Consider an imaginary bacterial cell that contains equal concentrations of 900 d
ID: 705951 • Letter: C
Question
Consider an imaginary bacterial cell that contains equal concentrations of 900 different enzymes in solution in the cytosol. Each protein has a molecular weight of 100,000. The cytosol has a specific gravity of 1.17 and contains 17.9% soluble protein by weight (the protein fraction consists entirely of enzymes) Calculate the molar concentration of each enzyme in this cell. Number Assume that the bacterial cell is a cylinder (diameter 1.00 um, height 2.00 um). Calculate the number of molecules of a single enzyme in the cell. Number molecules cellExplanation / Answer
ans)
1)Given: molecular weight of each protein = 100,000 g/ mol
The cell contains equal concentrations of 900 different enzymes in solution in the cytosol
The cytosol specific gravity =1.17 g/ml
17.9 % soluble protein by weight
Molar concentration of each enzyme:
The concentration of total protein in the cytosol = (1.17 gm/ml x 0.179)/100,000 g/mol
= 2.09 x 10^-6 mol/ml
= 2.09 x 10^-3 mol/L
Thus, for 1 enzyme in 800, the enzyme concentration = (2.09 x 10^-3 mol/L)/900
= 2.32x10^-6 M
2)
Find the radius:
r=1/2*diameter =1.0 µm/2 = 0.5 µm
h = 2.0 µm
Volume of bacterial cytosol = ? r^2h = 3.14 x (0.50µm)^2 x 2.0 µm = 1.6 µm^3
1.6 x 10^-12cm^3 = 1.6 x 10^-12 ml = 1.6 x 10^-15L
Moles of enzyme =
Molarity*volume (L) = moles
2.32 x10^-6 M x 1.6 x 10^-15L = 3.712 x 10^-21 moles
Convert moles to molecules by multiplying with Avogadro constant (6.022 x 10^23)
3.712 x 10^-21 moles*(6.022 x 10^23 molecules/mole) = 2.23 x 10^3 molecules/ cell