BCHM218_M02_OL-BL_onQ_CompanionGuide

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MOLECULAR BIOLOGY | BCHM 218 M2 PAGE 1 BCHM 218 MODULE 02 D N A REPLICATION, REPAIR, AND RECOMBINATION Please note: This course was designed to be interacted and engaged with using the online modules. This Module Companion Guide is a resource created to complement the online slides. If there is a discrepancy between this guide and the online module, please refer to the module. How can you help protect the integrity and quality of your Queen’s University course? Do not distribute this Module Companion Guide to any students who are not enrolled in B CH M 218 as it is a direct violation of the Academic Integrity Policy of Queen’s University. Students found in violation can face sanctions. For more information, please visit https://bhsc.queensu.ca/current- students/academic-calendar/academic-regulations-and-university-policies MOLECULAR BIOLOGY

MODULE 02 COMPANION GUIDE BCHM 218 MOLECULAR BIOLOGY | BCHM 218 M02 PAGE 2 TABLE OF CONTENTS INTRODUCTION ..................................................................................................................................................... 8 A. D N A Replication, Repair, and Recombination ......................................................................................... 8 B. Introduction ................................................................................................................................................... 9 C. Learning Outcomes .................................................................................................................................... 10 D. Module Assignments .................................................................................................................................. 11 D.1 Problem-Based Activities ..................................................................................................................... 12 E. Course Icons ................................................................................................................................................ 13 F. Module Outline ............................................................................................................................................ 14 SECTION 01: The Beginnings of Replication Enzymology ............................................................................... 15 1.1 Section 01: The Beginnings of Replication Enzymology ....................................................................... 15 1.2 The Eukaryotic Cell Cycle .......................................................................................................................... 16 1.3 Cell Cycle .................................................................................................................................................... 17 1.4 Arthur Kornberg ........................................................................................................................................ 18 1.5 Beginnings of Replication Enzymology ................................................................................................... 19 1.6 Kornberg: Purification of “Polymerase” .................................................................................................. 20 1.7 Purification of “Polymerase” Experiment ............................................................................................... 21 1.8 Kornberg: Requirements of the “Polymerization Reaction” ................................................................. 22 1.9 Requirements of the “Polymerization Reaction” Results ..................................................................... 23 1.10 Fo r Your Interest: Kornberg’s Publications .......................................................................................... 24 1.11 Kornberg's Experiments ......................................................................................................................... 25 1.12 Section 01 Summary ............................................................................................................................... 26 SECTION 02: D N A Replication .......................................................................................................................... 27 2.1 Section 02: D N A Replication .................................................................................................................. 27 2.2 Introduction to D N A Replication ........................................................................................................... 28 2.3 Video: D N A Replication ........................................................................................................................... 29 2.4 General Principles of D N A Replication ................................................................................................. 30 2.5 1) D N A Replication is Semiconservative ............................................................................................... 31 2.6 2) Replication is Initiated at Specific Sites .............................................................................................. 32 2.7 3) Replication is Typically Bidirectional ................................................................................................... 33 2.8 4) Replication is Continuous on the Leading Strand and Discontinuous on the Lagging Strand .... 34 2.9 Semidiscontinuous Replication ............................................................................................................... 35 2.9.1 Key Convention: Directionality of D N A Synthesis ........................................................................ 36

MODULE 02 COMPANION GUIDE BCHM 218 MOLECULAR BIOLOGY | BCHM 218 M02 PAGE 3 2.10 5) R N A Primers are Needed to Start D N A Replication .................................................................... 37 2.11 6) Nucleases, Polymerases, and Ligases .............................................................................................. 38 2.12 Sealing Nicks with D N A Pol I ................................................................................................................ 39 2.13 Activity: General Principles of D N A Replication ................................................................................. 40 2.14 D N A Polymerase Structure .................................................................................................................. 42 2.15 Conformations of D N A Polymerase I .................................................................................................. 43 2.16 Quality Control ........................................................................................................................................ 44 2.17 Incorporation of Incorrect Bases .......................................................................................................... 45 2.18 The Polymerase Active Site: Closed Form ............................................................................................ 46 2.18.1 Recall: Reading Reaction Mechanisms .......................................................................................... 48 2.18.2 Question: Polymerase I ................................................................................................................... 49 2.19 The Atomic Structure of the Polymerase Active Site .......................................................................... 50 2.20 The D N A Polymerase Reaction ............................................................................................................ 51 2.21 Translocation of D N A Polymerase ...................................................................................................... 52 2.22 Processivity .............................................................................................................................................. 53 2.23 What if Incorrect Base Pairing Occurs? ................................................................................................ 54 2.24 Effects of Mispairing ............................................................................................................................... 55 2.25 3’ - 5’ Exonuclease Activity of D N A Pol .................................................................................................. 56 2.26 Activity: Proofreading ............................................................................................................................. 57 2.27 3’ - 5’ Exonuclease Active Site .................................................................................................................. 59 2.28 Polymerase I ............................................................................................................................................ 60 2.29 Nick Translation by Pol I (5- 3’ Exonuclease) ......................................................................................... 61 2.29.1 The Process of Nick Translation by Pol I ....................................................................................... 62 2.30 Question: Exonucleases of D N A Polymerase .................................................................................... 63 2.31 Section 02 Summary ............................................................................................................................... 64 SECTION 03: D N A Replication in the Model Organism E. Coli ...................................................................... 65 3.1 Section 03: D N A Replication in the Model Organism E. coli ............................................................... 65 3.2 Introduction to E. coli Replication ............................................................................................................ 66 3.3 E. coli Contains Five D N A Polymerases ................................................................................................. 67 3.4.a E. coli D N A Polymerases ...................................................................................................................... 69 3.4.b E. coli D N A Polymerases ...................................................................................................................... 70 3.5 Stages of E. coli Replication ...................................................................................................................... 71 3.6 1) Initiation of Replication ........................................................................................................................ 72

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MODULE 02 COMPANION GUIDE BCHM 218 MOLECULAR BIOLOGY | BCHM 218 M02 PAGE 4 3.7 Structural Elements of the E. coli Replication Origin ............................................................................. 73 3.8 Activation of the E. coli Replication Origin (oriC) ................................................................................... 74 3.8.1 1) Generating the Open Complex .................................................................................................... 75 3.8.2 2) Activation of the Replication Origin ............................................................................................. 76 3.8.3 3) Assembly of the E. coli Replication Forks .................................................................................... 77 3.8.4 4) Replication Initiation and Leading Strand Synthesis ................................................................. 78 3.8.5 5) Lagging Strand Synthesis in E. coli ............................................................................................... 79 3.9 Initiation of Replication ............................................................................................................................ 80 3.10 EUKARYOTES VS E COLI ........................................................................................................................... 81 3.11 EUKARYOTES VS E COLI ........................................................................................................................... 82 3.12 Assembly of Bacterial Replication ......................................................................................................... 83 3.13 Components of Eukaryotic Initiation .................................................................................................... 84 3.14 2) E. coli Replication: Elongation by the Replisome ............................................................................. 85 3.15 The E. coli ‘Replisome’ ............................................................................................................................. 86 3.16 The ‘Replisome’ ........................................................................................................................................ 87 3.17 D N A Polymerase III Holoenzyme ........................................................................................................ 88 3.17.1 β Clamp ............................................................................................................................................. 89 3.17.2 Pol III Core ........................................................................................................................................ 90 3.18 β Clamps Facilitate Processivity ............................................................................................................ 91 3.19 Experiment: Evidence of β Clamp Function ......................................................................................... 92 3.20 Experiment: Requirement of Pol III for a β Clamp .............................................................................. 93 3.21 E. coli β Clamp Loader ............................................................................................................................ 94 3.22 E. coli β Clamp Loader Mechanism ....................................................................................................... 95 3.23 Question: β Clamp Function .................................................................................................................. 96 3.24 The Trombone Model of the Replication Fork ..................................................................................... 97 3.25 The Trombone Model: Lagging Strand Synthesis ............................................................................... 98 3.26 Clamp Recycling by D N A Pol I and Ligase .......................................................................................... 99 3.27 Question: The Trombone Model ........................................................................................................ 100 3.28 Question: D N A Polymerases ............................................................................................................. 101 3.29 3) E. coli Replication: Termination ...................................................................................................... 102 3.30 Termination of E. coli Chromosome Replication .............................................................................. 103 3.31 Activity: Ter sites .................................................................................................................................. 104 3.32 The Tus-Ter System and Replication Speed ...................................................................................... 105

MODULE 02 COMPANION GUIDE BCHM 218 MOLECULAR BIOLOGY | BCHM 218 M02 PAGE 5 3.33 Minimizing Competition Between Transcription and Replication in the E. coli Chromosome ... 106 3.34 Resolution of Catenated Daughter Chromosomes .......................................................................... 107 3.35 Topoisomerase Mechanism of Action ............................................................................................... 108 3.36 Eukaryotic Termination: The End Replication Problem ................................................................... 109 3.37 Eukaryotic Termination: Solving the End Replication Problem ...................................................... 110 3.38 Eukaryotic Termination: Telomerase ................................................................................................. 111 3.39 Question: Replicating the Ends of Linear D N A ............................................................................... 112 3.40 Section 03 Summary ............................................................................................................................ 113 SECTION 04: D N A Repair ............................................................................................................................... 114 4.1 Section 04: D N A Repair (1) .................................................................................................................. 114 4.2 D N A Repair (2) ...................................................................................................................................... 115 4.3 Karyotypes: Normal vs Breast Cancer Cell .......................................................................................... 116 4.4 Aneuploidy and Chromosomal Translocations .................................................................................. 117 4.5 D N A Lesions: Double-Strand Breaks ................................................................................................. 119 4.6 Deinococcus radiodurans: A Master of D N A Repair ........................................................................ 120 4.7 The Deinococcus radiodurans Genome .............................................................................................. 121 4.7.1 For Your Interest: Other Organisms with Extraordinary Capacity for D N A Repair ............... 122 4.8 Repair of Chromosomal Double-Strand Breaks ................................................................................. 123 4.9 Recombinational D N A Repair ............................................................................................................. 124 4.10 Step 5: Completing Repair of a D S B ................................................................................................. 126 4.11 D S B R: Resolution of Holliday Intermediates .................................................................................. 127 4.12 Bacterial D N A Repair ......................................................................................................................... 128 4.13 Repairing Double-Strand Breaks ........................................................................................................ 129 4.14 Formation and Repair of Double-Strand Breaks .............................................................................. 130 4.15 Section 04: Summary ........................................................................................................................... 131 SECTION 05: D N A Recombination ................................................................................................................ 132 5.1 Section 05: D N A RECOMBINATION (1) ............................................................................................... 132 5.2 D N A Recombination (2) ....................................................................................................................... 133 5.3 Homologous Recombination ................................................................................................................ 134 5.4 Review: Eukaryotic Meiosis ................................................................................................................... 135 5.5 Eukaryotic Meiosis: Overview ............................................................................................................... 137 5.6 Eukaryotic Meiosis: Crossing Over ....................................................................................................... 138 5.7 Initiation of Meiotic Recombination ..................................................................................................... 139

MODULE 02 COMPANION GUIDE BCHM 218 MOLECULAR BIOLOGY | BCHM 218 M02 PAGE 6 5.7.1 Spo11 Transesterification (Cleavage) Reaction ........................................................................... 141 5.8 Processing of S p o11-Mediated D S B s .............................................................................................. 142 5.9 Meiotic Recombination is Completed by a Classic D S B R Pathway ............................................... 143 5.10 Resolution of Holliday Intermediates ................................................................................................ 144 5.10.1 Crossover vs Non-crossover ........................................................................................................ 145 5.11 Catalysis of Holliday Intermediate Resolution: Ruv A B Complex .................................................. 146 5.12 Catalysis of Holliday Intermediate Resolution: Ruv C ...................................................................... 147 5.13 Products of Holliday Intermediate Resolution ................................................................................. 148 5.14 Resolution of Holliday Intermediates ................................................................................................ 149 5.15 Section 5 Summary .............................................................................................................................. 150 SECTION 06: Site-Specific Recombination ..................................................................................................... 151 6.1 Section 06: Site-Specific Recombination (1) ........................................................................................ 151 6.2 Site-Specific Recombination (2) ............................................................................................................ 152 6.3 Site-Specific Recombination (3) ............................................................................................................ 153 6.3.1 For Your Interest: Transposition ................................................................................................... 154 6.4 Biotechnological Applications of Site-Specific Recombination ......................................................... 155 6.5 Cre-Lox P and F l p-F R T Transgenes ................................................................................................... 156 6.6 Outcomes of Cre-Lox P and F l p-F R T Mediated Recombination ................................................... 157 6.7 Brainbow: Applying Cre-Lox P Site-Specific Recombination ............................................................. 158 6.8 The ‘Brainbow’ Technique ..................................................................................................................... 159 6.9 Brainbow Outcome ................................................................................................................................ 161 6.10 Brainbow ............................................................................................................................................... 162 6.11 A Gene’s Function Can Often Be Elucidated by Its Absence ........................................................... 163 6.12 1) Zinc Finger Nucleases ...................................................................................................................... 164 6.13 Applications of Z F Ns .......................................................................................................................... 165 6.14 2) TALENs .............................................................................................................................................. 166 6.15 Limitations of Z F Ns and TALENs ...................................................................................................... 167 6.16 3) CRISPR/Cas ....................................................................................................................................... 168 6.17 Video: CRISPR/ Cas9 ............................................................................................................................. 169 6.18 Activity: CRISPR/Cas Mechanism ........................................................................................................ 170 6.19 CRISPR/Cas9 System for Genomic Engineering ................................................................................ 171 6.20 CRISPR/Cas9 Genomic Engineering: Nuclease Domains ................................................................. 172 6.21 Genomic Engineering: Inactivation vs Mutation .............................................................................. 173

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MODULE 02 COMPANION GUIDE BCHM 218 MOLECULAR BIOLOGY | BCHM 218 M02 PAGE 7 6.22 CRISPR/Cas9 Genomic Engineering: Nuclease Domains ................................................................. 174 6.22.1 Endogenous CRISPR/Cas System ................................................................................................ 175 6.22.2 Synthetic CRISPR/Cas System ...................................................................................................... 176 6.23 Review: Technologies for Cutting Genes (1) ..................................................................................... 177 6.24 Review: Technologies for Cutting Genes (2) ..................................................................................... 178 6.25 Review Activity: CRISPR/Cas9 .............................................................................................................. 179 6.26 Inducing Mutations with CRISPR/Cas9 .............................................................................................. 180 6.27 Summary of CRISPR/Cas ..................................................................................................................... 181 6.28 Section 06 Summary ............................................................................................................................ 182 CONCLUSION .................................................................................................................................................... 183

MODULE 02 COMPANION GUIDE BCHM 218 MOLECULAR BIOLOGY | BCHM 218 M02 PAGE 8 INTRODUCTION A. D N A REPLICATION, REPAIR, AND RECOMBINATION

MODULE 02 COMPANION GUIDE BCHM 218 MOLECULAR BIOLOGY | BCHM 218 M02 PAGE 9 B. INTRODUCTION Module 02 will help you apply what you learned about protein and D N A structure in Module 01 to the process of D N A replication, repair, and recombination of genetic material. D N A replication is the synthesis of two daughter D N A molecules that are identical to the parental D N A. This process is highly regulated to limit deleterious mutations, and to ensure that dividing daughter cells receive identical copies of the replicated genome. In multicellular organisms, loss of control over replication can lead to cancer - uncontrolled cell division that eventually kills the entire organism. If you are interested in learning about the role of D N A replication in cancer, consider taking B MED 380, Evolutionary Biology of Cancer , which discusses this topic in detail. The double helix structure of D N A is so elegant and simple - you might think D N A replication would reflect this simplicity. However, this process is far from simple. The interplay of numerous replicative proteins follows a complex choreography to produce two identical D N A molecules from one parent molecule.

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MODULE 02 COMPANION GUIDE BCHM 218 MOLECULAR BIOLOGY | BCHM 218 M02 PAGE 10 C. LEARNING OUTCOMES By the end of this module, you will be able to: • Describe the structure and function of D N A polymerases. • Differentiate between the mechanisms of D N A replication in eukaryotes and prokaryotes. • Compare and contrast the various processes of D N A repair. • Discriminate between mechanisms of homologous recombination and techniques for site-specific recombination in order to better understand current gene-editing technologies.

MODULE 02 COMPANION GUIDE BCHM 218 MOLECULAR BIOLOGY | BCHM 218 M02 PAGE 11 D. MODULE ASSIGNMENTS ACTIVITIES THROUGHOUT THE MODULE Note that your responses within the module will NOT be graded. However, they are recorded in the module and can be viewed by your instructor The assessments listed must be completed by the end of Module 2. Module 2 - Quiz Problem-Based Activities - Refer to D.1 Problem Based Activities Biochemical Basis of Disease Assignment

MODULE 02 COMPANION GUIDE BCHM 218 MOLECULAR BIOLOGY | BCHM 218 M02 PAGE 12 D.1 PROBLEM-BASED ACTIVITIES SUB-PAGE OF D. MODULE ASSIGNMENTS PROBLEM-BASED ACIVITIES 1/1 You are required to complete four problem-based activities (PBAs) within your tutorial sessions. In each activity, you will be faced with a problem or research scenario that you must solve by applying module content to the situation presented. These scenarios are intended to simulate real life research applications in the field of molecular biology. Each P B A activity is worth 5%, for a total of 20% of your final grade. More details about each of the four P B As will be released on the course page closer to the due date.

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MODULE 02 COMPANION GUIDE BCHM 218 MOLECULAR BIOLOGY | BCHM 218 M02 PAGE 13 E. COURSE ICONS 4 OF 5 Throughout BCHM 218, there are several icons you should look out for. Some live in the sidebar. Select them for a description of what they signify. All the other icons live inline with the slide body. The following is a summary of every inline icon. Select them to see how they function. Big Buttons These elements control some dynamic content on the slide. Note that the blue signifies the display area. Hover Over Definitions When you hover over words or phrases that are formatted this way , a definition will appear. Circle Button This element toggles selected material or highlights related material on a slide. Inline Circle Button The circle at the end of this paragraph is similar in function to the Circle Button, but exists inline with the text body. It is used to highlight components of figures as they associate with the text body. Note the doctor. Toggle This control simply toggles between two views.

MODULE 02 COMPANION GUIDE BCHM 218 MOLECULAR BIOLOGY | BCHM 218 M02 PAGE 14 F. MODULE OUTLINE 5 OF 5 Section 1: The Beginnings of Replication Enzymology – Refer to 1.1 Section 2: D N A Replication – Refer to 2.1 Section 3: D N A Replication in the Model Organism E.coli – Refer to 3.1 Section 4: D N A Repair – Refer to 4.1 Section 5: D N A Recombination – Refer to 5.1 Section 6: Site-Specific Recombination – Refer to 6.1

MODULE 02 COMPANION GUIDE BCHM 218 MOLECULAR BIOLOGY | BCHM 218 M02 PAGE 15 SECTION 01: THE BEGINNINGS OF REPLICATION ENZYMOLOGY 1.1 SECTION 01: THE BEGINNINGS OF REPLICATION ENZYMOLOGY

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MODULE 02 COMPANION GUIDE BCHM 218 MOLECULAR BIOLOGY | BCHM 218 M02 PAGE 16 1.2 THE EUKARYOTIC CELL CYCLE 1 OF 11 You will begin your discussion of D N A replication with a refresher on the eukaryotic cell cycle. There are four stages of the cell cycle. Most of the cell’s time is spent in the G0 phase waiting to divide. When the cell starts the division process, most of its time is spent in the G1 phase, and it spends a slightly shorter time in the S (synthesis) phase. The cell spends very little time in the last two phases: the G2 phase, and M phase. Click each of the four phases in the figure for an overview of each phase. 1. G0/G1 Phase: Gap 0 and Gap 1. Cells are not actively dividing in this phase - it is the first growth phase of the cell cycle. Note: G 0 is not technically part of the cell cycle, but occurs when cells are not dividing - cells in this stage are often termed ‘quiescent.’ 2. S Phase: Synthesis. The cell replicates its entire genome in preparation for division. 3. G2 Phase: Gap 2. The second growth phase. The last chance for cells to grow before division. 4. M Phase: Mitosis. The phase where cell division occurs - the cell divides into two identical daughter cells. Reference: Cox, M. M., Doudna, J. A., & O'Donnell, M. (2012). Molecular Biology: Principles and Practice (2nd edition) . New York, N Y: W.H. Freeman and Co.

MODULE 02 COMPANION GUIDE BCHM 218 MOLECULAR BIOLOGY | BCHM 218 M02 PAGE 17 1.3 CELL CYCLE Activity: Drop Down Selection Please see the online learning module for the full experience of this interaction. 2 OF 11 Select the number of chromosomes (n) to the correct phase of the cell cycle. Note: Think the number of chromosomes at the end of each phase. Phase of the Cell Cycle Number of Chromosomes G1 Phase S Phase G2 Phase M Phase Feedback: Phase of the Cell Cycle Number of Chromosomes G1 Phase 2n: cells are diploid, containing two copies of each chromosome S Phase D N A replication is occurring - chromosomes are duplicated G2 Phase 4n: cells are tetraploid, containing four copies of each chromosome M Phase 2n: each daughter cells contains a complete set of chromosomes Present Icon Hint: ‘n’ is the number of unique chromosomes present at that step. In humans n = 23. You will learn more about the human chromosome in Module 03.

MODULE 02 COMPANION GUIDE BCHM 218 MOLECULAR BIOLOGY | BCHM 218 M02 PAGE 18 1.4 ARTHUR KORNBERG 3 OF 11 To start your study of D N A replication, you will learn about a scientist who contributed a great deal to the understanding of replication: Arthur Kornberg. Kornberg and his team made several critical discoveries with respect to the proteins that generate the D N A polymer, and how this process occurs. Learn a few of Kornberg’s discoveries: • Kornberg and his team developed an assay for measuring the amount of D N A synthesis that is occurring using bacterial cell extracts, which are prepared from lysing the bacteria to release all of the contents, including all of the proteins. • They discovered that several heat-stable factors (i.e. not proteins, which are generally denatured at high temperatures) are needed for the D N A synthesis reaction. They identified these to be nucleoside triphosphates. Using a multi-step fractionation of the bacterial lysate , they purified the protein that is now known as D N A polymerase I (Pol I). Definition: Lysate: Recall from Modules 01 that a lysate is the aqueous substance isolated following rupture of the cell membrane, which contains all contents of the cell.

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MODULE 02 COMPANION GUIDE BCHM 218 MOLECULAR BIOLOGY | BCHM 218 M02 PAGE 19 1.5 BEGINNINGS OF REPLICATION ENZYMOLOGY 4 OF 11 As you previously learned, cells go through a distinct cell replication cycle, with very little of their time actually spent in mitosis. The work of Arthur Kornberg and his team was instrumental to our understanding of how cells replicate their full complement of genomic D N A. This marked the ‘beginnings of biotechnology’ - the exploitation of biological information for other purposes. In other words, they were able to take genetic information and use it for technological purposes rather than biological ones. You will now explore some of Kornberg’s team’s accomplishments in more detail.

MODULE 02 COMPANION GUIDE BCHM 218 MOLECULAR BIOLOGY | BCHM 218 M02 PAGE 20 1.6 KORNBERG: PURIFICATION OF “POLYMERASE” 5 OF 11 One of Kornberg’s publications , Purification of “Polymerase,” involved a multistep process in which protein concentration and specific enzyme activity (D N A polymerization) were monitored. Kornberg and his team performed a multistep purification from bacterial lysate and monitored the activity of the proteins in the fractions for their capability to perform a certain function - in this case, D N A polymerization. The general steps of the multistep purification have been provided. You should recall the procedures mentioned from previous modules in the course. For more context, read the full article. Step I was to break the bacteria by sonication . Steps II through V removed D N A and R N A from the mixture. Step VI used salt (ammonium sulfate) to precipitate some proteins, and leave others in the solution, based on the properties of the proteins (as we talked about in Module 01). Finally, Step VII used a form of ion exchange chromatography to give a more purified product. Magnifying Glass Icon: If you’re interested, you can read the full paper. Definition: Sonication : The act of applying sound energy to agitate particles in a sample. Page Link: https://proxy.queensu.ca/login?url=http://dx.doi.org/10.1016/s0022-2836(64)80089-9

MODULE 02 COMPANION GUIDE BCHM 218 MOLECULAR BIOLOGY | BCHM 218 M02 PAGE 21 1.7 PURIFICATION OF “POLYMERASE” EXPERIMENT 6 OF 11 At each stage they measured the protein content in m g / m l. This was done using spectroscopy, S D S - PAGE, and additional assays to measure protein concentration in solution. They also measured the specific activity of the fraction, which means the ability of the enzyme to perform its function (i.e. polymerization of D N A). Specific activity increases with a more pure protein (in this case, polymerase). Recall your discussion of specific activity from Section 04 of Module 01. The table shows the steps that Kornberg and his colleagues took to purify polymerase. It is important to notice that with each purification step, the specific activity of the substance increases, suggesting to the team that their purification is working. Magnifying Glass Icon: If you’re interested, you can read the full paper. Page Link: http://www.jbc.org.proxy.queensu.ca/content/233/1/163 Reference: Lehman, I. R., Bessman, M. J., Simms, E. S., & Kornberg, A. (1958). Enzymatic synthesis of deoxyribonucleic acid. I. preparation of substrates and partial purification of an enzyme from escherichia coli . The Journal of Biological Chemistry, 233 (1), 163-170. Retrieved from http://www.jbc.org.proxy.queensu.ca/content/233/1/163.

