Exam 1 Bio 209 Review (Drexel University) Fall 2015 PDF

lOMoARcPSD|34711042 Exam 1 bio 209 review Cell, Mol & Dev Bio I (Drexel University) Scan to open on Studocu Studocu is not sponsored or endorsed by any college or university Downloaded by Mimi Kreger ([email protected]) ...

lOMoARcPSD|34711042 Exam 1 bio 209 review Cell, Mol & Dev Bio I (Drexel University) Scan to open on Studocu Studocu is not sponsored or endorsed by any college or university Downloaded by Mimi Kreger ([email protected])  lOMoARcPSD|34711042 Biology 209 CMDB I Fall 2015 Midterm Exam 1 Study Guide The outline below doesn’t necessarily describe 100% of what you’ll need to know to do well on the first midterm exam, but it should give you a good idea of what’s important. THIS IS NOT A LEGAL DOCUMENT! There will be ~50-60 questions on the exam. Most will be multiple-choice with a few true/false and matching questions. The questions involve a good level of detailed understanding of the lectures/notes. Make sure that you understand details of things like amino acid & nucleic acid structures & facts, examples of concepts covered, and specific terms covered in class. Remember, I won’t test on material in the book not covered in the lectures. Amino acids & Proteins:  For all 20 amino acids: know complete structures, full names, three-letter names, single letter code names (i.e. Aspartic acid = Asp = D), chemical characteristics  Know peptide bond structure, chemical features associated with this bond o Tend to be planar and rigid due to double bond character  Know given examples of covalent & non-covalent forces that stabilize protein structure and function (same for nucleic acids as well!) o Non-covalent: h-bonding, hydrophobic effect, van der waals, electrostatic o Porteins produced from amino acids (peptide bonds) o Nucleic acis produced from nucleotides (phosphdiester bond) o Covalent bonds like togehr individual subunits  Polarity of polypeptides (N- & C-termini, direction of protein synthesis) o Read from left to right, from amino N terminus to carboxyl C terminus  Positive on N terminus and negative on C terminus  The four levels of protein structure and examples of each covered in class: know the details of alpha-helices, beta-pleated sheets, etc. o Primary: linear sequence of amino acids  Determined by the linkage of peptide bonds o Secondary: certain repetitive structures in proteins  how the linear sequence folds upon itself  determined by backbone interactions (hydrogen bonds)  A helix and beta sheet  both stabilized by hydrogen bonds o Tertiary: how it looks in 3D space  the shape  many different folds within polypeptide  depends on distant interactions (stabilized by hydrogen bonds, van deer wals, hydrophobic packing, disulfide bridge)  Disulfide bridge between the sulfurs of cys  on exterior of cell because favor the oxidative state o Quaternary: when you have two or more polypeptides interact o A-helix:  Tightly coiled Downloaded by Mimi Kreger ([email protected])  lOMoARcPSD|34711042  Rod-like arrangement of amino acids  Backbone consists of repeating amino group N-Ca-carbonyl C  Amino acid R groups radiate outward  usually hydrophobic  Stabilized by extensive H-bonding between the NH and CO groups  Mostly right-handed  energetically favored o B-pleated sheet:  Extended polypeptide chains  Form sheets via H-bonding between NH and CO groups of different chains  Adjacent chains can run parallel, antiparallel, or mixed  R groups either stick up or down from top or bottom of the sheet  Definition & importance of protein domains o Folded proteins form functional structures called domains o Protein domains: distinct regions of a protein that have specific functions  Some of the features governing protein-protein interactions, protein assemblies, covalent modification and the importance of these things in how proteins function o Proteins assemble into larger complexes  noncovalent bonds with quaternary structure o Protein assemblies: exact structure depends on orientation of binding sites on each subunit  called polymer  Rigid structures: collagen  higher order of structure, 3 helices that are interwoven (strong)  Elastic structures: elastin  need proteins that are flexible  have a crosslink of disulfide bond and form disulfide bridge  Disulfide bond is a covalent bond that helps stabilize protein structure  seen in proteins that are outside the cell (because outside of cell is oxidative) o Some protein subunits can form sheets or tubes through multiple points of contacts o Interactions between two proteins