Lab 2 Enzyme Catalysis 684lw

Lab Report |1

Observing Enzyme Catalysis and Measuring Rate of Reactions

AP Biology

Fall 2008 Enzymes are globular proteins that act as biological catalysts for biochemical processes within a living cell. Certain spontaneous chemical reactions may occur so slowly as to be

Lab Report |2 imperceptible. Catalysts act as chemical agents that speed up the rate of a reaction without being consumed in the reaction. In an enzyme-catalyzed reaction, the substance to be acted upon, the substrate, bins onto an enzyme. Enzymes have active sites, in which only a particular substrate can bind onto. This specificity allows for a great number and variation of enzymes. The substrate fits into the active site, and in response, the enzyme modifies it shape ever so slightly to maintain a firmer grip on its substrate. This temporary union allows for a reduction in the activation energy required to activate the reaction. Each reaction has an energy barrier, which requires it to overcome to break the existing bonds and form new ones. This reduction in activation energy allows the reaction to occur at a much quicker rate. Enzymes are never used nor changed in the reaction. Therefore, they can be recycled to break down additional substrate molecules. Being proteins, enzymes have a highly organized structure which dictates its specificity for substrates. The primary structure of an enzyme includes its amino acid arrangement. This arrangement then controls the secondary and tertiary structures, which are intermolecular bonds forming between amino acids and their side chains, creating a three-dimensional conformation. If the shape of the enzyme is denatured or interrupted, then the enzyme will not function properly. In addition to observing enzyme catalysis, this lab examines how different factors affect enzyme activity. Some of these variables include salt concentration, pH, temperature, enzyme and substrate concentration, and inhibitors. If salt concentration is near 0, then polar amino acid side chains will attract to each other, thus denaturing the shape of the protein. However, if salt concentration exceeds a maximum limit, normal interaction of these side chains will be blocked completely, and the enzyme still won’t function properly. Therefore, it is necessary for enzymes to be placed in an intermediate salt concentration, such as human blood or cell cytoplasm.

Lab Report |3 The pH of an enzyme’s environment measures the acidity, or hydrogen ion concentration, in a solution. If the pH is lowered, the side chains will begin picking up hydrogen ions from its environment, which might disrupt its function. Likewise, as pH is raised, enzymes will lose hydrogen ions and lose its active shape. Hence, maintaining a neutral pH is necessary for most enzymes. Of course, there are also exceptions with stomach enzymes, such as pepsin (pH 2). Temperature is also another factor affecting enzyme functionality. Usually, as the temperature increases, enzyme activity will also consequently increase. However, a temperature optimum will eventually be reached, where anything above that denatures the protein. Most enzymes are functional around 40-50˚ C, although some enzymes can tolerate even greater temperature ranges. Enzyme concentration and substrate concentration also play a role in enzymatic activity. The more enzymes available, the quicker the reaction will occur until the substrate is all used up. More substrates will also mean quicker activity, until the enzyme is fully saturated so that it cannot continue increasing its activity. Activators and inhibitors interact with an enzyme so that its activity is altered. If a molecule increases the rate of reaction, it is an activator, whereas an inhibitor decreases or stops the activity. These substances regulate how fast an enzyme acts. Inhibitors work by unfolding or destabilizing bonds, thus denaturing the enzyme. Some inhibitors block or change the shape of the active site. This lab focuses on one particular enzyme – catalase. Catalase has four polypeptide chains, each composed of more than 500 amino acids, all interacting with each other to produce a unique structure and function. Catalase plays an indispensable role in a living cell by preventing accumulation of toxic levels of hydrogen peroxide formed as a byproduct of metabolic pathways.

