Exam 3 study guide

Question 1

1. Describe the three primary mechanisms used by bacterial RNA polymerase to find a specific promoter. (Ch 17)

Answer 1

Slide: How Does RNAP Find Promoters?

Option 1
* Random diffusion and nonspecific binding to short sequences
* Rapid dissociation of enzyme and repositioning
* Very, very inefficient mechanism

Option 2
* Nonspecific binding to genome and then movement along genome to specific promoter(s)
* Sliding
* Intersegment transfer
* Intradomain association and dissociation (hopping)

Question 2

2. What triggers the transition of a closed bacterial RNA polymerase holoenzyme complex to an open complex? What structural changes accompany this transition? (Ch 17)

Answer 2

Slide: Holoenzyme Structure - Initiation

1. When the holoenzyme slides into a promoter it transitions to a closed binary complex
   a. Closed=DNA duplex
   b. Covers approximately 75 bp of DNA
   From -55 to +20
   c. DNA begins to interact with RNAP subunits other than sigma
   d. Reversible
2. If the sigma factor interacts strongly with the promoter, the holoenzyme transitions to an unstable open complex
   a. DNA duplex is bent 90° in order to place it into the active site
   b. Promoter is denatured from -11 to +3 with assistance from sigma
   c. Transcription bubble increases to 22-24 nt in length
   d. Jaws close around downstream sequence
   e. Transition to open state is irreversible

Question 3

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? (Ch 17)

Answer 3

Several rNTPs are incorporated into the transcript, forming the ternary complex
• Additional rNTPs are added to the transcript while the RNAP remains tightly bound to the promoter
o RNAP is not able to move down the template
• RNAP pulls upstream DNA into the active site via a “scrunching” mechanism
• Scrunching creates considerable stress that results in release of these short transcripts
o Abortive initiation transcripts of 15-20 nucleotides
• Energy of successive abortive initiation events is used to eventually break RNAP free from the promoter and transition to elongation

The RNAP transitions into a ternary elongation complex
• Sigma factor is released
• Bubble returns to 10-12 nt in length
• RNAP coverts into the core enzyme
o Core enzyme only covers 30-40 bp of template DNA

Question 4

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 subsequent transition to an open complex. (Ch 17)

Answer 4

Slide: Holoenzyme Structure - Initiation

1. When the holoenzyme slides into a promoter it transitions to a closed binary complex
   a. Closed=DNA duplex
   b. Covers approximately 75 bp of DNA
   From -55 to +20
   c. DNA begins to interact with RNAP subunits other than sigma
   d. Reversible
2. If the sigma factor interacts strongly with the promoter, the holoenzyme transitions to an unstable open complex
   a. DNA duplex is bent 90° in order to place it into the active site
   b. Promoter is denatured from -11 to +3 with assistance from sigma
   c. Transcription bubble increases to 22-24 nt in length
   d. Jaws close around downstream sequence
   e. Transition to open state is irreversible

Sigma Factor N-terminal domain of sigma is autoinhibitory
* Normally masks the DNA binding domain of sigma
* Prevents sigma from nonspecifically binding to and blocking promoters
* Swings out of the way once sigma binds to the core RNAP
o N-terminal domain also blocks the DNA-binding domain of the holoenzyme until an open complex is formed

—A nonspecific interaction occurs between sigma2 and only the phosphodiester backbone in the closed binary RNAP complex

—A specific interaction between sigma2 and base pairs of the -10 and discriminator facilitates the melting that leads to the irreversible transition to the open RNAP complex
—Melting begins with the flipping of two specific bases by sigma2

