BMB 462 Lecture 34 Outline of Last Lecture I Overview of Translation II Determining the Genetic Code III Specifics of the tRNA code IV tRNA pairing with mRNA V Charging tRNAs with amino acids Outline of Current Lecture I II III IV V VI VII Ensuring Specificity of tRNA Charging Proofreading by aminoacyl synthetases Ribosome structure and assembly The ribosome as a ribozyme Initiation of translation Initiation complex in Eukaryotes The 1st step of elongation Current Lecture Concepts to remembers from previous courses lectures I Ensuring Specificity of tRNA Charging a 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 acid will be incorporated into the polypeptide b There are 3 mechanisms to insure the correct amino acid is added to its tRNA i Enzyme substrate interactions 1 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 synthetases a 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 escapes the 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 assembly a 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 of the 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 to 5 turns of DNA IV The ribosome as a ribozyme a Experimentation with Puromycin a drug inhibitor of translation showed that the active 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 translation a Translation much like replication or transcription has 3 phases Initiation Elongation and Termination i Initiation 1 There is a specific tRNA responsible for recognizing the first AUG codon In bacteria that is a formylated methionine The NFormylmethionine 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 process starts 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 the P site of the 30S ribosomal subunit since IF 1 is blocking the A site 1 IF 2 is a GTPase which will eventually hydrolyze GTP
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