Chapter 15

Question 0

mRNA open reading frame ORF

Answer 0

Open reading frame (ORF): continuous, non-overlapping string of codons (3 nucleotides)
in the protein-coding region of an mRNA.
Translation starts at the 5’end of the ORF and proceeds one codon at a time to the 3’ end.
In bacteria, the start codon is usually 5’-AUG-3’, but 5’-GUG-3’ and sometimes 5’-UUG-3’ are also used. Eukaryotic cells always use 5’-AUG-3’ as start codon. Stop codon: UAA, UAG and UGA.

Question 1

mRNA recruitment of tanslation machinery

Answer 1

Prokaryotic mRNAs have a ribosome-binding site that recruits the translational machinery
• RBS (ribosome-binding site): short sequence upstream of the start codon. Also referred to as a Shine-Dalgarno sequence.
• The RBS, located 3-9bp on the 5’ side of the start codon (seq: 5’- AGGAGG-3’), is complementary to a sequence located near the 3’ end of one of the ribosomal RNA (16s rRNA). The core of the region the 16s rRNA has the sequence 5’-CCUCCU-3’. The extent of complementarity and spacing between the RBS and the start codon has a strong effect on how actively a particular ORF is translated. High complementarity and proper spacing favors translation.
• Some prokaryotic ORFs, not first ORFs, lack a strong RBS, but are nonetheless actively translated. In these cases, the ORFs usually overlaps. 5’-AUGA-3’ (contains a start codon and stop codon). Called translational coupling. In this situation, the translation of the downstream ORF requires translation of the upstream ORF.
17-6

Question 2

tRNA overview/structure

Answer 2

Transfer RNAs are adaptors between codons and amino acids
• There are many types of tRNA molecules, but each is attached to a specific amino acid, and each recognizes a particular codon or codons.
• tRNA molecules are between 75 and 95 ribonucleotides in length.
• All tRNAs have the 3’ terminal sequence 5’-CCA-3’ (CCA adding
enzyme: synthesize RNA without a template).
• Presence of several unusual bases in their primary structure. (pseudouridine (ψU), dihydrouridine (D), hypoxanthine). This leads to improved tRNA function. Cells lacking these modifications show reduced rate of growth.

Question 3

tRNA secondary structure

Answer 3

Resembled a clover leaf. It has single-stranded and Double-stranded regions.
A. Acceptor stem (arm): attach amino acid. double strand, and 3-end protrusion.
B. ψU loop. 5’-TψUCG-3’
C. D loop: has dihydrouridine.
D. anti-codon loop: contains the anticodon,
a three nt long sequence that is responsible for
recognizing the codon by base pairing with
the mRNA.
E. Variable loop. 3-21 bases

Question 4

tRNA charging steps

Answer 4

Two steps of tRNA charging: attachment of amino acid to tRNA
• Charging requires an acyl linkage between the carboxyl group of the amino acid and the 2’ or 3’ hydroxyl group of the adenosine nucleotide that protrude from the acceptor stem at the 3’ end of the tRNA. (High energy bond, release energy during hydrxolysis. Significant for protein synthesis).
tRNA charging:
tRNA synthetases
Class I enzyme: attaches the amino
acid to the 2’-OH of the tRNA. (monomeric)
Class II enzyme: attaches the amino
acid to the 3’-OH of the tRNA. (dimeric or tetrameric)

Question 5

Mech of aminoacyl-tRNA synthetase

Answer 5

Each aminoacyl-tRNA synthetase attaches a single amino acid to one or more tRNAs. The same tRNA synthetase is responsible for charging all tRNAs for a particular amino acid.
Most organisms have 20 different tRNA synthetases, but this is not always the case. Some
bacteria lack the tRNA synthetase for Glutamine. The use tRNA synthetase for glutamate, and 17-13 a second enzyme to convert tRNA (glu) to tRNA (gln).

Question 6

How to ensure accuracy regarding aminoacyl-RNA synthetase

Answer 6

tRNA synthetases recognize unique structural features of cognate tRNAs
• The challenge for aminoacyl-tRNA synthetases: (A) they must recognize the correct ste of tRNAs for a particular amino acid, and (B) they must charge all these iso- accepting tRNAs with correct amnio acid.
• The specificity determinants are clustered at two distance sites on the molecule: the acceptor stem and the anticodon loop.
• In some cases, the discriminator base in the acceptor stem can determine the specificity of tRNA from one synthetase to another. (by mutation study)
• Serine is specified by 6 codons, so the synthetase for serine must rely on determinants that lie outside of anti-codon.

