Enzyme Kinetics 2014

Transcription

1 V 41 Enzyme Kinetics 2014 Atkins Ch.23, Tinoco 4 th -Ch.8 Enzyme rxn example Catalysis/Mechanism: E + S k -1 ES k 1 ES E is at beginning and k 2 k -2 E + P at end of reaction Catalyst: No consumption of E (ES): enzyme-substrate complex Intermediate Enzyme stabilize (lower E a ) transition state (between S and ES) by binding S and can promote bond formation/breaking Same as catalyst more efficient, selective - not general rate enhancements can be tremendous Mechanism above looks like typical rapid equilibrium but there may be more steps especially reverse k 2 Note: typically think of enzymes as proteins bind substrate (S) to specific site can modify S to activate for reaction Models: a) Lock and key substrate fits specific enzyme site b) Induced fit enzyme modifies structure to fit substrate Idea in site easier to: stretch bond / exchange charge /deform / add atom/ rearrange electron density etc. means stabilize TS, lower E a than solution specific dramatic efficiency Other enzymes: Ribozyme RNA catalyze Others carbohydrates s R- 41

2 V 42 ex: 2H 2 O 2 cat 2H 2 O + O 2 if no cat., slow in solution General catalyst: 1 st order: r ~ [H 2 O 2 ] [catalyst] - typical inorganic catalyst: Fe or HX, increase rate by ~ Ezymatically two behaviors (orders) 1 st order: r ~ [H 2 O 2 ] 1 low H 2 O 2 conc. enz = catalase r~ k 1 [enz][h 2 O 2 ] enhance rate ~ th order: r ~ [H 2 O 2 ] 0 high H 2 O 2 conc. r ~ k 0 [enz] max. rate (10 7 molec. sec -1 cat -1 ) - enzyme limit H 2 O 2 H 2 O + ½ O 2 (s) exergonic: G = -103 k J mol -1 most of this is enthalpy H 0 = k J mol -1 Activation high: E a = 71 k J mol -1 slow solution reaction since: r < 4 x 10-8 Ms -1 1 st order pre-exponential A < 1 x 10 5 s -1 E a = diagram miss intermed. ES with Fe or HBr E a ~ 45 k J mol -1 - chemical catalysis with Catalase E a ~ 8 k J mol -1 - enzymatic A ~ 1.6 x 10 8 M -1 s -1 - entropy barrier also lower e S/R Maximum reaction velocity max with inc. [S] m = k [E 0 ] Catalytic constant turnover number: k = max /[E 0 ] s -1 {# of active site limited} - max depend on [E 0 ] reaction 0 th order in [S], but 1 st order in [E 0 ] 42

3 V 43 Michaelis-Menten Analyze mechanism / get observed rate law Since enzyme so efficient work at very low concentration (actual value enzyme conc. may be unknown) Goals mechanism should fit these observed properties: a) = -ds/dt = dp/dt inc. linear w/[e] (double E double ) b) = k [S] if [E] constant 1 st order in substrate, low [S] c) max at high [S] (becomes zero order) goes through intermediate enzyme-substrate complex E + S ES P + E Mechanism: E + S E S k 1 k -1 k 2 k -2 E S E + P initial rate ignore k -2 no P initially 0 = (dp/dt) 0 = k 2 [ES] if steady state: d(es)/dt = k 1 [E][S] - k -1 [ES] - k 2 [ES] = 0 [ES] = k 1 [E][S]/(k -1 + k 2 ) Could plug in and solve mechanism but do not know E note: conc. are free E, S not initial E 0, S 0 can also picture as equilibrium, conc. ratio is ~const. K M = (k -1 + k 2 )/k 1 = [E][S]/[ES] - K M reflect ES dissociation 43 E+S T.S. ES E enz a Barrier between S and ES, E a enz, control rate, lower than TS E a E+P

4 V 44 (solve for [ES], plug into initial rate eqn. get regular 2 nd order) 0 = k 2 {k 1 /(k -1 + k 2 )}[E][S] ~ k 2 [E][S]/K M Note 1: if assume rapid equilibrium? Build up ES (slow P)? 0 = k 2 K e [E][S] K e = [ES]/[E] [S] = k 1 /k -1 = k 2 (k 1 /k -1 )[E][S] same form, but denom. miss k 2 Note 2: free enzyme hard to determine (protein conc. problem?): [E] 0 = [E] + [ES] [S] 0 = [S] + [ES] [S] initially, since [ES] low, [S]=[S 0 ] works for initial rate: equate [S] ~ [S 0 ] and [P] ~ 0 Substitute these into steady state result: [ES] = {k 1 /(k -1 + k 2 )}{[E 0 ] [ES]}{[S]} recall: K M = (k -1 + k 2 )/k 1 ES on both sides, rearrange: [ES]{1+ [S]/ K M } = [E 0 ][S]/ K M [ES] = {k 1 /(k -1 + k 2 )}[E 0 ][S] let K M = (k -1 + k 2 )/k 1 1+{k 1 /(k -1 + k 2 )}[S] mult. top/bottom by K M [ES] = [E 0 ][S]/{K M + [S] } now [ES] indep. [E], just [E 0 ] Product only in 2 nd step, depend on [ES], substitute: = k 2 [ES] = k 2 [E 0 ][S]/{K M + [S]} right form, low-high [S] = max /{(K M /[S]) + 1} div. by [S], max = k 2 [E 0 ] 44

