Per disputationem veritatem quaerimusSurface area to volume ratio
Surface area and exchange 3.3.1
- Surface-area-to-volume ratio: as any linear dimension increases by a factor , surface area increases by but volume by — a fixed geometric consequence, not a biological tendency, and it applies regardless of what the organism is made of.
- Every specialised exchange surface is shaped by the same three variables: surface area maximised by folding/subdividing/projecting (alveoli, gill lamellae, villi, root hairs), diffusion distance minimised (typically one cell thick), and a concentration gradient actively maintained (ventilation, or a transport system carrying material to or from the surface).
- Volume grows faster than surface area as an organism gets larger, so its surface-area-to-volume ratio falls.
- A large or metabolically active organism cannot rely on diffusion across its outer body surface alone — the distance is too great and the surface too small relative to the tissue volume served.
- A small, flattened body shape (a flatworm) keeps every cell close to the surface and keeps SA:V relatively high, but this has a geometric limit as size and metabolic demand increase.
- Beyond that limit, an organism instead develops specialised, localised exchange surfaces.
- Without an actively maintained concentration gradient, local concentrations would simply run down to equilibrium regardless of surface area or diffusion distance — ventilation or a transport system is as essential to an exchange surface's function as its area or thickness.
Worked examples
Worked example 3.3.1 · 5 marks
Four organisms are modelled as cubes of side length 2 mm, 20 mm, 60 mm and 200 mm.
Calculate the surface-area-to-volume ratio of each, and state what the trend implies for the largest organism if it depended on diffusion across its outer surface alone.
Show worked solution
Using:
for a cube of side : at ,
At ,
At ,
At ,
The ratio falls continuously as side length increases.
Since metabolic demand scales with the volume of respiring tissue while a plain outer surface's exchange capacity scales with surface area, diffusion across the outer surface alone could supply only a tiny fraction of what the 200 mm organism's tissue volume would require.
Mark scheme · 5 marks
- Calculates SA:V = 3 for the 2 mm cube 1 mark
- Calculates SA:V = 0.3 for the 20 mm cube 1 mark
- Calculates SA:V = 0.1 and 0.03 for the 60 mm and 200 mm cubes 1 mark
- States metabolic demand scales with volume while surface exchange scales with area 1 mark
- Explains diffusion alone could not supply enough for the largest organism's tissue volume 1 mark
Do not count matching words alone — ask whether your answer actually makes the same claim.
Worked example 3.3.1 · 4 marks
Two related species of shrew have almost identical body shapes but different sizes: species A has a mass of 3 g, species B has a mass of 12 g.
Predict, with reasoning, which species would be expected to have the higher metabolic rate per gram of body mass, and explain the specific adaptation very small mammals like shrews typically show as a consequence.
Show worked solution
Species A (the smaller shrew) would be expected to have the higher metabolic rate per gram of body mass.
A smaller body has a larger surface-area-to-volume ratio, so proportionally more heat is lost per gram of body mass.
To compensate and maintain a stable core body temperature, the smaller animal must generate more heat per gram through metabolism.
This is why very small mammals like shrews characteristically need to eat almost continuously relative to their body size.
Mark scheme · 4 marks
- States species A (smaller) has the higher metabolic rate per gram 1 mark
- Explains a smaller body has a larger surface-area-to-volume ratio 1 mark
- Links this to proportionally greater heat loss per gram of mass 1 mark
- Explains the need to eat almost continuously to offset this heat loss 1 mark
Do not count matching words alone — ask whether your answer actually makes the same claim.
Gas exchange
The human gas exchange system 3.3.2
- Air passes through a branching airway system of decreasing diameter — trachea → bronchus → bronchioles → alveoli — with ventilation (breathing movements) continually renewing the air inside the alveoli.
- At the alveolar epithelium, both the alveolar epithelium and the adjacent capillary endothelium are each just one cell thick, minimising diffusion distance between air and blood.
- Continuous blood flow through the capillaries constantly carries oxygenated blood away and brings deoxygenated blood in, maintaining the concentration gradients that keep diffusion happening in both directions (O₂ into blood, CO₂ out) — the same area/distance/gradient pattern seen throughout this topic.
