Under most in vitro assay conditions, the enzyme is used in catalytic amounts[10^(-12) to 10^(-8) M]. Estimate the concentration of an enzyme in a living cell. Assume that-
i) fresh tissue is 80% water and all of it is intracellular.
ii) the total soluble protein in a cell represents 15% of the wet weight.
iii) all the soluble proteins are enzymes.
iv) the average molecular weight of a protein is 150,000 and
v) about 1000 different enzymes are present.
Concentration of enzyme is estimated by the followingformula: V=m/M M - average molecular weight of a protein = 150,000
We should calculate m - mass. The real concentration of total enzyme is 0.8*0.15 = 0.12 or12 %. Mass of total enzyme is 12 % of the total cell mass, whichis 10^-9 on average. Thus, mass of the total enzyme is 0.83*10^-10. Mass of enzyme we investigate is the total mass of enzymesdivided by their number: 0.83*10^-10 /1000= 0.83*10^-13 Therefore, concentration of enzyme is 0.83*10^-13 / 150,000 = 5* 10^-19.