Question 7

Aminoacyl-tRNA formation accuracy

Answer 7

Aminoacyl-tRNA formation is very accurate
• Selecting the right amino acid is more challenging than recognizing the appropriate tRNA.
• Frequency of misincorporation:
<0.1%.
Some amino acids (like cysteine and tryptophan) differ substantiallyin size, shape and chemical groups.
(easy to distinguish)
Tyr can form hydrogen bond, phenylalanine cannot. Valyl-tRNA synthetase can sterically exclude isoleucine from
It catalytic pocket because isoleucine is larger than valine.
But valine should slip easily into the isoleucyl-tRNA synthetase amino-acid binding site. The extra mythelene group provide
-2 to 03 kcarl/mole of energy. This small difference In free energy will make binding to isoleucine ~100-fold more likely than binding to valine.
How is this additional level of fidelity achieved?

Question 8

Aminoacyl-tRNA proofreading mech

Answer 8

Some aminoacyl-tRNA synthetase use an editing pocket to charge tRNA with high accuracy
• Proofreading to increase fidelity.
• For example, in addition to its catalytic pocket (for adenylylation), isoleucyl tRNA synthetase has a nearby editing pocket that allows it to proofread the product of adenylylation. AMP-Valine can fit into the editing pocket, where it is hydrolyzed and released as free valine and AMP.
• So, two discrimination steps: one is the initial binding and adenylylation of the amino acid (discrimination factor of ~100), the other is the editing of the adenylylated amino acid (discrimination factor of ~100). Overall, less than ~0.01% error rate.

Question 9

Ribosome discrimination between correctly and incorrectly charged tRNAs

Answer 9

The ribosome is unable to discriminate between correctly and incorrectly charged tRNAs
• Genetic and biochemical evidence
1. Mutation in the anticodon. tRNA synthetases frequently don’t rely on interaction with the anti-codon to recognize the cognant tRNA. Hence, a subset of tRNA can be mutated in their anticodons but still be charged with their usual cognate amino acid. However, the mutated tRNA delivers amino acid to the wrong codon. That means ribosome does little to prevent an incorrectly charged tRNA from adding an inapproproate amino acid to the growing polypeptide.
2. Biochemical experiment nicely illustrates the point that ribosome recognizes tRNA and not the amino acid that it is carrying.

Question 10

Ribosome

Answer 10

The ribosome: a macromolecular machine that directs the synthesis of proteins
Ribosome: at least 3 RNA molecules, and more than 50 different proteins, speed: only 2-20 amino acid per second.
In prokaroytes, the transcription machinery and the translation machinery are located in the same compartment. Thus the ribosome can commence translation of the mRNA as it emerges from the RNA Polymerase.
In eukaryotes, transcription is in the nucleus, whereas translation is in
the cytoplasm.

Question 11

Ribosomal subunits

Answer 11

The large subunit: the peptidyl transferase center, which is responsible for the formation of peptie bonds.
• The small subunit: decoding center, in which charged tRNAs read or decode the codon units of the mRNA.
• Size: bacteria 30S and 50S; eukaryotic ribosome: 40S and 60S.
• Based on the velocity of their sedimentation after aultracentrifugation. S: Sverdberg (inventor of ultracentrifuge)

Question 12

The large and small subunits undergo association and dissociation during each cycle of translation (The ribosome cycle)

Answer 12

Binding of the mRNA and initiator to the small ribosome subunit
Recruits a large subunit Protein synthesis (initiation)
Ribosome translocation from one codon to the next. (translation elongation)
Complete polypeptide release and ribosome dissociation

Question 13

Poly some

Answer 13

Although a ribosome can synthesize only one polypeptide at a time, each mRNA can be translated simultaneously by multiple ribosomes. A mRNA bearing multiple ribosomes are known as a polyribosome or a polysome. A 1000-base ORF can bind to more than 10 ribosomes.

Question 14

The peptidyl transferase reaction

Answer 14

The peptidyl transferase reaction
• Two substrates: an aminoacyl-tRNA and a peptidyl-tRNA.
• The aminoacyl-tRNA is attached at its 3’ end to the carboxyl group of the amino acid.
• The reaction to form a new peptide bond is called
the peptidyl transferase reaction.
So the bond between the aminoacyl-tRNA and the amino acid is not broken during the formation of the next peptide bond. Instead, the bond between the peptide-tRNA and the growing peptide is broken. This reaction takes place without the simultaneous hydrolysis of a nucleotide triphosphate. Why? High energy acyl bond from the charging reaction. One ATP is needed for each tRNA charging reaction.

Question 15

Ribosome binding sites for tRNA

Answer 15

The ribosome has three binding sites for tRNA
• To perform the peptidyl transferase reaction, the ribosome must be able to bind at least two tRNAs simultaneously.

