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Cells and Organims

by: Amanda Huang

Cells and Organims Bio 0013

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Amanda Huang
Cells and Organisms

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Covers: Protein Structure,Glycolysis, Cell Cycle, DNA, Mitosis, Meiosis, RNA, Gene Expression and more 48 pages of detailed notes and pictures.
Cells and Organisms
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Date Created: 10/15/15
Lecture 2 CHEMICAL BONDS WATER CARBON CHEMISTRY O 00 O 00 O 00 O 00 O 00 O 00 O gt Organism living being that has I Energy Cells genes replication and evolution gt Cells compartment bound by plasma membrane and contain chemical in an aqueous solution gt Cell theory all organisms are made from cells and cells come from preeXisting ones 4 atoms makes up organisms C H O N other abundant P S Cl Na Mg STRUCTURE AFFECTS FUNCTION gt Cohesion attraction between same molecules gt Adhesion attraction between diff molecules Elements with diff of neutrons are isotopes gt For unstable isotopes gt Radioactive isotope What is the structure of an atom gt Electrons move in orbitals gt Orbitals grouped into electron shells gt ATOM MOST STABLE WHEN OUTER SHELLED FILLED PORLARITY OF WATER MAKES IT A GREAT SOLVENT What is the role of water in chemical rxns gt Water is both a weak acid and base 12 13 H3 GH How does covalent bonding hold molecules together gt Attractive force overcomes repulsive force gt Not always nonpolar some atoms hold e in much tights covalent bonds electronegativity function of valence shells amp shielding of valence e by shells of inner I MiX of ionization energy to remove an e and electro af nity energy released when e is added I Tendency to attract e to itself Increase up and to the right I Electronegativity OgtNgtCH gt Carbon forms tetrahedral structures orbitals hybridize to form 4 sp hybrid orbitals How does ionic bonds hold molecules together Transfer of e I Buffers restrict damaging changes in pH helps maintain homeostasis How does a spontaneous rxn occur gt Entropy and change in potential energy products held more tightly they have lower potential energy Thermal energy kinetic energy of molecular motion How to measure gt Liquid Scintillation Counting radioactive material to soln with chemical that becomes excited high energy returns to ground state by emitting photon of light gt Autoradiography Xray form or liquid photographic emulsion over slice of tissue with radioactive material beta particles will be emitted I trace biological molecules if they contain radioisotopes I cells take up radiolabel as if it were the normal form of the element because CHEMICAL REACTIVITY is based on atomic NOT atomic mass Speci c heat amount of energy to raise temp of 1 gram of a substance by 1 degree Celsius O 0 Functional Group Hydroxyl alcohol Carbonyl Carboxyl Amino Sulfhydryl Phosphate Methyl O 00 Lecture 3 FUNCTIONAL GROUPS AMINO ACIDS Structural Organlc Molecule wlth Example Formula Functlonal Group H H OH Carbohydrates HCCOH I I H H Ethanol ii I if C Carbohydrates HCCH I H Acetaldehyde O H O C Fats HCc OH OH H Acetic acid H ii T H H i H Alanine H H S H Proteins HO C C S H H H Bmercaptoethanol 039 OIH OI T i O P 039 Nucleic H C C C O P O acids O H H Oquot Glycerol phosphate H O O H I II II I C H Fats O39 C C C H I H Hydroxyl HIGHLY POLAR doesn t dissociate makes compounds more soluble through H bonding with water Carbonyl ketones and especially aldehydes are VERY reactive to form larger molecules Carboxyl acts as an acid Amino acts as a base Sul iydryl in proteins can form disulfide SS bonds that contribute to protein structure cystine covalently link to form disul de bond Phosphate When phosphate groups are linked together it breaks OP releasing LARGE amounts of energy H ALL PROTEINS ARE FROM fi39XKi iNo ACID SUBUNITS 20 standard 2 odd gt Amino Acid structure I ALL amino acids have CENTRAL CARBON ATOM C alpha that bonds to I Amino acids are distinguished by R groups Hydrogen Amino Carboxyl quot i39 H N C C H I o R Rgroup variant O 00 O 00 O 00 O 00 O 00 1 What are the properties of Amino Acids gt IONIZATION H H H N cO H3NO cCio 2 I OH o R R O Nonionized Ionized gt Stereoisomers I Isomers are molecules with same molecular formula but different structure Structuralgeometric optical isomers I OPTICAL ISOMERS any atom bonded to 4 UNIQUE PARTS form a CHIRAL CENTER Enantiomers are a pair of stereoisomers differ only in 3dimensional orientation same structural formula I Stereocenter is an atom bearing groups such that an interchanging of any two groups leads to a stereoisomer I ALL amino acids except GLYCINE have optical isomers I Only Lamino acids are found in PROTEINS used by cells in the body Side chains R groups differ in size shape reactivity and interactions with water CH Rgroups have low electronegatiVity so they are NONPOLAR also hydrophobic Polar side chain and electrically charged side chains acids amp bases How do amino acids link to form proteins gt Condensation monomer in water out gt Hydrolysis water in monomer out break bond gt Dipeptide 2 amino acids joined by l peptide gt Tripeptide 3 amino acids joined by 2 peptide Carbon skeletons don t write C or H s gt Angled bonds because tetrahedral 50gt AA proteins AAlt50 peptide oligopeptide All levels of structures are based on primary structure VVV H H o N c d H T 00 H 39 o H N C H GYLCINE H CH OH ALANINE Alphahelix Lecture 4 PEPTIDE BONDS PROTEIN STRUCTURE Primary sequence STARTS at amino end 3 Basic gt Proteins have 4 levels of structure gt Peptide bond is planar and restricts folding gt Secondary structure is stabilized by Hbonds between atoms in the peptide backbone gt Tertiary structure is stabilized by interactions among R groups a Rotation around CN bond is restricted because peptide bond has double bond character due to e resonancediffer in position of e e jumps from CO to CN gt Sometimes double sometimes single bond gt SO no rotation allowed around peptide bond gt ANGLES THAT ALLOW ROTATION Torsion dihedral angles or the backbone 3 Ramachadran s Plot valleys of energetically favorable conformations a Secondary Structure in Proteins gt Stabilized by Hbonds gt In alphahelix minimized interactions of R groups and R groups project out gt In betasheets R groups above or below gt Proteins 2nd structure increases its stability each H bond is week but large of H bonds stabilizes 3 Tertiary Structures amino acid Rgroups gt R groups interact with other side chains or with the peptidebonded backbone causing the polypeptide to fold in precise shape OTHER FORCES hydrophovic interactions ionic bond disul de covalent HydrophobiclntmIcllom 3 What are the principles of protein folding gt Fold back to become compact globular structures cquot Nevquotmm 3mm gt Polar folding places most charge amino acids on the 3 4 3 chm mam surface of the protein to interact with water 3333quotquot gt Nonpolar residues after peptide bond the amino acid c 0 CH O CH that is left face toward the interior of the proteins an mum M6 creating nonpolar amp hydrophobic interior gt EXCEPTIONS Integral Membrane spanning 0 proteins not compact Cquot CH 0quot1 6392 mquot0 C 0quot Ionic bond I Hydrophobic parts buried in lipid bilayer 3 Hydrophilicity XaXis is amino acid position yaXis is hydrophilicity a Quaternary Structure some proteins contain several polypeptide subunits the bonding of 2 polypeptide gt HOW Noncovalent interactions or covalent bonding interaction of disulfide bridges Lecture 5 CHEMICAL RXNS AND ENZYMES 339 FORM IS FUNCTION for a protein gt Anfinsen s experiment 1957 I Question is folding important And how does a protein fold I Started With Ribonuclease protein that degrades RNA I Then added active enzymes Betamercaptoethanol breaks disulfide bonds amp Urea breaks Hydrogen bonds I This DENATURES breaks apart proteins gt Some proteins Will spontaneously refold I Inactive enzyme slowly remove reagents9 ACTIVE ENZYME 339 TERTIARY BASED ON PRIMARY STRUCTURES amino acid sequence gt Sometimes proteins need help folding I CHAPERONES multiple large proteins that use ATP to help fold proteins 339 HOW to study proteins Electrophoresis method of separating complex mixtures of macromolecules according to their migration in an electric eld gt Macromolecules sorted by sieving properties of gel matrix gt NATIVE PROTEINS migrate based on SIZE AND CHARGE not helpful gt INSTEAD use SDSPAGE uses DENTAURED PROTEINS I SDS breaks hbond I SORTS ONLY ON SIZE I TWO subunit Will separate into TWO bands With addition of BME breaking disulfide bonds 339 Proteins are STRUCTURALLY FUNCTIONALLY diverse TABLE 33 A Summary 0 Pr tein Functims 339 CHEMICAL RXNS TRANSFER ENERGY Protein type Role in cell or organism gt Energy ability to do work supply heat Antiboclies and complement Defense destruction of gt STORED ENERGY POTENTIAL E We Sfiff 39 g gt MOVEMENT E KINETIC E or THERMAL E ngi fg39rigzzge m 8 M Vemem measured as heat from molecules Enzymes Cata39yze Chemica39 reactions zo ALL RXNS PROCEED in the direction of LOWER Pept de quot es AESSSdi nii Z39itTQZ c il es of FREE E R Sfeliecilimrca signastmm 339 changeG Gproducts Greactants fegtgfnige and inmate 339 Gibbs freeenergy tells Whether rxn is spontaneous or requires Structural proteins Pigziudeessgg qogrffcrt il and energy such as hair feathers gt changeGlt 0 SPONTAENOUS exergonic JEESZSZ39SSiZE39SZLSES gt changeG gt0 ENDERGONIC 33335 2 03 substances gt changeG changeH TchangeS 5933 mmogkawmsm 9999 WWW gt changeG is both enthalpy H and entropy S v Defining spontaneous Rxn gt Products have lower potential energy than reactants gt Lower potential energy H enthalpy e held more tightly gt Transfer of e to more electronegative atoms decreases potential energy gt Excess energy given off as heat Endothermic rxn must absorb heat Hgt0 Exothermic rxn must release heat Hlt0 Chemical Equilibria gt Important biological rxn are often reversible gt Rxn in a CLOSED system will eventually reach equilibrium I Concentration Will NOT change I changeG0 at equilibrium I equilibrium will change if you add reactants shift R or products shift L Electrons have the greatest potential energy in the outermost electron shells O 00 0 3rd Electron shells v IW quotum m M In 0 O O O 00 gt Spontaneous chemical rxn MAY occur slowly due to complex Rxn path ll quot quot ABc Transition state A 3c TranSItIon state 3 a Activation gt c 0 00 a quot energy 9 5 3 A BC i ADD 5 E 39th 9 c f 3 WI u Reactants AG 00 0 CATALYST g 2 BC quot enzyme A3 c 339 ENZYME 2 Products u Reactants 4739 AG does I i I Progress ofreaction not change Products Progress of reaction ENZYMES 339 What are enzymes highly SPECIFIC catalysts gt Reactant glucose SUBSTRATE binds to a specific region ACTIVE SITE of the hexokinase ENZYME where amino acids help the substrate reach the transition state and form the product INITIATION reactants bind at AS in a speci c orientation I Forms the enzymes substrate Shape changes amp enzyme bring components to the transition state TRANSITION interactions between enzyme and substrate lowers glucose activation energy TERMINATION products have lower affinity for active site amp released Enzyme is unchanged after rxn 339 When can the substrates not bind gt Competitive inhibition prevents the substrate from binding to the enzyme I Regulatory molecules SUBSTRATE ANALOGUES binds to the AS the substrate cannot bind gt Allosteric regulation changes enzyme s shape to activateinactivate it I Regulatory molecules binds to enzyme a shape change that aids OR prevents 339 How do we know which amino acids make up the active site gt Genetic methods I Margaret Dayhoff thought to compere sequence of enzyme from different species See residues that are shared between y and human Residues amino acid is crucial for the enzyme gt Mutational studies change amino acid sequence and look for changes in enzyme activity gt Physical crystallography gt Chemical Methods determining which amino acids are in the active site I Use reagents with specific amino acid Does it destroy catalytic activity of protein I Does the substrate or a competitive inhibitor protect against the action of a reagent V VV V A MODEL OF ENZYME ACTION A B C A B c Substrates Transition state Enzyme 39l Initiation Reactants bind 2 Transition state 3 Termination Products to the active site in a SpeCi c facilitation Interactions have lower affinity for active orientation between enzyme and site and are released Enzyme substrate lower the activation is unchanged after the reaction energy required Figure 324 Biological Science 2e 2005 Pearson Prentice Hall Inc Lecture 6 LIPID STRUCTURE ND MEMBRANE PROPERTIES gt gt gt O O a Structure of Steroid 4 member fused rings Lipids have major hydrocarbon component Usually nonpolar and hydrophobic higher CHO ration than carbs Fatty acid carboxyl group attached to hydrocarbon chain yi39f lb