Lab Report |4 The primary reaction that catalase catalyzes is the decomposition of hydrogen peroxide to produce water and oxygen gas: 2 H2O2 → 2 H2O+ O2 The decomposition of catalase is a spontaneous reaction, although incredibly slow. Catalase speeds up the reaction considerably. In this experiment, a rate for this reaction will be determined. The purpose of this lab is to measure the effects of changes in temperature, pH, enzyme concentration, substrate concentration on reaction rates of an enzyme-catalyzed reaction. This experiment examines how catalase reacts with hydrogen peroxide and how environmental factors affect the rate of this reaction. This experiment also applies titration and laboratory skills to find how much hydrogen peroxide is initially present in a solution, how much hydrogen peroxide is remaining and decomposed in a solution, and how catalase plays such an important role in living cells.

MATERIALS AND METHODS In this experiment, the decomposition of H2O2 was measured as follows: A purified catalase extract was added to the substrate (H2O2 ) in a beaker, which catalyzes the formation of H₂O and O₂. Before all of the H₂O₂ was converted to H₂O and O₂, the reaction was stopped by adding sulfuric acid (H₂SO₄) to denature the enzyme, thus halting any enzymatic activity. After the reaction was stopped, the amount of H₂O₂ was measured by titration. Potassium permanganate (KMnO₄) was used for this procedure. By adding the potassium permanganate, the following reaction occurred: 5 H2O2+ 2 KMnO4+ 3 H2SO4→K2SO4+ 2 MnSO4+ 8 H2O+5 O2

Lab Report |5 Once all the H₂O₂ had reacted, any more KMnO₄ added would be in excess and therefore did not decompose. The addition of excess KMnO₄ caused the solution to have a permanent pink or brown tint. After this had been obtained, no more KMnO₄ was added. The amount of KMnO₄ added was proportional to the measure of the amount of H₂O₂ remaining, therefore allowing us to calculate how much H₂O₂ had been remaining, and how much had been decomposed. This, however, is only a general overview of the detailed experiment. The following explains in further detail. To observe a general reaction of catalase, 10 mL of 1.5% (0.44 M) H₂O₂ was added into a 50-ml glass beaker. Into there, a 1 mL extract of fresh catalase was added. To demonstrate factors of enzymatic activity, 5 mL of catalase was boiled and then cooled. About 1 mL of this was added to another 10 mL of H₂O₂. Furthermore, catalase was exposed to both an acid (HCl) and basic (NaOH) environment to see how it functioned. This part of the lab, Exercise 2A, gave an insight on how catalase functioned. Exercise 2B involved determining the amount of H₂O₂ initially present in a 1.5% solution. This amount is known as the base line and is the basis for future procedures. Taking a 1.5% solution, 10 mL of H₂O₂ was placed in a clean glass beaker. Instead of enzyme solution, 1 mL of distilled water was added. Then, 10 mL of H₂SO₄ was added, forming a 21 mL solution. After mixing well, a 5 mL sample was extracted into another beaker for titration. A magnetic stirring was placed into the beaker, which was in turn placed on a hot plate. Using a burette, KMnO₄ was added one drop at a time. After measuring the KMnO₄ used, the amount of H₂O₂ was calculated. Another trial of this procedure was conducted to avoid errors. The following part of the lab called for examining how H₂O₂ decomposed spontaneously. About 15 mL of H₂O₂ was placed in a beaker. It was stored uncovered at room temperature for approximately 24 hours. From the remaining solution, a 5 mL extract was titrated to find the

Lab Report |6 amount of H₂O₂ spontaneously decomposed. This was also repeated several times to ensure accuracy. The final portion of the lab observed the rate of enzyme-catalyzed H₂O₂ decomposition. To observe the rate of reaction, the amount of substrate disappearing was measure over certain periods of time. For this procedure, 7 clean cups were needed. In each of these cups, 10 mL of 1.5% H₂O₂ was added by a syringe. Along with that, 1 mL of catalase was added into each cup one at a time, using another clean syringe. For example, one cup was left alone with catalase for 10 seconds. After 10 seconds, all enzymatic activity was halted with the addition of 10 mL of sulfuric acid, which was added by yet another clean syringe. Another cup had the 1 mL catalase extracted for 30 seconds. This procedure was consistent for 0, 10, 30, 60, 90, 120, and 180 seconds of catalase activity. From each cup, a 5 mL sample was removed to be titrated. This allowed us to find out how much H₂O₂ had been remaining after that set time. After this procedure was completed for all 7 times, it was repeated to ensure maintaining proper results.

