Diffusion  

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• Diffusion = movement of particles from high → low concentration

   until balanced (equilibrium).  

• No energy needed — happens automatically.

 

Examples:  

• Oxygen into cells during respiration  

• Carbon dioxide out during gas exchange in lungs

• Urea from cells into blood (for kidney removal)

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Surface Area to Volume Ratio (SA:V)  

• Small cells = high SA:V → fast diffusion → no special systems needed  

• Large organisms = low SA:V → need exchange organs

 

Calculation Example (Cube):  

Side length = 2 cm  

• Surface area = 6 × (2×2) = 24 cm²  

• Volume = 2×2×2 = 8 cm³  

• SA:V = 24:8 = 3:1

 

Compare:  

• 10:1 → smaller organism (better exchange)  - can rely on diffusion alone

• 3:1 → larger organism (needs lungs/gills) - diffusion isn’t quick enough so needs specialised exchange surface

 

Exchange Surfaces in Nature:

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“How is X adapted for exchange?” — Answer Formula  

1. Large surface area due to [villi / alveoli / etc.]  

2. Thin membrane (one cell thick) → short diffusion distance  

3. Good blood supply (animals) → maintains concentration gradient  

4. Ventilation (lungs) → keeps air moving

 

OSMOSIS  

 

Osmosis = movement of water from a dilute (weaker)

solution to a concentrated (stronger) solution across

a partially permeable membrane until balanced.

Common 3 marker on the exam.

 

• Only water moves — no other particles.  

• Partially permeable membrane = lets water

   through but blocks larger molecules.

   An example of one of these is a cell membrane.

 

Analogy:  

Imagine two rooms connected by a door that only lets water droplets through, not people. Water flows from the crowded room (concentrated) to the empty one (dilute) until both are the same.

 

Required Practical 3: Test how different salt/sugar solutions affect potato mass.

 

There is a common calculation you need to be able to do in the exam in relation to this practical:

 

% Mass Change Calculation  

(Final mass − Initial mass) ÷ Initial mass × 100  

 

Example:  

Potato starts at 5 g → ends at 6 g after soaking in water  

(6 − 5) ÷ 5 × 100 = +20% gain

 

ACTIVE TRANSPORT  

 

• Active transport = movement of particles from low → high concentration (against the gradient).  

• Requires energy from respiration. 

• Used when cells need substances in higher amounts than outside.

Imagine it’s like a cyclist going up a hill, he would be required to pedal and have lots of energy to pedal against the steep gradient.

 

Examples:  

• Root hair cells absorb mineral ions (e.g., nitrate) from soil  

• Small intestine absorbs glucose from digested food into blood

 

Diffusion vs Osmosis vs Active Transport — Summary Table

 

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