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Biology - Movement in and out of cells (Diffusion, Osmosis, and Active Transport)

Grade 9IGCSE

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 down a concentration gradient, as a result of their random movement. This process is passive and allows substances like O2O_2 and CO2CO_2 to cross membranes.

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The rate of diffusion is influenced by several factors: surface area (larger AA increases rate), temperature (higher TT increases kinetic energy), concentration gradient (steeper gradient increases rate), and distance (dd, shorter distance increases rate).

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Osmosis is the net movement of water (H2OH_2O) molecules from a region of higher water potential (dilute solution) to a region of lower water potential (concentrated solution), through a partially permeable membrane.

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In plant cells, water moving in by osmosis creates turgor pressure, making the cell 'turgid'. If water leaves the cell, the vacuole shrinks and the cell membrane pulls away from the cell wall, a process called 'plasmolysis'.

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Active transport is the movement of particles through a cell membrane from a region of lower concentration to a region of higher concentration using energy from respiration in the form of ATPATP. This process requires specific carrier proteins in the membrane.

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Importance of active transport: Ion uptake by root hair cells (e.g., Mg2+Mg^{2+} for chlorophyll or NO3−NO_3^- for proteins) and glucose uptake by epithelial cells of the villi in the small intestine.

📐Formulae

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

Percentage Change in Mass=Final Mass−Initial MassInitial Mass×100\text{Percentage Change in Mass} = \frac{\text{Final Mass} - \text{Initial Mass}}{\text{Initial Mass}} \times 100

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}}

💡Examples

Problem 1:

A potato cylinder has an initial mass of 2.50 g2.50\text{ g}. After being placed in a concentrated sucrose solution for 2 hours, its final mass is 2.15 g2.15\text{ g}. Calculate the percentage change in mass and explain the movement of molecules.

Solution:

Percentage Change=2.15−2.502.50×100=−14.0%\text{Percentage Change} = \frac{2.15 - 2.50}{2.50} \times 100 = -14.0\%

Explanation:

The mass decreased by 14.0%14.0\%. This occurred because the sucrose solution had a lower water potential than the potato cells. Consequently, water (H2OH_2O) moved out of the potato cells by osmosis across the partially permeable cell membranes.

Problem 2:

Explain how a root hair cell takes up nitrate ions (NO3−NO_3^-) from the soil where the concentration of ions is lower than inside the cell.

Solution:

Low [NO3−]soil→High [NO3−]cell\text{Low } [NO_3^-]_{\text{soil}} \rightarrow \text{High } [NO_3^-]_{\text{cell}}

Explanation:

Since the ions are moving against the concentration gradient (from low to high concentration), the cell must use active transport. This process requires energy in the form of ATPATP and involves specific carrier proteins located in the cell membrane to pump the NO3−NO_3^- ions into the cytoplasm.