Elasticity coefficient

Given the definition of the elasticity coefficient in terms of a partial derivative, it is possible, for example, to determine the elasticity of an arbitrary rate law by differentiating the rate law by the independent variable and scaling. For example, the elasticity coefficient for a mass-action rate law such as:

${\displaystyle v=k\ s_{1}^{n_{1}}s_{2}^{n_{2}}}$

where ${\displaystyle v}$ is the reaction rate, ${\displaystyle k}$ the reaction rate constant, ${\displaystyle s_{i}}$ is the ith chemical species involved in the reaction and ${\displaystyle n_{i}}$ the ith reaction order, then the elasticity, ${\displaystyle \varepsilon _{s_{1}}^{v}}$ can be obtained by differentiating the rate law with respect to ${\displaystyle s_{1}}$ and scaling:

${\displaystyle \varepsilon _{s_{1}}^{v}={\frac {\partial v}{\partial s_{1}}}{\frac {s_{1}}{v}}=n_{1}\ k\ s_{1}^{n_{1}-1}s_{2}^{n_{2}}{\frac {s_{1}}{k\ s_{1}^{n_{1}}s_{2}^{n_{2}}}}=n_{1}}$

That is, the elasticity for a mass-action rate law is equal to the order of reaction of the species.

For example the elasticity of A in the reaction ${\displaystyle 2A\rightleftharpoons C}$ where the rate of reaction is given by: ${\displaystyle v=kA^{2}}$, the elasticity can be evaluated using:

${\displaystyle \varepsilon _{a}^{v}={\frac {\partial v}{\partial a}}{\frac {a}{v}}={\frac {2kaa}{ka^{2}}}=2}$

Elasticities can also be derived for more complex rate laws such as the Michaelis–Menten rate law. If

${\displaystyle v={\frac {V_{\max }s}{K_{m}+s}}}$

then it can be easily shown than

${\displaystyle \varepsilon _{s}^{v}={\frac {K_{m}}{K_{m}+s}}}$

This equation illustrates the idea that elasticities need not be constants (as with mass-action laws) but can be a function of the reactant concentration. In this case, the elasticity approaches unity at low reactant concentration (s) and zero at high reactant concentration.

A. The slope of the reaction rate versus the reactant concentration scaled by both the reactant concentration and reaction rate yields the elasticity. If the log of the reaction rate and the log of the reactant concentration is plotted, the elasticity can be read directly from the slope of the curve. Curves were generated by assuming v = S/(2 + S)

For the reversible Michaelis–Menten rate law:

${\displaystyle v={\frac {V_{\max }/K_{m_{1}}(s-p/K_{\text{eq}})}{1+s/K_{m_{1}}+p/K_{m_{2}}}}}$

where ${\displaystyle V_{\max }}$ is the forward ${\displaystyle V_{max}}$, ${\displaystyle K_{m_{1}}}$ the forward ${\displaystyle K_{m}}$, ${\displaystyle K_{eq}}$ the equilibrium constant and ${\displaystyle K_{m_{2}}}$ the reverse ${\displaystyle K_{m}}$, two elasticity coefficients can be calculated, one with respect to substrate, S, and another with respect to product, P. Thus:

${\displaystyle \varepsilon _{s}^{v}={\frac {1}{1-\Gamma /K_{\text{eq}}}}-{\frac {s/K_{m_{1}}}{1+s/K_{m_{1}}+p/K_{m_{2}}}}}$

${\displaystyle \varepsilon _{p}^{v}={\frac {-\Gamma /K_{\text{eq}}}{1-\Gamma /K_{\text{eq}}}}-{\frac {p/K_{m_{2}}}{1+s/K_{m_{1}}+p/K_{m_{2}}}}}$

where ${\displaystyle \Gamma }$ is the mass-action ratio, that is ${\displaystyle \Gamma =p/s}$. Note that when p = 0, the equations reduce to the case for the irreversible Michaelis–Menten law.

As a final example, consider the Hill equation:

${\displaystyle v={\frac {V_{\max }(s/K_{s})^{n}}{1+(s/K_{s})^{n}}}}$

where n is the Hill coefficient and ${\displaystyle K_{s}}$ is the half-saturation coefficient (cf. Michaelis–Menten rate law), then the elasticity coefficient is given by:

${\displaystyle \varepsilon _{s}^{v}={\frac {n}{1+(s/K_{s})^{n}}}}$

Note that at low concentrations of S the elasticity approaches n. At high concentrations of S the elasticity approaches zero. This means the elasticity is bounded between zero and the Hill coefficient.

