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Add or subtract as indicated. (89 38') - (28 58')

Trigonometry | ISBN: 9780495108351 | Authors: Charles P McKeague ISBN: 9780495108351 200

Solution for problem 2.1.85 Chapter 2.2

Trigonometry

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Trigonometry | ISBN: 9780495108351 | Authors: Charles P McKeague

Trigonometry

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Problem 2.1.85

Add or subtract as indicated. (89 38') - (28 58')

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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 C-G and A-T 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 100-200bp long)  AT Rich Regions  DnaA Box Regions  Roughly five spots present  GATC Methylation Site  The Overall Process o 1. Prepping  A. Methylation  B. DnaA-Box/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  -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 β-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 Short 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  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 H-Bonds 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 H-Bonding 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. Εpsilon-Subunit 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 Alpha-Unit  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. DNAa-Box 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 in-tact, broke down nucleotides  The centrifuged and took the DNA Pellet o Findings 32  Complete the template, found radioactivity present ( P)  Without the template present, all nucleotides broken down and no radioactivity was present 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 C-G and A-T matching  Discovery of the Process o There Were Three Old Models to Explain DNA Replication Copies 15 o Cultured E. coli, specifically in a heavy Nitrogen medium 14N) 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 100-200bp 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 DnaA-Box/ATP Complex Proteins bind to the DnaA-Box 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 DnaA-Box Region  This binding causes strain in the AT-Rich 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. AT-Rich Regions  About 13bp Long  There are only two Hydrogen Bonds present, so they are more susceptible to separation  Very close to DnaA-Box 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 DnaA-Box 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 path-paver 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 10-12bp long RNA primer, so DNA Polymerase does not immediately and directly act  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  Hollow Enzyme, So Contains Several Subunits Polypeptide (10 Total)  Removes RNA  These subunits give the functionality, and the Primers (made by others merely aid these functioning DNA Primase) o – catalysis and Substitutes o β – clamp them with DNA o γ – clamp loader Nucleotides o ε – exonuclease  Each Subunit has a Specific Function o – Catalyzes Phosphodiester Bonds o β – Allows Polymerase III to hold onto the template; vital subunit o γ – Loads onto the clamp; important in lagging strand o ε – 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 1,000-2,000bp Fragments in Prokaryotes, 100-200 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) DnaA-Box Proteins DnaA-Box Sequences Form a Protein/ATP Complex HU and IHF also help DnaA Bind 5-6 Box Regions Present Each Box 9bp Induces Separation DnaA-Box Proteins create friction on the DNA Strand After the separation occurs, DNA-ATP complex becomes DNA-ADP Complex Protein Complex formed of about 20 Proteins These regions are next to AT-Rich regions which are susceptible to unwinding under stress 3) DnaC DnaB-Helicase Helicase is recruited to continue opening up the strands This is a 6-Protein 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 10-12bp 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 100-200bp Long in Eukaryotes, 1000-2000 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’

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Chapter 2.2, Problem 2.1.85 is Solved
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Textbook: Trigonometry
Edition:
Author: Charles P McKeague
ISBN: 9780495108351

The full step-by-step solution to problem: 2.1.85 from chapter: 2.2 was answered by , our top Math solution expert on 01/02/18, 08:55PM. This textbook survival guide was created for the textbook: Trigonometry, edition: . Trigonometry was written by and is associated to the ISBN: 9780495108351. Since the solution to 2.1.85 from 2.2 chapter was answered, more than 250 students have viewed the full step-by-step answer. The answer to “Add or subtract as indicated. (89 38') - (28 58')” is broken down into a number of easy to follow steps, and 10 words. This full solution covers the following key subjects: . This expansive textbook survival guide covers 58 chapters, and 3545 solutions.

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Add or subtract as indicated. (89 38') - (28 58')