Lab 6:  Enzymes – Catalase Function

OBJECTIVES

• Practice and apply hypothesis testing.
• Practice experimental design.
• Gain a better understanding of enzymes and some conditions that affect enzyme activity and the rate of an enzyme-catalyzed reaction.
• Understand these terms: enzyme, enzyme activity, active site, substrate, enzyme-substrate complex, product, denature, competitive inhibition, and noncompetitive inhibition.

INTRODUCTION

Enzymes are proteins that increase the rate of chemical reactions.  This process is called catalysis.  All cells use enzymes for metabolism, the sum of all chemical and physical reactions that a cell uses to break down nutrients to produce energy and to build up important molecules needed for cell function.  Enzymes increase the rate of a reaction by lowering activation energy but are not themselves consumed in the reaction.  Thus, one enzyme can catalyze repeated rounds of the same reaction.

As with all proteins, an enzyme’s function depends completely on its shape.  Enzymes have a very complex three-dimensional structure consisting of one or more polypeptide chains folded to form an active site – a specific area into which the substrate (material to be acted on by the enzyme) will fit. The shape of the enzyme determines which reaction it will catalyze, because only one substrate will fit into its active site.

Changes in temperature, pH, and/or the addition of certain ions or molecules may affect the structure of an enzyme’s active site and thus the ability of the enzyme to catalyze the reaction (the enzyme’s “enzyme activity”).  Changing an enzyme’s shape so that it is no longer biologically active is called denaturation.  Enzyme inhibitors can occupy the active site (competitive inhibitors) or change the shape of the active site by binding elsewhere (noncompetitive inhibitors). Hence, these factors also affect the rate of the reaction in which the enzyme participates.  The rate of an enzymatic reaction can also be affected by the relative concentrations of enzyme and substrate.

Salt concentration: If the salt concentration is very low or zero, the charged amino acid side-chains of the enzyme will stick together. The enzyme will denature and form an inactive precipitate. If, on the other hand, the salt concentration is very high, normal interaction of charged groups will be blocked, new interactions occur, and again the enzyme will precipitate. An intermediate salt concentration such as that of blood (0.9%) or cytoplasm is optimum for most enzymes.

pH: pH is a logarithmic scale that measures the acidity or H+ concentration in a solution. As the pH is lowered, an enzyme will tend to gain H+ ions, and eventually enough side chains will be affected so that the enzyme’s shape is disrupted. Likewise, as the pH is raised, the enzyme will lose H+ ions and eventually lose its active shape. Many enzymes have an optimum in the neutral pH range and are denatured at either extremely high or low pH. Some enzymes, such as those which act in the human stomach where the pH is very low, will have an appropriately low pH optimum. A buffer is a compound that will gain or lose H+ ions so that the pH changes very little.

Temperature: All 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 increasing temperature. However, if the temperature of an enzyme-catalyzed reaction is raised still further, an optimum is reached: above this point the kinetic energy of the enzyme and water molecules is so great that the structure of the enzyme molecules starts to be disrupted. The positive effect of speeding up the reaction is now more than offset by the negative effect of denaturing more and more enzyme molecules. Many proteins are denatured by temperatures around 40-50C, but some are still active at 70-80C, and a few even withstand being boiled.

In this exercise you will study the enzyme catalase, which accelerates the breakdown of hydrogen peroxide (a common end product of oxidative metabolism) into water and oxygen, according to the summary reaction:

          2H₂O₂ + Catalase —-> 2H₂O + O₂ + Catalase

Catalase is found in animal and plant tissues, and is especially abundant in plant storage organs such as potato tubers, corms, and in the fleshy parts of fruits. You will isolate catalase from potato tubers and measure its rate of activity under different conditions.

The fact that one of the products of the above reaction (oxygen) forms a gas, it is convenient to observe the conversion from reactants to products. A small piece of filter paper can be soaked in a solution of catalase and immersed in a solution of hydrogen peroxide. It will immediately sink to the bottom of the vessel. As the catalase converts the hydrogen peroxide to water and oxygen, some of the oxygen gas accumulates on the disk, changing buoyancy of the disk and floating to the surface. The elapsed time from immersion to floatation is thus proportional to the rate of the reaction (the time required to produce sufficient product (O₂) to float the disk). This allows us to explore the effect of a number of variables on the rate of a reaction.

