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Cell Biology - Membrane Transport (Diffusion, Osmosis, Active Transport)

Grade 9IB

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

🔑Concepts

Simple Diffusion: The passive net movement of particles from a region of higher concentration to a region of lower concentration ([high][low][high] \rightarrow [low]) down a concentration gradient. It occurs with small, non-polar molecules like O2O_2 and CO2CO_2.

Facilitated Diffusion: The passive movement of specific molecules (like glucose or ions such as ClCl^-) across a cell membrane via specific transmembrane integral proteins (carrier or channel proteins).

Osmosis: The net movement of water molecules (H2OH_2O) from a region of higher water potential (hypotonic solution) to a region of lower water potential (hypertonic solution) through a partially permeable membrane.

Active Transport: The movement of substances against a concentration gradient ([low][high][low] \rightarrow [high]). This process requires energy in the form of ATPATP and involves specific protein pumps, such as the Na+/K+Na^+/K^+ pump.

Factors Affecting Transport Rate: The rate of exchange is influenced by the concentration gradient (ΔC\Delta C), temperature (TT), and the available surface area (SASA).

Surface Area to Volume Ratio (SA:VSA:V): As a cell grows, its volume increases at a cubic rate while surface area increases at a squared rate. A high SA:VSA:V ratio is essential for efficient diffusion of nutrients and waste products.

📐Formulae

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

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

Surface Area of a Cube=6s2\text{Surface Area of a Cube} = 6s^2

Volume of a Cube=s3\text{Volume of a Cube} = s^3

💡Examples

Problem 1:

A piece of potato with an initial mass of 4.0 g4.0\text{ g} is placed in a concentrated salt solution. After 30 minutes, its mass is 3.2 g3.2\text{ g}. Calculate the percentage change in mass and explain the movement of water.

Solution:

Percentage Change=3.24.04.0×100=20%\text{Percentage Change} = \frac{3.2 - 4.0}{4.0} \times 100 = -20\%

Explanation:

The potato lost 20%20\% of its mass. This occurred because the external salt solution was hypertonic relative to the potato cells. Consequently, water (H2OH_2O) moved out of the potato cells by osmosis, down the water potential gradient.

Problem 2:

Calculate the surface area to volume ratio for a cubic cell with a side length of 2 \mum2\text{ \mu m} and determine if it is more efficient than a cell with a side length of 4 \mum4\text{ \mu m}.

Solution:

For s=2s=2: SA=6(22)=24 \mum2SA = 6(2^2) = 24\text{ \mu m}^2, V=23=8 \mum3V = 2^3 = 8\text{ \mu m}^3. Ratio =248=3= \frac{24}{8} = 3. For s=4s=4: SA=6(42)=96 \mum2SA = 6(4^2) = 96\text{ \mu m}^2, V=43=64 \mum3V = 4^3 = 64\text{ \mu m}^3. Ratio =9664=1.5= \frac{96}{64} = 1.5.

Explanation:

The smaller cell (s=2s=2) has a higher SA:VSA:V ratio (3:13:1) compared to the larger cell (1.5:11.5:1). Therefore, the smaller cell is more efficient at transporting materials across its membrane relative to its metabolic needs.