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OCR GCSE Biology A: Transport in plants: water uptake, transpiration and translocation (J247) – Revision Notes

Condensed recall notes on plant transport for OCR GCSE Biology A (J247), B2.2f to B2.2j, with formulas, a self-test and common mark-losing slips.

Subject
Biology
Level
GCSE
Topic
Transport in plants: water uptake, transpiration and translocation
Updated

Aligned to OCR GCSE Biology (J247), Version 4.0 (August 2026), for first assessment in 2018. Official specification .

Syllabus page (what it covers and how it is assessed): OCR GCSE Biology.

Syllabus points this page covers

J247

  • B2.2 The challenges of size

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Quick recall for OCR GCSE (9-1) Biology A (J247), statements B2.2f to B2.2j (plant transport, part of topic B2). For explanations, read the study guide, then test yourself with the practice questions. Course hub: OCR GCSE Biology. Printable checklist. A quick check of where you stand: free diagnostics. None of these statements is Higher Tier only.

Key terms

Term Meaning
Root hair cell Root surface cell with a long extension; takes in water and mineral ions
Osmosis Water moving through a partially permeable membrane down its concentration gradient
Active transport Movement against a concentration gradient using energy from respiration
Transpiration Loss of water vapour from the plant, mainly through stomata
Transpiration stream Movement of water up the xylem, replacing what is lost
Stoma (stomata) Pore in the leaf, controlled by two guard cells
Translocation Movement of dissolved sugars in the phloem from sources to sinks
Potometer Apparatus measuring the rate of water uptake by a shoot

B2.2f: Uptake in the root

  • Water: osmosis, from the dilute soil water into the root hair cell.
  • Mineral ions: mostly active transport, against the gradient.
  • Adaptations: long thin hair (large surface area), thin wall, big vacuole, many mitochondria, carrier proteins.
  • Then: water and ions pass inwards to the xylem.

B2.2g and B2.2h: Transpiration, translocation and the vessels

Transpiration in order:

  1. Water evaporates from mesophyll cell surfaces into leaf air spaces.
  2. Vapour diffuses out through open stomata, down a gradient.
  3. Water is pulled up the xylem to replace it, carrying mineral ions.

Stomata: guard cells turgid, pore open; guard cells flaccid, pore closed. Open in light for gas exchange, and shut to limit water loss.

Xylem Phloem
Carries Water, mineral ions Dissolved sugars
Direction Up only Both ways
Cells Dead, hollow, no end walls, lignin in walls Living sieve tubes, perforated end walls, companion cells
Adaptation reasoning Open tube means easy flow; lignin stops collapse Perforations let sap pass; companion cells supply energy
Process Transpiration stream Translocation (sources to sinks)

Ringing evidence: bark with phloem removed gives a swelling above the cut and starved tissue below; the leaves stay turgid because xylem is intact.

B2.2i: Factors and rate

Factor Rate of water uptake Reason to write
Light intensity up Up Stomata open wider, more vapour escapes
Temperature up Up Faster evaporation and diffusion, steeper gradient
Air movement up Up Vapour swept away, gradient kept steep
Humidity up Down Gradient smaller

Write the cause (stomata or gradient) in every explanation, and say that uptake follows loss.

B2.2j: Potometer and calculations

Method in steps

  1. Cut the shoot while submerged and join it to the apparatus underwater.
  2. Make sure every joint is airtight, and wipe the leaves.
  3. Let in a single bubble and note where it starts.
  4. Time the bubble over a fixed interval, note the distance, then average repeats.
  5. Alter one factor only; hold the rest steady.
Quantity Formula
Rate of bubble movement distance moved ÷ time (mm per minute)
Cross-sectional area of capillary π × radius²
Volume rate of uptake rate of bubble movement × area (mm³ per minute)
Percentage change (change ÷ starting value) × 100
Stomatal density mean number per field ÷ area of field (π × radius²)

Worked reminder. A bubble moves 63 mm in 7.0 minutes in a capillary of radius 0.40 mm.

rate   = 63 / 7.0 = 9.0 mm per minute
area   = pi x 0.40^2 = 0.503 mm^2
volume = 9.0 x 0.503 = 4.5 mm^3 per minute

Worked reminder (two conditions). Rate rises from 4.0 to 7.0 mm per minute: increase = 3.0, so 3.0 ÷ 4.0 × 100 = 75% increase.

Worked reminder (percentage mass loss). A shoot in a tube of water with oil on top loses 2.7 g from a starting mass of 130.0 g.

percentage loss = 2.7 / 130.0 x 100 = 2.1% (2 s.f.)

