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Notes · BiologyUK · A-Levels

Organisms exchange substances with their environment

As organisms get larger their surface-area-to-volume ratio falls, so simple diffusion across the body surface is no longer enough and specialised exchange surfaces and mass-transport systems evolve. This chapter covers gas exchange in animals and plants, digestion and absorption, the transport of oxygen by haemoglobin, the heart and cardiac cycle, and the movement of water and sugars through plants.

6 sections·~19 min reading time·3 competencies·Level Foundation 1 · Standard 2 · Advanced 3

T·0333 / 8
Exam profile
AO1 · Describe exchange surfaces and mass-transport systems in animals and plantsAO2 · Apply surface-area-to-volume, cardiac-output and rate calculations and interpret dissociation curves and pressure tracesAO3 · Analyse potometer, dissociation and cardiac data and evaluate the evidence for cohesion-tension and mass flow
Operators:describeexplaincalculatecompareanalyseevaluatesuggest

basic level

AS-Level covers surface area to volume ratio, exchange surfaces, gas exchange, digestion and absorption, and mass transport in animals and plants.

higher level

The full A-Level expects quantitative interpretation of dissociation curves and cardiac pressure traces and evaluation of the transport theories.

Depth

Reading depth: In depth

Text

Text size: Standard

Contents · 6 sections▾
  1. Organisms exchange substances with their environment
    • 01Surface area to volume ratio and exchange surfaces○
    • 02Gas exchange in animals and plants◐
    • 03Digestion and absorption◐
    • 04Haemoglobin and oxygen transport●
    • 05The heart and the cardiac cycle●
    • 06Mass transport in plants: transpiration and translocation●
§ 01

Surface area to volume ratio and exchange surfaces#

●○○FoundationLPAQA 7402 3.3.1LPDfE GCE Biology - surface area and exchange

SA:V ratio against size

Function graph, SA:V = 6/x = 6/x, 2 marked pointsGraph of SA:V = 6/x, decreasing, on the interval x from 1 to 6123456123456small: 6:1large: 1:1SA:V = 6/xSA:V ratiocube side length / arbitrary …
Fig. 1For a cube of side x, the SA:V ratio is 6/x - it falls steeply as the organism gets larger.

Key points

As an object gets larger, its volume increases faster than its surface area, so the surface-area-to-volume (SA:V) ratio falls. For a cube of side length xxx the surface area is 6x26x^26x2 and the volume is x3x^3x3, so the ratio is 6/x6/x6/x - it halves when the side doubles. This simple geometry has a profound biological consequence: a small organism has a large SA:V and can exchange gases and heat across its body surface, but a large organism cannot rely on its surface alone.
A low SA:V ratio also means a large organism has a relatively small surface for heat loss, and a longer diffusion distance from the surface to the centre. Large or active organisms therefore need specialised exchange surfaces (lungs, gills) and mass-transport systems (a circulation) to supply their cells and remove waste fast enough to meet a high metabolic rate.
Efficient exchange surfaces share the same adaptations, which can be predicted from the diffusion equation (Fick's law): a large surface area, a thin exchange surface (a short diffusion distance), and a steep concentration gradient maintained by a good blood supply or ventilation. The rate of diffusion is proportional to (surface area x concentration difference) / diffusion distance, so improving any of these three factors raises the rate.
These principles run through the whole chapter: the alveoli, gills, insect tracheae and leaf mesophyll all show a large, thin surface with a maintained gradient, and the circulatory system exists precisely because diffusion alone cannot serve a large body. Recognising the same three adaptations in an unfamiliar exchange surface is a reliable source of application marks.
SA:V ratio (cube)=6x2x3=6x\text{SA:V ratio (cube)} = \dfrac{6x^2}{x^3} = \dfrac{6}{x}SA:V ratio (cube)=x36x2​=x6​

Surface area to volume ratio

The ratio falls as size (xxx) rises; small organisms have a large SA:V and can exchange across the body surface.

rate of diffusion∝surface area×concentration differencediffusion distance\text{rate of diffusion} \propto \dfrac{\text{surface area} \times \text{concentration difference}}{\text{diffusion distance}}rate of diffusion∝diffusion distancesurface area×concentration difference​

Fick's law (proportionality)

Explains why exchange surfaces are large, thin and kept at a steep gradient.

