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Exchange surfaces and gas exchange

No living organism exists in a vacuum. To stay alive, every cell must continuously take in useful substances from its surroundings and get rid of waste materials. In this topic, we will explore why size matters when it comes to exchange, how multicellular organisms have evolved specialized exchange surfaces, and how we calculate the efficiency of gas exchange.


What you'll learn

  • Why organisms need to exchange substances with their environment
  • How to calculate and compare surface area to volume ratios (SA:V)
  • How the human alveoli are adapted for efficient gas exchange
  • The factors affecting the rate of diffusion and how to use Fick's Law (Separate Biology / Higher Tier only)

1. Why do organisms need to transport substances?

Cells are constantly busy carrying out metabolic reactions (such as aerobic respiration) to release energy. To support these reactions, substances must move into and out of cells across the cell membrane.

Substances that must be taken in:

  • Oxygen: Required for aerobic respiration by almost all living cells.
  • Dissolved food molecules: For example, glucose (for respiration) and amino acids (to build proteins).
  • Water: The medium in which all metabolic reactions take place.
  • Mineral ions: Such as sodium, potassium, and calcium in animals, or magnesium and nitrates in plants.

Substances that must be removed (excreted):

  • Carbon dioxide: A waste product of aerobic respiration that becomes toxic if it builds up.
  • Urea: A poisonous waste product formed in the liver from the breakdown of excess amino acids, which must be transported to the kidneys for excretion.
Definition

Excretion

Excretion is the process by which waste products of metabolism (such as carbon dioxide and urea) are removed from an organism.


2. Surface Area to Volume Ratio (SA:V)

Why can an amoeba (a single-celled organism) survive simply by letting gases diffuse across its cell membrane, while a human needs a complex pair of lungs and a circulatory system? The answer lies in the surface area to volume ratio (SA:V).

Definition

Surface Area to Volume Ratio (SA:V)

The surface area to volume ratio is a mathematical comparison of the total outward-facing area of an object (the surface area) to the amount of space it occupies on the inside (its volume).

Single-Celled Organisms

A single-celled organism has a very large surface area compared to its tiny volume. Because its body is so small, the distance from the outer membrane to the very center of the cell is tiny.

  • Substances can diffuse directly into and out of the cell quickly enough to meet its needs.
  • No specialized exchange surfaces or transport systems (like blood vessels) are needed.

Multicellular Organisms

As an organism gets larger, both its surface area and its volume increase—but they do not increase at the same rate. Volume increases much faster than surface area.

Surface Area to Volume Ratio

As a result:

  • The SA:V ratio decreases as size increases.
  • The diffusion distance becomes too great. If a multicellular organism relied on simple diffusion across its skin, oxygen would take days to reach the cells in its core, and those cells would die.
  • Therefore, multicellular organisms require specialized exchange surfaces (like lungs or gills) with huge surface areas, and transport systems (like the circulatory system) to deliver substances directly to cells.
Key Idea

The Big Takeaway

The larger an organism is, the smaller its surface area to volume ratio (SA:V) becomes. This is why large, multicellular organisms cannot rely on simple diffusion alone; they require specialized exchange surfaces and internal transport systems.

Example

Calculating surface area to volume ratio

An organism's body can be modeled as a cube. Calculate the surface area to volume ratio of a model multicellular organism represented by a cube of side length 3 cm. Express your answer in the form X:1X : 1X:1.

  1. Calculate the surface area of one face of the cube:
Area of one face=3 cm×3 cm=9 cm2 \text{Area of one face} = 3\text{ cm} \times 3\text{ cm} = 9\text{ cm}^2 Area of one face=3 cm×3 cm=9 cm2
  1. Calculate the total surface area of the cube (since a cube has 6 identical faces):
Total Surface Area=9 cm2×6=54 cm2 \text{Total Surface Area} = 9\text{ cm}^2 \times 6 = 54\text{ cm}^2 Total Surface Area=9 cm2×6=54 cm2
  1. Calculate the volume of the cube:
Volume=3 cm×3 cm×3 cm=27 cm3 \text{Volume} = 3\text{ cm} \times 3\text{ cm} \times 3\text{ cm} = 27\text{ cm}^3 Volume=3 cm×3 cm×3 cm=27 cm3
  1. Write down the ratio of surface area to volume:
SA : V=54:27 \text{SA : V} = 54 : 27 SA : V=54:27
  1. Simplify the ratio to the form X:1X : 1X:1 by dividing both sides of the ratio by the volume (27):
5427:2727=2:1 \frac{54}{27} : \frac{27}{27} = 2 : 1 2754​:2727​=2:1
Tip

Simplifying Ratios

In biology exams, always simplify your SA:V ratio so that the "volume" side is equal to 1. To do this, simply divide both the surface area and the volume by the volume value.


3. Alveoli: Gas Exchange in the Lungs

In humans, the specialized exchange surfaces responsible for gas exchange are the alveoli (singular: alveolus). These are tiny, microscopic air sacs located at the very ends of the bronchioles inside your lungs.

Oxygen diffuses from the air inside the alveoli into the blood capillary network that wraps around them. At the same time, carbon dioxide diffuses from the blood plasma into the alveoli to be exhaled.

