Enzyme reaction example of Catalysis, simplest form: E + P at end of reaction No consumption of E (ES): enzyme-substrate complex Intermediate

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1 V 41 Enzyme Kinetics Enzyme reaction example of Catalysis, simplest form: k 1 E + S k -1 ES E at beginning and ES k 2 k -2 E + P at end of reaction No consumption of E (ES): enzyme-substrate complex Intermediate Enzyme stabilize (lower E a ) transition state by binding S Can promote bond formation/breaking same as catalyst efficient, selective - usually 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: stretch bond / exchange charge /deform / add atom/ rearrange electron density etc. 41

2 V 42 means transition state lower E a than in solution specific dramatic efficiency Other enzymes: Ribozyme RNA catalyze Others carbohydrates ex: 2H 2 O 2 cat 2H 2 O + ½ O 2 slow in solution General catalyst: 1 st order: [H 2 O 2 ] [catalyst] typical- inorganic catalyst: Fe or HX, increase rate by ~ Ezymatically two behaviors (orders) 1 st order: ~[H 2 O 2 ] 1 low conc. catalase enhance r~ k 1 [cat][h 2 O 2 ] rate ~ th order: ~[H 2 O 2 ] 0 high conc. maximum rate (10 7 molecules sec -1 cat -1 ) r ~ k 0 [cat] 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 -1-1st ord. E a = 42

3 V 43 pre-exponential A < 1 x 10 5 s -1 with Fe or HBr E a ~ 45 k J mol -1 with Catalase E a ~ 8 k J mol -1 A ~ 1.6 x 10 8 M -1 s entropy barrier also lower Maximum reaction velocity υ max with inc. [S] υ m = k [E 0 ] Catalytic constant turnover number: k = υ max /[E 0 ] s -1 {active site determined} - υ max depend on [E 0 ] Michaelis-Menten Analyze mechanism / get rate law Since enzyme so efficient works very low concentration (actual value enzyme conc. may be unknown) Goals mechanism should fit these: a) υ = -ds/dt = dp/dt inc. linear w/[e] (double E double υ) b) υ = k [S] if [E] constant 1 st order in substrate c) υ υ max at high [S] (becomes zero order) goes through intermediate enzyme-substrate complex E + S ES P + E 43

4 T.S. V 44 Mechanism: E + S E S k 1 k -1 k 2 k -2 E S E + P ES initial rate ignore k -2 no P initially E+S υ 0 = (dp/dt) 0 = k 2 [ES] E+P 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 ) note: conc. are free E, S can also picture as equilibrium, conc. ratio is const. K M = (k -1 + k 2 )/k 1 = [E][S]/[ES] --reflect ES dissociation (plug in regular 2nd order) υ 0 = k 2 {k 1 /(k -1 + k 2 )}[E][S] ~ k 2 [E][S]/K M Note 1: what if assume rapid equilibrium?? υ 0 = k 2 K e [E][S] K e = [ES]/[E] [S] = k 1 /k -1 = k 2 (k 1 /k -1 )[E][S] miss denom. balance 2 k s Note 2: free enzyme hard to determine (protein conc.?): [E] 0 = [E] + [ES] [S] 0 = [S] + [ES] [S] since [ES] low works for initial rate: equate [S] ~ [S 0 ] and [P] ~ 0 Substitute these inot steady state result: [ES] = {k 1 /(k -1 + k 2 )}{[E 0 ] [ES]} {[S]} recall: K M = (k -1 + k 2 )/k 1 44

5 V 45 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: υ = k 2 [ES] = k 2 [E 0 ][S]/{K M + [S]} υ max = k 2 [E 0 ] υ = υ max /({K M /[S]} + 1) divide thru by [S] Analysis invert rate equation: 1/υ = (1/υ max )(K M /[S] + 1) Lineweaver-Burk plot: 1/υ vs. 1/[S] slope: K M /υ max x intercept: -1/K M, y intercept: 1/υ max recall υ max = k 2 [E 0 ] avoids need to have value E 0 integrate: (K M /[S] + 1)d[S] = -υ m dt K M ln {[S]/[S 0 ]} + [S] - [S 0 ] = -υ m t Behavior: a) low [S] υ (υ max /K M )[S] K M /[S] >> 1 1st order in S b) high [S] υ υ max K M /[S] << 1 0th order in S - turnover: k cat = υ max /E 0 = k 2 45

6 Interpret if [S] = K M υ = υ 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 ] V 46 How about Product if P build up need consider k -2 normally since consider initial rates could ignore k if include: ES 2 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 ) 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 } before just one term Also recall [E 0 ] = [E] + [ES] = [E] {1 + [S]/K M + [P]/K' M } Substitute in rates: υ = 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 cancel 2 nd term υ = [E 0 ] {(k 2 [S]/K M k -1 [P]/K' M )/(1 + [S]/ K M + [P]/K' M )} Now if υ max = k 2 [E 0 ] υ' max = k -1 [E 0 ] then υ = (υ max [S]/K M υ' max [P]/K' M )/(1 + [S]/K m + [P]/K' M ) (note: this took me many tries and help to get right form) 46

7 V 47 Point if include the Product, still interpretable a) Beginning of reaction [P] = 0 υ = (υ m [S]/K M )/(1 + [S]/K M ) = υ m /(K M /[s] + 1) same as MM b) as product builds up reaction will slow (-υ /K M ) Lineweaver-Burk Eadie-Hofstee (mult. by υ m υ 0 ) 1/υ 0 = K M /υ m 1/[S] + 1/υ m υ 0 = -K M υ 0 /[S] + 1/υ m Plot 1/υ vs. 1/S Plot υ 0 vs. υ 0 /S compress high S value spreads high S values a. Lineweaver-Burk plot is b. Eadie-Hofstee plot gives Two plots analyze same data set hydrolysis CBZ Gly Trp by carboxypeptidase vary [S] : mm 47