Question

If 6 molecules of a particular protein were present in an E. coli cell, what would be the nanomolar concentration of this protein? Assume an E. coli cell to be a cylinder 2.2 micrometers long and 1.04 micrometer in diameter. Report your answer to the nearest tenth of a unit.

Answer 1

volume = pi*(r)^2 * h
            = 3.14* (1.04/2 *10^-6)^2 * (2.2*10^-6)
            = 1.87*10^-18 m^3
             = 1.87*10^-15 L

Number of moles of protein = number of molecules / Avogadro's number
           = 6 /(6*10^23)
            = 10^-23 moles

Concentration = number of moles / volume
          = 10^-23 / (1.87*10^-15)
           = 5.35*10^-9 M
            = 5.35 nM

Answer: 5.35 nanoMolar