Question

Components of \(\boldsymbol{E}\). coli \(E\). coli cells are rod-shaped, about 2 \(\mu \mathrm{m}\) long, and \(0.8 \mu \mathrm{m}\) in diameter. E. coli has a protective envelope \(10 \mathrm{~nm}\) thick. The volume of a cylinder is \(\pi r^{2} h\), where \(h\) is the height of the cylinder. a. What percentage of the total volume of the bacterium does the cell envelope occupy? b. E. coli is capable of growing and multiplying rapidly because it contains some 15,000 spherical ribosomes (diameter \(18 \mathrm{~nm}\) ), which carry out protein synthesis. What percentage of the cell volume do the ribosomes occupy? c. The molecular weight of an \(E\). coli DNA molecule is about \(3.1 \times 10^{9} \mathrm{~g} / \mathrm{mol}\). The average molecular weight of a nucleotide pair is \(660 \mathrm{~g} / \mathrm{mol}\), and each nucleotide pair contributes \(0.34 \mathrm{~nm}\) to the length of DNA. Calculate the length of an \(E\). coli DNA molecule. Compare the length of the DNA molecule with the cell dimensions. Now, consider the photomicrograph showing the single DNA molecule of the bacterium \(E\). coli leaking out of a disrupted cell (Fig, 1-31b). How does the DNA molecule fit into the cell?

Short answer

a. 5.14% b. 4.56% c. The DNA is 1600 µm long, much longer than the cell, fitting by folding and supercoiling.

Step-by-step solution

Calculate Total Volume of E. coli

First, calculate the total volume of the E. coli bacterium treating it as a perfect cylinder. The formula for the volume of a cylinder is \( V = \pi r^2 h \). - The length (height) \( h \) of the cylinder is given as \( 2 \mu m = 2000 \ nm \) (convert to nm for consistency).- The radius \( r \) of the cylinder can be determined from the diameter \( 0.8 \mu m = 800 \ nm \), so \( r = 400 \ nm \). \[ V_{\text{total}} = \pi \times (400 \ nm)^2 \times 2000 \ nm = \pi \times 160,000 \ nm^2 \times 2000 \ nm \]\[ V_{\text{total}} = \pi \times 320,000,000 \ nm^3 \approx 1,005,310,381 \ nm^3 \]

Calculate Volume of E. coli Cell Envelope

The cell envelope has a thickness of 10 nm. To find its volume, first calculate the total volume of the bacterium including this layer with a radius of \( r = (400+10) \ nm = 410 \ nm \). \[ V_{\text{outer}} = \pi \times (410 \ nm)^2 \times 2000 \ nm = \pi \times 168,100 \ nm^2 \times 2000 \ nm \]\[ V_{\text{outer}} = \pi \times 336,200,000 \ nm^3 \approx 1,056,976,729 \ nm^3 \]Now calculate the volume occupied by just the cell envelope by subtracting the total E. coli volume from this outer volume:\[ V_{\text{envelope}} = V_{\text{outer}} - V_{\text{total}} \]\[ V_{\text{envelope}} = 1,056,976,729 \ nm^3 - 1,005,310,381 \ nm^3 \approx 51,666,348 \ nm^3 \]

Calculate Percentage Volume of Cell Envelope

To find the percentage of the total volume occupied by the envelope:\[ \text{Percentage} = \left(\frac{V_{\text{envelope}}}{V_{\text{total}}}\right) \times 100\% \]\[ \text{Percentage} = \left( \frac{51,666,348}{1,005,310,381} \right) \times 100\% \approx 5.14\% \]

Calculate Volume Occupied by Ribosomes

Calculate the volume of one ribosome, treating it as a sphere, with radius \( r = \frac{18 \ nm}{2} = 9 \ nm \):\[ V_{\text{ribosome}} = \frac{4}{3} \pi r^3 = \frac{4}{3} \pi \times (9 \ nm)^3 \]\[ V_{\text{ribosome}} = \frac{4}{3} \pi \times 729 \ nm^3 \approx 3053.63 \ nm^3 \]There are 15,000 ribosomes:\[ V_{\text{ribosomes otal}} = 15000 \times 3053.63 \ nm^3 \approx 45,804,450 \ nm^3 \]

