krit.club logo

Form and function - Membranes and membrane transport

Grade 11IBBiology

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

🔑Concepts

•

The Fluid Mosaic Model: Membranes consist of a phospholipid bilayer with embedded proteins, cholesterol, and carbohydrates. Phospholipids are amphipathic, containing a hydrophilic phosphate head (PO43−PO_{4}^{3-}) and two hydrophobic fatty acid tails.

•

Membrane Proteins: Integral proteins span the bilayer, while peripheral proteins are attached to the surface. Functions include junction, enzymes, transport (channels/pumps), recognition, anchorage, and transduction (JETRAT).

•

Cholesterol: An amphipathic molecule (C27H46OC_{27}H_{46}O) found in animal cell membranes that modulates membrane fluidity and permeability to some solutes.

•

Passive Transport: Movement of substances down a concentration gradient (ΔC{\Delta}C) without ATPATP expenditure. Includes simple diffusion, facilitated diffusion (via channel or carrier proteins), and osmosis.

•

Osmosis: The net movement of water molecules across a selectively permeable membrane from a region of low solute concentration (high water potential) to a region of high solute concentration (low water potential).

•

Active Transport: Movement of substances against a concentration gradient (ΔC{\Delta}C) using energy derived from ATPATP hydrolysis. A primary example is the Sodium-Potassium pump (Na+/K+Na^{+}/K^{+} pump), which moves 3Na+3Na^{+} out and 2K+2K^{+} in per cycle.

•

Bulk Transport: The use of vesicles to move large molecules or quantities. Endocytosis (phagocytosis and pinocytosis) brings material in, while exocytosis releases material out, utilizing membrane fluidity.

•

Surface Area to Volume Ratio (SA:VSA:V): As a cell grows, its volume (VV) increases faster than its surface area (SASA), decreasing the SA:VSA:V ratio and limiting the efficiency of membrane transport.

📐Formulae

M=IAM = \frac{I}{A}

% 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

Surface Area to Volume Ratio (Sphere)=4πr243πr3=3r\text{Surface Area to Volume Ratio (Sphere)} = \frac{4\pi r^2}{\frac{4}{3}\pi r^3} = \frac{3}{r}

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

💡Examples

Problem 1:

A piece of potato tissue with an initial mass of 2.50g2.50 g is placed in a concentrated sucrose solution. After 2 hours, the final mass is 2.15g2.15 g. Calculate the percentage change in mass and identify the tonicity of the solution relative to the potato cells.

Solution:

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

Explanation:

The negative value indicates a loss of mass. Water moved out of the potato cells via osmosis, meaning the external sucrose solution was hypertonic (higher solute concentration) compared to the cytoplasm.

Problem 2:

A micrograph shows a cell membrane with a thickness of 3.5mm3.5 mm. If the magnification of the image is 500,000×500,000\times, calculate the actual thickness of the membrane in nanometers (nmnm).

Solution:

A=IM=3.5 mm500,000=0.000007 mmA = \frac{I}{M} = \frac{3.5 \text{ mm}}{500,000} = 0.000007 \text{ mm} Converting to nmnm: 0.000007 mm×106=7 nm0.000007 \text{ mm} \times 10^6 = 7 \text{ nm}

Explanation:

The actual thickness (AA) is found by dividing the image size (II) by the magnification (MM). Since 1 mm=106 nm1 \text{ mm} = 10^6 \text{ nm}, the result is 7 nm7 \text{ nm}, which is the standard thickness of a biological membrane.

Problem 3:

Calculate the SA:VSA:V ratio for a cubic model of a cell with a side length (ll) of 2μm2 \mu m versus a side length of 4μm4 \mu m.

Solution:

For l=2μml = 2 \mu m: SA=6×22=24μm2,V=23=8μm3,Ratio=3:1SA = 6 \times 2^2 = 24 \mu m^2, V = 2^3 = 8 \mu m^3, \text{Ratio} = 3:1 For l=4μml = 4 \mu m: SA=6×42=96μm2,V=43=64μm3,Ratio=1.5:1SA = 6 \times 4^2 = 96 \mu m^2, V = 4^3 = 64 \mu m^3, \text{Ratio} = 1.5:1

Explanation:

As the cell size doubles, the SA:VSA:V ratio is halved (3.03.0 to 1.51.5). This demonstrates why cells must remain small or develop specialized shapes (like microvilli) to maintain efficient transport rates.