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MODULE 02 COMPANION GUIDE BCHM 218 MOLECULAR BIOLOGY | BCHM 218 M02 PAGE 22 1.8 KORNBERG: REQUIREMENTS OF THE “POLYMERIZATION REACTION” 7 OF 11 Another of Kornberg's studies was determining which components are critical for D N A replication. He and his team deduced the requirements of the “Polymerization Reaction” by determining the maximal incorporation of deoxyribonucleotides into D N A under various conditions. They accomplished this by performing 8 different experiments. In each experiment, they removed one or more factors and used radiolabeled d T T P to monitor the effect on deoxyribonucleotide incorporation into the D N A.

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MODULE 02 COMPANION GUIDE BCHM 218 MOLECULAR BIOLOGY | BCHM 218 M02 PAGE 23 1.9 REQUIREME NTS OF THE “POLYMERIZATION REACTION” RESULTS 8 OF 11 Some of the results of Kornberg’s experiment “Requirements of the Polymerization Reaction” are shown in the table. In each experiment where a factor was removed, there was almost a complete loss of the incorporation of radiolabelled d T T P into D N A. Can you interpret what this means for the polymerization reaction? The conclusion reached by Kornberg and his team: RESULTS: This means the requirements of the polymerase reaction are: • All four nucleotides present in the cell, • An intact D N A template, and • A M g++ cofactor for the polymerase

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MODULE 02 COMPANION GUIDE BCHM 218 MOLECULAR BIOLOGY | BCHM 218 M02 PAGE 24 1.10 FOR YOUR INTEREST: KORNBERG’S PUBLICATIONS 9 OF 11 You may find it interesting to learn that although Kornberg’s work is regarded as important to understanding cell replication, the impact of Kornberg’s findings weren’t immediately apparent to the scientific community. His papers were declined by the Journal of Biological Chemistry when submitted in the fall of 1957. Among the critical comments were: • “It is very doubtful that the authors are entitled to speak of the enzymatic synthesis of D N A” • “Polymerase is a poor name” Eventually, they were published once the impact of the findings on the replication of D N A, and thus heredity in general, were understood. Magnifying Glass Icon : If you’re interested, you can read about Kornberg’s publications Page Link: http://www.jbc.org.proxy.queensu.ca/content/280/49/e46

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MODULE 02 COMPANION GUIDE BCHM 218 MOLECULAR BIOLOGY | BCHM 218 M02 PAGE 25 1.11 KORNBERG'S EXPERIMENTS Activity: Text Entry Please see the online learning module for the full experience of this interaction. 10 OF 11 Answer the questions about Kornberg’s experiments on D N A polymerase. 1 of 2: Describe how Korberg deduced the requirements of the polymerase reaction. 2 of 2: What was the key replication component that Kornberg purified? Feedback: 1. He observed that when he removed critical components of the replication reaction there was a near complete loss of enzyme activity, as measured by radiolabelled d T T P being incorporated into D N A. 2. D N A polymerase I

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MODULE 02 COMPANION GUIDE BCHM 218 MOLECULAR BIOLOGY | BCHM 218 M02 PAGE 26 1.12 SECTION 01 SUMMARY 11 OF 11 Arthur Kornberg and his team were the first to isolate the D N A polymerase enzyme. Among their other discoveries, they determined the components required for the ‘polymerase reaction’ - all four deoxyribonucleotides, a D N A template, and M g++ (a cofactor for the polymerase). Their discoveries, while underappreciated at the time, led to the field of replication enzymology. Throughout the rest of this module, you will learn about the mechanism of D N A replication and the structure and function of D N A polymerase.

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MODULE 02 COMPANION GUIDE BCHM 218 MOLECULAR BIOLOGY | BCHM 218 M02 PAGE 27 SECTION 02: D N A REPLICATION 2.1 SECTION 02: D N A REPLICATION

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MODULE 02 COMPANION GUIDE BCHM 218 MOLECULAR BIOLOGY | BCHM 218 M02 PAGE 28 2.2 INTRODUCTION TO D N A REPLICATION 1 OF 30 D N A replication is the synthesis of two daughter D N A molecules identical to the parental D N A. During division, the cell must completely duplicate its genetic material so each daughter cell will have a copy of the cellular genome. The process of D N A replication is how the stored genomic information is passed down to the next generation and, as such, it is essential for life and evolution. If you consider that each cell has approximately six billion base pairs in its diploid genome, you can appreciate that this is no small undertaking. The process of D N A replication is highly regulated and a cell will only divide if there is enough resources to form two new cells. This ensures that the cell does not waste any resources by beginning division and not being able to complete it. Although D N A replication is a complex process, with a variety of components, it can be condensed down to six key principles which you will learn about in the rest of this section. We will finish with an overview of the D N A polymerase enzyme.

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MODULE 02 COMPANION GUIDE BCHM 218 MOLECULAR BIOLOGY | BCHM 218 M02 PAGE 29 2.3 VIDEO: D N A REPLICATION 2 OF 30 In Section 01, you learned about an experiment where Kornberg and his team deduced the important components of D N A replication - all four nucleotides, a D N A template, and a M g 2+ cofactor for the polymerase. Please watch the video to learn about the process of D N A replication. Throughout this module you will learn about the various molecular components described in the video. As you watch: Pay special attention to the components of D N A replication you learned about in Section 01 and the tasks they perform during replication. Watch this Video: D N A Replication Page Link: https://www.youtube.com/watch?v=4jtmOZaIvS0

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MODULE 02 COMPANION GUIDE BCHM 218 MOLECULAR BIOLOGY | BCHM 218 M02 PAGE 30 2.4 GENERAL PRINCIPLES OF D N A REPLICATION 3 OF 30 There are six general principles of D N A replication which you will learn about this section. 1. D N A replication is semiconservative. 2. Replication is initiated at specific sites. 3. Replication is typically bidirectional. 4. Replication is semidiscontinous: continuous on the leading strand and discontinuous on the lagging strand. 5. R N A primers are needed to start D N A replication. 6. Nucleases, polymerases, and ligases replace the R N A primers with D N A and seal the remaining nick.

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MODULE 02 COMPANION GUIDE BCHM 218 MOLECULAR BIOLOGY | BCHM 218 M02 PAGE 31 2.5 1) D N A REPLICATION IS SEMICONSERVATIVE 4 OF 30 The first principle you will explore is that D N A replication is semiconservative. Watson and Crick hypothesized this method of replication based on their discovery of the structure of D N A, however, before this was scientifically proven there were two other D N A replication theories: conservative and dispersive. Learn about each theory of D N A replication: Conservative In the conservative theory of replication, it was believed that during D N A replication the parental strands remained together and acted as a template for the creation of new daughter strands. Image displaying the conservative theory of D N A replication, which results in one completely new helix and one completely old helix. Dispersive In the theory of dispersive replication, the parental strand and new daughter strand will be randomly mixed together creating strands having a patchwork of old and new D N A Image displaying the dispersive theory of D N A replication, which results in both D N A strands being a patchwork of old and new D N A. Semiconservative Based on the double-stranded nature of D N A, Watson and Crick hypothesized that the helix must be unwound for replication to occur. This method is described as a semiconservative replication, as each new D N A molecule conserves one strand from the parental D N A and the other strand is new. Image displaying the semiconservative theory of D N A replication, which results in strands that are half old D N A and half new D N A. This is the true method of D N A replication. Reference: Cox, M. M., Doudna, J. A., & O'Donnell, M. (2012). Molecular Biology: Principles and Practice (2nd edition) . New York, N Y: W.H. Freeman and Co.

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MODULE 02 COMPANION GUIDE BCHM 218 MOLECULAR BIOLOGY | BCHM 218 M02 PAGE 32 2.6 2) REPLICATION IS INITIATED AT SPECIFIC SITES 5 OF 30 The point where D N A synthesis starts is a specific sequence called the origin of replication. Replication always initiates at defined origins, with the participation of an origin recognition system. Circular bacterial chromosomes can have only one origin, whereas long, linear eukaryotic chromosomes are dotted with numerous origins. The point where the parental duplex separates and the daughter duplexes form is referred to as a replication fork , and the open D N A that is being replicated is referred to as a replication bubble . Click to toggle between an animated image and electron micrograph of the replication origin in a eukaryotic chromosome. Note: the multiple origins of replication. Diagram displaying the various origins of replication will continue to grow until they connect with each other creating the new D N A strand. Reference: Cox, M. M., Doudna, J. A., & O'Donnell, M. (2012). Molecular Biology: Principles and Practice (2nd edition) . New York, N Y: W.H. Freeman and Co.

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MODULE 02 COMPANION GUIDE BCHM 218 MOLECULAR BIOLOGY | BCHM 218 M02 PAGE 33 2.7 3) REPLICATION IS TYPICALLY BIDIRECTIONAL 6 OF 30 The replication fork is typically (but not always) bidirectional. This means it moves away from the origin of replication in both directions. Note in the figure, D N A replication is occurring both towards and away from the replication fork, but always in the 5’ - 3’ direction. Interesting Fact Icon: Bidirectional replication is the most common form of replication, but not completely universal. The E. coli Col E1 plasmid is an example of a circular plasmid with a defined origin that replicates in a single direction. The genomes of some organisms, such as bacteriophage Φ29 and adenovirus, are single, linear, double-stranded D N As with no internal origin. These genomes are replicated starting from their ends. Another form of replication, used by certain phages with a circular genome, involves a single replication fork that proceeds multiple times around the circular D N A to generate numerous copies of the genome.

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MODULE 02 COMPANION GUIDE BCHM 218 MOLECULAR BIOLOGY | BCHM 218 M02 PAGE 34 2.8 4) REPLICATION IS CONTINUOUS ON THE LEADING STRAND AND DISCONTINUOUS ON THE LAGGING STRAND 7 OF 30 Both daughter strands cannot be replicated in the same direction that the replication fork moves. This is because D N A is antiparallel and all known D N A polymerases synthesize D N A in the 5’ - 3’ direction. This led to the theory of a ‘leading’ strand and a ‘lagging’ strand. The leading strand creates one long D N A strand, whereas the lagging strand creates many smaller D N A strands beginning with a R N A primer. These short D N A segments are called Okazaki fragments. Learn about the leading and lagging strand: • The leading strand is synthesized continuously in the direction of the replication fork (5’ - 3’). • The lagging strand is synthesized discontinuously in the opposite direction of the replication fork, but still in the 5’ - 3’ direction. The strand must be reinitiated at intervals and synthesized in a series of fragments known as the Okazaki fragments. Interesting Fact Icon Lagging-strand Okazaki fragments are 1 to 2 kb long in bacteria, but are shorter - only 100 to 200 nucleotides - in eukaryotes. Reference: Cox, M. M., Doudna, J. A., & O'Donnell, M. (2012). Molecular Biology: Principles and Practice (2nd edition) . New York, N Y: W.H. Freeman and Co.

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MODULE 02 COMPANION GUIDE BCHM 218 MOLECULAR BIOLOGY | BCHM 218 M02 PAGE 35 2.9 SEMIDISCONTINUOUS REPLICATION 8 OF 30 This mode of replication, with one leading and one lagging strand, is described as semidiscontinuous because only one daughter strand is synthesized continuously (the leading strand); the other is made as a series of discontinuous fragments (the lagging strand). The “lagging” designation is based on the fact that some unreplicated single -stranded D N A is generated on the lagging strand by the moving replication fork, so the conversion to duplex D N A on this strand lags behind that of the leading strand. As you just learned, the semidiscontinuous mode of replication has a specific directionality - i.e. the 5’ - 3’ direction. Key Convention: Directionality of D N A Synthesis - Refer to 3.9.1 Key Convention: Directionality of D N A Synthesis

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MODULE 02 COMPANION GUIDE BCHM 218 MOLECULAR BIOLOGY | BCHM 218 M02 PAGE 36 2.9.1 KEY CONVENTION: DIRECTIONALITY OF D N A SYNTHESIS SUB-PAGE OF 2.9 SEMIDISCONTINUOUS REPLICATION KEY CONVENTION: DIRECTIONALITY OF D N A SYNTHESIS 1/1 The direction of synthesis by D N A polymerase refers to the direction in which each new nucleotide is chemi cally linked to the growing daughter strand. All known D N A polymerases function in the 5′ - 3′ direction, linking the α - 5′ - phosphate of a new d N T P to the 3′ - O H position of the nucleotide residue at the 3’ end of the chain. This covalent linkage is sho wn in the figure. The image shows how D N A is built. The incoming d N T P is partially composed of a triphosphate compound, once the new base is added onto the growing chain, the beta and gamma phosphates are released as pyrophosphate. Reference: Cox, M. M., Doudna, J. A., & O'Donnell, M. (2012). Molecular Biology: Principles and Practice (2nd edition) . New York, N Y: W.H. Freeman and Co.

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MODULE 02 COMPANION GUIDE BCHM 218 MOLECULAR BIOLOGY | BCHM 218 M02 PAGE 37 2.10 5) R N A PRIMERS ARE NEEDED TO START D N A REPLICATION 9 OF 30 Enzymes called “primases” synthesize primers when and where they are required. Both the leading and lagging strands require short 10-13 base pair R N A primers to begin the synthesis of D N A. The primer strands must be complementary to the template and contain a free 3’ - O H group. As you will discuss in the coming slides, once the synthesis is complete, the R N A primer is removed by nuclease action. The Okazaki fragments will be joined by ligase after the replication fork has passed. The primer is located on the top strand in this image. Without a primer it is not possible to add d N T Ps to the strand. Interesting Fact Icon Primers can also be composed of D N A. You will learn more about this in your disucssion of PCR in Module 03. Reference: Cox, M. M., Doudna, J. A., & O'Donnell, M. (2012). Molecular Biology: Principles and Practice (2nd edition) . New York, N Y: W.H. Freeman and Co.

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MODULE 02 COMPANION GUIDE BCHM 218 MOLECULAR BIOLOGY | BCHM 218 M02 PAGE 38 2.11 6) NUCLEASES, POLYMERASES, AND LIGASES 10 OF 30 Following replication, the R N A primers must be removed, which leaves small nicks or gaps within the synthesized D N A strand that need to be sealed together. Nucleases, polymerases , and ligases work together to replace the R N A primers with D N A to seal the remaining nick. The polyermase that does this process is D N A polymerase I (D N A Pol I). D N A Pol I is organized into three major domains: 1. The D N A polymerase 2. 3′ - 5′ proofreading exonuclease 3. 5′ - 3′ exonuclease Definitions: Nucleases: Enzymes that hydrolyze the internucleotide (phosphodiester) linkages of nucleic acids. Polymerases: Enzymes that catalyze template-dependent synthesis of D N A from its deoxyribonucleoside 5′ -triphosphate precursors. Ligases: Enzymes that create a phosphodiester bond between the 3′ end of one D N A segment and the 5′ end of another. Exonuclease: An enzyme that hydrolyzes only those phosphodiester bonds that are in the terminal positions of a nucleic acid.

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MODULE 02 COMPANION GUIDE BCHM 218 MOLECULAR BIOLOGY | BCHM 218 M02 PAGE 39 2.12 SEALING NICKS WITH D N A POL I 11 OF 30 At a nick in the D N A, Pol I degrades the R N A primer in the 5′ - 3′ direction, releasing d N M P s, and simultaneously extends the 3′ terminus with d N T Ps in the same direction. The net result is movement of the nick in the 5′ - 3′ direction along the D N A until all R N A is removed. D N A ligase can then enter and seal the fragments. You will learn more about the structure of D N A polymerase I in a few slides.

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MODULE 02 COMPANION GUIDE BCHM 218 MOLECULAR BIOLOGY | BCHM 218 M02 PAGE 40 2.13 ACTIVITY: GENERAL PRINCIPLES OF D N A REPLICATION Activity: Drop Down Selection Please see the online learning module for the full experience of this interaction. 12 OF 30 Using your knowledge of the six general principles of D N A replication, use the dropdown menu to match the principles with the correct definition. Bin : Bidrectional, Nucleases, Polymerases, and Lisages, Origin of Replication, Primers, Semiconservative, Semidiscontinuous Principle: Definition: The replicated D N A has one strand from the parent and one new strand. Replication begins at specific sites. The replication fork moves away from the origin of replication in both directions. Replication happens in a continuous manner on the leading strand and discontinuously on the lagging strand. Short R N A fragments are needed for both strands to begin replication. Enzymes are needed to repair the nicks or gaps in the newly replicated D N A strands. Feedback: Principle: Definition: Semiconservative The replicated D N A has one strand from the parent and one new strand. Origin of Replication Replication begins at specific sites.

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MODULE 02 COMPANION GUIDE BCHM 218 MOLECULAR BIOLOGY | BCHM 218 M02 PAGE 41 Bidirectional The replication fork moves away from the origin of replication in both directions. Semidiscontinuous Replication happens in a continuous manner on the leading strand and discontinuously on the lagging strand. Primers Short R N A fragments are needed for both strands to begin replication. Nucleases polymerases and lisages Enzymes are needed to repair the nicks or gaps in the newly replicated D N A strands.

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MODULE 02 COMPANION GUIDE BCHM 218 MOLECULAR BIOLOGY | BCHM 218 M02 PAGE 42 2.14 D N A POLYMERASE STRUCTURE 13 OF 30 Looking at a crystal structure of D N A Pol I can explain a lot about its function. The figure shows a bacterial polymerase, but the 3D structure is remarkably well-conserved for eukaryotic polymerases. The structure of D N A Pol I resembles a right hand, with three separate domains: fingers, thumb, and palm. Note: there is also a 3’ - 5’ exonuclease domain that assists in the proofreading function, which will be discussed later in the module. Diagram of D N A polymerase. D N A polymerase is shaped like your right hand, with the fingers and thumb curving up and around the D N A strand. An overview of each domain: Fingers Domain The ‘fingers’ domain of the protein is where the d N T Ps enter. Thumb Domain The ‘thumb’ domain hold s the template D N A strand in place so that it can incorporate many nucleotides without disruption. Palm Domain The D N A template strand (shown in blue) lays along the palm domain, which contains the active site of the enzyme. The template threads through the polymerase, which attaches d N T Ps to the growing synthesized strand (shown in red). Reference: Cox, M. M., Doudna, J. A., & O'Donnell, M. (2012). Molecular Biology: Principles and Practice (2nd edition) . New York, N Y: W.H. Freeman and Co.

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MODULE 02 COMPANION GUIDE BCHM 218 MOLECULAR BIOLOGY | BCHM 218 M02 PAGE 43 2.15 CONFORMATIONS OF D N A POLYMERASE I 14 OF 30 There are two conformations of D N A Pol I: open and closed. The polymerase undergoes a conformational change every time it incorporates a nucleotide into the growing D N A strand. Toggle between the two conformations of D N A Pol I. New d N T P is held in the fingers of polymerase I. When it is in the open form it looks like an open palm with the thumb curved up. When polymerase I goes into its closed form the fingers rotate 40 degrees towards the palm to allow the new d N T P to be added. Pol I, open form: D N A Pol I begins in what is known as the open form . In the open form, the free 3’ - O H of the strand being synthesized (shown in red in the figure) is sitting in the active site, and the fingers domain has contacted a d N T P. Pol I, closed form: In order for a chemical reaction to take place, the d N T P has to be brought in proximity to the growing strand and attached at the ‘insertion site’. This is accomplished by the enzyme adopting the closed form - where the fingers domain approaches the active site. Reference: Cox, M. M., Doudna, J. A., & O'Donnell, M. (2012). Molecular Biology: Principles and Practice (2nd edition) . New York, N Y: W.H. Freeman and Co.

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MODULE 02 COMPANION GUIDE BCHM 218 MOLECULAR BIOLOGY | BCHM 218 M02 PAGE 44 2.16 QUALITY CONTROL 15 OF 30 How does the polymerase incorporate the correctly paired nucleotide into the growing strand? The accuracy of the polymerase relies on shape recognition involving H- bonding, van der Waals forces, and ionic bonding interactions, all concepts you learned about in Module 01. A-T and G-C base pairs have shapes that fit well into the closed form of the active site. Incorrect base pairs do not fit well, and prevent complete closure of the polymerase around the D N A.

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MODULE 02 COMPANION GUIDE BCHM 218 MOLECULAR BIOLOGY | BCHM 218 M02 PAGE 45 2.17 INCORPORATION OF INCORRECT BASES 16 OF 30 From an enzymatic perspective, the incorporation of incorrect bases is 10 to 1000 fold slower than incorporation of correct bases. Further, correct base pairing and base stacking of the newly added nucleotide provides additional stabilization that favours incorporation of the correct base. An incorrect fit results in dissociation of the d N T P and binding of a new one. Normally, the polymerase is able to distinguish the correct nucleotide by this method, with one in every 10 4 to 10 5 nucleotides added incorrectly. In the event of a misincorporated nucleotide, the polymerase has a way to remove it, which you will learn about shortly.

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MODULE 02 COMPANION GUIDE BCHM 218 MOLECULAR BIOLOGY | BCHM 218 M02 PAGE 46 2.18 THE POLYMERASE ACTIVE SITE: CLOSED FORM 17 OF 30 The closed conformation of the fingers domain in Pol I allows conserved, functionally important residues to contact the sugar and phosphate moieties of the incoming nucleotide. Each of these conserved components has an important role in the stability and efficiency of the D N A synthesis reaction. Read an introduction to the three conserved components and their role in D N A synthesis stability: 1. Asp Residues Once the correct fit is found, the reaction proceeds by using two essential Asp residues in the active site (recall these are acidic residues, and will therefore have a negative charge at physiological p H), in order to coordinate two divalent M g 2+ ions. As you will see next, the amino acids do not play a direct role in catalysis - the metal ions do it all. 2. First Magnesium Ion The first M g 2+ ion acts to deprotonate the 3’ O H group of the growing strand, in order to generate a 3’O - nucleophile. This nucleophile is able to attack the alpha-P on the incoming d N T P, in order to form a phosphodiester bond and release pyrophosphate. 3. Second Magnesium Ion The second M g 2+ ion is responsible for binding the negatively charged phosphate groups on the incoming d N T P, and facilitating the rapid release of the P P i leaving group after addition of the nucleotide, so that the next nucleotide can enter the active site. The magnesium ions are located on either side of the two asparagine residues. The metal ions are coordinated by the negatively charged Asp and phosphate oxygens, which act to stabilize the nucleotide in the active site. Definitions: Moieties: Specific groups of atoms within a molecule that are responsible for characteristic chemical reactions of that molecule. Interesting Fact Icon Reading Reaction Mechanisms - Refer to 2.18.1 Recall: Reading Reaction Mechanisms References: Cox, M. M., Doudna, J. A., & O'Donnell, M. (2012). Molecular Biology: Principles and Practice (2nd edition) . New York, N Y: W.H. Freeman and Co. Helmenstine, P. A. (n.d.). Definition of Moiety. Retrieved August 22, 2017, from

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MODULE 02 COMPANION GUIDE BCHM 218 MOLECULAR BIOLOGY | BCHM 218 M02 PAGE 47 https://www.thoughtco.com/definition-of-moiety-605357

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MODULE 02 COMPANION GUIDE BCHM 218 MOLECULAR BIOLOGY | BCHM 218 M02 PAGE 48 2.18.1 RECALL: READING REACTION MECHANISMS SUB-PAGE OF 2.18 THE POLYMERASE ACTIVE SITE: CLOSED FORM RECALL: READING REACTION MECHANISMS 1/2 In order for you understand how an active site works, you need to remember how to read reaction mechanisms from your basic chemistry class. This information is provided as a refresher for those who need it. Chemical reaction mechanisms, which trace the formation and breaking of covalent bonds, are followed using dots and curved arrows - this is known as ‘electron pushing’. Curved arrows represent the movement of electron pairs, and non-bonded electrons important to the reaction mechanism are designated by a pair of dots. Many reactions involve an electron-rich atom (a nucleophile ) reacting with an electron-deficient atom (an electrophile ). The figure shows some common nucleophiles and electrophiles used in biochemical reactions. Common nucleophiles include: negatively charged oxygen (which can be either an unprotonated hydroxyl group or an ionized carboxylic acid), negatively charged sulfhydryl, carbanion, uncharged amine group, imidazole, and hydroxide ions. Common electrophiles include: the carbon atom of a carbonyl group (the more electronegative oxygen of the carbonyl group pulls electrons away from the carbon), protonated imine group (activated for nucleophilic attack at the carbon by protonation of the imine), the phosphorus of a phosphate group, and a proton. Reference: Cox, M. M., Doudna, J. A., & O'Donnell, M. (2012). Molecular Biology: Principles and Practice (2nd edition) . New York, N Y: W.H. Freeman and Co.

