Injury, Recovery and Death in Relation to Conductivity and Permeability

During exposure to NaCl the reaction R — >8 — >T occurs. The value of S may be easily calculated by employing formula (1) and substituting the appropriate constants. We thus obtain The value of R at the start in sea water is taken as 1,041.77 and that of S as 2.7. In the solution of NaCl the values of KR (the velocity constant of the reaction R — >S) and K s (the velocity constant of the reaction S — > T) are taken as 0.04998 and 0.02856 respectively (see Table V). Hence the value27 of 8 at the end of 15.9 minutes is 447.26. When the tissue is replaced in sea water 8 is rapidly converted into A so that the total value of the latter becomes 447.26 + 2,027.96 == 2,475.22.

On replacing the tissue in sea water A = 2,475.22 and M = 70.69. The resistance due to A and M after any given time T in sea water is obtained by modifying formula (1) which becomes 27 In general the greater the rise in recovery the greater the value of 8, while the greater the fall the less the value of 8. replacement in sea water. The velocity constants K A and K M have the normal values in sea water, 0.0036 and 0.1080 respectively. Hence the resistance at the end of 10.6 minutes is 87.44.

We must likewise remember that on replacing the tissue in sea water the reactions 0 — > S — >• A recom- mence and produce a certain amount of A; this breaks down to form M, which in turn decomposes. The resulting amount of M may be easily calculated. It will be recalled that in sea water all processes are so adjusted that the amount of M remains constant; it is evident that if the reactions 0 — >S — > A were suddenly to stop, allow- ing A — >M — >B to continue, the amount of M would diminish. At the start the total resistance is 100. If 0 should stop producing this would diminish and we may call the loss of resistance L. Now if 0 were producing normally it would just replace this loss, so as to keep the resistance constant at 100: hence the amount pro- duced from 0 in any given time will be equal to the loss L which would occur in that time if 0 were to stop producing.

When tissue is exposed to a solution of NaCl, 0 di- minishes according to the scheme N — >• 0 — >P. Assum- ing that at the start N = 89.1 and 0 = 90 we find28 that the value of 0 after any given time (Ts) of exposure to a solu- 28 This value of 0 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 0 remains after any given time of exposure to sea water. It is assumed that in sea water any change in the amount of 0 is so small as to be negligible. This might' be due to the fact that 0 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).

tion of NaCl may be obtained by changing the constants in formula (1) thus: in which KN (the velocity constant of the reaction N — > 0) and KQ (the velocity constant of the reaction 0 — >P) have the values 0.03 and 0.0297 respectively (see Table V, page 98). We find by this formula that at the end of an exposure of 15.9 minutes the value of 0 + 10 is 92.57 ; hence it can produce only (92.57 — 10) -r- (100 — 10) = 0.917 as much of M in any given time as it could produce if it were intact.29 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 JTA = 0.0036 and K M= 0.1080 (these are the normal val- ues 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 0 could produce in 10.6 minutes if intact. This value may be called L and expressed as follows :

in which K A = 0.0036 and K M = 0.1080 (these are the nor- mal values in sea water) and T R is the time which has elapsed since the tissue was replaced in sea water. 38 In other words, if 8, T and A were completely removed, 0 could raise the level of M to 100 10 = 90 in the course of time. But if, for example, half of 0 is lost the remainder can raise the level of M to only one-half its former value; i.e., to 45. The recovery formula may therefore be expressed as follows :

Using this formula, we may find the resistance at any given time after replacement in sea water. A series of Recovery in Sea Water after Exposure of 15.9 Minutes to 0.52M NaCl. values so obtained is given in Table VI. It will be seen that they are in good agreement with the experimental values. The calculated and observed values are also plotted in Fig. 41, in which the abscissas represent the time in the solution of NaCl plus the time of recovery in sea water (in the case just discussed this would amount to 15.9+10.6=26.5 minutes).

