Henri-Michaelis-Menten kinetics of reversible enzymic reactions, and the determination of rate constants from kinetic constants

Abstract

The Michaelis constants derived for two reversible uni-reactant - uni-product reaction models, given originally by Haldane, are corrected. In the direction starting with the reactant having the lower binding constant, the steady state is one in which the enzyme-reactant intermediate has a concentration approximating the final equilibrium concentration., Consequently, the Haldane relationship is generally invalid, and kinetic analyses to validate the use of the kinetic constant ratio (kcat/Km) as a measure of specificity are also generally invalid. This correction of Michaelis constants is pertinent to attempts to back calculate rate constants from experimental values: the Michaelis constant used must be correct.

Introduction

Haldane^1,2 applied the now-called (quasi) steady-state assumption introduced earlier by Briggs and Haldane ³ to the reversible reactions shown in Figure 1, and derived for both models, and for the reactions in both directions, Henri-Michaelis-Menten equations. These, for a reaction starting with reactant A, apply to initial steady state velocities, have the form shown in equation 1, and identified the kinetic constants, $k_{cat}^A$ and $K_m^A$ , in terms of rate constants. (The superscript ^A denotes the starting reactant is A, and the subscript o indicate initial concentrations). The derivation required the assumption that the concentration of B was negligible at the entry to the steady state so that the concentration of A was sensibly measured by its starting concentration, a_o, and therefore in equation 1, db/dt is an initial steady-state velocity. Briggs and Haldane ³ solved the rate equation for X in an irreversible form of model 1 (with k- ₂  = zero) by noting that dx/dt was zero at a maximum concentration of X. A maximum is a necessary condition in a reaction in which x starts and ends at zero. Haldane^1,2 assumed that this procedure was sufficient in his treatment of reversible reactions, but about sixty years ago evidence appeared to show that it was not. An analysis of model 1 by Morales and co-authors^4,5 showed that x would pass through a maximum, before descending to its final equilibrium value (x_equ), only when k₁ > k-₂. Miller and Alberty ⁶ pointed out that the rate equation for X has an exact solution when k₁ = k-₂, and showed that in the pre-steady state x followed a simple exponential course, rising monotonically to x_equ. Using a method of numerical approximation, they showed that, when k₁ > k-₂, x proceeded through a maximum, and also gave one example (with k-₂ = 10k₁) of x rising monotonically to equilibrium. Walter and Morales ⁷ made computer-aided calculations (although for reactions with different equilibrium constants), and showed that when k₁ <k-₂, and also when k₁ = k-₂, x rises monotonically to x_equ. A more recent report ⁸ has shown one similar result. These observations do not in any way invalidate the usefulness of the steady state assumption, but it should have been recognized that Haldane's use of it gave results which were only conditionally correct. When starting the reaction with A, and if k₁ < k-₂ (model 1) or k₁ < k-₃ (model 2) his structures of $K_m^A$ , written in terms of rate constants, are incorrect because he had not found the correct steady state concentrations of x and y. His results are correct only when the binding constants are equal.

Figure 1.

Two models for(A)—uni-product (B) reaction catalysed by an enzyme E. The corresponding small case letters represent concentrations,and the letters k are rate constants.

Since roughly the middle of the last Century the view has been held that experimentally-derived kinetic constants should be viewed as purely empirical measurements which under defined conditions characterize the enzyme. Their values are now obtained directly, using computer-aided numerical methods, ⁹ from primary experimental data (reactant concentrations measured at intervals) rather than from secondarily derived reaction rates. None the less, the derivations of Haldane still appear in current texts and in the literature, and furthermore so do other relationships which are based on his work. Haldane's own relationship^1,2 between the kinetic constants of the forward and reverse reactions and the equilibrium constant of the reaction, and kinetic demonstrations that the ratio of kinetic constants, k_cat/K_m, is a good measure of enzymic specificity^10,11 are not generally valid. Recently, it has also been demonstrated ¹² that methods of numerical integration can be used to obtain rate constants from a measured K_m. Such an exercise requires the correct structure of the K_m, written in terms of the rate constants of the particular kinetic model used. In the case cited, an irreversible form of model 1 was used with the correct expression for the K_m, but for, any extension of the procedure to reversible reactions, the present work should be viewed as a cautionary note.

