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Movement in and out of Cells - Diffusion

Grade 12IGCSEBiology

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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Diffusion is the net movement of particles from a region of their higher concentration to a region of their lower concentration (downdown a concentration gradient) as a result of their random movement.

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The energy for diffusion comes from the kinetic energy of random movement of molecules and ions.

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Substances move into and out of cells by diffusion through the cell membrane. Only small, non-polar molecules like O2\text{O}_2 and CO2\text{CO}_2 can diffuse directly through the phospholipid bilayer.

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Several factors influence the rate of diffusion: the surface area of the membrane, the temperature (TT), the concentration gradient (ΔC\Delta C), and the diffusion distance (dd).

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The Surface Area to Volume ratio (SAV\frac{SA}{V}) is a critical factor for cell size; as a cell increases in size, its SAV\frac{SA}{V} decreases, which limits the efficiency of nutrient and gas exchange.

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In multicellular organisms, specialized surfaces (like the alveoli in lungs or root hair cells in plants) have adaptations to increase the rate of diffusion, such as being only one cell thick to minimize the diffusion distance (dd).

📐Formulae

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}}

Surface Area to Volume Ratio=Surface AreaVolume\text{Surface Area to Volume Ratio} = \frac{\text{Surface Area}}{\text{Volume}}

Net Movement=Molecules Moving In−Molecules Moving Out\text{Net Movement} = \text{Molecules Moving In} - \text{Molecules Moving Out}

💡Examples

Problem 1:

A cube-shaped cell has a side length of 2 \mum2\text{ \mu m}. Calculate its surface area to volume ratio and explain how doubling the side length to 4 \mum4\text{ \mu m} affects the efficiency of diffusion.

Solution:

For side s=2 \mums = 2\text{ \mu m}: SA=6s2=6(2)2=24 \mum2SA = 6s^2 = 6(2)^2 = 24\text{ \mu m}^2. V=s3=23=8 \mum3V = s^3 = 2^3 = 8\text{ \mu m}^3. SAV=248=3 \mum−1\frac{SA}{V} = \frac{24}{8} = 3\text{ \mu m}^{-1}. For side s=4 \mums = 4\text{ \mu m}: SA=6(4)2=96 \mum2SA = 6(4)^2 = 96\text{ \mu m}^2. V=43=64 \mum3V = 4^3 = 64\text{ \mu m}^3. SAV=9664=1.5 \mum−1\frac{SA}{V} = \frac{96}{64} = 1.5\text{ \mu m}^{-1}.

Explanation:

When the side length doubles, the SAV\frac{SA}{V} ratio is halved (from 33 to 1.51.5). This decrease in ratio means there is less surface area available for the diffusion of substances like O2\text{O}_2 relative to the volume of the cell that requires those substances, making diffusion less efficient for larger cells.

Problem 2:

Explain the movement of CO2\text{CO}_2 in a leaf during the daytime when photosynthesis is occurring rapidly.

Solution:

During the day, the concentration of CO2\text{CO}_2 inside the leaf is low because it is used in the chloroplasts for photosynthesis. The concentration of CO2\text{CO}_2 in the atmosphere is higher. Therefore, CO2\text{CO}_2 diffuses downdown the concentration gradient from the air, through the stomata, and into the spongy mesophyll cells.

Explanation:

This is a biological application of diffusion where a concentration gradient (ΔC\Delta C) is maintained by the metabolic consumption of the solute (CO2\text{CO}_2) within the cell.

Diffusion - Revision Notes & Key Diagrams | IGCSE Grade 12 Biology