INJURY AND RECOVERY 115 



decomposition of ; the amount of this may be found as 

 follows : The loss of A in sea water under normal condi- 

 tions 35 in 10.2 minutes is 



/ -(0.0036) 10.2 \ 



Loss = 2700- I 2700e 1 = 97.26 



and this could be completely replaced by if were 

 intact. But since has diminished 36 from 90 to 50.86 

 it can supply only 97.26 (50.86 -v- 90) = 54.95. This must 

 be added to A giving A 5 = A + 54.95. The value of 

 A 5 must be substituted for 2313 in formula (9). This 

 enables us to calculate the fall of resistance after the last 

 recovery (of 10.2 minutes). Fig. 44 shows the values 

 so obtained and also the observed values. 



77. Alternate Exposure to CaCl 2 and Sea Water. 



When the tissue of Laminaria is transferred from 

 sea water to a solution of CaCl 2 (of the same conductivity 

 as sea water) the resistance rises and then falls as 

 shown in Fig. 45. When it is replaced in sea water the 

 resistance falls (much more rapidly than if left in the 

 solution of CaClo) and eventually becomes stationary. 

 This fall of resistance may be spoken of as recovery, 

 since it may be regarded as analogous to the rise of 

 resistance which occurs when tissue is transferred from 

 NaCl to sea water. 



85 



The principle upon which this formula is based is explained on page 

 103 in discussing the loss of M and its replacement by 0. In the present 

 case the effect of S is negligible since the amount of 8 in sea water is 

 only 2.7. 



36 This is calculated as follows : at the beginning of the fourth exposure 

 =68.10 10 = 58.10. If its value were 90 it would lose 11.23 during 

 an exposure of 20.4 minutes to NaCl. Since = 58.10 the loss will be 

 11.23 (58.10-1-90) =7.24: subtracting this from 58.10 we have 50.86. 



