Chapter 6B Essays

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Chapter 6B Essays
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  1. 1) What features give polynucleotides a wider range of catalytic activities?
    a. Many ribozymes work by positioning metal ions at their active sites.
  2. 2) What is the process of RNA that allowed it do what it does today?
    a.       It could act as a template for the formation of another of itself; and, it is catalytic. Compartmentalization allowed barring of parasitic RNA molecules from the system and easier ability for things to remain in the neighborhood of RNA. They also had the property of amphiphilicity, one part hydrophobic and another hydrophilic. This allowed the formation of the first membrane-boudned cells. 
  3. 1)      What is the process of protein synthesis evolution? 
    a.       Some short peptides were synthesized by peptide synthetase enzymes; and, RNA could have catalyzed this by peptide bond formation. Other RNAs could have served as crude templates to direct the nonrandom polymerization of a few different amino acids. 
  4. 1)      What evidence supports the idea that RNA developed before DNA? 
    • a.       Chemical differences since deoxyribose is harder to make
    • b.      The double-helical structure for ehbancement of DNA stablility. 
  5. 1)      How are tRNAs constructed? 
    a.       They are constructed in such a wa that they require accurate base-pairing only at the first two positions of the codon and can tolerate a mismatch (wobble) at the third position
  6. 1)      What is the covalent modification that tRNAs undergo? 
    a.       They are synthesized by RNA polymerase III. They have introns that must be spliced out differently than pre-mRNA splicing in that it is a cut-and-paste mechanism catalyzed by proteins instead of formation of a lariat. This trimming and splicing requires the precursor tRNA to be correctly folded in its cloverleaf configuration.  Nearly 1 in 10 nucleotides of each mature tRNA is an altered version of a standard G, U, C, or A. 
  7. 1)      Explain how each amino acid goes to its appropriate tRNA molecule. 
    a.       It involves aminoacyl-tRNA synthetases, which covalently couple each amino acid to its appropriate set of tRNA molecules; there can be 20 synthetaes in eucaryotes and less in prokaryotes (in prokaryotes, one synthease places an amino acid on two different types of tRNA and a second enzyme modifies it. 
  8. Explain the reaction of the tRNA synthetases in terms of energy.
    • a.       The reaction for this uses ATP hydrolysis to produce a high-energy bond between tRNA and the amino acid. The steps are:
    •                                                               i.      The energy of ATP hydrolysis is used to attach each amino acid to its tRNA molecule in a high-energy linkage. The amino acid is first activated through the linkage of its carboxyl group directly to an AMP moiety, forming an adenylated amino acid; the linkage of the AMP is driven by the hydrolysis of the ATP molecule that donates the AMP.
    •                                                             ii.      Without leaving the synthetase enzyme, the AMP-linked carboxyl group on the amino acid is then transferred to a hydroxyl group on the sugar at the 3’ end of the tRNA molecule. This transfer joins the amino acid by an actigated ester linkage to the tRNA and forms the final aminoacyl-tRNA molecule. 
  9. 1)      Explain editing by tRNA synthetases. 
    • a.       It must select the correct amino acod, which is done in two steps:
    •                                                               i.      First, the correct amino acid has the highest affinity for the active site pocket of its synthetase and is favored over the other 19.
    •                                                             ii.      Second, after covalent linking to AMP, the tRNA binds the synthetase, trying to force the amino acid into a second pocket in the synthease, which is very precise. Once the amino acid is in the correct site, it is hydrolyzed from the AMP and is released from the enzyme.
    • b.      It must also recognize the correct set of tRNAs, and extensive structural and chemical complementarity between the synthetase and the tRNA allows the synthetase to probe various features of the tRNA.
    • Most tRNA synthetases directly recognize the matching tRNA anticodon, containing three adjacent nucleotide binding pockets. Others use the nucleotide sequence of the acceptor stem as the key determinant
  10. 1)      What is the fundamental reaction of protein synthesis?
    a.       It is the formation of the peptide bond between the carboxyl group and the free amino group of the incoming amino acid. Throughout the process, the growing carboxyl end of the chain remains activeated by its covalent attachment to a tRNA. Each addition disrupts the high-energy covalent linkage, but immediately replaces it with an identical linkage on the most recently added amino acid. So, each amino acid added carries with it the activation energy for the addition of the next amino acid rather than the energy for its own addition. 
