IB Biology · Theme B Form and function · Organisms
B3.1 Gas exchange
Every organism exchanges gases; the bigger it is, the harder diffusion alone has to work. Good exchange surfaces are permeable, thin, moist and large, with gradients kept steep. Lungs and leaves solve the same problem; haemoglobin then fine-tunes delivery at HL.
Compiled from the IB Biology guide (first assessment 2025, updated May 2026 for 2028) and our question bank ·
Specialist review in progress
· How these pages are made
Assessed in Paper 1A (multiple choice), Paper 1B (data-based) and Paper 2 (short and extended response). IB Biology guide (first assessment 2025, updated May 2026 for 2028).
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B3.1.1 Gas exchange gets harder as organisms get bigger
Gas exchange is diffusion of gases across a surface between organism and environment.
All organisms do it; a unicell exchanges across its plasma membrane.
As size grows, the surface area-to-volume ratio falls: doubling the side halves it.
The diffusion distance from centre to exterior grows too; diffusion time rises steeply with distance.
Students often think a larger organism has a larger ratio because it has more surface. In fact volume grows faster, so surface per unit volume falls.
Students often think gas exchange means breathing with lungs or gills. In fact every organism exchanges gases, however it does it.
B3.1.2 What a good exchange surface looks like
Permeable to oxygen and carbon dioxide, which pass through bilayers freely.
A thin tissue layer, so the diffusion distance is short.
Moist, so gases dissolve in the film before they cross the membranes.
A large surface area, made by folding or subdividing, so much gas crosses at once.
Students often think moisture is only there to keep cells alive. In fact gases must dissolve in it before they can cross the membranes.
Students often think a bigger organ simply has more area. In fact area comes from subdivision: millions of alveoli, or gill filaments and lamellae.
B3.1.3 Keeping the gradient steep
Net diffusion continues only while a concentration gradient exists; equilibrium stops it.
Ventilation brings fresh air to lungs or water over gills, keeping the outside oxygen-rich.
Dense capillary networks and continuous blood flow replace oxygenated blood with deoxygenated blood.
Ventilation moves the medium; gas exchange is the diffusion itself.
Students often think a gradient, once set up, lasts by itself. In fact diffusion erodes it; ventilation and blood flow must rebuild it constantly.
Students often think blood should linger in capillaries to fill with oxygen. In fact lingering blood reaches equilibrium; moving blood keeps loading.
B3.1.4 How mammalian lungs are adapted
Millions of alveoli give a huge surface area; each wall is one cell thick.
Extensive capillary beds surround each alveolus, so the diffusion distance is very short.
A branched network of bronchioles packs all those alveoli into the thorax.
Surfactant lowers the surface tension of the water film, so alveoli do not collapse.
Students often think surfactant is a mucus for trapping dust. In fact it reduces surface tension so alveoli stay open and expand easily.
Students often think exchange happens all along the airways. In fact only alveolar walls are thin enough; bronchi and bronchioles just conduct air.
B3.1.5 How the lungs are ventilated
Lungs have no skeletal muscle; pressure changes made by the thorax move air.
Inspiration: the diaphragm flattens, external intercostals lift the ribs; volume up, pressure down, air in.
Quiet expiration is mostly passive: muscles relax and the lungs recoil.
Forced expiration: internal intercostals pull ribs down; abdominal muscles push organs up against the diaphragm.
Students often think lungs suck air in and squeeze it out themselves. In fact muscles change thorax volume; the lungs follow.
Students often think the diaphragm pushes up to force air out. In fact it relaxes into a dome; only the abdominal muscles push, during forced expiration.
B3.1.6 Measuring lung volumes
Tidal volume: air moved in one relaxed breath, about 0.5 dm³ at rest.
Inspiratory reserve and expiratory reserve: the extra air beyond a normal breath in or out.
Vital capacity: deepest inhalation then fullest exhalation; equals tidal volume plus both reserves.
A spirometer traces volume against time; a bag or inverted water-filled container also works.
Students often think vital capacity is just tidal volume plus inspiratory reserve. In fact it includes the expiratory reserve too.
Students often think vital capacity is the total lung volume. In fact air always remains after a full exhalation, so the lungs hold more.
B3.1.7 How a leaf is adapted for gas exchange
The waxy cuticle blocks water and gases, so gases use the stomata.
The epidermis protects and secretes the cuticle; most stomata are on the lower surface.
Spongy mesophyll cells border interconnected air spaces; their moist surfaces are the exchange surface.
Guard cells swell and curve apart to open the pore; veins bring water, remove sugars.
Students often think guard cells contract like muscle. In fact they open the stoma by becoming turgid; their uneven walls bend them apart.
Students often think veins carry gases like blood vessels. In fact xylem brings water, phloem takes sugar; gases move by diffusion only.
B3.1.8 Tissues in a leaf section
A plan diagram shows tissue outlines and positions, with no individual cells.
