Escherichia coli cells are about \(2 \mu \mathrm{m}\) (microns) long and \(0.8 \mu \mathrm{m}\) in diameter. a. How many \(E\). coli cells laid end to end would fit across the diameter of a pinhead? (Assume a pinhead diameter of \(0.5 \mathrm{mm}\).) 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 an \(E\). colicell? What is the surface-to volume ratio of an \(E .\) colicell? d. Glucose, a major energy-yielding nutrient, is present in bacterial cells at a concentration of about \(1 \mathrm{m} M\). What is the concentration of glucose, expressed as \(\mathrm{mg} / \mathrm{mL}\) ? 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 \(E\). colicell 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 \(40 \mathrm{kD},\) what is its concentration, expressed as \(\mathrm{mg} / \mathrm{mL} ?\) f. \(\operatorname{An} E .\) coli cell contains about 15,000 ribosomes, which carry out protein synthesis. Assuming ribosomes are spherical and have a diameter of \(20 \mathrm{nm}\) (nanometers), what fraction of the \(E .\) colicell 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 \(0.34 \mathrm{nm}\) 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 nucleotide 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.)

Step by step solution

01

Calculate the number of E. coli cells fitting across a pinhead

To do this, first convert the diameter of the pinhead to micrometers since the length of E. coli is given in micrometers: 0.5 mm = 500 micrometers. Then, divide the diameter of the pinhead by the length of the E. coli cell: 500 micrometers / 2 micrometers = 250 cells.

02

Calculate the volume of an E. coli cell

The formula for the volume of a cylinder is \(V=\pi r^{2} h\). Here, \(r\) is the radius of the E. coli cell which is half the diameter, and \(h\) is the height, or length, of the cell. So, \(V = \pi * (0.4 micro m)^2 * 2 micro m = 1.005 micro m^3\).

03

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

The formula for the surface area of a cylinder is \(A = 2 \pi r (r + h)\). Plugging in the values, we get \(A = 2 * \pi * 0.4 micro m * (0.4 micro m + 2 micro m) = 5.65 micro m^2\). The surface to volume ratio is Surface Area/Volume = \(5.65/1.005 = 5.62 micro m^{-1}\).

04

Find the concentration of glucose in E. coli and the number of glucose molecules

First, to convert the glucose concentration to mg/mL: 1 M = 1 mole/L = 180 g/L = 180,000 mg/L since the molecular weight of glucose is 180 g/mole. Therefore, the concentration is 1 mM = 180 mg/L = 0.18 mg/mL. To find the number of glucose molecules, know that 1 M = 1 mole/L = \(6.023 * 10^{23}\) molecules/L. Therefore, 1 mM = \(6.023 * 10^{20}\) molecules/L. Multiply this by the volume of the cell (from Step 2) to find the number of glucose molecules in a cell.

05

Calculate the molar concentration of a protein and its concentration in mg/mL

To find the molar concentration, assume the cell contains one molecule of protein and divide this by the Avogadro's number (this gives us the number of moles) and the cell's volume (which gives us the volume in liter). Since the molecular weight of the protein is 40 kD = 40,000 Da = 40,000 g/mole, its concentration in mg/mL is obtained by multiplying the molar concentration by the molecular weight and converting g to mg and L to mL.

06

Calculate the fraction of E. coli volume occupied by ribosomes

Assuming ribosomes are spherical, find their volume using the formula \(V=4/3 \pi r^3\). Since each E. coli cell contains about 15,000 ribosomes, find the total volume occupied by the ribosomes by multiplying the volume of one ribosome by 15,000. The fraction of the E. coli volume occupied by ribosomes is obtained by dividing this by the volume of an E. coli cell calculated in Step 2.

07

Find the length and number of nucleotide pairs in E. coli chromosome and calculate the different proteins it can encode

Given that the mass of the E. coli chromosome is \(3*10^9\) daltons, divide this by the average molecular weight of a nucleotide pair to find the number of nucleotide pairs in the chromosome. Multiply this by the length per nucleotide pair to get the total length. To calculate the number of different proteins the E. coli can encode, find the number of amino acids that the nucleotide pairs can encode (dividing the number of nucleotide pairs by 3, as each 3 nucleotides encode 1 amino acid). Lastly, divide this by the average number of amino acids per protein (360) to get the number of different proteins.

Unlock Step-by-Step Solutions & Ace Your Exams!

• Full Textbook Solutions

  Get detailed explanations and key concepts

• Unlimited Al creation

  Al flashcards, explanations, exams and more...

• Ads-free access

  To over 500 millions flashcards

• Money-back guarantee

  We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!