RELATION OF TOXINES TO ANTITOXINES. 43 



a represents the number of toxic units and /3 the value D 1. 

 If z represent the amount of toxones, 200 - z represents the 

 amount of toxines and toxoids, on the assumption of 200 units 

 of combination. Thus, the formula for the value L of every 

 mixture of the poisons becomes 



L = (200 - z) Toxine-Toxoid + z Toxone, 

 all of which enter into combination with the antitoxine. Hence, 



in order to liberate one toxine it is necessary to add -^-r - , 



^ 200 z 



of which ^- - a represents the proportion of toxine. And 



this is, therefore, the amount of the poison solution, expressed 

 in toxic units, which must be added to (I.E.) + L to obtain a 

 mixture in which the whole of the toxones are free, so that 

 the addition of a single toxic unit produces L + ; the amount 



- a is thus equivalent to D 1 -- /?. 



To find the amount of toxone in a poison bouillon, we have 

 thus an equation containing one unknown value : 



whence is derived 



a + j8 



With the aid of this formula EHRLICH has calculated the 

 proportion of toxones in the poisons examined by him, and 

 has found that in the case of toxones, too, the values stand in 

 a very simple relationship towards 100 e.g., 100, 50, 25, or 33, 

 66, &c. Owing to these simple relationships it is possible, by 

 determining the values of a and ]8, always on the basis of the 

 assumption of 200 saturation units, to reproduce the immunity 

 unit, which was previously only an empirical unit of measure- 

 ment, since it is possible by this means to determine the 

 proportion of toxine and toxone i.e., in the case of fresh 

 poisons which do not contain toxoids, to explain their whole 

 constitution. Most of the poisons in the fresh condition appear 

 to consist of 100 parts of toxine and 100 parts of toxone. 



-z) = az-, 200/8 - 0s= 02; o z + |8 z = 200 |8; z (a + /3)=200j3; 

 200/3 



