Investigate suitable temperature using repeated seed counts
Calculate means and recognise the best tested condition without claiming a universal optimum.
Section 16.3 · Sexual reproduction in plants
Exam scope and assessment notes
Core and Extended · Biology 0970, 2026–2028. Use the stated specimen features and evidence. Diagrams are schematic. Written practice is self-checked; human subject-teacher review, supervised practical competence and cumulative assessment remain pending. Cambridge states that IGCSE 0970 has the same subject content as IGCSE 0610, with 9–1 grading. This StudyForA pathway maps the 2026–2028 syllabus only. This 9–1 graded qualification is available only in a limited number of administrative zones. Confirm that your centre can enter you for this code and exam series with Cambridge International. Cambridge has published a separate syllabus for 2029. StudyForA has not reconciled that version yet; this page supports exams in 2026, 2027 and 2028 only.
Before you begin
- Recall gametes, nuclear fusion, cellular respiration and the distinction between reproduction and growth.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
Which temperature has the highest mean in the supplied record?
What you will learn
- Control hydration and oxygen across temperatures
- Compare repeated timed germination results
Investigate suitable temperature using repeated seed counts
Invented three-repeat day-5 counts, 20 seeds per group: means at 5/15/25/35°C are 1/13/17/9. The best tested condition is 25°C; no universal optimum is established. Original teaching data.
| Temperature | Repeat counts | Mean count | Mean percentage |
|---|---|---|---|
| 5°C | 1, 2, 0 | 1 | 5% |
| 15°C | 12, 14, 13 | 13 | 65% |
| 25°C | 17, 18, 16 | 17 | 85% |
| 35°C | 8, 10, 9 | 9 | 45% |
| Feature | Meaning |
|---|---|
| Variable | Temperature |
| Critical controls | Moisture, oxygen, seed stock and time |
| Mean percentages | 5%, 65%, 85%, 45% |
| Conclusion | 25°C best among these tested conditions |
Control hydration and oxygen across temperatures
To investigate temperature, allocate matched viable seed groups to several stated temperatures while keeping moisture, oxygen availability, species and batch, starting total, timing and relevant additional requirements the same. Check moisture throughout because evaporation can differ between temperatures. Equal water supplied initially does not guarantee equal moisture after several days.
Repeat each temperature treatment using independently prepared groups. In the invented record, three groups of 20 seeds are tested at each temperature and germination is counted by day 5. At 5°C the counts are 1, 2 and 0; at 15°C they are 12, 14 and 13; at 25°C they are 17, 18 and 16; at 35°C they are 8, 10 and 9.
Compare repeated timed germination results
Mean counts are 1, 13, 17 and 9 respectively. Divide each mean by 20 and multiply by 100 to obtain mean germination percentages of 5%, 65%, 85% and 45%. The 25°C treatment has the highest mean among the four tested temperatures. The decline at 35°C shows that hotter is not automatically better in this record.
Do not claim the exact biological optimum is 25°C for all species or even every untested temperature for this seed stock. A temperature between the tested values could produce another result, and day-5 emergence does not measure all later growth. The repeated record supports a suitable-temperature explanation under its stated controls; supervised practical work and broader testing remain separate evidence.
Pause and explain
What does this comparison establish most directly?
Worked example
Find the mean of the 25°C counts 17, 18 and 16.
Show the worked solution
- Add the three counts: 51.
- 51 ÷ 3 = 17.
Answer Mean 17 of 20, or 85%.
Try a topic test
Choose Easy, Normal or Difficult. Save your answers and open worked review after finishing.
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.
Calculate the mean count and percentage for 15°C: 12, 14 and 13 of 20.
Give me a hint
Average the counts before using the common denominator.
Compare my reasoning
- (12 + 14 + 13)/3 = 13.
- 13/20 × 100 = 65%.
Mean 13 of 20; 65% by day 5.
Look for these in your work
- Compute the mean
- Use 20 as the starting total per group
Compare the mean percentages at 25°C and 35°C and identify the best tested treatment.
Give me a hint
Use means 17 and 9 of 20.
Compare my reasoning
- 25°C: 85%; 35°C: 45%.
- 85 − 45 = 40 percentage points.
25°C has the higher mean: 85% versus 45%, a 40-percentage-point difference. It is best among the four tested treatments.
Look for these in your work
- Calculate both percentages
- Use the tested-condition boundary
Why should moisture be monitored, and why is an exact universal optimum unsupported?
Give me a hint
Consider evaporation and untested temperatures.
Compare my reasoning
- Temperature can change evaporation and hence hydration.
- Only four temperatures for one stock at one time were tested.
Monitor moisture to avoid confounding temperature with hydration. The best tested result does not establish an exact optimum between values or a universal optimum across species.
Look for these in your work
- Name the hydration control
- Limit the optimum claim
Common mistakes
- Calling the highest tested result a universal exact optimum.
- Forgetting moisture can vary through evaporation.
What can you explain now?
What does this comparison establish most directly?
Compare with the explanation
25°C performs best among the tested conditions by day 5
Untested temperatures and other species remain outside the record.
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Make it stick
Compute repeat means and qualify conclusions about a suitable temperature.
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.