1. Enzyme Kinetics

2. Enzymes Are Uniquely Powerful Catalysts
Enzymes are proteins that can accelerate biochemical reactions by factors of 105 to 1017! This is much higher than chemical catalysts.
Enzymes can be extremely specific in terms of reaction substrates and products.
Enzymes catalyze reactions under mild conditions (e.g., pH 7.4, 37ºC).
The catalytic activities of many enzymes can be regulated by allosteric effectors.

3. Chemical Kinetics

4. Irreversible First-Order Reactionsk
A  B
v = d[B]/dt = -d[A]/dt = k[A]
(k = first-order rate constant (s-1))
Change in [A] with time (t):
[A]= [A]o e –kt or
[A]/[A]o = e –kt
ln([A]/[A]o) = –kt
([A]o = initial concentration)

5. Reversible First-Order Reactionsk1
A B
v = -d[A]/dt = k1[A] - k-1[B]
At equilibrium: k1[A]eq - k-1[B]eq = 0
[B]eq/[A]eq = k1/k-1 = Keq  
k-1

6. Second-Order Reactionsk
2A  P
v = -d[A]/dt = k[A]2
Change in [A] with time:
1/[A] = 1/[A]o + kt
A + B  P
v = -d[A]/dt = -d[B]/dt = k[A][B]
(k = second-order rate constant (M-1s-1)) k
Note: third-order reactions rare, fourth- and higher-order reactions unknown.

7. Free Energy Diagrams Keq = e –∆Gº/RT For A A‡ [A]‡/[A]o = e –∆Gº‡/RT 
Keq = equilibrium constant
[A]‡ = concentration of molecules having the activation energy
[A]o = total concentration of A
–∆Gº‡ = standard free energy change of activation (activation energy)

8. Relationship of Reaction Rate Constant to Activation Energy and Temperature: The Arrhenius Equation
k = A e -Ea/RT
Reaction rate constant (k) determined by activation energy (Ea or ∆Gº‡, applying transition state theory) and temperature (T) and proportional to frequency of forming product
(A or Q = kBT/h, where kB = Boltzmann’s constant, h = Planck’s constant):
k = (kBT/h) e -G°‡/RT
k = Q e -G°‡/RT
G = H - T S, so:
k = Q e S°‡/R  e -H°‡/RT
k = Q e -H°‡/RT
(where Q = Q e S°‡/R)
So: ln k = ln Q - H°‡/RT
L-malate  fumarate + H20
ln k

9. Relation of Equilibrium Constant to Activation Energy
Keq = k1/k-1
Keq = (Q e -G1°‡/RT)/(Q e -G-1°‡/RT)
Keq = e -(G1°‡ - G-1°‡)/RT
∆G° = G1°‡ - G-1°‡
Keq = e –∆G°/RT
Equilibrium constant Keq says nothing about rate of reaction, only free energy difference between final and initial states. The activation energy barrier opposes reaction in both directions

10. Effect of a Catalyst on Activation Energy
Catalysts do not affect GA (initial) or GB (final) and so do not affect overall free energy change (∆G° = GB - GA) or equilibrium constant Keq.
Equilibrium concentrations of A and B still determined solely by overall free energy change.
Catalysts only affect ∆G°‡, lowering the activation energy.
They accelerate both the forward and reverse reaction (increase kinetic rate constants k1 and k-1).

11. Intermediate States in Multistep Reactions

12. Enzyme Kinetics

13. The Effect of Substrate Concentration on Reaction Velocity

14. Michaelis-Menten Kinetics (1)
E = enzyme, S = substrate,
ES = enzyme-substrate complex,
P = product
v = k2[ES]
(Note: k2 also referred to as kcat)
[Enzyme]total = [E]t = [E] + [ES]
How to solve for [ES]?
1. Assume equilibrium, if k-1 >> k2:
KS = k-1/k1 = [E][S]/[ES] or
2. Assume steady state:
d[ES]/dt = 0
(Michaelis and Menten, 1913)
(Briggs and Haldane, 1925)

15. The Steady State in Enzyme Kinetics

16. Michaelis-Menten Kinetics Continued (2)
Rate of formation of ES complex = k1[E][S]
Rate of breakdown of ES complex = k-1[ES] + k2[ES]
Because of steady state assumption:
k1[E][S] = k-1[ES] + k2[ES]
Rearranging: [ES] = (k1/(k-1 + k2))[E][S]
Substituting Michaelis constant = KM = (k-1 + k2)/k1) = KS + k2/k1:
[ES] = ([E][S])/KM
So: KM[ES] = [E][S]

17. Michaelis-Menten Kinetics Continued (3)
Substituting [E] = [E]t - [ES]:
KM[ES] = [E]t[S] - [ES][S]
Rearranging: [ES](KM + [S]) = [E]t[S]
So: [ES] = [E]t[S]/(KM + [S])

18. Michaelis-Menten Kinetics Continued (4)
Now we can substitute for [ES] in the rate equation
vo = k2[ES].
But first note that the velocity in v = k2[ES] we use is the initial velocity, vo, the velocity of the reaction after the pre-steady state and in the early part of the steady state, i.e., before ~10% of substrate is converted to product. This is because at this stage of the reaction, the steady-state assumption is reasonable ([ES] is still approximately constant). Also, since not much P has yet accumulated, we can approximate the kinetics for even reversible reactions with this equation if we limit ourselves to vo.

