, RPB2, RPB3, RPB11, RPB6.
RNAP III - RPC1, RPC2, RPC5, RPC9, RPB6.

Question 7

Which subunits make up the pincers of RNA pol’s crab claw?

What is the active site called and how does it work?

Answer 7

Beta and beta’ in bacterial enzyme and RPB1 and RPB2 in eukaryotic enzyme.

The active centre cleft. According to a two-metal ion catalytic mechanism. The active site contain a highly bound Mg2+ ion and the second Mg2+ ion is brought in with each new nucleotide in the addition cycle and released with the pyrophosphates.

Question 8

Explain how initiation works.

Answer 8

A promoter in the DNA sequence bind RNA polymerase and undergoes structural changes required for initiation to proceed. The DNA around transcription point unwinds –> base pairs are disrupted, producing a “transcription bubble” of single stranded DNA. The new ribonucleotides are added in the 3’ direction. Only one of the DNA strands acts as a template.

Question 9

Explain how elongation works.

Answer 9

Once RNA pol has synthesised a short stretch of RNA it shifts into elongation. RNA pol unwinds the DNA in front and renewals it behind; it dissociates the growing RNA chain from the template as it moves along.

Question 10

Explain what is termination.

Answer 10

Once the polymerase has transcribed the length of the gene, it stops and releases the RNA product. In some cases well defined sequences trigger termination.

Question 11

Describe the steps of initiation.

Answer 11

Closed complex - initial binding of the polymerase to the promoter.

Open complex - DNA strands separate over a distance of 13 bp to form the transcriptional bubble.

Initial transcribing complex - incorporation of the first 10 or so ribonucleotides is a inefficient process, and at that stage, the enzyme often releases short transcripts and then begins synthesis again.

Question 12

What is the role of factor sigma?

Answer 12

This factor converts the core enzyme into the form that initiates only at the promoter. This form of the enzyme is called RNA polymerase holoenzyme.

Question 13

What is the predominant sigma factor called in E. coli? Which characteristics do polymerase containing this factor share?

Answer 13

Sigma 70. Two conserved sequences, each of 6 nucleotides, separated by 17-19 nucleotides. At around -10 and -35.

Question 14

What does “strength of a promoter” mean?

Answer 14

How many transcripts it initiates in a given time. Influenced by how well the promoter binds polymerase initially. how efficient it supports isomerization, and how readily the polymerase can then escape.

Question 15

What is an UP-element?

Answer 15

DNA element that binds RNA polymerase in strong promoters and increases polymerase binding by providing an additional specific interaction between the enzyme and the DNA.

Question 16

What is an extended -10 element ?

Answer 16

Comprises a standard -10 element with and additional short sequence element at its upstream end.

Question 17

What are the sigma 70 regions?

Answer 17

Region 1-4. Region 2 and 4 recognise -10 and -35 elements, respectively. Region 2 is involved in melting of the single stranded DNA.

Question 18

What is the role of the two helices in within region 4 of sigma 70?

What is the purpose of this binding?

Answer 18

They are called the helix-turn-helix. One of these helices inserts itself into the major groove and interacts with the bases int he -35 region. The other lies across the top of the groove, making contact with the DNA backbone.

Provides binding energy to secure polymerase to the promoter.

Question 19

How is melting driven through binding interactions between sigma and the single stranded DNA?

Answer 19

Two bases (A11 and T7) in the non-template strand are flipped out and inserted into pockets within the sigma protein where they make favourable contact that stabilises the unwound state of the promoter region.

Question 20

By what and how is the UP-element recognised?

Answer 20

It is not recognised by the sigma subunit, instead it is recognised by a carboxyl-terminal domain of the alpha subunit called the alphaCTD, which is connected to the alphaNTD by a flexible linker. Thus, although the alphaNTD is embedded in the body of the enzyme, the alphaCTD can reach the upstream element.

Question 21

In which position does melting occur, in respect to the transcription start site?

Answer 21

-11 and +2

Question 22

What is isomerization?

Answer 22

The transition from a closed to an open complex –> “melting”.

Question 23

Name the 5 channels of RNA pol and describe their function.

Answer 23

The NTP-uptake channel allows ribonucleotides to enter the active centre. The RNA-exit channel allows the growing RNA chain to leave the enzyme. The remaining three channels allow DNA entry from the enzyme, as follows. The downstream DNA enters the active centre cleft in double-stranded form through the downstream DNA channel. The non-template strand exit the active centre cleft though the non-template-strand (NT) channel and travels across the surface of the enzyme. The template strand exits through the template-strand (T) channel.

Question 24

What are the two structural changes seen in the enzyme upon isomerization from closed to open form?