Gas exchange in fish and insects 3.3.2
- Countercurrent system: water flows over the gill lamellae in the direction opposite to blood flow within them — the arrangement that keeps water more oxygen-concentrated than the adjacent blood along the whole length of the lamella, not just at one end.
- Fish gills consist of many filaments, each bearing many thin lamellae, giving a very large surface area for exchange.
- The countercurrent advantage is not about any single point exchanging gas faster; it is about what happens at BOTH ends of the lamella. Where water enters (highest O₂ concentration), it meets blood about to leave that is already well oxygenated — but the water is still more concentrated, so a gradient favouring further diffusion remains. Where water leaves (lowest O₂ concentration), it meets blood that has only just entered and is still very low in O₂ — so again the water remains more concentrated than the blood it meets.
- A parallel (concurrent) arrangement, by contrast, would let the two concentrations equalise partway along the lamella, after which no further net diffusion could occur — the countercurrent design sustains a favourable gradient along almost the whole exchange surface.
- Insects instead deliver gases directly to tissues via a tracheal system: spiracles (external openings, which can close to reduce water loss) lead to tracheae and then to tracheoles, extending close enough to respiring cells that O₂ and CO₂ diffuse directly between air and cell — no blood pigment is involved at all.
- In larger or more active insects, body movements can help ventilate the tracheal tubes, and fluid is drawn back from the tracheole ends during high activity, bringing air even closer to the tissues that need it fastest.
Gas exchange in leaves and single-celled organisms 3.3.2
- In a dicotyledonous leaf, CO₂ and O₂ diffuse in and out through stomata (pores controlled by guard cells) into a network of connected air spaces within the spongy mesophyll, which has a large internal surface area for gas exchange.
- Stomata present a genuine trade-off: opening them allows CO₂ uptake for photosynthesis, but also allows water vapour to diffuse out — stomatal closure limits water loss at the cost of also limiting CO₂ uptake.
- Once inside the leaf, diffusion to individual mesophyll cells is rarely the limiting step — the limiting factor is usually getting gases across the stomata in the first place.
- A single-celled aerobic organism living in water relies entirely on diffusion across its own cell-surface membrane: its large surface area relative to its small volume and the short diffusion distance to any point in the cell make a specialised exchange surface unnecessary, while its own respiration continuously maintains the diffusion gradients by using up O₂ and producing CO₂.
Ventilation mechanics 3.3.2
- Inspiration: external intercostal muscles contract (ribs move up and out) and the diaphragm contracts and flattens — thoracic volume increases, alveolar pressure falls below atmospheric pressure, and air enters.
- Quiet expiration: external intercostals relax (ribs move down and in) and the diaphragm relaxes and domes upward — elastic recoil reduces thoracic volume, alveolar pressure rises above atmospheric pressure, and air leaves.
- Forced expiration additionally recruits internal intercostal muscles and abdominal muscles, actively pulling the ribs down and compressing the abdomen to expel air faster and more completely than quiet, passive expiration alone.
- Pulmonary ventilation rate (PVR): the total volume of air moved into (or out of) the lungs per minute, , where tidal volume is the volume of air moved in a single normal breath and breathing rate is the number of breaths per minute.
- The entire mechanism works by creating a pressure difference: changing thoracic (and so lung) volume changes alveolar pressure relative to atmospheric pressure, and air always flows from higher to lower pressure — muscles change volume, and volume change is what actually moves the air.
- PVR can rise either by breathing more deeply (greater tidal volume) or by breathing faster (higher breathing rate), or both together — exercise typically increases both factors at once, so PVR rises by more than either factor would alone.
- Quiet expiration is normally a passive process, relying on elastic recoil of the lungs and thoracic wall rather than active muscle contraction — only forced expiration (e.g. during exercise) actively recruits additional muscles.
- Ventilation is what actively maintains the concentration gradient across the alveolar exchange surface — the third of the three variables (area, distance, gradient) that govern every exchange surface in this topic, alongside the large surface area and short diffusion distance covered in the previous sections.