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MODULE 02 COMPANION GUIDE BCHM 218 MOLECULAR BIOLOGY | BCHM 218 M02 PAGE 49 2.18.2 QUESTION: POLYMERASE I Activity: Text Entry Please see the online learning module for the full experience of this interaction. SUB-PAGE OF THE POLYMERASE ACTIVE SITE: CLOSED FORM POLYMERASE I 2/2 Answer the question about the reaction mechanism of Pol I. Thinking back to the active site of Pol I, what would act as the electrophile and what would be the nucleophile? Feedback: In order for the reaction to take place, the first M g2+ would be the electrophile, which attacks the 3’ - O H group to generate the O- nucleophile. This O- can then perform a nucleophilic attack on the β -phosphate. The second M g2+ and the Asp residue are present to stabilize the nucleotide and facilitate the reaction.

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MODULE 02 COMPANION GUIDE BCHM 218 MOLECULAR BIOLOGY | BCHM 218 M02 PAGE 50 2.19 THE ATOMIC STRUCTURE OF THE POLYMERASE ACTIVE SITE 18 OF 30 The figure shows the atomic structure of a D N A polymerase active site, along with the position of the reactants involved in nucleotidyl transfer. The active site of D N A polymerase. Both of the oxygen atoms in both of the asparagine residues coordinate the magnesium ions. One of the ions forms bonds with the oxygen atoms in the beta phosphate and gamma phosphate. Whereas the other ion forms a bond with the 3 prime end of the growing strand and another bond with the alpha phosphate. The 3′ - O H of the primer strand terminus is positioned for an in-line attack on the α -phosphate, followed by the release of pyrophosphate. Oxygen atoms are shown in red, and the metal ion (A & B) coordination is represented as dotted lines. Although there is very little similarity between the amino acid sequences of the various polymerases, they seem to have adopted a common mechanism for nucleotidyl transfer. There are two metal ions (usually magnesium ions) in the active site that are ligated by the conserved acidic residues of the palm. In all cases, one metal ion contacts the phosphates of the bound nucleotide and the other metal ion is situated between the α - phosphate of the nucleotide and the 3′ -end of the primer strand. References: Doublié, S., Sawaya, M. R., & Ellenberger, T. (1999). An open and closed form for all polymerases. Structure, 7(2), R31-R35. doi: 10.1016/S0969-2126(99)80017-3.

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MODULE 02 COMPANION GUIDE BCHM 218 MOLECULAR BIOLOGY | BCHM 218 M02 PAGE 51 2.20 THE D N A POLYMERASE REACTION 19 OF 30 The structure of D N A polymerase and its active site are an integral part of the polymerase reaction, your next topic of study. When the template strand is inserted in polymerase I and new D N A is being synthesized there are two defined sites. The insertion site is where new d N T Ps are added and the postinsertion site is just before the insertion site. Learn about the steps in the D N A polymerase reaction: 1. D N A polymerases require several components for the polymerase reaction to proceed: a primer strand and a template strand (i.e. a primed template), as well as d N T Ps complementary to the template strand. The d N T Ps are added to the 3’ - O H group of the primer strand, causing the primer strand to grow in the 5’ - 3’ direction. 2. Once the d N M P is incorporated into the growing strand, Pol I will move forward to the new 3’ terminus to incorporate another d N T P. Therefore, the D N A polymerase active site must be divided into at least two distinct sites: these are called the postinsertion site and the insertion site . 3. The insertion site contains the template strand nucleotide that will pair with the incoming d N T P. The postinsertion site contains the 3’ - O H group of the primer strand. The 3′ - O H group of the primer strand then attacks the phosphodiester bond that connects the α - and β - phosphates of the incoming d N T P, resulting in the addition of a d N M P to the primer strand and the release of pyrophosphate. 4. After a d N M P is added to the primer strand, the new terminal base pair occupies the insertion site and must be translocated to the postinsertion site, allowing the next template nucleotide and a new d N T P to occupy the insertion site. Translocation occurs by sliding or dissociation of the enzyme from the D N A, followed by rebinding with the terminal base pair in the postinsertion site. Reference: Cox, M. M., Doudna, J. A., & O'Donnell, M. (2012). Molecular Biology: Principles and Practice (2nd edition) . New York, N Y: W.H. Freeman and Co.

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MODULE 02 COMPANION GUIDE BCHM 218 MOLECULAR BIOLOGY | BCHM 218 M02 PAGE 52 2.21 TRANSLOCATION OF D N A POLYMERASE 20 OF 30 The polymerase is able to slide along the template strand because of several positively charged Lys residues located within the palm domain, which interact with the negative phosphodiester backbone via electrostatic interactions. This frees the insertion site for the next incoming nucleotide. The strength of these interactions and the process of translocation give the polymerase its property of processivity - its ability to catalyze consecutive reactions without releasing its substrate. Some polymerases require high processivity (i.e. when they need to synthesize templates many nucleotides long) and others require low processivity (i.e. those polymerases responsible for synthesizing the few bases between two Okazaki fragments). Later in this module you will learn about another factor that increases processivity - the beta clamp. Definition: Processivity: The ‘processivity number’ is the average number of nucleotides incorporated before the enzyme dissociates from the D N A.

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MODULE 02 COMPANION GUIDE BCHM 218 MOLECULAR BIOLOGY | BCHM 218 M02 PAGE 53 2.22 PROCESSIVITY 21 OF 30 Processivity can increase the efficiency of D N A polymerization it avoids the need for polymerase to locate and rebind a 3′ primer strand terminus. Pol I has a relatively low processivity number whereas Pol III, like most D N A polymerases that replicate chromosomes (i.e. replicases), has a processivity number that is quite high. As you will see later in this module, this results from direct association of the polymerase with a protein ring that encircles the D N A. Diagram displaying polymerase slides along the D N A strand in a movement called translocation. References: Cox, M. M., Doudna, J. A., & O'Donnell, M. (2012). Molecular Biology: Principles and Practice (2nd edition) . New York, N Y: W.H. Freeman and Co.

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MODULE 02 COMPANION GUIDE BCHM 218 MOLECULAR BIOLOGY | BCHM 218 M02 PAGE 54 2.23 WHAT IF INCORRECT BASE PAIRING OCCURS? 22 OF 30 Sometimes, the wrong d N T P base gets incorporated into the growing D N A chain. When this occurs, the incorrect base pairs do not properly fit within the closed formation of the active site, as shown in the figure. Incorrect base pairing can have negative effects, such as mutations and disease, or can result in positive changes, such as adaptation and evolution. However, in the majority of cases, the polymerase will sense and remove an incorrectly paired nucleotide before translocation happens. You will now learn about how the polymerase recognizes these incorrect base pairs and substitutes them with the correct nucleotide. Diagram displaying correct compared to incorrect base pairing. The incorrect base pairs shown are T = G, C = A, and A = G. Reference: Cox, M. M., Doudna, J. A., & O'Donnell, M. (2012). Molecular Biology: Principles and Practice (2nd edition) . New York, N Y: W.H. Freeman and Co.

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MODULE 02 COMPANION GUIDE BCHM 218 MOLECULAR BIOLOGY | BCHM 218 M02 PAGE 55 2.24 EFFECTS OF MISPAIRING 23 OF 30 After adding a complementary nucleotide to the growing D N A chain, D N A polymerase is now in a position to use this new base as the site of the subsequent addition. However, it only makes an addition if the previously added base is firmly base paired. Failure to get everything lined up will mean D N A Pol cannot proceed - the key chemical groups just won't be in the right places for catalysis to occur. What happens in this case? Somewhere in the course of evolutionary history, the polymerase domain was joined with a non- specific exonuclease. This exonuclease function was incorporated into the polymerase as a separate domain - the 3’ - 5’ exonuclease domain mentioned earlier in the section. Click the button to show the location of the 3’ - 5’ exonuclease domain . 3’ - 5’ Exonuclease Domain The structure of D N A polymerase. The 3’ to 5’ exonuclease site is located in the “palm”. Definition: Exonuclease: An enzyme that works by cleaving nucleotides from the end of a chain one at a time. Reference: Cox, M. M., Doudna, J. A., & O'Donnell, M. (2012). Molecular Biology: Principles and Practice (2nd edition) . New York, N Y: W.H. Freeman and Co.

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MODULE 02 COMPANION GUIDE BCHM 218 MOLECULAR BIOLOGY | BCHM 218 M02 PAGE 56 2.25 3’ - 5’ EXONUCLEASE ACTIVITY OF D N A POL 24 OF 30 Once the incorrect base pair is sensed, the polymerase will ‘fray’ the double stranded D N A by 4 nucleotides - disrupting the hydrogen bonds between 3 correct and 1 incorrect base pairs. This places the incorrect d N M P in the 3’ - 5’ exonuclease active site, which will cleave the incorrect d N M P from the growing polypeptide chain. Animated diagram of Pol “fraying” the double - stranded D N A to place one strand in the 3’ to 5’ exonuclease site. The 3’ to 5’ site is located in the “palm”. Interesting Fact Icon: D N A pol produces errors at a rate of 1 in 104 to 1 in 105 nucleotides by incorrect base selection. The 3’ to 5’ exonuclease activity reduces this to about 1 in 102 to 1 in 103 bases. Reference: Cox, M. M., Doudna, J. A., & O'Donnell, M. (2012). Molecular Biology: Principles and Practice (2nd edition) . New York, N Y: W.H. Freeman and Co.

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MODULE 02 COMPANION GUIDE BCHM 218 MOLECULAR BIOLOGY | BCHM 218 M02 PAGE 57 2.26 ACTIVITY: PROOFREADING Activity: Drop Down Selection Please see the online learning module for the full experience of this interaction. 25 OF 30 The mechanism you just learned about is called proofreading, and it is a function possessed by most polymerases. Using the knowledge learned in this section and the figure, place the steps of proofreading in the correct order. Diagram of the steps in proofreading. First, polymerase mispairs a C with a T. Second, polymerase repositions the mispaired 3’ terminus into the 3’ to 5’ exonuclease site. Third, exonuclease hydrolyzes the mispaired C, releasing d C M P. Fourth, the 3’ terminus repositions back to the polymerase insertion site. Fifth, polymerase incorporates the correct nucleotide, A. • Polymerase mispairs a d N T P - a cytosine is added to the primer strand and paired with thymine, instead of the correct d N T P adenine. • 3’ - 5’ exonuclease hydrolyzes the mispaired cytosine. • The 3’ - O H terminus repositions back to the polymerase insertion site. • Polymerase incorporates the correct nucleotide (adenine) into the primer strand. • Polymerase repositions the mispaired 3’ - O H terminus into the 3’ - 5’ exonuclease site by fraying the strand by 4 base pairs. Feedback: 1. Polymerase mispairs a d N T P - a cytosine is added to the primer strand and paired with thymine, instead of the correct d N T P adenine. 2. Polymerase repositions the mispaired 3’ - O H terminus into the 3’ - 5’ exonuclease site by fraying the strand by 4 base pairs. 3. 3’ - 5’ exonuclease hydrolyzes the mispaired cytosine. 4. The 3’ - O H terminus repositions back to the polymerase insertion site. 5. Polymerase incorporates the correct nucleotide (adenine) into the primer strand. Reference:

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MODULE 02 COMPANION GUIDE BCHM 218 MOLECULAR BIOLOGY | BCHM 218 M02 PAGE 58 Cox, M. M., Doudna, J. A., & O'Donnell, M. (2012). Molecular Biology: Principles and Practice (2nd edition) . New York, N Y: W.H. Freeman and Co.

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MODULE 02 COMPANION GUIDE BCHM 218 MOLECULAR BIOLOGY | BCHM 218 M02 PAGE 59 2.27 3’ - 5’ EXONUCLEASE ACTIVE SITE 26 OF 30 The 3’ - 5’ exonuclease catalyzes the removal of the incorrect d N M P using a similar “two metal ion mechanism” as the polymerase active site during D N A synthesis. Illustration of the 3’ to 5’ exonuclease site of. Learn the function of each metal ion: • One M g 2+ ion deprotonates an H 2 O molecule to form an O H - nucleophile, which can act to attack the phosphate and mediate hydrolysis of the d N M P that is incorrect, leaving a free 3’ O H. • The other M g 2+ ion promotes departure of the incorrectly-paired d N M P leaving group from the growing strand. Interesting Fact Icon: Why do these enzymes using magnesium-based cofactors? Observing the particular coordination geometry of the two metal ions, Steitz and Steitz proposed that metal ion A reduces the p K a of a water molecule (or 3 O H) and assists nucleophilic attack via the formation of a hydroxide ion, and metal ion B stabilizes the pentacovalent intermediate and facilitates leaving of the 3′ oxyanion group. Reference: Cox, M. M., Doudna, J. A., & O'Donnell, M. (2012). Molecular Biology: Principles and Practice (2nd edition) . New York, N Y: W.H. Freeman and Co.

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MODULE 02 COMPANION GUIDE BCHM 218 MOLECULAR BIOLOGY | BCHM 218 M02 PAGE 60 2.28 POLYMERASE I Activity: Text Entry Please see the online learning module for the full experience of this interaction. 27 OF 30 Answer the question about proofreading and the polymerization reaction. Proofreading is not simply the reverse of the polymerization reaction. Now that you have learned about both proofreading and the polymerization reaction, identify one significant difference between these two processes. Think about the bases being added/released by D N A pol. Feedback: In proofreading, the mismatched nucleotide is released as d N M P not a d N T P, which was what initially was incorporated in the polymerization reaction. It is important to note that in proofreading, the same enzyme that makes D N A also degrades it. Recall that D N A Pol I has an active site for the polymerase reaction, one for the 3’ - 5’ exonuclease reaction, and yet another one for the 5’ - 3’ exonuclease reaction, which you will learn about in the coming slides).

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MODULE 02 COMPANION GUIDE BCHM 218 MOLECULAR BIOLOGY | BCHM 218 M02 PAGE 61 2.29 NICK TRANSLATION BY POL I (5-3 ’ EXONUCLEASE) 28 OF 30 Most classes of D N A polymerase have a 3’ - 5’ exonuclease, however Pol I is unique in that it also has a 5’ - 3’ exonuclease. This means that it can excise either R N A or D N A in the direction of polymerase movement - the 5’ - 3’ exonuclease will degrade an R N A or D N A strand in the duplex, while the polymerase simultaneously adds d N T Ps behind it. One of the most relevant examples of this function is the removal of the primer between Okazaki fragments. The unique 5′ - 3′ exonuclease activity found in Pol I reflects the enzyme’s role in D N A repair. Pol I performs a host of clean-up functions during replication, recombination, and repair, which require the trimming of single-stranded D N A ends and the removal of R N A primers or D N A lesions. You will learn more about these functions in Sections 04-06 of this module. Learn more about the process of nick translation: Nick Translation - Refer to 2.30.1 The Process of Nick Translation by Pol I Present Icon: Exonucleases that digest a D N A strand from the 3′ terminus are called 3′ - 5′ exonucleases, because the strand shortens at the 3′ end while the 5′ end remains intact. In contrast, 5′ - 3′ exonucleases digest D N A from the 5′ terminus, while the 3′ end rema ins intact.

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MODULE 02 COMPANION GUIDE BCHM 218 MOLECULAR BIOLOGY | BCHM 218 M02 PAGE 62 2.29.1 THE PROCESS OF NICK TRANSLATION BY POL I SUB- PAGE OF NICK TRANSLATION BY POL I (5′ - 3′ EXONUCLEASE) THE PROCESS OF NICK TRANSLATION BY POL I 1/1 Diagram of nick translation showing the 5’ to 3’ exonuclease site excising nucleic acid s, releasing N M Ps or d N M Ps. At this same time, d N T Ps are added to the other side of the nick, releasing P Pi.nAn overview of nick translation: 1. At a nick - in the figure, this is the gap between lagging strand fragments - Pol I degrades the R N A pr imer in the 5′ - 3′ direction. 1. This releases d N M Ps, and simultaneously extends the 3′ terminus with d N T Ps in the same direction. 2. The net result is movement of the nick in the 5′ - 3′ direction along the D N A until all R N A (purple) is removed. 3. D N A ligase (which isn’t shown in the figure) can then seal the fragments once Pol I dissociates. Note: The 5’ - 3’ exonuclease uses a two - metal ion hydrolysis reaction very similar to that of the 3’ - 5’ exonuclease. Reference: Cox, M. M., Doudna, J. A., & O'Donnell, M. (2012). Molecular Biology: Principles and Practice (2nd edition) . New York, N Y: W.H. Freeman and Co.

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MODULE 02 COMPANION GUIDE BCHM 218 MOLECULAR BIOLOGY | BCHM 218 M02 PAGE 63 2.30 QUESTION: EXONUCLEASES OF D N A POLYMERASE Activity: Text Entry Please see the online learning module for the full experience of this interaction. 29 OF 30 You have now learned about both the 3’ - 5’ and 5’ - 3’ exonuclease sites present in D N A polymerases. Answer the questions about the D N A polymerase exonucleases. 1 of 3: How does D N A polymerase deal with mispairing? 2 of 3: Compare and contrast a 3’ - 5’ exonuclease to a 5’ - 3’ exonuclease. 3 of 3: Can you think of a common example of when nick translation would be used? Feedback: 1. A mispaired 3’ terminus frays by four nucleotides to allow it to be inserted into the 3’ → 5’ exonuclease site, where the incorrect base pair can be removed. The strand is then repositioned back into the polymerase insertion site, where the correct base pair can be added 2. D N A polymerase contains two exonuclease domains - a 3’ - 5’ and a 5’ - 3’ exonuclease. Both of these domains hydrolyze the phosphodiester bonds within nucleic acids, removing nucleotides in the terminal position of the polypeptide chain. The 3’ -5 exonuclease assists in the proofreading function of D N A pol, catalyzing the removal of incorrect d N M Ps from the 3’ end of D N A. The 5’ - 3’ exonuclease is essential for D N A repair (particularly double strand breaks), and degrades R N A / D N A a t the 5’ end while the D N A pol adds d N T Ps to the strand. 3. One of the most relevant examples of nick translation by Pol I is the removal of the primer between Okazaki fragments on the lagging strand.

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MODULE 02 COMPANION GUIDE BCHM 218 MOLECULAR BIOLOGY | BCHM 218 M02 PAGE 64 2.31 SECTION 02 SUMMARY 30 OF 30 In Section 02, you learned about the six key principles of D N A replication and the structure and function of D N A polymerase. There are a few key takeaways from this section: • A d N T P that correctly base pairs to the template strand is incorporated onto the primer strand with the release of pyrophosphate (P Pi). • D N A polymerases are inherently very accurate, and are made even more accurate by a proofreading 3′ - 5′ exonuclease. • Pol I also contains a 5′ - 3′ exonuclease that degrades D N A while, at the same time, the polymerase synthesizes D N A, in the process of nick translation. • Like the polymerase active site, exonuclease sites use a two metal ion mechanism for removal of incorrect d N M Ps.

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MODULE 02 COMPANION GUIDE BCHM 218 MOLECULAR BIOLOGY | BCHM 218 M02 PAGE 65 SECTION 03: D N A REPLICATION IN THE MODEL ORGANISM E. COLI 3.1 SECTION 03: D N A REPLICATION IN THE MODEL ORGANISM E. COLI

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MODULE 02 COMPANION GUIDE BCHM 218 MOLECULAR BIOLOGY | BCHM 218 M02 PAGE 66 3.2 INTRODUCTION TO E. COLI REPLICATION 1 OF 39 Now that you have the foundational knowledge of D N A replication, you will turn your attention to replication in Escherichia coli (E.coli) specifically, with additional comparison to eukaryotic replication. Molecular biologists often look to E. coli when studying the basic mechanisms of molecular biology, including replication, because it is very well understood when compared to other organisms. E. coli is an ideal model because of its simplicity and rapid growth. Throughout this section, you will learn about the D N A polymerases found in E. coli , the three stages of their replication process, and their replication machinery. You may recall this information from M I C R 271 : Introduction to Microbiology , or you may wish to take this course for a more in-depth look at replication in microorganisms.

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MODULE 02 COMPANION GUIDE BCHM 218 MOLECULAR BIOLOGY | BCHM 218 M02 PAGE 67 3.3 E. COLI CONTAINS FIVE D N A POLYMERASES 2 OF 39 You will begin your exploration of E. coli replication by learning about E. coli’s five D N A polymerases, which are involved in a variety of cellular processes. Read a brief introduction to the function of each polymerase. Pay special attention to which classes have 5’ - 3’ and 3’ - 5’ exonuclease activity. TABLE 11-1 The Five D N A Polymerases of E. coli D N A Polymerase Number of Subunits Mass ( k D a) Gene(s) Function 3’ - > 5’ Exonuclease? 5’ - > 3’ Exonuclease? Pol I 1 103 pol A Okazaki fragment processing and D N A repair Yes Yes Pol II 1 88 pol B Translesion synthesis Yes No Pol III 3 167 d n a E, hol E, d n a Q Chromosome replication Yes No Pol IV 1 40 din B Translesion synthesis No No Pol V 2 69 umu C, umu D Translesion synthesis No No Pol I mainly functions in processing Okazaki fragments and in the repair of damaged D N A, rather than in chromosome replication. The function of Pol II is not fully understood, however it is thought to be involved in D N A repair. In E. coli, Pol III is the polymerase that replicates the chromosome. Pol III is sometimes referred to as the replicase .

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MODULE 02 COMPANION GUIDE BCHM 218 MOLECULAR BIOLOGY | BCHM 218 M02 PAGE 68 Pol IV and Pol V were discovered much later than the other polymerases. They are different in that they lack a 3′ - 5′ proofreading exonuclease and thus often incorp orate the wrong nucleotide. These low- fidelity polymerases are produced in cells when the D N A sustains damage that stalls the replication fork (discussed later in this section). The reduced accuracy of Pol IV and Pol V enables them to insert an incorrect nucleotide opposite a damaged template base. Although this results in an error, it gets the replication fork moving again. The ability of the replication fork to move past a damaged site is a matter of life and death, and all cells - bacterial, archaeal, and eukaryotic alike - contain these error- prone “ translesion ” D N A polymerases.

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MODULE 02 COMPANION GUIDE BCHM 218 MOLECULAR BIOLOGY | BCHM 218 M02 PAGE 69 3.4.A E. COLI D N A POLYMERASES Activity: Drop Down Selection Please see the online learning module for the full experience of this interaction. 3a OF 39 Use what you have learned about E. coli polymerases to complete the activity. Select the “functions” of each Pol to the correct locations. D N A Polymerase: Function: Pol I Pol II Pol III Pol IV and V Feedback: D N A Polymerase: Function: Pol I Repair of damaged D N A Pol II D N A repair Pol III Chromosomal replication Pol IV and V Enables replication fork to move past a damaged site

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MODULE 02 COMPANION GUIDE BCHM 218 MOLECULAR BIOLOGY | BCHM 218 M02 PAGE 70 3.4.B E. COLI D N A POLYMERASES Activity: Drop Down Selection Please see the online learning module for the full experience of this interaction. 3b OF 39 Use what you have learned about E. coli polymerases to complete the activity. Select the “functions” of each Pol to the correct locations and select whether or not each polymerase contains 3’ - 5’ exonuclease activity, 5’ - 3’ exonuclease activity, or both. D N A Polymerase: Function: 3’ - 5’ Exonuclease: 5’ - 3’ Exonuclease: Pol I Repair of damaged D N A Pol II D N A repair Pol III Chromosomal replication Pol IV and V Enables replication fork to move past a damaged site Feedback: D N A Polymerase: Function: 3’ - 5’ Exonuclease: 5’ - 3’ Exonuclease: Pol I Repair of damaged D N A Activity Activity Pol II D N A repair Activity Pol III Chromosomal replication Activity Pol IV and V Enables replication fork to move past a damaged site

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MODULE 02 COMPANION GUIDE BCHM 218 MOLECULAR BIOLOGY | BCHM 218 M02 PAGE 71 3.5 STAGES OF E. COLI REPLICATION 4 OF 39 E. coli replication takes place in three main stages: 1. Initiation of Replication 2. Elongation by the Replisome 3. Termination You will start by examining initiation of replication in E. coli because it is understood in great detail and will serve as a model of the events involved in initiation of replication in eukaryotic cells. You will finish off this section by learning about elongation by the replisome and termination of E. coli replication.

