Chapter 1, Problem 9P

Answers to all problems are at the end of this book. Detailed solutions are available in the Student Solutions Manual, Study Guide, and Problems Book.

The Dimensions of Prokaryotic Cells and Their Constituents Escherichia coli cells are about 2 μm (microns) long and 0.8 μm in diameter. (Section 1.5)

a. How many E. coli cells laid end to end would fit across the diameter of a pinhead? (Assume a pinhead diameter of 0.5 mm.)

b. What is the volume of an E. coli cell? (Assume it is a cylinder, with the volume of a cylinder given by $V=\pi r^{2}h$ , where $\pi$ = 3.14.)

c. What is the surface area of ail E coli cell? What is the surface-to- volume ratio of an E coli cell?

d. Glucose, a major energy-yielding nutrient, is present in bacterial cells at a concentration of about 1 mM. What is the concentra¬tion of glucose, expressed as mg/mL? How many glucose molecules are contained in a typical E. coli cell? (Recall that Avogadro’s number $=6.023\times {10}^{23}$ .}

e. A number of regulatory proteins are present in E. coli at only one or two molecules per cell. If we assume that an E. coli cell contains just one molecule of a particular protein, what is the molar concentration of this protein in the cell? If the molecular weight of this protein is ${40}^{23}$ , what is its concentration, expressed as mg/mL?

f. An E coli cell contains about 15,000 ribosomes, which carry out protein synthesis. Assuming ribosomes are spherical and have a diameter of 20 nm (nanometers), what fraction of the E. coli cell volume is occupied by ribosomes?

g. The E coli chromosome is a single DNA molecule whose mass is about $3\times {10}^9$ daltons. This macromolecule is actually a linear array of nucleotide pairs. The average molecular weight of a nucleotide pair is 660, and each pair imparts 0.34 nm to the length of the DNA molecule. What is the total length of the E. coli chromosome? How does this length compare with the overall dimensions of an E. coli cell? How many nucleotide pairs does this DNA contain? The average E. coli protein is a linear chain of 360 amino acids. If three nucleo¬tide pairs in a gene encode one amino acid in a protein, how many different proteins can the E. coli chromosome encode?

(The answer to this question is a reasonable approximation of the maximum number of different kinds of proteins that can be expected in bacteria.)

Interpretation Introduction

To determine:

The number of E. coli cells per a pinhead.

Introduction:

Escherichia coli is a bacterium which lives in the intestine of healthy animals. E. coli genome consists of just one circular DNA.

Explanation of Solution

Given information:

Length of a cell $=2\mu m$

Diameter of pinhead $=0.5mm$

Formula used:

Number of E. coli per pinhead = Diameter of the pinhead/ Length of a cell

Calculation:

Number of E. coli cells that can be fitted to a pinhead $=0.5\text{ mm}/(2\times {10}^{-3}\text{mm})=250$

Number of E. coli cells that can be fitted to a pinhead $=250$

Interpretation Introduction

To determine:

The volume of an E. coli cell.

Introduction:

Escherichia coli is a bacterium which lives in the intestine of healthy animals. E. coli genome consists of just one circular DNA.

Explanation of Solution

Given information:

Length of a cell $=2\mu m$

Diameter of a cell $=0.8\mu m$

Formula used:

  $V=\pi r^{2}h$

  $V=$Volume

  $r=$Radius

  $h=$Height

Calculation:

  $\begin{array}{l}V=\pi r^{2}h\\ =3.14\times {(}^0\times (2\times 10^{-6}m)\\ =1{\text{×10}}^{\text{-18}}{\text{m}}^{\text{3}}{\text{=1×10}}^{\text{-15}}\text{L}\end{array}$

The volume of an E. coli cell $=1\times {10}^{\text{-18}}{\text{m}}^{\text{3}}$

Interpretation Introduction

To determine:

The surface area and surface to volume ratio of an E. coli cell.

Introduction:

Escherichia coli is a bacterium which lives in the intestine of healthy animals. E. coli genome consists of just one circular DNA.

Explanation of Solution

Given information:

Length of a cell $=2\mu m$

Diameter of a cell $=0.8\mu m$

Formula used:

Surface area = (Surface area of a circle $\times$2) + Surface area of the side of the cylinder

Surface area = $2\pi r^2+2\pi rh$

  $r=$Radius

  $h=$Height

Surface to volume ratio = Surface area/ Volume

Calculation:

Surface area of an E. coli cell

  $\begin{array}{l}=(2\times 3.14\times {(}^0)+(2\times 3.14\times (0.4\times 10^{-6}m)\times (2\times 10^{-6}m))\\ =6\times 10^{-12}{m}^2\end{array}$

Surface to volume ratio of an E. coli cell $=\frac{(6\times 10^{-12}{m}^2)}{(1\times 10^{-18}{m}^3)}=6\times 10^6{m}^{-1}$

Thus, surface area of an E. coli cell $=6\times {10}^{-12}{\text{m}}^{\text{2}}$

And, surface to volume ratio of an E. coli cell $=6\times {10}^6{\text{m}}^{-1}$

Interpretation Introduction

To determine:

The concentration of glucose in mg/mL and the number of glucose molecules in an E. coli cell

Introduction:

Escherichia coli is a bacterium which lives in the intestine of healthy animals. E. coli genome consists of just one circular DNA.

