HI 



and blood (free from antitoxin) contains pr. cc. 0,23: 

 10.23 units (as unit is chosen the amount of toxin, con- 

 tained in 1 cc. of a 1% solution of tetanolysin). When, 

 in another experiment, x cc. must be added to obtain 

 the same degree of colour, then the amount of toxin 

 pr. cc. is x : (10 + x). 



As in both experiments the same degree of colour 

 resulted, the amount of free toxin must be the same in 

 the two cases (a correction for the variations of the 

 amount of blood is not necessary, cfr. p. 15) \\'/. 

 0,23:10,23, the amount of toxin combined is conse- 

 quently the difference between the added and the free 

 toxin ' = x:(10 + x) - - 0.23 : 10.23. The amount ofanti 

 toxin combined is the same. As we now have added 

 n cc. antitoxin solution to 2 cc. 2% lysin, cc. anti- 

 toxin corresponds to every unity of toxin added. Then 

 in each cc. of solution ^ . ^ . p parts of antitoxin 

 are present supposed that 1 cc. of the used solution of 

 antitoxin is equivalent to p cc. of 1 % tetanolysin. 

 From this amount the amount of antitoxin just now 

 determined is subtracted, and then the value of the amount 

 of free antitoxin is found. Thus we get the equation: 



M 3 _\l!l_ JL_ / x 0,23 v | r /_x_ 0,23^ 



10,23/14 " 10+x P \10-i-x ~ ~ 10,23/J \10+x " " 10,23J 



For the determination of the values of p and K. 

 representing the two unknown, as many equations (12) 

 can be set up as we have experiments concerning the 

 loxicity of mixtures of toxin and antitoxin. By means 

 of successive approximation first the most probable va- 

 lue of p is found, and after that the most probable 

 value of K. The values calculated from the above data 

 as the most probable are: - -0,0(59: p 14,,").") K 

 0.115. 



60 



