enzyme kinetics and the michaelis-menten equation
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Enzyme Kinetics and theMichaelis-Menten EquationAndrew Biaglow , Keith Erickson & ShawneeMcMurranPublished online: 16 Feb 2010.
To cite this article: Andrew Biaglow , Keith Erickson & Shawnee McMurran(2010) Enzyme Kinetics and the Michaelis-Menten Equation, PRIMUS: Problems,Resources, and Issues in Mathematics Undergraduate Studies, 20:2, 148-168, DOI:10.1080/10511970903486491
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Enzyme Kinetics and theMichaelis-Menten Equation
Andrew Biaglow, Keith Erickson, and Shawnee McMurran
Abstract: The concepts presented in this article represent the cornerstone of classi-
cal mathematical biology. The central problem of the article relates to enzyme
kinetics, which is a biochemical system. However, the theoretical underpinnings
that lead to the formation of systems of time-dependent ordinary differential equations
have been applied widely to any biological system that involves modeling of popula-
tions. In this project, students first learn about the general balance equation, which is a
statement of conservation within a system. They then learn how to simplify the
balance equation for several specific cases involving chemically reacting systems.
Derivations are reinforced with a concrete experiment in which enzyme kinetics are
illustrated with pennies. While a working knowledge of differential equations and
numerical techniques is helpful as a prerequisite for this set of activities, all of the
requisite mathematical skills are introduced in the project, so the methods would also
serve as an introduction to these techniques. It is also helpful if students have some
basic understanding of chemical concepts such as concentration and reaction rate, as
typically covered in high school or college freshman chemistry courses.
[Supplementary materials are available for this article. Go to the publisher’s online
edition of PRIMUS for the following free supplemental resource(s): Appendices and
Sample Solution]
Keywords: Difference equations, elementary differential equations, Euler’s method,
linearization, parameter estimation, enzyme catalysis, Michaelis-Menten kinetics,
mass balance, law of mass action, conservation of mass.
The views expressed herein are those of the author(s) and do not reflect the position
of the United States Military Academy, the Department of the Army, or the
Department of Defense.
Address correspondence to Andrew Biaglow, Department of Chemistry and Life
Sciences, United States Military Academy, West Point, NY, 10996. E-mail:
PRIMUS, 20(2): 148–168, 2010
Copyright # Taylor & Francis Group, LLC
ISSN: 1051-1970 print / 1935-4053 online
DOI: 10.1080/10511970903486491
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1. SUMMARY OF PROBLEM
This lesson introduces Michaelis-Menten kinetics to model enzyme-cata-
lyzed reactions. After a brief introduction to enzyme catalysis and chemical
kinetics, students will derive the Michaelis-Menten equation and perform a
series of experiments to estimate the relevant parameters. They will also
linearize the equation to facilitate parameter estimation. This effort will
introduce students to mathematical tools used in settings in which enzymes
play a key role such as pharmaceutical production.
1.1. Requisite Mathematical and Scientific Background
Students should possess a familiarity with elementary first order differential
equations such as those presented in a first semester of calculus. Students
should also be familiar with numerical integration techniques, Euler’s
method in particular. This lesson presumes no background in science other
than a secondary-level course in biology or chemistry. The background
material includes all necessary definitions and explanations. However, stu-
dents having undergone at least one semester of college chemistry may
demonstrate more facility with the material.
2. SCENARIO
2.1. An Introduction to Enzyme Kinetics and the Michaelis-Menten
Equation
Humans and other organisms experience thousands of biochemical reactions
every minute on which they depend to maintain homeostasis and sustain life.
The rate at which such chemical reactions occur can determine if an organ-
ism survives. A catalyst is a substance that has the ability to accelerate a
chemical reaction without being changed or consumed in the process.
Without the help of catalysts, many of the reactions would occur at rates
too slow to be useful to an organism’s cells. Organisms depend on catalysts
to speed up necessary reactions.
For example, think of some sugar sitting out in a sugar bowl. The sugar
can sit for days, months, or even years without showing any significant signs of
oxidation. However, the oxidation of glucose is one of the processes that
provides energy to a cell. Without a catalyst to speed up the reaction time,
the glucose in an organism would oxidize too slowly to provide any benefit.
Humans use catalysts every day when preparing food, brewing beer, making
wine, and even when removing stains from laundry. Industries use catalysts in
the production of chemicals and pharmaceuticals. Penicillin is used to treat
Enzyme Kinetics 149
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bacterial infections because it is a substance that inhibits certain chemical
reactions that are necessary for the survival and growth of bacteria.
