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Competitive Inhibition, S and I Bind to the Same Site K E + S ES 2 E + P K-2 K E + I 3 EI K-3

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Competitive Inhibition, S and I Bind to the Same Site K E + S ES 2 E + P K-2 K E + I 3 EI K-3 VI–48 Complex Reactions/Inhibition – 2014 – Notes 6 e.g. two enzyme substrate complexes (reactant & product) H H - fumarase OOC H - - C C + H2O OOC C C COO H COO- H OH fumarate (F) malate (M) k k1 2 k3 E + F E F k E M k E + M k-1 -2 -3 Analysis gives initial rate (messy to work out, mixes rapid equilibrium & steady state, do forward reaction only, let k-3~0): k2k3 (k2 k2 k3 )[E0 ][F] VF[F] F = = F k1k2 k1k3 k2k3 K [F] [F] M k1k2 k2 k3 result appears simple, same as Michaelis Menton form: MAX[S]/(KM+[S]) MAX = VF, but constants complex usual initial rate, experiment not separate intermediate Aside: if do Relaxation measurement sense reverse rxn – analysis just do backwards, let "VR" be max. rev. rate R R = VR[M]/(K M + [M] ) from R = d[F]/dt = k-1[EF] net rate (F-R) - did this considering Product, M R R F R R R = (VFK M[F] – VRK M[M])/(K MK M + K M[F] + K M[M]) at beginning M = 0 -- gives back original form R F at equilibrium: VFK M[F]e = VRK M[M]e R F Ke = [M]e/[F]e = VFK M / VRK M To parse out all k1, k2 … etc. is difficult since observations combine Need to isolate/detect intermediates or vary conditions to get individual rate constants (ex. H/D exchange) 48 VI–49 Inhibition Sometimes path involves multiple intermediates but still apparently simple (like above) k k k E + S ES 2 ES' 3 ES'' 4 E + P 0 = kcat[E0][S]/(KM + [S]) k Differencek1 is kcat and KM combine3 all ki above k if onek-1 step (i) is slow (rate- 3limit) kcat = ki turnover depends on slow step KM always like (apparent) dissociation constant for all enzyme bound species: [E][S]/[ES] if k-1 >> ki then KM = k-1/k1 true dissociation const. Now many enzymes bind several different substrates but only one gives product Some substrates called inhibitor, I, bind tightly but have no reaction remove E0 - not avail. for S Competitive inhibition, S and I bind to the same site k E + S ES 2 E + P k-2 E + I EI This agrees with books and notes, reduces amount of E0 available to create ES, depend on KI –dissoc. const. of EI 49 VI–50 Competitive inhibition, parallel mechanism E + S ES E + P E + I EI k1 k3 k k-1 -3 k2 k-2 – Inhibitor (- - -) comparison to No Inhibitor (____) -- – Competitive inhibition – plot rate 0 vs. [S] and 1/0 vs. 1/[S] see comparison. – Inhibitor raises slope of L- B plot (dec. rate, 1/0 less) but same y intercept 1/m – change Km not m – half velocity, 0 = m/2, intercept on 0 vs. [S] plot = Km Two paths for E0, result: less available E for making P 50 VI–51 do steady state on [ES] again -- still product only from k2 step -1 = k2[ES] [ES] = k1[E][S]/(k-1 + k2) = KM [E][S] --as M-M but now [E0] = [E] + [ES] + [EI] reduce effect. enzyme, i.e [E] (EI ties up some of E0 - prevents formation ES, lowers E) let KI = [E][I]/[EI] KM = [E][S]/[ES] --2 dissoc const./paths [E0] = [E](1 + [I]/KI) + [ES] -- subst [EI] in [E0] eq. k1 not substitute for ES because will solve for it k-1 [I] reduces [E] + [ES] avail. -- consider E ES equil. Parameter for Inhibitor effect: = (1+[I]/KI) = (1+ [EI]/[E]) -- Note > 1, inc. [I] inc. , [E] = ([E0]–[ES])/(1+[I]/KI) = ([E0] – [ES])/ = [E] -1 -1 sub [E] into [ES]: [ES] = KM [E][S] = KM [S]([E0]–[ES])/ rearrange - [ES](KM+[S])=[E0][S] [ES]=[E0][S]/(KM+[S]) rate like before: 0 = k2[E0][S]/(KM + [S]) = MAX/(KM/[S] + 1) now inc. , decrease 0, but KM = K'M means 1/ vs. 1/S has diff. slope (KM/MAX) but same intercept i.e. change Km not m - L-B plot intersect on y-axis at 1/m - high KI, dissoc EI, = (1+[I]/KI) ~1, lots E, get same m 1/ = (K’M/ MAX)(1/[S]) + 1/MAX fit half rate K'M As [I] increases I << KI ~ 1 ordinary M - M I >> KI ~ [EI]/[E] >> 1, so [EI] >> [E] 51 VI–52 Three forms of inhibition are discussed in various texts: Competitive non-competitive uncompetitive E+S ES E+P E+S ES E+P E+S ES E+P E+I EI E + I EI ES+I ESI ES+I ESI k1 k3 k k-1 -3 as described above here both EI and ESI here no EI form ( = 1) (if ’, simple case) but ESI form (' > 1) _____ ______ Change with inhibitork2 ( ) compared to normal enzyme ( ) The alternative to kcompetitive-2 are non-competitive