Molecular Biology I
Molecular Biology I PCB 3522
University of Central Florida
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Popular in Biology Molecular Cell & Dev
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Date Created: 10/22/15
Molecular Biology II block 2 5 1 Weak and strong bonds determine macromolecular structure 23 c Polymerization Reactions 0 catalyzed linking together of activated building blocks 0 DNA RNA and polypeptides protein activated nucleotides nucleoside triphosphates activated amino acids20 aminoacyltRNA polymerization requires breakage of multiple high E bonds for addition of each building block 0 What specifies order of above molecules not enzymes nonrepetitive heteropolymers enzymes specify order for homopolymers ex starch cellulose and repetitive heteropolymers like GluMur GluMurGluMur cell wall Nonrepetitive specified by template central dogma o Central dogma of Molecular biology 0 DNA 9 RNA 9 Polypeptide 0 Now RNAe DNA reverse transcriptase Ex Retrovirus o Importance of Weak Bonds for Structure of Macromolecules 0 Order of the constituent building blocks determines their geneticbiochemical function specified by template 0 Primary information of RNA DNA amp proteins is the order of covalently linked building blocks but 0 Only after they form extensive additional weak bonds internally do they adopt their characteristic shapes that carry out function 0 Weak bonds play critical role in determining the structurefunction of DNARNAProtein o Importance of Weak Interactions 0 Weak interactions described in Chap 3 Hydrogen bonds ionic Van der Waals hydrophobic interactions 0 Direct proteins to form critical binding sites and DNA to assume double helical structure o Destruction of these interactions by disruption of weak bonds urea heat detergent without destruction of strong covalent bonds destroys activity of most Ex Ribonuclease treatment with urea HigherOrder Structures are Determined by Intra and Intermolecular Interactions 0 DNA Can Form a Regular Helix Examine Basic DNA Structure 0 Recall Xray diffraction studies of Franklin Borgon will cover amp Box 62 Chapter 5 Basic Protein DNA and RNA Structure 0 Cover structure of macromolecules and forces that control shape Emphasis on diverse structures of proteins Final sections focus on interactions between proteins and nucleic acids An activity central to many processes covered later in text Control of protein function Structure of DNA Details in Ch 6 Both Internal amp External Non Covalent Bonds Stabilize Structure 0 DNA molecules form a regular helix Two polynucleotide strands twisted in form of helix Nucleotide Building Blocks 3 5 phosphodiester linkage covalent Two antiparallel polynucleotide strands One strand 5 93 amp other 3 95 Complementary strands base pairing A T amp G C Two strands held together by H bonds between opposite base pairs purine pyrimidine Nucleotide Building Blocks 0 Nucleotide BSP 0 Base attached to sugar via an N glycosidic bond to C 1 on sugar N1 pyrimidine N9 purine o 5 C on sugar linked to phosphoric acid via ester bond 0 O 0 Both involve removal of water dehydration o Purine 2 rings Pyrimidines 1 o Tautomerization Keto to enol PICTURE o Keto form with carbonyl o enol form with hydroxyl o Tautomers Source of Errors 0 a amino in eq w imino o b keto in eq w enol 0 Two alternative tautomeric states in equilibrium with each other but equilibrium lies in direction of conventional structures Amino Keto 0 Basis of spontaneous mutations Ex Thymine enolGuanine o 3 5 Phosphodiester Linkage 0 Removal of water between the 3 OH on the sugar of one nucleotide and the 5 phosphate of the next nucleotide Recall AG 05 kcalmole o Pyrophosphate is released AG hydrolysis 7 kcal 0 Overall AG 65 kcal o Antiparallel 5 93 other 3 9 5 PICTURE o Hydrogen Bonds between base pairs on opposite strands o Ato T 2 H bonds 0 G to C 3 gtGC content gtTm 0 Almost all surface atoms in sugar and phosphate groups form bonds to water molecules 0 Position and length of H bonds between base pairs shown in FIGURE 51 o Stabilization of Double Helix o Purine amp Pyrimidine bases found in center of DNA allows base stacking Flat surfaces of bases stack creating shared pipi electrons between bases and limiting contact with water base stacking less satifactory if only one polynucleotide chain Because pyrimidines are shorter than purines single stranded DNA would result in unfavorable exposure of hydrophobic surface between adjacent bases 0 Disruption of double helix would bring the hydrophobic purine and pyrimidines into greater contact with water unfavorable 0 Double stranded DNA molecules contain a large number of weak bonds arranged so that most of them cannot break at same time If a bond breaks it can reform before additional bonds are broken thus multiple breaks rare so that double helices as short as ten remain stable o Breaking of Terminal Base Pairs in DNA By Random Thermal Motion 0 Typically when strands come apart without reforming this typically starts at one end and proceeds inward o This is because interactions between bases at the end of DNA are the least supported by adjacent interactions 0 At end they have only one neighboring base pair to secure interaction 0 Once some bonds break at termini they can reform or have additional breaks former favored at physiological temperatures o Ordered Collections of 2 bonds become less stable as T rises 0 As temperature rises above physiological temperatures the simultaneous breakage of several weak bonds is more likely After a significant number have broken it loses original form denaturation o Tm value for melting of DNA strand separation At Tm 50 of bonds broken and 50 bound Half way point Tm value goes up with GC content Why o RNA Forms a Wide Variety of Structures Details Chapter 614 0 Some like mRNA function as transient carriers of genetic information and are constantly associated with proteins thus they do not have an independent unique tertiary structure 0 Other RNAs fold into unique tertiary structures o Intramolecular interactions between distinct regions lead to formation of specific elements of secondary structure primarily between bases of RNA and include traditional base pairing unusual base pairing and hydrophobic base stacking Importance of 2 Hydroxyl 0 RNA differs from DNA in that the ribose sugar backbone carries a 2 OH group In folded structure of RNA molecules these 2 OH groups often participate in interactions that stabilize structure The binding of divalent metal ions ex Mg2 Mn2 Ca2 to RNA is often important to the formation of a stable folded conformation These ions can shield the negative charge of the RNA backbone allowing regions of the molecule to pack more tightly together Precisely folded compact nature of RNA Tertiary Structure 0 Illustrated by highresolution structures of some important RNA molecules 0 Ex TransferRNA Structures reveal that base stacking plays a major role in RNA conformation next slide Base stacking energetically favorable 7276 base pairs involved in base stacking Short base paired helical regions of RNA stack on top of one another to form longer discontinuous helical regions Then these regions of stacked helices pack against each other via additional tertiary interactions Transfer RNA Example 145 0 Cloverleaf a representation of tRNA structure exhibit a characteristic and highly conserved pattern of single amp double stranded regions 2 structure 0 Actual 3 dimensional structure using Xray crystallography reveals Lshaped tertiary structure o b Lbase stacking like DNA H bonds between bases in different helical regionsnonWatsonCrick interactions between bases amp SP backbone c ribbon representation o Chemical Features of Protein Building Blocks Amino Acids o 20 Amino Acid building blocks vary by R group 0 Common structural features Carboxyl group amp amino Alpha Carbon linked to a Hydrogen n The carboxyl group a Primary amino group n R group that is characterized by c Size shape chem composition o Amino Acids D amp L forms 0 Alpha carbon is a chiral carbon except for glycine R 0 Four different groups on alpha carbon thus assymmetric o Leads to optical isomers D amp L forms Mirror immages Enatiomers o R Groups All Participate in van der Waals Contacts That Are Dependent only on the Proximity of the Atoms Involved Rather Than Their Specific Chemical Makeup 0 Neutralnonpolar R side chains composed of simple C chains or aromatic rings that make principally hydrophobic contacts 0 Neutralpolar Include hydroxyl sulfhydral amide and imidazole also basic that make primarily H bonding interactions 0 Acidic Charged acidic carboxylates COOHe C0039 0 BasicLysine Arginine and Histidine Charged basic primary and secondary amines Lysine NH3 Arginine NHzCNH2 Histidine least basic of the three imidazole ring loses proton at pH 6 except after incorporated into polypeptide approx pKa of 7 Strongly polar thus all found more on exterior surface of polypeptide ie hydrated by H20 0 All 4 groups above have side chains that participate in Van der Waals contacts These depend more on proximity amp not chemical makeup Peptide Bond Formation PICTURE Primary Sequence 0 O O O 0 Sequence designation awkward to write full names Ex pentapeptide Glutamic acidGlycineAlanineLysineLeucine GluGlyAlaLysLeu E G A K L By Convention N amino terminus to left C carboxy terminus to right Peptide bond links amino acids 0 Peptide Bond primary covalent linkage between amino acids in polypeptides Water removed between the carboxyl group on one amino acid and the primary amine of a 2nd amino acid Linkage partial double bond fig 32 previous notes handout Because it involves more than one pair of electrons rotation around linkage limited All other linkages single rotate freely but steric interference between adjacent peptides limit Orientation of Adjacent Planar Peptide Bonds Limit Rotation O O Orientation of planar peptide bonds can be described by two bond angles Within proteins these angles constrained by the need to maximize the formation of 2 bond angles among functional groups within peptide backbone while minimizing steric interference o Angles of rotation about CaN and CaCO Bonds 0 Shaded planes of the peptide bonds PICTURE Four Levels of Protein Structure 0 Four Levels Primary linear sequence of amino acids held together by peptide bonds Secondary interactions between parts that make up polypeptide backbone major elements alpha helix amp beta pleated sheet hydrogen bonding Tertiary overall conformation stabilized by weak interactions final folding in 3D space a Formed from sequential interactions typically weak of increasingly distant amino acids Quaternary proteins consisting of multiple polypeptides manner in which they associate Molecular Structure 0 Primary sequence 9 Secondary local folding 9 Tertiary longrange folding 9 Quaternary multimeric organization 9 supramolecular largescale assemblies Functions Supramolecular o examples regulation signaling transport catalysis movement structure Flagellum o Filament individual flagellin chains 0 Fli proteins 0 Rings L P MS C MS amp C rings located within cell membrane amp cytoplasm C MS and P rings are motorbasal body Mot proteins surround MS amp C amp generate torque amp proton motive force provides the energy ABC Three Separate Proteins Needed for Transport of solute o Periplasmic binding protein PICTURE o Membranespanning transporter carrier 0 ATP hydrolyzing protein enzyme Primary Structure Determines 2 3 amp if gt 1 Polypeptide Chain 4 Denaturation of Ribonuclease 0 eg gentle denaturation urea will cause it to denature then cooling will refold it o Harsh high temperature will permanently misfold protein o Primary Structure Determines Eventual Tertiary Structure 0 Experiment with Ribonuclease Enzyme Subject to gentle conditions like urea that leads to denaturation interferes with weak chemical interactions H bonding but not covalent bonds Led to denaturation or unfolding When denatured Ribonuclease was restored to conditions that allowed formation of weak interactions enzyme rapidly regained 3D structure and RNA cleaving ability If excessively harsh conditions then irreversibly denatured and does not assume proper 3D conformation hydrophobics aggregate 0 Conclusion Primary 2 9 3 Tertiary interacts with other Tertiary polypeptides to form 4 Some require special help to achieve proper folding and enlist aid of special proteins Chaperonins or Molecular Chaperons prevents improper folding or aggregation hydrophobic Escherichia coli 0 Look at levels 2 3 in more detail 4 summary in next 2 o Quarternary Structure 4 0 Quarternary Structure is when a protein consists of multiple polypeptide chains known as protein subunits o Manner in which these subunits associate with one another is referred to as 4 amp includes 4 types weak bonds amp disulfide 0 Each subunit contains 1 2 amp 3 2 identical subunits homodimer nonidentical subunits heterodimer o Example 4 ATPase 0 Simple Example Insulin 1 0c1 a Two disulfide SS Hemoglobin 2 0c 2L3 o More Complex ATPase enzyme in membrane of bacteria PICTURE Alpha Helices and Beta Sheets Are the Common Forms of Secondary Structure 0 Alpha Helix PICTURE Most stable arrangement Righthanded helix repeating every 54 A along helical axis Favorable angles that accommodate a regular pattern of H bonding between carbonyl and imino groups Precise geometry each turn of helix has 36 amino acids Many amino acid sequences can adopt alpha helical symmetry because it is stabilized by contacts between nearly universal backbone atoms carbonyl and imino groups Proline is helixbreaking residue cyclical structure cannot participate as donor in H bonding Glycine tyrosine amp serine rarely found in alpha helix although their structures do not prevent it Side chains project away from helix as alpha helices constructed exclusively through backbone contacts Places them is good position to interact with another region of protein or another macromolecule Ex DNA or RNA 26 Beta Sheet is a Highly Extended Form of Polypeptide Backbone 0 Alignment of polypeptide in this extended form to allow for H bonds between carbonyl group of one beta strand and NH groups on adjacent strand 0 Typically a region of beta sheet is composed of 46 separate stretches of polypeptide each forming an individual beta strand each 810 amino acids in length 0 In beta sheet adjacent amino acids are related by a rotation of 180 thus respective side groups emerge from opposite sides Beta Sheets Held Together by Bonds PICTURE O a Beta sheet shown from above Oxygen and nitrogens of the backbone are fully hydrogenbonded b Beta sheet from side view Illustrates the location of side groups which alternate between emerging from above or below the plane of Beta sheet Representation emphasizes pleated sheet shape Ex antiparallel Two Types of Beta Sheets 0 0 Parallel Beta Sheet adjacent chains run in the same amino to carboxyl direction Antiparallel Beta Sheet Adjacent chains run in opposite directions In both all peptide groups lie approximately in the plane of the sheet Beta Sheets Twist in a RightHanded Manner Along Length 0 0 Most individual strands of beta sheets tend to be twisted along their length in a righthanded manner Thus instead of flat sheets of protein regions of beta sheet tend to curve to generate a compact protein molecule Thioredoxin protein To Fold Properly Favorable Interactions Must Be Maximized O O O For folding both backbone and side chains must adopt conformations that maximize favorable interactions Alpha helix and beta sheet are both stable conformations of backbone but for each side chain to make the max of weak bonds proteins adopt more varied shapes 3D conformations of polypeptide chains are thus compromises between tendency of backbones to form either alpha or beta and the tendency of side groups to twist the backbone into less regular configurations to maximize strength of 2 bonds of side chains Regular amp Irregular Features of Protein Structure Irregular Configurations in backbone green allow maximum formation of 2 structuresbeta amp alpha Burial of Hydrophobic Groups 0 One of strongest influences on protein folding is burial of hydrophobic nonpolar side chains into core of polypeptide s structure 0 Leads to prediction that in aqueous solutions polypeptides with large numbers of nonpolar side groups will tend to internalize the nonpolar residues and be more stable than polypeptides containing mostly polar groups 0 The decrease in free energy is small if Hbonds disrupted in a polar molecule held together by large of internal H bonds because polar groups can then H bond to H20 0 If we disrupt nonpolar groups typically a much greater loss in free energy because disruption inserts nonpolar groups into water o Specific Conformation of a Protein Results From Its Pattern of H Bonds 0 A portion of energy stabilizing protein comes from hydrophobic interactions Stabilization has no directional component 0 But Specific conformation of protein determined largely by H bonds H bonds require precise distances and angles fig 39 previously discussed In general all Hbond donors and acceptors within protein interior have suitable mates Failure to make an H bond in protein interior is energetically costly a few kcal H bond 0 Necessity of satisfying all the Hbond donors amp acceptors on backbone 2residue drives formation of large sections of alpha helix and beta pleated sheets 0 The only way that a polypeptide can transverse the non aqueous interior of a protein and satisfy Hbonding necessity is through formation of regular secondary structures side chains do not have enough donors and acceptors to do that Thus 0 All proteins contain significant regions of alpha helices amp beta sheet Polypeptide Chain Folding PICTURES 0 Even proteins that are composed of only alpha helices and beta pleated sheets adopt a variety of structures A alpha helices myoglobin amp lamba N terminal B beta sheets GFP amp gamma crystalline C Comparison of N terminal domain only alpha of lambda repressor with C terminal only beta o TurnsLoops 0 Some polypeptide sections are less regular to allow for turns at ends of alpha helices and individual strands of beta sheets Turns are loops of amino acids that link alpha helices and beta strands but no 2 Generally very short to minimize the number of unfulfilled Hbonds that accompany them 0 Less regular structure of loops critical for formation of binding sites for Small molecules Active sites of enzymes 0 Surfaces involved in proteinprotein interactions o Many Polypeptides Interact With One Another Through Supercoiling of Alpha Helices Around Each Other 0 Occurs only when nonpolar side chains along each alpha helix are arranged so that their side groups contact the other helix 0 Twisting of helices around each other reflects nonintegral 36 residuesturn nature of alpha helix 0 Allows side groups to pack neatly together only when the alpha helices interact at 18 angle from parallel If alpha helices remained rigid could stay in contact for only few residues o Coiledcoils Leucine Zipper of DNA Binding Proteins PICTURE o By supercoiling in lefthanded direction neatly packed and highly stable coiledcoils created 0 In leucine zipper family of DNA binding proteins two subunits come together to form a dimer through use of coiledcoil region 0 Region called leucine zipper because of repeating appearance of leucine and other amino acids with aliphatic side group valine or isoleucine o 2 turns 7 aa length aliphatic at 1St amp 4th position of 7 Positions such that aliphatic are on one side of helix 1 amp 4 on same face Two faces in two adjacent helices are packed against each other burying the hydrophobic away from water Leucine Zipper Combines Dimerization amp DNA binding sufaces within a single structural unit 0 Leucine zipper bound to DNA at right Fig 177 0 Two large at helices one from each monomer form both dimerization and DNA binding domain later 0 Two helices interact to form coiled coil that holds monomers together further down helices separate enough to interact with DNA Zinc Finger Domain 0 Alpha helix on left is recognition domain 0 Presented to DNA by 3 sheet on right 0 Zinc coordinated by 2 his residues in X helix amp 2 cys residues in 3 sheet 0 Arrangement stabilizes structure amp essential for DNA binding HelixLoopHelix 0 Long 0c helix involved in both DNA recognition and in combination with second shorter 0c helix dimerization alpha helix that binds DNA basic aa 0 Dimerization 2 0c helical regions 0 Separated by a flexible loop that allows them to pack together Most Proteins are Modular Containing Two or Three Domains 0 Single Proteins larger than about 20000 Daltons fold into 2 or more domains Smallest 11000 Da 100 residues Most single subunits2000070000 Da 0 Domain describes a structure that appears independent from the rest 0 Usually formed form a continuous amino acid sequence Box 52 read 0 Compact locally folded region of 3 structure Multiple domains common in larger globular proteins 5 2 Weak and strong bonds determine macromolecular structure c Wed 28 Dr Borgon I Protein Domains 0 small of domains account for variety of protein structures 0 limited to othelix Bsheet and a few other variants 310 helix Bhelix 0 20000 different proteins much fewer domains 0 why do we mainly see ohelix and Bsheet because they are efficient they fold properly 0 RNA world hypothesis RNA only has 4 nucleotides but amino acids have 20 o Protein Folding 0 want to maximize interactions until protein forms most stable conformation 0 Example of 3 structure motif small fragment 50 or so amino acids 808 888 greek key coiled coil maximizes hydrophobic interactions some small motifs are functional ex leucine zipper 0 There s a temperature which causes vibration and these proteins moving around and will eventually maximize those interactions in nanoseconds o Domain examples 0 Named by founder or from fraction from chromatography for example p53 was names after gel but its really 47kDa o SH2 interact with phosphotyrosine o SH3 interact with proline rich sequence H for homology SRC homology domain 2 and 3 because they look similar but specific purposes 0 WD40 Picture propeller conformation not every domain does the same thing but something very similar 0 EF Hand interacts with Ca so if a similar domain is present suspect Ca2 binding 0 Armadillo repeat 0 SNARE for vesicle trafficking 0 Death domain in pathway for cell death 0 Bromo acetylated lysine o Chromo methylated lysine regulate histones by acetylation and methylation then recognize sites with a protein that has a chromo or bromo domain 0 Example domains that recognize methylated lysine don t bind to everyone there are parts next to it to choose a specific domain o Most proteins are modular with 2 or 3 domains 0 proteins range from 1002000 amino acids 0 many proteins especially eukaryotic proteins are modular 0 Average size is larger in humans 9 indicates multiple domains bacteria can turn an operon 9 express 5 different proteins independently from one another CoA synthesis bacteria do it in 2 steps but humans have the domains in one proteins 9 regulate macromolecular structure better o Protein Domains 0 proteins are multidomain having sections that appear independent amino acids in active site can often be far apart folding 9 brings them close can cleave domains with proteses 9 often get a stable feature still functional or fuse proteins 0 Usually a continous chain as opposed to active sites 0 Usually stable on their own if cleaved or generated recombinantly tags ex GST Compartmentalization serve a function specialization tailored toward a specific function or stabilization enhance solubility structure ie scaffolding Advantageous in folding time n domains offer the advantage each domain can form independently 9 then can bundle 9 bundles can then fold thus reducing time to fold reduce large combinations of structures possible Ex LeucyltRNA synthetase Catalytic domain editing domain make sure corrent amino acid is there leucinespecific domain will cleave off ones other than leucine zincbinding Many homologues are synthesized separately in bacteria operons bacteria PPAT and DPCK 2 proteins Humans coenzyme A synthase 1 proteins multiple domains No operon needed cAbl tyrosine kinase involved in cell domain has SH2 SH3 other domains Pyruvate kinase and its domains 0 PEP ADP 9 pyruvate ATP Fructose 16bisphosphate binds to a domain step 3 phosphofructokinase F6P 9 FIGBP Step 10 Pyruvate kinase interacts FIGBP even though it s a few steps away what is the purpose it activates it too much PEP 9 stops whole system so FIGBP does feedforward activation to turn it on and make more room for PEP Phosphofructokinase 2Fructose bisphosphatase 2 0 adds phosphate takes away F6P e9 F1GBP 0 depending on glycolysis or gluconeogenesis negative feedback via ATP and citrate o Regulated by phosphorylated 32 no P 9 kinase P 9 phosphatase Fatty Acid Biosynthesis O O O O O O O o Acyl carrier protein ACP attaches to coenzyme A via Ser 41 phosphodiester bond CoA is bound to growing fatty acid At tip of CoA is aceter group so all of it is to scaffold