Aquot t v Phospholipids gt Dehydration rxns join fatty acids to glycerol creates ESTER LINKAGE I Phospholipids are AMPHIPATHIC cause they have both polar and nonpolar amphipathic do not dissolve in water I Phospholipid shield fatty acids tails from aqueous polar solvent gt Double bonds cause kinks in phospholipid tail gt Hydrocarbon side chains can be saturated no double bonds or unsaturated H 5H I TRANS double bond allows close packing saturated trans fats Dehydration reactions join fatty acids 3 tend to form solids MORE ENERGY trans double bond LESS t g39y emquot 0 Fatty at H 13 H energy than saturated bond 1 CW 1 I CIS double bonds can not pack closely unsaturated form oils H3C CH3 2 3 gt Art1 c1almembrane experlments 1 H20 CC I How rapidly can diff solutes cross membrane if at all when H H 0 Diff phospholipids are used to make the membrane 0 Proteins or other molecules are added to membrane Glycerol I SIZE and CHARGE affect rate of diffusion across membrane M gt Longer tails decrease uidity and permeability gt Increasing cholesterol content decreases uidity and permeability m gt Unsaturated fatty acid tails cis double bonds increase uid and permeability gt How to test membrane uidity FRAP Florescence Recovery After Photo bleaching I Bleach kill of section I Membrane is a uid layer time of recovery is a measure of uidity a Li id micelles b Li id bila ers p p y a R Water f a Permeability scale b Size and charge affect the rate of diffusion E E No water cmsec across a membrane Phospholipid amp u l 1quot quot High permeability bilayer quot39 WV 7 quot quot Phosphate a quot 6 W m n f 7 179 557 g A o a Salazar quot2233 a Wof jg fk Lar e unchar ed T g g Glucose sucrose 39 polar molecules r a F 39 3 0 CH1 Wolm Low permeability CH2 Figure 69 Biological Science 2e 2005 Pearson Prentice Hall Inc kink a typical unsaturated fat Lecture 7 ENDOMEMBRANE SYSTEM amp THE SECTOREY PATHWAY 1 In 1972 SingerNicholson suggest the FLUIDMOSAIC 3 4 5 6 7 8 Fluidmosaic model Cell exterior 39 MODEL a Integral proteins through membrane b Used freeze fracture to confirm this theory c Are membraneproteins xed in the plane 0 the membrane or lateral diffusion i Frye and Edidin experiment 1970marked surface protons and used SV4O a virus that allows cells to fuse into one giant cell 1 Saw the different proteins mixed Phospholipid bilayer Membrane proteins Cell interior Figule 619h Biological Srienu 2 2005 Pearson Pvenlire Hall lnr Is permeability a problem a PASSIVE TRANSPORT NO energy required WHY Going with the concentration gradient hi9 low i Channel proteins facilitate diffusion allow small molecules to move across membrane 1 Ion channel gramicidin antibiotic that attacks bacteria 2 Aquaporins allow transport of water 3 Gated channels open in response to binding of neurotransmitter or other signaling molecules ii TRANSPORT PROTEINS facilitate diffusion LARGER molecules 1 GLUT4 insulinregulated transport protein b ACTIVE TRANSPORT AGAINST concentration gradient i Active transport by membrane pumps movement against concentration gradient using ATP ATP stores energy in bonds pumps cleave bonds 9 release energy 1 Sodiumpotassium pump 2 Sodiumglucose cotransporter 3 ABC ATP binding cassette alpha helices trans membrane domains with nucleotide associating within cytosol a Use ATP to drive transport in and out of cells b Helps excrete toxins drugs other harmful molecules c Cancer patient tumor cells have increased expression of ABC transporters chemotherapy drugs gone before they could effect the tumor cell d Lot of these in bloodbrain barrier How to organize the uid mosaic membrane a The endomembrane system helps localize proteins and lipids to the correct organelles i Rough ER site of PROTEIN SYNTHESIS for those entering secretory pathway ii Golgi Apparatus site of protein PROCESSING and GLYCOSYLATION iii Smooth ER site of PHOSPHOLIPID SYNTEHSIS b RER system of channels bound on each side by a lipid bilayer Signal hypothesis proteins which enter the secretory pathway have sequence of amino acids at their Nterminus that directs the polypeptide to the ER and guides it during translation into the lumen of the ER a insertion of the signal sequence through the membrane needs GTP hydrolysis SRP and SRP receptor bind and hydrolyzed GTP to power changes that open the translocon amp inserts signal sequence into the translocon Either secrete protein OR bind it to uid membrane How to know this process of ER exocytosis Pule Chase Experiments PROTEIN SORTING AND VESICLE TRANSPORT Smooth endoplasmic reticulum Scrambalse and Flipase 21 Take phospholipids synthesized on cytosol side and ips into the organelle b Also can adjust Which phospholipids on Which side THE SIGNAL HYPOTHESIS Miocene Cytosol 1 Signal sequence 2 Ignal sequence 3 Signal recognl Ion 4 Protein synthesis 5 Protein synthesis is synthesized by binds to signal particle binds to continues Protein is complete Signal ribosome recognition particle receptor in ER enters ERSRP is sequence is SRP membrane released removed Figure 728 Biological Science 2Ie 2005 Pearson Prentice Hall Inc THE SECRETORY PATHWAY A MODEL RNA Rough ER 1 Protein enters a quot clsface of V Golgi apparatus GOIQI apparatus 394 L l 3 l III 3924 tTlquot ll 5 Protein is Wad from tall PROTEIN SORTING AND VESICLE TRANSPORT I In the endomembl system proteins bou for lysosomes or roug Lumen of 39 ER are given differen Golgifappauatus carbohydrate quottagsquot Proteins bound for secretion have builti export signal 2 Proteins are sortec in the Golgi when the bind to different g receptors 3 Transport vesicles bud off the trans face the Golgi and travel ti their destinations Cytosol To plasma membrane Transport Return to the ER for secretion vesicles 4 Proteins on vesicle surface interact with receptors at destinati 5 Vesicle delivers contents Figure 731 Biological Science 2e 2005 Pearson Prentice H Lecture 8 amp 9 Carbohydrate Structure Polysaccharides ECM l Carbohydrates contain a single carbonyl group either an ALDEHYDE at C1 or a KETONE group internally Glucose Galactose a Hexoses have 6 c pentoses have 5 c H O H o b Glucose and Galactose are STRUCTURALLY different 1 f c MonooligOpolysaccharides HZCIOH HZC OH d Naturally occurring sugars all have the C5 OH projecting to the Ho 3c H Ho 3c H right and are termed Dsugars There are 8 2A3 diff H C OH quot04 H combinations of the chiral orientations at carbons 2 3 and 4 H5L0H Hsc0H Different e All naturally occurrlng ketoses also have the C5 hydroxyl or1ented H6c0H configuration Hec0H to the right and are termed Dketoses J ofhvdroxylgroups 2 In soln glucose is found normally in CYCLIC form 39quotquotquot quotquotquot 39 a WHY E rich O of the C5 OH lies near the e poor Cl carbon share its e with it forming a new covalent bond b Anomeric C when C bonded to TWO oxygen reducing end ONLY possible on anomeric C Ring forms of glucose 6 C fHZOH 31331 3233239ime H 5 H We 1carbon resulting in a c H quot Ic When this reaction quotquot9 mm Ho 33 239 OH 3125 33 23 1quot is 6 formed and two H 5H20H6H q39jGIucg 36 are cJ 1CH HO 3H 3 0 Alpha ifthe c1 OH I I lies below the plane H OH GCHZOH of the rlng 39 5 7 Or H4 0 Beta if the C4 OH HoC IDH T CH lihabcrve the plane 3 2 hisediiction is H 0 ve im 0 an BGlucose 64 quot w p n t d Draw 1t as a r1ng for convenience but ACTUALLY puckered shape e Aldoses and ketoses have REDUCING power can donate e and become oxidized OILRIG i Benedict s test to detect reducing sugars requires free aldehyde group 9 0 c 2Cu2 5H0 4393quot 31120 3H20 R H R O 1 3 a Non reducing end ii only under basic conditions once hydroxyl group makes another rxn ANOMERIC C IS FIXED can no longer return to aldehyde formation Glycosidic bonds involves the anomeric C and an alcohol condensation rxn BRANCHED POLYSACCHARIDES MORE SOLUBLE By branching MORE NON REDUCING ENDS build faster and use up more quickly Glycosidic bonds involve the anomeric C and an alcohol 5 Maltose a disaccharide a Polymerization reaction aGlucose 601on aGlucose 6CH20H O T 15 139 1 4 39 a C S H C I 0H 3 l H Ho 339c 2 H Ho 339c Z39C 3c 239c I ethanol pr 0 I I O I I O I The hydroxyl groups from the lcarbon and 4 arbon react to produce an u 14gycosidic linkage and water b Sugars are often drawn in simplified form for Iegibility CHIOH CH20H CHon CH20H CHon CHon CHzOH H20H CHon wwwwa OH OH OH OH OH OH OH OH OH polysaccharides Reducing Flume 54 Blologkal Sdence lie 3913 2005 Pearson Prentice Hall Inc Thin nAl AA AAA AHAII TkA nl 1AAAAA kAA mnA A AiAnbh nnlib Aldoses and ketoses have reducing power they can donate electrons and become oxidized O 0 b 2012 5H0 ii H R o CuJO 3Ho Polarity M to identify a reducing sugar gt Find all anomeric c in rings these are the C s bound to two 0 s cant reopen already bonded to something else Does the anomeric C have a OH attached this will be a reducing end quot20quot quot20quot quot20quot quot20quot quot20quot quot20quot quot10 quot20quot 10quot Does the anomeric C have a 0R attached this will not be a reducing end CH 0 cannotreopen on on on on on on on on uh Figures4b Biological some21 critic gquot 7005 PeannnPrentIrtHall hr 9 lquot 2 W N 8 glycosldic bonds in tl 0 9on l reducrng unit lcquot H I 5c 0 HO rm H OH 39 0H OH H I H Bet 14 2 H c it in quot Beta o Ho 1 3 ou Reducing em 0 o 39 quota H But once the hydroxyl group of the anomeric CH20H tlt Jan 9 free grOUpSgt 18 CU OH H c sl Ho 3 H carbon IS substituted It IS no longer a reducing H c o H H o c c group can t reform the straight chain cllt r 20 m quot J p 0 H on CgtH J RC o Jug rlyntanomerlc C In each carbohydrate 3 f H on fnzon u cl 0 ee Aldeyhde Group H lc HO IH i H c 2c l in TABLE 5 Polysocchurides Differ in Structure Polysocchuride Chemical Structure Threedimensional Structure Starch aGlucose a Glucose quot20H CPIon o H F 0 H 1 a14 glycosidic linkage M Unbranched helix amylase Table 51 pan 1 Biological Science 2 539 Branched helices amylopectin C 2005 Pearson Prentic Hall Inc 130 residues TABlE Sl Polysuccliuritles Ditter in Structure continued hemicd Structure citGlucose CHzOtl Polysucchoride Threedimensional Structure Glycogen atGlucose 2 CH20H OH OH a14glycosidic linkage storage of energy in animals 3941 u I tb 5 can b 39 i x x 35 H dj 39C 1 ACE 1 I Q A x J I Aquot 1 O39 tO r85 Aid we a e g l L L 0 A J J t Cyl 0 1 alpha 16 amp Highly branched helices SUMMARY TABLE 51 Polysaccharides Differ in Structure animals alpha 1 4 cellulose beta 1 4 Polysaccharide Chemical Structure Cellulose 13Gucosc B Glucose CH30H 0n used for l Structural 0 support in I I cellwalls of 1 1 V slants and 7 v t 7 many algae 0H 1 h 1 CH20i4 CHIOH H l 4 x l 1 1 r on r 39 CH20H Threedimensional Structure Parallel strands joined by hydroge bonds c 29 Puma EEK1sz uc neighbors making it REALLY strong excludes H20 makes it INSOLUBLE we can39t digest it 1 09 tons of cellulose 1 09 tons of chitin alternating orientation or glucose means it can Hbond to its IABLE 5 Polysocchnrides Differ in Structure continued Chemical Structure Nacetyl Nacetyl glucosamtne glucosamine 81 4 glycosidic linkage Polysocclioride Chitin CH20H Hydrogen bond with Table 51 port 4 Biological Science 2 modified glucose not linked to hydroxyl group but Nacetyl adjacent strand forms here Threedimensional Structure Parallel strands joined by hydrogen bonds also excludes H20 c 1005 Parson Prentice Hall Inc used for EXOSKELETONS MRI 5 Polysoccliurides Differ in Structure continued Polysaccharide Chemical Structure Peptidoglycan Nacetyl N acetyl muramicacld glucosamlne B l 4glycosidic linkage CPIon CH2 H The amino acid chain forms peptid bond with same a chain of if 2 amino acids on 39 D adjacent strand o uses peptide bond I abla 51 part 5 Biological Science 210 teedimensional Structure I I Parallel strands joined by peptide bonds Used in BACTERIA MEMBRANES C 2005 Pearson Prentice Hall Inc Transpeptidasegt crosslink bacteria to gethercan be blocked by anitbiotics cause linkages to break down and cause bacteira to lyse Peptidoglycan Monomer NAM NAG 9H3 CHZOH H NlIc0 H O H O H H 0 I H NHO CHZOH quotseencw CH3 LAlanlne D Glutamlc acld pentnpeptlde mesa Diamhopimellc acid DAlanlne DAlanlne Peptidoglycan Monomer NAM NAG 9H3 CHZOH H NHCSO H o H O H H 0 9 H NHf0 CHZOH H3cCHco CH3 LAlanlne D Ghtamlne pentapeptide LLyslne DAlanine DAlanine E coli peptidoglycan Suars can be modified with amine Nacetyl amine acid and manv other nrnung Staphylococcus aureus peptidoglycan Cl IUI llllllg purysaccnal 165 are UlyCUSGIIIIIIUQIYCCIIS UHUSI GAGs are polymers of disaccharide building blocks repeating units H0 0quot HO OH OH coy o co co HO no 0 o o 0 0 o o no H O H0 0 OH HO Achl HO AcuH H0 Adm ill II He sugar Eamne can He sum Ft various positions 0020 050 0800f ego 090 03910 yo H0 0900quot am m pv hmo g a on f N quot M quot0 mquot adds negative 5 in vitreous humor cartilage synovial fluids mucose charge which SUCKS up watel I chondroitin alternatin glucuronic acid and NAcetylalactosamine