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MODULE 02 COMPANION GUIDE BCHM 218 MOLECULAR BIOLOGY | BCHM 218 M02 PAGE 72 3.6 1) INITIATION OF REPLICATION 5 OF 39 The first stage of E. coli replication is initiation. You will first take a look at the structural elements required for the E. coli replication origin, before examining the general steps in activation this origin. You will continue your discussion of the initiation stage with a look at regulation of D N A replication, as the initiation phase is the primary point of control. Finally, you will compare the E. coli initiation stage to eukaryotic initiation.

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MODULE 02 COMPANION GUIDE BCHM 218 MOLECULAR BIOLOGY | BCHM 218 M02 PAGE 73 3.7 STRUCTURAL ELEMENTS OF THE E. COLI REPLICATION ORIGIN 6 OF 39 Recall from the second principle of D N A replication, that the origin of replication is the specific site where D N A unwinds and replication is initiated, generating replication forks . Depending on the organism, there may be one or multiple origins of replication. E. coli has a single origin of replication, referred to as oriC , which is 245 base pairs long. D N A sequences of known E.coli origins share two common features. Diagram of E. coli origin in duplex D N A. On the left are the three 13-mer repeats, together called the D N A unwinding element, with a 5’ to 3’ directionality. On the right are the D n a A 9 -mer sites. R1 and R3 have a directionality opposite to that of the 13-mer repeats, while the directionality of R2 and R4 is in the same direction. Learn about these two features. 1. D n a A 9-mer sites: Four copies of a nine-nucleotide (9-mer) consensus sequence (called R sites, for “repeat”) to which the bacterial initiator protein , D n a A binds. 2. A=T rich 13-mer repeats: To one side of the D n a A 9-mer sites are three A=T-rich direct repeats (sequences repeated with the same directionality) of 13 b p each. These repeats, referred to as the D N A unwinding element, unwind readily (but not spontaneously) upon binding of the initiator. Many origins of replication contain A=T-rich repeats that probably function in a similar way. Interesting Fact Icon: All of the D N A replicated from a particular origin of replication is defined as a replicon (i.e. the total length of D N A replicated from one origin). For bacteria, which generally have only one origin, the entire chromosome is the replicon. In eukaryotes, it is the section of D N A replicated from one of its many origins. Reference: Cox, M. M., Doudna, J. A., & O'Donnell, M. (2012). Molecular Biology: Principles and Practice (2nd edition) . New York, N Y: W.H. Freeman and Co.

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MODULE 02 COMPANION GUIDE BCHM 218 MOLECULAR BIOLOGY | BCHM 218 M02 PAGE 74 3.8 ACTIVATION OF THE E. COLI REPLICATION ORIGIN (ORIC) 7 OF 39 The activation of the oriC and the assembly of the bacterial replication forks occurs in five steps. Review each step in the activation of the oriC. 1. Generating the Open Complex - Refer to 3.8.1 1) Generating the Open Complex 2. Activation of the Replication Origin - Refer to 3.8.2 2) Activation of the Replication Origin 3. Assembly of the E. coli Replication Forks - Refer to 3.8.3 3) Assembly of the E. coli Replication Forks 4. Replication Initiation and Leading Strand Synthesis - Refer to 3.8.4 4) Replication Initiation and Leading Strand Synthesis 5. Lagging Strand Synthesis - Refer to 3.8.5 5) Lagging Strand Synthesis in E. coli

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MODULE 02 COMPANION GUIDE BCHM 218 MOLECULAR BIOLOGY | BCHM 218 M02 PAGE 75 3.8.1 1) GENERATING THE OPEN COMPLEX SUB-PAGE OF ACTIVATION OF THE E. COLI REPLICATION ORIGIN ( ORIC ) 1) GENERATING THE OPEN COMPLEX 1/1 D n a A is the initiator protein at the E. coli oriC . D n a A is a member of the A A A+ protein family and, like most A A A+ proteins, binds and hydrolyzes A T P (although the turnover is very slow). Upon binding at the oriC , D n a A oligomerizes and wraps the D N A around the oligomer complex like a scarf, putting ‘strain’ on the D N A strand. In the presence of A T P, D n a A will destabilize the A=T-rich 13-mer repeats of the oriC , forming a s s D N A bubble. Formation of this bubble is also facilitated by H U , a small, basic, histone-like protein. Because the D n a A – A T P- oriC -H U complex forms a bubble at the origin, it is referred to as the open complex. The purpose of this first step is to generate the open complex so the replication machinery can bind at the A-T rich region to begin replication. D n a A oligomerizes upon binding the origin, and wraps the D N A around the oligomer. H U facilitates open complex formation. Definitions: Initiator protein : A protein that binds specific sites in an origin of replication and serves as a nucleation site for the assembly of protein complexes necessary to initiate replication. A A A+ proteins: A family of proteins with A T Pase activity that share a common structural domain, called the A A A domain, which assist in protein and D N A conformational changes. Reference: Cox, M. M., Doudna, J. A., & O'Donnell, M. (2012). Molecular Biology: Principles and Practice (2nd edition) . New York, N Y: W.H. Freeman and Co.

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MODULE 02 COMPANION GUIDE BCHM 218 MOLECULAR BIOLOGY | BCHM 218 M02 PAGE 76 3.8.2 2) ACTIVATION OF THE REPLICATION ORIGIN SUB-PAGE OF ACTIVATION OF THE E. COLI REPLICATION ORIGIN (ORIC) 2) ACTIVATION OF THE REPLICATION ORIGIN 1/1 Now that the D N A is in the open complex and the single stranded regions are exposed, two hexamers of the D n a B helicase assemble - one on each strand. A second protein , D n a C, is required for loading D n a B onto the s s D N A in the open complex. D n a C is a helicase loading protein that requires A T P for binding and is also part of the A A A+ protein family. D n a C pries open the hexameric ring of D n a B and slips it onto the s s D N A at the bubble. The A T P- bound form of D n a C binds tightly to D n a B and represses its helicase activity. Upon hydrolysis of A T P, D n a C is ejected from D n a B, which assembles it onto s s D N A at the bubble. This protein-D N A assembly at oriC is called the prepriming complex. D n a C loads D n a B onto each strand, forming the prepriming complex . Definition: Prepriming Complex: The complex of proteins assembled at oriC in E. coli at an early stage of replication fork assembly. Reference: Cox, M. M., Doudna, J. A., & O'Donnell, M. (2012). Molecular Biology: Principles and Practice (2nd edition) . New York, N Y: W.H. Freeman and Co.

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MODULE 02 COMPANION GUIDE BCHM 218 MOLECULAR BIOLOGY | BCHM 218 M02 PAGE 77 3.8.3 3) ASSEMBLY OF THE E. COLI REPLICATION FORKS SUB-PAGE OF ACTIVATION OF THE E. COLI REPLICATION ORIGIN (ORIC) 3) ASSEMBLY OF THE E. COLI REPLICATION FORKS 1/1 D n a B binds A T P, which enables it to translocate and unwind D N A, dislodging the D n a A protein. Unwinding happens outwards in both directions, widening the replication bubble. This unwinding generates positive supercoil stress in the D N A ahead of the replication fork, and this stress must be removed by the action of a topoisomerase enzyme. The newly unwound D N A in the replication bubble is also coated with single-stranded binding proteins (S S B) to protect it. Before D N A synthesis can begin, R N A primers must be synthesized by primase. Primase cannot interact with D n a B until the bubble grows to around 100 to 200 base pairs in size. D n a B helicase expands the replication bubble, and primase forms R N A primers at each replication fork. Definition: Topoisomerase: An enzyme that catalyzes alterations in D N A topology, introducing and/or removing positive or negative supercoils in closed, circular duplex D N A. Reference: Cox, M. M., Doudna, J. A., & O'Donnell, M. (2012). Molecular Biology: Principles and Practice (2nd edition) . New York, N Y: W.H. Freeman and Co.

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MODULE 02 COMPANION GUIDE BCHM 218 MOLECULAR BIOLOGY | BCHM 218 M02 PAGE 78 3.8.4 4) REPLICATION INITIATION AND LEADING STRAND SYNTHESIS SUB-PAGE OF ACTIVATION OF THE E. COLI REPLICATION ORIGIN (ORIC) 4) REPLICATION INITIATION AND LEADING STRAND SYNTHESIS 1/1 Once the R N A primer has been made, it directs loading of the β clamp and assembly of the leading strand Pol III holoenzyme . You will learn about the Pol III holoenzyme and the β clamp in more detail in the second stage of E. coli replication, elongation by the replisome. R N A primers direct beta clamp loading and assembly of the leading-strand Pol III core complexes. Core complexes advance the replication to opposite D n a B helicases. Definitions: β clamp: A component of the E. coli D N A polymerase III holoenzyme. The ring-shaped homodimer encircles and slides along the duplex D N A ahead of the Pol III core to which it is attached, greatly enhancing the processivity of D N A synthesis. Holoenzyme: The active form of an enzyme. An enzyme that a co-factor but is not bound to it is called an apoenzyme, and a holoenzyme is the apoenzyme bound to its co-factor. References: Cox, M. M., Doudna, J. A., & O'Donnell, M. (2012). Molecular Biology: Principles and Practice (2nd edition) . New York, N Y: W.H. Freeman and Co. Holoenzymes. Retrieved from https://www.nature.com/subjects/holoenzymes.

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MODULE 02 COMPANION GUIDE BCHM 218 MOLECULAR BIOLOGY | BCHM 218 M02 PAGE 79 3.8.5 5) LAGGING STRAND SYNTHESIS IN E. COLI SUB-PAGE OF ACTIVATION OF THE E. COLI REPLICATION ORIGIN (ORIC) 5) LAGGING STRAND SYNTHESIS IN E. COLI 1/1 Each Pol III holoenzyme extends its R N A primer until both Pol complexes connect with the D n a B helicase travelling in the same direction on the other side of the replication bubble. The replication fork moves rapidly - producing more s s D N A for primase to act upon as it goes. Priming is followed by clamp loading and synthesis of the lagging strand by a second Pol III enzyme in the complex, generating multiple Okazaki fragments discontinuously, along with the synthesis of the leading strand simultaneously. Therefore, there are four simultaneously functioning Pol III holoenzymes at each replication bubble. Replication continues to the end of the template or until another replication fork from an adjacent origin of replication is reached. This is how two bidirectional replication forks can be extended, and the D N A replicated, starting at oriC. Continued D n a B helicase activity allows lagging-strand priming and Okazaki fragment synthesis. Definition: Clamp loading: The placement of the beta clamp onto the D N A. Reference: Cox, M. M., Doudna, J. A., & O'Donnell, M. (2012). Molecular Biology: Principles and Practice (2nd edition) . New York, N Y: W.H. Freeman and Co.

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MODULE 02 COMPANION GUIDE BCHM 218 MOLECULAR BIOLOGY | BCHM 218 M02 PAGE 80 3.9 INITIATION OF REPLICATION Activity: Text Entry and Multiple Choice Please see the online learning module for the full experience of this interaction. 8 OF 39 Answer the questions about the first stage of E. coli replication, replication initiation. 1 of 3: What is the purpose of generating the open complex in the initiation of replication? 2 of 3: Do A A A+ proteins require A T P? 3 of 3: What are the components of the prepriming complex? Feedback: 1. The purpose of generating the open complex is to allow the replication machinery to bind at the A-T rich region to begin replication. 2. Yes 3. The prepriming complex includes D n a A bound to the D N A and D n a B helicases stabilizing the single strands of D N A at the replication bubble.

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MODULE 02 COMPANION GUIDE BCHM 218 MOLECULAR BIOLOGY | BCHM 218 M02 PAGE 81 3.10 EUKARYOTES VS E COLI 9 OF 39 Now that you’ve explored initiation in E. coli, you will now learn about the initiation step in eukaryotes. Eukaryotes have much greater D N A content than E. coli , and slower replication forks. As a result, eukaryotes need multiple origins of replication on each chromosome to allow for the complete replication of D N A in the 24-hour division time of, for example, the human cell. Origins of replication are 10-40 kilobase pairs apart along each chromosome, and multiple replication forks eventually meet to yield the two daughter chromosomes. “Firing” (activation) of the origins are under tight control, similar to those in E. coli , as is reinitiation at a single origin that has already been duplicated – eukaryotic origins fire once per cell cycle.

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MODULE 02 COMPANION GUIDE BCHM 218 MOLECULAR BIOLOGY | BCHM 218 M02 PAGE 82 3.11 EUKARYOTES VS E COLI Activity: Text Entry Please see the online learning module for the full experience of this interaction. 10 OF 39 Answer the question about the eukaryotic cell cycle. Recall from Section 01 that eukaryotes have defined phases of the cell cycle. In what phase(s) does (1) D N A replication and (2) separation of the duplicated chromosomes occur? Feedback: Replication occurs in the S phase of the cell cycle and separation of the duplicated chromosomes in M phase. Assembly of the replication protein complexes occurs late in the G1 phase and marks origins that will be used for replication during the S phase. This separation of events in the cell cycle is critical to the exquisite coordination that eukaryotes require to duplicate their long, linear chromosomes.

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MODULE 02 COMPANION GUIDE BCHM 218 MOLECULAR BIOLOGY | BCHM 218 M02 PAGE 83 3.12 ASSEMBLY OF BACTERIAL REPLICATION Activity: Text Entry Please see the online learning module for the full experience of this interaction. 11 OF 39 Answer the questions about the initiation of E. coli replication. Describe the steps involved in the assembly of bacterial replication forks at oriC. Feedback: • The A T P on D n a B allows it to translocate and unwind D N A, dislodging the D n a A protein. • Unwinding generates positive supercoil stress in the D N A ahead of the replication fork, and this stress must be removed by the action of a topoisomerase. • Newly unwound D N A is coated with single-stranded binding proteins (S S B) to protect it. • R N A primers are synthesized by primase.

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MODULE 02 COMPANION GUIDE BCHM 218 MOLECULAR BIOLOGY | BCHM 218 M02 PAGE 84 3.13 COMPONENTS OF EUKARYOTIC INITIATION 12 OF 39 There are many proteins involved in the initiation stage of eukaryotic replication, and many of these resemble proteins from bacterial replication. A simple eukaryote, Saccharomyces cerevisiae (brewer’s yeast) , is commonly used to study the assembly of replication forks in eukaryotes. An overview of replication initiation in S. cerevisiae: 1. S. cerevisiae has well-defined replication origins referred to as A R S (autonomously replicating sequences). A R S are 100 to 200 base pairs long and contain four common components: a highly conserved A sequence (for initiator binding), and B1, B2, and B3 elements (A T-rich). 2. The eukaryotic initiator is a heterohexamer called the origin recognition complex (O R C). The O R C has subunits with actions similar to D n a A, and A T P is required for O R C binding to the origin. 3. After O R C binds to D N A, the C d c 6 protein (also an A A A+ protein) binds to O R C (like H U in E. coli replication). The O R C – C d c 6 complex then loads the M c m 2-7 complex onto D N A . The M c m 2-7 helicase (D n a B-like) is a circular heterohexamer that binds one molecule of C d t 1 (D n a C-like), which is required before the O R C-C d c 6 complex can load the M c m 2-7 complex onto the D N A. These events occur only in the G 1 phase of the cell cycle. The resulting complex of O R C, C d c 6, C d t 1, and M c m 2-7 is referred to as the prereplication complex (pre R C). 4. Following the orchestrated addition of many proteins involved in replication and several ‘checkpoints’ to ensure that replication should proceed, the replication complex (R C) can begin to duplicate the D N A in S phase. Visual representation of the steps described. Definition: Cox, M. M., Doudna, J. A., & O'Donnell, M. (2012). Molecular Biology: Principles and Practice (2nd edition) . New York, N Y: W.H. Freeman and Co.

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MODULE 02 COMPANION GUIDE BCHM 218 MOLECULAR BIOLOGY | BCHM 218 M02 PAGE 85 3.14 2) E. COLI REPLICATION: ELONGATION BY THE REPLISOME 13 OF 39 Now that you’ve learned about the initiation of replication in both prokaryotes and eukaryotes, we will now move on to the second stage of replication using the same prokaryotic example, E coli. The second stage of E. coli replication is elongation by the replisome. Advancement of the replication fork requires the coordinated action of many different proteins. You will now take a closer look at the protein complex associated with one replication fork in prokaryotic replication, using the Pol III replicase to visualise how replication can happen simultaneously on both strands. Definition: Pol III replicase: Recall that this is the main polymerase involved in chromosomal replication in prokaryotes.

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MODULE 02 COMPANION GUIDE BCHM 218 MOLECULAR BIOLOGY | BCHM 218 M02 PAGE 86 3.15 THE E. COLI ‘REPLISOME’ 14 OF 39 The replisome is a multiprotein complex that promotes D N A synthesis at the replication fork. In E. coli, the replisome is composed of the Pol III holoenzyme, D n a B helicase, and primase. The system is even more complex in eukaryotes. An overview of the replisome components: Pol III Holoenzyme A 22-subunit D N A polymerase III complex found in E. coli responsible for D N A replication. It consists of three Pol III cores, three β sliding clamps, and a clamp loading complex. D n a B Helicase An enzyme in E. coli that opens the replication fork during D N A replication. Primse An enzyme that catalyzes the formation of R N A oligonucleotides used as primers by D N A polymerase. Animated diagram of the replisome. Reference: Cox, M. M., Doudna, J. A., & O'Donnell, M. (2012). Molecular Biology: Principles and Practice (2nd edition) . New York, N Y: W.H. Freeman and Co.

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MODULE 02 COMPANION GUIDE BCHM 218 MOLECULAR BIOLOGY | BCHM 218 M02 PAGE 87 3.16 THE ‘REPLISOME’ 15 OF 39 D n a B helicase connects to the Pol III core through the τ subunits of the clamp loader within the Pol III holoenzyme. Without this connection to polymerase, D n a B helicase is slow - it has low processivity - unwinding about 35 base pairs per second (b p/s). Upon connection of D n a B to the Pol III holoenzyme, unwinding proceeds at a rate of approximatel y 700 b p/s. The three τ subunits of the τ complex clamp loader bind three Pol III cores, and these same τ subunits also bind the D n a B helicase. This figure shows one Pol III core associated with the leading strand and two with the lagging strand. The leading-strand Pol III- β clamp complex moves continuously with D n a B helicase, while the lagging - strand Pol III- β clamp complex repeatedly moves on and off the D N A to extend the multiple R N A primers made by primase. Diagram of the replisome. Definitions: τ: The Greek letter tau. Clamp loader: A multiprotein complex that catalyzes the assembly of the sliding clamps onto D N A. Reference: Cox, M. M., Doudna, J. A., & O'Donnell, M. (2012). Molecular Biology: Principles and Practice (2nd edition) . New York, N Y: W.H. Freeman and Co. Clamp-loader. Retrieved from http://www.medilexicon.com/dictionary/17994.

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MODULE 02 COMPANION GUIDE BCHM 218 MOLECULAR BIOLOGY | BCHM 218 M02 PAGE 88 3.17 D N A POLYMERASE III HOLOENZYME 16 OF 39 The figure shows 20 of the 22 subunits that form the Pol III holoenzyme. The C- termini of the τ subunits protrude from the clamp loader and bind the Pol III cores. Each Pol III core also attaches to a β sliding clamp. Diagram of the D N A Pol III holoenzyme. Shown are three Pol III cores each associated with a beta sliding clamp and with the τ subunit of the clamp loader. Learn about the structures and functions of the β clamp and Pol III core: β clamp - Refer to 3.20.1 β Clamp Pol III core - Refer to 3.20.2 Pol III Core Reference: Cox, M. M., Doudna, J. A., & O'Donnell, M. (2012). Molecular Biology: Principles and Practice (2nd edition) . New York, N Y: W.H. Freeman and Co.

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MODULE 02 COMPANION GUIDE BCHM 218 MOLECULAR BIOLOGY | BCHM 218 M02 PAGE 89 3.17 .1 Β CLAMP SUB-PAGE OF D N A POLYMERASE III HOLOENZYME β CLAMP 1/1 Learn about the structure and function of the β clamp: • β clamps are homodimers, shaped like a ring that encircles the D N A duplex. • The function of the β clamp is to hold the polymerase onto the D N A while sliding along the duplex. This converts the Pol III core from a distributive enzyme, which moves on and off D N A as it works, to a processive enzyme that stays attached to D N A during repetitive cycles of d N M P incorporation. Diagram of the D N A Pol III holoenzyme. Shown are three Pol III cores each associated with a beta sliding clamp and with the τ subunit of the clamp loader. Reference: Cox, M. M., Doudna, J. A., & O'Donnell, M. (2012). Molecular Biology: Principles and Practice (2nd edition) . New York, N Y: W.H. Freeman and Co.

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MODULE 02 COMPANION GUIDE BCHM 218 MOLECULAR BIOLOGY | BCHM 218 M02 PAGE 90 3.17.2 POL III CORE SUB-PAGE OF D N A POLYMERASE III HOLOENZYME POL III CORE 1/1 Learn about the structure and function of the Pol III core: • The crystal structure of one Pol III core shows similarity to the Pol I structure with palm (contains active site), thumb (holds onto D N A), and finger (brings in d N T Ps) domains. • The Pol III alpha subunit is the main replicative subunit, and it recruits the ɛ subunit for 3’ - 5’ proofreading activity. Recall: this complex moves in the 5’ - 3’ direction along the D N A (in thi s case, from the bottom of the screen to the top of the screen). Structure of the Pol III core, resembling that of Pol I. The thumb can be seen holding the D N A to the palm. Reference: Cox, M. M., Doudna, J. A., & O'Donnell, M. (2012). Molecular Biology: Principles and Practice (2nd edition) . New York, N Y: W.H. Freeman and Co.

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MODULE 02 COMPANION GUIDE BCHM 218 MOLECULAR BIOLOGY | BCHM 218 M02 PAGE 91 3.18 Β CLAMPS FACILITATE PROCESSIVITY 17 OF 39 The Pol III core itself is capable of D N A synthesis at a slow rate, but D N A synthesis by the Pol III holoenzyme is exceedingly rapid, nearly 1 k b/s. The β clamps and clamp loader help maintain contact between the Pol III core and D N A, making the Pol III core highly processive ( ∼ 100 k b per binding event). Having three D N A polymerases in one holoenzyme assembly facilitates the coordinated synthesis of the leading and lagging strands at the replication fork. The β sliding clamps are assembled onto bo th D N A strands by the single clamp loader. The holoenzyme is held together by weak interactions including H-bonds and van der Waals forces. This ensures that each subunit is available in one place to perform its replicative function quickly and efficiently.

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MODULE 02 COMPANION GUIDE BCHM 218 MOLECULAR BIOLOGY | BCHM 218 M02 PAGE 92 3.19 EXPERIMENT: EVIDENCE OF Β CLAMP FUNCTION 18 OF 39 Experiments have demonstrated that the Pol core synthesize D N A more efficiently when attached to a β clamp, and that the core can transfer between templates. A purified Pol III holoenzyme ( just one β clamp and Pol III core are shown in the figure for simplicity) is assembled onto a primed donor s s D N A circle (M13mp18), then the Pol III-D N A is mixed with two competing acceptor D N A circles of different sizes, each containing a primer, but only one of which includes a β clamp. Learn about the two reactions: In Reaction A , the larger M13Gori D N A has a β clamp and the other (φX174) D N A does not. In Reaction B, smaller φX174 D N A has a β clamp and the other (M13Gori) does not. Replication is initiated using radioactive [32P]d N T Ps and timed. Representation of the reaction set- up. The M13Gori D N A is longer than the φX174 D N A. Both are in circular form. Reference: Cox, M. M., Doudna, J. A., & O'Donnell, M. (2012). Molecular Biology: Principles and Practice (2nd edition) . New York, N Y: W.H. Freeman and Co.