the acetyle group 0 Fatty Acid synthase has many domains each performs different reaction ACP sits in middle and linker is CoA and it carries CoA around the clock Until it gets chopped off around 16 or 18 because it can reach the cleavage site 0 Triclosan Put in antibacterial soap 9 inhibits fatty acid synthesis enoylCoA reductase in bacteria o Friday 210 Multidomain Protein Origins 0 Gene duplications exons shuffling retrotransposons unequal crossing over DNA polymerase slippage combined synthesis fusion of multiple genes followed by divergence 0 Gene Example if two sequences are similar enough 9 misalignment and crossing over 9 gene duplication and deletion doesn t mean its functional still needs the appropriate promoter and transcription factors 9 many duplications result in a dead end duplications and divergence Mutation in one and another 9 genes will eventually deviate 9 domains with similar function but slight variance example tissue plasminogen activator duplications of EGF F and K 9 exon shuffling 9 another duplication of K 9 produce multidomain protein Killer immunoglobulin very similar TM IC EC domains but deviations like sequence length occur Secretory Calciumbinding phosphoprotein n ancestor protein duplicated 9 fish protein 9 more duplications and variations 9 human protein Lymphocyte signaling Blue domain Pro rich SH2 u look the same but some binding partners differ n deviations indicate that there was an ancestral gene 9 diverge into specialized domains similar things bind to different partners Big picture if domains already functional duplications are more likely than starting from scratch o Hemoglobin can trace it back to cytochrome b5 variations of cytochrome 9 variations in Hb n in our own genome there are genes for variations of alpha and beta chains o turned on at different times eg fetus adults o some are pseudogenes dead end don t do anything 0 Example Tyrosine kinase and VEGFR Different amounts and different duplications but you can trace the tyrosine kinase back to an ancestral gene Vascular endothelial growth factor receptor also common ancestor 9 duplications and divergence o oCatulin 9 oCatenin 9 Vinculin chr 9 5 and 10 respectively alignments are similar oCatenin went through some duplications 9 vinculin a slight changes to adjust binding but structures are very similar 0 Gene Duplications within proteins coiled coil repeats People who solved the sequence identified repeats u when the structure was solved repeats corresponded to duplications each repeat is a helical bundle but slightly changed o eg a four helical bundle duplicated 9 similar overlap but with a kink specialization How you build a protein up from scratch a stable helical bundle 9 repeats over time Line up humans rats and chickens have insertions that fruit flies and worms don t a Notice that insertions are all on the same side thus indicates potential binding partners or membrane binding a line up splice variant 9 also another repeat a 68 amino acid duplication alternate splice greatly alters protein function and actin viscosity o determined with falling ball viscosity o II Quaternary Structures 0 O O O O O O Naming dimer trimer tetramer planar tetramer etc large numbers can have different symmetries n eg hexamer can be a trimer of dimers or other six subunit arrangement homodimer or heterodimer same or different subunits icosahedron in viral structures a eg bacteriophage may only need to synthesize one protein to make this structure pseudoheptameric structure 4 heptameric rings stacked Box 52 why synthesize subunits of quaternary structures Book reduce mistakes discard faulty subunits prior to incorporation Speed of translation 1000 amino acids 1 minute to synthesize 250 x 4 subunits 15 seconds to synthesize parallel processing 1 or 2 units are efficient enough as in hemoglobin so gene space not wasted fewer transcription factors needed Proper folding would a large macromolecule fold properly or is it easier to build subunits stable functional subunits would fold more efficiently 9 more likely chance for a large protein to aggregate Allosterykinetics kinetics are greatly affected by quarternary structures Cooperativity cross talking for example the fructokinasephosphatase eg if one subunit of hemoglobin is bound to Oz 9 others are likely to bind 0 Increase chance of coming into contact with substrates 0 Control actin exists as a monomer G and multimer F multimer can form some stable backbone structures 0 Book counterpoint Largest Known Protein GroEL O O O O O O 0 Titan connectin passive muscle elasticity resting tension acts like a spring 26926 amino acids single chain 2 domains 299345139 Da 3000 kDa 1 microm in length C132983H211861N36149O408835693 longeSt word in the English language 18 aasec 24 mintues to synthesize likely only 3 6 aasec in eukaryotes so up to 25 hours to synthesize associates with cytoskeleton during synthesis very few genetic disorders during synthesis very few genetic disorders related to this protein indicating its importance Essential that it is synthesized correctly If faulty subunits are so common how is this so stable 840 kDa 7000 amino acids 14 subunits 60kDa each 500 amino acids when heat shock protection is essential time is a factor Synthesis of single chain 65 minutes 390 seconds Synthesis of 14 separate chains 30 seconds due to parallel processing of ribosomes or 13X faster Time is vital since proteins needed to be refolded after a heat shock from high temperatures More examples proteasome is a pseudoheptameric ring insulin hexamer actin filament oactinin spectrin is a trimer that is like a skeleton for your cell ATP Synthase many subunits spins 9 change conformation 9 as they move they synthesize ATP O RuvA used in recombination binds to DNA and shuffles RAD51 repairs doublestranded breaks c Mon 213 ATP Synthase Video o III Protein Interactions O O recognition structurally based complementarity weak forces multiplicity o Redblue Picture 0 Color representation shows acidic negative asp glu red and positive basic his lys arg blue eg DNA binding protein would have positive charges to interact with DNA Examples acidic pocket basic ladder alternating acidicbasic domains pack together nicely help packing or activation Cdc34 s acidic tail occupies Cull s basic canyon can gain info from looking at surface of protein o IV ProteinNucleotide and Cofactor Interactions O O O O Rossmann fold NADflavin binding alternates helix and sheet w sheets parallel Dinucleotide fold ATPNAD binding parallel Bsheet with ohelices on both sides oBo P Loop Walker A Motif Phosphatebinding loop found in ATP synthase ATP binding cassette PICTURES ATP bound to various proteins RecA green RAD51 cyan 39 quot11500 30 is cut off for sequence level Yet structurally they have almost perfect homology for DNA repair 9 Won t find mutation deadend mutation not something passed on o Nucleotide Binding 0 O Motifs Walker A Ploop GXXXXGKTS and Walker B Use base stacking interactions o if you can find the sequence of a motif eg walker B 9 then likely that function Nucleotide binding o PICTURES ATP bound to an enzyme 2200 0 pocket next to it o ACP has a Ser32 that binds to phosphate of CoA final step of pathway put ATP next to growing CoA and has a phosphate transferring to the sugar Covalent interaction interaction between ACP and CoA o PICTURE Heme in a single chain of hemoglobin o Heme binds to Fe and 02 protein acts as scaffolding o Succinate Dehydrogenase 0 Proteins transfer the electrons o SDHA holds FADHZ 9 takes electrons from succinate and transfers it down the chain of the protein 9 electrons cross into the membrane o V ProteinDNA Interactions most important 0 Transcriptional regulation 0 Packing cells 10microns contain 2m of DNA 0 Recognition Conformation specific n Ionic or hydrogen bonds with phosphate backbone or through interacalation of bases with tyrosine or tryptophan SSBs Sequence specific a Major groove via ohelix insertion or other mechanisms a 67 of eukaryotic proteins o ConformationSpecific Binding 0 Single strand during replication dangerous because DNA may be permanently lost a double strand has complement o SSB with DNA bind and protect don t want it to be sequence specific base stacking interactions with ssDNA o Sequence specific binding Major groove matches the width and depth of an ohelix Multiple interactions improve specificity Major groove has many hydrogen bond donors and acceptors with which amino acids can interact o ohelices have a dipole moment that gives them a charge which can interact with the phosphate backbone slightly negative on O and slightly positive on H on HN o Major Groove Hydrogen Bond DonorsAcceptors 0 proteins can distinguish between GC and CG AT and TA are distinguished by other factors AT and TA ADA GC AAD CG DAA o Atoms can have more than one donor or acceptor so some that are involved in H bonding can also be involved in the groove interactions o Proteins that interact with DNA and how they do it o Helixloophelix Ca2 interacts with consensus sequence and loop around loops can stick into major groove 0 Leucine zipper shape of helices fit perfectly with major groove and is sequence specific interacts with multiple bases and downstream at another major groove 0 Zinc finger motif betabetaalpha has a beta strand inserted in the major groove 0 CAP bound to cAMP regulatory and DNA interacting in multiple places double specificity o Lambda repressor helixturnhelix stabilization helix stabilizes and recognition helix fits into groove Protein dimerizes o TFIIDTBPTATA Element Beta sheets in minor groove multiple interactions ensure its in the TATA box 0 O O o Nonspecific Interactions to scan DNA 0 O O O O O 0 Some proteins slide along DNA for replication repair and recombination requires ATP Proteins use positivelycharged side chains to interact with phosphate backbone It concentrates regulatory proteins to DNA allowing them to search for their appropriate target instead of floating freely in the cell This may cause affinity for nonspecific DNA sites Needs to be nonspecific to ensure replication of 3 billion bases pairs but still needs to make sure it is DNA Occasionally sequence specific will bind to a nonspecific site ProteinRNA interactions Stabilization processing unstable important to regulate RNA levels continuously degraded but proteins can stabilize it eg spliceosome Recognition 2 hydroxyl not present in DNA induces A form helix No ohelical insertion into grooves RNArecognition motif RRM 8090 amino acids in a four stranded antiparallel beta sheet and two ohelices that pack against it contacts RNA via Arg or Lys salt bridge to phosphodiester backbone base stacking eg NusA for transcription termination if it recognizes the right sequence Wed 215 Spliceosomal proteinRNA complex Don t maintain ds very long so interactions usually base stacking are typically different than Hbonging or intercalation of double helix Take RNA and bring exons close together to allow them to be connected beta sheets arranged antiparallel with side chains for base stacking VII Protein Modification Activation and regulation Apoptosis video 0 O O O ligandbinding causes conformational changes turn proteins on or off Amino acids sugars Other proteins Amino acid modifications phosphorylation Recruitment location Proteolytic cleaveage mechanical physical tugging or stress to cause it to activate Allostery changes at one location on a protein affect another location or other proteins in the quaternary structure 1 bind to receptor 2 Transduction 3 Response Pathways Picture integrins involved in cell attachment caspase involved in apoptosis indicates a death pathway Lac Repressor regulation add inducer eg IPTG 9 change conformation and then can no longer interact with DNA DNA binds to multiple places Aspartate Transcarbamylase open state synthesizes pyrimidines inhibitory state when CTP binds to it inhibits pyrimidine synthesis negative feedback Allostery O O ties into quaternaery structure an activator will cause the sites to bind an inhibitor will shut it down affects enzyme kinetics sigmoidal curve cooperativity standard MichaelisMenten is usually for single subunit proteins Pyruvate kinase activated by FIGBP o Phosphofructokinase is activated or inhibited my ATP depending on concentrations and AMP and FZGBP to ensure that it functions only when it should 0 ATPGTP Hydrolysis how ATP hydrolysis affects conformation eg EFTu with GTP bound u when hydrolyzed the phosphate is no longer in the way the switch helix moves 0 Cdk activation by Cyclin Book diagram control system of cell is like a computer ie input 9 and 9 output need multiple inputs met to achieve an output Has T loop PSTAIRE Partially active Critical glutamate moved into position looped move away from entrance to active site Active CDK activating kinase CAK phosphorylates T loop o Amino acid modification PICTURE 0 eg ubuitination when multiple ubiquitins are added then the protein is targeted for degradation 0 eg phosphorylation at Tyr Ser Thr o methylation and acetylation of histones acetyl allows bromo domain to bind o Phosphorylation 0 phosphate high energy group Changes charge decreases hydrophobicity of regions affects interations eg 2 proteins may not interface unless there is a phosphate 0 Up to 30 of proteins phosphorylated 0 Most cellular pathways 0 Kinase adds phosphate 500 human known 2 of genes drug targets O O O O O Phosphatase removes phosphate Amino acids that can be phosphorylated Ser Thr Tyr A single phosphate can activate a protein senses everything with the state of DNA Has many phosphorylation sites affects proteins it can interact with and determines which pathways it activates cAbl phosphorylation O O has SH2 for phosphotyrosines 9 keeps it in an inactive state its own domain is keeping it inactive remove phosphate 9 SH2 breaks free 9 activation of cAbl Glycosylation O O O O O O Serine threonine asparagine up to 50 of the mass of a protein proper folding in ER Secretion If protein doesn t have right amount of sugars 9 ER targets it for destruction eg if the proper amino acids are not in the surface Very difficult to crystallize a sugar Ubiquitination O O O O Ubiquitin binds to lysine Degradation proteasome targeting E1 binds to ubiquitin 9 transfers it to E2 9 transfers it to target protein with E3 ligating enzyme how cell knows when its time to get rid of protein Mechanical and conformational activation Mechanotransduction Cytoskeletal proteins Tension can cause proteins to spring open usually works handinhand with ligands and phosphorylation Usually used to take a EC signal So if you have collagen 9 integrins bind to that 9 proteins on inside talin and vinculin bind to that 9 tension 9 open up 9 signal transduction collagen 9 integrin 9 talin 9 vinculin 9 actin Integrin Actiation o Proteinprotein interactions ECM talin o Ligands PIPZ 0 Phosphorylation 0 Mechanical and conformational activation 0 has alpha and beta beta has IC tail Talin activation 0 proteinprotein interactions integrin o Ligands PIPZ 0 Phosphorylation FAK PKC 0 Mechanical and conformational activation Vinculin activation 0 proteinprotein interactions head talin alphaactinin paxillin VASP vinexin ponsin Arp23 profilin actin o Ligands PIPZ 0 Phosphorylation PKC serinethreoninetyrosine kinases o Mecahnical and conformational activation Vinculin changes conformation with different binding proteins PICTURES Talin has to change conformation to bind to protein because they overlap o 9 tension theory tension on membrane 9 proteins pull apart 9 helices move 9 then protein can bind Videos Vinculin movement and The Inner Life of the Cell 6 The Structures of DNA and RNA I History of the DNA structure 0 Known to be part of the chromosome Original isolation very rough which destroyed DNA Thought to be scaffolding Phoebus Levene Tetranucleotide Hypothesis 1912 bases were believed to be in equimolar quantities in the chromosomes 0 O O Frederick Griffith working on pneumonia discovers transformation 0 Oswald Avery later showed DNA was the genetic material via transformation but it was disputed because of conflict with the tetranucleotide hypothesis Nucleotides boring and bland Proteins exciting and unique 0 HersheyChase Experiment in 1952 bacteriophage 32P vs 355 o Erwin Chargaff impressed by Avery s results 9 abandoned previous work and was first to note nucleotides were not equimolar which disproved the tetranucleotide hypothesis AT GC PurinePyrimidine ratio via paper chromatography and UV spectrophotometry Explained theories to WatsonCrick in 1952 Initial DNA structures William Astbury took Xray diffractions pictures of DNA in the 1930s a misinterpreted u said phosphate and base were in the same plane a not disproved until 1950 with the structure of cytidylate by Sven Furberg which showed the base and sugar at a right angle Astbury also developed the idea of the alphahelix and betasheet he named both Inspired Linus Pauling with regards to H bonding o Linus Pauling discovered alphahelix structure used 15 angstorm reflections first model involved drawing amino acids on paper and folding it His work worried WatsonCrick Made 3 chain DNA model a problem phosphates would repel each other o Watson Crick and Franklin 0 James Watson 0 O Birdwatcher from Chicago PhD at 21 Worked on bacteriophage at Indiana University with Salvador Luria who discovered restriction enzymes Postdoc at Cambridge 1951 where the first Xray crystallography group was set up by Lawrence Bragg Max Perutz and John Kendrew were present structures of hemoglobin and myoglobin Heard Maurice Wilkins talk about DNA from Xray diffraction in 1951 0 Francis Crick Designed naval mines for the British Navy Joined Cambridge as a graduate student at 33 1949 Worked closely with Frederick Sanger insulin sequencing and under Max Perutz Crick realized protein sequence must be encoded in genes Both believed DNA was the genetic material but few others did Coined the term central dogma in 1958 o Rosalind Franklin Studied chemical structures in Paris brought in by Maurice Wilkins King s College in London to study some pure DNA samples Confusion as to whether she was working for Wilkins or on her own 0 October 1951 Watson meets Crick 0 Watson and Crick s 1St Structure Tried to model other people s data didn t take notes overconfidence Heard Franklin in 1951 where Like Pauling s idea on the ahelix they ignored the side chains Fit with Rosalind Franklin s images so they thought so Franklin Wilkins and Gosling came to take a look Franklin knew magnesiums would be bound by water and not available and that the hydrophilic phosphates should be on the outside Told to not build anymore models by Bragg as they were incompetent o Rosalind Franklin 0 O Captured images of both A and B Form DNA A form very low on water convinced her that A DNA was not helical not B DNA Franklin s mistake did not like model building preferred using the Patterson function as she was a crystallographer not a chemist o Photo 51 O O O O O O 0 Image from DNA obtained by Maurice Wilkins from Rudolph Singer Switzerland Wilkins gave the DNA to a graduate student Raymond Gosling and Rosalind Franklin chemist was later brought in This DNA was provided to Franklin and the images were shown to Watson by Wilkins Franklin and Wilkins did not get along Rosalind Franklin had written a summary of her results for an institute report in December Perutz gave Crick a copy In that report Crick read for the first time that the crystalline form of DNA was based on a facecentered monoclinic unit cell with a twofold axis of symmetry Water content of crystal Molecule looks the same rightside up or upside down Indicated antiparallel structure This is also the same space group of hemoglobin the molecule that Crick was working on as the subject of his PhD thesis so he recognized the implications facecentered monoclinic unicell with a twofold axis of symmetry It provided critical evidence in support of the WatsonCrick model for Bform DNA 4th spot missing due to where chains crossed DNA would be zig zags Spacing shows 34 nm periodicity 4 diamondshaped areas result from backbone and 10 lines can be counted representing the 10 repeats of the backbone o Building the model 0 Photo confirmed 2 nm helical structure stacked bases 034 nm apart 0 Initially TT AA CC GG Makes some sense identical information Chain would buckle Did not fit with Chargaff s rules Jerry Donohue Caltech but was studying at Cambridge with WatsonCrick Watson was using enol thymine and guanine not keto form o DNA Structure 0 Nobel Prize 1962 to Watson Crick Wilkins although 24 scientists contributed to structure Franklin had died and prizes are not given posthumously 0 Structure unchanged in over 50 years 0 Opened the field of Molecular Biology 0 Actual paper say that 3chain doesn t work but their model had xray evidence talked about the base pairing angle tautomer forms base pairs etc specific pairing we have postulated immediately suggests a possible copying mechanism Structure was perfect except for one H bond o Watson s The Double Helix 0 Very unfair to Rosalind Franklin 0 Called her Maurice Wilkins assistant o the best home for a feminist was in another person39s lab 0 Said that Dorothy Hodgkin was the best crystallographer in England insulin cholesterol penicillin B12 structures Nobel Prize in Chemistry WatsonCrick looked at her data without her knowledge Wilkins Franklin refused to look at Watson s data and she pointed out how little they knew about the chemistry of DNA Watson noted that Franklin was about to strike him when he was lecturing her on helices in her lab one day Epilogue basically tried to right the wrongs and paint Franklin in a better light Once described by E O Wilson as the most unpleasant human being I had ever met Other controversies Disputes with Venter Draft completed before she knew about WatsonCrick s structure 0 She used the Patterson function Awesome Chargaff quotes II DNA Structure 0 O O O O O O 0 Double helix Sugarphosphate backbone charge Majorminor groove Deoxyribose Base pairing Antiparallel 34 A 105 bpturn 2nm wide Nucleotide synthesis PICTURE O O 0 built up from amino acids Pyrimidine amino acids 9 make UTP 9 make CTP l39l39P Purines amino acids 9 IMP 9 ATPGTP Confo rmations O 0 most in amino and keto form imino form changes to DAA 9 ADA not very frequent but can cause problems tautomer pairs TG and CA Sugar Base nucleoside O 0 Sugar Base phosphate nucleotide 3 OH makes a nucleophilic attack to phosphate 2122012 123700 PM 2122012 123700 PM N L 4 UI 0 gt1 9 Types of Mutations Transitions a Pyrimidine to pyrimidine b Purine to purine more common Transversions a Pyrimidine to purine b Purine to pyrimidine Missense Point Mutation a A single change in a base leads to an amino acid substitution Nonsense Point Mutation a A change in bases causes a truncation of amino acids b Most damaging Single Nucleotide Polymorphisms SNP a A change in a base pair between sister chromatids Depurination a Loss of a purine base to form an apurinic site b Hydrolytic DNA Damage Depyriminidination a Loss of a pyrimidine to form an apyrimidinic site b Hydrolytic DNA damage Deamination a Cytosine deaminates to Uracil i Pairs with A instead of G b 5methylcytosine 5mC deaminates to Thymine i Pairs with A instead of G c Adenine deaminates to Hypoxanthine i Pairs With C instead of T d Guanine deaminates to Xanthine i Pairs with C but with only 2 Hbonds Strand Slippage a Occurs when a strand of DNA contains a multitude of the same repeats b Microsatellite DNA 10 Tautomeriz ation a Tautomer of A binds with C instead of T b Escapes the Proofreading exonuclease 11 Base Analogs a 5bromouracil 5bC i Pairs with A or G ii Analog of T b Aminopurine i Pairs with T or C ii Analog of A 12 Alkylation a Guanine forms to 06methylguanine i Pairs with T instead of C b Examples include i Nitosoguanidine ii Methyl methane sulfonate iii Ethyl methane sulfonate l3 Oxidation a 0239 H202 OH reactive oxygen39s b Guanine forms 78dihydro8oxoguanine oxoG i Pairs with C when oxoG is in the anti con guration ii Pairs with A when oxoG is in the syn con guration iii Most common mutation found in human cancers 1 Causes an AT gt CG transversion l4 yRadiation a Leads to reactive oxygen s b Causes dsDNA breaks 15 UV Radiation a Light at wavelengths approximately 260nm b Causes thymine thymine and thyminecytosine dimers i Between the bases a cyclobutane ring forms ii Halts polymerase and synthesis c Simplest damage and most instantaneous 1 Chapter 8 The Replication of DNA 2 The Chemistry of DNA Synthesis a DNA Synthesis Requires Deoxynucleoside Triphosphates and a Primer Template Junction i For the synthesis of DNA to proceed two key substrates must be present 1 N Four deoxynucleoside triphosphates they have three phosphoryl groups that are attached via the 539hydroxyl of the 239deoxyribose dGTP a b dATP c dTTP d dCTP i The innermost phosphoryl group is called the ot phosphate and the middle and outermost are referred to as the B and y Particular arrangement of singlestranded DNA ssDNA and double stranded DNA dsDNA a This is called a primertemplate