Investigation 13 T215 Big Idea Interactions 4 investigation 13 enZYMe aCtivitY* How do abiotic or biotic factors influence the rates of enzymatic reactions? Experiment to Demonstrate the Action of Amylase on Starch Solution Plan ==== Scientific knowledge: Substances called catalysts speed up many chemical reactions. Factors affecting enzyme activity. Enzymes are sophisticated catalysts for biological processes. These practicals (and the practicals at intermediate level) give you. Plot the enzyme activity versus pH. From this curve, what is the optimal pH? Explain why enzyme activities depend on the pH. Similarly plot the enzyme. They are biochemical catalysts meaning they lower the activation energy needed for a biochemical reaction to occur. Because of enzyme activity, cells can carry out complex chemical activities at relatively low temperatures. The substrate is the substance acted upon in an enzyme- catalyzed reaction, and it can bind reversibly to the active site of the enzyme. The active site is the portion of the enzyme that interacts with the substrate so that any substrate that blocks or changes the shape of the active sit effects the activity of the enzyme. The result of this temporary union is a reduction in the amount of energy required to activate the reaction of the substrate molecule so that products are formed. The following equation demonstrates this process: E + S . Therefore, the enzyme is not changed in the reaction and can be recycled to break down additional substrate molecules. Several factors can affect the action of an enzyme: salt concentration, p. H of the environment, temperature, activations and inhibitors. If salt concentration is close to zero, the changed amino acid side chains of the enzyme molecules will attract one another. The enzyme will then denature and form an inactive precipitate. Denaturation occurs when excess heat destroys the tertiary structure of proteins. This usually occurs at 4. If salt concentration is high, the normal interaction of charged groups will be blocked. The p. H scale is a logarithmic scale that measures the acidity or H+ concentration in a solution and runs from 0 to 1. Amino acid side chains contain groups such as –COOH that readily gain or lose H+ ions. As the p. H is lowered an enzyme will tend to gain H+ ions, disrupting the enzyme’s shape. If the p. H is raised, the enzyme will lose H+ ions and eventually lose its active shape. Reactions usually perform optimally in neutral environments. Chemical reactions generally speed up as the temperature is raised. More of the reacting molecules have enough kinetic energy to undergo the reaction as the temperature increases. However, if the temperature goes above the temperature optimum, the conformation of the enzyme molecules is disrupted. An activator is a coenzyme that increases the rate of the reaction and can regulate how fast the enzyme acts. It should be clear that if you can increase reaction rates by increasing temperature you can decrease reaction rates by lowering the temperature.The 8 step cancer self-treatment program presented here provides alternative treatments to prevent & remove cancer and build your health. A Selected Chronological Bibliography of Biology and Medicine — Part II Ca. 1810 — 1884 Compiled by James Southworth Steen, Ph.D. Delta State University. A BBC Bitesize secondary school revision resource for Standard Grade Biology on enzymes and aerobic respiration: synthesis, degradation, lock and key. It also makes the active site a better fit for the substrate. An inhibitor has the same power of activator regulation but decrease the reaction rate. An inhibitor also reduces the number of S- S bridges and reacts with the side chains near activation sites, blocking them. The enzyme used in this lab is catalase. It has four polypeptide chains that are each composed of more than 5. One catalase function is to prevent the accumulation of toxic levels of hydrogen peroxide formed as a by- product of metabolic processes. Many oxidation reactions that occur in cells involve catalase. The following is the primary reaction catalyzed by catalase, the decomposition of hydrogen peroxide to form water and oxygen: 2 H2. O2 . Catalase speeds up the reaction notably. The direction of an enzyme- catalyzed reaction is directly dependent on the concentration of enzyme, substrate, and product. For example, lots of substrate with a little product makes more product. Another example is lots of product with a little enzyme forms more substrate. Much can be learned about enzymes by studying the kinetics of enzyme- catalyzed reaction. 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. Hypothesis: Enzyme catalase, when working under optimum conditions, noticeably increases the rate of hydrogen peroxide decomposition. Materials: Exercise 2. AThe materials needed for exercise 2. A of the lab are: 3. L of 1. 5% (0. 4. M) H2. O2, a 5. 0- m. L glass beaker, 6 m. L of freshly made catalase solution, a test tube, boiling water bath, 1 cm. Record the results. Transfer 1. 0 m. L of 1. H2. O2 into a 5. 0- m. L beaker and add 1 m. L of the cooled, boiled catalase solution. Again record the results. To demonstrate the presence of catalase in living tissue, cut 1 cm of liver, macerate it, and transfer it into a 5. L glass beaker containing 1. L of 1. 5% H2. O2. Add 1 m. L of H2. O. Add 1. 0 m. L of H2. SO4 (1. 