Chapter 13, Problem 15P

Interpretation Introduction

(a)

To predict:

The equation should be determined for $\upsilon$which is the rate of the enzyme-catalyzed reaction $S\to P$in the terms of modified Michaelis- Menten model.

Introduction:

The Michaelis- Menten Equation relates reaction velocity with substrate concentration where they form enzyme-substrate complex which further reacts irreversibly to form a product and free enzyme.

$E+S\rightleftharpoons ES\to E+P$

  $\upsilon =\frac{d\left[P\right]}{dt}=\frac{V_{\max }\left[S\right]}{K_m+\left[S\right]}$

Where,

• $V_{\max }$= maximum velocity

  $K_m$

• = Michaelis- Menten constant

  $\left[S\right]$

• = substrate concentration

Explanation of Solution

Normally, simple Michaelis-Menten equation represents one equation in equilibrium and another is going to complete.

$E+S_{k_{-1}}^{k_1}\to ES\stackrel{k_2}{\to }E+P$

After modification in Michaelis-Menten equation, two equations in equilibrium can be represented as below:

$E+S_{k_{-1}}^{k_1}\to E{S}_{k2}^{k_1}\to E+P$

In the simple Michaelis-Menten equation, there is an assumption that $\left[ES\right]$remains constant throughout whole reaction. We can use this same assumption for modified equation which can also indicates that rate of ES formation must equal the rate of ES disappearance.

$k_1\left[E\right]\left[S\right]+k_{-2}\left[E\right]\left[P\right]=k_{-1}\left[ES\right]+k_2\left[ES\right]\mathrm{......}(1)$

The rate of reaction is the rate of production of P which can be represented by the forward $k_2$reaction as below.

$\upsilon =k_2\left[ES\right]\mathrm{.....}(2)\text{}\text{ where rate can be determined by }\left[ES\right].$

With the help of equation-(1), we can solve $\left[ES\right]$as below.

$\begin{array}{l}{k}_1\left[E\right]\left[S\right]+k_{-2}\left[E\right]\left[P\right]=k_{-1}\left[ES\right]+k_2\left[ES\right]\\ \left[E\right]\left(k_1\left[S\right]+k_{-2}\left[P\right]\right)=\left(k_{-1}+k_2\right)\left[ES\right]\\ \frac{\left[E\right]\left(k_1\left[S\right]+k_{-2}\left[P\right]\right)}{\left(k_{-1}+k_2\right)}=\left[ES\right]\mathrm{....}(3)\end{array}$

The total enzyme concentration and the concentration of enzyme bound to the substrate can be used for calculation of the concentration of free enzyme as below.

$\left[E\right]=\left[E_r\right]-\left[ES\right]\mathrm{....}(4)$

In the equation (3), substituting values and rearranging for solving $\left[ES\right]$as follow.

$\begin{array}{l}\frac{\left[E\right]\left(k_1\left[S\right]+k_{-2}\left[P\right]\right)}{\left(k_{-1}+k_2\right)}=\left[ES\right]\mathrm{....}(3)\\ \frac{\left(\left[E_r\right]-\left[ES\right]\right)\left(k_1\left[S\right]+k_{-2}\left[P\right]\right)}{\left(k_{-1}+k_2\right)}=\left[ES\right]\\ \frac{\left(\left[E_r\right]\left(k_1\left[S\right]+k_{-2}\left[P\right]\right)\right)}{\left(k_{-1}+k_2\right)}-\frac{\left[ES\right]\left(k_1\left[S\right]+k_{-2}\left[P\right]\right)}{\left(k_{-1}+k_2\right)}=\left[ES\right]\\ \frac{\left(\left[E_r\right]\left(k_1\left[S\right]+k_{-2}\left[P\right]\right)\right)}{\left(k_{-1}+k_2\right)}=\left[ES\right]+\frac{\left[ES\right]\left(k_1\left[S\right]+k_{-2}\left[P\right]\right)}{\left(k_{-1}+k_2\right)}\mathrm{....}(5)\\ \left[ES\right]=\frac{\frac{\left(\left[E_r\right]\left(k_1\left[S\right]+k_{-2}\left[P\right]\right)\right)}{\left(k_{-1}+k_2\right)}}{\frac{\left[ES\right]\left(k_1\left[S\right]+k_{-2}\left[P\right]\right)}{\left(k_{-1}+k_2\right)}}\end{array}$

