Exam 3 Study Guide

Question 1

Describe the three primary mechanisms used by bacterial RNA polymerase to find a specific promoter?

Answer 1

1. Random diffusion and nonspecific binding.
2. Sliding
3. Hopping

Both Sliding and Hopping have nonspecific binding and then movement to the promoter.

Question 2

What triggers the transition of a closed bacterial RNA polymerase holoenzyme to an open complex?

What structural changes accompany this transition?

Answer 2

When a sigma factor interacts strongly with the promoter, the holoenzyme transitions to an unstable open complex.
—> DNA duplex is bent 90 degrees to enter the active site.
—> Promoter is denatured from -11 to +3.
—> Transcription bubble increases to 22-24 nucleotides in length.
—> Jaws close around downstream sequence.

Opening is irreversible!

Question 3

How are the properties of the ternary bacterial RNA holoenzyme polymerase complex that undergoes rounds of abortive initiation related to the transition to elongation?

Answer 3

The energy of successive abortive initiation events is used to break RNA polymerase free from the promoter and transition to elongation.

The ternary complex is formed by several rNTPs. As these rNTPs are added to the transcript, the RNA polymerase remains tightly bound to the promoter and cannot move, this causes scrunching. Stress increases and causes the releases of short transcripts, 15-20 nucleotides.

Question 4

Describe how specific and nonspecific interactions between the sigma factor and the DNA duplex in the promoter region facilitate initial binding of the closed holoenzyme and the subsequence transition to an open complex?

Answer 4

The holoenzyme slides into a promoter, the DNA duplex closes.
Though once the sigma factor interacts strongly with the promoter the holoenzyme transitions to an unstable open complex.

The sigma factor binds to the core polymerase.

N-terminal domain of the sigma factor blocks the DNA-binding domain of the holoenzyme until an open complex is formed.

Question 5

What is the TBP?
Which positioning factor contains TBP for each of the eukaryotic RNA polymerases?

Answer 5

TBP is TATA-binding protein.
RNA polymerase 1: SL1 binds in conjunction with UBF
RNA polymerase 2: TF2D is responsible for binding
RNA polymerase 3: TF3B binds next to TF3C.

Question 6

How does TBP affect DNA structure upon binding the promoter and why is this an important function of TBD required by all RNA polymerases?

Answer 6

TBP binds to the minor groove of DNA, this causes the DNA to bend by 80 degrees. This causes transcription factors bound upstream to be in close proximity with RNA polymerases bound downstream.

Question 7

What are the general difference between transcription initiation complex assembly at TATA-containing and TATA-less RNA polymerase II promoters.

Answer 7

The same transcription factors are required, TF2D binds to the INR.
Some TATA-less promoters lack unique transcription start sites.

Question 8

What is the CTD of RNA polymerase II and how is modification of its structure related to the transition to elongation and subsequent RNA processing?

Answer 8

C-terminal Domain.

Phosphorylation of CTD tail of RNA polymerase II is required for promoter and transcription factor release.
CTD is also involved in mRNA processing, phosphorylated CTD serves as a recognition site for capping, tailing, and splicing enzymes.

Question 9

How do activators, repressors, coactivators, corepressors, and the mediator protein interact at enhancer regions to influence assembly of the general transcription factors and RNA polymerase II to the gene control region?

Answer 9

Activators and repressors have effects on transcription, either activation or repression.
Coactivators and corepressors interact with repressors and activators but do not directly bind DNA.
Mediator: Allows thee transcriptional regulators, factors, and RNA polymerase II to assemble at the promoter. Composed of 30+ subunits.

Question 10

How do elongation factors work to facilitate continued transcription by RNA polymerase II?

Answer 10

RNA polymerase II requires regulators to transition to elongation.
1. Recruit chromatin remodeling complexes to release chromatin that is blocking the RNA polymerase II movement.
2. Interacts with RNA polymerase II via a coactivator to unpause enzyme.
3. Act as or recruit elongation factors.

Elongation factors decrease the likelihood that RNA polymerase will dissociate from the DNA during elongation.

Question 11

Describe the differences between how intron definition and exon definition are used to define the 5’ and
3’ splice sites for a splicing event. What components are associated with each method of definition?
Why would a species/cell use intron definition versus exon definition?

Answer 11

Intron definition: 5’ and 3’ are simultaneously recognized by components of E complex. Uses U1 and then U2AF as nascent mRNA emerges from RNA polymerase II.

Exon definition: Takes advantage of the presence of small exons of a consistent size. Introns are long and variable. Many sequences in introns resemble true splice sites. The paired recognition of splice sites flanking an intron is quite inefficient.

