Is it possible to derive the Michaelis-Menten equation under conditions where the product formation is reversible

Question

Text books etc generally derive the Michaelis-Menten equation for the irreversible case i.e $$\ce{E + S <=> ES -> E +P}$$ I can't see how to do it for the reversible case i.e $$\ce{E + S <=> ES <=> E +P}\;$$ Is it possible?

The Michaelis-Menten equation is based on the explicit assumption that substrate is at a much higher concentration than the products. This is in fact what you have written and is true at the start of a reaction. At later stages it is generally not true and therefore the Michaelis-Menten equation no longer holds. If you ever do a biochemistry kinetics class you will always plot the initial rate of an enzymic reaction. — David, Feb 29, 2016 at 23:42
Now I understood what you mean to say. I was thinking you were referring to the first reaction of enzyme-substrate binding. Okay, there is no particular reversible conversion model that I have come across with but you can build a model based on a few assumptions. "Substrate concentration is much greater than the enzyme concentration" is true for the QSSA approximation. The original model used an equilibrium approximation which assumes that binding/unbinding reaction is much faster than the conversion. — WYSIWYG, Mar 1, 2016 at 5:48

David · Accepted Answer · 2016-03-01 11:22:46Z

No it is not possible to derive the equation for the reversible case. This is because the Michaelis-Menten equation is derived on the explicit assumption that the substrate concentration is much greater than the product concentration so that the back reaction does not occur. This is true at the start of a reaction in the lab when you add enzyme to your substrate. At later times when product builds up these conditions no longer hold and the M-M equation breaks down.

It is possible to derive a different equation for the reversible case (as some other posters do) but this begs the question of the importance of the Michaelis-Menten equation for biology. The poster seems to think that it is some universal truth, important in its own right or because it models the reaction in the living cell. I think this misses the point, which I think is as follows.

It originally was a prediction from, and hence a test of, the hypothesis that enzymes have a single active site at which the substrate binds and is converted to products.
It allowed two fundamental aspects of this enzyme model to be quantified: (i) the binding affinity of the substrate for the active site, as the inverse Km, and (ii) the catalytic efficiency ('smartness') as the turnover number or kcat, which is derived from Vmax (and enzyme concentration).
These enzyme constants can be used to address biological questions, for example whether an enzyme capable of a particular reaction in the test tube is able to account for the observed reaction rates in a tissue. Or whether the substrate you find the enzyme can use is likely to be the physiological substrate in a tissue.
It turns up enzymes that do not exhibit M-M kinetics, implying that some aspect of the model does not apply to them. This can be of mainly chemical interest (ping-pong kinetics and the like) or of physiological importance, as in allosteric kinetics (S-shaped rather than hyperbolic curve) shown by multi-subunit enzymes where the individual subunits show co-operativity, allowing more finely tuned regulation.
And if you do want to model reactions within cells, the kinetic constants can be of use. However as one is normally dealing with a chain of reactions in an open system the mathematics are complex (and nothing to do with the equation for a single reversible reaction in the test tube).

Most biochemistry students (and many teachers) would be hard pressed to explain why they should learn the derivation of the Michaelis-Menten equation. The answer is partly so they see that it is based on a particular model of the enzyme, but also so that they are aware of the assumptions on which it is based. Hence they need to be aware that the Michaelis-Menten equation is only true for initial rates of reaction, otherwise their determination of the practically and biologically useful Km and Vmax will be incorrect.

Well, I disagree. I think the reversible case is almost always considered in enzyme kinetics, especially for more complex mechanisms, and then simplifying assumptions applied to the full rate law. The equation for the reversible Michaelis-Menten equation is derived, for example, by Cornish-Bowden 'Fundamentals of Enzyme Kinetics', p 31 (of the first edition). See Cook and Cleland for more details. — user338907, Mar 1, 2016 at 2:00
The Reversible Michaelis-Menten Equation (after Cornish-Bowden, Fundamentals of Enzyme Kinetics)$$ v = {{{{V_{s}^f}[s]\over{K_{m}^s}}-{{V_{p}^f}[p]\over{K_{m}^p}}}\over{1 + {{[s]}\over{K_{m}^s}} + {{[p]}\over{K_{m}^p}}}}$$ — user338907, Mar 1, 2016 at 2:53
$ {V_{s}^f}$ is the maximum velocity for s (substrate) in the 'forward' direction, $ {V_{p}^r}$ is the maximum velocity for p (product) in the 'reverse' direction, $ {{K_{m}^s}}$ is the Michaelis Constant for s, and $ {{K_{m}^p}}$ is the Michaelis Constant for P. Set [p] to zero to get 'classical' Michaelis-Menten equation. — user338907, Mar 1, 2016 at 2:54
It is always possible to derive some mathematical model given the right assumptions. You perhaps mean to say that the reversible case would not be compatible with the MM model. Well, of course because MM model assumes irreversible conversion. The assumptions are different altogether. — WYSIWYG, Mar 1, 2016 at 5:51
@David Late to this thread, but it depends on whether you're referring to steady state (Briggs-Haldane) or MM equilibrium assumptions. It's perfectly ok to derive the reversible form using the steady state assumption. It's done all the time. For whole pathway models the reversible form is a decent approximation and far more realistic that the irreversible version. — rhody, Oct 4, 2018 at 22:39

