Enzymes and Enzyme Kinetics - Mechanism/Inhibition - CMBE

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Enzymes and Enzyme Kinetics - Mechanism/Inhibition - CMBE

E. Molecular Mechanism of Enzyme Catalysis1. ChymotrypsinAn example of base and nucleophiliccatalysis via a charge relay system.Serine is often involved as a strongnucleophile in proteases.


Aspartate102CCHHistidine57Serine195CH 2CHC CHOCHH N N2H OCH NO –ON CHH NR 1H R 2HC CH 2Protein Stabilized byE + SHydrogen Bonding


CHCH 2Histidine57Aspartate102CHC CHOCHH N N2HC CH 2 CH OOC––ON CHE + SR 1H R 2Serine195


CHCH 2Histidine57Aspartate102CHC CHOCHH N N2HC CH 2 CH OOC––ON CHES 1–R 1H R 2Serine195


Aspartate102OCHCH 2CCHHistidine57CHSerine195HC CH 2CHO ––NCHNHNOCCH 2OR 1HR 2ES 2 + P 1


CHCH 2Histidine57Aspartate102CHC CHOCHH N N2HC CH 2 CH OOC––O OH CHSerine195ES 2 + H 2 OR 2


CHCH 2Histidine57Aspartate102CHC CHOCHH N N2HC CH 2 CH OOC–O OH CHES 3–R 2Serine195


Aspartate102OCHCH 2HC CH 2 C H N NOC–HHE + P 2CCHHistidine57HOOCCH 2OCHSerine195R 2


Chemical Reaction ModelE + SES 1ES 2 + P 1ES 2 + H 2 OES 3E + P 2


2. Carboxylation with biotinCarboxylation involves the addition ofcarbon dioxide to an organicbiochemical. Several carboxylases usebiotin (Vitamin H) and two active sites.


BiotinOPyruvatecarboxylase–OOC CH 2 CH 2 CH 2 CH 2HNNH+SHC CH 2 CH 2 CH 2 CH 2 NH 3Lysine


Oketo formHNOHC CH 2 CH 2 CH 2 CH 2 N CCH 2 CH 2 CH 2 CH 2SNHHtautomerismOHReactionsiteureido formHNOHC CH 2 CH 2 CH 2 CH 2 N CCH 2 CH 2 CH 2 CH 2SNH


This enzyme has twoactive sites.biotinlysinePyruvate Carboxylase


CO 2 bindingoccurs at N atomHCO 3 -ATPPyruvate CarboxylaseS atom seems to beimportant in binding stability


Carboxybiotin is not asstable in first active siteADPOCOHPO 42-Pyruvate Carboxylase


PyruvateH atom bonded to other N atominteracts via H-bonding to aminoacid on enzymePyruvate Carboxylase


Biotin is not as stablein second active siteOxaloacetatePyruvate Carboxylase


pyruvate carboxylase (protein) + biotin = active enzymeapoenzyme + cofactor = holoenzymePyruvate Carboxylase


3. Site-directed mutagenesis of aspartateaminotransferaseBy altering the gene encoding a proteinat one location, the amino acid residueat one location in a protein may bechanged to another amino acid residue.If this one change is in the vicinity ofthe active site, it can have dramaticeffects on the reaction.


OO+H 3 NCCHCH 2–O+H 2 ONADP + NADPH 2+CCCH 2–OO+ NH 4+CO–CO–OOAspartateEnzyme also shows activity for glutamatebut not for arginine. Why?


+H 3 NOCCH–ONADP + NADPH 2+CH 2CH 2+H 2 OCH 2NHCNH 2NH 2+Arginine


Enzyme has an arginine residue atposition 292. The positive charge onthe side chain (of arginine 292) causessubstrate specificity as a result of ionpairing between it and substrateaspartate.


Arginine 292––+Reaction ProceedsAspartateAspartate Aminotransferase


–++No ReactionArginineAspartate Aminotransferase


The gene encoding for aspartateaminotransferase was altered so thatArginine 292 was replaced by Aspartate292. The effect was 1) increase inactivity of enzyme towards arginine by6 times, and 2) decrease in activity ofenzyme towards aspartate by 700,000times.


–+–Some ReactionProceedsArginineModified Aspartate Aminotransferase


F. Inhibition of Enzyme Catalysis1. Irreversible Inhibitors:• Bind irreversibly at active site• Destroy part of active site• By degrading a portion (or all!) of theenzyme present, these inhibitors lower theeffective concentration of the enzymeExamples• Iodoacetate attacks sulfhydryl groups• di-isopropylphosphofluoridate (DPF) is anerve gas which attacks serine.


• Bind reversibly at active siteReversible Inhibitors:• Can be removed from the enzyme, enablingit to recover its activity (somehow).


2. Competitive InhibitionA chemical which competes with the substratefor the active site. Instead of reacting, theinhibitor forms a “dead-end complex.”Competitive inhibitors are often structurallysimilar to intended substrate.E + Sk 1ESk 2E + Pk -1E + Ik 3k -3EINote: some of the rate constantshave different nomenclature fromthe book.


