Exam III Study Guide
Exam III Study Guide CELL 2050
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Joseph Merritt Ramsey
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This 56 page Study Guide was uploaded by Joseph Merritt Ramsey on Monday April 11, 2016. The Study Guide belongs to CELL 2050 at Tulane University taught by Dr. Meenakshi Vijayaraghavan in Winter 2016. Since its upload, it has received 70 views. For similar materials see Genetics in Science at Tulane University.
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Date Created: 04/11/16
March 16, 2016 Chapter 11: DNA Replication Chargaff’s Rule is Important to Consider o Complementarity is what allows Replication to occur o So replication depends on the CG and AT matching Discovery of the Process o There Were Three Old Models to Explain DNA Replication Copies o Cultured E. coli, specifically in a heavy Nitrogen medium ( N) Heavy Isotope, swapped the Medium to Light Isotope ( N) 14 Bacterial Replication o Bacteria have circular DNA with a single origin of replication o Components of oriC (About 100200bp long) AT Rich Regions DnaA Box Regions Roughly five spots present GATC Methylation Site The Overall Process o 1. Prepping A. Methylation must take place to loosen up the strands Occurs on Adenines This is the preliminary requirement This is the first thing that is looked for, that Adenines on both strands (Fully Methylated) are Methylated In the GATC Methylation Sites B. Then DnaABox/ATP Complex Proteins bind to the DnaABox Sequences This continues to add strain and opens up the DNA Each Box is 9 Nucleotides Long 20 Proteins Recruited ATP Complex gives the proteins and affinity for the DnaABox Region This binding causes strain in the ATRich region trying to wrap around the complex, opening the strand HU (Histone Like) and HIF (Histone Like Integration Factor) also help binding and recruitment C. ATRich Regions About 13bp Long There are only two Hydrogen Bonds present, so they are more susceptible to separation Very close to DnaABox Sequence This is the First Event, the break occurs in the AT Rich Region D. DnaC protein recruits DnaB Protein (Helicase), breaking the hydrogen bonds ATP dependent action This continues the action started by the DnaABox Proteins One Helicase binds to one strand, another binds to the other o Forms the Replication Forks Requires ATP Hydrolysis for energy E. The Strands need to remain unwound, so Single Strand Binding Protein comes in to stabilize the single strands This way they can act as a template to make daughter strands F. Gyrase (Topoisomerase II) acts as a sort of pathpaver ahead to loosen Positively Supercoiled regions o 2. Synthesizing A. DNA Primase This is the First Protein that comes in in the Synthesis process Like RNA Polymerase Makes 1012bp long RNA primer, so DNA Polymerase does not immediately and directly act o DNA Polymerase does not have the innate capacity to initiate replication B. DNA Polymerase II, IV, and V have repair functions I and III have normal replication functions DNA Polymerase III acts first and works with the Continuous Leading Strand first followed by the Lagging Strand with the creation of Okizaki Fragments But it Never Starts the Synthesis Polymerase I Polymerase III Single Polypeptide Hollow Enzyme, So Contains Several Subunits (10 Removes RNA Total) Primers (made by These subunits give the functionality, and the others DNA Primase) and merely aid these functioning Substitutes them o ▯– catalysis o ? – clamp with DNA Nucleotides o ? – clamp loader o ? – exonuclease Each Subunit has a Specific Function o ▯– Catalyzes Phosphodiester Bonds o ? – Allows Polymerase III to hold onto the template; vital subunit o ? – Loads onto the clamp; important in lagging strand o ? – exonuclease activity to fix incorrect nucleotide addition Does the most work replicating the DNA o 3. Polymerase Action It can never start synthesis Two Drawbacks 1. Cannot Initiate Synthesis – DNA Primase Helps 2. Can Only Add Nucleotides in 5’ to 3’ Direction – need the 3C Hydroxyl Group o Catalyzes Diester Bond Formation in 5’ of incoming nucleotide and 3’ on existing The above Polyermases act in very specific manners to create DNA copies A. Polymerase III o Continues to add Nucleotides to the DNA Primer made from DNA Primase o Leading Strand moves towards the replication fork o On the Lagging side, sections Loop and Unloop B. Polymerase I Looping in the Lagging Looping occurs with 1,0002,000bp Fragments in Prokaryotes, 100200 in Eukaryotes o The created fragments are called Okizaki Fragments Looping causes it to look like Synthesis