|Lab 2 Enzyme Catalysis|
The human body produces many things to keep it alive and healthy. Enzymes are proteins produced by living cells. Enzyme-catalysis binds with the active site of an enzyme, reducing the amount of energy needed to have a reaction with the substrate. Catalysis is a substance that lowers reaction energy and allows the reaction to take place in less time and at lower temperatures. Without catalysis people would die from poisons that the body produces, but would not be able to break down. Catalysis does not break down during a reaction so can be used reversibly (E is enzyme, S is substrate, and P is Product):
Even with catalysis, enzyme reactions can be affected by many factors: salt concentration, pH, temperature, substrate concentration, enzyme concentration, activators, and inhibitors. Salt concentration affects the enzyme if it is to high or to low. Not enough salt will cause the enzyme side chains to become attracted to each other and denaturalizes it. Too much salt blocks the action site of the enzyme. The pH of a substance is determined by the amount of hydrogen (H+) in it. The scale ranges from 0-14: 0-7 is acidic, 7 is neutral, and 7-14 is basic. If the pH is too basic, the enzyme gains (H+) and eventually denaturalizes. If the pH is too acidic, then the enzyme loses (H+) and becomes denaturalized. The ideal pH is between 6 and 8. Temperature affects the kinetic energy that causes the reaction to speed up or slow down. In general, the higher the temperature the faster the molecular reaction. If the enzyme raises to a temperature above its optimum level, the tertiary structure of the protein is destroyed (denaturing it). Most enzymes denaturalize around 40-50°C. The law of mass action states the direction of enzyme-catalyzed reaction is dependent on conservation of enzyme/substrate/product. For example, when the substance has high substrate and low product, the substrate is used and more products are made. When the product is high and enzyme low, the reaction reverses and produces more substrate. If the product is immediately metabolized or removed from the cell, there will be no substrate formed. Activators increase the rate of a reaction while inhibitors slow the rate of the reaction. Inhibitors unfold an enzyme or reduce the –S-S- chains that stabilize the enzyme’s structure. Some enzyme inhibitors are poisons like potassium cyanide and curare.
Even without catalase, a reaction will still occur, but slower. The study of kinetics helps to determine the amount of product or substrate formed.
Chemical reactions occur over periods of time. The first three minutes of the reaction, the rate of change stays about the same. After a while, when there is less substrate, the reaction slows down and the rate of change becomes less. To compare the change of kinetic energy between reactions, a common point must be obtained. The first part of the reaction is called the initial rate of change. The initial rate of any enzyme-catalyzed reaction can be determined by the characteristics of the enzyme molecule. This is always the same for any enzyme and substrate at the same temperature and pH, but the substrate must have an excessive amount.
Chemical reactions can be studied by measuring the disappearance of the substrate, the rate of appearance by the product, or measuring the release or the absorbence of heat. For example, hydrogen peroxide (H2O2) is converted to water (H2O) and oxygen (O2) gas. Catalysis speeds up the reaction and sulfuric acid (H2SO4) stops the reaction by lowering the pH and denaturalizing the enzyme. Potassium permanganate (KmnO4) measure the presence of H2O2:
5 H2O2 + 2 KmnO4 +3 H2SO4 → K2SO4 + 2 MnSO4 + 8 H2O + 5 O2
After a certain amount of KmnO4 is added and the substance reaches a permanent brown or pink, no more KmnO4 should be added because it can mot be broken down.
Under perfect conditions, the rate of enzyme-catalysis should denature most of the hydrogen peroxide in a short amount of time.
In Part 1, 10 mL of 1.5% of H2O2, a 50 mL beaker, and 1 mL of catalysis are needed. In Part 2, 5mL of catalysis, a water bath, and 10 mL of 1.5% H2O2 is needed. In Part 3, a 1 cm³ of liver, 50 mL beaker, and 10 mL of 1.5% H2O2. For all three parts, safety goggles, lab aprons, pencil, paper, erasers, and paper towels are needed.
To do this experiment, 10 mL of 1.5% H2O2, 1 mL of water, 10 mL of H2SO4, 50 mL beaker, 25 mL beaker, 5 mL syringe, and KmnO4 are needed. Safety goggles, lab aprons, pencil, paper, erasers, and paper towels are also needed.
