Potato Osmosis: When One Replicate Changes the Answer
A Form Six student studies water relations in potato tissue, using cylinders bored from a single tuber with a 5 mm borer and trimmed to 40 mm. She blots each cylinder, records its mass, and places three cylinders in each of six sucrose solutions (0.0, 0.2, 0.4, 0.6, 0.8 and 1.0 mol dm⁻³). After 30 minutes she blots and reweighs them. Mean percentage change in mass: 0.0 +12.4; 0.2 +5.1; 0.4 -0.8; 0.6 -6.9; 0.8 -14.2; 1.0 -22.0. At 0.4 mol dm⁻³ the three replicates gave -4.6%, -4.5% and +6.7%. Room temperature was 20 °C. She is asked to state the water potential of the tuber tissue, to say how confident she can be in the value, and to explain why her classmate's suggestion, that she should simply look at the cells under a microscope instead, may or may not give a better answer.
1. Estimate the sucrose concentration at which the tissue reaches incipient plasmolysis and hence the water potential of the tuber. Show your working and state every assumption you must make.
Read the worked answer
Locating the zero point. Net change in mass is zero when the bathing solution has the same water potential as the tissue, which is the definition of incipient plasmolysis for the bulk tissue. The curve crosses zero between 0.2 (+5.1%) and 0.4 (-0.8%). Linear interpolation: 0.2 + 0.2 × 5.1/(5.1 + 0.8) = 0.2 + 0.173 = about 0.37 mol dm⁻³.
Convert to water potential. Using ψs = -iCRT with i = 1 for sucrose, C = 370 mol m⁻³, R = 8.31 J mol⁻¹ K⁻¹ and T = 293 K: ψs = -(370 × 8.31 × 293) = -9.0 × 10⁵ Pa, so ψ ≈ -0.90 MPa.
The outlier matters. If the +6.7% replicate is a procedural error and only -4.6 and -4.5 are valid, the mean at 0.4 becomes -4.55% and the zero crossing moves to 0.2 + 0.2 × 5.1/(5.1 + 4.55) = 0.31 mol dm⁻³, giving ψ ≈ -0.75 MPa. The honest answer is therefore a range of about -0.75 to -0.90 MPa, not a single figure, because the estimate depends on how the anomalous datum is handled and because the change in mass near zero is small compared with weighing error.
Assumptions. Sucrose does not cross the membrane appreciably in 30 minutes; the cells behave as ideal osmometers with a reflection coefficient of 1; at incipient plasmolysis the pressure potential is zero so ψcell = ψsolution; the tissue is uniform (it is not: pith, cortex and vascular tissue differ); equilibrium is reached in 30 minutes; temperature is constant at 293 K.
2. Decide what she should do with the +6.7% replicate, justify the decision, and propose two likely technical causes plus two changes to the method.
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Decision. Do not silently delete it. A datum may be excluded only if a documented procedural fault is identified; here no fault was recorded, so the correct action is to report the mean of all three, flag the variability, and repeat the treatment. Deleting an inconvenient point because it changes the sign is fabrication, not standardisation.
Why this point is fragile. At 0.4 mol dm⁻³ the genuine change is only about -4.5%, very close to zero. Near the zero point a small systematic error is enough to reverse the sign, so this is exactly where replication and technique matter most.
Likely causes. (i) Uneven blotting: residual surface water on that cylinder adds mass and can convert a small loss into an apparent gain. (ii) The cylinder was cut from a different region of the tuber (pith or vascular tissue rather than storage parenchyma) or had a slightly different diameter, so its surface area to volume ratio and its tissue water potential differed. Other plausible causes are bruising or a cracked cylinder leaking solutes, the cylinder floating and not being fully immersed, or a weighing or transcription error.
Improvements. Increase replication to six cylinders per solution and allocate cylinders to treatments at random from one pooled batch; standardise blotting (roll once on dry filter paper for a fixed time with fixed pressure) and weigh immediately; seal tubes with cling film to prevent evaporation; and use finer concentration steps between 0.25 and 0.45 mol dm⁻³, because the whole answer depends on where the curve crosses zero in that narrow window.
3. Her classmate says she should forget masses and simply look at the cells under a microscope. Compare the two methods for locating incipient plasmolysis and say which estimate you would trust.
Read the worked answer
Mass-change method. It measures the whole cylinder, so it averages all cell types; it is objective (a balance reading), needs no staining, but takes 30 minutes per trial, depends on careful blotting, and is insensitive close to the zero point as the anomalous replicate shows.
Microscopic method. A thin strip of epidermis (red onion or rhodamine B-stained tissue) is mounted in each sucrose solution and examined after 5-10 minutes; plasmolysis is seen directly as the protoplast pulling away from the wall. It is fast, needs only small volumes, and shows what is actually happening to membranes rather than inferring it from a mass.
But it has its own bias. The first cells to plasmolyse are those with the least negative water potential, i.e. the most turgid cells, so first plasmolysis occurs at a lower sucrose concentration than the mean tissue value. The usual convention, scoring incipient plasmolysis when about 50% of cells are plasmolyed, corrects for this but is subjective and observer-dependent. Epidermal cells are also not the storage parenchyma that dominates the tuber's mass, so the two methods measure different populations of cells.
Judgement. Use both. If only one, the microscopic method gives a defensible value for a defined cell type and avoids the outlier problem, but the mass method is more reproducible across a class. Agreement within about 0.1 mol dm⁻³ would be evidence that both are sound; disagreement of the size produced by the outlier should prompt a repeat rather than a choice.