Estimating the isotonic point of potato tissue after a fire alarm
A student is asked to estimate the sucrose concentration that is isotonic with potato tissue. Six cylinders, each 4.0 cm long, are cut from the same potato with a cork borer, blotted dry, weighed, and each placed in a labelled beaker containing 20 cm³ of sucrose solution of a different concentration. After 30 minutes the cylinders are blotted and reweighed. The results, as percentage change in mass, are:
| Sucrose / mol dm⁻³ | 0.0 | 0.2 | 0.4 | 0.6 | 0.8 | 1.0 |
|---|---|---|---|---|---|---|
| Change in mass / % | +8.2 | +1.1 | −3.6 | −9.4 | −12.8 | −17.5 |
The student's partner notices that the 0.2 mol dm⁻³ beaker was left for 55 minutes, not 30, because the class was evacuated for a fire alarm and the cylinder was only blotted and reweighed afterwards. All other conditions were the same. The student plots a graph and reads off the concentration giving zero change.
1. Calculate the zero-change concentration using only the 0.0 and 0.4 mol dm⁻³ results, then using the 0.0 and 0.2 mol dm⁻³ results. Which estimate is wrong, and in which direction?
Read the worked answer
Using 0.0 and 0.4: the change falls from +8.2 % to −3.6 %, a fall of 11.8 percentage points over 0.4 mol dm⁻³. The zero point is therefore at 0.4 × (8.2 ÷ 11.8) ≈ 0.28 mol dm⁻³.
Using 0.0 and 0.2: the change falls from +8.2 % to +1.1 %, a fall of 7.1 points over 0.2 mol dm⁻³, giving 0.2 × (8.2 ÷ 7.1) ≈ 0.23 mol dm⁻³.
The second estimate is the wrong one, and it is an underestimate. The 0.2 mol dm⁻³ cylinder stayed in its solution for 55 minutes instead of 30, so it lost more water than it should have and its mass change was pushed downwards (its true 30-minute value would have been more positive than +1.1 %). A more positive value there drags the line to the right, so the true crossing lies above 0.23 and close to 0.28 mol dm⁻³.
The danger is that +1.1 % does not look like an outlier - it still fits the sequence - so the error is easy to miss unless the timing is checked. Any statement that the potato sap is more dilute than it really is would be wrong.
2. Describe what has happened to the cells in the 1.0 mol dm⁻³ cylinder, and predict what would be seen if that cylinder were then moved to distilled water for 30 minutes. Justify your prediction.
Read the worked answer
In 1.0 mol dm⁻³ sucrose the external solution has a lower water potential than the cell sap. Water leaves the cells by osmosis across the partially permeable membrane, the vacuole and protoplast shrink away from the cell wall (plasmolysis), and the tissue becomes limp and shorter - which is why it lost 17.5 % of its mass.
Prediction in distilled water: water enters the cells by osmosis because distilled water has a higher water potential than the cell contents; the cells become turgid, the tissue firms up, lengthens and gains mass rapidly at first, then more slowly as the cellulose walls resist further expansion.
Justification of the limit: the mass is unlikely to return exactly to the starting figure. Solute leaked out of the cells into the sucrose solution, so the sap is more dilute than before, and prolonged plasmolysis can damage membranes. Turgor is restored, but the original mass and length may not be.
3. Suggest one change that would make the estimate more precise, and explain why it would help.
Read the worked answer
Use more concentrations clustered around the expected crossing - for example 0.20, 0.25, 0.30, 0.35 and 0.40 mol dm⁻³ - or run five cylinders at each concentration and use the mean.
The zero-change point is read from the slope between the two points that straddle it. The further apart those points are, the more the answer depends on assuming the line is straight, and the more a single odd cylinder can shift it. Points close to the crossing reduce that extrapolation, so the estimate is fenced in much more tightly.
Replicates help for a different reason: they reduce the effect of one cylinder behaving unusually (as happened with the 55-minute cylinder), because a mean is less disturbed by a single value. Either change also makes it easier to spot an anomaly. Fixing the time exactly - a water bath and a timer, with the blottings done by the same person - removes the kind of variable that caused the problem in the first place.