Carbonated beverages can destroy digestive enzymes thereby interfering with
digestion, respiration, and nutrient absorption. Hence, the study of catalytic
reactions, and substances that can inhibit such reactions, has many applications.
Enzymes are biological catalysts made mostly from protein molecules.
Enzymatic catalysis can increase a reaction rate by a factor of 106 to 1018 [3].
The study of reaction rates in chemical reactions that are catalyzed by
enzymes is called enzyme kinetics and can help us understand the catalytic
process of an enzyme. In most cases, an enzyme converts one chemical,
called the substrate, into another, referred to as the product. In general,
during a reaction, the product concentration goes through three distinct
phases as a function of time. The first phase is a very brief transition period
during which the rate of product accumulation increases. During the second
phase, the increase of product concentration is nearly constant. In the final
phase, as the substrate is depleted, the rate of product accumulation decreases
and the product concentration reaches a limit. In this investigation we will
concentrate on the first two phases.
One important focus in enzyme kinetics is the analysis of the effect of
substrate concentration on the initial rate of an enzyme-catalyzed reaction.
Empirical data indicates that the relationship between the initial reaction rate
and the substrate concentration can be described by a rectangular hyperbolic-
like curve similar to that shown in Figure 1.
In the early 1900s, Leonor Michaelis, a German biochemist and physi-
cian, and Maud Menten, a Canadian medical scientist, investigated this
phenomenon and developed a quantitative model for it that now bears their
names [11].
Figure 1. Initial reaction rate versus substrate concentration.
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A quantitative description of enzyme kinetics follows from the develop-
ment of chemical kinetics. The foundations of the analysis are the mass
balance equation and the law of mass action. The mass balance equation
describes the law of conservation of mass and states that mass that enters a
system must either leave the system or accumulate within the system. That is,
Accumulation ¼ Input� Output:
The equation can be applied to any conserved quantity such as total
mass, total energy, or population of a particular species. In the study of
chemical kinetics, we often consider the balance on a particular type of
molecule that is participating in a chemical reaction. In a chemical reaction,
any accumulation of mass of a reactant or product must be balanced by input,
output, generation or consumption of that reactant or product. Thus the
modified mass balance equation must be written as
Accumulation ¼ Inputþ Generation� Output� Consumption:
It follows that the rate of accumulation can be described by the follow-
ing equation
Rate of
accumulation¼ Rate of
inputþ Rate of
generation� Rate of
output� Rate of
consumption(1)
The law of mass action states that the rate of a chemical reaction is
proportional to the product of the concentrations of reactants. A single
molecule is a very small mass. For example, a single molecule of glucose,
C6H12O6, has a mass of 180.16 atomic mass units, or 2.9916 � 10-22 grams.
Since analytical methods typically cannot detect mass changes on this scale,
chemists use concentrations to characterize reactions. Concentrations are
typically written in mass or moles per volume. A ‘‘1 molar’’ concentration
of glucose in water contains 1 mole of glucose molecules per liter of solution,
where a mole is 6.022 � 1023 molecules. Concentrations are denoted with
square brackets. So, for our glucose solution, we say that [C6H12O6] ¼ 1 M ¼1 mol/L. The appearance of a volume term in the concentration units is not a
problem in the mass balance, as long as the volume of the system remains
constant during the reaction. This will be the case for most biological
systems, which can be thought of as dilute solutions in water.
As a simple example, consider a chemical reaction in a ‘‘closed’’ system in
which a reactant, A, is converted to a product, B. By a closed system we mean
that neither reactants nor products can escape the system, nor will new reac-
tants or products enter the system from an outside source. Let us also suppose
that the reaction is irreversible, meaning that the products do not react to
reform reactants. Such a reaction is often denoted as follows:
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A �!k B:
The parameter k is the proportionality constant related to the ‘‘speed’’
with which the reaction occurs. Specifically, the law of mass action tells us
that the rate of consumption of A is proportional to the concentration of A, or
Rate of consumption of A ¼ k � ½A�:
Likewise, the rate of generation of B is also proportional to the concen-
tration of A:
Rate of generation of B ¼ k � ½A�:
Let us consider the mass balance. Since the system is closed, the rate of
change of input and output in the mass balance equation will be 0. Since the
reaction is irreversible, the rate of consumption of B is 0. Using Equation (1),
it then follows from a mass balance on B that
Rate of accumulation of B ¼ Rate of generation of B:
It is convenient to define the rate of accumulation of B as the time
derivative of the concentration of B. This means that we are writing a mass
balance on a differential time scale, as opposed to days or minutes. Using the
law of mass action, the mass balance can then be written as
d½B�dt¼ k½A�: (2)
The rate of generation of B is equal to the rate of consumption of A. Note
that for every B molecule that is produced, exactly one molecule of A is
consumed. This is shown in the chemical equation by the coefficients of A
and B, which are both equal to one. This means that
d½B�dt¼ � d½A�
dt: (3)
The mass balance equation then reduces to
d½A�dt¼ �k½A�: (4)
The mass balance on A can also be obtained directly by setting the rate of
accumulation of A equal to the negative of the rate of consumption of A and
applying the law of mass action to the consumption term, as discussed above.
