Chapter 13

Question 0

Template and non template strands

Answer 0

Gene Expression Requires Base Sequences
 The strand that is actually transcribed (used as the template) is termed the template strand
 The opposite strand is called the coding strand or the sense strand (non template strand)
 The base sequence is identical to the RNA transcript
 Except for the substitution of uracil in RNA for thymine in DNA

Question 1

Difference between transcription and DNA replication

Answer 1

Difference between transcription and DNA replication
• RNA Polymerase that catalyzes RNA synthesis can initiate transcription de novo, it doesn’t need a primer.
• The RNA product doesn’t remain base-paired to the template DNA strand. This displacement is critical for RNA to perform its function. Also, a cell can synthesize a large numbers of transcripts from a single gene in a short time.
• Transcription, although very accurate, is less accurate than replication (1 in 10,000 in error rate, compared to 1 in 10 million for replication). Lack of extensive proofreading mechanism for transcription, although two forms of proofreading mechanisms for RNA synthesis do exist.
• Replication: permanent. Transcription: transient.
• Replication copies the whole genome, while transcription only copy certain parts of the genome. The choice of what part of genome to transcribe and how extensively, can be regulated.

Question 2

Bacterial RNA pols and their subunits

Answer 2

RNA Polymerase Structure
By 1969 SDS-PAGE of RNA polymerase from E. coli had shown several subunits
– 2 very large subunits are β (150 kD) and β’ (160 kD) – Sigma(σ)at70kD
– Alpha (α) at 40 kD – 2 copies present in holoenzyme – Omega(w)at10kD
• Was not clearly visible in SDS-PAGE, but seen in other experiments
• Not required for cell viability or in vivo enzyme activity • Appears to play a role in enzyme assembly

Question 3

Bacterial RNA pols subunits’ functions/steps of transcription

Answer 3

Steps of transcription by RNA Polymerase
Transcription occurs in three stages
Initiation: Promoter is the sequence that initially binds the RNA Polymerase. Structural changes of the complex after binding (required for initiation to proceed). Then DNA unwinds (~13bp DNA separation), and initiation of transcription from 5’ to 3’ orientation. Because RNA Polymerase binds promoters in a defined orientation, the same strand is always transcribed from a given promoter. The transcription starting site is designated as +1 position.
Elongation: it starts after the first stretch of RNAs (~10bases). RNA synthesis. DNA unwinding in front and re-annealing behind. RNA dissociation with the template. Proofreading.
Termination: Specific, well-characterized sequences trigger termination: stop transcription, polymerase and RNA

Question 4

Bacterial promoter elements

Answer 4

Features of bacterial promoters
Two conserved sequences (- 10 and -35), each of 6 nucleotides, separated by a non-specific stretch of 17-19 nucleotides.
Extended -10 element
compensates for the absence of a -35 region.
A discriminator element
influences the stability of the complex between the enzyme and the promoter.

Question 5

Promoter strength

Answer 5

Promoter Strength
• Consensus sequences:
– -10 box sequence approximates TATAAT – -35 box sequence approximates TTGACA
• Mutations that weaken promoter binding:
– Down mutations
– Increase deviation from the consensus sequence
• Mutations that strengthen promoter binding:
– Up mutations
– Decrease deviation from the consensus sequence

Question 6

Sigma factor

Answer 6

Sigma as a Specificity Factor
• Core enzyme without the σ subunit could not transcribe viral DNA, yet had no problems with highly nicked calf thymus DNA
• With σ subunit, the holoenzyme worked equally well on both types of DNA

Question 7

Function of regions of sigma (regions 1-4)

Answer 7

Role of region 1 appears to be in preventing σ from binding to DNA by itself. Region 1.1 is highly negative charged, and acts as a molecular mimic of DNA. When not bound to DNA, σ region 1.1 lies within the active center cleft of the holoenzyme, blocking the path. In the open complex, region 1.1 shift some 50A and is now found on the outside of the enzyme. Region 1.2 recognizes the discriminator
• Region 2 is the most highly conserved of the four. There are four subregions – 2.1 to 2.4. 2.4 recognizes the promoter’s -10 box. The 2.4 region appears to be a-helix.
• Region 3 is involved in both core enzyme and DNA binding
• Region 4 is divided into 2 subregions. This region seems to have a key role in promoter recognition. Subregion 4.2 contains a helix-turn- helix DNA-binding domain and appears to govern binding to the -35 box of the promoter

Question 8

Function of α and σ subunits

Answer 8

Recruits RNA Polymerase core enzyme to the promoter
The αCTD and the αNTD is connected by a flexible linker. The αCTD element can reach the upstream element even when is further upstream of the -35 element.