hyaluronan alternating glucuronic acid and NAcetylglucosamine High Molecular Mass Polysaccharides Energy Storage Structural Support Starch Cellulose Alpha Glycogen Chitin Beta Peptidoglycan wee HydrationISupport Complex oligosaccharides can be attached to proteins and lipids glycosamlnoglycans can make glyCOSIdIc linkage between sugar and hydroxyl group Glycoprotein 0 ide Inside Outside of cell V w a 3 o Mm umuunmt wraith va tiltWWWWWI39L239 tumult u to K of cell at up wagu Idvnq 1r can Murm wiMMImmemmmmnmm they ttached to asparagine Nlinked or serine or threonine O linked side chair hey can also be attached to lipids to form glycolipids CH3CH3gt HzCCH2gt saturated reduced ATP consists of three phosphate groups ribose and adenine m unsaturated l o o 0 Ram I u n u o pov o39 o ltn en ampT Q5 39 w Adenine Lecture 10 Metabolism Energy Redox and Glycolysis quot Pquot quot939 quot quot on oquot quot 139 Metabolism synthesisdegradation rxns in living cells WNW Ribose gt Energy used to take oxidized carbon and convert to reduced C CH O 60 4RD Zip gt Reduced C from Glucose is oxidized9 release energy as ATP 5 13 5 3 glycole 00 REDUCTION AND OXIDATION OF CARBON 01110088 4m gt Energy from sun used to remove e from water 9 C02 to make glucose photons 2Pymvate photosynthems CO2 gain 6 mm mm 30 gt Through glycolysis citric acid electron transport chain and oxidative TS hop phosphyorylation e from O2 to water QB ox39p 39 ENERGY RELASED in form of ATP m 6H 0 30m 339 1 Oxidation by direct transfer of e 2 2 gt A transfers e to B A is being oxidized A B0 6 Ao B 339 2 Oxidation of C by Rxn with Oxygen CH4 2 02 OCO 2 H201 gt carbon oxidized as e further away from C in CO2 than in CH4 I the fewer H s attach to C the MORE OXIDIZED is the Carbon 339 3 By transfer of a HYDRIDE ion H equivalent to H 2 e gt often Y or X is a cofactor work with enzyme in a Enzyme chem rxn picking up e from substrate 9 on to XH2 Y X YH2 another acceptor red ox ox red v OVERVIEW OF CELLULAR RESPIRATION How do we capture energy from the oxidation of glucose cs ancient pathway 10 of total energy Glucose 06 Glyc ysls PROCESS OVERVIEW OF CELLULAR RESPIRATION t ADP NADH NADH NADH noquot age I j 39 l ADP Energy Investment Stage m c F t 16 b P uses ATP Glucose Man 02 ttmususedtoproduceAYP mxluoose M 02 quot0 ruc ATP C02 AWhIOTP A I39P C 3P 6 gigglms imagining latrich Cycle 45ctgrgn11 1rzaur M P 13 AD Stage in WWI Mb aw I N Energy Recovery Stage What comes out W 2C3P2 ATP d R d d E C6 gt 2x CS an e uce v 2C3P gt20quot3Myruvate v 2x CS NADH is an electron carrier it has reducing power quotmph 7 HI TV Pique 93 Blnlngkal Smnm Zln R d d Reduction 9 we H H 0 NM C NH o H 0 2r NH oxidation Electron carrier cw W1 c 2005 Pearson Petunia NaIIan NAD QH2 oxid NADH Q H red 39339 OVERVIEW OF GLYCOLYSIS IVER r quot 39 Allioreacuonsof m 00 Drum m 939 39 0C0quot LW tun e saunacut Out w l i has new 2 no 00 OH 139 Glucose Fructou Fru ou sphosphate 63pth 1 6 bls osphate C ADP quot H ketose Glycolysis begins with an 39 39quotquot quot quot quot glyceraldehyde 3P G3P nu nr 0 o Q 572 n n Oh M W I m l Iut h U The 2quot indicates that glucose 2 quotAD 2 0 has been split Into two 3carbon sugars 2 P 2 1 2 039 039 039 l I 0 0 0 I I 910quot gt uco Q gt froQ euro0 mow 39 cu urlng the energy payoff phase 4 A re produced for a net gain of 2 MP CMWMOMHA vacuum debuuiu o 0 v How is glycolysis regulated What if don t need more ATP gt Step 3 is catalyzed by PFKase phosphofructokinase is HIGHLY EXERGONIC I Products of steps 1 amp 2 can be put to other pathways but NOT Fructose bisphosphate I PFK is regulated by feedback inhibition 0 ATP bind to enzyme s active site regulatory site 9 at this site ATP 9 ADP and phosphate group transferred to create fructose bisphosphate O at HIGH concentration level ATP binds on secondary site of enzyme and the enzyme ATP synthase panicles SUMMARY OF GLUCOSE OXIDR 39 inler membrane space Ribosome cnstae Granules 0 Inner membrane uter membrane DNA 2MP Lecture 11 Metabolism TCA Cycle and Respiration Cellular respiration If electron acceptor I such as oxygen KREBS CYCLE ELECTRON TRANSPORT AND GLYCOLYSIS OXIDATIVE PHOSPHORYLATION is present Glucose gt Pyruvate lt ATP 4 Is NOT present v Mitochondria can vary in shape dynamic in shape gt Aerobic exercise increases of mitochondria in muscle I In cardiac muscle cell mito 3x as much cristae than in a liver cell mitochondria gt How did mito come about Why does it have two layers two bilayers I Mitochondria and chloroplasts evolved from microbes engulfed by nucleated ancestral cells 0 Bacteria membrane inner membrane Cell membrane outer membrane 0 Mitochondrial cells have their OWN DNA necessary for replicating mitochondria circular DNA gt Mitochondrial DNA comes ONLY FROM MOM not dad v Outer membrane has LARGE pores Inner membrane is impermeable to ions Transport proteins carry pyruvate into the matrix where it can enter the citric acid cycle FERMENTATION To enter the Krebs cycle pyruvate is rst oxidized to 602 and acetyl CoA 00 Process of Cell Respiration oo gt glycolysis glucose 9 pyruvate generate new 2 ATP 2 quotW quotquot quot 5 quot39 quotquot 2 20 H0 EC00 gt w oxygen pyruvate 9 Acetyl CoA then into Krebs Cycle DH 5 coo 030 All reducing power will be harnessed for ATP synthesis in 353339 MW carryimmbsyde cl ffwmu OXIDATIVE PHOSPHORYLATION each glucose will yield C6gt 2C3gt02 in end around 30 ATP molecules gt2 coz IKrebs cycle citric acid cycle TCA tricarboxylic acid cycle All 3 39ODllOn 04 the pyruvate broken down to 002 xvebuyumwmm mrtmhondrul motnL coo outside cristae GLYCOLYSIS PYRUVATE PROCESSING 00 AND KREBS CYCLE ATPATP I KNOW if 0 NADH 0quot 3339m going forward Glucose C compounds 400 MD 39 being oxrdlzed m r 9quot to generate e 200 adding any quot quot 2 carriers H 39 39 substrate will ico E id f m WM 300 speed up Mmquot foo m if or NADH b k39 d i 1 39 rea mg own n E 6 40 In each ofthese drops Of pyruvate W Swdmu energy is transferred ti Wmquot energystoring molecucs m u swammm t zooswenmuuu 400 ATP NADH and FADH 3 Cgt 2C 002 20 4Cgt 2002 40 39339 Feedback inhibition 600 1 Enzyme that converts acetyl CoA citrate 2 NADH binds to active site Joe M 7 quotquot quot m 3 ATP to allosteric regulatory site 4 Pyruvate Acetyl CoA by ATP VVVV 00 Fermentation regenerate NAD from NADH In the absence 0f oxygen yeast regenerate Lactic acid fermentation occurs in humans NAD by reducing pyruvate to ethanol 2A0 ZA I P Krebs need Glucose 02 Alcohol fermentation occurs in yeast 2ADP ZATP I g Wig 2NAD 2 NADH 2Pyruvate 139 U GIucose gt gt gt in J t quot g Iquot V LDH Nointermediate Lquot pymvale wedenwmenasezrg m quot 2 w W Wage zuctate T co f g ADH 3 5mm aloom39 dehydrogengiietvlaldehvde 2 02 Dl erentlated tlssue Proliferatlve Thmor r ssue V I fermentation 3902 Glucose Glucose Glucose 02 Pyruvate 02 Pyruvate Pymvate 5 w Lactate Lam Meme tumor oesn t C02 002 wantt give Oxidatlve Anaerobic Aerobic quotp 0 bonds phosphorylation glycolysis glycolysls rath rs 36 mol ATP 2 mol ATP Warburg effect build in 5 mol glucose mol glucose 4 mol ATPmol glucose 39 g Inste ot Vendor Helden et el 2009 Science 324 1029 energ Cellular respiration quotWWW 39 moumuspommo at cm cucoursus quot 33mm quot 0mm mosmomrunou an 02 gt 30 ATPglucose quotohmquot m mutton such as cum WW no 02 gt 30 ATP15 glucose gure We Selene 10 c MWMRO HOIIJN Warburg effect cancer cells do NOT use cellular respiration NO oxidative phosphoylation after therapy the tumors are no longer sucking in the glucose no longer malignant ATP synthase panicles inler membrane space Ribosome 395 Lecture 12 Electron Transport and Oxidative Phosphorylation 1 Process a energy released in moving an e from one molecule to another with HIGHER e affinity USED to pump protons against gradient pH GRADIENT and CHARGE Gradient b 6 transport and proton pumping couple if protons cannot be PUMPED OUT then e will not move down the ETC 0 if p gradient builds too much then not enough energy is released in moving 6 339 NADH Inner membrane Outer membrane 2 Heme part of protein NOT of the polypeptide chain 0sz a Atoms that hold a chemical group that is held by a protein a i b Each of the cytochrome has a slightly different heme to carry its 6 3 w 33439 3 Proton gradient in inner membrane space 3 magma a Tries to move p into mitochondrial matrix by going through the shaft F0 unit w W quotquot5quotquot b Binding of a p to subunit turns shaft one notch g quot quotquotquotquotquotquot c Rotor mechanism serves as a paddle as it is turned by F0 unit E i Changes conformation of F1 unit catalytic subunits f mu huoinma aviv 1 Allows binding of ADP P and catalyze ATP and release it M d E ow into the system from sugars via NADH or FADHZ MW e E ow OUT of the system by being transferred to 02 to make H20 m 331mm 4 What happens if the inner membrane is leaky Q quotN m h a Chemical uncoupler used as weightloss aids DNP can kill 0 nmmmm 0 i DNP carries protons from intermembrane space to the matrix ii Facilitated transport NOT active iii Let energy gradient dissipate so get Glucose and INSTEAD of STORING AS mmnormsvmse ENERGY RELEASE AS HEAT b Naturally do this through brown fat when we need HEAT i Mitochondria are abundant and have UCP uncoupling protein in their inner membrane so 6 transport results in release of heat RATHERN than ATP c Ionophores like valinomycin lipid soluble K carrier STOP ATP synthesis by decreasing membrane potential dissipating charge BUT NOT pH 5 Gradient drives other processes in addition to ATP synthesis The quotRhoquot quot605Wquot i The F0 unit is the base the F1 unit is the knob chain oltlturs in the inner quot 27 membrane of the quot SSS mitochondrion 2 membranes of cristaei Kt304 quotquot39 ELECTRONTRANSPORTCHAIN H m nan m mm m m m m 4 quot lnner membrane l 39 u 39 u v 39 39 g l I I I I 39 0 a39 Mitochortdrial matrix higher affinity Lecture 13 Nucleic Acids and DNA Packaging 39339 Frederick Griffith 1928 studied bacteria that cause pneumonia gt Studied rough R strain and smooth S strain gt 4 studies R strain S Strain Heatkilled S strain R strain heatkilled S strain I R strain live I S straindie I Heat killed 5 strain live I R strain heatkilled S strain die I Transformation r strain transformed into S strain doesn t not know if its DNA protein RNA that transforms 139 Avery MacLeod McCarty 1944 started with heatkilled Streptococcus cells removed lipid and carbohydrates gt At end of experiment DNA TRANSFORMING FACTOR 39139 Martha Chase amp Alfred Hershey 1952 studied T bacteriophage virus that infects bacteria to replicate itself gt What is the hereditary material gt Prediction of DNA hypothesis radioactive DNA will be located within pellet gt Prediction of Protein hypothesis Radioactive protein will be located within pellet gt Conclusion viral genes consist of DNA Viral coats of protein v Franklin Crick Watson and Wilkins proposed a structural model for doublestranded DNA 1953 gt Franklin and Wilkins use Xray crystallography to deduce structure of DNA 39339 The DNA double helix Bform DNA most common gt What type of interactions promotes the stability of the secondary structure of DNA giolzz m mb39e quot9quotquot I Hbond and hydrophobic interaction between bases counteracts the negative of the outer 39SUQarPhosphate backbone faces structure outwards base pairs gt 5 end phosphate group NOT bound to anythmg Nitrogenous bases It gt 3 end hydroxyl end NOT bound to anythmg face inwards ALWAYS go 59 3 Wham new 95221 00 Hbond occurs between bases basis for base pairing Wide major groove gt Guanine and Cytosine 3 Hbond MORE narrow minor groove STABLE because MORE HBONDS gt Adenine and Thymine 2 Hbond gt Relative Stability GCgtATgt AU v Structure of DNA provided physical basis for deoxyrib se L 5 C sugar v Chargaff s Rules gt AGCT 1 39339 RNA differs from DNA in 3 aspects a Nucleotide nucleoside c Nitrogenous bases 13 V r w 1 NM 0 o gt 1 R1bose sugar 2 Carbon has an OH o x N M m gt Urac1l mstead of thymme 1 0 0 Yr L o I o N o 5 1 quothas gt RNA doesn tnormally form a doublehehcal him Mme WWW Mmmmw 970w 2 4 structure BUT 1t can have s1gn1f1cant secondary mm MM IUgI39 quot structure 1 o b Sugars N quot F on oquot O 39 Guanine is Adenine A Riboso i Dooxyribou Purine Thl ahorl RNA utmnd l39 toldAd un min m mum rrqlnn39 In mm doublv trumiLd hvlu v39 Tea39ahymena ribozyme can carry out chemical quotenzymatic I reactions The DNA double helix phosporhous Nucleic acid polymerization can be viewed as a simple 5 340m DNA condensation reaction 39 Sugarphosphate backbone faces mm outwade base pairs Nitrogenous bases face inwards complete 7 Righthanded helix 33mm 7 e um mm 3 4 nm 9 uses energy in monc linkages to drive nuc polymerization rxn to in prevent in reversi 39 Wide major groove narrow minor groove Dolmen bases 034 I Distance I ribose quotquot minor But nucleotides in the cell exist as nucleoside triphosphates 1 Human genome 6 billion base pair Length of DNA in each cells is about 