user338907 · Accepted Answer · 2016-03-15 10:00:53Z

The Reversible Michaelis-Menten Equation

An equation for the reversible form of the Michaelis-Menten equation was first derived by Haldane (1932, p 81), and the form shown in Eqn (1) was first given by Peller and Alberty (1953).

$$ v = { {{{V_{max}^f}\over{K_{m}^s}}\ s\ -{{V_{max}^r}\over{K_{m}^p}}\ p }\over{1 + {{s}\over{K_{m}^s}} + {{p}\over{K_{m}^p}}}}\ \ \ \ \ (1)$$

$v$ is velocity
$s$ is substrate concentration
$p$ is product concentration
$K_{m}^s$ is the Michaelis Constant for $s$
$K_{m}^p$ is the Michaelis Constant for $p$
$V_{max}^f $ is the maximum velocity in the 'forward' direction
$V_{max}^r $ is the maximum velocity in the 'reverse' direction

Setting $p$ to zero gives the more familiar form of the Michaelis-Menten equation (where $v_i$ is now the initial velocity)

$$ v_i = { {{V_{max}^f} s }\over{K_{m}^s + s}}\ \ \ \ \ (2)$$

Derivation of the Rate Law

How can Eqn (1) be derived?

This has been done by (among others) Cornish-Bowden (2004, pp 48 -52), and the derivation given below closely follows this source (with some help from Wolfram Mathematica).

Let's consider the following form of the reversible single-substrate, single-product enzyme mechanism

A substrate $s$ may reversibly bind to $E$ (the 'free' enzyme) to give $ES$. This in turn may be reversibly converted to $EP$, and reversible dissociation of product gives back the original form of the 'free' enzyme together with $p$.

The complexes $ES$ and $EP$ are sometimes called 'central complexes'

However, in order to simplify the algebra, I will derive the rate law for the following simplified (and unrealistic) mechanism containing a single central complex.

We need to be very, very careful here. This simplifying assumption will not change the overall form of the kinetic constant form of the rate law, that is both mechanism (3) and (4) will give rise to Eqn (1), but it will make the definition of kinetic constants simpler.

We must not assume that these simplified definitions necessarily apply to any realistic case. The definitions for $V_{max}^f$ (Eqn 10) and $V_{max}^r$ (Eqn 11), for example, contain a single rate constant but those for the mechanism of Eqn (3) are more complex (Eqns 15 & 16).

Thus statements such as 'for unit enzyme concentration, $k_{3,2}$ equals the maximum velocity in the forward direction' are fraught with difficulties. (It probably is valid to argue, at least for the type of mechanisms considered here, that 'for unit enzyme concentration, the maximum velocity cannot exceed $k_{3,2}$').

But we digress. Let's return to the derivation.

Let $e_o$ be the total enzyme concentration
Let $x$ equal the concentration of $ES$
The concentration of $E$ is therefore ($e_o$ - $x$)

In addition, let's make two fundamental assumptions (see here for a great elucidation of the steady-state method in enzyme kinetics).

Very soon after the reaction is initiated, a steady-state is established so that the rate of formation of $x$ equals the rate of breakdown of $x$. This is the steady-state assumption. (What happens in the pre-steady-state period does not concern us here).
The concentration of $e_o$ is so low that formation of $ES$ and $EP$ may be ignored in calculating substrate and product concentrations (this also greatly simplifies the algebra)

The following differential equation, which describes the rate of breakdown of $x$, may now be written:

$$ {dx\over dt} = {k_{1,2}\ (e_o -x)\ s + k_{3,2}\ (e_o -x)\ p - (k_{2,1} + k_{2,3})\ x = 0}\ \ \ \ \ \ \ (5)$$

Let's solve for $x$. I am going to be lazy here and allow Wolfram Mathematica to do the work.

(A more detailed version of the Mathematica notebook may be found here).

The above command produces the following output (how cool is that?)