Question 5

1. What is the TBP? Which positioning factor contains TBP for each of the eukaryotic RNA polymerases? (Ch 18)

Answer 5

Slide: TBP

• Each class of eukaryotic RNAP is assisted by a positioning factor that contains TBP and other components
• TATA-binding protein was originally identified as a protein that binds to the TATA box in RNAP II promoters
o Subsequently determined to also be a part of positioning factors for RNAP I and RNAP III
o Does not actually recognize the TATA box for RNAP I and RNAP III promoters
o Some RNAP II promoters lack TATA boxes but still require TBP
• TFIID
• Positioning factor required by RNAP II
• Also contains 14 subunits called TAFs
o TBP associated factors
• Multiple TFIID variants contain different combinations of TAFs
• Different TFIID variants are tissue-specific
• The positioning factor recognizes the promoter in different ways for different RNAPs
• RNAP III
o TFIIIB binds next to TFIIIC
• RNAP I
o SL1 binds in conjunction with UBF
• RNAP II
o TFIID is solely responsible for binding
• TBP binds to the minor groove of DNA
o External surface of TBP remains available for TAF interactions
• Nucleosomes also bind in the minor groove
o Nucleosome and TBP binding are mutually exclusive
• Upon binding, TBP bends the DNA by 80°
o Brings transcription factors bound upstream in close proximity with RNAP bound downstream

Question 6

2. How does TBP affect DNA structure upon binding the promoter and why is this an important function of TBD required by all RNA polymerases? (Ch 18)

Answer 6

• TBP binds to the minor groove of DNA
o External surface of TBP remains available for TAF interactions
• Nucleosomes also bind in the minor groove
o Nucleosome and TBP binding are mutually exclusive
• Upon binding, TBP bends the DNA by 80°
o Brings transcription factors bound upstream in close proximity with RNAP bound downstream

Question 7

3. What are the general differences between transcription initiation complex assembly at TATA-containing and TATA-less RNA polymerase II promoters? (Ch 18)

Answer 7

Slide: Transcription Initiation Complex

1. TBP subunit of TFIID directs transcription factor to TATA box
   —May be enhanced by upstream elements acting through a coactivator
   —Also recognizes the INR element and the DPE
2. TFIIB binds
   —Is recruited to the promoter along with RNAP II TFIIA also binds for some promoters
3. TFIIF binds
   —Is recruited to the promoter along with RNAP II
   —Large subunit contains DNA helicase activity
   —Small subunit has some homology to bacterial sigma factor regions that bind core polymerase
4. RNAPII is recruited to the promoter
   —TFIIB binds near RNA exit site and may influence switch from abortive initiation to promoter escape
   —TFIIB also inserts into the active site of RNAP II and assists TFIID with stabilization of promoter melting
5. TFIIH binds
   —10 subunits, almost as large as RNAP II
   —Kinase that phosphorylates the CTD of RNAP II
   —Interacts with RNAP II downstream of start site
   —Involved in promoter escape
   —Involved in nucleotide excision repair pathways
6. TFIIE binds
   —Extends region covered by the apparatus to +30
   Have now formed the transcription initiation complex

Slide: Transcription Initiation Complex- No TATA box

o Same transcription factors required as with TATA box- containing promoters
o INR supplies the positioning element instead of TATA box
o TFIID binds to the INR via interactions with TAFs
o Other TAFs also recognize the DPR element downstream of start
o Some TATA-less promoters lack unique transcription start sites
o Initiation occurs somewhere within a cluster of possible start sites
o Transcription factors play a role similar to that of the bacterial sigma factors
o Allow the basic polymerase to recognize promoters
o Have evolved more independence and far more variety than sigma factors

Question 8

4. 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? (Ch 18)

Answer 8

Slide: Transition to Elongation

o Phosphorylation of CTD tail of RNAP II is required for promoter and transcription factor release
o CTD tail contains 52 tandem repeats of a 7 amino acid sequence
o Phosphorylation is facilitated by a kinase complex that includes TFIIH and Cdk9
o TFIIH phosphorylates serines in the fifth position of each repeat
o Cdk9 is also involved in cell cycle control
o RNAP II changes conformation
o Disengages from general transcription factors
o Tightens interactions with DNA
o Acquires new proteins that increase RNAP II processivity
o CTD is also involved in mRNA processing
o Phosphorylated CTD serves as a recognition site for capping, tailing, and splicing enzymes
o Once initiated, the RNAP II can pause for long periods of time before moving to elongation

