All design choices and example data
What should be varied?
- Vary external sucrose concentration between comparable tissue treatments. The intended comparison changes external solution concentration. This is the suitable option for the stated question.
- Use progressively thinner tissue pieces as sucrose concentration increases. Thickness changes surface area relative to volume and may alter the extent of water exchange in the fixed interval.
- Use tissue from a different plant source for each sucrose concentration. Tissue composition could then vary with concentration. That would confound the intended comparison.
Which calculation gives percentage mass change?
- (Final mass − initial mass) ÷ final mass × 100. Percentage change uses initial mass as its reference. Using final mass changes the denominator and the result.
- Final mass ÷ initial mass × 100. That gives final mass as a percentage of initial mass. Subtract the initial mass first to calculate percentage change.
- (Final mass − initial mass) ÷ initial mass × 100. Subtract initial mass to find the change, then use initial mass as the reference. A loss gives a negative result. This is the suitable option for the stated question.
Which control approach is useful?
- Weigh each piece straight from solution without removing surface liquid. Surface solution contributes to the reading. Its amount may vary, so use a consistent blotting method.
- Use comparable tissue dimensions, exposure time and consistent blotting. These reduce differences unrelated to external concentration. This is the suitable option for the stated question.
- Leave the larger pieces immersed for longer to compensate for their size. Changing both dimensions and exposure time prevents a comparable concentration test. Standardise both instead.
Matched example
Pieces are measured after consistent blotting.
Calculation questions and worked method
Calculate percentage mass change at 0.40 mol dm⁻³, using the mean final mass.
Mean = (2.03 + 2.04 + 2.05) ÷ 3 = 2.04. Percentage change = (2.04 − 2) ÷ 2 × 100 = 2%.
Estimate the zero-mass-change concentration by local linear interpolation, in mol dm⁻³.
The adjacent bracket is 0.4 mol dm⁻³ (2%) to 0.6 mol dm⁻³ (-3%). Local interpolation = 0.4 − (2 × 0.2 ÷ -5) = 0.48 mol dm⁻³. This estimate does not establish identical solute composition.
Surface solution remains on pieces
This example adds surface-fluid mass, increasing with concentration. It illustrates a measurement bias rather than predicting tissue water uptake.
Calculation questions and worked method
Calculate percentage mass change at 0.40 mol dm⁻³, using the mean final mass.
Mean = (2.09 + 2.1 + 2.11) ÷ 3 = 2.1. Percentage change = (2.1 − 2) ÷ 2 × 100 = 5%.
Estimate the zero-mass-change concentration by local linear interpolation, in mol dm⁻³.
The adjacent bracket is 0.6 mol dm⁻³ (1.5%) to 0.8 mol dm⁻³ (-2%). Local interpolation = 0.6 − (1.5 × 0.2 ÷ -3.5) = 0.68571 mol dm⁻³. This estimate does not establish identical solute composition.
Evaluation prompt
Explain how the different blotting assumptions change the apparent zero-mass-change estimate. Suggest an improvement and explain why even a well-measured zero crossing does not establish identical solute composition inside and outside the tissue.
Example explanation and self-check points
Surface fluid raises the measured final mass in the biased example, moving its apparent zero crossing towards a higher external concentration. Consistent blotting, tissue dimensions and exposure time make the comparison more defensible. A local zero crossing estimates a condition with no net mass change under the method used; it does not show that the tissue and solution contain identical solutes. Separate pieces can reveal variation, but pieces from one tissue source do not represent every plant.
- I link retained surface fluid to a shifted apparent zero crossing.
- I propose consistent blotting with comparable tissue and exposure conditions.
- I limit the inference from zero mass change and from pieces sharing one tissue source.
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Original fixed datasets and reasoning tasks. Full requirements: AQA practical assessment · OCR A specification. These activities do not replace laboratory work or assessment of practical competence.
