Ink LearningBiologyPracticalsExam boards

BiologyExchange surfaces and gas exchange › Why nothing large can breathe through its skin

Why nothing large can breathe through its skin

An amoeba has no lung, no gills and no blood, and manages perfectly well. Scale it up to the size of a mouse and it would suffocate inside a minute. The reason is arithmetic rather than biology, and once you have the arithmetic the rest of the unit is a set of answers to it.

Before this Diffusion and Fick's law · Membrane structure and permeability

Before you start

A large animal needs lungs because it has more cells to supply than a small one. That sounds obvious and it points at the wrong quantity. An elephant does need more oxygen per second than a mouse, but it also has far more skin to collect it through, so the totals alone settle nothing. The problem is that the two totals do not grow at the same rate. Make an animal twice as long and its volume goes up eight times while its surface goes up four, so every gram of tissue now has half as much surface working for it. Size does not add demand faster than it adds supply; it adds demand faster than it adds surface, which is a different and much sharper problem.

What you should be able to do

The arithmetic that forces the problem

Take a cube of side L. Its surface area is 6L², its volume is L³, and the ratio between them is 6L² ÷ L³, which simplifies to 6 ÷ L. Nothing about that expression is biological. It says that the ratio of surface to volume depends on nothing except size, and that it falls as size rises, and it falls without limit.

A cube of side 1 cm has a ratio of 6. A cube of side 2 cm has a ratio of 3. A cube of side 10 cm has a ratio of 0.6. The 10 cm cube has a hundred times the surface of the 1 cm cube and a thousand times the contents, so each cubic centimetre inside it is served by a tenth as much outside. Every organism that has ever grown past a millimetre or so has had to deal with that, and the ways of dealing with it are what this unit is about.

surface area to volume ratio = surface area ÷ volumeFor a cube this comes to 6 ÷ side; for a sphere it comes to 3 ÷ radius. Both fall as the object grows.

Size is not the only variable, though, and the exam question that separates candidates is usually the one about shape. Volume can be held fixed while the ratio is changed a long way, simply by flattening the body out.

The sphere is the shape with the least surface for a given volume, which is why a hibernating dormouse curls into a ball. Flattening does the opposite, and it is the trick a flatworm, a gill lamella and a leaf all use.

The sphere at 2.4 and the sheet at 8.8 contain identical amounts of material. What differs is how much of that material is close to the outside. This is why a tapeworm can be metres long and still have no specialised exchange surface: it is under a millimetre thick, so nothing inside it is ever far from its own outside.

Working with the ratio

A cube-shaped organism has sides of 0.1 mm. A second organism of the same shape has sides of 1.0 mm. Calculate both ratios, and state how many times greater the demand per unit of surface is in the larger one.

Work in the same units throughout. For the small one, surface = 6 × 0.1² = 0.06 mm² and volume = 0.1³ = 0.001 mm³, giving a ratio of 60 mm⁻¹. The shortcut 6 ÷ L gives the same answer in one step.

For the large one, 6 ÷ 1.0 = 6 mm⁻¹.

The ratio has fallen by a factor of ten, so each unit of surface is now supplying ten times as much tissue. The larger organism needs ten times the flux across every square millimetre of its surface to stay level, and there is no guarantee diffusion can supply it.

Marks are lost here for dropping the units. A ratio of surface to volume has units of one over length, so it is mm⁻¹ or cm⁻¹, and it changes if you switch between them. Quoting '60' with no unit and no statement of the length unit used is not a measurement.

Fick's law: three terms, three things to change

The rate at which anything crosses an exchange surface by diffusion depends on three quantities, and the relationship between them is usually named after Adolf Fick. At A-level it is used as a proportionality, not as an equation with a constant to substitute into.

rate of diffusion ∝ (surface area × difference in concentration) ÷ thickness of the surfaceThe first two multiply the rate. The third divides it, which is why 'thin' appears in every mark scheme about exchange surfaces.

All three terms are properties of one physical surface, and an organism can work on any of them. Folding raises the area; a blood supply that carries the substance away holds up the difference; a wall one cell thick keeps the denominator small.

Read the law as a list of design instructions and the features of every exchange surface fall out of it. A large area means more molecules crossing per second. A large difference in concentration means each molecule crosses faster, which is why a surface that is not ventilated or not supplied with blood soon stops working: the gradient flattens as the substance accumulates on one side. A thin surface shortens the crossing.

