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DEMO Practical-Psi 28OCT2026 Investigating Water Potential of Potato Tissue

7 minutes ago
5 min read


You can see a potato cylinder swell or shrink, but you cannot see its water potential. In this practical, we measure it indirectly. Potato cylinders are soaked in sucrose solutions of different strengths, weighed before and after, and the results are plotted on a graph. The concentration at which the potato neither gains nor loses mass tells us the water potential of the potato tissue.

Watch the full practical demonstration on YouTube: https://youtu.be/eYYlF9V7KKg. The practical manual is here: Practical-Psi: Investigating Water Potential of Potato Tissue.


Why Psi?

Psi (Ψ, pronounced "sigh") is the Greek letter used for water potential in plant physiology. Water potential is the energy state of water compared with pure water. Pure water at atmospheric pressure is the reference, set at 0 MPa (megapascal). Dissolving something in water, such as sugar or salt, lowers its water potential below zero. Across a membrane, water always moves from higher water potential to lower water potential. For example, water moves from −0.2 MPa towards −0.8 MPa, because −0.2 is the higher value.


From cells to tissue

In school, you learned how a single cell behaves in hypotonic, isotonic and hypertonic solutions: a plant cell becomes turgid, flaccid or plasmolysed. This practical takes that idea one step further, to the tissue level, and turns it into a number we can calculate.


What you do

  1. Label and fill. Prepare sucrose solutions of 0.0, 0.1, 0.2, 0.3, 0.4 and 0.5 M (mol per litre), using the same volume in every container.

  2. Cut equal cylinders. Use a cork borer to cut potato cylinders of the same diameter and length (for example, 3 cm). The starting length is a constant variable, so keep it the same for every cylinder.

  3. Blot and weigh. Gently blot off surface liquid without squeezing, then record the initial mass to 0.01 g and the initial length in cm.

  4. Soak for 1 hour. Place the cylinders in the solutions at room temperature, and record the temperature.

  5. Blot, reweigh and remeasure. Blot the same way as before, record the final mass and final length, and note the firmness (firm, little change or soft).

  6. Photograph. Lay all the cylinders side by side with a ruler or coin in the frame for scale.

  7. Calculate and plot. Work out the percentage change in mass and plot it against sucrose concentration.


Always know your three types of variables: what you change (sucrose concentration), what you measure (percentage change in mass, with length and firmness as visual observations) and what you keep constant (cylinder size, solution volume, soaking time and temperature).


See it, then calculate it


What you can see. Length and firmness show the direction of water movement. Cylinders in hypotonic solutions take up water, swell and feel firm. Cylinders in hypertonic solutions lose water, shrink and feel soft. Record these in the Visual Observation Table, and keep your photograph as evidence.

What you cannot see. No matter how closely you look, you cannot see the water potential of the tissue. It has to be calculated from the change in mass, which is why you plot the graph and find the isotonic point.

Why mass, not length? Length changes in potato are small. A 3% change on a 3 cm cylinder is 3 × 0.03 = 0.09 cm, which is less than the smallest ruler division (0.1 cm). The balance reads to 0.01 g, so mass gives far more precise data for calculating water potential.

Check that both agree. Observations show the direction of change, not its size: a cylinder that looks unchanged near the isotonic point may still have gained or lost a little water. Swollen cylinders should gain mass and shrunken ones should lose mass. If they disagree, report it and discuss why, such as uneven blotting or a measuring error.


The calculation, step by step


Percentage change in mass = (final mass − initial mass) ÷ initial mass × 100

Example: a cylinder weighs 2.00 g before and 2.15 g after soaking in 0.0 M (pure water). The change is 2.15 − 2.00 = 0.15 g. Then 0.15 ÷ 2.00 = 0.075, and 0.075 × 100 = +7.5%.


Solute potential of each solution: Ψs = −iCRT, where:

  • i = 1 (sucrose does not split into ions);

  • C = sucrose concentration in mol per litre;

  • R = 0.0831 L·bar per mol per K (the gas constant);

  • T = temperature in kelvin (°C + 273), for example 22 °C = 295 K.

Example for 0.3 M sucrose: 1 × 0.3 × 0.0831 × 295 = 7.35 bar. Since 1 bar = 0.1 MPa, 7.35 × 0.1 = 0.735, so Ψs = −0.735 MPa.


Sample results

  • 0.0 M: +7.5% (solute potential 0.00 MPa)

  • 0.1 M: +4.0% (−0.245 MPa)

  • 0.2 M: −0.8% (−0.490 MPa)

  • 0.3 M: −3.4% (−0.735 MPa)

  • 0.4 M: −5.2% (−0.981 MPa)

  • 0.5 M: −7.7% (−1.226 MPa)


The curve crosses 0% change at about 0.19 M. This is the isotonic point, where there is no net water gain or loss. Here, the water potential of the potato tissue equals the solute potential of that solution, because the open solution has zero pressure potential:

1 × 0.19 × 0.0831 × 295 = 4.66 bar. Then 4.66 × 0.1 = 0.466, so the water potential of the potato tissue is about −0.47 MPa.


Three lessons from this practical

1. Pure water is not isotonic to the potato. At 0.0 M the potato gains mass, because pure water (0 MPa) has a higher water potential than the potato cells. Pure water is hypotonic relative to the tissue, not isotonic.

2. Use artificial intelligence (AI), but check it. AI can plot your curve and help you interpret it, but its first answer is often wrong. In the video, the first graph simply connected the dots instead of drawing a line of best fit. You need the fundamentals to spot such errors. Report the prompt you used in your lab report, and keep your own library of prompts for future work.

3. Know what r and R² mean. The correlation coefficient, r, shows how strongly two variables follow a straight line. Values close to +1 or −1 mean a strong relationship. A negative r means that as one variable goes up, the other goes down, as percentage change in mass does when sucrose concentration rises. The coefficient of determination, R², is r multiplied by itself. For example, 0.938 × 0.938 = 0.88, which means 88% of the variation in mass change is explained by sucrose concentration.


Report checklist

  • A completed Visual Observation Table (length, firmness and solution type) and your photograph with a scale.

  • A completed results table, including the solute potential of each solution.

  • A graph of percentage change in mass against sucrose concentration, with a line or smooth curve of best fit and the isotonic point marked.

  • Your estimated water potential of the potato tissue, with the calculation shown.

  • The AI prompt you used to produce your graph.

  • If your curve does not cross zero between 0.0 and 0.5 M, report that honestly; do not force an answer.


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