Enzyme
Catalysis
Introduction:
In general,
enzymes are proteins produced by living cells, they act as catalysts in
biochemical reactions. A catalyst affects the
rate of a chemical reaction. One consequence of enzyme
activity is that cells can carry out complex chemical activities at relative
low temperatures. In an enzyme-catalyzed reaction, the
substance to be acted upon ( the substrate = S ) binds reversibly to the
active site of the enzyme (E). One
result of this temporary union is a reduction in the energy required to
activate the reaction of the substrate molecule so that the products (P)
of the reaction are formed.
In summary: E + S ---> ES --> E + P
Note
that the enzyme is not changed in the reaction and can even be recycled to
break down additional substrate molecules. Each enzyme
is specific for a particular reaction because its amino acid sequence is unique
and causes it top have a unique three-dimensional structure. The
active site is the portion of the enzyme that interacts with the
substrate, so that any substance that blocks or changes the shape of the active
site affects the activity of the enzyme. A description
of several ways enzyme action may be affected follows:
1. Salt
Concentration. If the salt concentration is close
to zero, the charged amino acid side chains of the enzyme molecules will
attract to each other. The enzyme will denature and
form an inactive precipitate. If, on the other hand,
the salt concentration is too high, normal interaction of charged groups will
be blocked, new interactions will occur, and again the enzyme will precipitate. An intermediate salt concentration such as that of human
blood (0.9% ) or cytoplasm ins the optimum for many
enzymes.
2. pH. Amino acid side chains contain groups such as - COOH and
NH2 that readily gain or lose H+ ions. As
the pH is lowered an enzyme will tend to gain H+ ions, and
eventually enough side chains will be affected so the enzyme's shape is
disrupted. Likewise, as the pH is raised, the enzymes
will lose H+ ions and eventually lose its active shape. Many of the enzymes function properly in the neutral pH
range and are denatured at either an extremely high or low pH.
Some enzymes, such as pepsin, which acts in the human stomach where the
pH is very low, have a low pH optimum.
3. Temperature. Generally, chemical reactions speed up as the temperature
is raised. As the temperature increases, more of the
reacting molecules have enough kinetic energy to undergo the reaction. Since enzymes are catalysts for chemical reactions, enzyme
reactions also tend to go faster with increase temperature. However,
if the temperature of an enzyme-catalyzed reaction is raised still further, a temperature
optimum is reached; above this value the kinetic energy of the enzyme and
water molecules is so great that the conformation of the enzyme molecules is
disrupted. The positive effect of speeding up the
reaction is now more than offset by the negative effect of changing the
conformation of more and more enzyme molecules. Many
proteins are denatured by temperatures around 40-50 degrees C, but some are
still active at 70-80 degrees C, and a few even withstand boiling.
4. Activation's and
Inhibitors. Many molecules other than the
substrate may interact with an enzyme. If such a
molecule increases the rate of the reaction it is an activator, or if it
decreases the reaction rate it is an inhibitor. These
molecules can regulate how fast the enzymes acts. Any
substance that tends to unfold the enzyme, such as an organic solvent or
detergent, will act as an inhibitor. Some inhibitors
act by reducing the -S-S- bridges that stabilize the enzyme's structure. Many inhibitors act by reacting with the side chains in or
near the active site to change its shape or block it. Many
well known poisons such as potassium-cyanide and curare are enzyme inhibitors
that interfere with the active site of critical enzymes.
The enzyme used in
this lab, catalase, has four polypeptide chains, each composed of more than 500
amino acids. This enzyme is ubiquitous in aerobic
organisms. One function of catalase within cells is to
prevent the accumulation of toxic levels of hydrogen peroxide formed as a
by-product of metabolic processes. Catalase might also
take part in some of the many oxidation reactions that occur in the cell.
2H2O -------> 2 H2O
+ O2 (gas )
In the absence of
catalase, this reaction occurs spontaneously, but very slowly.
Catalase speeds up the reaction considerably. In
this experiment, a rate for this reaction will be determined.
Much can be learned about enzymes by studying the kinetics of
enzyme-catalyzed reactions. For example, it is
possible to measure the amount of product formed, or the amount of substrate
used, from the moment the reactants are brought together until the reaction has
stopped. If the amount of product formed is measured
at regular intervals and this quantity is plotted on a graph, a curve like the
one that follows is obtained.
Figure 2.1 Enzyme Activity