• A-site: the binding site for aminoacylated-tRNA
• P-site: the binding site for peptidyl-tRNA
• E-site: the binding site for the tRNA that is released after the growing peptide chain has been transferred to the aminoacyl-tRNA (E is for “exiting”).

Each tRNA-binding site is formed at the interface between the large and the small subunits of the ribosome.
• In this ways the bound tRNA can span the distance between the peptidyl transferase center in the large subunit and the decoding center in the small subunit.
• The 3’ ends of the tRNAs that are coupled to the amino acid or to the growing peptide chain are adjacent to the large subunit. The anticodon loops of the bound tRNAs are located adjacent to the small subunits

Question 16

Factors involved during the initiation of translation

Answer 16

For translation to be successfully initiated:
The ribosome must be recruited to the mRNA.
A charged tRNA must be placed into the P-site of the ribosome.
The ribosome must be precisely positioned
over the start codon. (critical for the reading frame)

Prokaryotic mRNAs are initially recruited to the small subunits by base pairing to rRNA
For ideally positioned RBSs, The small subunit is positioned on the mRNA such that the start codon will be in the P-site when the large subunit joins the complex.

Question 17

Translation initiation mechanism involving tRNA

Answer 17

A specialized tRNA charged with a modified methionine binds directly to the prokaryotic small subunit
Translation initiation is the only time a tRNA is binding to the P-site without previously occupying the A-site. The event requires a special tRNA (initiator tRNA) , which base pairs with the start codon: usually AUG or GUG.
AUG and GUG have a different meaning when they occur within an ORF, where they are read by tRNA for methionine and valine respectively.
Although the initiator tRNA is first charged with a methionine, a formyl group is rapidly added to the methionine amino group by a separate enzyme (Met-tRNA transformylase). Thus rather than valine or methioine, the initator tRNA is coupled to N-formyl methionine. The charged tRNA is refered to fMet- tRNAfMet.
In prokayotes, a enzyme known as deformylase removes the formyl group form the amino terminus during or after the synthesis of the polypeptide chain. In fact, many mature prokayotic protein do even start with methionine; aminopeptidases often remove the amino-terminal methionine as well as one-two additional amino acid

Question 18

Three initiation factors direct the assembly of an initiation complex that contains mRNA and the initiator tRNA

Answer 18

Translation initiation factors: IF1, IF2 and IF3
IF1: prevent tRNAs from binding to the portion of the
small subunit that will be part of the A-site. (binds to the A-site) IF2: is a GTPase that interacts with three key components of the intiation machinery: the small subunit, IF1 and the charged tRNA. It facilitates the association of fMet-tRNA with the small subunit and prevents other charged tRNAs from associating with the small subunit.
IF3: binds to the small subunit and blocks it from re-associating with large subunit. (occupy the future E-site)
With the binding of three initiation factors, the small subunit is
preparing to bind to the mRNA and the initiator tRNA.
The last step of initiation involves the association with the large
subunit to create a 70S initiation complex. In this step, IF3 is released, and the large subunit is free to bind to the small subunit. This interaction stimulates the GTPase activity of IF2•GTP. IF2•GDP has reduced affinity for the ribosome and the initiator tRNA, leading to the release
of IF2•GDP as well as the IF1 from the ribosome.

Question 19

Prokaryotic initiation and eukaryotic initiation comparisons and contrasts

Answer 19

Prokaryotic initiation and eukaryotic initiation
Similarity:
Both use a start codon, and a dedicated initiator tRNA, both use initiator factors to form complex with the small ribosomal subunit that assembles on the mRNA before addition of the large subunit.
Difference:
The small subunit is already associated with an initiator tRNA when it is recruited to
the capped 5’ end of the mRNA.
The small subunit scans along the mRNA in a 5’ to 3’ direction until it reaches the first 5’-AUG-3’. (explains why the vast majority of eukaryotic RNAs encode a single polypeptide.
More auxiliary factors are needed.

Question 20

PolyAntail in translation initiation

Answer 20

Translation Initiation factors hold eukaryotic mRNAs in circles.
The presence of a Poly-A tail contributes to the efficiency of
eukaryotic translation.
Mediated by eIFG binding directly to the 3’ end of mRNA and the Poly-A binding protein.
Result in the circular configuration of the mRNA, several rounds of translation.

Question 21

uORF

Answer 21

Sometimes the first ORF is in a poor sequence context, resulting in its frequent bypass.
• In other cases, short upstream ORFs (uORFs, less than 10 AAs) are translated. This allows interaction between initiation factors (eg. eIFG and eIF3) that tether the 40S subunit to the mRNA to be retained after termination. They can scan the next AUG after binding to a new initator tRNA complex.