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The corresponding reaction is referred to as a first-order reaction because
Equation (4) is linear in [A]; that is, the exponent on [A] is one. In practice,
reaction order is determined empirically.
The rate of accumulation can have a negative or positive value. If the
rate of accumulation is negative, then this simply means that the mass
decreases during the time interval over which we are measuring the rate.
The resulting mass balance on A, given by Equation (4), has a solution of
the form
½A� ¼ ½A�0e�kt; (5)
where [A] is a function of time t, and [A]0 is the concentration of A at time
zero. The concentration of B is determined by taking the initial amount of B,
described by [B]0, and adding the amount of B formed by the reaction,
½B� ¼ ½B�0 þ1 mol B produced
1 mol A reactedð½A�0 � ½A�Þ:
The difference [A]0 – [A] describes the amount of A that reacted, and [A]
is given by Equation (5).
The usefulness of the method is apparent because we can now predict the
concentrations of A and B at all future times, as long as we know the initial
concentrations of A and B, and the rate constant k. Often the concentration is
measured in the laboratory as a function of time, and by using curve-fitting
procedures, we can extract a value for k that can be used for making such
predictions.
A numerical solution to Equation (4) is also possible. Euler’s method
provides one elementary technique for obtaining a numerical solution. For
the application of Euler’s method to this equation, we start by assuming that
the differentials in Equation (4) can be replaced with very small discrete
changes, that is, d[A] � �[A] and dt � �t. Thus, we assume
d½A�dt� �½A�
�t::
Equation (4) then becomes
�½A��t¼ �k � ½A�:
Assuming we have a table of concentrations, then for the j and j þ 1
entries, �[A] ¼ [A]jþ1 – [A]j. Making this substitution into the above
equation and solving for [A]j þ 1 gives the difference equation for [A]:
�½A��t¼½A�jþ1 � ½A�j
�t¼ �k � ½A�j ) ½A�jþ1 ¼ �k � ½A�j ��t þ ½A�j: (6)
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From Equation (3), the difference equation for [B] is derived in the same
fashion giving
½B�jþ1 ¼ k � ½A�j ��t þ ½B�j: (7)
Equations (6) and (7) can be plotted in a spreadsheet provided we know k,
[A]0 and [B]0. The results shown in Figures 2 and 3 were created with Excel using
parameter values of k ¼ 1 sec-1, [A]0 ¼ 1.00 mol/L, and [B]0 ¼ 0.00 mol/L.
Figure 2. Euler’s method solution for [A] and [B] compared to exact solutions.
Figure 3. Plot of Euler’s method solution and exact solutions for [A] and [B].
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Notice in Figures 2 and 3 that there is significant difference between the
Euler’s method solution and the exact solution. This difference occurs natu-
rally in numerical methods and is a function of the time increment �t. In our
calculation we used a time increment of �t ¼ 0.2 sec. As �t gets closer to 0,
the amount of error will also decrease. Since a spreadsheet was used for this
calculation, much smaller values of �t are possible. For example, with �t ¼0.01, the numerical solution is nearly indistinguishable from the actual
solution depicted by the solid curves in Figure 3.
The mass balance rate equation, Equation (1), can be adapted to fit many
systems as long as the boundaries of the system are defined along with the
‘‘input’’ and ‘‘output.’’ For example, the equation can be employed to
describe the process of balancing the amount of money in a checking
account. The rate at which money accumulates, measured in dollars per
unit time, is equal to the rate at which new money is generated, via interest,
plus the rate at which we deposit money minus the rate at which we spend
money from our account. The predator-prey model provides another applica-
tion of the balance equation. The rate at which deer accumulate in a given
region, for example, measured in number of deer per unit time, is equal to the
rate at which new deer are generated, described by the difference between
birth rate and death rate (to include hunting), plus the rate at which deer
migrate into the area minus the rate at which deer migrate out of the area.
As an example of an enzyme catalyzed reaction we consider nutrient
uptake by a cell. Cells need nutrients to grow and develop, but how does the
nutrient enter the cell? Let us consider the case of a microorganism such as
bacteria growing in a petri dish [1]. Nutrient molecules form the substrate, S.
