How many progress-curves do we need to estimate \(K_m\) and \(V_{max}\)?
Suppose we have a Michaelian enzyme whose kinetic parameters we want
to determine. To this end, a conventional approach is to find the
initial rates at different concentrations of the substrate, to obtain,
in this way, a dataset relating substrate concentration to rate. In this
approach, a progress-curve for each initial substrate concentration
assayed need to be built in order to calculate the initial rate as the
slope of the tangent line to the curve at time 0.
For instance, let’s assay our enzyme at substrate concentrations 1,
2, 4, 8, 12, 20 and 30 mM.
At [S] = 1 mM:
At [S] = 2 mM:
At [S] = 4 mM:
At [S] = 8 mM:
At [S] = 12 mM:
At [S] = 20 mM:
At [S] = 30 mM:
Note that, in the current example, we have needed 7 progress-curves
to collect the following dataset:
| [S] (mM) |
1.0 |
2.0 |
4 |
8.0 |
12.0 |
20.0 |
30.0 |
| v (mM/min) |
8.1 |
14.4 |
23 |
32.1 |
36.7 |
41.3 |
43.9 |
Afterwards, we can proceed to estimate \(K_m\) and \(V_{max}\), for instance, by fitting our
data to the Michaelis-Menten model:
dir.MM(Sv, unit_v = "mM/min")
#> $parameters
#> Km Vm
#> 5.028 51.678
#>
#> $data
#> S v fitted_v
#> 1 1 8.1 8.572993
#> 2 2 14.4 14.706318
#> 3 4 23.0 22.896766
#> 4 8 32.1 31.733497
#> 5 12 36.7 36.418605
#> 6 20 41.3 41.296148
#> 7 30 43.9 44.260021
Although we have succeeded in our aim (estimating the kinetic
parameters) we have needed 7 progress-curves. Wouldn’t be nice to be
able get \(K_m\) and \(V_{max}\) from one single progress-curve?
Well, that is precisely what the integrated Michaelis-Menten equation
allow us.
Integrating the Michaelis-Menten equation
Suppose that, starting from an initial concentration of substrate 1
mM, we monitor their evolution over time, obtaining the following
data:
| 0.00 |
1.000 |
| 0.05 |
0.593 |
| 0.10 |
0.338 |
| 0.15 |
0.188 |
| 0.20 |
0.103 |
| 0.25 |
0.056 |
| 0.30 |
0.030 |
| 0.35 |
0.016 |
Assuming that the substrate concentration drops as it is converted
into a product by the direct unidirectional reaction catalyzed by our
enzyme (this is only true when the reverse reaction is negligible, for
example in hydrolysis reactions), and that the product has no effect on
the catalyzed reaction rate, then we can relate the variation in the
concentration of substrate over time integrating the velocity
equation.
\[\begin{equation}
\tag{1}
v = \frac{-d[S]}{dt} = V_{max} \frac{[S]}{[S] + K_m}
\end{equation}\]
\[\begin{equation}
\tag{2}
\int_{S_o}^{S_t} \frac{[S] + K_m}{[S]} d[S] = -\int_{0}^t V_{max} dt
\end{equation}\]
\[\begin{equation}
\tag{3}
\frac{1}{t} \ln \frac{S_o}{S_t} = - \frac{1}{K_m} \frac{(S_o - S_t)}{t}
+ \frac{V_{max}}{K_m}
\end{equation}\]
Equation (3) can be recognized as the equation of a line where the
dependent variable is \(\frac{1}{t} \ln
\frac{S_o}{S_t}\) and the independent variable is \(\frac{(S_o - S_t)}{t}\). Thus, if the data
given in the table above are conveniently transformed, they can be fit
to a line by least square.
The function int.MM() from the package renz takes
as argument a dataframe where the first column is time and the second
column is [S], and returns both the kinetic parameters and a dataframe
with the transformed variable:
Some final considerations
In addition to saving us work, since a single progress-curve is
enough to estimate the kinetic parameter, this method has an additional
advantage: since the method does not employ initial rates, it avoid the
bias introduced by underestimating initial rates (see doi:
10.1016/0307-4412(85)90008-1). Nevertheless, before using this approach,
we have to make sure that the rate of the inverse reaction is negligible
and that the product of the reaction does not affect the direct reaction
rate.