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.
4. Irreversible First-Order Reactions
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)
k
5. Reversible First-Order Reactions
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
k1
k-1
6. Second-Order Reactions
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
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
[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).
13. The Effect of Substrate Concentration on
Reaction Velocity
14. Michaelis-Menten Kinetics (1)
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)
E = enzyme, S = substrate,
ES = enzyme-substrate complex,
P = product
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]
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.
22. Multistep Reactions
E + S ES ES E + P
vo = kcat[E]t[S]/(KM + [S])
k2 k3
k1
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.
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
E
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.)
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
45. Effects of Mixed (Noncompetitive)
Inhibitor on Enzyme Kinetics
Kapp
M = (1 + [I]/KI)KM/(1 + [I]/KI)
(= KM, when KI = KI, which is often the case.)
Vapp
max = Vmax/(1 + [I]/KI) < Vmax
k1
k-1
•Not the same as uncompetitive inhibition.
•In mixed inhibition, inhibitor can bind E or
ES.
47. a = 1 + [I]/KI
a = 1 + [I]/KI
(For mixed inhibitor,
generally, ~ KM)
48. Irreversible Inhibition
k1
k-1
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]
Slope = -kobs
