Estimating the Product Inhibition Constant from Enzyme Kinetic Equations Using the Direct Linear Plot Method in One-Stage Treatment

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Estimating the Product Inhibition Constant from Enzyme Kinetic Equations Using the Direct Linear Plot Method in One-Stage Treatment catalysts Article Estimating the Product Inhibition Constant from Enzyme Kinetic Equations Using the Direct Linear Plot Method in One-Stage Treatment Pedro L. Valencia 1,* , Bastián Sepúlveda 2, Diego Gajardo 2 and Carolina Astudillo-Castro 3 1 Department of Chemical and Environmental Engineering, Universidad Técnica Federico Santa María, P.O. Box 110-V, Valparaíso 2390123, Chile 2 Department of Mathematics, Universidad Técnica Federico Santa María, Valparaíso 2390123, Chile; [email protected] (B.S.); [email protected] (D.G.) 3 School of Food Engineering, Pontificia Universidad Católica de Valparaíso, Valparaíso 2360100, Chile; [email protected] * Correspondence: [email protected] Received: 15 June 2020; Accepted: 8 July 2020; Published: 1 August 2020 Abstract: A direct linear plot was applied to estimate kinetic constants using the product’s competitive inhibition equation. The challenge consisted of estimating three kinetic constants, Vmax, Km, and Kp, using two independent variables, substrates, and product concentrations, in just one stage of mathematical treatment. The method consisted of combining three initial reaction rate data and avoiding the use of the same three product concentrations (otherwise, this would result in a mathematical indetermination). The direct linear plot method was highly superior to the least-squares method in terms of accuracy and robustness, even under the addition of error. The direct linear plot method is a reliable and robust method that can be applied to estimate Kp in inhibition studies in pharmaceutical and biotechnological areas. Keywords: direct linear plot; median method; product inhibition; kinetic constants; non-parametric; distribution-free method 1. Introduction The direct linear plot (DLP) is a graphic method to estimate kinetic constants from enzymatic reactions based on the median as a statistic. The method was developed by Eisenthal and Cornish-Bowden in 1974 [1]. The accuracy and robustness of this method were tested during the estimation of Vmax and Km from the Michaelis–Menten equation [1,2]. As in this article, all of the recent studies regarding the application of DLP are based on the estimation of these two parameters from the Michaelis–Menten equation. The application of DLP to equations with more than two parameters has not been studied until now [3]. This median-based method was applied herein to estimate three kinetic constants from the substrate-uncompetitive inhibition equation. The results indicated that DLP was not only applicable to equations with more than two parameters, but it was reliable and robust when compared to the least-squares (LS) method. In the present article, we explored the application of DLP to the product-competitive inhibition equation, which is a different type of problem, as will be exposed. Some attempts to estimate inhibition constants from enzyme kinetics using DLP have involved estimating apparent kinetic constants and using secondary plots to estimate the values of these inhibition constants. The application of DLP to estimate inhibition constants was first proposed by Eisenthal and Cornish-Bowden [1]. The literature contains outdated examples of this application because the use of DLP to estimate inhibition constants in one-stage has not been assessed [4–8]. In the literature, DLP has been used to estimate the apparent kinetic Catalysts 2020, 10, 853; doi:10.3390/catal10080853 www.mdpi.com/journal/catalysts Catalysts 2020, 10, x FOR PEER REVIEW 2 of 11 Catalysts 2020, 10, 853 2 of 10 assessed [4–8]. In the literature, DLP has been used to estimate the apparent kinetic constants Vmax and Km; however, studies have used secondary plots to estimate the inhibition constant Kp. The constants Vmax and Km; however, studies have used secondary plots to estimate the inhibition constant present article studied the application of DLP to estimate the product inhibition constant in one-stage Kp. The present article studied the application of DLP to estimate the product inhibition constant in one-stagetreatment, treatment, avoiding avoidingthe estimation the estimation of apparent of apparent kinetic constants kinetic constants and secondary and secondary plots. As plots. the proper As the propercharacterization characterization of inhibition of inhibition in pharmaceutical in pharmaceutical and biotechnological and biotechnological applications applications is a major is concern, a major concern,the importance the importance of this proposal of this proposal is that it isopens that itup opens the possibility up the possibility of using ofa reliable using a and reliable robust and method robust to estimate product inhibition constants. The purpose of this study is to estimate Vmax, Km, and Kp