enzyme kinetics
DESCRIPTION
enzymeTRANSCRIPT
Enzyme Kinetics
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.
Chemical Kinetics
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
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
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.
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)
Relationship of Reaction Rate Constant to Activation Energy and
Temperature: The Arrhenius Equationk = 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 + H20ln k
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
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).
Intermediate States in Multistep Reactions
Enzyme Kinetics
The Effect of Substrate Concentration on Reaction Velocity
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]
or2. 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
The Steady State in Enzyme Kinetics
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]
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])
Michaelis-Menten Kinetics Continued (4)
Now we can substitute for [ES] in the rate equationvo = 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.
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)
A Lineweaver-Burk (Double Reciprocal) Plot
An Eadie-Hofstee Plot
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.
Catalytic Efficiency (kcat/KM )
“Perfect enzyme”
Diffusion-controlled limit: 108-109 M-1s-1
Substrate preferences of chymotrypsin
pH-Dependence of Enzyme Activity
Enzyme-Catalyzed Bisubstrate Reactions: Two Examples
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
Ping Pong reaction
Sequential binding
Ternarycomplex
Indicative of ternary complex formation and a sequential mechanism
Indicative of a Ping Pong mechanism
Enzyme Inhibition
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.)
Competitive Inhibition
Effects of Competitive Inhibitor on Enzyme Kinetics
KappM = KM(1 + [I]/KI) > KM
Vappmax = Vmax
KI (inhibitor dissociation constant) = koff/kon
= 1 + [I]/KI
A Substrate and Its Competitive Inhibitor
HIV Protease Inhibitors
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]/(KM + [S]))/(Vmax[S]/(KM + [S])) = (KM + [S])/(KM + [S])vI = initial velocity with inhibitorv0 = initial velocity without inhibitor= 1 + [I]/KI
When vI/v0 = 0.5, [I] = IC50 = KI(1 + [S]/KM)
If measurement made when [S] << KM, IC50 = KI
Uncompetitive Inhibition
Effects of Uncompetitive Inhibitor on Enzyme Kinetics
KappM = KM/(1 + [I]/KI) < KM
Vappmax = Vmax/(1 + [I]/KI) < Vmax
= 1 + [I]/KI
Mixed (Noncompetitive) Inhibition
Effects of Mixed (Noncompetitive) Inhibitor on Enzyme Kinetics
KappM = (1 + [I]/KI)KM/(1 + [I]/KI)
(= KM, when KI = KI, which is often the case.)
Vappmax = Vmax/(1 + [I]/KI) < Vmax
k1k-1
•Not the same as uncompetitive inhibition.•In mixed inhibition, inhibitor can bind E or ES.
= 1 + [I]/KI
= 1 + [I]/KI
= 1 + [I]/KI
= 1 + [I]/KI
(For mixed inhibitor, generally, ~ KM)
Irreversible Inhibition
k1
k-1
k2E + 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
Irreversible Inhibition by Adduct Formation
(diisopropylfluorophosphate)
Irreversible Inhibition of Chymotrypsin by TPCK
(N-tosyl-L-phenylalanine chloromethylketone)