(a) Problem-solving and application
- Syllabus
- 2024
- Topic
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- Level
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Learning objectives
Turn a practical context into a solvable problem
Solving a practical problem means extracting the relevant variables and evidence, choosing a valid comparison or relationship, and producing an answer with units and a context check.
| Move | What to do |
|---|---|
| 1 state the target | identify exactly what must be found: difference, rate, percentage, optimum or predicted outcome |
| 2 map the variables | identify the changed condition and the measured response; ignore details that do not affect the target |
| 3 align the evidence | compare like with like—same time, unit, sample basis or body-mass basis |
| 4 choose an operation | difference = final − initial; rate = change ÷ time; percentage change = change ÷ original × 100 |
| 5 calculate transparently | show selected values, substitution, arithmetic and units |
| 6 check the context | confirm sign, size and unit are biologically plausible and answer the stated target |
In a seedling table at day 20, dry mass is 8.5 g with fertiliser and 6.8 g without it. The matched difference is 1.7 g; relative to the no-fertiliser value, the increase is 1.7 ÷ 6.8 × 100 = 25%.
Do not compare unmatched days or switch denominators silently. A numerical pattern is an answer to the calculation; explaining why it occurs requires relevant biological knowledge and belongs to the next reasoning step.
Apply biological knowledge to a practical observation
Applying knowledge in a practical context means selecting the biological principle that controls the measured outcome and building an explicit causal link from condition to observation.
| Reasoning move | Example of a complete link |
|---|---|
| identify the observation | fertilised seedlings have greater dry mass at the same time |
| select relevant biology | nitrate supports amino-acid and protein synthesis; magnesium supports chlorophyll |
| connect mechanism to measurement | mineral supply can increase growth and photosynthesis, producing more biomass and therefore greater dry mass |
| respect measurement choice | dry mass excludes variable water content, so it better represents accumulated biological material |
| transfer to another context | a mouse has a larger surface-area-to-volume ratio than a human, loses heat faster per gram and needs a higher respiration rate per gram to maintain temperature |
A strong explanation has the form condition → biological process → effect on the measured variable. Use the context's scale and units: total oxygen use and oxygen use per gram answer different questions.
Naming a topic is not application. 'Fertiliser helps growth' or 'surface area matters' is incomplete until the mechanism is connected to the actual result, measurement basis and direction of change.