From top: upper epidermis, palisade mesophyll, spongy mesophyll, lower epidermis.
Veins sit within the mesophyll: xylem towards the upper surface, phloem towards the lower.
Palisade cells are elongated, packed, chloroplast-rich, placed to catch light from above.
Students often draw a few cells in a plan diagram. In fact it shows regions only; drawing cells is a different kind of diagram.
Students often put phloem above xylem in a vein. In fact xylem is uppermost; phloem lies beneath it.
B3.1.9 Transpiration comes with gas exchange
Transpiration is loss of water vapour, mainly through open stomata.
Stomata open to admit carbon dioxide; while open, vapour diffuses out down its gradient.
Rate rises with temperature, wind and light; it falls with humidity.
Wind strips the humid layer at the leaf surface; light opens the stomata.
Students often think plants transpire on purpose to pull water up. In fact it is an unavoidable side effect of letting carbon dioxide in.
Students often think humidity increases transpiration. In fact it shrinks the vapour gradient, so the rate falls.
B3.1.10 Counting stomata
Stomatal density is stomata per unit area, usually per mm².
Make a leaf cast with clear varnish, peel it with tape, and mount it.
Count stomata in several high-power fields, take the mean, then divide by one field's area.
Counts vary from place to place; replicates reveal this and make the estimate reliable.
Students often total all counts and divide by one field's area. In fact you divide the mean count per field by that area.
Students often think differing replicates mean the method failed. In fact they show natural variability, which is exactly why replicates are taken.
B3.1.11 Haemoglobin, adult and foetal (HL) HL
2028 guide: scope reduced — Reported (unverified secondary source) that the 2028 guide no longer requires myoglobin. The 2025 guidance for this statement names only foetal and adult haemoglobin, cooperative binding and allosteric binding of carbon dioxide, so no object in this bank mentions myoglobin; all items on this statement are valid under both guides. Candidates sitting May/Nov 2026 or 2027 exams still need the fuller 2025 scope.
Haemoglobin has four subunits, each with a haem group whose iron binds one oxygen.
Cooperative binding: one oxygen bound changes the shape, raising the other haems' affinity.
Carbon dioxide binds allosterically, away from the haems, lowering affinity for oxygen.
Foetal haemoglobin has higher affinity, so it loads oxygen where maternal haemoglobin unloads it.
Students often think each haem binds independently. In fact the first binding raises the affinity of the rest.
Students often think foetal haemoglobin has lower affinity to release oxygen faster. In fact it is higher, so it can take oxygen from maternal blood.
B3.1.12 The Bohr shift (HL) HL
More carbon dioxide lowers haemoglobin's affinity for oxygen: the curve shifts right.
Carbon dioxide binds allosterically and lowers pH via carbonic acid; both change the conformation.
So at any given oxygen concentration, more oxygen dissociates from haemoglobin.
Active tissues make more carbon dioxide, so they receive more oxygen.
Students often think affinity rises where more oxygen is needed. In fact it falls, and that is what releases the oxygen.
Students often think tissues get extra oxygen only because they used it up. In fact carbon dioxide itself lowers affinity, adding to the low-oxygen effect.
B3.1.13 Reading an oxygen dissociation curve (HL) HL
From 2028 this is supplied in the Biology data booklet — you need to recognise and interpret it, not reproduce it from memory.
The curve plots percentage saturation against oxygen concentration, not time.
It is S-shaped because of cooperative binding: slow start, steep middle, plateau near full.
The steep region matches respiring-tissue concentrations, so a small fall releases much oxygen.
A curve to the left means higher affinity; to the right, lower.
Students often read the curve as saturation over time along the circulation. In fact the x-axis is oxygen concentration.
Students often think the plateau is where oxygen is delivered. In fact delivery happens where saturation falls, in the steep part.
Diagnostic a bearings check, not a test
10 questions, one per part of the topic where we can. Answer them, then see which statements you own and which to read.
1 Why does gas exchange become a greater challenge as organisms increase in size?
Answer and reasoning
The surface area-to-volume ratio decreases, and the distance from the centre of the organism to its exterior increases. — Volume grows faster than surface area, so a large organism has proportionally less surface across which to exchange gases, and its inner cells are too far from the surface for diffusion alone to supply them. Both effects make specialized gas-exchange surfaces and transport systems necessary.
The surface area-to-volume ratio increases, so far more gas has to cross each unit of the exchange surface. — A student who reasons that a bigger organism has 'more surface' picks this. Total surface area does increase, but the ratio of surface to volume decreases, so there is less surface per unit of respiring tissue, not more.
Only animals with lungs or gills exchange gases, so the challenge arises once an organism is big enough to have them. — A student who equates gas exchange with breathing picks this. Gas exchange is a vital function in all organisms, including unicellular organisms and plants; the challenge grows with size because the surface area-to-volume ratio falls and the diffusion distance rises, not because small organisms do not exchange gases.