  11. 1)      Explain synthesis of RNAs.
    • a.       Eukaryotic ribosome subunits are assembled at the nucleolus, when newly transcribed and modified rRNAs associate with ribosomal proteins that are transported into the nucleus.
    • b.      When not synthesizing proteins, the two subunits of the ribosome are separate. They join on an mRNA near the 5’ end to initiate synthesis, then pulling the mRNA through. When a stop codon is encountered, the ribosome releases the polypeptide chain and separates. 
  12. 1)      Compare and contrast eukaryotic and prokaryotic ribosomes.
    • a.       They both have one large and one small subunt that fit together to form a complete ribosome.
    •                                                               i.      The small provides framework for tRNA and mRNA matching; and, the large subunit catalyzes the formation of the peptide bonds that link the amino acids together into a polypeptide chain. 
  13. 1)      What is the structure of ribosomes? 
    • a.       It contains four binding sites for RNA molecules: one is for the mRNA and three (A, P, and E) are for tRNAs, which is held tightly at the A and P sites only if the aticodon pairs with the codon on mRNA.
    • b.      It is 2/3 RNA and 1/3 protein. The proteins areon the surface and fill in the gaps and crevices on folded RNA. Some proteins send out extended regions of polypeptide chain that pepetrate short distacnes into holes in the RNA core. It functions to stabilize the NA core, while permitting changes in rRNA conformation that are necessary for this RNA to catalyze efficient protien synthesis. The proteins also aid in assembly of rRNAs to make up the core.
    • c.       Both the A, P, and E sites, as well as catalytic site are formed by RNA. 23S rRNA forms a highly structured pocket that precisely orients the two reactants and thereby accelerates their covalent joining. In addition, the tRNA in the P site contributes to the active site, supplying a functional OH group that participates directly in catalysis. 
  14. 1)      What are the steps of elongation? 
    • tRNA binding
    • peptide bond formation
    • large subunit translocation
    • small subunit translocation
  15. Explain tRNA binding
    •                                                               i.      Elongation begins at a point where amino acids have already been linked and there is a tRNA in the P site fo the ribosome, covalently linked to the end of the growing polypeptide.
    •                                                             ii.      Step 1: tRNA carrying the next amino acid in the chain binds to the A site by base pairing with the mRNA codon so the P and A site contain adjacent bound tRNAs
  16. Explain peptide bond formation.
    •                                                               i.      the carboxyl end of the polypeptide chain is released from the tRNA at the P site (by breaking the bodn between tRNA and amino acid) and joined to the free amino group of the amino acid linked to the tRNA at the A site, forming a new peptide bond
    • 1.       this is catalyzed by peptidyl transferase in the large subunit
  17. Explain large subunit and small subunit translocation
    •                                                               i.      as a result of these last two steps, the entire ribosome moves three nucleotides along mRNA and is positioned to start the next cycle
    •                                                             ii.      Step Three: the large subunit moves relative to the mRNA held by the small subunit, shifting the acceptor stems of the two tRNAs to the E and P sites of the large subunit
    •                                                           iii.      Step Four: another series of conformational changes moves the small subunit and its bound mRNA exactly three nucleotides, resetting the ribosome so it is ready to receive the next aminoacyl tRNA
  18. 1)      What feature of elongation makes it efficient?
    a.       Two elongation factors, which enter and leave the ribosome during each cycle, each hydrolyzing GTP to GDP and undergoing conformational changes in the process: Ef-Tu and EF- G (in bacteria) and EF1 and EF2 (in eucaryotes)
  19. 1)      How does EF-Tu (EF1 in eucarytotes) increase accuracy of translation? 
    • a.       First, as it escorts an incoming aminoacyl-tRNA to the ribosome, EF-Tu checks whether the tRNA-amino acid match is correct (which may be a result of EF-Tu’s discrimination among different amino acid0tRNA combinations
    • b.      Second, EF-Tu monitors the initial interaction between the anticodon of an incoming aminoacyl-tRNA and the codon of the mRNA in the A site
    •                                                               i.      Aminoacyl-tRNAs are bent when GTP bound Ef-Tu is bound, preventing incorporation fo the amino acid. Matching hydrolyzes GTP and releases EF-Tu.