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MODULE 02 COMPANION GUIDE BCHM 218 MOLECULAR BIOLOGY | BCHM 218 M02 PAGE 93 3.20 EXPERIMENT: REQUIREMENT OF POL III FOR A Β CLAMP 19 OF 39 Aliquots of the reaction mixtures are analyzed on an agarose gel, followed by autoradiography. In both cases, the original donor plasmid is replicated, and then the acceptor D N A with a β clamp is replicated in preference to the acceptor D N A without a β clamp (looking at the amou nt of product formed over time). The result suggests that Pol III replicates D N A strands that have a β clamp more efficiently than those lacking a clamp. Further studies confirmed that Pol III hops from one clamp to another, leaving clamps behind on the D N A as it does so. Gel showing the results of the experiment. In reaction A, the M13Gori D N A bands are closest to the top of the gel. The M13m p18 bands are just below it. The φX174 bands ran the furthest down the gel, but are extremely faint. In reaction B, the order of bands is the same, however this time M13Gori produced very faint bands, while the M13m p18 bands and the φX174 bands are quite dark. Reference: Cox, M. M., Doudna, J. A., & O'Donnell, M. (2012). Molecular Biology: Principles and Practice (2nd edition) . New York, N Y: W.H. Freeman and Co.

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MODULE 02 COMPANION GUIDE BCHM 218 MOLECULAR BIOLOGY | BCHM 218 M02 PAGE 94 3.21 E. COLI Β CLAMP LOADER 20 OF 39 The β sliding clamp does not assemble onto D N A by itself - it requires the multiprotein clamp loader to open and close the ring around the D N A. The clamp loader within the Pol III holoenzyme is the τ complex (composed of τ 3 (τ trimer), δ, and δ’). Animated visual representation of the beta clamp docking underneath the clamp loader (γ complex). There is a gap between two of the 5 units in the clamp loader where this occurs. Reference: Cox, M. M., Doudna, J. A., & O'Donnell, M. (2012). Molecular Biology: Principles and Practice (2nd edition) . New York, N Y: W.H. Freeman and Co.

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MODULE 02 COMPANION GUIDE BCHM 218 MOLECULAR BIOLOGY | BCHM 218 M02 PAGE 95 3.22 E. COLI Β CLAMP LOADER MECHANISM 21 OF 39 The clamp loader uses the energy of A T P binding to open the β sliding clamp. In this operation, the clamp loader binds to one of the flat surfaces of the clamp and forces it to open. In the absence of A T P, the clamp loader cannot bind the β clamp, beca use the subunits are oriented in a way that blocks their interaction with the clamp. Learn how the clamp loader works: 1. A T P binding to the τ subunits induces a conformational change that enables the clamp loader to bind and open the clamp. 2. Binding of A T P is needed for the clamp loader to bind D N A. 3. A T P hydrolysis causes the clamp loader to revert to the form that cannot bind the β clamp or D N A, thereby ejecting the clamp loader and allowing the clamp to close around D N A. Note: The clamp-loading activity is specific to a primed site, because the D N A must bend out of a gap in the side of the clamp loader, and only single-stranded D N A, not double-stranded D N A, has the flexibility to make this sharp bend. Visual depiction of the steps described. Reference: Cox, M. M., Doudna, J. A., & O'Donnell, M. (2012). Molecular Biology: Principles and Practice (2nd edition) . New York, N Y: W.H. Freeman and Co.

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MODULE 02 COMPANION GUIDE BCHM 218 MOLECULAR BIOLOGY | BCHM 218 M02 PAGE 96 3.23 QUESTION: Β CLAMP FUNCTION Activity: Text Entry Please see the online learning module for the full experience of this interaction. 22 OF 39 Answer the question, using your knowledge of β clamps. Explain 1) the experiment that was used to demonstrate the function of the β clamp, and 2) what the results were. Feedback: In this experiment, purified Pol III holoenzyme is assembled onto a primed donor s s D N A circle, and then the Pol III-D N A is mixed with two competing acceptor D N A circles of different sizes, each containing a primer, but only one of which includes a β clamp. Replication is initiated using [32P] d N T Ps and timed. Then, samples from the reaction mixtures at various timepoints are analyzed on an agarose gel and subjected to autoradiography. In both cases, the original donor plasmid is replicated, and then the acceptor D N A with a β clamp is replicated, followed by the acceptor D N A without a β clamp. This can be seen on the gel because the acceptor plasmids are different sizes, and Pol III has amplified the plasmid with the β c lamp.

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MODULE 02 COMPANION GUIDE BCHM 218 MOLECULAR BIOLOGY | BCHM 218 M02 PAGE 97 3.24 THE TROMBONE MODEL OF THE REPLICATION FORK 23 OF 39 You will now turn your attention to an inherent issue with the replisome: how can the lagging strand Pol III synthesize D N A in the opposite direction to replication fork movement, yet remain tethered to the replisome? This question can be answered using the trombone model. This model describes D N A replication on the lagging strand, with its repeated cycles of loop growth and disassembly, which is analogous to the movement of slide on a trombone. Continue to the next slide to learn how lagging strand synthesis occurs. Reference: Music Arts. Jupiter Brass Instruments . http://www.musicarts.com/Jupiter,Brass-Instruments.mac

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MODULE 02 COMPANION GUIDE BCHM 218 MOLECULAR BIOLOGY | BCHM 218 M02 PAGE 98 3.25 THE TROMBONE MODEL: LAGGING STRAND SYNTHESIS 24 OF 39 The lagging- strand polymerase extends the 3′ terminus of an Okazaki fragment in the opposite direction to fork movement, yet this polymerase is part of the replisome and thus moves with the fork. The opposing directions result in formation of a D N A loop for each Okazaki fragment. As multiple Okazaki fragments are synthesized, loops repeatedly grow and are released, analogous to the movement of a trombone slide as the instrument is played. For simplicity, the third Pol III core involved in lagging-strand synthesis is not shown in action in the figure. When the lagging-strand polymerase finishes an Okazaki fragment, it must dissociate from the D N A in order to transfer to a new R N A primer and extend the next fragment. It does so by detaching from the β clamp, leaving it behind on the D N A. Once released, the lagging -strand Pol III core can associate with a new R N A- primed site on which a new β clamp has been assembled. This process results in a bui ldup of β clamps and R N A primers on replicated D N A. On the next slides, you will learn how this buildup is dealt with. Visual representation of the trombone model for lagging strand synthesis. Reference: Cox, M. M., Doudna, J. A., & O'Donnell, M. (2012). Molecular Biology: Principles and Practice (2nd edition) . New York, N Y: W.H. Freeman and Co.

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MODULE 02 COMPANION GUIDE BCHM 218 MOLECULAR BIOLOGY | BCHM 218 M02 PAGE 99 3.26 CLAMP RECYCLING BY D N A POL I AND LIGASE 25 OF 39 The resolution of β clamp buildup can be loosely divided into four steps. Learn the steps involved in the resolution of β clamp build -up: 1. Once the Pol III core dissociates from the β clamp following synthesis of the Okazaki fragment, the β clamp site acts to attract Pol I which is able to remove the R N A primer using its 5’ - 3’ exonuclease activity (nick -translation). 2. Pol I then dissociates from the D N A, l eaving behind a s s D N A break. The β clamp site now acts to attract the D N A ligase enzyme which is able to seal the break by forming a phosphodiester bond using the free 3’ O H and 5’P groups. 3. The unoccupied β clamp is then opened and unloaded (step 4) by excess δ subunits of the clamp loader. Visual representation of the steps described. Reference: Cox, M. M., Doudna, J. A., & O'Donnell, M. (2012). Molecular Biology: Principles and Practice (2nd edition) . New York, N Y: W.H. Freeman and Co.

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MODULE 02 COMPANION GUIDE BCHM 218 MOLECULAR BIOLOGY | BCHM 218 M02 PAGE 100 3.27 QUESTION: THE TROMBONE MODEL Activity: Text Entry Please see the online learning module for the full experience of this interaction. 26 OF 39 Answer the question about the trombone model of lagging strand replication. In one sentence, describe what is meant by the “trombone model”? Feedback: D N A loops repeatedly grow and collapse on the lagging strand, allowing the replisome to move towards the replication fork - like the movement of a trombone slide.

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MODULE 02 COMPANION GUIDE BCHM 218 MOLECULAR BIOLOGY | BCHM 218 M02 PAGE 101 3.28 QUESTION: D N A POLYMERASES Activity: Text Entry Please see the online learning module for the full experience of this interaction. 27 OF 39 Researchers who pioneered the use of D N A polymerases for D N A sequencing, P C R, and site- directed mutagenesis were awarded Nobel prizes for their ground-breaking achievements, but D N A polymerases play key roles in numerous additional indispensable methods including D N A labeling, cloning, whole genome amplification, and diagnostic techniques. Answer the question about D N A polymerases. What types of modifications have been made to D N A polymerases that have enabled or improved their usefulness in research applications? Text for Hint: For example the ability to utilize non- standard nucleotides and to replicate “difficult” D N A templates with simple repeats or sequences that are excessively rich in A+T or G+C base pairs. Or design of various fusion polymerases to enhance processivity. D N A polymerases from several organisms have been subjected to extensive engineering in order to remove or curb unwanted activities and to modify or enhance others that are required for optimal performance in vitro. Feedback: While there are many answers to this question, one example you will learn about in Module 03 is the use of polymerases in P C R, where advances in fidelity, speed, and processivity have improved genetic analysis, cloning, and diagnostics. Combining desirable protein domains from several D N A polymerases into a single engineered enzyme can increase both speed and processivity during P C R. Novel D N A polymerases can also be created by swapping domains from different D N A polymerases (e.g. those found in hot springs) to select for hybrid polymerases with desirable P C R properties. Reference: Gardner, A. F., & Kelman, Z. (2014). D N A polymerases in biotechnology. Frontiers in Microbiology , 5 , 659. http://doi.org/10.3389/fmicb.2014.00659

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MODULE 02 COMPANION GUIDE BCHM 218 MOLECULAR BIOLOGY | BCHM 218 M02 PAGE 102 3.29 3) E. COLI REPLICATION: TERMINATION 28 OF 39 The third and final stage of E. coli replication is termination. Once replication has started, there is no stopping it - the chromosome or cell is committed to division. In both bacteria and eukaryotes, replication forks meet head-on, at which point termination occurs. You will first learn about termination in E. coli cells, which have a specialized mechanism for preventing head-on collisions between two D N A polymerases, and between a D N A polymerase and an R N A polymerase that is transcribing the D N A. Then, you will learn about termination in eukaryotic cells, which have the additional problem of replicating the ends of linear chromosomes. Evolution has provided a solution to this problem in the form of a novel D N A polymerase - called telomerase - that is specialized for this purpose. A double stranded D N A forming three replication bubbles, with two replication forks each. The new D N A is synthesized on either side of the bubble until the replication forks meet. The end result is two double stranded D N A duplexes. Reference: Cox, M. M., Doudna, J. A., & O'Donnell, M. (2012). Molecular Biology: Principles and Practice (2nd edition) . New York, N Y: W.H. Freeman and Co.

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MODULE 02 COMPANION GUIDE BCHM 218 MOLECULAR BIOLOGY | BCHM 218 M02 PAGE 103 3.30 TERMINATION OF E. COLI CHROMOSOME REPLICATION 29 OF 39 In E. coli, a region located halfway around the circular chromosome from oriC contains two clusters of 23 base pair sequences called Ter sites . The two clusters of Ter sites are oriented in opposite directions. The monomeric Tus protein tightly binds to a Ter site and blocks advance of the replication fork by stopping D n a B helicase. A fascinating property of the Tus-Ter complex is that its fork-blocking activity is directional (denoted by the arrowheads). Replication forks are blocked when approaching a Tus-Ter complex from one direction (the nonpermissive direction), but not when approaching from the opposite (permissive) direction. Diagram of the Tus-ter system. Both a clockwise and counterclockwise replication fork originate from the oriC. The clockwise fork will first encounter Ter H, Ter I, Ter E, Ter D, and Ter A, which all face in the clockwise direction -- and thus it would be allowed to proceed -- since these are the counterclockwise trap. The clockwise fork would then encounter in the following order: Ter C and Ter B, and if it made it past there, Ter F, Ter G, and Ter J. These are called the clockwise fork trap and are oriented in the counterclockwise direction. Replication forks are not allowed to travel past Ter sites oriented in the opposite direction. Definition: Tus: ‘Tus’ stands for termination utilization substance. Reference: Cox, M. M., Doudna, J. A., & O'Donnell, M. (2012). Molecular Biology: Principles and Practice (2nd edition) . New York, N Y: W.H. Freeman and Co.

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MODULE 02 COMPANION GUIDE BCHM 218 MOLECULAR BIOLOGY | BCHM 218 M02 PAGE 104 3.31 ACTIVITY: TER SITES Activity: Fill in the Blanks Please see the online learning module for the full experience of this interaction. 30 OF 39 Diagram of the Tus-ter system. Both a clockwise and counterclockwise replication fork originate from the oriC. The clockwise fork will first encounter Ter H, Ter I, Ter E, Ter D, and Ter A, which all face in the clockwise direction -- and thus it would be allowed to proceed -- since these are the counterclockwise trap. The clockwise fork would then encounter in the following order: Ter C and Ter B, and if it made it past there, Ter F, Ter G, and Ter J. These are called the clockwise fork trap and are oriented in the counterclockwise direction. Replication forks are not allowed to travel past Ter sites oriented in the opposite direction. Using what you learned on the previous slide and the diagram provided, fill in the blanks with clockwise or counter-clockwise in the paragraph . Ter D and Ter A block progress of the of the _____ fork, but allow _____ replication to proceed through. Whereas, Ter B and Ter C block progress of the _____ fork, but allow ____ replication to proceed through. Feedback: Ter D and Ter A block progress of the of the counterclockwise fork, but allow clockwise replication to proceed through. Whereas, Ter B and Ter C block progress of the clockwise fork, but allow counterclockwise replication to proceed through. Reference: Cox, M. M., Doudna, J. A., & O'Donnell, M. (2012). Molecular Biology: Principles and Practice (2nd edition) . New York, N Y: W.H. Freeman and Co.

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MODULE 02 COMPANION GUIDE BCHM 218 MOLECULAR BIOLOGY | BCHM 218 M02 PAGE 105 3.32 THE TUS-TER SYSTEM AND REPLICATION SPEED 31 OF 39 The Tus-Ter mechanism works regardless of the speed of replication in either direction. Learn how the Tus-Ter system accommodates the speed of the replication forks. Equal Speed: Replication forks of equal speed meet at the same time in the terminus region of the chromosome. Visual depiction of replication forks meeting at the terminus region. Unequal Speed: Replication forks moving at unequal speeds still meet in the terminus region, since replication will be stalled in one direction until it is completed in the other direction. The diagram shows the result when the clockwise fork has replicated more rapidly than the counterclockwise fork. Visual depiction of the clockwise fork proceeding faster than the counterclockwise fork, and being stopped by the Ter cluster. Reference: Cox, M. M., Doudna, J. A., & O'Donnell, M. (2012). Molecular Biology: Principles and Practice (2nd edition) . New York, N Y: W.H. Freeman and Co.

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MODULE 02 COMPANION GUIDE BCHM 218 MOLECULAR BIOLOGY | BCHM 218 M02 PAGE 106 3.33 MINIMIZING COMPETITION BETWEEN TRANSCRIPTION AND REPLICATION IN THE E. COLI CHROMOSOME 32 OF 39 Actively replicating bacteria are also growing and metabolizing, and therefore actively transcribing R N A from promoters all over the chromosome. This means that collisions between R N A polymerase and replication forks are inevitable. Studies have shown that codirectional collisions do not impede forks, whereas head-on collisions can cause a fork to pause or stall. Most transcripts in bacteria are oriented in the same direction as replication , and therefore most collisions are codirectional, provided that the forks do not proceed more than halfway around the chromosome. As you just learned, the Tus-Ter system prevents a replication fork from extending much beyond the halfway point around the chromosome, and thus ensures that the fork always moves in the same direction as transcription. Perhaps this system evolved to prevent replication forks from going too far around the circular chromosome, where the direction of transcription would result in head-on collisions! The majority of transcription is in the same direction as replication. Replication forks are blocked by Ter sites, preventing them from moving into the transcription taking place in the opposite direction. Reference : Cox, M. M., Doudna, J. A., & O'Donnell, M. (2012). Molecular Biology: Principles and Practice (2nd edition) . New York, N Y: W.H. Freeman and Co.

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MODULE 02 COMPANION GUIDE BCHM 218 MOLECULAR BIOLOGY | BCHM 218 M02 PAGE 107 3.34 RESOLUTION OF CATENATED DAUGHTER CHROMOSOMES 33 OF 39 When replication of the circular chromosome is completed, it results in the attachment or tangling of the two daughter strands of D N A at the site of termination. These catenated daughter chromosomes are then resolved (unlinked) into two separate chromosomes by the action of a type II topoisomerase enzyme. You will learn about how topoisomerases work on the next slide. Visual representation of the development and resolution of catenated daughter chromosomes. Reference: By Catherinea228 - Own work, CC BY-SA 4.0, https://commons.wikimedia.org/w/index.php?curid=59320465

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MODULE 02 COMPANION GUIDE BCHM 218 MOLECULAR BIOLOGY | BCHM 218 M02 PAGE 108 3.35 TOPOISOMERASE MECHANISM OF ACTION 34 OF 39 Learn about how topoisomerase works: 1. The type II topoisomerase protein in prokaryotes is a heterodimer, consisting of two A T Pase domains and two cleavage core domains, linked by a scaffolding domain. The enzyme is able to pull one of the pieces of catenated d s D N A into its cleavage site, and through the use of A T P, it can generate a double-strand break in this D N A fragment. 2. The mechanism for generating this double stranded break is through the use of a Try residue in the catalytic site. Tyr gives up its proton to form a nucleophile, which is able to attack the phosphate in a phosphodiester bond (this occurs on both strands). This temporarily covalently links the D N A strand to the protein forming a phosphotyrosyl linkage, in what is known as a 5’ adduct. Once the daughter chromosomes are separated, a hydrolysis reaction reforms the phosphodiester bonds. Visual depiction of the mechanism described. Visual representation of the reaction. Reference: Vos, S. M, Tretter, E. M., Schmidt, B. H., & Berger, J. M. (2011). All tangled up: how cells direct, manage, and exploit topoisomerase function. Nature Reviews Molecular Cell Biology 12, 827-841. doi: 10.1038/nrm3228.

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MODULE 02 COMPANION GUIDE BCHM 218 MOLECULAR BIOLOGY | BCHM 218 M02 PAGE 109 3.36 EUKARYOTIC TERMINATION: THE END REPLICATION PROBLEM 35 OF 39 The replication of linear chromosome ends poses a unique problem. At the end of a chromosome, after the leading strand has been completely extended to the last nucleotide, the lagging strand has a single-stranded D N A gap that must be primed and filled in. Visual depiction of why linear chromosomes present a problem for the replication of their ends. After Okazaki fragment processing, there is a gap at the 5’ end of each strand. If nothing was done about this, the second generation will be missing the information that was contained in those gaps. Break this problem down: 1. The problem arises when the R N A primer at the extreme end is removed for replacement with D N A. Since there is no 3′ terminus for D N A polymerase to extend from, this single - stranded gap cannot be converted to duplex D N A. 2. The genetic information in the gap will be lost in the next round of replication, and repeated rounds will cause the ends to progressively shorten until the genes near the ends are entirely lost. Reference: Cox, M. M., Doudna, J. A., & O'Donnell, M. (2012). Molecular Biology: Principles and Practice (2nd edition) . New York, N Y: W.H. Freeman and Co.

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MODULE 02 COMPANION GUIDE BCHM 218 MOLECULAR BIOLOGY | BCHM 218 M02 PAGE 110 3.37 EUKARYOTIC TERMINATION: SOLVING THE END REPLICATION PROBLEM 36 OF 39 As you will learn about in more detail in Module 03, the ends of the linear eukaryotic chromosomes are called telomeres and are composed of repeats of a unique sequence. The telomerase reverse transcriptase (T E R T) carries a tightly bound, non-coding telomerase R N A (T R). The T R contains about 1.5 telomere repeat units, which match the telomere repeat sequence, that it uses as a template to extend the 3′ terminus of the telomere beyond what was replicated. The T E R T – T R holoenzyme is referred to as telomerase . This reaction occurs in S phase, as part of the chromosome duplication process. You will now look at the reaction cycle of Tetrahymena telomerase as an example of the action of telomerase.

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MODULE 02 COMPANION GUIDE BCHM 218 MOLECULAR BIOLOGY | BCHM 218 M02 PAGE 111 3.38 EUKARYOTIC TERMINATION: TELOMERASE 37 OF 39 The reaction cycle of Tetrahymena (a protozoa) telomerase is shown here as an example of the action of telomerase. Reveal the steps of this process: 1. First, at the 3′ -terminal end of a linear D N A, three nucleotides of the telomere anneal to three R N A nucleotides in telomerase. 2. Then the telomerase extends the 3′ end of the single -stranded D N A by the length of one telomere repeat-six nucleotides in the case of Tetrahymena . Telomerase is different from other D N A polymerases in that it carries its own template and synthesizes single-stranded D N A. 3. After adding a six-nucleotide repeat, telomerase separates the R N A - D N A hybrid and repositions on the telomere for extension of the next 6-mer repeat. Telomerase acts processively, synthesizing many telomere repeats in one telomerase-binding event. 4. The telomerase- extended 3′ single -stranded D N A terminus is then converted to duplex D N A by the same priming and polymerization machinery used in chromosome replication. Because the telomere D N A is a simple repeat that does not encode a biomolecule, the cell can tolerate a certain amount of variability in final telomere length. 5. It is important to note, however, that the D N A product is not completely duplex. The 3′ terminus of a new telomere still has single-stranded D N A, due to the same R N A primer- removal problem discussed earlier, and these are bound and protected by telomere D N A- binding proteins so that they aren’t subjected to D N A repair mechanisms. Visual depiction of the steps described. Reference: Cox, M. M., Doudna, J. A., & O'Donnell, M. (2012). Molecular Biology: Principles and Practice (2nd edition) . New York, N Y: W.H. Freeman and Co.

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MODULE 02 COMPANION GUIDE BCHM 218 MOLECULAR BIOLOGY | BCHM 218 M02 PAGE 112 3.39 QUESTION: REPLICATING THE ENDS OF LINEAR D N A Activity: Text Entry Please see the online learning module for the full experience of this interaction. 38 OF 39 Answer the question about replication termination in eukaryotes. 1. What problem is encountered when replicating the ends of linear D N A? How is it solved? 2. You have now learned about all three steps of replication. Which is the primary control point for replication? Why? Feedback: 1. When the R N A primer at the extreme end of the strand is removed for replacement with D N A, it leaves no 3′ terminus for D N A polymerase to extend from, resulting in a single -stranded gap. If this problem was not addressed, the ends of the D N A would progressively shorten until the genes near the ends are entirely lost. This problem is solved by telomerase, which extends the telomeres (chromosome ends), preventing the loss of important genetic information. 2. Regulation occurs primarily at the initiation step because, once replication has begun, the cell is committed to division.

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MODULE 02 COMPANION GUIDE BCHM 218 MOLECULAR BIOLOGY | BCHM 218 M02 PAGE 113 3.40 SECTION 03 SUMMARY 39 OF 39 In Section 03, you learned about D N A replication in E. coli , frequently comparing it to eukaryotic replication . D N A replication occurs in 3 stages: initiation, elongation by the replisome, and termination. During the initiation stage in E. coli , the assembly of replication forks at the oriC occurs in a series of four steps. There are several mechanisms of regulation that occur during this stage, including D N A methylation, A T P turnover by D n a A, the activity of the H d a protein, and R N A polymerase activity. Elongation occurs via D N A polymerase III - the Pol III core connects to the β clamp, which encircles the D N A for processive D N A synthesis. Finally, termination of E. coli replication occurs halfway around the circular chromosomes from the oriC . Bidirectional replication forks meet head-on within a terminus region bordered on both sides by Ter sites able to bind Tus proteins. You will now turn your attention to how cells handle damaged D N A as you explore the topic of D N A repair.

