Example 1
Find SA:V for a cube with side 2 micrometres.
Area =6×22=24 micrometres2; volume =23=8 micrometres3. SA:V=3 micrometres−1.
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Year 12 · SACE Stage 2
Membranes and exchange
Classify transport and calculate surface-area-to-volume ratios.
Method and worked examples
The fluid mosaic model describes a phospholipid bilayer with mobile proteins and other components. Small non-polar molecules can cross it; many ions and polar solutes require transport proteins. Passive movement follows a concentration gradient for uncharged solutes, but an electrochemical gradient for ions, combining concentration and electrical effects. Active transport uses energy to oppose the relevant gradient. Aquaporins are water channels.
Osmosis is net water movement across a selectively permeable membrane toward lower water potential; for otherwise comparable solutions this is toward greater non-penetrating solute concentration. Endocytosis brings material into vesicles and exocytosis releases vesicle contents. A cube of side a has surface area 6a2, volume a3 and SA:V=a6. Smaller cells therefore have more exchange surface per unit volume.
Example 1
Example 2
A membrane passes water but not sucrose. Outside sucrose concentration is greater. Predict initial net movement.
Water moves out of the cell toward the more concentrated non-penetrating solute solution; the cell tends to lose volume.
Watch out: Facilitated diffusion uses a channel or carrier but moves down a gradient without direct energy input.