Proceeding in this manner with different times of exposure we obtain the series of recovery curves shown in Fig. 41. The number attached to each curve denotes the Fio. 42. — Curves showing the rise and fall of net electrical resistance in Laminaria agardhii in 0.278 M CaCh (single curve which rises and falls) and recovery in sea water (descending curves). The figure attached to each recovery curve denotes the time of exposure (in min- utes) to the solution of CaCh. In the recovery curves the experimental results are shown by the dotted lines, the calculated results by the unbroken lines. The observed points repre- sent the average of eight or more experiments. Probable error of the mean less than 10%

time of exposure to the solution of NaCl. The observed results are plotted as dotted lines, the calculated values as unbroken lines. It will be seen that the agreement is satisfactory throughout. In general the greater the number of experi- ments which were averaged to obtain the result the nearer it approached to the calculated curve. Let UB now consider the behavior of tissues transferred from a solution of 0.278 M CaCl2 (which has the conduc- tivity of sea water) to sea water. In such a solution the resistance rises and then falls. If tissue is allowed to remain in the solution for a short time and is then replaced in sea water the resistance falls rapidly, as shown in Fig. 42. This fall of resistance may be regarded as analogous to the rise of resistance which occurs in the experiments with NaCl and the term recovery may be used in both cases. It is evident from the figure that ..as the exposure to the solution of CaCl2 lengthens the level which is reached as the result of recovery gets lower. This is pre- cisely what happens in the experiments with NaCl. It would therefore appear as though the same mechanism of recovery were involved. If this is so the same method of calculation should enable us to predict recovery in both cases. This is found to be true. Using the same formulas which have already been employed in the experiments with NaCl we are able to predict the course of the curves obtained in experiments with CaCL. This is rather strik- ing in view of the fact that the two sets of curves differ so fundamentally in appearance.

In calculating the curves for CaCl2 the constants given in Table V (page 98) are employed. The results are shown as unbroken lines in Fig 42 (the dotted lines show the experimental results). It is evident that the agree- ment is very satisfactory. Some assistance in picturing the reactions which occur during exposure is afforded by Fig. 43, which shows the curve of 0 + 10 in NaCl (unbroken line) and in CaCL (dotted line). These curves are plotted from the calcu-

lated values; the observed values are shown as points; it will be observed that they lie fairly close to the calculated curve. The figure also shows the calculated values of $; in this case no observed values are given because such values cannot be very precisely determined. This is owing to the fact that the value of S affects only Fia. 43. — Curves showing the values of O + 10 in NaCl (upper unbroken line) and in CaC (upper dotted line); also the values of S in NaCl (lower unbroken line) and in CaCh (lower dotted line). The ordinates give the values of O; these must be multiplied by 6.75 to obtain the values of S. The observed points represent the average of eight or more experiments; probable error of the mean less than 10% of th* mean.

the speed of recovery (not the final level attained) and as the speed is variable the only satisfactory procedure is to assume such values of K _ and K _ as cause the closest approximation to the observed speed of recovery. When these values have been found the value of 8 can readily be calculated. The results of these calculations are plot- ted in Fig. 43. In this figure the ordinates give the values of 0 + 10 : these must be multiplied by 6.75 to obtain the values of 8. In all curves the value of 8 at the start is 2.7 (the value of

in sea water30 ) ; this appears on the ordinate in the figure as 2.7 -f- 6.75 = 0.4. The curves rise to a maximum and then fall to zero. The curves for 0 + 10 start at 100 and fall to 10 (since the base line is taken as 10, just as in the curve of M). It is found that the rate of recovery is approximately the same in all cases ; this applies to the experiments with CaCl2 as well as with those in NaCl. In general it may be said that it usually requires about 60 minutes for the curve to complete nine-tenths of the total rise or fall which occurs in recovery.

If the theory here developed is sound it should also enable us to predict the behavior of tissue transferred from one toxic solution to another. In order to put this to a test a variety of experiments was made in which the tissue was exposed to several solutions in succession. The procedure may be illustrated by a typical experi- ment, the results of which are shown in Fig. 44. 30 The normal value of 8 in sea water is taken as 2.7 which is exceed- ingly small as compared with the amount of 0. The amount of 8 which is produced from 0 in each unit of time is relatively large, but 8 is so rapidly transferred into A that its amount in sea water never becomes greater than 2.7.