PROCEDURE

Part A : Extraction of Catalase

1. Peel a fresh potato tuber and cut the tissue into small cubes. Weigh out 50 g of tissue.
2. Place the tissue, 50 ml of cold distilled water, and a small amount of crushed ice in a pre-chilled blender.
3. Homogenize for 30 seconds at high speed. From this point on, the enzyme preparation must be carried out in an ice bath.
4. Filter the potato extract, then pour the filtrate into a 100-ml graduated cylinder. Add cold distilled water to bring up the final volume to 100 ml. Mix well. This extract will be arbitrarily labeled 100 units of enzyme per ml (100 units/ml). (NOTE – this will be enough catalase for all lab groups)

Materials:

Work in groups of 4.

1. Tape or permanent marker for labeling
2. Five 50 ml beakers or large test tubes.
3. Stock 3% H₂O₂
4. Catalase solution
5. No2 Whatman filter paper
6. Hole punchers
7. Beaker with water
8. Water baths at the following temperatures (4, 25, 37, 60^oC). These will be beakers on hot plates with temperature monitored by a thermometer.

PART B: Effect of Substrate Concentration

Before beginning, make a hypothesis and prediction about the scientific questions we are asking.

Q1. How does substrate concentration affect the speed at which catalase converts H₂O₂ into water and oxygen?

Ha (Alternative hypothesis); H0 (Null hypothesis)

Write your scientific prediction (remember to use if—then—).

What are independent and dependent variables in this experiment?

Methods

1. Place 20 ml stock 3% H₂O₂ in one of the beakers or test tubes and label. Make sure your label identifies your group and concentration of H₂O₂.
2. Dilute stock 3% H₂O₂ to make 1.5%, 0.75%, 0.38% and 0.19%. You should figure out how to do this by serial dilution of your 3% H₂O₂ stock. All beakers or test tubes should contain 10 ml except for the last beaker containing 0.19% H₂O₂.
3. Make 25 small filter paper disks with a hole punch.
4. With a forceps, grab a disk and immerse it in catalase solution for 3 seconds. BE PRECISE WITH THE TIME IT IS IMMERSED IN THE ENZYME. Immediately drop it into one of the solutions of H₂O₂ and start your timer. If you look closely, you will see bubbles of O₂ forming on the disk. Stop your timer the moment the paper disk rises to the surface. Remove the paper disk and repeat the process four more times.
5. Repeat step 4 for each of the concentration of H₂O₂.
6. Record your data in a table.

DATA ANALYSIS

Graph your results by hand in your lab notebook. Plot the independent variables on X axis, dependent variable on Y axis. Use the averages of your replicates for the graph.

PART C: The effect of temperature on enzyme function and the reaction

Before beginning, make some hypothesis and predictions about the scientific questions we are asking.

Q1. How does temperature affect the speed at which catalase converts hydrogen peroxide into water and oxygen?

Ha (Alternative hypothesis)

H0 (Null hypothesis)

Write your scientific prediction here (remember to use if—then—)

What are independent and dependent variables in this experiment?

Methods

1. Prepare 5 test tubes with 10 ml each of 1.5% H₂O₂solution, and label them to identify your group and temperature. Keep one of your test tubes immersed in water at room temperature, the others in water baths at 4^oC (on ice), 37^oC, and 60^oC, respectively.
2. Make 25 more filter paper disks.
3. Measure the reaction rate as you did in part 1, by dipping each filter paper into catalase for 3 seconds, then recording the time it takes for a disk to rise to the top of each test tube after dropped to the bottom. As in part 1, repeat this procedure with 5 filter paper per test tube.
4. Record your data in a table.

DATA ANALYSIS

Graph your results by hand in your lab notebook. Plot the independent variables on X axis, dependent variable on Y axis. Use the averages of your replicates for the graph.

Post-lab Questions – Answer these questions in your lab notebook

1. What were your conclusions regarding the effect of substrate concentration on catalase activity?
2. What were your conclusions regarding the effect of temperature on catalase activity? Identify the optimum temperature on catalase activity, in other words, at what temperature was the amount of product produced by the enzyme-catalyzed reaction greatest?
3. If an enzyme were isolated from an organism, such as a clam, that lived in seawater that averages 14°C, what would you predict would be the optimal temperature for that enzyme, and why?
4. Define enzyme denaturation in terms of protein structure.