Question 9

5. How do activators, repressors, coactivators, co-repressors, 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? (Ch 18)

Answer 9

Slide: Enhancers

o Transcriptional regulators bind to enhancer regions to influence the assembly of the general transcription factors and RNAP II to the gene control region
o The whole expanse of DNA involved in initiating and regulating transcription
o Enhancers are cis-regulatory sequences located a variable distance from core promoter
o Upstream, downstream, and within an intron
o Can be located over a 100,000 bp region
o Regulators that bind enhancers can be classified by their potential effect on transcription
o Activators versus repressors
o Other regulators called coactivators and co-repressors also interact with activators and repressors
o But do not usually directly bind DNA
o At any time, a cell contains a mix of regulators of different strengths
o Interactions between regulators are too weak to assemble in solution
o Cis-regulatory regions can “crystallize” the regulators at the gene control region
o May involve the formation of a biomolecular condensate
o Overall effect of an enhancer on transcription is determined by the specific combination of bound factors
o Because of the contextual nature of classifying activators based on their effect on transcription, we can more accurately classify them based on their function
o True activators that bind specific DNA elements and the basal machinery at the promoter
 Usually via coactivators
o Chromatin remodeling activators that recruit chromatin modification
 enzymes and remodeling complexes
o Architectural modifying activators that bend DNA in order to bring factors bound apart on linear duplex into close proximity
o Can also classify repressors into similar groups based on their function

Slide: Mediator
o A large protein complex that allows the transcriptional regulators, general transcription factors, and RNAP II to assemble at the promoter
o Correctly positions TFIIH near the tail of RNAP II, which facilitates CTD phosphorylation

Question 10

6. How do elongation factors work to facilitate continued transcription by RNA polymerase II? (Ch 18)

Answer 10

Slide: Elongation Factors

o RNAP II often pauses after the completion of initiation and requires additional regulators to transition to elongation
o Common in humans, where a significant fraction of genes have a paused polymerase approximately 50 bp downstream of the start site
o Poised gene
o These new regulators act in three ways to facilitate elongation
o Recruit chromatin remodeling complexes to release chromatin that is blocking RNAP II movement
o Interacts with RNAP II via a coactivator to unpause enzyme
o Act as or recruit elongation factors
o Elongation factors decrease the likelihood that RNAP will dissociate from the DNA during elongation
o Major function is to help RNAP move through nucleosomes
o Pry the DNA away from the histone core
o Reduce innate “stickiness” of RNAP for nucleosomes

Question 11

1. 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? (Ch 19)

Answer 11

Slide: Splice Site Recognition

o Intron definition
o 5’ and 3’ splice sites are simultaneously recognized by components of E complex
 Sequential deposit of U1 and then U2AF as nascent mRNA emerges from RNAP II
o Used for splicing of small, single-intron genes in unicellular eukaryotes

o Exon definition
o Takes advantage of presence of small exons of a consistent size
 Introns are long and variable in multicellular eukaryotes
 Many sequences in introns resemble true splice sites
 The paired recognition of splice sites flanking an intron is generally quite inefficient
o U2AF binds to the 3’ splice site
o U1 binds to the 5’ splice site at the beginning of the next intron
 Bridges the exon
o Sequential deposit of U2AF and then U1 as nascent mRNA emerges from RNAP II
o Complexes are switched to link across the introns

Question 12

2. 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? (Ch 19)

