Cambridge O Level Biology 5090: Human gas exchange (5090) – Practice Questions

These are original questions written for Marlbridge, for revision and practice on this content. They are not reproduced past-paper questions, and they do not replicate the exam’s exact structure, question count or mark tariffs – examination boards hold copyright in their own papers. Use these alongside the official past papers from your board or school.

These questions cover Topic 9, Human gas exchange, of the Cambridge O Level Biology 5090 syllabus for examination in 2026, 2027 and 2028: every learning outcome in section 9.1 (9.1.1 to 9.1.9), including the inspired-air and physical-activity investigations. The syllabus is not tiered, so every question applies to every candidate.

Revise first with the study guide and the revision notes. The course hub and the printable checklist list the other topics, and the free 10-minute diagnostics help you decide what to revise next. A calculator is useful for questions 4, 5, 8, 9 and 11.

Questions

1. State the approximate percentages of nitrogen and of carbon dioxide in atmospheric air. [2]

2. Identify each structure of the gas exchange system from its description.

(a) A wider region at the top of the windpipe, containing the vocal cords. [1] (b) A tube in the neck, held open by rings of cartilage. [1] (c) Narrow, branching tubes inside each lung. [1] (d) Tiny air sacs, each covered by a network of blood vessels. [1]

3. Describe two features of human gas exchange surfaces, and explain how each one increases the rate of gas exchange. [4]

4. A person takes in 500 cm³ of air in each breath. Inspired air contains 21% oxygen and expired air contains 16% oxygen.

(a) Calculate the volume of oxygen in one inspired breath. [1] (b) Calculate the volume of oxygen absorbed from one breath, assuming 500 cm³ is also breathed out. [2] (c) The person breathes 14 times per minute. Calculate the volume of oxygen absorbed per minute. [1] (d) Explain why expired air contains less oxygen than inspired air. [2]

5. A student compares inspired and expired air. Two test-tubes each contain 10 cm³ of hydrogencarbonate indicator. Inspired air is drawn through tube A and expired air is breathed out through tube B.

(a) State the colour change in the indicator when the carbon dioxide concentration increases. [1] (b) Tube B changed colour after 25 s and tube A after 350 s. Calculate how many times faster tube B changed than tube A. [2] (c) State two variables, other than the volume of indicator, that should be kept the same. [2] (d) Describe a simple test to show that expired air contains more water vapour than inspired air. [1] (e) State one safety precaution for this investigation. [1]

6. Explain how the ribs, intercostal muscles and diaphragm cause air to enter the lungs. [5]

7. During a hard cough, air is forced out of the lungs.

(a) Name the muscles that contract to pull the ribs down and in. [1] (b) Explain why the external and internal intercostal muscles are described as antagonistic. [2]

8. A student’s resting breathing rate was 14 breaths per minute. She ran for four minutes, then her breathing rate was measured every minute.

Time after running / min	0	1	2	3	4	5
Breathing rate / breaths per minute	38	31	24	18	15	14

(a) Calculate the percentage increase in breathing rate from rest to time 0. Give your answer to one decimal place. [2] (b) Describe the change in breathing rate after running. [2] (c) Explain why breathing rate increased during running. [4] (d) Suggest how the depth of breathing could be measured. [1]

9. At rest a man breathes 12 times per minute with 0.5 dm³ per breath. During cycling he breathes 30 times per minute with 2.4 dm³ per breath. Calculate the volume of air breathed per minute in each case, and how many times larger it is during cycling. [3]

10. Explain how goblet cells, mucus and ciliated cells protect the gas exchange system. [4]

11. This question is about the alveoli.

(a) The barrier between alveolar air and the blood in a capillary is 1.5 µm thick. Calculate its length, in mm, on a photomicrograph at ×2000. [2] (b) Describe how oxygen moves from the air in an alveolus into a red blood cell. [3] (c) Explain why carbon dioxide diffuses in the opposite direction. [2] (d) A student wants to show the composition of atmospheric air as a pie chart. Calculate the angle for carbon dioxide and suggest why a pie chart shows it poorly. [2]

Answers

1. Nitrogen about 78% [1]; carbon dioxide about 0.04% [1]. [2] Examiner insight: The two gases are separate points; writing 4% for carbon dioxide confuses atmospheric air with expired air and loses that mark.

2. (a) Larynx [1] (b) Trachea [1] (c) Bronchioles [1] (d) Alveoli [1] [4] Examiner insight: “Bronchi” and “bronchioles” are different structures, so a name with the wrong ending is a wrong answer, not a spelling slip; learn the endings exactly.

3. Any two pairs. Large surface area [1], so more gas can diffuse at the same time [1]. Thin surface [1], so the diffusion distance is short [1]. Also accept: good blood supply, which keeps a steep concentration gradient by carrying oxygen away; good air supply, which keeps oxygen concentration high in the alveoli. [4] Examiner insight: A feature listed with no explanation usually gains only part of the credit for that pair; “thin, so short diffusion distance” is safer than “thin, so diffusion is quicker”.