They enter the membrane of a bacterial cell by attaching to an enzyme,
specifically, a membrane-bound receptor. The receptor along with its
attached nutrient molecule is referred to as an enzyme-substrate complex.
(See Figure 4.) The product in the reaction would be any nutrient molecules
Figure 4. Schematic of an enzyme that facilitates transport into the cell.
Enzyme Kinetics 155
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that are successfully captured by the cell. Note that the reaction E þ S$ ES,
where E, S and ES respectively denote the amounts of enzyme, substrate, and
enzyme-substrate complex, is reversible since an occupied receptor may lose
a nutrient molecule before it is captured by the cell.
In their development of what is now referred to as Michaelis-Menten
kinetics, Michaelis and Menten made two simplifying assumptions when
constructing the reaction scheme for an enzyme-catalyzed reaction such as
the one just described. The first is that the enzymatic step, or formation of
products, is irreversible. Alternatively, one may assume that only initial rate
before enough product has accumulated for a back reaction to occur will be
considered. Thus the reaction scheme will be represented by
E þ S �!k1ES; (8)
ES �!k�1E þ S; (9)
ES �!k2E þ P; (10)
where P denotes the amount of product formed and k1, k-1, and k2 denote rate
constants in the corresponding reaction directions. (For a nice interactive simula-
tion of this reaction process see [13].) Note that the reaction described by
Equation (9) is the reverse of the reaction described by Equation (8), and we
remind ourselves of this by denoting the rate constants by k-1 and k1 for the
reverse and forward reactions, respectively. A classical representation of a reactive
catalyst like that described by Equations (8) through (10) is shown in Figure 5.
We can now apply the concept of mass balance and the law of mass
action to develop the following system of differential equations that describe
the accumulation of all four substances. Beginning with a mass balance on
enzyme E, we have
Figure 5. Classical representation of a reactive catalyst.
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Rate ofaccumulation
of E
¼ Rate ofinput
of E
þ Rate ofgeneration
of E
� Rate ofoutput
of E
� Rate ofconsumption
of E
:
As before, we shall assume that the system is closed, the law of mass
action applies, and that the rate of accumulation of E is given by the time
derivative of the concentration of E. Thus,
Rate ofaccumulation
of E
¼ Rate ofgeneration of E
by reaction ð9Þ� Rate of
consumption of E
by reaction ð8Þ;
which can be written
d½E�dt¼ k�1½ES� � k1½E�½S�: (11)
The mass balance on S proceeds in a similar fashion to give
d½S�dt¼ k�1½ES� � k1½E�½S�: (12)
The mass balance on the ES complex is similar except that it contains
one generation term from reaction (8) and two consumption terms from
reactions (9) and (10):
Rate ofaccumulation
of ES
¼ Rate ofgeneration of ESby reaction ð8Þ
� Rate ofconsumption of ES
by reaction ð9Þ� Rate of
consumption of ESby reaction ð10Þ
;
which can be written
d½ES�dt¼ k1½E�½S� � k�1½ES� � k2½ES�: (13)
Finally, for product P, we have:
Rate ofaccumulation
of P
¼ Rate ofgeneration of Pby reaction ð10Þ
;
or
d½P�dt¼ k2½ES�: (14)
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The exact solution of Equations (11) to (14) is beyond the scope of this
work. The numerical solution using Euler’s method is obtained by writing
difference equations for each equation as follows:
½E�jþ1 ¼ ðk�1½ES�j � k1½E�j½S�jÞ�t þ ½E�j; (15)
½S�jþ1 ¼ ðk�1½ES�j � k1½E�j½S�jÞ�t þ ½S�j; (16)
½ES�jþ1 ¼ ðk1½E�j½S�j � k�1½ES�j � k2½ES�jÞ�t þ ½ES�j; (17)
½P�jþ1 ¼ k2½ES�j�t þ ½P�j: (18)
The spreadsheet solution of Equations (15) through (18) using Euler’s
method is shown in Figure 6. The different phases of the product concentra-
tion versus time were discussed earlier, and we can see them in Figure 6. The
‘‘first phase’’ of the product concentration versus time profile is seen as the
‘‘lag’’ in [P] between 0 and approximately 5 ns. From approximately 5 ns to
about 50 ns, [P] increases in a roughly linear fashion, which corresponds to
the ‘‘second phase’’ mentioned above. After about 50 ns, [P] starts to level
off as the system enters the ‘‘third phase.’’
An interesting effect is observed in the results plotted in Figure 6. The
concentration of enzyme-substrate complex [ES] increases rapidly before ,5
ns, and stays nearly constant between ,5 ns and ,50 ns. After ,50 ns, the
[ES] drops to zero. This leads us to the second major simplifying assumption
made by Michaelis and Menten.
