Question

The inside of a cell is extremely crowded. We will try to gain some insight for exactly how crowded by considering some numbers that are estimates of the situation in the common bacterium E. coli. One can estimate the total number of proteins present in a cell in several ways. For example, one can measure the dry mass of a cell (the amount of material left once the cell is dried out), analyze the composition to determine what fraction is amino acids, estimate the average molecular weight of a protein, and use all these numbers to calculate the total number of protein molecules present. Alternatively, one can use experimental analysis to determine the approximate amount of proteins present in a sample consisting of a known number of cells. All of these methods agree that a typical E. coli bacterial cell contains about 3 x 10⁶ proteins.

a) We can estimate the volume of an E. coli cell to be about 1 µm³, that is, equivalent to a cube of 1 µm on each side. Convert that volume to liters, and assuming there are 3 x 10⁶ total proteins present, calculate the total molar concentration of proteins contained in a bacterial cell.

b) Take the reciprocal of the molar concentration obtained in part a, and convert units to find the number of cubic nanometers per protein molecule.

c) One way to think about the volume per protein molecule found in part b is that if you divided up the total volume of the cell into tiny cubes, with each cube having the volume found in part b, then on averge, each cube would contain one protein. The length of each side of one of these cubes would just be the cube root of the volume per protein molecule. Further, the length of each side of the cube is an estimate of the center-to-center separation distance of each protein molecule. From the volume per protein obtained in part b, use this approach to estimate the center-to-center distance between protein molecules (on average) in an E. coli cell.

d) Averaged across a cell's inventory of proteins, the average protein molecular weight is about 30000 grams per mole (i.e., 30 kiloDaltons). For proteins, an average mass density is found of about 1.4 g/cm³. Use these numbers, after appropriate unit conversions, to calculate the average volume of a protein molecule in cubic nanometers. Compare this number to the value obtained in part b, and estimate what fraction of the total volume of a cell is filled with proteins.

Answer 1

E. coli bacterial cell contains about 3 x 106 proteins.

a) We can estimate the volume of an E. coli cell to be about 1 µm3, that is, equivalent to a cube of 1 µm on each side. Convert that volume to litres, and assuming there are 3 x 106 total proteins present, calculate the total molar concentration of proteins contained in a bacterial cell.

b) Take the reciprocal of the molar concentration obtained in part a, and convert units to find the number of cubic nanometers per protein molecule.

c) One way to think about the volume per protein molecule found in part b is that if you divided up the total volume of the cell into tiny cubes, with each cube having the volume found in part b, then on average, each cube would contain one protein. The length of each side of one of these cubes would just be the cube root of the volume per protein molecule. Further, the length of each side of the cube is an estimate of the center-to-center separation distance of each protein molecule. From the volume per protein obtained in part b, use this approach to estimate the center-to-centre distance between protein molecules (on average) in an E. coli cell.

d) Averaged across a cell's inventory of proteins, the average protein molecular weight is about 30000 grams per mole (i.e., 30 kiloDaltons). For proteins, an average mass density is found of about 1.4 g/cm3. Use these numbers, after appropriate unit conversions, to calculate the average volume of a protein molecule in cubic nanometers. Compare this number to the value obtained in part b, and estimate what fraction of the total volume of a cell is filled with proteins.