Answer 12

Slide: Spliceosome Assembly

4. Upon release of U4, U6 pairs more extensively with U2
   o U4 sequesters the U6 snRNA until it is needed
5. This step creates the entirety of the active site
   o U2 is already paired with the branch point
   o U6 now pairs with intronic sequence downstream of the 5’ splice site
   o U2-U6 pairing brings 5’ splice site in close contact with branch point
   o Assisted by interactions between U5 and upstream exon

o ATP hydrolysis reactions are used to break specific RNA- RNA base pairs
o Breaking of specific base pairs is required to make others that are specifically required for the sequential assembly of the spliceosome
o If the initial correct base pairs do not form, then ATP hydrolysis will not occur, and spliceosome assembly will not proceed
o Examples
 Specific U6 pairing with U4 is broken by ATP hydrolysis and replaced by specific pairing with U2

Question 13

3. Describe the two models of RNA polymerase II termination and how they are related to tail formation. (Ch 19)

Answer 13

Slide: RNAP II Termination

o RNAP II continues transcription for hundreds of nucleotides after RNA is cleaved
o Two factors lead to RNAP II termination
o 1. Allosteric changes
 Binding of cleavage factors and subsequent RNA cleavage leads to a conformational change in RNAP II
 Conformational change makes the enzyme less processive and more likely to dissociate from the DNA
o 2. Exonuclease torpedo
 RNA cleavage produces an uncapped 5’ RNA end which is eventually bound by a 5’ –> 3’ exonuclease
* Exonuclease is carried on RNAP II?
 The exonuclease degrades the RNA 5’ –> 3’
 When the exonuclease reaches RNAP II it destroy the RNA-DNA hybrid
 RNAP II dissociates
o RNAP I and RNAP III both terminate at specific terminator sites
o RNAP III looks for a discrete poly-T sequence in the template strand
o RNAP I requires
o Accessory terminator proteins that recognize one of two terminator sequences
o Cleavage of the nascent RNA

Question 14

1. Describe the general differences between the two pathways of poly(A) removal-dependent degradation. (Ch 20)

Answer 14

Slide: Poly(A) Removal-Dependent Degradation

1) 5’ –> 3’ decay pathway
a) Digestion of the poly(A) tail down to 10-12 nt
b) Lsm1-7 decapping enhancer binds to short poly(A) tail
c) Lsm1-7 activates the decapping reaction on the 5’ end
d) Removal of the cap produces a 5’ monophosphorylated RNA
e) 5’ –> 3’ Xrn1 exonuclease rapidly degrades the mRNA

• The 5’ cap is usually resistant to decapping while it is being translated because it is bound to a cytoplasmic cap-binding protein
  o Cap-binding protein is also a component of the eukaryotic translation initiation eIF4F complex
• Translation and decapping machineries compete for the cap
  o Initiation of translation involves an interaction between the PABPs and the eIF4F initiation complex at the 5’ end
  o Removal of the tail and release of PABPs destabilizes the eIF4F-cap interaction
  o Cap is more exposed to decapping enzymes

1) 3’ –> 5’ decay pathway
a) Digestion of the poly(A) tail down to 10-12 nt triggers exosome action
b) The exosome is a multiprotein complex that contains a 3’5’ exonuclease
c) The exosome degrades the mRNA from the 3’ end

Question 15

1. Describe the biochemical steps of tRNA charging by aminoacyl-tRNA synthetase. (Ch 22/23)

Answer 15

Slide: Aminoacyl-tRNA Synthetase

o Aminoacyl-tRNA synthetases are the family of enzymes that load tRNAs with the correct amino acid
1. An amino acid reacts with ATP to form an aminoacyl adenylate intermediate
o Energy of hydrolysis is trapped in the mixed anhydride linkage of the adenylate
o Pyrophosphate is released
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

2. How do aminoacyl-tRNA synthetases differentiate between different amino acids? How do they differentiate between different tRNAs? (Ch 22/23)