Diffusion
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 their random motion. It is passive and requires no metabolic energy.
Exchange surface
A surface across which substances move between an organism and its environment, or between two compartments within it.
Diffusion distance
The thickness of the barrier a substance has to cross, sometimes called the length of the diffusion pathway.
Concentration gradient
The difference in concentration between two points, divided by the distance between them.

One thing the law does not say is anything about time. It gives a rate, and a rate is only useful if the distance involved is short enough for that rate to matter. That is where the real limit on body size sits.

Why diffusion runs out of road

Diffusion is not a fixed-speed process like a car on a motorway. A molecule wandering randomly makes progress in proportion to the square root of the time it has been wandering, which turns around to say that the time it needs rises with the square of the distance. Ten times further takes a hundred times as long.

The numbers are for oxygen in water at body temperature. The first two rows are the scale a cell works at and diffusion handles them easily. By the fourth row it has become useless, and there is nothing an organism can do about it except stop relying on it.

Those figures explain a lot at once. A bacterium 2 micrometres across is supplied throughout in under a millisecond. A human egg cell, around 120 micrometres, is at the far edge of what diffusion will serve, and it is one of the largest cells in the body. A block of tissue a centimetre thick could not be supplied by diffusion in any useful time at all, which is why every large organism has both a specialised exchange surface and a transport system to carry substances the rest of the way.

Small organisms escape all of this. An Amoeba or a Paramecium has a very high surface area to volume ratio and no part of its cytoplasm far from the outside, so oxygen diffuses in across the cell surface membrane fast enough to meet demand and carbon dioxide diffuses out the same way. No lung, no gill, no blood, no problem. It is worth saying explicitly in an answer that the whole body surface is the exchange surface, because that is the marking point.

Flatworms take the other route. A planarian may be a centimetre long but it is a fraction of a millimetre thick, so its flattened shape keeps every cell within diffusion distance of the outside while allowing the animal to be big enough to see. Length is cheap; thickness is what costs.

Metabolic rate complicates the picture in a way worth knowing. A small mammal has a high ratio, so it loses heat quickly, so it has to respire faster per gram to stay warm, so its oxygen demand per gram is higher than an elephant's. Geometry that helps with supply hurts with heat, and a shrew ends up with both a high demand and a short life expectancy. Questions on this often pair a table of body masses with a table of breathing rates and ask you to explain the trend.

What every specialised exchange surface has in common

Once an organism is too large for its own outside to serve it, the answer is always the same in outline: grow a dedicated surface, make it enormous by folding it, keep it thin, and keep something moving on both sides of it so the gradient never flattens.

FeatureWhich term of Fick's law it servesExample
Large surface area, usually by foldingRaises the area on top of the fractionAlveoli, gill lamellae, root hairs, villi
A wall one cell thickCuts the thickness underneath the fractionSquamous alveolar epithelium; a single layer on a gill lamella
A blood supply, or its equivalentRemoves the substance and holds the difference upAlveolar capillary network; the blood in a gill filament
Ventilation, or another way of renewing the mediumRenews the supply on the outside and holds the difference upBreathing; the buccal pump of a fish; abdominal pumping in insects
A permeable, usually moist surfaceLets the substance cross at all; gases dissolve before diffusingThe alveolar lining fluid; the moist walls of mesophyll cells

That table is a better revision tool than a list of adaptations, because it forces you to say why. An answer that states an alveolus is thin scores nothing on its own. An answer that states an alveolus is one cell thick, so the diffusion pathway is short, so the rate of diffusion is higher, scores the marks it is asked for.

The last row hides a cost that the third lesson of this unit returns to. A surface thin enough and wet enough for oxygen to cross is also thin enough and wet enough for water to leave, which is why the gas exchange surface of a land animal is folded away inside the body rather than hung out in the air, and why a plant spends a good deal of its structure managing the same trade-off.

TRY IT — Predicting from the ratio

A spherical single-celled organism has a radius of 20 micrometres. A second species, also spherical, has a radius of 60 micrometres. Both rely on diffusion across the cell surface membrane. Using the ratio, explain which is more likely to be limited by oxygen supply, and suggest one change of shape that would help the larger one.

Check your answer

For a sphere the ratio is 3 ÷ radius. The small one gives 3 ÷ 20 = 0.15 per micrometre; the large one gives 3 ÷ 60 = 0.05 per micrometre, a third as much.