A large organism uses more oxygen in total; the distance gases must diffuse to reach its cells makes no difference. — A student who thinks diffusion is fast over any distance picks this. Diffusion time rises steeply with distance, and the distance from centre to exterior grows with size, so distance is one of the two reasons the challenge increases.
2 Why is the surface of a gas-exchange organ kept moist?
Answer and reasoning
The moisture is not needed for exchange; it only keeps the cells of the surface from drying out and dying. — A student who pictures gases crossing the surface as gases picks this. Keeping cells alive is a benefit, but the moisture is essential to exchange itself: gases can only cross the plasma membranes in solution.
The moisture forms a sticky layer that traps dust and pathogens before they reach the exchange surface. — A student who confuses the moist film with the mucus of the airways picks this. Trapping particles is the job of mucus in the trachea and bronchi; the film on the exchange surface is there so that gases can dissolve.
Oxygen and carbon dioxide dissolve in the film of water before diffusing across the cell membranes. — Gases cross plasma membranes only in solution. The thin film of moisture lets oxygen dissolve and then diffuse into the cells and blood, and lets carbon dioxide dissolve and diffuse out; moisture is therefore a property of every gas-exchange surface.
The water is absorbed into the blood across the surface, carrying the dissolved oxygen in with it. — A student who pictures oxygen being taken in with water picks this. The film of water stays on the surface; only the dissolved oxygen diffuses across the membranes, down its concentration gradient, into the blood.
3 A fish rests in still, well-oxygenated water. What keeps the oxygen concentration gradient across its gill surface steep?
Answer and reasoning
The water around the gills contains dissolved oxygen, so the gradient is maintained by the water itself. — A student who thinks gills need no ventilation picks this. The layer of water in contact with the gill surface is depleted of oxygen within moments, and diffusion through still water is very slow, so the fish must pump fresh water over the gills.
Once oxygen has diffused into the blood the gradient is fixed and does not need to be maintained further. — A student who believes a gradient looks after itself picks this. Diffusion destroys the gradient that drives it, so it must be continuously renewed by ventilation and blood flow.
Blood is held in the gill capillaries until it is saturated, so that every red blood cell loads fully. — A student who thinks blood loads more if it lingers picks this. Blood that stays approaches equilibrium with the water and net diffusion stops; continuous flow replaces it with deoxygenated blood, keeping the gradient steep.
Water is pumped over the gills while blood flows continuously through their dense capillaries. — Ventilation with water keeps the outer side of the gill surface rich in oxygen, and continuous blood flow through the dense network of capillaries keeps the inner side low in oxygen by carrying oxygenated blood away. Together they maintain the concentration gradient.
4 What is the function of surfactant in the alveoli of a mammalian lung?
Answer and reasoning
It traps dust and microorganisms in a sticky layer so that they can be swept out of the lungs. — A student who assumes every airway secretion is mucus picks this. Trapping particles is the role of mucus in the trachea and bronchi; surfactant is a phospholipid layer with a physical role at the alveolar surface.
It lowers the surface tension of the moisture lining the alveoli, so they do not collapse and inflate easily. — Water has a high surface tension that would pull the walls of each tiny alveolus inward and collapse it. Surfactant, secreted onto the moist lining, reduces this surface tension so the alveoli stay open and expand with little effort during inspiration.
It is the film of water absorbed into the blood, carrying the oxygen dissolved in it across the alveolar wall. — A student who pictures oxygen entering with absorbed water picks this. The moist film stays on the alveolar surface and only the dissolved oxygen diffuses across; surfactant is a phospholipid layer on that film that lowers its surface tension.
It dries the alveolar lining so that gases cross the wall as gases rather than having to dissolve first. — A student who thinks moisture hinders gas exchange picks this. The lining must stay moist because gases cross membranes only in solution; surfactant reduces the surface tension of that moisture rather than removing it.
5 Which statement correctly defines vital capacity?
Answer and reasoning
The maximum volume that can be inhaled in one breath from the end of a normal exhalation. — A student who pictures 'capacity' as filling up from the resting position picks this. This is tidal volume plus inspiratory reserve; vital capacity also includes the expiratory reserve that can be forced out below a normal breath.
The maximum volume that can be exhaled after a normal, resting inhalation. — A student who focuses on the exhalation into the spirometer and forgets the maximal inhalation first picks this. This is tidal volume plus expiratory reserve, which leaves out the inspiratory reserve.
The maximum volume of air that can be exhaled after the deepest possible inhalation. — Vital capacity runs from the peak of a maximal inhalation to the end of a maximal exhalation, and equals tidal volume plus inspiratory reserve plus expiratory reserve. It is measured by inhaling fully and then exhaling fully into a spirometer.
