THE CONSTITUENTS OF DIPHTHERIA TOXIN 495 



toxin reaches just up to the 100 limit and shows no trace of any 

 gradual decline. This follows from the determination of the L t dose,. 

 as can be seen from the following analysis. 



Given a poison in which, in the L dose, the hemitoxin zone reaches- 

 exactly to 100, how large will the L t dose be? Lt, i.e., the amount 

 of poison which on the addition of 200 combining units still leaves 

 1 L. D. free, will be reached when 200 equivalents of hemitoxin are 

 present. We shall therefore have to multiply the L dose of the 



202 

 poison by - ^ in order to obtain the Lt dose. If we carry out this 



multiplication we obtain an Lt dose of 0.253, which agrees very well 

 with the value actually found, 0.25 cc. 



Thus the important fact is demonstrated that in this case the 

 neutralization of the diphtheria poison by antitoxin proceeded exactly 

 the same as the neutralization of a strong acid by a strong base. 

 Here then the course of the reaction is represented by a straight line 

 and not by a curve. 



Further evidence for the view 7 that in this poison the hemitoxin 

 extended right up to the limit 100 is furnished by phase II. Here 

 we see a simultaneous increase of the L t dose and a decrease of the 

 toxicity manifesting themselves by the fact that the L. D. increases 

 from 0.0025 to 0.003 cc., so that the number of L. D. contained in the 

 L dose has decreased from 50 to 42. 



This increase cf the Lt dose amounted to about 0.26 cc. and from 

 it, by means of the simple calculation already mentioned, it can be 

 shown that toxoid formation took place in the end zone of the toxin, 

 the "tritotoxoid zone," as I term it. 



Let us assume that the end zone (which before as well as after the 



second phase extended to 100) contains a toxoid mixture of toxicity 

 instead of the hemitoxin. In order to reach the Lt dose in this 



210 



case we must multiply the L dose by and not by \ -, as was 



uu zoo 



the case wdth hemitoxin. On carrying out this calculation, L being 

 0.125, we get ^^i- = 0.2625 = L t . 



In the determination made at that time I actually found the L t 

 dose to be 0.26, but noted "a little over." That the tritotoxoid zone 



possessed a toxicity of was showm by the subsequent analysis by 

 means of partial neutralization, for near the end, a zone of 18-20 