Question 9

Polymerase/Promoter Binding

Answer 9

Polymerase/Promoter Binding
• Holoenzyme binds DNA loosely at first
• Complex loosely bound at promoter = closed promoter complex, dsDNA in closed form
• Holoenzyme melts DNA at promoter forming open promoter complex - polymerase tightly bound

Question 10

Local DNA Melting at the Promoter

Answer 10

Local DNA Melting at the Promoter
• From the number of RNA polymerase holoenzymes bound to DNA, it was calculated that each polymerase caused a separation of about 10 bp
• In another experiment, the length of the melted region was found to be 12 bp
• Later, size of the DNA transcription bubble in complexes where transcription was active was found to be 17-18 bp

Question 11

Experiment to locate melted promoter

Answer 11

DMS: dimethyl sulfate.
S1 nuclease: endonuclease that specifically cut single-stranded DNA (use mild condition). S1 Nuclease degrades single-stranded nucleic acids, releasing 5’-phosphoryl mono- or Oligonucleotides. S1 Nuclease also cleaves dsDNA at the single-stranded region caused by a nick, gap, mismatch or loop. S1 Nuclease exhibits 3’-phosphomonoesterase activity.
RNA Polymerase melts DNA in the -9 to +3 region of the T7 A3 promoter by Methylation S1 assay
Methylation-S1 assay: R: RNA Polymerase S: S1 nuclease

Question 12

Sigma initiates transcription

Answer 12

• In this first experiment stimulation by σ appears to cause both initiation and elongation
• Or stimulating initiation by σ provides more initiated chains for core polymerase to elongate
• Further experiments by the same group proved that σ does not stimulate elongation. (use antibiotic rifamycin to block bacterial transcription initiation)
[14C]ATP label RNA transcript
[γ-32]ATP and [γ-32]GTP label the first position of the RNA (because the first position has all three phosphates (αβγ

Question 13

Transition to the open complex and melting of promoter

Answer 13

Transition to the open complex involves structural changes in the RNA Polymerase and in the promoter DNA
In the transition, σ factor undergoes isomerization, a spontaneous conformational change in the DNA- enzyme complex (does not require ATP hydrolysis), to a more energetically favorable form.
Two bases in the non-template strand of the -10 element (A11 and T7) flip out from their base-stacking interactions and instead insert into the pockets within the σ protein where they make more favorable interactions. By stabilizing the single-stranded form of the -10 element, these interaction drives melting of the promoter region.

Question 14

Channels in and out of open complex

Answer 14

Channels into and out of the open complex
Template strand (gray), nontemplate strand (orange). Four regions of σ are shown.
There are 5 channels into the enzyme. NTP uptake channel (not shown); RNA-exit channel for growing RNA to leave the enzyme;
Three channels allow DNA entry and exit from the enzyme: downstream DNA enter the active center cleft through the downstream DNA channel (between the pincers).
Within the cleft, the DNA strands separate from position +3. The non-template exits the active center through the non-template (NT) channel. The template strand follows a path through the active center cleft and exits through the template-strand (T) channel.
The double helix re-forms at -11 in the upstream DNA behind the enzyme.

Question 15

Summary of sigma and alpha factors

Answer 15

The σ-factor allows initiation of transcription by causing the RNA polymerase holoenzyme to bind tightly to a promoter
• This tight binding depends on local melting of the DNA to form an open promoter complex and is stimulated by σ
• The σ-factor can therefore select which genes will be transcribed

Question 16

Transcription initiation by RNA pol

Answer 16

Transcription is initiated by RNA Polymerase without the need for a primer
• DNA Polymerase needs RNA primer to synthesize new DNA;
• RNA Polymerase can initiate a new RNA chain on a DNA template and thus does not need a primer. This requires that the DNA template and the initiating ribonucleotide to be brought into the active site.
• RNA Polymerase starts most transcripts with an A, and that ribonucleotide binds the template nucleotide (T) with only two two hydrogen bonds.
• The enzyme has to make specific interactions with one or all of the DNA template strand, initiating ribonucleotide, and the second- ribonucleotide—holding one or all rigidly in the correct orientation to allow chemical attack on the incoming NTP.

