Question

A spherical intervertebral disc cell has an outer diameter of 20 um with a nuclear diameter (sphere) of 4 um. Protein X is present at a concentration of 35 mg/mL in its cytoplasm. There are some organelles in the intervertebral disc cell that have a cumulative volume of 0.15 picoliters. Determine the number of protein molecules within the cell if the molecular weight of the protein is 150,000 daltons (g/mol). (Note: Avogadro's number is 6.022 x 1023 molecules/mol).

Answer 1

The number of protein X molecules in the cell is 5.7 × 10⁸ proteins.

Step-by-step explanation:

First, let´s calculate the volume of the cytoplasm:

Volume of cytoplasm = volume of the cell - volume of the nucleus - volume of the organelles

The volume of a sphere of radius "r" is:

4/3 · π · r³

The volume of the cell is:

4/3 · π · (10 um)³ = 4.2 × 10³ um³ = 4.2 × 10 ⁻⁹ ml = 4.2 pl

The volume of the nucleus is:

4/3 · π · (2 um)³ = 33.5 um³ = 3.35 × 10⁻¹¹ ml = 0.0335 pl

Then, the volume of the cytoplasm is:

volume of cytoplasm = 4.2 pl - 0.0335 pl - 0.15 pl = 4.02 pl = 4.02 × 10⁻⁹ ml

The mass of protein present in the cytoplasm will be:

4.02 × 10⁻⁹ ml · 35 mg/ml = 1.41 × 10⁻⁷ mg = 1.41 × 10⁻¹⁰ g

The number of moles of the protein in the cytoplasm will be:

1.41 × 10⁻¹⁰ g · (1 mol / 150000 g) = 9.4 × 10⁻¹⁶ mol.

The number of proteins will be:

9.4 × 10⁻¹⁶ mol · (6.022 × 10²³ molecules / mol) = 5.7 × 10⁸ molecules.

The number of protein molecules is 5.7 × 10⁸.