To do this exercise, safety goggles, lab aprons, pencil, paper, erasers, paper towels, about 20 mL of 1.5% H2O2, 1 mL of H2O, 10 mL of H2SO4, 50 mL beaker, 25 mL beaker, 5 mL syringe, and KmnO4 are needed.
To do this experiment, about 60 mL of 1.5% H2O2, 6 mL of catalysis, 60 mL of H2SO4, 12 cups labeled 10, 30, 60, 120, 180, and 360 seconds, six cups labeled acid, a black marker, and a timer are needed. Safety goggles, lab aprons, pencil, paper, erasers, and paper towels will be needed, also.
Methods: * Remember to wear the goggles and apron. *
In Part 1, transfer 10 mL of 1.5% H2O2 a 50 mL glass beaker and add 1 mL of freshly made catalase to the solution. Remember to keep the catalase solution on ice at all times. Record the observations made. In Part 2, transfer 5 mL of the purified catalase extract to a test tube and place it in a boiling water bath for five minutes. Next, transfer 10 mL of 1.5% H2O2 into a 50 mL beaker and add 1 mL of the cooled, boiled catalase solution. Again record the results. In Part 3, cut 1 cm of liver, transfer it into a 50 mL glass beaker containing 10 mL of 1.5% H2O2, and mix it. Record the results.
To form a baseline for this experiment, put 10 ml of 1.5% H2O2 into a clean glass beaker. Add 1 mL of H2O and then add 10 mL of H2SO4 (1.0 M). Be careful when using acid. Mix this solution well. Remove a 5 mL sample and place it into another beaker. Assay for the amount of H2O2 as follows. Place the beaker containing the sample over white paper and use a 5 mL syringe to add one drop of KMnO4 at a time to the solution until it becomes a persistent pink or brown color. Gently swirl the solution after adding each drop. Record all results.
To determine the rate of spontaneous conversion of H2O2 to H2O and O2 in an uncatalyzed reaction, put about 20 mL of 1.5% H2O2 in a beaker. Store it uncovered at room temperature for approximately 24 hours. Put 10 ml of 1.5% H2O2 into a clean glass beaker (using the uncatalyzed H2O2 that set out). Add 1 mL of H2O and then add 10 mL of H2SO4 (1.0 M). Be careful when using acid. Mix this solution well. Remove a 5 mL sample and place it into another beaker. Assay for the amount of H2O2 as follows. Place the beaker containing the sample over white paper and use a 5 mL syringe to add one drop of KMnO4 at a time to the solution until it becomes a persistent pink or brown color. Gently swirl the solution after adding each drop. Record all results.
If a day or more has passed since Exercise B was performed, it is necessary to reestablish the baseline. Repeat the assay from Exercise B and record the results. Compare with other groups to check that results are similar. To determine the course of an enzymatic reaction, how much substrate is disappearing over time must be measured. The first thing to be done is to set up the cups labeled with times and acid. Add 10 mL of H2SO4 to each of the cups marked acid. Then put 10 mL of 1.5% H2O2 into the cup marked 10 sec. Add 1 mL of catalase extract to this cup. Swirl gently for 10 seconds (use the timer for accuracy). At 10 seconds, add the contents of one of the acid filled cups. Remove 5 mL and place in the second cup marked 10 sec. Assay the 5 mL sample by adding one drop of KMnO4 at a time until the solution turns a pink or brown. Repeat the above steps except allow the reactions to proceed for 30, 60, 120, 180, and 360 seconds, respectively. Use the times corresponding with the marked cups. Record all results and observations.