is moving towards replication fork o But after it is unwound synthesis is away from fork Primase Initiates the Looping Process itself The looping causes a small delay Reaction of DNA Polymerase Prepping Actor Acts On 1 Methylation Adenine ) Must occur in order for replication to start This is the preliminary requirement GATC Region 2 DnaABox Proteins DnaABox Sequences ) Form a Protein/ATP Complex HU and IHF also help DnaA Bind 56 Box Regions Present Each Box 9bp Induces Separation DnaABox Proteins create friction on the DNA Strand After the separation occurs, DNAATP complex becomes DNAADP Complex Protein Complex formed of about 20 Proteins These regions are next to ATRich regions which are susceptible to unwinding under stress DnaC DnaBHelicase 3 ) Helicase is recruited to continue opening up the strands This is a 6Protein Subunit Creates a replication fork ATP dependent Process 4 Single Strand Binding Proteins Unwound DNA Strands ) Binding needs to occur to keep the strands separated Small Proteins that Specifically Bind to Single Stranded DNA 5 Gyrase/Topoisomerase II ) Continues to loosen the DNA ahead as well Synthesizing 1 DNA Primase ) Has the same function as RNA Polymerase Makes a 1012bp Long strand of RNA Primer This initiates the Synthesis, not Polymerase III Primase also acts to initiate the Looping Process 2 DNA Polymerase III Leading ) Leading Strand Starts adding on at the open 3’ Locations Can go continuously Follows the primer Strands 3 DNA Polymerase III Lagging ) Lagging Strand Has to add through a Looping Process because addition cannot occur in the 5’ location After the Loop is formed and coded, it flips back These loops are about 100200bp Long in Eukaryotes, 10002000 In Prokaryotes Polymerase III Action Why Towards the 5’ End? The 3’ end is the available end and is the only one that can be added onto, so it needs to be the open side Restrictions of DNA Polymerase 1. Cannot Initiate Synthesis 2. Can Only Add in 5’3’ March 18, 2016 Chapter 11: DNA Replication Chargaff’s Rule is Important to Consider o Complementarity is what allows Replication to occur o So replication depends on the CG and AT matching Discovery of the Process o There Were Three Old Models to Explain DNA Replication Copies o Cultured E. coli, specifically in a heavy Nitrogen medium ( N) 14 Heavy Isotope, swapped the Medium to Light Isotope ( N) Bacterial Replication o Bacteria have circular DNA with a single origin of replication o Components of oriC (About 100200bp long) AT Rich Regions DnaA Box Regions Roughly five spots present GATC Methylation Site The Overall Process o 1. Prepping A. Methylation B. DnaABox/ATP Complex Proteins C. DNA Helicase D. Single Strand Binding Protein E. Gyrase (Topoisomerase II) o 2. Synthesizing A. DNA Primase B. DNA Polymerase II, IV, and V have repair functions I and III have normal replication functions o 3. Polymerase Action DNA Polymerase Actors A. Polymerase III B. Polymerase I o DNA Polymerase I removes the RNA Primer Cap and adds in the 5’ to 3’ direction o It latches on to the second Primer nucleotides and removes from the first primer when it adds DNA nucleotides o Primer removed so synthesis can be completed o Works in conjunction with Polymerase III Reaction of DNA Polymerase Role of the ▯ Subunit o They come in as Triphosphates o Pyrophosphate given up in formation of Diester Bond Dehydration Synthesis o Results in the Phosphate adjacent to the sugar being the one that binds to the next Nucleotide o This catalysis is done by the ▯Subunit Importance of the ?Subunit o Holds DNA Polymerase III in place in order to react, works up to 500,000bp Most significant for continuity o Mutations in the Beta Subunit result in significantly less efficient replication, falls off every 10bp and goes from 750bp/sec to 20bp/sec o 4. DNA Ligase Ligase stitches things together Powered by NADPH Acts about every 500,000bp, When Poly III Hops off Acts after Poly I has removed the Primers Catalyzes the one Covalent that is needed between Long Strand from III and Sh+rt Strand of I Prokaryotes use NAD , Eukaryotes ATP Broken down into Nicotinamide Monoamine and Adenosine o 5. Termination The replication forks eventually come together in Bacteria But the Ter Sequence also plays a role They are present in both directions, the clockwise movement and the anticlockwise movement So Two Sequences present Ter Does not actually cause termination Tus binds to the Ter Sequence and triggers termination on the opposite side 1 Tus (Termination Utilization Substance) is sufficient to stop replication (one continues to the termination portion) o Tus