31 This is corrected from 20 minutes (as previously explained) in order to make it conform to the standard curve. replaced in sea water and left for 10.4 minutes32 the value of T R (in formulas (6) and (7) ) is 10.4 and the value of L is found to be 1.33. FIG. 44. — Curves showing the net electrical resistance of Laminaria agardhii in NaCl 0.52 M and in sea water. Unbroken line, calculated values; broken line, observed values. Average I of ten or more experiments; probable error of the mean less than 10% of the mean.

Substituting these values in formula (1) we find that when the tissue has been replaced in sea water the resis- tance at the end of 10.4 minutes is 83.49. Proceeding in this manner we calculate the resistance at various intervals after replacement in sea water and obtain the first (calculated) recovery curve shown in Fig. 44. It is evident that it is in fairly good agreement with the observed values. After 200 minutes in sea water (during which the resistance rose to 87.10% and remained practically constant) the tissue was replaced in the solution of NaCl. In the course of 21.2 minutes33 the resistance fell from

32 This is corrected from 10 minutes (as previously explained). 33 The actual time was 20 minutes: the manner in which the corrected figure is obtained is explained in a subsequent paragraph. 87.10 to 64.18. It was then replaced in sea water. The recovery curve may be calculated as before, the only differences being as follows : 1. On replacing the tissue in sea water the destruction of 0 (by the reactions A* — >0 — >P) ceases (or becomes negligible) ; hence the value of 0 at the beginning of the second exposure (if equilibrium has been reached) is that of the observed resistance less 10, or 87.10-

77.10. We find by means of formula (5) that when 0 at the start equals 90 it loses 11.95 during an exposure of 21.2 minutes to the solution of XaCl, but as it only equals 77.10 at the start the loss will be 11.95 (77.10 -r- 90) = 10.23. Subtracting this from 77.10 gives 66.87, the value of 0 at the end of the second exposure, and adding 10 (since the base line is 10) makes 76.87, the level to which the resistance should rise after the second exposure.

2. At the start of the first recovery34 S is rapidly con- verted into A, but is partially restored during the subsequent stay in sea water and at the beginning of the second exposure equals 2.7 (0-1-90) in which 0 has the value given above (77.10). 3. During exposure to XaCl the value of R diminishes from Pitl to Rl according to the formula period in sea water the speed of recovery will fall off somewhat with each successive exposure. 5. The value of A is obtained by multiplying by 30 the resistance observed at equilibrium (less 10). This is based upon the f ollowing considerations :

Just before the beginning of the second exposure A and M are assumed to be in equilibrium in sea water, in which case as much of A must decompose in any minute as of M (otherwise M would not remain constant). But the amount of A which decomposes in 1 minute is AK A and of M is MKMj and since KM is 30 times as great as KA it follows that A = 30 M. At the beginning of the second exposure M= 87.10 — 10= 77.10 and A= (77.10) In order to ascertain how the resistance would change during the second exposure if it conformed to the standard curve previously employed, we may employ the formula

in which JE^ = 0.018, KM= 0.540 and T£=time the tis- sue has remained in the solution of NaCl. Comparing the values thus obtained with the observed resistance after an exposure of 20 minutes we find that if the time is multiplied by 1.06 (making it 21.2 minutes) the observed resistance (64.18) agrees with the standard curve. This figure is therefore adopted. The value of TE in formulas These data were employed in calculating the second recovery curve and the results are shown in Fig. 44. The third recovery curve was calculated in the same fashion.