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MODULE 02 COMPANION GUIDE BCHM 218 MOLECULAR BIOLOGY | BCHM 218 M02 PAGE 114 SECTION 04: D N A REPAIR 4.1 SECTION 04: D N A REPAIR (1)

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MODULE 02 COMPANION GUIDE BCHM 218 MOLECULAR BIOLOGY | BCHM 218 M02 PAGE 115 4.2 D N A REPAIR (2) 1 OF 14 So far in this course, you’ve learned about the structure and function of D N A, how all the geneti c information is packaged within a single cell, and how it is replicated to form two new cells. D N A is essential to all life, and when errors and/or damage occurs to this molecule, it can have major consequences. There are a collection of processes that function to recognize and correct errors within D N A molecules, referred to as D N A repair. In this section, you will learn about double strand breaks within D N A and the process of recombinational D N A repair.

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MODULE 02 COMPANION GUIDE BCHM 218 MOLECULAR BIOLOGY | BCHM 218 M02 PAGE 116 4.3 KARYOTYPES: NORMAL VS BREAST CANCER CELL 2 OF 14 A karyotype is a visualization of the number and appearance of the chromosomes within a cell, where each chromosome pair is labelled with a different colour. The figure shows the karyotype of a normal human cell and a H C C1937 human breast cancer cell. You will note the karyotype of the breast cancer cell has a number of problems and varies drastically from the normal cell karyotype, suggesting a major issue with the cell repair machinery. Karyotype of a normal human cell. Karyotype of a human breast cancer cell. References: Levy, S., Sutton, G., Ng, P., Feuk, L., Halpern, A., Walenz, B., . . . Venter, J. (2007). The diploid genome sequence of an individual human. Plos Biology, 5 (10), 2113-2144. doi:10.1371/journal.pbio.0050254 Grigorova, M., Staines, J. M., Ozdag, H., Caldas, C., & Edwards, P. A. W. (2004). Possible causes of chromosome instability: Comparison of chromosomal abnormalities in cancer cell lines with mutations in B R C A1, B R C A2, C H K2 and B U B1. Cytogenetic and Genome Research, 104 (1-4), 333-340. doi:10.1159/000077512

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MODULE 02 COMPANION GUIDE BCHM 218 MOLECULAR BIOLOGY | BCHM 218 M02 PAGE 117 4.4 ANEUPLOIDY AND CHROMOSOMAL TRANSLOCATIONS 3 OF 14 Problems within a cancerous cell are often the result of faulty recombination and D N A repair. Two prime examples of problems that occur within a cancer cell include aneuploidy and chromosomal translocations. Learn about aneuploidy and chromosomal translocations: Aneuploidy Aneuploidy is the presence of an abnormal number of chromosomes in a cell. In the figure, you can see double the chromosome number for almost every chromosome in this cell. Virtually all cancer cells are aneuploid as a result of mitotic segregation defects or chromosomal non- disjunction. Alt Text: Karyotype of a human breast cancer cell. Chromosomal Translocations A chromosomal translocation is an abnormality caused by rearrangement of parts between non- homologous chromosomes. These can be direct switching of material between two chromosomes (shown in the figure), or large scale deletions and insertions. These types of chromosomal abnormalities are generally caused by errors during homologous recombination or double-strand break repair - two processes you will learn about in this section and Section 05. Chromosomal translocation between chromosome Definitions: Recombination: The rearrangement of genetic material, especially by crossing over in chromosomes. Non-disjunction: The failure of one or more pairs of homologous chromosomes or sister chromatids to separate normally during and after mitosis. Non-homologous: Chromosomes that are not members of the same pair. References: Grigorova, M., Staines, J. M., Ozdag, H., Caldas, C., & Edwards, P. A. W. (2004). Possible causes of chromosome instability: Comparison of chromosomal abnormalities in cancer cell lines with mutations in B R C A1, B R C A2, C H K2 and B U B1. Cytogenetic and Genome Research, 104 (1-4), 333-340. doi:10.1159/000077512

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MODULE 02 COMPANION GUIDE BCHM 218 MOLECULAR BIOLOGY | BCHM 218 M02 PAGE 118 By Courtesy: National Human Genome Research Institute - [1] (file), Public Domain, https://commons.wikimedia.org/w/index.php?curid=935175

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MODULE 02 COMPANION GUIDE BCHM 218 MOLECULAR BIOLOGY | BCHM 218 M02 PAGE 119 4.5 D N A LESIONS: DOUBLE-STRAND BREAKS 4 OF 14 One of the most common and dangerous D N A lesions is a double-strand break (D S B) . A D S B is a break in the phosphodiester backbone on both strands of D N A at the same site. These lesions usually arise during replication, when the replication fork encounters a single-strand break in the template strand, which then becomes propagated into a D S B in both the template and daughter strands. Broken D N A ends make it impossible for D N A replication to continue. D S Bs can also arise as a result of exposure to U V or gamma radiation. The result goes beyond a simple mutation that may or may not affect cellular function – D S Bs, if not repaired, typically lead to cell death.

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MODULE 02 COMPANION GUIDE BCHM 218 MOLECULAR BIOLOGY | BCHM 218 M02 PAGE 120 4.6 DEINOCOCCUS RADIODURANS: A MASTER OF D N A REPAIR 5 OF 14 Understanding the mechanisms that some bacteria use to repair D N A lesions following irradiation provides information that scientists can use to better understand the repair of D S Bs, and perhaps generate therapies to aid repair in people exposed to radiation, such as astronauts or pilots. One such bacteria is Deinococcus radiodurans, a ball-shaped bacteria that can withstand tremendous amounts of radiation because it regenerates its D N A every 12-24 hours. Treatment of D. radiodurans with high levels of ionizing radiation can produce hundreds of genomic double-strand breaks, but the genome is re-assembled accurately before initiation of the next cycle of cell division. This organism can live in many harsh environments because of its strange mechanism of D N A replication. Interesting Fact Icon: The energy deposited by electromagnetic radiation is measured in rads or grays (1 G y = 100 rads). For a human cell, a dose of 2 to 5 G y is lethal. In the 1950s, it became clear that some organisms were surprisingly resistant to radiation. For example, in efforts to use radioactive sources to sterilize food, some sealed food samples were spoiled by bacteria even after exposure to gamma radiation at levels up to 4,000 G y. The culprit was a pink, non-spore-forming, nonmotile bacterium eventually named Deinococcus radiodurans. Reference: Tiempos Exponentiales. (2009). Retrieved from http://vonneumannmachine.files.wordpress.com/2009/04/d_radiodurans.jpg

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MODULE 02 COMPANION GUIDE BCHM 218 MOLECULAR BIOLOGY | BCHM 218 M02 PAGE 121 4.7 THE DEINOCOCCUS RADIODURANS GENOME 6 OF 14 Not only does Deinococcus radiodurans replicate its D N A rapidly, but it also contains many copies of its cellular D N A to ensure there is always a backup template for repair, and it contains high levels of manganese - thought to be important in the repair process. The Deinococcus genome consists of four circular D N A molecules, all generally present in multiple copies. After γ irradiation, the cells stop growing and D N A repair begins. Overlapping D N A fragments are spliced together, and the entire genome is accurately reconstituted within a few hours. The cells begin to grow and divide again as if nothing had happened. It is perhaps the most remarkable feat of D N A repair we know of so far. For Interest - Refer to 4.7.1 For Your Interest: Other Organisms with Extraordinary Capacity for D N A Repair Sub-Page Repair of double stranded breaks in the four circular D N A molecules of Deinococcus after gamma irradiation. Manganese is present in the cell. Reference: Cox, M. M., & Battista, J. R. (2005). Deinococcus radiodurans [mdash] the consummate survivor. Nat Rev Micro , 3 (11), 882-892. https://doi.org/10.1038/nrmicro1264. (https://search-proquest-com.proxy.queensu.ca/docview/224638057?pq-origsite=summon)

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MODULE 02 COMPANION GUIDE BCHM 218 MOLECULAR BIOLOGY | BCHM 218 M02 PAGE 122 4.7.1 FOR YOUR INTEREST: OTHER ORGANISMS WITH EXTRAORDINARY CAPACITY FOR D N A REPAIR SUB-PAGE OF 4.7 THE DEINOCOCCUS RADIODURANS GENOME FOR INTEREST 1/1 For several decades, D. radiodurans was considered the most radiation-resistant organism known. However, recent research has revealed many microbial species that are highly resistant to radiation, some of them more so than D. radiodurans . In addition, these highly resistant species are found in various unrelated genera, indicating that this phenotype has evolved independently many times. A few species have been found in hot springs with high radon backgrounds, where exposure to chronic low levels of radiation has forced adaptations to permit more efficient D N A repair. However, the most reliable source of bacteria with extreme radiation resistance is the desert environment, where the major selective pressure is not radiation but desiccation. When water disappears for an extended period of time, most metabolic processes, including generation of the A T P needed for D N A repair, ceases. When desiccation gives way to conditions favorable for growth, the extraordinary capacity of these organisms for rapid genomic reconstitution and repairing D S Bs returns.

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MODULE 02 COMPANION GUIDE BCHM 218 MOLECULAR BIOLOGY | BCHM 218 M02 PAGE 123 4.8 REPAIR OF CHROMOSOMAL DOUBLE-STRAND BREAKS 7 OF 14 The capacity for the enzymatic repair of D S Bs is inherent to every free-living organism. Recombinational D N A repair is, to a large extent, directed at the repair of D S Bs. Recombinational D N A repair refers to a group of recombination processes directed at the repair of D N A strand breaks or cross-links, especially at inactivated replication forks. The repair of double-strand breaks by recombinational D N A repair requires the presence of another, undamaged, homologous double-stranded D N A. In a diploid cell, that double-stranded D N A is either the second copy of the chromosome or the sister chromatid present immediately after D N A replication. This second D N A molecule guides the repair process by providing a template for the restoration of genetic information that might otherwise be lost as missing nucleotides at the site of the break. You will now learn about the mechanism of recombinational D N A repair.

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MODULE 02 COMPANION GUIDE BCHM 218 MOLECULAR BIOLOGY | BCHM 218 M02 PAGE 124 4.9 RECOMBINATIONAL D N A REPAIR 8 OF 14 Use the numbers to navigate through the first 4 steps in the process of D N A repair by recombination. STEP 1 After a D S B, first, the broken D N A ends are processed by helicases and nucleases, with the 5′ -ending strands selectively degraded to create 3′ single -stranded extensions, or overhangs, at the site of the break. This single-stranded D N A is coated and protected by single-strand binding proteins (S S Bs). Note the orientation of each strand in the figure. Diagram of the process of recombinational D N A repair. First, a double-strand break undergoes end processing, creating the 3’ overhangs. Helicases an d nucleases participate in this step. Second, the first strand invasion takes place through the action of recombinase. Third, the second strand invasion occurs by recombinase. Fourth, D N A polymerase extends the strands. STEP 2 Second, the 3′ single -stranded extensions invade the homologous chromosome, in a process catalyzed by a ubiquitous class of enzymes called recombinases (RecA proteins) which replace the S S Bs. This D N A strand invasion is the quintessential step of homologous recombination and all processes related to it. The invading strand displaces one strand of the intact homologous chromosome and base pairs with the other. The structure created by strand invasion by the 3′ single -stranded extension is sometimes referred to as a D-loop (displacement loop). STEP 3 A second strand invasion takes place, similar to the first. Note: For the pathway shown in the figure, the two strand invasions produce two D N A crossover points, where strands originating separately from the two participating D N A molecules are base paired to each other. STEP 4 The defining reaction is the replicative extension of the invading strand by D N A polymerases. Importantly , the use of 3′ ends for the invasion step has the important consequence that these ends can also act as primers for D N A synthesis.

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MODULE 02 COMPANION GUIDE BCHM 218 MOLECULAR BIOLOGY | BCHM 218 M02 PAGE 125 D N A polymerase-mediated extension of the invading strands after they are paired lengthens them in a manner that faithfully restores any information lost at the site of the break, using the invaded homologous chromosome as the template. Lightbulb There is evidence that D N A repair synthesis is initiated by Pol III and then elongated by with either Pol I alone or by Pol III, with Pol I filling gaps that arise from excision of damaged bases. Definition: Recombinases: Restriction endonucleases and D N A ligases combined into a single efficient package. Reference: Cox, M. M., Doudna, J. A., & O'Donnell, M. (2012). Molecular Biology: Principles and Practice (2nd edition) . New York, N Y: W.H. Freeman and Co.

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MODULE 02 COMPANION GUIDE BCHM 218 MOLECULAR BIOLOGY | BCHM 218 M02 PAGE 126 4.10 STEP 5: COMPLETING REPAIR OF A D S B 9 OF 14 Once extension by the polymerase has occurred, there are two possible pathways to complete the recombinational D N A repair process in the 5th and final step. Learn about each process: A) Synthesis-dependent strand annealing (S D S A): The now-lengthened invading strands can simply be displaced by the action of helicases and then anneal to each other. Any remaining gaps can be filled by D N A polymerases. D N A ligases complete the repair by ligating the ends, re-forming the two d s D N A chromosomes. B) Double-strand break repair (D S B R): Replication is completed by ligating the strands while they are still linked. A four-branched crossover junction, with all D N A strands intact such that each branch is a segment of duplex D N A, is called a Holliday intermediate (or junction) - represented by ‘X’ shaped structures formed by the extension of the D-loops. Specialized endonucleases present in all cells, called Holliday junction resolvases, recognize and cleave the Holliday intermediates in one of two ways outlined in the figure. The synthesis-dependent strand annealing pathway. The strands dissociate and anneal, then replication and ligation are completed. Reference: Cox, M. M., Doudna, J. A., & O'Donnell, M. (2012). Molecular Biology: Principles and Practice (2nd edition) . New York, N Y: W.H. Freeman and Co.

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MODULE 02 COMPANION GUIDE BCHM 218 MOLECULAR BIOLOGY | BCHM 218 M02 PAGE 127 4.11 D S B R: RESOLUTION OF HOLLIDAY INTERMEDIATES 10 OF 14 As you learned on the previous slide, D S B R results in the formation of Holliday intermediates. Holliday intermediates are recognized and cleaved in one of two ways. Learn about these two methods: Method 1 If both Holliday intermediates are cleaved at the sites labeled X or both are cleaved at the sites labeled Y, there is ‘non - crossover’ - which means the genetic material between the cleavage sites will be exchanged, but the chromosomes will not (notice the regions flanking the repair site are the same colour). Method 2 A genetic ‘crossover’ occurs i f one Holliday junction is cleaved at the X sites and the other at the Y sites, such that the genetic material outside the site of repair is now from two separate chromosomes (notice the strands flanking each repair site are different colours). In either case, the breaks left behind by the cleavage are then sealed by D N A ligase. Diagram of double-strand break repair pathway. After completion of replication and ligation, Holliday intermediates are generated. These can be resolved in two ways. In option 1, cleavage occurs at both “X” sites, and this X by X resolution results in a non-crossover event. In option 2, cleavage occurs at the X sites for one Holliday junction and the Y sites for another. This is called in X by Y resolution, resulting in a crossover event. Reference: Cox, M. M., Doudna, J. A., & O'Donnell, M. (2012). Molecular Biology: Principles and Practice (2nd edition) . New York, N Y: W.H. Freeman and Co.

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MODULE 02 COMPANION GUIDE BCHM 218 MOLECULAR BIOLOGY | BCHM 218 M02 PAGE 128 4.12 BACTERIAL D N A REPAIR Activity: Multiple Choice Please see the online learning module for the full experience of this interaction. 11 OF 14 Answer the question about bacterial D N A repair. Which D N A polymerase(s) is/are involved in bacterial recombinational D N A repair? • Pol III • Pol II • Pol I • Pol V • A D N A polymerase is not involved in bacterial recombinational D N A repair. Feedback: Pol I

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MODULE 02 COMPANION GUIDE BCHM 218 MOLECULAR BIOLOGY | BCHM 218 M02 PAGE 129 4.13 REPAIRING DOUBLE-STRAND BREAKS Please see the online learning module for the full experience of this interaction. 12 OF 14 As you learned in this section, there are two pathways for completing the repair of a double-strand breaks by recombination: synthesis-dependent strand annealing (S D S A) and double-strand break repair (D S B R) Use the dropdown menus to identify the correct order of steps in the D S B R pathway. • Broken D N A ends are processed to create 3’ single -stranded overhangs. • The 3’ single -stranded extensions invade the homologous chromosome, in a process catalyzed by recombinases. • Replicative extension of the invading strand by Pol III occurs. • The strands are ligated while still linked. • Holliday junction resolvases recognize and cleave one Holliday intermediate at the X site, and the other at the Y site. • The breaks left behind by the cleavage are sealed by D N A ligase. Feedback: 1. Broken D N A ends are processed to create 3′ single -stranded overhangs. 2. The 3′ single -stranded extensions invade the homologous chromosome, in a process catalyzed by recombinases. 3. Replicative extension of the invading strand by Pol III occurs. 4. The strands are ligated while still linked. 5. Holliday junction resolvases recognize and cleave one Holliday intermediate at the X site, and the other at the Y site. 6. The breaks left behind by the cleavage are sealed by D N A ligase.

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MODULE 02 COMPANION GUIDE BCHM 218 MOLECULAR BIOLOGY | BCHM 218 M02 PAGE 130 4.14 FORMATION AND REPAIR OF DOUBLE-STRAND BREAKS Activity: Text Entry Please see the online learning module for the full experience of this interaction. 13 OF 14 Answer the questions about double-strand breaks. 1 of 2: What are two situations in which D S Bs commonly arise? 2 of 2: What is the name of the structure created by strand invasion by the 3′ single - stranded extension during D S B repair? Feedback: 1) D S Bs most commonly arise as the result of replication or exposure to UV or gamma radiation. 2) The structure is called a D-loop (or displacement loop).

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MODULE 02 COMPANION GUIDE BCHM 218 MOLECULAR BIOLOGY | BCHM 218 M02 PAGE 131 4.15 SECTION 04: SUMMARY 14 OF 14 D N A repair refers to a collection of processes that function to recognize and correct errors within D N A molecules. Recombinational D N A repair is a mechanism for the repair of single-strand and double- strand breaks, both of which can arise during replication. It requires the presence of an undamaged, homologous D N A strand - either a sister chromatid or a second copy of the chromosome. Double-strand breaks are the most deleterious D N A lesions, and repairing them requires the generation of 3’ single strand extensions at both of the broken D N A ends. There are two mechanisms through which recombinational D N A repair can be completed: • Synthesis-dependent strand annealing (S D SA) • Double-stranded break repairs (D S B R) - which involves the production of a Holliday intermediate

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MODULE 02 COMPANION GUIDE BCHM 218 MOLECULAR BIOLOGY | BCHM 218 M02 PAGE 132 SECTION 05: D N A RECOMBINATION 5.1 SECTION 05: D N A RECOMBINATION (1)

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MODULE 02 COMPANION GUIDE BCHM 218 MOLECULAR BIOLOGY | BCHM 218 M02 PAGE 133 5.2 D N A RECOMBINATION (2) 1 OF 14 In Section 04, you learned about recombinational D N A repair. In this section, you will learn about D N A recombination in other cellular processes. D N A recombination is the exchange of segments within D N A strands to produce new nucleotide sequence arrangements. Recombination typically occurs between regions of similar sequence by breaking and rejoining D N A segments. D N A recombination is essential for generating genetic diversity and for maintaining genome integrity. In this section, you will learn about homologous recombination and in Section 06, you will learn about a second type of recombination called site-specific recombination. Reference: D N A recombination - Latest research and news | Nature. (n.d.). Retrieved August 23, 2017, from https://www.nature.com/subjects/dna-recombination

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MODULE 02 COMPANION GUIDE BCHM 218 MOLECULAR BIOLOGY | BCHM 218 M02 PAGE 134 5.3 HOMOLOGOUS RECOMBINATION 2 OF 14 Homologous recombination (H R) almost certainly began as a D N A repair process, but has evolved into a broader mechanism that allows populations of organisms to genetically adapt more quickly to their environment. H R refers to recombination between two D N A molecules of similar sequence (i.e. homologous chromosomes or sequences with small allelic differences), occurring in all cells. This process takes place during meiosis and mitosis in eukaryotes, and during the repair of D S Bs in all organisms. In order to understand how H R works, it is important to understand the process of meiosis. You will begin your study of homologous recombination with a brief review of meiosis.

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MODULE 02 COMPANION GUIDE BCHM 218 MOLECULAR BIOLOGY | BCHM 218 M02 PAGE 135 5.4 REVIEW: EUKARYOTIC MEIOSIS 3 OF 14 Recall from your previous biology courses that homologous recombination is an integral part of the pairing of homologous chromosomes during meiosis and occurs between non-sister chromatids during the first stage of meiosis I. Answer the questions regarding meiosis. Please see the online learning module for the full experience of this interaction. 1 of 6: In what phase of meiosis are each pair of homologous chromosomes replicated to create four chromosome copies (two copies, or chromatids per chromosome)? 2 of 6: In what phase of meiosis do homologous sister chromatid pairs align? 3 of 6: In one round of meiosis, the four sets of chromosomes are segregated through two cell divisions (meiosis I and meiosis II) to create how many gametes?

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MODULE 02 COMPANION GUIDE BCHM 218 MOLECULAR BIOLOGY | BCHM 218 M02 PAGE 136 4 of 6: Are the gametes haploid or diploid? 5 of 6: In meiosis, what segregates at the first cell division? What about the second cell division? 6 of 6: What provides the physical link between sister chromatids? Diagram showing the stages of meiosis. Diploid cells enter interphase then proceed through S phase into Meiosis I, where they split into two cells. The two diploid cells enter Meiosis II and then split a final time to create four haploid gametes or spores. Feedback: 1) S Phase 2) The first meiotic prophase (prophase I). 3) Four 4) Haploid 5) Homologous chromosomes segregate in the first cell division, and sister chromatids segregate in the second. 6) Proteins called cohesins provide the link. Definition: Meiosis: A special type of cell division necessary for sexual reproduction in eukaryotes. The cells produced by meiosis are gametes or spores. Reference: Cox, M. M., Doudna, J. A., & O'Donnell, M. (2012). Molecular Biology: Principles and Practice (2nd edition) . New York, N Y: W.H. Freeman and Co.

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MODULE 02 COMPANION GUIDE BCHM 218 MOLECULAR BIOLOGY | BCHM 218 M02 PAGE 137 5.5 EUKARYOTIC MEIOSIS: OVERVIEW 4 OF 14 The segregation of homologous chromosomes during meiosis I is an event unique to meiosis. The chromosomes to be segregated are not related by a recent replication event, and thus, there needs to be some molecular device to ensure that only homologous chromosomes are linked at this stage. Find out how cells accomplish this: 1) The eukaryotic answer to this problem is homologous recombination. Recombinational crossovers provide the accurate alignment of homologous chromosomes at the metaphase plate during meiosis. 2) Many of the same enzymes are involved in both recombinational D N A repair during replication and meiotic recombination. The process again begins with D S Bs, but in meiosis I, the breaks are programmed events. Continue to the next slide to learn about the crossovers between homologous chromosomes. Diagram showing the stages of meiosis. Diploid cells enter interphase then proceed through S phase into Meiosis I, where they split into two cells. The two diploid cells enter Meosis II and then split a final time to create four haploid gametes or spores. Reference: Cox, M. M., Doudna, J. A., & O'Donnell, M. (2012). Molecular Biology: Principles and Practice (2nd edition) . New York, N Y: W.H. Freeman and Co.

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MODULE 02 COMPANION GUIDE BCHM 218 MOLECULAR BIOLOGY | BCHM 218 M02 PAGE 138 5.6 EUKARYOTIC MEIOSIS: CROSSING OVER 5 OF 14 As you just learned, homologous recombination results in a crossing over between equivalent parts of chromosomes to help accurately align homologous pairs at the metaphase plate during meiosis I. Crossing over has two roles in meiosis. Learn the roles of crossing over in meiosis: 1) To create the physical link essential for proper chromosomal segregation. 2) Following segregation, the sister chromatids (now daughter chromosomes) are no longer identical. Parts of each set of paired chromatids have been exchanged with the homologous chromosome, generating genetic diversity. The first stage of meiosis I is when the homologous chromosomes are aligned. The crossovers of recombination are visible in prophase I. During this phase, the homologous chromosomes partially separate, but they are still held together at joints called chiasmata - these are likely the actual crossover sites between chromatids of homologous chromosomes (where recombination occurs).