Question 22

IRES: internal ribosome entry site

Answer 22

IRES: internal ribosome entry site
• A more example of initiation at sites downstream from the most 5’-proximal AUG are internal ribosomal entry sites (IRES). IRESes are RNA sequences that function like prokayoritc RBSs. The recrut the small subunit to bind and initiate even in the absence of a 5’ cap. The are often encoded in viral mRNAs that lack a 5’ cap and have a need to exploit the sequence of their genome maximally.

Question 23

IRES mechs

Answer 23

Two mechanisms: 1. Recruit eIFG
2. Cricket paravirus virus
The 5’-UTR of the mRNA forms
a complex RNA structure that mimics a tRNA bound to the P
-site of the 40S ribosomal subunits. This way it bypass all the initiation factors and an initiation tRNAs.
This leads to the hypothesis that early in evolution all mRNAs have such IRESs and that initiation factors evolved later to make translation more efficient and versatile.

Question 24

Elongation

Answer 24

Aminoacyl-tRNAs are delivered to the A-site by elongation factor EF-Tu
Once a tRNA is aminoacylated, EF-Tu binds to
the tRNA’s 3’ end, masking the coupled amino acid.
EF-Tu binds and hydrolyzes GTP. EF-Tu can only bind to an aminoacyl-tRNA when it is associated with GTP. When EF-Tu hydrolyzes its bound GTP, any associated aminoacyl-tRNA is released.
EF-Tu GTPase is active when it is associated with the factor-binding center, the same domain in the large subunit that activates the IF2 GTPase when the large subunit joins the initiation complex.
EF-Tu only interacts with factor-binding center after tRNA enters the A-site and a correct codon-anticodon match
is made.
At this point, EF-Tu hydrolyzes its bound GTP, and is released form the ribosome.

Question 25

Ribosome as a ribozyme

Answer 25

Once the correctly charged tRNA has been placed in the A-site and has rotated into the peptidyl transferase center, peptide-bond formation takes place. This reaction is catalyzed by RNA, specifically by the 23S rRNA components of the large subunit.
• Early evidence came from the experiment when a large subunit that have been largely stripped of its protein was still able to direct peptide bond formation.
• It is also supported by structural studies. (A-site: green, P site: red). There is no amino acid within the 18A of the active site.
Amino terminals of L27 protein does reach into the active site. Mutations in the region of the protein still have 30-50% of the protein synthesis.
How does the 23SrRNA catalyze the peptide-bond formation?
17-50

Question 26

Peptide bond formation: shuttle model

Answer 26

Proton shuttle model (2’-OH as the shuttle): the 2’-OH donates a hydrogen to the 3’-OH of the peptidyl-tRNA and accepts a proton from the attacking amino group of the amino acid attached to the A site tRNA. This electron movement drives peptide-bond formation.

Question 27

Ribosome translocation during elongation

Answer 27

Peptide-bond formation initiates translocation in the large subunit
• Once the peptidyl transferase reaction has occurred, the tRNA in the P-site is dacetylated (no longer attached to an amino acid), the growing peptide chain is linked to the tRNA in the A-site.
• For a new round of peptide chain reaction to occur, the P-site tRNA must move to the E-site and the A-site tRNA must move to the P-site.
• At the same time, the mRNA must move by three nucleotides to expose the next codon.

Question 28

Factors involved in translation termination

Answer 28

Termination of translation
• There is no terminating tRNA to recognizing the stop codon.
• Stop codon is recognized by proteins called relsease factors (RFs).
• Two classes of RFs:
• Class I RFs recognize the stop codons and trigger hydrolysis of the peptide chain form the tRNA in the P-site. In prokaryotes, RF1 recognizes UAG, RF2 recognizes UGA, UAA is recognized by both RF1 and RF2. In eukaryotes, eRF1.
• Class II RFs stimulate the dissociation of the class I factors from the ribosome after release of the polypeptide chain. RF3, or eRF3 (regulated by GTP binding).
Short regions of Class I release factors recognize stop codons and trigger release of the peptidyl chain
• Experiments to swap RF1 and RF2 short coding regions identified a three-amino-acid sequence that is critical for release factor specificity. (peptide anticodon)
• 3D structure confirms the binding of RF1 to the A-site of the ribosome.

Question 29

Euk translation termination and ribosome recycling

Answer 29

Difference:
1.eRF1 recognizes all stop codons 2.eRF3 delivers eRF1 to the
ribosome
3.There is no evidence of ribosome
recycling factors in eukaryotic cells, nor does the eEF2 ( the eukaryotic EF-G) participate in ribosome recycling.
4. eRF1 (in conjunction with an ATPase called Rli1) participates in
ribosome recycling.