RESULTS For Exercise 2A, an enzyme-catalyzed reaction was observed. The enzyme in the reaction was catalase. The substrate acted upon was hydrogen peroxide, which was broken down into water and oxygen gas. The release of the gas was proven as oxygen by a glowing splint test. When boiling catalase, the reaction does not occur as quickly, since the function of the enzyme has been altered. Potatoes also have catalase stored in them, as observed when bubbles were formed as pieces of potato were added to hydrogen peroxide. The initial amount of H₂O₂ in a 1.5% solution had to be calculated, which acted as the basis for the rest of the experiment. In order to find this, the molarity of 2% KMnO₄, which was used as the titrant, had to be measured as well. The following calculations were made:

Lab Report |7 2% KMnO₄ = 2 grams KMnO₄100 grams H₂O = 2 grams KMnO₄100 mL H₂O = 20 grams KMnO₄1000 mL H₂O 20 grams KMnO4 1 mole KMnO₄ 1 L 158.04 grams KMnO₄ = 0.127 moles KMnO₄1 L 0.127 M (moles/liter) = 2% KMnO₄ The baseline measurement was calculated as follows: 3.70 mL KMnO₄ used to titrate 1.5% H₂O₂ 3.70 mL KMnO₄ 0.127 moles 1000 mL KMnO₄ = 0.000470 moles KMnO₄ .000470 moles KMnO₄ 5 moles H₂O₂ 2 moles KMnO₄ = 0.00118 moles of H₂O₂ or 1180 μmoles Baseline = 1180 μmoles For Exercise 2C, the spontaneous conversion of H₂O₂ into H₂O and O₂ was observed in an uncatalyzed reaction. The following calculations were made: 0.100 mL KMnO₄ used to titrate the solution 0.100 mL KMnO₄ 0.127 moles 1000 mL KMnO₄ = 0.0000127 moles KMnO₄ 0.0000127 moles KMnO₄ 5 moles H₂O₂ 2 moles KMnO₄ = 0.0000318 moles or 31.8 μmoles of H₂O₂ remains Amount of H₂O₂ spontaneously decomposed = Baseline - H₂O₂ remaining 1148 μmoles = 1180 μmoles – 31.8 μmoles After leaving H₂O₂ uncovered overnight, approximately 1148 μmoles or 97% of the initial amount was spontaneously decomposed without the aid of catalase. For Exercise 2D, data was obtained by manipulating the time for catalase activity to see how time would correlate to H₂O₂ decomposition. The following calculations were used to determine the moles of H₂O₂ remaining from the catalyzed reaction, and how much had been decomposed by the catalase:

Lab Report |8 0 seconds – 3.70 mL of KMnO₄ was used as titrant 3.70 mL KMnO₄ 0.127 moles KmnO₄ 5 moles H₂O₂ 1000 mL KMnO₄ 2 moles KMnO₄ = 0.00118 moles H₂O₂ remaining = 1180 μmoles remaining after 0 seconds of catalase activity Baseline - H₂O₂ remaining = H₂O₂ decomposed 1180 μmoles - 1180 μmoles = 0 μmoles 0 μmoles H₂O₂ decomposed after 0 seconds of catalase activity

10 seconds – 3.00 mL of KMnO₄ used as titrant 3.00 mL KMnO₄ 0.127 moles KmnO₄ 5 moles H₂O₂ 1000 mL KMnO₄ 2 moles KMnO₄ = 0.000953 moles H₂O₂ remaining = 953 μmoles remaining after 10 seconds of catalase activity Baseline - H₂O₂ remaining = H₂O₂ decomposed 1180 μmoles - 953 μmoles = 227 μmoles 227 μmoles H₂O₂ decomposed after 10 seconds of catalase activity

