54 TOXINES AND ANTITOXINES. 



ratios between toxine and antitoxine follow a definite curve 

 closely resembling that which represents the relation between the 

 decomposed part of a substance and the products of such decom- 

 position. There is thus a condition of equilibrium between the 

 free toxine and antitoxine, on the one hand, and the compound 

 of the two, on the other, so that all three components are present. 

 This deviates from EHRLICH'S fundamental view in just the same 

 way as we pointed out above in the case of the compound of 

 lysine with the cell. For EHRLICH assumes that, in the case of 

 diphtheria virus, there is a firm combination between toxine and 

 antitoxine, so that only the actual excess of one component is 

 active. We shall return to this point presently. 



ARRHENIUS and MADSEN have strengthened the results of these observa- 

 tions by theoretical calculations of the values of G and x. G is obtained 

 from the equation : 



Free Toxine Free Antitoxine _ / Toxine - Antitoxine V 2 



Vol. Vol. \ Vol. ; 



The amount of free and combined toxine can be calculated in a complicated 

 way. The amount of toxine present in 1 c. c. of a 1 per cent, solution is 

 taken as the unit of measurement. Now, the original mixture of toxine 

 with 10 c.c. of blood (without antitoxine) contained 0'23 : 10'23 units 

 per c.c. Suppose that, in an experiment with antitoxine, it is necessary to 

 add x c. c. in order to obtain the same shade of colour i. e. , to ensure that 

 the same amount of free lysine is present. Then the amount of toxine 

 that has combined with the antitoxine is equal to the difference between 

 the amounts of added and free toxine = x : (10 + x) - 0"23 : 10'23, and 

 obviously as great as the combined quantity of antitoxine. The quantity 

 of antitoxine added (n) is distributed over 4 c.c. of the lysine solution, and 



N 

 hence each unit of toxine corresponds to c.c. of antitoxine. If, now, 



the ratio of antitoxine to toxine in c.c. be represented by p i.e., 1 c.c. of 

 antitoxine solution saturating p c.c. of a 1 per cent, solution of lysine 

 it follows that 



. --pr . p = the amount of antitoxine per c.c. 



n 10 + x J 



From this must be deducted the known quantity of antitoxine in order to 

 obtain the actual amount of free toxine: Hence, by interpolation of these 

 values into the first equation, there results : 



0-23 

 (ll - } RF23 



K and p can be calculated approximately from the mean results of the 

 saturation experiments, and the values thus obtained in 12 determinations 

 being 



K = 0-115 p = 14-55, 



i.e., 1 c.c. of the 0'0025 per cent, antitoxine solution employed neutralises 



( x 0-23 \"| _ v( x _0'23\ 2 

 VlO+JB 10-22/J VlO + tf 10'23/ ' 



