Answer to Question #151011 in Biochemistry for Erfan

Question

Estimate the concentration of an enzyme in a living cell. Assume that fresh tissue is 80% water and all of it is intracellular, the total soluble protein in a cell represents 15% of the weight, all the soluble proteins are enzymes, the average weight of a protein is 150000 and about 1000 different enzymes are present.

Answer 1

Let’s use 1 g of cell as the basis.

Mass of total protein is 0.15 g

Mass of water = 0.8 g

V(water) = 0.8 mL

C(protein) $= \frac{0.15}{0.8} = 0.1875 \;g/mL$

MM(protein) = 150000 g/mol

C(protein) $= \frac{0.1875}{150000} = 0.125 \times 10^{-5} \;mol/mL$

C(enzyme) $= \frac{0.125 \times 10^{-5}}{1000} = 0.125 \times 10^{-8} \;mol/mL = 0.125 \times 10^{-5} \;mol/L$

Answer: $0.125 \times 10^{-5} \;M$