Question

To approximate the actual concentration of enzymes in a bacterial cell, assume that the cell contains equal concentrations of 1,000 different enzymes in solution in the cytosol and that each protein has a molecular weight of 100,000. Assume also that the bacterial cell is a cylinder (diameter 1.0 m, height 2.0 m), that the cytosol (specific gravity 1.20) is 20% soluble protein by weight, and that the soluble protein consists entirely of enzymes. Calculate the average molar concentration of each enzyme in this hypothetical cell.

Answer 1

2.4 × 10⁻⁶ M

Step-by-step explanation:

Molar concentration = no of moles of solute (enzymes)/volume of solvent(cystosol)

Let V = volume of solvent = volume of cystosol = πd²h/4 where d = diameter of cell = 1.0 m and h = height of cell = 2.0 m

Concentration of enzymes, C = n/V where n = no of moles of enzymes

Also, n = m/M where m = mass of all enzymes and M = molecular weight of protein = 100,000

Note that the solvent for the enzymes is cystosol, since they are dissolved in it.

Density of cystosol,ρ = specific gravity of cystosol in g/cm³ = 1.20 g/cm³.

Now, ρ = m'/V where m = mass of cystosol and V = volume of cystosol

m' = ρV

m' = 1.20V

Given that 20 % of the cystosol is protein and all protein are enzymes, the mass of enzymes is thus m = 0.2m' = 0.2 × 1.20V = 0.24V

C = n/V

= m/MV

= 0.24V/100000V

= 2.4 × 10⁻⁶ mol/cm³

= 2.4 × 10⁻³ mol/dm³

= 2.4 × 10⁻³ M

This is the concentration of all the 1000 enzymes.

So, the concentration of one enzyme is 2.4 × 10⁻³/1000 M = 2.4 × 10⁻⁶ M