BMB 251 1st Edition Lecture 20 Outline of Last Lecture I. ClickersII. RNA polymerasesIII. General transcription factorsa. TFIIDb. TFIIBc. TFIIHIV. Transcriptional activator V. MediatorsVI. RNA modificationsa. 5’ end cappingb. 3’ end polyadenylationVII. RNA splicing VIII. snRNA/snRNPa. SpliceosomeOutline of Current Lecture IX. ClickersX. Splicing and Polyadenylation via C-terminal tailXI. RNA polymerase I, II, and IIIXII. Nucleolus XIII. TranslationXIV. Genetic codeCurrent Lecture- Clicker Question 1: You are studying a mutation in a certain gene. You sequence the geneand find a mutation in the splice site at the junction of exon 1 and intron 1. What do you predict you will find when you sequence the mutant mRNAs?o Intron 1 will be retained in the mRNA- Clicker Question 2: How is rRNA cleavage different from mRNA splicing?o The cleaved fragments are not re-attached covalently They both occur in the nucleus They do require protein complexes, but they are different for rRNA and mRNA. These reactions do not occur spontaneously, they need enzymes to put substances on the rRNA and recognize sequences for cleavageThese 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.- C-terminal tail of RNA polymerase allows capping and splicing components to transfer directly onto RNA as soon as it emerges from the enzymeo 5’ end of pre-mRNA is almost immediately capped and spliceosome assembles on RNA to splice as it’s transcripto Similar mechanism allows polyadenylation on 3’ end of gene Position of 3’ end is determined by signal encoded in the genome signals are transcribed into RNA recognized by series of RNA-binding proteins and RNA-processing enzymes Two of the more important binding proteins: CstF (cleavage stimulation factor) and CPSF (cleavage and polyadenylation specificity factor) proteinstravel with RNA polymerase tail and are transferred to 3’ end processing sequence Once they are bound, additional proteins assemble to create the 3’ end of mRNA and poly-A-binding proteins create the poly-A tail on the 3’ end **The cleavage and poly-A signals encoded in the DNA shows where the end of RNA will be - Mature mRNAs are coated with accessory proteins required for exportation out of the nucleuso Nuclear pores will only open up to let specific sequences in/out of nucleuso mRNAs are ALWAYS circularized conformation is needed to recruit a ribosome o The proteins added to the mRNA help bind the cap with the poly-A tail, which is needed for protein synthesis- There are three specialized RNA polymerases in eukaryoteso RNA polymerase I: transcribes 5.8S, 18S, and 28S rRNA geneso RNA polymerase II: transcribes all protein-coding genes, snoRNA genes, miRNA genes, siRNA genes and most snRNA geneso RNA polymerase III: transcribes tRNA genes, 5S rRNA genes some snRNA genes and genes for other small RNAs- The proteins in ribosomes are strictly scaffolding in building ribosomal structure. All catalysis occurs from RNA side chains rather than protein side chains (RNA makes up 2/3 of ribosome)- Nucleolus: site where the proteins that are synthesized in cytosol are imported back intothe nucleus and bind with rRNA to create the small and large subunits of ribosome these HUGE ribosomal structures are then exported back out of the nucleus into the cytosol to carry out more protein synthesis- DNA and RNA are chemically similar DNA serves as a direct template of RNA synthesisvia complementary base pairing - On the other hand, RNA and proteins are chemically different. Translation of informationmust occur to create proteins of RNA translated through rules known as the genetic code because there is no direct one-to-one correspondence between nucleotides and amino acids- mRNA is read in consecutive groups of three, called codons, and there are 64 different combinations of these pairings of nucleotides can be (4x4x4 = 64)- 20 amino acids, 64 possibilities of codons genetic code is redundant o Only Met and Trp amino acids have one codon coding for them; the rest have at least two codons per amino acido Having some redundancy allows the system to repel some mutations. Having 6 codons for one amino acid (in some cases) allows room for silent mutations to occur, which changes the mRNA but not eh amino acid coded for - RNA sequence could technically be read in any one of three reading frames, depending on the starting position of decoding processo This would give three completely different polypeptideso Only one is correct, though, and synthesizes the desired proteino Ribosome and RNA polymerase must start in the correct reading frame or else complete garbage will be
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