The larger species therefore has a third of the surface per unit of volume, while its oxygen demand goes with volume, so it is the one more likely to be limited by supply. Its centre is also three times further from the surface, and since diffusion time goes with the square of distance, oxygen takes about nine times as long to reach it.

Flattening or elongating the cell would raise the ratio without losing any volume, and would shorten the longest internal distance at the same time. Both matter, and an answer that mentions only the ratio has answered half the question.

In the exam

Check yourself

A student claims that a large organism could avoid needing a gas exchange system by simply having a very large surface area. Evaluate this claim, using the ratio and the behaviour of diffusion over distance.

Answer

The claim is not wrong so much as incomplete. A large surface area on its own is no use; what matters is the surface area relative to the volume it has to supply, because demand rises with volume.

An organism can raise its ratio by changing shape, and flatworms and tapeworms do exactly that. So the strategy is genuinely available, and some large organisms use it.

It runs out for a second reason, though, which is distance. Even with plenty of surface, any cell buried deep inside is a long way from it, and diffusion time rises with the square of distance: a centimetre takes hours. So a body that stays thin can manage without an exchange system, and a body that is thick cannot, however large its surface is.

A complete answer therefore separates the two limits: the ratio limits how much can cross, and the diffusion distance limits how quickly it reaches the middle. Large active animals fail both tests, which is why they have a specialised exchange surface and a transport system to finish the delivery.

Questions

Written to the command words the boards use. Try them on paper before opening a scheme: the marks go to points made, not to length.

Question 14 marks

A cube-shaped organism has sides of 0.5 mm. A second organism of the same shape has sides of 2.0 mm. Calculate the surface area to volume ratio of each, giving a unit, and calculate how many times greater the smaller organism's ratio is.

Mark scheme
  1. M1 ratio is surface area divided by volume, which for a cube simplifies to 6 divided by the length of a side
  2. A1 smaller organism: 6 ÷ 0.5 = 12 mm⁻¹, with the unit given as one over length
  3. M1 larger organism: 6 ÷ 2.0, worked in the same length unit as the first
  4. A1 larger organism gives 3 mm⁻¹, so the smaller ratio is four times greater

Question 24 marks

A flatworm 0.3 mm thick supplies all of its cells with oxygen by diffusion across its body surface. A block of tissue one centimetre thick cannot. Explain the difference.

Mark scheme
  1. B1 surface area rises with the square of length while volume rises with the cube, so the surface area to volume ratio falls as an organism gets thicker
  2. B1 demand for oxygen goes with the volume of respiring tissue, so each unit of surface has more tissue to supply in the larger body
  3. B1 diffusion time rises with the square of the distance, so ten times further takes about a hundred times as long
  4. A1 over a centimetre diffusion would take hours and could never meet demand, whereas the flatworm's flattened shape keeps every cell a fraction of a millimetre from the outside

Question 33 marks

A sphere and a flattened sheet contain identical volumes of tissue. The sphere has a surface area to volume ratio of 2.4 and the sheet one of 8.8. Compare the two shapes as exchange surfaces.

Mark scheme
  1. B1 the sheet's ratio is about 3.7 times the sphere's, even though the two contain the same volume of material
  2. B1 the sheet therefore has far more surface serving each unit of volume, while the sphere is the shape with the least surface possible for a given volume
  3. B1 the centre of the sphere is also much further from the nearest surface than any point in the sheet, so the diffusion distance is longer in the sphere as well

Question 43 marks

A hibernating dormouse curls its body into a tight ball. Suggest how this reduces the rate at which it loses heat to its surroundings.

Mark scheme
  1. B1 curling brings the body closer to a sphere, which is the shape with the smallest surface area for a given volume
  2. B1 the surface area to volume ratio is therefore lowered without any change in the volume of the animal
  3. A1 heat is lost across the body surface, so less surface per unit of body volume means a slower rate of heat loss

Question 52 marks

State the relationship known as Fick's law, naming all three of the quantities on which the rate of diffusion across a surface depends.

Mark scheme
  1. B1 rate of diffusion is proportional to the surface area multiplied by the difference in concentration
  2. B1 and inversely proportional to the thickness of the surface, so a thinner barrier gives a faster rate

Worth remembering

← The cell cycle: growth, copying, and one division that changes nothing · Lungs: seventy square metres, folded into a chest →