Potato water potential and a second potato sample
A student investigates water potential in cylinders cut from a single potato. Six cylinders, each 4.0 cm long and 0.8 cm diameter, are blotted, weighed, and placed in sucrose solutions from 0.0 to 0.50 mol dm−3 for 60 minutes. The percentage change in mass is: 0.00 M +6.8%; 0.10 M +4.9%; 0.20 M +1.4%; 0.30 M −1.1%; 0.40 M −3.9%; 0.50 M −7.2%. The student estimates the solute potential of the sucrose solutions as −350, −620, −950, −1300 and −1750 kPa for 0.10, 0.20, 0.30, 0.40 and 0.50 mol dm−3 respectively. A second potato, bought from a different shop, gives a zero change at about 0.16 mol dm−3. The student concludes that the first potato has a water potential of about −780 kPa and that the second potato is 'more concentrated'.
1. Use interpolation to estimate the sucrose concentration and water potential at which the first potato shows zero change. Show your working.
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Zero mass change lies between 0.20 M (+1.4%) and 0.30 M (-1.1%). The total change across the interval is 1.4 - (-1.1) = 2.5 percentage points. The zero point is therefore 1.4 / 2.5 = 0.56 of the way from 0.20 to 0.30 M:
- sucrose concentration = 0.20 + 0.56 × 0.10 = 0.256 mol dm−3, about 0.26 M.
The solute potentials are -620 kPa at 0.20 M and -950 kPa at 0.30 M, a change of -330 kPa over 0.10 M. Interpolating:
- water potential = -620 - (0.56 × 330) = -805 kPa.
The student's -780 kPa is close but should be about -805 kPa from these data. The estimate assumes a linear change between the two solutions and that 60 minutes was enough to reach equilibrium, so it is an approximation.
2. The student says the second potato is 'more concentrated'. Explain whether this is correct, using the 0.16 mol dm−3 zero point.
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A zero change at 0.16 M means the external sucrose solution has the same water potential as the potato tissue. Interpolating between 0.10 M (-350 kPa) and 0.20 M (-620 kPa): 0.16 is 0.06/0.10 = 0.60 of the interval, so potential ≈ -350 - (0.60 × 270) = -512 kPa. The second potato therefore has a water potential of about -510 kPa, compared with about -805 kPa for the first. A less negative water potential means a more dilute solution of solutes in the tissue, not a more concentrated one. The student has reversed the interpretation: the second potato needs a weaker sucrose solution to balance it, so its tissue is less concentrated and has a higher water potential. The comparison also ignores possible differences in turgor pressure, but at zero net water movement the external potential matches the total tissue water potential.
3. Explain two reasons why the estimated water potential may not be reliable, and state how the method could be improved.
Read the worked answer
Two limitations:
- Equilibrium may not be reached. A 4.0 cm × 0.8 cm cylinder is relatively thick; 60 minutes may not allow uniform water movement to the centre. Use thinner or shorter cylinders, or leave longer, and confirm constant mass.
- Mass change is not a direct measure of water potential. If sucrose or other solutes enter the cells, the mass change includes solute as well as water, and the balance point is shifted. Use a non-permeating solute such as mannitol, or measure plasmolysis rather than mass.
Other errors include uneven blotting, evaporation during weighing, and variation in surface area-to-volume ratio. Improve by using replicates of identical cylinders, blotting in a standard way, using a sealed container and a balance reading to 0.001 g, and plotting a calibration curve with smaller concentration intervals. A pressure chamber or the Chardakov method could give an independent check.