The Sweet Potato Null Point: Estimating Water Potential from Mass-Change Data
A sixth-form student investigates water relations in sweet potato (Ipomoea batatas). Cylinders are cut with a 5 mm cork borer from a single tuber, trimmed to 30 mm, blotted and weighed to 0.01 g. Batches of three are placed for 30 minutes in 0.0, 0.2, 0.4, 0.6 and 0.8 mol dm⁻³ sucrose at 28 °C, then reblotted and reweighed. Mean percentage changes in mass are +14.2, +7.8, +1.1, −6.5 and −13.9 per cent. In 0.4 mol dm⁻³ the three replicates change by −6.1, −6.9 and −18.4 per cent; the last cylinder was left uncovered on the bench for four minutes before weighing. A classmate repeats the work using a tuber stored for three weeks at 8 °C and finds the concentration giving no change in mass to be 0.25 mol dm⁻³. The teacher requires the tissue water potential to be reported in MPa to one decimal place, with an assessment of how far the data support that precision.
1. Estimate the sucrose concentration at which the tissue would show no net change in mass, and explain what that concentration does and does not tell you about the tissue's water potential.
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The null point lies between 0.2 and 0.4 mol dm⁻³. Over that interval the mean change falls by 7.8 + 6.5 = 14.3 percentage points, so zero change occurs at 0.2 + (7.8/14.3) × 0.2 ≈ 0.31 mol dm⁻³.
At that concentration there is no net water movement, so the water potential of the bathing solution equals the water potential of the tissue - that is the definition being exploited. What the value is not is a pressure. 0.31 mol dm⁻³ is a concentration, and converting it requires either Ψ = −iCRT (about −0.8 MPa at 28 °C) or, better, a calibration curve from an osmometer. Interpolating a straight line between five points also assumes linearity that the real curve does not have near the null point.
A further limitation: the null point is where the intact tissue balances the solution. It is not necessarily incipient plasmolysis, which is scored microscopically as the concentration at which about 50% of cells show plasmolysis. The two are related but not interchangeable.
2. Decide whether the −18.4% replicate should be excluded from the mean at 0.4 mol dm⁻³, and explain the effect of that decision on the final answer.
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Two questions must be separated: is the point statistically extreme, and is there a technical reason for it?
Statistically, Dixon's Q = gap/range = (18.4 − 6.9)/(18.4 − 6.1) = 11.5/12.3 ≈ 0.93. For n = 3 the 95% critical value is about 0.97, so the value cannot be rejected on statistics alone.
However, rejection is defensible when a procedural fault is identified. The cylinder sat uncovered on the bench, so it lost water by evaporation in addition to osmosis; the measured mass loss is therefore inflated. On that basis exclusion is justifiable - but it must be reported with the reason, not simply deleted to tidy the curve.
The consequence is real. Retaining the value gives a mean of −10.5% at 0.4 mol dm⁻³ and moves the null point to about 0.29 mol dm⁻³, shifting the quoted water potential by roughly 0.05 MPa. That is larger than the one-decimal-place precision the teacher requested, which shows the request is over-precise for a five-point, three-replicate design.
3. Explain why the classmate's null point differs from the student's, and propose changes that would make the estimate more trustworthy.
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The two tubers are different biological material. Storage at 8 °C allows starch to be hydrolysed to sugars, which lowers (makes more negative) the tissue's solute potential; a stored tuber should therefore need a higher sucrose concentration to balance it, not the lower 0.25 mol dm⁻³ observed. That contradiction suggests other causes: solutions made up inaccurately or allowed to evaporate (raising their concentration and so lowering the apparent null point), a different surface-to-volume ratio from imprecise trimming, or simply tuber-to-tuber variability.
A single tuber is itself a limitation: it gives good internal consistency but says nothing about the species.
Improvements: more concentrations around the suspected null point (0.2, 0.25, 0.3, 0.35, 0.4), more replicates per concentration, a sealed humid chamber to prevent evaporation, a water bath to hold temperature constant, standardised blotting with a fixed rolling protocol, and equilibration long enough (often 45–60 minutes) for the tissue to reach equilibrium. Precision in mol dm⁻³ should then be converted to MPa using an independent calibration rather than assumed.