$$db/dt=k_{cat}^A{e}_o{a}_o/(a_o+K_m^A)$$ (1)

In Section 2, I give for model 1, when k- ₂ > k₁, the correct expressions for the steady state concentration, the initial (steady state) velocity of the reaction, and $K_m^A$ . Noting that expressions for kinetic constants in terms of rate constants are derived from rate equations entirely by algebraic approximation, in Section 3 I shall continue with these methods to show for model 2 that, when k- ₃ > k₁ and X and Y are in equilibrium, the concentration of y in the pre-steady state also rises monotonically to its equilibrium concentration. This then leads to the correct equations for the initial steady-state rate of the reaction and for the K_m. In addition, in Section 4 I also give an alternative treatment of model 2 which indicates errors in Haldane's^1,2 deductions. I shall restrict my observations to the condition, usual for the characterization of an enzyme, that the reactant concentration is several orders of magnitude greater than that of the enzyme, a condition which may not always occur in vivo or where enzymes are used in vitro.

Quasi-Equilibrium in the steady state, and the correct Michaelis constants for model 1

Morales and co-authors^4,5 pointed out that, for x to pass through a maximum value when dx/dt = zero, d²x/dt ² must be negative at the maximum (which is the value used as the steady-state concentration). They showed that, when starting with A, this required that k₁ > k-₂. The implication of this is that, when k-₂ > k₁, x rises only to x_equ. (The analysis of Morales and co-authors is essentially repeated in appendix i, where I use their approach as part of the analysis of model 2). The demonstration that, in the pre-steady state, x rises monotonically to x_equ is provided by the work of Miller and Alberty, ⁶ Walter and Morales ⁷ and Tzafriri and Edelman. ⁸ When the pre-steady state is brief (Briggs and Haldane's “in the first instant of the reaction”, ^{3 )} and if the product concentration is negligible at the entry into the steady state, then beginning with A and when k- ₂ > k₁, the logical expression for the initial (steady state) velocity, $v_i^A$ , is then given by equation 2.

$$v_i^A=db/dt=k_2{x}_{equ}=k_2{a}_o{e}_o/(a_o+k{-}_1/k_1+k_2/k{-}_2)$$ (2)

The expression for x_equ is stated in reference, ⁶ and its derivation is indicated in appendix ii. Equation 2 leads to the identity $K_m^A=k{-}_1/k_1+k_2/k{-}_2$ . Only when k₁ = k- ₂ is this identical with that, (k- ₁  + k₂)/k₁, derived by Haldane.^1,2 For the reaction starting with B, and with $k{-}_2>k_1,K_m^B$ is that given by Haldane, (k-₁ + k₂)/k-₂. The correctly derived Michaelis constants, and those obtained by Haldane's approach, are given in Table 1. It may be noted that Haldane's results are correct only when the binding constants k₁ and k- ₂ (for model 1), and k₁ and k- ₃ (for model 2) are identical.

Table 1.

The Michaelis constants of reactant A for model 1 when k-₂ > k₁, and for model 2 when k-₃ > k₁ and X and Y are in equilibrium.

	Model 1	Model 2
Correct expression	k-1/k1 + k2/k-2	k-1k-2 + k1k2k3/k-3 k1(k-2 + k2)
Following Haldane*	(k-1 + k2)/ k1	k-1k-2 + k2k3 k1(k-2 + k2)

*The expression for model 2 can be found from the general K_m given by Haldane by approximation, using the condition that k-₂ and k₂ are very large. Alternatively, it is obtained from the value of y obtained when the rate equations for X and Y are equated to zero, the results are added, and x = k-₂ y/k₂ is substituted. Setting a = a_o and b = zero, then leads to a supposed steady state value of y.

