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Updated: Aug 19, 2026

Hot Biological Catalysis: Isothermal Titration Calorimetry to Characterize Enzymatic Reactions
Published on: April 4, 2014
Evolution of enzyme catalytic power. Characteristics of optimal catalysis evaluated for the simplest plausible
Abstract:
1. Evolutionary changes in the structure of an enzyme that provide an increase in its K(m) value are considered. Provided that K(m) increases as a result of increases in the forward rate constants of the catalysis relative to the reverse rate constants, the enzyme catalyses the conversion of a fixed concentration of its substrate more rapidly when its structure provides that K(m)>[S] than when K(m)<[S]. 2. Catalytic efficiency of enzymes is discussed in terms of the simplest plausible model, the Haldane [(1930) Enzymes, Longmans, London] reversible three-step model: [Formula: see text] The rate equation for the forward reaction of this model (formation of P) may be written in the simple form: [Formula: see text] K(eq.) is the equilibrium constant (=[P](eq.)/[S](eq.)), and k(cat.)=V/[E](T), where [E](T) is the total enzyme concentration. 3. To assess the effectiveness of an enzyme, it is necessary only to determine the extent to which the constraints of a particular kinetic mechanism permit v(2) (v when K(m)>>[S]) to approach v(d) (the diffusion-limited rate). 4. The value of the optimal rate of catalysis (v(opt.), the maximal value of v(2)) is dictated by the equilibrium constant for the reaction, K(eq.); v(2)=v(d)/a, where [Formula: see text] when k(+1) is assumed equal to k(-3), and v(opt.)=v(d)/a(min.). When K(eq.)>/=1, it is necessary that k(+2)>>k(-1) for a to take its minimum value, a(min.); when K(eq.)<<1, it is necessary only that k(+2)>>K(eq.).k(-1), i.e. a can equal a(min.) even if k(+2)
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