stops the same side (binding to clockwise side allows counterclockwise to proceed) Ligase comes in after the Tus initiated termination Stiches normal spots together Sometimes Ligase has errors, but Topoisomerase II comes in to help fix o Makes a cut and Ligase comes in to “Remend” DNA Replication Complexes (they form to increase efficiency) o 1. Primosome Works to break HBonds and Lay Primers This process occurs within a unit Done by Helicase and DNA Primase o 2. Replisome With the activity of Polymerase III, the unit gains another functionality Becomes the Replisome Does not include Ligase and Poly I o 3. DNA Polymerase III Unit The protein is dimerized, so they move along together Proofreading Mechanisms o Fidelity – retaining the correct DNA sequence o Three Manners I. Inherent Instability of Mismatched Bases Improper HBonding creates mismatches that twist the DNA out of shape Torsion and turning occurs Can occur at a rate of 1/1000bp II. ▯Subunit Specificity Catalysis cannot occur if the match isn’t exact Only allows matched base pairs to bind together Has induced fit model The mismatched pairs cannot fit into the Alpha Domain Subunit The Induced Fit changes error rate from 1/100,000 to 1/1,000,000 III. ?psilonSubunit has a Proofreading Function Compared to Poly I (Removes and Synthesizes in 5’ to 3’) Exonuclease Polymerase III Subunit digests 3’ to 5’ and adds the repair in 5’ to 3’ o Addition is the same direction, but digestion is different Still a part of Polymerase III, just located behind the AlphaUnit So no the error rate is changed to 1/100,000,000 How Does the Cell Coordinate Replication with Division? – Components are Needed o 1. Methylation (Adenine Mutation) After the replication two strands are not methylated So both strands are not methylated (Fully Methylated) Results in Hemimethylated strands o 2. DNAaBox Proteins Similarly, the box protein complexes go from ATP to ADP complexes No more affinity is present, so fewer complexes The necessary concentration of Box proteins is no longer present immediately after replication either Because there are more DNAa Boxes (mores strands) 20 DnaA Proteins to bind with the 5 Boxes Present o 3. Cell Growth and the Dam Gene When the cell grows to a certain extent and reaches a particular level This occurs on Adenine for GATC Methylated Regions Dam (DNA Adenine Methyltransferase) acts to Methylate Adenine Confirmation of Triphosphate Theory – Arthur Kornberg o Background In vitro experiment Wanted to see how the Nucleotides formed He theorized the Triphosphates were recruited to form Nucleotides o Set Up and Procedure Took all the necessary proteins and components to necessary to synthesize (Complete System) as well as a DNA template Did a control without the template as well Then labeled the Triphosphates to be able to observe them Add the end of the replication period (30min), Perchloric Acid was added to destroy the remaining bases Kept the strand intact, broke down nucleotides The centrifuged and took the DNA Pellet o Findings Complete the template, found radioactivity present ( P)2 Without the template present, all nucleotides broken down and no radioactivity was present March 30, 2016 Chapter 11: DNA Replication Chargaff’s Rule is Important to Consider Discovery of the Process Bacterial Replication The Overall Process o 1. Prepping A. Methylation B. DnaABox/ATP Complex Proteins C. DNA Helicase D. Single Strand Binding Protein E. Gyrase (Topoisomerase II) o 2. Synthesizing A. DNA Primase B. DNA Polymerase II, IV, and V have repair functions I and III have normal replication functions o 3. Polymerase Action DNA Polymerase Actors A. Polymerase III B. Polymerase I Reaction of DNA Polymerase Role of the ▯Subunit Importance of the βSubunit o 4. DNA Ligase Acts about every 500,000bp, When Poly III Hops off + Prokaryotes use NAD , Eukaryotes ATP o 5. Termination But the Ter Sequence also plays a role Tus binds to the Ter Sequence and triggers termination Ligase comes in after the Tus initiated termination Sometimes Ligase has errors, but Topoisomerase II comes in to help fix Catenanes DNA Replication Complexes (they form to increase efficiency) o 1. Primosome (HBonds and Primers) o 2. Replisome (With Poly III) o 3. DNA Polymerase III Unit Proofreading Mechanisms o Fidelity – retaining the correct DNA sequence I. Inherent Instability of Mismatched Bases II. ▯Subunit Specificity III. ΕpsilonSubunit has a Proofreading Function How Does the Cell Coordinate Replication with Division? – Components are Needed o 1. Methylation (Adenine Mutation) Results in Hemimethylated strands o 2. DNAaBox