EnzymeII

k1 [Et ][S ] [Et ][S ] ⇒ [ES ] = = (k1 [S ] + (k −1 + k 2 )) [S ] + (k −1 + k 2 )
k1
[Et ][S ] [S ] + K m
•
Rate of production formation (rate law), v = k2[ES]. So:
k-1
Kd = = −1 Reversible reaction, dissociation constant is [ ES ] k1 So small Km means tight substrate binding; high Km means weak substrate binding. • Km equals to the substrate concentration at which v=1/2vmax [ E ][S ] k
⇒ ⇒ k1 [Et ][S ] − k1 [ES ][S ] = (k −1 + k 2 )[ES ] k1 [Et ][S ] = [ES ](k1 [S ] + ( k −1 + k 2 ))
k1 →[ES ] = K m =( k −1 + k 2 )
At steady state, the forward rate should equal to the reverse rate:
E+S
k1
K-1
ES
k2
E+P
Rate Law in Enzyme Catalyzed Reactions
E+S
k1 K-1
ES
k2
E+P
• Rate law still applies in enzyme catalyzed reactions. • The forward velocity, or rate, vf is,
The catalytic efficiency
• • • • Name for kcat/Km An estimate of "how perfect" the enzyme is kcat/Km is an apparent second-order rate constant It measures how the enzyme performs when S is low The upper limit for kcat/Km is the diffusion limit - the rate at which E and S diffuse together
1) 2) 3) State steady assumption; Initial velocity assumption; Rate law.
Initial Velocity Assumption
E+S
k1 K-1
ES
k2 K-2
E+PFra Baidu bibliotek
• In the beginning of the reaction, there is very little product, or [P] is small. So the amount of [ES] contributed by E+P is negligible. • Thus, the MM equation concerns the reaction rate that is measured during early reaction period. • In which case, the enzyme catalyzed reaction can be modified to: •
Determining the VMAX and KM : LineweaverLineweaver-Burk plot
KM 1 1 + 1 = • V [S] VMAX VMAX
Understanding Vmax
• • • • The theoretical maximal velocity Vmax is a constant Vmax is the theoretical maximal rate of the reaction - but it is NEVER achieved in reality To reach Vmax would require that ALL enzyme molecules are tightly bound with substrate Vmax is asymptotically approached as substrate is increased
The dual nature of the MichaelisMenten equation
Combination of 0-order and 1st-order kinetics • When S is low, the equation for rate is 1st order in S • When S is high, the equation for rate is 0-order in S • The Michaelis-Menten equation describes a rectangular hyperbolic dependence of v on S!
• Km is a constant derived from rate constants
Km = k −1 + k 2 k1
• Km is, under true Michaelis-Menten conditions, an estimate of the dissociation constant of E from S, because k1 at equilibrium, k1[ E ][S ] = k −1[ ES ] E+S ES
v f = k1 [E ][S ]
• The reverse velocity or rate, or the rate of disappearance vd is,
v d = k −1 [ES ] + k 2 [ES ] = (k −1 + k 2 )[ES ]
• At steady state, there is no accumulation of [ES], thus:
[Et ] = [ES ] + [E ]
•
⇒ k1 ([Et ] − [ES ])[S ] = (k −1 + k 2 )[ES ] ⇒ k1 [Et ] = k1 [ES ][S ] + (k −1 + k 2 )[ES ]
v f = k1 [E ][S ] = k1 ([Et ] − [ES ])[S ]
You need to know how this is derived
E+S
• •
k1 K-1
ES
k2 K-2
E+P
•
This is the complete chemical formula for an enzyme-catalyzed (E) reaction of substrate, S and product, P; Mechaelis-Menten equation describes the relationship between reaction rate and substrate concentration. It can explain the saturation behavior in catalyzed reactions as shown in the previous slide. Mechaelis-Menten equation is derived based on the following three conditions:
v= k 2 [Et ][S ] [S ] + K m
Notes on the MM Equations
• The rate of production formation can usually be measured experimentally by monitoring the progress curve of production formation. • The maximum rate can be reached at saturating substrate concentration, or when [S] ∞
Enzyme Kinetics
• The rate of unimolecular reaction is proportional to the concentration of the reactant. Thus rate is linearily dependent on [A].
A→ P
v[ S ]→∞ k [E ] = 2 K t = k 2 [Et ] = v max 1 + m [S ]
Enzyme-catalyzed rate is saturated
So MM equation can be re-written as:
v=
v max [S ] [S ] + K m
Understanding Km
v
v=−
d [ A] = k [ A] dt
[A]
•
But if this reaction is catalyzed by an enzyme, the rate shows saturation behavior.
Why?
v
A Enzyme→ P
[A]
The Mechaelis-Menten Equation
The turnover number
A measure of catalytic activity • kcat, the turnover number, is the number of substrate molecules converted to product per enzyme molecule per unit of time, when E is saturated with substrate. • If the M-M model fits, k2 = kcat = Vmax/Et • Values of kcat range from less than 1/sec to many millions per sec
Enzymes II -Michaelis-Menten Kinetics
Leonor Michaelis (1875-1949)
Maud Menten (1879-1960)
Introduction to Enzyme Kinetics
• Kinetics concerns with the rates of chemical reaction. Enzyme kinetics addresses the biological roles of enzymatic catalysts and quantify the remarkable function of biological enzymes; • Enzyme kinetics information can be exploited to control and manipulate the course of metabolic events. Pharmaceuticals or drugs are often special inhibitors targeted at a particular enzyme. Thus the science of pharmacology relies on such information
v f = vd
Derivation of Michaelis-Menten Equation
• We need one more condition, that is, the total enzyme concentration, [Et] is the sum of that of enzyme-substrate complex, [ES], and that of free enzyme, [E]: