Chapter 6, Problem 31RE

MATHEMATICAL You do an enzyme kinetic experiment and calculate a $V_{\text{max}}$ of 100 $\text{μmol}$ of product per minute. If each assay used 0.1 mL of an enzyme solution that had a concentration of 0.2 mg/mL, what would be the turnover number if the enzyme had a molecular weight of 128,000 g/mol?

Interpretation Introduction

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.

$\text{E }+\text{ S }\rightleftharpoons \left[ES\right]\to \text{ E }+\text{ P}$

The rate of production of product from the enzyme–substrate complex is known as, ${\text{k}}_{\text{Cat}}$ (turnover number). The maximum rate of the enzymatic reaction is, ${\text{V}}_{\text{max}}$(maximum velocity).

Answer to Problem 31RE

$\begin{array}{c}\left[{\text{E}}_{\text{T}}\right]=\text{ 1}\text{.56 }\times {\text{ 10}}^{\text{-10}}\text{ mol}\\ {\text{k}}_{\text{Cat}}={\text{ 10,700 sec}}^{\text{-1}}\end{array}$

Explanation of Solution

Given information:

${\text{V}}_{\text{max}}$ of the solution $=100\text{ μmol}$ of product per minute.

Volume of an enzyme solution $=0.1\text{ mL}$.

Molecular weight of an enzyme $=128.000\text{ g/mol}$.

Concentration of enzyme solution $=0.2\text{ mg/mL}$.

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 $V_{\text{max}}$ and $\left[E_{\text{T}}\right]$ is needed for the calculation of the turnover number of an enzyme.

The turnover number $\left(k_{\text{Cat}}\right)$ of a reaction can be calculated as:

$k_{\text{Cat}}=\frac{V_{\text{max}}}{\left[E_{\text{T}}\right]}$

$\left[E_{\text{T}}\right]$ is the total number of enzymes present in the reaction and $V_{\text{max}}$ is the maximum rate of a reaction that is catalyzed by an enzyme.

0.2 mg/ml is the density of an enzyme present in 0.1 ml of solution. Since,

$\begin{array}{l}\text{density (mg/mL) }=\frac{\text{mass (mg)}}{\text{volume (mL)}}\\ \text{upon rearranging the expression,}\\ \text{mass (mg) }=\text{ density (mg/mL) }\times \text{ volume (mL)}\end{array}$

Substitute 0.2 mg/mL for density and 0.1 mL for volume in the above expression for calculating the mass:

$\begin{array}{c}\text{mass (mg)}=0.2\left(\frac{\text{mg}}{\cancel{\text{mL}}}\right)\times 0.1\cancel{\text{mL}}\\ \text{mass (mg)}=0.02\text{ mg}\end{array}$

Now, for calculating the Enzyme concentration (density) in 1 ml of solution,

$\text{density (mg/mL) }=\frac{\text{mass (mg)}}{\text{volume (mL)}}$

Substitute 0.02 mg for mass and 1 mL for volume,

$\begin{array}{c}\text{density (mg/mL) }=\frac{\text{mass (mg)}}{\text{volume (mL)}}\\ \text{density (mg/mL) }=\frac{\text{0}\text{.02 mg}}{1\text{ mL}}\\ \text{density (mg/mL) }=\text{ 0}\text{.02 mg/mL}\end{array}$

Conversion of 0.02 mg into g is as follows:

$\begin{array}{c}1000\text{ mg }=\text{ 1 g}\\ \text{1 }=\frac{1\text{ g}}{1000\text{ mg}}\\ \text{0}\text{.2 mg }=\frac{1\text{ g}}{1000\cancel{\text{mg}}}\times \text{ 0}\text{.02 }\cancel{\text{mg}}\\ =\text{ 0}\text{.02 }\times {\text{ 10}}^{\text{-3}}\text{ g}\end{array}$

$128,000\text{ g/mol }(128\times {10}^3\text{g/mol})$ is the given enzyme concentration, hence the calculation of total enzyme concentration $\left[{\text{E}}_{\text{T}}\right]$ is done as follows:

128x 10³ g of enzyme present in 1 mole of solution.

This implies, 1 g of enzyme present in $\frac{1}{128\times {\text{ 10}}^{\text{3}}\text{ g/mol}}$ a mole of solution.

Thus, the concentration of $\left[E_{\text{T}}\right]$($0.02\times {\text{ 10}}^{\text{-3}}\text{ g}$ of enzyme) is as follows:

$\begin{array}{c}\left[E_{\text{T}}\right]=\frac{0.02\times \text{ 10-3 }\cancel{\text{gm}}}{128\times \text{ 10-3 }\frac{\cancel{\text{gm}}}{\text{mol}}}\\ \left[E_{\text{T}}\right]=\text{ 0}\text{.000156 }\times {\text{ 10}}^{\text{-6}}\text{ mol}\\ \left[E_{\text{T}}\right]=\text{ 1}\text{.56 }\times {\text{ 10}}^{\text{-10}}\text{ mol}\end{array}$

The value of total concentration $\left[E_{\text{T}}\right]$ is 1.56x10^-10 mole.

$V_{\text{max}}$ of this reaction is $100\text{ μmol/min}$, the conversion of $\text{μmol/min}$ into mol/sec is as follows:

Since, $1\text{ mol }=\text{ 1000000 μmol}$ and 1 min =60 sec, hence, the conversion of $100\text{ μmol/min}$ is

$\begin{array}{c}{V}_{\text{max}}=\frac{100\cancel{\text{μ mol}}}{1\cancel{\min }}\times \frac{1\text{ mol}}{1000000\cancel{\text{μ mol}}}\times \frac{1\cancel{\min }}{60\text{ sec}}\\ V_{\text{max}}=\frac{100\text{ mol}}{1000000\times \text{ 60 sec}}\end{array}$

Recall the expression for the turnover number of an enzyme:

$k_{\text{Cat}}=\frac{V_{\text{max}}}{\left[E_{\text{T}}\right]}$

Substitute 1.56 x 10^-10 mol for $\left[E_{\text{T}}\right]$ and $\frac{100\text{ mol}}{1000000\times \text{ 60 sec}}\text{ for }{V}_{\text{max}}$,

Hence the calculation is as follows:

$\begin{array}{c}{k}_{\text{Cat}}=\frac{100\cancel{\text{mol}}}{1000000\times \text{ 60 sec }\times \text{ 1}\text{.56 }\times {\text{ 10}}^{\text{-10}}\cancel{\text{mol}}}\\ k_{\text{Cat}}={\text{ 10,700 sec}}^{\text{-1}}\end{array}$

Conclusion

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 $1.56\times {\text{ 10}}^{\text{-10}}\text{ mol}$ and their turnover number is $10,700{\text{ sec}}^{\text{-1}}$.