junction b The primertemplate junction has two functions i The template provides the ssDNA that directs the addition of each complementary deoxynucleotide 1 Primer is complementary to but shorter than the template ii Provides only the information necessary to pick which nucleotides are added b DNA is Synthesized by Extending the 339 End of the Primer i The chemistry of DNA synthesis requires that the new chain grows by extending the 339 end of the primer ii The phosphodiester bond is formed in an SNZ reaction in which the hydroxyl group at the 339 end of the primer strand attacks the ot phosphoryl group of the incoming nucleoside triphosphate 1 The leaving group is pyrophosphate iii The template strand directs which of the four nucleoside triphosphates is added c Hydrolysis of Pyrophosphate Is the Driving Force for DNA Synthesis i The addition of a nucleotide to a growing polynucleotide chain of length n is indicated by the following reaction XTP XMPn 9 XMPn1 PP a The AG for the reaction is 35 kcalmol b The driving reaction is the release of the pyrophospate P9 2Pi a The AG for the reaction is 7 kcalmol 3 Mechanism of DNA Polymerase a DNA Polymerases Use a Single Active Site to Catalyze DNA Synthesis i The synthesis of DNA is catalyzed by DNA polymerase 1 DNA polymerase uses a single active site to catalyze the addition of any of the four deoxynucleoside triphosphates ii The DNA polymerase monitors the ability of the incoming nucleotide to form an AT or GC base pair rather than detecting the exact nucleotide that enters the active site 1 Only when a correct base pair is formed are the 339OH of the primer and the 6 phosphate of the incoming nucleoside triphosphate in the optimum position for catalysis to occur a This is an example of kinetic selectivity i An enzyme favors catalysis using one of several possible substrates by dramatically increasing the rate of bond formation only when the correct substrate is present DNA polymerases have the ability to distinguish between ribonucleoside and deoxyribonucleoside triphosphates rNTPs and dNTPs 1 Although rNTPs are present approximately 10fold higher concentration in the cell they are incorporated at a rate that is more than 1000fold lower than dNTPs a This discrimination is mediated by the steric exclusion of rNTPs form DNA polymerase active site i The nucleotidebinding pocket is too small to allow the presence of a 239OH on the incoming nucleotide 1 This space is occupied by two amino acids that make van der Waals contacts with the sugar ring ii Changing these amino acids to other amino acids with smaller side chains results in a DNA polymerase with significantly reduced discrimination between dNTPs and rNTPs b DNA Polymerases Resemble a Hand That Grips the PrimerTemplate Junction iv Structures reveal that the DNA substrate sits in a large cleft that resembles a partially closed right hand There are three domains in conjunction to the quothandquot analogy 1 Thumb 2 Fingers 3 Palm Palm Domain 1 Composed ofa Bsheet 2 Contains the primary elements of the catalytic site a This region of DNA polymerase binds two divalent metal ions Typically Mgzi or ani that alter the chemical environment around the correctly basepaired dNTP and the 339OH of the primer i One metal ion reduces the affinity of the 339OH for its hydrogen 1 This generates a 339O39that is primed for the nucleophilic attack of the ot phosphate of the incoming dNTP ii The second metal ion coordinates the negative charges of the B and yphosphates of the dNTP and stabilizes the pyrophosphate by joining the primer and the incoming nucleotide 3 Monitors the base pairing of the most recent added nucleotides a This region of the polymerase makes extensive hydrogen bond contacts with base pairs in the minor groove of the newly synthesized DNA i These contacts are not basespecific but only form if the recently added nucleotides are correctly basepaired ii Mismatched DNA in this region interferes with these minor groove contacts and dramatically slows catalysis Finger Domain 1 Important for catalysis 2 Several residues located within the fingers bind to the incoming dNTP a Once a correct base pair is formed between the incoming dNTP and the template the finger domain moves to enclose the dNTP Squot This closed form of the polymerase hand stimulates the catalysis by moving the incoming nucleotide in close contact with the catalytic metal ions 3 Contains an OHelix which has the following amino acids to bind to the incoming dNTP the Ohelix will bend 40 and close the domain on the dNTP allowing proper bonding a Arginine i Binds to the yphosphate upon closure of the finger b Lysine i Binds to the Bphosphate upon closure of the finger c Tyrosine i Binds by van der Waals interactions to the nucleotide being added 4 The finger domain associates with the template region leading to a nearly 90 turn of the phosphodiester backbone between the first and second bases of the template a This bend serves to expose only the first template base after the primer at the catalytic site and avoids any confusion concerning which template base should pair with the next nucleotide to be added v Thumb Domain 1 The thumb domain is NOT intimately involved in catalysis a The thumb interacts with the DNA that has been most recently synthesized This serves two purposes i It maintains the correct position of the primer and the active site ii Helps to maintain a strong association between the DNA polymerase and its substrate 1 This association contributes to the ability of the DNA polymerase to add many dNTPs each time it binds a primertemplate junction DNA Polymerases are Processive Enzymes i Catalysis by DNA polymerase is rapid 1 Capable of adding as many as 1000 nucleotides per second 2 Speed of synthesis is largely due to the processive nature of DNA polymerase a Processivity is a characteristic of enzymes that operate on polymeric substrates i Defined as average number of nucleotides added each time the enzyme binds a primertemplate junction ii The rate of DNA synthesis can be dramatically increased by adding multiple nucleotides per binding event 1 It is the initial binding of polymerase to the primertemplate junction that is ratelimiting for DNA synthesis iii Increased processivity is facilitated by the ability of DNA polymerases to slide along the DNA template iv Once bound to a primertemplate junction DNA polymerase interacts tightly with much of the doublestranded portion of the DNA in a sequencenonspecific manner 1 These interactions include a Electrostatic interactions between the phosphate backbone and the quotthumbquot domain 57 O b Interactions between the minor groove of the DNA and the palm domain Exonucleases Proofread Newly Synthesized DNA A major limit to DNA polymerase accuracy is the occasional approx 1 in 105 times flickering of the bases into the quotwrongquot tautomeric form Proofreading of DNA synthesis is mediated by nucleases that remove incorrectly basepaired nucleotides 1 This nuclease was identified in the same polypeptide as the DNA polymerase termed proofreading exonuclease a Capable of degrading DNA starting from a 339 DNA end In the rare event that an incorrect nucleotide is added to the primer strand the proofreading exonuclease removes this nucleotide from the 339 end of the primer strand The removal of mismatched nucleotides is facilitated by the reduced ability of DNA polymerase to add a nucleotide adjacent to an incorrectly basepaired primer 1 Mispaired DNA alters the geometry of the 339OH and the incoming nucleotide because of poor interactions with the palm region a This altered geometry reduces the rate of nucleotide addition in must the same way that addition of an incorrectly paired dNTP reduces catalysis As for processive DNA synthesis proofreading occurs without releasing the DNA from the polymerase the process is outlined below When a mismatched base pair is present in the polymerase active site the primertemplate junction is destabilized creating several base pairs of unpaired DNA The DNA polymerase active site binds such a mismatched template poorly but the exonuclease active site has a tenfold higher affinity of singlestranded 339 ends Newly unpaired 339 end moves from the polymerase active site to the exonuclease active site a Incorrect nucleotide is removed by the exonuclease Primertemplate junction then rebinds to polymerase active site and DNA synthesis continues N W P 4 The Replication Fork Both Strands of DNA Are Synthesized Together at the Replication Fork a Junction between the newly separated template strands and the unreplicated duplex DNA is called the replication fork 1 The fork moves continuously toward the duplex region of unreplicated DNA leaving two ssDNA templates that direct the formation of two daughter DNA duplexes There are two strands of newly synthesized DNA 1 Leading strand a The DNA strand that is continuously synthesized due to the 339OH end 2 Lagging Strand a Template directs the DNA polymerase to move in the opposite direction of the replication fork Although the leadingstrand DNA polymerase can replicate its template as son as it is exposed synthesis of the lagging strand must wait for the movement of the replication form to expose a substantial length of template before it can be replicated iv The resulting short fragments of new DNA formed on the lagging strand are called Okazaki fragments and vary in length from 1000200 nucleotides in bacteria and from 100400 nucleotides in eukaryotes 1 Shortly after being synthesized Okazaki fragments are covalently joined together to generate a continuous intact strand of new DNA 2 Okazaki fragments are therefore transient intermediates in DNA replication b The Initiation of a new Strand of DNA Requires an RNA Primer Squot i How then are new strands of DNA synthesis started 1 The cell takes advantage of the ability of RNA polymerases to do what DNA polymerases cannot start new RNA chains de novo from the beginning a Primase is a specialized RNA polymerase dedicated to making short RNA primers on an ssDNA template b These primers are subsequently extended by DNA polymerase ii Although both the leading and lagging strands require primase to initiate DNA synthesis the frequency of primase function on the two strands is dramatically different 1 Each leading strand requires only a single RNA primer 2 Synthesis of the lagging strand means that new primers are needed for each Okazaki fragment a Because a single replication fork can replicate millions of base pairs synthesis of the lagging strand can require thousands of Okazaki fragments and their associated RNA primers iii Unlike the RNA polymerases involved in messenger RNA mRNA transfer RNA tRNA and ribosomal RNA rRNA synthesis primase does not require an extensive DNA sequence to initiate synthesis of a new RNA primer 1 Instead primases prefer to initiate RNA synthesis using an ssDNA template containing a particular primer iv Primase activity is dramatically increased when it associates with another protein that acts at the replication fork called DNA helicase 1 This protein unwinds the DNA at the replication fork creating an ssDNA template that can be acted on by primase RNA Primers Must Be Removed to Complete DNA Replication To complete DNA replication the RNA primers used for initiation must be removed and replaced with DNA ii Removal of the RNA primers can be thought to as a DNArepair event and this process shares many of the properties of excision DNA repair iii To replace the RNA primers with DNA an enzyme called RNase H recognizes and removes most of each RNA primer 1 This enzyme specifically degrades RNA that is basepaired with DNA a HENCE the H in its name which stands for hybrid DNARNA hybrid 2 RNase H removes all of the RNA primer except the ribonucleotide directly linked to the DNA end a This is because RNase H can cleave only bonds between two ribonucleotides iv Removal of the RNA primer leaves a gap in the dsDNA that is an ideal substrate for DNA polymerasea primertemplate junction d e DNA polymerase fills this gap until every nucleotide is basepaired leaving a DNA molecule that is complete except for a break in the backbone between the 339OH and 539phosphate of the repaired strand a This quotnickquot in the DNA can be repaired by an enzyme called DNA ligase i DNA ligases use highenergy cofactors such as ATP to create a phosphodiester bond between an adjacent 539phosphate and 339 OH v ONLY after all RNA primers are replaced by DNA and the associated nicks are sealed is DNA synthesis complete DNA Helicases Unwind the Double Helix in Advance of the Replication Fork i DNA polymerases are generally poor at separating the two basepaired strands of duplex DNA 1 At the replication fork a second class of enzymes DNA Helicases catalyze the separation of the two strands of duplex DNA a These enzymes bind to and move directionally along ssDNA using the energy of nucleoside triphosphate usually ATP hydrolysis to displace any DNA strand that is annealed to the bound ssDNA b DNA helicases that act at replication forks are hexameric proteins that assume the shape of a ring ii Like DNA polymerases DNA helicases act processively 1 Each time they associate with substrates they unwind multiple base pairs of DNA 2 The ringshaped hexameric DNA helicases found at replication forks exhibit high processivity because they encircle the DNA a Release of the helicase from its DNA substrate therefore requires the opening of the hexameric protein ring which is rare iii Each DNA helicase moves along ssDNA in a defined direction 1 This property is a characteristic of each DNA helicase call its polarity a DNA helicase can either have a polarity of i 539 9 339 1 Lagging strand 2 Allows the DNA helicase to proceed toward the duplex region of the replication fork ii 339 9 539 1 This direction is always defined according to the strand of DNA bound rather than the strand that is displaced iv The energy provided for helicases is provided by ATP hydrolysis DNA Helicase Pulls Single Stranded DNA Through a Central Protein Core i How does a hexameric DNA helicase use the energy of ATP hydrolysis to move along the DNA 1 Determination of the atomic structure of a viral hexameric helicase bound to a singlestranded DNA substrate provides some clues a In the structure ssDNA is encircled by the sit subunits of the helicase i Each subunit has a quothairpinquot protein loop that binds a phosphate of the DNA backbone and its two adjacent ribose components 1 Loops are found in a righthanded spiral b The atomic structure is only a single snapshot each of the sex different subunits is at a different stage in the DNA translocation process r Q i Together the interactions of the different subunits with DNA and ATPADP reveal the coordinated movement of these proteins hairpins can pull the ssDNA through the central pore of the helicase 2 Mechanism a A subunit first binds the ssDNA at the top of the structure and the DNA binding loop moves through successive conformations toward the bottom bringing the bond DNA long with it i Each of these conformations is associated with a different nucleotidebound state 1 Top conformation is an ATPbound state 2 Middle conformation is an ADPbound state 3 Bottom lacks a bound nucleotide ii A single subunit binds hydrolyzes and releases ATP cycling through the above conformations ii In addition to translocating along ssDNA a helicase must also displace the complementary strand to cause DNA unwinding 1 In the case of the hexameric helicase the structure of the central channel shows that strand displacement must occur for DNA to pass through the channel a At its narrowest point the central channel has a diameter of 13A large enough to fit ssDNA SingleStranded DNABinding Proteins Stabilize ssDNA Prior to Replication i After the DNA helicase has passed the newly generated ssDNA must remain free of base pairing until it can be used as a template for DNA synthesis 1 The stabilize the separated strands ssDNAbinding proteins designated SSBs rapidly bind to the separated strands a Binding of one SSB promotes the binding of another SSB to the immediately adjacent ssDNA i This is called cooperative binding occurs because SSB molecules bound to immediately adjacent regions of ssDNA also bind to each other 1 This SSBSSB interaction strongly stabilizes SSB binding to ssDNA and makes sites already occupied by one or more 53 molecules preferred SSB binding sites ii Cooperative binding ensures that ssDNA is rapidly coated by SSB as it emerges from the DNA helicase 1 Once covered with 53 ssDNA is held in an elongated state that facilitates its used as a template for DNA or RNA primer synthesis iii SSB interacts with ssDNA in a sequenceindependent manner 1 SSBs primarily contact ssDNA through electrostatic interactions with the phosphate backbone and stacking interactions with the DNA bases 2 In contrast to sequencespecific DNA binding proteins SSBs make few if any hydrogen bonds to the ssDNA bases Topoisomerases Remove Supercoils Produced by DNA Unwinding at the Replication Fork i As the strands of DNA are separated at the replication fork the dsDNA in front of the replication fork becomes increasingly positively supercoiled I This accumulation of supercoils is the result of DNA helicase eliminating the base pairs between the two strands If the DNA strands remain unbroken there can be no reduction in linking number a Thus as the DNA helicase proceeds the DNA must accommodate the same linking number within a smaller and smaller number of base pairs b For the DNA in front of the replication fork to remain relaxed one DNA link must be removed approximately every 10bp of DNA unwound ii The problem is most clear for the circular chromosomes of bacteria but it also applies to eukaryotic chromosomes 1 Because eukaryotic chromosomes are no closed circles they could in principle rotate along their length to dissipate the introduced supercoils this is NOT the case a It is simply not possible to rotate a DNA molecule that is millions of base pairs long each time one turn of the helix is unwound iii The supercoils introduced by the action of the DNA helicase are removed by topoisomerases that act on the replicated dsDNA in front of the replication fork 1 These enzymes do this by breaking either one or both strands of the DNA without letting go of the DNA and passing the same number of DNA strands through the break 2 Topoisomerases act as a quotswivelasequot that rapidly dissipates the accumulation of supercoils ahead of the replication fork Replication Fork Enzymes Extend the Range of DNA polymerase Substrates i On its own DNA polymerase can only efficiently extend 339OH primers annealed to ssDNA templates 1 The addition of primase DNA helicase and topoisomerase dramatically extends the possible substrates for DNA polymerase a Primase i Primase provides the ability to initiate new DNA strands on any piece of ssDNA ii The use of primase also imposes a requirement for the removal of the RNA primers to complete replication b DNA helicase i Strand separation by DNA helicase and dissipation of positive supercoils by topoisomerase allow DNA polymerase to replicate dsDNA ii Both DNA helicase and topoisomerase perform their functions without permanently altering the chemical structure of DNA or synthesizing any new molecule 1 DNA helicase breaks only the hydrogen bonds that hold the two strands of DNA together without breaking any covalent bonds 2 Topoisomerases break one or more bonds of DNA39s covalent bonds each bond broken is precisely reformed before topoisomerase releases DNA a Instead of altering the chemical structure of DNA the action of these enzymes results in a DNA molecule with an altered conformation i These conformational alterations are essential for the duplication of the large dsDNA molecules that are the foundation of both bacterial and eukaryotic chromosomes N iii The proteins that act at the replication fork interact tightly but in a sequence independent manner with the DNA 1 These interactions exploit the features of DNA that are the same regardless of the particular base pair a Negative charge and structure of the phosphate backbone b Hydrogen bonding residues in the minor groove c Hydrophobic stacking interactions between the bases 5 The Specialization of DNA Polymerases a DNA Polymerases Are Specialized for Different Roles in the Cell i The central role of DNA polymerases in the efficient and accurate replication of the genome requires that cells have evolved multiple specialized DNA polymerases ii Example E coli 1 DNA polymerase III DNA Pol III is the primary enzyme involved in the replication of the chromosome a DNA Pol III is generally found to be part ofa larger complex that confers very high processivitya complex known as the DNA Pol III holoenzyme iii DNA Pol is specialized for the removal of the RNA primers that are used to initiate DNA synthesis 1 DNA polymerase has a 539 exonuclease that allows DNA Pol to removed RNA or DNA immediately upstream of the site of DNA synthesis a Unlike DNA Pol III holoenzyme DNA Pol is not highly processive adding only 20100 nucleotides per binding event i These properties are ideal for RNA primer removal and DNA synthesis across the resulting ssDNA gap ii The 539 exonuclease of DNA Pol I can remove the RNADNA linkage that is resistant to RNase H iv Because both DNA Pol and DNA Pol III are involved in DNA replication both of these enzymes must be highly accurate 1 Both proteins carry an associated proofreading exonuclease v Eukaryotic cells also have multiple DNA polymerases with a typical cell having more than 15 1 Of these three are essential to duplicate the genome a DNA Pol 6 b DNA Pol E c DNA Pol otprimase i Each is composed of multiple subunits vi DNA Pol otPrimase Specifically involved in initiating new DNA strands Foursubunit protein complex consists of a two subunit DNA Pol 01 and two subunit primase a After the primase synthesizes an RNA primer the resulting RNA primertemplate junction is immediately handed off to the associated DNA Pol 01 to initiate synthesis 3 Because of its relative lox processivity it is rapidly replaced by highly processive DNA Pol 6 and DNA Pol E a The process of replacing the three is called polymerase switching and results in three different DNA polymerases functioning at the eukaryotic replication fork Nl Recent studies indicate that DNA Pol 6 and DNA Pol e are specialized to synthesize different strands at the replication fork 1 DNA Pol 8 making the leading strand 2 DNA Pol 6 making the lagging strand b Sliding Clamps Dramatically Increase DNA Polymerase Processivity High processivity at the replication fork ensures rapid chromosome duplication One key to the high processivity of the DNA polymerases that act at replication forks is their association with proteins called sliding DNA clamps 1 These proteins are composed of multiple identical subunits that assemble in the shape of a quotdoughnutquot a The hole in the center of the clamp is large enough to encircle the DNA double helix and leave room for a layer of one or two water molecules between the DNA and the protein i These properties allow the clamp proteins to slide along the DNA without dissociating from it 2 Sliding DNA clamps also bind tightly to DNA polymerases at replication forks a The resulting complex between the polymerase and the sliding clamp moves efficiently along the DNA template during DNA synthesis How does the association with the sliding clamp change the processivity of the DNA polymerase In the absence of the sliding clamp a DNA polymerase dissociates and diffuses away from the template DNA on average once every 20100bp synthesized In the presence of the sliding clamp the DNA polymerase still disengages its active site from the 339OH end of the DNA frequently but the association with the sliding clamp prevents the polymerase from diffusing away from the DNA a By keeping the DNA polymerase in close proximity to the DNA the sliding clamp ensures that the DNA polymerase rapidly rebinds the same primertemplate junction increasing the processivity of the DNA polymerase Once an ssDNA template has directed synthesis of its complementary DNA strand the DNA polymerase must release from the completed dsDNA and the sliding clamp to act at a new primertemplate junction 1 This release is accomplished by a change in the affinity between the DNA polymerase and the sliding clamp that depends on the bound DNA a When a DNA polymerase reaches the end of an ssDNA template Okazaki fragment the presence of dsDNA in its active site results in a change in conformation that reduces the polymerase39s affinity for the sliding clamp and the DNA Once released from a DNA polymerase sliding clamps are not immediately removed from the replicated DNA 1 Instead other proteins that must function at the site of recent DNA synthesis to perform their function interact with the clamp proteins Enzymes that assemble chromatin in eukaryotic cells are recruited to the sites of DNA replication by an interaction with the eukaryotic sliding DNA clamp PCNA b Okazaki fragment repair also