Potato Osmosis Practical: Which Reading Is Wrong, and Can It Prove a Drink Is 'Isotonic'?
A Form 5 student investigates the water potential of potato tissue. She cuts 30 cylinders of equal length and diameter from one tuber, blots them, weighs them, and places five cylinders in each of six sucrose solutions: 0.0, 0.2, 0.4, 0.6, 0.8 and 1.0 mol dm⁻³. After 25 minutes she removes the cylinders, blots them and reweighs. Mean percentage change in mass: 0.0 = +13, 0.2 = +6, 0.4 = −3, 0.6 = −14, 0.8 = +2, 1.0 = −22. She plots the graph and reads off the point where the line crosses zero. Her teacher says one result must be discarded and the experiment repeated. Meanwhile a drinks company has heard about the method and wants to advertise a new sports drink as 'isotonic to human cells' on the basis of the same potato test.
1. Identify the reading that should be discarded, describe the most likely procedural error behind it, and estimate the sucrose concentration of the potato cell sap. Show your working.
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Which reading is wrong. The 0.8 mol dm⁻³ result of +2 %. The trend is clean everywhere else: as sucrose concentration rises, mass change falls steadily (+13, +6, −3, −14, ... −22). A gain in mass in a strongly hypertonic solution breaks the trend, so it is an outlier, not a biological finding. Note that it is also decisive: plotting it drags the crossing point from about 0.3 to somewhere near 0.7 mol dm⁻³.
Most likely cause. The cylinders in that beaker were not blotted properly before the second weighing, so surface water added mass and hid the actual loss. Other realistic causes: wrong solution made up, cylinders left much longer than 25 minutes, or cylinders cut with a different surface area.
Estimate of cell sap concentration. Use the two readings that bracket zero change: +6 % at 0.2 mol dm⁻³ and −3 % at 0.4 mol dm⁻³. Assuming a straight line, the zero crossing is at 0.2 + (6/(6+3)) × 0.2 = 0.2 + 0.133 ≈ 0.33 mol dm⁻³.
At that external concentration there is no net movement of water, so the water potential of the potato cell sap is roughly equal to that of a 0.33 mol dm⁻³ sucrose solution. This is an estimate, so it should be quoted to about one decimal place (0.3 mol dm⁻³) and confirmed by repeating with narrower intervals.
2. Explain why the cylinders gained mass in 0.0 mol dm⁻³ and became floppy in 1.0 mol dm⁻³, using the terms water potential, osmosis, turgor pressure and cell wall.
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In distilled water (0.0 mol dm⁻³). The external solution has a water potential of zero, higher than the water potential of the potato cell sap, which is negative because of dissolved sugars and ions. Water therefore moves by osmosis down the water potential gradient, from the solution into the cells, across the partially permeable cell membrane. The vacuole expands and pushes the cytoplasm against the cell wall. The wall is strong but slightly elastic, so it resists and builds up turgor pressure, which opposes further water entry until the gradient is balanced. Mass rises by 13 %, and the tissue feels firm.
In 1.0 mol dm⁻³ sucrose. The external water potential is much lower than that of the cell sap, so water leaves the cells by osmosis. The vacuole shrinks, the cytoplasm pulls away from the wall, and the cell becomes plasmolysed. With no turgor pressure holding the tissue rigid, the cylinder becomes floppy and its mass falls by 22 %.
The reasoning to hold on to is that the direction of water movement depends on the difference in water potential, not on the type of solution, and that turgor pressure is what makes plant tissue firm.
3. The company wants to claim that its drink is 'isotonic to human cells' using this potato method. Explain why the potato results cannot support that claim, and suggest what evidence would be needed instead.
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Why potato tissue is the wrong model. The method measures the water potential of potato cell sap, which is a mixture of starch, sugars and ions inside a vacuole, and it depends on the cell wall. Human cells have no vacuole, no cell wall and a different internal solute concentration. A potato cylinder in a solution can reach equilibrium by building turgor pressure against its wall; a human cell cannot, so water continues to enter until the membrane bursts.
The specific error in the claim. A drink that is isotonic for potato tissue (around 0.3 mol dm⁻³ sucrose) is not automatically isotonic for blood plasma, which is equivalent to about 0.9 % sodium chloride (roughly 0.15 mol dm⁻³ of particles). The two numbers are not interchangeable, and they were measured on different tissue.
Evidence that would actually be needed. Measure haemolysis of red blood cells in a series of dilutions of the drink: in a hypotonic drink the cells swell and burst and the supernatant turns red; in an isotonic drink they keep their shape; in a hypertonic drink they shrink and crenate. Alternatively measure the drink's osmolality with a freezing-point osmometer and compare it with plasma. Until then, the advertising claim is not supported by the data.