Potato Tissue and Sucrose: Finding the Isotonic Point
A student investigates osmosis in potato tissue. She cuts 24 cylinders, each 4.0 cm long and 0.8 cm in diameter, from one large potato. After blotting, she records each cylinder's initial mass. She places groups of four cylinders into six beakers containing sucrose solutions of 0.0 M, 0.2 M, 0.4 M, 0.6 M, 0.8 M, and 1.0 M. After 30 minutes, she removes the cylinders, blots them gently, and reweighs them. The mean percent changes in mass are +8.1%, +3.4%, -0.6%, -5.2%, -10.8%, and -14.7%, respectively. One cylinder from the 0.6 M beaker has a split in its skin, but its mass is included in the mean. Room temperature is 22 degrees C, and the beakers are covered to limit evaporation. The student's lab partner says the potato cells must have been dead in the 1.0 M solution because the cylinders lost so much mass, and that the cell walls should have prevented any shrinking.
1. Estimate the sucrose concentration that is isotonic to the potato cell sap. Use the data to justify your estimate, and explain the cellular process that produces the pattern across all six solutions.
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The isotonic concentration is near 0.4 M, probably between 0.35 M and 0.40 M. At 0.4 M the mean mass change is only -0.6%, closest to zero. At 0.2 M the tissue gained +3.4%, so the 0.2 M solution had a higher water potential than the cell sap and water entered the cells. At 0.6 M the tissue lost -5.2%, so the 0.6 M solution had a lower water potential and water left the cells. Interpolating between +3.4% at 0.2 M and -0.6% at 0.4 M puts the zero-change point at about 0.37 M.
The pattern is explained by osmosis. The plasma membrane and tonoplast are selectively permeable, so water moves down its water-potential gradient. In dilute solutions, water enters the vacuole, turgor pressure rises, and mass increases. In concentrated solutions, water leaves the vacuole, the protoplast shrinks away from the cell wall (plasmolysis), and mass decreases. The cell wall is strong but permeable; it limits expansion, not water movement.
2. The partner claims the 1.0 M cells were dead and that the cell wall should have prevented shrinking. Evaluate both claims using evidence from the experiment and your knowledge of cell structure.
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The claim that the 1.0 M cells were dead is not supported. A graded response across the six concentrations is exactly what living, selectively permeable membranes produce: the higher the external sucrose concentration, the greater the water loss. If most cells were dead, membranes would lose selective permeability, solutes would leak, and the mass changes would be erratic rather than steadily more negative. The split skin in one 0.6 M cylinder is mechanical damage to that cylinder, not evidence that all cells in the 1.0 M beaker were dead.
The claim that the cell wall should have prevented shrinking is incorrect. The cell wall is rigid but not impermeable, and it does not regulate water movement. In a hypertonic solution, water leaves the vacuole and cytoplasm, and the plasma membrane pulls away from the wall. The tissue as a whole can shrink and lose mass even though the wall remains. The wall mainly resists bursting when water enters.
3. Predict how the results would change if the potato cylinders were boiled for 10 minutes before being placed in the 0.0 M and 1.0 M solutions. Explain your prediction.
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Boiling would destroy the selective permeability of the cell membranes by denaturing membrane proteins and killing the cells. In the 0.0 M solution, water could still enter, but solutes would also leak out of damaged cells, so the mass gain would likely be smaller and more variable than the +8.1% seen in living tissue. In the 1.0 M solution, water could still leave, but solutes diffusing out would reduce the concentration gradient, so the mass loss would probably be less than the -14.7% seen in living tissue. The clean, graded relationship between sucrose concentration and mass change would be lost because the membranes would no longer control what crosses.