Final answer:
The spherical cell with a surface area-to-volume ratio of approximately 1.2 is more efficient at exchanging nutrients and wastes compared to the cubed cell, which has a ratio of approximately 0.86, due to having a higher surface area relative to its volume.
Step-by-step explanation:
To determine which cell exchanges nutrients and wastes with its environment more efficiently, we compare the surface area-to-volume ratios of a spherical cell to a cubed cell. A spherical cell with a diameter of 5 µm (meaning a radius of 2.5 µm) has a surface area of 4π(2.5µm)2 and a volume of (4/3)π(2.5µm)3. Calculating these, we find a surface area of approximately 78.54 µm2 and a volume of approximately 65.45 µm3, resulting in a surface area-to-volume ratio of roughly 1.2.
In contrast, a cube-shaped cell with a side length of 7µm has a surface area of 6 x (7µm)2 and a volume of (7µm)3. The resulting surface area is 294 µm2, and the volume is 343 µm3, leading to a surface area-to-volume ratio of approximately 0.86.
Therefore, the spherical cell has a higher surface area-to-volume ratio and is more efficient in terms of exchanging nutrients and wastes compared to the cubed cell. This is because a larger surface area-to-volume ratio allows for more surface area per unit of volume for diffusion to occur.