The quasi - steady states of model 2, with X and Y in equilibrium, and the correct K_m

The conservation equation for E in model 2 is e_o = e + x + y, and its first differential is de + dx + dy = 0. There is insufficient information to provide an exact relationship between dx and dy in the pre-steady state: it is only in the quasi- steady state itself that x, y and e each become (quasi) constant. I have therefore given an analysis of a special case of model 2, when k₂ and k₋₂ are sufficiently large to maintain X and Y in equilibrium, a condition which allows an exact relationship between dx and dy in the pre-steady state.

In appendix i it is shown, for a reaction starting with A and with X and Y in equilibrium, that the concentration y proceeds through a maximum value only when k₁ > k- ₃ . The analysis essentially follows that of Morales and co-workers.^4,5 I then extend the analysis to show that when k- ₃ > k₁, d²y/dt² is always negative and rises monotonically to zero at equilibrium. Consequently, the only steady-state concentration available is that when y approaches the equilibrium value, y_equ, and the initial (steady state) velocity would logically be written as $v_i^A=k_3{y}_{equ}$ (as usual, b is assumed to be negligible at the beginning of the steady state, thus giving negligible back reaction, and allowing a to be written as a_o). The equilibrium concentration is given by equation 3, or with the expansion of the equilibrium constant K_equ (K_equ = k₁k₂k₃/k-₁k-₂k- ₃ ), by equation 4 (appendix ii). The initial velocity is then given by equation 5. If numerator and denominator are divided by k₁(k- ₁  + k₂) to give an equation with the form of equation 1, the $K_m^A$ predicted is (k-₁k- ₂  + k₁k₂ k₃/k-₃)/ k₁(k- ₁  + k₂)

$$y_{equ}=e_o{a}_o/\{a_o(1+k{-}_2/k_2)+k_3(1+K_{equ})/k{-}_3{K}_{equ}\}$$ (3)

$$y_{equ}=k_1{k}_2{e}_o{a}_o/\{k_1{a}_o(k{-}_2+k_2)+k{-}_{1}k{-}_2+k_1{k}_2{k}_3/k{-}_3\}$$ (4)

$$v_i^A=k_3{y}_{equ}=k_1{k}_2{k}_3{e}_o{a}_o/\{k_1{a}_o(k{-}_2+k_2)+k{-}_{1}k{-}_2+k_1{k}_2{k}_3/k{-}_3\}$$ (5)

For the condition k- ₃ > k₁, Haldane's $K_m^B$ is correct. For the condition that k₁ = k- ₃ , the terms in b disappear from the equation for (dx/dt + dy/dt) (equation (8) in appendix i), and following Miller and Alberty ⁶ the variables can be separated, and the rate equation integrated to give y without approximation (Starting with both A and B, it rises monotonically to y_equ).

A general view of model 2

An interesting general but incomplete analysis of model 2 is obtained by comparing the equilibrium concentration of Y (equation 3) with the maximum value of y (y*), obtained from Haldane's general steady-state equation for y^1,2 by setting b to zero. The condition that y* > y_equ reduces to k₁a_o <k₂ (k₁/k- ₃ - 1) - k- ₁ . (The algebra is elementary and has been omitted here). The left hand side of the inequality is positive, and so the right hand side is also positive, and therefore k₁/k- ₃ >1 + k- ₁ /k₂. When k₂ >>k- ₁ , as for example when X and Y are in equilibrium, the condition for a maximum in the progress curve of Y reduces to k₁/k-₃> 1, that given in section 3 for the same equilibrium. Although the inequality obtained by comparing y* with y_equ provides limited general information, it is a further indication that Haldane's method of simply equating dx/dt and dy/dt to zero should generally be treated with caution.