30 seconds – 3.10 mL KMnO₄ used as titrant 3.10 mL KMnO₄ 0.127 moles KmnO₄ 5 moles H₂O₂ 1000 mL KMnO₄ 2 moles KMnO₄ = 0.000985 moles H₂O₂ remaining = 985 μmoles remaining after 30 seconds of catalase activity Baseline - H₂O₂ remaining = H₂O₂ decomposed 1180 μmoles - 985 μmoles = 195 μmoles 195 μmoles H₂O₂ decomposed after 30 seconds of catalase activity

Lab Report |9 60 seconds – 2.50 mL KMnO₄ used as titrant 2.50 mL KMnO₄ 0.127 moles KmnO₄ 5 moles H₂O₂ 1000 mL KMnO₄ 2 moles KMnO₄ = 0.000795 moles H₂O₂ remaining = 795 μmoles remaining after 60 seconds of catalase activity Baseline - H₂O₂ remaining = H₂O₂ decomposed 1180 μmoles - 795 μmoles = 385 μmoles 385 μmoles H₂O₂ decomposed after 60 seconds of catalase activity

90 seconds – 2.50 mL KMnO₄ used as titrant 2.50 mL KMnO₄ 0.127 moles KmnO₄ 5 moles H₂O₂ 1000 mL KMnO₄ 2 moles KMnO₄ = 0.000795 moles H₂O₂ remaining = 795 μmoles remaining after 90 seconds of catalase activity Baseline - H₂O₂ remaining = H₂O₂ decomposed 1180 μmoles - 795 μmoles = 385 μmoles 385 μmoles H₂O₂ decomposed after 90 seconds of catalase activity

120 seconds – 2.00 mL KMnO₄ used as titrant 2.00 mL KMnO₄ 0.127 moles KmnO₄ 5 moles H₂O₂ 1000 mL KMnO₄ 2 moles KMnO₄ = 0.000635 moles H₂O₂ remaining = 635 μmoles remaining after 120 seconds of catalase activity Baseline - H₂O₂ remaining = H₂O₂ decomposed 1180 μmoles - 635 μmoles = 545 μmoles 545 μmoles H₂O₂ decomposed after 120 seconds of catalase activity

L a b R e p o r t | 10 180 seconds -- 2.20 mL KMnO₄ used as titrant 2.20 mL KMnO₄ 0.127 moles KmnO₄ 5 moles H₂O₂ 1000 mL KMnO₄ 2 moles KMnO₄ = 0.000698 moles H₂O₂ remaining = 698 μmoles remaining after 180 seconds of catalase activity Baseline - H₂O₂ remaining = H₂O₂ decomposed 1180 μmoles - 635 μmoles = 482 μmoles

0 sec

10 sec

30 sec

60 sec

90 sec

120 sec

180 sec

Baseline (mL)

3.70

3.70

3.70

3.70

3.70

3.70

3.70

Baseline (μmoles)*

1180

1180

1180

1180

1180

1180

Final Reading (mL)

12.90

37.00

16.00

9.20

37.00

24.80

9.20

34.00

12.90

6.70

1180 18.5 0 16.0 0

35.00

22.60

3.70

3.00

3.10

2.50

2.50

2.00

2.20

0.00

0.70

0.60

1.20

1.20

1.70

1.50

0

227

195

385

385

545

482

Initial Reading (mL) Amount of KMnO₄ Consumed (mL) Amount of H₂O₂ Decomposed (mL) Amount of H₂O₂ Decomposed (μmoles)**

482 μmoles H₂O₂ decomposed after 180 seconds of catalase activity

Table 1. This table shows an overview for all of the results for this part of the lab. The manipulated variable was the time that the catalase was allowed to catalyze. The responding variable would be the amount of H₂O₂ remaining and decomposed from the reactions. *Calculations for this measurement are shown in the results for 2B (see Lab Report 7) **Calculations for these measurements are shown above (see Lab Report 8, 9, 10)

L a b R e p o r t | 11

Figure 1. This scatter plot depicts how catalase activity affected the decomposition of hydrogen peroxide. A line of best fit was added to show how catalase activity is supposed to catalyze H₂O₂ breakdown, and to explain that there were slight errors in the obtained data. However, the line of best fit is not completely accurate itself, as shown that it starts at 150 μmoles for 0 seconds of catalase activity, which is false.