Worked example

Comparing SA:V ratios

Calculate the surface-area-to-volume ratio of a cube of side 1 cm and a cube of side 3 cm, and comment.

  1. 01Side 1 cm

    SA = 6×12=66 \times 1^2 = 66×12=6 cm2; V = 13=11^3 = 113=1 cm3; ratio = 6:1.

  2. 02Side 3 cm

    SA = 6×32=546 \times 3^2 = 546×32=54 cm2; V = 33=273^3 = 2733=27 cm3; ratio = 54:27=2:154:27 = 2:154:27=2:1.

    5427=2⇒2:1\frac{54}{27} = 2 \quad\Rightarrow\quad 2{:}12754​=2⇒2:1
  3. 03Comment

    The larger cube has a much smaller SA:V ratio, so diffusion across the surface cannot supply its greater volume - a specialised exchange surface is needed.

Result: 6:1 for the small cube and 2:1 for the large cube.

Exam focus

  • Calculate SA:V ratios for cubes or other shapes and explain the consequence for exchange and heat loss.
  • Relate the three adaptations of an exchange surface (large area, thin, steep gradient) to Fick's law.

Typical mistakes

  • Stating SA and V but not forming the ratio, or getting the ratio the wrong way up.
  • Saying a large organism has a large surface area 'so exchange is easy' - it is the ratio to volume that matters.

Active revision

Calculate the SA:V ratio of a cube of side 2 mm and of side 4 mm, and use them to explain why a large organism needs a specialised gas-exchange surface.

Active recall

Recall the key points — then reveal.

Sources: GCE AS and A level subject content for the sciences (Department for Education) · AQA A-level Biology 7402 specification (AQA)

§ 02

Gas exchange in animals and plants#

●●○StandardLPAQA 7402 3.3.2LPDfE GCE Biology - gas exchange

Counter-current flow in a fish gill

Counter-current exchangeSchematic diagram with 3 elements, water flow (100% -> 40% O2), blood flow (30% -> 90% O2), gradient maintained along the whole lamellawater flow (100%-> 40% O2)blood flow (30%-> 90% O2)gradientmaintained alon…
Fig. 2Water and blood flow in opposite directions, so an oxygen gradient is kept across the whole lamella.

Key points

Gas-exchange surfaces all balance the same trade-off: they must be large, thin and moist to maximise diffusion, yet a large moist surface also loses water. Different organisms resolve this in different ways. In insects, air enters through spiracles into a system of tracheae and fine tracheoles that deliver oxygen directly to the tissues; the spiracles can close to limit water loss, and abdominal movements ventilate the system in active insects.
Fish use gills with a counter-current exchange system. Water flows over the gill lamellae in the opposite direction to the blood flow beneath them. Because the two flow in opposite directions, a concentration gradient for oxygen is maintained across the entire length of the lamella, so blood can absorb oxygen even from water that already has a fairly low oxygen content. If the flows ran in the same direction (concurrent), the two would reach equilibrium halfway along and no more oxygen could be absorbed - the counter-current arrangement is far more efficient.
In plants, gas exchange occurs mainly through the stomata on the leaf underside; carbon dioxide diffuses in and oxygen out through the air spaces of the spongy mesophyll, whose large internal surface area and short diffusion path make exchange efficient. The stomata are opened and closed by guard cells, allowing the plant to control gas exchange and, unavoidably, water loss.
Xerophytes are plants adapted to reduce water loss in dry conditions, and their adaptations illustrate the same exchange-versus-water trade-off from the opposite direction. They include sunken stomata (trapping humid air and reducing the gradient), a thick waxy cuticle, rolled leaves, hairs (trichomes) trapping moist air, and a reduced surface area such as spines. Each adaptation lowers the water-potential gradient driving evaporation, at the cost of some gas exchange.
Worked example

Why counter-current beats concurrent flow

Compare oxygen uptake in counter-current and concurrent gill systems, explaining the difference in terms of concentration gradients.

