BIOMG 3320 Group 6 (Lecture 14-15-16)

  1. Why is protein synthesis is more complex than DNA replication and synthesis
    • No specific affinity between amino acids and purine/pyrimidine bases in RNA
    • Polymerization of 20 amino acids (instead of 4 nucleotides)
    • DNA replication is 200-1000nt/second; translation is 2-20 amino acids/second
    • Protein synthesis can account for up to 90% of the chemical energy used by a cell for biosynthetic reactions.
  2. Explain in lab protein synthesis
    • Small proteins (peptides) can be chemically synthesized in vitro. However, peptide bonds don’t form spontaneously. The amino-group is a good nucleophile, but the hydroxyl is not a good leaving group.
    • Chemical peptide synthesis has low efficiency, and a maximum of 100 amino acids can be synthesized in a few days.
    • Biological protein synthesis, on the other hand, produce proteins of up to 4,000 amino acids. Can synthesize 100 amino acid proteins in 5 seconds.
    • Protein synthesis is efficient in vivo, because tRNA activates the carboxyl group.
  3. What are the primary components of the translational machinery?
    • mRNAs: information
    • tRNAs: physical interface mRNA codon/amino acid. Can be aminoacyl-tRNA (tRNA charged with a single amino acid). Or a peptidy-tRNA (a tRNA charged with the growing polypeptide chain)
    • aminoacyl-tRNA synthetases: couple tRNA to amino acid
    • Ribosome: the machine that coordinates mRNA/tRNA recognition and catalyzes peptide bond formation. The ribosome positions the reactants to facilitate the nucleophilic attack that characterizes peptide bond formation.
  4. What is the peptidyl transferase reaction?
    • A peptide bond being formed between the C-terminal amino acid of the growing polypeptide chain and the incoming amino acid.
    • The polypeptide chain is thus transferred from the old peptidyl-tRNA to the aminoacyl-tRNA, which is now called the new peptidyl-tRNA.
  5. Synthesizing aminoacyl-tRNAs
    • Aminoacyl-tRNA syntheses are enzymes that attach the correct amino acid to their tRNAs.
    • The reaction occurs in two steps:
    • The amino acid is activated through adenylylation. This step requires ATP.
    • tRNA replaces the adenyl.
    • One aminoacyl-tRNA synthetase will only charge one amino acid, to multiple tRNAs.
    • Amino acid + tRNA + ATP -> aminoacyl-tRNA + AMP + 2P
  6. How to ensure fidelity in tRNA charging?
    • It’s very important that the correct set of tRNAs for a particular amino acid. This is because the ribosome doesn’t have the mechanism to check whether the tRNA is coupled with the right amino acid.
    • Mischarging is, thus, very rare (less than 0.005%)
    • Shape and size of the amino acid binding site (synthesis site) in the aminoacyl-tRNA synthetase. One major issue with this is that some amino acids are very similar. Based only on the synthesis site, valine would be attached to isoleucine tRNAs 1% of the time.
    • Editing pocket increases accuracy. The editing pocket is small, and the correct amino acid can’t get in. If it does get in, it is hydrolyzed. The editing site proofreads the aminoacyl tRNA products.
  7. Composition of the Prokaryotic ribosome
    • 70S
    • A large subunit (50S): broken down into ribosomal RNA and proteins
    • A small subunit (30S): broken down into ribosomal RNA and proteins
    • Subunits are named according to the velocity of their sedimentation under centrifugation.
  8. Composition of the Eukaryotic ribosome
    • 80S
    • A large subunit (60S): broken down into ribosomal RNA and proteins
    • A small subunit (40S): broken down into ribosomal RNA and proteins
  9. Ribosome structure
    • rRNA + protein
    • Large subunit: contains the peptidyl transferase center
    • Small subunit: Contains the decoding center (where charged tRNAs "decode" the mRNA codons. mRNA goes through small subunit.
  10. What are the tRNA binding sites in the ribosome?
    • A site: binds aminoacylated tRNA (incoming tRNA)
    • P site: binds peptidyl-tRNA (tRNA holding polypeptide chain)
    • E site: binds the exiting tRNA that is released.
  11. Explain the ribosome cycle
    • Initiation: The small ribosomal subunit engages with the mRNA and allows for the initiator tRNA to be recruited.
    • Elongation: The large ribosomal subunit engages with this unit. Multiple rounds of peptide chain formation occurs.
    • Termination: The large and small subunits disengage.
    • The ribosome must be assembled and disassembled every time a single protein is formed.
    • One ribosome can only synthesize one polypeptide at a time.
    • Polysomes are multiple ribosomes that assemble on the same mRNA molecule.
  12. Translation initiation in bacteria
    • Ribosome binding site (RBS) in mRNA< which is a sequence upstream of the start codon AUG, base pairs with the small subunit of the 16s rRNA.