3 meter a How does DNA fit into a eukaryotic nucleus i HISTONES basic charge and DNA charged 1 DNA spontaneously folds up around histones 2 How to open up chromatin chain Acetylation of histones a Add acetyl groups to hiStOIleS One way to promote chromatin opening acetylation of histones lysine residue normally basic charged is removed c phosphate charged interacts 1 f my ionic relation With charge quot m 39 39 i 39 i 2 lin tillklllll n u Illll ll39l lle Jump 39l tquot ll39i pullnull rv39wuq mi 1h r m quotmam m not monophosphate Condensed chromatin hiStone histones Simon I acellylransferase 39 eace y ase remove charge on d ONLY 1n eukaryotes nd remove acetyl histones h1stones group and restore 3 RubinsteinTaybi Syndrome RTS Stone 0 former state HDAC caused by changes 1n genes that code for HATS Acetyl group Decondensed chromatin n h39smne 350q391zA quot393 Lecture 14 The Cell Cycle and its Regulation Nucleosomes other DNA packaging proteins chromatin Chromosome contiguous stretch of chromatin DNA packaged using DNA into chromosomes in eukaryotic cells CELL CYCLE O O 00 o 9 O 00 O 0 cell leiSlOl l cycle begins ilTliiOSlS Iv m if f u quot 7 cell prepares to dlvde cell grow gt gt gt During INTERPHASE centrosomes duplicate G1SG2 chromatin relaxed I G1 4 unreplicated chromosomes time for cell to increase in size and organelles I S and G2 4 replicated chromosomes each consisting of two sister cells Prepare for M Phase mitosis PROPHASE I Chromosomes condensed I Form microtubules made up of alpha and beta dimers form and end I end microtubules grow faster I each centrosome composed of centrioles positioned at 90 degree angles I mitotic spindles attach to chromosomes and draw them away MITOSIS I Prefer Plain Milk and Tea I PROPHASE Chromosomes condense and mitotic spindles begin to form I PROMETAPHASE nuclear envelope breaks down spindle bers connect to chromosomes at kinetochore I METAPHASE chromosomes complete migration to middle of cell I ANAPHASE sister chromatids separate I Kinetochore microtubules shorten and motor proteins pull chromo to opp poles I TELOPHASE nuclear envelop reform mitotic spindle disappear cleaved due to actin myosin ring Cell Division Begins cytoplasm divides when actinmyosin ring cause plasma membrane to pinch in Cell Division is complete 2 daughter cells cell deClCeS whether to continue 39epllamp of DNA Chromosome MUST be copied before cells can divide unreplicated chromosome replicated chromosome condensed replicated chromosome Alma Howard Stephen 1953 When in eukaryotic cell division does DNA synthesis occur gt Pulse Chase Experiment I Feed radioactive thymidine deoxynucleoside sugar base I After 30 min wash unincorporated T out of cell culture I Exposed to photographic emulsion Conclude chromosomes are replicated during interphase or S synthesis phase How long does this phase last Figuring out the length of S phase I engh ul 8 phase M p39wase ca 15 wquot rudiuact v39 1quotquot y u A u u y I M I Pulse 7 in i Chase fquot so if quot I Waited 2 4 6 16 hours then exposed to photographic Wquot j emulsion I n 12 us 20 Tlmc since and 01 rhyml c pulun may 3910 M pl39ase radloaclu I Cells asynchronous every cells is in a different stage of the cycle I Figuring out length of S phase What makes the chromosomes move gt Microtubule depolymerizatin pulls chromosomes towards the poles What controls entry into mitosis gt CONCLUSION mitosis promoting factor MPF is found in the cytoplasm of mitotic cells DIVISION M Parent cell DIVISION M Parent cell Sister chromatids 4 unre icatedmhfom 4 replicated chromosome 0 romosomes are Showquot each consisting of two si partially condensed to make chromatids them visible IGS62 GSG DIVISION M 32 OF INTERPHASE PROPHASE PROMETAPHASE Centrosomes Chromatin Early mitotic Aster Fragments Kinetochore of nuclear Centromere envelope with centriole pairs duplicated spindle centromere r r Nonkinel 39 quot V micro Nucleolus Nuclear Plasma Chromosome consisting Kinetocl envelope membrane of two sister chromatids microtul At start of mitosis replica Chromosomes METAPHASE ANAPHASE TELOPHASE AND CYTOKIt condense Metaphase 7 V7 Cleavage Nuc INTERPiIASE G1 s Ga 7 7 398 quot 39quotquot 39 39 quot39 During mitosis sister 39 quot 39 chromatids separate no daughter cells are formed by cytokinesis 39 Nuclear Centrosome at Daughter envelope ipim e one spindle pole chromosomes forming KlDO OChore Klmeochore Pm fibers A I Number of 39 e A Mlcrotubuie chromosomes 4 4 4 4 4 3 4 39 per cell 5555555quotf5 55 5555 s 39 39 connectmn at Pius end Minus a Number centromere is orDNA 4 broken quotquot19 molecules per cell 0 Number of hromosome movement chromatids 4 8 8 8 8 4 a Cycltn concentration regulates MPF concentration quot 39 I 39 39 39 39 39 cyclin A thus a cyclindependent 37 W quotquot amp39n WMPF ig39mm 3quot l MPF Cdk MPF is inactive until cdk loses its inhibitory phosphate MPF component concentration Then ls b Activated MPF has an array of aftecta mto Activated MPF active entry Into mrtosrs a n i Cyclin Cdk with phOWOWtw39 W m awn Lecture 15DNA SYNTHESIS 39 Many types of cancer cells have a shorter cell cycle compared to noncancerous cells gt Why I Mitosis is faster NO I G2 or S phase is shorter NO I Thymidine pulechase 9 Peak between 412 hours 9 G2 4 hrs S8hrs M lhr Gl 7hrs O For normal cells same length of G2 and S I Cancer cells have lost ability to STOP dividing I Most cells stop in G1 and depending on signalsgrowth condition they go to S phases 0 0r GXISt o b Activated MPF has an array of effects Coancoer cells DO NGT STOP 1n Gl Active MPF v MPF mitotic cyclin cyclindependent klnase cdk 939 quotquotm gt 2 phosphate groups are added to the cdk during G2 lamina mg nuclear 00W meltdown I one promoteskinase activ1ty Wm mm I one inhibits kinase act1v1ty I y mmm W W when the inhibitory phosphate is removed MPF gggggsbg mg39 e mm th becomes active 3323quot ooncenitmtlonydicflfniyc gt How to deactivate MPF Negative Feedback I Deactivated by enzyme activated during mitosis I Protease destroys cyclin 00 Checkpoints prevent cell from progressing inappropriately through the cell cycle Mphase checkpoints inactivating Pass checkpoints if t 3 1 chromosomes have 4 attached to spindle quot apparatus 2 chromosomes have properly segregated and MPF is absent Pass checkpoint if chromosomes have replicated successfully 0 DNA is undamaged activated MPF is present G checkpoini passed ATP transcription factor ADP 4 Inactivating 5 Phosphorylated 6 E2F trigger phosphate is Rb releases 52F production of removed and Sphase prote active Cdk phosphorylates Rb G1 cyclin s a molecular brake G1 CDK I lnactivating 1 phosphate Go quots l v I Activating Pass checkpo39nt 39f39 Mature cells do not g a target phosphate cequot s39ze 393 adequte pass this checkpoint Growquot 0 nutrients are suffucnent they enter 30 state actors 0 social sngnals are present I DNA is undamaged v Rb IS a molecular brake 139 G1 checkpoint G cyclin and E2F 1 Growth lactora 2 Growth factors 3 Cyclln binds v Fluorescnce ub1qu1t1nation cell cycle 1nd1cator mmmm other cause increase in to Cdk Cdk is cells cyclin and E2F phosphorylated FUCCI concentrations Rb inactivates 82F or cell itself by binding to it In eukaryotes the cell cycle is a way to separate the processes of DNA synthesis and mitosis Also time for cells to prepare for divisionBUT in bacteria cell division and DNA replication are NOT always separate Possible because bacteria don t have nuclei How is new DNA synthesized gt Semiconservative conservative or dispersive 1 it the hydrogen bonds between complementary base pairs are broken the DNA helix can separate 2 Each strand 01 DNA can serve as a template tor the iormation ot a new strand Free nucleotides attach according to complementary base pairing 1 st theory semi conservation mo 5 3 When the M We to i phoephete Dec secondary stni restored 2nd theory the DNA strand would separate then switch in and out At the end one s COMPLETELY new and old conservation model 3rd theory DNA strand replicate in short stretches dispersive theory gt MeselsonStahl Experiment Q J I Chromosome 2 Chromosome is located mid replicates cell 3 Chromosomes FtsZ ring 5 F pull apart ring constricts oi Ftsz protein Membrane forms and cell wall intold Grew ecoli in N15 soln then transferred to Nl4 allowed to divide once collected samples then allowed to divide again in Nl4 Semiconservative The strands separate and each gives rise to a new complementary strand Conservative Doublestranded DNA gives rise to a new 39r39 ex x Generation 0 Hquot K i x 1 xx Oeneretion 1 39 H I 39 15quot I N K r I r x L quot ammoquot a 2 E E E 1 Hybrid N Hybrid N rm 139 quot After 2 generations Alter 2 generations 1 burdenle DNA 1 highdensity DNA 1 2 intermediatedenslty DNA 314 lowdensity DNA Dispersive Doublestranded DNA replicated over short doublestranded structure stretches 39 2 Hybrid ngld 932 After 2 generations All Intermediatedensity DNA gt RESULT semiconservative model each parental strand is used to synthesize a complementary strand 0 3 What happens during DNA synthesis gt Deoynucleaoside triphosphates are added to template DNA polymerase enzymes that catalyzes this 9 have MULTIPLE points of origin gt Lagging and Leading strands gt This causes strain on DNA strand as you unwind it Okazaki fragments Of39g39n Direction of l 39 Direction of fork movement lork movement Lagging Origin Lagging M A 339 Okazaki iragmer ts Bacteria only have ONE point of origin eukaryotic chromosomes Solution DNA synthesis is continuous on one strand the leading strand but discontinuous on the lagging strand discovered by Reii Okazaki r lirn puu strands I I r Litwmdi39iu a I and rrpl r quotinn 2 It Syri MES zed tnt Legging Discomu DNA 3y I39 39 quot Leading Continue DNA yn I39l rV yv u Leadin strand quotquot g Sliding Clamp quotVquot Vquot quot quotquotu quotNINJNquot39MVMWVFMWM g mmm v nmmmmmnquot Mr lw lw w qulw qw 39 quot 39 39quotquot39quot5 5 3 DNA polymerase Last Okazaki fragment Lagging strand 5 ererw lwrMquot quot39llllllllllllllll39llquot 7 v vuvvu In 1 quotrun vvru a repeated sequence TTAGGG in r r 7 Telomerase is an enzyme th an RNA template to add DNA nucleotides to the ends of vvv TYI39AOOOYYAOOGYYAOOO SSBPS Hencase 3 DNA polymerase synthesizes the last Okazaki lragment in lagging strand 39 39 39 T 39 39 quot chromosomes nationalism mmquotmwcmmmma 3 v NWNWNWNWNWNWnNNN s W a 203 mncmmnenevcncmmronwcg39 WWawnwawawaw39wwawawawawawaw its J 3 7 IM in succes Ww wqw wqwqmm 37735 3 Unreplicatedend quotWW Lecture 16 DNA replication and repair 7 Telomeres 00 DNA replication in eukaryotes same idea different protein names l39 fxxfsszyga DON T NEED TO KNOW quot39 quot 39 quotquot gt Bacteria s chromosome is circular but MOST eukaryotes have linear chromosome gt Issue What happens when the replication fork reaches the end of a LINEAR chromosome I Primer is removed but then NO DNA synthesis on lagging strands chromosome is shortened 70 nonorage gt Solution 2 50 y 3 139 Cells with short telomeres stop dividing 5 quot 398 39 H5 gt Late 1980 s Calvin Harvey showed that telomere shortening can cause 2 40 a 2 cells to exit the cell cycle enter into G0 and become senescent f39iTi 3939f i11392quot Telomere clock 53 30 39 c gt Telomerase is expressed in germline produces sperm and egg cells 39 3 but not in most somatic cells gobo 6000 7000 8000 9000 whoonboo I Thought that most somatic cells have a limited number of divisions InmHIomm numbu am gt Could aging by slowed down if telomere clock can be turned off I Become like cancer gt Cancer cells require the ability to produce telomerase cancer cells considered immortal I Company in Cali developed drug that acts as a competitive inhibitor of telomerase binds to telomerase RNA therefore can t extend cancer DNA strand v What happens when mistakes are made during replication gt Predict error 1 10100k bases polymerized I Humans over 6 billion base pairs in genome gt 1st line of defense some DNA polymerase have proofreading ability I DNA polymerase adds mismatched THEN detects and removes it I Howstructure of polymerase III 39 I I Exonuclease removes base from growing chain quotI Q 4 llllil Ill I where exonuclease is Q I gt 2nd line of defense Mismatch repair a x k i I Destabilization when mismatch occurs that allows it to move a DNA polymerase adds a mismatched base 5 OH 339 TTCTGTCCATCGCT 3 b but detects the mistake and corrects it E S39rrmmw TTCTGTCCATCGCT 3 if mistake NOT eliminated by exonuclease how does a cell KNOW which base 0 seo strand is the right strand Mismatch repairs uses 3 proteins 0 recognize and remove mismatches on the newly synthesized DNA strand MutSscans binds kinks it MutL MutH nicks I In E coli adenines are methylated by enzymes after DNA is synthesized Takes a slow or no DNA synthesis b removal of mismatched nucleotidelts mispaired f thumb last base 3 palr x 4 sum 1 he m A l 1 r t 392 r 39 51 ns rs gttgam 5 l 5 3 