Proteins o 3. Cell Growth and the Dam Gene Confirmation of Triphosphate Theory – Arthur Kornberg o He theorized the Triphosphates were recruited to form Nucleotides o Set Up and Procedure Took all the necessary proteins and components to necessary to synthesize (Complete System) as well as a DNA template Add the end of the replication period (30min), Perchloric Acid was added to destroy the remaining bases The centrifuged and took the DNA Pellet o Findings Complete the template, fεound radioactivity present ( P) Without the template present, all nucleotides broken down and no radioactivity was present Mutants and Quest for Discovering Protein Function o How could the function of various replication proteins be discovered and determined? If mutants were observed, the functionality could be determined Allowed them to discover DNA Polymerase II and III in Bacteria o Conditional Mutants The mutations had to be screened One by one, Brute Force Screening to what mutations were present They couldn’t induce a specific locus to mutate Specifically observed a particular type that were sensitive to the environment (Temperature Sensitive, ts) These would act in Permissive Temperatures o Experimental Procedure I. Mutagenic Agent (usually Radiation) II. Agar Plating III. RepliPlating They touch the bacterial colony onto a disk They then grow that colony and simultaneously touch two other plates in order to replicate IV. Permissive vs. NonPermissive Temperature o Experimental Findings Rapid Stop Slow Stop Vital to Replication Important in the Second process, causes Round, so Replication immediate stop halts after one go Polymerase I,III, Not enough DnaABox Helicase, Primase Proteins All are vital to DnaC cannot recruit Replication Helicase Dam does not function properly o Importance of Understanding 1) Identification of Proteins Function 2) Mapping of Given Genes 3) Genetic Cloning (cDNA) could now Occur Eukaryotic Replication Overview o Prokaryotic Comparisons (Highlight Similarities in Replication Processes) 1. OriC Locations – one in prokaryotes, various in Eukaryotes (every 100,000bp) 2. DNA Nature – linear with histones vs. naked circular 3. Cell Cycle Regulation Checkpoints – protein checkpoints to push forward, complex methods o Still utilize many of the same proteins as well: 1. Gyrase 2. Single Stranded Binding Proteins 3. A form of Primase Process o I. Prepping 1) Origin of Replication (ARS) 50bp Long in Eukaryotes Known as Autonomously Replicating Sequences (ARS) Element Contain: o 1] These sequences have ATRich Regions in them o 2] They also have a protein binding sequence analogous to DnaABox Regions 2) PreReplication Complex (with the Origin Recognition Complex) The PreRC is made of 14 Proteins This occurs during the G1 Phase A particular complex within the PreRC determines initial recognition o Origin Recognition Complex (ORC) o Made of 6 Proteins o Forms an ATP complex and binds to ARS (analogous to DnaAProteins) This binding initiates the S Phase (leads to MCM) 3) MCM Helicase MCM Helicase initiates the replication process (this is the beginning of S Phase) This first break is in the AT Regions of the ARS MCM Helicase must be present to form replication forks (wherever it is present, replication forks form) o II. Primer Formation 4) Polymerase ▯Action (with Primase) Works in conjunction with Primase to make DNA Polymerase / ▯ Primase Complex Forms a 10bp RNA Primer and 2030bp DNA Strand This is the Primer Stage Theory is that ▯only has a 2030bp capacity o III. Synthesizing 5) Secondary Polymerase Action After the Alpha/Primer Complex has acted, Alpha falls off and either ? (Delta) or ? (Epsilon) Leading o ? works in the leading strand Lagging o ? works in the lagging strand o Needs to have a motile capacity in order to function in Lagging Strand o Works on 100200bp Long Fragments (more origins so no needs for long strands) 6) Primer Removal DNA Polymerase I worked in Prokaryotes o In Eukaryotes, ?Polymerase works always, leading and lagging, with Flap Endonuclease Works by creating peripheral “flaps” that can be removed Lagging Process o I. Moves up the fragment and “butts into” and “pushes off” a 2bp Long fragment (the next primer upfield) This process occurs on the whole primer o II. Flap Endonuclease comes and removes the small strand o III. Delta then continues to add DNA o IV. As a longer strand accumulates, DNA Helicase (DNAII) comes in and removes (5+bp) Leading Process o I. PCNA (Proliferating Cell Nuclear Antigen) allows for Delta Polymerase separation and reattachment Analogous to ? Clamp in Prokaryote Polymerase III This proved