Consider an imaginary bacterial cell that contains equal concentrations of 1300
ID: 1026865 • Letter: C
Question
Consider an imaginary bacterial cell that contains equal concentrations of 1300 different enzymes in solution in the cyíosol. Each protein has a molecular weight of 100,000. The cytosol has a specific gravity of 1 .22 and contains 19.5% 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 , height 2.00 ). Calculate the number of molecules of a single enzyme in the cell. Number molecules cellExplanation / Answer
Ans. # Part A: Note: Since the volume or mass of the cell has NOT been mentioned in the first part, the calculation would not be possible.
So, data from part 2 is taken to present the solution.
# Part 1: Step 1: Volume of a cell (cylinder) = (pi) r2 l
Where, r = radius of the cell = 1 um / 2 = 0.5 um
l = length of the cell = 2 um
Now,
Volume, V = (3.14159) x (0.5 um)2 x 2.0 um = 1.5708 um3
= 1.5708 x (10-12 cm3) ; [1 um3 = 10-12 cm3]
= 1.5708 x 10-12 cm3 = 1.5708 x 10-12 mL
= 1.5708 x 10-15 L
# Step 2: Specific gravity of 1.22 means that the density of bacterial cell is 1.22 g/ cm3.
Mass of the cell =Volume x Density
= 1.5708 x 10-12 cm3 x (1.22 g/ cm3)
= 1.9164 x 10-12 g
# Step 3: Given, % mass of enzyme in cell = 19.5 %
# Total mass of enzyme = 19.5 % x (1.9164 x 10-12 g) = 3.7369 x 10-13 g
# Mass of one enzyme = Total mass of enzyme / Number of enzyme molecules
= 3.7369 x 10-13 g / 1300 enzyme
= 2.8746 x 10-16 g/ enzyme
# Moles of one enzyme = Mass / Molar mass
= 2.8746 x 10-16 g / (100000 g/ mol)
= 2.8746 x 10-21 mol
# Now,
Molarity of one enzyme = Number of moles / Volume of cell in liters
= 2.8746 x 10-21 mol / (1.5708 x 10-15 L)
= 1.83 x 10-6 M
# Part 2: Moles of one enzyme = Mass / Molar mass ; [see above]
= 2.8746 x 10-16 g / (100000 g/ mol)
= 2.8746 x 10-21 mol