5 V 45 Analysis invert rate equation: 1/ = (1/ max )(K M /[S] + 1) Independent of [E 0 ] Lineweaver-Burk plot: 1/ vs. 1/[S] slope: K M / max x intercept: -1/K M, y intercept: 1/ max recall max = k 2 [E 0 ] avoids expressing value for E 0 integrate M-M: = -d[s]/dt (K M /[S] + 1)d[S] = - m dt K M ln {[S]/[S 0 ]} + [S] - [S 0 ] = - m t complex conc./time-dep. Behavior: a) low [S] ( max /K M )[S] K M /[S] >> 1 1st order in S b) high [S] max K M /[S] << 1 0th order in S - turnover: k cat = max /E 0 = k 2 Interpret if [S] = K M = max /2 K M small E bind S tightly (k -1 +k 2 <<k 1 ) or low [E] not much [S] needed to saturate [E 0 ] S does not come off ES easily How about Product? if P build up need consider k -2 normally since consider initial rates could ignore k 2 if include: ES P + E do a steady state: k -2 d(es)/dt = 0 = k 1 [E][S] k -1 [ES] k 2 [ES] + k -2 [E][P] [ES] = (k 1 [E][S] + k -2 [E][P])/(k -1 + k 2 ) 45

6 V 46 let: K M = (k -1 + k 2 )/k 1 and K' M = (k -1 + k 2 )/k -2 [ES] = [E] {[S]/K M + [P]/K' M } was one term w/o [P] Also recall [E 0 ] = [E] + [ES] [E 0 ] = [E] {1 + [S]/K M + [P]/K' M } Substitute in rates: = k 2 [ES] k -2 [E][P] = [E] {k 2 ([S]/K M + [P]/K M ) k -2 [P]} = [E 0 ] {k 2 ([S]/K M + [P]/K M ) k -2 [P])/(1 + [S]/K m + [P]/K M )} recall: k -2 K m = (k -1 + k 2 ) k -2 K m [P]/K m = (k -1 + k 2 )[P]/K m Substitute this for 3 rd term numerator, k 2 part cancels 2 nd term = [E 0 ] {(k 2 [S]/K M k -1 [P]/K' M )/(1 + [S]/ K M + [P]/K' M )} Now if max = k 2 [E 0 ] (max forward) ' max = k -1 [E 0 ] (max rev rxn) then = ( max [S]/K M ' max [P]/K' M )/(1 + [S]/K m + [P]/K' M ) (note: this took me many tries and help to get right form) Point if include the Product, still interpretable a) Beginning of reaction [P] = 0 = ( m [S]/K M )/(1 + [S]/K M ) = m /(K M /[S] + 1) same as MM b) as product builds up reaction slows (- /K M ) dominate 46

7 V 47 Example: (from Atkins) Carbonic anhydrase catalyze hydration of CO 2 in red blood cells to give bicarbonate (HCO 3 - ) CO 2 + H 2 O HCO H + At ph 7.1, K, and [E 0 ] = 2.3 nmol L -1 invert both values: Plot results as Lineweaver-Burk: 40 1/ Slope = 40 Intercept = 4 30 MAX 1/intercept 20 mmol L -1 s -1 K M = slope/intercept = 10 mmol L turnover: k = MAX [E 0 ] =1.09 x 10 5 s -1 [CO 2 ] ½ (at ½ MAX ) = K M 0 10 x (1/[CO 2 ]) 47

8 V 48 Alternate way of looking at data: Lineweaver-Burk Eadie-Hofstee (mult. by m 0 ) 1/ 0 = (K M / m )(1/[S]) + 1/ m 0 = -K M { 0 /[S]} + m Plot 1/ vs. 1/S Plot 0 vs. 0 /S compress high S value spreads high S values slope: K M / m int: 1/ m slope:-k M,int: m a.lineweaver-burk plot: b. Eadie-Hofstee plot: Two plots analyze same data set hydrolysis CBZ Gly Trp by carboxypeptidase vary [S] : mm 48

9 V 49 Another look at lock-key, from Atkins, Chymotrypsin cuts peptide bonds, via tetrahedral TS releases amine and separate carboxylic acid, Site is Ser---His---Asp Catalytic triad creates stabilized tetrahedral intermediate after proton transfer to amide, create alkoxide (Ser) which is reactive to break peptide, negative charge stabilized by N-H groups from other peptides or sidechains to form oxoanion hole 49