or another molecule type and a protein involve specific non-covalent interactions that determine the specificity of binding  success depends on the number of interactions o Non-covalent bonds mediate specificity of binding between molecules o Function of proteins regulated by covalent modification  Covalently linking certain functional groups  modify certain amino acids  Add phosphate group (- charge), then changes non-covalent interactions (change shape, then change function)  Kinetic properties of enzymes covered in class o Increase the rate of biological reactions without altering reaction equilibria (rate increased in both directions) Downloaded by Mimi Kreger ([email protected])  lOMoARcPSD|34711042  Products with less energy than reactants: exergonic (releasing energy)  Products with more energy than reactants: endergonic (absorbing energy)  Remain exer or endo with or without enzyme present o Decrease activation energy of a reaction  AE: energy barrier that needs to be overcome in order for a reaction to reach completion o Accelerate reactions through stabilization of transition states (reduce size of transition state) o Enzyme active site  Enzyme will orientate substrate in such a way that substrate molecules will reach the transition state sooner  More stable with less free energy  Relationships b/w reaction equilibria, standard free energy change, reaction rates, stabilization of transition states o All chemical reactions can be described in terms of equilibrium, which is the degree to which a reaction will process forward to yield a product or backwards to the starting point o Equilibrium constant expressed as K (concentration of products over reactants)  no unit  K > 1 exergonic  favoring left to right  K < 1 endergonic  favoring right to left o Reaction equilibria are linked to the standard free energy change  Energy states of reactants and products remain unchanged in uncatalyzed vs. enzyme catalyzed reactions  Free energy = amount of energy to do some type of work  Prefect equilibrium then 0 standard free energy change  When K >1, standard free energy change becomes negative (less energy then you started with  exergonic)  When K < 1, standard free energy values are positive (end up with more energy then when you started  endergonic) o Reaction rates are linked to the activation energy  Enzymes increase reaction rate by decreasing activation energy o When an enzyme binds to a substrate there is not always a perfect fit, but has enough contact points that your binding is going to be fairly specific  when enzyme binds to substrate it has to accomdate the transition state o When enzyme binds perfectly to a substrate -> it is too stable have such an energy barrier to overcome (increase in activation energy)  Reaction progression plots for exergonic, endergonic, “perfect equilibrium” reactions  Differences b/w enzyme-catalyzed reactions vs reactions taking place in the absence of an enzyme Downloaded by Mimi Kreger ([email protected])  lOMoARcPSD|34711042  Features associated with the enzyme active site and how these attributes enable enzymes to possess catalytic activity o Enzymes stabilize transition points by: enough points of contact with the substrate, but needs to accommodate the transition state, active site of enzyme is largely hydrophobic environment, but hydrophilic amino acids inside so active sites are embedded deep within the enzymes o The catalytic site is relatively small compared with the rest of the enzyme  enzymes are so big relative to the active site b/c some enzymes are controlled by other things (protein allistary)  have enzyme and allistaric site then its going to transmit information to enzyme that will relate in shape change so it can accommodate  Allistaric regulator: change shape and activity in some way  Create regulatory sites, and active sites are very exclusive (create selective environment) o Substrates bound to enzymes by multiple weak non-covalent interactions  M & M “saturation” kinetics: what this means, what it says about how enzymes work o Constant concentration of enzyme  the rate of a reaction increases with increasing substrate until a maximal velocity is achieved o At some critical substrate concentration you don’t see an increase in reaction velocity o This saturation effect is an important distinction vs. unacatalyzed reactions…why  Factors affecting enzyme activity: examples given in class o Transcriptional, translational, posttranscriptional, and posttranslational control o Transcriptional: regulate mRNA turnover o Translation: