152 Mr. W. Sutherland. [July 26, 



accounts for the powerful toxicity of copper towards those lower plants to 

 whose protoplasm it gains ready access. 



The large molecular mass of about 40,000 found for globulin accounts for 

 the fact emphasised by Hardy that the polybasic acids H 2 S0 4 , H3PO4, 

 combine by molecules and not by equivalents with globulin. The eight 

 NH 2 radicles are so widely distributed over the large molecule that only one 

 can come within the reach of an acid molecule. Thus only one OH radicle 

 in H 2 S0 4 and H3PO4 can operate. In the same way, since globulin contains 

 only two COOH radicles, these are probably far apart, and a divalent atom 

 like Ba can replace only one H. It would be possible for Ba to unite two 

 globulin molecules through a COO radicle in each, and for S0 4 to unite two 

 through an NH 2 radicle in each. Probably the colloidal state of a globulin 

 suspension is due to the ordinary ineffective doublet of trivalent nitrogen 

 becoming effective, thus enabling the globulin molecules to grip each other 

 by means of the eight free NH 2 radicles in each. Solution in acid occurs 

 when the acid satisfies the extra valencies of the NH 2 radicles, thus 

 breaking their hold on one another. The semplar G of globulin escapes into 

 the freedom of solution as part of the molecule G(HCl)s. In edestin, 

 T. B. Osborne has found what he calls the monohydrochloride insoluble, 

 while the dihydrochloride is soluble. The explanation of this is that in the 

 monohydrochloride sufficient junctions amongst the semplars are still left 

 intact to keep the colloidal monohydrochloride from going into solution. 

 The equivalent weights, 14,300 to 7000 given by Osborne for edestin, are 

 1/3 and 1/6 of the 40,000 found above for globulin. 



As to solution of colloidal globulin in bases, we have seen that in the case 

 of Hardy's ox globulin it was not sufficient to supply merely enough NaOH 

 to form the easily dissociable globulin salt, but twice as much was required 

 for complete solution. Here the extra NaOH is required along with OH and 

 H due to the first half of the NaOH to break the colloidal junctions between 

 the semplars. How it does this as efficiently as four times the equivalent 

 HC1 forms an interesting chemical question. The probable mode of 

 operation of neutral salts in dissolving colloidal globulin has already been 

 discussed. 



Summary. 



By expressing the experimental results of Hardy and Mellanby in simple 

 formulae, it is shown that the solution of globulin and its precipitation take 

 place under simple conditions of chemical equilibrium. For example, if p is 

 the fraction of a globulin suspension dissolved in a salt solution whose 

 concentration is the fraction q of C that is required just to dissolve the whole 