Answer 24

The pincers at the front of the enzyme clamp down on the downstream DNA. There is a major shift in the position of the amino-terminal region of the sigma (1.1). When not bound sigma 1.1 lies within the active centre cleft of the holoenzyme, blocking the path that, in the open complex is followed by the template DNA strand. In the open complex this region shifts 50 Å and is now found on the outside of the enzyme, allowing the DNA access to the cleft.

Question 25

What does the sigma region 3/4 do?

Answer 25

It interacts with the template strand, organising it in the correct conformation and location to allow intiation.

Mimics RNA

Question 26

Explain the 3 models of abortive transcription.

Answer 26

1. Transient excursion - transient cycles of forward and reverse translocation of RNA pol.
2. Inchworming - flexible element within the polymerase that allows a module at the front of the enzyme, containing the active site, to move downstream, synthesising a short transcript before aborting and retracting tot he body of the enzyme still at the promoter.
3. Scrunching - DNA downstream from the stationary promoter-bound, polymerase is unwound and pulled into the enzyme.

Question 27

What is promoter escape?

Answer 27

Breaking of all interactions between polymerase and promoter elements and between polymerase and any regulatory proteins operating at the given promoter.

Question 28

Which part of the sigma factor is involved in abortive initiation?

Answer 28

The 3/4 linker (mimics RNA). This region of sigma lies in the middle of the RNA exit channel in the open complex, and for an RNA chain to be made longer than 10 nucleotides this region of sigma must be ejected from hat location, a process that can take the enzyme several attempts.

Question 29

Explain how RNA polymerase elongation works.

Answer 29

During elongation, the enzyme adds one nucleotide at a time to the growing RNA transcript. Polymerase uses a step mechanism: the enzyme steps forward as a molecular motor, advancing in a single step a distance equivalent to a base pair for every nucleotides it adds to the growing RNA chain. The size of the bubble remains constant throughout elongation.

Question 30

Which two proofreading functions does RNA polymerase perform? And explain how they work.

Answer 30

Pyrophosphorolytic editing and hydrolytic editing.

Pyrophosphorolytic editing: The enzyme uses its active site in a simple back-reaction, to catalyse the removal of an incorrectly inserted ribonucleotide, by reincorporation of PPi.

Hydrolytic editing: the enzyme backtracks by one to more nucleotides and cleaves the RNA.

Question 31

Which factors stimulate hydrolytic editing?

Answer 31

Cre factors. They function as elongation stimulating factors –> they ensure that polymerase elongates efficiently and help overcome “arrest” at sequence that are difficult to transcribe.

Question 32

What is the equivalent to Cre factors in eukaryotic RNA polymerase?

Answer 32

TFIIS

Question 33

Which class of proteins join polymerase in the elongation phase and promotes the processes of elongation and termination?

Answer 33

Nus proteins.

Question 34

What is TRCF?

Answer 34

Protein with ATPase activity. It binds double-stranded DNA upstream of the polymerase and uses the ATPase motor to translocate along the DNA until it encounters the stalled RNA polymerase. The collision pushes polymerase forward, either allowing it to restart elongation or, more often, causing dissociation of the ternary couples of RNA polymerase, template DNA, and RNA transcript. This terminates transcription by the enzyme, but it makes way for repair enzymes and for another RNA polymerase.

Question 35

Which two forms does transcription termination occur in bacteria? And describe this process?

Answer 35

Rho is a protein (hexameter) that binds to the naked (non translating) RNA chain and travels on it. When it reaches DNA/RNA hydrid, it dissolves the complex.

Rho-dependent and Rho-independent.

Rho-dependent: Have ill-defined RNA elements called ruts sites and for them to work requires the action of the Rho factor. The energy derived from ATP hydrolysis powers the translocase/helicase activity of Rho to unwind RNA/DNA duplex.

Rho-independent: they need no other factors to work. Consists of two sequence elements: a short inverted repeat followed by a stretch of about eight A:T base pairs. These elements do not affect the polymerase until they have been transcribed-that is, they function in the RNA rather than in the DNA. When polymerase transcribes an inverted repeat sequence, the resulting RNA can form a stem-loop structure –> hairpin –> termination.

Question 36

Describe the Rho protein?

Answer 36

Ring-shaped protein with six identical subunits, bind to single-stranded RNA as it exits the polymerase. The protein has also an ATPase activity, and once attached to the transcript, Rho uses the energy derived from ATP hydrolysis to induce termination.

Question 37

Which models of Rho terminations exist?