Worked examples
Worked example 3.3.2 · 4 marks
In a countercurrent gill exchange system, water enters with a partial pressure of oxygen of 20 kPa and leaves with 6 kPa.
Blood entering the lamella (opposite direction to the water) has a partial pressure of oxygen of 3 kPa and leaves with 14 kPa.
Calculate the percentage of the oxygen originally present in the water that diffused into the blood, and explain why a favourable gradient for diffusion into the blood exists at both the point where water enters and the point where it leaves.
Show worked solution
Percentage of oxygen removed from the water:
At the point where water enters (20 kPa), it meets blood about to leave, already at 14 kPa; since , oxygen can still diffuse into blood.
At the point where water leaves (6 kPa), it meets blood that has only just entered, at 3 kPa; since , oxygen can still diffuse into blood here too.
The water is always at a higher partial pressure than the blood alongside it, maintaining a favourable gradient along the whole exchange surface.
Mark scheme · 4 marks
- Calculates 70% of the oxygen removed from the water 1 mark
- States water at entry (20 kPa) still exceeds the exiting blood (14 kPa) 1 mark
- States water at exit (6 kPa) still exceeds the entering blood (3 kPa) 1 mark
- Explains a favourable gradient is maintained along the whole exchange surface, unlike parallel flow 1 mark
Do not count matching words alone — ask whether your answer actually makes the same claim.
Worked example 3.3.2 · 3 marks
At rest, a person has a tidal volume of 0.4 dm³ and a breathing rate of 14 breaths per minute.
During moderate exercise, their tidal volume rises to 1.6 dm³ and their breathing rate rises to 22 breaths per minute.
Calculate their pulmonary ventilation rate at rest and during exercise, and calculate the factor by which it increases.
Show worked solution
At rest:
During exercise:
Factor of increase:
PVR rises by a larger factor than either variable alone, because it depends on the product of the two, and their combined effect is multiplicative, not merely additive.
Mark scheme · 3 marks
- Calculates resting PVR = 5.6 dm³ min⁻¹ 1 mark
- Calculates exercise PVR = 35.2 dm³ min⁻¹ 1 mark
- States the increase (≈6.3×) is larger than either individual factor's own increase, since PVR is their product 1 mark
Do not count matching words alone — ask whether your answer actually makes the same claim.
Digestion and absorption
The digestive tract, carbohydrate and protein digestion 3.3.3
- Carbohydrate: amylase (salivary then pancreatic) hydrolyses starch to maltose and short chains in the lumen → membrane-bound disaccharidases (maltase, sucrase, lactase) finish the job at the epithelial surface, producing glucose, fructose or galactose immediately before absorption.
- Protein: endopeptidases (e.g. pepsin, trypsin) cleave peptide bonds within the chain, creating new ends → exopeptidases release amino acids from those ends → membrane-bound dipeptidases complete the final hydrolysis step at the epithelial surface.
- Carbohydrate and protein digestion both follow the same 'work inward from the whole molecule, finish at the membrane' pattern — a lumen-based stage first (using secreted enzymes), then membrane-bound enzymes completing the final hydrolysis step immediately before absorption.
- The liver produces bile (stored in the gallbladder) and the pancreas secretes both digestive enzymes and bicarbonate (to neutralise acidic stomach contents entering the duodenum) — both empty into the duodenum, the first section of the small intestine.
- Different membrane-bound disaccharidases are each specific to one substrate: maltase acts on maltose (from starch digestion), sucrase on sucrose (glucose + fructose), lactase on lactose (glucose + galactose) — the same general 'membrane-bound enzyme finishes the job' pattern, applied to three different everyday sugars.
Villus structure and absorption 3.3.3
- The ileum's villi and microvilli greatly increase surface area, its epithelium (one cell thick) minimises diffusion distance, and its dense capillary network maintains the concentration gradient by continuously carrying absorbed material away — the same three-variable pattern (area, distance, gradient) that governs every specialised exchange surface in this topic.
- Monosaccharides and amino acids are absorbed by sodium-coupled (co-transport): a basal Na⁺/K⁺ pump actively maintains a low intracellular Na⁺ concentration using ATP; Na⁺ then diffuses back into the epithelial cell down its electrochemical gradient through a luminal cotransporter, and this movement is coupled to glucose or amino acids being carried in against their own concentration gradient.
- Galactose can use the same glucose cotransporter; fructose instead enters by facilitated diffusion, not cotransport — a genuine difference between the three monosaccharides worth keeping distinct.
- Visking tubing (a partially permeable artificial membrane, allowing small molecules through but not large ones) can model this absorption process in the laboratory: a mixture of a large molecule (e.g. starch) and a small one (e.g. glucose) is sealed inside a length of Visking tubing, which is then placed in a beaker of water or an appropriate test-reagent solution. Only the small molecule diffuses out through the tubing's pores into the surrounding solution (detected there, e.g. glucose giving a positive Benedict's test in the surrounding water after time), while the large molecule remains trapped inside — modelling how the gut wall allows small digestion products through while excluding large, undigested molecules.
- Glucose and amino acids then leave the epithelial cell into the tissue fluid, and so into the blood, by facilitated diffusion — the whole absorption pathway uses active transport, facilitated diffusion and cotransport together, exactly as in the general cotransport mechanism covered under Cells.
Lipid digestion and absorption 3.3.3
- Emulsification: bile salts physically break large lipid droplets into many smaller droplets (a mixed micelle), without breaking any chemical bonds — increasing the total surface area available for lipase to act on.
- Lipase hydrolyses the ester bonds in triglycerides, releasing fatty acids and monoglycerides.
- Fatty acids and monoglycerides (lipid-soluble) diffuse directly into the epithelial cell, are reassembled into triglycerides at the smooth ER, packaged with protein at the Golgi into lipoprotein chylomicrons, and leave the cell by exocytosis into the lacteal (part of the lymphatic system), not directly into the blood capillaries.
- Bile emulsification is a purely physical process — no chemical bonds are broken — but it substantially raises the rate at which lipase can act, simply by exposing far more lipid surface area to the enzyme.
- Lipid absorption is routed differently from monosaccharide/amino acid absorption specifically because the products are lipid-soluble: they can diffuse directly across the epithelial cell's membranes without needing a transport protein, but are too large once reassembled into chylomicrons to enter the narrower blood capillaries, so they enter the more permeable lymphatic lacteal instead — lymph eventually carries them to the bloodstream via a route separate from the hepatic portal vein that carries absorbed sugars and amino acids.
Worked examples
Worked example 3.3.3 · 4 marks
Bile salts are added to a mixture of lipid droplets suspended in water.
Without bile salts, one large lipid droplet forms; with bile salts, many much smaller droplets (micelles) form instead.
Explain why this change increases the rate at which lipase can digest the lipid, and explain why bile salts themselves are not digestive enzymes.
Show worked solution
Lipase can only act at the surface of a lipid droplet.
Breaking one large droplet into many much smaller micelles dramatically increases the total surface area of lipid exposed for a given total volume, so far more lipase molecules can access and act on the lipid simultaneously.
Bile salts are not digestive enzymes because they do not catalyse any chemical reaction breaking a bond within the lipid molecules — they act purely physically (emulsification), speeding up the separate enzyme-catalysed reaction but not carrying it out themselves.
Mark scheme · 4 marks
- States lipase can only act at a droplet's surface 1 mark
- Explains smaller droplets give a much greater total surface area for the same volume 1 mark
- Links this to more lipase molecules acting simultaneously 1 mark
- Explains bile salts act physically (emulsification), not catalysing any bond-breaking reaction themselves 1 mark
Do not count matching words alone — ask whether your answer actually makes the same claim.
Worked example 3.3.3 · 4 marks
| Time | Iodine test (outside water) | Benedict's test (outside water) |
|---|---|---|
| Before (0 min) | Negative (orange-brown) | Negative (stays blue) |
| After (30 min) | Negative (orange-brown) | Positive (brick-red precipitate) |
A mixture of starch and glucose is sealed inside a length of Visking tubing, which is then placed in a beaker of distilled water for 30 minutes.
