A moving-bubble potometer measures water uptake by a shoot. It does not directly collect the water vapour lost from leaves. This distinction is the starting point for reliable worked examples of potometer calculations: calculate the measured uptake first, then explain when it can be used as an estimate of transpiration.
All datasets here are constructed classroom examples. Use the supplied readings and dimensions; do not assemble apparatus or carry out an unsupervised practical from this lesson. Real apparatus preparation and checks belong with a teacher or technician.
What does the bubble tell you?
As water enters the shoot from a correctly functioning sealed apparatus, a marker bubble moves along the capillary. Its change in position gives a distance. Dividing distance by time gives a linear rate, such as millimetres per minute. To find a volume rate, you also need the capillary’s internal cross-sectional area.
Transpiration is the loss of water vapour from the aerial parts of a plant, mainly through stomata. Uptake often tracks this loss reasonably well over suitable periods, but a shoot can store or use some absorbed water. Changing tissue water content and delays in adjustment mean uptake and loss need not match instantaneously.
Check which kind of instrument a question describes. A mass-loss arrangement on a balance records a different quantity from a moving-bubble potometer. Do not transfer the units or limitations of one measurement to the other without reading the question.
Worked example 1: from two positions to volume rate
A capillary has a uniform internal radius of 0.50 mm. In four minutes the bubble moves from the 14 mm position to the 62 mm position. Assume the observed displacement represents uptake and use π = 3.142.
- Distance moved = final position − initial position = 62 − 14 = 48 mm.
- Linear rate = 48 ÷ 4 = 12 mm min−1.
- Internal area = πr2 = 3.142 × 0.502 = 0.7855 mm2.
- Volume taken up = area × distance = 0.7855 × 48 = 37.704 mm3.
- Volume rate = 37.704 ÷ 4 = 9.426 mm3 min−1, about 9.43 mm3 min−1.
The same result follows from area × linear rate: 0.7855 × 12 = 9.426. Because 1 cm3 = 1,000 mm3, this is 0.009426 cm3 min−1. Divide by 1,000 when changing cubic millimetres into cubic centimetres.
Two common errors give misleading answers. Using 62 mm treats the final position as the displacement and ignores where the bubble started. Using the diameter in πr2 makes the calculated area four times too large. Here the diameter would be 1.00 mm; the formula needs its half, 0.50 mm.
Worked example 2: compare repeated environmental readings
The same shoot and capillary are compared in still and moving air. The supplied record says temperature, light, humidity, leaf area and measurement duration were kept similar, with time allowed for the shoot to adjust before recording. Each reading lasts five minutes.
| Condition | Trial 1 / mm | Trial 2 / mm | Trial 3 / mm | Mean / mm | Mean linear rate / mm min−1 |
|---|---|---|---|---|---|
| Still air | 18 | 20 | 22 | 20 | 4 |
| Moving air | 36 | 40 | 44 | 40 | 8 |
For still air, mean displacement = (18 + 20 + 22) ÷ 3 = 20 mm, giving 4 mm min−1. Moving air gives a mean of 40 mm, or 8 mm min−1. The increase relative to still air is (8 − 4) ÷ 4 × 100 = 100%. Uptake doubled in this constructed comparison.
Air movement can remove humid air next to a leaf, helping maintain a water-vapour gradient between the leaf and its surroundings. This is a plausible explanation for greater transpiration and associated uptake. State the measurement accurately: the apparatus recorded faster water uptake, supporting greater transpiration under the controlled conditions.
The repeats also reveal variation: still-air displacements span 18–22 mm; moving-air displacements span 36–44 mm. Their ranges do not overlap, but three trials are still limited evidence. If all still-air readings occurred first, a gradual change in shoot condition could affect the comparison. Alternating treatment order in a properly supervised investigation can help check that possibility.
Worked example 3: faster movement need not mean more water
Two instruments have different internal radii. Apparatus A has radius 0.20 mm and a bubble speed of 10 mm min−1. Apparatus B has radius 0.40 mm and a speed of 4 mm min−1. Use π = 3.142.
