What should be varied?
- Vary external sucrose concentration between comparable tissue treatments.
- Use progressively thinner tissue pieces as sucrose concentration increases.
- Use tissue from a different plant source for each sucrose concentration.
AQA RP3 context · practical reasoning
Investigate the effect of external sucrose concentration on the mass change of comparable plant-tissue pieces.
Each piece starts at 2.00 g. All pieces spend the same stated interval in solution. The example uses separate pieces from comparable tissue; it does not represent independent plants. Final masses are measured with a consistent blotting method.
Questions and answers · Matched example · 3 readings per condition.
Pieces are measured after consistent blotting.
| Sucrose concentration / mol dm⁻³ | Reading 1 | Reading 2 | Reading 3 |
|---|---|---|---|
| 0 | 2.23 | 2.24 | 2.25 |
| 0.2 | 2.13 | 2.14 | 2.15 |
| 0.4 | 2.03 | 2.04 | 2.05 |
| 0.6 | 1.93 | 1.94 | 1.95 |
| 0.8 | 1.83 | 1.84 | 1.85 |
Percentage change = (mean final mass − initial mass) ÷ initial mass × 100. Estimate the zero crossing by local linear interpolation between the adjacent opposite signs. Give numerical answers to about three significant figures.
Use the alternate example below alongside the data above.
| Sucrose concentration / mol dm⁻³ | Reading 1 | Reading 2 | Reading 3 |
|---|---|---|---|
| 0 | 2.23 | 2.24 | 2.25 |
| 0.2 | 2.16 | 2.17 | 2.18 |
| 0.4 | 2.09 | 2.1 | 2.11 |
| 0.6 | 2.02 | 2.03 | 2.04 |
| 0.8 | 1.95 | 1.96 | 1.97 |
Vary external sucrose concentration between comparable tissue treatments. The intended comparison changes external solution concentration.
(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.
Use comparable tissue dimensions, exposure time and consistent blotting. These reduce differences unrelated to external concentration.
Mean = (2.03 + 2.04 + 2.05) ÷ 3 = 2.04. Percentage change = (2.04 − 2) ÷ 2 × 100 = 2%.
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 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.
Original authored examples, not real experimental predictions, official marks or a laboratory protocol.