Chapter 1, Problem 1P

(a)

Summary Introduction

To determine: The size of a typical eukaryotic cell with cellular diameter of $50\mu \text{m}$ if the cell is magnified $10,000$ times.

Introduction:

The eukaryotic cells are more complex than the prokaryotic cells and they vary based on the structure and functions. The eukaryotes have been evolved from the unicellular organisms in the timeline of evolution.

Explanation of Solution

The diameter ($\text{D}$) of the given eukaryotic cell is $50\mu \text{m}$. The power of magnification is $10,000$ times. The diameter of the magnified cell is as follows:

$\begin{array}{c}\text{D}=50\mu \text{m}\\ \text{D}\text{(After}\text{magnification)}=50\mu \text{m}\times {10}^4\\ =500\times {10}^3\mu \text{m}\\ =\text{500}\text{mm}\end{array}$

Conclusion

The size of a typical eukaryotic cell with cellular diameter of $50\mu \text{m}$ if the cell is magnified $10,000$ times is $\underset{\_}{500mm}.$

(b)

Summary Introduction

To determine: The number of molecules of actin in a myocyte of diameter $50\mu \text{m}$ if the diameter of actin molecules is $\text{3}\text{.6}\text{nm}$ and cell has no other cellular components.

Introduction:

The eukaryotes have been evolved from the unicellular organisms in the timeline of evolution. The second phase of evolution has been started when the levels of multicellularity diversify the organisms into algal species, fungi, plants, and animals. Myocyte is known as a muscle cell.

Explanation of Solution

The diameter ($\text{D}$) of the given myocyte cell is $50\mu \text{m}$. The radius of this cell is half the diameter and is equal to $25\mu \text{m}$. The radius of cell with unit in (meter) is $25\times {10}^{-6}\text{m}$.

The diameter ($\text{D}$) of the given actin molecule is $\text{3}\text{.6}\text{nm}$. The radius of this actin molecule is half the diameter and is equal to $\text{1}\text{.8}\text{nm}$. The radius of actin with unit in (meter) is $1.8\times {10}^{-9}\text{m}$.

The volume of the given myocyte is as follows:

$\begin{array}{c}{\text{V}}_{(\text{Myocyte})}=\frac{4}{3}\pi r^3\\ =\frac{4}{3}\pi {(25\times {10}^{-6}\text{m})}^3\\ =6.54\times {10}^{-14}{\text{m}}^3\end{array}$

The volume of the given actin is as follows:

$\begin{array}{c}{\text{V}}_{(\text{Actin})}=\frac{4}{3}\pi r^3\\ =\frac{4}{3}\pi {(1.8\times {10}^{-9}\text{m})}^3\\ =2.44\times {10}^{-26}{\text{m}}^3\end{array}$

The number of actin molecules in the given myocyte is as follows:

$\begin{array}{c}\text{N}=\frac{{\text{V}}_{(\text{Myocyte})}}{{\text{V}}_{(\text{Actin})}}\\ =\frac{6.54\times {10}^{-14}{\text{m}}^3}{2.44\times {10}^{-26}{\text{m}}^3}\\ =2.7\times {10}^{12}\end{array}$

Conclusion

The number of molecules of actin in a myocyte (if the diameter of actin molecules is $\underset{\_}{3.6nm}$ and cell has no other cellular components) is $\underset{\_}{2.7\times 10^{12}}.$.

(c)

Summary Introduction

To determine: The number of molecules of mitochondria in a liver cell of diameter $50\mu \text{m}$ if the diameter of action molecules is $\text{1}\text{.5}\mu \text{m}$ and cell has no other cellular components is.

Introduction:

The eukaryotic cells are more complex than the prokaryotic cells and they vary based on structure and functions. The eukaryotes have been evolved from the unicellular organisms in the timeline of evolution. Mitochondrion is part of a eukaryotic cells and the energy storage of any cell.

Explanation of Solution

The diameter ($\text{D}$) of the given liver cell is $50\mu \text{m}$. The radius of this cell is half the diameter and is equal to $25\mu \text{m}$. The radius of cell with unit in (meter) is $25\times {10}^{-6}\text{m}$.

The diameter ($\text{D}$) of the given mitochondria is $\text{1}\text{.5}\mu \text{m}$. The radius of this mitochondria is half the diameter and is equal to $\text{0}\text{.75}\text{nm}$. The radius of mitochondria with  in meter is $0.75\times {10}^{-6}\text{m}$.