  1. 01Concurrent flow

    Water and blood flow the same way; their oxygen concentrations equalise about halfway along, after which no net diffusion occurs, so the blood leaves only about 50% saturated.

  2. 02Counter-current flow

    Water and blood flow in opposite directions, so at every point the water still has a higher oxygen concentration than the adjacent blood - a gradient exists along the entire lamella.

  3. 03Conclusion

    Diffusion of oxygen into the blood continues all along the lamella, so the blood can reach a much higher saturation (around 80-90%).

Result: Counter-current flow maintains a gradient along the whole lamella, giving far higher oxygen uptake.

Exam focus

  • Explain how counter-current flow maintains a diffusion gradient along the whole gill lamella and why it is more efficient than concurrent flow.
  • Relate each xerophytic adaptation to a reduced water-potential gradient or a smaller exchange surface.

Typical mistakes

  • Saying counter-current flow gives a 'bigger' gradient rather than one maintained along the whole length.
  • Describing xerophyte adaptations without linking them to the reduced rate of transpiration.

Active revision

Explain why a fish gill using a counter-current system can absorb more oxygen from the water than one in which water and blood flow in the same direction.

Active recall

Recall the key points — then reveal.

Sources: GCE AS and A level subject content for the sciences (Department for Education) · AQA A-level Biology 7402 specification (AQA)

§ 03

Digestion and absorption#

●●○StandardLPAQA 7402 3.3.3LPDfE GCE Biology - digestion and absorption

Digestion of starch to glucose

Carbohydrate digestion and absorptionGraph, starch → maltose, maltose → glucose, glucose → into blood (co-transport with Na+)starchmaltoseglucoseinto blood (co-transport withNa+)amylasemaltase(membrane-bou…Na+/glucoseco-transport
Fig. 3Two enzymes in sequence, the second membrane-bound, then absorption by sodium-glucose co-transport.

Key points

Digestion is the hydrolysis of large, insoluble food molecules into small, soluble ones that can be absorbed. Carbohydrates are digested by amylase (starch to maltose) and then by membrane-bound disaccharidases such as maltase (maltose to glucose); proteins by endopeptidases and exopeptidases and membrane-bound dipeptidases (to amino acids); and lipids by lipase, aided by bile salts. The recurrence of membrane-bound enzymes on the epithelial cells is a favourite examined detail.
Lipid digestion depends on emulsification. Bile salts emulsify lipids into tiny droplets, greatly increasing the surface area for lipase to act on; lipase hydrolyses triglycerides to monoglycerides and fatty acids, which combine with bile salts to form micelles. Micelles carry these products to the epithelial cell surface, where they are released and diffuse across the membrane (being lipid-soluble) before being reassembled into triglycerides.
Absorption of the products occurs across the epithelium of the ileum, whose villi and microvilli give an enormous surface area, with a rich blood supply maintaining steep gradients. Amino acids and monosaccharides are absorbed, and here the co-transport mechanism is central: glucose and amino acids are absorbed against their concentration gradient by co-transport with sodium ions.
The glucose co-transport mechanism links back to the Cells chapter. Sodium ions are actively pumped out of the epithelial cell into the blood, lowering the sodium concentration inside the cell; sodium then diffuses in from the gut lumen through a co-transporter protein, and this movement drags glucose in with it, even against the glucose gradient. The glucose then leaves the cell into the blood by facilitated diffusion. This indirect use of ATP (to pump sodium) is a neat and frequently examined idea.
Worked example

The pathway of glucose absorption

Describe, in sequence, how sodium ions enable glucose to be absorbed against its concentration gradient in the ileum.

