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) |
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| 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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| AP LAB PAGE |