The volume of the given liver is as follows:

$\begin{array}{c}{\text{V}}_{(\text{Liver}\text{cell})}=\frac{4}{3}\pi r^3\\ =\frac{4}{3}\pi {(25\times {10}^{-6}\text{m})}^3\\ =6.54\times {10}^{-14}{\text{m}}^3\end{array}$

The volume of the given mitochondria is as follows:

$\begin{array}{c}{\text{V}}_{(\text{Mitochondria})}=\frac{4}{3}\pi r^3\\ =\frac{4}{3}\pi {(0.75\times {10}^{-6}\text{m})}^3\\ =1.76\times {10}^{-18}{\text{m}}^3\end{array}$

The number of mitochondria molecules in the given myocyte is as follows:

$\begin{array}{c}\text{N}=\frac{{\text{V}}_{(\text{Myocyte})}}{{\text{V}}_{(\text{Mitochondria)}}}\\ =\frac{6.54\times {10}^{-14}{\text{m}}^3}{1.76\times {10}^{-18}{\text{m}}^3}\\ =36,000\end{array}$

Conclusion

The number of mitochondria in a liver cell (if the diameter of action molecules is $\underset{\_}{1.5\mu m}$ and cell has no other cellular components) is $\underset{\_}{36,000}.$

(d)

Summary Introduction

To determine: The number of molecules of glucose be present in spherical eukaryotic cell if the cellular concentration is $\text{1}\text{mM}$.

Introduction:

Glucose is known as a simple sugar molecule which possess a molecular formula  ${\text{C}}_{\text{6}}{\text{H}}_{\text{12}}{\text{O}}_{\text{6}}$. It is referred as the blood sugar.

Explanation of Solution

The volume of the given eukaryotic cell is $6.54\times {10}^{-14}{\text{m}}^3$. The volume of the same cell is as follows:

$\begin{array}{c}1{\text{m}}^3={(100)}^3{\text{cm}}^3\\ 6.54\times {10}^{-14}{\text{m}}^3=6.54\times {10}^{-14}{\text{m}}^3\times {(100)}^3{\text{cm}}^3\\ =6.5\times {10}^{-8}{\text{cm}}^{\text{3}}\end{array}$

One cubic centimeter is equal to one millimeter. Therefore, the volume is $6.5\times {10}^{-8}\text{mL}$.

The number of molecules in one liter of $\text{1}\text{mM}$ solution of glucose is as follows:

$\begin{array}{c}\text{N}=\frac{0.001\text{mol}}{1000\text{mL}}\times \frac{6.022\times {10}^{23}\text{molecules}}{\text{1}\text{mol}}\\ =6.022\times {10}^{17}\text{molecules}/\text{mL}\end{array}$

The total number of glucose in given cell is as follows:

$\begin{array}{c}N=\text{Number}\text{of}\text{molecules}\times \text{Volume}\text{of}\text{cell}\\ =6.022\times {10}^{17}\text{molecules}/\text{mL}\times 6.5\times {10}^{-8}\text{mL}\\ =3.9\times {10}^{10}\text{molecules}\end{array}$

Conclusion

The number of molecules of glucose be present in spherical eukaryotic cell if the cellular concentration is $\text{1}\text{mM}$ is $\underset{\_}{3.9\times 10^{10}}.$

(e)

Summary Introduction

To determine: The number of glucose molecules in hexokinase molecule if the hexokinase concentration is $20\mu \text{m}$.

Introduction:

Glucose is known as a simple sugar molecule which possess a molecular formula  ${\text{C}}_{\text{6}}{\text{H}}_{\text{12}}{\text{O}}_{\text{6}}$. It is circulated in an organism’s body and is referred as the blood sugar. Hexokinase refers to an enzyme that helps in phosphorylating the hexoses.

Explanation of Solution

The concentration of hexokinase molecule is $20\mu \text{m}$ and can be written as $\text{0}\text{.00002}\text{M}$. The concentration of glucose molecule is $\text{1}\text{mM}$ and can be written as $\text{0}\text{.001}\text{M}$. The ratio of the concentration of glucose and concentration of hexokinase is as follows:

$\begin{array}{c}\text{Ratio}=\frac{\text{Concentration}\text{of}\text{glucose}}{\text{Concentration}\text{of}\text{hexokinase}}\\ =\frac{\text{0}\text{.001}\text{M}}{\text{0}\text{.00002}\text{M}}\\ =\frac{50}{1}\end{array}$

One molecule of hexokinase will have $50$ molecules of glucose.

Conclusion

The number of glucose molecules in hexokinase molecule (if the hexokinase concentration is $20\mu \text{m}$) is $\underset{\_}{50}.$