Instead of waiting for the establishment of equilib- rium we may replace the tissue in NaCl after it has been for a short time in sea water. During the fourth recovery, after the tissue had been 10.2 minutes in sea water and the resistance had risen to 54.92%, it was replaced in sea water : the subsequent fall in resist- ance was calculated by means of formula (9). For the value 77.1 in this formula we must substitute the observed resistance less 10, or 55.89 — 10 = 45.89 ; and in place of 2313 we must substitute the present value of A. We assume that at the beginning of the fourth exposure to NaCl equilibrium had been reached in sea water: hence as the resistance was 68.10 the value of A (which we call A, ) is, A1 = 30 (68.10 — 10) . During the fourth exposure to NaCl (lasting 20.4 minutes) the value of Al diminished to A2 according to the formula

On replacing the tissue in sea water A2 was augmented by the conversion of 8 into A. The value of S is found according to formula (3) in which T Eis equal to the total time of exposure (20.8 + 21.2 + 20.8 + 20.4=83.2). We may call this 8^ Hence the value of A immediately after replacement in sea water is As = A2-{- 8lt During the subsequent 10.2 minutes in sea water A3 diminished to A4 according to the formula decomposition of 0 ; the amount of this may be found as follows : The loss of A in sea water under normal condi- tions35 in 10.2 minutes is

and this could be completely replaced by 0 if 0 were intact. But since 0 has diminished36 from 90 to 50.86 it can supply only 97.26 (50.86 -v- 90) = 54.95. This must be added to A giving A5 = A± + 54.95. The value of A5 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. When the tissue of Laminaria is transferred from sea water to a solution of CaCl2 (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.

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. Recovery after exposure to CaCl2 may be calculated in precisely the same manner as recovery after exposure to NaCl. The only difference is that in formulas (2), (3), (5), (8) and (9) we must employ for the velocity con- stants (KN, K0, KR, Ks, KA and KM) the values given for CaCl2 in Table V, page 98. In formulas (6) and (7) the values of the velocity constants are always the same ( = 0.0036 and JTM= 0.1080) since these are the

FIG. 45. — Curves showing the net electrical resistance of Laminaria aaardhii in CaCh 0.27SM and in sea water. Unbroken line, calculated values; broken line, observed values. Average of ten or more experiments; probable error of the mean less than 10% of the mean. Results of such calculations are shown in Fig. 45 together with the observed values. It seemed desirable to test the theory further by vary- ing the experiments in the manner shown in Fig. 46. The calculations are made as already explained. It will be

noticed that in this and in some other experiments the resistance rises rather more rapidly in CaCl2 than the calculations would lead us to expect. This is due to the FIG. 46. — Curves showing the net electrical resistance of Laminaria aaardhii in NaCl 0.52 M , m CaCl-2 0.278 M and in sea water. Unbroken line, calculated values; broken line, observed values. Average of ten or more experiments; probable error of the mean less than 10% fact that the " standard curve " for CaCl2, which was based upon previous experiments made under different conditions, seems to be a little too low for the present material.

A series of experiments was made to determine the effect of CaClo followed directly by NaCl. The results are shown in Fig. 47. The rise in CaCl2 during the first 91.8 minutes is calculated in the usual manner. In order to calculate the subsequent drop in NaCl we must substi- tute for 77.1 in formula (9) the value of M; i.e., the observed resistance (less 10) at the beginning of exposure to NaCl. In place of 2313 we must substitute the value of A, which is Al =2700e -(o.ooism.s

During the exposure of 60.6 minutes to NaCl the value of A changes from A, to A2 = A,e -f FIG. 47. — Curves showing the net electrical resistance of Laminaria agardhii in NaCl 0.52 M, in CaCh 0.278 M and in sea water. Unbroken line, calculated values; broken line, observed values. Average of ten or more experiments; probable error of the mean less than 10% This value must be substituted for A in formula (7) in calculating the recovery in sea water.

In finding the value of 8 (by means of formula (3)) we must remember that during the 91.8 minutes in CaCl2 the value of R (which at the start is RQ = 1041.77) dimin- ishes from R0 to R1 according to the formula ing the 60.6 minutes in NaCl R1 diminishes to R2 accord- ing to the formula We must also bear in mind that 0 diminishes during the exposure. Since this process is 6 times as rapid in NaCl as in CaCl2 we may consider 91.8 minutes in CaCl2 to be equivalent to 91.8 -f- 6 = 15.3 minutes in NaCl and the total exposure to be equivalent to 60.6 + 15.3 = 75.9 minutes in NaCl.37 The value of 0 may then be found by means of formula (5).