Biology  by Bradford
IB Biology 2025 · Theme B · Form and function

B3.1 Gas exchange. Practice questions with markscheme.

48 original IB-style questions on B3.1, written from the 2025 guide: 21 multiple-choice, 13 short-answer, 6 data-based, 3 extended-response part, 3 drawing, 2 labelling. Below is a 20-mark standard-level practice paper built from them, ready to hand out as a class quiz or homework, or to sit yourself and mark against the scheme. Print it, project it, or build a fresh one on the same topic.

What the guide asks for

10 statements at SL and HL, 3 additional higher level.

  1. B3.1.1SL / HL Gas exchange as a vital function in all organisms
  2. B3.1.2SL / HL Properties of gas-exchange surfaces
  3. B3.1.3SL / HL Maintenance of concentration gradients at exchange surfaces in animals
  4. B3.1.4SL / HL Adaptations of mammalian lungs for gas exchange
  5. B3.1.5SL / HL Ventilation of the lungs
  6. B3.1.6SL / HL Measurement of lung volumes
  7. B3.1.7SL / HL Adaptations for gas exchange in leaves
  8. B3.1.8SL / HL Distribution of tissues in a leaf
  9. B3.1.9SL / HL Transpiration as a consequence of gas exchange in a leaf
  10. B3.1.10SL / HL Stomatal density
  11. B3.1.11HL Adaptations of foetal and adult haemoglobin for the transport of oxygen
  12. B3.1.12HL Bohr shift
  13. B3.1.13HL Oxygen dissociation curves as a means of representing the affinity of haemoglobin for oxygen at different oxygen concentrations

In the bank for B3.1

  • 21 multiple-choice
  • 13 short-answer
  • 6 data-based
  • 3 extended-response part
  • 3 drawing
  • 2 labelling
  • 6 higher level only

Every question is original and tagged to a guide statement. See the whole bank →

Make your own

The practice paper

Take it on screen → Build a fresh paper Paper code BbB-EAAAQAAAABQAOrN-
Biology · topic quiz
Standard level · topic practice, not an exam format
30 minutes20 marks

Paper code: this paper was generated, so the code is its recipe. Enter it at biologybybradford.com/exam-maker to rebuild this exact paper and its markscheme.

Covers B3.1 Gas exchange
Name:
1.

A plant in dry soil closes its stomata during the afternoon. Which statement describes the cost and the benefit of this response?

[1]
  1. Water loss by transpiration is reduced, and photosynthesis is unaffected because oxygen can still leave the leaf
  2. Uptake of carbon dioxide is increased, but the leaf can no longer be cooled by evaporation
  3. Water loss by transpiration is reduced, but the uptake of carbon dioxide for photosynthesis is also reduced
  4. Water uptake by the roots is increased, but the plant can no longer release oxygen
2.

Most stomata occur on the lower surface of (dorsiventral) leaves. Which advantage does this give?

[1]
  1. More light reaches the guard cells there
  2. The shaded lower surface is cooler, so less water is lost
  3. CO₂ is denser than air and collects beneath the leaf
  4. Guard cells cannot open on the upper epidermis
3.

Which properties are needed by all gas-exchange surfaces?
I. Large surface area
II. Thin barrier giving a short diffusion distance
III. Maintained concentration gradients

[1]
  1. I and II only
  2. I, II and III
  3. II and III only
  4. I and III only
4.

Stomatal density was determined from leaf casts taken from ten trees growing at different distances from a woodland edge, and the mean density for each tree was plotted against distance. The points are scattered but show a downward trend. Describe how a line of best fit should be drawn on this scatter graph, and state one conclusion that may not be drawn from it.

[3]
5.

All organisms must exchange gases with their environment. Explain why gas exchange becomes a greater challenge as the size of an organism increases.

[3]
6.

The diagram shows a transverse section of a dicotyledonous leaf. Identify the structures labelled I–IV.

[4]
waxy cuticleupper epidermisIIIIIIphloemIVlower epidermisair spaceTransverse section of a dicotyledonous leaf (cells simplified)
I.
II.
III.
IV.
7.

Stomatal density was measured on the upper and lower surfaces of leaves of four plant species growing in different habitats. The rate of transpiration per unit leaf area was measured for each species under the same conditions of light, temperature and humidity.

