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What determines the efficiency of diffusion throughout the model "cells" ?

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1. The most effective cube size that we tested, in terms of maximizing diffusion, was the 1cm^3 agar cube.
2.  The smallest “cell”, 1cm^3, was the most effective at maximizing diffusion because of the lower surface area and smaller volume. This makes less work for the vinegar and less area to cover. Even if the rate of diffusion was the same, the smallest would still be the most efficient because the vinegar must diffuse less area.
3. Cells do not grow to increase surface area because when the surface area to volume ratio gets too small, the cell can no longer grow and function. Therefore, if the cell grows beyond a certain limit and that ratio gets too small, not enough material will be able to diffuse across the membrane fast enough to accommodate the increased volume.
4. Cube C, ratio 4:1, would be the most efficient at maximizing diffusion due to a higher surface area and lower volume.
5. Your body adapts surface area-to-volume ratios to help the exchange of gases by
6. Small cells, like bacteria, are not able to grow to the size of a small fish because bacteria are single celled organisms and have cell walls restricting any further growth. Also, growth would require more energy that the cells do not have.
7. The advantages of large organisms being multicellular is the organisms can diffuse more efficiently because they have multiple small cells that can diffuse faster than one large cell.

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If different-sized model cells are diffused by vinegar solutions then smaller cells will diffuse cells faster due to the higher surface area to volume ratio of smaller cells.

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Rate of Diffusion vs. Ratio of SA to Volume

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Rate of Diffusion vs. Ratio of SA to Volume