Potato Osmosis and the Isotonic Point
Fifty potato cylinders were cut to 3.0 cm and blotted to 1.00 g. Five were placed in each of five sucrose solutions (0.0, 0.2, 0.4, 0.6, 0.8 mol dm^-3) for 30 minutes. Lengths were measured to the nearest 0.1 cm. Median percentage length changes were +6%, +1%, -4%, -9% and -14%. In 0.4 mol dm^-3 the five replicates changed by -3%, -4%, -4%, -5% and -22%. The laboratory warmed from 21°C to 23°C during the practical, and the 0.8 mol dm^-3 cylinders were noticeably limp. The student concludes that potato tissue has a water potential of -0.4 MPa because the 0.4 mol dm^-3 solution caused shrinkage. No direct water-potential measurement was made, and the student assumes sucrose concentration is proportional to water potential.
1. Using the median data, estimate the sucrose concentration that is isotonic to the potato tissue. Explain your reasoning in terms of water potential and show your working.
Read the worked answer
The median data cross zero between 0.2 and 0.4 mol dm^-3. Using linear interpolation: the change is +1% at 0.2 and -4% at 0.4, so zero change is at 0.2 + (1/(1+4)) × 0.2 = 0.24 mol dm^-3. At this external concentration there is no net water movement, so the water potential of the external sucrose solution equals the water potential of the potato cell sap. The student's -0.4 MPa is not supported: 0.4 mol dm^-3 is a concentration at which the tissue already loses length, not the isotonic point. Also, sucrose concentration is not linearly proportional to water potential, so converting 0.24 mol dm^-3 to a pressure requires a calibration curve. The estimate is for the tissue as a whole, not one cell, and should be checked with replicate cylinders.
2. The -22% replicate is suspicious. Evaluate whether the student should exclude it, and describe one check that would help decide.
Read the worked answer
The -22% value is an outlier: the other four replicates are -3 to -5%, and the median is -4%. It should not be deleted simply because it is inconvenient; first check the record for a cracked cylinder, incomplete blotting, a different starting length, or a solution mix-up. If the raw data show a procedural error, exclude it and state the reason; if not, repeat that treatment. Reporting the median is more robust than the mean when one value is extreme. A useful check is to weigh the cylinders before and after, because a large mass loss with only a small length change suggests the cylinder split or lost a piece. The 2°C rise is unlikely to explain a 17-percentage-point difference, since temperature affects the rate of osmosis more than the final equilibrium length.
3. Explain why the 0.8 mol dm^-3 cylinders were limp and why measuring length may underestimate water loss compared with measuring mass.
Read the worked answer
In 0.8 mol dm^-3 the external solution is hypertonic to the cell sap. Water leaves the cells by osmosis, the protoplast shrinks away from the cell wall (plasmolysis), and turgor pressure falls. The tissue becomes limp because the cells are no longer pressing against the cell walls. Length is a one-dimensional measure: a cylinder can shrink in diameter as well as length, and a split or curved cylinder changes length unpredictably. Mass measures total water lost and is less affected by shape, provided the surface is blotted consistently. However, mass also includes the cell walls and any solutes taken up, so the two measurements answer slightly different questions.