Alveolus and Capillary Cross-Section

Key Adaptations of the Alveoli

To make diffusion as rapid and efficient as possible, alveoli have several key structural adaptations:

  1. Huge Surface Area: There are millions of alveoli in each lung, creating a massive total surface area (about the size of a tennis court!) for diffusion to occur.
  2. Extremely Thin Walls: Both the alveolar wall and the capillary wall are only one cell thick. This means the diffusion distance is incredibly short (less than a micrometer), allowing gases to cross almost instantly.
  3. Moist Lining: The inside of the alveolus is coated with a thin layer of moisture. Gases must dissolve in this liquid layer before they can diffuse across the cell membranes.
  4. Rich Blood Supply: Each alveolus is surrounded by a dense network of capillaries. Blood is constantly moving, carrying oxygenated blood away and bringing deoxygenated blood back. This maintains a steep concentration gradient.
  5. Ventilation (Breathing): By constantly breathing in (bringing in high O2O_2O2​ air) and breathing out (removing high CO2CO_2CO2​ air), we maintain a high concentration of oxygen inside the alveolus compared to the blood.
Common Mistake

Cell walls vs. One cell thick

Do not say that alveoli have "cell walls". Animals do not have cell walls! Instead, write that the alveolar wall is one cell thick. This is a crucial distinction that examiners look for.


4. Factors Affecting the Rate of Diffusion (Separate Biology)

The rate at which a substance diffuses across a membrane is determined by physical properties. There are three primary factors:

  • Surface Area: The larger the area available for molecules to pass through, the more molecules can diffuse at once. (Rate of diffusion is directly proportional to surface area).
  • Concentration Difference: The greater the difference in concentration between the two sides of the membrane, the faster the particles will net-migrate down their gradient. (Rate of diffusion is directly proportional to concentration difference).
  • Diffusion Distance: The thicker the membrane, the longer it takes for particles to travel across it. (Rate of diffusion is inversely proportional to membrane thickness).

5. Fick's Law of Diffusion (Separate Biology & Higher Tier)

We can bring all of these variables together into a single mathematical relationship known as Fick's Law.

Definition

Fick's Law

Fick's Law states that the rate of diffusion is directly proportional to the surface area and concentration difference, and inversely proportional to the thickness of the membrane.

Rate of Diffusion∝Surface Area×Concentration DifferenceThickness of Membrane \text{Rate of Diffusion} \propto \frac{\text{Surface Area} \times \text{Concentration Difference}}{\text{Thickness of Membrane}} Rate of Diffusion∝Thickness of MembraneSurface Area×Concentration Difference​

The symbol ∝\propto∝ means "is proportional to". This formula is incredibly useful because it allows us to calculate how changing one or more of these variables will scale the rate of diffusion.

  • If you double the surface area, the rate of diffusion will double.
  • If you double the concentration difference, the rate of diffusion will double.
  • If you double the thickness of the membrane, the rate of diffusion will be halved.
Example

Applying Fick's Law

A patient has a lung condition that damages their alveoli. The condition causes the total surface area of their lungs to decrease by 50%50\%50% (halved), and causes the thickness of their alveolar membranes to double due to scarring.

Assuming the concentration gradient of oxygen remains unchanged, calculate the factor by which the rate of oxygen diffusion changes.

  1. Write down Fick's Law:
Rate∝Surface Area×Concentration DifferenceThickness of Membrane \text{Rate} \propto \frac{\text{Surface Area} \times \text{Concentration Difference}}{\text{Thickness of Membrane}} Rate∝Thickness of MembraneSurface Area×Concentration Difference​
  1. Assign an arbitrary value of 1 to all initial variables for comparison:
    • Initial Surface Area = 1
    • Initial Concentration Difference = 1
    • Initial Membrane Thickness = 1
    • This gives an initial rate index of:
1×11=1 \frac{1 \times 1}{1} = 1 11×1​=1
  1. Determine the new values based on the changes described in the question:

    • New Surface Area = 0.50.50.5 (decreased by 50%50\%50%)
    • New Concentration Difference = 1 (remains unchanged)
    • New Membrane Thickness = 2 (doubled)
  2. Substitute these new values into Fick's Law to calculate the new relative rate:

New Relative Rate=0.5×12=0.25 \text{New Relative Rate} = \frac{0.5 \times 1}{2} = 0.25 New Relative Rate=20.5×1​=0.25
  1. Compare the new rate to the original rate. The rate of oxygen diffusion has decreased to 0.250.250.25 times (or a quarter) of its original value.

Exam technique

In the exam

  1. Never forget units: If asked to calculate SA:V ratio, make sure to write it as a ratio (e.g., 3:13:13:1). Do not leave it as a fraction.
  2. Structure-function link: When describing alveoli adaptations, always link the structure to its function. Do not just write "alveoli have thin walls". Write "alveoli have thin walls, which are one cell thick, to provide a short diffusion distance for faster gas exchange."
  3. Analyze the math: In Fick's Law questions, look carefully at whether a variable is on the top of the fraction (directly proportional) or the bottom (inversely proportional) to predict how it affects the rate.

Self review

Check yourself

  • Why can single-celled organisms rely on simple diffusion for gas exchange, while humans cannot?
  • List three structural adaptations of human alveoli that maximize the rate of gas exchange.
  • [Separate Biology] State Fick's Law of diffusion. If the concentration difference of a gas across a membrane is tripled, what happens to the rate of diffusion?
Recap questions

1 of 5

A cube-shaped organism has side length 2 cm. What is its surface area to volume ratio in the form X:1X:1X:1?

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To stay alive, every living cell must continuously take in useful substances and remove metabolic waste. Helpful substances include oxygen, glucose, and water. Waste substances that must be excreted include carbon dioxide and urea.

Single-celled organisms have a very large surface area compared to their volume. Because they are tiny, the distance from their outer membrane to the center of the cell is incredibly small.

Substances can diffuse directly into and out of single-celled organisms fast enough to sustain life. They do not require specialized exchange surfaces or complex internal transport networks.

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Which substance do most living cells take in for aerobic respiration?

Exchange surfaces and gas exchange Revision Guide

  1. GCSE
  2. /Biology
  3. /Exchange surfaces and gas exchange