Question 12

How do U2, U4, and U6 interact within the spliceosome and how do these components interact with
each other to control the triggering and specificity of the catalytic reactions?

Answer 12

U2: snRNP binds to the branch point. Displaces BBP/SF1 and U2AF. Requires ATP hydrolysis. Forms the A complex.
U4: Release along with U1 results in forming the B2 complex.
U6: Once U4 releases, U6 pairs with U2. U4 is hiding U6 from U2 until U4 is released.

Question 13

Describe the two models of RNA polymerase II termination and how they are related to tail formation.

Answer 13

Allosteric: Binding of cleavage factors causes a conformational change that lessens the processivity, leading to a higher chance of dissociation from DNA.
Exonuclease Torpedo: RNA cleavage produces and uncapped 5’ RNA end which is susceptible to degradation by nucleases. The exonuclease eventually reaches the RNA polymerase II and destroys the RNA-DNA hybrid, causing RNA polymerase II to dissociate.

Question 14

Describe the general differences between the two pathways of poly(A) removal-dependent degradation.

Answer 14

5’->3’ decay pathway: digestion of the poly A tail down to 10-12 nucleotides. LSM1-7 decaying enhancer binds to short poly A tail. LSM1-7 activates the decaying reaction on the 5’ end. Removal of the cap produces a 5’ monophosphorylated RNA. 5’->3’ Xrn1 exonuclease rapidly degrades the mRNA.
The cap is usually resistant to decapping while being translating.

3’->5’ decay pathway: Digestion of the poly A tail down to 10-12 nucleotides triggers exosome action. The exosome is a multi protein complex that contains a 3’->5’ exonuclease. The exosome degrades the mRNA from the 3’ end.

Question 15

Describe the biochemical steps of tRNA charging by aminoacyl-tRNA synthetase.

Answer 15

Aminoacyl-tRNA syntheses are the family of enzymes that load tRNAs with amino acids.
1. An amino acid reacts with ATP to form an aminoacyl adenylate intermediate.
2. The 2’-OH or 3’-OH of the terminal 3’ nucleotide in the tRNA attacks the carbonyl carbon of the adenylate.
3. An aminoacyl-tRNA and AMP is formed.

Question 16

How do aminoacyl-tRNA synthetases differentiate between different amino acids? How do they
differentiate between different tRNAs?

Answer 16

Each tRNA synthetase should be specific for the amino acid. All tRNAs share the same general structure but differ at nucleotide positions of the four arms. They use both direct (nucleotide differences) and indirect (phosphodiester) methods.
There are discriminators in shape of different amino acids.

Question 17

What are the differences between kinetic and chemical proofreading by aminoacyl-tRNA synthetases?

Answer 17

Kinetic Proofreading: tRNAs that match the sequence for the synthetase properly align their amino acid acceptor stem with the ATP amino acid, quickly triggers aminoacylation reaction. Misalignment will not trigger aminoacylation reaction, dissociates much faster than it can react.
Chemical Proofreading: 9 different tRNA syntheses are able to proofread and correct errors once incorrect amino acid has bound to enzyme. 1 error every 40,000 reactions.

Question 18

How is chemical proofreading related to pre-transfer and post-transfer editing?

Answer 18

Pre-transfer editing and post-transfer editing are the two forms of chemical proofreading.
Pre-transfer editing: Incorrect aminoacyl-AMP is hydrolyzed after tRNA binding, but before charging has occurred.
Post-transfer editing: Amino acid is hydrolyzed from aminoacyl-tRNA after tRNA charging. Uses an editing active site in the synthetase enzyme that is separate from the synthetic/loading active site.

Question 19

Describe how the integrated double-sieve mechanism is used by some aminoacyl-tRNA synthetases to
prevent the wrong amino acid from participating in the tRNA charging reaction.

Answer 19

The double sieve is based on relative sizes of the synthetic and editing sites. The first sieve is the synthetic site since it is larger than the editing site. Amino acids larger than correct amino acids will be excluded from the synthetic site, loading will not occur.
The second sieve is the editing site, amino acids smaller than the correct amino acid will fit. The incorrect amino acid will then be hydrolyzed and removed in the editing site.

Question 20

Describe how IF-1, IF-2, and IF-3 work together with the components of the translational machinery to
initiate transcription in bacteria. What role does each component play in the process?

Answer 20

IF-3 helps the 30S subunit bind to the initiation sites on the mRNA. IF-2 aids binding of the initiator tRNA to the complex. IF-1 binds to the 30S subunits at the A site and prevents aminoacyl-tRNAs from binding prematurely.
All initiation factors are then released and the 50S subunit joins to form the full ribosomal structure.

