DOC PREVIEW
IUB BIOL-L 211 - Exam 3 Study Guide
Type Study Guide
Pages 11

This preview shows page 1-2-3-4 out of 11 pages.

Save
View full document
View full document
Premium Document
Do you want full access? Go Premium and unlock all 11 pages.
Access to all documents
Download any document
Ad free experience
View full document
Premium Document
Do you want full access? Go Premium and unlock all 11 pages.
Access to all documents
Download any document
Ad free experience
View full document
Premium Document
Do you want full access? Go Premium and unlock all 11 pages.
Access to all documents
Download any document
Ad free experience
View full document
Premium Document
Do you want full access? Go Premium and unlock all 11 pages.
Access to all documents
Download any document
Ad free experience
Premium Document
Do you want full access? Go Premium and unlock all 11 pages.
Access to all documents
Download any document
Ad free experience

Unformatted text preview:

BIOM 211 Exam # 3 Study Guide Lectures: 20 - 28Lecture 20 (October 20)Transcription in EukaryotesDescribe the major differences between transcription in eukaryotes and transcription in prokaryotes.Eukaryotes lack operons (which are only found in prokaryotes). DNA is compacted by histones into nucleosomes, which makes transcription more complicated in eukaryotes than prokaryotes.Additionally, eukaryotic transcription occurs in the nucleus, while prokaryotes lack a nucleus altogether. Transcription and translation take place in different areas of the cell (nucleus vs. outside the nucleus). Again, prokaryotes lack the nucleus so everything occurs in the same area of the cell. RNA polymerase II is the dominant RNA polymerase for eukaryotic transcription. There are neither sigma factors nor -35/-10 regions in eukaryotes. Multiple transcription factors are necessary for eukaryotic transcription. Finally, processing of mRNA requires the addition of the poly-A tail, splicing is required to remove introns (which are not present in prokaryotic mRNA), and 5' capping of the transcript.What is the difference between exons and introns?Exons are coding regions of DNA in eukaryotes, and introns are non-coding regions of DNA. What is the promoter?The promoter is the location where general transcription factors bind and recruit RNA polymerase.What are regulatory sequences and where are they located?Regulatory sequences are variable sequences that bind regulatory proteins. They can be locatedupstream or downstream of a promoter, and they are generally tens to hundreds of kilobases long. They control whether a gene is turned on or off.What are the key elements of the core promoter and what do they do?The TATA box, BRE, and Inr are they key elements of the core promoter, which are required in order for RNAP to initiate transcription. The TATA box binds specific transcription factors. BRE is the recognition element for the transcription factor TFIIB. Inr is the initiator element, sandwiched at the transcriptional start site.What are the differences between the five main types of RNA polymerases?Pol I transcribes specialized RNA-encoding genesPol II transcribes nearly all protein encoding genes, has multiple subunits, is shaped like a claw, and catalyzes RNA synthesis (most important for this unit)Pol III works in conjunction with Pol I and will be discussed in the futurePol IV and V are present in plants and not relevant to this courseWhat are the three general steps of transcription initiation in eukaryotes?1. Binding of RNA Pol II to promoter in closed complex2. Open complex forms (about 13 base pairs long)3. Initial transcription and promoter escapeHow is the preinitiation complex assembled?Pol II binds with six general transcription factors and additional proteins at the promoter. First, TFIID binds the TATA box, aided by TFIIA. TFIIB binds BRE. TFIIF stabilizes protein-protein interactions, and is recruited to the Inr element by the action of TFIID, TFIIA, and TFIIB. TFIIE joins the complex and recruits TFIIH. TFIIH uses the energy from ATP hydrolysis to mediate promoter melting.*Note that "TF" stands for "Transcription Factor"Additionally, the mediator complex is a component of the preinitiation complex that mediates interactions between Pol II and activator proteins as well as Pol II and general transcription factors. Activators are also present and bind enhancers upstream of the promoter to help recruit RNA Pol II to the promoter (similar to prokaryotes).Describe the two steps of promoter escape.Step 1: The promoter meltsTFIID holds the upstream promoter in place. TFIIH binds downstream of RNAP and uses ATP hydrolysis to feed dsDNA into polymerase and separate the strands. Abortive initiation thenoccurs prior to step 2.Step 2: Pol II is PhosphorylatedPol II has a tail (carboxyl-terminated domain: CTD). The tail comes to the promoter in an unphosphorylated state. TFIIH phosphorylates the tail in order to help Pol II escape to elongation.What are elongation factors?Elongation factors stimulate transcript elongation. Some help displace initiation factors, while others stimulate Pol II proofreading activity.What challenge to histones present to eukaryotic elongation?Histones are physical roadblocks in the form of nucleosomes that inhibit transcription.Lecture 21 (October 22)Transcription in Eukaryotes 1.5What does FACT stand for, and what does it do?FACT stands for "Facilitates Chromatin Transcription), and it allows elongation to proceed despite chromatinWhat is the main heterodimer component of FACT, what does each component do, and why is itcalled a heterodimer?The main heterodimer component of FACT is composed of Spt16 and SSRP1. Spt16 binds to the H2A and H2B dimer and removes one dimer ahead of Poll II to allow elongation to continue. Afterward, the dimer is replaced. Meanwhile, SSRPI binds to the H3 and H4 tetramer. A heterodimer simply means that a dimer (two piece complex) consists of two different (hetero) componentsWhat is the purpose of SPT6 and what is its relation to the FACT heterodimer?SPT6 helps to reassemble the nucleosome, but it is not part of the FACT complex.What is the purpose of phosphorylated carboxyl-terminated domain (CTD)?The CTD tail recruits RNA processing factors to Pol I.Describe the formation of the 5' cap on the RNA transcript.The 5' cap contains a methylated guanine on the 5' end of the RNA strand. To form the cap, RNAtriphosphatase first removes a gamma phosphate from the 5' end of RNA. The beta and gammaphosphates are removed from guanine. Guanylyltransferase transfers a guanine (makes sense- "guanylyl"- guanine; "transferase"- transfers) to the 5' beta phosphate of the RNA strand (note: it transfers to the beta because the gamma has been removed). Then, methyltranferase (you guessed it- transfers methyl groups) adds methyl groups to the guanine.Why is the 5' cap so important?The 5' cap on RNA prevents degradation. Additionally, the methyl group later serves as a signal to the ribosome to bind for translation.Describe the process of polyadenylation of the 3' end of the RNA transcript.The CTD tail recruits the enzymes necessary for polyadenylation. After the Poly-A tail is transcribed, those enzymes are transferred to the RNA strand. The Poly-A Signal is the transcribed trigger that sends the polyadenylation enzymes over to RNA. CstF (Cleavage stimulation Factor) stimulates cleavage. After mRNA cleavage, the shorter piece is degraded. CPSF (Cleavage and


View Full Document

IUB BIOL-L 211 - Exam 3 Study Guide

Type: Study Guide
Pages: 11
Documents in this Course
Load more
Download Exam 3 Study Guide
Our administrator received your request to download this document. We will send you the file to your email shortly.
Loading Unlocking...
Login

Join to view Exam 3 Study Guide and access 3M+ class-specific study document.

or
We will never post anything without your permission.
Don't have an account?
Sign Up

Join to view Exam 3 Study Guide 2 2 and access 3M+ class-specific study document.

or

By creating an account you agree to our Privacy Policy and Terms Of Use

Already a member?