Question

What is the surface-area-to-volume ratio for the cuboidal cell that removes waste via diffusion most efficiently?

A) (6 μ m⁻¹)
B) (3 μ m⁻¹)
C) (2 μ m⁻¹)
D) (1 μ m⁻¹)

Answer 1

The most efficient cuboidal cell for waste removal by diffusion is the one with the highest surface-area-to-volume ratio, which is 6 μm⁻¹.

Step-by-step explanation:

The efficiency of a cell in removing waste through diffusion greatly depends on its surface-area-to-volume ratio. For a cuboidal cell, the surface area (SA) is calculated by the formula 6 × S², and the volume (V) by S³, where S is the length of one side of the cube.

Let's calculate the surface-area-to-volume ratio (SA:V) for different sizes of cuboidal cells:

• If S = 1 μm, SA = 6 μm² and V = 1 μm³, then SA:V = 6 μm²:1μm³ = 6 μm⁻¹.
• If S = 2 μm, SA = 24 μm² and V = 8 μm³, then SA:V = 24 μm²:8μm³ = 3 μm⁻¹.
• If S = 3 μm, SA = 54 μm² and V = 27 μm³, then SA:V = 54 μm²:27μm³ = 2 μm⁻¹.
• If S = 4 μm, SA = 96 μm² and V = 64 μm³, then SA:V = 96 μm²:64μm³ = 1.5 μm⁻¹.

A higher surface-area-to-volume ratio means that the cell has more surface area available for diffusion compared to its volume. As seen from the calculations above, a cuboidal cell with a higher SA:V ratio will be more efficient in removing waste. Therefore, the answer is that the most efficient cuboidal cell for waste removal by diffusion has a surface-area-to-volume ratio of 6 μm⁻¹ (Answer A).