interact with sliding clamp proteins Sliding clamp proteins are a conserved part of the DNA replication apparatus derived from organisms as diverse as viruses bacteria yeast and humans N m c Sliding Clamps Are Opened and Placed on DNA by Clamp Loaders i The sliding clamp is a closed ring in solution but most open to encircle the DNA double helix 1 A special class of protein complexes called sliding clamp loaders catalyze the opening and placement of sliding clamps on the DNA 2 These enzymes couple ATP binding and hydrolysis to the placement of the sliding clamp around primertemplate junctions on the DNA 3 The clamp loader also removes sliding clamps from the DNA when they are no longer in use 4 These enzymes alter the conformation of their target sliding clamps like DNA helicases and topoisomerases but NOT its chemical composition ii What controls when sliding clamps are loaded and removed from the DNA 1 2 Loading of a sliding clamp occurs anytime a primertemplate junction is present in a cell a These DNA structures are formed not only during DNA replication but also during several DNA repair events b A sliding clamp can only be removed from the DNA if it is NOT bound by another protein A sliding clamp that is bound to a DNA polymerase is not subject to removal from the DNA a Sliding clamps will only remove from the DNA once all of the enzymes that interact with them have completed their function 6 DNA Synthesis At The Replication Fork a When an ssDNA region of DNA is broken there is a complete break in the chromosome that is much more difficult to repair than an ssDNA break in a dsDNA region b How do two DNA polymerases remain linked at the replication fork while synthesizing DNA on both the leading and lagging template strands i A model that explains this coupling proposes that the replication machinery exploits the flexibility of DNA 1 Iquot As the helicase unwinds the DNA at the replication fork the leadingstrand template is exposed and acted on immediately by the leadingstrand DNA r 39 which 39 39 a 39 strand of 39 y DNA The laggingstrand template is not immediately acted on by DNA polymerase it is spooled out as ssDNA that is rapidly bound by SSB a Intermittently primase interacts with the DNA helicase and is activated to synthesize a new RNA primer on the laggingstrand template b The resulting DNARNA hybrid is recognized as a primer template junction by the sliding DNA clamp loader and a sliding clamp is assembled at this site c When the laggingstrand DNA polymerase completes the previous Okazaki fragment this polymerase is released from the template i Because this polymerase remains tethered to the leadingstrand DNA polymerase and the sliding clamp loader it is in an ideal position to bind the RNA primertemplate junction immediately after the addition ofa sliding clamp c Interactions Between Replication Fork Proteins Form the E coli Replisome i The combination of all of the proteins that function at the replication fork is referred to as the replisome 7 Initiation of DNA Replication a b C Specific Genomic DNA Sequences Direct the Initiation of DNA Replication i The specific sites at which DNA unwinding and initiation of replication occur are called origins of replication 1 Can contain one or many depending on the organism The Replicon Model of Replication Initiation i 1963 Francois Jacob Sydney Brenner and Jacques Cuzin proposed a model to explain the events controlling the initiation of replication in bacteria 1 All the DNA replicated from a particular origin of replication as a replicon a IE E coli cells have only one origin of replication the entire chromosome is a single replicon 2 This model proposed two components that controlled the initiation of replication a Replicator b Initiator 3 Replicator is defined as the entire set of cisacting DNA sequences that is suf cient to direct the initiation of DNA replication a This contrasts the origin of replication which is the site on the DN where the DNA is unwound and DNA synthesis initiates i Although the origin of replication is always part of the replicator sometimes particularly in eukaryotic cells the origin of replication is only a fraction of the DNA sequences required to direct the initiation of replication replicator 4 The initiator protein specifically recognizes a DNA element in the replicator and activates the initiation of replication a All initiator proteins select the sites that will become origins or replication although they recognize DNA using different DNAbinding motifs b All fthe known initiator proteins are regulated by ATP binding and hydrolysis and share a common core ATPbinding motif related to but distinct from that used by sliding DNA clamp loaders 5 The initiator protein is the ONLY sequencespecific DNAbinding protein involved in the initiation of replication a The remaining proteins required for replication initiation do not bind to DNA sequence specifically Replicator Sequences Include Initiator Binding Sites and Easily Unwound DNA i The DNA sequences of known replicators share two common features They include a binding site for the initiator protein that nucleates the assembly of the replication initiation machinery They include a stretch of ATrich DNA that unwinds readily but NOT spontaneously a Unwinding of DNA at replicators is controlled by the replication initiation proteins and the action of these proteins is tightly regulated in most organisms The single replicator required for E colichromosomal replication is called MC 1 Two repeated motifs are critical for oriC function a The 9mer motif is the binding site for the E coli initiator DnaA and is repeated five times at oriC N b 13mer motif repeated three times is the initial site of ssDNA formation during initiation Although the specific sequences are different the overall structures of replicators derived from many eukaryotic viruses and the singlecell eukaryote Saccharormyces cerevisae are similar Replicators functioning in multicellular eukaryotes are not well understood 1 Identification and characterization have been hampered by the lack of genetic assays for stable propagation of small circular DNA comparable to those used to identify origins in singlecell eukaryotes and bacteria 8 Binding and Unwinding Origin Selection and Activation by the Initiator Protein a Initiator proteins always perform at least two different functions during the initiation of replication These proteins bind a specific DNA sequence within the replicator Initiator proteins interact with additional factors required for replication initiation thus recruiting them to the replicator Some but not all initiator proteins perform a third function they distort or unwind a region of DNA adjacent to their binding site to facilitate the initial opening of the DNA duplex E coli initiator protein 1 DnaA binds the repeated 9 mer elements in oriCand is regulated by ATP 2 When bound to ATP NOT ADP DnaA also interacts with DNA in the region of the repeated 13mer repeats of uric The formation of ssDNA at a site in the chromosome is NOT sufficient for the DNA helicase and other replication proteins to assemble 1 DnaA recruits additional replication proteins to the ssDNA formed at the replicator including the DNA helicase In eukaryotic cells the initiator is a sixprotein complex called the origin recognition complex ORC 1 Function of ORC is best understood in yeast cells 2 It recognizes a conserved sequence found in yeast replicators called the A element as well as a second lessconserved Bl element 3 Like DnaA ORC binds and hydrolyzes ATP a ATP binding is REQUIRED for sequencespecific DNA binding at the origin and ATP hydrolysis is required for ORC to participate in the loading of the eukaryotic DNA helicase onto the replicator DNA 4 Unlike DnaA binding of ORC to yeast replicators does not lead to strand separation of the adjacent DNA a ORC is REQUIRED to recruit either directly or indirectly all of the remaining replication proteins to the replicator b ProteinProtein and ProteinDNA Interactions Direct the Initiation Process Once the initiator binds to the replicator the remaining steps in the initiation of replication are largely driven by proteinprotein interactions and proteinDNA interactions that are sequenceindependent 1 The end result is the assembly of two replisomes Initiation E coli Mechanism 1 After the initiator DnaA has bound to MC and unwound the 13mer DNA the combination of ssDNA and DnaA recruits a complex of two proteins a DNA helicase DnaB Squot b Helicase Loader DnaC i BOTH proteins are present in six copies within the complex ii The DNA helicase is inactive in the helicasehelicase loader complex to prevent it from functioning at inappropriate sites 2 Once bound to the ssDNA at the origin the helicase loader directs the assembly of its associated DNA helicase around the ssDNA recall ssDNA passes through the middle of the DnaB helicase39s hexameric protein ring a This process is analogous to the assembly of sliding DNA clamps around a primertemplate junction 3 DnaC is an ATPutilizing AAA protein 4 Upon completion of the above task the helicase loader is released allowing the helicase to become active a One helicase is loaded onto each of the two separated ssDNA strands at the origin and the orientation of these two helicases is such that they will proceed toward each other as they move with a 539 9 339 polarity along their associated ssDNAs Proteinprotein interactions between the helicase and other components of the replication fork described above direct the assembly of the rest of the replication machinery 1 Helicase recruits DNA primase to the origin of DNA resulting in the synthesis of an RNA primer on each strand of the origin 2 The DNA Pol III holoenzyme is brought to the origins through interactions with the primertemplate junction and the helicase a Once the holoenzyme is present sliding clamps are assembled on the RNA primers and the leadingstrand polymerases are engaged 3 As new ssDNA is exposed by the action of the helicase it is bound by SSB and DNA primase synthesizes the first laggingstrand DNA synthesis a At this point two replication forks have been assembled and initiation of replication is complete Eukaryotic Chromosomes Are Replicated Exactly Once per Cell Cycle i Chromosomal DNA replication occurs only during the S phase of the cell cycle 1 During this time all of the DNA in the cell must be duplicated exactly once ii Incomplete replication of any part of a chromosome causes inappropriate links between daughter chromosomes iii Addition of even one or two more copies of critical regulatory genes can lead to catastrophic defects in gene expression cell division or the response to environmental signals 1 It is critical that every base pair in each chromosome can be replicated once and only once each time a eukaryotic cell divides iv The need to replicate the DNA once and only once is a particular challenge for eukaryotic chromosomes because they each have many origins of replication 1 Origins are typically separated by approx 30kb so even a small eukaryotic chromosome may have more than ten origins and a large human chromosome may have thousands v Whether an origin is activated to cause its own replication or replicated by a replication fork derived from an adjacent origin it must be inactivated until the next round of cell division 1 If these conditions were NOT true the DNA associated with an origin could be replicated twice in the same cell cycle Prereplicative Complex Formation is the First Step in the Initiation of Replication in Euka ryotes Initiation of replication in eukaryotic cells involves two events that occur at distinct times in the cell cycle 1 Replicator selection 2 Origin activation Replicator Selection 1 The process of identifying sequences that will direct the initiation of replication and occurs in 61 prior to S phase a This process leads to the assembly of a multiprotein complex at each replicator in the genome 2 Mediated by the formation of prereplicatve complexes preRCs a Composed of four separate proteins that assemble in an ordered fashion at each replicator Origin activation 1 Occurs after cells enter S phase and triggers the replicatorassociated protein complex to initiate DNA unwinding and DNA polymerase recruitment The separation of replicator selection and origin activation is different from the situation in prokaryotic cells where the recognition of replicator DNA is intrinsically coupled to DNA unwinding and polymerase recruitment 1 The temporal separation of these two events during the eukaryotic cell cycle ensures that each chromosome is replicated only once during each cell cycle Prereplicative Complex Formation Mechanisms 1 Recognition of the replicator by the eukaryotic initiator ORC 2 Once ORC is bound it recruits two helicase loading proteins Cdc6 and Cdt1 a Together ORC and the loading proteins recruit the eukaryotic replication fork helicase the Mcm 27 complex i Both ORC and the Cdc6 protein are members of the AAA family of proteins like DnaC and the subunits of the sliding clamp loaders b Studies of the preRC assembly suggest that these proteins use ATP binding and hydrolysis to load the ringshaped Mcm27 complex around dsDNA c Consistent with the Mcm27 complex encircling dsDNA formation of the preRC does not lead to the immediate unwinding of origin DNA or the recruitment of DNA polymerases i The preRCs that are formed during 61 are only activated to unwind DNA and initiate replication after cells pass from the 61 to the S phase of the cell cycle 3 PreRCs are activated to initiate replication of two protein kinases a Cdk cyclindependent kinase b de Dbf4dependent kinase i Protein kinases are proteins that covalently attach phosphate groups to target proteins ii Each of these kinases is inactive in 61 and is activated only when cells enter S phase e f iii Once activated these kinases target the preRC and other replication proteins iv Phosphorylation of these proteins results in the recruitment of many additional replication proteins to the origin and the initiation of replication v These new proteins include the three DNA polymerases and a number of other proteins 1 Pol 6 2 Pol E 3 Pol 01 1 6 and E associate first followed by 01 1 This order ensures that all three DNA polymerases are present at the origin prior to the synthesis of the first RNA primer vi Only a subset of proteins that assemble at the origin go on to function as a part of the eukaryotic replisome 1 In addition to the three DNA polymerases the Mcm complex and many of the factors required for DNA polymerase recruitment become part of the replication fork Similar to the E coli DNA helicase loader DnaC the other factors such as Cdc6 and Cdtl are released or destroyed after their role is complete PreRC Formation and Activation are Regulated to Allow only a Single Round of Replication During Each Cell Cycle i How do eukaryotic cells control the activity of hundreds or even thousands of origins of replication such that not even one is activated more than once during a cell cycle 1 Answer lies in the tight regulation of the formation and activation of preRCs by cyclindependent kinases Cdks a Cdks have two contradictory roles in regulating preRC function i They are required to activate preRCs to initiate DNA replication ii Cdk activation inhibits the formation of new preRCs The tight connection between preRC function Cdk levels and the cell cycle ensures that the eukaryotic genome is replicated only once per cell cycle i Cdk levels are low during 61 whereas elevated levels of Cdk are present during the remainder of the cell cycle S 62 M 1 Thus there is only ONE opportunity for preRCs to form During 61 and only ONE opportunity for those preRCs to be activated during S 62 and M ii PreRCs are disassembled after they are activated or after the DNA to which they are bound is replicated 1 These exposed replicators are then available for new preRC formation and rapidly bind to ORC The elevated levels of Cdk activity in S 62 and M phase cells prevent the association of other members of the preRC complex with ORC a It is only when cells segregate their chromosomes and complete cell division that Cdk activity is eliminated and new preRC complexes can form Similarities Between Eukaryotic and Prokaryotic DNA Replication Initiation N 57 N i Similar 1 Recognition of the replicator by the initiator protein a The initiator protein in combination with one or more helicase loading proteins assembles the DNA helicase on the replicator 2 The helicase generates a region of ssDNA that can act as a template for RNA primer synthesis 3 Once primers are synthesized the remaining components of the replisome assemble through interactions with the resulting primertemplate junction ii Different 1 Regulation of replication 2 Unlike in eukaryotic cells rapidly dividing bacterial cells initiate replication more than once per cycle 3 Eukaryotic cells focus regulation on the initial association of the MCM helicase with the DNA a Bacterial cells focus regulation on the binding of the DnaA initiator protein to DNA 9 Finishing Replication a U Squot Completion of DNA replication requires a set of specific events i Events are different for circular vs linear chromosomes 1 Circular 39 the 39 39 39 fork machinery can replicate the entire molecule but the resulting daughter molecules are topologically linked to each other 2 Linear chromosome cannot be completed by the replication fork machinery Type II Topoisomerases Are Required to Separate Daughter DNA Molecules i After replication of a circular chromosome is complete the resulting daughter DNA molecules remain linked together as catenanes 1 General term for two circles that are linked links on a chain ii To segregate these chromosomes into separate daughter cells the two circular DNA molecules must be disengaged from each other or decatenated 1 This separation is accomplished by the action of type II topoisomerase a These enzymes have the ability to break a dsDNA molecule and pass a second dsDNA molecule through its break b This reaction can easily decatenate the two circular daughter chromosomes by breaking one and passing the second through the break iii Although the importance of this activity for the separation of circular chromosomes is most clear the activity of Topo II is also critical to the segregation of large linear molecules 1 Although there is no inherent topological linkage after the replication of a linear molecule the large size of eukaryotic chromosomes necessitates the intricate folding of the DNA into loops attached to a protein scaffolding LaggingStrand Synthesis is Unable to Copy the Extreme Ends of Linear Chromosomes i The requirement for an RNA primer to initiate all new DNA synthesis creates a dilemma for the replication of the ends of linear chromosomes called the end replication problem 1 This is NOT observed during the duplication of the leadingstrand template 2 A single internal RNA primer can direct the initiation of a DNA strand that can be extended to the extreme 539 terminus of its template Q a The requirement for multiple primers to complete lagging strand synthesis means that a complete copy of its template cannot be made b Even if the end of the last RNA primer for Okazaki fragment synthesis anneals to the final base pairs of the laggingstrand template once this RNA molecule is removed there will remain a short region of unreplicated ssDNA at the end of the chromosome i This means that each round of DNA replication would result in the shortening of one of the two daughter DNA molecules ii Organisms sole the end replication problem in a variety of ways 1 One solution is to use a protein instead ofan RNA as the primer for the last Okazaki fragment at each end of the chromosome a In this situation the quotpriming proteinquot binds to the laggingstrand template and uses an amino acid to provide an RNA primer b By priming the last lagging strand the priming protein becomes covalently linked to the 539 end of the chromosome c Terminally attached replication proteins of this kind are found at the end of the LINEAR chromosomes of certain species of BACTERIA 2 Most eukaryotic cells use telomeres a The ends of eukaryotic chromosomes are called telomeres and are generally composed of headtotail repeats of a TGrich DNA sequence i For example the human telomeres consist of repeats 539TTAGGG 3 Although many of these repeats are double stranded the 339 end of each chromosome extends beyond the 539 end as ssDNA i This unique structure acts as a novel origin of replication that compensates for the end replication problem 1 The origin recruits a specialized DNA polymerase called telomerase Telomerase is a Novel DNA Polymerase That Does NOT Require an Exogenous Template i Telomerase is a remarkable enzyme that includes multiple protein subunits AND an RNA component 1 It acts to extend the 339 end of its DNA substrate a BUT unlike most DNA polymerases telomerase does NOT need an exogenous DNA template to direct the addition of new dNTPs i Instead the RNA component of telomerase serves as the template for adding the telomeric sequence to the 339terminus at the end of the chromosome ii Telomerase specifically elongates the 339OH of a particular ssDNA sequence using its own RNA as a template iii As a result of this mechanism the newly synthesized DNA is single stranded ii The key to telomerase39s unusual function is revealed by the RNA component of the enzyme called telomerase RNA or TER 1 Depending on the organism TER varies in size from 1501300 bases 2 The sequence of the RNA in ALL organisms a short region that encodes approximately 15 copies of the complement of the telomere sequence a This region of the RNA can anneal to the ssDNA at the 339 end of the telomere 57 5 f i Annealing occurs in such a way hat a part of the RNA template remains singlestranded creating a primertempate junction that can be acted on by telomerase One of the protein subunits of telomerase is a member of a class of DNA polymerases that use RNA templates called reverse transcriptases TERT or telomerase reverse transcriptase iii The characteristics of telomerase are in some ways distinct and in other way similar to those of other DNA polymerases 1 Different a Inclusion ofan RNA component b Lack of a requirement for an exogenous template c Ability to use an entirely ssDNA substrate to produce an ssDNA product d Telomerase must have the ability to displace its RNA template from the DNA product to allow repeated rounds of templatedirected synthesis i This means that telomerase includes an RNADNA helicase activity 2 Similar a Telomerase requires a template to direct nucleotide addition i Can only extend a 339OH end of DNA b Uses the same nucleotide precursors c Can only act as a processive manner i Adding many sequence repeats each time it binds to a DNA substrate Telomerase Solves the End Replication Problem by Extending the 339 End of the Chromosome When telomerase acts on the 339 end of the telomere it extends ONLY one of the two strands of the chromosome ii How is the 539 end extended 1 This is accomplished by the laggingstrand DNA replication machinery 2 By providing an extended 339 end telomerase provides additional template for lagging strand replication machinery By synthesizing and extending RNA primers using the telomerase extended 339 end as a template the cell can effectively increase the length of the 539 end of the chromosome as well iii Even after the action of the laggingstrand machinery has acted there remains a short ssDNA region at the end of the chromosome 1 The action of telomerase and the laggingstrand replication machinery ensures that the telomere is maintained at sufficient length to protect the end of the chromosome from shortening TelomereBinding Proteins Regulate Telomerase Activity and Telomere Length i Proteins that bind to the doublestranded regions of the telomere or other proteins bind to the telomere and act as weak inhibitors of telomerase activity 1 When ere are relatively few copies of the telomere sequence repeat few of these proteins are bound to the telomere and telomerase can extend the 339 OH end of the telomere As the telomere becomes longer more of the telomerebinding proteins accumulate and telomerase can extend the 339OH end of the telomere W N g a This is negative feedback loop mechanism longer telomeres inhibit telomerase Proteins that recognize the singlestrand form of the telomere can also modulate telomerase activity 1 In S cervisiae cells the Cdcl3 protein binds to the singlestranded regions of the telomere a Studies show this protein recruits telomerase to the telomeres i Therefore Cdcl3 is a positive activator of telomerase The human protein that binds to singlestranded teomeric DNA POTl acts in the opposite manner N a An inhibitor of telomerase activity i In vitro studies show that POTl binding to singlestranded telomere DNA inhibits telomere activity ii Cells that lack this protein show dramatically increased telomere DNA length b This protein interacts indirectly with the doublestrand telomere binding proteins