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Movement in and out of Cells - Diffusion

Grade 11IGCSEBiology

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

🔑Concepts

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Diffusion is defined as 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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Diffusion is a passive process, meaning it does not require energy input from the cell; the energy for movement is derived from the kinetic energy of the molecules or ions.

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Living organisms obtain many requirements through diffusion, such as O2O_2 for respiration and CO2CO_2 for photosynthesis, and remove waste products like CO2CO_2 from respiration.

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The cell membrane is a partially permeable barrier that allows small molecules like H2OH_2O, O2O_2, and CO2CO_2 to pass through while restricting larger molecules like starch or proteins.

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The rate of diffusion is influenced by several factors: the steepness of the concentration gradient, the temperature (higher temperature increases kinetic energy), the surface area of the membrane, and the diffusion distance.

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The Surface Area to Volume ratio (SA:VSA:V) is critical; as an object (or cell) increases in size, its volume increases much faster than its surface area, leading to a decrease in the efficiency of diffusion.

📐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=Total Surface AreaTotal Volume\text{Surface Area to Volume Ratio} = \frac{\text{Total Surface Area}}{\text{Total Volume}}

💡Examples

Problem 1:

A cell has a concentration of glucose at 2 mmol/dm32\text{ mmol/dm}^3 inside the cytoplasm and is surrounded by an extracellular fluid with a glucose concentration of 10 mmol/dm310\text{ mmol/dm}^3. Describe the movement of glucose if the membrane is permeable to it.

Solution:

Glucose will move into the cell via diffusion until an equilibrium is reached.

Explanation:

Since the external concentration (10 mmol/dm310\text{ mmol/dm}^3) is higher than the internal concentration (2 mmol/dm32\text{ mmol/dm}^3), a concentration gradient exists. Glucose molecules move down this gradient from the higher concentration to the lower concentration.

Problem 2:

Compare the diffusion of O2O_2 at 10∘C10^\circ\text{C} and 40∘C40^\circ\text{C}. Which temperature results in a faster rate?

Solution:

The rate of diffusion will be significantly higher at 40∘C40^\circ\text{C}.

Explanation:

At higher temperatures, particles possess more kinetic energy (EkE_k). This causes the O2O_2 molecules to move faster, leading to more frequent and rapid net movement across a space or membrane.

Problem 3:

Explain why single-celled organisms, such as Amoeba, do not need specialized gas exchange surfaces like lungs.

Solution:

They have a very large Surface Area to Volume ratio (SA:VSA:V).

Explanation:

Because the organism is so small, its surface area is large enough relative to its volume to allow O2O_2 to diffuse into the cell and CO2CO_2 to diffuse out fast enough to meet the metabolic demands of the entire cytoplasm.

Diffusion - Revision Notes & Key Diagrams | IGCSE Grade 11 Biology