Question 21

Describe the scanning model of initiation of translation in eukaryotes

Answer 21

Small subunit binds to the 5’ cap and begins to move 5’->3’ down the mRNA. As it moves, the small subunit can melt some secondary structures of the mRNA. The small subunit stops when it recognizes the start codon and flanking sequences at -4 and +1 (Kozak sequence).

Question 22

What are the steps of elongation in prokaryotic translation? How are EF-Tu, EF-Ts, and GTP involved
in elongation?

Answer 22

1. Entry of an aminoacyl-tRNA into the A site is mediated by EF-Tu. EF-Tu-GTP binds an aminoacyl-tRNA, escorts it to the ribosome.
2. The anticodon end of the ternary complex moves into the A site of the 30S subunit and base pairs with the codon.
3. The EF-Tu-GTP end of the ternary complex binds to the factor binding center of the large subunit.
4. The factor binding center simulates EF-Tu-GTP hydrolysis.
5. The aminoacyl end of the aminoacyl-tRNA moved to face the P site tRNA.
6. EF-Tu-GDP is released.
   EF-Ts mediates the regeneration of EF-Tu-GTP
   –EF-Ts binds EF-Tu-GDP and displaces GDP
   –GTP binds to the EF-Ts-TU complex, causing it to dissociate into EF-Tu-GTP and EF-Ts.

Question 23

Describe the proofreading function of EF-Tu. How is this related to correct binding of the aminoacyl-
tRNA into the A site, interaction between EF-Tu and the factor binding center, and the peptidyl
transferase reaction? How is GTP involved in proofreading?

Answer 23

1. When there is a correct base-pairing between the first two positions of the codon and anticodon, 16S rRNA forms nonspecific hydrogen bonds with the minor groove of the A site tRNA. Correctly base paired tRNAs will dissociate slowly but the incorrectly base paired tRNAs will dissociate quickly.
2. EF-Tu-GTP only interacts with the factor binding center if the aminoacyl-tRNA fully enters the A site, an incorrect aminoacyl-tRNA will dissociate before EF-Tu-GTP interacts with the factorbinding center and promotes GDP hydrolysis (which commits the amino acid to incorporate into the growing polypeptide).
3. The aminoacyl-tRNA initially binds to the A site with the amino acid oriented away from the P site. When EF-Tu-GTP is hydrolyzed to EF-Tu-GDP the A site tRNA rotates towards the P site. Correct base pairs handle this rotational strain, incorrect base pairs will break and dissociate.

Question 24

How is the 23S rRNA involved in the peptidyl transferase process?

Answer 24

23S rRNA positions aminoacyl end of aminoacyl-tRNA near the end of the peptidyl-tRNA, assisted by a kink between the P and A sites. This kink maximizes distance between tRNAS at mRNA end, while minimizes distance between tRNAs at peptidyl-aminoacyl end.

Question 25

What is the “hybrid state” and how is it and other changes in ribosomal configuration related to the
translocation process?

Answer 25

Upon transfer of the polypeptide to the A site, the tRNAs move into a hybrid state, this is when the anticodons of the tRNA remain in their pre-peptidyl transfer positions. The 3’ end of the A site tRNA is bound to the polypeptide and prefers to bind in the P site, the 3’ end of the P site tRNA has now been deacylated and prefers to bind in the E site.

Question 26

Describe the similarities and differences between the peptidyl transferase process with an aminoacyl-
tRNA in the A site versus RF1/RF2 in the A site.

Answer 26

Termination codons are recognized by class 1 protein release factors; RF1 recognizes UAG and UAA, RF2 recognizes UGA and UAA.
*****

Question 27

What are the differences between the lacOC, lacI-
, lacI S, and lac -d mutations?

Answer 27

lacOC: Constitutive mutation, always on. Destroys repressor binding site. Cis-acting
lacI-: Constitutive trans acting. Synthesizes defective repressors that won’t bind to operator.
lacI S: Uninducible trans acting. Cannot be turned off.
lac -d: Dominant negative. Blocks repressor action.

Question 28

Describe how the lac repressor binds to the lac operator structure. What effect does repressor binding
have on the DNA in the region surrounding the repressor binding site and on RNA polymerase
holoenzyme binding?

Answer 28

The repressor binds to an inverted repeat in the operator, in the major groove of DNA.
There are 3 operators, 2 of which need to be bound to induce repression. A loop in the DNA forms between the two binding sites and blocks the RNA polymerase.

Question 29

How are the differential affinities of the active versus inactive lac repressor related to its location in the
cell and ability to bind the operator?

Answer 29

Active repressor has low affinity for random DNA, all DNA sequences in the E. coli genome can serve as low affinity sites. Repressors are almost always bound. Though operators have high affinity so an active repressor is usually bound to the operator, 96% of the time.