in human cells i As telomeres increase in length more POTl is recruited thereby increasing the likelihood that it binds to the ssDNA ends of the telomere and inhibits telomerase TelomereBinding Proteins Protect Chromosome Ends i In addition to regulating telomerase function telomerebinding proteins also play a crucial role in protecting the ends of chromosomes 1 Ordinarily in a cell the presence of a DNA end is considered the sign of a doublestranded break in DNA which is targeted by DNA repair machinery 2 The most common outcome of this repair is to initiate recombination with other DNA in the genome a Whereas this response is appropriate for random DNA breaks it would be disastrous for the telomeres to participate in the same events i Attempts to repair telomeres in the same manner as double stranded DNA breaks would lead to chromosome fusion events which eventually result in random chromosome breaks ii What protects the telomeres from this fate 1 The simple answer is that the proteins bound at the telomere distinguish telomeres from other DNA ends in the cell 2 Elimination of these proteins leads to the recognition of the telomeres as normal DNA breaks a It is possible that protection is conferred simply by coating the telomere with binding proteins 3 Studies of the structure of human telomere have yielded a different response a Telomeres isolated from human cells were observed by SEM and found to form a loop rather than a linear structure Subsequent analysis indicated that this structure called a tloop was formed by the 339ssDNA end of the telomere invading the dsDNA region of the telomere i It has been proposed that by forming a tloop the end of the telomere is masked and cannot be recognized as a normal DNA end 57 1 Purified TRFZ is capable of directing tloop formation with purified telomere DNA iii The Tloop structure may also be relevant to telomere length control 1 Just as the loop structure may protect the telomere from DNA repair enzymes it is also likely that telomerase cannot recognize this form of the telomere as it lack an obvious singlestrand 339 end a It has been proposed that as telomeres shorten the would have an increasingly difficult time forming the tloop thereby allowing increased access to the 339 end of the telomere 1 Chapter 7 Genome Structure Chromatin and the Nucleosome 2 Introduction Within the cell DNA is associated with proteins and each DNA molecule and its associated protein is called a chromosome b Packaging of the DNA into chromosomes serves several important functions a C d e g h k Chromosome is a compact form of the DNA that readily fits inside the cell Packaging the DNA into chromosomes serves to protect the DNA from damages Only DNA packaged into a chromosome can be transmitted efficiently to both daughter cells when a cell divides iv Chromosome confers an overall organization to each molecule of DNA Half of the molecular mass of a eukaryotic chromosome is protein i In eukaryotic cells a given region of DNA with its associated proteins is called chromatin 1 And the majority of the associated proteins are small basic proteins called histones Other proteins called nonhistone proteins are also associated with eukaryotic chromosomes These proteins include the numerous DNAbinding proteins that regulate the transcription replication repair and recombination of cellular DNA The protein component of chromatin performs another essential function compacting the A Huma i Most n cell contains 3x109bp per haploid set of chromosomes DNA laid out end to end would be approximately 1m in length compaction in human cells is the result of the regular association of DNA with histones to form structures called nucleosomes Assoc of the i Altera intera Proka It is le I Formation of the nucleosome is the first step in a process that allows the DNA to be folded into much more compact structures that reduce the linear length by as much as 10000fold iation of the DNA with histones and other packaging proteins limits the accessibility DNA This can interfere with the proteins that direct the replication repair recombination and transcription of the DNA tions to individual nucleosomes allow specific regions of the chromosomal DNA to ct with other proteins ryotic cells are required to compact their DNA ss clear how prokaryotic DNA is compacted Bacteria have no histones or nucleosomes 3 Genome Sequence and Chromosome Diversity a Chromosomes Can Be Circular or Linear Traditional view is that prokaryotic cells have a single circular chromosome and eukaryotic cells have multiple linear chromosomes Most studied prokaryotes do indeed have single circular chromosomes there are now numerous examples of prokaryotic cells that have multiple chromosomes linear chromosomes or even both All eukaryotic cells has multiple linear chromosomes 1 Number of chromosomes typically varies from 2 to less than 50 but can reach thousands protozoa Tetrahymena 57 Squot O D iv Circular and linear chromosomes each pose specific challenges that must be overcome for maintenance and replication of the genome 1 Circular chromosomes require topoisomerases to separate the daughter molecules after they are replicated a Without the two daughter molecules would remain interlocked or catenated with each other after replication 2 The ends of linear chromosomes have to be protected from enzymes that normally degrade DNA ends and present a different set of difficulties during DNA replication Every cell Maintains a Characteristic Number of Chromosomes i Prokaryotic cells typically have only one complete copy of their chromosomes that is packaged into a structure called the nucleoid 1 When prokaryotic cells are dividing rapidly portions of the chromosome in the process of replicating are present in two and sometimes even four copies 2 Also carry one or more smaller independent circular DNAs called plasmids Plasmids typically are not essential for bacterial growth b They carry genes that confer desirable traits to the bacteria antibiotic resistance Plasmids are often present in many complete copies per cell ii Majority of eukaryotic cells are diploid 1 Contain two copies of each chromosome iii Two copies of a given chromosome are called homologs 1 One is derived from each parent iv Not all cells in a eukaryotic organism are diploid 1 Subset of eukaryotic cells are either haploid or polyploid a Haploid cells contain a single copy of each chromosome and are involved in sexual reproduction i SpermEgg cells b Polyploid cells have more than two copies of each chromosome i Megakaryocytes are specialized polyploid cells approx 28 copies of each chromosome Genome Size is Related to the Complexity of the Organism i Genome size is defined as the length of DNA associated with one haploid complement of chromosome 1 Varies substantially between different organisms ii Table 72 1 Genome size in Mb increases between prokaryotes and eukaryotes 2 The number of genes increases between prokaryotes and eukaryotes 3 Gene density genesMb DECREASES between prokaryotes and eukaryotes The E coIiGenome s Composed Almost Entirely of Genes i Great majority of the single chromosome of the bacteria E coli encodes proteins or structural RNAs ii Critical element of the E coli genome is not a gene or a sequence that regulates gene expression 1 Instead the E coli origin of replication is dedicated to directing the assembly of the replication machinery More Complex Organisms Have Decreased Gene Density 1 Squot What explains the difference in genome sizes of organisms of apparently similar complexity 1 Largely related to gene density a Average number of genes per megabase or genomic DNA b When gene densities are compared it becomes clear that different organisms use the geneencoding potential of DNA with varying efficiencies Inverse correlation between organism complexity and the gene density i Less complex the organism the higher the gene density d Gene density is consistently lower and more variable in eukaryotic cells Equot f Genes Make up Only a Small Proportion of the Eukaryotic Chromosomal DNA viii Two factors contribute to the decreased gene density observed in eukaryotic cells 1 Increases in gene size 2 Increases in the DNA between genes a Intergenic sequences Major reason that gene size is larger in more complex organisms is not that the average protein is bigger or that more DNA is required to encode the same protein but proteinencoding genes in eukaryotes frequently have discontinuous protein coding genes Introns 1 Interspersed nonproteincoding regions 2 Introns are removed from the RNA after transcription a Process called RNA splicing Approximately 5 of the average human proteinencoding gene directly encodes the desired protein 1 The other 95 is made up of introns An increase in the amount of intergenic sequences in more complex organisms is responsible for the remaining decreases in gene density Intergenic DNA is the portion of a genome that does NOT encode proteins or structural RNAs 1 More than 60 of the human genome is composed of intergenic sequences and must of the DNA function is unknown There are two kinds of intergenic DNAs 1 Unique 2 Repeated Unique Intergenic DNAs 1 Approximately 14 of the intergenic DNA is unique 2 One contributor to an increase in such is an increase in regions of the DNA that are required to direct and regulate transcription a Regulatory sequences 3 As an organism becomes more complex and encodes for more genes regulatory sequences required to coordinate gene expression also grow in complexity and size 4 Unique regions of the human intergenic DNA also include many apparently nonfunctional relics a Nonfunctional mutant genes b Gene fragments i Above two arise from simple random mutagenesis or mistakes in DNA recombination c Pseudogenes i Arise from the action of an enzyme called reverse transcriptase 1 This enzyme copies NA into doublestranded DNA 2 Only expressed by certain types of viruses that require this enzyme to reproduce It is clear that there are likely functions of the unique intergenic regions that are no understood a Recent identification of microRNAs i miRNA ii Act to regulate the expression of other genes by altering either the stability of the product mRNA or its ability to be translated iii Estimated that human cells may have more than 400 miRNAs g The Majority of Human ntergenic Sequences Are Composed of Repetitive DNA i Almost half of the human genome is composed of DNA sequences that are repeated many times in the genome 1 Two general classes of repeated DNA a Microsatellite DNA b GenomeWide repeats 2 Microsatellite DNA a Composed of very short lt13 bp tandemly repeated sequences b Most common sequences are dinucleotide repeats CACACACACACACACACACACACA i Repeats arise from difficulties in accurately duplicating the DNA and represent approximately 3 of the human genome 3 Genomewide repeats a Much larger approximately greater than 100bp in length and many are larger than 1kb b Can be found either as single copies dispersed throughout the genome or as closely spaced clusters c Common feature that they all are forms of transposable elements i Sequences that can move from one place in the genome to another ii During transposition the element moves to a new position in the genome often leaving the original copy behind iii Movement of transposable elements is a relatively rare event in human cells 1 Over long periods of evolutionary time they have become so successful they now constitute 45 of the human genome U39I 4 C39 l rquot quot and g a Eukaryotic Chromosomes Require Centromeres Telomeres and Origins of Replication to be Maintained during Cell Division i There are several important DNA elements in eukaryotic chromosomes that are NOT genes and are NOT involved in regulating the expression of genes 1 Elements include origins of replication that direct the duplication of the chromosomal DNA 57 2 Centromeres that act as quothandlesquot for the movement of chromosomes into daughter cells 3 Telomeres that protect and replicate the ends of linear chromosomes Origins of Replication are the sites at which the DNA replication machinery assembles to initiate replication 1 Typically found 3040 kb apart through the length of each eukaryotic chromosome Centromeres are required for the correct segregation of the chromosomes after DNA replication Two copies of each replicated chromosomes are called sister chromosomes and during cell division thy must be separated with one copy going into each of the two daughter cells 2 Centromeres direct the formation of an elaborate protein complex called a kinetochore a Interacts with the centromere DNA and with protein filaments microtubules that pull the sister chromosomes away from each other and into the two daughter cells 3 It is critical that each chromosome have only ONE centromere a In the absence of a centromere the replicated chromosomes segregate randomly resulting in daughter cells that either have lost a chromosome or have two copies of a chromosome In the case of two or more 57 I the attached to filaments pulling in opposite directions this can lead to chromosome breakage are iv Telomeres are located at the ends of linear chromosomes 1 Bound by a number of proteins 2 The proteins provide two important functions a Telomeric proteins distinguish the natural ends of the chromosomes from sites of chromosome breakage and other DNA breaks in a cell i DNA ends are the sites of frequently recombination and DNA degradation proteins that assemble at telomeres form a structure that is resistant to both events b Telomeres act as a specialized origin of replication that allows the cell to replicate the ends of the chromosomes 3 Telomeres facilitate end replication through the recruitment ofan unusual DNA polymerase called telomerase 4 In contrast to most of the chromosome a portion of the telomere is maintained in a singlestranded form Eukaryotic C39 l rquot quot and g g 39 Occur in Separate Phases of the Cell Cycle During cell division the chromosomes must be duplicated and segregated into the daughter cells In bacterial cells these events occur simultaneously 1 As the DNA is replicated the resulting two copies are separated into opposite sides of the cell Eukaryotes duplicate and segregate their chromosomes at distinct times during cell division C iv The events required for a single round of cell division are collectively known as the cell cycle mitotic cell division 1 The mitotic cell division cycle can be divided into four phases a 61 b S c Gz d M 2 S Phase a Chromosome replication occurs resulting in the duplication of each chromosome b Each chromosome of the duplicated pair is called a chromatid i Two chromatids of a given pair are called sister chromatids c Sister chromatids are held together after duplication through the action of a molecule called cohesion i The process that holds them together is called sisterchromatid cohesion 1 Tethered state is maintained until the chromosomes segregate from one another 3 M Phase a Mitosis b Three key steps i Each pair of sister chromatids is bound to a structure called the mitotic spindle 1 This structure is composed of long protein fibers called microtubules that are attached to one of the two 39 39 39 g 39 g centers in animal cells spindle pole bodies in yeasts 2 Microtubuleorganizing center are located on opposite sides of the cell forming quotpolesquot toward which the microtubules pull the chromatids 3 Attachment of the chromatids to the microtubules is mediated by thequot 39 quot Iat each ii Cohesion between the chromatids is dissolved by proteolysis of cohesion 1 Before cohesion is dissolved it resists the pulling forces of the mitotic spindle iii Sisterchromatid separation 1 In the absence of counterbalancing force of chromatid cohesion the chromatids are rapidly pulled toward opposite poles of the mitotic spindle 2 Cohesion between the sister chromatids and attachment of sister chromatid kinetochores to opposite poles of the mitotic spindle play opposing roles that must be carefully quot Ifor 39 g 39 to occur properly Chromosome Structure Changes as Eukaryotic Cells Divide i As chromosomes proceed through a round of cell division their structure is altered numerous times ii There are two main states for chromosomes 1 Chromosomes are in their most compact form as cell segregate their chromosomes a The process that results in this compact form is called chromosome condensation i In this state the 39 are 39 39 quot U 39 one another greatly facilitating the segregation process During phases of the cell cycle when chromosome segregation is not occuring collectively referred to as interphase the chromosomes are significantly less compact a At these stages the chromosomes are likely to be highly intertwined resembling spaghetti 3 DNA replication requires the nearly complete disassemny and reassembly of the proteins associated with each chromosome 39 from Iquot d SisterChromatid Cohesion and Chromosome Condensation Are Mediated by SMC Proteins 5 i The key proteins that mediate sisterchromatid cohesion and chromosome condensation are related to one another ii The structural maintenance of chromosome SMC proteins are extended proteins that form defined pairs by interacting through lengthy coiledcoil domains 1 Together with nonSMC proteins they form multiprotein complexes that act to link two DNA helices together iii Cohesion is an SMC protein containing complex that is required to link the two daughter DNA duplexes together after DNA replication 1 Basis for sister chromatid cohesion a Structure of cohesion is thought to be a large ring composed of two SMC proteins and two nonSMC proteins iv The 39 39 that I 39 requires a related SMC containing complex called condensin 1 Condensin shares many of the features of the cohesion complex suggesting that it too is ringshaped 2 Induces chromosome condensation Mitosis Maintains the Parental Chromosome Number i Mitosis occurs in several stages also 1 Prophase 2 Metaphase 3 Anaphase 4 Telophase ii Prophase 1 Chromosomes condense into the highly compact form required for segregation 2 At the end of prophase the nuclear envelope breaks down and the cell enters metaphase iii Metaphase 1 Mitotic spindle forms and the kinetochores of sister chromatids attach to the microtubules 2 Proper chromatid attachment is only achieved when the two kinetochores of a sisterchromatid pair are attached to microtubules emanating from opposite microtubuleorganizing centers r Q m This type of attachment is called bivalent attachment and results in the microtubules exerting tension on the chromatid pair by pulling the sisters in opposite directions i Tension exerted by bivalent attachment is opposed by sister chromatid cohesion and results in all the chromosomes aligning in the middle of the cell between the two microtubuleorganizing centers metaphase plate ii Chromosome segregation starts only after All sisterchromatid pairs have achieved bivalent attachment b quot 39 of both 39 quot to 39 39 39 from the same microtubuleorganizing center of attachment of only one chromatid of the pair is known as monovalent attachment i Does not result in tension ii Monovalent attachment can lead to both copies of a chromosome moving into one daughter cell iv Anaphase 1 Chromosome segregation is triggered by proteolytic destruction of the cohesion molecules resulting in the moss of sisterchromatid cohesion 2 Stage signifies at which the sister chromatids separate and move to opposite sides of the cell 3 Once the two sisters are no longer held together the tension causes them to part v Telophase 1 All the nuclear envelope reforms around each set of segregated chromosomes 2 Cell division can be completed by physically separating the shared cytoplasm of the two presumptive cells a Separation of the shared cytoplasm is known as cytokinesis During Gap Phases Cells Prepare for the Next Cell Cycle Stage and Check That the Previous Stage is Completed Correctly i The remaining two phases of the mitotic cell cycle are gap phases 61 a Occurs prior to DNA synthesis 2 G2 a Occurs between S phase and M phase ii Gap phases of the cell cycle provide time for the cell to accomplish two goals 1 To prepare for the next phase of the cell cycle 2 Check that the previous phase of the cell cycle has been completed appropriately a IE prior to entry into S phase most ells must reach a certain size and level of protein synthesis to ensure that there will be adequate proteins and nutrients to complete the next round of DNA synthesis If there is a problem with a previous step in the cell cycle cell cycle checkpoints stop the cell to provide time for the cell to complete that 57 step Meiosis Reduces the Parental Chromosome Number i A second type of eukaryotic cell division is specialized to produce cells that have half the number of chromosomes than the parental cells ii Meiotic cell cycle 1 61 2 S 3 Elongated G2 iii S Phase 1 Each chromosome is replicated and the daughter chromatids remain associated as in the mitotic S phase 2 Cells that enter meiosis must be diploid and thus contain two copies of each chromosome prior to DNA replication one derived form each parent 3 After DNA replication these homologs related sisterchromatid pairs pair with each other and recombine 4 The most significant difference between the mitotic and meiotic cell cycles occurs during chromosome segregation a Unlike mitosis which there is a single found of chromosome segregation chromosomes participating in meiosis undergo two founds of segregation known as meiosis and meiosis H i Each stages have segregation events including prophase metaphase anaphase with the corresponding roman numeral iv Metaphase Meiosis I 1 Homologs attach to opposite poles of the microtubulebased spindle a This attachment is mediated by the kinetochore b Monovalent attachment occurs in meiosis v Meiosis 1 Very similar to mitosis big difference is that the round of DNA replication does no precede this segregation event a Instead a spindle is formed in association with each of the two newly separated sister chromatid pairs vi Metaphase 1 Spindles attach in a bivalent manner to the kinetochores of each sister chromatid pair 2 Cohesion that remains at the centromeres after meiosis critical to oppose the pull of the spindle vii Anaphase 1 Second round of chromosome segregation 2 Initiated by the elimination of centromeric cohesion 3 At this point there are four sets of chromosomes in the cell each of which contains a single copy of each chromosome h Different Levels of Chromosome Structure Can be Observed by Microscopy i Outside of mitosis chromosomal NA is less compact in the electron microscope two states of chromatic are observed 1 Fibers with a diameter of 30nm a More compact version of chromatin that is frequently folded into large loops reaching out from a protein core or scaffolding 2 Fibers with a diameter of 10nm a Less compact form of chromatin that resembles a regular series of quotbeads on a stringquot i Beads are nucleosomes and these proteinDNA structures play a critical role in regulating the structure and function of chromosomes 5 The Nucleosome a 57 Nucleosomes Are the Building Blocks of Chromosomes i The majority of the DNA in eukaryotic cells is packed into nucleosomes 1 Each nucleosome is composed of a core of eight histone proteins and the DNA wrapped around them ii The DNA between each nucleosome is called linker DNA iii By quot39 Uinto 39 Ais 39 iv Core DNA 1 DNA most tightly associated with the nucleosome 2 Is wound approximately 165 times around the outside of the histone octamer like thread around a spool 3 The length of DNA associated with each nucleosome can be determined using nuclease treatment a The approximate 147bp length of this DNA is an invariant feature of nucleosomes in all eukaryotic cells b Length of linker DNA is variable Typically 2060bp and each eukaryote has a characteristic average linker DNA length 1 The difference in average linker DNA length is likely to reflect the differences in the larger structures formed by nucleosomal DNA in each organism rather than differences in the nucleosomes themselves V In any cells there are stretches of DNA that are not packaged into nucleosomes 1 These are regions of DNA engaged in gene expression replication or recombination a Not bound by nucleosomes these sites are typically associated with nonhistone proteins that are either regulating or participating in these events Histones Are Small Positively Charged Proteins i Histones are by far the most abundant proteins associated with eukaryotic DNA ii Eukaryotic cells commonly contains five abundant histones 39 sixfold I rr 1 H1 H2A 3 HZB 4 H3 5 H4 a Histones H2A HZB H3 and H4 are the core histones i Two copies of each of these histones form the protein core around which nucleosomal DNA is wrapped ii Are present in equal amounts in the cell iii Core histones are also relatively small proteins ranging in size from 11 to 15 kDa b Histone H1 is not part of the nucleosome core particle i Instead it binds to the linker DNA and is referred to as a linker histone ii Half as abundant as the core histones