Samples of the water outside the tubing are tested with iodine solution and with Benedict's solution both before the experiment starts and after 30 minutes.
Predict and explain the results of both tests at both time points.
Show worked solution
Before the experiment starts, the water outside the tubing contains neither starch nor glucose, so both tests give a negative result.
After 30 minutes, the iodine test remains negative — starch is too large a molecule to pass through the pores of the partially permeable Visking tubing.
The Benedict's test is now positive — glucose is small enough to diffuse freely through the tubing's pores, down its concentration gradient, so some has diffused out into the surrounding water by 30 minutes.
This models how the gut wall allows small digestion products through while excluding large, undigested molecules.
Mark scheme · 4 marks
- States both tests are negative before the experiment starts 1 mark
- States the iodine test remains negative after 30 minutes 1 mark
- Explains starch is too large to pass through the tubing's pores 1 mark
- States the Benedict's test becomes positive, explained by glucose diffusing out down its concentration gradient 1 mark
Do not count matching words alone — ask whether your answer actually makes the same claim.
Mass transport in animals
The circulatory system and heart structure 3.3.4.1
- Double circulatory system: blood passes through the heart twice per complete circuit — once around the pulmonary circuit (heart–lungs–heart), once around the systemic circuit (heart–body–heart) — restoring pressure between the two, rather than just changing route.
- The double circulatory system's advantage over a single circuit is pressure, not just route: passing through the heart a second time restores blood pressure after the lower-pressure pulmonary circuit, before it is sent at high pressure around the larger, higher-resistance systemic circuit — supporting faster bulk delivery than a single pass through the heart could sustain.
- Within the heart, wall thickness matches the pressure a chamber must generate: the atria (thin) feed the ventricles; the right ventricle (thicker) drives the shorter, lower-pressure pulmonary circuit; the left ventricle (thickest of all) drives the longer, higher-pressure systemic circuit around the whole body.
- The atrioventricular valves (tricuspid on the right, bicuspid/mitral on the left) prevent backflow from ventricle to atrium; the semilunar valves (pulmonary and aortic) prevent backflow from the arteries back into the ventricles — every valve opens and closes purely in response to the pressure difference across it, never cued by anything else.
- The coronary arteries branch from the aorta itself, supplying the heart's own muscle with oxygenated blood — the heart cannot rely on the blood passing through its own chambers to supply its own tissue.
The cardiac cycle 3.3.4.1
- Cardiac output: the volume of blood pumped by one ventricle per minute, .
- Stroke volume: the volume of blood ejected by one ventricle in a single heartbeat.
- Ventricular filling: both AV valves open, semilunar valves closed — pressure is below that of the atria, so ventricular volume increases.
- Atrial systole: AV valves still open, semilunar still closed — atrial contraction raises atrial pressure, completing ventricular filling.
- Isovolumetric contraction: both valve types closed — ventricular pressure rises sharply while volume stays unchanged (an enclosed, incompressible chamber).
- Ventricular ejection: AV valves closed, semilunar valves open — ventricular pressure now exceeds arterial pressure, so blood is ejected and volume decreases.
- Isovolumetric relaxation: both valve types closed again — ventricular pressure falls while volume stays unchanged, until it drops low enough for the AV valves to reopen and the cycle repeats.
- Every transition in the cycle is driven by a pressure comparison, and every valve simply opens or closes in response to whichever side of it has higher pressure at that instant — there is no separate 'signal' telling a valve when to move.
- The two isovolumetric phases exist specifically because both valve types are momentarily closed at once: with nowhere for blood to go, the ventricle's volume cannot change even though its pressure is changing rapidly — a genuinely distinct phase from either filling or ejection.
- Cardiac output rising during exercise reflects both a higher heart rate and, often, a higher stroke volume — the two factors multiply together, so a change in either alone changes cardiac output proportionally.
