No. 474] 



SAP FLOW IN MAPLE 



443 



5 gms. Sufficient sugar is passing out through the y membranes 

 to equal the difference of 10 gms. pressure. The sugar now in 

 chamber a would tend to exert a pressure of 15 gms. toward 

 membrane x and a pressure of 10 gms. against membrane y, which 

 would cause a reverse pressure of 10 gms. offsetting the pressure 

 caused by the cell. Only if the solute could be carried away as 

 soon as excreted could this mechanism work to produce pressure, 

 but the assumption of such freedom for the solute would neces- 

 sitate a still greater freedom for the solvent, which would make 

 the existence of two unconnected reservoirs impossible. It is 

 obvious therefore that no pressure can be produced by this method. 

 In order to obtain pressure it is necessary to assume unequal 

 permeability of the membrane in the two directions. In B we 

 may suppose each end membrane to show a tendency to 5 gm. 

 pressure on its left-hand, side and 15 gms. pressure on its right. 

 Then water would pass from the cell to chamber a under 10 gms. 

 pressure, and from this chamber to the next cell under 10 gms. 

 again, and so on. This arrangement would also account for the 

 passage of sugar from one cell to another, which could not be 

 explained by the first method. 



If, as there seems some reason to believe, the two membranes 

 on either side of the cell wall act as one owing to the numerous 

 plasma connections between them, then, as may be seen from C, 

 (Fig. 3), the assumption of unequal permeability of the same 

 membrane in opposite directions is the only one that will account 

 for the phenomenon. It seems, therefore, that in any case we must 

 assume not simply that the membrane at one end of the cell is 

 more permeable than that at the other, but that each end membrane 

 is more permeable in the direction toward the bark than toward 

 the pith. 



I see no reason why the pressure produced by the various cells 

 should not be accumulative, that is, if we have three cells each 

 producing a 10 gms. pressure we might expect a pressure of 30 

 gms. at the end of the series, or perhaps even double that if it is 

 assumed that the cell wall forms a distinct chamber. For example, 

 in B (Fig. 3) sap from cell I is forced into chamber a under 10 gms. 

 pressure. Chamber a forces sap into cell II under a pressure of 

 10 gms. also; but, disregarding friction, cell I would have forced 



