INJURY AND RECOVERY 99 



see that the permanent injury is an index of the 

 condition of 0. 



We may now calculate the curve of recovery 23 after 

 exposure to a solution of NaCl 0.52 M. We assume that 

 when the tissue is transferred from sea water to the 

 solution of NaCl the reactions >8 >A cease and 

 that the velocity constant K A of the reaction A >M 

 increases from 0.0036 to 0.0180 while the velocity constant 



K .of the reaction M >B increases from 0.1080 to 0.540. 



M. 



We may then calculate the resistance in the solution of 

 NaCl after any length of exposure by means of the 

 formula 



/ KA V -KATE -KMTR\. -K M T E , , , 

 Resistance = 2,700! = _ II e -e l+goe +10 (i) 



in which T E is the time of exposure in minutes, and e is 

 the basis of natural logarithms. 10 is added in the for- 

 mula because the base line is taken as 10 (not as 0) for the 

 reason that the resistance sinks to 10 (as shown in Fig. 28) 

 when the tissue dies. 



We assume that when the tissue is replaced in sea 

 water the reactions >S > A recommence and that 

 the values of K A and JT^ become 0.0036 and 0.1080 respec- 

 tively, while the other velocity constants likewise acquire 

 the values which they normally have in sea water. Under 

 these conditions M will be formed faster than it is decom- 

 posed and the resistance will rise. 



The fact that the rise does not reach as high a level 

 after a long exposure as after a short one indicates that 

 during the exposure gradually diminishes ; we assume 

 that this takes place by the reactions 



N >O >P 



We likewise assume that during exposure to the solu- 



23 



See Chapter II. 