Answer 16

Slide: Aminoacyl-tRNA Synthetase

o Each tRNA synthetase should be selective for a specific group of tRNAs and a specific amino acid
1.Detecting tRNA differences
o All RNAs share the same general tertiary structure, but differ at nucleotide positions of the four arms
 Changes in the nucleotide sequences
 Subtle differences between the shape of the L-shaped arms
o tRNA synthetases discriminate between tRNAs using both direct (nucleotide differences) and indirect (phosphodiester) methods
 Most common discriminators are in the anticodon loop and amino acid acceptor arm
2. Detecting amino acid differences
o Primary discriminator is shape of different amino acids
o But amino acids are very small, and some are very similar in structure
o Those that are similar in structure have different binding efficiencies and free energies
o The two detection methods work together to produce the optimal induced fit between enzyme, tRNA, and amino acid

Question 17

3. What are the differences between kinetic and chemical proofreading by aminoacyl-tRNA synthetases? (Ch 22/23)

Answer 17

Slide: Aminoacyl-tRNA Synthetase Proofreading

*Kinetic proofreading
o tRNAs that match the specific nucleotide sequence combination for the synthetase
 Properly align their amino acid acceptor stem with the ATP and amino acid in the active site
 Quickly trigger aminoacylation reaction
o Incorrect tRNAs
 Misalignment of acceptor stem in active site
 Will not quickly trigger aminoacylation reaction
 Dissociates much faster than it can react

*Chemical proofreading
o Some tRNA synthetases have great difficulty in distinguishing between amino acids with a similar structure
 Isoleucyl-tRNA synthetase cannot effectively distinguish isoleucine from valine using shape of amino acid binding site
 Unable to prevent significant levels of valine-tRNAIle synthesis without proofreading
o Nine different tRNA synthetases are able to proofread and correct errors once incorrect amino acid has bound to enzyme
o Analogous to the 3’ –> 5’exonuclease proofreading function of DNA polymerases

*Overall accuracy is 1 error for every 40,000 charging reactions

Question 18

4. How is chemical proofreading related to pre-transfer and post-transfer editing? (Ch 22/23)

Answer 18

Slide: Chemical Proofreading

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

Slide: Post-Transfer Editing

o The post-transfer editing pathway can be thought of as an integrated double-sieve
o Based on relative sizes of the synthetic and editing sites
o The synthetic site is larger than the editing site
o The first sieve is the synthetic site
o Amino acids larger than correct amino acid will be excluded from the synthetic site
o Loading will not occur
o The second sieve is the editing site
o Amino acids smaller than the correct amino acid will fit into the synthetic site and the editing site
o The incorrect amino acid will then be hydrolyzed and removed in the editing site
o The correct amino acid can fit into the synthetic site, but not the editing site
o Will be correctly charged and retained in an aminoacyl-tRNA
o The synthetic and editing sites are located a considerable distance across the enzyme from each other
o The amino acid is covalently bonded to the tRNA in the aminoacyl-tRNA
o The aminoacyl-tRNA is able to form an extended structure that can “reach” the editing site
o Before charging, the tRNA forms a hairpin structure that is unable to reach the editing site
o Pre-transfer editing cannot use the editing site

Question 19

5. 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. (Ch 22/23)

Answer 19

Slide: Post-Transfer Editing

The post-transfer editing pathway can be thought of as an integrated double-sieve
Based on relative sizes of the synthetic and editing sites
The synthetic site is larger than the editing site

The first sieve is the synthetic site
Amino acids larger than correct amino acid 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 into the synthetic site and the
editing site
The incorrect amino acid will then be hydrolyzed and removed in the editing site

The correct amino acid can fit into the synthetic site, but not the editing site
Will be correctly charged and retained in an aminoacyl-tRNA
The synthetic and editing sites are located a considerable distance across the enzyme from each other
The amino acid is covalently bonded to the tRNA in the aminoacyl-tRNA
The aminoacyl-tRNA is able to form an extended structure that can “reach” the editing site
Before charging, the tRNA forms a hairpin structure that is unable to reach the editing site
Pre-transfer editing cannot use the editing site