Now let's define the velocity equation in terms of $x$, taking the reverse reaction into account

$$ {dp\over dt} = {k_{1,2}\ x\ -\ k_{3,2}\ (e_o -x)\ p}\ \ \ \ \ \ \ \ \ (6)$$

We now need to substitute for $x$ in the above equation. Again, using Mathematica

Using substitution

the following output is obtained:

That is, the rate-constant form of the rate law may be expressed as follows:

$$ {dp\over dt} = { {(k_{1,2}\ k_{2,3}\ s - {k_{2,1}\ k_{3,2}\ p)\ e_o}}\over{{k_{1,2}\ s} + k_{2,1}+ k_{2,3}+{k_{3,2}\ p}}} \ \ \ \ \ \ \ (7)$$

The 'trick' now is to define kinetic constants in such a way that the more useful kinetic-constant form of the rate law may be obtained.

Let's define the kinetic constants as follows

$$K_{m}^s = {{k_{2,1}\ +\ k_{2,3}}\over{k_{1,2}}}\ \ \ \ (8)$$

$$K_{m}^p = {{k_{2,1}\ +\ k_{2,3}}\over{k_{3,2}}}\ \ \ \ (9)$$

$$ k_{cat}^f = {{V_{max}^f}\over{e_o}} = k_{2,3}\ \ \ \ (10)$$

$$ k_{cat}^r = {{V_{max}^r}\over{e_o}} = k_{2,1}\ \ \ \ (11) $$

Dividing Eqn.(7) 'above-and-below' by ($ {k_{1,2}\ + k_{2,3}}$) gives the following expression:

Substituting the values of the kinetic constants into the above expression leads logically to the following kinetic-constant form of the rate law:

$$ v = { {({{k_{cat}^f}\over{K_{m}^s}}\ s\ -{{k_{cat}^r}\over{K_{m}^p}}\ p)\ e_o }\over{1 + {{s}\over{K_{m}^s}} + {{p}\over{K_{m}^p}}}}\ \ \ \ \ (12)$$

Kinetic Constants for the More Realistic Mechanism

The definition of kinetic constants for the more realistic mechanism of Eqn (3) are as follows: (see here for the Mathematica derivation).

In general, different mechanisms may give rise to an identical rate law.

The following mechanism will also give rise to Eqn (1).

In fact, adding any number of central complexes to the mechanism of Eqn (3) will give rise to a rate law in the form of Eqn (1).

An Extension to the Michaelis-Menten Mechanism

Here is a mechanism that won't give rise to a rate law as shown in Eqn (1):

In the above mechanism, the form of the 'free' enzyme generated upon product release ($F$) is different from that which combines with substrate ($E$), and emphasizes that we may need to take their inter-conversion into account in any complete description of a single-substrate, single-product enzyme (see Taraszka & Alberty, 1964).

The rate law for this mechanism contains an extra kinetic constant, here called $K_{sp}$ $$ v = { {({{k_{cat}^f}\over{K_{m}^s}}\ s\ -{{k_{cat}^r}\over{K_{m}^p}}\ p)\ e_o }\over{1 + {{s}\over{K_{m}^s}} + {{p}\over{K_{m}^p}} + {{s p}\over{{K_{sp}}}}}}\ \ \ \ \ (19)$$

This mechanism could, for example, describe membrane transport where $F$ represents the carrier on the inside of the membrane and $E$ represents the carrier on the outside. But it also could apply to any single-substrate, single-product enzyme. This again emphasizes the danger of over-simlification.

The rate law of Eqn (19) is well reviewed by Darvey (1972), and the pdf is freely available to all.

Final Thoughts ...

As the number of enzyme species increases, obtaining the rate-constant form of the rate law becomes difficult, as the equations very quickly accumulate many terms. Before the days of Mathematica and the like, enzymologists used the schematic method of King and Altman (1956), or one of its many variants, to obtain the rate-constant form of the rate law.

So what are the advantages of deriving the kinetic constant form of the full rate law?

If the reaction is demonstrably reversible, it may studied in both directions
For a multi-substrate enzyme, the product inhibition equations are easily written down.
It allows a relationship (or relationships) between the kinetic constants and the equilibrium constant to be obtained. For the reversible Michaelis-Menten mechanism of Eqns (3) & (4), the Haldane relationship is as follows:

$$ K = {{k_{cat}^f\ k_{m}^p }\over{k_{cat}^r\ k_{m}^s }}\ \ \ \ (20)$$

where $K$ is the equilibrium constant for the reaction (Haldane, 1930)

References

Cook, P.F. & Cleland, W.W (2007) Enzyme kinetics and mechanism Garland Science.
Cornish-Bowden (2004) Fundamentals of Enzyme Kinetics 3rd Edn. Portland Press (London)
Darvey, I. G. (1972) The application of product inhibition studies to one-substrate one-product enzymic reactions. Biochem J. 128, 383 - 387 [pdf]
Haldane, J.B.S (1930) Enzymes Longmans (London)
King, E. L. & Altman, C. (1956). A schematic method of deriving the rate laws for enzyme-catalyzed reactions. J. Phys. Chem. 60, 1375 - 1378
Peller, L. & Alberty, R. A. (1959). Multiple intermediates in steady-state enzyme kinetics. I. The mechanism involving a single substrate and product. J. Am. Chem. Soc. 81, 5907 - 5915
Segel, I. H. (1975) Enzyme Kinetics. Behavior and Analysis of Rapid Equilibrium and Steady-State Systems Wiley
Taraszka, M. & Alberty, R. A. (1964). Extensions of the steady-state rate law for the fumarase reaction. J. Phys. Chem. 68, 3368 - 3373.

