Download Exercise 7. Enzyme Kinetics

Exercise 7. Enzyme Kinetics
Objectives
This experiment is designed to familiarize the student with basic enzyme kinetics and
analysis of initial rate data. The enzyme chosen for study is -galactosidase, a large enzyme that
is fairly stable and easily assayed. The substrate is chosen such that a colored product results
upon hydrolysis. This product can then be measured spectrophotometrically. An understanding
of the Michaelis-Menten equation is important.
Pre-lab
1) Prior to lab, come ready with a table listing all 18 reactions, and the amount of enzyme
solution (0.25 g/ml), Z buffer, and ONPG solution (0.5 mg/ml) that each cuvette will
contain.
2) In the procedure text, the Michaelis Menten equation is derived assuming a rapid
equilibrium step. Show the derivation for an alternative form, instead assuming a pseudosteady state. Experimentally, how could one test verify the accuracy of this assumption if
direct observation of the intermediate is not available?
3) What order is the MM equation with respect to enzyme? How might this change if the
enzyme were immobilized on a solid support instead of in solution and why?
4) Consider an enzymatic reaction product that is an inhibitor of the free enzyme. Where
would you expect to find such a system and why?
Background
Consider the following simple mechanism for an enzyme (E)-catalyzed reaction in which
a substrate (S) is converted to a product (P):
K
k
s
2
E S 

E  P 
E  S 
Eq (1)
In the first step, the enzyme and substrate reversibly bind to form an enzyme-substrate complex.
In the second step, the substrate is converted to a product and released from the enzyme. (The
kinetics of product release are neglected here.) If we assume that the first step is in equilibrium,
the following expression can be obtained:
Eq (2)
Ks E S  E S
    
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The constant Ks is the ratio of the reverse (k-1) and forward (k1) rate constants for step one, also
known as the dissociation constant. In a closed system, the rate of product formation is then
derived from the second step to give the following expression:
   k E S 
dP
2
dt
Eq (3)
The ES complex is not easily measured and is often a very short-lived intermediate. Therefore,
we need to obtain an expression for [ES] in terms of other, measurable variables. Eq (2)
expresses the complex concentration in terms of [S] and [E], the free enzyme concentration.
However, the free enzyme concentration is also difficult to determine. Remembering that the
enzyme only participates in the reaction as a catalyst, a mass balance around the enzyme yields:
E   E  E S
0
Eq (4)
where [E]0 is the initial, or total enzyme concentration. Now, Eqs (2) and (4) can be combined to
give an expression for [ES] in terms of the substrate concentration and the total enzyme
concentration. This expression can be inserted into Eq (3) to give expression for the rate of
production formation, or the velocity of the reaction:
v
   k E S  V S 
dt
K  S  K  S 
dP
2
max
0
s
m
Eq (5)
where Vmax is the maximum reaction velocity and Km is the Michaelis constant. Note that Km is
equal to the dissociation constant, Ks, for the simple mechanism of Eq (1) (but this is not always
the case for enzyme kinetics). The final expression given in Eq (5) is known as the MichaelisMenten equation for enzyme kinetics. The parameters Vmax and Km give important insight into
the behavior of the enzyme for a particular set of experimental conditions. These parameters can
be determined graphically, given the reaction velocity as a function of the substrate concentration
and the initial enzyme concentration. Vmax is further defined as:

