13 part 2 Enzyme kinetics
酵素動力學
溫鳳君0993b303
姜喆云0993b039
心
智
圖
課程重點 part 2
• Kinetics are the branch of science concerned
with the 反應速率.
• Enzyme kinetics are to determine the 最大反
應速率and binding affinities for substrates
and inhibitors.
1.一級反應(First-order reaction) :unimolecular
reaction A→P
V=-[ A]/dt= k[A]
2.二級反應(second-order reaction)=
Bimolecular reaction
A + B →P + Q或2 A → P + Q
註:很少三級反應
Increase chemical reaction rate
(a)升高溫度 fromT1 to T2
(b) 加催化劑
Transition state是在energy profile最高點
The reaction rate is 按比例 to the
concentration of reactant molecules with the
transition-state energy
• Free energy of activation(G‡)
the energy required to raise the average
energy of 1 mol of reactant to the transition
state energy
• Decreasing G‡ increases the reaction rate
• Enzyme did change kinetics!!!!!!!!
The Michaelis-Menten Equation
• 假設形成ES complex
• 假設ES complex is in rapid equilibrium with
free enzyme
• Breakdown of ES to form products is
assumed to be slower than
1) formation of ES and
2) Breakdown of ES to re-form E and S
E+S ES
E+P
Km
• 根據M-M assumption，假設E+S ES
E+P
• 根據Briggs & Haldane assumed that Vf = Vd
under steady-state!
推導出
• Km= Sum of dissociation rate constants
ES synthesis rate constant
Km’s meaning
• Km和ES binding capacity 有關
• Km和反應速率有關
• 當最大速率的一半時，Km會和基質濃度相
同
kcat
• A measure of catalytic activity
• the activity of one molecule of enzyme
• 當M-M model fits, k2 = kcat = Vmax/Et
The Ratio kcat/Km
• Define the catalytic efficiency of enzyme
• The upper limit for kcat/Km is the diffusion
limit -the rate at which E and S diffuse
together
問題
• 1.
Under what conditions can a bisubstrate enzymecatalyzed reaction with a double-displacement
mechanism be treated kinetically as if it is a singlesubstrate reaction? (I.e., under what conditions does
its Michaelis-Menten equation approximate that for a
single-substrate reaction?)
A:
The Michaelis Menten equation for a bisubstrate
reaqction with a double displacement mechanism
reduces to the equation for a single substrate reaction
when the concentration of the second substrate is
extremely large relative to the binding affinity for that
substrate, as is the case when that substrate is water.
• 2. Explain mathmematically how a value for Km can be
obtained from the Vo vs
So graph when Vo = 1/2 Vmax.
When Vo = Vmax/2, then Vmax/2 = Vmax
So ,
Km + So
cancelling Vmax,
1/2 = So
or Km + So = 2So
Km + So
or Km = So at Vo = (value of) Vmax/2