rate and amount of protein produced o Ex. Of posttranslational (protein already synthesized)  Covalent modification  Involves formation of covalent bond  targeting or modifying amino acids that are part of a structure  Phosphorylation (take phosphate form something else and transfer it to a protein, utilize ATP as phosphate source and it generates ADP and phosphate goes to one of amino acids) and diphosphorylation (remove phosphate group  do the opposite) o Phosphorylation adds negative charge  It is reversible  Allosteric regulation  Positive: enzyme enhanced and negative: activity of enzyme decreased (ex. Of feedback inhibition, way of Downloaded by Mimi Kreger ([email protected])  lOMoARcPSD|34711042 regulating levels of synthesized endproduct) not always activation  Protein sequestration (location) (proteolytic modification)  Ex. Digestive enzymes b/c a lot are produced by organs in which you don’t want to have them active for instance in the pancreas (not active until undergoes proteolysis) o Secreted from one site and released into another site and acted on by other proteses that lead to their activation o Inactive forms have –ogen ending  Can’t go back  Control where the enzyme is activated and when it is DNA, Chromosomes, & Genomes:  Proof that demonstrates DNA as the genetic material of an organism o Most DNA is located in chromosomes, RNA/proteins distributed throughout the cell o Precise correlation between amount of DNA and # of chromosomes o In diploid organisms, somatic cells 2x DNA as germ cells o Molecular composition of DNA unchanged throughout cells of organism, while composition of RNA/protein are variable in different cell types o DNA more stable than RNA or proteins  Griffith’s and Avery, MacLeod, & McCarty’s experiments: what they demonstrated, the importance of their work to understanding DNA function o Griffith: discovery of bacterial transformation  Discovered that bacteria can take pieces of DNA from outside (transformation)  Once they knew bacteria could take up DNA then you could look at the effects of what incorporated DNA does to a cell  Sets stage for future experimenters to determine that DNA is genetic material  DNA is transforming principle  Took rough strain cell and he took heat killed bacteria from smoother to see if DNA isolated from smooth cells was capable of changing non-virulant cells o Avery, MacLeod, McCarty  Used Griffith’s streptococcus assay  Replicate Griffith’s work and more finely tune it so they could formally demonstrate transformation principle  Used process of exclusion to show that DNA is in fact the transforming principle  Essentially the same experiment as the previous one, but pre treated them with enzymes that destroy different categories of macromolecules  only when DNase you didn’t see any smooth colonies Downloaded by Mimi Kreger ([email protected])  lOMoARcPSD|34711042  Know all features of the Watson & Crick Model of DNA structure covered in class, forces that aid in DNA stability, principles of H-bonding in DNA basepairing, etc. o DNA is double stranded stabilized by a number of forces (h-bonding) o Every building bloack has 3 parts: sugar phosphate + base = nucleotide o Bases face inward thus have the opportunity to form base pairs via h- bonding o Forces that stabilze DNA:  H-bonding between bases  Base stacking effect: bases within each strand are right on top of another and bases are perpendicular to the backbone and are hydrophobic (hydrophobic effect)  Negatively charged phosphate groups repel each other and give rise to helical structure  phosphate groups farther a part achieve lowest free energy (most stable) o A=T (2 H-bonds) and C=_ G(3 H–bonds) o 2 strands have different ends  polarity  one strand going one wat and the other going the opposite way  5’ end refers to the 5’ phosphate group  3’ end refers to the 3’ hydroxyl group o Watson & Crick Model:  Right handed double helix  Bases are flat, perpendicular to sugar/phosphate backbone  ~10 bases per turn of helix  major and minor grooves  bases h-bond to a complementary base in 2nd strand  purines (A/g) h-bond to pyrimidines (T/C)  2 strands are antiparallel  Chargaff’s Principle o Studied DNA and found that these relationships help for all DNA  [Thymine] = [Adenine]  [Cytosine] = [Guanine]  [T] + [C] = [A] + [G]  fixed realtionships between T & A, G & C  Know structures of all nucleotides: be able to recognize and give complete/correct names of these, differences b/w DNA & RNA o Macromolecules are polymers  Monomer subunits are called nucleotides  DNA has 2’ H  RNA has 2’ OH o Formation of a polypeptide chain is joinging nucleotides with phosphodiester linkages Downloaded by Mimi Kreger ([email protected])  lOMoARcPSD|34711042  Covalent bond between 3’ OH group of a deoxyribose sugar and 5’ carbon of ribose ring structure o In RNA uracail replaces thymine o RNA single stranded and folds back on self  Four different levels of DNA condensation/packaging, importance to the cell/organism. Know different models proposed to explain higher levels of DNA packaging covered in class o DNA synthesis occurs during interphase  DNA not condensed and can’t tell where individual chromosomes are o Chromosomal condensation occurs during M phase  In M phase DNA chromosomes are visible and well defined and look at different regions of sister chromatids  To achieve high condensation  multineme  Chromosomes contain a lot of protein and even a small amount of RNA  called Chromatin: complex of DNA with these components  Protein component has two classes: basic (+)  histones and acidic (-)  group of proteins called nonhistone chromosomal proteins  One level of condensation: part of what forsm the chromatins involved these groups of 8 histones that re referred to as the nucleosomes (basic unit of eukaryotic chromosome structure) o Levels of DNA condensation:  1) DNA double helix (2 nm)  2) DNA + histones “beads on a string” (10 nm)  histones are basic (positively charged)  interact with negatively charged phosphates  electrostatic interactions between histones and phospahtes to create stability  arragned in set of 8 histones and 4 kinds o fhistones than arrange into octmetric histone core  DNA protected because wound around the nucleosome  3) solenoid or zigzag (30nm)  4) 30 nm fiber organized into loops via chromatin “scaffold”  Unineme vs multineme: what terms mean, etc. o Unineme model: one sement or piece of double stranded DNA per chromosome o Multinwmw: have multiple double helices running through the chromosome Downloaded by Mimi Kreger ([email protected])  lOMoARcPSD|34711042  Chromatin: definition, types, components, chromatin remodeling, importance in regulation of gene expression o Chromatin = complex of DNA + histones + nonhistone proteins  Heterochromatin: more condense compacted  largely transcriptionally silent  low gene expression  Euchromatin: more open, accessible  more open the DNA is then it is going to have greater transcriptional activity  active gene expression  Type of chromatin determines the transcription activity o Chromatin can undergo changes/modifications  some are reversible o Conseuqences of chromatin modification include changes in gene expression o Gene expression can be controlled/regulated which has implications for organismal development, activation/deactivation of cell functions o Non histone proteins:  Large number of different proteins (heterogeneous)  Composition of fraction varies from cell type to cell type in the same organism  Function is regulationg expression of different sets of genes o Solenoid model:  Supported by electron microscopy and x-ray diffraction studies  ~6 nucleosomes per turn  30 nm fiber made up of nucleosome discs stacked on edge in the shape of a helix  linker DNA buried in center of helix, but never passes through the axis of fiber o Zigzag model:  30 nm fiber made up of zigzag pattern of nucleosomes formed through binding of histone H1  model supported by biophysical studies showing a spring like property of isolated 30 nm fibers  in this model linker DNA passes through central axis of fiber  longer linker DNA favors this configuration  since the length of the linker DNA varies from species to species both models may be correct o Final level of condensation: a scaffold of nonhistone proteins condense the 30 nm chromatin fiber into loops or supercoild domains  Structural features of histones, nucleosomes, nucleosome assembly, interactions with DNA o 5 classes of histone proteins: H1, H2a, H2b, H3, and H4 o unique amino acid composition o found in all eukaryotes chromosomes in amounts similar to DNA o specific molar ratios observed:  1 H1 :2 H2a : 2 H2b : 2 H3 : 2 H4 Downloaded by Mimi Kreger ([email protected])  lOMoARcPSD|34711042 o form an octamer called nucleosome core o N-terminal tails of histones exposed (sensitive to proteases) o Tails required for formation of 30 nm fiber  Assemble so end terminals hang outward from the nucleosome core o Nucleosome assembly:  Formation of H32 * H42 tetramer  Tetramer binds to DNA  DNA-H3H4 complex recruits 2 copies of H2a * H2b dimer to complete nucleosome