Hey.. While you are at it, can you also try obtaining the rate as purely a function of substrate (by using the conservation relation). I haven't used mathematica. — WYSIWYG, Mar 1, 2016 at 17:28

WYSIWYG · Accepted Answer · 2016-03-15 05:10:14Z

The derivation for the equation mentioned by TomD in the comments:

$$ v = {{{{V_{s}^f}[s]\over{K_{m}^s}}-{{V_{p}^f}[p]\over{K_{m}^p}}}\over{1 + {{[s]}\over{K_{m}^s}} + {{[p]}\over{K_{m}^p}}}}$$

can be found here. This formula assumes a three step reaction. This equation uses product as a variable in the equation. Ideally, since there is no formation/degradation of the reactants/products (closed system), the product can be represented as a function of the other variables ($\ce{[S0] -[ES] -[S]}$). In short, the product formation rate can be simply represented as a function of the substrate concentration.

What I am going to show is not a full derivation but one approach that you can start with.

For this two step reversible reaction:

$$\ce{E + S <=>[k_1][k_2] I <=>[k_3][k_4] E + P}$$

You would have the system of ODEs:

$ \dfrac{d\ce{[E]}}{dt}=-k_1\ce{[E][S]} + (k_2+k_3)\ce{[I]} -k_4\ce{[E][P]}\\ \dfrac{d\ce{[I]}}{dt}=\quad k_1\ce{[E][S]} - (k_2+k_3)\ce{[I]} +k_4\ce{[E][P]}\\ \dfrac{d\ce{[S]}}{dt}= -k_1\ce{[E][S]} + k_2\ce{[I]} \\ \dfrac{d\ce{[P]}}{dt}= -k_4\ce{[E][P]} + k_3\ce{[I]} $

Assuming QSSA for $\ce{[I]}$ (and hence also $\ce{[E]}$) and setting $\ce{[E]} = \ce{[E0] -[I]}$, you'll get:

$$\ce{[I]}=\frac{\ce{[E0]}(k_1\ce{[S]}+k_4\ce{[P]})}{k_2+k_3+k_1\ce{[S]}+k_4\ce{[P]}}$$

Since $\ce{[P]}=\ce{[S0] -[I] -[S]}$, the equation for $\ce{[I]}$ shown above, will become quadratic when you substitute this relation. This expression can be finally plugged into the fourth ODE to obtain the expression for the product formation rate, purely in terms of the substrate.

If you keep product as a variable then you can substitute the above expression for $\ce{[I]}$ in the fourth ODE.

$$V=-k_4\ce{[P]}\left(\ce{[E0]} - \frac{\ce{[E0]}(k_1\ce{[S]}+k_4\ce{[P]})}{k_2+k_3+k_1\ce{[S]}+k_4\ce{[P]}}\right) +k_3\frac{\ce{[E0]}(k_1\ce{[S]}+k_4\ce{[P]})}{k_2+k_3+k_1\ce{[S]}+k_4\ce{[P]}}$$

Some algebraic rearrangements will yield:

$$V=\ce{[E0]}\frac{k_3k_1\ce{[S]}+k_4k_2\ce{[P]}}{k_2+k_3+k_1\ce{[S]}+k_4\ce{[P]}}$$

Dividing both numerator and denominator by $k_2+k_3$ yields:

$$V=\ce{[E0]}\frac{\frac{k_3k_1}{k_2+k_3}\ce{[S]}+\frac{k_4k_2}{{k_2+k_3}}\ce{[P]}}{1+\frac{k_1}{k_2+k_3}\ce{[S]}+\frac{k_4}{k_2+k_3}\ce{[P]}}$$

This is the form of equation mentioned by TomD (just create new constants from the old ones).

For some more details have a look at this paper by Keleti^[1] and the other papers citing it.

^[1] Keleti, T. "Two rules of enzyme kinetics for reversible Michaelis‐Menten mechanisms." FEBS letters 208.1 (1986): 109-112.

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Is it possible to derive the Michaelis-Menten equation under conditions where the product formation is reversible