iii One molecule of H1 can associated with a nucleosome iv Histone H1 is approximately 21 kDa Consistent with their close association with the negatively charged DNA molecule the histones have a high content of positively charged amino acids 1 At least 20 of the residues in each histone are either LYSINE K or ARGININE R The protein core of the nucleosome is a discshaped structure that assembles in an ordered fashion only in the presence of DNA 1 Without DNA the core histones form intermediate assemblies in solution 2 Conserved region found in every core histone histonefold domain mediates the assembly of these histoneonly intermediates HistoneFold Domain 1 Compose of three ot helical regions separated by two short unstructured loops 2 This domain mediates the formation of headtotail heterodimers of specific pairs of histones Histones H3 and H4 first form heterodimers that then come together to form a tetramer with two molecules each of H3 and H4 H2A and HZB form heterodimers in solution but not tetramers The assembly of a nucleosome involves the ordered associated of these building blocks with DNA 1 The H3H4 tetramer binds to DNA 2 Two H2AHZB dimers join the H3H4 DNA complex to form the final nucleosome Core histones each have an aminoterminal extension because it lacks a defined structure and is accessible within the intact nucleosome 1 The accessibility can be detected with treatment of nucleosomes with the protease trypsin specifically cleaves proteins after positively charged amino acids a Treatment of nucleosomes with trypsin rapidly removes the accessible aminoterminal tails of the histones but cannot cleave the tightly packed histonefold regions 2 The exposed aminoterminal tails are not required for the association of DNA with the histone octamer as DNA is still tightly associated with the nucleosome after protease treatment Instead the tails are the sites of extensive modifications that alter the function of individual nucleosomes i These modification include 1 Phosphorylation 2 Acetylation and methylation on serine lysine and arginine residues m c The Atomic Structure of the Nucleosome d Histones Bind Characteristic Regions of DNA within the Nucleosome i High resolution threedimensional structure of the nucleosome core particle has provided many insights into nucleosome function 1 The high affinity of the nucleosome for DNA 2 The distortion of the DNA when bound to the nucleosome 3 Lack of DNA sequence specificity twofold Although not perfectly has an H axis of symmetry called the dyad axis 5 1 This can be visualized by thinking of the face of the octamer disk as a clock with the midpoint of the 147bp of DNA located at the 1200 position a This places the ends of the DNA just short of 1100 and 100 b The line drawn from 1200 and 600 through the middle of the disk defines the dyad axis The H3H4 tetramers and the H2AHZB dimers each interact with a particular region of the DNA within the nucleosome 1 Of the 147bp of DNA included in the structure the histonefold regions of the H3H4 tetramer interact with the central 60bp 2 The aminoterminal region of H3 most proximal to the histonefold region forms a fourth othelix that interacts with the final 13bp at each end of the bound DNA To picture the nucleosome with a clock face the H3 H4 tetramer formsthe top half of the histone 1 Histone H3H4 tetramers occupy a key position in the nuclosome by binding the middle AND both ends of the DNA The two H2AHZB dimers each associated with approximately 30bp of DNA on either side of the central 60bp of DNA bound by H3 and H4 1 Using the clock face the DNA associated with H2AHZB is located from approximately 500 to 900 on either face 2 Together the two H2AHZB dimers form the bottom part of the histone octamer located across the disc from the DNA ends The extensive interactions between the H3H4 tetramer and the DNA help to explain the ordered assembly of the nucleosome 1 H3H4 tetramer association with the middle and ends of the bound DNA would result in the DNA being extensively bent and constrained making the association of H2AHZB dimers relatively easy 2 In contrast the relative short length of DNA bound by H2AHZB dimers is not sufficient to prepare the DNA for H3H4 tetramer binding Many DNA SequenceIndependent Contacts Mediate the Interaction Between the Core Histones and DNA A closer look at the interactions between the histones and the nucleosomal DNA reveals the structural basis for the binding and bending of the DNA within the nucleosome Fourteen distinct sites of contact are observed one for each time the minor groove of the DNA faces the histone octamer The association of DNA with the nucleosome is mediated by a larger number approx 40 of hydrogen bonds between the histones and the DNA 1 The majority of these hydrogen bonds are between the proteins and the oxygen atoms in the phosphodiester backbone near the minor groove of the DNA 2 Only seven hydrogen bonds are made between the protein side chain and the bases and all of these are made in the MINOR groove of the DNA A large number of these hydrogen bonds provides the driving force to bend DNA 1 A typical sequencespecific DNAbinding protein only has approx 20 hydrogen bonds with DNA V 2 The highly basic nature of the histones further facilitates DNA bending by masking the negative charge of the phosphates that ordinarily resists DNA bending The finding that all of the sites that contact between the histones and the DNA involve either the minor groove or the phosphate backbone is consistent with the nonsequencespecific nature of the association of the histone octamer with DNA 1 NEITHER THE PHOSPHATE BACKBONE NOR THE MINOR GROOVE IS RICH IN BASESPECIFIC INFORMATION 2 Of the seven hydrogen bonds formed with the bases in the minor groove NONE IS WITH PARTS OF THE BASES THAT DISTINGUISH BETWEEN GC AND AT f The Histone AminoTerminal Tails Stabilize DNA Wrapping Around the Octamer In structure of the nucleosome also tells something about the histone aminoterminal tails 1 Four HZB and H3 tails emerge from between the two DNA helices a In each case their path of exit is formed between two adjacent minor grooves making a quotgapquot between the two DNA helices just big enough for a polypeptide chain The HZB and H3 tails emerge at approximately equal distances from each other around the octamer disk at 100 and 1100 for the H3 tail 400 and 800 for the HZB tail Instead of emerging between the two DNA helices the H2A and H4 amino terminal tails emerge from either quotabovequot or quotbelowquot both DNA helices a These tails are also distributed around the face of the nucleosome with the HZA tails emerging at 500 and 700 and the H4 tails at 300 and 900 By emerging both between an don either side of the DNA helices the histone tails can be though of as the grooves of a screw a This is because it directs the DNA to wrap around the histone octamer disc in a Lhanded manner i Lefthanded nature of the DNA wrapping introduces NEGATIVE supercoils in the DNA The parts of the tails most proximal to the histone disc also make some of the many hydrogen bonds between the histones as the DNA as they pass by the 57 N W 4 DNA Wrapping of the DNA Around the Histone Protein Core Stores Negative Superhelicity Each nucleosome added to a covalently closed circular template changes the linking number of the associated DNA by approximately 12 The DNA that is packed into nucleosomes would become negatively supercoiled if nucleosomes were removed from the DNA remainder of the DNA is kept relaxed by topoisomerases 1 Nucleosomes can be viewed as storing or stabilizing negative superhelicity Why would a cell want to maintain a stockpile of negative superhelicity 1 Initiation of DNA replication transcription and recombination 2 Favors DNA unwinding a Removal of a nucleosome allows increased access to the DNA but also facilitates DNA unwinding of nearby DNA sequences If nucleosomes store negative superhelicity in eukaryotic cells what serves the equivalent function in prokaryotic cells 1 The entire genome is maintained in a negatively supercoiled state a This is accomplished by a specialized topoisomerase called gyrase that has the ability to induce negative superhelicity into relaxed DNA by reducing the linking number i Gyrase requires the presence of ATP to introduce negative supercoils 1 In the absence of ATP gyrase can only relax DNA 2 Not all bacteria need to maintain their DNA in a negatively supercoiled state a Bacteria that prefer to grow at very high temperatures must expend energy to prevent their DNA from unwinding i Contain topoisomerase known as reverse gyrase 1 Increases the linking number of relaxed DNA in the presence of ATP 2 Counteracts the effect of thermal denaturation that would ordinarily result in many regions of the genome being unwound 6 HigherOrder Chromatin Structure a 57 Heterochromatin and Euchromatin i Early studies of chromosomes divided chromosomal regions into two categories 1 Euchromatin a Characterized by dense staining with a variety of dyes and a more condense appearance b Euchromatic regions showed higher levels of gene expression c Euchromatic nucleosomes are found to be in much less organized assemblies 2 Heterochromatin a Staining poorly with dyes and having a relatively open structure b Heterochromatic regions of chromosomes had very limited gene expression c Heterochromatic regions are of 39 39 DNA 39 39 39 into higherorder structures that result in a barrier to gene expression ii Heterochromatin 1 Heterochromatic regions show little gene expression 2 Associated with particular chromosomal regions including the telomere and the centromere a Important for the function of both of these key chromosomal elements iii The difference in heterochromatin and euchromatin structure is how the nucleosomes in these chromosomal regions are or are not assembled into larger assemblies Histone H1 Binds to the Linker DNA Between Nucleosomes i Once nucleosomes are formed the next step in the packaging of DNA is the binding of histone H1 ii H1 interacts with the linker DNA between nucleosomes further tightening the association of the DNA with the nucleosome This can be detected by the increased protection of the nucleosomal DNA from micrococcal nuclease digestion Beyond the 147bp protected by the core histones addition of the histone H1 to a nucleosome protects an additional 20bp of DNA from micrococcal nuclease digestion l N Squot iii Histone H1 has the unusual property of binding two distinct regions of the DNA duplex 1 These two regions are part of the single DNA molecule associated with a nucleosome Sites of H1 binding are located asymmetrically relative to the nucleosome Site Locale a One of the two regions bound by H1 is the linker DNA at one end of the nucleosome b Second site of DNA binding is in the middle of the associated 147bp i The additional DNA protected from the nuclease digestion is restricted to linker DNA on only one side of the nucleosome 4 H1 binding increases the length of the DNA wrapped tightly around the histone octamer iv H1 binding produces a more defined angle of DNA entry and exit form the nucleosome This results in the nucleosomal DNA taking on a distinctly zigzag appearance Angles of entry and exit observed vary substantially depending on conditions a Salt concentrations b pH c Presence of other proteins 3 If we assume that these angles are approximately 20 relative to the dyad axis this would result in a pattern in which nucleosomes would alternate on either side ofa central region of linker DNA bound to histone H1 Nucleosome Arrays Can Form More Complex Structures The 30nm Fiber i Binding of H1 stabilizes higherorder chromatin structures ii When salt concentrations increased the addition of histone H1 results in the nucleosomal DNA forming a 30nm fiber 1 Can be observed in vivo 2 Represents the next level of DNA compaction 3 Incorporation of DNA into this fiber makes the DNA less accessible to many DNAdependent enzymes a RNA polymerases iii There are two models for the structure of the 30nm fiber 1 Solenoid model a Nucleosomal DNA forms a superhelix containing approximately six nucleosomes per turn b Helical pitch ofapproximately 11nm i Approximately diameter of the nucleosome disc suggesting that the 30nm fiber is composed of nucleosome discs stacked on edge in the form ofa helix 1 Flat surfaces on either face of the histone octamer disc are adjacent to each other and the DNA surface of the nucleosomes forms the accessible surface of the superhelix 2 Linker DNA is buried in the center of the superhelix but never passes through the axis of the fiber 1 The linker DNA circles around the central axis as the DNA moves from one nucleosome to the next WN NH 2 Zigzag Model 0 5 a Based on the Zigzag pattern of nucleosomes formed upon H1 addition i 30nm fiber is a compacted form of these Zigzag nucleosome arrays ii Studies CONFIRM this model iv Unlike the solenoid model the Zigzag conformation requires the linker DNA to pass through the central axis of the fiber in a relatively straight form 1 Longer linker DNA varies between different species the form of the 30nm fiber may not always be the same and both models may be correct The Histone AminoTerminal Tails Are Required for the Formation of the 30nm Fiber i Core histones lacking their aminoterminal tails are incapable of forming 30nm fibers ii The most likely role of the tails is to stabilize the 30nm fiber by interacting with adjacent nucleosomes 1 This is supported by the 3D crystal structure of the nucleosome which shows that each of the aminoterminal tails of H2A H3 and H4 interacts with the adjacent nucleosome cores in the crystal lattice iii Recent studies indicate that the interaction between the positively charged amino terminus of histone H4 and a negatively charged region of the histonefold domain of histone HZA is particularly important for 30nm fiber formation 1 The residues of HZA that interact with the H4 tail are conserved across many eukaryotic organisms but are not involved in DNA binding or formation of the histone octamer a One theory is that these regions of HZA are conserved to mediate internucleosomal interactions with the H4 tail Further Compaction of DNA involves Large Loops of Nucleosomal DNA i The packaging of DNA into nucleosomes and the 30nm fiber results in the compaction of the linear length of DNA by approximately 40 fold 1 This is still insufficient to fit 12 meters of DNA into a nucleus approximately 1039 5m across a Additional folding of the 30nm fiber is required to compact the DNA further i Although unclear one model suggests that the 30nm fiber forms loops of 4090kb that are held together at their bases by a proteinaceous structure referred to as the nuclear scaffolding ii Two classes or proteins that contribute to the nuclear scaffolding have been identified 1 Topoisomerase Topo II a Abundant in both scaffolding preparations and purified mitotic chromosomes b Treating cells with drugs that result in DNA breaks at the sites of Topo DNA binding generates DNA fragments that are approx 50kb in size c The presence of Topo II at the bottom of each loop would ensure that the loops are topologically isolated from each other 2 SMC Proteins a These proteins are key components of the machinery that condenses and holds sister chromatids together after chromosome duplication b Serve to enhance their functions by providing an underlying foundation for their interactions with chromosomal DNA for nuclear scaffolding f Histone Variants Alter Nucleosome Function The core histones are among the most conserved eukaryotic proteins the nucleosomes formed by these proteins are very similar in all eukaryotes H2AX is a variant of HZA that is widely distributed in eukaryotic nucleosomes 1 When chromosomal DNA is broken H2AX located adjacent to the break is phosphorylated which is then specifically recognized by DNA repair enzymes leading to their localization at the site of DNA damage CENPA is associated with nucleosomes that include centromeric DNA Replaces H3 subunits in the nucleosome 2 Incorporated into the kinetochore that mediates attachment of the chromosome to the mitotic spindle 3 Includes a substantially extended aminoterminal tail region It is unlikely that incorporation of CENPA changes the core structure of the nucleosome a Instead the extended tail ofCENPA is likely to generate novel binding sites for other protein components of the kinetochores P 7 Regulation of Chromatin Structure The Interaction of DNA With the Histone Octamer Is Dynamic a 57 vi It is critical that nucleosomes can be moved or that their grip on the DNA can be loosened to allow access to particular regions of DNA DNA is inherently dynamic Like all interactions mediated by noncovalent bonds the association of any particular region of DNA with the histone octamer is not permanent 1 Any individual region of the DNA will transiently be released from tight interaction with the octamer now and then The dynamic nature of DNA binding to the histone core structure is important 1 This is because many DNAbinding proteins strongly prefer to bind to histone free DNA As a result of intermittent spontaneous unwrapping of DNA from the nucleosome a protein can gain access to its DNAbinding sites with a probability of 1 in 50 to 1 in 100000 1 This is dependent on where the binding site is within the nucleosome a The more central the binding site the less frequently accessible Association of H1 and incorporation of nucleosomes into the 30nm fiber will also alter probabilities of protein gaining access to the DNA NucleosomeRemodeling Complexes Facilitate Nucleosome Movement The stability of the histone octamerDNA interaction is influenced by large protein 39 called 39 39 quot U I 39 These multiprotein complexes facilitate changes in nucleosome location or interaction with the DNA using the energy of ATP hydrolysis There are two basic types of nucleosome changes mediated by these enzymes 1 Sliding 2 Transferred Sliding 1 All nucleosomeremodeling complexes can catalyze the quotslidingquot of DNA along the surface of the histone octamer Transferred 1 Subset of nucleosomeremodeling complexes can catalyze a more extreme change in which a histone octamer is quottransferredquot from one DNA helix to another vi Each of these multisubunit enzymes contains an ATPhydrolyzing subunit that is capable of moving in a directional manner also called translocation on double stranded DNA 1 In the context of a nucleosomeremodeling complex these enzymes are though to bind tightly to the histone octamer and translocate the DNA relative to the histones 2 A DNA molecule that is sliding across a histone octamer can be viewed as binding to the octamer in many different energetic equivalent states and the nucleosomeremodeling complex is allowing DNA to access these different states more easily vii There are multiple types of nucleosomeremodeling complexes in any given cell 1 They can have as few as two subunits or more than ten 2 Each of these complexes contains a similar ATP hydrolyzing subunit that catalyzes the DNA movement as described by figure 736 viii Figure 736 Notes 1 The model proposes that the ATPhydrolyzing subunit of the nucleosome remodeling complex binds the nucleosomal DNA two helical turns from the central dyad 2 Using the ATPdependent DNA translocating activity the nucleosome remodeling complex first pulls the DNA from the nearest linker domain into the nucleosome a This breaks the five histoneDNA contacts between the ATPhydrolyzing subunit and the linker DNA i The broken contacts then reform with the translocation DNA leaving too much DNA associated with the nucleosome next to the ATPhydrolyzing subunit 1 To relive this strain the model proposes that the additional sequences move like a quotwavequot across the surface of the histones breaking one contact at a time until all the contacts have reformed with the appropriate amount of DNA between them at which point the excess DNA is no longer present within the histoneassociated DNA and the nucleosome has shifted its position of the DNA 3 In essence ATPase domain lifts DNA off the nucleosome in positions 15 position 3 at 1200 position 5 at 900 then 6gt7gt8 ix Although the ATPhydrolyzing subunit is similar the different nucleosomeremodeling complexes the other subunits associated with each complex modulate their function 1 IE These complexes can include subunits that target them to particular chromosomal locations In other instances nucleosomeremodeling complexes are localized by subunits that bind to specific modifications of the histone aminoterminal tails via CHROMOBROMO domains c Some Nucleosomes Are Found in Specific Positions nucleosome Positioning i There are occasions when restricting nucleosome locations positioning nucleosomes is beneficial N 0 viii Positioning a nucleosome allows the DNAbinding site for a regulatory protein to remain in the accessible linker DNA region 1 In many instances such nucleosomefree regions are larger to allow the binding sites for multiple regulatory proteins to remain accessible Nucleosome positioning can be directed by DNAbinding proteins or particular DNA sequences 1 In the cell one frequent method involves competition between nucleosomes and DNAbinding proteins Just as many proteins cannot bind to DNA within a nucleosome binding ofa protein to the DNA can prevent the subsequent association of the core histones with that stretch of DNA If two such DNAbinding proteins are bound to sites closer than the minimal region of DNA required to assemble a nucleosome approx 150bp the DNA between the proteins will remain nucleosome free 1 Binding of additional proteins to adjacent DNA can further increase the size of a nucleosome free region 2 In addition to this inhibitory of Motel 39 I 39 39 positioning some DNAbinding proteins interact tightly with adjacent nucleosomes leading to 39 39 quot 39 quot 39 quot adjacent to these proteins The second method of nucleosome positioning involves particular DNA sequences that have a high affinity for the nucleosome Because DNA bound in a nucleosome is bent nucleosomes preferentially form on DNA that bends easily 1 ATrich DNA has an intrinsic tendency to bend toward the minor groove a ATrich DNA is favored in positions in which the minor groove faces the histone octamer 2 GC rich DNA has the opposite tendency and is therefore favored when the minor groove is facing away from the histone octamer a EACH nucleosome will try to maximize this arrangement of ATrich and GC rich sequences Recent studies of nucleosome positioning in the yeast S cerecisiae suggest that as many as 50 of tightly positioned nucleosomes can be attributed to preferential binding of the histone core to the sequences they include 1 Such sequences are NOT required for nucleosome assembly and the action of other proteins including chromatinremodeling and transcription factors can move nucleosomes from such preferred positions Mechanisms of nucleosome positioning influence the organization of nucleosomes in the genome 1 Despite this many nucleosomes are not tightly positioned r J u Modification of the AminoTerminal Tails of the Histones Alters Chromatin Accessibility When histones are isolated from cells their aminoterminal tails are typically modified with a variety of small molecules Lysines in the tails are frequently modified with acetyl or methyl groups at least one arginine is also known to be methylated Serine are subject to modification with phosphate It has been proposed that the modification of histone tails forms a quothistone codequot that can be read by proteins involved in gene expression and other DNA transactions 1 This model proposes that in addition to the sequence of the genome the nucleosome associated with a particular gene and how they are modified will also strongly influence whether an associated gene is expressed v Both the type of modification and the particular site modified are important for understanding this code 1 IE Acetylation of Lysines at positions 8 and 16 of the histone H4 amino terminal tail is associated with the start sites of expressed genes a Acetylation of Lysines 5 and 12 is NOT it marks newly synthesized H4 molecules that are ready to be deposited onto DNA as part of a new nucleosome 2 Methylation of Lysines 4 36 or 79 of histone H9 is associated with expressed genes a Methylation of Lysines 9 or 27 of the same histone is associated with transcriptional repression 3 Phosphorylation of the aminoterminal tail of histone H3 is commonly observed in the highly condensed chromatin of mitotic chromosomes vi How does histone modification alter nucleosome function 1 One change is that acetylation and phorphorylation each acts to reduce the overall positive charge of the histone tails acetylation of lysine neutralizes its positive charges a The loss of positive charge reduces the affinity of the tails for the negatively charged backbone of the DNA 2 Modification of the histone tails affects the ability of nucleosome arrays to form more repressive higherorder chromatin structure vii Histone aminoterminal tails are