3m 4 quot13 g exonuclease R W it cut active site c resume DNA synthesis L r i time so newly synthesized strand of DNA is unmethylated for several minutes scans m 7 a 3 Y MutL t 539 U n39inal lggsinlck gt A CH3 1 c I binds mismatch leL r quot i 3quot I 39 539 x wh 5 J D MU H i 39 CH quotwk lexonuclease 3 r I 39 m 139 539 CH3 CH3 DNA polymerase I 3 i I 339 39 I 7 sum 5 mismatch repaired 1 s gt Is mismatch repair really important 39 mmwm I Heredity colon cancer is cause by mutation of MutH and MutL 5 c Cquot 39 gt Other lines of defense repair processes outside of S phase 1393 ma I EX DNA damage by UV light 3 6 A T A as quot39 I Lesions Thymine dimer covalent bonds between adjacent thymine 5 3 3 I How can thymine dimers be repaired by nucleotide excision repair I ZNucleotide G A c c A A T A a excisionexonucll 3 5 3 539 393 3 Nucleotide o A c c A A A a replacementch 339 539 Repaired damage r 39i 5 3 o A c c A A A a quot quotquot ading strandlsynthesis 3 s Lagging strand synthesis discontinuous 4 relieves strain by breaking 2 one of the phosphodiester A is 39 MWMMW quotNA PMquot bonds helps it unwind Toaisomerag relieves twisting forces RNA Iquot t r 393quotstrard A v 39 xx xquot 1Prlmersdded Wmquot U 39H quot7V5 xquot x A xx 39 39 7 3 s 39l quot Topoisomsnse A 5 1 SSBPS Helicase39 Iquot 39 7 v Helicase opens double helix Primsse Singlestrand DNAbinding proteins SSB stabilize single strands M is opened unwound and primed 2 quot3 quotgmen lregment synthesized I n M w 3 E Sliding ciamg holds DNA polymerase in place n 5 M A quot 5 I DNA 0 se lll works in 539 s 339 direction synthesizing leading strand M RNA primer snain cl Lesdingstrand 39 39 4 A xa 2nd Okazelrl moment 3 Second trsgment 1quot 0mm mgmm I quot39 synthesized 39 539 A A x M g M inquot I removes RNA replace w DNA polymerase I we 4 Primer replaced DNA p quot quotquot quot a 5 3 uili 3 5 e II39 in p s DNAllgsse doses 3 C l 5 cap closed a g p 3 s w V I39VT v v v v v v v y v v v v 0 Heliease unwinds Hbonds to unwind SSBP stabilizes single stand rimase adds RNA primer DNA polymerase III synthesizes leading stand Sliding clamp keeps DNA polymerase attached Topoisomerase relieves tensions The products are separated by pillary gel electrophoresis the identity of the ddNTPs is determined using a uorescence detectc and the sequence of the synthesized DNA is determined L y 739 11quot LECTURE 17 APPLICATIONS OF DNA SYNTHESIS IN MLECULAR BIOLOGY p H gt What is needed for DNA replications gt Cycle 1 Denature 93 quot Output i1quotii39rquotlquot39lquotll b 4 Separate fragments detector 5 Read output 539 Primers not incluoed sine wasn39t synthe 5 G Y A A Y c c G 3 synthesized strand Helicase primase DNA pol dNTP topoisomerase q mow j if a SSB proteins DNA ligase telomerase eukaryotes 6 5 CMquot quot clamp DNA pol I removes primer in bacteria M template Sequence of nontemplate DNA 0 Underlined ONLY things you need to create DNA in a test tube Kary Mullis incented PCR polymerase chain reaction in 1983 PCR makes million of copies of a short DNA sequence 50 base pair 5 kilobases Can make copies even if DNA sequence is part of a complex mixture ex human The Sanger method logic behind DNA sequencing genome ONE CATCH have to know the sequence of DNA that you re trying to amplify How to design primers for PCR 0 PCR primers must be located on either side of the target sequence on OPPOSITE strands must be in 5 9 3 sequence DNA can be made singlestranded by DNA helicases or heat disrupt H bonds How does PCR work 0 Multiple cycles of O Denaturation of DNA at high temperature INSTEAD of helicase O Annealing primers cool down 9 DNA synthesis by polymerase mid temperature Applications of PCR any time when you have very little DNA and need to make copies for furthers analysis Detection of mutations screening for inherited disorders Bioterrorism prevention 0 Prenatal sex determination 0 Screening for genetically modified foods 0 Forensics Paternity testing Primers 21 539 39 quot 339 39 539 39 xa dNTPs O 1 Start with a solution containing W synthesized primers and an abundant supply of the four dNTPs 3 2 Heating loads to denaturation of the doublestranded DNA pairing DNA to be sequenced is mixed with a single primer DNA polymerase dNTP chainterminating dideoynucleoside triphosphates ddNTPs are included in low concentration p P P P P 5 0 Base 5 CH2 0 Base Normal A ddN dNTP V tern extends 3 8 H syntl DNA strand 1 W quot 0 Cu gquot l 1 Incul 390 reactloi I mixture Template DNA 3 5 COAAVCAVVAV chain tern a 3 Collect on strands that v are produced I h lA l when ddNTP are added to DNA polymer prevents formation of a phosphodiester linage with an incoming nucleotide 1 Anneal primers and synthesize new DNA Cycles 23 Repeat denaturation annealing and DNA synthesis steps 5 339 5 3 cycle 2 m 7 W 539 3r 5 r W 3 i 5 I 3 5 rm 5W7 W 31 5 3 5 3 Primer annealing 4 Extension m During incubation Tag polymerase uses dNTPs to synt esue complementary At ggglgr temperaturesI the primers bind to the template DNA by Denaturation 5 Repeat cycle cycle 3 alter cycle 3 that desired product appears 5 Repeat cycle again of three steps 24 again doubling the copies of DNA up to 2030 times to produce millions of copies of template DNA complementary base DNA strand starting at the primer Haplold n l Hlplold glmotu n 23 Dlplold 2n 39 ovumquot j 3337 Lecture 18 Meiosis 139 Karyotype connection between DNA and Genetics in gt Show and type of chromosomes present in cell A ozo Humans 23 haploid 46 diploid 39 139 Ploidy of choromosomes of each type that are present in a cell gt Unusual karyotypes I I MIOIII Ind 139 What IS dlfferent about a cancer cell karyotype 312337 quotquot39 quotquot quotquot gt no longer diploid PLOIDY OFF not homogenous in color 2 chromosomes come together and get stuck Y chromosome is missing I 2 checkpoints during mitosis checkpoint at What is crossing over metaphase line up properly and anaphase separate pairo properly Crossing over occurs when iair Of SiSter Chromatids homo39ogou gt dur1ng me1051s chromosome exchange among each other 333337 337337 improper crossmg over heme mquot he39 l i o How do organlsms produce progeny W1th dlff tra1ts crgs39fs39fgggfg l39gzemw doublestranded DNA break gt In order to produce hap101d gametes of chromosomes 39 39P39 P 39Se939 93 quot f chromosomes durlng MUST be halved melos1s mews39s 39 synaptonema 00 What is crossing over quot quotequot gt Crossmg over occurs When nonSlster chromatlds physically exchange parts between each 0th I later dammed erosns gt Held together at Synaptonema complex Homologous chromosomes 1n replicated gt Separatlon oi homologs CAN T occur separate untll connectlon at chlasma 13 broken 2n replicated 39 Summary of meiosis diploid 2n 2n replicated reduction in ploidy DNA replication x quotmoss 39 I 4 Metaphase 1 n migrate to meta 6 Telophase I and replicated plate 2 39 c okinesis p opposite sides of cell cg mmowms moquot m reduction m synapSIs of homologous opposite sides of coll Meiosis II DNA content pairing of M and F then coll divides 2n replicated Q Q g 1 DNA replication 39quot quot quot DNA replication 2n replicated 2n replicated o metaphase separated homologs 1n replicated 1 separate sisters 0 gzrrsate 391 Q Q Q s to Tolophase II and Y J K n Meiosis II Sister chromatids separate metaphase aid 8 Metaphase ll Clvomosomos line Cytokinesls upatmiddloolcoll Chromosomosrnoveto muons plate moving to opposite opposite sides of coil old at coll than coll divides Lecture 19 Sex determination and mistakes in meiosis 9 9 disadvantages to sexual reproduction gt asexual vs sexual reproduction there are only 12 as many childproducing offspring in the sexual pop only female can have a child gt if disadvantageous then why exists I lst idea purifying selection hypothesis sexual if 2 parents and one has a mutational gene it is likely to be offset by the Spennatogenesis Oogenesis 2 stem cells telomerase active Il39 Spemalogonium an I 39 Dogoriuum uan I May drama by mitoys to norm more mammoan I Mtosrs and l Mutasusand l clmerenuauon 39 clumenuanon 39 Prim Primaw 00er 2n Meuosasl V P Meiosisl 3quot Secondary a Seconoar 00 l spermatocyte nl 1 if Meiosis II J k f Mantels lIl i i t n quot arent 90an to Q Q 3 Q sne39mm 3 Q 03m 9 9 9 become Vn 39 l l l 39 L pow bodies W mature eggs 0 O O J tune sperm ll Mamie egg cell ovum in ellls n3 4 hap oid gametes 1 mature cell and 3 polar bodies tha eventually dismtegrate normal gene and if male and female mate only 2 or 4 of kids will inherit the mutational gene 2Ild idea changing environment hypothesis outcrossing is favored exposure to pathogens favored outcrossing I 9 v NONDISJUNCTION when chromosomes don t segregate properly in environments where evolving pathogens are present sexual reproduction by a Nondisjunuion in meiosis I amclcs ngotcs 39iisiosm 39fquot has if NONDIBJUNCTION 1 I a 4 gametes n 1 IirJ aneuploid 39 ll 39 139 quotPM 1921 quot 39 39 l K was n b Nondlsjunuion in moioslscl llmuu J wows HEIOSIS u 39 oniy2 gametes gtlt l aneuploid 1 Mom I an Why t We run on menu ligaltv i T 133033 Monosomlc thvutbhupmmdcou Wot Ham v on too my at on too kw taill I aneuploud NOT normal amount gametes N M I d l orma upon r 2n 139 Down s syndrome can be cause by an extra copy of chromosome 21 trisomy 21 gt Frequency of trisomies INCREASES as a function of the mother s age gt WHY Eggs are arrested in prophase I of meiosis until ovulation I May allow more time for damage to cohesins 99 after prophase I Misregulation of cohesion breakdown can lead to nondisjunction can t add cohesion gt What are cohesins Current model is that several proteins form ringlink structure around 2 DNA molecules 9 beginning of anaphase Cohesins CM mitotic cells o o 0 I n 0 n I 39l u o quotm quot n 0 mm t4 n ID early lalc mcmphasc anaphase prophase prophase chromosome acuon of W condensation scpann 39 C COhCSIl COlnpleX 1 NOW 39 l agelrer A I J saplm cc u pvt Nat L39c Hal 3 l I Formed during S phase when DNA replicates rings open sequentially during mitosis w centromeric cohesins degraded at Dcutlostmnd mu manner During meiosis l cohesins at the tips of chromosomes are lost During meiosis ll cohesins at the centromeres are lost 3mm hvcmnl J l39lquotquotln Cmmwwv A II lt C mesaer 39 itth quot1 L LII CI cohesion act3 rIIOlFZISIZFIE am gt 39 gt loss 01 adiosinn at ccnlrcmcies rrx mug v Sex determination in humans gt Regions at a tip of the chromosomes on Y chromosome that recombine w X chromosome Sexdetermining region Y SRY gene I Psuedoautosomal regions regions on Y chromosome that can recombine w X SRY transferred to X9 XX male SRY lost from Y9 XY female SRY encodes a protein that causes other genes that determine maleness to be expressed Sex determination vary for diff organism XXXO sex determination grasshoppers Females are XX males are XO ZZZW sex determination birds Males are 22 homogametic Females are ZW heterogametic Haplodiploidy bees wasps ants Sex based on of chromosome sets in nucleus I Males haploid n femals diploid 2n 39339 Environmental sexdetermination gt Temperature during embryonic development snakes and lizards gt Position in group mollusk s position in a stack of limpets 139 How does meiosis contribute to genetic diversity gt 1 independent assortment of traits gt 2 crossing over during meiosis promotes recombination between diff chromosomes I additional genetic diversity I allows biologists to gure out location of disease genes 39V39Vo VVV During meiosis I tetrads can line up two different ways before the homologs separate ewe I X 999 Brown eyes Green eyes Brown eyes Green eyes Black hair Red hair Red hair Black hair The red and blue chromosomes can line up in different ways 39 during metaphase 1 3 LA 9 anAn A11 393 luAMm A A n E EA lgil Hr parent Dominant allele Recessive alle for seed shape for seed shapi C hromosomes replicate R Lecture 20 Mendelian genetics 39339 A diploid organism can produce 2n chromosomes n n Allelrsegzjfelgate 39 r gt For humans 84 million gametes that can be produced AX l gt of offspring with diff genetic combinations is 7 X 1013 genetically Mewsquot distinct offspring gt crossing over increase diversity even more v Mendel s work Gametes Dominant allele lor H m 1Y9 us gt Why was it so successful r f jjzfm m ther I Idea model garden peas Could control mating Easy to study quot quotquot 39 quotMeiosis gt Started by cross between 2 true breeding plants homozygous Femmegamems gt Year 1 monohybrid crosses R X rr Homozygous 3 r gt Year 2 monohybrid crosses with F1 plants Rr X Rr 3932 in 0 gt Year 3 monohybrid crosses with F2 plants Rr X Rr Me39os39sg m Offspring genotypes All Rr hetero Offspring phenotypes All round set 0 v Genes eXist in diff versions called alleles gt Genotype refers to the combination of alleles for 1 genes gt Diff alleles occupy the same locus on homologous chromosomes XS Results of monohybrid cross illustrate Mendel s principles of segregation a result of