A new mathematical model for the enzymatic kinetic resolution of racemates

Abstract

A mathematical model is presented for the kinetic resolution of racemates. It takes all intermediate binding steps into account and assumes that such steps are reversible. The model describing dynamics of the chiral reaction products consists of two nonlinear differential equations. With this model, the enantioselectivity of enzyme has been studied. Mathematical and numerical simulation of the model show that there are several ways to control the enantiomeric ratio (E) but the affinity and the binding rates of the intermediate enzyme complex to the racemic substrates are the key steps for the enzyme enantioselectivity.

Appendix: Derivation of the model equations

The system of eight differential equations in Eqs. (2.4)–(2.11) and the mass conservation equations in Eqs. (2.1)–(2.3) describe dynamic evolution of the reaction network in Scheme 1. To simplify this model further, we assume that this system is at quasi-steady state and all the enzyme complexes stay constant over the course of the reaction. Therefore,

$$\begin{aligned}&\frac{d[{ EC}]}{dt}=0=k_1 [E][C ]-k_2 [{EC}]-k_3 [{EC}]+k_4 [{ EE}][P]\end{aligned}$$

(5.1)

$$\begin{aligned}&\frac{d[{ EE}]}{dt}=0=k_3 [{EC}]-k_4 [{ EE}][P]-k_5 [{ EE}][{DS} ]+k_6 [{EDS}]\nonumber \\&\qquad -k_{11} [{ EE}][{LS}]+k_{12} [{ ELS}]\end{aligned}$$

(5.2)

$$\begin{aligned}&\frac{d[{ EDS}]}{dt}=0=k_5 [{ EE}][{ DS}]-k_6 [{ EDS}]-k_7 [{ EDS}]+k_8 [{ EDP}]\end{aligned}$$

(5.3)

$$\begin{aligned}&\frac{d[{ EDP}]}{dt}=0=k_7 [{ EDS}]-k_8 [{ EDP}]-k_9 [{ EDP}]+k_{10} [E ][{ DP}]\end{aligned}$$

(5.4)

$$\begin{aligned}&\frac{d[{ ELS}]}{dt}=0=k_{11} [{ EE}][{ LS}]-k_{12} [{ ELS}]-k_{13} [{ ELS} ]+k_{14} [{ELP}]\end{aligned}$$

(5.5)

$$\begin{aligned}&\frac{d[{ ELP}]}{dt}=0=k_{13} [{ ELS} ]-k_{14} [{ ELP}]-k_{15} [{ ELP} ]+k_{16} [E][{ LP}] \end{aligned}$$

(5.6)

By solving Eq. (5.6) for [ELP], we get

$$\begin{aligned}{}[ELP]=\alpha _0 [ELS] + \alpha _1 [E][LP] \end{aligned}$$

(5.7)

where,

$$\begin{aligned} \alpha _0 ={k_{13} }/{\left( {k_{14} + k_{15} } \right) }\quad \text{ and } \quad \alpha _1 ={k_{16} }/{\left( {k_{14} + k_{15} } \right) } \end{aligned}$$

(5.8)

After plugging [ELP] into Eq. (5.5) and solving it for [ELS] gives

$$\begin{aligned}{}[ELS]=\alpha _4 [LS][EE] + \alpha _5 [E][LP] \end{aligned}$$

(5.9)

Here $\alpha _4$ and $\alpha _5$ are defined in terms of individual rate constants as,

$$\begin{aligned} \alpha _4&= \frac{k_{11} (k_{14} + k_{15} )}{k_{12} (k_{14} + k_{15} ) + k_{13} k_{15} } \nonumber \\ \alpha _5&= \frac{k_{14} k_{16} }{k_{12} (k_{14} + k_{15} ) + k_{13} k_{15} } \end{aligned}$$

(5.10)

From Eq. (5.4), [EDP] becomes

$$\begin{aligned}&[EDP]=\alpha _2 [EDS] + \alpha _3 [E][DP]\end{aligned}$$

(5.11)

$$\begin{aligned}&\alpha _2= {k_7 }/{({k_8 + k_9})}\quad \text{ and } \quad \alpha _3 ={k_{10} }/{( {k_8 + k_9 })} \end{aligned}$$

(5.12)

After substituting $[EDP]$ into Eq. (5.3) and solving it [EDS], we obtain

$$\begin{aligned}&[EDS]=\alpha _6 [DS][EE] + \alpha _7 [E][DP]\end{aligned}$$

(5.13)

$$\begin{aligned}&\alpha _6 = \frac{k_5 (k_8 +k_9 )}{k_6 (k_8 +k_9 )+k_7 k_9 }\quad \text{ and }\quad \alpha _7 =\frac{k_8 k_{10} }{k_6 (k_8 +k_9 )+k_7 k_9 } \end{aligned}$$