Equations:123456____ dSdt____ dEdt____ dESdt____ dIdt____ dEIdt____ dPdt= -k 1 [E][S] + k -1 [ES]= -k1 [E][S] + k -1 [ES] + k 2 [ES] - k 3 [E][I] + k -3 [EI]= k 1 [E][S] - k -1 [ES] - k 2 [ES]= -k 3 [E][I] + k -3 [EI]= k 3 [E][I] - k -3 [EI]= k 2 [ES]


From definition of K I :[EI]=_____[E][I]K IInserting into Equation 3:[E] = [E 0 ] - [ES] -_____[E][I]K ISolving for [E]:[E] =__________([E 0 ] - [ES])1+___ [I]( K I)Notes:if I = 0, then E = E 0 -ESif K I , k -3 and [EI] 0compare to results without inhibition


Insert equation for [E] into steady-state assumption (1):_______________k 1 [S]([E 0 ] - [ES])0 = -k -1 [ES] - k 2 [ES]1+___ [I]( K I)Solving for [ES]:[ES] =__________________[E 0 ][S]_____ (k -1 + k 2 )[S] + 1+___ [I]k 1( K I)


Insert expression for [ES] into Diff Eqn 6:____ dPdt=k 2 [ES]____ dPdt=__________________k 2 [E 0 ][S]_____ (k -1 + k 2 )[S] + 1+___ [I]k 1( K I)____ dPdt=_______________V MAX [S][S] + K M 1+___ [I]( K I)Note:Strictly, [I] is concentration ofinhibitor not associated withenzyme. [I] is not the totalconcentration of inhibitor!


____ dPdt=_______________V MAX [S][S] + K M 1+___ [I]( K I)Effects on Michaelis-Menten Parameters:1) V MAX is unchanged( K I2) K M is increased by 1+___ [I])( 1+___ [I]K )APPM = K M K I


Competitive Inhibition:V MAX5Reaction VelocityV MAX2432100 100 200 300 400 500APPK M K MSubstrateConcentration


Eadie-Hofstee Plot for competitive inhibition:Increasing [I]V 0V 0 /S 0


Method 2Computer Solution____ dSdt____ St=S i+1 –S i _______t____ dEdt____ Et=_______ E i+1 –E it____ dESdt____ dIdt____ dEIdt____ dPdt____ ESt____ It____ EIt____ Pt====__________ES i+1 –ES it_______ I i+1 –I it_________ EI i+1 –EI it_______ P i+1 –P it


So, based on 6 derivatives...S i+1==(-k 1 E i S i + k -1 ES i )t + S iE i+1ES i+1(-k 1 E i S i + k -1 ES i + k 2 ES i - k 3 E i I i + k -3 EI i )t + E i= (k 1 E i S i -k -1 ES i -k 2 ES i )t + ES iI i+1 =EI i+1 =(-k 3 E i I i + k -3 EI i )t + I i(k 3 E i I i -k -3 EI i )t + EI iP i+1= k 2 ES i t + P ik 3 = 10 L/gmink -3 = 40 /min


*Program 2Simple Enzyme Kinetics with Competitive Inhibition*/#include#includemain(){FILE *fout;double km1,km3,k2; // units = 1/mindouble kp1,kp3; // units = L/g mindouble Snew,Pnew,Enew,ESnew; // units = g/Ldouble Sold,Pold,Eold,ESold; // units = g/Ldouble Iold,Inew,EIold,EInew; // units = g/Ldouble deltatime; // units = minutesdouble currenttime; // units = minutesdouble endtime; // units = minutesint count=0;// Define Some Constantskp1 = 30.;km1 = 160.;k2 = 110.;kp3 = 10.;km3 = 40.;endtime = 30;deltatime=0.00001;// Set Initial ConditionsSold = 10.;Pold = 0.;Eold = 0.00875;ESold = 0.;Iold = 10.;EIold = 0.;Again, small value for t. And,like Program 1, we have3 million iterations.Use an initial value for I and EI.


Calculationsfout=fopen("a:ex3b1.dat","w");currenttime = 0.;do{Snew = (-kp1*Eold*Sold+km1*ESold)*deltatime+Sold;Enew = (-kp1*Eold*Sold+km1*ESold+k2*ESoldkp3*Eold*Iold+km3*EIold)*deltatime+Eold;ESnew = (kp1*Eold*Sold-km1*ESold-k2*ESold)*deltatime+ESold;Inew = (-kp3*Eold*Iold+km3*EIold)*deltatime+Iold;EInew = (kp3*Eold*Iold-km3*EIold)*deltatime+EIold;Pnew = k2*ESold*deltatime+Pold;count++;if (count==25000){fprintf(fout,"%7.2f %7.4f %7.4f %6.4f\n",currenttime,Snew,Pnew,(Pnew-Pold)/deltatime);count=0;}}Sold=Snew;Eold=Enew;ESold=ESnew;Pold=Pnew;EIold=EInew;Iold=Inew;currenttime+=deltatime;}while(currenttime


Figure 3Simple Enzyme Kinetics with Competitive InhibitionConcentration [g/L]10864dP/dtS 0= 10 g/L k 1= 30 L/gminE 0= 8.75 mg/L k -1= 160 min -1k 2= 110 min -1 k 3= 10 L/gmin1.00.9k -3= 40 min -1 PS0.8 I 0= 00.7 I 0= 10 g/L0.60.50.4dP/dt20.30.200 5 10 15 20 25 30Time [min]0.1


3. Noncompetitive InhibitionA chemical which binds with enzyme on site(s)other than the active site. The inhibitor can alterenzyme affinity to substrate: the product will notform from enzyme complexed with inhibitor.