  1. 01Pump sodium out

    Sodium ions are actively transported (using ATP) from the epithelial cell into the blood, lowering the sodium concentration inside the cell.

  2. 02Sodium co-transport

    Sodium diffuses from the gut lumen into the cell through a co-transporter protein, and glucose is carried in with it, even against the glucose gradient.

  3. 03Glucose to blood

    Glucose then moves from the cell into the blood by facilitated diffusion down its gradient.

Result: A sodium gradient set up by active transport drives glucose uptake by co-transport.

Exam focus

  • Describe the roles of bile salts and micelles in lipid digestion and absorption.
  • Explain glucose absorption by sodium co-transport, making clear that ATP is used to pump sodium, not glucose directly.

Typical mistakes

  • Saying bile salts digest lipids - they emulsify them (increasing surface area); lipase does the hydrolysis.
  • Describing glucose co-transport as active transport of glucose directly, rather than co-transport driven by a sodium gradient.

Active revision

Explain how a molecule of glucose in the lumen of the ileum ends up in the blood, referring to the role of sodium ions and ATP.

Active recall

Recall the key points — then reveal.

Sources: GCE AS and A level subject content for the sciences (Department for Education) · AQA A-level Biology 7402 specification (AQA)

§ 04

Haemoglobin and oxygen transport#

●●●AdvancedLPAQA 7402 3.3.4.1LPDfE GCE Biology - transport of oxygen

Oxygen dissociation curves

Function graph, fetal (higher affinity) = 100*x^2.8/(2.5^2.8 + x^2.8); adult = 100*x^2.8/(3.5^2.8 + x^2.8); Bohr shift (high CO2) = 100*x^2.8/(5^2.8 + x^2.8)Graph of fetal (higher affinity), roots at x = 0, y-intercept at y = 0, increasing, on the interval x from 0 to 14, Graph of adult, roots at x = 0, y-intercept at y = 0, increasing, on the interval x from 0 to 14, Graph of Bohr shift (high CO2), roots at x = 0, y-intercept at y = 0, increasing, on the interval x from 0 to 14246810121420406080100fetal (higheraffinity)adultBohr shift (highCO2)% saturation of haemoglobinpartial pressure of oxygen / …
Fig. 4Higher affinity shifts the curve left (fetal); the Bohr effect (high CO2) shifts it right, releasing more oxygen.

Key points

Haemoglobin is a globular protein with quaternary structure - four polypeptide chains, each with a haem group that binds one oxygen molecule, so one haemoglobin carries up to four oxygens. Loading (association) occurs where the partial pressure of oxygen is high (the lungs) and unloading (dissociation) where it is low (respiring tissues). Its affinity for oxygen - how readily it binds - is the key variable, and it is not constant.
The oxygen dissociation curve, a graph of percentage saturation against the partial pressure of oxygen, is S-shaped (sigmoidal). This shape arises from cooperative binding: the binding of the first oxygen changes the shape of the haemoglobin, making it easier for the next to bind, so the curve is shallow at first, then steep, then levels off as the molecule saturates. The steep middle section is biologically important because a small fall in oxygen partial pressure in the tissues causes a large release of oxygen.
The Bohr effect describes how a higher carbon dioxide concentration shifts the curve to the right, meaning haemoglobin has a lower affinity for oxygen and releases it more readily. Actively respiring tissues produce more carbon dioxide, which lowers the pH; this reduces haemoglobin's affinity so that more oxygen is unloaded exactly where it is most needed - an elegant example of structure serving demand.
Different haemoglobins are tuned to their environment. Fetal haemoglobin has a higher affinity for oxygen than adult haemoglobin (its curve lies to the left), so it can load oxygen from the mother's blood at the placenta, where the adult haemoglobin is unloading. Organisms living where oxygen is scarce (such as at high altitude or in mud) also tend to have high-affinity haemoglobins. Interpreting and comparing these shifted curves is a core AO3 skill.
Worked example

Reading a dissociation curve

Using the adult curve, at a tissue partial pressure of oxygen where the saturation is 40% and a lung partial pressure where it is 96%, calculate the percentage of oxygen unloaded between the lungs and the tissues.

