126 INJURY, RECOVERY, AND DEATH 



concentration is the same or not. Thus, if a solution of 

 A 0.05 M is just as toxic as a solution of B 0.1 M, mixtures 

 of the two will give a horizontal straight line (as in Fig. 

 49) provided their effects are additive. 



Emphasis should be laid upon the fact that the method 

 of mixing two equally toxic solutions eliminates disturb- 

 ances due to variations of osmotic pressure. If a mole- 

 cule of A is twice as toxic as a molecule of B, a solution 

 of A 0.05 M will be just as toxic as a solution of B 0.1 M, 

 provided there are no other factors to be considered. But 

 if the osmotic pressure of the 0.05 M solution of A is less 

 than that of the 0.1 M solution of B, there will in many 

 cases be better growth in the 0.05 M solution of A. In 

 order to make the solution of A appear equally toxic with 

 the solution of B, the concentration of A must be some- 

 what increased, say to 0.055 M. We thus compensate for 

 the variation in osmotic pressure, and this compensation 

 is not destroyed when the 0.055 M solution of A is mixed 

 with the 0.1 M solution of B. If the effects of the salts 

 are additive, we must therefore get a horizontal straight 

 line, as shown in Fig. 49. 



It is evident that this straight line furnishes a quanti- 

 tative criterion of antagonism. All that is necessary is 

 to determine what concentrations of A and B are equally 

 toxic, mix these solutions in various proportions, and 

 determine the amount of growth. The antagonism in 

 any mixture may then be expressed in a very simple 

 manner. In the curve LKM (Fig. 49) the antagonism 

 in a mixture in which the molecules are 50% A and 

 50% B may be expressed as KJ-^JE. JE is the 

 growth which would have been obtained if the effect of 

 the salts had been additive (that is, if there had been no 

 antagonism, but each salt had produced its effect inde- 



