Chapter 1, Problem 18Q

Summary Introduction

To calculate: The volume, surface area, and surface-to-volume ratio for a bacterial and animal cell.

Explanation of Solution

It is given that,

Diameter of the bacteria cell=1 µm

Diameter of the animal cell=15 µm

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

$\begin{array}{c}\text{Radius}\text{of}\text{the}\text{bacterial}\text{cell=}\frac{\text{1}}{\text{2}}{\text{×10}}^{\text{-6}}\text{m}\\ \text{=0}{\text{.5×10}}^{\text{-6}}\text{m}\\ \text{Radius}\text{of}\text{the}\text{bacterial}\text{cell=}\frac{\text{15}}{\text{2}}{\text{×10}}^{\text{-6}}\text{m}\\ \text{=7}{\text{.5×10}}^{\text{-6}}\text{m}\end{array}$

$\begin{array}{c}\text{Volume}\text{of}\text{a}\text{sphere=}\frac{\text{4}}{\text{3}}{\text{πr}}^{\text{3}}\\ \text{Surface}\text{area}\text{of}\text{a}{\text{sphere=4πr}}^{\text{2}}\end{array}$

$\begin{array}{c}\text{Volume}\text{of}\text{a}\text{spherical}\text{bacterial}\text{cell=}\frac{\text{4}}{\text{3}}\text{π}{\left(\text{0}{\text{.5×10}}^{\text{-6}}\right)}^{\text{3}}\\ \text{=}\frac{\text{4}}{\text{3}}\text{×}\frac{\text{22}}{\text{7}}\text{×}{\left(\text{0}\text{.5}\right)}^{\text{3}}\text{×}{\left({\text{10}}^{\text{-6}}\right)}^{\text{3}}\\ \text{=0}{\text{.5237×10}}^{\text{-18}}\\ \simeq \text{5}{\text{.24×10}}^{\text{-19}}{\text{m}}^{\text{3}}\end{array}$

$\begin{array}{c}\text{Surface}\text{area}\text{of}\text{a}\text{spherical}\text{bacterial}\text{cell=4π}{\left(\text{0}{\text{.5×10}}^{\text{-6}}\right)}^{\text{2}}\\ \text{=4×}\frac{\text{22}}{\text{7}}\text{×}{\left(\text{0}\text{.5}\right)}^{\text{2}}\text{×}{\left({\text{10}}^{\text{-6}}\right)}^{\text{2}}\\ \text{=3}{\text{.14×10}}^{\text{-12}}{\text{m}}^{\text{2}}\end{array}$

$\begin{array}{c}\text{Volume}\text{of}\text{a}\text{spherical}\text{animal}\text{cell=}\frac{\text{4}}{\text{3}}\text{π}{\left(\text{7}{\text{.5×10}}^{\text{-6}}\right)}^{\text{3}}\\ \text{=}\frac{\text{4}}{\text{3}}\text{×}\frac{\text{22}}{\text{7}}\text{×}{\left(\text{7}\text{.5}\right)}^{\text{3}}\text{×}{\left({\text{10}}^{\text{-6}}\right)}^{\text{3}}\\ \text{=1767}{\text{.656×10}}^{\text{-18}}\\ \simeq \text{1}{\text{.77×10}}^{\text{-15}}{\text{m}}^{\text{3}}\end{array}$

$\begin{array}{c}\text{Surface}\text{area}\text{of}\text{a}\text{spherical}\text{animal}\text{cell=4π}{\left(\text{7}{\text{.5×10}}^{\text{-6}}\right)}^{\text{2}}\\ \text{=4×}\frac{\text{22}}{\text{7}}\text{×}{\left(\text{7}\text{.5}\right)}^{\text{2}}\text{×}{\left({\text{10}}^{\text{-6}}\right)}^{\text{2}}\\ \text{=7}{\text{.07×10}}^{\text{-10}}{\text{m}}^{\text{2}}\end{array}$