and uncompetitive. The difference is how an inhibitor interacts with the enzyme-substrate complex, and prevents product forming Non-competitive: (derive Engel - pp 712-13) E + S ES E + P E + I EI ES + I ESI 52 VI–53 0 = k2[E0][S]/{[S](1+[I]/KIS)+Km(1+[I]/KI)} = MAX [S]/(’[S] + KM ) where ’ = (1+[I]/KIS), = (1+[I]/KI), KIS = [ES][I]/[EIS] Again: [I]=0 ’=1 (MM), as [I] inc. reaction slows, less P In Lineweaver Burk, both slope and intercept change 1/ = (KM/ MAX)(1/[S]) +k 3 ’/MAX k-3 Here - if ’>1 are independent, ≠ ’ i.e. KI≠ KIS then measured KM will be different with inhibitor thus they do not have to intercept at x-axis at 1/ KM However, most books show intercept on x-axis (called -1/KM ) between inhibited and no inhibitor for this, or = ’. Why? Tinoco makes comment that they describe “simplest case” – true Engel just asserts they intercept and KM is not changed Atkins points out this is a special case, i.e. ’>1 but, = ’ Why do this? if I binds to a site away from the S binding site, then it is likely that S does not affect the binding constant of I if true then, KI = KIS and thus = ’ makes x-intercept be KM Uncompetitive: drop E + I EI = 1, ' > 1, So in this case the inhibitor cannot bind enzyme alone, needs the substrate first. Without this step KM = KM’ ( = 1) slope not changed, but intercept change (’/MAX) Same equation, but Lineweaver-Burk plots are parallel As Tinoco notes, this is a special case 53 VI–54 Plot for non-competitive inhibition (iff ’ >1) KM same, x-intercept, or 0 = ½ MAX but slope changes max reduced –non-competitive inhibition See shape change: two slopes, two y-intercept 54 VI–55 Inhibition hugely important – most biochemistry processes need enzymes – need specific reaction to go under control – need operate with low concentration enzyme – need fast response to S perturbation - on/off control – bio feedback/signaling– gone wrong – “sick” Drugs/pharma intercede inhibit example dihdrofolate reductase (See Engel pp. 713-14) DHFR reduces 7,8 dihydrofolate (DHF) to 5,6,7,8- tetrahydrofolate, a step in the biosynthesis of thymidine. It is inhibited by methotrexate (MTX) – see structures: Shape of MTX looks like DHF, but amine function is methylated, prevents use of N-H in reduction mechanism Used in cancer therapies, less thymidine, less cell division 55 VI–56 With MTX need more substrate to get to half MAX, or KM’ = KM > KM -- slows reaction, steeper slope 1/ L-B: 1/ = (KM/ MAX)(1/[S]) + ’/MAX Problem: Use the data below to determine the inhibition mechanism for DHFR and the maximum rate. If KM for DHFR is 0.10 mM, what is KI? adding more inhibitor, see slower rate: MTX competitive inhibitor, slope change, same intercept fit data at left: Slope= 0.535 s Intercept =0.091 mM-1 -1 MAX =11 mM s KM’ = 5.89 mM KI = 3.4 pM MTX – binds strongly! --remember KI and KM dissociation const, smaller value, stronger binding 56 VI–57 e.g. succinic dehydrogenase Tinoco---p417-8 – – SH1 – – a) OOC–CH2–CH2–COO OOC–CH=CH–COO succinate fumarate – – if add malonate: OOC–CH2–COO has similar shape, 2 –COO- groups – binds active site but cannot dehydrogenate (not form C=C) so stays on enzyme: EI forms, ties up E inhibit reaction of succinate: less ES like case (a), uses same site — Competitive inhibition b) Triosephosphate dehydrogenase - Cys in active site if alkylate: TPD S H + I CH2CNH2 TPD S CH2C NH2 O O This enzyme now lost to reaction inhibition (no reversal) but KM same, max reduced (recall slope Km/max) like case (c), sort of kills off the enzyme – non-competitive ' > 1 see figure above 57 VI–58 Allosterism binding substrate to one site/subunit affect binding/release at 2nd or 3rd … e.g. hemoglobin – 4 subunits, each bind O2 cooperate MWC model each subunit has 2 forms different binding efficiency R – relaxed – -binds better T – tense - - no substrate: L = [T0]/[R0] equilibrium Sigmoidal shape Solve mechanism use pre-equilibrium: max L = [S]/KR L = [T0]/[R0] L = 0 no T max/(1 + ) this is M–M L increase sigmoidal shape in rate vs. [S] This is like inhibitor, reduces amount of R available to bind S binding S convert T R max rate constant analogous to increasing enzyme [E0], only R binds KR increase: denom (1+ L) ~ 1 and maxmax[S]/KR So inc. KR dec. rate – it dissociates E-S complex 58 VI–59 59 .
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