Enzyme Kinetics of Acid Phosphatase
Introduction
Enzymes are organic catalysts, proteins that increase the rates of chemical
reactions. The portion of a system’s total energy that is available to do work is free
energy. Chemical reactions involve transition states, whose free energy is higher than
the total energy of either the reactants or the products. The difference between the free
energy of the reactants and the free energy of the transition state is called the
activation energy of the reaction. Enzymes reduce the activation energy of the
reactions they catalyze. Because they are neither consumed nor modified in the
chemical reactions they affect, small quantities of enzyme can mediate the formulation
of large quantities of product.
The enzyme we will work with this week is a phosphatase, an enzyme that
catalyzes the removal of phosphate groups (PO4-2) from organic molecules. They are
found throughout the body. These enzymes play a role in many important biological
processes, including the removal of phosphates from DNA and ATP. Given that
phosphates attach to many kinds of molecules, we can use “R” to indicate any one of
these:
e.g. R-PO4
(a molecule with a phosphate group)
The particular enzyme we will work with today is an enzyme found in many cells. This
version is found in wheat germ cells. Since it works best under the acid conditions of the
small intestine, its common name is acid phosphatase. It brings a substrate
molecule into contact with a water molecule, catalyzing the removal of a -PO4 group
from, and the addition of an -OH group to, the substrate molecule. An example of its
function is as follows:
phosphatase
enzyme
R-PO4
+ H2O

R-OH
+
H3PO4
substrate
product
product
In this particular reaction, the phosphate group forms the substrate upon which the
enzyme acts and the R-OH and H3PO4 are the products. This is a form of hydrolytic
reaction (i.e. hydrolysis), in which water is split (and is not considered a substrate).
If we have a means to measure the accumulation of R-OH, we can follow the progress
(rate) of the reaction. Nitrophenyl phosphate (i.e. our R- PO4) is colorless, while
Nitrophenol (i.e. R-OH) is yellow. Therefore, the reaction proceeds as follows:
Acid
phosphatase
Nitrophenyl phosphate + H2O

Nitrophenol +
H3PO4
substrate
product
product
COLORLESS
YELLOW
The progress of the reaction is monitored by a change in color. As the reaction
proceeds to the right, the solution becomes more yellow.
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How do we measure the color of a solution?
As white light strikes a "yellow" solution it appears that color because the solution is
absorbing the purple, blue and red light, while allowing the yellow light to transmit
through and out of the solution. The color of light as perceived by the human visual
system appears as it does based on the wavelength (in nanometers, 10-9 meters) of
the light. Yellow light has a wavelength of approximately 570 nm, while purple light is
about 405 nm, blue 500 nm and red 620 nm. A solution appears yellow because,
relative to the white light striking it, it contains the same amount of yellow light, and less
purple, blue and red light. Therefore, the difference between a colorless and yellow
solution is not how much yellow light comes out (they both have about the same
amount), but rather how much light of other wavelengths (purple, blue, red) are
absorbed (the yellow solution has more).
If we want to measure “how yellow” a solution is, we do this by monitoring how much of
the other wavelengths of light the solution absorbs. We will measure the absorbance of
purple light (wavelength = 405 nm) in this experiment. The more yellow the solution, the
more purple light is absorbed. The more colorless, the less purple light is absorbed.
We will use a spectrophotometer to precisely measure how much light of 405 nm a
solution absorbs as a precise measure of how much product is present.
For many solutions, absorbance is directly proportional to the concentration of pigment
in the solution. This relationship is represented by the Beer-Lambert Equation: A= ε•l•c
where A is the absorbance, ε is the extinction coefficient for p-nitrophenol (18.8 x 103
liter mol-1cm-1, l is the path of light length (1 cm) and c is the concentration.
You will take advantage of this relationship in this experiment. Use the absorbance and
coefficient values to calculate the concentrations. These data will be used to monitor
the accumulation of product as the basis for estimating the reaction rate.