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MODULE 02 COMPANION GUIDE BCHM 218 MOLECULAR BIOLOGY | BCHM 218 M02 PAGE 139 5.7 INITIATION OF MEIOTIC RECOMBINATION 6 OF 14 To understand how cells generate diversity during meiosis, it is helpful to look at how yeast cells perform H R, as they are well studied and extremely proficient at performing genetic recombination. Learn about initiation meiotic recombination in yeast. 1. Recall that following the premeiotic S phase replication cycle, homologous chromosomes are brought together. As the cell enters meiosis, D S Bs are intentionally introduced at multiple locations along one chromatid of each chromatid pair. The breaks are not random, and yet are not entirely predictable. Breakage occurs at chromosomal ‘hotspots’ more frequently than other sites in the genome. A protein called Spo11 catalyzes the formation of these D S Bs. Diagram showing the stages of D N A replication. Diploid cells enter interphase then proceed through S phase into Meiosis I, where they split into two cells. The two diploid cells enter Meiosis II and then split a final time to create four haploid gametes or spores. 2. The protein Spo11 is found in all eukaryotes. It acts as a dimer and uses an active-site Tyr residue as a nucleophile in a transesterification reaction. Each subunit cleaves one D N A strand, with the phosphodiester bond replaced by a covalent 5′ -phosphotyrosyl linkage. The reaction halts at this point. A dozen or more additional proteins may cooperate in the formation of an active Spo11 complex on the D N A, and in processing the D N A after it is cleaved, which you will learn about next. Learn about the Spo11 Transesterification Reaction - Refer to 5.7.1 Spo11 Transesterification (Cleavage) Reaction Sub-Page Diagram of the Spo11 transesterfication reaction. Shows the intact duplex with two Spo-11 dimers. Each D N A strand is cleaved and then the phosphodiester bond replaced by a covalent 5′ - phosphotyrosyl linkage. This stage is called the transient covalent intermediate. Definition: Transesterification Reaction: The process of exchanging the organic group of an ester with the organic group of an alcohol. References:

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MODULE 02 COMPANION GUIDE BCHM 218 MOLECULAR BIOLOGY | BCHM 218 M02 PAGE 140 Cox, M. M., Doudna, J. A., & O'Donnell, M. (2012). Molecular Biology: Principles and Practice (2nd edition) . New York, N Y: W.H. Freeman and Co.

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MODULE 02 COMPANION GUIDE BCHM 218 MOLECULAR BIOLOGY | BCHM 218 M02 PAGE 141 5.7.1 SPO11 TRANSESTERIFICATION (CLEAVAGE) REACTION SUB-PAGE OF 5.7 INITIATION OF MEIOTIC RECOMBINATION LEARN ABOUT THE SPO11 TRANSESTERIFICATION REACTION 1/1 Details of the Spo11 transesterification (cleavage) reaction are shown in the figure. The active site Tyr- residue acts as a nucleophile, attacking the phosphodiester bond and replacing it with a 5’ - phosphotyrosyl linkage. Its mechanism is similar to the type II topoisomerase catalytic mechanism you learned previously. Spo11 is closely related to eukaryotic type II topoisomerases.

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MODULE 02 COMPANION GUIDE BCHM 218 MOLECULAR BIOLOGY | BCHM 218 M02 PAGE 142 5.8 PROCESSING OF S P O11-MEDIATED D S B S 7 OF 14 Once the Spo11-mediated D S Bs have formed during the initiation phase, there are a few key proteins that process the broken ends of the D S Bs. Learn about this process: 1. To remove S p o11 from the D N A and initiate nucleolytic degradation of the 5′ -ending strand, a complex of proteins consisting of M r e11-Rad50-X r s2 binds to each S p o 11 complex and cleaves the D N A by several base pairs on the 3′ side of S p o11. This liberates the linked protein along with a short segment of the attached D N A strand. 2. The nuclease S a e2 then degrades the D N A a bit more. Other enzymes, including the helicase S g s1 and the nucleases Exo1 or D n a2 , have been implicated in the more extended degradation of the 5′ ends to create long 3′ single -stranded extensions which can now be processed in a mechanism similarly to D S B repair. Diagram of the remainder of the processing of S p o11-mediated double strand breaks. From the transient intermediate, an M r e-Rad50-X r s2 complex cleaves the D N A on the 3’ side of S p o11. This leaves 3’ overhangs with a free O H. S a e2 then degrades the D N A to make the 3’ extensions even longer. Finally, S g s1, D n a2, and Exo1 further degrade the 5’ ends to create 3’ single -strand extensions. Reference: Cox, M. M., Doudna, J. A., & O'Donnell, M. (2012). Molecular Biology: Principles and Practice (2nd edition) . New York, N Y: W.H. Freeman and Co.

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MODULE 02 COMPANION GUIDE BCHM 218 MOLECULAR BIOLOGY | BCHM 218 M02 PAGE 143 5.9 MEIOTIC RECOMBINATION IS COMPLETED BY A CLASSIC D S B R PATHWAY 8 OF 14 Now that 3’ overhangs are present (which are able to scan and locate regions of homologous Watson - Crick base pairing) the process of meiotic recombination proceeds the same way as D S B repair (using some specific proteins in eukaryotes). Review the steps of this process. As you read, try to compare the process to D S B repair. 1. The single-stranded regions are first bound by R P A , the eukaryotic single-stranded D N A- binding protein. Aided by mediator proteins, two Rec A-class recombinases called D m c1 ( d isrupted m eiotic c D N A) and Rad51 are loaded onto the 3′ extensions on either side of the double-strand break. Note: D m c1 and Rad51 are the eukaryotic counterparts to Rec A (the Rec proteins used for sensing D S Bs in prokaryotes). They exhibit sequence and structural homology to Rec A and, like Rec A, they form extended nucleoprotein filaments on the D N A. 2. The site is now set up for recombination. Sequential cycles of binding, sampling and release of d s D N A - through transient Watson- Crick base pairing between the s s D N A and the complementary strand of the duplex partner - are performed until homology is detected. 3. A stable D N A joint is then formed by intertwining of the s s D N A with its complement from the homologous target. Recall from Section 04 that recombinases are involved in this step. Now polymerase and ligase can complete the process of repair. The same diagram that was used to show the process of recombinational D N A repair. First, a double-strand break undergoes end processi ng, creating the 3’ overhangs. Helicases and nucleases participate in this step. Second, the first strand invasion takes place through the action of recombinase. Third, the second strand invasion occurs by recombinase. Fourth, D N A polymerase extends the strands. Reference: Cox, M. M., Doudna, J. A., & O'Donnell, M. (2012). Molecular Biology: Principles and Practice (2nd edition) . New York, N Y: W.H. Freeman and Co.

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MODULE 02 COMPANION GUIDE BCHM 218 MOLECULAR BIOLOGY | BCHM 218 M02 PAGE 144 5.10 RESOLUTION OF HOLLIDAY INTERMEDIATES 9 OF 14 Recall the stable, X-shaped joint resulting when one strand of each parental duplex crosses over into the other duplex is called a Holliday intermediate. These linked homologous chromosomes can exchange genetic material by resolving Holliday Intermediates. You discussed this in Section 04, but will now go into the mechanism in more detail. There are two possible fates, only one of which leads to a crossover, but both involve the exchange of genetic material - the non-crossover exchange of genetic material is known as gene conversion . What determines whether or not resolution will result in crossover versus non-crossover products? It is dependant on whether resolution involves cleaving the crossover strands (horizontal arrows in the figure) or the template strands (vertical arrows). Non-Crossover vs. Crossover - Refer to 5.11.1 Crossover vs Non-crossover Sub-Page Diagram of the two possible ways Holliday intermediates can be resolved. In the first option, the template strands are cleaved in the same way, resulting in a non-crossover product of chimeric chromosomes. In the second option, the Holliday junctions are cleaved at different sites (one junction is cleaved at the template strands, the other at the crossover strands), resulting in a crossover product. Reference: Cox, M. M., Doudna, J. A., & O'Donnell, M. (2012). Molecular Biology: Principles and Practice (2nd edition) . New York, N Y: W.H. Freeman and Co.

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MODULE 02 COMPANION GUIDE BCHM 218 MOLECULAR BIOLOGY | BCHM 218 M02 PAGE 145 5.10.1 CROSSOVER VS NON-CROSSOVER SUB-PAGE OF 5.11 RESOLUTION OF HOLLIDAY INTERMEDIATES NON-CROSSOVER VS. CROSSOVER 1/1 Cleaving the crossover strands leads to completion of replication and segregation of two monomeric chromosomes in daughter cells - shown in figure (b). Cleaving the template strands creates chimeric chromosomes - shown in figure (a). As a general rule: • If both junctions are cleaved the same way then non-crossover products will be generated. • In contrast, when the two Holliday junctions are cleaved using different sites, then cross-over products are generated. Diagram of the two possible ways Holliday intermediates can be resolved. In the first option, the template strands are cleaved in the same way, resulting in a non-crossover product of chimeric chromosomes. In the second option, the Holliday junctions are cleaved at different sites (one junction is cleaved at the template strands, the other at the crossover strands), resulting in a crossover product. Definitions: Monomeric: A molecule that can combine with others of the same type to form a polymer. Chimeric: A fusion of two kinds of molecules. Reference: Cox, M. M., Doudna, J. A., & O'Donnell, M. (2012). Molecular Biology: Principles and Practice (2nd edition) . New York, N Y: W.H. Freeman and Co.

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MODULE 02 COMPANION GUIDE BCHM 218 MOLECULAR BIOLOGY | BCHM 218 M02 PAGE 146 5.11 CATALYSIS OF HOLLIDAY INTERMEDIATE RESOLUTION: RUV A B COMPLEX 10 OF 14 The recombination intermediate is resolved by nicking a strand in each duplex followed by ligation. The processing of Holliday intermediates is facilitated by a complex called Ruv A B ( r epair of U V damage). Up to two Ruv A protein tetramers bind to a Holliday intermediate and form a complex with two Ruv B hexamers. Learn about the two Ruv A B proteins: Ruv A Protein Ruv A protein is a Holliday junction-specific D N A binding protein that recognizes the structure of the D N A junction and keeps it in a ‘box - like’ state. Ruv B Protein The donut-shaped Ruv B hexamers surround two of the four arms of the Holliday intermediate. Ruv B is a D N A translocase, related in structure and function to hexameric D N A helicases. The D N A is propelled outward through the hole in the donut-shaped Ruv B, away from the junction, in a reaction coupled to A T P hydrolysis. The result is very rapid movement of the position of the Holliday intermediate. This can be thousands of base pairs in a few seconds.

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MODULE 02 COMPANION GUIDE BCHM 218 MOLECULAR BIOLOGY | BCHM 218 M02 PAGE 147 5.12 CATALYSIS OF HOLLIDAY INTERMEDIATE RESOLUTION: RUV C 11 OF 14 Now that the double Holliday intermediate is stabilized, and the Ruv A B complex has moved the Holliday intermediate away from the region of damaged D N A, it recruits Ruv C , a Holliday intermediate resolvase. Ruv C replaces one of the Ruv A tetramers at the junction and cleaves strands in opposing arms of the Holliday intermediate to resolve it into viable chromosomal products. It accomplishes this by nicking two strands with the same polarity. These nicks are later sealed into a different arrangement by D N A ligase. You will explore this step further later in this section.

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MODULE 02 COMPANION GUIDE BCHM 218 MOLECULAR BIOLOGY | BCHM 218 M02 PAGE 148 5.13 PRODUCTS OF HOLLIDAY INTERMEDIATE RESOLUTION Please see the online learning module for the full experience of this interaction. 12 OF 14 Answer the question about the products of a Holliday intermediate resolution. As you learned earlier in this section, D N A can be cleaved at either site 1 or site 2 (the template and crossover strands, or just the template strands). In order to best see how the chromosomes are resolved, it is easiest to first draw in the 5’ and 3’ ends of the Holliday intermediates, and then look at the resulting products. Would the products be crossover or non-crossover if the Holliday intermediate was cleaved at site 1? At site 2? Feedback: If the Ruv C resolvase nicks the Holliday intermediate at site 1, a crossover product would result. If it cleaves at site 2, there would be a non-crossover product. The two sites where the Holliday intermediate can be cleaved, resulting in either a crossover or non-crossover product. Diagram showing the stages of meiosis. Diploid cells enter interphase then proceed through S phase into Meiosis I, where they split into two cells. The two diploid cells enter Meiosis II and then split a final time to create four haploid gametes or spores. Reference: Cox, M. M., Doudna, J. A., & O'Donnell, M. (2012). Molecular Biology: Principles and Practice (2nd edition) . New York, N Y: W.H. Freeman and Co.

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MODULE 02 COMPANION GUIDE BCHM 218 MOLECULAR BIOLOGY | BCHM 218 M02 PAGE 149 5.14 RESOLUTION OF HOLLIDAY INTERMEDIATES Please see the online learning module for the full experience of this interaction. 13 OF 14 Using the diagram provided, answer the question about the result of Holliday intermediate resolution. A Holliday intermediate is formed between two chromosomes at a point between two genes, A and B , as shown below. The two chromosomes have different alleles of the two genes ( A and a ; B and b ). The two cut sites to resolve a Holliday intermediate - vertical (X) and horizontal (Y). Where would the Holliday intermediate have to be cleaved (points X and/or Y) to generate a chromosome with an A b genotype? Feedback: The Holliday intermediate would have to be cleaved at the Y sites. Reference: Cox, M. M., Doudna, J. A., & O'Donnell, M. (2012). Molecular Biology: Principles and Practice (2nd edition) . New York, N Y: W.H. Freeman and Co.

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MODULE 02 COMPANION GUIDE BCHM 218 MOLECULAR BIOLOGY | BCHM 218 M02 PAGE 150 5.15 SECTION 5 SUMMARY 14 OF 14 • The repair of D S Bs by homologous recombination requires use of a homologous chromosome and multiple enzymes. • During meiosis, recombination generates crossovers that create a physical link between homologous chromosomes just before the first meiotic cell division. • Meiotic recombination events are initiated at programmed double-strand breaks. • Meiotic recombination in eukaryotes makes an important contribution to the generation of genetic diversity in a population. • Some meiotic recombination events do not generate crossovers but result in a more subtle exchange of genetic information known as gene conversion . • Mitotic recombination is rarer but is also initiated at D S Bs.

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MODULE 02 COMPANION GUIDE BCHM 218 MOLECULAR BIOLOGY | BCHM 218 M02 PAGE 151 SECTION 06: SITE-SPECIFIC RECOMBINATION 6.1 SECTION 06: SITE-SPECIFIC RECOMBINATION (1)

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MODULE 02 COMPANION GUIDE BCHM 218 MOLECULAR BIOLOGY | BCHM 218 M02 PAGE 152 6.2 SITE-SPECIFIC RECOMBINATION (2) 1 OF 27 To conclude this module, you will now turn your attention to site-specific recombination. Site-specific recombination is a type of genetic recombination that occurs only at specific sequences. This section can be loosely divided into three sections: • First, you will learn the definition of site-specific recombination and transposition. • Then, you will explore some biotechnological applications of site-specific recombination. • Finally, you will look at three site-specific recombination technologies which scientists use to cut at specific sequences in vitro and in vivo . These are: zinc finger nucleases, TALENs, and CRISPR-Cas9.

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MODULE 02 COMPANION GUIDE BCHM 218 MOLECULAR BIOLOGY | BCHM 218 M02 PAGE 153 6.3 SITE-SPECIFIC RECOMBINATION (3) 2 OF 27 You will begin your lesson on site-specific recombination by gaining an understanding of what it is, as well as its outcomes. Learn about site-specific recombination and its effects. Definition Site-specific recombination is a precise and predictable process in which D N A is rearranged between two specific sequences. It involves the movement of specialized nucleotide sequences, called mobile genetic elements , between non-homologous sites. This procedure is carried out by recombinases, and can result in the insertion, deletion, or inversion - depending on the arrangement of the recombination sites - of a particular D N A segment. The reactions are often tied to genomic replication, but they have been recruited for other purposes as well -- which you will see later in this Module. Effects Site-specific recombination can alter gene order - which generally wouldn’t happen during homologous recombination because you exchange the same parts. Both eukaryotes and prokaryotes use this process to regulate gene expression and to increase the organism's genetic repertoire. Site-specific recombination can give rise to spontaneous mutations in organisms. The relics of site- specific recombination (repeated D N A sequences that can act as mobile genetic elements) can be found in many vertebrate chromosomes (~45% of human genome - and in plants this is even higher). For Interest - Refer to 6.3.1 For Your Interest: Transposition Sub-Page Definition: Mobile Genetic Elements: D N A sequences that can move around the genome, changing either their number of copies or location. They include transposable elements, plasmids, and bacteriophage elements. Reference: Mobile elements. Retrieved from https://www.nature.com/subjects/mobile-elements.

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MODULE 02 COMPANION GUIDE BCHM 218 MOLECULAR BIOLOGY | BCHM 218 M02 PAGE 154 6.3.1 FOR YOUR INTEREST: TRANSPOSITION SUB-PAGE OF 6.3 SITE-SPECIFIC RECOMBINATION FOR INTEREST 1/1 Another type of recombination system that, like site-specific recombination, brings together D N A sites without extensive homology is transposition. Originally discovered by Barbara McClintock, transposons are a kind of molecular parasite. They are nucleic acid sequences that insert themselves into the genome of another organism and are replicated passively every time a cell divides. Some transposons in the human genome are active, moving at a low frequency, but most are inactive, evolutionary relics altered by mutations. Through the activity of enzymes called transposases , these genetic elements can move from one location on a chromosome to another. This process is known as transposition and can lead to the redistribution of other genomic sequences and has played a major role in human evolution. Both site-specific recombination and transposition generally involve bringing together D N A sites without extensive homology. Reference: Barbara McClintock. (2013). Retrieved from http://atlaswikigr.wikifoundry.com/page/Barbara+McClintock.

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MODULE 02 COMPANION GUIDE BCHM 218 MOLECULAR BIOLOGY | BCHM 218 M02 PAGE 155 6.4 BIOTECHNOLOGICAL APPLICATIONS OF SITE-SPECIFIC RECOMBINATION 3 OF 27 Transgenic organisms can be produced. Creation of these organisms takes advantage of site-specific recombinase enzymes to edit the genome. These systems can be (and have been) used to activate a particular gene, insert a new gene into a cell at a chosen location, replace one gene with another gene or an altered version of the same gene, delete a gene, or alter the linear structure of an entire chromosome. This is particularly useful for making knock-out and knock-in mice. Two systems that use site-specific recombination for genome editing are: • Cre-Lox • F l p-F R T You will learn about these systems and their applications in the following slides. Definitions: Knock-out: The generation of a mutant organism in which the function of a particular gene has been completely eliminated. Knock-in: A desired gene is inserted into a specific locus in the target genome via homologous recombination. References: Knockout. Retrieved from http://www.biology-online.org/dictionary/Knockout. Minikel, E. (2012). The difference between knock-in and transgenic mice. Retrieved from http://www.cureffi.org/2012/11/13/the-difference-between-knock-in-and-transgenic-mice/.

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MODULE 02 COMPANION GUIDE BCHM 218 MOLECULAR BIOLOGY | BCHM 218 M02 PAGE 156 6.5 CRE-LOX P AND F L P-F R T TRANSGENES 4 OF 27 In the Cre-Lox P and F l p-F R T systems, Lox P and F R T are specific directional sequences that are placed into the genome. Note: they are not naturally occurring in the genome. Cre and F l p are the recombinase enzymes that recognize the Lox P and F R T sites, respectively. These Cre and F l p recombinases are also not naturally occurring in eukaryotes, but they work to cleave or invert the intervening sequence when engineered into the cells of any organism. Diagram showing both strands of the recombination sites from lox P and F R T; the inverted 13 b p repeats are binding sites for the recombinases, named Cre and F l p, respectively. The inverted repeats are separated by an asymmetric core sequence. The cleavage and exchange events occur at or near the ends of the core sequence. Reference: Cox, M. M., Doudna, J. A., & O'Donnell, M. (2012). Molecular Biology: Principles and Practice (2nd edition) . New York, N Y: W.H. Freeman and Co.

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MODULE 02 COMPANION GUIDE BCHM 218 MOLECULAR BIOLOGY | BCHM 218 M02 PAGE 157 6.6 OUTCOMES OF CRE-LOX P AND F L P-F R T MEDIATED RECOMBINATION 5 OF 27 There must be two Lox P or F R T sites in the genome for the system to work, and the overall outcome of a site-specific recombination reaction depends on the location and relative orientation of the recombination sites within the genomic D N A in which they reside. As you will see, the orientation of the Lox P or F R T sites is extremely important. Learn about the importance of the orientation of the Lox P or F R T sites. 1. If the sites are inverted (facing towards each other) the recombinase enzyme will invert the intervening sequence - changing its orientation on the D N A. 2. If the sites are oriented in the same direction, the recombinase can cleave out the intervening sequence, leaving behind one perfectly re-formed Lox P or F R T site. However, if there is a foreign piece of D N A that contains homology to the region outside of the Lox P or F R T sites, and also contains two matching sites, the intervening sequence of the foreign D N A can be inserted, and used to replace the endogenous sequence in between the sites - this is not shown in the figure. Diagrammatic representations of recombination on a single strand of D N A resulting in 1) inversion, and 2) deletion and insertion. Two recombination sites (F R T sites) flank a length of D N A to be recombined. A deletion results in 2 separate products: a linear D N A with none of the original genes, and a circular one with the three original genes and one F R T site. An insertion is the opposite of a deletion. Reference: Cox, M. M., Doudna, J. A., & O'Donnell, M. (2012). Molecular Biology: Principles and Practice (2nd edition) . New York, N Y: W.H. Freeman and Co.

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MODULE 02 COMPANION GUIDE BCHM 218 MOLECULAR BIOLOGY | BCHM 218 M02 PAGE 158 6.7 BRAINBOW: APPLYING CRE-LOX P SITE-SPECIFIC RECOMBINATION 6 OF 27 Using site-specific recombination, molecular biologists have determined a method to precisely modify the sequence of D N A. A prime example of this is the Brainbow technology. The Brainbow mouse is an example where individual neurons in the brain are visualized using fluorescent proteins. This process combines transgenic technology and site-specific recombination, and has been used to map all the neurons in a mouse brain. The use of site-specific recombination to make genomic alterations always requires some genetic engineering. Recombination target sites must be installed where a biotechnologist wants them to be, and the corresponding recombinase must also be present in the same cell and at the right time.

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MODULE 02 COMPANION GUIDE BCHM 218 MOLECULAR BIOLOGY | BCHM 218 M02 PAGE 159 6.8 THE ‘BRAINBOW’ TECHNIQUE 7 OF 27 The Brainbow method makes use of green fluorescent protein (G F P) and site-specific recombination to trace the path of the countless neurons that make up the brain. Learn how the Brainbow method works: Step 1 Inserted into the mouse genome is a gene cassette (a structured set of genes arranged for a particular biotechnological purpose) with several copies of G F P variants that encode proteins fluorescing with different colors, such as red (R F P), orange (O F P), yellow (Y F P), and cyan (C F P). Variants of the loxP target site for the Cre recombinase are engineered between the genes. Step 2 The different lox sites differ in their core sequences, so lox1 reacts only with lox1, and lox2 only with lox2, etc. Step 3 In this case, the cassette utilizes three or more different lox sites so that Cre-mediated recombination results in different patterns of G F P variant expression, and thus a different color, in each neuron! The loxP sites are arranged so that only one of the three possible recombination events can occur in a particular cassette, and each event will result in gene expression of one of the four G F P variants. Step 4 The cassette also includes a promoter that directs gene expression only in neurons - the promoter will only allow expression of the most proximal G F P variant. Transgenic mice have been engineered that have several of these cassettes, with the potential for expressing one G F P variant from each cassette in a given neuron. The engineered mice with G F P cassettes are homozygous for these cassettes, and they pass them on to all their progeny. Separately, a second population of homozygous transgenic mice is engineered to express the Cre recombinase, again from a promoter directing gene expression transiently, and only in developing neurons. Alt text: Animated diagram of how Brainbow works. The arrangement of the original D N A segment is (from left to right): promoter, lox 1, lox 2, lox 3, O F P, lox 1, R F P, lox 2, Y F P, lox 3, C F P. Given this arrangement, a lox 1 by lox 1 recombination will result in a red neuron, a lox 2 by lox 2 recombination will produce a

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MODULE 02 COMPANION GUIDE BCHM 218 MOLECULAR BIOLOGY | BCHM 218 M02 PAGE 160 yellow neuron, and a lox 3 by lox 3 recombination will produce a cyan neuron. Reference: Cox, M. M., Doudna, J. A., & O'Donnell, M. (2012). Molecular Biology: Principles and Practice (2nd edition) . New York, N Y: W.H. Freeman and Co.