Question 17

Transcription initiation 2

Answer 17

Transcription initiation was assumed to end as RNA polymerase formed 1st phosphodiester bond
• Carpousis and Gralla found that very small oligonucleotides (2-6 nt long) are made without RNA polymerase leaving the DNA
• RNA Polymerase produces and releases short RNA transcripts of <10 nucleotides (abortive transcripts) before escaping the promoter, entering the elongation phase.

Question 18

Transcription initiation mechanism

Answer 18

During initial transcription, RNA Polymerase remains stationary and pulls downstream DNA into itself. Several hypotheses have been proposed
• The polymerase cannot move enough downstream to make a 10-nt transcript without doing one of three things:
- transient excursion: moving briefly downstream and then snapping back to the starting position
- inchworming: stretching itself by leaving its trailing edge in place while moving its leading edge downstream
- scrunching: compressing the DNA without moving itself (appears to be correct)

Question 19

FRET technology

Answer 19

Fluorescence Resonance Energy Transfer
• To answer this question Fluorescence Resonance Energy Transfer (FRET) was used as it relies on two fluorescent molecules that are close enough together to engage in transfer of resonance energy. When the two molecules move apart, the efficiency of this energy transfer will decrease
• FRET allows the position of σ relative to a site on the DNA to be measured without using separation techniques that might displace σ from the core enzyme
FRET experiments support that σ factor remains with the core after promoter clearance

Question 20

Transcription elongation and proofreading

Answer 20

The elongating Polymerase is a processive machine that synthesizes and proofreads RNA
• After transcription initiation is accomplished, core polymerase continues to elongate the RNA. Nucleotides are added sequentially, one after another in the process of elongation.
• Double-stranded DNA enters the front of the enzyme between the pincers;
• At the opening of the catalytic cleft, the strands separate to follow different paths through the enzyme before exiting via their respective channels and re-forming a double helix behind the elongating polymerase.
• Ribonucleotides enter the active site through their defined channel and are added to the growing RNA chain under the guidance of the template DNA strand.
• Only 8 or 9 nucleotides of the growing RNA chain remain base-paired to the DNA template at any giving time; the remainder of the RNA chain is peeled off and directed out of the enzyme through the RNA exit channel.

Question 21

Template and transcript within the RNA Polymerase elongation complex (One step mechanism)

Answer 21

A. untranslocated Polymerase (9-base pair stretch between the template DNA and the RNA chain);
B. Forward translocated polymerase (+1): the enzyme translocates one base forward)
C. NTP bound the template
D. Reversetranslocated polymerase (-1) (for hydrolytic editing, proofreading mechanism)
The enzyme steps forward as a molecular motor, advancing
in a single step a distance equivalent to a base pair for
every nucleotide it adds to the

Question 22

Pausing and proofreading during transcription

Answer 22

Pausing and Proofreading
• RNA polymerase frequently pauses, or even backtracks, during elongation
• Pausing allows ribosomes to keep pace with the RNA polymerase, and it is
the first step in termination.
• Two mechanisms of proofreading on the growing transcript:
A. Backtracking (Polymerase backtrack) aids proofreading by extruding the 3’-end of the RNA out of the polymerase, where misincorporated nucleotides can be removed by an inherent nuclease activity of the polymerase, stimulated by auxiliary factors (GreA and GreB). (Hydrolytic editing).
B. Pyrophosphorylytic editing: the RNA Polymerase uses is active site, in a simple back-reaction, to catalyze the removal of an incorrectly inserted ribonucleotide, by reincorporation of PPi. The enzyme can then incorporate another ribonucleotide in its place in the growing RNA chain.

