Potato Osmosis: Finding the Isotonic Point from Mass Change
Potato cylinders of equal size are blotted, weighed, and placed for 30 min in sucrose solutions. Each cylinder initially has mass 2.00 g. The percentage mass changes are: 0.0 M, +6.5%; 0.2 M, +2.1%; 0.4 M, −1.4%; 0.6 M, −7.8%; 0.8 M, −11.2%. One cylinder at 0.6 M was left in the solution for 2 h by mistake and lost 18.0% of its mass. A student estimates the isotonic point by joining the 0.0 M and 0.8 M points with a straight line, obtaining about 0.30 M; another uses the 0.2 and 0.4 M points and obtains about 0.32 M. The potato tissue has cell walls and living cells. The lab temperature is 22 °C.
1. Explain the mass changes in terms of water potential and identify the best estimate of the isotonic concentration. Why is interpolation between 0.2 M and 0.4 M more defensible than using the endpoints?
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At 0.0 M the external solution has the highest water potential, so water enters the potato cells and mass rises 6.5%. As sucrose concentration rises, external water potential falls; above the isotonic point water leaves and mass falls. The zero crossing lies between +2.1% at 0.2 M and −1.4% at 0.4 M. Linear interpolation gives 0.2 + 0.2 × (2.1/(2.1 + 1.4)) = 0.32 M. This interval is better than the 0.0–0.8 M line because the response is not linear over the whole range: at 0.0 M turgor pressure and cell-wall elasticity limit uptake, and at 0.8 M cells may be plasmolyzed or damaged. The 0.2–0.4 M points bracket zero with small changes, so local interpolation is more defensible.
2. The 0.6 M cylinder left for 2 h lost 18.0%. Does this change the isotonic estimate? Explain what it does show.
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The 18.0% loss after 2 h should not be used in the 30-min isotonic estimate. It shows that at 0.6 M the tissue had not reached a stable mass in 30 min, or that longer exposure caused additional water loss and possibly cell damage. The isotonic point is defined by no net water movement, not by the largest possible loss. Excluding that point, the best estimate remains about 0.32 M. However, the result warns that equilibration time may depend on concentration; a fixed 30-min protocol is essential, and a time-course at 0.6 M would test whether the 30-min value was still changing.
3. A student says the 0.0 M cylinders should burst because they gain water. Use cell structure and osmosis to evaluate this claim.
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The 0.0 M cylinders gain water but do not burst because potato cells are plant cells with rigid cellulose cell walls. As water enters, turgor pressure rises and the cell's internal water potential increases until it matches the external pure water potential; further net inflow stops. The wall bears the tension. Animal cells without walls could swell and lyse. The observed 6.5% mass gain is therefore turgor-driven swelling, not uncontrolled bursting.