Potato Osmosis: Where Is the Isotonic Point?
Students determine the water potential of potato tissue using sucrose solutions. They cut 12 cores, record initial masses, soak four cores in each of three concentrations (0.0, 0.3, and 0.6 M) for 30 minutes, and calculate percent change in mass. The class means are +12% at 0.0 M, +1% at 0.3 M, and −9% at 0.6 M. A student concludes that 0.3 M is isotonic and that the potato has a water potential of −0.3 MPa because the sucrose concentration equals the solute potential. The teacher notices that the 0.3 M cores came from one potato and the 0.6 M cores came from another, that cores were blotted inconsistently, and that the 0.3 M mean of +1% is an average of +4%, +2%, −1%, and −2%. The class has a balance and a pressure chamber but no osmometer.
1. Evaluate the student's conclusion that 0.3 M is the isotonic point. Refer to the replicate data and the experimental design in your answer.
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The 0.3 M claim is not well supported. The mean of +1% hides replicates ranging from +4% to −2%, so the zero point is uncertain; a mean near zero can result from averaging cores that gained and lost water. The 0.3 M cores came from one potato and the 0.6 M cores from another, so potato-to-potato variation is confounded with concentration. Inconsistent blotting adds error to the mass measurements. Interpolating between the 0.3 M mean of +1% and the 0.6 M mean of −9% puts the zero mass change at about 0.33 M, not exactly 0.3 M, and the scatter makes any single value provisional. A better estimate would use several cores per concentration from the same potato, consistent blotting, a longer incubation, and a regression of percent mass change on concentration. The student is right that the isotonic sucrose concentration is where net water movement is zero, but the data only bracket that point; they do not pin it down to 0.3 M.
2. Using the interpolated isotonic concentration, estimate the potato tissue's water potential. Show the calculation and explain why the student's value of −0.3 MPa is dimensionally incorrect.
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At the isotonic point the potato's water potential equals the bathing solution's water potential. The solution is open to the atmosphere, so its pressure potential is zero and its water potential equals its solute potential: ψs = −iCRT. For sucrose, i = 1; C ≈ 0.33 mol/L; R = 0.0831 L·bar·mol⁻¹·K⁻¹; T ≈ 295 K. ψs ≈ −(1)(0.33)(0.0831)(295) ≈ −8.1 bar ≈ −0.81 MPa. The student's −0.3 MPa incorrectly treats molarity as megapascals; 0.3 M is a concentration, not a water potential. Inside the potato at equilibrium, ψ = ψs + ψp = −0.81 MPa; at incipient plasmolysis ψp ≈ 0, so ψ ≈ ψs. The estimate depends on the interpolated 0.33 M, so it carries the uncertainty in the mass-change data. A pressure chamber could measure the balance pressure directly and provide an independent check.
3. Describe two changes to the protocol that would most improve the reliability of the isotonic-point estimate, and explain why each matters.
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First, use cores from a single potato for all concentrations, or block the experiment by potato so each concentration is represented in every potato. Potato tubers differ in solute content and tissue age, so using one potato for 0.3 M and another for 0.6 M confounds concentration with tuber identity. Second, standardize blotting and incubation: blot each core the same way for the same time, use a larger number of replicates, and incubate long enough for equilibrium. Inconsistent blotting changes the initial mass and the surface water film, which directly affects percent change. Increasing replicates narrows the confidence interval around each mean and makes the zero-crossing easier to locate. A third improvement would be to use a finer concentration series and fit a line rather than interpolate between two widely spaced points. These changes reduce random error and remove a systematic confound, making the isotonic point more trustworthy.