Answer 37

Rho pushes polymerase forward relative to the DNA and RNA, resulting in termination in a manner analogous to termination by TRCF; Rho pulls RNA out of the polymerase, resulting in termination; Rho induces a conformational change in polymerase, causing the ezyme to terminate.

Question 38

How is Rho directed to work on particular RNA transcripts?

Answer 38

There is specificity in the sites it binds, optimally, these sites consists of stretches of 40 nucleotides that do not fold into a secondary structure. Rho fails to bind to any transcript that is being translated. In bacteria, transcription and translation are highly coupled-translation initiates on growing RNA transcription as soon as they start exiting polymerase, while they are still being synthesised. Rho typically terminates only those transcripts still being transcribed beyond the end of a gene or operon.

Question 39

How do hairpin structures induce termination?

Answer 39

By either pushing polymerase forward relative to the DNA and RNA, wrestling the transcript from polymerase, or inducing a conformational change in polymerase. It works as an terminator when it is followed by a stretch A:U base pairs. A:U are the weakest of all base-pairs.

Question 40

Name the differences in transcription between eukaryotes and bacteria.

Answer 40

Bacteria: one RNA polymerase, one initiation factor (sigma)
Eukaryotes: at least three RNA polymerases, several initiation factors –> general transcription factors (GTFs).

Question 41

What is the eukaryotic core promoter?

Answer 41

Minimal set of sequences elements required for accurate transcription initiation by pol II machinary (typically 40-60 nucleotides long) extending either upstream or downstream from the transcription start site.

Question 42

Which elements are found in the core promoter site?

Answer 42

TFIIB recognition element (BRE), the TATA element (or box), the initiator (Inr), the downstream promoter element.(DPE, DCE, MTE).

Question 43

Which elements make up the regulatory sequences?

Answer 43

Promoter proximal elements, upstream activator sequence (UASs), enhancer, and a series of other elements called silencers, boundary elements, and insulators.

Question 44

What is the preinitiation complex?

Answer 44

The complete set of general transcription factors and polymerase, bound together at the promoter and poised for initiation.

Question 45

Where does the preinitiation complex formation begin?

Answer 45

At the TATA elements.

Question 46

Which element recognises the TATA element and where does it bind?

Answer 46

TFIID. Binds to the TATA DNA sequence called TBP (TATA-binding protein). TBP cause the minor groove to be flattened by 80 degrees and intercalates two Phe residues the base stack.

Question 47

What is role of TFIID?

Answer 47

Is a critical factor in promoter recognition and preinitiation complex establishment.

Question 48

In which orders does the preinitiation complex assemble?

Answer 48

TBP binding distorts the TATA sequence. The resulting TBP-DNA complex provides a platform to recruit other general transcription factors and polymerase itself to the promoter. In vitro, these proteins assemble at the promoter in the following order: TFIIA, TFIIB, TFIIF together with polymerase, and then TFIIE and TFIIH. Formation of the preinitiation complex containing these components is followed by promoter melting.

Question 49

In eukaryotes which two steps are required for promoter escape?

Answer 49

1. ATP hydrolysis

2. Phosphorylation of the polymerase

Question 50

Explain how phosphorylation of polymerase II occurs?

Answer 50

The large subunit of Pol II has a carbonyl-terminal domain (CTD), which is referred to as the “tail”. The CTD contains a series of repeats of the heptapeptide sequence: Tyr-Ser-Pro-Thr-Ser-Pro-Ser (52 repeats in human). Proline is the only amino acid where the side chain is covalently linked to the backbone –> takes 40 X longer to add than other amino acids.

Each repeat contains sites for phosphorylation by specific kinases, including a subunit of TFIIH. The form of Pol II recruited to the promoter initially contains an unphosphorylated tail, but the species found in the elongation complex bears multiple phosphoryl groups on its tail. Addition of these phosphates helps polymerase to shed the majority of the general transcription factors used for initiation.

Question 51

Why is TBP’s binding unorthodox?

Answer 51

It uses an extensive region of Beta sheet to recognise the minor groove of the TATA element. The reason is linked to the need of the protein to distort the local DNA structure.

Question 52

What happens to DNA upon TBP binding?

Answer 52

Causes the minor groove to be widened to an almost flat conformation, it also bends the DNA by an angle of 80 degrees. Specificity is imposed by two pairs of phenylalanine side chains that intercalate between the bas pairs at either end of the recognition sequence and drive the strong bend in the DNA.

Question 53

What is the role of TFIIB?

Answer 53

It enters the preinitation complex after TBP. It also contacts Pol II in the preinitiation complex. These protein appear to bridge the TATA-bound TBP and polymerase . Structural studies suggests that segments of TFIIB inserts into the RNA-exit channel and active centre cleft of Pol II in a manner analogous to the sigma region 3/4 linker in the bacterial case.