Initial 0 to 10

10 to 30

30 to 60

60 to 90

90 to 120

120 to 180

22.7

-1.6

6.3

0

3.3

-1.05

Rates (μmoles/second)

Table 2. This table shows the rate of catalysis as was obtained from the results. The rate is highest in the beginning, but soon slows down and there are signs that it will level off. Because of some errors in the data, some rates are negative and do not make sense, although the rate is supposed to be decreasing over time.

DISCUSSION In Exercise 2A of this experiment, the degradation of H₂O₂ was observed by enzyme catalysis. The enzyme involved in this reaction was catalase, a very important enzyme in living cells, as discussed in the Introduction. The substrate acted upon was hydrogen peroxide (H₂O₂), which was broken down to form water (H₂O) and oxygen gas (O₂). The gas released by the reaction was proved to be oxygen by the glowing splint test. If it was oxygen, then the glowing splint would turn back into a flame because of the presence of oxygen. Since this did happen, the gas was shown to be oxygen. To demonstrate the effects of temperature change on enzymatic function, 5 mL of purified catalase was extracted into a test tube and placed in a hot water bath. This boiled catalase was cooled, and added to a test tube of 1.5% H₂O₂. However, very few oxygen bubbles formed, unlike that when unboiled catalase was used. The explanation for this is because temperature affects the enzymatic activity by

L a b R e p o r t | 12 changing the enzyme’s molecular conformation. This denaturization changes the shape of the enzyme’s active shape, and it no longer functions properly, as modeled by the boiled catalase. Another test tube was used to test if potato contained any catalase. As H₂O₂ was added into the test tube with the small piece of potato, oxygen bubbles formed and water filled the bottom of the test tube, showing that potato cells have catalase. Next, a piece of potato was mashed to see if it would have any effect on catalase activity. As it was added to the hydrogen peroxide in another test tube, the number of bubbles increased since more catalase was exposed to the H₂O₂. The pieces of potato were also subjected to an acid and base. Although both denatured the enzyme, few bubbles formed in the NaOH solution, whereas no reaction occurred in the HCL solution. This part of the experiment showed that strong acids and high heat denature enzymes, thus altering their function. If the potato had been boiled, the reaction would not have occurred as quickly as it had without any disruptions in environment. The baseline assay was calculated by finding how much KMnO₄ reacted with the H₂O₂ initially present in the 1.5% hydrogen peroxide. Since the amount of KMnO₄ consumed is proportional to the amount of H₂O₂, the baseline was calculated, in both milliliters and moles. For every 5 moles of H₂O₂ in the beaker, 2 moles of KMnO₄ reacted, hence a 2.5:1 ratio of hydrogen peroxide to potassium permanganate.

L a b R e p o r t | 13 In Exercise 2C of the experiment, the spontaneous reaction of H₂O₂ was observed. After left overnight, only 0.10 mL of KMnO₄ was needed before all of the H₂O₂ had been reacted with. This allowed us to conclude that very little H₂O₂ had remained, and most of it had been disintegrated. Nearly 97% of the initial hydrogen peroxide had decomposed spontaneously, showing that although catalase is not needed, it occurs very slowly. The main part of this experiment was to observe the rate of reactions for catalase-catalyzed reactions, allowed to occur for different times. The data table and graph present every piece of data obtained from this part of the experiment, although there were a few random errors, which will be discussed later on. Figure 1 shows how catalase activity correlated to the breakdown of H₂O₂. Although the values may not be increasing linearly, there is a positive correlation in the data. The independent variable is time, in seconds, of catalase activity, while the dependent variable is the amount or H₂O₂ consumed, in micromoles. The final part of the experiment was to analyze the rate of these reactions, and how this rate changed over the time of catalase activity. As Table 2 shows, the rate is highest from initial time to 10 seconds, with an average of 22.7 μmoles per second. The next rate is a bit off, since negative values are obtained from the data. However, it is certain that the rate is decreasing over time of catalase activity. It lowers to 6.3 μmoles per second, and then to 3.3 μmoles per second, showing that the reactions is slowing down. Despite some errors while obtaining data, the reaction rate is generally decreasing.

