Potato Osmosis and an Outlier in the Sucrose Series
At a school laboratory in Zomba, a candidate investigates osmosis in potato tissue. Five cylinders are cut from the same potato tuber, each 30 mm long and 5 mm in diameter. Each is blotted dry and weighed; all start at 2.00 g. They are placed separately in 0.0, 0.2, 0.4, 0.6 and 0.8 mol/dm3 sucrose solutions for 30 minutes at room temperature. After removal, each cylinder is blotted and reweighed. The final masses are 2.18 g, 2.10 g, 1.78 g, 1.90 g and 1.64 g respectively. The candidate expects a smooth trend, but the 0.6 mol/dm3 result does not fit. The teacher confirms the balance is working and all cylinders were left for the same time. The candidate must decide whether to use the 0.6 result when estimating the concentration of the potato cell sap.
1. Calculate the percentage change in mass at 0.2 and 0.4 mol/dm3. Use these values to estimate the sucrose concentration of the potato cell sap, stating one assumption.
Read the worked answer
At 0.2 mol/dm3: ((2.10 - 2.00) / 2.00) x 100 = +5.0%. At 0.4 mol/dm3: ((1.78 - 2.00) / 2.00) x 100 = -11.0%.
The tissue gains mass in 0.2 and loses mass in 0.4, so the isotonic point lies between them. Assuming a linear change between the two readings, the zero point is 5/(5+11) = 0.3125 of the way from 0.2 to 0.4: 0.2 + (0.3125 x 0.2) = 0.2625, about 0.26 mol/dm3.
Assumption: at the isotonic point there is no net water movement, and the tissue behaves as a single osmotic compartment over the 30-minute period. The estimate is only approximate because the relationship need not be linear.
2. The 0.6 mol/dm3 cylinder finished at 1.90 g. Explain why this value is inconsistent with the rest of the series and describe how the candidate should check it.
Read the worked answer
Osmosis predicts that a more concentrated sucrose solution has a lower water potential, so water loss should be greater. The 0.6 mol/dm3 cylinder should therefore weigh less than the 1.78 g cylinder in 0.4 mol/dm3, not more.
A mass of 1.90 g suggests either that the cylinder was not blotted consistently, that the solution was weaker than labelled, that a different tuber was used, or that the immersion time was shorter. Surface water left on the cylinder would add mass and could explain the result.
The candidate should repeat the 0.6 treatment with a fresh cylinder from the same tuber, freshly prepared solution, the same blotting method and the same timing, and check the labels. If the repeat still disagrees, the point should be excluded from the estimate and the reason reported.
3. Explain the direction of water movement in the 0.8 mol/dm3 solution and contrast the firmness of the cylinder in distilled water.
Read the worked answer
In 0.8 mol/dm3 sucrose, the external solution has a lower water potential than the potato cell sap. Water leaves the cells by osmosis through the partially permeable cell surface membrane and cell wall. The vacuole shrinks, the protoplast pulls away from the cell wall (plasmolysis), and the tissue loses turgor, so the cylinder becomes limp and shorter.
In distilled water the external water potential is higher than the cell sap, so water enters the vacuole by osmosis. The vacuole expands, presses the cytoplasm against the cell wall, and the tissue becomes turgid and firm. The cell wall prevents bursting.
The difference is therefore explained by the direction of the water potential gradient, not by active transport of sucrose.