The total volume of air in the lungs when they are inflated as fully as possible. — A student who reads 'capacity' as the whole contents of the lungs picks this. The lungs cannot be emptied completely by exhaling, so the total lung volume is larger than the vital capacity that a spirometer can measure.
6 Which adaptation of a leaf allows carbon dioxide to reach the photosynthesizing cells inside it?
Answer and reasoning
The cuticle and epidermis are permeable to gases, so carbon dioxide diffuses in across the whole surface of the leaf. — A student who treats the whole leaf surface as the exchange surface picks this. The waxy cuticle is almost impermeable and the epidermal cells are tightly packed, so gases enter only through the stomata.
The veins carry carbon dioxide into the leaf from the stem and roots and take the oxygen produced away. — A student who models plant veins on blood vessels picks this. Xylem delivers water and phloem removes sugars; gases are not transported in veins but move through the leaf by diffusion.
The palisade mesophyll lies just beneath the upper epidermis, so its cells absorb the gas from the outside air. — A student who assumes the photosynthetic cells must be the exchange surface picks this. Palisade cells are covered by the epidermis and cuticle; they receive carbon dioxide from the air spaces of the spongy mesophyll beneath them.
Air spaces between the spongy mesophyll cells connect to the stomata, so the gas diffuses to every cell. — Carbon dioxide enters through the stomata and diffuses through the interconnected air spaces of the spongy mesophyll, dissolving in the moist surfaces of the mesophyll cells, including the palisade cells above. The air spaces give the leaf a large internal exchange surface.
7 Which statement about a plan diagram of a transverse section of a dicotyledonous leaf is correct?
Answer and reasoning
It shows a few representative cells of each tissue in outline, including their nuclei and chloroplasts. — A student who assumes every microscope drawing shows cells picks this. Individual cells belong in a high-power drawing; a plan diagram shows each tissue as a labelled region bounded by lines.
It shows the outline and position of each tissue, with the palisade mesophyll directly beneath the upper epidermis. — A plan diagram is a low-power drawing of tissue regions, not cells. From the upper surface down it shows upper epidermis, palisade mesophyll, spongy mesophyll and lower epidermis, with veins in the mesophyll; the palisade layer lies immediately under the upper epidermis where light enters.
It shows each vein with phloem above the xylem, because the phloem lies nearer to the upper surface. — A student who transfers the stem arrangement, phloem outside xylem, picks this. In a leaf vein the xylem lies toward the upper surface and the phloem toward the lower surface.
It shows the spongy mesophyll directly beneath the upper epidermis, above the palisade mesophyll. — A student who has learnt the layers as a list without their positions picks this. The palisade mesophyll lies under the upper epidermis and the spongy mesophyll below it, above the lower epidermis.
8 A student counts the stomata visible in five different fields of view on the lower epidermis of a leaf: 22, 26, 19, 27 and 21. The area of one field of view at high power is 0.20 mm². What is the stomatal density?
Answer and reasoning
115 per mm² — The mean count is (22 + 26 + 19 + 27 + 21) ÷ 5 = 23 stomata per field. Dividing by the area of a field, 23 ÷ 0.20 mm², gives 115 stomata per mm².
575 per mm² — A student who pools the counts and divides the total, 115, by the area of one field picks this. The 115 stomata were counted over five fields, so either the mean count (23) is divided by 0.20 mm², or the total by 1.0 mm².
110 per mm² — A student who thinks one count is enough uses only the first field: 22 ÷ 0.20. The counts vary from 19 to 27 because the leaf is variable, which is why the replicates are averaged before the density is calculated.
4.6 per mm² — A student who multiplies the mean count by the area, 23 × 0.20, picks this. Density is a number per unit area, so the count is divided by the area; 4.6 stomata per mm² is far below any real leaf.
9 What is meant by cooperative binding of oxygen to haemoglobin? HL
Answer and reasoning
Each of the four haem groups binds oxygen independently, with the same affinity whether or not the others are occupied. — A student who pictures four separate sockets picks this. Independent binding would give a simple saturating curve; the haem groups interact through the protein, so binding at one raises the affinity of the rest.
Oxygen and carbon dioxide share the haem groups, taking turns to bind as their concentrations change in the blood. — A student who thinks carbon dioxide competes with oxygen at the haem groups picks this. Carbon dioxide binds allosterically, at sites on the polypeptide chains, and cooperative binding concerns oxygen alone.
Binding of oxygen to one haem group changes the conformation of the protein so the remaining haem groups bind oxygen more readily. — Haemoglobin has four subunits, each with a haem group. When the first oxygen binds, the shape of the whole molecule shifts and the affinity of the other haem groups rises, so the second, third and fourth oxygen molecules bind progressively more easily. This is the basis of the S-shaped dissociation curve.
Haemoglobin molecules bind to one another once loaded, so oxygen is carried through the blood in molecular clusters. — A student who reads 'cooperative' as cooperation between separate molecules picks this. The cooperation is within one molecule, between its four subunits; haemoglobin molecules do not join together.
