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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 Grade 11 Notes & Examples | IB Biology