processivity to allow it to work Also present for Epsilon, but the notes action is the on and off falling of Delta o II. Then the above process occurs on the single location in the leading strand Polymerase Types ▯ ? ? ? Also the Replaces Alpha Replaces Alpha Replication of catalyzing type in lagging in the Leading Mitochondrial Works with strand Strand DNA Primase to Removes Primer form initial primer Repair o Types – Works in conjunction with cell checkpoints (such as p53) I. Polymerase ? can remove bp mismatches Base excision repair – removes incorrect bases to be replaced Works as replication is occurring II. Lesion Replicating Polymerases Continues elongation if repair does not end up occurring o Involved in continuing replication of replication Interestingly, can also detect mismatched bp in the parent strand and allows for a correction in the daughter without altering daughter o Recognition 1. HBonding Mismatch 2. Induced Fit and DNA Alignment 3. DNA Tortion Nucleosome Replication o Histone need to be replicated as well (H2A, H2B, H3, H4 from the Octamer especially) o Also follows a Semiconservative Model (so old and new histones form) Telomere Replication o When the Delta Polymerase comes to the end, what happens with the Telomeres? The loops are 100200bp, so what if the remaining end is not exact? The template isn’t long enough – 3’ Overhang remains Telomerase acts (loses functionality with age, so Telomere length reduction) o Telomeres The remaining part is known as the 3’ Overhang These are usually about 1216bp Long Telomeres have numerous Repeat Sequences comprised of GGG and TT Repeats Normal Polymerase function presents some problems: Cannot flip/loop Still has to code 5’ to 3’ o Telomerase Process TERTS (Telomere Reverse transcriptase) is a part of Telmoerase There are two of these proteins present RNA is used as a primer to extend Template Strand TERTS has an internal 9bp RNA Complementary Strand that codes a DNA fragment Attaches at first three Codes six Repeats It is then moved to the 3’ end since replication and looping cannot occur there Ligase stiches the translocated portion together Forming a template long enough to form a loop which can then be replicated April 1, 2016 Chapter 12: Gene Transcription and RNA Modification Prokaryotic Transcription Overview of Transcription o Only time that DNA can be accessed for the information to be expressed o Players Involved 1) DNA – codes for RNA, has signals for starting/stopping 2) Proteins – recognize, regulate, modify transcribed RNA o DNA does not change through the process – a copy is made identical to one of the strands o What is a Gene? A unit of heredity DNA sequences that codes A transcriptional unit (not necessarily structural) o Some Terminology Important to consider the terminology based on the directionality of the transcriptional synthesis Central Dogma o DNA RNA Proteins o Classes 1. Eukaryotes Follows the Dogma Polycistronic RNA cannot occur because organelles perform the specific tasks (which is the purpose of Polycistronic) 2. Prokaryotes Also follow the Central Dogma But can also have Polycistronic RNA – can code for several types of proteins o Consider, a given vitamin or mineral needs to be processed o In Prokaryotes, a single long strand of mRNA can be transcribed for various proteins in the metabolic pathway This can occur in mitochondria and chloroplasts as well due to their evolutionary nature 3. Outliers Viruses are organisms with RNA as their genetic material o So RNA goes straight to DNA with Reverse Transcriptase This is important to note that the RNA in viruses does not act like a protein, can only perform the same functions o Viroids are infectious RNA’s, but they only affect plants (no domains) o Prions are infectious Proteins that unbind other proteins, more complex than simply RNA, because they have domains Components of Transcription o Transcription Sequences DNA and RNA each have specific sequences needed for Transcription This process occurs in G1 (Transcription), relative to the S Phase (Replication) DNA RNA Starting Starting o Promoter Sequence o Prokaryotes site for RNA Ribosomal Binding Site at the 5’ Polymerase Binding Start Codon – specifies the first amino acid in the sequence, modified AUG in o Transcription Sequence also Prokaryotes (NFormyl Meth Unit) present in the DNA o Eukaryotes Have 3’ and 5’ Modifications (Cap and Stopping o Terminator Tail), so just a start Codon is needed Sequence (AUG, Methionine) (different than Prokaryotes) Regulatory o Also has regulatory Always starts, even if Meth not sequences that alter present