10 During vigorous exercise a muscle produces large amounts of carbon dioxide. How does this affect the delivery of oxygen to the muscle? HL
Answer and reasoning
Carbon dioxide lowers haemoglobin's affinity for oxygen, so more oxygen dissociates at the oxygen concentration in the muscle. — This is the Bohr shift. Carbon dioxide binds allosterically to haemoglobin and lowers blood pH; both change the protein's conformation and reduce its affinity for oxygen. The dissociation curve shifts to the right, so a larger fraction of the oxygen carried is released in the tissue producing the most carbon dioxide.
Haemoglobin's affinity for oxygen rises in the working muscle so that it can hold on to more oxygen for delivery to the muscle cells. — A student who thinks greater need means greater affinity picks this. Higher affinity would keep oxygen bound to haemoglobin; the Bohr shift lowers the affinity so that more oxygen is released to the working muscle.
Only the low oxygen concentration in the working muscle causes extra unloading; carbon dioxide itself has no effect on haemoglobin. — A student who applies only diffusion down a gradient picks this. Low oxygen concentration does promote unloading, but carbon dioxide additionally lowers haemoglobin's affinity, so at the same oxygen concentration more oxygen dissociates.
Carbon dioxide binds to the haem groups in place of oxygen, forcing the oxygen molecules off the haemoglobin. — A student who generalizes from carbon monoxide picks this. Carbon dioxide does not bind at the haem groups; it binds allosterically to the polypeptide chains, which changes the protein's shape and lowers its affinity for oxygen.
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13 more questions. Every wrong answer here is a real misconception, and you see why it is wrong straight away.
1 Three cubes are used to model organisms of different sizes. Their surface area-to-volume ratios are: side 1 mm, 6 mm⁻¹; side 2 mm, 3 mm⁻¹; side 4 mm, 1.5 mm⁻¹. Which conclusion is supported by these values?
Answer and reasoning
The largest cube has the most surface per unit of volume, because its total surface area is the greatest. — A student who confuses total surface area with the ratio picks this. The 4 mm cube does have the largest total area, but the data show it has the least surface per unit volume: 1.5 mm⁻¹ against 6 mm⁻¹ for the smallest cube.
Each doubling of the side halves the ratio, leaving proportionally less surface to supply each unit of volume. — From 6 to 3 to 1.5 mm⁻¹, the ratio halves each time the side doubles, because surface area rises fourfold while volume rises eightfold. A larger organism therefore has less exchange surface for each unit of respiring tissue, which is why gas exchange becomes more challenging with size.
The ratio should have stayed constant as the cubes were scaled up, so the values must contain a measurement error. — A student who thinks area and volume scale by the same factor picks this. Area scales with length squared and volume with length cubed, so the ratio must fall; the values are exactly what the geometry predicts.
Only the fall in the ratio matters; the distance from the centre to the surface has no effect on diffusion. — A student who believes diffusion is equally fast over any distance picks this. The larger cube's centre is further from its surface, and diffusion time rises steeply with distance, so distance is a second, independent reason larger organisms need transport systems.
2 Suppose that blood flow through the gill capillaries of an anaesthetized fish is stopped while water continues to flow over the gills, and that oxygen uptake across the gills falls almost to zero within a minute. Which explanation is best supported?
Answer and reasoning
With no blood flow, oxygen built up in the stationary blood until its concentration matched the water's, so the gradient was lost. — This is the correct explanation. Once the blood in the capillaries was as rich in oxygen as the water, there was no concentration gradient and net diffusion stopped; continuous blood flow normally removes oxygenated blood and keeps the gradient steep.
Stopping blood flow cannot affect the gradient across the gill; the water flowing over the gills must have run out of oxygen. — A student who thinks the gradient is fixed once established picks this. Fresh water continued to flow, so the outer side of the gradient was maintained; it was the inner side, the blood, that came to equilibrium when flow stopped.
Blood held still in the capillaries should load more fully, so the fall must have been caused by damage to the gill surface. — A student who believes stationary blood absorbs more picks this. Stationary blood equilibrates within seconds and then absorbs nothing; the result is exactly what continuous flow normally prevents, not evidence of damage.
Oxygen enters with water absorbed into the blood, and once the blood stopped moving no more water could be taken up. — A student who pictures oxygen entering dissolved in absorbed water picks this. Water is not absorbed at the gill; dissolved oxygen diffuses across the membranes on its own, and it stopped because the gradient was lost.
3 Mammalian lungs contain a branched network of bronchioles ending in millions of alveoli. How does this branching contribute to gas exchange?
Answer and reasoning
It multiplies the area of airway wall across which gases are exchanged into the blood as air travels inward. — A student who thinks the whole airway tree exchanges gases picks this. Bronchiole walls are too thick and too far from capillaries for exchange; the branching matters because of the number of alveoli it delivers air to.