Question 23

RNA Polymerase can become arrested and need removing

Answer 23

RNA Polymerase can become arrested and need removing
• One common cause of the arrest is a damaged DNA strand.
• The consequence of arrest can be catastrophic. To deal with this situation, the cell has machinery that removes the arrested polymerase and at the same time recruits repair enzymes (Uvr(A)BC). The repair followed is called trancription-coupled repair.
• Both Polymerase removal and repair enzyme recruitment are performed by a single protein TRCF.
• TRCF has ATPase activity. It binds double-stranded DNA upstream of the polymerase and use 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 complex 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 Polymerease.

Question 24

Termination of transcription

Answer 24

When the polymerase reaches a terminator at the end of a gene it falls off the template and releases the RNA
• There are 2 main types of terminators
– Intrinsic terminators function with the RNA polymerase by itself
without help from other proteins
– Other type depends on auxiliary factor called rho (ρ), these are rho or ρ-dependent terminators

Question 25

Roh-independent termination

Answer 25

Rho-Independent Termination
• Intrinsic or rho-independent termination depends on terminators of 2 elements:
– Inverted repeats followed immediately by
– T-rich region in the nontemplate (sense) strand of the gene
• An inverted repeat predisposes a transcript to form a hairpin structure due to complementary base pairing between the inverted repeat sequences

Question 26

Rho independent termination mech

Answer 26

Transcription Termination
RNA Polymerase is not shown in here.
Bacterial terminators act by:
A. RNA-DNA hybrid formation
B. hairpin formation
C. This causes transcription to pause
a string of U’s incorporated just downstream of hairpin to destabilize the hybrid and the RNA falls off the DNA template
How the hairpin disrupts the transcribing polymerase is not clear, but the weak interaction between the transcript and the template DNA (Us in the transcript and As in the template) appear to make the release of the transcript easier.

Question 27

Rho dependent termination

Answer 27

Rho-Dependent Termination
• Rho-dependent terminators have rather ill-defined RNA elements called rut sites, and for them to work requires the action of Rho factor. Rut site is an unstructured, cytosine-rich sequence on the mRNA known for which a consensus sequence has not been identified. Rho-dependent termination occurs downstream of translational stop codons and consists of rut site and a downstream transcription stop point.
• Rho, which is a ring-shaped protein with six identical subunits, binds to single-stranded RNA as it exits the polymerase.
• The protein also has ATPase activity, once attached to the transcript, Rho uses the energy from ATP hydrolysis to induce termination.
• The precise mechanism of termination remains to be determined. Models: (A) Rho pushes polymerase forward relative to the DNA and the RNA (like TRCF); (B) Rho pulls RNA out of the Polymerase, resulting in termination; (C) Rho induces a conformational change in polymerase, causing enzymes to terminate (has evidence to support).

Question 28

Rho transcription termination factor

Answer 28

The Rho transcription termination factor
Hexamer: six monomers form an open ring. The ring is not flate, the six subunit is further down in the plane of the page than the first.
The RNA transcript on which Rho acts is believed to bind along the bottom of each subunit and then thread through the middle of the ring.

Question 29

Rho attributes

Answer 29

Rho decreases RNA elongation rate but not initiation
Rho Affects Chain Elongation
• There is little effect of rho or ρ on transcription initiation, if anything it is increased
• The effect of rho or ρ on total RNA synthesis is a significant decrease
• This is consistent with action of rho or ρ to terminate transcription forcing time-consuming reinitiation

Question 30

Rho release mech

Answer 30

Rho releases the RNA Product from the DNA template
RNA co-sedimented with DNA
In the absence of rho.
By contrast, the transcripts made in the presence of Rho sedimented at a much slower rate, independent of the DNA.
Rho (means Release in Greek)
Rho Releases Transcripts from the DNA Template
• Compare the sedimentation of transcripts made in presence and absence of ρ
– Without ρ, transcripts cosedimented with the DNA template – they hadn’t been released
– With ρ present in the incubation, transcripts sedimented more slowly – they were not associated with the DNA template
• It appears that ρ serves to release the RNA transcripts from the DNA template

Question 31

Mechanism of Rho

Answer 31

No string of T’s in the ρ-dependent terminator, just inverted repeat to hairpin
• (a). Rho binds to the RNA Polymerase;
• (b). Binding to the growing transcript (via rho loading site in the RNA), ρ follows the RNA polymerase
• (c) It catches the polymerase as it pauses at the hairpin. Then Rho tightens the RNA loop and irreversibly trapped the elongation complex.
• (d). Releases transcript from the DNA- polymerase complex by unwinding the RNA-DNA hybrid (RNA-DNA helicase activity)