meiosis How can you determine whether an organism is homozygous dominant or heterozygous gt R could RR or Rr gt Use a test cross where you cross homozygous recessive individual 00 Shortcut gt Probability I For mutually exclusive events ADD the probability I For independent events that occur together or consecutively MULTIPLY the probabilities I Probability that fournucleotide DNA sequence in Ecoli will be 5 GATC 3 O 025A4 1256 Mendel observed a 31 ratio with nuusmmaartwu m ummmv a 9Ver F1 selfcross that he af rmed b During menosus I tetrads can line up two differe W W ways before the homologs separate arc ranc mg enzyme F 5 H I o o en me involved in 39 quot 39 W O 0 gene of unknown function gene of unknown function h 39 transcription factorthat quot 1 39 39 quot X X l regulates expression of chalcone s nthase 7 a pigmentgene tj 39 39 39 jugs 39 gene of unknown function gt 3 in Purple White Purple White 39 39 Tall Dwarf onverts an inactive form 39 394 k Dwarf Tall fgibberellin plant growth 9 I39 3 ormone to active form n a 1 I KEIIPI MII I39OSS W39IIXQ39 eyed female crossed wilh reamed male 39 P generation White even 52 d ryed female mah e to V convention Red eyed Whue eyed a male male 5 F2 generation 1 of each 3quot 1 39rquot 139 39Y xiv 39 39r M39l r39 r Iquot 39 WAR Lecture 21 Sex linkage and pedigrees O 00 O 00 O O 00 0 Thomas Hunt Morgan 18661945 Drosophila melanogaster gt Short generation time cheap to raise gt Mated whiteeyed male y w a redeyed female I Result all progeny had red eyes gt Then mated F1 progeny I Result ratio of red eyes white eyes 31 I EXCEPT no FEMALES had white eyes gt Did reciprocal cross Morgan mated whiteeyed female w a redeyed male I Result ALL females had red eyes all males had white eyes I Then interbred the F1 progeny Sex chromosomes I Result 12 progeny had white eyes irregardless of sex 9 White did NOT act like most recessive alleles male Female 39 Grasshopper X0 XX gt Morgan 5 proposal Drosophlia XY XX I Red allele W or is dominant to white w Human XY XX I The w gene is on the X chromosome Xlinked Bird 22 ZW I Genes on the X chromosome do not have alleles on the Y chromosome I Males are HEMIZYOUS only 1 member of a pair for Xlinked gene Xlinked dominant traits Affect both males and females affected male MUST have affected mother Xlinked recessive traits appear more often in males and are not passed from father to son Ylinked traits ONLY affects males Autosomal Recessive appear with equal frequency in both sexes and seem to skip generations gt Werner Syndrome missing helicase causes rapid aging I 23 because can cancel out the double recessive from Punnett square because she is either homozygous dominant or heterozygous Autosomal dominant traits appear with equal frequency in both sexes and do NOT skip generations Lecture 22 Genetic linkage O 00 Gene 1 Gene 1 7 Gene 2 Crossing over occurs frequently between genes that are far apart Gene 3 4 XVI 3x W Crossing over is rare between genes that are close together IIIIIIIIIIII xw xquot V Meiosis II I How do inheritance patterns change when 2 genes are on the same 2 8 F chromosome E 5quot Q gt Two rather than four types of gametes m 1 xwY xwy xWV xWY Crossing over can result in separation of linked chromosomes gt Recombinant usually smaller THAN parental must be dealing with genetic linkage If genes are completely linked NO crossing over Gametes AB AB ab ab If genes are linked but crossing over between nonsister chromatids can occur during meiosis I prophase Gametes AB ab parental aB Ab recombinant INCREASED CROSSOVER FREQUENCY CORRELATES WITH INCREASED DISTANCE BETWEEN GENES discovered by Alfred Sturtevant recombinants o total progeny X 100 gt recombination frequency is also expressed in map units mu which are sometimes called centiMorgans CM single nucleotide polymorphisms SNPs single nucleotide changes after sequencing many human genomes millions of genetic diff have now been identified gt SNPs can be used as positional markers if they are located outside of genes gt SNPs can also be mutations that affect phenotypes if they change the expression of genes or the info that they contain gt How do clinicians use them I Probability of getting certain disease Recombinant chromosomes Recombination frequency Carrier has the recessive allele but not shown therefore heterozygous Colorblindness and Hemophilia are Xlinked recessive traits Lecture 23 Special topics in Genetics 9 00 Delormmzhon of ABO Blood Group by Multiple Alleles Product of Product of Product of IA allehlle l3 allele i allele 39808 galactosamine 931901086 l K r J Glycoproteins in plasma membrane of red blood cells Recessive lethality codominance multiple alleles gene interactions IAIB BIB Genotype IAIA IAi IBi ii 1ncomplete dommance guant1tat1ve Blood type a g g a g o traits and polygenic inheritance Recessive lethal alleles gt Y acts as a dominant allele for fur color but a recessive allele for DEATH Codominance Human ABO blood type is an ex of both multi and codominant alleles gt 3 alleles IA IE I IA 1B codominant I is recessive I Human blood types are determined by glycoproteins gt 1 gene 3 diff alleles encode enzymes adds sugar to glycoproteins on RBC S Y dominant for yellow fur color but RECESSIV allele for death P generation Yellow f x Cametes JD mm F generation Yellow lt N d Dead Yellow Nonyellow Itrr My quotmy Conclusion YY mice die and so 23 of progeny are Yy yellow V3 of progeny are yy nonyellow Product of Product of Product of Phenotype IA allele Ia allele i allele Genotype Blood Group nee Blood ells M ivquot quot quot fggamme galactose l39l39ml39r A 31221198 A A or O REC m9 l r or m I tg moan B m N B or O Glycoprotelns v in plasma 307 N membrane of l l39 u w 3 AB Nquot u red blood cells antigen u MAMquotBN O I i O Genotype I M Ml BIB 1 Bi ii antlgen A A8 A 8 0 Blood typo I 139 Gene Interaction two genes interaction to control one trait segregating independently O O 00 00 as bb cc M 88 CC pamma39 pureline while pureline red x mam l How many possrblc mm 1 rod genotypes 4x4x4 64 h 48 80 CC 41 of genotypes in a generation pmquot meam fed monohybrid CI39OSS l kenfertilization F2 f 20 most look IIKO Fl parents A 900039 390quot out wrdo range am i 1 Shonowpos gone 80cc Q88 23 g 2 Mbch 25 X3 4 AaBbcc 2 AaBBcc 4 MBch i 6 154 lanBBcc 2Aabbcc1Mbbc 2 1 1 Aabbcc 4 WC 2 aaBBCc 4 418060 MBBCc unlinked 2 mo led reu x Y39y C39e x Y yC39 W C no le T I i I hm I been Orenge CreeI Yoyc J qu39C39 MHZ MeyyC39 39myytt 2 unhnked genes each w 2 allele WOMiluednlen enc l Incomplete dominance blending of traits usually seen when pigment is involved Quantitative traits traits determined by the interaction of many diff genes polygenic gt Can be in uenced by environmental factors Parental generation x RR 1 IT 0 blending of trails F1 generenon 39 l R39 Selllenilizalio 39 F2 generation quot O F l I 0 MRI l4Rr 14Rr l4rr Purple Lavender While Whlle Pink 0 Red Kernel color Q 2 aaBba l aabbCC 2 aaBbCC 1 aaeacc 2 Aaasc sabbcc 2 sabch 4 1338ch 8 Me 4 AaBBCc 2 MBDCC o 1 2 3 5 6 Wm Number 0 red pigment alelos A 8 039 Cl in WWW AABBCC rod 121 1 1 1 LeCture 2439 Genomics and mediCine Enzymeactivity Fullenzymatic 50enzymatic Noenzyma1 I Family hypercholesterolemia patients only produce 12 as much of quotquotm Wquot Wquot W LDL low density lipoprotein Fewer LDL particles are taken up quoti i mquot I jy me 396 Ty39 Phe 1 by cells via endocytosis elevated blood levels of cholesterol Phenom NORMAL atr g qg iet m AUTO SOMAL DOMINANT gt Phenotypically dominant if present will show I Applying genetic principles to human diseases gt The severity of hyperchlo is dependent upon environmental factor I Eat low protein diets to lead a normal life for those that are hetero High as phenylalanine concentration in vivo I pre v 2 case studies gt 1 Huntington s Disease a genetic disorder characterized by neurodegeneration and impairment of motor amp cognitive functions autosomal dominant I in HD patients basal ganglia responsible for motor movement is most Murrow WW affected L childhood I How was it discovered Nancy Wexler my I gene IT15 which codes for 3k amino acid protein called Huntington gene has a large of CAG trinucleotide repeats at its 5 end I Huntington s disease shows GENETIC ANTICIPATION as the disease is passed from generation to generation the severity of the disease increase and the age of MM onset decreases I WHY 9 What cases the repeat expansions Perhaps faulty DNA replication or repair gt 2 Charcot MarieTooth Disease inherited neurological disorder that mainly affect the peripheral nervous system and results in muscle weakness and atrophy I James Lupski has CMT had his genome sequence as well as his family members analyzed discovered 2 causative mutations for CMT I R954X mutation changes an arginine amino acid in SH3TC2 protein into a stop codon I Y169H mutation changes a tyrosine amino acid in the protein into a histidine amino acid I SH3TC2 protein is required for proper myelination of axons in the nervous system Personal genomics branch of genomic concerned w sequencing amp analysis of the genome of an individual gt Uses SNP analysis chips typically covering 002 of the genome partial or full genome sequencing I once genotypes are known individual genotype can be compared with published literature to determine likelihood of trait expression Phenylalanine oxidation rate in vitro Manifestations Mild Grandparent vParent I Grar O 00 Lecture 25 Molecular basis behind the blue phenotype v mutant allele responsible for the blue skin tone is a mutant version of a gene that codes for the enzyme NADH diaphorase found in red blood cells gt in the presence of oxygenop Lecture 26 How Genes Work Central dogma v Understanding how gene expression occurs techniques that allow us to conduct this type of research DNA Q1 Ni f r 3 mRNA i mRNA 1 Ribosome Protein v Gene expression is the process of translating the info in DNA into functioning molecules within the cell v Phenotype outward physical manifestation of internally coded inheritable info blood types blood levels gt Sometimes by changing the environment lifestyle temperature diet can affect the phenotype 00 What do genes do gt Beadle amp Tatum worked with Neurospora crass bread mold I Figure out what something does by making it defective KO Lossoffunction alleles nult strategy I Led of one gene one enzyme hypothesis I Auxotroph organism that can t synthesize what is needed for growth Using genetics to examine metabolic pathways The onegene oneenzyme hypothesis Using genetics to probe metabolic pathways Neurospora crassa Xravsx Precursor thine ltrulline gt Arginins Neuros am my Meau Enzyme 1 Enzyme 2 W Enzyme 3 Supplement type V None Ornithine only 1 Citrulline only 39 Argininel L3 3 Wildtype 39 minimal COMBINE medium 7 arg 1 no growth GROWTH GROWTH GROWT nladlum quot 39 Mutant type arg2 no growth no growth GROWTH GROWT check for auxotrophic arg 3 no growth no growth no growth GROWT Mmm quot quot mutants t I medium Precursor gt lthinel gt gt Arglnlne LN Enzyme 1 E e 2 Enzyme 3 Minimal auxotroph V r medium 2132 Precursor gt ithine r gt Citrulline Arginine Singleamino EEEE E EEEEEEE medium 7 1 acid arq1 argz arg3 4 a S t Z lt g i 5 g 3 a V v What is the central dogma of molecular biology The central dogma summarizes the ow of info in cells It states that DNA codes for RNA which codes for proteins v Many genes code for RNA molecules that do NOT function as mRNAs carries info out of nucleus and are NOT translated into proteins tRNA rRNA microRNA gt Sometimes info ows in the opp direction from RNA back to DNA reverse transcriptase gt George Gamow s triplet code gt Nirenberg and P Leder had to figure out the amino acid coded for by the 3base code 39339 THE GENETIC CODE It is redundant degenerate almost all amino acids are encoded by more than one codon It is unambiguous one codon ONLY codes for one amino acid It is nearly universal with few exceptions all codons specify the same amino acids in all organisms It is conservative the rst two bases are usually identical when multiple codons specify the same amino acid Most common start codon is AUG I 3 termination codons UAGUUUUGA RNA polymerase polymerize nucleotides into strands of RNA Transcription process of copying hereditary info in DNA to RNA Translation use info in nucleic acids to synthesize proteins V V VVV O O O 00 00 00 W amino acid equence b Phenotype 9 39 Phenytalamnc Phc Tyrosine Tyr 32 Cystelno Cys U 39 Senne Ser quot UUA Lcucmc Lcu UCi UAA Stop codon UGA Stop codon UU UC UN Stop codon uGl Tryptophan Trp Lecture 27 Mutations and cut cc CAL39 H d H CGL o o o m I Sh Ine ls overv1ew of transcription 0 CU LeucmelLeu cc prolmelpm CA cm ArgmlnctArg lt quot Glutammc Glu o v Mutation any permanent m CU cc CA cm l a AUU ACl AG 1 change in an organism 5 DNA gg AU WWW AC MK AsparagmoMsn my SennelSer E AUA Ac Threonlne Thr AAA AGA39 I Methionine Met Lysmc Lys Arglnlne Arg AUu AC AA AGL