Vmax  kcat E
0
Eq (6)
where kcat is the catalytic rate constant (or turnover number). For this simple mechanism, kcat is
equal to k2, the first order rate constant for the second step of Eq (1). More complicated reaction
mechanisms may also yield a rate expression in the Michaelis-Menten form; kcat may then be a
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function of several rate constants, each representing a different step in the reaction sequence.
The Michaelis constant, Km, is also a more complex function of rate constants in such cases.
In this exercise, we will investigate the kinetics of -galactosidase, a large enzyme
(monomer molecular weight 116,000) that is fairly stable and easily assayed. The enzyme galactosidase hydrolyzes -D-galactosides, such as lactose. It can be easily assayed with
chromogenic substrates, which contain a reactant that releases a colored product when
hydrolyzed. One such material is ONPG (o-nitrophenyl--D-galactopyranoside). This colorless
compound is converted to galactose and o-nitrophenol in the presence of -galactosidase. Onitrophenol is yellow and can be measured by its absorption at 420 nm. The reaction is halted by
the addition of concentrated sodium carbonate, which shifts the pH to 11, inactivating the galactosidase.
The amount of o-nitrophenol produced is proportional to the amount of enzyme present
under conditions of excess ONPG, permitting an assay of the amount of enzyme present in a
sample. In this exercise however, we will determine the kinetics of hydrolysis by varying the
ONPG concentration and measuring the initial rate of formation of o-nitrophenol.
PROCEDURE
1. A 0.5 g/ml solution of -galactosidase in Z buffer will be given to you. Stock ONPG
solution (0.5 mg/ml) will also be provided for use as the substrate. Z buffer contains a small
amount of 2-mercaptoethanol, a chemical which is toxic and can be absorbed through the
skin. It also smells terrible! Wear gloves at all times during this exercise and be careful
not to spill reagents.
2. You will be running 18 reactions simultaneously. Three will be controls, and the rest will be
your enzymatic reactions. In this experiment, you will be varying the substrate concentration
by changing the proportion of substrate solution added to the reaction mixtures. There will
be a total of 13 different substrate concentrations. There will also be two additional reactions
run at high substrate concentration in which the enzyme concentration is varied.
3. In one cuvette you will have 1.0 ml of Z buffer equilibrated to the assay temperature (all
assays will be run at room temperature), to which is added 0.2 ml of ONPG solution. Later,
100 L of the enzyme solution will be added to start the reaction, for a total reaction volume
of 1.3 ml. This is your 1X substrate concentration.
4. Into subsequent cuvettes, add appropriate amounts of Z buffer and ONPG solution, such that
you arrive at the following substrate concentrations: 0.1X (ten-fold dilution), 0.2X, 0.3X,
0.4X, 0.5X, 0.6X, 0.8X, 1.2X, 1.4X, 1.6X, 2X, and 3X. (Keep in mind that the amount of
enzyme added to these reactions will be the same--100 L.) The total volume for each of
these will be 1.3mL
5. Make three control cuvettes at three substrate concentrations: 1X, 0.2X, and 2X. These
cuvettes should contain 100 L of Z buffer in place of enzyme. For example, the 1X control
should contain 1.1 ml Z buffer and 0.2 ml ONPG solution.
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6. The final two cuvettes should contain 0.2X and 2X of the enzyme solution and 3X of the
ONPG solution. Add the Z buffer and substrate first. All cuvettes should now contain the
correct proportion of Z buffer and ONPG solution.
8. As quickly as possible, add the appropriate amounts of enzyme solution to the cuvettes. Stir
each well with your pipette tip and start the timer after the addition of the first.
9. You will be adding 150uL of Na2CO3 to each cuvette to halt the reaction. 0.1X thru 0.8X
OPNG, and also the reactions from step 6. should be stopped after 2 minutes. 1X thru 3X
OPNG should be stopped after 10 minutes. Mix the Na2CO3 thoroughly with your pipette tip
10. Blank the spectrophotometer with a cuvette containing 1.3 ml Z buffer plus 0.15 ml Na2CO3
solution. Read the OD at 420 nm for all 18 cuvettes. One student should take measurements
while the other records the data.
Guideline for Analysis & Conclusions Section
The extinction coefficient of o-nitrophenol is 4500 M-1cm-1; the molecular weight of ONPG is
301.3. Use concentration units of M (mol/L) and time units of seconds in your calculations.

Explain the mechanism of this enzyme for hydrolysis of ONPG. Be specific with respect to
the important amino acids on the enzyme, the role of crucial salts and water.

Why would one want to stop those reactions with lower OPNG earlier than those with higher
OPNG?

Prepare Lineweaver-Burke and Eadie-Hofstee plots for the data obtained during the lab.
Use these plots to determine values of Vmax and Km. Also perform a non-linear regression
using mathematical software (Mathcad, KaleidaGraph, etc.) to determine another set of
values for vmax and Km.

Create a velocity vs. concentration plot showing the data you obtained and the three curves
generated by incorporating your values for vmax and Km into the Michaelis-Menten equation.
Remember, data is plotted as x’s or dots on the graph, and equations are represented by lines.

Comment on the determined value of KM. What values (range) would you expect for an
enzyme such as -galactosidase for this substrate relative to (e.g.) lactose?

Determine the value of kcat from the concentration of the enzyme in your solutions and the
value of vmax. What assumptions did you make concerning the integrity of the enzyme
preparation?
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EQUIPMENT AND REAGENTS
Z buffer:
16.1 gm Na2HPO47H2O
5.5 g NaH2PO4.H2O
0.75 g KCl
0.246 g MgSO47H2O
2.7 ml 2-mercaptoethanol
Adjust to pH 7.0 and bring to 1.0 liter with distilled water
Do not autoclave.

-galactosidase, 0.5 g/ml in Z buffer
o-nitrophenyl--D-galactosopyranoside (ONPG), 0.5 mg/ml
1.0 M Na2CO3 solution
Spectrophotometer (OD at 420 nm)
Spectrophotometer cuvettes, 1 cm width (path length)
REFERENCES
Bailey, J.E., and Ollis, D.F. (1986) Biochemical Engineering Fundamentals, 2nd Ed., pp. 89108, McGraw Hill, Inc., New York.
Blanch, H.W., and Clark, D.S. (1996) Biochemical Engineering, pp. 1-1-18, Marcel Dekker,
Inc., New York.
Fogler, F.S. (1992) Elements of Chemical Reaction Engineering, 2nd Ed., pp. 359-364, Prentice
Hall, Englewood Cliffs, NJ.
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