Final answer:
The immune cell with a radius of 8 µm would have the highest efficiency for exchanging nutrients and wastes with its environment due to its higher surface area-to-volume ratio compared to the larger or differently shaped cells listed.
Step-by-step explanation:
The cell that would have the highest exchange rate of metabolites with its environment based on the surface area-to-volume ratio would be the immune cell (sphere) with a radius of 8 µm. To determine this, one must calculate the surface area-to-volume ratio for each cell. The surface area of a sphere is given by the formula 4πr² and the volume by ⅓πr³. For the immune cell (sphere):
- Surface Area = 4π(8²) = 804.25 µm²
- Volume = ⅓π(8³) = 2144.66 µm³
- Surface Area-to-Volume Ratio = Surface Area / Volume = 804.25 / 2144.66 = 0.38 approximately
Since the immune cell has a smaller radius than the oocyte cell, and much smaller dimensions in comparison to the rectangular cells, it would have a higher surface area-to-volume ratio, making it more efficient in exchanging materials with its environment. It is recognized that as cells grow larger, there is a decrease in the surface area-to-volume ratio, which makes diffusion less efficient in larger cells compared to smaller ones.
Therefore, in terms of efficiency for the exchange of nutrients and wastes with the environment, the spherical immune cell with a smaller radius is superior relative to the other cell types provided because of its favorable surface area-to-volume ratio.