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IGCSE · Cambridge (CIE) · 0970

Calculate a potometer uptake rate

Convert bubble distance into displaced water volume using capillary area and elapsed time.

Transpiration pathway · 0970 (9–1) · 8.3
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 distance change, area and rates.

Work on paper. Try each question before revealing help, and explain why your method works.

Quick readiness check

Which distance is the measured change?

What you will learn

  • Use change in bubble position rather than final position
  • Calculate displaced volume per time with units
Apparatus model, not to scale: a leafy shoot is sealed to a water-filled capillary. Bubble movement is a water-uptake proxy. For area 0.02 cm², travel 15 cm in 3 min gives 0.10 cm³/min. Teacher-supervised preparation and leak checks are needed.A potometer follows wateruptakeLeafy shoot in a sealed tubeSealWater-filled capillaryBubbleTravel towards shoot: 15 cmArea × travel ÷ time0.02 × 15 ÷ 3 = 0.10 cm³/min
Apparatus model, not to scale: a leafy shoot is sealed to a water-filled capillary. Bubble movement is a water-uptake proxy. For area 0.02 cm², travel 15 cm in 3 min gives 0.10 cm³/min. Teacher-supervised preparation and leak checks are needed.

Distance becomes volume

For a uniform capillary, displaced volume equals its cross-sectional area multiplied by bubble travel distance. If area is in cm² and travel in cm, the volume is in cm³. Use final position minus initial position, not the final ruler reading alone.

Average uptake rate is displaced volume divided by elapsed time. A calibrated apparatus may provide volume markings directly; then use the volume change. The method must be chosen from the supplied apparatus information.

Keep the proxy visible

With area 0.02 cm², bubble positions 2 and 17 cm give travel 15 cm. Volume is 0.02 × 15 = 0.30 cm³; over 3 min the average water-uptake rate is 0.10 cm³/min.

This is water uptake by the shoot, not an exact direct transpiration measurement. Leaks, stored water, water used by the plant and changing conditions can cause differences. Reject a negative movement or zero duration rather than hiding an invalid reading.

Pause and explain

Which calculation gives volume rate?

Put the idea to work

Worked example

Calculate uptake for area 0.02 cm², positions 2 to 17 cm, time 3 min.

Show the worked solution
  1. Travel = 17 − 2 = 15 cm.
  2. Volume = 0.02 × 15 = 0.30 cm³; rate = 0.30 / 3.

Answer Average uptake = 0.10 cm³/min.

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.

1 · Guided

Find travel from 2 to 17 cm, then volume for area 0.02 cm².

Give me a hint

Subtract first, multiply second.

Compare my reasoning
  1. 17 − 2 = 15 cm.
  2. 0.02 × 15 = 0.30 cm³.

Travel 15 cm; volume 0.30 cm³.

Look for these in your work

  • Use the change
  • Calculate cm³
2 · Independent

Area is 0.01 cm², travel is 20 cm and time is 4 min. Calculate uptake rate.

Give me a hint

Volume is area times distance.

Compare my reasoning
  1. Volume = 0.01 × 20 = 0.20 cm³.
  2. Rate = 0.20 / 4 = 0.05 cm³/min.

Average uptake = 0.05 cm³/min.

Look for these in your work

  • Calculate volume
  • Divide by time with the rate unit
3 · Transfer

Area is 0.02 cm² and travel 60 mm in 2 min. Calculate the rate and explain its limit.

Give me a hint

Convert 60 mm to cm before multiplying.

Compare my reasoning
  1. 60 mm = 6 cm; volume = 0.02 × 6 = 0.12 cm³.
  2. 0.12 / 2 = 0.06 cm³/min; this measures uptake, not direct vapour loss.

0.06 cm³/min water uptake; it is a proxy rather than an exact transpiration rate.

Look for these in your work

  • Convert to 6 cm
  • Calculate rate and identify the proxy

Common mistakes

  • Using 17 cm instead of the 15 cm change.
  • Multiplying cm² by mm without converting the distance.
Recall without your notes

What can you explain now?

Which calculation gives volume rate?

Compare with the explanation

Area × distance change ÷ time

The resulting units are volume per time.

After trying it yourself, choose your next review. This is your self-assessment.

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Make it stick

Write a rate formula with units, then state what the apparatus actually measures.

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.

Keep learning Biology (9-1)

Course scope and assessment

780 sequenced lessons provide 2340 guided, independent and application problems in characteristics, classification, cells, organisation, specimen size, membrane transport, biological molecules, enzymes, plant and human nutrition, plant transport, circulatory systems, the heart, blood vessels, blood, disease transmission, body defences, immunity, gas exchange in humans, respiration energy uses and investigations, aerobic respiration, anaerobic respiration and recovery, excretion in humans, coordination and response, sense organs, hormones, homeostasis, tropic responses, drugs, asexual reproduction, sexual reproduction, sexual reproduction in plants, sexual reproduction in humans, sex hormones in humans, sexually transmitted infections, chromosomes, genes and proteins, mitosis, meiosis, monohybrid inheritance, variation, adaptive features, selection, energy flow, food chains and webs, nutrient cycles, population, community, ecosystem, population-growth phases, food supply, habitat destruction, pollution, conservation, biotechnology and genetic modification. 61 of 61 syllabus sections have linked teaching and staged practice. The earlier inheritance overview remains introductory, and the monohybrid pathway develops the relevant Core and Supplement requirements. This is not full programme completion or guaranteed exam readiness. Written practice is self-checked. This is the separate 9–1 qualification 0970; Cambridge states that its 2026–2028 subject content is otherwise the same as 0610. 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. The 2029 syllabus is separately published at https://www.cambridgeinternational.org/Images/764352-2029-syllabus.pdf and is not yet mapped here. This course supports exams in 2026, 2027 and 2028 only.

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