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Cell Biology - Membrane Structure

Grade 11IBBiology

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

🔑Concepts

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The Fluid Mosaic Model: Proposed by Singer and Nicolson in 1972, describing the membrane as a fluid bilayer of phospholipids with embedded proteins.

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Phospholipid Structure: Phospholipids are amphipathic molecules containing a hydrophilic (polar) phosphate head and two hydrophobic (non-polar) fatty acid tails. This results in the spontaneous formation of a bilayer in H2OH_2O.

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Membrane Proteins: Categorized into Integral proteins (transmembrane, amphipathic) and Peripheral proteins (bound to the surface, polar). Their functions include Transport, Receptors, Anchorage, and Enzymatic activity.

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Cholesterol: A steroid (C27H46OC_{27}H_{46}O) found in animal cell membranes. It regulates membrane fluidity by preventing the tails from crystallizing at low temperatures and restricting molecular motion at high temperatures.

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The Davson-Danielli Model: An early 'protein-lipid sandwich' model (19351935) which was falsified by evidence from freeze-fracture electron microscopy and fluorescent antibody tagging.

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Membrane Fluidity: Dependent on the ratio of saturated to unsaturated fatty acid tails. Unsaturated tails have 'kinks' due to C=CC=C double bonds, increasing fluidity.

📐Formulae

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

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

SA:V=Surface AreaVolumeSA:V = \frac{\text{Surface Area}}{\text{Volume}}

Width of Plasma Membrane≈7.5×10−9 m\text{Width of Plasma Membrane} \approx 7.5 \times 10^{-9} \text{ m}

💡Examples

Problem 1:

An electron micrograph shows a plasma membrane with a measured thickness of 5 mm5\text{ mm}. If the magnification of the image is ×500,000\times 500,000, calculate the actual thickness of the membrane in nanometers (nmnm).

Solution:

Actual Size (AA) = 5 mm500,000=0.00001 mm\frac{5\text{ mm}}{500,000} = 0.00001\text{ mm}. To convert to nmnm: 0.00001×106=10 nm0.00001 \times 10^6 = 10\text{ nm}.

Explanation:

Using the magnification formula A=IMA = \frac{I}{M}, we divide the image size by the magnification factor. Since 1 mm=1,000,000 nm1\text{ mm} = 1,000,000\text{ nm} (or 106 nm10^6\text{ nm}), the final result is 10 nm10\text{ nm}.

Problem 2:

Contrast the solubility of the two regions of a phospholipid molecule in relation to H2OH_2O.

Solution:

The phosphate head is polar and therefore hydrophilic (soluble in H2OH_2O), while the hydrocarbon tails are non-polar and hydrophobic (insoluble in H2OH_2O).

Explanation:

The 'amphipathic' nature is critical for the formation of the bilayer where tails face inward, away from the aqueous environment of the cytoplasm and interstitial fluid.

Membrane Structure - Revision Notes & Key Diagrams | IB Grade 11 Biology