L a b R e p o r t | 14 Initially, the catalase is fresh and accepts as many substrates (H₂O₂) as possible into its active sites. This makes up for the exceptionally high rate of reaction. Eventually, the enzyme becomes saturated with substrates, and cannot continue to increase its rate of catalysis, so therefore cannot increase. As less substrate molecules become available, reaction rate slows down until it eventually levels off or reaches 0, when all substrate sources are depleted. This is when the rate is lowest, near the end of the reaction; in this case it’s supposed to be from 120 to 180 seconds. The inhibiting effect of sulfuric acid (H₂SO₄) is mainly because it lowers the pH of the enzyme. Sulfuric acid is a very strong acid, so it denatures the protein. If the pH is lowered, the side chains will begin picking up hydrogen ions from its environment, which might disrupt its conformation, ultimately inhibiting its function. The secondary and tertiary structures of the enzyme become altered, thus changing the behavior of the enzyme. Lowering the temperature will have a negative effect on an enzyme, since kinetic energy is needed to start the reaction. If it becomes too cold, the enzyme will denature and it will no longer function properly. This would restrict enzymatic activity, just as a temperature too high would. To test the effect of varying pH, I would create a controlled experiment where the only thing that is different is the pH level, which would be the independent value. There would be three test tubes, each with 5 mL fresh catalase. The first test tube would be left alone, while one of the other two will contain an acid, while another will have a base. About 10 mL hydrogen peroxide would be added to each of these test tubes to observe the reaction. This could be repeated in a similar fashion for temperature and enzyme concentration, by changing only one thing among the three test tubes. For example, the test tubes would be set up in the same fashion

L a b R e p o r t | 15 as described, except pH would all be consistent. The test tubes would vary in temperature: freezing, room temperature, and boiling. This also applies to enzyme concentration. Some errors were recognized in the results, which slightly messed up some of the calculations. However, these errors were too minor to greatly impact the results too profoundly and were considered acceptable, as they seemed to make sense most of the time. These errors could have been caused for multiple reasons. The catalase might have mixed with ice water, it might have come into with another substance, been contaminated, etc. Titration measurements may have slightly incorrect, if too much or too little KMnO₄ was used. Dirty or used syringes could have caused the problem or a wide assortment of other minor errors could have caused this random occurrence in the data. However, the data made sense for the most part, and the objective of the lab was reached.

SUMMARY Enzyme catalysis was observed in order to analyze how changes in temperature, pH, enzyme concentration, and substrate concentration affected an enzyme-catalyzed reaction. This experiment analyzed the rate of enzyme-catalyzed reactions and observed the correlation between catalase activity and products formed. It was found out that the rate of an enzymecatalyzed reaction starts off rapidly, decreases, and levels off or completely stops, and can be further affected by environmental factors, which play a crucial role in regulating enzymes and metabolic processes.

References

L a b R e p o r t | 16

Campbell, Neil A. 2005. Biology/Neil A. Campbell, Jane B. Reece. 7th ed. San Francisco, CA: Pearson Education, Inc. p. 80-5, 150-7

“Enzyme Catalysis.” 25 October 2008. http://kvhs.nbed.nb.cal/gallant/biology/labenz.html

Goldberg, Deborah T. 2007. Barron’s AP Biology. 2nd ed. New York, NY: Barron’s Educational series, Inc. p 409.

“LabBench Activity – Enzyme Catalysis.” 28 October 2008. http://www.phschool.com/science/biology_place/labbench/lab2/concepts.html

Pack, Phillip E. 2007. Cliffs AP Biology. 3rd ed. Hoboken, NJ: Wiley Publishing, Inc. p 260.