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

Transition state是在energy profile最高點 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 enzyme-catalyzed reaction with a double-displacement mechanism be treated kinetically as if it is a single-substrate 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