Blood vessel structure 3.3.4.1
- Arteries: thick, muscular, elastic wall around a relatively narrow lumen — built to withstand high pressure; elastic recoil helps maintain flow between heartbeats.
- Arterioles: a smooth-muscle layer around a narrow lumen — constriction and dilation of this muscle controls how much blood flows into a given capillary bed.
- Capillaries: a wall just one cell thick (endothelium only), with no muscle or elastic layer — minimising diffusion distance; their narrow lumen also slows flow, giving more time for exchange across their very large total surface area.
- Veins: a wide lumen with a relatively thin wall, plus valves in many veins — since blood pressure has fallen very low by this point, veins rely on their valves (preventing backflow) and compression by contracting skeletal muscle to keep blood moving back toward the heart.
- Each vessel type's structure follows directly from the pressure and function it must support — arteries need strength and elastic recoil for high, pulsing pressure; capillaries need thinness for exchange, not strength; veins need one-way valves rather than a thick wall, since their internal pressure is already low.
Tissue fluid formation and return 3.3.4.1
- At the arteriole end of a capillary bed, hydrostatic pressure is relatively high and drives filtration: water and small solutes (but not most plasma proteins, which are too large) are forced out of the capillary into the surrounding tissue fluid.
- Plasma proteins remaining in the blood lower the blood's own water potential relative to the tissue fluid, generating an osmotic force that opposes filtration and draws water back in — hydrostatic pressure and this osmotic (oncotic) effect act in opposite directions throughout the capillary bed.
- Near the arteriole end, hydrostatic pressure dominates and fluid moves out; near the venule end, hydrostatic pressure has fallen (lost along the capillary) while the osmotic effect stays roughly constant, so the balance shifts and more fluid re-enters the capillary.
- Not all filtered fluid is reabsorbed directly into the capillary — the remainder is drained by lymphatic capillaries and eventually returned to the blood via veins near the heart, rather than being lost from the circulation altogether.
- Tissue fluid itself is what actually bathes body cells directly, providing the real medium for exchange of O₂, nutrients and CO₂ between blood and cells — blood in the capillary never contacts a body cell directly.
Haemoglobin and oxygen transport 3.3.4.1
- Bohr effect: rising CO₂ concentration (and so falling pH) shifts the oxygen dissociation curve to the right — at a given partial pressure of oxygen, saturation is lower, so more oxygen is released exactly where CO₂ (and so respiration) is highest.
- Haemoglobin is a quaternary protein: 4 globin chains, each carrying one haem group with a central Fe²⁺ that binds one O₂ molecule reversibly — one haemoglobin molecule can carry up to four O₂ molecules.
- Cooperative binding: binding one O₂ changes haemoglobin's shape, increasing the affinity of its remaining binding sites — producing the characteristic sigmoid (S-shaped) oxygen dissociation curve.
- The curve's shape gives ready loading at high pO₂ (in the lungs, where the curve is already near its plateau) and ready unloading at low pO₂ (in respiring tissue, where the curve is at its steepest, most sensitive part).
- The sigmoid shape is a direct consequence of cooperative binding: the first O₂ binds relatively reluctantly (few sites already occupied, no affinity boost yet), but each subsequent binding becomes easier, and the last one becomes harder again as available sites run out — producing the characteristic S-curve rather than a simple rising line.
- Because CO₂ concentration is highest exactly where tissue is respiring fastest, the Bohr effect automatically increases oxygen delivery precisely where demand is greatest, without needing any separate signalling mechanism.
- A haemoglobin variant with HIGHER oxygen affinity than the reference curve shown loads oxygen at a lower pO₂ than usual — useful for organisms living in low-oxygen environments, but such a variant would also need a correspondingly favourable arrangement to unload that oxygen again at the tissues, or higher affinity alone would actually impair oxygen delivery.
Mass transport in plants
Xylem and phloem structure 3.3.4.2
- Xylem: dead, hollow, lignified vessel elements joined end to end with the cross-walls between them broken down, forming a continuous tube — transports water and mineral ions, one direction only, roots to leaves.