  1. 01Read the two saturations

    In the lungs the haemoglobin is 96% saturated; in the tissues it falls to 40%.

  2. 02Find the difference

    Percentage unloaded = 96 - 40.

    96%−40%=56%96\% - 40\% = 56\%96%−40%=56%
  3. 03Interpret

    56% of the haemoglobin's oxygen capacity is released to the tissues - the steep middle of the curve makes this large release possible for a modest fall in partial pressure.

Result: 56% of the oxygen carried is unloaded to the tissues.

Exam focus

  • Explain the sigmoidal shape using cooperative binding, and interpret loading in the lungs versus unloading in the tissues.
  • Describe and explain a left shift (higher affinity, e.g. fetal) and a right shift (Bohr effect, high CO2) and their advantages.

Typical mistakes

  • Saying a right shift means haemoglobin carries less oxygen overall, rather than that it releases oxygen more readily (lower affinity).
  • Confusing loading (lungs, high pO2) with unloading (tissues, low pO2).

Active revision

Fetal haemoglobin has a dissociation curve to the left of the mother's. Explain how this allows the fetus to obtain oxygen from the maternal blood at the placenta.

Active recall

Recall the key points — then reveal.

Sources: GCE AS and A level subject content for the sciences (Department for Education) · AQA A-level Biology 7402 specification (AQA)

§ 05

The heart and the cardiac cycle#

●●●AdvancedLPAQA 7402 3.3.4.1LPDfE GCE Biology - the heart

The cardiac cycle sequence

The cardiac cycleGraph, atrial systole: atria contract → ventricular systole: AV valves shut, SL valves open, ventricular systole: AV valves shut, SL valves open → diastole: SL valves shut, heart refills, diastole: SL valves shut, heart refills → atrial systole: atria contractatrial systole:atria contractventricularsystole: AVvalves shut, SL…diastole: SLvalves shut,heart refills
Fig. 5Valves open and close according to the pressure differences across them.

Key points

The heart is a double pump: the right side pumps deoxygenated blood to the lungs and the left side pumps oxygenated blood to the body. The left ventricle has a thicker muscular wall than the right because it must generate a higher pressure to force blood all around the body, whereas the right pumps only to the nearby lungs at lower pressure (which protects the delicate capillaries). Atrioventricular valves (bicuspid and tricuspid) and semilunar valves ensure a one-way flow.
The cardiac cycle is the sequence of contraction (systole) and relaxation (diastole) in one heartbeat. In atrial systole the atria contract, pushing the last of the blood into the ventricles; in ventricular systole the ventricles contract, raising ventricular pressure, closing the atrioventricular valves (the first heart sound) and opening the semilunar valves so blood is ejected into the aorta and pulmonary artery; in diastole all chambers relax, the semilunar valves close (the second heart sound) and the heart refills.
Valves open and close according to pressure differences, and this is the key to interpreting a cardiac pressure graph. A valve opens when the pressure behind it exceeds the pressure in front, and closes when the pressure in front exceeds that behind. So, for example, the semilunar valve opens only when ventricular pressure rises above aortic pressure, and closes when the ventricle relaxes and aortic pressure again exceeds it. Being able to read valve events off such a graph is a routine examination task.
The output of the heart is quantified by cardiac output, the volume of blood pumped by one ventricle per minute, given by cardiac output = heart rate x stroke volume, where stroke volume is the volume pumped per beat. This links directly to exercise physiology: during exercise both heart rate and stroke volume rise, increasing cardiac output to deliver more oxygenated blood to the muscles.
cardiac output=heart rate×stroke volume\text{cardiac output} = \text{heart rate} \times \text{stroke volume}cardiac output=heart rate×stroke volume

Cardiac output

Volume of blood pumped by one ventricle per minute; both factors rise during exercise.

