Question

4 mark for review which one of the following cells is likely to be the most efficient at eliminating waste by diffusion? a 1 μm 1 μm 1 μm b 2 μm 2 μm 2 μm c 4 μm 1 μm 2 μm d 6 μm 2 μm 3 μm

Explanation:

Cells with a higher surface - area - to - volume ratio are more efficient at eliminating waste by diffusion. Calculate the surface - area - to - volume ratio for each cell. For a cube of side length $s$, the volume $V = s^3$ and surface area $SA=6s^2$, so the ratio is $\frac{SA}{V}=\frac{6s^2}{s^3}=\frac{6}{s}$. For a rectangular prism with length $l$, width $w$ and height $h$, $V = l\times w\times h$ and $SA = 2(lw+lh + wh)$.

• Option A: Cube with $s = 1\ \mu m$. Volume $V_A=1^3 = 1\ \mu m^3$, surface area $SA_A = 6\times1^2=6\ \mu m^2$, ratio $\frac{SA_A}{V_A}=6$.
• Option B: Cube with $s = 2\ \mu m$. Volume $V_B = 2^3=8\ \mu m^3$, surface area $SA_B=6\times2^2 = 24\ \mu m^2$, ratio $\frac{SA_B}{V_B}=3$.
• Option C: Rectangular prism with $l = 4\ \mu m$, $w = 2\ \mu m$, $h = 1\ \mu m$. Volume $V_C=4\times2\times1=8\ \mu m^3$, surface area $SA_C=2(4\times2 + 4\times1+2\times1)=28\ \mu m^2$, ratio $\frac{SA_C}{V_C}=\frac{28}{8}=3.5$.
• Option D: Rectangular prism with $l = 6\ \mu m$, $w = 3\ \mu m$, $h = 2\ \mu m$. Volume $V_D=6\times3\times2 = 36\ \mu m^3$, surface area $SA_D=2(6\times3+6\times2 + 3\times2)=72\ \mu m^2$, ratio $\frac{SA_D}{V_D}=2$.

The cell with the highest surface - area - to - volume ratio is the most efficient at diffusion.

Answer:

A. A cube with side - length 1 $\mu m$