and influence the The Ribosome will travel back and rates of transcription forth over the mRNA before it “decides” which start codon Stopping o Stop Codon exists instead o Transcription Steps I. Initiation The strands have to separate to be accessible o Done by transcription factors and RNA Polymerase They bind and an Open Complex (unbinding) Forms II. Elongation The actual sequence is constructed III. Termination Process stopped Prokaryotes have RhoDependent and Rho Independent Pathways All steps involve DNA/RNA interaction, so when the DNA/RNA Hybrid is broken (separated), Termination has Occurred o This is the temporary connection between the Template and the mRNA o The Various Roles of RNA Transcripts – This is an important point to consider: RNA has numerous functions it can perform after transcription 1. Structural This is mRNA Codes for proteins, over 90% of genes 2. Standard tRNA and rRNA are also present in high concentrations Involved in the Transcribing portion 3. Atypical (Assessed by the Sedimentation Coefficient) 7S is the Signal Recognition Particle (SRP) (7 Sedimentation Coefficient + 6 Proteins, co translational recruitment of proteins to the ER) Spliceosomes (snRNA and snoRNA, Small Nuclear RNA) Nucleolus RNA (ribosomal subunit assembly) Viruses Steps Detailed o I. Initiation A. Promoter Sequence +1 Location on the DNA o The first nucleotide that is coded for information o Upstream units are Negative Units Has TTGACG at 35 and TATAAT at 10 crucial for the +1 location o These are a consensus o Binding first occurs on the 35 sequence o 1618 bp long o Upstream Promoter sequences essentially “point” to the +1 Location o These are the sequences to which RNA Polymerase Binds o Any sort of acetylation, methylation, or modification to promoter sequences drastically reduces transcription rates Alteration in the binding sequences results in the rate changing from Once Every 2sec to Once Every 10min o These sequences need to be of high fidelity B. RNA Polymerase II (HoloEnzyme, made of two components) Core Enzyme This process is a two Component process Subunits: Core Unit has Various Subunit Components (RNA ▯(x2) Polymerase) o ▯subunits allow for the addition and binding of ? (and Beta Prime) – the transcription factor (Sigma) Catalyzes the bond o ? and ?’ catalyze ester formation in both Eukaryotes and Prokaryotes o ? – Assembles the Core (Omega) ? Factor (this is the transcription factor) o Works with RNA Polymerase in order to: 1) Identify 35 Sequence 2) Cause formation of closed complex o Single subunit that allows the Core to function o Induces the opening up of the strands o Once initiation done, though, Sigma Released o Has a Motif (supersecondary structure) on the ▯turn that binds to 35 location on the major groove C. Haloenzyme Binding It binds loosely and beings to “scan” the DNA for a Promoter Sequence Then the Closed Complex forms o It binds, but nothing is open to Transcript yet o This occurs when the ? Factor and DNA Poly II bind at the 35 Promoter Region o This Closed Complex creates a tension in the TATATT Region at 10 Shortly afterwards, the Open Complex Forms o The strain placed on the molecule from the Closed Complex causes the adjacent AT rich region to unravel, forming the Open Complex o This is the first opening up o So this is the crux of initiation D. Short RNA Strand Synthesis In the Open Complex, a small strand of mRNA is formed (810bp) This synthesis causes the release of the ? Factor and the End of Initiation The Core Enzyme continues to move and transcribe – goes to elongation o II. Elongation The actual sequence is constructed in this step From +1 Onwards The Transcription Bubble forms as copying occurs This is any unwound section open for transcription About 17bp long Codes at a rate of 43 Nucleotides Hz o III. Termination This is marked by the breaking of the DNA/RNA hybrid Prokaryotes rely on Rho Pathways Rho separates them by breaking the HBonds Rho Dependent Rho Independent Rho acts like Helicase Intrinsically occurs Has Requirements to Bind: because of the nature of the o I. Rho Utilization Sequence (Rut) mRNA sequence in DNA (allows for Rho Binding) This form also has a Stem o II. Alternating GC Rich Regions to Loop that interferes with form Stem Loop the Ribosome Process Sequences Present o RNA Stem Loop forms downstream o 1. Alternating GC Rich This occurs because of the GC o 2. 