Start codon 33 g 39 Aspartic acnd Asp I Valme Val Alanine Ala 39 39 W thcme Gly GUquot C39C 0 quot Glutamic acid Glu 06 GUN CC GA quot GG gt Point mutation single base change Name De nition Example Consequence Original DNA sequence of nontemplate coding strand Original polypeptide Silent Change in nucleotide that does not change amino acid specified by codon Change in genotype but no change in phenotype Usually neutral with respect to tness Missense Change in nucleotide that Change in primary structure of protein replacement changes amino acid specified may be bene cial neutral or by codon deleterious Nonsense Change in nucleotide that Premature termination polypeptide results in early stop codon is truncated Usually deleterious l Frameshift Addition or deletion of a Reading frame is shifted massive nucleotide 7 missense Usuall deleterious a m m y gt Chromosomelevel mutations inversions translocations deletions duplications poly and aneuploidy is diff of chromosome copies v Step 1 Transcription of a gene production of a RNA short lived version of the info on the DNA gt Step 2 sequence of nucleotides in mRNA is translated into a sequence of amino acids protein v Transcription Tc process where RNA is synthesized from a DNA template DNA 539 i y gt Requires NTPs w ribose as sugar ribonucleoside triphosphate 3 5v gt NO primer needed for RNA to being Tc gt Transcription in PROKARYOTES 5 I DNA template strand that is read by the enzyme 0 I Promoter region of DNA that is start site for transcription I Termination signal for RNA polymerase stop Tc I RNA polymerase enzyme that Makes RNA NTPs I Catalyzes the formation of a phosphodiester bond between the growing 3 end of the RNA chain and the incoming ribonucleoside triphosphate I Sigma factors detachable protein subunit must bind to the polymerase I Sigma binds to promoter region STILL DS not SS I RNA polymerase and sigma form a holoenzyme 139 Which DNA strand is used as a template to synthesize RNA gt Either strand depends on the location of the promoter v Promoters a region of DNA to which the transcription machinery holoenzyme will bind in order to initiate transcription TATAAT at 10 box and TTGACA at 35 box gt Consensus sequences best match when a large of similar sequences are compared gt For promoters DNA sequences that are closer to the consensus sequence are often stronger promoters v Tc initiation complex RNA polymerase sigma promoter site neg pos Upstream Downstream Slgma Jontemplate trand I Promoterl RNAcoding region 1 gt WAS s 3 5 empate Transcription Terminator Transcription and start site termination site RNA transcript 539 339 Core enzyme v Transcription in Eukaryotes gt 1 Eukaryotes have 3 polymerases RNA polymerase I II and III I Each pol transcribes only certain types of RUNA I RNA pol II is the ONLY pol that transcribes proteincoding genes gt 2 Promoters in eukaryotes are most diverse gt 3 Instead of a sigma protein eukaryotic RNA pol recognizes promoters used a group of proteins basal transcription factors gt 4 Termination of eukaryotic proteincoding genes involve a short sequence polyadenylation signal poly A signal I in eukaryotes transcription ends a variable distance from the polyA signal as the RNA polymerase has to fall off the DNA template 00 Transcription and translation occurs concurrently in bacteria as there is nonnuclear envelope to separate the two processes gt Initiation factors bind 5 cap of mRNA together With the small ribosome subunit 139 In bacteria RNA folds back to form short double helix loop hair pin loop Worm quot393quot cleus zp NAMature mRNA quot W Mature mRNA Wanslation Ribosom in cytosol Protein Lecture 28 RNA processing overview of translation 0 v Transcription factors tells RNA polymerase Where transcription starts 139 What does the complex look like gt Promoter TATA promoterproximal element Basal transcription complex RNA polymerase 11 others 00 Eukaryotic RNA polymerase Name of Enzyme RNA polymerase I RNA pol I RNA polymerase II RNA pol ll RNA polymerase III RNA pol Ill Type of Gene Transcribed Genes that code for most of the large RNA molecules rRNAs found in ribosomes Proteincoding genes produce mRNAs also genes that code for RNAs that function in ribosome assembly and in processing and regulation of mRNAs Genes that code for transfer RNAs tRNAs for one of the small rRNAs found in ribosomes and for noncoding RNAs ncRNAs also genes that code for RNAs that function in ribosome assembly and in processing and regulation of mRNAs Promoter proximal element v Unlike prokaryotes eukaryotes have stretches of intervening bases that DON T code for a product then combine separated sequences into a Whole gt Some of the primary RNA sequence is spliced out of the nal transcript gt Regions that are part of final mRNA exons and regions that are removed are introns gt RNA splicing snRNP small nuclear RNAS snRNAs amp a complex of proteins I Groups of snRNPs spliceosomemultipart complex I Spliceosome catalyzes the looping out and removal of introns introns destroyed 139 Eukaryotic transcripts add 5 cap and a 3 polyA tail during processing cap and tail help gt Prevents mRNA from being degraded by enzymes present in the cell amp serve to help regulate amount of protein translated Transcription in Bacterial and Eukaryotes Point of Comparison RNA polymerases Promoter structure Proteins involved in contacting promoter RNApror essing Bacteria One Typically contains a 35 box and a 10 b0x moiiierent versions of sigma bind to litterth promoters None Eukaryotes Other basal transcription factors Basal transcription factors associated with TBP b PROCESS snRNPs EDIT mRNA WITHIN THE NUCLEUS 1 snRNPs bind 339 2 snRNPs assemble Spliceosome 3 3 lntron is cut aftquot 3 Edited mRNA 4 lntron is released exons join together 5 Exon Exon ach produces a different class of RNA transcription start site Mansibasal transcription factors Complex and variable often includes a TATA box ab0ut 30 from the Extensive several processing steps occur in the nucleus before RNA is exported to the cytosol for translation I1 Enzymecatalyzed addition of 5 cap 2 Splicing intron removal by Spliceosome 3 Enzymecatalyzed addition of 3 polyA tail 0 v TRANSLATION Tl process whereby protein is synthesized from a mRNA template gt involves ribosome proteins and rRNAs mRNA template start site for Tlinitation factors and terminateion signal stop Tlrelease factors aminoacyl transfer RNAs tRNA modi ed guanine V methylguanylate 5 cap PolyA tail zA A m G39P P39P 339 r v r J 5 untranslated Coding region 339 untranslated region region v How do we know that translation occurs at ribosomes gt Experimental evidence from pulsechase experiments I During pulse newly synthesized proteins are labeled with radioactivity I correlation between of ribosomes and rate which cells syntehsize proteins 139 Ribosome made of rRNAs and proteins gt Subunit Complex of RNA molecules and proteins I Contains at least one ribosomal RNA rRNA I rRNAs catalysts catalyze peptide bond formation gt How do ribosomes read the genetic code I Theory 1 amino acids interact directly w mRNA codons I Theory 2 adapter molecules hold amino acids and interact w mRNA codons I Transfer RNAs tRNAs interpreters of the genetic code 139 How does tRNA work gt CCA sequence at 3 end of tRNA is a site for amino acid attachment while triplet on the loop at end of structure serves as an anticodon set of 3 ribonucleotides that forms base pairs with the mRNA codon gt Tertiary structure is important as it maintains physical distance between anticodon and amino codon important in positioning within ribosome Amino acid Xray crystallography How does tRNA work Binding site for Eany model of tRNA function Amino acid ammo ac39d attached to 3 and Hydrogen bonds Binding site for mRNA codon called anticodon Serine codon codons are written 5 339 Serine anticodon Antleodon 5 mRNA mRNA ATP III 0 v Wobble hypothesis lSt two nucleotides are usually the same Aminoacyl IRMA synthetase smeatc to leucmo g AM 9 v How are amino acids attached to tRNAs Acumen gt Aminoacyl tRNA synthetases has a specific amino acid 20 tRNA synthetases 2332 and l tRNA for each amino acid I Amino acid activated during hydrolysis of ATP 122 m gt tRNA synthetase binds to tRNA Mquot gt activated amino acid transfers to the tRNA producing an aminoacyl tRNA 1 Q N 00 TRANSLATION IN 3 STEPS gt Initiation the small subunit binds mRNA and moves to the AUG codon translation start site gt Elongation amino acids are joined together and the ribosome moves to the next codon gt Termination When a stop codon on mRNA is encountered the completed polypeptide is released and the ribosome disengages Diagram of ribosome during translation olypeptldc grows in amino 0 carboxyl direction amino acids in green Peptide bond formation occurs here Aminoacyl tRNA subunit The E site The P site holds The A site holds a tRNA the tRNA with holds an that will exit growing polypeptide aminoacyl attached tR INITIATING TRANSLATION IN BACTERIA Large subunit of ribosome 5 The large subunit of the ribosome binds positioning the fmet aminoacyl tRNA in the P site Step 1 How does translation get started I Bacteria Start Shine sequence binding Also called the Ribosome site 539 Initiation 55 sequence factor Small subunit of ribosome Initiation factor ribozyme Peptide bond formed amino acid from tRNA in P site is transferred to tRNA in A site Incoming aminoacyl tRNA binds in A site New aminoacyl tRNA Peptide bond formation enters Into A Site Translation termination INITIATING TRANSLATION IN BACTERIA Aminoacyl tRNA Small subunit of ribosome An aminoacyl tRNA carrying formylmethionine fMet base pairs with the start codon of the mRNA Translocation tRNA attached to protein moves into P site Ellame Translocation A release factor protein recognizes a stop codon and lls the A site protein is hydrolyzed by the release factor Bond linking the tRNA to the The ribosome separates from the mRNA and the small and large subunits dissociate ll n l I H x 39 I V I I Il 39 PU3I l I Life span stablllty of mRNA Translation rate p Protein activation 39 h39b39ti l truer modi catio 32 tzlfigtion Protein Lecture 29 Control of Gene Expression in Prokaryotes Rib s m I Roles bacteria play in our lives our quot 39Vquot 5quot RquotA quot quotquot quot39s gt Medicine diseasecausing bacteria health m a a promoting bacteria Transcrl tlfnal Translational Posttranslatior gt Biotechindustry commercial value to manufacture con r0 contrOI contm39 products degrade pollutants Y Y gt Environment making soil snow formation EFFICIENT USE FAST I OF RESOURCES I Why are bacteria so successful Great reproduction very adaptable gt gt gt gt gt I Gene regulation many levels of control Transcriptional control regulatory proteins affect RNA pol ability to bind to a promoter and initiate transcription Translational control regulatory molecules alter length of time mRNA survives OR affects translation initiation or elongation Posttranslational control prevents proteins being activated by chemical modification Procon for each step saving energyrecourses vs speed Constitutive expression action is constant I Lots ON9 some none OFF I Ecoli uses a Wide variety of foods for ATP production via glycolysis and ETC VVVV Not membrane bound nucleus but circular chromosome Why does Ecolo prefer glucose to lactose Lactose MORE energy to process Lactose acts as an inducer triggers transcription of certain gene For Ecoli cells to use lactose I 1 Sugar get into organism membrane proteinGlactoside permease I 2 Inside cell break down catabolism the disaccharide sugar into monosaccharides glucose and galatctose via enzyme Betagalactosidase Replica plating revealed 3 types of mutations constitutive mutants produce product ALL the time lacI I lacZ gene encodes for Betagalactosidase I lacY gene encodes for transporter protein I W mutation Galactoside permease can t transport lactose into cell I lacI gene encodes for repressor protein binds to the operon 32131 protein I W mutant lacI gene expression occurs With or Without lactose repressor protein binds to operator site part of operon I lactose binds to lacI induces allosteric regulation so DNA transcription occurs Iac Operon gt l Regulatory sequences gtlt Structuralgenes gt I DNA DJ 86 pm 0 lacZ lacY IacA I Operator Structural gene PromOtef for for Bgalactosidase structural genes Structural gene for Bgalactoside transacetylase Promoter for regulatory gene Structural gene for Bgalactoside permease Regulatory gene I Mapping the location of the 3 genes gt VVV Repressor protein lactose present Repressor present normal Iacl gene Lactose present binds to repressor Lactoserepressor complex cannot bind to DNA Transcription occurs lactose acts as inducer WW Iacl39 Lactoseremessor lactose YRA SCRIPYION BEGINS I Repressor for bacteria sometimes one promoter can regulate 2 genes operon I need both functional LacZ and LacY for lactose to