3. Noncompetitive InhibitionE + SE + Ik 1k -1k 3ESk 2E + Pk -1 ____K S =k1EIk -3EI + SES + Ik 4k -4k 5EISK S´ =EISk -5k -4____k 4


a. “Pure” NoncompetitiveSubstrate has the same affinity for enzyme as ithas for the enzyme-inhibitor complex.K S = K S´Presence of inhibitor lowers the effectiveconcentration of ES. Some of the enzymesubstrateis busy in unproductive reactions.


Quasi-steady state assumption...____ dPdt=1) K M is unchangedV_______________MAX [S]/( 1+___ [I]) K I[S] + K M( K I2) V MAX is decreased by 1+___ [I])k -3 ____K I =[EI] k 3____ [E][I]______[ES][I] k -5= ____[EIS] k 5V MAXAPP= V MAX/( 1+___ [I]) K I


Effect of noncompetitive inhibitor is to reducethe maximum reaction velocity. From thereaction equations, note that increasing [S] isnot very effective at increasing reaction ratebecause more S will also go towardsunproductive reactions.


Eadie-Hofstee Plot for noncompetitive inhibition:Noncompetitive inhibition can be“worse” than competitive sinceadding more substrate will notget back to V MAXIncreasing [I]V 0V 0 /S 0


. “Mixed” NoncompetitiveK S K S´Substrate has a different affinity for enzyme thanfor the enzyme-inhibitor complex.Both V MAX and K M are altered by inhibitorconcentration. Inhibition lies betweennoncompetitive and uncompetitive.


4. Uncompetitive InhibitionA chemical which binds with enzyme-substratecomplex but not with the enzyme itself. Thepresence of the inhibitor lowers the effective ESconcentration.


Uncompetitive InhibitionE + Sk 1ESk 2E + Pk -1ES + Ik 3EISk -3


Quasi-steady state assumption...____ dPdt=V MAXAPP__________[S]KAPPM + [S]( K I1) K M is decreased by 1+( K I2) V MAX is decreased by 1+___ [I])___ [I])K I ______[ES][I][EIS]=k -3____k 3/( 1+___ [I]V)APPMAX = V MAX KAPPM = K M 1+K I/( K I___ [I])


Eadie-Hofstee Plot for uncompetitive inhibition:Increasing [I]V 0V 0 /S 0


5. Substrate InhibitionIdentical to uncompetitive inhibition except thatthe substrate is the actual inhibitor. A specialcase of uncompetitive inhibition.


Substrate InhibitionE + Sk 1ESk 2E + Pk -1ES + Sk 3ESSk -3


Quasi-steady state assumption...____ dPdt=V MAXAPP__________[S]KAPPM + [S]Note: since chemical reaction is thesame as uncompetitive inhibition, thesolution is the “same” except [S] = [I]K I ______[ES][S][ESS]=k -3____k 3/( 1+___ [S]V)APPMAX = V MAX KAPPM = K M 1+K I/( K I___ [S])


Simplification...V =____ dPdt=______________V MAX [S]K M + [S] + [S] 2 /K IInitially, increasing substrate concentration increases rate.Eventually, increasing substrate concentration decreases rate.Thus, a maximum reaction rate exists.V 0S 0Maximumreaction rate


A maximum v is found for S when dv/dS = 0____ dvdS=_____________________________________V MAX (K M + [S] + [S] 2 /K I ) - V MAX [S](1 + 2[S]/K I )(K M + [S] + [S] 2 /K I ) 2= _________________V MAX( K(K M + [S] + [S] 2 /K I ) 2 M + [S] + [S] 2 /K I -[S]-2[S] 2 /K I )= _________________V MAX( K M -[S] 2 /K I )(K M + [S] + [S] 2 /K I ) 2


____ dvdS = 0 when ( K M -[S] 2 /K I ) = 0( K M -[S] 2 /K I ) = 0S MAX = K I K M


Comparison of common types of inhibition:V MAXK MCompetitiveNoncompetitive– –UncompetitiveFor given values of V MAX , K I , K M and [S]:V (noncompetitive) < V (uncompetitive)


Figure 4Comparison of Enzyme Inhibition4.03.5No InhibitionCompetitivedP/dt3.02.52.0Vmax/2UncompetitiveNoncompetitiveSubstrate1.51.00.50.0V MAX = 5.05 min -1K Km m = 10.1 g/LK I = 24.4 g/L[I] = 15.15 g/L0 5 10 15 20 25 30Substrate Concentration [g/L]

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