BMB 251 1st Edition Lecture 19 Outline of Last Lecture I. ClickersII. Central dogmaIII. RNAIV. Transcriptiona. RNA polymeraseb. mRNAc. rRNAd. tRNAe. snRNAf. RNA holoenzymeg. RNA core enzymeh. Sigma factorV. Promoters/terminatorsOutline of Current Lecture VI. ClickersVII. RNA polymerasesVIII. General transcription factorsa. TFIIDb. TFIIBc. TFIIHIX. Transcriptional activator X. MediatorsXI. RNA modificationsa. 5’ end cappingb. 3’ end polyadenylationXII. RNA splicing XIII. snRNA/snRNPa. SpliceosomeCurrent Lecture- Clicker Question 1: Unlike DNA polymerase, RNA transcription with RNA polymerase does not need:o Helicase, ligase, primers or SSB proteins (the answer was “all of the above”)- Whereas prokaryotes have one type of RNA polymerase, eukaryotic nuclei have three: RNA polymerase (I), (II), and (III)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.o Structurally similar, share common subunits, but they transcribe different types of geneso RNA polymerase (I) and (III) transcribe genes encoding tRNA, rRNA, and various other small RNAso RNA polymerase (II) transcribes most genes, such as ones encoding proteins- Whereas prokaryotic RNA polymerase only require sigma factor to begin transcription, eukaryotes require additional proteins called general transcription factors (GTF)- GFT: help position eukaryotic RNA polymerase correctly at promoters, help pull DNA double helixapart and release RNA polymerase from promoter when transitioning into elongation modeo **Designated as TFII (for transcription factor for polymerase II) and are needed at almostevery promoter of genes in which RNA polymerase (II) transcribeso TFIID: binds to TATA box (subunit TBP –for TATA binding protein- recognizes it), which is located about 25 nucleotides upstream from transcription start site. This signals the startof the active promoter to be detected within an entire genome. Other GTF’s and polymerase (II) then assemble and form the transcription initiation complex TBP bends the DNA and forms two kinks in the double helix, which serve as a landmark within the genome that the promoter is active and ready for the binding of RNA polymerase (II)o TFIIB: recognizes the BRE element in DNA (located right before TATA box) and accurately positions RNA polymerase (II) at the transcription start siteo TFIIH: unwinds DNA so RNA polymerase (II) can gain access to template strand - Clicker Question 2: Which eukaryotic promoter sequence element is most analogous to sigma factor binding site in prokaryotic genes?o TATA box- basically unchanging and very conservedo BRE recognition element in eukaryotes differs greatly from the -35 sequence seen in prokaryoteso The initiator (Inr) and downstream promoter sequence (DPE) are unique and specific to eukaryotes- RNA polymerase (II) then disassociates from promoter and enters elongation phase- Transcriptional activators: gene regulatory proteins which bind to specific sequences in DNA (sometimes several thousands of nucleotide pairs away from promoter) and help RNA polymerase, GTF’s and mediator to assemble at promotero No promoter in eukaryotes alone can start/activate transcription, which is why an activator is needed - Mediator: required in transcription initiation, which allows activator proteins to communicate properly with RNA polymerase and GTF’s - Chromatin-modifying proteins: allow better access to DNA in chromatin for RNA polymerase- DNA supercoiling: conformation DNA adopts in response to superhelical tension - Unlike in prokaryotes, there are many more modifications and splicing to be made to RNA beforetranslation can occur- Both ends of mRNA are modified by capping on the 5’ end and polyadenylation on the 3’ end  allows cell to assess that both ends are present and message is intacto After the first 25 nucleotides are transcribed, cap consisting of guanine nucleotide is added to 5’ end of RNA 3 unique steps to create cap: Phosphatase removes PO4 from 5’ end, GMP molecule is added and methyl group is added to the guanosine o 3’ end has anywhere from 150-250 adenines added to it Without the A’s, mRNA would be unstable  The amount of time mRNA stays in the cytoplasm is proportional to length of added fuse because it goes away with time and mRNA needs it to continue to “survive” in the cytoplasm- Clicker Question 3: Where are protein coding sequences foundo Exonso **In order to reach the ribosome, sequence has to leave the nucleus and only coding sequences (exons) can do this- RNA splicing removes noncoding intron sequences, allowing only the coding exon sequences to exit the nucleuso As soon as the 5’ side is synthesized, enzymes attach and get it ready to be spliced; as soon as the 3’ end is finished being synthesized, these factors bind and immediately begin splicing (as the rest of RNA is being transcribed); this occurs in the following four steps:o 2’ –OH in adenine nucleotide within intron attacks 5’ splice site and cuts the sugar-PO4 backbone of RNA, leaving a free 3’ –OH end on the exono 5’ end of the cut intron loops back and covalently bonds with the adenine baseo Released 3’ –OH end of exon reaction with start of next exon, releasing the entire intron in the form of a lariato Released intron sequence is degraded in the nucleus- Clicker Question 4: Why can’t DNA undergo splicing?o It has no 2’ –OH groups- snRNA: involved in major form of pre-mRNA splicing and are complexed with seven protein subunits to form snRNP (small nuclear ribonucleoprotein which form the core of the spliceosome; spliceosome: large assembly of RNA and protein molecules to carry out pre-mRNA splicing)- Exon junction complex (EJC): site marked by proteins where successful splicing event