Exam 2 PDF - Chapter 6 Enzymes

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USF Health College of Nursing

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enzymes biochemistry biology biological processes

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This document appears to be notes on enzymes, likely from a biology class. It covers topics such as enzyme types, cofactors, and activation energy. No exam board or year is specified.

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CHAPTER 6-ENZYMES that can speed the rate of rxn Enzyme : protein catalyst ↳ reactants rxns also called substrates...

CHAPTER 6-ENZYMES that can speed the rate of rxn Enzyme : protein catalyst ↳ reactants rxns also called substrates in enzyme ↳ highly specific cofactor : molecule required enzyme activity for ↳ typically a metal (mg Fe zn). , molecule or devived from a cofactor that is an organic Coenzyme : A specific Vitamin (Heme] A functional enzyme Holoenzyme : ↳ Apozyme (protein) + cofactor = holoenzyme cofactor binds think non-covalent interactions Binding Sites : Where substrate + , that catalyze hydrolysis of a peptide bond Proteases : Enzyme chemical bond by addition of water (very common Hydrolysis : breaking Six classes of enzymes : 1) Oxiductases : Catalyze oxidation-reduction rxns within a molecule 2) Isomers : more functional groups groups between molecules 3) Transferases : more function 4) Hydrolases : enzymes break bonds by adding water bonds 5) Lyases : break bonds wout using water forming double wh ATP 6) Ligases enzymes : join 2 molecules together, usually State Activation Energy > Lower transition - Endergonic · · pluse products ↳ Wienzyme Ordinate AG less than O = Spontaneous exergonic &G : more than 0 = nonspontaneous : exergonic Catalyst (enzymes) REDUCE the · AE & G Catalyst DO NOT afrect · AG = 0 · AG determines Spontaneity , not rate · Atequilbrilum affect equilbrilum or A G* * Enzymes Do NOT -G Large Keg = product Fav = · · Standard AG = 1 Go products concentration at equilbrilum keg = reactants AG = AG RTInQ T = 298k energy Change + Free R = 8. 3145/mol K run from hours to seconds reduce the time required for a - Enzymes can the AE , binding WI Catalyst of ↳ enzymes speed up rxns by lowering active site get to a * Tinkenymikeashortcutrace-theyhelpmolecules (t) rate (5) (conc ] = increase equal to [Conch increase. Rate is , Stabilize the transition state · Enzymes to reach trans state 1) no enzyme-high A3 to sub-not good 2) complimentary sub-binding to trans- Lowers of Complimentary 3) energy state trans state making trans ealser to reach EXRXn : Proline Racemase to or molecule similar ↳ transition State analog mimics a : the trans state - allows molecule to bind in active site common mode of forCompetite Inhib - binding proteins typically have high specficity-only certain ligands bind to ligand trans state ↳ binding site: complementary surfaces are preformed DLock and key complementary : active site is perfect idea > the - enzymes shape for substrate ↳ rigid Es complex As thesubstratebindsthetransisti s ligand binding (common 2) induced fit : conformational Changes upon site changes Idea - enzymes active bit binds Shape when substrate M abC # To Facilitate binding and state formation of trans AGE RXn Spontaneity reactant products H: "rxn involves breaking down FBP Into DHAP and G3P glycosis during 2) Asked to calculate G under given concen. Inside the cell and determine if the rxn 3Q occurs spontaneously. 3) Given - Standard 1G (AG) = + 23 8. k)/mol > - 23800j/mo Temp = 298k Gascontent (1) = 8. 3145/mol. concentrations : FBP 1 S X =. 10 - 4 M DHAP = 4 3 X10 - 3 M 4) formulator 1G AG. RT en Q. = + > - G3p 9 6x10 SM - = A G. A Calculate Q : rxn Quotient (a) is calculated using the concentrations of - the products and reactants Q d = - = 25 Step 2: Plug values into the AG Formula 3 a - = 2 752. x 10 AG = 23 800 , + (8. 31451mo) (698k) In (2. 572x10 - 4) Step 3 Calculate the natural : e(ena) en (6. 732x10 - % = - 12. 807 Step 4 Calculate the second (8 :. 314)x(298)x( - 12. 