The diffusion rate for essential molecules limits cell size because of the \( \mathrm{SA}: \mathrm{V} \) of the cell. Small molecules can move through the cell cytoplasm via simple diffusion. \( S A: V \) is directly related to the efficiency of diffusion through the cell. True False SA:V can be calculated by multiplying the surface area of the cell by the
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Bonus Knowledge
The statement is true. The surface area to volume ratio (\( SA:V \)) is crucial because as a cell grows, its volume increases faster than its surface area. This limits the cell’s ability to exchange materials efficiently, making diffusion less effective for larger cells. This is why many cells are microscopic in size, allowing for a greater \( SA:V \) ratio and more efficient nutrient intake and waste removal. To calculate the \( SA:V \), you can find the surface area of the cell (for example, a cube has a surface area of \( 6a^2 \) where \( a \) is the length of a side) and divide it by the volume (for a cube, it’s \( a^3 \)). So, the formula becomes \( SA:V = \frac{6a^2}{a^3} = \frac{6}{a} \). This simplicity shows how decreasing the cell size dramatically increases the \( SA:V \), enhancing diffusion efficiency!