It lets the lungs grow larger, and a larger organ must have a bigger surface for gas exchange. — A student who equates size with surface area picks this. Enlarging an organ raises its volume faster than its surface; the large area comes from subdivision into millions of alveoli, not from the lungs being big.
It packs a very large alveolar surface, each part close to a capillary bed, into the volume of the thorax. — Repeated branching ends in millions of alveoli, each thin-walled and wrapped in capillaries. The combined alveolar surface of tens of square metres is fitted into a chest cavity of a few litres, giving the high surface area that gas exchange needs.
It houses the muscle that contracts to pump air in and out of the alveoli found at the end of every branch. — A student who thinks the lungs move air by their own muscle picks this. Air is moved by the diaphragm and intercostal muscles changing the volume of the thorax; the bronchioles conduct air, they do not pump it.
4 Which sequence of events draws air into the lungs during inspiration?
Answer and reasoning
Air entering the lungs inflates them, and the inflating lungs push the ribs outward and the diaphragm downward. — A student who reverses cause and effect, as in the balloon model, picks this. The ribs and diaphragm move first, lowering the pressure in the lungs; air flows in because of that pressure difference.
The muscular walls of the lungs contract, widening the alveoli and sucking air in through the trachea. — A student who thinks the lungs are muscular picks this. Lungs contain no skeletal muscle; the thorax is enlarged by the diaphragm and external intercostal muscles, and the lungs expand passively with it.
The diaphragm relaxes and rises into a dome, and the ribs move up so that air is drawn in to fill the space. — A student who pictures the diaphragm's dome as its active state picks this. A relaxed, domed diaphragm reduces thorax volume and belongs to expiration; in inspiration the diaphragm contracts and flattens.
The diaphragm and external intercostals contract, thorax volume rises and lung pressure falls. — Contraction of the diaphragm and external intercostal muscles increases the volume of the thorax. The pressure inside the lungs falls below atmospheric pressure, and air flows in down the pressure gradient.
5 A person breathes out as hard as possible. What is the role of the abdominal muscles in this forced expiration?
Answer and reasoning
They contract, pushing the abdominal organs up against the relaxed diaphragm so that the volume of the thorax decreases. — Contraction of the abdominal muscles raises the pressure in the abdomen and forces the organs upward against the diaphragm, which has relaxed into its dome. Together with the internal intercostal muscles pulling the ribs down and in, this reduces the thorax volume and expels air.
They relax so that the diaphragm can rise passively, and so they have no active role in forcing the air out. — A student who thinks the abdominal muscles work only during inspiration picks this. They are relaxed during inspiration; in forced expiration they contract, and it is their contraction that pushes the diaphragm upward.
They contract together with the diaphragm, which pushes upward to squeeze the air out of the lungs from below. — A student who thinks the diaphragm contracts to push upward picks this. A muscle cannot push; the diaphragm relaxes in expiration and is moved up by the abdominal organs beneath it.
They have no role, because the lungs themselves contract to expel the air once the person decides to breathe out. — A student who believes the lungs are muscular picks this. The lungs cannot contract; forced expiration depends on the abdominal and internal intercostal muscles reducing the volume of the thorax.
6 A student's spirometer readings are: tidal volume 0.5 dm³, inspiratory reserve volume 3.0 dm³, expiratory reserve volume 1.2 dm³. What is the student's vital capacity?
Answer and reasoning
4.2 dm³ — A student who thinks each reserve already includes the tidal volume adds only the two reserves: 3.0 + 1.2. The reserves are the extra volumes beyond a normal breath, so the tidal volume of 0.5 dm³ must be added as well.
4.7 dm³ — Vital capacity is the maximum volume that can be exhaled after the deepest inhalation, which is tidal volume + inspiratory reserve + expiratory reserve = 0.5 + 3.0 + 1.2 = 4.7 dm³.
3.5 dm³ — A student who takes vital capacity to be the deepest breath in from rest adds tidal volume and inspiratory reserve only: 0.5 + 3.0. The expiratory reserve of 1.2 dm³ can also be exhaled and is part of the vital capacity.
1.7 dm³ — A student who measures only the exhalation after a normal breath in adds tidal volume and expiratory reserve: 0.5 + 1.2. Vital capacity must start from a maximal inhalation, so the inspiratory reserve of 3.0 dm³ is included.
The guard cells take up water and swell, and their unevenly thickened walls make them bend apart. — When guard cells gain water they swell. Because the wall next to the pore is thicker than the outer wall, the swelling bends each cell outward, away from its partner, and the pore between them opens; loss of water reverses this and closes the stoma.
The guard cells contract like muscle cells, pulling the two edges of the pore apart from one another. — A student who supplies a muscle mechanism for anything that opens and closes picks this. Plant cells have no contractile machinery; guard cells change shape through changes in turgor.