107) term = - 29 , 708 mo maximum -...... [substrate] Vmax = Kcat[E]T Step s : Calculate AG Kcat = Turnover number (# of substrate convened time) to product (per unit (EX 000/min) 92X103 : 57 ( 31730 2646) , -G = 23 800, + -. = - 7. ↳ speed of 1 enzyme molecule per minute 1) m = Michealis - Mentor constant Step 6 : Interpret results Grelates to substrate binding affinity AG is negative rxn is spontaneous &man CHAPTER 7 : ENZYME KINETICS REGULATION complex Enzymes Form an enzyme-substrate where rxn occurs (substrate binds ↳ Active site-site on enzyme + substrate Es > - enzyme + product enzyme or disappearance of a product time wh appearance in concentration over Rate-change Rate = (constant) Rate observations for enzyme Kinetics : Drate-dependent on sub at low concen. 2) rate-independent on sub at high concern ↳ Vmax reached - - - zero First second 1st order Low concentration substrate = Zero order concentrations substrate = High catalyzedxn depenis Michaelis-Menton Equationrateofenzyme : Vo = Umax Vmax : maximum rate based on km + [S] concentration of enzyme Low Substrate Km = greater than substrate rate dependent on conc of. sub High substrate km = less than substrate = rate independent on conc of sub ↳ approach umax at high [C] why does anenzyme reach umax and not go higher? - enzyme becomes saturated wh substrate , all enzymes must the have a substrate to bind with How could the Umax potentially be increased ? Add more enzyme - Kcat Turnover number > # or substrate per unit time = - ES = "Steady state" = Overall [ES], not changing overtime M-M Kinetics Important assumptions for substrate than enzyme 1) You need a lot more 2) Enzyme is "Steady State" meaning rate of Es formation ↳ rate of ES Formation = rate OF ES > Conc -. of Es , not changing 3)Initial rates only , revene xn not occuring km = measure of substrate affinity" "binding ↳ relates to substrate binding to enzyme (formation + breakdown Vmax measure of enzyme rate = ↳ dependent on enzyme conc. km = [S] required to reach zUmax rxn toreach Zumax · higher km = more CS] to get · have to use km3 Umax together to analyze comparision of enzymes + substrates specificity constant = atmatgarfinity substrates > products Kat : at saturation how - many (km) ↳ not afrected by sub affinity rate ↳ measure enzyme Kcat/km : How well does the enzyme > - sub > - product ↳ dependent on enzyme rate (kat) + km Sequential RXNS : both A , B bind to enzyme before any rxn occurs ↳Ternary Complex Forms (E + 2 [s]) D ordered : A must bind to encyme than B can bind A , B binds first then other binds 2) Random : Either at the same time in a sequence , one after the other - Overall both subs bind to enzyme P1 leaves , B binds Double Replacement (pingpong) : A binds + modifies enzyme , modified by enzyme , p2 leaves dif sites then exchange parts between , leading to 2 products overall the enzyme binds to 2 sub at - not active site ; increases or decreases energy Enzyme Regulation : 1) Allosteric Reg binds to enzyme : , or deactivated by phosphorylation 2) Reversible covalent mod : enzyme activated : adds phosphate ↳ Phosphorylation enzyme Kinase 3) Proteolytic Activation : becomes active when hydrolyzing occurs same function 4) Isozymes : Multiple Forms of enzyme , amount of enzyme present 5) Expression controling : ↳ prevent RNA trans Isozymes fetal : hemaglobin mod : phosphorylation + acetylation z common covalent a Phosphorylation : Occursatahydroxyetherkaseaddaphosat At's w/ a hydroxyl : Thr , Ser , Tyr Why phosphorylate ? · electrostatic interactions make rxn faster , ATP is the cellular energy ↳ alter , Allosteric Regulation at a site other than the active site Allosteric enzyme : activated or inhibited by regulators of the the initial rxn that leads to the rest - Allosteric enzymes regulate pathway ↳ think of them as a "Start button" for a series , once you push start , the rest happens curve - demonstrate a sigmodial inhibition : The end product of the metabolic pathway Inhibits (slows down) an I Feedback Xs production the pathway to prevent enzyme early in When theres enough product It binds to the enzyme b tells the pathway - , to slow down or stop saving energy ,. An intermediate or early product in pathway activates 2) FeedForward Activation : more product (speeds up) an enzyme to promote produces molecule that activates an early step in the pathway a - An to go faster enlyme