The Rogue Potato Cylinder: Deciding What the Osmosis Data Mean
Five groups in a Grade 11 practical each cut six 4.0 cm potato cylinders from the same tuber and weighed them before soaking for 30 minutes in sucrose solutions of 0.0, 0.2, 0.4, 0.6 and 0.8 mol/dm3. Mean percentage mass changes were +6.2, +2.1, -3.4, -9.8 and -15.1. Group 3's 0.4 mol/dm3 cylinder, however, gained 1.2%, while its other two replicates lost 3.1% and 3.7%. The teacher must submit one class set of results and decide whether to include Group 3's figure. A student asks whether adding a metabolic inhibitor would change the outcome, since 'the potato is alive'.
1. Estimate the sucrose concentration at which net water movement into the cylinders is zero. Show your method and state precisely what that value represents.
Read the worked answer
Interpolate between the two concentrations that bracket zero change: 0.2 mol/dm3 (+2.1%) and 0.4 mol/dm3 (-3.4%).
- Total change across the interval = 2.1 + 3.4 = 5.5 percentage points.
- Fraction of the interval from 0.2 mol/dm3 = 2.1 / 5.5 = 0.38.
- Zero point = 0.2 + (0.38 × 0.2) = 0.276 ≈ 0.28 mol/dm3.
Interpretation: at about 0.28 mol/dm3 the external solution is isotonic with the cell sap, so there is no net water movement and the tissue is at incipient plasmolysis. It estimates the water potential of the vacuolar sap, not of the whole tuber, and it is valid only because the response is close to linear (the 0.6 and 0.8 points fall on the same trend). It is a class mean, so a rogue replicate inflates the uncertainty; a graph with the line of best fit and error bars is a more defensible presentation than a single calculated number.
2. Should Group 3's 1.2% value be included in the class mean? Give two procedural explanations and describe one check that would distinguish between them.
Read the worked answer
It should be excluded from the mean and reported separately as an anomalous result, because including it shifts the 0.4 mol/dm3 mean from about -3.4% to -1.9% and moves the interpolated isotonic point toward 0.31 mol/dm3, distorting the conclusion.
Two plausible explanations: (a) the cylinder was not blotted consistently, so surface water was still on it when weighed, adding mass; (b) the cylinder was cut from a different radial region of the tuber - the inner pith and the outer vascular ring differ in solute concentration - or the timing was wrong (a shorter soak gives less water loss).
Distinguishing check: include a distilled-water (0.0 mol/dm3) cylinder handled by the same student in exactly the same way. If that cylinder also gains more than the class mean, the fault is technique (blotting/weighing), not the tissue. A better fix is to standardise blotting (roll once on dry paper, weigh within 30 seconds) and to cut all cylinders from the same radial zone, then repeat in triplicate.
3. Would 1 mmol/dm3 sodium azide (a respiratory inhibitor) alter the 30-minute mass changes? Justify your prediction and state one way the result could be misleading.
Read the worked answer
For 30 minutes, little change is expected: osmosis is passive movement of water down a water-potential gradient across a membrane, and it does not require ATP. The initial rate of water movement would be essentially the same, so the 0.0 mol/dm3 cylinders would still gain and the 0.8 mol/dm3 cylinders would still lose mass.
Over hours, however, the inhibitor would matter: ATP-dependent ion pumps that load solutes into the vacuole fail, so sap solute concentration falls, sap water potential rises (becomes less negative), and the tissue loses the ability to hold water. Cylinders in water would gain less, and the isotonic point would shift to a lower sucrose concentration. The misleading outcome: azide also damages membranes and can be cytotoxic, so at longer times the cylinders may lose mass for reasons unrelated to osmosis, and the results would be erratic rather than showing a clean shift.