  20. 1)      How is the correctness of the codon-aticodon match assessed? 
    a.       The rRNA in the small subunit of the ribosome forms a series of hydrogen bonds with the codon-anticodon pair that allows determination of its corectlness. rRNA flds around the pair, and its final closure triggers GTP hydrolysis, thus releasing EF-Tu. 
  21. 1)      After Ef-Tu dissociates, what is the other opportunity for correction? 
    a.       After GTP hydrolysis,there is a short time delay as the amino acid carried by the tRNA moves into position on the ribosome. It is a shorter delay for correct pairs. Moreover, incorrectly matched tRNAs dissociate more rapidly than those correctly bound because their interaction with the codon is weaker. Incorrect tRNAs leave the ribosome without being used. 
  22. 1)      What is important about inititiation of translation?        
    a.       The site at which protein synthesis begins on the mRNA is crucial since it sets the reading frame for the whole length of the message. An error of one nucleotide would cause misreading of the rest. The initiation step is also important because for most genes it is the last point at which the cell can decide whether the mRNA is to be translated and the protein synthesized. 
  23. 1)      How does translation begin
    • a.       It begins with AUG and a special tRNA for starting it called the initiator tRNA. In eucaryotes, the initiator tRNA-methionine complex is first loaded intot he small ribosomal subunit along with additional proteins called eukaryotic initiation factors, or eIFs.
    • b.      Next, the small ribosomal subunit binds to the 5’ end of an mRNA molecule, which is recognized by virtue of its 5’ cap and its two bound initiation factors, eIF4E (which directly binds the cap) and eIF4G.
    • c.       The small ribosomal subunt then moves forward along the mRNA, searching for the first AUG. Additional initiation factors that act as ATP-powered helicases facilitate the ribosome’s movement through RNA secondary structure.
    • d.      Once AUG is found, the initiation factors dissociate, allowing the large ribosomal subunit to assemble with the complex and complete the ribosome. Initiation tRNA is still in the P-site, leaving the A-site vacant. 
  24. 1)      What is leaky scanning? 
    a.       When scanning ribosomal subunits skip the first AUG due do substantial diference in recognition sites of the consensus recognition sequence. This produces two or more proteins, differing in their N-termini, from the same mRNA molecule. 
  25. 1)      How does initiation in bacteria occur? 
    a.       Because bacterial mRNAs have no 5’ cap to signal the ribosome where to begin searching for the start of translation, there is a specific ribosome-binding site (Shine-Dalgarno sequence) that is located a few nucleotides upstream of the AUG at which translation is to begin. This nucleotide sequence (5’-AGGAGGU-3’) forms base pairs with the 16S rRNA of the small ribosomal subunit to position the initiating AUG codon int eh ribosome. A set of translation initiation factors orchestrates this interaction, as well as the subsequent assembly of the large ribosomal subunit to complete the ribosome. This process enables a bacterial ribosome to directly assemble on a start codon that lies in the interior of an mRNA molecule, as long as a ribosome-binding site precedes it by several nucleotides. This makes bacterial mRNAs polycistronic.
  26. Explain termination.
    a.       Stop codons (UAA, UGA, UAG) signal the end of the protein-coding message. Proteins known as release factors bind to any ribosome with a stop codon positioned in the A site, forcing the peptidyl transferase in the ribosome to catalyze the addition of a water molecule instead of an amino acid to the peptidyl-tRNA. This reaction frees the carboxyl end of the growing polyeptide chain from its attachment to a tRNA molecule, and since only this attachment normally holds the growing polypeptide to the ribosome, the completeled protein chain is immediately released into the cytoplasm. The ribosome then releases the mRNA and separates into the large and small subunits, which can assemble on this or another mRNA molecule to begin a new round of protein synthesis. 
  27. 1)      How is translation different in bacteria? 
    a.       Because bacterial mRNA does not to be processed, ribosomes attach to the free end of the bacterial mRNA molecule and start translating it even before the transcription of that RNA is complete, following closely behind the RNA polymerase as it moves along DNA
  28. 1)      What is interesting about the coding of amino acids? 
    a.       The standard colde allows cells to make proteins using only 20 amino acids. However, bacteria, archae, and eucaryotes have a 21st amino acid that can be incorborated into a growing polypeptide chain through translation recoding called selenocysteine, which is essential for the efficient function of a variety of enzymes and contains a selenium atom in place of sulfur of cysteine. 
