Question

An almost cube-shaped animal cell (side length 10 μm) uses 109 ATP molecules per minute. You can assume that the cell replaces the ATP used with a glucose oxidation reaction with the reaction equation 6O2 + C6H12O6 → 6CO2 + 6H2O, and that complete oxidation of one glucose produces 30 ATP molecules.

a) How much oxygen does the cell consume per minute?
b) How long does it take for a cell to use its own volume of oxygen? (One mole of gas = 22.4 liters)

Answer 1

a)

Given that:

Number of ATPs generated per minute = 109

Number of ATPs produced from complete oxidation of one glucose molecule = 30

Therefore; number of glucose molecules oxidised per minute = 109 / 30 = 3.63

From the reaction stoichiometry:

1 molecule of glucose requires 6 molecules of oxygen for complete oxidation.

3.63 molecule will require = 3.63 X 6 / 1 = 21.8 molecules

Rounding off to next largest integer; 22 molecules

Therefore;

The cell consumes 22 molecules of O2 per minute.

b)

Side length of cell = 10 um = 10^-5 m

Volume of cell = (10^-5)³ = 10^-15 m³

1 m³ = 1000 L

Therefore;

Volume of cell = 10^-12 L

Given that;

22.4 L = 1 mol

10^-12 L = 10^-12 X 22.4 / 1 = 2.24 X 10^-11 mol

Also;

1 mole contains = 6.022 X 10²³ molecules of O₂

Therefore;

2.24 X 10^-11 mol contains = 2.24 X 10^-11 X 6.022 X 10²³ = 1.35 X 10¹³ molecules of O2

Molecules of O2 to be used = 1.35 X 10¹³ molecules

Molecules of O2 used per minute = 22 molecules / min

Time required = 1.35 X 10¹³ / 22 = 6.1 X 10¹⁰ minutes

Therefore;

It will take 6.1 X 10¹⁰ minutes for a cell to use its own volume of oxygen.