Study the solid line
of the graph of this reaction. At time 0 there is no
product. As time progresses the production of product
increases at a steady rate. After a period of time
this rate slows down and at a certain point the reaction rate is very slow.
General
Procedure:
In this experiment the
disappearance of the substrate, H2O2, is measured as
follows:
1. A purified catalase
extract is mixed with substrate ( H2O2) in a beaker. The enzyme catalyzes the conversion of H2O
to H2O and O2 (gas ).
2. Before all the H2O2
is converted to H2O and O2 , the reaction is stopped by
adding sulfuric acid ( H2SO4 ). The
sulfuric acid lowers the pH, denatures the enzyme, and thereby stops the
enzyme's catalytic activity.
3. After the reaction
is stopped, the amount of substrate (H2O2) remaining in
the beaker is measured. To measure this quantity,
potassium permanganate is used. Potassium permanganate
(KMnO4), in the presence of H2O2 and H2SO4
reacts as follows:
5 H2O2 +
2 KMnO4 + 3 H2SO4 --------------> K2SO4
+ 2 MnSO4 + 8 H2O + 5 O2
Note that H2O2
is a reactant for this reaction. Once all the H2O2
has reacted, any more KMnO4 added will be in excess and will not be
decomposed. The addition of excess KMnO4
causes the solution to have a permanent pink or brown color. Therefore,
the amount of H2O2 remaining is determined by adding KMnO4,
until the whole mixture stays a faint pink or brown, permanently. Add no more KMnO4 after this point.
Figure 2.2 The General Procedure for the above exercise and
Exercise 2C.
The figure below represents the complete Exercise 2C.

Exercise
2A: Test of Catalase Activity:
1. To observe
the reaction to be studied, transfer 10 mL of 1.5% (0.44M) H2O2
into a 50 ml glass beaker and add 1 mL of freshly made catalase solution. The bubbles coming from the reaction mixture are oxygen,
which results from the breakdown of H2O2 by catalase. Be sure to keep the freshly made catalase solution on ice
at all times.
a. what is the
enzyme in this reaction? ____________________________________________________
b. What is the
substrate in this reaction? ___________________________________________________
c. What is the
product in this reaction? ____________________________________________________
d. How could you
show that the gas evolved is oxygen ? _____________________________________
2. To demonstrate the
effect of boiling on enzymatic activity, transfer 5 mL of purified catalase
extract to a test tube and place it in a boiling water bath for five minutes. Transfer 10 mL of 1.5% H2O2 into a
50 mL glass beaker and add 1 mL of the cooked, boiled catalase solution. How does the reaction compare to the one using the unboiled catalase? Explain the
reason for this difference.
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3. To demonstrate the
presence of catalase in living tissue, cut 1 cm3 of potato or liver,
macerate it, and transfer it into a 50 mL beaker containing 1.5% H2O2
. What do you observe? What
do you think would happen if the potato or liver was boiled before being added
to the H2O2?
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Exercise
2B: The Baseline Assay:
To determine
the amount of H2O2 initially present in a 1.5% solution,
one needs to perform all the steps of the procedure without adding catalase to
the reaction mixture. This amount is known as the
baseline and is an index of the initial concentration of H2O2
un solution. In any series of experiments, a baseline
should be established first.
Procedure
for Establishing Baseline:
1. Put 10 mL
of 1.5% H2O2 into a clean glass beaker.
2. Add 1 mL of H2O
( instead of enzyme solution).
3. Add 10 mL of H2SO4
(1.0 M) Use Extreme care in Handling Acids.
4. Mix well.
5. Remove a 5 mL
sample. Place this 5 mL sample in another beaker,
and assay for the amount of H2O2 as follows: Place the
beaker containing the sample over white paper. Use a
burette or 5 mL pipette to add potassium permanganate a drop at a time to the
solution until a persistent pink or brown color is obtained. Remember
to gently swirl the solution after adding each drop. Record
your data below.
Baseline calculations
Final Reading of burette ________ mL
Initial reading of burette ________mL
Baseline (Final -Initial) _________mL KMnO4
Figure 2.4: Proper Reading of a Burette

Exercise 2C: An
Enzyme-Catalyzed Rate of H2O2 Decomposition
Refer to figure 2.2 to
complete this section and record the data in Table 2.1 below.
Table 2.1
|
Potassium
Permanganate (ml) |
Time
(Seconds) |
|||||
|
|
10 |
30 |
60 |
120 |
180 |
360 |
|
A. Baseline |
|
|
|
|
|
|
|
B. Final Reading |
|
|
|
|
|
|
|
C. Initial Reading |
|
|
|
|
|
|
|
D. Amount of KMnO4
consumed (B-C) |
|
|
|
|
|
|
|
E. Amount of H2O2 used (A-D) |
|
|
|
|
|
|
Graph the data for
enzyme-catalyzed H2O2 decomposition.
Graph Title:
___________________________________________________________________
Graph 2.1

Analysis of
Results:
1. Explain the inhibiting effect of sulfuric acid on the function of catalase. Relate this to enzyme structure and chemistry.
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2. Predict the
effect lowering the temperature would have on the rate of enzyme activity. Explain you prediction.
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3.
Design a controlled experiment to test the effect of varying pH, temperature,
or enzyme concentration.
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