Potato Cores in Sucrose: Osmosis or Active Transport?
Students cut six potato cores of equal size and mass (5.0 g) from the same potato. They blot each core, record its mass, and place it in a labeled beaker containing 50 mL of sucrose solution for 30 minutes. The concentrations are 0.0 M, 0.2 M, 0.4 M, 0.6 M, 0.8 M, and 1.0 M. After 30 minutes, they remove the cores, blot them, and record final masses: 5.8 g, 5.4 g, 5.0 g, 4.6 g, 4.2 g, and 3.9 g. The students conclude that water moves by active transport because the cores gained mass in distilled water. They also note that the 0.6 M core became limp and the 1.0 M core became very limp, while the 0.0 M core became firm. One student asks whether the 0.4 M solution is isotonic to potato cells.
1. Is 0.4 M sucrose isotonic to the potato cells? Use the mass data and explain net water movement at 0.4 M and at 0.8 M.
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Yes. The 0.4 M core starts at 5.0 g and ends at 5.0 g, so there is no net change in mass. That means water enters and leaves at equal rates; the solution is isotonic to the potato cell contents. At 0.8 M, the core loses mass (5.0 g to 4.2 g), so the solution is hypertonic. Water moves out of the cells by osmosis, from the higher water concentration inside the cells to the lower water concentration in the sucrose solution. The cells lose turgor and become limp, which is why the 0.8 M and 1.0 M cores were flaccid. At 0.0 M, the core gains mass because distilled water is hypotonic; water enters the cells and they become firm. The 0.4 M result is the best estimate of isotonicity among the tested concentrations, though a narrower range could refine it.
2. The students conclude that water movement is active transport because the cores gained mass in distilled water. Evaluate this conclusion and explain the correct mechanism.
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The conclusion is incorrect. Active transport uses energy to move substances against their concentration gradient, usually through protein pumps. Water movement in this experiment is osmosis, a passive process: water diffuses across the selectively permeable membrane from an area of higher water concentration (lower solute concentration) to an area of lower water concentration (higher solute concentration). In distilled water, the potato cells have a higher solute concentration than the surrounding water, so water enters and the mass increases; no ATP is required. The fact that mass changes does not indicate active transport. If the cells were dead, water would still enter by diffusion, which is another reason the result does not show active transport. The students should also note that osmosis is a type of diffusion.
3. Predict what would happen if the 1.0 M core were moved to distilled water for 30 minutes. Explain using cell structures, and state one limit on your prediction.
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The 1.0 M core had lost water and become very limp, meaning its cells were plasmolyzed: the vacuole shrank and the plasma membrane pulled away from the cell wall. If placed in distilled water, water would enter the cells by osmosis because the distilled water has a higher water concentration than the cell contents. The vacuole would expand, pushing the plasma membrane against the cell wall, and the tissue would become firm again (turgid). The cell wall, made of cellulose, limits expansion and prevents bursting, which is a key difference from animal cells. However, if the cells had been severely damaged or killed by the 1.0 M treatment, they might not recover normal selective permeability. Also, the final mass would depend on how long the core was left in distilled water and whether it reached equilibrium.