23.6 Enzymes Three principal features of enzyme-catalyzed reactions: 1. For a given initial concentration of substrate, [S] 0, the initial rate of product formation is proportional to the total concentration of enzyme, [E] For a given [E] 0 and low values of [S] 0, the rate of product formation is proportional to [S] 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, v max.

Michaelis-Menten mechanism E + S → ES k 1 ES→ E + Sk 2 ES→ P + E k 3 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

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

With the definition of K M and v max, we get The above Equation can be rearranged into: Therefore, a straight line is expected with the slope of K M /v max, and a y- intercept at 1/v max 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, k cat, 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 k cat /K M,

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 K M /υ max, a y-intercept at 1/υ max, and an x-intercept at −1/K M. The value of k 3 is then calculated from the y-intercept and eqn. However, the plot cannot give the individual rate constants that appear in the expression for K M.

Example: The enzyme carbonic anhydrase catalyses the hydration of CO 2 in red blood cells to give bicarbonate ion: CO 2 + H 2 O →HCO 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. [CO 2 ]/(mmol L -1 ) rate/(mol L -1 s -1 )2.78x x x x10 -4 Determine the catalytic efficiency of carbonic anhydrase at 273.5K Answer: Make a Lineweaver-Burk plot and determine the values of K M and v max from the graph. The slope is 40s and y-intercept is 4.0x10 3 L mol -1 s v max = = 2.5 x10 -4 mol L -1 s -1 K M = (2.5 x10 -4 mol L -1 s -1 )(40s) = 1.0 x mol L -1 k cat = = 1.1 x 10 5 s -1 ε = = 1.1 x 10 7 L mol -1 s -1

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

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.

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