required to form the 30 nm fiber and modification of the tails modulates this function 1 Consistent with the association of some types of acetylated histones with expressed regions of the genome acetylation of the H4 aminoterminal tail interferes with the ability of nucleosomes to be incorporated into the repressive 30nm fiber viii Formation of the 30 nm fiber is facilitated by an interaction between the positively charged H4 aminoterminal tail and the negatively charged surface of the HZA histonefold domain 1 Acetylation interferes with this association by altering the charge of the H4 tail e Protein Domains in NucleosomeRemodeling and Modifying Complexes Recognize Modified Histones i Modified histone tails can also act as recruit specific proteins to the chromatin ii The following protein domains are responsible to recognize modified forms of histone tails Bromo a Interact with acetylated histone tails 2 Chromo 3 TUDOR 4 PHD plant homeodomain a The previous three Chromo TUDOR PHD interact with methylated histone tails iii SANT domain 1 Interact preferentially with unmodified histone tails r 2 Example the protein HP1 contains a chromodomain that will bind to methylated lysine 9 of histone H3 but not to any other site of histone methylation iv How do the domains that recognize modified histones alter the function of the associated nucleosomes 1 One way is that modified histones recruit enzymes that will further modify adjacent nucleosomes a Many of the enzymes that acetylate histone tails called histone include 39 that recognize the same histone modifications that they create v Modified histones can also recruit other proteins that act on chromatin 1 Many nucleosomeremodeling complexes include one or more subunits with domains that recognize modified histones allowing modified histones to recruit these enzymes Specific Enzymes Are Responsible for Histone Modification i Histone acetyltransferases catalyze the addition of acetyl groups to histones whereas histone deacetylases remove these modifications 1 A number of different histone acetyltransferases and deacetylases have been identified and are distinguished by their abilities to target a different subset of histones or in some cases specific lysines in one histone ii Histone methyltransferases add methyl groups to histones and histone demethylases remove these modifications 1 Histone methyltransferases and demthylases always target only one of the many lysines or arginines on a specific histone iii Because these different modifications have different effects on nucleosome function the modification of a nucleosome with different histone acetyltransferases or methyltransferases can modulate chromatin structure and influence a wide array of DNA transaction iv Like the nucleosomeremodeling complex counterparts these modifying enzymes are part of large multiprotein complexes 1 Additional subunits play important roles in recruiting these enzymes to specific regions of the DNA v Similar to the U these 39 can be with transcription factors bound to DNA or directly with specifically modified nucleosomes The recruitment of these enzymes to particular DNA regions is responsible for the distinct patterns of histone modification observed along the chromatin and is a major mechanism for modulating the levels of gene expression along the eukaryotic chromosome and 39 quot U Work Together to Increase DNA Accessibility i The combination ofaminoterminal tail quotquot 39 and 39 39 quot U can dramatically change the accessibility of the DNA 1 The protein complexes involved in these modifications are frequently recruited to sites of active transcription ii Modification of aminoterminal tails can reduce the ability of nucleosome arrays to form repressive structures creating sites that can recruit other proteins including nucleosome remodelers a 1 Remodeling of the nucleosomes can then further increase the accessibility of the nucleosomal DNA to allow DNAbinding proteins access to their binding sites 8 Nucleosome Assembly 39 Are 39 39 39 39 quot 39 After DNA Replication i The 39 quot 39 ofa 39 requires rquot 39 of the DNA AND the V reassembly of the associated proteins on each daughter DNA molecule 1 The latter process is tightly linked to DNA replication to ensure that the newly replicated DNA is rapidly packaged into nucleosomes Although the replication of DNA requires the partial disassembly of the nucleosome the DNA is rapidly repackaged in an order series of events 1 DNA binding ofan H3H4 tetramer 2 Two H2AHZB dimers associate to form the final nucleosome 3 H1 joins this complex last presumably during the formation of higherorder chromatin assemblies To duplicate a chromosome at least half of the nucleosomes on the daughter chromosomes must be newly synthesized Two questions 1 Are all the old histones lost and only new histones assembled into nucleosomes 2 If not how are the old histones distributed between the two daughter chromosomes a The fate of the old histones is a particularly important issue given the effects that histone modification can have on the accessibility of the resulting chromatin If the old histones were lost completely then chromosome duplication would erase any quotmemoryquot of the previously modified nucleosomes i In contrast if the old histones were retained on a single chromosome that chromosome would have a distinct set of modifications relative to the other copy of the chromosome In experiments that differentially labeled old and new histones it was found that the old histones are present on both of the daughter chromosomes 1 Mixing is not entirely random a H3H4 tetramers and H2AHZB dimers are composed of either all new or all old histones i As the replication fork passes nucleosomes are broken down into their component subassemblies H3H4 tetramers appear to remain bound to one of the two daughter duplexes at random and are never released from DNA into the free pool of histones H2AHZB dimers are released and enter the local pool available for new nucleosome assembly 2 The distributive inheritance of old histones during chromosome duplication provides a mechanism for the propagation of the parental pattern of histone modification a By this mechanism old modified histones will tend to rebind one of the daughter chromosomes at a position near their previous position on the parental chromosome 57 N 57 Assembly of Nucleosomes Requires Histone quotChaperonesquot i The assembly of nucleosomes is not a spontaneous process ii The majority of the histones aggregate in a nonproductive form iii For correct nucleosome assembly it was necessary to raise salt concentrations to high levels gt1M NaCl then slowly reduce the concentration over many hours 1 This occurs in vitro and NOT in vivo iv Studies of nucleosome assembly under physiological salt concentrations identified factors required to direct the assembly of histones onto the DNA 1 These factors are negatively charged proteins that form complexes with either H3H4 tetramers or H2AH23 dimers and escort them to sites of nucleosome assembly a They act to keep histones from interacting with the DNA nonproductively these factors have been referred to as histone chaperones v How do the histone chaperones direct nucleosome assembly to sites of new DNA synthesis 1 Studies of the histone H3H4 tetramer chaperone CAFI reveal a likely answer a Nucleosome assembly directed by CAFI requires that the target DNA be replicating b Studies of CAFI dependent assembly have determined that the mark is a ringshaped sliding clamp protein called PCNA i The above will be discussed in detail in Chapter 8 9 Chapter Table Details a 57 Squot 0 5 Table 71 Variation in Chromosome Makeup in Different Organisms Page 137 i Chromosome Form 1 Prokaryotes are circular 2 Eukaryotes are linear ii Number of Chromosomes increase with the complexity of the organisms iii Genome size increases with the complexity of the organism Table 72 Comparison of the Gene Density in Different Organisms39 Genomes Page 139 i Genome size increases in organism complexity ii Number of genes increases in organism complexity iii Gene density DECREASES in organism complexity Table 73 Contribution of ntrons and Repeated Sequences to Different Genomes i Average number of introns per gene in E coli is ZERO very low numbers in eukaryotes Percent of repetitive DNA is significantly larger in eukaryotes highest in homo sapiens Table 75 General Properties of the Histones Page 159 i Core histones are lighter in weight 1100015400 gmol than linker histone 20800 ii Linker histones have greater percentage of lysine and arginine than core 32 20 24 Table 77 Histone Modifications Page 183 i H2A 1 Phosphorylation 2 Acetylation ii H2B r 1 Phosphorylation 2 Acetylation iii H3 1 Phosphorylation 2 Methylation 3 Acetylation iv H4 1 Phosphorylation 2 Methylation 3 Acetylation Table 79 Properties of Histone Chaperones i CAF1 1 4 subunits 2 Binds histones H3H4 3 Interacts with sliding clamp ii HIRA 1 4 Subunits 2 Binds histones H3H4 3 Does not interact with sliding clamp iii RCAF 1 1 subunit 2 Binds histones H3H4 3 Does not interact with sliding clamp iv NAP1 1 1 subunit 2 Binds histones H2AH2B 3 Does not interact with sliding clamp 10 Chapter Figure Details a 57 Squot 0 Figure 72 Comparison of Chromosomal Gene Density for Different Organisms Page 140 i Largest subunit of RNA polymerase in eukaryotic cells is RNA polymerase M ii As the organism complexity increases the number of genes decreases Figure 73 Schematic of RNA Splicing Page 141 i Transcription of prem RNA is initiated at an exon and proceeds in a general direction ii This transcript is then processed by splicing to remove noncoding introns to produce mRNA Figure 74 Organization and Content of the Human Genome i Out of the 3200Mb of the human genome only 48Mb are composed of genes ii Most of the genome is comprised of intergenic DNA Figure 75 Processed Pseudogenes Arise from Integration of ReverseTranscribed mRNA Page 144 i When reverse transcriptase is present in a cell mRNA molecules can be copied into doublestranded DNA 1 In rare instances these DNA molecules can integrate into the genome creating pseudogenes Pseudogenes have the common characteristic of lacking introns 1 This is because introns are rapidly removed from newly transcribed RNAs iii Pseudogenes lack the appropriate promoter sequences to direct their transcription as these are not part of the mRNA from which they are derived 5 r Figure 76 Centromeres Origins of the Replication and Telomeres are Required for Eukaryotic Chromosome Maintenance Page 145 i Each eukaryotic chromosome includes 1 Two teomeres a Located at each end of the chromosome 2 One centromere a Can be found anywhere on chromosome 3 Many origins of replication a Can be found at numerous locations across the length of the chromosome Figure 77 More or Less Than One Centromere Leads to Chromosome Loss or Breakage i No centromere leads to random segregation of an entire chromosome ii More than one centromere leads to chromosomal breakage Figure 78 C Size and C Vary T quot Among Different Organisms i The increase in organism complexity yields an increase in the size of the centromere 1 In humans it ranges from 240 kb to several Mb Figure 710 Eukaryotic Mitotic Cell Cycle i Four stages 1 S a Chromosomalreplication a Establishes that the level of nutrients is sufficiently high to allow the completion of cell division a Allow the cell to prepare for next cell division phase a Chromosomalsegregation Figure 711 Events ofS Phase i Two main chromosomal events occur during S Phase 1 DNA replication copies each chromosome completely 2 After replication sisterchromatid cohesion is established by ringshaped cohesion molecules Figure 712 Events of M Phase Mitosis Phase i Three major events Two kinetochores of each linked sisterchromatic pair attach to opposite poles of mitotic spindle 2 Once all kinetochores are bound to opposite poles sisterchromatid cohesion is eliminated by destroying cohesion ring 3 After cohesion is quot 39 39 sister 39 39 39 are D g Ito opposite poles of the mitotic spindle Figure 713 Changes in Chromatin Structure i Chromosomes are maximally condensed in M phase and decondensed throughout the rest of the cell cycles 1 Decondensed stages are referred to as interphase Figure 714 Model for the Structure and Function of Cohesions and Condensins 3 3 9 3955 Q E Iquot C lt i Cohesions and condensins are ringshaped protein complexes that include two SMC proteins that play important roles in bringing distant or different regions of DNA together Figure 720 Assembly ofa Nucleosome i Initiated by the formation of a H32H42 tetramer ii Tetramer then binds to doublestranded DNA iii H32H42 tetramer bound to DNA recruits two copies of the H2AHZB dimer to complete structure Figure 721 AminoTerminal Tails of the Core Histones are Accessible to Proteases i Treatment of nucleosomes with limiting amounts of proteases that cleave after basic amino acids specifically removes the aminoterminal quottailsquot leaving the histone core intact Figure 723 Interactions of the Histones with Nucleosomal DNA i H3H4 bind the middle and ends of the DNA ii H2AHZB bind 30bp of DNA on one side of the nucleosome Figure 727 Histone H1 binds Two DNA Helices i H1 binds to the linker DNA at one end of the nucleosome and the central DNA helix of the nucleosome bound DNA Figure 729 Histone H1 Induces Tighter DNA Wrapping Around the Nucleosome i Histone H1 can interact with each nucleosome inducing a tighter coil Figure 730 Two Models for the 30nm Chromatin Fiber i Solenoid 1 Linker DNA does not pass through the central axis of the superhelix and that the sides and entry and exit points of the nucleosomes are relatively inaccessible ii Zigzag 1 Linker DNA frequently passes through the central axis of the fiber and the sides and even the entry and exit points are more accessible Figure 733 Alteration of Chromatin by Incorporation of Histone Variants i Incorporation of CENPA in place of Histone H3 is proposed to act as a binding site for one or more protein components of the kinetochore Figure 734 Model for Gaining Access to NucleosomeAssociated DNA i The ability of sequencespecific DNAbinding proteins to bind nucleosomes suggest that unwrapping of the DNA from the 39 is I quot 39 for 39 quot39 o the DNA 1 DNA sites closest to the entry and exit points are the most accessible and sites closest to the midpoint of the bound DNA are least accessible Figure 735 Nucleosome Movement Catalyzed by NucleosomeRemodeling Activities i Nuclosome movement by sliding along a DNA molecule exposes sites for DNA building proteins ii Nucleosome movement can occur by transfer of the nucleosome from one strand of DNA to another Figure 736 Model for Nucleosomal DNA Sliding Catalyzed by NucleosomeRemodeling Complexes i ATPhydrolyzing subunit of the nucleosomeremodeling complex binds the nucleosomal DNA two helical turns from the central dyad 5 gtlt lt E The nucleosomeremodeling complex using the ATP dependent DNA translocating activity first pulls the DNA from the nearest linker domain into the nucleosome a This breaks the five histoneDNA contacts between the ATPhydrolyzing subunit and the linker DNA The broken contacts then reform with the translocated DNA leaving too much DNA associated with the nucleosome next to the ATP hydrolyzing subunit i To relieve this strain the model proposes that the additional sequences move like a quotwavequot across the surface of the histones breaking one contact at a time until all the contacts have reformed with the appropriate amount of DNA between them Figure 737 Two Modes of DNABinding Protein Dependent Nucleosome Positioning i Association of many DNAbinding proteins with DNA is incompatible with the association of the same DNA with the histone octamer 1 Because a nucleosome requires MORE THAN 147bp of DNA to form if two such factors bind to the DNA less than this distance apart the intervening DNA can not assemble into a nucleosome ii A subset of DNAbinding proteins have the ability to bind to nucleosomes 1 Once bound to DNA such proteins will facilitate the assembly of nucleosomes immediately adjacent to the protein39s DNA binding site Figure 738 Nucleosomes Prefer to Bind Bent DNA i Because the DNA is bent severely during association with the nucleosome DNA sequences that position nucleosomes are intrinsically bent 1 AT base pairs have an intrinsic tendency to bend toward the minor groove and GC base pairs have the opposite tendency a Sequences that alternate between AT and GC rich sequences with a periodicity of 5 bp will act as preferred nucleosomebinding sites Figure 741 Chromatin Remodeling and HistoneModifying Complexes Work Together to Alter Chromatin Structure i Sequences specific DNA binding proteins typically recruit these enzymes to specific regions of a chromosome ii The DNAbinding protein first recruits a histone acetyltransferase that modifies the adjacent nucleosomes increasing the accessibility of the associated DNA by locally converting the chromatin fiber from the 30nm fiber to the most accessible 10 nm form 1 Allows the binding of a second DNAbinding protein that recruits a nucleosome remodeling complex iii quot 39 of the 39 39 quot U complex facilitates the sliding of the adjacent nucleosomes which allows the binding site for a third DNAbinding protein to be exposed 1 This could be the binding site for the TATAbinding protein at a start of transcription Figure 742 Inheritance of Histones After DNA Replication i As the chromosome is replicated histones that were associated with the parental chromosome are differently distributed 1 Histone H3H4 tetramers are randomly transferred to one of the two daughter strands but do not enter into the soluble pool of H3H4 tetramers 57 a Newly synthesized H3H4 tetramers form the basis of the nucleosomes on the strand that does not inherit the parental tetramer HZA and HZB dimers are released into the soluble pool and complete for H3H4 association with newly synthesized HZA and HZB a As a consequence on average every second H3H4 tetramer on newly synthesized DNA will be derived from the parental chromosome N b These tetramers will include all of the modifications added to the parental nucleosomes c The H2AHZB dimers are more likely to be derived from newly synthesized protein aa Figure 743 Inheritance of Parental H3H4 Tetramers Facilitates the Inheritance of Chromatin States i As a chromosome is replicated the distribution of the parental H3 H4 tetramers results in the daughter chromosomes receiving the same modification as the parent 1 The ability of these modifications to recruit enzymes that perform the same modifications facilitates the correct propagation of the same state of modification to the two daughter chromosomes bb Figure 744 Chromatin Assembly Factors Facilitate the Assembly of Nucleosomes i After the replication fork has passed chromatin assembly factors chaperone free H3H4 CAFI tetramers and H2AHZB NAPl dimers to the site of newly replicated DNA 1 Once at the newly replicated DNA these factors transfer their histones contents to the DNA a CAFI is recruited to the newly replicated DNA by interactions with DNA sliding clamps i These ringshaped auxiliary replication factors encircle the DNA and are released from the replication machinery as the replication fork moves 1 Chapter 9 The Mutability and Repair of DNA 2 Replication Errors and Their Repair a The Nature of Mutations 0quot 0 i iii The simplest mutations are switches of one base for another two kinds 1 Transitions a Pyrimidine to pyrimidine b Purine to purine 2 Transversions a Pyrimidine to purine b Purine to pyrimidine Mutations that alter a single nucleotide are called point mutations Extensive insertions and deletions and gross rearrangements of chromosome structure cause drastic changes in DNA 1 Insertion of a transposon which typically places many thousand nucleotides of foreign DNA in the coding or regulatory sequences of a gene or by the aberrant actions of cellular recombination processes The overall rate at which new mutations arise spontaneously at any given site on the chromosome ranges from about 10396 to 10391 per round of DNA replication One kind of sequence that is particularly prone to mutation are repeats of simple di tri or tetranucleotide sequences known as DNA microsatellites 1 Stretches of CA repeats are found at many widely scattered sites in the chromosomes of humans and some other eukaryotes 2 The replication machinery has difficulty copying such repeats accurately and undergoes slippage Some Replication Errors Escape Proofreading 1 Within the proofreading exonuclease some misincorporated nucleotides become a mismatch between the newly synthesized strand and the template strand 1 Three dilTerent nucleotides can be misincorporated opposite each of the four kinds of nucleotides in the template strand T G or C opposite a T in the template for atotal of 12 possible mismatches TT TG TC etc The error will only become permanent of the DNA undergoes a SECOND round of synthesis 1 Upon second S phase the error will be corrected but the incorrect base pairs originally will have caused a mutation Mismatch Pair removes Errors that Escape Proofreading i Mismatch repair system increases the accuracy of DNA synthesis by an additional 23 orders of magnitude Proofreading exonuclease increasing it originally by 100 orders This system faces two challenges 1 It must scan the genome for mismatches a Mismatches are transient eliminated following a second S phase the mismatch repair system must rapidly find and repair mismatches 2 The system must correct the mismatch accurately a It must replace the misincorporated nucleotide in the newly synthesized strand and not the correct nucleotide in the parental strand iii In E coli mismatches are detected by a dimer of the mismatch repair protein MutS l MutS scans the DNA recognizing mismatches from the distortion in the DNA backbone 2 MutS embraces the mismatchcontaining DNA inducing a pronounced kink inthe DNA and a conformational change in MutS itself 3 The key to the speci city is that DNA containing a mismatch is more readily distorted than properly base paired DNA 4 MutL is recruited by MutS a MutL activates MutH i Enzyme that causes an incision or nick on one strand near the site of the mismatch l Nicking is followed by the action ofa specific helicase Uer and one of three exonucleases 5 How does E coli mismatch repair system know which of the two mismatched nucleotides to replace a E 001139 tags the parental strand by transient hemimethylation i The E coli enzyme Dam methylase methylates A residues on both strands of the sequence 539 GATC3 That sequence is widely distributed along the entire genome occuring once every 256bp attaches the CH3 on the T ii When a replication fork passes through DNA that is methylated at GATC sites on both strands fully methylated DNA the resulting daughter DNA duplexes will be hemimethylated iii Ifthe new strand is lacking a methyl group it is then recognized for repair 6 MutH binds to hemimethylated sites but its endonuclease activity is latent a Only when it is contacted by MutL and MutS located at a nearby mismatch which can be within a distance of a hundred or more base pairs does MutH become activated b How Recent evidence indicates that the MutSMutL compleX leaves the mismatch and moves along the DNA contour to reach MutH at the site of hemimethylation Once activated MutH selectively nicks the unmethylated strand so only newly synthesized DNA in the vicinity of the mismatch is removed and replaced i Methylation is therefore a quotmemoryquot device that enables the E coli repair system to retrieve the correct sequence from the parental strand if an error has been made during replication 7 Different exonucleases are used to remove ssDNA between the nick created by MutH and the mismatch depending on whether MutH cuts the DNA on the 539 or the 339 side of the misincorporated nucleotide 0 iv a If the DNA is cleaved on the 539 side of the mismatch then exonuclease VII or RecJ which degrades DNA is a 539 gt3 direction removes the stretch of DNA from the MutHinduced cut through the misincorporated nucleotide If the nick is on the 3 side of the mismatch then the DNA is removed by exonuclease I which degrades DNA in a 339 gt5 direction i DNA Pol III lls in the missing sequence Eukaryotic cells also repair mismatches and do so using homologs to MutS called MSH proteins for MutS homolo gs and MutL called MLH and PMS Eukaryotes have multiple MutSlike proteins with different speci cities l a One is speci c for simple mismatches whereas another recognizes small insertions or deletions resulting from quotslippagequot during DNA replication 2 Evidence that mismatch repair has a critical role in higher organisms came from the discovery that a genetic predisposition to colon cancer is due to a mutation in the genes for human homologs of MutS speci cally the MSH2 homolog and MutL Eukaryotic cells lack MutH and E coli39s clever trick of using hemimethylation to tag the parental strand LA 5 3 DNA Damage a Most bacteria lack Dam