23.6 Enzymes

1. 23.6 Enzymes Three principal features of enzyme-catalyzed reactions: • For a given initial concentration of substrate, [S]0, the initial rate of product formation is proportional to the total concentration of enzyme, [E]0. • For a given [E]0 and low values of [S]0, the rate of product formation is proportional to [S]0. • For a given [E]0 and high values of [S]0, the rate of product formation becomes independent of [S]0, reaching a maximum value known as the maximum velocity, vmax.

2. Michaelis-Menten mechanism E + S → ES k1 ES → E + S k2 ES → P + E k3 The rate of product formation: To get a solution for the above equation, one needs to know the value of [ES] Applying steady-state approximation Because [E]0 = [E] + [ES], and [S] ≈ [S]0

3. Michaelis-Menten equation can be obtained by plug the value of [ES] into the rate law of P: • Michaelis-Menten constant: KM can also be expressed as [E][S]/[ES]. • Analysis: 1. When [S]0 << KM, the rate of product formation is proportional to [S]0: 2. When [S]0 >> KM, the rate of product formation reaches its maximum value, which is independent of [S]0: v = vmax = k3[E]0

4. With the definition of KM and vmax, we get The above Equation can be rearranged into: Therefore, a straight line is expected with the slope of KM/vmax, and a y-intercept at 1/vmax when plotting 1/v versus 1/[S]0. Such a plot is called Lineweaver-Burk plot, • The catalytic efficiency of enzymes Catalytic constant (or, turnover number) of an enzyme, kcat, is the number of catalytic cycles (turnovers) performed by the active site in a given interval divided by the duration of the interval. • Catalytic efficiency, ε, of an enzyme is the ratio kcat/KM,

5. A Lineweaver–Burk plot is a plot of 1/υ against 1/[S]0, and according to eqn it should yield a straight line with slope of KM/υmax, a y-intercept at 1/υmax, and an x-intercept at −1/KM. The value of k3 is then calculated from the y-intercept and eqn. However, the plot cannot give the individual rate constants that appear in the expression for KM.

6. Example: The enzyme carbonic anhydrase catalyses the hydration of CO2 in red blood cells to give bicarbonate ion: CO2 + H2O → HCO3- + H+The following data were obtained for the reaction at pH = 7.1, 273.5K, and an enzyme concentration of 2.3 nmol L-1.[CO2]/(mmol L-1) 1.25 2.5 5.0 20.0rate/(mol L-1 s-1) 2.78x10-5 5.00x10-5 8.33x10-5 1.67x10-4Determine the catalytic efficiency of carbonic anhydrase at 273.5K Answer:Make a Lineweaver-Burk plot and determine the values of KM and vmax from the graph. The slope is 40s and y-intercept is 4.0x103 L mol-1s vmax = = 2.5 x10-4 mol L-1s-1 KM = (2.5 x10-4 mol L-1s-1)(40s) = 1.0 x 10-2 mol L-1 kcat = = 1.1 x 105s-1 ε = = 1.1 x 107 L mol-1 s-1

7. Mechanisms of enzyme inhibition • Competitive inhibition: the inhibitor (I) binds only to the active site. E + I ↔ EI • Non-competitive inhibition: binds to a site away from the active site. It can take place on E and ES E + I ↔ EI ES + I ↔ ESI • Uncompetitive inhibition: binds to a site of the enzyme that is removed from the active site, but only if the substrate is already present. ESI ↔ ES + I • The efficiency of the inhibitor (as well as the type of inhibition) can be determined with controlled experiments

9. Lineweaver–Burk plots characteristic of the three major modes of enzyme inhibition: (a) competitive inhibition, (b) uncompetitive inhibition, and (c) non-competitive inhibition, showing the special case α = α′ > 1.

10. Autocatalysis • Autocatalysis: the catalysis of a reaction by its products A + P → 2P The rate law is = k[A][P] To find the integrated solution for the above differential equation, it is convenient to use the following notations [A] = [A]0 - x; [P] = [P]0 + x One gets = k([A]0 - x)( [P]0 + x) integrating the above ODE by using the following relation gives or rearrange into with a=([A]0 + [P]0)k and b = [P]0/[A]0