Chapter 2, Problem 1AQ

Summary Introduction

To calculate:

The surface-to-volume ratio of spherical cells of different sizes (2 µm and 15 µm) and mention the consequences of this difference in the cell function.

Concept introduction:

In the field of microbiology, the shape of a cell is termed as morphology. The size and shape of microorganisms are immediately revealed upon examination using a microscope. Archaeal and bacterial cells varies in size ranging from 0.2 µm to >700 µm in diameter. Several properties of a cell, including its evolution and growth rate are controlled by its surface area to cell volume (S/V) ratio.

Explanation of Solution

Surface area to cell volume (S/V) ratio:

$\begin{array}{l}{\text{Surface area = 4πr}}^{\text{2}}\\ \text{Volume = }\frac{\text{4}}{\text{3}}{\text{πr}}^{\text{3}}\\ \text{Therefore, }\frac{\text{Surface}}{\text{Volume}}=\frac{3}{\text{r}}\end{array}$

Diameter of the spherical cell = 15 µm

$\text{Radius}\text{= }\frac{\text{Diameter}}{\text{2}}$= 7.5 µm

Surface area to cell volume (S/V) ratio is $\frac{\text{Surface}}{\text{Volume}}=\frac{3}{7.5}=0.4$

Diameter of the spherical cell = 2 µm

$\text{Radius}\text{= }\frac{\text{Diameter}}{\text{2}}$ = 1 µm

Surface area to cell volume (S/V) ratio is $\frac{\text{Surface}}{\text{Volume}}=\frac{3}{1}=3$

Consequences of the differences in surface to volume ratio: The ratio of surface area to cell volume increases as the size of a cell decreases. In comparison to the cells of larger sizes, cells of smaller sizes possess increased surface area relative to the volume of the cell. Several properties of a cell, including its evolution and growth rate are controlled by its S/V ratio. Due to the presence of larger surface area, metabolic waste products and nutrients pass easily into and out of the smaller cells compared to the larger cells. The rate of exchange of wastes and nutrients facilitates the fast growth of a cell. Compared to larger cells, smaller cells with a higher S/V ratio grow faster.