ENZYME KINETICS 



where E, S, and P represent enzymatic site, substrate, and 

 product, respectively. The form of the equation for the steady- 

 state velocity v obtained from this mechanism is the same 

 whether it is assumed that ES is in rapid equilibrium with E and 

 S or in a steady state. 



" 1 + ^s/(S) ^-"^ 



In this equation V is the maximum steady-state velocity which is 

 equal to kz (E)^, and K^ is the Michaelis constant which is equal 

 to {k2 + k^/ky. The mere fact that the rate data for so many 

 enzymatic reactions fit this equation, however, does not prove 

 that they follow mechanism (2). All the rate equation tells us 

 is the composition of the activated complex for the reaction in 

 terms of the predominant forms of the reactants in the solution. 

 Thus for the first step in mechanism (2) the activated complex 

 (not to be confused with the enzyme-substrate compound ES) 

 contains one substrate molecule per enzymatic site. Since this 

 activated complex is built up from a molecule of enzyme 

 and a molecule of substrate, the over-all reaction rate will be 

 proportional to the concentrations of both E and S if this is 

 the rate-determining step. At high substrate concentration 

 a complex of composition ES already exists in solution and so 

 the reaction rate will be independent of substrate concentra- 

 tion if the second step is rate-determining. 



As pointed out by King (9) the requirements of rate equation 

 (3) that the velocity be first order in (S) at low (S) and zero order 

 at high (S) may be met by several other mechanisms. Two 

 examples are 



E + S . ES (4) 



E + S > E + P 



in which the ES complex does not yield product and 



E ^=± E' (5) 



E + S » E' + P 



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