Pressure changes in the left heart

Left-heart pressures over one cycleLine chart: pressure / mmHg by time / s, Data: left ventricle · 0.0: 6; left ventricle · 0.1: 8; left ventricle · 0.2: 75; left ventricle · 0.3: 120; left ventricle · 0.4: 100; left ventricle · 0.5: 20; left ventricle · 0.6: 6; left ventricle · 0.7: 6; left ventricle · 0.8: 6; aorta · 0.0: 82; aorta · 0.1: 80; aorta · 0.2: 75; aorta · 0.3: 120; aorta · 0.4: 105; aorta · 0.5: 95; aorta · 0.6: 90; aorta · 0.7: 86; aorta · 0.8: 83; left atrium · 0.0: 5; left atrium · 0.1: 11; left atrium · 0.2: 6; left atrium · 0.3: 4; left atrium · 0.4: 6; left atrium · 0.5: 10; left atrium · 0.6: 12; left atrium · 0.7: 8; left atrium · 0.8: 50204060801001200.00.10.20.30.40.50.60.70.8pressure / mmHgtime / sleft ventricleaortaleft atrium
Fig. 6Representative pressures (mmHg): the semilunar valve opens where ventricular pressure exceeds aortic.
Worked example

Calculating cardiac output

At rest a person has a heart rate of 72 beats per minute and a stroke volume of 70 cm3. Calculate their cardiac output in cm3 min-1 and in dm3 min-1.

  1. 01Apply the equation

    Cardiac output = heart rate x stroke volume = 72×7072 \times 7072×70.

    72×70=5040 cm3 min−172 \times 70 = 5040\ \text{cm}^3\,\text{min}^{-1}72×70=5040 cm3min−1
  2. 02Convert units

    5040 cm3=5.04 dm35040\ \text{cm}^3 = 5.04\ \text{dm}^35040 cm3=5.04 dm3 (since 1 dm3 = 1000 cm3).

Result: Cardiac output = 5040 cm3 min-1 = 5.04 dm3 min-1.

Exam focus

  • Read valve-opening and valve-closing events off a cardiac pressure graph using the rule that a valve opens when the pressure behind exceeds that in front.
  • Calculate cardiac output, stroke volume or heart rate from the other two.

Typical mistakes

  • Saying valves open and shut 'to let blood through' rather than because of a pressure difference.
  • Mixing up the units of cardiac output (cm3 min-1) with stroke volume (cm3).

Active revision

A person has a stroke volume of 75 cm3 and a heart rate of 68 beats per minute. Calculate their cardiac output in dm3 per minute.

Active recall

Recall the key points — then reveal.

Sources: GCE AS and A level subject content for the sciences (Department for Education) · AQA A-level Biology 7402 specification (AQA)

§ 06

Mass transport in plants: transpiration and translocation#

●●●AdvancedLPAQA 7402 3.3.4.2LPDfE GCE Biology - mass transport in plants

A potometer

PotometerSchematic diagram with 7 elements, leafy shoot, capillary tube (scale), air bubble, water reservoir, bubble moves as water is taken up, distance moved in time tleafy shootcapillary tube(scale)air bubblewater reservoirbubble moves aswater is taken …distance movedin time t
Fig. 7As the shoot transpires, water is drawn up and the air bubble moves along the capillary tube.