Question 32

Rho summary

Answer 32

Summary
• rho-independet terminator revealed two important features (using the trp attenuator as a model):
1 - an inverted repeat that allows a hairpin to for at the end of the transcript
2 - a string of T’s in the nontemplate strand that results in a string of weak rU-dA base pairs holding the transcript to the template strand
• Rho-dependent terminators consist of an inverted repeat, which can cause a hairpin to form in the transcript but no string of T’s

Question 33

Euk transcription

Answer 33

Eukaryotic transcription
• Eukaryotic transcription process is highly similar to that in bacteria.
• Bacteria have only one RNA Polymerase, but all eukaryotes have at least
three different ones.
• Bacteria require only one additional initiation factor (σ), but several initiation factors are required for efficient and promoter-specific initiation in eukaryotes. They are called general transcription factors (GTFs).
• In vitro, the general transcription factors are all that are required, together with Pol II, to initiate transcription on a DNA template. In vivo, however, the DNA template in eukaryotic cells is incorporated into nucleosomes. Additional factors are required, including DNA-binding regulatory proteins and chromatin-modifying enzymes.

Question 34

Forms of euk RNA polymerases

Answer 34

There are at least three RNA polymerases operating in eukaryotic nuclei
– RNA Pol I transcribes large ribosomal RNA genes (28S, 18S and 5.8S rRNA)
– RNA Pol III transcribes tRNA genes, some small nuclear RNA genes and 5S rRNA gene.
– RNA Pol II transcribe rest of nuclear genes, espcially the protein coding genes.
– RNA Pol IV and V are found in plants, where they transcribe small interfering RNAs.
• Ribosomal genes are different from other nuclear genes
– Different base composition from other nuclear genes (GC content 60%. The rest of the genes: 40%)
– Unusually repetitive (hundreds to 20,000 copies)
– Found in different compartment, the nucleolus

Question 35

Euk core promoter elements

Answer 35

Core Promoter Elements
• The core promoter is modular and can contain almost any combination of the following elements:
– TATA box
– TFIIB recognition element (BRE)
– Initiator (Inr)
– Downstreampromoterelement(DPE) – Downstream core element (DCE)
– Motif ten element (MTE)
• At least one of the four core elements is missing in most promoters
• TATA-less promoters tend to have DPEs
• Promoters for highly specialized (luxury) genes tend to have TATA boxes, while many housekeeping genes (constitutively active) and developmentally regulated genes (i.e., homeotic genes) have TATA- less promoters.

Question 36

Euk class II promotors

Answer 36

Class II promoters are recognized by RNA polymerase II
• Considered to have two parts:
– Core promoter - attracts general transcription factors and RNA polymerase II at a basal level and sets the transcription start site and direction of transcription
– Proximal promoter - helps attract general transcription factors and RNA polymerase and includes promoter elements upstream of the transcription start site (also as part of the regulatory elements)

Question 37

Euk regulatory seqs

Answer 37

Regulatory sequences
• Other sequences that are required for accurate and efficient transcription in vivo:
1. promoterproximalelement
2. Upstreamactivatorsequence
3. Enhancer
4. Silencer
5. Boundaryelement
6. Insulator
• All these regulatory elements bind regulatory proteins (activators and repressors), which help or hinder transcription from the core promoter.
• Some regulatory elements can be located many or even hundreds of kilobases from the core promoter on which they act.

Question 38

Euk enhancers and silencers

Answer 38

Enhancers and Silencers
• These are position- and orientation-independent DNA elements that stimulate or depress, respectively, transcription of associated genes
• Are often tissue-specific in that they rely on tissue- specific DNA-binding proteins for their activities
• Some DNA elements can act either as enhancer or silencer depending on what is bound to it

Question 39

Transcription Initiation by RNA Pol II in Euks

Answer 39

Transcription Initiation by RNA Pol II
RNA Pol II forms a preinitiation complex with general transcription factors at the promoter:
A. TATA elements is recognized by TFIID through its component TBP (TATA-binding protein).
B. TBP distorts the TATA sequence, and the resulting TBP-DNA complex provides a platform to recruit other general transcription factors. In vitro, the order is TFIIA, TFIIB and TFIIF, and then TFIIE and TFIIH.
C. Formation of the preinitiation complex is followed by promoter melting, which requires ATP hydrolysis mediated by TFIIH.

