Potato Osmosis: When to Reject an Anomalous Result
A student cut five potato cylinders, each 4.0 cm long and 2.00 g, from the same tuber. Each was blotted, weighed, and placed in a labelled beaker containing 0.0, 0.2, 0.4, 0.6 and 0.8 mol/dm³ sucrose solution for 30 minutes. After removal, the cylinders were blotted and reweighed. The final masses were 2.36 g, 2.18 g, 2.00 g, 1.84 g and 1.62 g respectively. The 0.4 mol/dm³ cylinder was recorded as 2.00 g, but the student noticed that its beaker had been left uncovered overnight and the solution level had fallen. A second student repeated the 0.4 mol/dm³ test using a freshly prepared solution and obtained 1.94 g. The teacher asked the class to decide whether the original 0.4 mol/dm³ result should be kept or rejected, and to explain what the results show about potato tissue.
1. Calculate the percentage change in mass for the 0.2 and 0.8 mol/dm³ cylinders. Use all the data to identify the solution closest to isotonic for the potato tissue, and justify your choice.
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For 0.2 mol/dm³: ((2.18 − 2.00) ÷ 2.00) × 100 = +9%. For 0.8 mol/dm³: ((1.62 − 2.00) ÷ 2.00) × 100 = −19%. The smallest net change is at 0.4 mol/dm³: the original was 0% and the repeat was about −3%. A solution is closest to isotonic when there is little net water movement, so 0.4 mol/dm³ is the best estimate. It is slightly hypertonic on the repeat because the mass fell a little, but it is much closer than 0.2 (+9%) or 0.6 (−8%). The original 0% should not be treated as exact because that reading is suspect.
2. Should the original 0.4 mol/dm³ result be kept or rejected? Support your decision with evidence and explain how the uncovered beaker could have changed the result.
Read the worked answer
Reject or flag the original 0.4 value as anomalous and use the repeat. An uncovered beaker left overnight would lose water by evaporation, making the sucrose solution more concentrated. A more concentrated external solution has a lower water potential, so the potato should lose more water and fall below 2.00 g, not stay exactly at 2.00 g. The repeat gave 1.94 g, a small loss, which fits a solution just above the isotonic point. Keeping 2.00 g would falsely suggest exact isotonicity and would distort the estimate. The original reading may also have come from incomplete blotting or a misread balance, so it is not reliable.
3. Explain why the 0.8 mol/dm³ cylinder became shorter and flaccid. Use water potential and plasmolysis in your answer.
Read the worked answer
The 0.8 mol/dm³ sucrose solution has a lower (more negative) water potential than the potato cell sap. Water therefore moves out of the vacuole and cytoplasm by osmosis, across the partially permeable cell membrane, down the water potential gradient. As cells lose water, the vacuole shrinks and turgor pressure falls. The protoplast pulls away from the cell wall, which is plasmolysis. With less turgor, the tissue cannot support itself, so the cylinder becomes flaccid and shorter. The 19% mass loss supports this.