Gas exchange surfacesEdexcel A-Level Biology A: Revision notes
Section 1
Why organisms need gas exchange surfaces
Cells need oxygen for aerobic respiration and must remove carbon dioxide. Gases cross the exchange surface by diffusion.
The surface area to volume ratio (SA:V) falls as organisms get bigger. For a cube of side 2 cm, SA = 6 × 2² = 24 cm² and V = 2³ = 8 cm³, a ratio of 3 : 1. For a cube of side 4 cm, SA = 96 cm² and V = 64 cm³, a ratio of 1.5 : 1.
A single-celled organism has a large SA:V and a short diffusion distance, so its cell surface is enough. A large animal has a small SA:V and a high demand for oxygen, so it needs a specialised gas exchange surface with a large surface area.
Do not write that large organisms 'have less surface area'. They have a smaller surface area compared with their volume.
Section 2
Properties of gas exchange surfaces
Efficient gas exchange surfaces share three properties:
- a large surface area, so more molecules can cross at once
- a thin surface (short diffusion distance), often one cell thick
- a steep concentration gradient, maintained by ventilation (moving air or water) and blood flow (removing the diffusing gas)
Gas exchange surfaces are also moist, so gases dissolve and can cross the membrane.
Section 3
Fick's law of diffusion
Fick's law states:
rate of diffusion ∝ (surface area × concentration difference) ÷ thickness of the exchange surface
So the rate increases with a larger area or a larger concentration difference, and decreases with a greater thickness.
Worked example: surface A has area 12 cm², concentration difference 8 and thickness 0.2 mm, so rate ∝ 12 × 8 ÷ 0.2 = 480. Surface B has area 12 cm², difference 6 and thickness 0.1 mm, so rate ∝ 12 × 6 ÷ 0.1 = 720. B is 1.5 times faster.
Doubling the thickness halves the rate (if the other factors are constant), because the rate is proportional to 1/thickness.
Show the substitution into the formula in a calculation. A relative rate has no units.
Section 4
Structure of the mammalian lung
Air flows through the trachea, bronchi and bronchioles to the alveoli, which are the gas exchange surfaces. The alveoli are surrounded by a capillary network.
- Many alveoli (about 300 million) give a very large surface area (about 70 m²).
- The alveolar wall is squamous epithelium, one cell thick, and the capillary wall is also one cell thick, so the diffusion distance is very short.
- Ventilation (breathing) keeps the oxygen concentration high and carbon dioxide low in the alveolar air.
- Blood flow removes oxygen and brings carbon dioxide, keeping the gradient steep.
- The moist lining lets gases dissolve.
Section 5
Applying Fick's law to the lungs
Lung diseases can be explained by Fick's law.
- Pulmonary fibrosis: scar tissue thickens the alveolar walls, so the diffusion distance increases and the rate falls.
- Emphysema: alveolar walls break down, forming fewer, larger air spaces, so the surface area falls.
- Pneumonia: fluid in the alveoli increases the distance and reduces the area in contact with air.
- High altitude: the inhaled air contains less oxygen, so the concentration difference is smaller.
- Exercise: faster breathing and blood flow keep the concentration gradient steeper, increasing the rate.
That's the notes covered.
Carry on to the next subtopic.
Exam questions on Gas exchange surfaces
- Single-celled organisms such as Amoeba obtain oxygen by diffusion across their cell surface membrane, but large animals have specialised gas exchange surfaces such as lungs. Students use agar cubes to model how the surface area to volume ratio changes as an organism gets larger.Explain why a large mammal needs a specialised gas exchange surface, but an Amoeba does not.2 marks
- Fick's law states that the rate of diffusion is proportional to (surface area × concentration difference) ÷ thickness of the exchange surface. In a healthy adult the alveolar walls have a thickness of 0.5 μm. In a disease called pulmonary fibrosis, scar tissue thickens the alveolar walls.During vigorous exercise, a person's breathing rate and heart rate both increase. Explain how this increases the rate of diffusion of oxygen into the blood.2 marks
- A student models diffusion across an exchange surface using Fick's law, where rate of diffusion is proportional to (surface area × concentration difference) ÷ thickness. Surface A has an area of 12 cm², a concentration difference of 8 arbitrary units and a thickness of 0.2 mm. Surface B has an area of 12 cm², a concentration difference of 6 arbitrary units and a thickness of 0.1 mm.Calculate the relative rate of diffusion for surface A and for surface B, and state how many times faster diffusion is across B than across A.3 marks
Written by the Exaim team, led by Shaun Daswani (Head of Upper Secondary, Improve ME Institute; MSc Financial Mathematics, Imperial College London; BSc, UCL) and Jason Daswani (operational lead, Improve ME Institute; LSE).