core  Has a twofold axis of symmetry  H1 will bind to linker DNA and will stabilize the structure o Histone-DNA interactions  14 sites of contact; one for each time minor groove of DNA faces histon octomer  ~ 40 H-bonds involed (majority occur between proteins of core and oxygens of phosphodiester bonds within minor groove)  force of so many H-bonds drives bending of DNA around nucleosome core  basic amino acids of cord neutralize negatively charged phosphates of DNA backbone  allows for bending of DNA without repulsive forces between phosphates on inside of bend  Many points of contact  noncovalent points of interaction  Structural/sequence features associated with centromeres & telomeres, experimental approaches used to visualize these structures o Telomeres: beginning/end of chromosomes  3 functions: prevent degradation, prevent fusion ends with other DNA molecules, facilitate faithful replication of linear end of DNA  Need telomerase to maintain the ends  without it then they would gradually shorten  Cancerous and germ cells no shortening with age moralization of cell  Normal somatic cells  shorten with age  End of telomere  Single stranded overhang of the 3’ end  Overhang length organism dependent  G rich : allows for non WC base pairing, hoogstein bping, G-quartet, akin to tying a loose knot in the end or Velcro  Structure 1: G- quartet  Folding back on itself like Velcro, 4 bases involved in this structure  Structure 2: t-loop  3’ overhang is going to fold back and base pair with some portion of the chromosomes further back Downloaded by Mimi Kreger ([email protected])  lOMoARcPSD|34711042  This folding and base pairing is going to be stabilized by an array of proteins o Centromere: middle of chromosomes where sister chromatids interact with one another  Allows one copy of each duplicated and condensed chromosome to be pulled into each daughter cell when a cell divides  Kinetochore forms at the centromere and attaches the duplicayed chromosomes to the mitotic spindle  allowing them to be pulled apart  During anaphase spindle fiber attaches to centromere and separate sisters to opposite ends of cell to become daughter chromosomes  Constricted region of metaphase chromosome  Centromere provide important function during mitosis: kinetochores attach to centromere  Attachement point of spindle fibers DNA Replication:  Semiconservative mode of DNA replication and how this makes sense given what you know about the W&C model of DNA structure o Requirments of replication: fast and accurate  have several systems of DNA repair enzymes to help ensure that if there is a mistake thay can fix the errors o Semiconservative mode of replication: each parental strand serves as a template for new daughter strands  End up with a hybrid (mix of new and old)  Direction of synthesis: 5’ to 3’ direction for each incoming nucleotide, the first phosphate group is what is going to be covalently bonded to the 3’ OH group from the previous  Triphosphate important because when you break the first one it gives you energy to create phosdiester linkages  Primer: short piece of RNA (eventually replaced by DNA)  Primerase: lay down short track of RNA o DNA polymerases catalyze the synthesis of DNA  Pyrophosphate provides energy for inclusion of nucleotide  2 template strands are exposed and read by polymerase and then synthesize new strands  DNA polymerases: know what reactions they catalyze, different prok & eukaryotic types emphasized in class, cofactor requirements, roles in DNA synthesis and proofreading of sequences o DNA polymerases are proteins that catalyze covalent addition of nucleotides to pre-exisiting DNA  Requires: Mg2+, dNTPs, primer DNA, template DNA Downloaded by Mimi Kreger ([email protected])  lOMoARcPSD|34711042  Primer initially going to be an RNA molecule and provide a 3’ OH then release 2 phosphate groups and attach phosphate group to the 3’ position on the strand o Polymerase in Proks:  Pol 1: RNA primer removal, DNA repair  Catalyzes formation of phosphodiester bond  5’ to 3’ exonuclease activity: along with RNase H removes RNA primer  cut on adjacent strand, cut sites are staggered  3’ to 5’ exonuclease activity: removes misincorporated nucleotides  at cleveage sites you are going to remove sections of that primer  remove incorrect base and replace with correct base  Pol 2: DNA repair  Pol 3: chromosomes replication  replaices entire bacterial genome, true replicase  Pol 2, 4, 5 : replication of damaged DNA o Polymerase in Euks: (13 in total)  Pol alpha: primer synthesis during DNA replication  