106 I. M. KLOTZ 



that a protein tyrosine chain is involved in the interaction with the (CH3)2N — , 

 and if we assume that the removal of the anionic charge followed by disappear- 

 ance of binding indicates that a positively charged group of the protein is in- 

 volved, then we can make a model of a particular molecule as shown in Fig. 

 12 and, from a scale model, get the distance between (CH;i)2N — and — C00~ 

 substituents. 



In the case of the para dye its maximum extension is 12.8 A. (Fig. 14). In 

 the meta dye, the distance between (CH3)2N — and — C00~ depends on the ex- 

 tension but it can vary from about 10.5 to 12.8 Angstroms, so at maximum ex- 

 tension it too can cover a distance of about 12 A. On the other hand, in the 

 ortho dye, the maximum distance you can reach from (CH3)2N — to — C00~ 

 is only approximately 8.5 A. So, as an alternative explanation, it would seem, 

 then, that perhaps the binding to the protein was at two points and that these 

 two exterior substituents were involved. In the para and meta molecules the 

 exterior substituents can span the distance between the two side chains of the 

 protein which are involved, but with the ortho molecule that distance cannot be 

 spanned and, as a result, the electrostatic bond being stronger, there is only a 

 bonding here and the (CH3)2N — group is too far away to give anomalous spec- 

 trophotometric interaction, or to give the bond at this particular point. 



If this explanation is true, then we felt we ought to be able to argue in an- 

 other direction. If the reason the ortho dye molecule does not interact with the 

 human serum albumin to give the particular special spectrum is because it is 

 too short, then it ought to be possible to separate these groups by too long a 

 distance and have the same phenomenon occurring as it occurs here. 



We could put in a methylene group between — C00~ and the ring and we 

 did (Peticolas (1954)), and found that the methylene compound acted exactly 

 like the regular para dye. Perhaps that was not quite far enough apart, however. 

 So we put in two methylene groups, but that, too, acted like a regular para dye. 

 That is not too puzzling, though, for if you make the model (Fig. 15) you will 

 find that while, of course, at maximum extension the distance is approximately 

 14.5 A, there is a methylene group with free rotation and you can rotate the 

 anionic substituent around and you will get an extension which is only approxi- 

 mately 13 A (Fig. 15). 



That, then, says that what we need is a rigid bond between the ring and the 

 anionic substituent, and as is shown in Fig. 16 and 17, bottom, clearly we should 

 use a cinnamic acid in which case we cannot obtain rotation. So in this case 

 (Fig. 17, bottom) we should have too long a distance between the (CH3)2N — 

 group and the C00~ group, approximately 14.5 A and, consecjuently, this dye 

 too should not be able to span the distance between protein side chains. Thus 

 this molecule should show a spectrum not like the para compound of Fig. 17 

 (top) but like the ortho substituted one, which, indeed, it does. 



I would like to show you, then. Fig. 16 with models indicating the compari- 

 son of the regular compound (top) which does span the distance of the side 



