AS Biology 9700: Water Potential and Reaching Full Turgor

A disc of potato tuber tissue, cell P, is flaccid (has no pressure potential) before an experiment. Cell P has a solute potential (ψs) of −700 kPa and a pressure potential (ψp) of 0 kPa. Cell P is then placed into a beaker of distilled water, which has a water potential of 0 kPa.

(a) Using the equation ψ = ψs + ψp, calculate the water potential of cell P before it is placed in the distilled water. Show your working. [2]

(b) Compare this value with the water potential of the distilled water, and predict and explain the direction of net water movement between cell P and the distilled water immediately after cell P is placed in it. [3]

(c) As cell P takes in water, its pressure potential rises because the cell wall resists the expansion of the protoplast. Explain why the net movement of water into cell P eventually stops, and state the water potential of cell P at this point. [3]

Part (a): Calculating the initial water potential of cell P

The water potential of a plant cell is the sum of its solute potential and its pressure potential:

ψ = ψs + ψp

Substituting the given values for the flaccid cell:

ψ = (−700 kPa) + (0 kPa) = −700 kPa

So, before being placed in the distilled water, cell P has a water potential of −700 kPa.

Part (b): Predicting the direction of net water movement

Pure distilled water contains no solutes, so its solute potential is 0 kPa and it has no pressure potential either, giving it a water potential of 0 kPa. Comparing the two values:

Cell P: ψ = −700 kPa
Distilled water: ψ = 0 kPa

0 kPa is higher (less negative) than −700 kPa. Water always moves by osmosis from a region of higher (less negative) water potential to a region of lower (more negative) water potential, down the water potential gradient. Since cell P has the more negative water potential, there is a net movement of water out of the distilled water, into cell P.

Part (c): Why net water entry eventually stops, and the water potential at that point

As water continues to enter cell P by osmosis, the protoplast (cytoplasm and cell-surface membrane) swells and pushes outward against the rigid cell wall. The cell wall does not stretch indefinitely. It resists this expansion and pushes back on the protoplast, generating a pressure potential (ψp) that increases from 0 kPa as more water enters. Because the cell’s overall water potential is ψs + ψp, and ψs stays constant at −700 kPa (assuming negligible dilution of the cell sap), a rising ψp makes the cell’s water potential become less negative over time, narrowing the gap between the cell’s water potential and the distilled water’s water potential of 0 kPa.

Net water entry stops once there is no longer a water potential gradient between the cell and the distilled water. That is, once the cell’s water potential rises to equal the distilled water’s water potential of 0 kPa. This happens when ψp has risen to +700 kPa, since ψ = ψs + ψp = (−700) + (700) = 0 kPa. At this point, cell P is described as being at full turgor: the cell wall’s inward push exactly balances the tendency of water to enter by osmosis, so although water molecules continue to move across the membrane in both directions, there is no further net movement.

Final answers

(a) ψ = (−700) + (0) = −700 kPa.
(b) Cell P’s water potential (−700 kPa) is more negative than the distilled water’s (0 kPa), so water moves, by osmosis, out of the distilled water and into cell P.
(c) As water enters, the cell wall generates a rising pressure potential, making the cell’s water potential less negative. Net entry stops when the cell’s water potential rises to equal the distilled water’s water potential (0 kPa), which occurs when ψp = +700 kPa (full turgor).