Most biological reactions proceed at very slow rates under normal conditions. These
rates are too slow for cells to grow and maintain normal cellular functions. The rates of
these reactions can be increased by raising the temperature of the environment or
lowering the activation energy of the reaction. Most biological systems have a low
tolerance for high temperatures. In addition, temperature regulation requires a high
input of energy into a system. Therefore, biological systems have evolved enzymes.
Enzymes are biological catalysts. Their function is to lower the activation energy of
specific chemical reactions, thereby allowing the reactions to proceed at rates faster
than under standard conditions. Without their presence in the cell, the multitude of
chemical reactions that make life possible would not proceed at the required rate.
Enzymes accelerate the velocity of almost all reactions that occur in biological systems,
including catabolic and anabolic reactions.
The rate of an enzyme reaction is referred to as the velocity of the reaction. It is most
common to express the velocity in amount of product formed per minute. The initial
reaction velocity, vo, of an enzymatic reaction varies with the substrate concentration,
[S]. This can be illustrated by plotting vo vs. [S]. This plot is known as a MichaelisMenten plot. It was named for two pioneers in the field of enzyme kinetic analysis.
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Figure 1. Michaelis-Menten plot for an enzyme-catalyze reaction.
From this plot, an equation can be derived to describe the characteristics of an enzyme
catalyzed reaction. The equation is called the Michaelis-Menten equation. The common
form of the equation is
v0 = initial reaction velocity. The v0 is expressed as amount of product produced per unit
time. The units of v0 are M/min or moles/min.
Vmax = maximal reaction velocity; achieved when all the enzyme active sites are
occupied with substrate molecules. This condition is called substrate saturation.
[S] = substrate concentration
Km = Michaelis constant = (k2 + kp)/k1
Two important values, Vmax and Km can be determined from the graph shown in figure 1.
The Vmax is the maximal rate at which an enzyme catalyzes a reaction. It is expressed
as the amount of product formed per minute. The units of Vmax are the same as for v0.
This value can be extrapolated from the plateau of the curve on the Michaelis-Menten
plot. The plateau region represents reaction conditions in which the concentration of
substrate far exceeds the concentration of enzyme active sites. Therefore, all active
sites will be occupied at any given time. The enzyme is said to be saturated with
substrate under these conditions. The Vmax is not a constant because it depends on the
amount of enzyme in the reaction.
The Km or Michaelis constant is the other very useful term that is derived from the
Michaelis-Menten equation. As is shown in the key to the Michaelis-Menten equation,
the Km is related to the rate constants of the reaction. In simple terms, it is the ratio of
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the rate at which ES forms (k1) to the rate at which ES dissociates (k2+kp). The
dissociation of ES is the sum of the rate that the substrate dissociates from the enzyme
before it undergoes the enzyme catalyzed reaction (k2) and the dissociation of product
following the reaction (kp). There are three important points to remember about Km:
1) The Km is a measure of the apparent affinity of the substrate for the enzyme. In other
words, it indicates how well the substrate fits into the binding site of the enzyme. The
substrate with the lowest Km value has the highest apparent affinity for the enzyme.
The "best" substrate is that which has the highest Vmax/Km.
2) Km is directly related to Vmax. It represents the substrate concentration at 1/2 the
maximal velocity (1/2 Vmax) of the enzyme.
3) The Km of an enzyme is a constant for every individual enzyme with a particular
substrate. Therefore, it serves as a fingerprint for the presence of a specific enzyme.
This is very useful for investigators who are determining whether an enzyme is present
or absent in a particular cell or tissue. The units of Km are molarity, M.
The fact that the Michaelis-Menten plot gives a curve makes it difficult to accurately
determine the Vmax and Km values from a graph. Therefore, the terms in the equation
are often rearranged and used to generate a plot that gives a straight line. This equation
is called the Lineweaver-Burk equation.
This equation is in the form, y = mx + b. By plotting 1/v0 versus 1/[S], this equation gives
a straight line. This is illustrated in figure 2. The intercept on the 1/v0 axis is 1/Vmax and
the intercept on the 1/[S] axis is -1/Km. The slope of the line on the plot is Km/Vmax.
Figure 2. Lineweaver-Burk plot for an enzyme-catalyzed reaction.