Discussion

Haldane first applied the steady state approximation of Briggs and Haldane ³ to reversible reactions in 1930, ¹ and his work was republished in 1965. ² He appears not to have considered the full requirement for a maximum, (for model 1 dx/dt = 0 and d²x/dt² < 0, and for model 2, dy/dt = 0 and d²y/dt² < 0), and it is probably this which led to the deficiency in his analyses. I have pointed out here that starting a reaction with A, and with k₁ < k- ₂ for model 1 and k₁ < k-₃ for model 2, dx/dt and dy/dt may approach zero only as the equilibrium concentrations of X and Y are approached: the second derivatives, d²x/dt ² and d²y/dt², rise monotonically to, but do not exceed, zero. In fairness to Haldane, he did state in the Introduction to the second edition of his book, ² written shortly before his death in 1964, why it was deliberately republished unchanged. It is also possible that he was not aware of the then recent work, with model 1, of Morales and co-workers,^4,5 Miller and Alberty ⁶ and Walter and Morales. ⁷

Although the final equilibrium concentration of an enzyme-substrate intermediate is, naturally, a possible solution to its rate equation, it is not revealed when, in the rate equations for models 1 and 2, the rate of change of their concentrations is simply equated to zero. The correct equations for initial rates of reaction lead to the correct analytical Michaelis constants for models 1 and 2, and these are given in Table 1.

In a steady state, the concentrations of X and Y might be in a constant ratio other than equilibrium, but an analytical solution of the rise of x and y during the pre-steady state is not possible for this condition, and it is for this reason that I have considered the special case of equilibrium (appendix i). A more general evaluation of model 2 can only be achieved by modern numerical methods. Sets of progress curves of Y could be generated, each set for a given equilibrium constant, and encompassing a reasonable range of rate constants compatible with both the equilibrium constant and the requirement that the rate constants do allow that, after the “first instant of the reaction”, the steady state is reached and b remains negligible

The condition that X and Y might be in equilibrium is not an unknown feature of multistep reactions, and has been proposed earlier based on numerical modelling (for example see. ¹³ This equilibrium essentially reduces model 2 to model 1, because X and Y can be considered as a single quantity. By ordinary kinetic means, when rates are measured from changes in reactant or product concentrations, X and Y are indistinguishable, and this equilibrium is one possible condition which would make sense of model 1. The latter model is a naïve kinetic view of a reaction mechanism: the intermediate X gives rise directly to two different products, A and B. Its continued use as a basis for numerical determinations of the catalytic constants ⁹ is only justified by the application of Ockhams razor: model 1 is the simplest to predict the observed kinetics. Haldane introduced model 2 based on experimental chemical evidence that a second intermediate may be formed after the initial binding of the reactant.

It must be concluded that, for reversible reactions, the derivation of kinetic constants in terms of rate constants must be made with caution. The application of the quasi-steady state concept requires more thought than simply equating in a rate equation dx/dt or dy/dt to zero, and if rate constants are to be back-calculated from experimental kinetic constants, the equation describing the kinetic constant must be valid. This discussion has been about single reactant-single product reactions, but it should be born in mind when multi-substrate reactions are considered.

There are other consequences of misidentified kinetic constants. The Haldane relationship,^1,2 relating the kinetic constants of a reversible reaction to the equilibrium constant, is correct for models 1 and 2 only when, respectively, k₁ = k-₂ and k₁ = k- ₃ . Although Bock and Alberty ¹⁴ made well-controlled kinetic measurements using fumarase (quite reasonably assumed to follow model 1), and obtained results consistent with the Haldane relationship, it has since been shown that the enzyme is a tetramer with four identical subunits, and extensive kinetic studies by Rose ¹⁵ have shown there are several possible kinetic pathways within each subunit. Mescam et al. ¹³ used numerical methods to determine the relative fluxes through three possible kinetic pathways of an 11-state reversible mechanism, which they adapted from the work of Rose, and their results indicate that, except for the binding steps, each pathway had intermediate stages in a state of quasi-equilibrium. The actual mechanism of fumarase is not described by either model 1 or 2, and so whatever is concluded from the results of Bock and Alberty, ¹⁴ those results are not a confirmation of the Haldane relationship.