The guard cells lose water and shrink, leaving a gap between them through which the gases can pass. — A student who reasons that smaller cells leave a bigger hole picks this. Loss of water makes the guard cells flaccid and closes the stoma; it is swelling that opens it.
The plant needs to lose water to pull the transpiration stream up the xylem, and that need triggers opening. — A student who treats transpiration as the purpose of stomata picks this. Stomata open when guard cells become turgid, in response to light and water availability, so that carbon dioxide can enter; the water loss and the pull on the xylem are consequences of opening, not its cause.
8 Why is transpiration described as a consequence of gas exchange rather than as a process the leaf carries out for its own sake?
Answer and reasoning
The plant opens its stomata in order to lose water, because transpiration is needed to pull water up through the xylem. — A student who treats the transpiration stream as the purpose of stomata picks this. Stomata open to admit carbon dioxide; the pull on the xylem is a useful consequence of the water loss, not the reason the stomata open.
Water vapour escapes across the cuticle and epidermis wherever gases are exchanged over the leaf surface. — A student who thinks gases and water cross the whole leaf surface picks this. The waxy cuticle is almost impermeable; both gas exchange and nearly all water loss take place through the stomata.
Stomata must be open for carbon dioxide to enter, and water vapour inevitably diffuses out through the same pores. — Photosynthesis requires carbon dioxide, which can only enter through open stomata. While they are open, water vapour from the saturated air spaces diffuses out down its concentration gradient to the drier atmosphere; the water loss is an unavoidable side effect of gas exchange.
Transpiration is evaporation that depends only on temperature, so it happens whether or not the stomata are open. — A student who applies the everyday model of evaporation picks this. Water evaporates inside the leaf and can only escape through open stomata, which is exactly why transpiration is tied to gas exchange.
9 A potometer was used to estimate transpiration by a leafy shoot at 20 °C. Water uptake in still air was 0.8 cm³ per hour; with a fan blowing air across the leaves, 1.9 cm³ per hour; in still air with a clear plastic bag enclosing the shoot, 0.3 cm³ per hour. Which conclusion is supported by these results?
Answer and reasoning
Moving air raised the rate because the fan warmed the leaves, and warmer leaves evaporate water faster. — A student who thinks wind acts only through temperature picks this. The temperature was held at 20 °C; moving air raised the rate by sweeping away the humid layer next to the leaves, steepening the water-vapour gradient.
The humid air inside the bag should have increased the rate, so the reading of 0.3 cm³ per hour must be a fault. — A student who believes humidity increases transpiration picks this. The bag trapped water vapour, which reduced the concentration gradient between the air spaces and the surrounding air, so the rate fell as expected.
The shoot cut down its transpiration inside the bag because it no longer needed to pull water up the stem. — A student who treats transpiration as something the plant does on purpose picks this. Transpiration is passive diffusion of water vapour; the rate fell because the humid air in the bag reduced the gradient, not because the plant chose to reduce it.
Moving air and low humidity both steepen the water-vapour gradient from the air spaces to the atmosphere. — The fan removed the layer of humid air at the leaf surface, and the bag did the opposite by trapping water vapour around the shoot. Both results are explained by the size of the water-vapour concentration gradient that drives diffusion out of the stomata.
10 Three students estimated the stomatal density of leaves from the same plant. Student A counted one field of view at high power and reported 90 per mm². Student B counted ten different fields, obtained values between 60 and 130 per mm², and reported the mean, 95 per mm². Student C counted the same single field as A five times, getting 90 per mm² each time. Which evaluation of these results is best?
Answer and reasoning
C's result is the most reliable, because repeating the count five times confirmed that no counting error had been made. — A student who sees repetition only as a check for mistakes picks this. Recounting the same field confirms the count of that field, but it samples only one region and says nothing about how the density varies across the leaf.
B's estimate is most reliable, because the spread shows the leaf varies and the mean averages it out. — Biological material is variable, so counts from different fields differ. Replicate counts in different fields reveal that variability and their mean is a more reliable estimate than any single field, however carefully or repeatedly it is counted.
B's method is the least reliable, because results that vary from 60 to 130 show that the counting was inconsistent. — A student who expects good replicates to be identical picks this. The spread reflects real variation in the leaf, not poor counting; it is exactly the variability that replicate fields are meant to capture and average.
A would have been more reliable using low power, because one larger field of view contains more stomata. — A student who applies 'bigger sample is better' to a single field picks this. Stomata are counted at high power because they can be resolved reliably there; reliability comes from several independent fields, not from one bigger one.
11 In the placenta, maternal blood and foetal blood are separated by a thin barrier. How is oxygen transferred from maternal blood to foetal blood? HL
Answer and reasoning
Foetal haemoglobin has a lower affinity for oxygen, so that it releases oxygen to the rapidly growing foetal tissues more easily than adult haemoglobin would. — A student who thinks about unloading rather than loading picks this. A lower affinity would prevent foetal haemoglobin taking oxygen from maternal blood in the first place; its affinity is higher, which is what allows the transfer.
