ON CHEMICAL REACTION RATES; ENZYMES 203 



Now this reaction can be very complicated, but for our purpose it will suf- 

 fice to consider one (the simplest) set of conditions, and examine how the 

 rate varies with changes in either enzyme or substrate concentration. First 

 we assume that v_ } is much faster than v 2 , and therefore that reaction 1 is 

 essentially at equilibrium, or that K = kjk , . 



The rate is then given by 



v 2 = k 2 [ES] 



where the square bracket again denotes concentration. Now the only prob- 

 lem remaining is to compute [ES] from the equilibrium constant. If the 

 initial concentrations of enzyme and substrate, respectively, are [£"] n and 

 [S] , the concentrations of free E and S are given by 



[E] = [E] - [ES] 

 and 



[S] = [S] - [ES] 



and therefore the equilibrium constant is given by 



K = \m 



Cq ([£]„ - [ES]) ([S] - [ES]) 



This becomes simpler if only the usual case is considered, namely that in 

 which the substrate concentration is much higher than the enzyme concen- 

 tration; for under this condition only a small fraction of the substrate mole- 

 cules will ever be tied up as complexes ES because there are so few enzyme 

 molecules with which the substrate can form a complex. Hence 



[S] - [ES] - [S} 



Rearrangement gives 



[ES] = 



K eq [E] [S] () [E] Q [S] 



i + /ysio K m + [s] 



if K m is defined as 1/A~, f] . This holds for any value of [S] at any time. 



Therefore the rate, v 2 , of the enzyme-catalyzed reaction (proportional to 

 the concentration of complexes) is: 



k 2 [E] [S] 



v 2 = -d[S]/dt = 



K m + [S] 



This is the rather celebrated Michaelis-Menten Equation, and describes the 

 rate as a function of initial substrate concentration under the particular con- 

 ditions we assumed. A plot of v 2 vs [5"] is shown in Figure 8-6 for both high 

 and low enzyme concentrations. The expression says that: (1) the rate is 