Question 20

6. 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? (Ch 22/23)

Answer 20

Slide: Bacterial Translation - Initiation

Initiation occurs at the start codon and Shine-Dalgarno sequence on mRNA
 Polypurine hexamer approximately 10 nt upstream of start codon
 5’-AGGAGG-3’
 Complementary to a portion of the 16S rRNA

30S subunit binds to mRNA first, aided by initiation factors

IF-3 stabilizes the free 30S subunit and must eventually be released to allow the 50S subunit to join the 30S-mRNA complex

A 30S subunit carrying several initiation factors binds to an initiation site on mRNA to form an
initiation complex
 IF-3 also 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 subunit 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
 rRNA interactions between subunits

Question 21

7. Describe the scanning model of initiation of translation in eukaryotes. (Ch 22/23)

Answer 21

Slide: Eukaryotic Translation - Initiation

The small ribosomal subunit in eukaryotes recognizes the 5’ cap of the mRNA and moves to the initiation site
 No Shine-Dalgarno sequence
Scanning model of eukaryotic initiation
 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
 Weak Kozak consensus can lead to “leaky scanning”

Question 22

8. What are the steps of elongation in prokaryotic translation? How are EF-Tu, EF-Ts, and GTP involved in elongation? (Ch 22/23)

Answer 22

Slide: Bacterial Translation - Elongation

Entry of an aminoacyl-tRNA into the A site is mediated by EF-Tu
The number of aminoacyl-tRNAs in the cell is approximately the same as the number of
EF-Tu molecules

1. EF-Tu-GTP binds an aminoacyl-tRNA
    Ternary complex
    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
    The aminoacyl end of the aminoacyl-tRNA is in the A site but not aligned facing the P site tRNA
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
7. 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
    EF-Ts is a GTP exchange factor

Question 23

9. 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? (Ch 22/23)

Answer 23

Slide: Bacterial Elongation - Proofreading

EF-Tu serves as a proofreader
II. EF-Tu-GTP only interacts with the factor binding center if the aminoacyl-tRNA fully enters the A site
–Only occurs if there is correct base pairing between the codon and anticodon
 An incorrect aminoacyl-tRNA will dissociate before EF-Tu-GTP interacts with the factor-binding center and promotes GTP hydrolysis
 Hydrolysis to EF-Tu-GDP would commit the incorrect amino acid to incorporation into
the growing polypeptide
III. 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 and the peptidyl transferase center
 Accommodation
–Correctly base paired tRNA will handle the rotational strain
–Incorrectly base paired tRNA will be unable to handle the rotational strain
 Base pairs will break
 tRNA will dissociate from A site

Question 24

10. How is the 23S rRNA involved in the peptidyl transferase process? (Ch 22/23)

Answer 24

Slide: Bacterial Translation - Elongation

Peptidyl transferase
 Process is a function of the protein and rRNA components of the large ribosomal subunit called the peptidyl transferase center
 Triggered when EF-Tu releases the aminoacyl end of A site tRNA
 23S rRNA positions aminoacyl end of aminoacyl-tRNA near the end of the peptidyl-
tRNA
–Assisted by a kink in mRNA between the P and A sites
–The kink maximizes distance between tRNAs at mRNA end to facilitate codon-anticodon pairing
–Also minimizes distance between tRNAs at peptidyl-aminoacyl end
 23S rRNA base pairs with CCA terminus of P and A site tRNAs
–Amino group of amino acid on aminoacyl-tRNA placed in close proximity to the carbonyl of the last amino acid added to the peptidyl-tRNA
–Proximity stimulates catalysis
 Entropic catalysis
 The rRNA and P site tRNA also directly participate in the reaction in an enzymatic
fashion
–2’-OH of A2451 in 23S rRNA
–2’-OH of P site tRNA and proton shuttle
 Substrate-assisted catalysis

Slide: Roles of rRNA in Translation

23S rRNA
 Interacts with 3’-CCA terminus of P site tRNAs in peptidyl transferase center
 Removing almost all proteins from the 50S subunit results in a 23S rRNA complex (with protein fragments) that still retains peptidyl transferase activity
–23S rRNA alone has a low level of peptidyl transferase activity
 Archaeal large subunit has only 23S rRNA in the peptidyl transferase center
 Directly or indirectly involved in removing proton from amino group of peptidyl-tRNA
–Proton shuttle?