Maternal red blood cells pass across the barrier into the foetal circulation and deliver their oxygen directly to the foetus. — A student who pictures the mother 'sharing her blood' picks this. The two circulations never mix; only dissolved substances such as oxygen diffuse across the placental barrier.
Maternal haemoglobin's affinity for oxygen rises in the placenta because the foetus needs oxygen, so it hands its oxygen on to the foetus. — A student who equates affinity with delivery picks this. A rise in affinity would make maternal haemoglobin hold on to oxygen; transfer depends on maternal haemoglobin releasing oxygen and foetal haemoglobin, with the higher affinity, binding it.
Foetal haemoglobin has a higher affinity for oxygen, so at the oxygen concentration in the placenta it loads oxygen that maternal haemoglobin releases. — Both haemoglobins experience the same oxygen concentration in the placenta. Maternal haemoglobin, with the lower affinity, unloads oxygen there; foetal haemoglobin, with its different subunits and higher affinity, binds that oxygen to a high saturation, so oxygen diffuses across the barrier into foetal blood.
12 Adult haemoglobin is about 20% saturated with oxygen at a partial pressure of oxygen of 2 kPa, 60% at 4 kPa, 90% at 8 kPa and 98% at 13 kPa. Which explanation accounts for this pattern? HL
Answer and reasoning
The four haem groups fill one after another at equal rates, so saturation rises in direct proportion to the oxygen concentration. — A student who thinks the haem groups bind independently picks this. The data are not proportional: saturation triples between 2 and 4 kPa but rises only slightly between 8 and 13 kPa, which is the signature of cooperative binding followed by saturation.
Saturation rises steeply from 2 to 8 kPa because each oxygen bound raises the affinity of the other haem groups. — At low oxygen concentration the first oxygen binds with difficulty, so saturation is low. Once bound, it changes the protein's conformation and the other haem groups bind more readily, giving the steep rise; above 8 kPa most haem groups are occupied and the curve levels off. This cooperative binding produces the S-shape.
The curve shows the order in which haemoglobin loads over time, from its arrival in the lungs to full saturation. — A student who reads the x-axis as time picks this. The axis is oxygen concentration; each value gives the saturation haemoglobin reaches at that concentration, wherever and whenever it occurs.
The plateau above 8 kPa is where haemoglobin releases most of its oxygen to tissues, because saturation is highest there. — A student who reads saturation as delivery picks this. The plateau corresponds to loading in the lungs; oxygen is released in the steep region, where a small fall in oxygen concentration causes a large fall in saturation.
13 Two oxygen dissociation curves are compared. Curve X lies to the left of curve Y: at a partial pressure of oxygen of 4 kPa, haemoglobin X is 75% saturated and haemoglobin Y is 60% saturated. What can be concluded? HL
Answer and reasoning
Haemoglobin X has the lower affinity for oxygen, because its curve lies further to the left on the axis. — A student who reads 'left' as 'less' picks this. Comparing at the same concentration, X is more saturated than Y, so X holds oxygen more readily; a left-shifted curve means higher affinity.
Haemoglobin X loaded before Y, because the curve further to the left represents an earlier time. — A student who reads the x-axis as time picks this. The axis is oxygen concentration, so the two curves describe how each haemoglobin responds to concentration, not when each was loaded.
Haemoglobin X has the higher affinity for oxygen, because it is more fully loaded at the same oxygen concentration as Y. — Affinity is compared at one oxygen concentration: X reaches a higher saturation than Y at 4 kPa, so it binds oxygen more readily. A curve further to the left always indicates higher affinity; foetal haemoglobin's curve lies to the left of the adult curve for this reason.
Haemoglobin X delivers more oxygen at 4 kPa, because a higher saturation means more oxygen is being released. — A student who confuses saturation with delivery picks this. Saturation is oxygen held, not oxygen given up; the higher affinity of X means it retains more oxygen at 4 kPa than Y does.
That was your twenty minutes. Real practice on B3.1 is past-paper questions marked against the mark scheme.
What the exam asks of B3.1
Paper 1A asks you to match a feature of a surface, lung or leaf to its purpose, or to read a spirometer trace. Paper 1B may give surface area-to-volume data, stomatal counts or a dissociation curve and ask you to calculate, then explain. Paper 2 uses *draw* for the leaf plan diagram, *outline* for the properties of exchange surfaces, and *explain* for ventilation, where you must name the muscle, the volume change and the pressure change in order. At HL, *explain* the S-shape and the Bohr shift in terms of conformation and affinity, and *compare* the foetal and adult curves.
Compiled from the IB Biology guide (first assessment 2025, updated May 2026 for 2028) and our question bank · Specialist review in progress. How these pages are made ·
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