Key points

Water travels up the xylem from root to leaf, driven ultimately by transpiration - the evaporation of water from the leaf surfaces through the stomata. The cohesion-tension theory explains the mechanism: water evaporating from the mesophyll lowers the water potential there, pulling water out of the xylem; because water molecules are cohesive (hydrogen-bonded to one another), they form a continuous column, so the loss at the top exerts tension that pulls the whole column up. Adhesion of water to the xylem walls helps, and the column is under negative pressure (tension), which is why a xylem vessel narrows slightly during rapid transpiration.
The rate of transpiration is affected by the environmental factors that alter the water-potential gradient or the rate of evaporation: light (opens stomata), temperature (raises the kinetic energy of water molecules), humidity (a humid atmosphere reduces the gradient) and air movement (wind removes saturated air, keeping the gradient steep). A potometer measures the rate of water uptake, which approximates transpiration, by tracking the movement of an air bubble along a capillary tube of known dimensions.
Sugars (mainly sucrose) are transported in the phloem from a source (where they are made or released, such as a photosynthesising leaf) to a sink (where they are used or stored, such as a root or growing fruit) - a process called translocation. The mass-flow hypothesis explains it: sucrose is actively loaded into the sieve tubes at the source, lowering the water potential there so that water enters by osmosis and raises the hydrostatic pressure; at the sink sucrose is removed, water leaves and the pressure falls; the resulting pressure gradient drives a mass flow of solution from source to sink.
Evidence for these theories is examined critically. Support for mass flow includes the observation that phloem sap is under pressure (it exudes when cut) and that removing a ring of bark (which contains the phloem) causes sugars to accumulate above the ring. Support for cohesion-tension includes the narrowing of tree trunks during the day when transpiration is fastest, and the fact that a broken water column stops flow. Evaluating such evidence - and recognising the limitations of each theory - is a distinctive A-Level demand.
rate of water uptake=distance moved×πr2time\text{rate of water uptake} = \dfrac{\text{distance moved} \times \pi r^2}{\text{time}}rate of water uptake=timedistance moved×πr2​

Potometer rate

rrr is the radius of the capillary tube; multiplying the distance the bubble moves by the cross-sectional area gives a volume.

Worked example

Calculating water uptake from a potometer

In a potometer the air bubble moves 60 mm in 5 minutes. The capillary tube has a radius of 0.5 mm. Calculate the rate of water uptake in mm3 min-1.

  1. 01Cross-sectional area

    Area = πr2=π×0.52=0.785\pi r^2 = \pi \times 0.5^2 = 0.785πr2=π×0.52=0.785 mm2.

  2. 02Volume taken up

    Volume = distance x area = 60×0.785=47.160 \times 0.785 = 47.160×0.785=47.1 mm3.

  3. 03Rate

    Rate = 47.1÷547.1 \div 547.1÷5.

    rate=47.15=9.4 mm3 min−1\text{rate} = \frac{47.1}{5} = 9.4\ \text{mm}^3\,\text{min}^{-1}rate=547.1​=9.4 mm3min−1

Result: About 9.4 mm3 min-1.

Exam focus

  • State the cohesion-tension and mass-flow theories and the evidence for and against each.
  • Calculate a rate of water uptake from potometer data and explain how a named factor changes the rate.

Typical mistakes

  • Saying water is 'pushed' up the xylem - it is pulled up under tension by transpiration (root pressure is minor).
  • Confusing the source and the sink in translocation, or forgetting that loading sucrose at the source requires active transport.

Active revision

Explain how an increase in wind speed and a decrease in humidity would each affect the rate of transpiration measured by a potometer.

Active recall

Recall the key points — then reveal.

Sources: GCE AS and A level subject content for the sciences (Department for Education) · AQA A-level Biology 7402 specification (AQA)

Contents

Section -- / 06

    • 01Surface area to volume ratio and exchange surfaces○
    • 02Gas exchange in animals and plants◐
    • 03Digestion and absorption◐
    • 04Haemoglobin and oxygen transport●
    • 05The heart and the cardiac cycle●
    • 06Mass transport in plants: transpiration and translocation●

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  • AQA A-level Biology 7402 specification

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