Explanation of Solution

Given information:

Glucose concentration $=1mM$

Molecular weight of glucose $=180.156g/mol$

Formula used:

Number of moles = Concentration $\times$Volume

Calculation:

Concentration of glucose $=(1\times 10^{-3}M)\times 180.156g/mol=0.18g/L=0.18mg/mL$

Moles of glucose in an E. coli cell $={\text{(1×10}}^{\text{-15}}{\text{L)×(1×10}}^{\text{-3}}\text{M})=1\times {10}^{-18}$mol

Number of glucose molecules in an E. coli cell

  $\begin{array}{l}=(1\times {10}^{-18}\text{mol})\times (6.023\times {10}^{23}\text{ molecules/mol})\\ =6\times {10}^5\text{ molecules}\end{array}$

Thus, concentration of glucose $\text{=0}\text{.18mg/mL}$

And, number of glucose molecules in an E. coli cell $=6\times {10}^5$molecules

Interpretation Introduction

To determine:

The molar concentration of the protein in the cell

Introduction:

Escherichia coli is a bacterium which lives in the intestine of healthy animals. E. coli genome consists of just one circular DNA.

Explanation of Solution

Given information:

Molecular weight of protein $=40kD$

Formula used:

Molar concentration = Moles/ Volume

Calculation:

Number of moles $=\frac{1molecule}{6.023\times 10^{23}molecules/mol}=1.7\times 10^{-24}mol$

Molar concentration $=\frac{1.7\times 10^{-24}mol}{10^{-15}L}=1.7\times 10^{-9}M=1.7nM$

Protein concentration $=(1.7\times 10^{-9}M)\times (40\times 10^{3}g/mol)=6.8\times 10^{-5}mg/mL$

Thus, molar concentration $=1.\text{7 nM}$

Protein concentration $=6.8\times {10}^{-5}\text{μg/mL}$

Interpretation Introduction

To determine:

The fraction of the E. coli cell volume which is occupied by ribosomes.

Introduction:

Escherichia coli is a bacterium which lives in the intestine of healthy animals. E. coli genome consists of just one circular DNA.

Explanation of Solution

Given information:

Number of ribosomes $=15000$

Diameter of ribosome $=20nm$

Formula used:

Volume of a ribosome (cylinder) $=\frac{4}{3}\pi r^3$

Fractional volume = Volume of ribosome/ Volume of cell

Calculation:

Volume of one ribosome $=\frac{4}{3}\times 3.14\times {(10\times {10}^{-9}m)}^3=4.2\times {10}^{-24}{m}^3$

Volume of 15,000 ribosome $=4.2\times {10}^{-24}{m}^3\times 15000=6.3\times {10}^{-20}{m}^3$

Fractional volume $=\frac{6.3\times {10}^{-20}{m}^3}{1\times {10}^{-18}{m}^3}=0.063$

Thus, fractional volume $=0.063$

Interpretation Introduction

To determine:

The total length of E. coli chromosome and compare this length with the overall dimensions of E. coli cells, number of nucleotide pairs contains in the DNA, and number of different proteins can be encoded by E. coli chromosome.

Introduction:

Escherichia coli is a bacterium which lives in the intestine of healthy animals. E. coli genome consists of just one circular DNA.

Explanation of Solution

Given information:

Mass of E. coli chromosome $=15000$

Average molecular weight of a nucleotide pair $=660g/mol$

Distance of a nucleotide pair $=0.34nm$

Number of amino acids is an average protein $=360$

Calculation:

Number of moles of base pairs in DNA

  $\begin{array}{l}=\frac{3\times {10}^{9}g/mol.DNA}{660g/mol.bp}\\ =4.55\times {10}^{6}mol.bp/mol.DNA\end{array}$

Length

  $\begin{array}{l}=(4.55\times {10}^{6}mol.bp/mol.DNA)\times (0.34nm/bp)\\ =1.55\times {10}^{-3}m=1.55mm=1550\mu m\end{array}$

Length DNA/ Length E. coli $=\frac{1550\mu m}{2\mu m}=775$

Number of base pairs per protein $=360\times 3=1080$

Number of different proteins $=\frac{4.55\times {10}^6}{1080}=4213$

Thus,

Total length of E. coli chromosome $=1.55mm$

Length DNA/ Length E. coli $=775$

Number of nucleotide pairs $=4.55\times {10}^6$

Number of different proteins $=4213$