The elasticities for a reversible uni-uni enzyme catalyzed reaction was previously given by:

${\displaystyle \varepsilon _{s}^{v}={\frac {1}{1-\Gamma /K_{\text{eq}}}}-{\frac {s/K_{m_{1}}}{1+s/K_{m_{1}}+p/K_{m_{2}}}}}$

${\displaystyle \varepsilon _{p}^{v}={\frac {-\Gamma /K_{\text{eq}}}{1-\Gamma /K_{\text{eq}}}}-{\frac {p/K_{m_{2}}}{1+s/K_{m_{1}}+p/K_{m_{2}}}}}$

An interesting result can be obtained by evaluating the sum ${\displaystyle \varepsilon _{s}^{v}+\varepsilon _{p}^{v}}$. This can be shown to equal:

${\displaystyle \varepsilon _{s}^{v}+\varepsilon _{p}^{v}={\frac {1}{1+s/K_{m_{1}}+p/K_{m_{2}}}}}$

Two extremes can be considered. At high saturation (${\displaystyle s>K_{m_{1}},p>K_{m_{2}}}$), the right-hand term tends to zero so that:

${\displaystyle \varepsilon _{s}^{v}\approx -\varepsilon _{p}^{v}}$

That is the absolute magnitudes of the substrate and product elasticities tends to equal each other. However, it is unlikely that a given enzyme will have both substrate and product concentrations much greater than their respective Kms. A more plausible scenario is when the enzyme is working under sub-saturating conditions (${\displaystyle s<K_{m_{1}},p<K_{m_{2}}}$). Under these conditions we obtain the simpler result:

${\displaystyle \varepsilon _{s}^{v}+\varepsilon _{p}^{v}\approx 1}$

Expressed in a different way we can state:

${\displaystyle ||\varepsilon _{s}^{v}||>||\varepsilon _{p}^{v}||}$

That is, the absolute value for the substrate elasticity will be greater than the absolute value for the product elasticity. This means that a substrate will have a great influence over the forward reaction rate than the corresponding product.[6]

This result has important implications for the distribution of flux control in a pathway with sub-saturated reaction steps. In general, a perturbation near the start of a pathway will have more influence over the steady state flux than steps downstream. This is because a perturbation that travels downstream is determined by all the substrate elasticities, whereas a perturbation downstream that has to travel upstream if determined by the product elasticities. Since we have seen that the substrate elasticities tends to be larger than the product elasticities, it means that perturbations traveling downstream will be less attenuated than perturbations traveling upstream. The net effect is that flux control tends to be more concentrated at upstream steps compared to downstream steps.[7][8]

The table below summarizes the extreme values for the elasticities given a reversible Michaelis-Menten rate law. Following Westerhoff et al.[9] the table is split into four cases that include one 'reversible' type, and three 'irreversible' types.

Approximate values of elasticities under various conditions

Equilibrium State	Saturation Levels	Elasticities
Near Equilibrium	All degrees of saturation	${\displaystyle \varepsilon _{s}^{v}\gg 1,\varepsilon _{p}^{v}\ll -1}$
Out of Equilibrium	High Substrate, high product	${\displaystyle \varepsilon _{s}^{v}\approx -\varepsilon _{p}^{v}}$
Out of Equilibrium	High Substrate, low product	${\displaystyle \varepsilon _{s}^{v}\approx 0,\varepsilon _{p}^{v}\approx 0}$
Out of Equilibrium	Low Substrate, high product	${\displaystyle \varepsilon _{s}^{v}\approx 1,\varepsilon _{p}^{v}\approx -1}$
Out of Equilibrium	Low Substrate, low product	${\displaystyle \varepsilon _{s}^{v}\approx 1,\varepsilon _{p}^{v}\approx 0}$

Elasticities coefficient can also be computed numerically, something that is often done in simulation software.[10]

For example, a small change (say 5%) can be made to the chosen reactant concentration, and the change in the reaction rate recorded. To illustrate this, assume that the reference reaction rate is ${\displaystyle v_{o}}$, and the reference reactant concentration, ${\displaystyle s_{o}}$. If we increase the reactant concentration by ${\displaystyle \Delta s_{o}}$ and record the new reaction rate as ${\displaystyle v_{1}}$, then the elasticity can be estimated by using Newton's difference quotient:

${\displaystyle \varepsilon _{s}^{v}\simeq {\frac {v_{1}-v_{o}}{\Delta s_{o}}}{\frac {s_{o}}{v_{o}}}={\frac {v_{1}-v_{o}}{v_{o}}}/{\frac {s_{1}-s_{o}}{s_{o}}}}$

A much better estimate for the elasticity can be obtained by doing two separate perturbations in ${\displaystyle s_{o}}$. One perturbation to increase ${\displaystyle s_{o}}$ and another to decrease ${\displaystyle s_{o}}$. In each case, the new reaction rate is recorded; this is called the two-point estimation method. For example, if ${\displaystyle v_{1}}$ is the reaction rate when we increase ${\displaystyle s_{o}}$, and ${\displaystyle v_{2}}$ is the reaction rate when we decrease ${\displaystyle s_{o}}$, then we can use the following two-point formula to estimate the elasticity:

${\displaystyle \varepsilon _{s}^{v}\simeq {\frac {1}{2}}{\frac {v_{1}-v_{2}}{s_{1}-s_{o}}}\left({\frac {s_{o}}{v_{o}}}\right)}$