methylase and are also unable to use hemimethylation to mark the newly synthesized strand How then does the mismatch repair system know which of the two strands to correct a 0quot Laggingstrand synthesis takes place discontinuously with the formation of Okazaki fragment that are joined to previously synthesized DNA by DNA ligase i Prior to the ligation step the Okazaki fragment is separated from previously synthesized DNA by a nick which can be though of as being equivalent to the nick created in E coli by MutH on the newly synthesized strand Recent results indicate that human homologs of MutS MSH interact with the sliding clamp component of the replisome PCNA and would thereby by recruited to the site of discontinuous DNA synthesis on the lagging strand i Interaction with the sliding clamp could also recruit mismatch repair proteins to the 3 growing end of the leading strand a DNA Undergoes Damage Spontaneously from Hydrolysis and Deamination i Mutations arise not only from errors in replication but also from damage to the DNA ii Some damage is caused by environmental factors such as radiation and so called mutagens which are chemical agents that increase the rate of mutation cytosine DNA also undergoes spontaneous damage from the action of water The most frequent and important kind of hydrolytic damage is deamination of vi 1 Under normal physiological conditions cytosine undergoes spontaneous deamination thereby generating the unnatural in DNA base Uracil Uracil preferentially pairs with Adenine and so introduces that base in the opposite strand upon replication rather than the G that would have been directed by C Adenine and Guanine are also subject to spontaneous deamination l Deamination converts Adenine to Hypoxanthine which hydrogen bonds to Cytosine rather than to Thymine 2 Guanine is converted to Xanthine which continues to pair with Cytosine although with only two hydrogen bonds DNA also undergoes depurination by spontaneous hydrolysis of the N glycosidic linkage and produces an abasic site deoxyribose lacking a base in the DNA All the above reactions result in alterations to the DNA that are unnatural Apurinic sites are unnatural and each of the deamination reactions generates an unnatural base 1 This allows changes to be recognized by the repair systems 2 This also suggests an explanation for why DNA has thymine instead of Uracil a If DNA naturally contained Uracil instead of Thymine then deamination of Cytosine would generate a natural base which the repair systems could not easily recognize The hazard of having deamination generate a naturally occurring base is illustrated by the problem caused by the presence of 5methylcytosine 5mC l Vertebrate DNA frequently contain 5mC in place of cytosine as a result of the action of methyltransferases This modified base has a role in the transcriptional silencing Deamination of 5mC generates Thymine which will not be recognized as an abnormal base and following a round of DNA replication can become fixed as a C to T transition 4 Methylated C s are hot spots for spontaneous mutations in vertebrate DNA LAN b DNA is Damaged by Alkylation Oxidation and Radiation i ii Alkylation 1 Methyl or ethyl groups are transferred to reactive sites on the bases and to phosphates in the DNA backbone 2 Alkylating chemicals include nitrosamines and N methylNInitroN nitrosoguanidine 3 One 0 the most vulnerable sites of alkylation is the oxygen of carbon atom 6 of guanine a The product of this methylation 06methylguanine mispairs with Thymine resulting in the change ofa GC base pair into an AT base pair when the damaged DNA is replicated Reactive Oxygen Species 0 H202 OH 1 Potent oxidizing agents are generated by ionizing radiation and by chemical agents that generate free radicals 2 Oxidation of Guanine generates 78dihydro8oxoguanine or oxoG C iii iv 9 OxoG is highly mutagenic because it can basepair with Adenine as well as Cytosine i If it pairs with Adenine during replication it gives rise to a GC to TA transversion which is one of the most common mutations found in human cancers 1 Perhaps the carcinogenic effects of ionizing radiation and oxidizing agents are partly caused by free radicals that convert Guanine to oxoG ii Nothing mentioned in book about the consequences of oxoG binding to Cytosine UV light 1 Radiation with a wavelength of approximately 260nm is strongly absorbed by the bases and the main consequence of which is the photochemical fusion of two pyrimidines that occupy adjacent positions on the same polynucleotide chain a In the case of two Thymine s the lsion is called a thymine dimer which comprises a cyclobutane ring generated by links between carbon atoms 5 and 6 of adjacent thymine39s b In the case of Thymine adjacent to a Cytosine the resulting fusion is a ThymineCytosine adduct in which the Thymine is linked via its carbon atom 6 to the carbon atom 4 of Cytosine i These linked bases are incapable of base pairing and cause the DNA polymerase to stop during replication Radiation l N L yradiation and Xrays ionizing radiation are particularly hazardous because they cause doublestranded breaks in the DNA which are difficult to repair a Ionizing radiation can directly attack ionize the deoxyribose in the DNA backbone b It can also attack indirectly by generating reactive oxygen species which in turn react with the deoxyribose subunits Since cells require intact chromosomes to replicate their DNA ionizing radiation is used therapeutically to kill rapidly proliferating cells in cancer treatment a Certain anticancer drugs Bleomycin cause breaks in the DNA Ionizing radiation and agents like Bleomycin that cause DNA to break are referred to as quotclastogenicquot agents a Greek klastos is broken Mutations Are Also Caused by Base Analogs and Intercalating Agents Mutations are also cause by two compounds i ii 1 2 Base Analogs a Compounds that substitute for normal bases Intercalating Agents a Compounds that slip between the bases Base Analogs l Structurally similar to proper bases but differ in ways that make them treacherous to the cell a They are similar enough to the proper bases to get taken up by cells and converted into nucleoside triphosphates ad incorporated into DNA during replication b Due to the structural di erences between these analogs and the proper bases the analogs basepair inaccurately leading to frequent mistakes during the replication process 2 5 br0m0uracil a One of the most mutagenic base analogs analog to Thymine b The presence of the bromo substituent allows the base to mispair with guanine via the enol tautomer i The keto tautomer is strongly favored over the enol iii Intercalating Agents 1 Flat molecules containing several polycyclic rings that bind to the equally at purine or pyrimidine bases of DNA just as the bases bind or stack with each other in the double helix Pro avin Acridine and Ethidium more commonly known as Ethidium bromide cause the deletion of addition of a base pair or even a few base pairs a When deletions or additions arise in a gene they can have profound consequences on the translation of its mRNA because they shi the coding sequence out of its proper reading frame 3 How do intercalating agents cause short insertions and deletions a One possibility in the case of insertions is that by slipping between the bases in the template strand these mutagens cause the DNA polymerase to insert an extra nucleotide opposite the intercalated molecule i Intercalation of one of these structures approximately doubles the typical distance between two base pairs b In the case of deletions the distortion to the template caused by the presence of an intercalated molecule might cause the polymerase to skip a nucleotide 4 Repair of DNA Damage a The most direct DNA repair system is a repair enzyme that simply reverses the damage i One more elaborate step involves excision repair systems in which the damaged nucleotide is not repaired but removed from the DNA ii The undamaged strand serves as a template for reincorporation of the correct nucleotide by DNA polymerase iii Two types exist 1 One involving the removal of ONLY the damaged nucleotide 2 The other involving the removal of a short stretch of ssDNA that contains the lesion b More elaborate is recombinational repair which is employed when both strands are damaged as when the DNA is broken N 0 2391 FD i One strand cannot serve as a template for the repair of the other ii Recombinational repair is known as double strand break repair iii In recombinational repair sequence information is retrieved from a second undamaged copy of the chromosome When progression of a replicating DNA polymerase is blocked by damaged bases a special translesion polymerase copies across the site of the damage in a manner that does NOT depend on the base pairing between the template and the newly synthesized DNA strands i THIS MECHANISM IS A SYSTEM OF LAST RESORT BECAUSE TRAN SLESION SYNTHESIS IS INEVITABLY HIGHLY ERRORPRONE MUTAGENIC Direct Reversal of DNA Damage i An example of repair by simple reversal of damage is photoreactivation 1 This directly reverses the formation of Pyrimidine dimers that result from UV irradiation The enzyme DNA photolyase captures energy from light and uses it to break the covalent bonds linking adjacent pyrimidines the damaged bases are mended directly ii Another example of direct reversal is the removal of the methyl group from the methylated base 06methylguanine l Methyltransferase removes the methyl group from the Guanine residue by transferring it to one of its own Cysteine residues 2 This is VERY COSTLY to the cell because the methyltransferase is not catalytic having once accepted a methyl group it can never be used again Base Excision Repair Enzymes Removed Damaged Bases by a BaseFlipping Mechanism i The most prevalent way in which DNA is cleansed of damaged bases is by repair systems that remove and replace the altered bases ii The two principal repair systems are 1 Base excision repair 2 Nucleotide excision repair Base Excision Repair 1 Uses an enzyme called glycosylase to recognize and remove the damaged base by hydrolyzing the glycosidic bond a The resulting abasic sugar is removed from the DNA backbone in a further endonucleolytic step i Endonucleolytic cleavage also removes apurinic and apyrimidinic sugars that arise by spontaneous hydrolysis b After the damaged nucleotide has been entirely removed from the backbone a repair DNA polymerase and DNA ligase restore an intact strand using the undamaged strand as a template 2 DNA glycosylases are lesionspecific and cells have multiple DNA glycoslyases with different specificities a A specific glycosylase recognizes Uracil generated as a consequence of deamjnation of Cytosine and another is responsible N iii for removing oxoG generated as a consequence of oxidation of Guanine b Atotal of 8 different DNA glycosylases have been identified in the nuclei of human cells iv How do DNA glycosylases detect damaged bases while scanning the genome 1 Evidence indicates that these enzymes diffuse laterally along the minor groove of the DNA until a specific kind of lesion is detected v But how is the enzyme able to act on the base it if is buried in the DNA helix 1 Xray crystallography studies reveal that the damaged base is ipped out so that it projects away from the double helix where it sits in the specificity pocket of the glycosylase The double helix is able to allow base ipping with only modest distortion to its structure and hence the energetic cost of base ipping is not favorable vi It is unlikely that glycosylases slip out every base to check for abnormalities as they diffuse along DNA vii What is the damaged base is not removed by base excision before DNA replication Does this inevitable mean that the lesion will cause a mutation 1 In the case of oxoG which has the tendency to mispair with Adenosine a failsafe system exists N a A dedicated glycosylase recognizes oxoGA base pairs generated by misincorporation of an Adenosine opposite an oxoG on the template strand b Glycosylase REMOVES THE ADENOSINE i The repair enzyme recognizes an Adenosine opposite an oxoG as a mutation and removes the undamaged but incorrect base pair 2 Another example of a failsafe system is a glycosylase that removes T opposite a G a Such a TG mismatch can arise by spontaneous deamination of 5mC which occurs frequently in the DNA of vertebrates b Because both T and G are normal bases how can the cell recognize which is the INCORRECT base i The glycosylase system assumes that the T in the TG mismatch arose from deamination of 5mC and selectively removes the T so that it can be replaced with a C f Nucleotide Excision Repair Enzymes Cleaved Damaged DNA on Either Side of the Lesion i Unlike base excision repair the nucleotide excision repair enzymes do NOT recognize any particular lesion 1 This system works by recognizing distortions to the shape of the DNA helix cause by variations such as thymine dimers presence of a bulky chemical adduct on a base etc a Such distortions trigger a chain of events that lead to the removal of a short singlestrand segment patch that includes the lesion b This removal creates a singlestrand gap that is lled in by DNA polymerase using the undamaged strand as a template and thereby restoring the sequence ii Nucleotide excision repair in E coli 1 Largely accomplished by four proteins a UvrA b UvrB c UvrC d Uer A complex of UvrA and UvrB scans the DNA with UvrA being responsible for detecting distortions to the helix Upon encountering a distortion UvrA exits the complex and UvrB melts the DNA to create a singlestranded bubble around the lesion UvrB recruits UvrC with UvrC creating two incisions a One located eight nucleotides away on the 5 side of the lesion b One located four to ve nucleotides away on the 339 side of the lesion These cleavages create a 1213 residuelong singlestranded DNA segment which is made accessible by the action of the DNA helicase N L 5 V39 Uer 6 DNA Pol I and DNA ligase ll in the resulting gap iii Principle of nucleotide excision repair in higher cells is much the same as that in E 001139 but the machinery for detecting excising and repairing the damage is more complicated involving 25 or more polypeptides l XPC which is responsible for detecting distortions to the helix similar to UvrA 2 The DNA is opened to create a bubble around the lesion a Formation of the bubble involves the helicase activities of the proteins XPA and SPD equivalent to UvrB and the singlestrand binding protein RPA The bubble creates cleavage on the 539 side of the lesion for a nuclease known as ERCClXPF and on the 3 side for the nuclease XPG XPG similar to UvrC 4 The resulting singlestranded DNA segment is 2432 nucleotides long 5 As in bacteria the DNA segment is released to create a gap that is lled in by the action of DNA polymerase and ligase iv UVR proteins are needed to mend damage from UV light mutants of the uvr genes are sensitive to UV light and lack the capacity to remove Thymine Thymine and ThymineCytosine adducts v Humans can exhibit a genetic disease called xeroderma pigmentosum Render af icted individuals highly sensitive to sunlight and results in skin lesions including skin cancer Seven genes referred to as XP genes have been identi ed in which mutations give rise to xeroderma pigmentosum a These genes correspond to proteins such as XPA XPC XPD XPF and XPG in the human pathway for nucleotide excision repair U N underscoring the importance of nucleotide excision repair in mending damage from UV light vi Not only is nucleotide excision repair capable of mending damage throughout the genome but it is also capable of rescuing RNA polymerase the progression of which has been arrested by the presence of a lesion in the transcribed templatestrand of a gene 1 This is known as transcription coupled repair and involves recruitment to the stalled RNA polymerase of nucleotide excision repair proteins a The significance is that it focuses repair enzymes on DNA genes being actively transcribed i In effect RNA polymerase serves as another damagesensing protein in the cell b Central to the transcriptioncoupled repair in eukaryotes is the general transcription factor TFIIH i Unwinds the DNA template during the initiation of transcription ii Subunits include the DNA helixopening proteins XPA and SPD iii Responsible for two separate functions 1 Its standseparating helicases melt the DNA around a lesion during nucleotide excision repair including transcriptioncoupled repair 2 Help to open the DNA template during the process of gene transcription g Recombination Repairs DNA Breaks by Retrieving Sequence Information From Undamaged DNA i Excision repair uses the undamaged DNA strand as a template to replace a damaged segment of DNA on the other strand ii How do cells repair doublestranded breaks in DNA in which both strands of the duplex are broken 1 This is done by the double strand break DSB repair pathway which retrieves sequence information from the sister chromosome a Because of its central role in general homologous recombination as well as in repair the DSB repair pathway is extremely important covered in Chapter 10 Molecular Biology II 2 DNA recombination also helps to repair errors in DNA replication a Consider A replication fork that encounters a lesion in DNA that has NOT been corrected by nucleotide excision repair b The DNA polymerase will sometimes stall attempting to replicate over the lesion Although the template strand cannot be used the sequence information can be retrieved from the other daughter molecule of the replication fork by recombination Chapter 10 Molecular Biology II iii Once the recombinational repair is complete the nucleotide excision system has another opportunity to repair the error 1 Consider Situation in which the replication fork encounters a nick in the DNA template Passage of the fork over the nick will create a DNA break repair of which can only be accomplished by the DSB repair pathway h DSBs in DNA are Also Repaired by Direct Joining of Broken Ends 1 iii A DSB is the most cytotoxic of all kinds of DNA damage 1 If not fixed a DNA break can have multiple deleterious consequences such as blocking replication and causing chromosome loss resulting in cell death or neoplastic transformation The DSB repair pathway relies on DNA sequence information in a sister chromosome to repair broken DNA molecules 1 This is effective because the sister chromosome provides atemplate for the precise restoration of the original sequence across the site of the break a In yeast cells DSB repair is the principal pathway by which breaks are mended What happens early in the cell cycle before two sister chromosomes have been generated by DNA replication 1 If a stillunreplicated chromosome suffers a break then no sister chromosome is present to serve as a template in the DSB repair pathway a Alternative system comes into play known as nonhomologous end joining or NHEJ NHEJ 1 Backup system in yeast but in higher cells it is the principal pathway by which breaks are repaired 2 The machinery for carrying out NHEJ protects and processes the broken ends and then joins them together a Because sequence information is lost from the broken ends the original sequence across the break is not faithfully restored during NHEJ and therefore NHEJ is mutagenic What is the mechanism that joins DNA ends together in NHEJ a The two ends of the broken DNA are joined to each other by misalignment between single strands producing from the broken ends b This misalignment is believed to occur by pairing between tiny stretches as short as lbp of complementary bases serendipitous microhomologies c Singlestrand tails are removed by nucleases and gaps are filled in by DNA polymerase 4 Seven components of the NHEJ pathway have been discovered in mammalian cells U a Ku70 b Ku80 c DNAPKcs d Artemis e XRCC4 f CemunnosXLF g DNA Ligase IV i i Ku70 and Ku80 are the most lndamental components of 1 They constitute a heterodimer that binds the DNA ends and recruits DNAPKcs which is a protein kinase ii DNAPKcs forms a complex with Artemis 1 Artemis is BOTH a 539 to 3 exonuclease and a latent endonuclease that is activated by phosphorylation by DNAPKcs 2 These activities process the broken ends and prepare them for ligation iii Ligation is carried out by Ligase IV in a complex with XRCC4 and CernunnosXLF 5 NHEJ is widely present in eukaryotic organisms but occurs less frequently in bacteria 6 Anomaly a Spores of the bacterium Bacillus subtilis or B subtilis produces a Kulike protein and a DNA ligase when it sporulates and packages the proteins into the mature spore b Ku and the ligase repair DNA breaks when the spore germinates i Mutant spores lacking these proteins are highly susceptible to dry heat a condition known to break DNA c Upon germination heated mutant spores are unable to resume grth because they are unable to rejoin the heatinduced breaks d Germinating spores rely on NHEJ instead on DSB repair e Spores only have ONE CHROMOSOME i Therefore they CANNOT rely on sister chromosomes to use as a template for repair of the break ii The chromosome is tightly coiled in a doughnutlike structure could hold the ends of DNA breaks in DNA in close proximity to each other 1 This close juxtaposition could facilitate correct rejoining of ends even if the chromosome has multiple breaks Translesion DNA Synthesis Enables Replication to Proceed Across DNA Damage 1 iii DNA is generally mended by excision followed by resynthesis using an undamaged template Sometimes DNA polymerase encounters a lesion that has been missed such as a pyrimidine dimer or an apurinic site 1 The polymerase must force itself to copy the DNA anyway or cease replication Even it cells cannot repair these lesions there is a failsafe mechanism that allows the replication machinery to bypass these sites of damage a process known as translesion synthesis 1 Highly error prone and likely to introduce mutations this mechanism is favored over apoptosis Translesion Synthesis l Catalyzed by a specialized class of DNA polymerases that synthesize DNA directly across the site of the damage a In E coll it is carried out by proteins UmuC and UmuD i UmuC is a distinct family of DNA polymerases found in many organisms known as the Y family of DNA polymerases ii An important feature of these polymerases is that although they are templatedependent they incorporate nucleotides in a manner that is independent of base pairing 1 This explains how the enzymes can synthesize DNA over a lesion on the template strand 2 Due to the fact that the enzyme is not reading the template strand it is highly errorprone 2 Due to a high rate of error similar to that of NHEJ this is a last resort system a Granted it enables the cell to survive but it induces a high level of mutagenesis i This causes translesion polymerase to NOT BE PRESENT in E 001139 under normal circumstances and only activated in response to DNA damage 3 The steps are as follows a DNA polymerase 111 along with the Sliding Clamp noted as the B clamp is synthesizing DNA until it hits a lesion b DNA Pol 111 along with the sliding clamp fall off and DNA Pol IV Din B or DNA Pol V UmuD39zC attaches and copies over the lesion c As soon as the lesion is copied DNA Pol IV or V fall off and DNA Pol III and the sliding clamp reattach and continue synthesizing 4 The genes encoding the translesion polymerase are expressed as part of a pathway known as the SOS Response a Damage leads to the proteolytic destruction of a transcriptional repressor LexA repressor that controls expression of genes involved in the SOS response including those for UmuC and UmuD and then inactive precursor for UmuD i The same pathway is also responsible for the proteolytic conversion of UmuD to UmuD Cleavage of LexA and UmuD are both stimulated by a protein called RecA which is activated by singlestranded DNA resulting from DNA damage i RecA is a dualfunction protein that is also involved in DNA recombination Discussed in Chapter 10 Molecular Biology 11 How atranslesion polymerase gains access to athe stalled replication machinery at the site of DNA damage 1 In mammalian cells entry into the translesion repair pathway is triggered by chemical modification of the sliding clamp also known as PCNA in eukaryotes which anchors the replicative polymerase to the DNA template 0quot N U 5 VI The chemical modi cation is the covalent attachment to the sliding clamp of a peptide known as ubiquitin in a process known as ubiquitination a Widely used in eukaryotic cells to mark proteins for various processes mostly degradation Once ubiquitinated the sliding clamp recruits a translesion polymerase which contains domains that recognize and bind to ubiquitin The translesion polymerase displaces the replicative polymerase from the 3 end of the growing strand and extends it across the site of damage UBIQUITINATION OF THE SLIDING CLAMP IS THEREFORE A DISTRESS SIGNAL THAT RECRUITS A TRAN SLESION PROLYMERASE TO RESCUE A REPLICATION MACHINERY THAT IS STALLED AT A SITE OF DNA DAMAGE
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