Question

question 7 of 15 predict which cell will experience the most rapid exchange of materials with the environment. a. cube - shaped cell with dimensions of 3 μm×3 μm×3 μm b. cube - shaped cell with dimensions of 2 μm×2 μm×2 μm c. spherical cell with radius of 3 μm d. spherical cell with radius of 2 μm

Explanation:

Step1: Recall surface - area to volume ratio concept

Faster material exchange is related to higher surface - area to volume ratio.

Step2: Calculate surface area of cube formula

For a cube of side length $s$, surface area $A_{cube}=6s^{2}$.

Step3: Calculate volume of cube formula

Volume of cube $V_{cube}=s^{3}$.

Step4: Calculate surface - area to volume ratio of cube

Ratio for cube $r_{cube}=\frac{6s^{2}}{s^{3}}=\frac{6}{s}$. For $s = 2\ \mu m$, $r_{cube}=\frac{6}{2}=3$; for $s = 3\ \mu m$, $r_{cube}=\frac{6}{3}=2$.

Step5: Calculate surface area of sphere formula

Surface area of sphere $A_{sphere}=4\pi r^{2}$.

Step6: Calculate volume of sphere formula

Volume of sphere $V_{sphere}=\frac{4}{3}\pi r^{3}$.

Step7: Calculate surface - area to volume ratio of sphere

Ratio for sphere $r_{sphere}=\frac{4\pi r^{2}}{\frac{4}{3}\pi r^{3}}=\frac{3}{r}$. For $r = 2\ \mu m$, $r_{sphere}=\frac{3}{2}=1.5$; for $r = 3\ \mu m$, $r_{sphere}=\frac{3}{3}=1$. The cube with side length 2 μm has the highest surface - area to volume ratio among the options, so it will have the most rapid material exchange.

Answer:

B. Cube - shaped cell with dimensions of 2 μm × 2 μm × 2 μm