    • This positions the AUG (start codon) of the mRNA within the P site of the small subunit.
    • Initiator tRNA (charged with fMet) is recruited and pairs with the AUG in the P site.
    • The large subunit is recruited.
  13. Explain initiation factors in the assembly of the initiation complex
    • IF3 prevents reassociation between small and large subunit.
    • IF1 prevents tRNAs from binding A site.
    • IF2 facilitates binding of fMet with the small subunit.
  14. At which step in translation do most errors occur?
    • Elongation: The addition of the correct aminoaceyl-tRNA.
    • Elongation is highly conserved
  15. Explain translation initiation in eukaryotes
    • Initiated tRNA (fMet) binds the P site on small subunit with the support from initiation factor eIF2-GTP. This is called the 43S preinitation complex.
    • Independently, the 5’cap recruits eIF4E, which recruits RNA helicase. Stimulates the processing of mRNA, the secondary structures are. Translation is not possible without this processing (there are exceptions).This is why this process is called 5’-cap dependent translation.
    • Once the mRNA is ready to be engaged, the small subunit scans the mRNA from 5’ to 3’ to find the AUG. Once the small subunit finds the start codon, the large subunit is recruited.
  16. Compare prokaryotic translation and eukaryotic translation
    • Eukaryotes also use start codon, dedicated initiator tRNA and initiation factors
    • How mRNA and start codon are recognized is different from prokaryotes.
    • Initiator tRNA is charged with methionine
    • The small subunit binds the initiator tRNA (43S) before recruitment of mRNA
    • The start codon is found by scanning downstream from the 5’ end of the mRNA.
  17. How do we find start sites in eukaryotes?
    • Done through ribosome profiling
    • We freeze the ribosomes in the start position.
    • Isolated polysomes are treated with RNAse 1, which digests free mRNA. The mRNA that is not digested will be inside ribosomes.
    • The sequence inside will be exactly 20 nt, which we can convert to DNA and sequence.
    • Ribosome profiling is revealing non-canonical start sites.
    • Translation can occur in many different AUGs. If the cell is treated differently, the AUG changes.
  18. Explain Internal ribosome entry sites (IRESs)
    • This is a mechanism of alternative initiation that facilitates 5’-cap independent translation.
    • IRESs are RNA sequences that recruit the small ribosomal subunit to bind and initiate even in the absence of a 5’cap.
    • IRESs bypass normal requirements for initiation of translation.
    • This is important for viruses, where when 5’ cap dependent translation is shut down by the cell in response to viruses, 5’ independent translation occurs.
    • During cell division, 5’ cap dependent translation is shut down. If a protein is needed during cell division, it is encoded by IRES mRNA.
  19. Translation elongation
    • Once the ribosome is assembled with the charged initiator tRNA in the P site, three key events must occur.
    • Binding of aminoacyl-tRNA to A site
    • Peptide bond formation
    • Translocation: The ribosome moves by three nucleotides. This frees up the A site for another tRNA to enter and puts the P site tRNA (now lacking an amino acid) into the E site.
    • ​​Unlike initiation to translation, the mechanism of elongation is highly conserved between prokaryotic and eukaryotic cells.
  20. What are the two elongation factors that make translation especially efficient and accurate?
    • EF-TU and EF-G in bacteria
    • EF1 and EF2 in eukaryotes.
  21. Explain Ef-Tu
    • Ef-Tu delivers aminoacyl-tRNAs to the A site.
    • Many tRNAs enter the A site, but are not stabilized. EF-Tu is crucial to stabilize the right tRNA. Ef-Tu increases the accuracy of translation.
    • This is done through the factor binding center. The Ef-Tu only interacts with the factor binding center after the correct tRNA enters the A site. The factor binding center activates the GTPase activity of Ef-Tu.
    • The Ef-Tu then leaves the A site.
    • If the correct aminoacyl-tRNA has not entered the A-site, the GTPase activity of Ef-Tu is not activated and the Ef-Tu prevents any interaction between the polypeptide chain and the wrong tRNA.
  22. How many incorrectly added amino acids would you expect to find in a protein of 140 thousand daltons?
    • 1 amino acid is 100 Da.
    • Error rate is 0.01-0.1%
    • Between 0 and 2.
  23. What is the error rate of DNA replication?
    10-9
  24. How does the cell try to ensure that the correct aminoacyl-tRNA is delivered?
    • Three mechanisms ensure correct pairing between tRNA and mRNA
    • 16S rNRA forms a series of hydrogen bonds with the codon-anticodon match is made. These bonds help stabilize the codon-anticodon pair.
    • GTPase activity of Ef-Tu is sensitive to the correct codon-anticodon pairing.