Further down , encouraging sites structure ; multiple active sites + mutiple regulatory have a quaternary - All allostenc enzymes T-state = Less active = Less Csb binding R-state = more active = more [s] binding [S] either Tor R D concerted Model : For one enzyme , ↳T = more Stable active ↳ R = enzymatically more to R- State [S] the other active sites In R-State as more T-state enzymes go Binding of traps · a : , more substrate can bind a time changes structure from R-T or T-R one at , 2) Sequential Model : each active site Changes in Homotrophic Efrect : same molecule binds to the enzyme affects its affinity , resulting In a sigmodial curve Ess makes it easier or harder for other ↳ The binding of one molecule of the the same ass to bind activity. molecules of , enchancing enzyme for others to bind of oneO2 molecule makes it easier EX : hemaglobin binding , Heterotrophic Efrect : Different molecules bind to the enzyme afrect its activity molecules can either increase/decrease ↳The binding of these different enzyme activity ↳ causes curve to shift , changing km curve be they don't Follow the typical Michealls-Menten Allosteric enzymes result in a sigmodial curve ↳ more sensitive rate known as threshold effect large effect on ↳ small effects can have a certain minimum level of Cs] activate the enzyme Threshold Effect : Only need a to before the increase Allosteric Regulators afrect RET when bound T-state = less active Inhibitors = Stabilize Stablize R-state = more active Activators = ATP = activator CTP = Inhibitor CHAPTER 8 : ENZYME MECHANISM + INHIBITION Catalytic stratigies modified active site has a nuc that is quickly covalently D common catalysis : run occur faster , more efficient donates or acceptsIt to help 2) Acid-base Catalysis enzyme : react together increasing chance they will approximation enzyme brings : 2 Css , 3) Catalysis by e-transfers bond formator help catalyze rxns by stabiling charge, , 4) Metal- Fon Catalysis : Metallons good inhibitors analogs very = mechs? Transition State Why study enzyme - Increase of temp = increase in rate (initally increases stability of protein/enzyme - Ligand/subbinding - Enzyme activity is pdependent side chains = protonated can result in actuation or deactivation - protonation/deprotonation of specific groups inhibitor at active site-presents (s] binding competitive Binding = of prevents product formation-un/non competitive inhibitor at allosteric site Binding of Competitive uncompetitive noncompetitive ⑧ substrate ~ im immigr Inhibitor binds to active Dinhibitor binds to " site allosteric site binds allosteric prevents s from binding 1) Inhibitor to he Site binds to E , not Es I * only to 2) binds to E or Es complex w/ equal affinity * does not care if substrate is bound or not be observed presence of inhibitors Km + Vmax can normal-umax in onconsi ~ apparentumax Binds Es only Noncompetitive : binds E or ES binds same site as sub uncompetitive : competitive : vmax = decreases umax = decease no affect umax km = stays same km = decreases Y- intercept = same km increases representation of data Double-Reciporical Plot : linear + slope = m Y-Intercept = umax umax Competitive uncompetitive Noncompetitive competite in - uncompetive ~ noncompetive ~ ↓ To # no inhibitor present Vo Tho inhibitor present ↳ no inhibitor Slope =m present vmax same Imax-samey-erhe a O & - [S] Vmax decreased Umax decreased Vmax = same : = km decreased km = unchanged = km = increased · Unable to interact Yare competiting · we · same y-intercept covalent bond to the enzyme Irreversible inhibitors : typically or attach a group to enzyme. Kinetics D Protease RXn + - proteases cleave proteins by hydrolysis slow (w/out catalyst) - hydrolysis is exergonic , shows > Biphasic Kinetics Chymotyrpsin - of Measuring rate · ↳ look for nucleophile + electrophile Oh + X-0 Acylation- Deacylation H + Ho-c"0 Ex 20 'R ↑ ↑ 'R R enzyme nuc electrophie acyl-enzyme Enzyme 2) Serine Protease - Catalytic Trend protease : hydrolysis of a peptide bond (nucleophie Asp-His-Ser ↑ alkoxde e G Ser-19S..... o - H ASP-102 His-57 H -.............. - 0 H H............. in 2 ~ :O 1: 0 acts as the nucleophil Chydroxyl · needs to remove the Proton to create a better nucleophle and