(5.14)

Solution of Eq. (5.1) for [EC] is

$$\begin{aligned}&[EC]=\alpha _8 [E] + \alpha _9 [EE]\end{aligned}$$

(5.15)

$$\begin{aligned}&\alpha _8\!=\!\frac{k_{1C} }{k_2\!+\!k_3 }\quad \text{ and } \quad \alpha _9\!=\!\frac{k_{4P} }{k_2\!+\!k_3 },k_{1C}\!=\!k_1 [C]\quad \text{ and } \quad k_{4P}\!=\!k_4 [P] \end{aligned}$$

(5.16)

By solving Eq. (5.2) for [EE] after substituting [EC] in Eq. (5.15), [EDS] in Eq. (5.13) and [ELS] in Eq. (5.9), we get

$$\begin{aligned}{}[EE]=\rho [E] \end{aligned}$$

(5.17)

Here $\rho$ is a function of the substrate and the product concentrations, which has the following form

$$\begin{aligned} \rho :=\rho \left( {[DP],[LP],[DS],[LS]} \right) =\frac{\Theta _1 + \Theta _2 [DP] + \Theta _3 [LP]}{\Theta _4 + \Theta _5 [DS] + \Theta _6 [LS]} \end{aligned}$$

(5.18)

Here, $\Theta _{\mathrm{i}}$’s $(i=1\ldots 6)$ are defined as

$$\begin{aligned} \Theta _1&= k_3 \alpha _8 =\frac{k_3 k_{1C} }{k_{2P} +k_3 } \nonumber \\ \Theta _2&= k_6 \alpha _7 =\frac{k_6 k_8 k_{10} }{k_6 (k_8 +k_9 )+k_7 k_9 }\nonumber \\ \Theta _3&= k_{12} \alpha _5 =\frac{k_{12} k_{14} k_{16} }{k_{12} (k_{14} + k_{15} ) + k_{13} k_{15} } \nonumber \\ \Theta _4&= k_4 -k_3 \alpha _9 =\frac{k_2 k_{4P} }{k_2 +k_3 } \nonumber \\ \Theta _5&= k_5 -k_6 \alpha _6 =\frac{k_5 k_7 k_9 }{k_6 (k_8 +k_9 )+k_7 k_9}\nonumber \\ \Theta _6&= k_{11} -k_{12} \alpha _4 =\frac{k_{11} k_{13} k_{15} }{k_{12} (k_{14} + k_{15} ) + k_{13} k_{15}} \end{aligned}$$

(5.19)

After replacing $[EE]$ in Eq. (5.15) by Eqs. (5.17), (5.15) becomes

$$\begin{aligned}{}[EC]=\left( {\alpha _8 + \alpha _9 \rho }\right) [E] \end{aligned}$$

(5.20)

Substituting $[EE]$ in Eq. (5.17) into Eq. (5.13) gives

$$\begin{aligned}{}[EDS]=\left( {\alpha _6 [DS]\rho + \alpha _7 [DP]}\right) [E] \end{aligned}$$

(5.21)

By substituting $[EE]$ in Eq. (5.17) into Eq. (5.9), we obtain

$$\begin{aligned}{}[ELS]=\left( {\alpha _4 [LS]\rho + \alpha _5 [LP]} \right) [E] \end{aligned}$$

(5.22)

By plugging [ELS] in Eq. (5.22) into [ELP] in Eq. (5.7), we get

$$\begin{aligned}{}[{ ELP}]=\left( {\alpha _0 \alpha _4 [{ LS}]\rho + \left( {\alpha _0 \alpha _5 + \alpha _1 } \right) [{ LP}]} \right) [E] \end{aligned}$$

(5.23)

When we replace [EE] in Eq. (5.13) by [EE] given by Eq. (5.15), we obtain

$$\begin{aligned}{}[{ EDS}]=\left( {\alpha _6 [DS]\rho + \alpha _7 [{ DP}]} \right) [E] \end{aligned}$$

(5.24)

Then substituting [EDS] in Eq. (5.24) into Eq. (5.11) gives us

$$\begin{aligned}{}[{ EDP}]=\left( {\alpha _2 \alpha _6 [DS]\rho +\left( {\alpha _2 \alpha _7 +\alpha _3 } \right) [DP]} \right) [E] \end{aligned}$$

(5.25)

