of by Socirg Leaming Cox Consider an imaginary bacterial cell that contains equa
ID: 539949 • Letter: O
Question
of by Socirg Leaming Cox Consider an imaginary bacterial cell that contains equal concentrations of 800 different enzymes in ution in the Cytosol. Each protein has a molecular weight of 100,000. The cytosol has a specific gravity of 1.23 and 20.5% soluble protein by weight (the protein fraction consists entirely of enzymes). Calculate the molar concentration of each enzyme in this cell. Assume that the bacterial cell is a cylinder (diameter 1.00 , height 2.00 m) Calculate the number of molecules of a single enzyme in the cell molecules Hint ® Give Up & View Solution O Check ExitExplanation / Answer
1)Given: molecular weight of each protein = 100,000 g/ mol
The cell contains equal concentrations of 800 different enzymes in solution in the cytosol
The cytosol specific gravity =1.23 g/ml
20.5% soluble protein by weight
Molar concentration of each enzyme:
The concentration of total protein in the cytosol = (1.23 gm/ml x 0.205)/100,000 g/mol
= 2.52 x 10^-6 mol/ml
= 2.52 x 10^-3 mol/L
Thus, for 1 enzyme in 800, the enzyme concentration = (2.52 x 10^-3 mol/L)/800
= 3.12 x10^-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
3.12 x10^-6 M x 1.6 x 10^-15L = 4.99 x 10^-21 moles
Convert moles to molecules by multiplying with Avogadro constant (6.022 x 10^23)
4.99 x 10^-21 moles*(6.022 x 10^23 molecules/mole) = 3.05 x 10^3 molecules/ cell
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