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MODULE 02 COMPANION GUIDE BCHM 218 MOLECULAR BIOLOGY | BCHM 218 M02 PAGE 161 6.9 BRAINBOW OUTCOME 8 OF 27 When a mouse with the G F P cassettes mates with a mouse expressing the Cre recombinase, their progeny are heterozygous for both the cassette and the recombinase genes. As these mouse embryos develop, the Cre recombinase is expressed early in the development of each neuron. Recombination events occur in some or all of the cassettes in a given neuron, leading to the expression of a particular set of G F P variants. Mixing several different G F P variants in a cell increases the number of potential colors. The outcome is unpredictable for each cell. However, only one set of recombination reactions occurs in each cell, and the end result imparts a distinctive color that is expressed for the life of that neuron. Neighbouring developing neurons go through the same recombination processes, but the number of possible outcomes is large and neighbouring cells rarely acquire the same color. The result is a rainbow-like array of fluorescent colors in the neural network - a brainbow! Researchers use the brainbow to trace the paths of the axons through the brain. Neurons displaying various fluorescent colours from the brainbow mouse. Reference: Cox, M. M., Doudna, J. A., & O'Donnell, M. (2012). Molecular Biology: Principles and Practice (2nd edition) . New York, N Y: W.H. Freeman and Co.

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MODULE 02 COMPANION GUIDE BCHM 218 MOLECULAR BIOLOGY | BCHM 218 M02 PAGE 162 6.10 BRAINBOW Activity: Text Entry Please see the online learning module for the full experience of this interaction. 9 OF 27 Answer the questions about Brainbow. In the cassette shown in the figure, three different lox sites are used. How are these three different sites prevented from recombining with each other? Feedback: The core sequence must be modified to prevent recombination between the different lox P sites. Alt text: Animated diagram of how brainbow works. The arrangement of the original D N A segment is (from left to right): promoter, lox 1, lox 2, lox 3, O F P, lox 1, R F P, lox 2, Y F P, lox 3, C F P. Given this arrangement, a lox 1 by lox 1 recombination will result in a red neuron, a lox 2 by lox 2 recombination will produce a yellow neuron, and a lox 3 by lox 3 recombination will produce a cyan neuron. Reference: Cox, M. M., Doudna, J. A., & O'Donnell, M. (2012). Molecular Biology: Principles and Practice (2nd edition) . New York, N Y: W.H. Freeman and Co.

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MODULE 02 COMPANION GUIDE BCHM 218 MOLECULAR BIOLOGY | BCHM 218 M02 PAGE 163 6.11 A GENE’S FUNCTION CAN OFTEN BE ELUCIDATED BY ITS ABSENCE 10 OF 27 One of the main methods molecular biologists use to understand gene function is to delete it. This allows you to examine the effects of an absence of a protein on the cell function, and thereby infer the gene product’s function in an unaltered cell. An increasingly common strategy within the lab is to cut a gene at a site that is functionally critical, producing a D S B which is not properly repaired, and thus alters the gene function. As you’ve seen, double-strand breaks are usually repaired by homologous recombination systems. There are three technologies that have been developed to cut genes at particular sites in vivo : 1. Zinc finger nucleases 2. TALENs 3. CRISPR/Cas systems The remainder of this section will cover how these technologies work. Reference: The Nation (2016, June 7). Gene editing technique could transform future. Retrieved from http://nation.com.pk/snippets/07-Jun-2016/gene-editing-technique-could-transform-future.

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MODULE 02 COMPANION GUIDE BCHM 218 MOLECULAR BIOLOGY | BCHM 218 M02 PAGE 164 6.12 1) ZINC FINGER NUCLEASES 11 OF 27 Zinc finger nucleases (Z F Ns) represent a kind of designer D N A cleavage system based on using a zinc finger-containing protein to recognize a specific site, and cleave it. A zinc finger is a protein domain characterized by a single atom of zinc coordinated to four Cys residues or to two His and two Cys residues. It consists of about 30 amino acids and folds into a characteristic β - α - β structure, aided by a bound Z n2+ ion. Amino acid residues within the α -helix structure of the zinc finger are able to contact three consecutive nucleotides within the major groove of D N A. This leads to binding with a substantial degree of selectivity. The structure of a zinc finger binding domain. Reference: Cox, M. M., Doudna, J. A., & O'Donnell, M. (2012). Molecular Biology: Principles and Practice (2nd edition) . New York, N Y: W.H. Freeman and Co.

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MODULE 02 COMPANION GUIDE BCHM 218 MOLECULAR BIOLOGY | BCHM 218 M02 PAGE 165 6.13 APPLICATIONS OF Z F NS 12 OF 27 Molecular biologists can stitch together several zinc fingers in tandem, creating structures that allow the specific recognition of almost any D N A sequence 9 to 18 base pairs long. When the zinc fingers are fused to a nonspecific nuclease domain (often derived from an enzyme called FokI) to create a Z F N, the D N A sequence to which the Z F N binds is cleaved at a site adjacent to the recognition sequence of the associated zinc fingers. An engineered zinc finger nuclease (Z F N) in complex with its target D N A. The cartoon shows a Z F N dimer bound to D N A, with the nuclease domain in the middle. Reference: Cox, M. M., Doudna, J. A., & O'Donnell, M. (2012). Molecular Biology: Principles and Practice (2nd edition) . New York, N Y: W.H. Freeman and Co.

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MODULE 02 COMPANION GUIDE BCHM 218 MOLECULAR BIOLOGY | BCHM 218 M02 PAGE 166 6.14 2) TALENS 13 OF 27 TALENs are similar to Z F Ns, except that the D N A binding is directed by a series of transcription activator-like effector (T A L E) domains. TALENs are similar in size to zinc fingers, but recognize single base pairs. Like zinc fingers, they can be linked together and fused to a nonspecific nuclease domain (FokI) to yield a T A L E nuclease, or TALEN. These enzymes can be expressed in a cell, and the resulting enzyme cleaves the target site in the genome to generate a double-strand break. TALENs (and Z F N) can be designed to inactivate genes or integrate specific foreign D N A sequences. These methods can even inactivate viral D N A that has integrated into the genome. Alt text: An engineered TALEN in complex with its target D N A. A cartoon shows a TALEN dimer bound to D N A, with the nuclease domain in the middle. Reference: Cox, M. M., Doudna, J. A., & O'Donnell, M. (2012). Molecular Biology: Principles and Practice (2nd edition) . New York, N Y: W.H. Freeman and Co.

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MODULE 02 COMPANION GUIDE BCHM 218 MOLECULAR BIOLOGY | BCHM 218 M02 PAGE 167 6.15 LIMITATIONS OF Z F NS AND TALENS 14 OF 27 A major limitation of both Z F Ns and TALENs is that they require modularity to recognize different D N A sequences. This means that you need a different designer enzyme for every D N A site you want to recognize. This requires a lot of protein engineering, which is an expensive process that can take weeks or even months.

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MODULE 02 COMPANION GUIDE BCHM 218 MOLECULAR BIOLOGY | BCHM 218 M02 PAGE 168 6.16 3) CRISPR/CAS 15 OF 27 A more robust and convenient approach to cutting genes in vivo has been derived from a kind of bacterial immune system called CRISPR/Cas - short for C lustered, R egularly I nterspaced S hort P alindromic R epeats, and C RISPR- A ssociated P roteins. Learn about the origin of the CRISPR/Cas system: 1. Viral D N A seized by the bacterial immune system upon initial infection is incorporated into the CRISPR regions (repeats shown as black diamonds in the figure) to become the spacers (shown as coloured boxes). These hybrid chunks of native (repeats) and foreign (spacers) sequences are then expressed in the form of short R N A molecules. These aptly named ‘guide R N As’ can form a larger ‘seek and destroy’ structure when conjoined with the nuclease Cas9. 2. This system can both protect the cell, but also record events of previous viral attacks by detecting foreign D N A. This is possible because the CRISPR system allows integration of those viral D N A fragments into the CRISPR locus, and then CRISPR R N A biogenesis happens, which forms interference complexes able to hybridize to the viral sequences - the Cas9 nuclease can then recognize this complex, cleaving and inactivating the viral D N A. 3. The CRISPR/Cas system has been adapted by biotechnology into a nucleotide-specific, R N A- guided endonuclease system. Diagram of the development of CRISPR immunity. First, the virus invades the bacterial cell. Second, a new spacer is derived from the virus and integrated into the CRISPR sequence. These two steps are called adaptation. Third, CRISPR R N A is produced. This is the production of CRISPR R N A stage. Finally, CRISPR R N A guides molecular machinery to target and destroy the viral genome. This stage is called targeting. Reference: Barrangou, R., & Marraffini, L. A. (n2014). CRISPR-Cas Systems: Prokaryotes Upgrade to Adaptive Immunity. Molecular Cell , 54 (2), 234-244. https://doi.org/10.1016/j.molcel.2014.03.011.

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MODULE 02 COMPANION GUIDE BCHM 218 MOLECULAR BIOLOGY | BCHM 218 M02 PAGE 169 6.17 VIDEO: CRISPR/ CAS9 16 OF 27 You have now learned about how the CRISPR/Cas system functions naturally in bacteria. Watch the video to review the natural CRISPR/Cas system and see how researchers are using CRISPR/Cas9 both in test tubes and in live cells. You will learn more about the use of CRISPR/Cas9 for genomic engineering later in this section. As you watch: Focus on the mechanism of the CRISPR/Cas system and how it can be used for gene editing. Page Link: https://www.youtube.com/embed/2pp17E4E-O8

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MODULE 02 COMPANION GUIDE BCHM 218 MOLECULAR BIOLOGY | BCHM 218 M02 PAGE 170 6.18 ACTIVITY: CRISPR/CAS MECHANISM Activity: Drop Down Selection Please see the online learning module for the full experience of this interaction. 17 OF 27 Using your knowledge of the CRISPR/Cas mechanism, use the drop down menu to select the words that complete the summary. Word Bin: spacers, guide R N As, Cas, harpin loop, trans-activating When a bacteriophage attacks a bacterium with the appropriately integrated CRISPR/Cas, the CRISPR sequences are transcribed to: R N A, and ____ are cleaved to form products called _____. The g R N A then forms a complex with Cas proteins and often a ____ CRISPR R N A. This R N A folds into a ___ structure, which is recognized by ___. The resulting complex binds specifically to the invading bacteriophage D N A, cleaving and destroying it. Feedback: R N A, and spacers are cleaved to form products called guide R N As . The g R N A then forms a complex with Cas proteins and often a trans-activating CRISPR R N A. This R N A folds into a hairpin loop structure, which is recognized by Cas . The resulting complex binds specifically to the invading bacteriophage D N A, cleaving and destroying it.

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MODULE 02 COMPANION GUIDE BCHM 218 MOLECULAR BIOLOGY | BCHM 218 M02 PAGE 171 6.19 CRISPR/CAS9 SYSTEM FOR GENOMIC ENGINEERING 18 OF 27 Work in many laboratories, including those of Jennifer Doudna and Emmanuelle Charpentier, has produced a streamlined CRISPR/Cas9 system that is increasingly robust. The guide R N As (g R N As) and trans-activating CRISPR R N A (tracr R N As) are typically fused into a what is called a single guide R N A (sg R N A). This does not exist in the endogenous bacterial system. The guide sequence can be programmed (by altering its sequence) to target almost any specific genomic site. Continue to the next slide to learn how CRISPR/Cas9 is used in genomic engineering. Reference: The Conversation. (2015, Marxh 26). Explainer: CRISPR technology brings precise genetic editing - and raises ethical questions. Retrieved from http://theconversation.com/explainer-crispr-technology- brings-precise-genetic-editing-and-raises-ethical-questions-39219/.

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MODULE 02 COMPANION GUIDE BCHM 218 MOLECULAR BIOLOGY | BCHM 218 M02 PAGE 172 6.20 CRISPR/CAS9 GENOMIC ENGINEERING: NUCLEASE DOMAINS 19 OF 27 The Cas9 protein has two separate nuclease domains: one domain cleaves the D N A strand paired with the sg R N A, and the other cleaves the opposite D N A strand. Learn how these two domains can be manipulated to achieve different outcomes: Option 1 Inactivating one domain creates an enzyme that cleaves just one strand, creating a single-strand break, called a nick. This will be repaired using homology-directed repair (H D R). Using a ‘copy -and- paste’ mechanism, D N A sequences similar to those around the D N A break site, if introduced by the researcher, can act as templates in repair (red in the figure). This reaction makes sure the template is precisely integrated into the repaired sequence. Option 2 On the other hand, if a D S B is introduced, a ‘quick -and- dirty’ way to repair is by non-homologous end joining (N H E J) which simply joins the loose ends - an error-prone process which often leads to the disruption of the gene. Note: The s g R N A is needed both to pair with the target sequence in the D N A and to activate the nuclease domains for cleavage. Diagram showing the diverging pathways for double and single stranded breaks in the CRISPR/Cas9 System. In both cases, electroporation introduces the plasmid and fragment into the cell. Inside the cell, the assembled CRISPR/Cas9 complex scans the genome for a target sequence. In the case of a double strand break, the CRISPR system repairs the break using non-homologous end joining. In the case of a single strand break (nick), the recombination fragment is used to repair the break via homology-directed repair. Reference: Cox, M. M., Doudna, J. A., & O'Donnell, M. (2012). Molecular Biology: Principles and Practice (2nd edition) . New York, N Y: W.H. Freeman and Co.

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MODULE 02 COMPANION GUIDE BCHM 218 MOLECULAR BIOLOGY | BCHM 218 M02 PAGE 173 6.21 GENOMIC ENGINEERING: INACTIVATION VS MUTATION 20 OF 27 Plasmids expressing the required protein and R N A components of CRISPR/Cas9 can be introduced into cells by electroporation. In cells from many organisms, targeted gene inactivation occurs in 10% to 50% of the treated cells. If a genomic change (mutation) rather than a simple gene inactivation is desired, it can be introduced by recombination using a D N A fragment encompassing the cleavage site and including the planned change, which can be introduced into the cell with the CRISPR/Cas9 plasmids. This recombination is generally more efficient if a nick rather than a double-strand break is introduced at the target site because it will use homologous ends of the inserted D N A fragment to be integrated into the genome. Diagram showing the diverging pathways for double and single stranded breaks in the CRISPR/Cas9 System. In both cases, electroporation introduces the plasmid and fragment into the cell. Inside the cell, the assembled CRISPR/Cas9 complex scans the genome for a target sequence. In the case of a double strand break, the CRISPR system repairs the break using non-homologous end joining. In the case of a single strand break (nick), the recombination fragment is used to repair the break via homology-directed repair. Reference: Cox, M. M., Doudna, J. A., & O'Donnell, M. (2012). Molecular Biology: Principles and Practice (2nd edition) . New York, N Y: W.H. Freeman and Co.

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MODULE 02 COMPANION GUIDE BCHM 218 MOLECULAR BIOLOGY | BCHM 218 M02 PAGE 174 6.22 CRISPR/CAS9 GENOMIC ENGINEERING: NUCLEASE DOMAINS 21 OF 27 The Cas9 protein is comprised of six domains: Rec I, Rec II, Bridge Helix, Ruv C, H N H (H i s-A s n-H i s), and P A M-Interacting. The P A M ( P rotospacer A djacent M otif) sequence is an N G G trinucleotide sequence, where N is any base. It is always located next to the spacer sequences and is absolutely necessary for Cas9 to be able to bind target D N A. The P A M sequence must immediately follow the target sequence, and distinguishes invading D N A from self D N A. Learn how PAM is involved in both the endogenous CRISPR/Cas system and the synthetic system: Endogenous CRISPR/Cas System - Refer to 6.22.1 Endogenous CRISPR/Cas System Sub-Page Synthetic CRISPR/Cas System - Refer to 6.22.2 Synthetic CRISPR/Cas System Sub-Page Visual depictions of the endogenous and synthetic CRISPR/Cas systems, showing the P A M sequence next to the target D N A. Reference: Jinek, M., Chylinski, K., Fonfara, I., Hauer, M., Doudna, J. A., & Charpentier, E. (2012). A Programmable Dual-R N A-Guided D N A Endonuclease in Adaptive Bacterial Immunity. Science , 337 (6096), 816-821. https://doi.org/10.1126/science.1225829

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MODULE 02 COMPANION GUIDE BCHM 218 MOLECULAR BIOLOGY | BCHM 218 M02 PAGE 175 6.22.1 ENDOGENOUS CRISPR/CAS SYSTEM SUB-PAGE OF 6.22 CRISPR/CAS9 GENOMIC ENGINEERING: NUCLEASE DOMAINS ENDOGENOUS CRISPR/CAS SYSTEM 1/1 In the bacterial system, the mature CRISPR R N A (c r R N A) base pairs to trans-activating c r R N A (tracr R N A), forming a dual-R N A structure that directs the CRISPR-associated protein Cas9 to introduce double-strand breaks in the target D N A adjacent to the P A M sequence. At sites complementary to the c r R N A-guide sequence, the Cas9 H N H nuclease domain cleaves the complementary strand, and the Cas9 Ruv C-like domain cleaves the noncomplementary strand. Visual depictions of the endogenous and synthetic CRISPR/Cas systems, showing the P A M sequence next to the target D N A. Reference: Jinek, M., Chylinski, K., Fonfara, I., Hauer, M., Doudna, J. A., & Charpentier, E. (2012). A Programmable Dual-R N A-Guided D N A Endonuclease in Adaptive Bacterial Immunity. Science , 337 (6096), 816-821. https://doi.org/10.1126/science.1225829

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MODULE 02 COMPANION GUIDE BCHM 218 MOLECULAR BIOLOGY | BCHM 218 M02 PAGE 176 6.22.2 SYNTHETIC CRISPR/CAS SYSTEM SUB-PAGE OF 6.22 CRISPR/CAS9 GENOMIC ENGINEERING: NUCLEASE DOMAINS SYNTHETIC CRISPR/CAS SYSTEM 1/1 In the engineered system, the dual-tracr R N A:c r R N A, when engineered as a single R N A chimera (s g R N A), also directs sequence-specific Cas9 d s D N A cleavage adjacent to the P A M sequence. Visual depictions of the endogenous and synthetic CRISPR/Cas systems, showing the P A M sequence next to the target D N A. Reference: Jinek, M., Chylinski, K., Fonfara, I., Hauer, M., Doudna, J. A., & Charpentier, E. (2012). A Programmable Dual-R N A-Guided D N A Endonuclease in Adaptive Bacterial Immunity. Science , 337 (6096), 816-821. https://doi.org/10.1126/science.1225829

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MODULE 02 COMPANION GUIDE BCHM 218 MOLECULAR BIOLOGY | BCHM 218 M02 PAGE 177 6.23 REVIEW: TECHNOLOGIES FOR CUTTING GENES (1) Please see the online learning module for the full experience of this interaction. 22 OF 27 Use what you have learned to choose whether each statement refers to Z F Ns, TALENs, CRISPR/Cas, or more than one of these technologies. Select your answer(s) by clicking in the appropriate box . Characters Z F N TALEN CRISPR Has a protein domain that fold s into a β - α - β structure. Several can be stitched together. Derived from a kind of bacterial immune system. The protein is fused to a nonspecific nuclease domain. Particular amino acid residues in the α helix contact about three consecutive nucleotides in the major groove of a D N A sequence. Feedback: Characters Z F N TALEN CRISPR Has a protein domain that folds into a β - α - β structure. Z F N Several can be stitched together. Z F N TALEN Derived from a kind of bacterial immune system. CRISPR The protein is fused to a nonspecific nuclease domain. Z F N TALEN Particular amino acid residues in the α helix contact about three consecutive nucleotides in the major groove of a D N A sequence. Z F N

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MODULE 02 COMPANION GUIDE BCHM 218 MOLECULAR BIOLOGY | BCHM 218 M02 PAGE 178 6.24 REVIEW: TECHNOLOGIES FOR CUTTING GENES (2) Please see the online learning module for the full experience of this interaction. 23 OF 27 Use what you have learned to choose whether each statement refers to Z F Ns, TALENs, CRISPR/Cas, or more than one of these technologies . Select your answer(s) by choosing the appropriate box. Characters Z F N TALEN CRISPR Able to recognize single base pairs. A different enzyme is required for every D N A site that you want to recognize, which is very costly. Requires a P A M sequence to direct cleavage. Can be used to integrate D N A into a particular locus. Uses a nuclease to cleave D N A. Feedback: Characters Z F N TALEN CRISPR Able to recognize single base pairs. TALEN A different enzyme is required for every D N A site that you want to recognize, which is very costly. Z F N TALEN Requires a P A M sequence to direct cleavage. CRISPR Can be used to integrate D N A into a particular locus. Z F N TALEN CRISPR Uses a nuclease to cleave D N A. Z F N TALEN CRISPR

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MODULE 02 COMPANION GUIDE BCHM 218 MOLECULAR BIOLOGY | BCHM 218 M02 PAGE 179 6.25 REVIEW ACTIVITY: CRISPR/CAS9 Activity: True or False Please see the online learning module for the full experience of this interaction. 24 OF 27 Complete the CRISPR/Cas9 review activity. For each of the following statements, select either True or False: 1 of 3: Single guide R N A sequences are present in the endogenous bacterial system. 2 of 3: Non-homologous end joining is more error-prone than homology-directed repair (H D R). 3 of 3: The P A M sequence must immediately precede the Cas9 target sequence. Feedback: 1) False; single guide R N A is a synthetic fusion of g R N A and tracr R N A sequences. 2) True; homology-directed repair uses a recombination fragment to allow for precise copying. 3) False; it must immediately follow the target sequence.

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MODULE 02 COMPANION GUIDE BCHM 218 MOLECULAR BIOLOGY | BCHM 218 M02 PAGE 180 6.26 INDUCING MUTATIONS WITH CRISPR/CAS9 Activity: Text Entry Please see the online learning module for the full experience of this interaction. 25 OF 27 Answer the question about how CRISPR/Cas9 is used. Describe how you would introduce a specific genomic change using CRISPR/Cas9? Feedback: A change can be introduced by recombination when a D N A fragment encompassing the cleavage site and including the planned change enters the cell with the CRISPR/Cas9 plasmids. This type of recombination is generally more efficient if a nick rather than a double-strand break is introduced at the target site.

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MODULE 02 COMPANION GUIDE BCHM 218 MOLECULAR BIOLOGY | BCHM 218 M02 PAGE 181 6.27 SUMMARY OF CRISPR/CAS 26 OF 27 Review the points summarizing the CRISPR/Cas system. These are some of the take-home points that you should focus your studying on. • Cas nuclease proteins have homology to enzymes that perform D N A repair functions. • With the help of a guide R N A, the nuclease can be targeted to generate d s D N A breaks almost anywhere in the genome (and it can be multiplexed using several different g R N As). • Cells have ways of repairing these d s D N A breaks (N H E J and homologous recombination), and these processes can also be taken advantage of to edit genomes. • Genes can be inactivated or altered with the CRISPR/Cas9 system, as a method of elucidating or manipulating gene function. • This has huge future potential for gene editing, but also requires major ethical considerations • Genes can be inactivated or altered with the CRISPR/Cas9 system, as a method of elucidating or manipulating gene function. • This has huge future potential for gene editing, but also requires major ethical considerations.

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MODULE 02 COMPANION GUIDE BCHM 218 MOLECULAR BIOLOGY | BCHM 218 M02 PAGE 182 6.28 SECTION 06 SUMMARY 27 OF 27 Site-specific recombination entails the precise cleavage and rejoining of D N A ends at specific and reproducible sites in the D N A. This technology has been adapted by molecular biologists to manipulate D N A sequences. This function can be carried out by three methods of cutting genes in vivo , including: • Z F Ns, • TALENs, and • CRISPR/CAS.

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MODULE 02 COMPANION GUIDE BCHM 218 MOLECULAR BIOLOGY | BCHM 218 M02 PAGE 183 CONCLUSION You have now completed Module 02. SECTION 1: THE BEGINNINGS OF REPLICATION ENZYMOLOGY – Refer to 1.1 SECTION 2: D N A REPLICATION – Refer to 2.1 SECTION 3: D N A REPLICATION IN THE MODEL ORGANISM E. COLI – Refer to 3.1 SECTION 4: D N A REPAIR – Refer to 4.1 SECTION 5: D N A RECOMBINATION – Refer to 5.1 SECTION 6: SITE SPECIFIC RECOMBINATION – Refer to 6.1

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BCHM218_M02_OL-BL_onQ_CompanionGuide

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