19. The Michaelis-Menten Equation
vo = k2[E]t[S]/(KM + [S]) or
vo = Vmax[S]/(KM + [S])
(since Vmax = k2[E]t when [S] >> KM)

20. A Lineweaver-Burk (Double Reciprocal) Plot

21. An Eadie-Hofstee Plot

22. Multistep Reactions E + S ES  ES  E + P  k1
k k3
E + S ES  ES  E + P
vo = kcat[E]t[S]/(KM + [S])  
k-1
kcat = empirical rate constant that reflects rate-determining component. Mathematically, for the reaction above,
kcat = k2k3/(k2 + k3).
However, k2 and k3 often very hard to establish with precision as individual rate constants.

26. Catalytic Efficiency (kcat/KM )
“Perfect enzyme”
Diffusion-controlled limit: M-1s-1
Substrate preferences of chymotrypsin

27. pH-Dependence of Enzyme Activity

28. Enzyme-Catalyzed Bisubstrate Reactions: Two Examples

29. Bisubstrate Reactions
S1 + S2 P1 + P2
A-X + B A + B-X (in transferase reactions)
Sequential binding of S1 and S2 before catalysis:
Random substrate binding - Either S1 or S2 can bind first, then the other binds.
Ordered substrate binding - S1 must bind before S2.
Ping Pong reaction - first S1  P1, P1 released before S2 binds, then S2  P2.  E  

30. Sequential binding
Ternary
complex
Ping Pong reaction

31. Indicative of ternary complex formation and a sequential mechanism

32. Indicative of a Ping Pong mechanism

33. Enzyme Inhibition

34. Types of Enzyme Inhibition
Reversible inhibition
(Inhibitors that can reversibly bind and dissociate from enzyme; activity of enzyme recovers when inhibitor diluted out; usually non-covalent interaction.)
Competitive
Mixed (noncompetitive)
Uncompetitive
Irreversible inhibition
(Inactivators that irreversibly associate with enzyme; activity of enzyme does not recover with dilution; usually covalent interaction.)

35. Competitive Inhibition

36. Effects of Competitive Inhibitor on Enzyme Kinetics
KI (inhibitor dissociation constant) = koff/kon
KappM = KM(1 + [I]/KI) > KM
Vappmax = Vmax

37. a = 1 + [I]/KI

38. A Substrate and Its Competitive Inhibitor

39. HIV Protease Inhibitors

40. Relationship of KI to Half-Maximal Inhibitory Concentration (IC50)
For a competitive inhibitor of an enzyme that follows Michaelis-Menton kinetics:
vI/v0 = (Vmax[S]/(KMa + [S]))/(Vmax[S]/(KM + [S])) = (KM + [S])/(KMa + [S])
vI = initial velocity with inhibitor
v0 = initial velocity without inhibitor
a = 1 + [I]/KI
When vI/v0 = 0.5, [I] = IC50 = KI(1 + [S]/KM)
If measurement made when [S] << KM, IC50 = KI

41. Uncompetitive Inhibition

42. Effects of Uncompetitive Inhibitor on Enzyme Kinetics
KappM = KM/(1 + [I]/KI) < KM
Vappmax = Vmax/(1 + [I]/KI) < Vmax

43. a = 1 + [I]/KI

44. Mixed (Noncompetitive) Inhibition

45. Effects of Mixed (Noncompetitive) Inhibitor on Enzyme Kinetics
Not the same as uncompetitive inhibition.
In mixed inhibition, inhibitor can bind E or ES. k1  
k-1
KappM = (1 + [I]/KI)KM/(1 + [I]/KI)
(= KM, when KI = KI, which is often the case.)
Vappmax = Vmax/(1 + [I]/KI) < Vmax

46. a = 1 + [I]/KI
a = 1 + [I]/KI

47. (For mixed inhibitor, generally, ~ KM)
a = 1 + [I]/KI
a = 1 + [I]/KI

48. Irreversible Inhibitionk1 k2
E + I E·I  E-I
Plot:
ln(residual enzyme activity) vs. time
If [I]>>[E], conditions are pseudo-first order and slope is -kobs (pseudo-first order inactivation rate constant)
kinact (second-order inactivation constant) = k1k2/k-1 = kobs/[I]  
k-1
Slope = -kobs

49. Irreversible Inhibition by Adduct Formation
(diisopropylfluorophosphate)

50. Irreversible Inhibition of Chymotrypsin by TPCK
(N-tosyl-L-phenylalanine chloromethylketone)