ENERGY OF ANTIBODY-ANTIGEN REACTIONS 135 



Kistiakowsky and his group (Boyd et al.^ 1941), and the second by 

 Steiner and Kitzinger (1956), gave —40 and —6 kcal per mole of 

 antibody, respectively. I doubt if this difiference v^as due to experi- 

 mental error ; more likely it should be traced to differences in features 

 between the two very different antibody-antigen systems used. 

 Steiner and Kitzinger 's result agrees better with values of AH° 

 calculated indirectly for other serological reactions, as we shall see 

 below. 



Free Energy from Equilibrium Measurements 



Various methods have been used to measure the equilibrium be- 

 tween free antibody and antigen and their compounds, or between 

 antibody and hapten and their compounds, including (i) equilibrium 

 dialysis, (ii) direct analyses of precipitates and supernatants, (iii) 

 electrophoretic and ultracentrifugal observations, and (iv) light scat- 

 tering. Details of the experimental procedures will have to be found 

 in the references cited. Here we may say merely that all are methods 

 of determining or calculating the concentrations at equilibrium of free 

 antibody, free antigen, free rapten, or compounds thereof. From such 

 measurements the equilibrium constant K and AF° can be calculated. 

 If measurements can be made at more than one temperature, AH° 

 and AS° can also be estimated. 



Of the above methods, (i) and (iv) are applicable only to simple 

 antibody-hapten systems, the former only to univalent hapten sys- 

 tems. Method (iii) can be applied to systems in which the antibody 

 is reacting with a protein, but application of the method may in some 

 cases disturb somewhat the very equilibrium which it is desired to 

 measure. Method (iv) does not disturb the equilibrium. 



In applying method (iii), allowance must be made for the fact 

 that antibody is divalent, at least usually, and protein antigens are 

 multivalent (Epstein, Doty, and Boyd, 1956). Therefore, if we 

 measure the equilibrium in which each antibody is combined with as 

 many antigen molecules as possible (two in the case of divalent anti- 

 body), in the presence of free antigen and the compound AG, where 

 A represents antibody and G represents antigen, our equilibrium 

 constant corresponds to 



(AG2)/(G) (AG) = K (2) 