$\begin{array}{c}\text{Surface}\text{area}\text{to}\text{volume}\text{ratio}\text{of}\text{animal}\text{cell=}\frac{\text{7}{\text{.07×10}}^{\text{-10}}{\text{m}}^{\text{2}}}{\text{1}{\text{.77×10}}^{\text{-15}}{\text{m}}^{\text{3}}}\\ \text{=3}{\text{.99×10}}^{\text{5}}{\text{m}}^{\text{-1}}\\ \simeq {\text{4×10}}^{\text{5}}{\text{m}}^{\text{-1}}\end{array}$

$\begin{array}{c}\text{Surface}\text{area}\text{to}\text{volume}\text{ratio}\text{of}\text{bacterial}\text{cell}\text{=}\frac{\text{3}{\text{.14×10}}^{\text{-12}}{\text{m}}^{\text{2}}}{\text{5}{\text{.24×10}}^{\text{-19}}{\text{m}}^{\text{3}}}\\ \text{= 0}{\text{.599×10}}^{\text{7}}{\text{m}}^{\text{-1}}\\ \simeq \text{0}{\text{.6×10}}^{\text{7}}{\text{m}}^{\text{-1}}\\ \simeq {\text{6×10}}^{\text{6}}{\text{m}}^{\text{-1}}\end{array}$

Thus, the volume and surface area of the bacteria cell is $5.24\times {10}^{-19}{m}^3$ and $3.14\times {10}^{-12}{m}^2$respectively and of animal cell is $1.77\times {10}^{-15}{m}^3$and $7.07\times {10}^{-10}{m}^2$respectively.

Summary Introduction

To explain: The change in the ratio of surface area to volume if the internal membranes of the cell are included which is 15 times the area of the plasma membrane in surface area calculation.

Explanation of Solution

The volume of the animal cell is 3375 times more than bacterial cell, but the surface area increases only 225 times more than bacterial cell. The surface area to volume ratio of the bacterial and animal cell when the surface area is 15 times that of normal changes to

$\begin{array}{c}\text{Surface}\text{area}\text{of}\text{bacterial}\text{cell}\text{including}\text{internal}\text{membrane}\text{=}\text{15×3}{\text{.14×10}}^{\text{-12}}{\text{m}}^{\text{2}}\\ \text{=47}{\text{.1×10}}^{\text{-12}}{\text{m}}^{\text{2}}\end{array}$

$\begin{array}{c}\text{Surface}\text{area}\text{to}\text{volume}\text{ratio}\text{of}\text{bacterial}\text{cell}\text{including}\text{internal}\text{membrane=}\frac{\text{47}{\text{.1×10}}^{\text{-12}}{\text{m}}^{\text{2}}}{\text{5}{\text{.24×10}}^{\text{-19}}{\text{m}}^{\text{3}}}\\ \text{=8}{\text{.98×10}}^{\text{7}}{\text{m}}^{\text{-1}}\\ \simeq {\text{9×10}}^{\text{7}}{\text{m}}^{\text{-1}}\end{array}$

$\begin{array}{c}\text{Surface}\text{area}\text{of}\text{animal}\text{cell}\text{including}\text{internal}\text{membrane}\text{=}\text{15×7}{\text{.07×10}}^{\text{-10}}{\text{m}}^{\text{2}}\\ \text{=106}{\text{.05×10}}^{\text{-10}}{\text{m}}^{\text{2}}\end{array}$

$\begin{array}{c}\text{Surface}\text{area}\text{to}\text{volume}\text{ratio}\text{of}\text{animal}\text{cell}\text{including}\text{internal}\text{membrane =}\frac{\text{106}{\text{.05×10}}^{\text{-10}}{\text{m}}^{\text{2}}}{\text{1}{\text{.77×10}}^{\text{-15}}{\text{m}}^{\text{3}}}\\ \text{=59}{\text{.91×10}}^{\text{5}}{\text{m}}^{\text{-1}}\\ {\text{6×10}}^{\text{6}}{\text{m}}^{\text{-1}}\end{array}$

The surface area of bacterial and animal cell changes to $9\times {10}^7{m}^{-1}$ and $6\times {10}^6{m}^{-1}$ respectively. Even after including the internal membranes, the surface area to volume ratio remains nearly the same. Thus, the internal membrane of eukaryotic cell helps the cells to grow bigger and yet maintain copious membrane area. Note, bacteria lack membrane-bound organelle and the calculation for bacterial cell is to gain an idea about the change in surface area to volume ration only.