Question 40

Function of polymerase tail

Answer 40

Promoter escape requires phosphorylation of Polymerase tail
• For polymerase escapes the promoter and enters the elongation phase, two steps occur which are not seen in bacteria:
(A) ATP hydrolysis, which is required for DNA melting; (B) Phosphorylation of the polymerase.
Carboxyl-terminal domain (CTD) or tail of the Pol II contains a series of repeats of the heptapeptide sequence: Tyr-Ser-Pro-Thr-Ser-Pro-Ser.

Question 41

Euk elongation factors

Answer 41

Elongation factors
• P-TEFb: is a kinase recruited to polymerase by transcriptional activators. Once bound to Pol II, this protein phosphorylates the serine residue at position 2 of the CTD repeats. This phosphorylation event correlates with elongation. P-TEFb also phosphorylates and activates SPT5, which itself is an elongation factor. P-TEFb also recruits another elongation factor TAT-SF1.
• SPT5: is comparable to bacterial elongation facotr NusG. SPT5/NusG factors bind to their respective RNA Polymerases at the tip of the clamp, overlapping with the region contacted by σ region 4 (in bacterial) and TFIIB (in eukaryotes). This overlapping and presumably mutually exclusive binding raises the possibility that displacing the initiation factors may be part of the function of these elongation regulation.
• ELL protein: binds to elongating polymerase and suppresses transient pausing by the enzyme. ELL was originally identified as the product of a gene that undergoes translocations in acute myeloid leukemia.
• TFIIS: Stimulates the overall rate of elongation by limiting the length of time that polymerase pauses when it encounters sequences that would otherwise tend to slow the enzyme’s progress. (similar to ELL). TFIIS also contributes to proofreading by polymerase. TFIIS stimulates an inherent RNase activity in polymerase, allowing an alternative approach to removing misincorporated bases through local limited RNA degradation. This is comparable to the hydrolytic editing in the bacterial case stimulated by GreB factors.

Question 42

Elongating RNA Polymerase dealing with histones

Answer 42

Elongating RNA Polymerase must deal with histones in its path
Chromatin impedes transcription, which leads to the identification of factors that facilitate transcription.
FACT (facilitate chromatin transcription): is a heterodimer of two well-conserved proteins, Spt16 and SSRP1.
Spt16 binds to H2A•H2B dimer, and SSRP1 binds to H3•H4 tetramer. FACT can both dismantle histones, by removing one H2A•H2B dimer (ahead of an transcribing RNA Polymerase), and reassemble them by restoring that dimer (after passing the nucleosome).

Question 43

Elongation factors and RNA processing

Answer 43

Elongating Polymerase is associated with a new set of protein factors required for various types of RNA processing
Once transcribed, eukaryotic RNA has to be processed in various ways before being exported to the cytoplasm for translation.
The RNA processing events include capping of the 5’ end of the RNA, splicing, and polyadenylation of the 3’ end of the RNA.
Elongation, termination of transcription and RNA processing are interconnected.
Elongation factor SPT5 also helps to recruit the 5’ capping enzyme to the CTD tail of the Polymerase (phosphorylated at serine position 5). The hSPT5 stimulates the 5’-capping enzyme activity. After capping, dephosphorylation of the serine-5 within the tail may be responsible for the dissociation of the capping machinery. And further phosphorylation at serine-2 causes recruitment of the machinery for RNA splicing.
The elongation factor TAT-SF1 recruits components of the splicing machinery to polymerase with a serine-2 phosphorylated tail.