- Phloem: living sieve tube elements joined through perforated sieve plates, each supported by an adjacent companion cell connected via plasmodesmata — transports sucrose and other organic solutes, in either direction depending on where sources and sinks currently are.
- Xylem vessel elements are dead at maturity, with no cytoplasm and a lignified wall (except at unlignified regions called pits, which allow lateral water movement) — being dead and hollow is exactly what makes them an efficient, low-resistance tube for one-way bulk water flow.
- Phloem sieve tube elements are alive but have lost their nucleus and most organelles, relying on an adjacent companion cell (connected via plasmodesmata) to provide the metabolic support (including ATP for active loading) that a sieve tube element cannot generate for itself.
Water transport and translocation 3.3.4.2
- Source (phloem): where sugar is made or mobilised — a photosynthesising leaf, or a reserve-releasing storage organ. Sink (phloem): where sugar is used or stored — a root, a growing region, a fruit; which tissue plays which role can change with the plant's needs.
- Cohesion-tension: evaporation of water from mesophyll cell walls (then diffusion out through stomata as transpiration) lowers water potential at the top of the xylem, pulling water upward under tension; hydrogen bonding between water molecules (cohesion) keeps the water column continuous all the way down to the roots.
- Mass flow (pressure-flow) hypothesis: at the source, sucrose is actively loaded into the sieve tube (cotransport with H⁺, requiring ATP) — water then follows by osmosis, raising hydrostatic pressure at the source. At the sink, sucrose is removed — water then leaves by osmosis, lowering pressure there. The resulting pressure difference drives mass flow of the whole solution along the sieve tube, from source to sink.
- The cohesion-tension mechanism starts with water LOSS at the leaf (transpiration), not with any pumping action at the root — the root's own uptake of water by osmosis from the soil is a consequence of this pull, not its cause.
- Adhesion between water molecules and the xylem vessel walls also contributes to keeping the water column moving, alongside cohesion between water molecules themselves.
- Loading sucrose into the phloem at the source is the one ATP-dependent step in the whole mass-flow mechanism — the actual bulk movement of solution along the sieve tube afterward is driven purely by the pressure gradient this loading (and unloading) creates, not by any further direct energy input along the way.
- Because different sieve tubes can each have their own independent source-to-sink pressure gradient, sap can be observed flowing in opposite directions simultaneously in different phloem sieve tubes within the same plant — entirely consistent with the mass-flow model, since each tube's flow direction depends only on where ITS OWN source and sink currently are.
Evidence for phloem transport 3.3.4.2
- Ringing: sugars accumulate and the stem swells just above the ring, while tissue below the ring is unaffected — consistent with downward phloem transport being physically interrupted at the ring, while xylem (left intact) continues supplying water upward as normal.
- Radioactive tracer: labelled sugars are subsequently detected specifically in the phloem and at sink tissues, confirming that photosynthetic products move through the phloem to reach sinks.
- Ringing experiment: remove a complete ring of bark and phloem from a woody stem, leaving the xylem inside intact, then observe over time.
- Radioactive tracer experiment: supply a leaf with ¹⁴CO₂, allow photosynthesis to fix it into labelled sugars, then track the radioactivity's movement to other parts of the plant over time.
- Both experiments identify the ROUTE sugars travel by (the phloem) and are consistent with the mass-flow hypothesis, but neither alone proves the pressure-flow mechanism's every detail is correct — they show sugars move through the phloem from source to sink, not the precise physical mechanism driving that movement.
- Observed transport speeds in some plants, and the simultaneous bidirectional sap transport described in the previous section, need the more careful explanation given there (each sieve tube having its own independent pressure gradient) rather than a single, plant-wide description of 'the' direction of flow.
Worked examples
Worked example 3.3.4 · 4 marks
A potometer's capillary tube has a radius of 0.4 mm.
An air bubble moves 45 mm along the tube in 15 minutes.
Calculate the rate of water uptake in mm³ per minute.
If moving the shoot into bright light increased the rate of water uptake by 40%, calculate the distance the bubble would be expected to move in the same 15 minutes under the brighter light.
Show worked solution
Volume taken up:
(3 s.f.).
