Calculate a potometer uptake rate
Convert bubble distance into displaced water volume using capillary area and elapsed time.
Core and Extended; Biology 0610, 2026–2028. Use the stated specimen features and evidence. Diagrams are schematic. Written practice is self-checked
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
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?
Worked example
Calculate uptake for area 0.02 cm², positions 2 to 17 cm, time 3 min.
Show the worked solution
- Travel = 17 − 2 = 15 cm.
- 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.
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
- 17 − 2 = 15 cm.
- 0.02 × 15 = 0.30 cm³.
Travel 15 cm; volume 0.30 cm³.
Look for these in your work
- Use the change
- Calculate cm³
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
- Volume = 0.01 × 20 = 0.20 cm³.
- 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
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
- 60 mm = 6 cm; volume = 0.02 × 6 = 0.12 cm³.
- 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.
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.
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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.