The kinetic constants ratio, k_cat /K_m, is often named the specificity constant, and this has a good intuitive basis, but it has been shown that k_cat/K_m is not always useful for comparing different enzymes which catalyse the reaction of a single substrate. ¹⁶ Theoretical analyses, to provide a basis for the use of k_cat /K_m, when a single enzyme is catalysing two competing reactions, have all depended on the use of kinetic constants deduced from rate equations by writing “dx/dt = 0”,^10,11 thus failing to consider quasi-equilibria as solutions. For reversible reactions, the use of the kinetic constant ratio, k_cat /K_m, is only substantiated if each reactant has a binding constant greater than that of its product, or if each has a binding constant less than its product. The validity of the kinetic constant ratio as a measure of specificity therefore requires that all binding constants are known, but I am not aware of results for which this information is available.

Conclusions

Earlier work with model 1^4–8 has shown that, in a reaction starting with A and when the binding constant of B is greater than that of A, the quasi-steady state of X is one in which its concentration is approaching its final equilibrium concentration. I have pointed out that this leads to a correction of the equation for the initial steady-state velocity, and for the structure, written in terms of rate constants, of $K_m^A$ (Table 1). Using the approach of Morales and co-authors,^4,5 I have also shown for model 2 that in a reaction starting with A, and when the binding constant of B is greater than that of A, the sum (x + y) of the concentrations of the enzyme-reactant intermediates does not proceed through a maximum greater than the equilibrium concentration. In the special case that X and Y are in equilibrium, I have then shown that y then rises monotonically to its equilibrium concentration, and pointed out that a close approach to this is the steady state concentration with which the initial velocity is to be calculated. Consequently, the structure of the K_m for this reaction is not that which would be derived by Haldane's approach of equating dx/dt and dy/dt to zero.

As a result of these observations, I suggest caution is used if kinetic constants are derived by a back-calculation from kinetic constants, and I have concluded that the Haldane relationship^1,2 (between the kinetic constants of the reactants and the equilibrium constant) is not generally valid. Furthermore, the validity of using the ratio of the kinetic constants, k_cat/K_m, as a specificity constant is not generally supported by the kinetic analyses^10,11 of the catalysis of two reactions in a single solution by a single enzyme.

Appendix

 

i. When k- ₃ > k₁, y rises monotonically to y_equ

Equations 6 and 7 are the rate equations for X and Y, respectively, in model 2.

$$dx/dt=k_{1}ae+k{-}_{2}y\text{–}x(k{-}_1+k_2)$$ (6)

$$dy/dt=k{-}_{3}be+k_{2}x\text{–}y(k{-}_2+k_3)$$ (7)

Because both dx/dt and dy/dt cannot become zero at exactly the same instant, ⁷ equating each of these in equations 5 and 6 to zero, in search of a condition when y has a maximum value, introduces an uncertainty. At the best, if dy/dt is exactly zero at a maximum, dx/dt must also be negligible. To avoid any possible ambiguity,note that the sum of equations 6 and 7 gives equation 8, and differentiation of the conservation equation for the enzyme (e_o = e + x + y) gives the exact differential de + dx + dy = 0.

$$dx/dt+dy/dt=(k_{1}a+k{-}_{3}b)e\text{–}k{-}_{1}x\text{–}{k}_{3}y$$ (8)

Now dx/dt + dy/dt = d(x + y)/dt, and it can easily be shown that the sum (x + y) can have a negative phase, that is it passes through a maximum, only when k₁ > k- ₃ . Here the interest is in the change of y with time, which requires a relationship between x and y, and one does occur when k₂ and k- ₂ are sufficiently large to maintain X and Y close to equilibrium. Using the equilibrium assumption, that is k₂x = k- ₂ y, then x = k- ₂ y/k₂, dx/dt = (k- ₂ /k₂) dy/dt, and thus (dx/dt + dy/dt) can be written as mdy/dt where m = (k- ₂  + k₂)/ k₂. Noting also that a = a_o –b, and writing (k- ₃ – k₁) = δ, a compact form of equation 8 becomes equation 9.

$$mdy/dt=(k_1{a}_o+\delta b)e\text{–}y(k_3+k{-}_{1}k{-}_2/k_2)$$ (9)

Partial differentiation of equation 9 gives equation 10.

$$m{d}^{2}y/d{t}^2=\delta edb/dt+(k_1{a}_o+\delta b)de/dt\text{–}(dy/dt)(k_3+k{-}_{1}k{-}_2/k_2)$$ (10)

When in the steady state de/dt and dy/dt are negligible, d²y/dt² can be negative only when, in the term δedb/dt, δ is negative, that is k₁ > k- ₃ .