  29. 1)      How are antibiotics used effectively against bacteria? 
    a.       Many antibiotics lodge in pockets of rRNAs and interfere with the smooth operation of the bacterial ribosome. Because they block specific steps in the processes that lead from DNA to protein, many are useful for cell biological studies. 
  30. 1)      Aside from time delays in protein translation, what else does accuracy in translation depend on? 
    a.       It depends on the expenditure of energy. Protien synthesis consumes more energy than anything. At least four high-energy phosphate bonds are split for each new peptide bond: two consumed in charging a tRNA molecule with an amino acid and two drive steps in the cycle of reactions occurring on the ribosome during synthesis itself. Extra energy is consumed each time an incorrect amino acid linkage is hydrolyzed by tRNA synthetase and each time that an incorrect tRNA enters the ribosome, triggers GTP hydrolysis, and is rejected. 
  31. 1)      How does the cell avoid translatin of damaged mRNAs?
    • a.       The 5’ cap and poly-A til are both recognized by translation-initiation machinery before translation begins. To help ensure that mRNAs are properly spliced before translation, the exon junction complex (EJC) [which is deposited on mRNA after splicing] stimulates the subsequent translation of mRNA.
    • b.      The most powerful is nonsense-mediated mRNA decay, which eliminates defective mRNAs before they can be translated. 
  32. 1)      Explain nonsense-mediated mRNA decay.
    a.       It begins as an mRNA molecule is being transported from the nucleus to the cytosol. As its 5’ end emerges from the nuclear pore, the mRNA is met by a ribosome, which begins to translate it. As translation proceeds, the exon junction complexes (EJC) bound to the mRNA at each splice site are displaced by the moving ribosome. The normal stop codon will be within the last exon, so by the time the ribosome reaches it and stalls, no more EJCs should be bound to the mRNA. If this is the case, the mRNA “passes insepection” and it released to the cytosol where it can be translated in earnest. However, if the ribosome reaches a premature stop codon and stalls, it senses that the EJCs remain and the bound mRNA molecule is rapidly degraded. In this way, the first round of translation allows the cell to test the fitness of each mRNA molecule as it exits the nucleus. 
  33. 1)      Why is nonsense-mediated mRNA decay important? 
    • a.       Evoluntionarily: allow eukaryotes to explore new genes formed by DNA mutations, rearrangements, or splicing.
    • b.      Cells of developing immune system: extensive DNA rearrangements often generate premature termination codons. Those that are toxic degrade the mRNAs
    • c.       Finally, they play a role in mitigating the symptoms of many inherited human diseases; nonsense decay eliminates the aberrant mRNA and prevents a potentially toxic protein from being made
  34. 1)      What is the bacterial equivalent tononsense decay? 
    a.       When a bacterial ribosome translates to the end of an incomplete RNA it stalls and does not release the RNA. Rescue comes in the form of a special RNA (called tmRNA), which enters the A site and is translated, releasing the ribosome. The special amino acid tag added to the C-terminus of the truncated protein signals to proteases that the entire protein is to be degraded. 
  35. 1)      Over evolutionary time, what has the amino acid sequence been selected for? 
    a.       The conformation that it adopts as well as its ability to fold rapidly.  Folding can begin immediately, as the protein spins out the ribosome, starting from the N-terminal end.. Within a few seconds of emerging from the ribosome, it forms a compact structre that has secondary features. This unusually dynamic and flexible state called a molten state is the starting point for a relatively slow process in which many side-chain adjustments occur that eventually form the correct tertiary structure.            
  36. 1)      What mechanisms recognize hydrophobic patches on proteins and minimize the damage they cause? 
    a.       Two depend on molecular chaperones, which bind to the patch and try to repair the defective protein by giving it another chance to fold. They also prevent protein aggregation by covering the patches. When refolding fails, a third mechanism is called into play that completely destroys the protein by proteolysis. The proteolytic pathway begins with the recognition of an abnormal hydrophobic patch on a protein’s surface, and it ends with the delivery of the entire protein to a protein destruction machine, a proteasome. 
