Answer to Question #151011 in Biochemistry for Erfan

Question #151011
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.
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Expert's answer
2020-12-14T05:38:15-0500

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"


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