The second simplifying assumption made by Michaelis and Menten was
that the concentration of the intermediate enzyme-substrate complex equili-
brates rapidly. In terms of our cell nutrient example, any new receptor-
nutrient complexes that form are balanced by complexes that break up by
forming either an empty receptor and product or converting back to an empty
receptor and substrate, so that receptors have an approximately constant
occupancy rate [1]. This means that we may assume that the derivative
d[ES]/dt in Equation (13) can be approximated by zero. This assumption is
often referred to as the quasi-equilibrium hypothesis or pseudo-steady-state
hypothesis (PSSH).
The PSSH was an extremely useful development in the study of chemi-
cal kinetics. While it is a fairly straightforward matter to solve the Michaelis-
Menten equations using numerical methods, computers were not readily
available prior to the 1960s. The use of the PSSH was therefore widespread
in the field of chemical kinetics, and its use has continued to this day in the
study of enzyme kinetics. Virtually all enzyme kinetics parameters available
in the literature were derived using this assumption and the linearization
methods that followed [6, 9, 10]. In the Requirement problems, we assume
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that Equations (11) – (14) hold, and, unless otherwise stated, we assume the
pseudo-steady-state hypothesis is valid.
The Michaelis-Menten equation (19) is derived using the PSSH (see
Requirement 2). This equation is given by
Figure 6. Euler’s method solution for [E], [S], [ES], and [P].
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V0 ¼Vmax½S�
KM þ ½S�(19)
where
V0 ¼dP
dt;Vmax ¼ k2½E�tot; and KM ¼
k�1 þ k2
k1
:
These parameters are discussed in more detail in Requirement 2.
2.2. Exploring Enzyme Kinetics Using Pennies
This experiment was adapted from an activity described in [4].
This activity involves an experiment in which students will have an
opportunity to empirically explore the relationship between V0 and [S].
2.2.1. Materials required
� 200 pennies
� 1 meter by 1 meter square table
� timer or stopwatch
� data collection sheet
� blindfold
In this experiment, we model an enzymatic reaction similar to the
process of nutrient uptake by a cell that was described earlier and illu-
strated in Figure 4. The pennies represent the substrate. Various concen-
trations of pennies will be placed on the table and spread out randomly.
The concentration of substrate is represented by the number of pennies per
square meter. The ‘‘product’’ will be the pennies that are picked up from
the table and placed in a cup. Without some sort of enzyme it is clear that
no product will be formed since the pennies cannot jump up and land in the
cup! In the experiment, fingertips will act as an enzyme by picking up
pennies and placing them in the cup. Initial reaction rate will be measured
by the number of pennies the enzyme can convert in a 30 second time
interval. (Note: In order to gather useful data it is important to have a large
enough table to spread out the pennies, especially for low concentrations.
It is also important for the ‘‘enzyme’’ to use only his or her fingertips, not
the entire palm, to locate and pick up pennies.)
2.2.2. Procedure
1. Designate one student to be the catalyst. The catalyst will be blindfolded
and will use his or her fingers from one hand to pick up pennies from a
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large square table and place them in a cup held in the other hand. However,
the catalyst must perform this task solely by touch—no peeking allowed!
Also, the catalyst is allowed to use only his or her fingertips to locate and
pick up each penny and pennies may be picked up only one at a time.
2. Have the catalyst stand or sit near the table with his or her eyes closed.
Spread twenty pennies out on the table. Try to spread the pennies out
randomly and evenly.
3. Set the timer for 30 seconds. Once the timer starts the catalyst may begin
to seek out pennies to place in the cup. When the timer stops, no more
pennies may be placed in the cup.
4. Record the amount of product formed for the given concentration by count-
ing the pennies in the cup. Use the data collection table provided in Figure 7.
5. Repeat steps 1 through 4 using 40, 60, 80, 100, 120, 140, 160, 180, and
200 pennies.
6. Use your results to address the questions in Requirement 1.
3. REQUIREMENTS
1. Use the results from the Penny Experiment to address the following three
questions.
a. Use Equation (19) to create a model for your data. Then plot both your
data and the model on the same axes.
b. Explain why there must be a maximum initial reaction rate for this
experiment.
c. What values does your model predict for Vmax and KM? Does the
predicted value of Vmax seem reasonable? Explain.
2. In this problem we derive the Michaelis-Menten equation and investigate
some of the properties of enzyme kinetics that it implies.
a. Let [E]tot denote the total concentration of enzyme. Then [E]tot is
equal to the sum of the free enzyme concentration, [E], and the
Figure 7. Data collection form for the penny experiment.
