Interpret repeated root and shoot response data
Calculate means and angular rates with the stimulus direction retained.
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
- Identify the root and shoot of a seedling; recall growth, cell structure and the variables in a controlled investigation.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
What unit fits angle change divided by hours?
What you will learn
- Read each experiment and its reference separately
- Calculate mean curvature and a mean angular rate
Interpret repeated root and shoot response data
Four separate invented six-hour experiments retain angle magnitude and stimulus-relative direction. An even-blue-light root control has mean magnitude 1°. Original teaching model.
| Part and stimulus | Readings / ° | Growth direction |
|---|---|---|
| Root · Gravity | 28°, 32°, 30° | Towards gravity |
| Shoot · Gravity | 38°, 42°, 40° | Away from gravity |
| Shoot · One-sided light | 24°, 30°, 36° | Towards source |
| Root · Stated blue light | 6°, 9°, 12° | Away from source |
Separate even-blue-light root control: 0, 1, 2°; mean magnitude 1°. Angle magnitudes do not supply the response sign.
| Feature | Meaning |
|---|---|
| Mean magnitudes | Root gravity 30°; shoot gravity 40°; shoot light 30°; root light 9° |
| Mean rates | 5; about 6.7; 5; 1.5°/h respectively |
| Root control | Mean 1°; treatment–control difference 8° |
| Scope | Separate invented experiments; interval averages, not universal effects |
Read each experiment and its reference separately
The table contains four separate invented experiments observed over six hours. Each row gives three seedling angle magnitudes in degrees and a separately labelled growth direction. Root gravity readings are 28, 32 and 30° towards gravity; shoot gravity readings are 38, 42 and 40° away from gravity. Shoot-light readings are 24, 30 and 36° towards the source, and the stated blue-light root readings are 6, 9 and 12° away from the source.
Calculate each mean by adding its three readings and dividing by three. The means are 30° for root gravity, 40° for shoot gravity, 30° for shoot light and 9° for root light. The numbers are magnitudes: the response sign comes from the named direction, not from a negative arithmetic sign. These are separate experiments, not four observations of the same individual seedling.
Calculate mean curvature and a mean angular rate
A mean angular rate over the supplied interval is mean angle change divided by elapsed time. The values are 5°/h, about 6.7°/h, 5°/h and 1.5°/h respectively. The unit is degrees per hour, not millimetres per hour, because no length change was measured. These interval averages do not show a constant rate at every moment.
In a separate even-blue-light control, root angle magnitudes are 0, 1 and 2°, giving a mean of 1°. The stated one-sided root model therefore has an 8° greater mean magnitude. Compare variation and method before drawing a strong conclusion: three invented readings do not prove a precise population effect or that all roots behave identically. Supplied data interpretation also does not establish practical competence.
Pause and explain
What supplies positive or negative phototropism in this table?
Worked example
Find the mean and six-hour mean angular rate for root-light readings 6, 9 and 12°.
Show the worked solution
- Mean is (6+9+12)/3 = 9°.
- Rate is 9/6 = 1.5°/h.
Answer Mean 9° away from the source; interval mean angular rate 1.5°/h.
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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 four mean angle magnitudes from the table.
Give me a hint
Add each row separately and divide by three.
Compare my reasoning
- Root gravity: 90/3 = 30; shoot gravity: 120/3 = 40.
- Shoot light: 90/3 = 30; root light: 27/3 = 9.
Root gravity 30°; shoot gravity 40°; shoot light 30°; root light 9°.
Look for these in your work
- Calculate all four means
- Retain degree units
Calculate the four mean angular rates over six hours.
Give me a hint
Divide each mean angle by six.
Compare my reasoning
- 30/6 = 5; 40/6 ≈ 6.7.
- 30/6 = 5; 9/6 = 1.5.
Root gravity 5°/h; shoot gravity about 6.7°/h; shoot light 5°/h; root light 1.5°/h.
Look for these in your work
- Use division by elapsed time
- Retain angular rate units
Compare mean root curvature in the one-sided model and even-blue-light control, then state one inference limit.
Give me a hint
The control mean is (0+1+2)/3.
Compare my reasoning
- Control mean is 1°; treatment mean is 9°, a difference of 8°.
- Small invented records do not establish a universal root response or precise population effect.
The treatment mean is 8° greater than the control mean. This small invented comparison does not establish a universal root response or a precise population effect.
Look for these in your work
- Calculate the mean difference
- Limit the inference
Common mistakes
- Using millimetres per hour for an angle measurement.
- Assigning response signs from angle magnitudes alone.
What can you explain now?
What supplies positive or negative phototropism in this table?
Compare with the explanation
Growth direction relative to the light source
The angle magnitude alone does not give the sign.
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Make it stick
Calculate angle means and interval rates, then classify responses using the labelled stimulus-relative direction.
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