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The Experiment
Reagents
p-nitrophenyl phosphate (PNPP), wheat germ acid phosphatase, Bovine Serum
Albumin, potassium hydroxide, sodium acetate, 0.1 M MgCl2 .
Waste Disposal
A bottle for the collection of wastes is in the hood. All excess stock reagents, unused
reaction solutions and solutions from the reaction vessel should be disposed of in this
bottle. The glassware can then be given a single small rinse into the waste container
before being placed in the wash basin in the sink.
Procedure
Note: All glassware MUST be washed with phosphate free soap.
Standard Curve for p-nitrophenol (Each group)
1. Obtain 6 test tubes and label them 1-6. Prepare dilutions of the p-nitrophenol (PNP)
as outlined in the following table. Mix well.
Table 1 Dilution Series from 60 uM stock p-nitrophenol
Test tube
1
2
3
4
5
6
60 uM PNP
(ml)
0
1
2
3
4
5
0.02 M NaOH
(ml)
6
5
4
3
2
1
Diluted PNP (uM)
0
10
20
30
40
50
2. Transfer each to a cuvette and read the absorbance at 405 nm. Use the sample
containing no PNP as the blank to adjust the spectrophotometer to zero absorbance. In
excel, use a scatter plot to plot A405 (ordinate) against the umoles of p-nitrophenol in the
sample. Insert a trend line in the graph. Is there a linear relationship between
concentration and absorbance as shown by the R2 value for the trend line? Does this
determination of p-nitrophenol obey the Beer-Lambert Law?
For the Class
1. Prepare 50 ml of a 1.0 M sodium acetate buffer solution, and adjust to pH 5.7.
2. Use a portion of this buffer to prepare 50.00 mL of 5mM sodium acetate buffer.
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3. Weigh accurately 0.28 g of p-nitrophenyl phosphate (PNPP) into a clean and dry 30
mL beaker. Dissolve the substrate in a 15 mL aliquot of double distilled water.
4. Weigh accurately 9 mg of wheat germ acid phosphatase into a 15mL beaker
previously charged with 20 mg of bovine serum albumin (BSA).
5. Dissolve the enzyme and BSA in 20 mL of 5 mM sodium acetate buffer [pH 5.7].
6. Obtain 7 test tubes and label them “1” to “7”. These are the stock tubes. Prepare
dilutions of the substrate as outlined in the following table.
Table 2: Dilution series from 0.05 M stock PNPP substrate
Test tube 0.05 M PNPP
(mL)
1
-2
0.10
3
0.25
4
0.50
5
1.00
6
2.50
7
~5.00
double distilled
water (mL)
~5.00
4.90
4.75
4.50
4.00
2.50
--
Diluted PNPP
(M)
0.0
0.0010
0.0025
0.0050
0.0100
0.0250
0.0500
Determine the effect of substrate concentration on enzyme activity (Each Group)
1. Obtain a total of 14 clean and dry test tubes. Label them 1-7 in 2 sets. The
experiment must be performed twice (14 reaction tubes are required) and the
absorbance values averaged.
2. Into the test tubes add 0.5 mL of 1.0 M sodium acetate (pH 5.7), 0.5 mL of 0.1 M
MgCl2 (magnesium chloride), 0.5 mL of p-nitrophenyl phosphate (from stock tubes 1 to
7) and 3.30 mL of deionized distilled water. So tubes 1 get the substrate from stock
tube 1, and so on for tubes 2-7
°
3. Equilibrate these solutions to 37 C in the water bath for approximately 5 minutes.
4. Set a timer to 5 min. When the second hand on a clock or watch is at 0 or 30, remove
tube 1 from the bath and add 200 µL of the wheat germ acid phosphatase, quickly mix
with the vortex and put it back in the bath. Start the timer.
5. Add 200 uL of enzyme to the other tubes at 0.5 minute intervals. If the appropriate
reagents have been added according to the above protocol there should be a total
volume of 5.00 mL in each reaction tube.