Question 44

5’ capping mechanism

Answer 44

The structure and formation of the 5’ RNA cap
5’ capping: unusual 5’- 5’ linkage between nucleotides
3 steps:
1. The γ-phosphate at the 5’ end of the RNA is removed by an enzyme called RNA triphosphatase;
2. The GMP moiety is added using guanylyltransferase. An enzyme-GMP complex is generated from GTP first with release of the β- and γ-phosphates of that GTP, and then the GMP from the enzyme is transferred to the β-phosphate of the 5’-end of the RNA;
3. The newly added guanine and the purine at the original 5’ end of the mRNA are further modified by the addition of methyl group by methyltransferase. The resulting 5’ cap structure subsequently recruits the ribosome to the mRNA for translation to begin

Question 45

Polyadenylation and termination

Answer 45

Polyadenylation and termination
The final RNA processing event, polyadenylation of the 3’ end of the mRNA, is intimately linked with the termination of transcription.
The Polymerase CTD tail is also involved in recruting some of the enzymes necessary for polyadenylation.
Once polymerase has reached the end of a gene, it encounters specific sequences that, after transcribed into RNA, trigger the transfer of the polyadenylation enzymes to the RNA, leading to four events:
1.cleavage of the message;
2.addition of many adenine residues to its 3’ end;
3.degradation of the RNA remaining associated with RNA polymerase by a 5’ to 3’ ribonuclease;
4.and subsequently, termination of transcription.

Question 46

Polyadenylation and termination mechanism

Answer 46

Polyadenylation and termination
The sequence that, once transcribed into RNA, trigger transfer of the factors to bind the RNA for polyadenylation are called Pol-A signals.
Two protein complexes are carried by the CTD of the Polymerase as it approaches the end of the gene: CPSF (cleavage and polyadenylation specific factor) and CSTF (cleavage stimulation factor). Once these two complexes are bound to the RNA, other proteins are recruited as well, leading to initially to RNA cleavage and then polyadenylation.
Polyadenylation is mediated by an enzyme called poly-A polymerase, which adds approximately 200 adenines to the RNA 3’ end produced by the cleavage. The enzyme uses ATP as precursor and adds the nucleotides using the same chemistry as RNA polymerase. But it does so without a template.
RNA Polymerase doesn’t terminate immediately after the RNA is cleaved and polyadenylated. Rather, it continues to move along the template, generating a second RNA molecule.

Question 47

RNA Pol I and RNA Pol III promoters that are recognized and the basic mechanism

Answer 47

RNA Pol I and Pol III recognize distinct promoters but still require TBP
RNA Pol I and III enzymes are related to Pol II and even share several subunits, but they initiate transcription from distinct promoters and transcribes distinct genes. These genes encodes specialized RNAs rather than proteins.
Each enzyme works with its own unique set of transcription factors. But TBP is universal.
Some of the other GTFs discussed in the Pol II case do have structurally and functionally equivalent components in the other systems. TFIIE-like subunits are found in Pol I and Pol III enzymes.

Question 48

Pol I and its promoter

Answer 48

Pol I transcribes just the rRNA genes
rRNA is expressed at far higher levels than any other genes, perhaps explaining why it has its own dedicated polymerase.
Pol I promoter: the core element (around the start site) and the UCE (upstream control element) element (100-150bp upstream).
Two factors: SL1 and UBP. SL1 comprises TBP and three TAFs specific for Pol I transcription. It binds to the core element. SL1 binds DNA only in the presence of UBF, which binds to UCE.

Question 49

Pol III and its promoter

Answer 49

Pol III promoters are found downstream form the transcription start site
Pol III promoters come in various forms, with the unusual feature of being located downstream from the transcription start site. Some consists of two regions, called BoxA and BoxB (tRNA genes), separated by a short element; Others contain BoxA and BoxC (5s rRNA gene); and still others contain a TATA element like those of Pol II.
TFIIIB and TFIIIC are required for tRNA gene transcription, and those plus TFIIIA for the 5srRNA gene.
In tRNA promoter, TFIIIC binds to the upstream of the start site, which recruits TFIIIB to the DNA upstream of the start site. This in turn, recruits Pol III to the start site for transcription.

Question 50

RNA splicing summary

Answer 50

RNA Splicing
• Coding sequence: a series of three-nucleiotide codons that specifies the linear sequence of amino acids in a polypeptide product (contiguous in bacteria and phage genes, but mosaic in eukaryotic genes
• Exon: any region of the primary transcripts that retains in a mature RNA (coding and non-coding exons: non- coding RNAs, microRNAs). Exon does not equal to coding sequence
• Intron: any region of the primary transcripts that is not in a mature RNA (noncoding introns)