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Enzyme Kinetics – 2014 – Atkins Ch.23, Tinoco 4th-Ch.
Enzyme rxn – example Catalysis/Mechanism :
E + S k-
k 1
ES E is at beginning and
ES
k 2
k-2 E + P at end of reaction
Catalyst: No consumption of E (ES): enzyme-substrate complex – Intermediate
Enzyme stabilize (lower Ea) 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 Ea than solution specific – dramatic efficiency Other enzymes: Ribozyme – RNA catalyze Others – carbohydrates …
R-
s
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 ~ 10^4 - 10^5
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 ~ 10^15
0 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^0298 = -103 k J mol- most of this is enthalpy H^0 = -94.6 k J mol- Activation high: Ea = 71 k J mol- slow solution reaction since: r < 4 x 10-8^ Ms-1^ – 1 st^ order pre-exponential A < 1 x 10^5 s-1^ diagram miss intermed. ES
with Fe or HBr Ea ~ 45 k J mol-1^ - chemical catalysis with Catalase Ea ~ 8 k J mol-1^ - enzymatic A ~ 1.6 x 10^8 M-1^ s-1^ - entropy barrier also lower – eS/R
Maximum reaction velocity max with inc. [S] m = k [E 0 ] Catalytic constant – turnover number: k = max/[E 0 ] s- {# of active site limited} - max depend on [E 0 ] reaction 0th^ order in [S], but 1st^ order in [E 0 ]
Ea=
(solve for [ES], plug into initial rate eqn. – get regular 2nd^ order) 0 = k 2 {k 1 /(k-1 + k 2 )}[E][S] ~ k 2 [E][S]/KM
Note 1: if assume rapid equilibrium? Build up ES (slow P)? 0 = k 2 Ke[E][S] Ke = [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: KM = (k-1 + k 2 )/k 1
ES on both sides, rearrange: [ES]{1+ [S]/ KM } = [E 0 ][S]/ KM
[ES] = {k 1 /(k-1 + k 2 )}[E 0 ][S] let KM = (k-1 + k 2 )/k 1 1+{k 1 /(k- 1 + k 2 )}[S] mult. top/bottom by KM
[ES] = [E 0 ][S]/{KM + [S] } now [ES] indep. [E], just [E 0 ] Product only in 2nd^ step, depend on [ES], substitute: = k 2 [ES] = k 2 [E 0 ][S]/{KM + [S]} right form, low-high [S] = max/{(KM/[S]) + 1} div. by [S],max = k 2 [E 0 ]
Analysis invert rate equation: 1/ = (1/max )(KM/[S] + 1) Independent of [E 0 ]
Lineweaver-Burk
plot: 1/ vs. 1/[S] slope: KM/max x intercept: -1/KM , y intercept: 1/max recall – max = k 2 [E 0 ] avoids expressing value for E 0
integrate M-M: = -d[S]/dt (KM/[S] + 1)d[S] = -mdt KMℓn {[S]/[S 0 ]} + [S] - [S 0 ] = -mt complex conc./time-dep.
Behavior: a) low [S] (max/KM)[S] KM/[S] >> 1 1st order in S b) high [S] max KM/[S] << 1 0th order in S - turnover: kcat = max/E 0 = k 2
Interpret if [S] = KM = max/ KM 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- normally– since consider initial rates – could ignore
if include: ES k k^2
P + E do a steady state:
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 )
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 3 -^ + H+
At pH 7.1, 273.5 K, and [E 0 ] = 2.3 nmol L-
invert both values:
Plot results as Lineweaver-Burk: 40 1/ Slope = 40 Intercept = 4 30 MAX1/intercept 20 mmol L-1^ s- KM = slope/intercept = 10 mmol L-
10 turnover: k = MAX[E 0 ] =1.09 x 10^5 s- [CO 2 ]½ (at ½ MAX) = KM 0 10 x (1/[CO 2 ])
Alternate way of looking at data: Lineweaver-Burk Eadie-Hofstee (mult. by m 0 ) 1/ 0 = (KM/m)(1/[S]) + 1/m 0 = -KM{ 0 /[S]} + m Plot 1/ vs. 1/S Plot 0 vs. 0 /S compress high S value spreads high S values slope: KM/m int: 1/m slope:-KM,int:m a.Lineweaver-Burk plot: b. Eadie-Hofstee plot:
Two plots analyze same data set – hydrolysis CBZ–Gly–Trp by carboxypeptidase vary [S] : 2.5 20 mM