Question 54

What is the role of TFIIF?

Answer 54

This factor associates with Pol II and is recruited to the promoter together with the enzyme. Stabilising the DNA-TFIIB complex and is required before TFIIE and TFIIH are recruited to the preinitiation complex.

Question 55

What is the role of TFIIE and TFIIH?

Answer 55

TFIIE consists of two subunits that has a role in recruitment of TFIIH. TFIIH is a helices that unwinds DNA.TFIIH controls the ATP-dependent transition of the preinitiation complex to the open complex. TFIIH has a subunit that functions as ATPase and another that is a portion kinase, with roles in promoter melting and escape.

Question 56

How does TFIIH mediate promoter melting?

Answer 56

It is believed that the TFIIH acts as an ATP-driven translator of double-stranded DNA.This subunit binds to DNA downstream from polymerase and feeds double-stranded DNA, with a right-handed threading, not the cleft of the polymerase. This action drives the melting of the DNA because the upstream promoter DNA is held in a fixed position by TFIID and the rest of the GTFs.

Question 57

Which additional proteins are required for in vivo transcription compared to in vitro?

Answer 57

Activators that help recruit polymerase to he promoter, stabilising its binding there. This is mediated through interaction between DNA-bound activators, chromatin-modifying and -remodeling factors, and parts of the transcription machinery.

Question 58

What is the function of hammerhead Ribozyme?

Answer 58

The hammerhead ribozyme is an RNA motif that catalyzes reversible cleavage and ligation reactions at a specific site within an RNA molecule. It is one of several catalytic RNAs (ribozymes) known to occur in nature.

Question 59

Give an overview of transcription.

Answer 59

RNA transcription goes only in 5’->3’ direction. RNA is transcribed from one strand only. Transcription initiation involves recognition of a particular sequence located upstream the gene called the promoter. DNA unwinding starts from the promoter after RNA polymerase binds on it. DNA unwinding is temporary in transcription. The primary transcript is unstable. Special proteins are involved in termination of transcription.

Question 60

How big is the RNA polymerase?

Answer 60

It covers a region between 75-80 bp, extending from -55 to +20 during initiation.

Question 61

What are consensus sequences? What are the two consensus sequences at bacterial promoter?

Answer 61

Sequences most often conserved at a particular site.

• 10 sequence (TATA box)
• 35 sequence

Question 62

Which are the common consensus sequences in eukaryotic promoters?

Answer 62

Element 
BRE (TFIIB recognition element)
TATA
INR (initiator element)
DPE (downstream promoter element)

Question 63

Which are the general transcription factors needed for RNA pol I in eukaryotes?

Answer 63

TFIID, TFIIH and TFIIE.

Question 64

Explain the terms activators, mediators and chromatin modifying enzymes.

Answer 64

Activators - attract RNA Pol to the promotor, role in gene expression.

Mediator - Big proteins, allow activator and RNA pol interaction.

Chromatin modifying enzymes - allow greater accessibility to DNA.

Question 65

How is promoter escape mediated in Eukaryotes?

Answer 65

Promoter escape requires phosphorylation of RNA pol CTD tail with hetapetide repeats (Tyr-Ser-Pro-Thr-Ser-Pro-Ser). Promoter escape happens when RNA pol II breaks its contact with the promoter.

Question 66

Which complex couples the Ribosome with the RNA pol?

Answer 66

The NusG - NusE complex.

Question 67

Explain 7-methyl guanosine capping?

Answer 67

The gamma phosphate is removed from the 5’ end by RNA triphosphatase. Then, guanyltransferase adds a GMP moiety to the 5’ beta phosphate. The last step is addition of the methyl group to the guanine residue at eh 7’ position. Capping proved a rare example of 5’->5’ linkage.

Question 68

Explain the process of polyadenylation and the torpedo model of termination.

Answer 68

CPSF and CSTF are poly-A signals. These factors once bound to RNA recruit other proteins that leads to RNA cleavage and then polyadenylation.

Polyadenylation is mediated by poly-A polymerase, which adds approximately 200 adenines.

Torpedo model
The free end of the second RNA is uncapped and thus can be distinguished from genuine transcripts. This new RNA is recognised by an RNase called (in yeast) Ra1 (in human hXrn2) that is located onto the end of the RNA by another protein (Rtt103) that binds the CTD of RNA polymerase. The Rat1 enzyme quickly degrades the RNA in a 5’-to-3’ direction, until it catches up to the still-transcribing polymerase from which the RNA is being spewed.