THE PROBLEM OF ENDOGENOUS METABOLISM 



445 



(a) Case 1: one enzyme and one substrate 



For noncompetitive inhibition, both factors in Eq. 9-2 are multiplied by 

 1 + [{I) IK,]; hence the intercept and slope will be increased by this term 

 and the curves will appear as in Fig. 9-3. This presents no particular prob- 

 lem unless the nonlinearity is interpreted as arising from another mech- 



Fig. 9-3. Effect of endogenous metabolism 

 on the determination of the inhibitor con- 

 stant. The intercepts of both the uninhi- 

 bited and the inhibited reaction curves are 

 unchanged but the slopes are altered; it is 

 thus impoi'tant that sufficiently high sub- 

 strate concentrations be used in order to 

 determine these intercepts. 



anism. On the other hand, the plot of l/v, against (I) will remain a straight 

 line, but the intercept and slope will each contain the term (S^) -f (S) in 

 place of (S) (see Fig. 5-3D). The fractional inhibition will be the same as 

 the inhibition on either the endogenous reaction or the exogenous reaction 

 alone, since each rate expression will be multiplied by K,'[(l) -f K^] (Eq. 

 3-16). Thus any plots using i will not be altered by the endogenous factor. 

 The fractional inhibition for a competitive inhibitor, however, will not be 

 the same for each individual reaction and the total reaction, because it de- 

 pends on the sul)strate concentration. That is, the inhibition found on the 

 total reaction will be less than would have occurred if only the added 

 substrate were i)resent. It may also be noted that if only the slope and 

 intercept values are used in the Ijv against (I) plot, an incorrect value for 

 K, may be calculated when only (S) is used instead of (S^) -{- (S) (see 

 Fig. 5-lD). 