Going further than Morales and co-authors, the question now for positive δ is, can y rise only to equilibrium with a positive dy/dt falling monotonically to zero at equilibrium: is d²y/dt² always negative, rising to zero? (Were dy/dt to have a phase in which it was positive but “negligible”, before rising to equilibrium, this phase could correspond to a steady state, but under these circumstances d²y/dt² would not always be negative). Because de/dt = - (dx/dt + dy/dt) = - mdy/dt, substitution in equation 10 then gives equation 11. The term {(k₁a_o + δb)m + (k₃ + k-₁k-₂/k₂)} is not a constant, but is always

$$m{d}^{2}y/d{t}^2=\delta edb/dt-dy/dt\{(k_1{a}_o+\delta b)m+{(k_3+k{-}_{1}k{-}_2/k_2)}_{}\}$$ (11)

positive, with a value proportional to b and a maximum at equilibrium. It will now be abbreviated with the letter n. Noting that e = e_o – (x + y) = e_o – my, and substituting these in equation 11 gives equation 12.

$$dy/dt=(k_1{a}_o+\delta b)e_o/m\text{–}y\{(k_1{a}_o+\delta b)/m+k_3+k{-}_{1}k{-}_2/k_2\}$$ (12)

Substitution of db/dt = (k₃y – k-₃be) into equation 12 and rearranging leads to equation 13. The factor

$$m{d}^{2}y/d{t}^2=-k{-}_3\delta b{e}^2-n[(k_1{a}_o+\delta b)e_o/m\text{–}y\{(k_1{a}_o+\delta b)/m+k_3(1-\delta e/n)+k{-}_{1}k{-}_2/k_2)\}]$$ (13)

(1 - δe/n) cannot at any stage of the pre-steady state significantly alter the value of k₃ and alter the sign of the term in y, so that d²y/dt² is always negative when k- ₃ > k₁, and y can only rise monotonically to y_equ.

ii. The equilibrium concentration of enzyme-reactant intermediates.

The equation x_equ = a_oe_o / (a_o + k- ₁ /k₁ + k₂/k- ₂ ) is stated in reference. ⁶ It is obtained using the same principles as outlined here for y_equ in model 2. In the following, the small case letter (except a_o and e_o) are now understood to represent equilibrium concentrations. There are two conservation equation for model 2: e_o = e + x + y and (when a_o >>e_o) a_o = a + b. There are the following equilibrium conditions: b/a = K_equ = k₁k₂ k₃/k-₁k-₂ k-₃, x = k- ₂ y/k₂, and e = k₃y/k- ₃ b. Now b = aK_equ = (a_o –b)K_equ, and so b = a_oK_equ/(1 + K_equ), and e = k₃y (1 + K_equ) /k- ₃ a_oK_equ. Hence, y_equ is given by y_equ = e_o / {1 + k- ₂ /k₂ + k₃(1 + K_equ)/ k- ₃ a_oK_equ}, which on rearrangement gives equation 3.

$$y_{equ}=e_o{a}_o/\{a_o(1+k{-}_2/k_2)+k_3(1+K_{equ})/k{-}_3{K}_{equ}\}$$ (3)

In expanded form, equation 3 leads to equation 4.

$$y_{equ}=k_1{k}_2{e}_o{a}_o/\{k_1{a}_o(k{-}_2+k_2)+k{-}_{1}k{-}_2+k_1{k}_2{k}_3/k{-}_3\}$$ (4)