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enzyme-substrate complex concentration, [ES]. Show that the rate of
product formation is given by
d½P�dt¼ k2½E�tot½S�
KM þ ½S�(20)
where KM, referred to as the Michaelis constant, is given by
KM ¼k�1 þ k2
k1
:
b. Use the equation from part (a) to create a plot of d[P]/dt against [S].
c. By analyzing a plot of d[P]/dt against [S] along with Equation (20), we
can see several key characteristics of Michaelis-Menten kinetics. First
we note that in enzyme kinetics it is customary to consider Equation
(20) in terms of the initial rate of product formation in order to mini-
mize reversible reactions and the inhibition of the enzyme by product.
This initial velocity, denoted V0, can be defined as the velocity d[P]/dt
measured before about 10% of the substrate has been converted to
product [7]. Our first observation is that for large values of [S], [S]
>> KM, the reaction reaches a maximum velocity, Vmax. This occurs
when all enzyme molecules are occupied by substrate molecules, an
event termed enzyme saturation. Explain why it follows that Equation
(20) may be rewritten as the Michaelis-Menten equation (19) introduced
in the reading. Recall, this equation is given by
V0 ¼Vmax½S�
KM þ ½S�:
d. Confirm that when the substrate concentration is equal to the Michae-
lis constant, the initial reaction rate is half of the maximum rate. (This
means that KM represents the substrate concentration at which half of
the enzyme active sites are occupied by substrate molecules [3].)
Illustrate this point on your graph from part (b).
e. For low substrate concentrations, i.e., [S] << KM, the relationship
between V0 and [S] should be nearly linear. How is this implied by
Equation (19)? What is the slope of this nearly linear relationship in
terms of Vmax and KM?
3. The rate constants k2 in Equation (10) describes the rate at which the
enzyme-substrate complex reacts to produce the product and to regenerate
the enzyme. This value is often called the turnover number of an enzyme
since it measures the number of ES complexes that are converted to
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product per enzyme molecule per unit time when the enzyme is fully
saturated with the substrate.
a. Show that the turnover number is given by
k2 ¼Vmax
½E�tot
: (21)
b. Table 1 gives parameter values for KM and k2 in some representative
enzyme-substrate systems [7]. Note that the Michaelis constant varies
considerably, not only between different enzymes, but also between
different substrates with the same enzyme. For each system
i. Find out what the purpose of each reaction is;
ii. Determine how many substrate molecules can be converted by one
enzyme molecule per minute.
4. In practice, it can be quite difficult to determine the parameters KM and
Vmax from a plot of V0 versus [S]. In 1934, Dean Burk, an American
biochemist, and his lab assistant, Hans Lineweaver, developed a way to
more easily analyze the parameters of the Michaelis-Menten equation by
taking the reciprocals of both sides of Equation (19) [10]. (For a short and
interesting history see [5].)
a. Show that the plot of this new equation, referred to as a Lineweaver-
Burk plot, describes a linear relationship and determine its slope and
intercepts.
b. Create a Lineweaver-Burk plot of the data you obtained from the
penny experiment. Find the line of best-fit for your data and use this
line to estimate the parameters KM and Vmax. How do these estimates
compare to those you obtained when answering Requirement 1(c) in
the penny experiment?
5. One of the disadvantages of the Lineweaver-Burk plot is that it com-
presses the data points at high substrate concentrations and emphasizes
Table 1. Values of KM and k2 for some enzyme-substrate systems
Enzyme Substrate Product KM (mol/L) k2 (s-1)
Carbonic Anhydrase CO2 HCO3- 0.012 1.0 � 106
Carbonic Anhydrase HCO3- H2CO3- 0.026 4.0 � 105
Catalase H2O2 H2O þ O2 0.025 1.0 � 107
Fumarase Fumarate Malate 5.0 � 10-6 800
Fumarase Malate Fumarate 2.5 � 10-5 900
Urease Urea CO2 þ NH3 0.025 1.0 � 104
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the points at lower concentrations where the data are less precise.
An alternative is the Eadie-Hofstee plot [6, 9]. The Eadie-Hofstee equa-
tion is given by
V0 ¼ �KMV0
½S� þ Vmax: (22)
a. Show how this equation can be derived from the Michaelis-Menten
equation.
b. Create an Eadie-Hofstee plot of your data from the penny activity.
Find the line of best-fit for your data and use this line to estimate KM
and Vmax. Compare your results to those you obtained when answering
Requirement 1(c) in the activity.