A’s uptake rate = 3.142 × 0.202 × 10 = 1.2568 mm3 min−1. B’s rate = 3.142 × 0.402 × 4 = 2.01088 mm3 min−1. Although B’s bubble moves more slowly, B records the greater volume rate.
Doubling the radius quadruples the area. Consequently, each millimetre of travel in B represents four times the water volume represented in A. Comparing bubble speed alone is valid for volume comparisons only when capillary areas are the same or appropriate calibration factors are supplied.
Evaluate a claim before explaining it
Suppose a report says that moving a shoot from a cool room to a sunny, windy outdoor position increased its uptake. The result is evidence of a change, but not an isolated wind effect. Light, temperature and humidity may also have changed. A persuasive answer names these confounders and proposes a comparison in which only the intended factor changes.
Other limitations are specific to the apparatus. A leak allows water or air to move by an unintended route, so bubble movement may no longer represent shoot uptake. An obstruction or an unintended air gap in the water pathway can interfere with movement. Do not automatically label an unusually fast reading “more transpiration” without checking the supplied integrity record.
Leaf area affects whole-shoot uptake. If one shoot has twice the leaf area, a larger overall rate may simply reflect more evaporating surface. Where area is supplied, rates per unit leaf area can improve comparison, but leaf age, stomatal behaviour and species still matter. Greater heat or wind does not guarantee unlimited increases: water stress can promote stomatal closure.
Measurement precision also limits the answer. With an uncertain position reading, a very short bubble displacement has a large relative uncertainty. Longer observation may make displacement easier to resolve, but an interval must not be so long that conditions drift. Report precision justified by the readings rather than many calculator digits.
Three practice tasks with solved answers
Practice 1: keep units consistent
A bubble travels 30 mm in 150 seconds through a capillary with area 0.40 mm2. Find the volume rate in mm3 s−1 and mm3 min−1.
Answer: Volume = 0.40 × 30 = 12 mm3. Rate = 12 ÷ 150 = 0.080 mm3 s−1. Multiplying by 60 gives 4.8 mm3 min−1.
Practice 2: use a diameter correctly
The internal diameter is 0.60 mm and bubble speed is 5 mm min−1. Calculate volume uptake using π = 3.142.
Answer: Radius = 0.30 mm. Rate = 3.142 × 0.302 × 5 = 1.4139 mm3 min−1, approximately 1.41. Substituting 0.60 as the radius would wrongly quadruple the answer.
Practice 3: normalise and qualify
Shoot P takes up 6 mm3 min−1 with 30 cm2 leaf area. Shoot Q takes up 9 with 60 cm2. Which has greater uptake per unit area?
Answer: P: 6 ÷ 30 = 0.20 mm3 min−1 cm−2. Q: 9 ÷ 60 = 0.15 in the same units. P is greater per area, although Q is greater overall. Neither figure alone is an exact direct transpiration measurement.
Connect root uptake, xylem transport and leaf water loss
A shoot potometer isolates one part of a longer water pathway. In an intact plant, water must first reach root surfaces, enter living tissues, pass into xylem and travel to leaves. Distinguish liquid-water movement from evaporation and diffusion of water vapour.
Root hairs increase the absorbing surface in contact with soil water. Their membrane is selectively permeable; water moves by osmosis when root-cell water potential is lower than the surrounding soil solution. Mineral-ion uptake is a separate process involving transport proteins and electrochemical gradients. Active uptake can help establish solute differences, but water itself is not pumped into the root by ATP-powered osmosis.
Xylem vessel elements and tracheids have lignified walls. Their conducting spaces provide a pathway for water and dissolved mineral ions. Transpiration lowers leaf water potential and generates tension in the xylem water column. Cohesion between water molecules helps transmit the pull; adhesion to walls also contributes. Xylem water transport does not require a living pumping cell at every point. Root pressure can contribute under some conditions, but it is not a universal explanation of transport to the tops of tall trees.