Calculate Percentage Volume of Ribosomes

Calculate the percentage of total cell volume occupied by ribosomes:\[ \text{Percentage} = \left(\frac{V_{\text{ribosomes otal}}}{V_{\text{total}}}\right) \times 100\% \]\[ \text{Percentage} = \left( \frac{45,804,450}{1,005,310,381} \right) \times 100\% \approx 4.56\% \]

Calculate Length of E. coli DNA

First, find the number of nucleotide pairs in the E. coli DNA. The number of nucleotide pairs is given by:\[ \text{Moles of DNA} = \frac{3.1 \times 10^9 \text{ g/mol}}{660 \text{ g/mol}} \approx 4.7 \times 10^6 \text{ nucleotide pairs} \]Each nucleotide pair contributes \( 0.34 \ nm \) to the length:\[ \text{Length of DNA} = 4.7 \times 10^6 \times 0.34 \ nm \approx 1.6 \times 10^6 \ nm \]

Compare DNA Length to Cell Dimensions

The length of the DNA molecule is approximately \( 1.6 \times 10^6 \ nm = 1600 \ \mu m \), which is significantly longer than the 2 \( \mu m \) length of the E. coli cell. The DNA molecule fits into the cell through extensive folding and supercoiling mechanisms.

Key concepts

Cell Volume Calculation

Understanding the cell volume calculation of an E. coli bacterium is crucial in comprehending its overall architecture and function. E. coli is modeled as a perfect cylinder to simplify volume calculations. The basic formula used here is the volume of a cylinder: \[ V = \pi r^2 h \] where \( r \) is the radius of the cylinder, and \( h \) is its height.

• Given that the height \( h \) is 2 µm, which can be converted to 2000 nm for consistency.
• The diameter of the E. coli cell is 0.8 µm, so the radius \( r \) is half of that, translating to 400 nm.

By plugging in these values, we achieve a volume of approximately \( 1,005,310,381 \) cubic nanometers.
This foundational understanding showcases the compact yet complex design of bacterial cells, allowing them to perform essential functions within a restricted space.

Ribosome Volume

Ribosomes are pivotal in protein synthesis, and their volumetric calculation in E. coli provides insights into spatial dynamics within the cell. They are approximated to spheres for simple calculations. The formula used is as follows:\[ V = \frac{4}{3} \pi r^3 \]where \( r \) is the radius. Each ribosome in E. coli has a diameter of 18 nm, thus making the radius 9 nm.

• The volume of a single ribosome is, therefore, approximately 3053.63 cubic nanometers.
• In total, there are 15,000 ribosomes in an E. coli cell.

Their cumulative volume reaches about 45,804,450 cubic nanometers.
When comparing this with the total cell volume, roughly 4.56% of the cellular space is dedicated to ribosomes, highlighting their essential role in the cell's rapid growth and protein synthesis capabilities.

DNA Length Calculation

The DNA length calculation provides fascinating insights into how such a long molecule fits inside a tiny bacterial cell. With the known molecular weight of E. coli DNA being 3.1 x 10⁹ g/mol and the average molecular weight of each nucleotide pair as 660 g/mol, we can deduce the number of nucleotide pairs.

• This corresponds to around 4.7 x 10⁶ nucleotide pairs.
• Each nucleotide pair extends the DNA by 0.34 nm.

Multiplying the number of pairs by their length contribution results in a total DNA length of approximately 1.6 x 10⁶ nm or 1600 µm.
This length vastly exceeds the 2 µm length of the E. coli cell. Such an extended DNA fits inside the cell through coiling and intricate packaging mechanisms, allowing for efficient genetic storage and transmission.

Bacterial Cell Envelope

The bacterial cell envelope of E. coli is an essential barrier that protects the cell from external threats. Calculating its volume offers insights into its relative space within the cell. The outer radius, including the envelope thickness, is calculated by adding 10 nm to the existing radius of the bacterium.

• This makes the new radius 410 nm for the entire bacterium.
• The outer cell volume is roughly 1,056,976,729 cubic nanometers.
• By subtracting the original cell volume, the envelope accounts for approximately 51,666,348 cubic nanometers.

This results in about 5.14% of E. coli's total volume being attributable to its protective envelope, demonstrating its critical role in cell integrity and interaction with the environment. Understanding the contributions of such layers elucidates how bacterial cells maintain their structure and functionality under different conditions.