Table 1 Catalysis Activity
|H 2O2 and Fresh Catalase||Small amount of bubbles|
|H 2O2 and Boiled Catalase||Little or no bubbling|
|Catalase with Liver||Very bubbly or reactive|
Table 2 Baseline Assay
|Final Reading of Burette||1.5 mL|
|Initial Reading of Burette||5.0 mL|
|Baseline (Final-Initial)||3.4 mL of KmnO4|
Table 3 The Uncatalyzed Rate of H2O2 Decomposition
|Final Reading of Burette||2.2 mL|
|Initial Reading of Burette||7.0 mL|
|Amount of KMnO4 Titrant||4.8 mL|
|H2O2 Spontaneously Decomposed||1.3 mL|
|Percentage Spontaneously Decomposed in 24 hours||62.9%|
Table 4 New Baseline
|Final Reading of Burette||1.4 mL|
|Initial Reading of Burette||5.0 mL|
|Baseline (Final-Initial)||3.6 mL of KmnO4|
Table 5 Catalyzed Rate of H2O2 Decomposition
|3.6 mL||3.6 mL||3.6 mL||3.6 mL||3.6 mL||3.6 mL|
B. Final Reading
|1.2 mL||1.4 mL||1.8 mL||1.9 mL||2.4 mL||2.8 mL|
C. Initial Reading
|5 mL||5 mL||5 mL||5 mL||5 mL||5 mL|
D. Amount of KmnO4 Consumed (B-C)
|3.8 mL||3.6 mL||3.2 mL||3.1 mL||2.6 mL||2.2 mL|
E. Amount of H2O2 Used (A-D)
|.2 mL||0 mL||.4 mL||.5 mL||1.0 mL||1.4 mL|
Graph 1 Affect of Time on Enzyme-Catalyzed H2O2 (Remaining amount)
(a) What is the enzyme in this reaction?
The enzyme in the reaction is catalase.
(b) What is the substrate in this reaction?
The substrate in the reaction is hydrogen peroxide.
(c) What is the product in this reaction?
The products in the reaction are water and oxygen gas.
(d) How could you show that the gas evolved is O2?
The formula 2 H202 + catalase →2 H2O + O2 proves that water and oxygen gas can only be produced.
How does the reaction compare to the one using unboiled catalysis? Explain the reason for the difference.
The boiled catalysis was not as reactive as the regular catalysis, because boiling the catalysis denatures it.
What do you observe? What do you think would happen if the liver was boiled before being added to the H2O2?
The liver has a high amount of catalase in it causing it to be very reactive when put with hydrogen peroxide. If the liver was boiled first, the catalysis would have been denatured and would not have reacted as much as previously.
1) From the formula described earlier recall that rate = G y/G x. Determine the initial rate of the reaction and the rates between each of the time points. Record the rates in the table below.
|Time Intervals (seconds)|
2) When is the rate the highest? Explain why.
The rate is the highest at initial to 10, because of the high concentration of catalysis.
When is the rate the lowest? For what reason is the rate low?
The rate is the lowest at 30 to 60 seconds, because the concentration of calase and the concentration of the product are beginning to balance each other out.
Explain the inhibiting effect of sulfuric acid on the function of the catalysis. Relate this to enzyme structure and chemistry.
The sulfuric acid changes the pH of the catalase function and causes it to denature. Most enzymes work in a range of 6 to 8 and by adding the acid, the pH drops too low.
Predict the effect of lowering the temperature would have on the rate of the enzyme activity. Explain your prediction.
Lowering the temperature would slow the reaction. If the temperature is lowered a great deal (below 40ºC) it will be denatured.
Design a controlled experiment to test the effect of varying pH, temperature, or enzyme concentration.
To test the effect of temperature on enzymes: put 5 mL of catalase in the freezer until it is completely frozen, add 1 mL of catalase to 10 mL of 1.5% H2O2 (which is in a 50 mL beaker). Watch and record results.
Error of Analysis:
Errors in this experiment could have come from inaccurate measurements and timing. Also the catalase not being frozen when received most likely affected the data.
Discussion and Conclusion:
The purpose of this lab was to show the decomposition of hydrogen peroxide under different circumstances. Exercise 2A showed the affects of catalysis (added to and from living cells) in hydrogen peroxide. In Exercise 2B, the baseline was determined for the experiment (3.5). In Exercise 2C, the natural decomposition of hydrogen peroxide was viewed and found to be slower that when catalyzed. In Exercise 2D, a new baseline was made (3.6) and the decomposition of hydrogen peroxide with a catalase, over six minutes, was discovered to decompose more rapidly than an uncatalyzed reaction over 24 hours.