---## Enzyme Kinetics: Understanding Turnover NumberThe active form of the enzyme alcohol dehydrogenase exists as a homotetramer (MW = 150 kDa) and features an identical active site on each of its four subunits. To study this enzyme, we performed a series of enzyme kinetic experiments with an enzyme sample concentration of 4.0 µg/mL. From these experiments, we determined a Vmax (maximum reaction velocity) of 2.442 x 10^-2 µmoles/min.### Problem Statement**Question:** What is the turnover number of this enzyme (in units of #/sec)?**Options:**- 76/sec- 305/sec- 152/sec- 4579/sec- 1526/sec### ExplanationThe turnover number (also called kcat) is a measure of the catalytic activity of an enzyme. It is defined as the number of substrate molecules converted into product by an enzyme molecule in a unit of time when the enzyme is fully saturated with substrate.### Calculation of the Turnover Number1. **Determine enzyme concentration in moles:** \[ \text{Enzyme concentration} = \frac{4.0 \, \text{µg/mL}}{150,000 \, \text{g/mol}} = \frac{4.0 \times 10^{-6} \, \text{g/mL}}{150,000 \, \text{g/mol}} = 2.67 \times 10^{-11} \, \text{mol/mL} \]2. **Convert moles to per minute Vmax:** \[ Vmax = 2.442 \times 10^{-2} \, \mu \text{moles/min} \]3. **Convert Vmax from per minute to per second:** \[ Vmax = \frac{2.442 \times 10^{-2} \, \mu \text{moles/min}}{60 \, \text{sec/min}} = 4.07 \times 10^{-4} \, \mu \text{moles/sec} \]4. **Calculate kcat:** \[ kcat = \frac{Vmax}{[\text{enzyme concentration}]} \] Since there are 4 active sites

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The active form of the enzyme alcohol dehydrogenase exists as a homotetramer (MW = 150 kDa), and has one identical active site located on each of its four subunits. You have performed a series of enzyme kinetic experiments using 0.025 mL of an enzyme sample that is 4.0 µg/mL and determined that the following Vmax = 2.442 x 102umoles/min What is the turnover number of this enzyme (in units of #/sec)?

The active form of the enzyme alcohol dehydrogenase exists as a homotetramer (MW = 150 kDa), and has one identical active site located on each of its four subunits. You have performed a series of enzyme kinetic experiments using 0.025 mL of an enzyme sample that is 4.0 µg/mL and determined that the following Vmax = 2.442 x 102umoles/min What is the turnover number of this enzyme (in units of #/sec)?

Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

Section: Chapter Questions

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

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