3’ End mRNA Uracil Rich Alternating Regions (so DNA has Adenine) They have an affinity for each Adenine of DNA and the other Uracil of RNA are already o The Stem Loop slows down weak, now Poly II slowed polymerase by blocking the NUSA has also been found Ribosome (touches lightly) and is known to further o So Rho can catch up from behind slow and stall the transcription process April 4, 2016 Chapter 12: Gene Transcription and RNA Modification Overview of Transcription Central Dogma o DNA RNA Proteins o Classes 1. Eukaryotes 2. Prokaryotes Polycistronic RNA 3. Outliers Viruses and Reverse Transcriptase Components of Transcription o Transcription Sequences DNA RNA Starting Starting o Promoter Sequence site o Prokaryotes for RNA Polymerase Ribosomal Binding Site at the 5’ Binding Start Codon – specifies the first o Transcription Sequence amino acid in the sequence, also present in the DNA modified AUG in Prokaryotes Stopping o Eukaryotes o Terminator Sequence Have 3’ and 5’ Modifications (Cap Regulatory and Tail), so just a start Codon is o Also has regulatory needed (AUG) (different than sequences that alter and Prokaryotes) influence the rates of The Ribosome will travel back and transcription forth over the mRNA before it “decides” which start codon Stopping o Stop Codon exists instead o Transcription Steps I. Initiation II. Elongation III. Termination o The Various Roles of RNA Transcripts – This is an important point to consider: RNA has numerous functions it can perform after transcription 1. Structural This is mRNA Codes for proteins, over 90% of genes 2. Standard tRNA and rRNA are also present in high concentrations Involved in the Transcribing portion 3. Atypical (Assessed by the Sedimentation Coefficient) 7S is the Signal Recognition Particle (SRP) (7S + 6 Proteins, cotranslational recruitment) Spliceosomes (snRNA and snoRNA) Nucleolus RNA Viruses Steps Detailed o I. Initiation A. Promoter Sequence Has TTGACG at 35 (1618bp long) and TATAAT at 10 crucial for the +1 location B. RNA Polymerase II (HaloEnzyme, made of two compenents) Core Enzyme Subunits: Has a Motif (supersecondary structure) that binds ▯(x2) to 35 location on the major groove ? (and Beta Prime) – Core Unit Various Subunit Components Catalyzes the bond o β and β’ catalyze ester formation in both Eukaryotes and Prokaryotes ? – Assembles the Core o ω – Assembles the Core σ Factor o Single subunit that allows the Core to function o Induces the opening up of the strands C. Haloenzyme Binding It binds loosely and beings to “scan” the DNA for a Promoter Sequence Then the Closed Complex forms o This occurs when the σ Factor and DNA Poly II bind at the 35 Promoter Region Shortly afterwards, the Open Complex Forms D. Short RNA Strand Synthesis In the Open Complex, a small strand of mRNA is formed (810bp) This synthesis causes the release of the σ Factor and the End of Initiation The Core Enzyme continues to move and transcribe o II. Elongation The Transcription Bubble forms as copying occurs, About 17bp long, Codes at a rate of 43 Nucleotides Hz o III. Termination This is marked by the breaking of the DNA/RNA hybrid Prokaryotes rely on Rho Pathways Rho separates them by breaking the HBonds Rho Dependent Rho Independent Rho acts like Helicase Intrinsically occurs Has Requirements to Bind: because of the nature of the o I. Rho Utilization Sequence (Rut) mRNA sequence in DNA (allows for Rho Binding) This form also has a Stem o II. Alternating GC Rich Regions to Loop that interferes with form Stem Loop the Ribosome Process Sequences Present o RNA Stem Loop forms downstream o 1. Alternating GC Rich This occurs because of the GC o 2. 3’ End mRNA Uracil Rich Alternating Regions (so DNA has Adenine) They have an affinity for each Adenine of DNA and the other Uracil of RNA are already o The Stem Loop slows down weak, now Poly II slowed polymerase by blocking the NUSA has also been found Ribosome (touches lightly) o So Rho can catch up from behind and is known to further slow and stall the transcription process Eukaryotic Transcription Overview o Complications and Differences with Eukaryotic DNA 1) Histone Association The Histones are attracted because of the charges (Agrinine, Lysin to the Phosphates) Histone modification is a way of allowing the unwinding of DNA from Histones o Acetylation essentially neutralizes the positive charges are Histone Tails o So it essentially blocks the DNA binding with the Histone and allows unwinding 2) Cellular Complexity Polycistronic mRNA is not seen in Eukaryotes Organelles exist which allow for specific, regional protein construction – no Polycistronic nuclear mRNA 3) Multicellularity All the sequences in every cell cannot function at all times There’s a complex balance in Eukaryotes – some genes are permanently upregulated, some permanently downregulated o RNA