E + S ES P + E k2k2 v o = k 2 (ES) Michaelis reasoned that If k 2 is the smallest rate constant, the overall velocity of the reaction is Problem: We cannot measure [ES] Solution: Substitute [ES] into equation

Evaluating and Using the Michaelis-Menten Equation

V max [S] K m + [S] v o = [S] variables vovo Dependent variable V max KmKm Constants [S] C + [S] v o =  pO2pO2 P 50 +pO2pO2 Rectangular Hyperbola (Dependent Variable) (Independent variable) Y axis X axis

V max [S] K m + [S] v o = V max [S] K m v o = V max [S] 2[S] v o = V max [S] [S] v o = S << Km S = Km S >> Km First order with [S] First order with [S] One-half Vmax One-half Vmax Zero Order with [S] Zero Order with [S]

V max = k 2 [E T ] V max is first order with enzyme vovo [S] E1 E2 E3 E4 V max Enzyme Slope = k cat When the velocity = V max k 2 = k cat When the velocity = V max k 2 = k cat

V max [S] K m vo=vo= V max 2 v o = S S S S S v o = V max Vmax 2 Km = [S] at one-half Vmax [S] = Km

Picture it this way LOW [S] HIGH [S]

TWO DEFINITIONS OF K M Rate Constant Definition Km =Km = k 2 + k -1 k1k1 Affinity No units Substrate Definition Km = [S] that gives 1/2 Vmax Km = [S] that fills half the sites on the enzyme Km has units of substrate concentration

Calculate the Km of an enzyme. When [S] is 2 micromolar, v o = 3 micromoles per minute. At saturation Vmax = 10 micromoles per minute. What does the answer tell you? V max [S] K m + [S] v o = 3  moles/min = 10  moles/min x 2  M Km + 2  M 3(Km + 2) = 20 3Km + 6 = 20 3Km = 14 Km = 4.7  M When the [S] is 4.7  M, the enzyme is half-saturated with [S] When the [S] is 4.7  M, [E free ] = [ES] Setup: Km holds the same literal meaning as pKa and P50

An enzyme has a Km of 10  M. At what [S] will the reaction be at Vmax? a) 20  Mb) 5  Mc) cannot tell Reason: The effect of S on velocity is hyperbolic, not linear An enzyme has a Km of 10  M. When [S] equals 5  M 15  moles of S are consumed per minute. At what [S] will the reaction be at Vmax a) 20  Mb) 45  Mc) cannot tell What is the Vmax of the above reaction a) 20  Mb) 45  moles/minc) cannot tell [S] VoVo

How close to Vmax will a reaction be when [S] = 1) Km 2) 10 Km 3) 100 Km VoVo Vmax = [S] Km + [S] Relative Max velocity 1)One-half Vmax 1/2 2) 90.9% Vmax10/11 3) 99% Vmax100/101

Based on the Kinetic analysis we can conclude: There are two phases of an enzyme-catalyzed reaction 1. Binding the substrate as determined by Km 2.Modifying the substrate and releasingthe product as determined by K 2 The two phases will become more apparent when we study inhibitors