In the equation (2), substituting values as below,

$\begin{array}{l}\upsilon =k_2\left[ES\right]\mathrm{.....}(2)\text{}\\ \upsilon =k_2\frac{\frac{\left(\left[E_r\right]\left(k_1\left[S\right]+k_{-2}\left[P\right]\right)\right)}{\left(k_{-1}+k_2\right)}}{\frac{\left[ES\right]\left(k_1\left[S\right]+k_{-2}\left[P\right]\right)}{\left(k_{-1}+k_2\right)}}\mathrm{....}(6)\end{array}$

The above equation (6) can be used for getting value of $V_{\max }{\text{ and two K}}_m$value. When all available enzymes is combined with the substrate then the maximum rate can be achieved.

In the equation (2), we can substitute value of $\left[E_T\right]for\left[ES\right]$as below.

$\begin{array}{l}{V}_{\max }=k_2\left[ES\right]\\ V_{\max }=k_2\left[E_T\right]\end{array}$

With the help of rate constants, we can determine the two value of $K_m$.

$\begin{array}{l}\frac{1}{K_m^S}=\frac{k_1}{k_{-1}+k_2}\\ \frac{1}{K_m^p}=\frac{k_{-2}}{k_{-1}+k_2}\end{array}$

In the equation (6), substituting values,

$\begin{array}{l}\upsilon =k_2\frac{\frac{\left(\left[E_r\right]\left(k_1\left[S\right]+k_{-2}\left[P\right]\right)\right)}{\left(k_{-1}+k_2\right)}}{\frac{\left[ES\right]\left(k_1\left[S\right]+k_{-2}\left[P\right]\right)}{\left(k_{-1}+k_2\right)}}\\ \upsilon =\frac{V_{\max }\left(\frac{\left[S\right]}{K_m^S}+\frac{\left[P\right]}{K_m^P}\right)}{1+\left(\frac{\left[S\right]}{K_m^S}+\frac{\left[P\right]}{K_m^P}\right)}\end{array}$

Interpretation Introduction

(b)

To predict:

The equilibrium constant should be determined for solved Michaelis-Menten equation when $\upsilon =0$.

Introduction:

The Michaelis- Menten Equation relates reaction velocity with substrate concentration where they form enzyme-substrate complex which further reacts irreversibly to form a product and free enzyme.

$E+S\rightleftharpoons ES\to E+P$

  $\upsilon =\frac{d\left[P\right]}{dt}=\frac{V_{\max }\left[S\right]}{K_m+\left[S\right]}$

Where,

• $V_{\max }$= maximum velocity

  $K_m$

• = Michaelis- Menten constant
• $\left[S\right]$= substrate concentration

Explanation of Solution

We have the solved Michaelis-Menten equation as below.

$\upsilon =\frac{V_{\max }\left(\frac{\left[S\right]}{K_m^S}+\frac{\left[P\right]}{K_m^P}\right)}{1+\left(\frac{\left[S\right]}{K_m^S}+\frac{\left[P\right]}{K_m^P}\right)}$

Now, we have $\upsilon =0$, so the above equation would be,

$\begin{array}{l}\upsilon =\frac{V_{\max }\left(\frac{\left[S\right]}{K_m^S}+\frac{\left[P\right]}{K_m^P}\right)}{1+\left(\frac{\left[S\right]}{K_m^S}+\frac{\left[P\right]}{K_m^P}\right)}\\ 0=\frac{V_{\max }\left(\frac{\left[S\right]}{K_m^S}+\frac{\left[P\right]}{K_m^P}\right)}{1+\left(\frac{\left[S\right]}{K_m^S}+\frac{\left[P\right]}{K_m^P}\right)}\\ 0=V_{\max }\left(\frac{\left[S\right]}{K_m^S}+\frac{\left[P\right]}{K_m^P}\right)\\ \left(\frac{\left[S\right]}{K_m^S}+\frac{\left[P\right]}{K_m^P}\right)=0\\ \frac{\left[S\right]}{K_m^S}=-\frac{\left[P\right]}{K_m^P}\end{array}$

Therefore the equilibrium constant would be,

$\begin{array}{l}\frac{\left[S\right]}{K_m^S}=-\frac{\left[P\right]}{K_m^P}\\ \frac{K_m^P}{K_m^S}=-\frac{\left[P\right]}{\left[S\right]}\\ \frac{\left[P\right]}{\left[S\right]}=-\frac{K_m^P}{K_m^S}\end{array}$