Polymerases Poly I – rRNA except for 5S Subunit Poly II – mRNA, structural genes, and Small Nuclear RNA of Spliceosomes Poly III – tRNA and the 5S Ribosomal RNA subunit A Look at RNA Polymerase o I. Clamp – The units work like a jaw, having to open up to take the DNA o II. Bridge – The DNA settles at the Bridge and makes a Right Handed turn Right turns occur at β’ and β (The catalytic units) o III. Rudder – Eukaryotes have a rudder that is 9bp ahead that breaks the template in order to dissociate This is only significant difference between Eukaryotes and Prokaryotes Steps for Structural Genes o I. INITIATION A Promotor region initiates the transcription process, similar concept to that of Prokaryotes 1. Start Site (+1) 2. TATA Box Region (Adjacent to 25) o A Core Promoter sequence region exists in Eukaryotes, the TATA Box and Start Site are the Core Promoter o A welldefined TATA Box free of mutations results in definitely having small amounts of transcription (Basal Transcription) 3. Regulatory Elements (Between 50 and 100bp) o There are enhancers and silencers (and they are sequences, not proteins o Present between 50 and 100 o Two Actors: A. Cis Next to, immediately upstream or within the chromosome Usually sequences B. Trans Opposite, away from, Different chromosome, factors that diffuse through cytoplasm and nucleus, and bind to trans Then the Closed Complex must form (All 5Transcription Factors and RNA II) I. RNA Polymerase II II. Transcription Factors of RNA Polymerase II (5 Proteins, General Transcription Factors (GTFs)) o 1] TFIID Recognizes the TATA Sequence After Promoter Region, 3 Huge protein complex Components for transcription: TATA Binding Protein (TBP) TBPAssociated Factors (TAFs) A. RNA Polymerase II Binds and moves to read the rest, then B. Transcription Factors reverses before “deciding” to bind Does so after DNA has opened up C. Mediator Complex o 2] TF2B Closed Complex: Brings in the RNA Poly II to bind to the Promoter Sequence Prokaryotes have the Forms the Bridge Sigma Complex o 3] TFIIF Comes in along with RNA Polymerase II o 4] TFIIE Eukaryotes have: Binds to RNA Polymerase II 1. All Five TFAs Initiates and maintains the Open Complex o 5] TFIIH Binds to RNA Polymerase II and Acts as Helicase to continue open up o II. ELONGATION How Does Basal transcription move into Elongation? RNA Polymerase II has a Carboxy Terminal Domain (CTD) When it is Phosphorylated, Elongation begins TFIIH also has ATPase and Protein Kinase Activity (hydrolyzes ATP and transfers Phosphate) That activity causes separation of TFIIB and RNA Polymerase II That release causes the release of others as well While TFIIH is Phosphorylating, it is still breaking HBonds So TFIIF remains attached to RNA Polymerase II III. Mediator Protein Complex o CoActivator, Large Complex o Appears to regulate the ability of TFIIH to Phosphorylate CTD o TFIIH and Mediator have the same function, so they ensure that Phosphorylation and Elongation occurs Elongation process o III. TERMINATION Initiation of Termination begins 5002000bp Downstream of the Poly Adenylation sequence (AAUAAA) This sequence is the signal to end, coming in at the 3’ End Once PolyA is made, Termination Downstream occurs Two Models 1. Allosteric Model o Factors play a role o At the core, Polyadenylation Sequence induces an instability o Elongation Factors are lost or Termination Factors bind 2. Torpedo Model o And exonuclease binds to the 5’ end of the mRNA o Starts digesting in the 5’ to 3’ direction and eventually catches up to RNA Polymerase II, which then falls off RNA Modification o There is CoLinear Process to Gene Expression in Prokaryotes Similar all the way from DNA to RNA to AA Sequence o Eukaryotes stray from CoLinearity They instead splice and remove and modify (the whole process is Trimming) PreRNA or Heterogeneous RNA is the same length as the Template Strand, so essentially the Coding Strand Then splicing and modification takes place on PreRNA to form the mRNA (removing introns) o Type of RNA Processing 1) rRNA Trimming occurs with rRNA Formed by RNA Polymerase I (except for 5S) Occurs in the Nucleolus Here, done by Endonucleases in the center 2) mRNA 5’ Capping and 3’ PolyA Tails Occur to stops nucleases from digesting 3) tRNA Made of huge precursors and various processes o 2 Endo and Exo Nucleases They have to undergo cleavage at the 5’ and 3’ ends 1. RNaseP o Acted on by RNase P (made of 375bp RNA and 20kDA Protein) o Forms the Acceptor Stem at the 3’ End, reading as 5’ CCA 3’ April 6, 2016 Chapter 12: Gene Transcription an
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