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6. When the 5 min time is over for tube 1 remove the tube from the bath and add 2.5
mL of 0.5 M KOH. Carefully mix the tube with the vortex (final volume is 7.5 mL). The
quenched solution should be set aside undisturbed, to allow the cloudy precipitate to
form and settle to the bottom of the reaction tube.
7. Remove the other tubes at 0.5 min intervals so each tube remains in the bath for 5
min after the enzyme has been added.
8. Repeat the assay for the second set of tubes of the different substrate concentrations
(“1” to “7”).
9. Set the spectrophotometer to a λ of 405 nm. Calibrate to zero absorbance with the
blank (tube 1) and measure the absorbance for each of the assay tubes. The precipitate
at the bottom of the reaction tube must be avoided at all cost as the particulate matter
will interfere with absorbance readings. The supernatant can be removed carefully for
analysis with the use of clean Pasteur pipettes to fill the cuvettes.
10. It would be wise to charge and equilibrate all 14 reaction tubes before starting the
first assay with the addition of the phosphatase enzyme. As there are 14 assays
initiated at 0.5-minute intervals, each requiring 5 minutes of incubation before
quenching, this experiment will require good cooperation and communication in each
lab group.
Determination of Vmax and Km for Acid Phosphatase
1. Average the absorbance values for each pair of tubes.
2. Use the Beer-Lambert Equation to determine the micromoles of p-nitrophenol
.
–1
produced per reaction mixture. Express the velocity as micromoles of product min by
dividing by the incubation time.
Prepare a Michaelis-Menten plot with the above data (substrate conc vs velocity).
Put the data in a scatter plot with data points and a line.
2. Estimate the Km and Vmax values from this plot. Express the values in the units of µM
and µM/min, respectively.
As the Michaelis-Menten plot results in a hyperbola, the resulting estimates of Km and
Vmax are not very precise. Better values for Km and Vmax are derived by taking the
double reciprocal form of the Michaelis-Menten ( a Lineweaver-Burk plot)
3. Prepare a Lineweaver-Burk plot of 1/v0 versus 1/[S].
Put the data into a scatter plot with data points only, no line. Insert a trend line
(regression line). Show the regression equation for the line.
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4. Determine the Km and Vmax from the Lineweaver-Burk plot regression line equation.
The regression line is in the form of an algebraic formula for a straight line.
Y=MX + B
Where y= dependent variable; m = slope of the line; x= independent variable; and
b= y-intercept.
The equation of the line is used to provide the Km and Vmax values for the
enzyme. The y-intercept is equal to 1/Vmax. Therefore, Vmax = 1/y-intercept value. The
slope of the regression line is equal to Km/Vmax, so Km = (Vmax)(Km/Vmax) Calculate and
record these values.
5. Are the Vmax and Km values different from those that you estimated from the
Michaelis-Menten plot? Why or why not?
Effect of pH on Enzyme Activity
1. Prepare two 1.0 M sodium acetate buffer solutions. Record the pH of one of them
and do not adjust it. Adjust the pH of the other solution to pH 4.
2. Repeat the experiment showing the effect of substrate concentration on enzyme
activity using these buffers instead of the pH 5.7 buffer.
3. Plot a Lineweaver-Burk plot of these data on a graph with the original experiment
data. What effect (if any) does changing the pH have on enzyme activity?
Inhibition by inorganic phosphate
1. Repeat the original experiment with the modification that 0.3 ml of 0.05 M sodium
phosphate (pH 5.7) is added to each tube and the volume is adjusted so that the final
volume 1s the same (3.0 ml water, not 3.3 ml).
2. Calculate the reaction velocity at each substrate concentration as before.
3. Plot a Lineweaver-Burk plot of these data on a graph with the original experiment
data. Do you detect inhibition? What kind?
Also include in the discussion:
1. A general description of the structure and function of enzymes.
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2. Which is the limiting factor in a reaction with low amounts of substrate: enzyme or
substrate? Why?
3. Which is the limiting factor in a reaction with high amounts of substrate: enzyme or
substrate? Why?
4. How do your answers to questions 2 and 3 relate to the control of enzyme reactions
in cells?