Pol beta: base excision repair  Pol omega: lagging strand DNA synthesis; nucleotide and base excision repair  Pol epsilon: leading strand DNA synthesis; nucleotide and base excision repair  a, d and/or e work together to replicate nuclear DNA  g replicates DNA in mitochondria  b, z, and h are nuclear DNA repair enzymes o All polyermases need primers (3’ OH), except RNA polymerases o DNA primase requires only a DNA template o Synthesis of DNA is going to require RNA primer  RNA primers generated by DNA primase  RNA primers excised and replaced by DNA  Know different proofreading systems that give rise to high fidelity DNA synthesis o 5’  3’ polymerization o 3’  5’ exonucleolytic proofreading o strand-directed mismatch repair  repairs erros missed by proofreading exonuclease of DNA polumerase  SDMMR must recognize and repair error in newly synthesized strand (NOT replace original base in parent strand)  In bacteria it scans and detects unmethylated A’s on new strand  older DNA is the more methylated itbecomes  In humans it detects nicks prior to DNA ligase sealing of okazaki fragments  these fragemnts are a sign of newly Downloaded by Mimi Kreger ([email protected])  lOMoARcPSD|34711042 synthesized DNA, do not want to correct parental strand because you will introduce a mutation  Recognizes mismatch, excises segment with mismatch, sunthezies new segment o polymerizing and editing function o new strand being synthesized is constanyl modified to makre sure it has the correct nucleotides  General features, similarities, differences b/w prok & euk DNA replication o DNA replication takes place at the origin  Proks only have a single origin of replication  simpler than in euks  AT rich areas are found where you have to separate DNA (beginning of replication)  In euks have multiple origins  takes time to replicate DNA and we have a large genome  so allows for faster replication times  Direction of DNA synthesis (uni or bi?) and evidence to support what we know o DNA replication is bidirectional: happens in both directions simultaneously  Use of denaturation mapping in phage lambda  Phage has AT rich regions (easy to separate), preferentially denature at high PH, use these bubbles as physical markers for tracking replication  If it were uni then would intitally only one bubble would duplicate  take time to see 2nd bubble duplicate  If bidirectional then both denaturation sites duplicate early on (why supports bi)  Terms covered: Leading vs lagging strands, continuous/discontinuous synthesis, Okazaki fragments o Both strands synthesized in 5’ to 3’ direction o Leading strand: strand that is continuously sunthesized o Lagging strand: discontinuously synthesized  Short stretched synthesized in 5’ to 3’ direction  Overall extension in 3’ – 5’ direction, but polymerase works in 5’ to 3’ direction  Short 5’ to 3’ stretches called okazaki fragments  must be joined together to make the discontinuous strand continuous  Fragments have own RNA primer that provide 3’ OH (extension of DNA strand)  Leading strand only needs one primer, while lagging strand depends on number of fragments Downloaded by Mimi Kreger ([email protected])  lOMoARcPSD|34711042  Primers will eventually be removed and filled in with DNA and end up with parental and new DNA  Reactions catalyzed by and importance of DNA primase, ligase, helicases, topoisomerases o DNA ligase catalyzes the covalent closure of nicks in DNA  links adjacent okazki fragemtns  Need phospideister linkage between two fragments of DNA  ligase does this and seals it off  Requires ATP or NAD  Ligase does not work to fll gaps (where bases are missing)  need polymerase and then ligase o DNA primase adds RNA primers for initiation of DNA synthesis o DNA ligase seals sugar  P backbone between okazaki fragments to form single strand of replicated DNA o Synthesis occurs 5’ to 3’ and not the other way around because incoming nucleotides need to be added on the 3’ group o DNA helicases unwind DNA using ATP  Move ahead of the polyermase  as DNA separated it has to be rotated  Powered by ATP hydrolysis  DNA helicases produce two single strands of DNA  H-bonding potential of bases unfulfilled so H-bond ot itself  Need to keep DNA extended for replication  SSBs  If helicases unwinding DNA, tangles could form, overwinding possible (positive supercoiling)  this makes it harder to separate the strands and reach a point where the helicases cant advance any further  Know components of prok & euk DNA replication that serve as accessory proteins to the major enzymes discussed (SSBs, clamp loader, clamp, etc.) o Single stranding DNA binding protein (SSB): keeps single stranded DNA extended  allows for each template strand to still be accessible and limits interchain H-bonds o The replication fork: “sliding clamp” holds DNA pol on template strand o In proks have single stranded DNA binding protein o Topoisomerases: catalyze transient breaks in DNA to prevent overwinding, positive supercoils  Topo 1: single stranded break  Removes supercoils one at a time  Transient break allows opposite sides of break to spin indepently around intact phosphodiester bond  Reseal break  Topo 2: introduces negative supercoils  2 – supercoil added  double standed break Downloaded by Mimi Kreger ([email protected])  lOMoARcPSD|34711042  in euks found where 2 DNA helices cross one another  breaks both strands of double helix and allows other double helix to pass through gate then reseals double stranded break  end up with two interlocked DNA circles o In euks: replication protein A keeps unwound strand in extended state, clamp is formed by PCNA proteins, instead of pol 1 you have combinations of ribonucleases to excise RNA primers once touve had extensions during synthesis (pol o fills gaps and DNA ligase seals)  PCNA protein act as clamp  DNA pol a:  Initiation at origins and priming okazaki fragments  Found complexed to DNA primase o As primase extends the polyermase can utlize the 3’ OH group to keep extending  DNA pol o or e:  Completes okazaki fragment synthesis  Responsible for processive (have an enzyme that can stay on track longer nd syntheize longer pieces of new DNA) synthesis of chromosomal DNA  Must interact with PCNA and replication factor C to be active  Contain 3’ to 5’ exonuclease activity o In proks: clamo formed by two beta subunits o DNA replication in Euks vs Proks  More complex  RNA primers shorter  Ozaki fragments short  DNA replication only in S phase of cell ycle  Multiple origins of replication per chromosomes  Multiple polymerases at replication fork  Nucleosomes  Telomeres  Centromeres  The DNA clamp in eukaryotes is PCNA (proliferating cell nuclear antigen), while its counterpart in bacteria is the β subunit of Pol III. Also bacteria have single-strand binding proteins (SSB), while eukaryotes have replication factor A (RF-A) to bind to single strand DNA.  Dispersive mechanism of nucleosome replication o Eukaryotic DNA wrapped around nucleosome  As cells enter the S phase there is a massiv burst in histone synthesis  need new histone proteins for DNA replication and no change in position of mucleosomes o Density transfer experiments: Downloaded by Mimi Kreger ([email protected])  lOMoARcPSD|34711042  Nucleosomes on progeny DNA  contain old and new histone complexes  Suggests that nucleosome duplication may occur through a dispersive mechanism  Nucleosome are essentially cut in half as the replication fork advances  When that happens the removal of 4 of 8 subunits allows for thr passing of the replication fork  New histones used to restore nucleosome  half old and hald new  Centromeres & Telomeres: General types of sequences associated with each, where found on the chromosome, issues related to replication of telomere lagging strand, importance of telomerase, relationships b/w telomere length and aging vs. cancers o Telomeres found on the ends of eukaryotic chromosomes  Problem synthesizing the lagging strand at telomeres  cant produce the terminal okazaki fragment  No primer to provide free 3’ OH  Have single stranded 3’ overhang  Ends of chromosomes are replicated by enzyme called telomerase which prevents the ends of chromosomes from becoming shorter during each replication  The nucleotide sequence at the terminus of the lagging strand is specified by a short RNA molecule present as an essential component of telomerase  Telomarse packs it own RNA primer to get around problem of 3’ overhang o Telomarse  Binds to G rich telomere overhang using internal RNA template  lengthens the overhang and essentially provides enough material so DNA polymerase can fill in the gaps  Adds single telomere repeats to parent strand  After several additions: RNA primer made and DNA polymerase synthesizes new strand o Cancer cells display increased telomerase activity and longer telomeres than normal cells  if cells maintain telomeres then greater replication potential o Human cells grow for a limited number of generations older cells have shorter telomere lengths o Progerias: rare human disease characterized by premature aging  Downloaded by Mimi Kreger ([email protected])