Question

refer to the table.

	diameter
cell b	2.0 μm
cell c	3.0 μm

how do the surface area - to - volume ratios compare for the cells listed in the table, and which cell will have the most rapid transport of materials into its interior? assume the cells are spherical.
the surface area - to - volume ratios increase from c to b to a, with cell c having the most rapid transport.
the surface area - to - volume ratios increase from a to b to c, with cell b having the most rapid transport.
the surface area - to - volume ratios increase from c to b to a, with cell a having the most rapid transport.
the surface area - to - volume ratios increase from a to b to c, with cell c having the most rapid transport.
the surface area - to - volume ratios increase from a to b to c, with cell a having the most rapid transport.

Explanation:

Step1: Recall surface - area and volume formulas for sphere

The surface - area formula of a sphere is $A = 4\pi r^{2}$, and the volume formula is $V=\frac{4}{3}\pi r^{3}$. The surface - area to volume ratio $R=\frac{A}{V}=\frac{4\pi r^{2}}{\frac{4}{3}\pi r^{3}}=\frac{3}{r}$.

Step2: Calculate ratios for each cell

For cell A with $r_A = 0.5\ \mu m$, $R_A=\frac{3}{0.5}=6$. For cell B with $r_B = 1\ \mu m$, $R_B=\frac{3}{1}=3$. For cell C with $r_C = 1.5\ \mu m$, $R_C=\frac{3}{1.5}=2$. So the ratios increase from C to B to A ($2<3<6$).

Step3: Understand the relationship between ratio and transport

A higher surface - area to volume ratio allows for more rapid transport of materials into the cell interior. Since cell A has the highest ratio, it has the most rapid transport.

Answer:

The surface area - to - volume ratios increase from C to B to A, with cell A having the most rapid transport.