If a hypothetical cell A with diffusion pressure deficit (DPD) 3 bars is connected to three cells B, C, D whose osmotic potential (OP) and total pressure (TP) are respectively 4 and 4, 10 and 6 and 7 and 2 bars, the flow of water will be

A and D to B and C

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A to B, C and D

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B to A, C and D

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C to A, B and D

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Solution

The correct option is C B to A, C and D
DPD = OP - TP.
DPD (A) = 3 bar.
DPD (B) = 4 - 4 = 0 bar.
DPD (C) = 10 - 6 = 4 bars.
DPD (D) = 7 - 2 = 5 bars.

Water always flows from lower DPD (Diffusion Pressure Deficit) to higher DPD. Since the DPD of cell B is the lowest (0) the water will flow from B to A and then to C.

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