Modeling: the Michaelis-Menten Model of Enzyme
Modeling: the Michaelis-Menten Model of Enzyme
Kinetics
In this component we see how the basic properties of linear and
rational functions can be used to analyze data from enzyme
kinetics in terms of the Michaelis-Menten Model. There are two
sections:
We begin by seeing how to find approximate linear models for data
that are not exactly linear.
At the end of the component on linear functions we gave the
following table of data adapted from a study of enzyme kinetics.
Here the variable s is the substrate (which has the units of mM
NPG, millimolar di-nitro-phenol) and the function v is
absorbance (which has units of OD, optical density) per second.
| s = Substrate | v = Absorbance per second
|
|
|
| 0.5 | 0.0030 |
| 1.0 | 0.0066 |
| 1.5 | 0.0102 |
| 2.0 | 0.0138 |
We'll show that the data are represented by a linear function and
find a formula for that function. Note that successive substrate
readings differ by 0.5, so the values of the variable s are
equally spaced. To test whether there is a linear model
representing the data, we see whether the function values v are
equally spaced. The first difference is 0.0066-0.0030 = 0.0036,
and it is easy to see that the other differences agree with this.
It follows that the data are linear, and the slope is
|
Slope = |
Change in function
Change in variable
|
= |
0.0036
0.5
|
= 0.0072 = 7.2×10-3. |
|
Here the
units of the slope are mM NPG per OD per second. Now we use the
point-slope form to find a formula:
|
v-0.0030 = 0.0072(s-0.5). |
|
Rearranging gives the slope-intercept form:
The graph of the line, along with the data points, is below.
Note that the initial value of the function (the same as the
vertical intercept of the line) is -0.0006 = -6×10-4.
This says that the absorbance rate will be nonzero (in fact,
negative) when the substrate reading is 0! This result does not
make sense physically, because we expect the absorbance rate to
approach 0 as the substrate level gets close to 0. We
conclude that the linear model is not valid for very small
substrate levels. In fact, as we will see when we discuss the
Michaelis-Menten Model, the data should be represented by a curve,
not a straight line, over a wide range of values.
The results and discussion above raise a question: Why do we
bother to use linear models if they are not always accurate? One
answer is that such models are easy to use and do represent many
data sets approximately over relatively short intervals. This
answer raises another question: If the data are only
approximately linear, how do we find a linear model? This is a
common problem. We should be suspicious about data tables like
the one above from enzyme kinetics-surely it's very unlikely
that the data will be exactly linear. In fact, the data on enzyme
kinetics were adjusted a bit to make them exactly linear. Here is
a more realistic table:
| s = Substrate | v = Absorbance per second
|
|
|
| 0.5 | 0.00319 |
| 1.0 | 0.00641 |
| 1.5 | 0.00972 |
| 2.0 | 0.01422 |
Checking differences in values of v shows that these data cannot
be represented exactly by a linear function. However, a plot of
the data points (see the graph below) shows that they nearly fall
on a straight line. The mathematical procedure of linear
regression gives the equation for a line that approximates the
data. We now turn to discuss this procedure.
There are many lines that might be used to approximate a data plot
that is nearly, but not exactly, linear. The line produced by
linear regression is the best approximation in a certain sense. A
formula for the regression line can be given in terms of the data,
but it is rather complicated. Usually scientists perform linear
regression using a calculator or a spreadsheet program on a
computer. For the data set above, this procedure gives the line
It is worthwhile to compare this line
with the line we found to fit the first data set (which was
slightly altered from this second set to be exactly linear):
A graph showing the regression line along
with the nearly linear data is shown below. The fit is fairly
good.
We have seen that the regression line provides an approximate
linear model for data that are nearly linear. Such models should
be used with care because they may give unreasonable results if
they are applied over large intervals. Now we turn to the question
of how to find a nonlinear model over a large interval in the
study of enzyme kinetics.
Recall that the Michaelis-Menten Model for enzyme kinetics
describes the relation between the initial reaction rate v and
the initial concentration of the substrate s. It has the form
where V (sometimes called Vmax)
and Km are positive constants whose meaning we discuss below.
This is a rational function of the type discussed in the component
on rational functions. As a function, its domain consists of all
real numbers s except for s = -Km; in practice, though, we
restrict our attention to nonnegative values of s. Its graph
passes through the origin (0,0) (the location of the horizontal
intercept and the vertical intercept). The graph has a vertical
intercept at s = -Km, which has no physical interpretation. It
has a horizontal intercept at v = V. An example of this type of
graph was given in the component on rational functions, where the
function f(x) = 5x/(x+2) was studied and
graphed. Its shape is significant: It is increasing and concave
down, which means that, as s increases, so does v, but at a
decreasing rate.
The interpretation of the constants V and Km is important.
Because V is the horizontal asymptote, the graph levels off at
that value, and we know that for large values of substrate
concentration s the initial rate v will be close to V. How
about Km? Notice what happens in the formula when we put in
s = Km:
|
v(Km) = |
VKm
Km+Km
|
= |
V
2
|
. |
|
This means
that at s = Km the initial rate v is half of the limiting value
V. For example, consider the function
v = 1.8s/(s+5). Here V = 1.8 and Km = 5. Thus
the initial rate v will be close to 1.8 when s is large, and
the initial rate v will be 1.8/2 = 0.9 when
s = 5.
Using this interpretation we can give one simple method of
estimating the constants V and Km if a fairly complete set of
data is given. Namely, we estimate the horizontal asymptote by
seeing where the function values level off, and this gives an
estimate of V. Once we have an estimate of V we estimate the
value of s that gives v = V/2, and this
gives an estimate of Km. This method is simple to apply, but
it does require a pretty complete data set.
There is another method for estimating the constants V and Km
in fitting a Michaelis-Menten Model to data. To explain this, we
start with the model
and note that
|
|
1
v
|
= |
s+Km
Vs
|
= |
1
V
|
+ |
Km
V
|
|
1
s
|
. |
|
This says that 1/v is a linear function of
1/s with slope [(Km)/ V]
and vertical intercept 1/V. This fact
suggests a way of estimating the constants V and Km if a set
of data points in the form (s,v) is given. First we invert the
variable and function values given to get data in the form
(1/s,1/v). (We have to discard any points where s = 0 or v = 0.)
Then we apply linear regression to the transformed data. The
vertical intercept of the regression line gives an estimate for
1/V, so we invert that intercept to get an
estimate for V. The slope of the regression line gives an
estimate for Km/ V, so multiplying the
slope by the estimate for V we just obtained gives an estimate
for Km. It should be noted that this method has its limitations
since inverting the data can lead to distortion with small
readings for s and v.
File translated from TEX by TTH, version 2.25.
On 08 Aug 2002, 15:15.