makes His a better bond accepts / Histidine making histidine Aspartate (ASP) : hydrogen Stabilize + initate w/histidine bond. Just base. helps reactive from Serine (nuc generates a highly proton Histidine (HIS) base removes a , : serine. alkoxide Ion on the Serine (ser) : nucleophie Alkoxide Ion (ser) attacks the peptide bond of the substrate * Residues 3) Mutagenesis of Catalytic Triad loss of activity · Mutation of amino acids result in significant - 105 or greater loss in activity ! - EachAn is essential/required Protease specificity acid that accomidate amino · protease specificity occurs by pockets in enzyme side chains. can be simple (one AA) or complex (multiple pockets · specificity Ocharged pocket a G At can fit and want to. interact or small A large pocket nonpolar where only a Side chain Fits. Specificity pockets can accomidate different Als bond) · sites for hydrolysis (the peptide by binding specific specific proteases identifying specific An side chains. CHAPTER 9 : HEMOGLOBIN proteins = Hemoglobin Red blood cells responsible = for Oxygen transport within the body Oxygen Transport in the body binds to red blood ces n the ↓ Lungs /hemoglobin · Oxygen enters blood from the lungs 02 CO2 via blood · Transported to tissues CO2 02 & ↓ +issue then also to the tissues cells then transfers , protein that binds Of In red blood Hemaglobin - carries (O2 back to longs. reserve Found muscle provides Oxygen Myoglobin - in , Hemoglobin > tissue 4 part protein Transport Oxygen longs - , · Contains 4 similar subunits · Tetramer : other subih its subunit affects Of binding on · Allosteric Protein : O2 binding one release "Cooperativity" Calostery) binding in O2 + ↳ displays Myoglobin in muscle cells · binds Oxygen one subunit binds only 102 only · , multiple ↳ not cooperative , needs Globins Require Heme to Bind Oxygen cofactor that contains an iron on has Heme ! Heme > a - Myoglobin iron can coordinate essential for protein activity cofactor : molecule that is , wid total of SIX atoms. Heme = organic molecule = coenzyme · myglobin binds the heme , in the heme there is an or where the iron can bind to both require home to bind to O2 Lone per subunit Hemoglobin 3 myglobin · Oxygen Binding · proximal histidine interacts directly with the iron of heme ↳ closest to heme · then O2 binds to home whiron in the Feat State · then the bound O2 stabilized by interaction W/ distal Histidine Myglobin (O2 Affiniti) saturation of protein Is Oxygen concentration (POL Curve Binding : ↳ shows Or binding us the amount of Oxygen present at low concentrations Myglobin-binds O2 tightly , even ↳ reserve Function = Oxygen Cooperativity of Hemaglobin indicates O2 binding and release is cooperative · Hemaglobin-Sigmodial curve Oz can have a large effect on binding release ↳ small changesIn curve = Cooperativity Hemaglobin = sigmodial w/high (lungs) Hemaglobin Function : 1) bind O2 at locations O2 w/low O2 (tissues 7 thisispossiblea 2) releaseO2 at locations cooperativity allows for fraction of Oz to be transfered from long > tissue large - · cooperativity a can be in the Tor R state Hemaglobin · Or binding afrects structure of hemoglobin on other submits · Allosteric effects that change binding affinity Tstate (tense) : Low O2 binding Rstate (relaxed) : high O2 binding Conversion · from T-R State Facilitates (increases) oxygen binding If we want to increase O2 binding > - Stabilize R state or destablize T state Stabilize T State or destabilize R state Of T and R State Binding Curves Estate R State = high affinity for 02 T Lobserved hemaglobin(sigmodial T state = Low affinity for 02....................... T - State · like allosteric proteins , hemaglobin shift between T , R state to vary Its own O2 affinity. · All subunits in concerted model are in Tor R · O2 binding offects nearby units in sequential 2 , 3-BPG Stabilizes the T-State Stabilizes the T-state of · hemaglobin helps , release O2 increase of 2. 3 BPG = more O2 released ↳ when you have an the T-State released when in the R-state · binds to hemoglobin only in , for O2 In womb , how do they get the O2 ? Bables rely on moms 1 Fetus relies on mother for O 2 2) Hemaglobin must bind Oxygen at the same poz that , the mothers releases O2 that maternal hemoglobin releases O2 ↳ fetus must bind O2 at the same poz (similar to tissue poz). ↳ retal hemaglobin bind O2 w/ must higher affinity to take from mother · Fetal hemaglobin does not bind 2 , 3-BPG as well as an adult hemaglobin. so less ↳ The 2. 