Incipient plasmolysis: sucrose versus urea
A student peels the lower epidermis of a red onion scale leaf and floats pieces in sucrose solutions at 27 °C. She uses 0.10, 0.20, 0.30, 0.40 and 0.50 M sucrose, waits 15 minutes, and counts cells showing incipient plasmolysis in five random fields per solution. The results are: 0% at 0.10 M, about 5% at 0.20 M, about 50% at 0.30 M, and nearly 100% at 0.40 M and 0.50 M. She takes R = 8.3 J mol−1 K−1 and 1 MPa = 1000 kPa. Her partner concludes that because 50% of the cells plasmolyse at 0.30 M, the cell sap must have an osmotic concentration of exactly 0.30 M and an osmotic potential of −0.30 MPa. The teacher asks her to calculate the water potential of the 0.30 M solution, to explain what 50% plasmolysis actually means, and to predict what would happen if the same peel were placed in 0.30 M urea instead of sucrose.
1. Calculate the water potential of the 0.30 M sucrose solution at 27 °C and state what it tells you about the cell sap at incipient plasmolysis. Correct the partner's two errors.
Read the worked answer
For a non-electrolyte, ψs = −iCRT with i = 1:
ψs = −(1)(0.30 mol L−1)(8.3 J mol−1 K−1)(300 K) = −747 J L−1 ≈ −747 kPa ≈ −0.75 MPa.
At incipient plasmolysis the protoplast is just about to pull away from the wall, so turgor pressure ψp = 0 and ψcell = ψs(cell). Water potential equilibrium with the bathing solution then gives ψs(cell sap) ≈ −0.75 MPa.
Two corrections. First, 0.30 M is a concentration, not a pressure; the corresponding potential is −0.75 MPa, not −0.30 MPa, so the partner has simply dropped the RT factor. Second, 50% plasmolysis is a population midpoint: the epidermal cells have a spread of sap concentrations, and 0.30 M is the concentration at which half of them have crossed the threshold. It is an estimate of the median sap potential, not an exact value for every cell.
2. Predict whether cells will be turgid or plasmolysed in 0.20 M sucrose and in 0.50 M sucrose, and calculate the turgor pressure in the 0.20 M case.
Read the worked answer
ψ of 0.20 M sucrose = −(0.20)(8.3)(300) = −498 kPa ≈ −0.50 MPa. ψ of 0.50 M sucrose = −(0.50)(8.3)(300) = −1245 kPa ≈ −1.25 MPa.
- 0.20 M: the external solution (−0.50 MPa) is less negative than the cell (−0.75 MPa), so water enters. It enters until ψcell equals −0.50 MPa. Since ψs(sap) ≈ −0.75 MPa, ψp = ψcell − ψs = −0.50 − (−0.75) = +0.25 MPa. The cell is turgid, with the wall pushing back at 0.25 MPa.
- 0.50 M: the external solution (−1.25 MPa) is more negative than the cell, so water leaves and the protoplast shrinks away from the wall. The cells are plasmolysed, and the degree is greater than at 0.40 M.
The direction of water movement is decided by the total water potential, not by solute concentration alone - the turgor term must be included whenever the cell is not at incipient plasmolysis.
3. Predict what happens if the peel is placed in 0.30 M urea rather than 0.30 M sucrose, and explain why urea cannot be used to estimate cell sap potential.
Read the worked answer
Urea is a penetrating solute; sucrose is effectively non-penetrating over this timescale.
Initially, 0.30 M urea has the same water potential as 0.30 M sucrose (−0.75 MPa), so the cells sit at incipient plasmolysis. Urea then crosses the membrane and enters the cell, making the internal ψs more negative. Water follows osmotically, ψp rises, and the cells become turgid again - that is, deplasmolysis. In 0.50 M urea the sequence is more pronounced: visible plasmolysis first, then recovery.
The judgement is that a penetrating solute destroys the equilibrium assumption used in the experiment. At incipient plasmolysis with sucrose, ψcell = ψs(solution) because the solute cannot enter; with urea, the internal concentration keeps changing, so the bathing concentration no longer reports the sap potential. This is why sucrose or mannitol, not urea or glycerol, is used for such determinations.