Plugging [ELP], [EDP], [ELS], [EDS], [EE] and [EC] into the conservation equation for the enzyme given by Eq. (2.1), and then solving it for [E] we get

$$\begin{aligned}&[E]\nonumber \\&\quad \qquad \!=\! \frac{[E]_0 \left( {\Theta _4 \!+\! \Theta _5 [DS] \!+\! \Theta _6 [LS]} \right) }{ \left( {\Omega _1 \!+\! \Omega _2 [LP] \!+\! \Omega _3 [DP]} \right) \left( {\Theta _4 \!+\! \Theta _5 [DS] \!+\! \Theta _6 [LS]} \right) \!+\!\left( {\Omega _4 \!+\! \Omega _5 [DS] \!+\! \Omega _6 [LS]} \right) \left( {\Theta _1 \!+\! \Theta _2 [DP] \!+\! \Theta _3 [LP]} \right) }\nonumber \\ \end{aligned}$$

(5.26)

where,

$$\begin{aligned} \Omega _1&= 1+\alpha _9 >0 \nonumber \\ \Omega _2&= \alpha _1 +(1+\alpha _0 )\alpha _5 >0 \nonumber \\ \Omega _3&= \alpha _3 +(1+\alpha _2 )\alpha _7 >0 \\ \Omega _4&= 1 + \alpha _8 >0 \nonumber \\ \Omega _5&= \alpha _6 \left( {1+\alpha _2 } \right) >0 \nonumber \\ \Omega _6&= \alpha _4 (1+\alpha _0 )>0\nonumber \end{aligned}$$

(5.27)

From Eqs. (5.25), (5.26) and (2.10), rate of change for [DP] becomes

$$\begin{aligned}&\frac{d[DP]}{dt}\nonumber \\&\quad \qquad \!=\! \frac{ V_{\max 1} [DS]\left( {\Theta _1 \!+\! \Theta _2 [DP] \!+\! \Theta _3 [LP]} \right) \!-\! V_{\max 2} [DP]\left( {\Theta _4 \!+\! \Theta _5 [DS] \!+\! \Theta _6 [LS]} \right) }{ \left( {\Omega _1 \!+\! \Omega _2 [LP] \!+\! \Omega _3 [DP]} \right) \left( {\Theta _4 \!+\! \Theta _5 [DS] \!+\! \Theta _6 [LS]} \right) \!+\!\left( {\Omega _4 [DS]\!+\! \Omega _5 \!+\! \Omega _6 [LS]} \right) \left( {\Theta _1 \!+\! \Theta _2 [DP] \!+\! \Theta _3 [LP]} \right) }\nonumber \\ \end{aligned}$$

(5.28)

Here $V_{\max 1}$ and $V_{\max 2}$ are defined in terms of rate constants as

$$\begin{aligned}&V_{\max 1} = k_9 \alpha _2 \alpha _6 [E]_0 =\frac{k_5 k_7 k_9 }{k_6 (k_8 +k_9 )+k_7 k_9 }[E]_0 >0 \nonumber \\&V_{\max 2} = \left( {k_{10} -k_9 \left( {\alpha _2 \alpha _7 + \alpha _3 } \right) } \right) [E]_0 =\frac{k_6 k_8 k_{10} [E]_0 }{k_6 (k_8 +k_9 )+k_7 k_9 } >0 \end{aligned}$$

(5.29)

Finally, form Eqs. (5.23), (5.26) and (2.11), rate of change for [LP] becomes

$$\begin{aligned}&\frac{d[LP]}{dt}\nonumber \\&\quad \qquad \!=\! \frac{ V_{\max 3} [LS]\left( {\Theta _1 \!+\! \Theta _2 [DP] \!+\! \Theta _3 [LP]} \right) \!-\! V_{\max 4} [LP]\left( {\Theta _4 \!+\! \Theta _5 [DS] \!+\! \Theta _6 [LS]} \right) }{ \left( {\Omega _1 \!+\! \Omega _2 [LP] \!+\! \Omega _3 [DP]} \right) \left( {\Theta _4 \!+\! \Theta _5 [DS] \!+\! \Theta _6 [LS]} \right) \!+\!\left( {\Omega _4 [DS] \!+\! \Omega _5 \!+\! \Omega _6 [LS]} \right) \left( {\Theta _1 \!+\! \Theta _2 [DP] \!+\! \Theta _3 [LP]} \right) }\nonumber \\ \end{aligned}$$

(5.30)

Here $V_{\max 3}$ and $V_{\max 4}$ are given by

$$\begin{aligned}&V_{\max 3} =k_{15} \alpha _0 \alpha _4 [E]_0 =\frac{k_{11} k_{13} k_{15} }{k_{12} (k_{14} + k_{15} ) + k_{13} k_{15} }[E]_0 >0 \nonumber \\&V_{\max 4} =\left( {k_{16} -k_{15} \left( {\alpha _0 \alpha _5 + \alpha _1 } \right) } \right) [E]_0 \text{= }\frac{k_{12} k_{14} k_{16} [E]_0 }{k_{12} (k_{14} +k_{15} )+k_{13} k_{15} } >0 \end{aligned}$$

(5.31)