Worked case 1: separate water uptake from mineral uptake
A theoretical intact-root record gives soil water potential −0.10 MPa and root-hair water potential −0.35 MPa. Water can move into the root hair because −0.10 is the higher value. The drop is 0.25 MPa. A learner reverses the arrow because 0.35 is numerically larger than 0.10 without its sign. Keep the signs: higher water potential here means less negative. If dry soil changes to −0.60 MPa while the root hair remains −0.35 MPa, the simple local model predicts net water movement outwards, not automatic entry “because roots need water”.
Suppose a neutral solute in a separate membrane model moves from lower to higher concentration with a supplied energy-coupled carrier. That active step can alter solute conditions, but it is not the movement of water by osmosis. Ion cases would also require the electrical gradient. The pressure units in the water-potential record cannot be compared as if they were solute concentrations.
Worked case 2: interpret linked evidence
| Condition | Uptake / cm³ in 30 min | Water-vapour loss / cm³ liquid equivalent in 30 min | Net retention / cm³ |
|---|---|---|---|
| Moderate humidity | 6.0 | 5.6 | 0.4 |
| Higher humidity | 3.0 | 2.7 | 0.3 |
Temperature, light, leaf area, root supply and air movement are matched. Higher surrounding humidity reduces the leaf-to-air vapour difference, supporting lower transpiration and associated uptake. Uptake rates are 6/30 = 0.20 and 3/30 = 0.10 cm³/min. They are not exactly the loss rates: the record retains 0.4 and 0.3 cm³ respectively. This connects the root–xylem pathway to the retained potometer lesson while preserving its measurement limitation. It does not prove that every species responds identically or that stomatal behaviour is unchanged in all real conditions.
Independent practice with answers
- Task: Soil is −0.20 MPa, root hair −0.50 MPa. Which way is the favourable osmotic movement? Answer: Into the root hair, from higher to lower water potential.
- Task: Name the state of water in xylem and as it leaves the stomatal air space. Answer: Mainly liquid in xylem; vapour diffuses out after evaporation from moist leaf surfaces.
- Task: Uptake is 8.0 cm³ and vapour loss is 7.4 cm³ over 40 min. Calculate uptake rate and net retention. Answer: 0.20 cm³/min and 0.6 cm³ retained under the given complete account.
- Task: Does cutting off the roots in a supplied potometer model prove roots normally play no role? Answer: No. It bypasses the intact-root part of the pathway. It cannot establish how soil water or ions enter a whole plant.
All records are interpretation tasks; no cutting or apparatus preparation is required. Background: plant water transport.
Frequently asked questions
Does a moving-bubble potometer measure transpiration directly?
No. It measures water uptake by the shoot. Under suitable conditions uptake can estimate transpiration, but storage, use and adjustment delays can cause differences.
Why subtract the two scale positions?
The required distance is the change in position. A final reading includes the starting offset unless the bubble began at zero.
When can I compare distance per minute alone?
When the capillaries have the same internal area, or when the question specifically asks for a linear rate. Different areas require volume calculations for a fair water-uptake comparison.
What if the question gives diameter?
Halve it before substituting into πr². Using diameter as radius gives an area, and therefore a calculated volume, four times too large.
Why can a leak invalidate the result?
The marker may respond to water or air entering or leaving somewhere other than the intended shoot pathway, so its motion no longer reliably represents uptake.
Does higher humidity normally increase water loss?
Higher humidity usually reduces the water-vapour gradient from leaf to air, tending to reduce transpiration when other factors are similar. Use the actual conditions and supplied observations.
Should I remove an anomalous reading?
Investigate and report it. A justified equipment failure may support exclusion, but removing a value merely because it disagrees with the expected trend biases the result.
Sources and further reading
Scientific background was checked against these educational references. The examples, tables, diagrams and practice answers on this page are original learning material.
- Science & Plants for Schools: investigating transpiration with a potometer
- Practical Biology: measuring water uptake with a potometer
Continue your revision
You can complete every question above without signing in. For a wider resource collection, visit the KCSE Biology topic-tests page. Its resource content currently requires login; check membership access and the listed contents before joining. This lesson does not promise a particular downloadable file.
WhatsApp Us Now