3 BPG Stabilizes T-state Restate would be favored BPG , "Babies bind O2 better , but not BPG" retal hemoglobin be 2 dif proteins same function Developemental Ex- > Isoenzyme > - , not bind BPG as well ? Different AA dif protein complementary Why mighfetal hemaglobin , CO2 and (H ] + are intrinsically related CH + ) Lower PH Stabilizes T-State Higher = = Higher CO2 = CO2 and lowers pl of blood · As our body uses energy , it generates ↳ O2 energy (cellular respiration) is essential part of regenerating to be released ↳ Lover the PH causes more O2 It increase of O2 release of CO2 and Bohr Effect : Increase ↳ T-state-stabilized = facitating O2 release ↳H so increase in Oz released + Stabilize the T-state , += protonation of His He Effect : (AKAPHefrect > - lower pH = higher H Answer the following questions regarding the reaction mechanism shown below for the peplide bond enzyme HIV-1 protease.-hydrolysis of a substrate 2 = = enzyme = Water ① ② What molecule is acting as a nucleophile in Step 2? how to determine new bond welectrophile Steps on : D nucleophile donates pair of e- to make 2) look at atoms In 52 , find where a huc attacks an electro Most. hydrolysis acts as the nuc. rxns , water or another e-rich molecule 3) In H20 is wh the carbonyl S2 , reacting Hz0 = nucleophile In what step(s) is Asp25’ acting as a Bronsted-Lowry acid? Bronsted-Lowry acid proton (H ) a Charge + : Donates a , look for Step 5 Write a reaction scheme for this reaction and determine whether it is a sequential or double Gidentify replacement reaction. If the mechanism Sequential : involves a step by step process, where reactants , enzyme , products one rxh occurs after another. Double-replacement : Involve 2 exchanges between reacting species. water Reactants : Peptide and Enzyme : Asp 25 , Asp25 Products : 2 peptide Fragment each step wh the enzyme facilitating - Sequential be the peptide bond is hydrolyzed in a series of steps Answer the following questions based on the rate plot shown below. Assume reaction velocity is units of mM/s. umax Steps to find Umax/km N Didentify Vmax is the 2) Identify km - km substrate conc. at half the 1lzumax Maximum Velocity (1 Umax) ↓ at the graph go Looking down half way of Vmax then go down to the X-axis to determine #. - 0. 09mg/mL What is the Km and Vmax for the uninhibited enzyme (white squares)? Vmax = 25mM/s km = 0. 09mg/m Find Kat If 10 nmol of enzyme were used in a 100 μL reaction, what would the kcat be? > - steps to enzyme = Ionmol 1) Formula For Umax substrate = 100z 2) calculate [E] T Umax = 25 mM/s 3) solve for kat 1) Vmax Ka + x[E] + 3) m kat = = [e]+ = & (10nmo)(maxom ! (00m)(e) kcat SmM = 4 [t]T = - = 1x10 [E]T = 10-Smmol 0 Imm - 7 1 ca+ = 250s. = - sloved down If the black diamonds represent inhibited, enzyme what type of inhibitor would you expect it to be? The black diamond show Vmax and km decreasing compared to unhibited enzyme. Vmax = ↓ = uncompetitive km = Slope y = Answer the following questions based on the rate plot shown below. to find Vmax + km steps D Find Umax from Intercept With the equation 2) solve for Umax 3) solve for km Umax - What is the Vmax and KM for the uninhibited enzyme? (Trendline: y = 2.2x + 16.5) 2 2 Umax = 0 060e Umax == 0 0600mm/min km y-intercept max. =. ,. 16 S 0 1333mM 2x0 0606 =. 2... 2 94X 10 5 y 2 2x + 16 5 y =. +. =.. 10 S 2 2 km = 0 1333 mm Y Intercept = Slope =... - + 10 5. (min/nm) Slope = =. + 0. 0606 x 10 kat = turnover # 1mm = 10 3 - equation Kat max 2) calculate specificity constant Vmax = katx[C]T = [E]T ↓ Vmax Kat = 0606x10-3M/m = 60 , 600 min Kat = 1x10 - 9M [E]T at 600min 60 , 600 min x108 M'min ↳ _ 60 , = = 4 55. km 0. 1333mM 0 1333x10. - 3M Based on the Vmax and Km values calculated, what would be the rate of the reaction at 15 mM substrate? Vo maxs = Vmax = 0. 0606mM/ min Vo 90 0 060uXIS km = 0 1333mm. =. = 0 1333 15 [S] = 1SmM. + vo = 0. 0001mm/min T-State R-State - Lower 02

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