be metabolize Negative control when a regulatory protein repressor brings to DNA and shuts Tc Positive control regulatory protein activator triggers Tc LacI exerts negative control on lacZ and lacY I Lac Operon genes OFF if amp lactose present xs Gene encodes protein product quot33233quot lac LacI Repressor protein always on39 LacZ Betagalactosidase lacZ llf null mulatm 3133 LacY Galactosrde permease use lactose lac Y T 1 do W 0 Incl lac Y RNA polymerase bound to promoter Lecture 30 Control of Gene Expression in Prokaryotes Part 2 transcription transcription How can the repressor protein LacI regulate Positive control when a regulatory protein called an activator binds to DNA and triggers Cleaves lactose Regulatory Negative control when a regulatory protein called a repressor binds to DNA and shuts down Membrane transport m transcription of the lac operon genes function 6 992235quot proteinim orts gt Operon set of coordinately regulated E bacterial genes that are transcribed fzgizggg g mi4 m M p lacy257fj ye together into one mRNA gt lac genes codes for a product that quotquotquot quotquotquotquot quot32 Lactose represses transcription of the c202 and Ecoquot Gamma lacY gene gt glucose easily gets in but if there is both there is a series of phosphorylation rxns that modi es an enzyme so lactose can t enter the cell I Phosphotransferase system PTS if E protein is phosphorylated IN or dephosphorylated BLOCKED that lets lactose in converts lactose glucose galactose a Partial diploids and the lac operon gt Creates partial diploids bacteria w 2 copies of just the lac operon amp associated genes gt Trans made in one area and operating somewhere else gt make is one area and woks on area right next to it What happens when a person says that they can t tolerate milk gt Answer nonfunctional enzyme lactase which normally I How to test if something works in cis versus trans Mechanism for genetic exchange for bacteria gt Transformation genetic material transferred as naked DNA a Partial diploid created via transformation of plasmid w can we test if something rks in cis vs trans add back the dna for Chromosome pennease llg alactosidase How do Ecoli use glucose over lactose as a food source when BOTH are available Inducer exclusionold textbook catabolite repression quotcatabolite repression inducer exclusion phosphotransferase system PTS diffusible CIS lactose glucose e 4439 t d39t S n Lacl 0 T quot 439Elv739739uu ru39 ltquot39 1quot 1quot39 3 u lt1Elt39E x A w IacY CIS partial diploi created via transforrnatin of plasmid Lacl Iacl somewhere else 1 TRANS on the chromosome not in wild type area I at recovered the M Q phenotype crs Lowno transcription mam awaaowuzu mz mow I 2 amp a Operator IacZ IacY Iacl Tool partial diploids rescue mutant phenotype PROTEINS Repressor Betagal Permease RNA polymerase XGal amp UNFG to detect Betagal Tefln r c39 MOLECULES glucose V lactose allolactose n In galactose W l 39 N H i 4 INDUCER EXCLUSION Lecture 31 CONTROL of GENE EXPRESSION EUKARYOTES part 1 I How do we know the DNA is the same in every somatic cell of an adult organism gt If the DNA in the nucleus is the same in each cell then the DNA in EACH somatic cell should be a able to create a new organism clone gt Types of CLONING I Molecular cloning recombinant DNA technology mVectors small dsDNA that are physically separate from and IN 4 can replicated independently of chromosomal DNA within a cell I Reproductive cloning duplicating the entire organism I A plant or animal which has the same genes as the original from which it was produced gt How can we do this experimentally Somatic Cell Nuclear Transfer I l The DNA needs proper environment in order to instruct the development of a new organism egg I 2 Mammalian embryos need 0 develop in the proper environment uterus of surrogate host gt SCNTSomatic Cell Nuclear Transfer SCNT embryos made by combining the DNA of somatic cell w an egg cell from which the nucles was removed I l Enucleate remove DNA from unfertilized egg I 2 Combine somatic cell nucleus DNA with enucleated egg cell I 3 Allow zygote to develop until blastocyst stage implantation stage and implant into mammalian surrogate uterus of the same species I Differential gene expression diff cell types arranging them into tissues and coordinating their activity to form the multicellular society individual I How is the expression of the DNA regulated histone modification DNA methylation chromatin remodeling complexes gt How is Chromatin altered I l Chromatinremodeling complexes needs ATP enzymes that form macromolecular machines I must occur before transcription I DNA w promoter must be released from tight interactions w proteins so RNA polymerase can make contact w promoter 9 Chromatinremodeling complexes cause nucleosomes to slide along DNA to open chromatin for gene transcription I 2 Histone modifications acetylation amp deacetylation of histones I 3 DNA methylation it s a sign of an inhibitor I Methylation of DNA ex Methylated globin promoter DNA methyltransferases add methyl groups CH3 to cytosine residues in DNA I Methylated sequences are recognized by proteins that trigger chromatin condensatlon Acetylation amp deacetylation of histones I O l Fetus needs fetal glob1n gene to be turned ON adult gene turned OFF methylated lx O 2 Adult needs adultglobin gene to be turned ON and fetal gene turned OFF methylated Unmethylated promoter DNA promoter double helix A rd Fetalglobm l adultglobin 3333 I gene gene for chemical 5 39 3 modification D NA 3 5 I a Histono tails prolruoe outward from a nocleosome ACTIVE inactive hismne histone acetyl T 39 deacetylases transferases QUESTION Why CAN39T you get TC from a methylated promoter HDAC HAT Unacelylmod hislones hemmed muck reduces the b Acetylation ol histone tails promotes loose chromatin pos39 Charge slructure lhal permits transcnptnon DNA 539 5 3 5 inactive ACTIVE 0 v DNA is NOT destiny epigenetics rewrites the rules of disease heredity and identity gt Epigenetie inheritance patterns of inheritance that are due to something other than difference in DNA sequences gt Ex Same genes but one mice had diabetes I High in methyl donors prevents the expression of the Agouti gene that produces mRNA I How is the RNA regulated Processing splicing capstails Sequestering Degradation PROCESS A MODEL OF TRANSCRIPTION INITIATION Enhancer Enhancer 39 Activator lt i n L Basal transcription tactors h n M r 7 I 5 J l2 lo t l l xiquot l I Promoterproximal elernentIPromoter RNA polymerase II Droteins bind that helps the reczulator r elements function better Mediator MM 0 v What regulatory regions are located in the DNA sequence that controls the transcription of a gene Promoter elements BOTH in pro amp eukaryotes gt other regions that can regulate transcription in eukaryotes Enhancers PPE I in prokaryotes Operator Promoter Basal Promoter sequence in common to most genes RNA polymerase makes initial contact with gene Tc Start site Enhancer Promoterproximal Element PPE sequence that is unique for this gene eIement Activate or repress Tc quotrom the basal promoter Promoterproximal To start site 9 0 v Experiments that show role of enhancers gt Assay to examine gene levels RNA Northern Blotting I Hybridization nucleic acid hybridization is the formation of a duplex between 2 complementary sequences I Can occur between 2 polynucleotide chains which have complementary bases 9 DNADNA DNARNA RNARNA I Annealing describe hybridization of 2 complementary molecules I Probe sequence of nucleic acids that I Can be labeled w a marker which allows identification I Will hybridize to another nucleic acid on nbonds3 mmmp g ems39 labeled probe 539 AGGACCGGAAAUUAACCGGGCCCGCUUCAGAUUCAAU 339 mRNA of intereg I hybridization w radioactive probe Incubate in soln w labeled probe cDNA Radioactive probe base pairs to the fragments containing complementary sequences in RNA sample located on the nylon Lecture 32 CONTROL OF GENE EXPRESSION EUKARYOTES part 2 O 00 O 00 O O O 00 00 0 RNA processing splicing captails Sequestering Degradation Enhancers regulatory sequences in the DNA POSTITIVE control Enhancers act over long distances In any orientation Can be located in introns or in untranscribed 5 or 3 sequences anking the gene Any types Of enhancers exist diff for diff genes Can work if moved tO new location near original gene Opposite function silencers NEGATIVE control How dO enhancers function gt TO regulate Tc Of a gene specific regulatory proteins transcriptional activators or repressors need tO bind tO right DNA sequences ex Enhancers silencers PPE VVVVVV Promoter Enhancer Enhi How can you test that enhancers are binding sites for regulatory proteins that activate transcription gt Can Bcells express a glObin mRNA if you add a Bcell enhancer sequence tO the glObin gene and insert this recombinant DNA construct into Bcells gt Conclusion Bcells cells contain a factor regulatory protein that interacts w this Bcell enhancer tO regulate Tc What s the use Using specific enhancers or promoter sequences you can examine where a gene is expressed Enhancer Tc Start site KltM xmm l Exon Intron Exon Exon Promoterproximal element You make a transgenic animal in the lab recombinant DNA e fty 3 w 39 w r Tissue speci c 1 enhancersequence GFP RNA x i 39F i quot 3 r A TO regulate Tc Of a gene you need BOTH specific regulatory proteins transcriptional activators or repressors amp specific DNA sequences enhancers silencers Proximal promoter element PPE Sequestered mRNA keep it in the nucleus Cap and tail essential for ability to T1 Alternative splicing of mRNAs make diff proteins from the same RNA template Most mammalian primary mRNAs contain introns and exons Cut and paste diff regions Of the RNA message in order tO generate diff proteins Importance allows for many splicing variants isoforms from just one gene sequence 0P cell type 1 V VVV l 139 1 quot 1 7quot makes it more efficient instead of TI in a chain tranglamg in 2 Circle Lecture 33 CONTROL OF GENE EXPRESSIONEUKARYOTES part 3 Northern blot analysis alternative RNA splicing gt If the probe is complementary to A exon amp A exon is expressed in both smooth and skeletal muscle there Will be a band on the gel in each lane Western blot analysis detect specific proteins gt Detects specific antigens separated by electrophoresis by use of labeled antibodies I Exploits principle of antibodies binding speci cally to antigens in biological tissues gt Didn t exhibit purple protein in petunias Cosuppression suppressed expression of BOTH the endogenous gene and transgene Posttranscriptional gene silencing PTGS mediated by doublestranded RNA RNA interference RNAi gt RNAi I Occurs in many organisms I Protects genome from some viruses I Important for regulating gene expression I New tool to learn how genes work KO I dsRNA is more ef cient than ssRNA I can be arti cially introduced lab or encoded in genome microRNA I microRNA expression regulated locationtime gt Process I 1 In the nucleus cell DNA sequences I transcribed into mRNA premRNA odd stem structures made at speci c times I 2 PremRNA moves into cytoplasm I 3Dicer complex cuts dsRNA into specific dsRNA sections I 4 RISC RNA induced silencing complex picks up dicer fragments and then cleaves the mRNA either cleaves its entirely or makes it so Tl machine can t attach9NO PROTEIN RNA interference Engineered I siRNAs enter in pathway nucleus here r 6 RNA cytoplasm quotquotquotquotquotquotquotquot quot Dicer small 19 nt duplex interferin 5 RNA pathway RNA Induced 2 m 3 overhangs siRNA K silencing complex 39 R39s I39 quot l l39ir 9l RISC quotIi xf vuh18v PremiRNA L J A39u 39 Antigenbinding site antigen Disulphide bridges Protein chains Detection in Western Blots microRNA pathway miRNA Kll l II I It Trans Iranspon out In Misfolded ModT t39 395 are t and lJquumn proleasonm attaChed WGV v How are proteins regulated Folding chemlcal modificatlons degradatlon sequestering Correctly gt most proteins need help folded I molecular chaperone family of proteins that assist in the correct folding of peptides I PREVENTS INCORRECT INTERACTIONS I capture unfolded polypeptides stabilize 089555 W intermediates prevent misfolded species from r accumulating assist in transport across membranes 3 Vesicle gt How do you remove a misfolded protein I Tags ubiquitin are attached to the proteins I Proteasome chews it up Go39gi gt What if protein folding goes wrong in an organism I Parkinson s disease Alzheimer s disease some types of cancer cystic fibrosis 1m disease Proteinaceous infectious particle PrP I Prion disease Mad cow brink shrinkage and deterioration occurs rapidly I If you ate a protein it cause your own proteins PrP to misfold ltIJI mu Ilulluphnlm r I la llnsmlu ulmr lum umulnmlnun Inu39l llnll I ll 6 lIIlIlIgt Il39i39 Slu c39l DiulHidr lrm lun II Iu39lun bond Sulr 1min hull39nuvn hundin


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