SpeciesHabitatStomatal density, upper surface / mm⁻²Stomatal density, lower surface / mm⁻²Transpiration rate / mg cm⁻² h⁻¹
Water lilyfloating leaves on a pond46003.6
Sunfloweropen field851753.5
Oakwoodland03402.8
Oleanderdry scrubland0900.9
(a)Calculate the total stomatal density (both surfaces) of the sunflower leaf.[1]
(b)Describe the relationship between total stomatal density and transpiration rate.[2]
(c)Explain why the water lily has stomata only on the upper surface of the leaf.[1]
(d)Explain how the low stomatal density of oleander is an adaptation to its habitat.[2]
(e)Outline how stomatal density could be measured for one of these leaves.[1]

Original practice questions © Biology by Bradford · CC BY-NC-SA 4.0 · Not affiliated with or endorsed by the International Baccalaureate Organization.
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Show the markscheme

Markscheme BbB-EAAAQAAAABQAOrN-

One mark per point; / separates alternative wording within a point, OR separates alternative answers, words in brackets are not required, underlined words are essential. OWTTE = or words to that effect.

1. B3.1.9 [1]
  • C: stomata are the route for both water vapour out and CO₂ in; closing them (abscisic acid) conserves water at the cost of photosynthesis;
2. B3.1.7 [1]
  • B: placing stomata away from direct sun lowers evaporation from the pores while still admitting CO₂;
3. B3.1.2 [1]
  • B: all three, plus moisture (gases dissolve before diffusing) and (in many animals) a good blood supply;
4. B3.1.10 [3 max]
  • draw a single straight line, or one smooth curve, following the overall trend of the points;
  • position it so the points are balanced about it, with roughly equal numbers above and below and the distances from the line as small as possible;
  • do not join the points dot to dot / do not draw a zigzag line passing through every point;
  • do not force the line through the origin unless the origin is itself a measured or justified point;
  • extend the line only across the range of distances actually measured;
  • may not be concluded: that distance from the edge causes the change in stomatal density, since a correlation alone does not establish cause;
  • may not be concluded: a stomatal density for a distance outside the range measured;

Award [1 max] for the conclusion that may not be drawn. Design move: name what the method could not see — the line is a summary of the data, not a licence to read beyond them.

5. B3.1.1 [3]
  • as an organism increases in size its volume increases faster than its surface area, so its surface area-to-volume ratio decreases, OWTTE;
  • the need for oxygen / production of carbon dioxide depends on the volume (number of respiring cells), whereas the exchange of gases depends on the surface area;
  • (so) the surface of a large organism is too small to supply all of its cells by diffusion across the body surface;
  • the distance from the centre of the organism to its exterior increases, so diffusion (which is slow over long distances) cannot deliver gases fast enough to inner cells;
  • (hence) large organisms need a specialized gas-exchange surface / organ and a transport system, OWTTE;

Do not accept 'larger organisms need more oxygen' without reference to surface area, volume or distance.

6. B3.1.7, B3.1.8 [4]
  • I, palisade mesophyll;
  • II, spongy mesophyll;
  • III, xylem;
  • IV, guard cell / stoma;
7. B3.1.10, B3.1.9
  • (a) [1]
    • 260 mm⁻²;
  • (b) [2]
    • generally the higher the stomatal density, the higher the rate of transpiration (oleander 90 mm⁻² and 0.9 mg cm⁻² h⁻¹; water lily 460 mm⁻² and 3.6 mg cm⁻² h⁻¹);
    • the relationship is not exact: oak has more stomata (340) than sunflower (260) but a lower transpiration rate, OWTTE;
  • (c) [1]
    • the lower surface is in contact with water, so stomata there could not exchange gases with the air / would be blocked by water;
  • (d) [2 max]
    • in a dry habitat water is scarce, so the plant must reduce water loss;
    • fewer stomata means less water vapour diffuses out of the leaf / lower transpiration (0.9 mg cm⁻² h⁻¹);
    • the trade-off is a lower rate of CO₂ uptake for photosynthesis, OWTTE;
  • (e) [1 max]
    • make an impression of the leaf surface (e.g. with clear nail varnish, peeled off with tape) and view it under a microscope, counting stomata in a field of view of known area;
    • repeat counts in several fields of view / leaves and calculate a mean per mm², OWTTE;

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