Estimating potato tuber water potential from mass changes in sucrose solutions
Potato tuber cylinders are cut with a 1.0 cm diameter cork borer and trimmed to 4.0 cm. Groups of six cylinders are blotted, weighed together, and placed in 20 cm3 of sucrose solution for 30 min at 22 °C. They are blotted again and reweighed. Mean percentage mass changes from five replicates are: 0.0 M, +6.4%; 0.2 M, +2.3%; 0.4 M, −1.9%; 0.6 M, −6.8%; 0.8 M, −11.2%. One 0.6 M replicate lost 14.9% and is suspected to have been cut thinner. The sucrose solutions have water potentials of 0 kPa (0.0 M), −540 kPa (0.2 M), −1120 kPa (0.4 M), −1760 kPa (0.6 M) and −2460 kPa (0.8 M). The student estimates the tuber water potential as −780 kPa and states that cells at 0.4 M are plasmolysed because the tissue lost mass.
1. Assess the student's estimate of −780 kPa for the tuber water potential.
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The zero mass change lies between 0.2 M and 0.4 M. Linear interpolation gives 0.20 + 0.20 × (2.3 ÷ (2.3 + 1.9)) = 0.310 M. Interpolating between −540 kPa and −1120 kPa: the fraction is (0.310 − 0.20) ÷ 0.20 = 0.55, so water potential ≈ −540 + 0.55 × (−580) = −859 kPa, about −860 kPa. The student's −780 kPa is therefore too high, meaning not negative enough. However, the estimate depends on assuming that percentage mass change is linear with water potential between 0.2 M and 0.4 M, which may not be true. The 0.4 M point is close to zero, so the zero crossing is not tightly bracketed. A better estimate would use narrower intervals, for example 0.25, 0.30 and 0.35 M, or plot a graph of percentage mass change against water potential and read the intercept. The anomalous 0.6 M replicate should also be checked, because it could distort the mean.
2. Is the claim that cells at 0.4 M are plasmolysed justified by the mass data? Explain your reasoning.
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The claim is not justified by mass data alone. A mean mass decrease of 1.9% shows that the tissue has lost water, so the external solution has a more negative water potential than the cell contents. Water leaves by osmosis, and turgor pressure falls. But plasmolysis is a specific observation: the protoplast shrinks away from the cell wall. Incipient plasmolysis is usually defined as the point at which 50% of cells show this separation. A small mass loss can occur while cells are still turgid or just flaccid, with the protoplast still pressed against the wall. Mass change also includes water from cut cells and intercellular spaces, and depends on the amount of tissue and surface area. To test the claim, the student should make a peel of the potato epidermis, mount it in 0.4 M sucrose, and examine under a microscope for protoplasts pulled away from cell walls. Without that observation, the correct conclusion is only that the tissue lost water at 0.4 M.
3. Explain how the anomalous 0.6 M reading should be handled and suggest one change that would improve the precision of the whole investigation.
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The 14.9% value should not simply be discarded. First check the raw data and the cylinder: a thinner or shorter cylinder has a larger surface area to volume ratio, so it loses a greater percentage of its mass. If there is a documented technical reason, such as a damaged cylinder or a different immersion time, it can be excluded, but the reason should be stated. Otherwise repeat that treatment. For precision, increase the number of replicates, for example to ten per concentration, and calculate individual percentage changes before finding the mean. Standardise the cylinders by mass as well as length and diameter, blot all cylinders in the same way for the same time, use sealed containers to prevent evaporation, and keep immersion time and temperature constant. Use smaller concentration intervals around the estimated zero point so that the water potential can be interpolated more accurately.