INFECTION AND RESISTANCE 



tity of the toxoid of weaker affinity. In consequence, as far as the 

 poisonous properties of the mixture are concerned, the addition of 

 toxin would not render the neutral mixture poisonous for guinea 

 pigs until the toxoids had been completely displaced from union with 

 antitoxin. Finally, after all the antitoxin had been bound to un- 

 changed toxin, further addition would then result in the presence of 

 free toxin and poisonous properties would again appear in the mix- 

 ture. Ehrlich at first spoke of the toxoids possessing weaker affinity 

 for antitoxin than the toxin itself as "epitoxoids." This conception 

 can be rendered clear by the following example : 



In the case cited above we had 

 TorMLD = 0.0025 c. c. 

 L+ = 0.25 c. c. 



Lo = 0.125 c. c. 



The difference = 0.125 = 50 M L D. 



Supposing that the toxoids (epitoxoids) present in the mixture 

 possessed an affinity for antitoxin less than that of toxin, the follow- 

 ing conditions might obtain : 



151 toxin-antitoxin + 49 epitoxoid-antitoxin = L . 

 Add 1 M L D or T and we have: 



152 toxin-antitoxin + 48 epitoxoid-antitoxin + 1 epitoxoid free. 

 Add 2 M L D or T and we have: 



153 toxin-antitoxin + 47 epitoxoid-antitoxin + 2 epitoxoid free until, 

 finally, adding 50 T, we get: 



200 toxin-antitoxin -f- 49 epitoxoid free + 1 toxin free = L+. 13 



Later experience led Ehrlich to abandon the opinion that the 

 epitoxoids were deterioration products of the toxin. He found that 

 the relation between L and L + which we have just outlined, was 

 demonstrable in the same way, in freshly prepared toxin nitrates, 

 in which there had been little time for toxoid formation. He further 



13 An example identical in significance with the one just given, but some- 

 what simpler in its arithmetical conditions, is here added for the sake of 

 permitting no possibility of unclearness. This example is taken from Ehr- 

 lich's own work. 



T = .01 c. c. of the toxin bouillon. 



L+ (neutral, of antitoxin unit yet killing 1 pig) = 2.01 c. c. or 201 T. 

 Lo (complete neutral, of 1 antitoxin unit) =1 c. c. or 100 T. 



Difference = 1.01 c. c. or 101 T 



100 toxin-antitoxin + 100 epitoxoid antitoxin = L ; 

 Add 1 T, and we have: 



101 toxin-antitoxin + 99 epitoxoid-antitoxin + 1 epitoxoid free; 

 Add 101 T, and we have: 



200 toxin-antitoxin +100 epitoxoid free + 1 T free = L+. 