Worked reminder (stomatal density). Counts per field of view: 6, 8, 7 and 7, with a field area of 0.20 mm².

mean    = (6 + 8 + 7 + 7) / 4 = 7
density = 7 / 0.20 = 35 stomata per mm^2

Practical work you may be asked about

Practical What to remember
Root hair cells under a light microscope Thin extension of the cell; stain if needed; draw with a sharp pencil and a scale
Position of xylem and phloem in root, stem and leaf sections Xylem and phloem sit in groups called vascular bundles; the arrangement differs between organs
Stomatal impression Paint clear nail varnish on the leaf, peel it off when dry, view under the microscope, count stomata in several fields
Potometer Rate from bubble movement, then repeat and average
Sucking water through a long straw Shows how powerful the pull of the transpiration stream is

How to build a full-mark explanation

  1. Name the factor and the direction of the change.
  2. Say what changes in the leaf: stomata open or close, or the gradient steepens or flattens.
  3. Say what that does to evaporation and diffusion of water vapour.
  4. Link to the result in the question, such as a faster bubble or a larger mass loss.

Example: “A fan raises the rate because moving air removes water vapour from beside the stomata. This keeps the concentration gradient steep, so vapour diffuses out faster and water is drawn up the xylem faster.”

Reading data and graphs

  • Plot the independent variable on the horizontal axis, with a sensible scale and labelled axes with units.
  • To describe a trend, quote two values with units from the table or graph.
  • If a rate flattens, say it has levelled off and suggest another factor now limiting it, such as stomata being fully open.
  • Check repeat readings for outliers before averaging, and say which result you left out.
  • A change of a percentage is a ratio to the starting value, so doubling is +100%.

Sources and sinks

A source is where sugar is made or released, such as a photosynthesising leaf. A sink is where sugar is used or stored, such as a growing root tip, a developing fruit or a storage organ. The direction of translocation depends on which is which, which is why phloem carries sugar both up and down. Use the words source and sink in an answer about translocation.

Must-know distinctions

  • Osmosis moves water; active transport moves ions. Only active transport needs energy.
  • Transpiration is a loss of water vapour; the potometer measures uptake, which approximates it.
  • Xylem is dead and one-way; phloem is living and two-way.
  • Guard cells control the pore. Stomata are the pores.

Quick self-test

  1. Name the process by which water enters a root hair cell.
  2. Why does a root hair cell contain many mitochondria?
  3. A bubble moves 45 mm in 9.0 minutes. Calculate the rate in mm per minute.
  4. The capillary radius is 0.40 mm. Use the rate in question 3 to find the volume of water absorbed each minute, to 2 decimal places.
  5. A shoot’s tube falls from 85.0 g to 83.3 g. Calculate the percentage mass lost.
  6. Nine stomata are counted in a field of view of area 0.12 mm². Calculate the stomatal density.
  7. A rate rises from 4.0 to 7.0 mm per minute. Calculate the percentage increase.
  8. State two ways the xylem is adapted for its function.
  9. What is the role of a companion cell?
  10. Explain why transpiration slows in still, humid air.

Answers

  1. Osmosis.
  2. They release energy by respiration for active transport of ions.
  3. 45 ÷ 9.0 = 5.0 mm per minute.
  4. Area = π × 0.40² = 0.5027 mm²; 5.0 × 0.5027 = 2.51 mm³ per minute.
  5. Loss = 1.7 g; 1.7 ÷ 85.0 × 100 = 2.0%.
  6. 9 ÷ 0.12 = 75 per mm².
  7. (7.0 − 4.0) ÷ 4.0 × 100 = 75%.
  8. Any two: dead hollow cells with no end walls; lignin in the walls to stop collapse and give support.
  9. It has many mitochondria and supplies energy to load sugars into the sieve tube.
  10. Vapour stays near the leaf, so the concentration gradient is small and diffusion out of the stomata is slow.

Where marks are usually lost

  • Saying “osmosis” for mineral ion uptake.
  • Describing transpiration as water leaving the roots or the xylem, rather than vapour leaving the leaf by stomata.
  • Writing that phloem carries “food” without saying dissolved sugars.
  • Calling a potometer a transpiration meter without noting that it measures uptake.
  • Leaving out “one variable changed, others controlled” in a method.
  • Dividing by the final value in a percentage change.
  • Giving the volume rate without multiplying by cross-sectional area, or mixing mm and cm.
  • Explaining a factor with “faster” and no mention of gradient or stomata.

Official syllabus

OCR, GCSE (9-1) Biology A (Gateway Science) J247 Specification, Version 4.0 (August 2026), first assessment 2018, statements B2.2f to B2.2j.

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