660 14. EFFECTS OF pH ON ENZYME INHIBITION 



It is assumed that only HES breaks down to form the product. In this case: 

 V = F™ ,^^^ (14-35) 



This equation may be rearranged to give: 



''" <^> (14.36) 



fhes" (S) + (afhe'lfhes'lK, 



This allows us to see that the apparent substrate constant, Kg = af^J'KJ 

 fhes" , is that expected from the above discussion; it must be remembered 

 that aKg is the substrate constant for the product-forming complex. It 

 is also observed that the apparent maximal rate, F,,/ = V^Jf/^gg" , is less 

 than the true rate because only a fraction of the enzyme combined with 

 substrate is in the HES form. If ES and HgES can also form product, the 

 expression for the rate will contain terms for them corresponding to Eq. 

 14-35. It is evident that this system will exhibit a maximum in the rate-pH 

 curve if HES exclusively forms product (or more rapidly than do ES or 

 HgES). 



An excellent presentation of the effects of substrate ionization is that 

 given by Frieden (1958). Equations are derived for various situations and 

 some methods for handling the experimental data are discussed. An in- 

 teresting possibility is brought up l)y Frieden when he suggests that the 

 form of the substrate that is not catalytically active may be an inhibitor; 

 i.e., if S is the form that is reacted by the enzyme, SH may not only fail 

 to react but may compete with S for the substrate site. 



Location of pH Optima 



The pH at which maximal enzymic activity occurs can be calculated by 

 finding the hydrogen ion concentration necessary to minimize the pH func- 

 tions in the denominator of the rate equation. For example, if E~ combines 

 with HS"^ (HE and S not contributing to the reaction). A',' = f/f/^/Kg, 

 and the rate will be highest when fjf/j is at its minimum value. Taking 

 the derivative of this product with respect to (H) and setting it equal to 

 zero: 



d{f,%.') 1 K^. 



dm K,^ (RY 



(14-37) 



where K^ and K^ are the ionization constants for the enzyme and substrate 

 respectively. Thus: 



pH,,, = l/2(pA„^, + pAJ (14-38) 



In such a simple case, the optimal pH lies midway between the pA^'s of the 