from method to estimate product inhibition constants. The purpose of this study is to estimate Vmax, Km, the competitive inhibition equation in just one stage of calculations, avoiding the apparent constants and Kp from the competitive inhibition equation in just one stage of calculations, avoiding the apparent constantsand secondary and secondaryplots. This plots.means Thisthe values means of the the values product of inhibition the product constants inhibition must constants be obtained must from be obtainedthe DLP method, from the which DLP method, has never which been has tried never before. been The tried problem before. Theis outlined problem as is follows. outlined as follows. The equation for competitive product inhibition involves three parameters—Vmax, Km, and Kp— The equation for competitive product inhibition involves three parameters—Vmax, Km, and Kp—and threeand variables—substratethree variables—substrate concentration, concentration, product concentration,product concentration, and initial reaction and initial rate (Equation reaction (1)).rate (Equation (1)). VmaxS v = (1) = K K + S + m P m Kp (1) + + This is a different type of problem compared to previous articles where the substrate-uncompetitive This is a different type of problem compared to previous articles where the substrate- inhibition equation involved three parameters—Vmax, Km, and KS—but only two variables—substrate uncompetitive inhibition equation involved three parameters—Vmax, Km, and KS—but only two concentration and initial reaction rate (Equation (2)) [3]. variables—substrate concentration and initial reaction rate (Equation (2)) [3]. V S v = max (2) = S2 Km + S + (2) + K+s TheThe experimental dataset dataset required required is istraditionally traditionally used used in the in initial the initial reaction reaction rate method, rate method, which whichdenotes denotes a series a seriesof initial of initial rates ratesvs. substrate vs. substrate conc concentrationsentrations at constant at constant product product concentrations, concentrations, as asindicated indicated in inthe the scheme scheme of ofFigure Figure 1.1 . P = P1 P = P2 P = Pm S v S v S v S1 v11 S1 v21 S1 vm1 … … … … … … Sn v1n Sn v2n Sn vmn FigureFigure 1.1. SchemeScheme ofof thethe traditionaltraditional experimentalexperimental designdesign based on thethe initialinitial reactionreaction raterate methodmethod toto estimateestimate inhibitioninhibition constantsconstants inin enzymeenzyme kinetics.kinetics. EquationEquation (1)(1) cancan bebe rearrangedrearranged toto EquationEquation (3)(3) inin thethe followingfollowing form:form: v P Km − + ij i = − (3) vijKm SjVmax + = vijSj (3) − Kp − The definition of vij is the rate of product concentration Pi and substrate concentration Sj. As the The definition of v is the rate of product concentration P and substrate concentration S . As the estimation of three parametersij is required, three data pairs andi three equations are neededj to solve estimation of three parameters is required, three data pairs and three equations are needed to solve the the matrix system in Equation (4). matrix system in Equation (4). 0 − 10 1 0 − 1 v S v P Km v S B ii0 −i0 ii0iCB ⎛ ⎞C =B −ii0 i0 C (4) B − CB C B − C B v S v P CB Vmax C = B v S C (4) B jj0 −j0 jj0jCB C B −jj0 j0 C B − CB K C B − C @ v S v P A@B ⎝ m ⎠AC @ v S A kk0 − k0 kk0 k Kp − kk0 k0 Likewise, the definition of vii’ = v(Pi, Si’), i.e., i values are between 1 and m (the number of product concentrations), and i’ values are between 1 and n (the number of substrate concentrations). Alternatively, the solution for Equation (4) is presented in Equation (5) in algebraic form. ( − + − + (− + )) = (5) − + ( −) + − Catalysts 2020, 10, 853 3 of 10 Likewise, the definition of vii’ = v(Pi, Si’), i.e., i values are between 1 and m (the number of product concentrations), and i’ values are between 1 and n (the number of substrate concentrations). Alternatively, the solution for Equation (4) is presented in Equation (5) in algebraic form. vii vjj vkk (Pk(Si Sj )+Pi(Sj Sk )+Pj( Si +Sk )) Vmax = 0 0 0 0 − 0 0 − 0 − 0 0 (Pj Pi)S v v +(Pi P )S v v +(Pj Pi)S v v − k0 ii0 jj0 − k j0 ii0 kk0 − i0 jj0 kk0 PiSj Sk vii (vkk vjj )+PkSi Sj vkk (vjj vii )+PjSi Sk vjj (vii vkk ) Km = 0 0 0 0 − 0 0 0 0 0 − 0 0 0 0 0 − 0 (5) (Pi Pj) S v v +v ( PiS v +P S v +PjS v P S v ) − k0 ii0 jj0 kk0 − j0 ii0 k j0 ii0 i0 jj0 − k i0 jj0 S S v (v v )+S S v (v v )+S S v (v v ) Km j0 k0 ii0 jj0 − kk0 i0 j0 kk0 ii0 − jj0 i0 k0 jj0 kk0 − ii0 K = p (Pi Pj)S v v +v ( PiS v +P S v +PjS v P S v ) − k0 ii0 jj0 kk0 − j0 ii0 k j0 ii0 i0 jj0 − k i0 jj0 The rank of the matrix in Equation (4) must be 3 for the system to have a solution. For the case when three product concentrations are taken from the same set, Pi = Pj = Pk, the system has no solution because the rank of the resulting matrix is 2. This problem can be easily checked by observing the denominators of Equation (5) when replacing Pi, Pj, and Pk with the same value of product concentration.
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