  37. 1)      How does a proteasome do its work? 
    • a.       The proteasome cap recognizes a substrate protein and subsequently translocates it into the proteasome core, where it is digested. At an early stage, the ubiquitin is cleaved from the substrate protein and is recycled. Translocation into the core of the proteasome is mediated by a ring of ATP-dependent proteins that unfold the substrate protein as it is threaded through the ring and into the proteasome core. The trheading reaction, driven by ATP hydrolysis, unfolds the targe proteins as they move through the cap, exposing them to the proteases lining the proteasome core. This keeps the entire substrate bound until all of it is converted into short peptides.
    • b.      The 19S cap also acts as regulated “gates” at the entrances to the inner proteolytic chamber, and they are responsible for bindin a targeted protein substrate to the proteasome. With few exceptions, the proteasomes act on proteins that have been specifically marked for destruction by the covalent attachment of a recognition tag formed from a small protein called ubiquitin. 
  38. 1)      How are denatured or misfolded proteins recognized? 
    a.       They are recognized and destroyed because abnormal proteins tend to present on their surface amino acid seuqnces or motifs that are recognized as degradation signals by a set of E3 molecules in the ubiquitin-proteasome system; these seuqences must be buried and inaccessible to normal counterparts of these proteins. However, a proteolytic pathway that recognizes and destroys abnormal proteins must be able to distinguish between completed proteins that have wrong conformations and the many growing polypeptides on ribosomes that are not normally folded. 
  39. 1)      Aside from tagging for elimination certain proteins, what is another mechanism of control ?
    • a.       Some proteins have short lifetimes and conditioned this way. One generation of ubiquitin lingases is turned on either by E3 phosphorylation or allosteric transition in an E3 protein caused by its binding to a specific small or large molecule
    • b.      Alternatively, in response either to intracellular signals or to signals from the environment, a degradation signal can be created in a protein, causing its rapid ubiquitylation and destruction by the proteasome. One common way to create a signal is to phosphorylate a specific site on a protein that unmasks a normally hidden degradation signal. Another way to unmask such a signal is by regulated dissociation of a protein subunit. Finally, powerful degradation signals can be created by cleaving a single peptide bond, provided that this cleavage creates a new N-terminus that is recognized by a specific E3 as a “destabilizing“ N-terminal residue. 
  40. 1)      How does the N-terminal type of degradation signal arise? 
    a.       It arises because of the “N-end rule,” which relates the lifetime of a protein in vivo to the identity of its N-terminal residue. These destabilizing N-terminal residues are recognized by a special ubiquitin ligase that is conserved from yeast to humans. 
  41. 1)      What happens to methionine? 
    a.       Special proteases, called methionine aminopeptidases, willremove the first methionine of a nascent protein, but will do so only if the second residue is also stabilizing according to the N-end rule. These substrates are formed by site-specific proteases. 
  42. 1)      How do normal humans get disease? 
    a.       The gradual decline of the cell’s protein quality controls permits normal proteins to form aggregates, which are released from dead cells and acculmulate in the ECM that surrounds the cells in a tissue, and even, damage tissues. This occurs especially in the nerve cells; and protein aggregates cause neurodegenerative diseases. 
  43. 1)      How does a protein aggregate sustatin itself? 
    a.       It must be resistant to proteolysis inside and outside of the cell. They form fibrils built from a series of polypeptide chains that are layered one over the other as a continuous stack of Beta sheets, called a cros-beta filament, a structure resistant to proteolysis that causes staining deposits called amyloids. This causes prion diseases. 
  44. 1)      What is significant about prion diseases? 
    a.       They can spread among organisms through consumption especially. A set of diseases in humans is caused by a misfolded aggregated form of PrP. Normal PrP is normally located on the outer surface of the plasma membrane. When converted, it forms protease-resistant, cross-beta fiaments and induces others to convert as well. This spreads rapidly from cell to cell in the brain, causing death in animals and humans. 
  45. 1)      What is a special feature of prions? 
    • a.       They form different types of aggregates from the same polypeptide chain. Each type can be infectious, forcing others to adopt the same abnormal structure.
    • b.      A positive role is used by fungi to establish different types of cells. 

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