4. (a) 0.21 × 500 = 105 cm³ [1] (b) Oxygen breathed out = 0.16 × 500 = 80 cm³ [1]; absorbed = 105 − 80 = 25 cm³ [1] (c) 25 × 14 = 350 cm³ per minute [1] (d) Oxygen diffuses from the alveoli into the blood [1] and is used by cells in aerobic respiration [1]. [6] Examiner insight: Error carried forward is often allowed in linked calculation parts, so show the method in (c) even if you doubt (b); give a unit with every volume.

5. (a) From red/orange to yellow [1] (b) 350 ÷ 25 [1] = 14 times [1] (c) Any two [2]: same concentration of indicator; same starting temperature; same person; same breathing pattern or rate; same size of tube. (d) Breathe onto a cold, dry mirror and look for condensation [1]. (e) Any one [1]: wear eye protection; never suck liquid into the mouthpiece; use a clean mouthpiece for each person. [7] Examiner insight: Vague variables such as “same conditions” or “same amount” earn nothing; name the quantity (volume, concentration, temperature).

6. External intercostal muscles contract [1]; ribs move up and out [1]. Diaphragm contracts and flattens [1]. Volume of the thorax increases [1], so pressure inside falls below atmospheric pressure and air moves into the lungs [1]. [5] Examiner insight: Volume and pressure are separate points here, and an answer that says the lungs “suck in” air without a pressure change cannot earn the last mark.

7. (a) Internal intercostal muscles [1] (b) They pull the ribs in opposite directions [1]; when one set contracts the other relaxes [1]. [3] Examiner insight: Saying the muscles “work against each other” is not enough for both marks; describe the opposite movements and the contract-relax pairing.

8. (a) (38 − 14) ÷ 14 × 100 [1] = 171.4% [1] (b) The rate falls after running [1], reaching the resting rate of 14 by 5 minutes [1]. (c) Muscles contract more, so respiration increases [1]. More carbon dioxide is produced, so its concentration in the blood rises [1]. The brain detects this rise [1] and increases the rate of breathing, supplying more oxygen and removing carbon dioxide faster [1]. (d) Use a spirometer (or measure chest expansion with a tape measure) [1]. [9] Examiner insight: In (b) quoting data with units (for example “returns to 14 breaths per minute after 5 minutes”) secures the second mark; describing only “it decreases” earns one.

9. Rest: 12 × 0.5 = 6.0 dm³ per minute [1]. Cycling: 30 × 2.4 = 72 dm³ per minute [1]. 72 ÷ 6.0 = 12 times larger [1]. [3] Examiner insight: The two volumes are credited separately from the ratio, so show both even though the question ends with the comparison.

10. Goblet cells secrete mucus [1]. Mucus traps pathogens and particles [1]. Cilia on ciliated cells beat [1] and move the mucus up to the throat, away from the alveoli, where it is swallowed [1]. [4] Examiner insight: Giving the trapping role to the cilia, or saying the cilia kill bacteria, contradicts the correct points and loses those marks.

11. (a) Image size = actual size × magnification = 1.5 × 2000 = 3000 µm [1] = 3.0 mm [1] (b) Oxygen dissolves in the moist lining of the alveolus [1]. It diffuses down a concentration gradient [1] across the thin alveolar wall and capillary wall into the blood and into the red blood cell [1]. (c) The concentration of carbon dioxide in the blood is higher than in the alveolar air [1], because cells produce it in respiration; it diffuses from higher to lower concentration [1]. (d) 0.04 ÷ 100 × 360 = 0.144° [1]; the sector is far too small to see or measure [1]. [9] Examiner insight: In (a) the unit conversion is its own step; an answer of “3000 mm” shows the right method but the wrong unit, so it cannot gain full credit.

Where marks are usually lost

• Writing 4% instead of 0.04% for carbon dioxide in atmospheric air.
• Naming a feature of the alveoli without linking it to diffusion distance, surface area or concentration gradient.
• Reversing the volume-pressure link when explaining breathing in.
• Saying the internal intercostals contract during quiet breathing in.
• Leaving out units on volumes or rates.
• Forgetting to convert µm to mm in magnification answers.
• Explaining faster breathing without carbon dioxide and the brain.
• Saying the cilia trap particles or move mucus down into the lungs.
• Naming vague controlled variables in the inspired-air investigation.

Next steps

• Revision notes for this topic
• Study guide for this topic
• Movement into and out of cells practice
• Cambridge O Level Biology course hub
• Printable checklist
• All free 10-minute diagnostics
• Book a free trial class

Official syllabus

Cambridge O Level Biology 5090 syllabus for examination in 2026, 2027 and 2028 (version 4), Topic 9: Human gas exchange, published by Cambridge Assessment International Education.