    • After Ef-Tu release, incorrectly matched tRNAs dissociate more rapidly.
  25. Explain EF-G
    • When the tRNA in the A-site is holding the peptide chain, the factor binding site is revealed.
    • EF-G can only bind to the ribosome when associated with GTP.
    • Binding of EF-G-GTP stimulates GTP hydrolysis at the factor binding center.
    • GTP hydrolysis changes the conformation of EF-G, so it can reach into the small subunit.
    • EF-G-GDP mimics the structure of a tRNA bound to Ef-Tu in order to assemble itself within the small subunit.
    • EF-G drives translocation by displacing the tRNA bound to the A site.
  26. Explain termination of translation
    • Translation elongation ends when one of the three stop codons enters the A site.
    • Release factors mimic a tRNA, with an anticodon with an affinity for stop codons.
    • RF1 and RF2 activate the hydrolysis of the polypeptide from the peptidyl-tRNA.
    • RF3 stimulates the dissociation of RF1 and RF2 from the ribosome.
  27. How do GTP-binding/hydrolyzing proteins coordinate translation?
    • Protein conformation changes depending on GDP or GTP binding.
    • GTP hydrolysis is used to monitor the completion of key steps.
  28. Inhibitors of protein synthesis as antibiotics
    • Translation is a strict sequence of steps, so impairing one step compromises the whole process.
    • 40% of antibiotics are inhibitors of translational machinery.
    • Tetracycline: targets the A site of the 30S subunit and inhibits the aminoacyl-tRNA from binding to the A site. Is not able to interact with eukaryotic ribosome, as it is specific to the structure of the prokaryotic ribosome.
  29. What is the step of translation that is usually regulated?
    Initiation is usually the step at which translation regulation occurs to minimize the energy wasted.
  30. Why is the regulation of translation more rapid than regulation of transcription?
    The ultimate goal of regulation is to reduce the number of protein produced. Transcription is further removed from protein synthesis.
  31. Explain the regulation of translation initiation in bacteria
    • RNA-binding protein can bind to the ribosome-binding site on mRNA when nutrient levels are low.
    • This simply occludes the mRNA from the 30S subunit trying to bind to it.
  32. Regulation of translation in eukaryotes
    • 4E-BP binds to the elF4E at 5’ end of mRNA and blocks the engagement of other factors such as RNA helicase to remove secondary structures, ultimately preventing 43S complex (small subunit) from binding to the RNA.
    • 4E-BP can be phosphorylated, and is then unable to bind to the RNA.
    • Phosphorylation is removable, completing the dynamic cycle that characterizes regulation.
    • 4E-BP phosphorylation is performed by mTOR kinase
    • mTOR is extremely important and very complex. It is a master regulator of nutrient availability among many things.
    • May in fact prove to be a target of anti-cancer drugs, as inhibition leads to less phosphorylation of translation-repressing factors, which ultimately starves cancer cells of translation.
    • When there is less mTOR, 4E-BP is not phosphorylated and binds to mRNA to block translation.
  33. Explain post-translational events
    • Polypeptide chain must be folded into proper protein conformation once translation is complete.
    • Chaperones and other proteins can aid in this process.
    • Proteins undergo covalent modification (post-translational modification, PTMs)
    • PTMs include phosphorylation, acetylation, methylation, ubiquitination (appending an entire small protein)
  34. Why doesn't amount of mRNA correlate with amount of protein?
    Post-translational modification.
  35. Explain protein electrophoresis: SDS-PAGE
    • SDS is sodium dodecyl sulfate
    • PAGE is polyacrylamide gel electrophoresis
    • SDS denatures extracted proteins, goal is to get them to become more linear so they migrate predictably. Heat helps with denaturation, mercaptoethanol reduces disulfide bridges.
    • SDS surrounds proteins, coats with with a uniform negative charge. Proteins will then migrate towards positive electrode.
    • Larger proteins migrate faster than smaller ones and proteins separate based on apparent molecular weight.
    • Gels can be stained to visualize protein bands after migration.
  36. Explain western blotting (immunoblotting)
    • Run SDS-PAGE
    • Transfer proteins to a membrane
    • Incubate with an antibody that recognizes your protein of interest.
    • Wash un-bound antibody off the membrane, then detect the remaining antibody bound to your protein of interest. Intensity indicates the amount of protein detected.
    • Only useful if you have an antibody for your protein of interest and you know what protein you’re going after.
  37. Explain mass spectrometry
    • Mass spectrometer can detect a fingerprint of each peptide, essentially sequencing each one.
    • Map the sequences acquired onto protein databases to determine the protein identities.
Author
pelinpoyraz
ID
358500
Card Set
BIOMG 3320 Group 6 (Lecture 14-15-16)
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Updated