W.. J. V. OSTERHOUT 151 



should stop producing this would diminish and we may call the loss 

 of resistance L. Now if O were producing normally it would just 

 replace this loss, so as to keep the resistance constant at 100: hence the 

 amount produced from O in any given time will be equal to the loss 

 L which would occur in that time if O were to stop producing. 



When tissue is exposed to a solution of NaCl, O diminishes accord- 

 ing to the scheme N -^O -^ P. Assuming that at the start N = 

 89.1 and = 90 we find^^ that the value of after any given time (T) 

 of exposure to a solution of NaCl may be obtained by changing the 

 constants in formula (1) thus: 



= 89.1 (^^) {e-^''^ - e-^0T)j _ ^^ - KoT ^ ,^ 



(5) 



in which Kj^r (the velocity constant of the reaction N -^0) and Kg 

 (the velocity constant of the reaction -^ P) have the values 0.03 

 and 0.0297 respectively (see Table I). 



We find by this formula that at the end of an exposure of 15.9 

 minutes the value of O is 92.57; hence it can produce only (92.57 — 

 10) -^ (100 — 10) = 0.917 as much of M in any given time as it could 

 produce if it were intact.^* The amount it could produce, if intact, 

 during recovery in sea water is easily found by subtracting from 100 

 the resistance obtained by means of formula (1), when K^ = 0.0036 

 and Km = 0.1080 (these are the normal values in sea water). Using 

 these values we find that at the end of 10.6 minutes the amount of 

 resistance, as given by formula (1), would be 98.55. Hence the loss 

 during that time would be 100 — 98.55 = 1.45, which is the amount 



^^ This value of is assumed merely for convenience in calculation without 

 reference to other assumed values. Its real value must be much greater than 

 that of A but it is not necessary to assign any definite real value to it, since the 

 only point of interest is to determine what per cent of remains after any given 

 time of exposure to sea water. It is assumed that in sea water any change in the 

 amount of is so small as to be negligible. This might be due to the fact that 

 is present in large amount and decomposes slowly or to the fact that it is formed 

 as rapidly as it decomposes (by the reactions N -^O — > P) . 



^^ In other words, if S, T, and A were completely removed, could raise the 

 level of M to 100 — 10 = 90 in the course of time. But if, for example, half of 

 is lost the remainder can raise the level of M to only one-half its former value; 

 i.e., to 45 + 10 = 55. 