Rate:
Since tube radius and time are unchanged, distance moved is directly proportional to volume taken up, so a 40% increase in rate corresponds to a 40% increase in distance:
Mark scheme · 4 marks
- Calculates volume taken up ≈ 22.6 mm³ 1 mark
- Calculates rate ≈ 1.51 mm³ min⁻¹ 1 mark
- States distance moved is proportional to volume taken up (radius/time unchanged) 1 mark
- Calculates the new distance as 63 mm 1 mark
Do not count matching words alone — ask whether your answer actually makes the same claim.
Worked example 3.3.4 · 4 marks
A patient's cardiac output is measured as 5.6 dm³ min⁻¹, with a heart rate of 70 beats per minute.
(a) Calculate the patient's stroke volume.
(b) During exercise, the patient's heart rate rises to 140 beats per minute and their stroke volume rises to 0.12 dm³.
Calculate their cardiac output during exercise, and state whether the increase in cardiac output is better explained by the change in heart rate or the change in stroke volume.
Show worked solution
(a) Stroke volume:
(b) During exercise:
Heart rate has doubled (factor of 2), while stroke volume increased by a smaller factor (:
) — proportionally, the change in heart rate contributes more to the rise in cardiac output.
Mark scheme · 4 marks
- Calculates stroke volume = 0.08 dm³ 1 mark
- Calculates exercise cardiac output = 16.8 dm³ min⁻¹ 1 mark
- States heart rate changed by a factor of 2, stroke volume by a factor of 1.5 1 mark
- Concludes heart rate's change contributes proportionally more to the rise in cardiac output 1 mark
Do not count matching words alone — ask whether your answer actually makes the same claim.
Worked example 3.3.4 · 4 marks
At a partial pressure of oxygen of 4 kPa, resting muscle tissue has a saturation of haemoglobin with oxygen of about 20%, but exercising muscle tissue (producing more CO₂, and so at a lower local pH) has a saturation of only about 8% at the same partial pressure of oxygen.
Explain this difference, and explain why it is beneficial to the exercising muscle.
Show worked solution
This is the Bohr effect: a rise in CO₂ concentration (and the corresponding fall in pH) shifts the oxyhaemoglobin dissociation curve to the right, reducing haemoglobin's affinity for oxygen at a given partial pressure.
Exercising muscle respires faster, producing more CO₂ and lowering local pH, shifting the curve rightward there specifically.
This is beneficial because a lower affinity means haemoglobin releases more of its bound oxygen into the tissue fluid, precisely where oxygen demand is highest.
Mark scheme · 4 marks
- Names the Bohr effect 1 mark
- States rising CO₂/falling pH shifts the dissociation curve to the right 1 mark
- Explains this reduces haemoglobin's oxygen affinity at a given partial pressure 1 mark
- Explains the benefit: more oxygen unloaded exactly where demand is highest 1 mark
Do not count matching words alone — ask whether your answer actually makes the same claim.
Worked example 3.3.4 · 4 marks
A woody stem is completely ringed, with a strip of bark and phloem removed all the way around, leaving the xylem beneath fully intact.
Predict what would be observed in the region of stem immediately above the ring, and immediately below it, over the following days, and explain your reasoning in terms of which tissue was interrupted.
Show worked solution
Above the ring, sugar produced by photosynthesis in leaves above will accumulate, since phloem — interrupted at the ring — can no longer carry it downward, typically causing swelling.
Below the ring, tissue would show a shortage of organic nutrients, since its supply via phloem from above is cut off.
Water transport is unaffected in either region, since xylem was left fully intact — this is exactly why ringing experiments are used as evidence that phloem, not xylem, carries organic solutes.
Mark scheme · 4 marks
- States sugar accumulates above the ring, since phloem transport is interrupted there 1 mark
- States tissue below the ring shows a nutrient shortage 1 mark
- States water transport is unaffected, since xylem remains intact 1 mark
- Explains this is evidence that phloem, not xylem, carries organic solutes 1 mark
Do not count matching words alone — ask whether your answer actually makes the same claim.