Question 30

Regulation of translation

Answer 30

Regulation of translation
• Transcription regulation vs translation regulation (more rapid change in protein levels).
• More on the initiation step (efficient).
• Protein or RNA binding near the ribosome-binding site negatively regulates batercial
translation initiation.
• Ribosomal proteins are translational repressors of their own synthesis.
• Global regulators of eukaryotic translation target key factors required for mRNA recognition and initiator tRNA ribosome binding.
• Spatial control of translation by mRNA-specific 4E-BPs.
• An Iron-regulated, RNA-binding protein controls translation of ferritin.
• Translation of the yeast transcriptional activator Gcn4 is controlled by short upstream ORFs and ternary complex abundance.

Question 31

Nonstop mediated decay

Answer 31

AAA codes for lysine. No stop coding leads to the addition of multiple lysines to the end of the protein.
The stalled ribosome is bound by two proteins related to eRF1 and eRF3 (Dom34 and Hbs1) that stimulate ribosome dissociation and release of the peptidyl-tRNA and mRNA. A second eRF3 related factor, Ski7, recruits a 3’ to 5’ exonuclease that degrades the “non-stop” mRNA. Also, endonucleus (Xm1) is cutting the uncapped RNA.
Proteins containing Polylysines in the C-terminals are not stable, and are degraded by proteases.

Question 32

No go decay

Answer 32

This mechanism recognizes ribosomes that are stalled on a mRNA.
This can occur as a stable mRNA secondary structure in the coding region of the mRNA or because of a stretch of codons for which there are few corresponding tRNAs in the cells.
The mechanisms are similar to nonstop-mediated decay.
Although indirectly, eukaryotic cells rely on translation as a mechanism to proofread their mRNAs.

Question 33

Eukaryotic cells degrade mRNA that are incomplete or have premature stop codons (A: nonsense-mediated mRNA decay)

Answer 33

Eukaryotic cells degrade mRNA that are incomplete or have premature stop codons (A: nonsense-mediated mRNA decay)
a) Translation of normal mRNA displaces all the exon-junction complexes; (b) if premature stop codon is present in the mRNA, then the ribosome is released before the displacement of all the exon-junction complexes. This results in the recruitment of Upf1, Upf2, and Upf3
to the ribosome, which activate decapping enzyme, deadenylation enzyme. The unprotected mRNA is rapidly degraded by 5’- to 3’ and 3’ to 5’ exonucleases.

Question 34

Energy consumption in peptide bond formation

Answer 34

A circle of peptide formation consumes two molecules of GTP and one molecules of ATP
• How many molecules of nucleoside triphosphate does it cost per round of peptide formation (setting aside the energetics of amino acid biosynthesis and the energetics of initiation and termination?)
• A. An ATP is consumed by the aminoacyl-tRNA snthetase in creating the high energy acyl bond that links amino acid and tRNA. The breakage of the high-energy bond drives the peptidyl transferase reaction that creates the peptide-bond.
• B. A GTP is consumed in the delivery of a charged tRNA to the A-site of the ribosome by EF-Tu, ensuring the correct codon-anticodon recognition. (Accuracy)
• C. A GTP is consumed in the EF-G mediated process of translocation. (order of events)
• Translation elongation is similar in prokaryotes and eukaryotes.

Question 35

Translational termination mechanism: Class I and the way in which GDP/GTP exchange and GTP hydrolysis control the function of the Class II releasing factor

Answer 35

Short regions of Class I release factors recognize stop codons and trigger release of the peptidyl chain
• Experiments to swap RF1 and RF2 short coding regions identified a three-amino-acid sequence that is critical for release factor specificity. (peptide anticodon)
• 3D structure confirms the binding of RF1 to the A-site of the ribosome.
Unlike other GTP-binding proteins, RF3 has a higher affinity to GDP than GTP.
• RF3 binding to the ribosome is dependent on the presence of a class I release factor.
• RF1 acts as a GTP exchange factor for RF3 (similar to EF-Ts for EF-Tu). After the Class I release factor stimulates polypeptide release, an change in the conformation of the ribosome and class I release factor stimulates RF3 to exchange its bound GDP for GTP.
• The binding of GTP to RF3 leads to the formation of a high affinity interaction with the ribosome that favors the rotated hybrid state as we discussed as a translational intermediate above.
• This changes also allow RF3 to associate with facto-binding center of the large subunit.
• This stimulates the hydrolysis of GTP, and release of RF3•GDP.