7 8 



DIFFERENCE BETWEEN L. 



before,.Jthe acutely lethal portion, and the whole is shaded hori- 

 zontally to show that it is completely neutralized by the i unit of 

 antitoxin. 



Now let us take 1*25 of the same solution and add to it i unit 

 of antitoxin. In this extra 0-25 c.c. of toxin (a quarter of 

 the original amount) there are 12-5 parts of prototoxoid, 25 of 

 toxin, and 12-5 of epitoxoid. There will now be 62-5 parts of 

 prototoxoid, 125 of toxin, and 62-5 parts of epitoxoid. The 200 parts 

 into which we imagine the unit of antitoxin is divided will now 

 neutralize the whole of the prototoxoid (62-5 parts), the whole of 

 the toxin (125 parts), and 12-5 parts of toxon. There will be 

 50 parts of epitoxoid left free, but no toxin. Hence, 1*25 c.c. of 

 the toxic solution is less than the L + dose. The result may 

 be represented thus : 



n-6 parts of epifoxotd 



62-5 parrs 



125 parrs. 

 FIG. 15. 



200 62-5 parrs 



Let us now imagine a third mixture of 1-33 c.c. of the toxic 

 solution and i unit of antitoxin. The 0-33 c.c. of toxin will 

 contain 16-6 c.c. of prototoxoid, 33-3 c.c. of toxin, and 16*6 c.c. of 

 toxon, and the total 1*33 c.c. will thus contain 66-6 c.c. of proto- 

 toxoid, i33'3 c.c. of toxin, and 66*6 c.c. of epitoxoid. The proto- 

 toxoid + toxin ( = 200 parts) will just absorb the whole of the unit 

 of antitoxin, leaving nothing but toxon free. Thus : 



66 6 parrs 



133-3 parrs 



FIG. 1 6. 



ZOO 66- 6 parrs 



Then, if i extra lethal dose of toxin be added to the above 

 mixture, it will find all the antitoxin utilized by substances with 

 a combining affinity as great as, or greater than, its own, and 

 will be left free. Hence, the L + dose is just greater than 

 1 33 c.c. 



All this follows from what has previously been said concerning 