0 M) using extreme caution. Mix this solution well. Remove a 5 m. L sample and place it into another beaker. Assay for the amount of H2. O2 as follows. Place the beaker containing the sample over white paper. Use a 5- m. L syringe to add KMn. O4 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 all results. Check with another group before proceeding to see that results are similar. Exercise 2. CTo determine the rate of spontaneous conversion of H2. O2 to H2. O and O2 in an uncatalyzed reaction, put about 2. L of 1. 5% H2. O2 in a beaker. Store it uncovered at room temperature for approximately 2. Record the results. Exercise 2. DIf a day or more has passed since Exercise B was performed, it is necessary to reestablish the baseline. Repeat the assay and record the results. To determine the course of an enzymatic reaction, how much substrate is disappearing over time must be measured. First, set up the cups with the times and the word acid up. Then put 1. 0 m. L of 1. H2. O2 into the cup marked 1. Add 1 m. L of catalase extract to this cup. Swirl gently for 1. Remove 5 m. L and place in the second cup marked 1. Assay the 5- m. L sample by adding KMn. O4 a drop at a time until the solution obtains a pink or brown color. Repeat the above steps except allow the reactions to proceed for 3. Use the times’ corresponding, marked cups. Record all results and observations. Results: Table 1: Test of Catalysis Activity. Experiment. Observations. Hydrogen Peroxide + Fresh Catalase. Bubbling in solution with the release of oxygen. Hydrogen Peroxide + Boiled Catalase. No reaction occurred. Hydrogen Peroxide + Liver. Much bubbling in solution with the release of O2. Table 2: Establishing a Baseline #1. Baseline Calculations (syringe contains KMn. O4)Readings. Final Reading of Syringe. LInitial Reading of Syringe. LBaseline. 3. 8. Table 3: Uncatalyzed H2. O2 Decomposition(Syringes Contain KMn. O4)Results. Final Reading of Syringe. LInitial Reading of Syringe. LAmount of H2. O2 Spontaneously Decomposed. LPercent of H2. O2 Spontaneously Decomposed in 2. Hours. 94. 3%Table 4: Establishing a Baseline #2. Baseline Calculations (syringe contains KMn. O4)Readings. Final Reading of Syringe. LInitial Reading of Syringe. LBaseline. 3. 5. Table 5: Time- Course Determination. Potassium Permanganate (m. L) 1. 03. 06. 01. Baseline. 3. 5. 3. Final Reading. 1. Initial Reading. 5. Amount of KMn. O4 Consumed. Amount of H2. O2 Used. Effect of Time on the Amount of H2. O2 Remaining after an Enzyme Catalyzed Reaction. Exercise 2. A: 1. What is the enzyme in this reaction? The enzyme in this reaction is the catalase solution. What is the substrate in this reaction? The substrate is hydrogen peroxide. What are the products in this reaction? The products are water and oxygen gas. How could you show that the gas evolved is oxygen? Referring to the equation 2. H2. O2 + Catalase solution. How does the reaction compare to the one using the unboiled catalase? With the boiled catalase, there was no sign of bubbling because the catalase was denatured by the heat and caused no reaction. What do you observe? I observe quite a bit of gas being released from the solution. What do you think would happen if the liver were boiled before being added to the hydrogen peroxide? I think that no signs of a reaction occurring would be shown. The catalase that occurs naturally within the liver would have been denatured. 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) Initial 0- 1. Rates. 37/1. 00- 3/2. When is the rate the highest? Explain why. The rate is the highest in the first ten seconds because the rate decreases as the concentration of the catalase decreases over time. When is the rate the lowest? For what reason is the rate low? The rate is lowest during the last time period of 3. The catalase concentration has been reduced and the product amount has increased, blocking the enzymes from reacting with the hydrogen peroxide. Explain the inhibiting effect of sulfuric acid on the function of the catalysis. Relate this to enzyme structure and chemistry. The sulfuric acid’s high concentration of H+ ions gives the acid a low p. H. Predict the effect of lowering the temperature would have on the rate of the enzyme activity. Explain your prediction. Enzymes generally only work at the between the temperatures of forty and fifty degrees Celsius. Design a controlled experiment to test the effect of varying p. H, temperature, or enzyme concentration. Part One (the effects of a strong acid on enzyme activity): Add 1. L of 1. 5- % hydrogen peroxide to a 5. L beaker, and add 1 m. L of catalase solution. Mix well and then add 1 m. L of (0. 5 M) HCl to the beaker. Observe the reaction and record the results. Part Two (the effects of a neutral solution on enzyme activity): Add 1. L of 1. 5- % hydrogen peroxide to a 5. L beaker, and add 1 m. L of catalase solution. Mix well and then add 1 m. L of pure water with a p. H of 7. 0. Observe the reaction and record the results. Part Three (the effects of a strong base on enzyme activity): Add 1. L of 1. 5- % hydrogen peroxide to a 5. L beaker, and add 1 m. L of catalase solution. Mix well and then add 1 m. L of (0. 5 M) Na. OH to the beaker. Observe the reaction and record the results. Error Analysis: Several errors could have occurred throughout the experiment. Miscalculations involving numbers and amounts of solutions would have a severe effect upon the results. Mathematical errors may also have of occurred. When the catalase arrived, it had melted.
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