Question 25

11. What is the “hybrid state” and how is it and other changes in ribosomal configuration related to the translocation process? (Ch 22/23)

Answer 25

Slide: Bacterial Translation - Elongation

9. Upon transfer of the polypeptide to the A site tRNA, the tRNAs move into their hybrid state
    The anticodons of the tRNAs remain in their pre-peptidyl transfer positions
    The 3’ end of the A site tRNA is now 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
   10.EF-G-GTP binds to the ribosome and stabilizes the hybrid state

Question 26

12. 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. (Ch 22/23)

Answer 26

Slide: Translation - Termination

1. The termination codons are recognized by class 1 protein release factors
    RF1 recognizes UAG and UAA
    RF2 recognizes UGA and UAA
    In eukaryotes there is only eRF1

Class 2 release factors help release the class 1 factor from the ribosome
 RF3 requires GTP for activity
 eRF3 in eukaryotes

All three release factors have structures similar to that of EF-Tu-tRNA and EF-G
 All bind GTP
 All bind to the factor binding center or the A site

Initiation
1. Initiation
 The steps of translation up to entrance of the first aminoacyl-tRNA into the A site

Question 27

1. What are the differences between the lacOC, lacI-, lacIS, and lac-d mutations? (Ch 24)

Answer 27

Slide: lac Operon Mutants

 LacOc is a constitutive cis-acting mutation
 LacI- is a constitutive trans-acting mutation
 LacIS is an uninducible trans-acting mutation
 LacI-d is a dominant negative trans-mutation when found in a heterozygous merozygote 
	 lacI-d/_
	 One mutant subunit in the tetramer will block repressor action

Lac Oc: Destroy binding site to repressor

Lac I-: Cannot bind operator, does not contribute to transcription on/off, affects DNA binding domain

Lac Is: Unable to bind allolactose, always active, always has the potential to bind to repressor

Lac I-d: Shut off activity whether homozygous or heterozygous mutant. Mutant allele has no activity on its own and can inactivate any normal copies. Merozygote: partial diploid. Blocks ability of repressor to bind to operator.

Question 28

2. 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? (Ch 24)

Answer 28

Slide: lac Repressor

 The repressor binds to an inverted repeat in the operator
 Binds more tightly to the upstream side of the repeat
 Within the major groove of DNA
 Inducer binding changes the structure of the repressor core so that it can no longer bind into the major groove
 When the repressor binds to O1-O2 or O1-O3, it induces a loop into the DNA between the two binding sites
 Repressor binding enhances RNAP binding to the promoter
 The RNAP is trapped and “stored” on the promoter by the loop structure
 Cannot initiate transcription

Question 29

3. 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? (Ch 24)

Answer 29

Slide: lac Repressor

 Active repressor has low affinity for random DNA
 A large number of low-affinity binding sites will compete just as well for a repressor as a small number of high-affinity operator sites
 All DNA sequences in the E. coli genome can serve as low-affinity sites!
 Almost all active repressor is bound to either low-affinity sites in the genome or the operator
 Less than 0.01% of active repressor is free in cell
 The difference in affinities between the low-affinity and high-affinity sites is large enough that active repressor is usually bound to the operator
 Operator is occupied by repressor 96% of the time
 Even with no inducer present, the operator is unoccupied 4% of the time
 Transcription is never fully “off”
 Basal (not zero) level of transcription allows cell to always be poised for operon induction
 Always some β-gal and permease present
 Solves the “lac paradox”