Estimate tissue water potential from plant-tissue mass changes
Changes in standardized plant-tissue mass across known solutions can locate the concentration at which there is no net water movement.
AS Level · Cambridge International Biology 9700, 2028–2030. These lessons explain membrane transport and investigation reasoning. Written practice is self-checked; it does not demonstrate supervised practical competence or replace teacher-reviewed assessment.
Before you begin
- Relate phospholipid and protein structure to their roles, and distinguish a concentration gradient from the direction of net movement.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
What does zero percentage mass change suggest in a plant-tissue osmosis investigation?
What you will learn
- Investigate plant tissue in solutions of differing water potential
- Estimate tissue water potential from a zero-net-change result using an appropriate calibration
Measure water movement consistently
Cut equal-sized plant-tissue cylinders, record initial masses and place them for equal times in a series of known solutions. Remove them, blot consistently and record final masses. Keep temperature, solution volume, tissue source and dimensions as constant as possible.
Calculate percentage mass change = (final mass − initial mass) ÷ initial mass × 100. Repeats help identify variability. Positive mass change indicates net water uptake; negative change indicates net water loss under the test conditions.
Use the isotonic point as an estimate
Plot percentage mass change against the known solution concentration or water potential. The point where percentage mass change is zero estimates the isotonic condition: no net water movement between tissue and solution.
Use a supplied calibration or known water potentials to estimate the tissue water potential at that point. Without a calibration linking solution concentration to water potential, report the isotonic concentration rather than inventing a numerical water potential.
Pause and explain
What additional information is needed to report tissue water potential from an isotonic concentration?
Worked example
A tissue cylinder changes from 2.00 g to 2.10 g in one solution. Calculate percentage mass change and interpret its sign.
Show the worked solution
- Change = 0.10 g.
- 0.10 ÷ 2.00 × 100 = 5%.
Answer Percentage mass change = (2.10−2.00) ÷ 2.00 × 100 = +5%. The tissue gained mass, consistent with net water uptake under these conditions.
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Test this topic →From guided practice to a new situation
Write your answer and reasoning first. Use a hint only if you are stuck. The model solution is for self-checking; your written work is not automatically marked.
A 1.50 g tissue sample becomes 1.44 g. Calculate percentage mass change.
Give me a hint
Use (final−initial) ÷ initial × 100.
Compare my reasoning
- Change = −0.06 g.
- −0.06 ÷ 1.50 ×100 = −4%.
Percentage mass change = (1.44−1.50) ÷ 1.50 ×100 = −4%. The sample lost 4% of its starting mass.
Look for these in your work
- Change sign correct
- Initial mass used as denominator
One solution causes +6% mass change and another causes −3%. What does the zero-change point between them estimate?
Give me a hint
Interpret where net uptake changes to net loss.
Compare my reasoning
- At the zero-change point, there is no net mass change.
- It estimates the isotonic solution condition for the tissue.
The solution at the zero percentage mass change estimates the isotonic condition, where there is no net water movement between tissue and solution under the method conditions.
Look for these in your work
- Zero point interpreted
- No-net-water-movement stated
A student reports tissue water potential as −0.35 MPa from an isotonic sucrose concentration but has no concentration-to-potential table or calibration. Explain the limitation.
Give me a hint
Separate the isotonic concentration from a numerical water potential.
Compare my reasoning
- The experiment locates the isotonic concentration.
- A calibration is needed to convert it to water potential.
The data may estimate the isotonic sucrose concentration, but a concentration-to-water-potential calibration or supplied reference is needed to justify −0.35 MPa. The numerical value should not be invented.
Look for these in your work
- Measured inference stated
- Missing calibration identified
Common mistakes
- Using final mass as the percentage denominator
- Failing to blot surface solution consistently
- Converting isotonic concentration into a water potential without a calibration
What can you explain now?
What does zero percentage mass change suggest in a plant-tissue osmosis investigation?
Compare with the explanation
No net water movement under those conditions
Equal water potentials give no net osmosis.
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Make it stick
How do mass changes identify an isotonic condition, and when can you report a numerical tissue water potential?
If you needed the solution, close it and solve the problem again from a blank page. A correct answer is stronger when you can explain the reason for each step.