6. In enzyme kinetics, there are two types of inhibitors: competitive and non-
competitive. A competitive inhibitor essentially binds the active site of the
enzyme, thus preventing the substrate from binding. A noncompetitive
inhibitor binds elsewhere affecting the active site in such a way that,
although the enzyme may still be able to bind with the substrate, its ability
to convert the substrate to product is diminished. Both types of inhibitor
have the overall effect of slowing product formation. Here we suggest two
ways in which the penny experiment may be modified in order to explore
the effects of inhibitors. Results of each experiment can be analyzed using
the techniques developed in Requirements 1, 4, and 5. Perform the penny
experiment with the suggested modification, then compare the new data
with the penny data collected for Requirement 1(a) of the penny activity.
In this manner you can assess the effects of the inhibitor on V0 and Vmax.
You can then create either a Lineweaver-Burk plot or an Eadie-Hofstee
plot of the data and use the results to provide a model for V0.
a. In order to simulate the effects of a competitive inhibitor, we will
replace some of the pennies in our previous experiment with quarters.
Each quarter will represent a substrate molecule whose active site is
occupied by an inhibitor. Thus, when the enzyme encounters a quarter,
it cannot bind with it. In other words, when a quarter is picked up it
must be put back down, rather than placed in the cup.
b. To simulate the effects of a non-competitive inhibitor the ‘‘enzyme’’
can either wear a surgical glove or tape the three middle fingers of his
or her hand together. In this way, the enzyme will still have the ability
to convert the substrate to product, but it will be more difficult to
do so.
7. Use a computer algebra system such as Mathematia or Maple to verify that
the solution to the mass balance equation for the first-order irreversible
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reaction A ! B is given by Equation (5). Make a plot of [A] versus t for
[A]0 ¼ 1 and k ¼ 0.1, 1, and 10 s-1.
8. Create an Euler’s method solution for Equations (11) to (14) for urease,
with k1 ¼ 4.0 � 10-3 s-1, k–1 ¼ 1.0 � 10-8 s-1, k2 ¼ 1.0 � 10-4 s-1, [E]0 ¼0.1 M, [S]0 ¼ 1.0 M, [ES]0 ¼ 0 M, and [P]0 ¼ 0 M. (M represents mol/L.)
a. Do your results confirm the pseudo-steady-state hypothesis? Why or
why not?
b. Increase [E]0 to 1.0 M and discuss what happens to [ES]. Is the
pseudo-steady-state hypothesis still valid?
4. INSTRUCTOR NOTES
4.1. Implementation
This module was designed to be conducted as an out of class project for
students who were familiar with elementary differential equations and
numerical solution methods. We believe that the penny experiment along
with its variations concerning substrate inhibitors (Requirement 6) should be
an integral part of the module. We recommend modeling the basic penny
experiment in class to facilitate some measure of uniformity in its imple-
mentation. However, doing all of the experiments in class would consume an
inordinate amount of class time. Students should not find it too difficult to
conduct the experiments outside of class after seeing the in-class sample
experiment.
4.2. Student Challenges
We believe that students who are unfamiliar with numerical integration,
Euler’s method in particular, would benefit from working through the
procedure by hand. Appendix A contains an Euler’s method worksheet
for Equations (2) and (4). The worksheet encourages students to work
through Euler’s method by hand in a simple case so that they will gain
a better understanding of how the numerical solutions are generated by a
spreadsheet.
4.3. Benefits
The implementation of the penny experiment to estimate the values Vmax and
KM helps demonstrate to students the roles of both experimentation and
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mathematics in deriving many of the parameter values used by scientists in
models of various phenomena. By comparing the results obtained from
various modeling strategies such as the least squares curve, Lineweaver-
Burk plots, and Eadie-Hofstee plots, students learn that there may be more
than one ‘‘best fit’’ model. Determination of which best fit model is ‘‘best’’
emphasizes the need for a solid understanding of both mathematics and
science.
We noticed another possible benefit of the lesson in the (unexpected)
construction and evaluation of the experiment described in the solution to
Requirement 6. First, a particular student group learned a valuable lesson
about choosing the ‘‘right experiment’’ after discovering that the experiment
they had designed was not really appropriate for addressing the given
requirement to investigate the effect of a competitive inhibitor on Vmax.
However, their effort was in no way futile since the students became self-
motivated to use methods of scientific inquiry to examine their results. The
students formulated and addressed questions such as the following: What
hypothesis was their experiment designed to test? Why did they design the
experiment in the way that they did? How could they improve the experiment
next time? What information could be gleaned from the data? Does the
model developed from the data seem intuitively reasonable? Based on their
data, the students formulated their own conjecture that the relationship
between product reaction rate, V0, and the percent of substrate occupied by
competitive inhibitors, r, would be given by
V0ðrÞ ¼ Vcð1� rÞ
where Vc represents the initial product reaction rate for a given concen-
tration of substrate with no inhibitors. Thus, these students were inspired to
do research beyond the scope of the initial lesson.
