BMB 462 Lecture 34 Outline of Last Lecture I. Overview of TranslationII. Determining the Genetic CodeIII. Specifics of the tRNA codeIV. tRNA pairing with mRNAV. Charging tRNAs with amino acidsOutline of Current Lecture I. Ensuring Specificity of tRNA ChargingII. Proofreading by aminoacyl synthetasesIII. Ribosome structure and assemblyIV. The ribosome as a ribozymeV. Initiation of translationVI. Initiation complex in EukaryotesVII. The 1st step of elongationCurrent LectureConcepts to remembers from previous courses/lectures:-I. Ensuring Specificity of tRNA Charginga. Accuracy in charging a tRNA is crucial for the fidelity of translation.i. If the wrong amino acid is charge to a tRNA, than the incorrect amino acidwill be incorporated into the polypeptide.b. There are 3 mechanisms to insure the correct amino acid is added to its tRNA.i. Enzyme-substrate interactions1. There are 20 different amino acid synthetases, 1 for each amino acid. The substrate for the synthetase is the tRNA. The tRNA provides context for the synthetase by having certain recognition points in the RNA sequence. The first filter, then, in synthetase recognition involves certain nucleotides in the RNA sequence.These notes represent a detailed interpretation of the professor’s lecture. GradeBuddy is best used as a supplement to your own notes, not as a substitute.a. Some recognition points, marked in orange in Figure 27-21,interact with one aminoacyl. Other sites, marked in green, can interact with one or more aminoacyl groups (thus making them more important in the process).b. Invariant nucleotides, marked in blue, are the same in all tRNA molecules and therefor cannot be used for discrimination by the synthetases. They still make contact though.c. The anticodon can also make contact with the synthetases.2. The second recognition filter is enzyme recognition of the amino acid itself.a. This can be problematic because some amino acids actually look quite similar, i.e. valine and isoleucine - the only main difference between the two is size (isoleucine has one more carboxyl group).II. Proofreading by aminoacyl synthetasesa. To recognize the amino acid itself, the synthetase proofreads the aminoacyl-AMP.Once AMP activates the aminoacyl, if it gets bound by the wrong enzyme (i.e. a Val-AMP is bound by an Ile-tRNA-synthetase), the synthetase checks the amino acid in the secondary active site. When it recognizes that it has bound the wrong aminoacyl-AMP, it actually hydrolyzes the bond and releases individual Val and AMP molecules.b. The third filter occurs if the aminoacyl binds to the incorrect enzyme and escapesthe proofreading of the secondary active site, and is bond to the wrong tRNA.i. The bond between the tRNA and the incorrect aminoacyl can still be recognize by a third proofreading step and be hydrolyzed.c. The 3 steps of proofreading allow the overall error rate to be reduced to 1 error per 104 amino acids incorporated. This strongly improves the fidelity of translation.i. This error rate is approximately the same error rate that DNA makes without its own proofreading mechanisms (which increase DNA's fidelity to about 1 error in every 109 nucleotides added).1. DNA replication has a lower error rate due to the fact that it must house the genetic information of the cell, whereas proteins are quickly turned over.III. Ribosome structure and assemblya. In both bacterial and eukaryotic ribosomes, there is a small subunit and a large subunit.i. The measure for subunit size is 'S', which stands for Svedberg unit or Sedimentation unit.ii. In bacteria, the large subunit is 50S and the small subunit is 30S; it's 60S and 40S for eukaryotes.1. Together, the whole ribosome in bacteria adds up to 70S, and in eukaryotes it is 80S. These values are smaller than an exact sum ofthe two subunits. This is because sedimentation (separation via centrifugation) depends on the weight and shape of your component.a. Something that is bulky and spread out may sediment at a slower during centrifugation than a molecule that is more compact or globular. Thus compacting the two subunits together allows them to sediment faster than would be expected by adding up the sedimentation units of the 2 separate subunits.b. In E. coli, there are approximately 15,000 ribosomes in a single cell, accounting for 1/4 of the cell's mass.i. 65% of a single ribosome is RNA and the remaining 35% is composed of proteins.1. For example, in a bacterial ribosome, there are only 2 rRNAs (the 5S and the 23S), but they are quite long, while the 36 proteins that make up the rest of the subunit are quite small. Thus, by mass, the ribosomal subunit is composed mostly of RNA.ii. The particle size of a ribosome is about 18nm in diameter. In comparison to one turn of DNA, which is 3.4nm, the ribosome roughly corresponds to5 turns of DNA.IV. The ribosome as a ribozymea. Experimentation with Puromycin, a drug inhibitor of translation, showed that theactive site is only formed by rRNA, making the ribosome a ribozyme; there is no protein found near the active site.i. Visualization of puromycin bound to the ribosome shows it completely surrounded by rRNA, with no proteins in proximity.V. Initiation of translationa. Translation, much like replication or transcription, has 3 phases: Initiation, Elongation, and Termination.i. Initiation1. There is a specific tRNA responsible for recognizing the first AUG codon. In bacteria, that is a formylated methionine. The N-Formylmethionine is charged onto a tRNA that is specific for it.a. Other AUGs read further downstream in the sequence are recognized by a different tRNA.2. Eukaryotes also have a special tRNA for the first methionine; it isn't formylated, but has a more complicate modification differentiating it from other Mets.b. Initiation begins with formation of the initiation complex in bacteria. This processstarts with the smaller 30S subunit and 2 initiation factors, IF-1 and IF-3.i. IF-1 prevents tRNAs from prematurely binding to the A-site by binding to and blocking the A-site.ii. IF-3 prevents premature assembly of the small and large ribosomal subunits.c. There is a Shine-Delgarno (SD) sequence just upstream of the first AUG that directs where the 30S subunit binds prior to initiation of translation.i. At this point, the 30S subunit is bound to the mRNA, IF-1, and IF-3.d. The second step involves binding of another initiation factor, IF-2, and the first tRNA comes in to bind the AUG.i. These two components are bound to
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