BCHM 3050 1st Edition Lecture 23 Outline of Last Lecture I Transcription in Eukaryotes II Eukaryotic RNA Polymerases III Eukaryotic Promoters IV Initiation of Eukaryotic Transcription V Enhancers VI RNA Processing VII 5 Cap of Eukaryotic mRNA VIII mRNAPolyadenylation IX mRNA Splicing Outline of Current Lecture I Introduction to Translation II Cracking of the Genetic Code III The Complete mRNA Genetic Code IV Codons vs Anticodons V The Wobble Hypothesis VI Formation of Aminoacyl tRNAs VII Requirements for Protein Synthesis VIII Ribosome Structure IX Formation of the Prokaryotic Initiation Complex X Peptide Elongation Current Lecture I Introduction to Translation 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 HIV is a virus that has genetic material it has RNA at first and once it gets into a host it makes the RNA into DNA and it integrates this DNA into the genome of the host b Translation the coded mRNA dependent synthesis of protein or the use of the genetic directions encoded in mRNA to synthesize proteins II Cracking of the Genetic Code a Bacterial extract contains all components co factors required for protein synthesis including tRNA s ribosomes amino acids ATP GTP except mRNA b Similar experiments were performed with other mRNA s e g poly A s polylysine poly C s poly proline other triplet combinations gave other peptides c Eventually all possible combinations of triplet tri nucleotide codons were tested and assigned their corresponding amino acid d Several scientists came to this same conclusion but Nirenberg and Matthaei won the race e Everything needed for translation in is the bacterial extract except mRNA which has to be made new everytime f The key difference between DNA polymerase and RNA polymerase is that RNA polymerase does not need a primer DNA polymerase requires a primer in order to start extending a DNA chain g 1st observation there is a code the nitrogenous bases code for specific amino acids h Then when they put 2 different alternating bases in a sequence they got 2 different amino acids and 2 different amino acid sequence combinations i 2 different base sequences UCU CUC UCU CUC CUC UCU CUC UCU j They determined that there is probably more than one combination that codes for a specific amino acid k Then they put 3 nucleotides together and they were able to figure out that 3 bases 1 amino acid III The Complete mRNA Genetic Code a The start AUG in eukaryotes codes for methionine in prokaryotes it codes for N formyl methionine b Note All other AUG codons elswhere in the transcript always code for methionine c Some amino acids are coded for by more than 1 triplet d Leucine is the most abundant amino acid in the body so it has the most amount of different triplet codes e Methionine AUG is the start codon and must be present in order for amino acids to be coded in the body start translating where there is an AUG and ignore everything that comes before this start codon f The stop codons end translation g The start and stop codons are where translation starts and where translation stops h Promoter and terminator are present for transcription and they start and stop transcription i Each amino acid has about 4 codons that code for it the third position of the codons varies while the first and second positions are the same between the 4 different codons the 3rd nitrogenous base is called a degenerative base j There is only 1 codon that codes for tryptophan so it is likely the least abundant amino acid k All of the codons are read 5 3 l IV The more codons that code for a specific amino acid the more abundant that amino acid is Codons vs Anticodons a Make sure you read the tRNAs 5 3 when trying to code for an amino acid b 5 CAU3 anti codon c 3 GUA5 mRNA codon d So the amino acid you code for is methionine AUG e 64 total codons and 20 total amino acids f V In order to find the total number of possible base combinations of an organism you must take the total number of bases and raise it to the power of the number of bases that have to be combined to make an amino acid so if you have 4 nucleotide bases and 3 together make a codon than 43 64 which gives you the total number of possible base combinations or codons The Wobble Hypothesis a wobble rules of base pairing will allow a minimum of 31 codons required to provide code combinations for the amino acids 32 codons including a stop VI Formation of Aminoacyl tRNAs a tRNA Charging i e synthesis of aminoacyl tRNA s is essentially irreversible because of the hydrolysis of pyrophosphate b There are 20 special enzymes that are specialized for the 20 amino acids VII Requirements for Protein Synthesis a mRNA transcript b Large and small ribosome subunits c Assorted initation and elongation factors d Assorted supply of aminoacyl tRNas e GTP energy i Need energy both ATP and GTP ii GTP serves are the primary energy source rather than ATP f Three Stages i Initiation ii Elongation iii Termination VIII Ribosome Structure a There are 3 main spots in the ribosome b Which codon is sitting in which spot tells you how translation will proceed c 2 tRNAs take up A and P sites and the two tRNAs transfer amino acids to form a peptide d The acceptor site accepts new tRNA all of the time e The 1st codon amino acid sits at the P Site f IX The 2nd and 3rd codons amino acids sit at the A Site the A Site has 2 amino acids Formation of the Prokaryotic Initiation Complex a Shine Delgarno sequence is a purine rich mRNA sequence just upstream from the AUG start codon b This is facilitated by base pairing with the 16s rRNA c This insures that protein synthesis starts at the true start codon and not at some internal methionine codon d Shine Dalgarno sequence is a purine rich mRNA sequence just upstream from the AUG start codon e This is facilitated by base pairing with the 16s rRNA f X This insures that protein synthesis starts at the true start codon and not at some internal methionine codon Peptide Elongation a EF Tu is a GTP binding Elongation Factor protein It carries aminoacyl tRNA s to the ribosome and positions them in the empty A site b GTP is hydrolyzed causing release of EF Tu GDP from the ribosome c GDP of EF Tu is replaced by GTP d EF Tu GTP picks up another aminoacyl tRNA and cycle repeats e Tu stands for temperature unstable Later Ts stands for temperature stable
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