4.4. Improvements or Alternate Implementations
Although the reading indicates that the Michaelis-Menten equations might be
applied to situations other than enzyme kinetics, students often fail to make
that connection. To this end, we have included three supplemental problems
in Appendices 2–4 with their respective solutions in Appendix 5. We hope
that such problems might clarify how the mathematical principles used to
describe the dynamics of enzyme kinetics and mass balance can be applied in
a wide range of problems. The three supplemental problems provided here
describe applications to the following: finance, the classic predator-prey
model, and chemical equilibria. We have also seen a reference to the
Michaelis-Menten equation being used to approximate the average weight
of an adult male Siberian tiger [14]. An investigation into this application
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might make a nice interdisciplinary mini-project for a math student interested
in biological applications, or vice versa.
5. SUPPLEMENTAL READING
Introductory chapters on protein and enzyme structure, as well as introduc-
tions to the Michaelis-Menten Equation can be found in general biochemistry
texts such as [12] and [2]. For a detailed treatment of the integrated form of
the Michaelis-Menton equation as well as a full implementation in Matlab,
see [8].
REFERENCES
1. Allen, L. J. S. 2006. An Introduction to Mathematical Biology, pp.
263–265. Upper Saddle River NJ: Prentice Hall.
2. Berg, J. M., J. L. Tymoczko, and L. Stryer. 2006. Biochemistry, 6th
Edition. New York: W.H. Freeman.
3. Chang, R. 2005. Physical Chemistry for the Biosciences, pp. 363–372.
Sausalito CA: University Science Books.
4. Chayoth, R., and A. Cohen. 1996. A simulation game for the study of
enzyme kinetics and inhibition. The American Biology Teacher. 58(3):
175–177.
5. Dagani, R. 2003. Straightening Out Enzyme Kinetics. Chemical and
Engineering News. 81(24). http://pubs.acs.org/cen/science/8124/8124jacs125.
html. Accessed on 8 Sept. 2008.
6. Eadie, G. S. 1942. The inhibition of cholinesterase by physostygmine
and prostigmine. Journal Biological Chemistry 146: 85–93.
7. Gladney, L. 1998. The Interactive Textbook of PFP. http://www.physics.
upenn.edu/courses/gladney/mathphys/java/sect5/subsection5_1_4.html.
Accessed on 8 Sept. 2008.
8. Helfgott, M., and E. Seier. 2007. Some mathematical and statistical
aspects of enzyme kinetics. Journal of Online Mathematics and Its
Applications. 7, Article ID No. 1611. http://mathDL.maa.org/mathDL/
4/?nodeId=1611&pa=content&sa=viewDocument. Accessed December
2009.
9. Hofstee, B. H. J. 1959. Non-inverted versus inverted plots in enzyme
kinetics. Nature. 184: 1296–1298.
10. Lineweaver, H., and D. Burk. 1934. The Determination of enzyme
dissociation constants. Journal of the American Chemical Society. 56:
658–666.
11. Michaelis, L., and M. L. Menten. 1913. Die Kinetik der Invertinwirkung
Biochemische Zeitschrift 49: 333–369.
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12. Nelson, D. L. and M. M. Cox. 2008. Lehninger Principles of
Biochemistry, 5th Edition. New York: W.H. Freeman.
13. Stieff, M., and U. Wilensky. 2001. NetLogo Enzyme Kinetics Model.
Evanston, IL: Center for Connected Learning and Computer-Based
Modeling, Northwestern University. http://ccl.northwestern.edu/netlogo/
models/EnzymeKinetics. Accessed December 2009.
14. Siberian Tiger. (2008, Aug. 12). In Wikipedia, The Free Encyclopedia.
http://en.wikipedia.org/w/index.php?title¼Siberian_Tiger&oldid¼231427177.
Accessed August 2008.
BIOGRAPHICAL SKETCHES
Andy Biaglow is a chemical engineer and a member of the faculty of the
Chemistry and Life Science Department at the United States Military
Academy. Andy spends his free time thinking about problems in heteroge-
neous catalysis and is an avid lover of accordion music and ancient coins.
Keith Erickson is a member of the mathematics faculty at Georgia Gwinnett
College. His background is in chemical engineering and bioengineering, and
he enjoys applying mathematics in a wide variety of subjects.
Shawnee McMurran serves on the mathematics faculty at California State
University, San Bernardino. Her background is in partial differential equa-
tions. Her recent research has focused on the history of mathematics, along
with her increasing involvement with mathematics education.
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