Concept explainers
MATHEMATICAL You do an enzyme kinetic experiment and calculate a
Interpretation:
The turnover number of the enzyme of the following enzyme-kinetic experiment is to be calculated.
Concept introduction:
In an enzymatic reaction, initially the substrate binds with the active site of an enzyme forming an enzyme–substrate [ES] intermediate that ultimately leads to the desired product.
The rate of production of product from the enzyme–substrate complex is known as,
Answer to Problem 31RE
Explanation of Solution
Given information:
Volume of an enzyme solution
Molecular weight of an enzyme
Concentration of enzyme solution
Turnover number of an enzyme represents the number of substrate moles, getting converted into product per enzyme molecule or per active site per unit time. The unit of the turnover number is sec-1 or min-1. The value of both
The turnover number
0.2 mg/ml is the density of an enzyme present in 0.1 ml of solution. Since,
Substitute 0.2 mg/mL for density and 0.1 mL for volume in the above expression for calculating the mass:
Now, for calculating the Enzyme concentration (density) in 1 ml of solution,
Substitute 0.02 mg for mass and 1 mL for volume,
Conversion of 0.02 mg into g is as follows:
128x 103 g of enzyme present in 1 mole of solution.
This implies, 1 g of enzyme present in
Thus, the concentration of
The value of total concentration
Since,
Recall the expression for the turnover number of an enzyme:
Substitute 1.56 x 10-10 mol for
Hence the calculation is as follows:
The turnover number of an enzyme is, the number of substrate molecules getting converted into product molecules by a single active site per unit time. The total concentration of enzymes is
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Chapter 6 Solutions
Biochemistry
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- BiochemistryBiochemistryISBN:9781305961135Author:Mary K. Campbell, Shawn O. Farrell, Owen M. McDougalPublisher:Cengage Learning