Chapter 1, Problem 48QAP

To determine

(a)

The volume of the cell.

Answer to Problem 48QAP

The volume of the cell including the membrane is $525.2{\text{μm}}^{\text{3}}$.

Explanation of Solution

Given info:

Diameter of the human cell is $1\text{0}\text{μm}$.

Convert micrometer into meters.

  $\begin{array}{c}1\text{0}\text{μm}\text{=}10\text{μm}\times \frac{{\text{10}}^{-6}\text{m}}{\text{1}\text{μm}}\\ =1\times {10}^{-5}\text{m}\end{array}$

The thickness of the cell membrane is $5\text{nm}$.

Convert nanometer into meters.

  $\begin{array}{c}5\text{nm}\text{=}5\text{nm}\times \frac{{\text{10}}^{-9}\text{m}}{\text{1}\text{nm}}\\ =5\times {10}^{-9}\text{m}\end{array}$

Formula used:

The formula to find the volume of the cell including membrane is,

$V_{cell+memberane}=\frac{4}{3}\pi R^3$

Equation-1

Here, $R$ is the radius of the cell enclosed by membrane.

Calculation:

First find the radius of the cell enclosed by membrane.

The radius of the cell enclosed by membrane is equal to sum of radius of cell plus thickness of membrane.

  $\begin{array}{c}R=r_{cell}+t\\ =\frac{1\times {10}^{-5}}{2}\text{m}+5\times {10}^{-9}\text{m}\\ \text{=}5.005\times {10}^{-6}\text{m}\end{array}$

Convert unit of radius from meter to micrometer.

  $\begin{array}{c}R\text{=}5.005\times {10}^{-6}\text{m}\times \frac{{10}^6\text{μm}}{1\text{m}}\\ =5.005\text{μm}\end{array}$

Next find the volume of the cell by substituting the value of radius in equation-1.

  $\begin{array}{c}{V}_{cell+membrane}=\frac{4}{3}\pi {\left(5.005\text{μm}\right)}^3\\ =525.17{\text{μm}}^{\text{3}}\text{Equation-2}\end{array}$

Conclusion:

Thus, form equation-2the volume of the cell including the membrane is $525.2{\text{μm}}^{\text{3}}$.

To determine

(b)

The volume of the cell membrane.

Answer to Problem 48QAP

The volume of the cell membrane is $1.57{\text{μm}}^{\text{3}}$.

Explanation of Solution

Given info:

The volume of the cell including the membrane is $525.2{\text{μm}}^{\text{3}}$.

Diameter of the human cell is $1\text{0}\text{μm}$.

Formula used:

The formula to find the volume of the cell membrane is found by subtracting the volume of the cell form the volume of the cell with membrane.

  $\begin{array}{c}{V}_{membrane}=V_{cell+membrane}-V_{cell}\\ V_{membrane}=V_{cell+membrane}-\left(\frac{4}{3}\pi r_{cell}{}^3\right)\text{Equation-3}\end{array}$

Here, $r$ is the radius of the cell enclosed by membrane.

Calculation:

First find the volume of the cell.

  $\begin{array}{c}{V}_{cell}=\frac{4}{3}\pi {\left(\frac{\text{10}\text{μm}}{2}\right)}^3\\ =523.6{\text{μm}}^{\text{3}}\text{Equation-4}\end{array}$

Next find the volume of the membrane using equation-4 and 3.

  $\begin{array}{c}{V}_{membrane}=525.17{\text{μm}}^{\text{3}}-523.6{\text{μm}}^{\text{3}}\\ =1.57{\text{μm}}^{\text{3}}\text{Equation-5}\end{array}$

Conclusion:

Thus, form equation-5the volume of the cell membrane is $1.57{\text{μm}}^{\text{3}}$.

To determine

(c)

The percentage of volume occupied by the cell membrane.

Answer to Problem 48QAP

The cell membrane occupies $0.3\%$of volume.

Explanation of Solution

Given info:

The volume of the cell including the membrane is $525.2{\text{μm}}^{\text{3}}$.

The volume of the cell membrane is $1.57{\text{μm}}^{\text{3}}$.

Formula used:

The formula to find the percentage of volume occupied by the cell membrane is,

  $\begin{array}{c}\text{Percentage}\text{of}\text{volume}\text{occupied}\text{bycell-memberane}=\frac{V_{\text{membrane}}}{V_{\text{cell+membrane}}}\times 100\%\\ \end{array}$

Equation-6

Calculation:

Substitute all known values in equation-6 to find the percentage of volume occupied by the cell membrane is,

  $\begin{array}{c}\text{Percentage}\text{of}\text{volume}\text{occupied}\text{bycell-memberane}=\frac{1.57{\text{μm}}^{\text{3}}}{525.2{\text{μm}}^{\text{3}}}\times 100\%\\ =0.298\%\\ \approx 0.3\%\text{Equation-7}\\ \end{array}$

Conclusion:

Thus, form equation-7cell membrane occupies $0.3\%$of volume.