Organisms Exchange Substances with their Environment
Organisms Exchange Substances with their Environment is the A-level Biology topic covering surface area to volume ratio, gas exchange in fish, insects and leaves, and digestion and absorption in mammals. It explains why larger, more complex organisms need specialised exchange surfaces and mass transport systems instead of relying on diffusion alone, and includes surface area and rate calculations.
Method
- Start every exchange-surface question by considering surface area to volume (SA:V) ratio: smaller organisms have a high SA:V ratio and can rely on diffusion, larger organisms have a low SA:V ratio and need specialised surfaces.
- Learn the four features that maximise any exchange surface: large surface area, short/thin diffusion pathway, steep concentration gradient (maintained by ventilation or blood flow), and permeability to the substance exchanged.
- For the fish gill countercurrent system, use given oxygen saturation data to show that a diffusion gradient is maintained along the whole length of the lamella, not just at one end.
- For insect tracheal systems and leaf gas exchange, link structure (spiracles, tracheoles, stomata, guard cells) to how gases move by diffusion or mass flow.
- For calculations, identify the correct formula (surface area, volume, or rate = quantity / time), substitute values carefully with consistent units, and round to the number of significant figures asked for.
- For digestion and absorption questions, trace the pathway of a named nutrient from digestion (naming the enzyme) through to absorption in the ileum (naming the transport mechanism, e.g. co-transport).
Worked example
A cube-shaped organism has sides of length 5 mm. Calculate its surface area, volume, and surface area to volume ratio in its simplest form, and state whether it could rely on diffusion alone across its body surface.
- Calculate the surface area: a cube has 6 faces, so surface area = 6 x (5 mm)^2 = 6 x 25 = 150 mm^2.
- Calculate the volume: volume = (5 mm)^3 = 125 mm^3.
- Write the ratio and simplify: 150 : 125 = 1.2 : 1 (divide both sides by 125).
- Final answer: SA:V = 1.2 : 1. This is a relatively low ratio, so this organism could not rely on diffusion alone across its whole body surface and would need a specialised exchange surface and a mass transport system.
Practice questions
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Q1What is meant by the term diffusion?Show answer
Answer: The net movement of particles from a region of higher concentration to a region of lower concentration, down a concentration gradient, as a result of random motion.
Q2Name the pores in a leaf's lower epidermis through which gas exchange occurs, and the cells that control their opening.Show answer
Answer: Stomata; guard cells.
Q3State two features that maximise the rate of diffusion across a gas exchange surface.Show answer
Answer: Any two of: large surface area, short/thin diffusion pathway, steep concentration gradient (maintained by ventilation/blood flow), permeable surface.
Q4A cube-shaped cell has sides of 2 mm. Calculate its surface area to volume ratio in its simplest form.Show answer
Answer: 3 : 1 (surface area = 6 x 4 = 24 mm^2; volume = 8 mm^3; 24 : 8 = 3 : 1)
Q5Explain why insects rely mainly on diffusion through their tracheal system rather than a mass transport (blood) system to deliver oxygen to tissues.Show answer
Answer: Air (and therefore oxygen) diffuses directly to respiring tissues through the tracheae and tracheoles, so a separate blood-based oxygen transport system is not needed; this limits insect body size because diffusion alone becomes too slow over larger distances.
Q6An agar cube of side length 3 cm, containing an alkali and phenolphthalein indicator, is placed in acid. Acid diffuses in from all six faces and the cube takes 270 seconds to turn fully colourless. Calculate the mean rate of diffusion into the cube, in cm per second, to 2 significant figures.Show answer
Answer: 0.0056 cm/s (greatest distance to the centre = half the side length = 1.5 cm; rate = 1.5 / 270 = 0.00556, rounds to 0.0056)
Exam-style questions
Written in the style of a A Level Science exam paper, with a full mark scheme.
State three factors, other than the distance travelled, that affect the rate of diffusion of a substance across a surface.
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Fish gills use a countercurrent system. At four equally spaced positions along a gill lamella (in the direction of water flow), the oxygen saturation of the water is 92%, 74%, 56% and 38%, and the oxygen saturation of the blood (flowing in the opposite direction) at the same four positions is 82%, 64%, 46% and 28%. Use these data to explain how the countercurrent arrangement maintains a diffusion gradient for oxygen along the whole length of the lamella.
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A student compared stomatal density between a xerophyte leaf (adapted to a dry habitat) and a mesophyte leaf (adapted to average water availability), using a light microscope with a circular field of view of diameter 0.4 mm. The xerophyte leaf had a stomatal density of 40 stomata per mm^2, and the mesophyte leaf had a stomatal density of 220 stomata per mm^2. Calculate the number of stomata expected in one full field of view for each leaf type. Use area of a circle = pi x r^2, and pi = 3.142. Give each answer to the nearest whole stoma.
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