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Biology - Movement in and out of Cells

Grade 10IGCSE

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.

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Factors that influence the rate of diffusion include: surface area, temperature, concentration gradient, and diffusion distance.

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Osmosis is the net movement of water 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 against the cell wall, making the cell turgid. Water moving out causes the cell to become flaccid and eventually undergo plasmolysis (where the cell membrane pulls away from the cell wall).

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In animal cells, which lack a cell wall, excessive water intake by osmosis can lead to lysis (bursting), while water loss leads to crenation (shriveling).

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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 (against a concentration gradient) using energy from respiration in the form of ATPATP.

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Active transport requires specific carrier proteins in the cell membrane to move molecules across.

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The surface area to volume ratio (SA:VSA:V) is critical for efficient exchange. As an organism increases in size, its SA:VSA:V ratio decreases, necessitating specialized exchange surfaces and transport systems.

📐Formulae

% change in mass=final mass−initial massinitial mass×100\% \text{ change in mass} = \frac{\text{final mass} - \text{initial mass}}{\text{initial mass}} \times 100

SA:V=Total Surface AreaTotal VolumeSA:V = \frac{\text{Total Surface Area}}{\text{Total Volume}}

Rate of Diffusion∝Surface Area×Concentration GradientDiffusion Distance\text{Rate of Diffusion} \propto \frac{\text{Surface Area} \times \text{Concentration Gradient}}{\text{Diffusion Distance}}

💡Examples

Problem 1:

A potato cylinder has an initial mass of 2.50 g2.50\text{ g}. After being placed in a 0.8 mol/dm30.8\text{ mol/dm}^3 sucrose solution for 22 hours, its final mass is 2.15 g2.15\text{ g}. Calculate the percentage change in mass and explain the movement of water.

Solution:

Percentage change=2.15−2.502.50×100=−14%\text{Percentage change} = \frac{2.15 - 2.50}{2.50} \times 100 = -14\%

Explanation:

The negative result indicates a loss in mass. Water moved out of the potato cells by osmosis because the sucrose solution had a lower water potential than the H2OH_2O inside the potato cells. The water moved through the partially permeable cell membranes down a water potential gradient.

Problem 2:

Compare the SA:VSA:V ratio of two cubes: Cube A with a side length of 1 cm1\text{ cm} and Cube B with a side length of 3 cm3\text{ cm}.

Solution:

For Cube A: SA=6×(12)=6 cm2SA = 6 \times (1^2) = 6\text{ cm}^2, V=13=1 cm3V = 1^3 = 1\text{ cm}^3. Ratio =6:1= 6:1. For Cube B: SA=6×(32)=54 cm2SA = 6 \times (3^2) = 54\text{ cm}^2, V=33=27 cm3V = 3^3 = 27\text{ cm}^3. Ratio =54:27=2:1= 54:27 = 2:1.

Explanation:

As the size of the cube increases, the surface area to volume ratio decreases (6:16:1 vs 2:12:1). This explains why larger organisms cannot rely solely on diffusion across their body surface to exchange nutrients and waste.

Problem 3:

Explain how root hair cells take up magnesium ions (Mg2+Mg^{2+}) from the soil when the concentration of ions in the soil is lower than inside the cell.

Solution:

The cell uses active transport.

Explanation:

Since the Mg2+Mg^{2+} ions are moving against a concentration gradient (from low to high concentration), the cell must use energy in the form of ATPATP generated from respiration. Specific carrier proteins in the root hair cell membrane bind to the ions and pump them into the cytoplasm.

Movement in and out of Cells - Revision Notes & Key Formulas | IGCSE Grade 10 Science