Biology 2180 Laboratory #3 Name Introduction Enzyme Kinetics and Quantitative Analysis Catalysts are agents that speed up chemical processes and the catalysts produced by living cells are called enzymes. Enzymes are proteins and each cell produces hundreds of them. enzymes accelerate the velocity of virtually all reactions that occur in biological systems including those involved in breakdown, synthesis and chemical transfers. In so doing, they are responsible for preforming essentially all the changes associated with life processes. The general expression used to describe an enzyme reaction is shown below. Enzyme Action Enzyme (E) + Substrate (S) E + S Enzyme-Substrate Complex (E-S) E S Enzyme (E) + Product (P) E + P1 P2
Three important features of enzyme reactions can be seen in the above diagram. 1.The term substrate refers to the compound that is acted upon by the enzyme. In general, enzymes exhibit a high degree of substrate specificity in that they usually catalyze only a single chemical reaction. Indeed, enzymes are often denoted by the name of the substrate that is acted upon and the suffix -ase. Thus, protein-splitting enzymes are proteases and enzymes that hydrolyze lipids are lipases, for example. 2.The enzyme binds to the substrate to form an enzyme-substrate complex. This interaction is responsible for the specificity of enzyme action since only those compounds that fit into the active site can be acted on by the enzyme. The binding of the substrate to the enzyme also serves to alter the substrate in such a way that the conversion of substrate to product is facilitated. 3. The enzyme is not destroyed during the reaction but rather is set free after the formation of the end product. Thus, the liberated enzyme is available to combine with more substrate to produce more product. This feature makes enzyme molecules exceedingly efficient in catalysis and explains how very small quantities of enzymes are sufficient for cellular processes. Measuring The RATE OF Enzyme Reactions An enzyme catalyzed reaction can be represented in a simplified form as follows: Substrate (S) Enzyme Product (P) The velocity or rate of the reaction can be determined by measuring the decrease in substrate concentration with time or, more commonly, by measuring the rate of appearance of product as shown in Figure 1-2. During the early part of the reaction illustrated in the figure, the amount of product formed in creases linearly with time. However, in the latter part of the reaction (20 minutes to 40 minutes), the rate of product appearance diminishes, and at 40 minutes, product is no longer formed. A number of reasons can account for the decline in reaction rate with time inducing the depletion of substrate or the breakdown of the enzyme. In any case, in order to describe this reaction, its velocity must be determined during an early time interval when the amount of product is increasing in a linear manner. The rate measured during this time is the slope of a straight line and is called the initial velocity of the reaction (Vo). The Vo in the example shown below is 10 nanomoles (nmoles) of product formed per minute. A mole is the molecular weight of a substance expressed in grams. thus, one nanomole (nmole) is the molecular weight of a substance expressed in ng (1ng=10g9). A one molar (1M) solution contains one mole of solute per liter of solution. Thus, a one nanomolar (1nM) solution contains one nmole of solute per liter.
Typical curve for an enzyme catalyzed reaction. 300 200 100 20 40 60 Time (minutes) The initial velocity, (Vo), is the slope of the linear portion of the curve. Therefore, in this example, the velocity during time points 1 to 2 can be calculated as: Vo = nmoles at time 2 - nmoles at time 1 time 2 - time 1 = 100-50 10-5 = 50 5 = 10 nmoles/min Repeat this calculation for time points 3 to 4 and for time points 7 to 8. How can you explain these different results?
Factors that Influence the Rate of an Enzyme Catalyzed Reaction Amount of Enzyme The Vo is directly proportional to the amount of active enzyme molecules as shown below. This feature is important for it enables one to determine the amount of an enzyme in an unknown sample. For example, if 1 µg of pure enzyme gives a Vo of 10 nmoles of product/min, a cell extract that yields a Vo of 20 nmoles product/min contains 2 µg of active enzyme. Initial velocity and Enzyme Concentration. 40 30 20 10 1.0 2.0 4.0 3.0 Amount of Enzyme (µg) Temperature Since enzymes are proteins, they are usually denatured and inactivated by temperatures above 50-70 C. On the other hand, increased temperature also speeds up chemical reactions. With a typical enzyme, the predominant effect of increased temperature up to about 45 C is to increase the enzyme catalyzed reaction rate. Above 45 C, thermal denaturation becomes increasingly important and destroys the catalytic function of the enzyme. At some temperature, a maximum reaction rate called the temperature optimum is observed, and this temperature is usually in the range found in cells (20 C-40 C). ph The activity of enzymes is also greatly influenced by acidity or alkalinity. Excess acidity or alkalinity generally causes denaturation and inactivation of enzymes, just as high
temperature causes heat denaturation. Most enzymes in plants and animals operate effectively at neutral ph. However, the ph optimum of the enzyme that you will study today is 4.5. In fact, the name of the enzyme, acid phosphatase, is derived in part from its low ph optimum. Substrate Concentration The binding of an enzyme to its substrate is an essential part of the enzyme catalyzed reaction. As diagramed below, at low substrate concentration the active site on the enzyme is not saturated by substrate and thus the enzyme is not working at full capacity. As the concentration of substrate increases, the sites are bound to a greater degree until at saturation, no more sites are available for substrate binding. At this saturating substrate concentration, the enzyme is working at full capacity and the maximum velocity ( Vmax ) of the reaction is seen. Diagram showing the Effects of Substrate Concentration on the Saturation of Enzyme Active Sites. Substrate Concentration S + E E-S E + P Low Half Saturated Fully Saturated This diagram shows that product ( P ) formation is directly related to the concentration of the enzyme-substrate complex (E-S) rather than the concentration of the substrate (S). The consequences of saturating the enzyme with substrate on the reaction rate is shown below. The initial velocity of the reaction (V0) increases in a hyperbolic manner as the substrate concentration is increased. This increase in reaction rate is proportional to the concentration of the enzyme-substrate complex. Thus, the Vmax occurs because the enzyme becomes saturated with substrate. The substrate concentration required to yield half the maximal velocity ( Vmax /2 ) can also be determined from Figure 1-5 and is an important constant in describing an enzyme.
This constant is known as the Michaelis constant and is abbreviated Km. Under conditions of defined temperature, ph and ionic strength, the Km approximates the dissociation constant of an enzyme for its substrate. The dissociation constant ( Kd ) of an enzyme is the concentration of substrate the yields half-saturation of the enzyme with substrate. Thus, the Kd and Km reflect the affinity of the enzyme for the substrate. For example, a Km of 0.2moles/literof substrate would indicate that the substrate binding site would be half saturated when the substrate is present at the concentration. Such an enzyme would have a low affinity for its substrate. In contrast, a Km of 10-7 moles/liter indicates that the enzyme has a high affinity for its substrate, since it is half saturated at this low concentration. Effect of Substrate Concentration on the reaction Rate. VMax VMax 2 V0 V0 = Vmax [S] Km + [S] Km Substrate Concentration [S] The Michaelis-Menten equation for the hyperbola graph on the left is given on the right. As shown here, the effects of substrate concentration on the velocity of an enzyme catalyzed reaction is described quantitatively by the Michaelis-Menten equation given above. For experimental determination of the Km and Vmax, the Michaelis-Menten equation is usually rearranged to a form that is equaivalent to the straight line equation (y=mx +b). One such rearrangement, first described by Lineweaver and Burk, involves taking the reciprocals of each side of the Michaelis-Menten equation as shown in the figure below. If a double reciprocal plot is made with 1/V0 on the vertical axis and 1/[S] on the horizontal axis, a straight line relationship exsists where the slope is km/vmax and the y intercept is 1/Vmax while the x intercept is -1/Km. Wheat Germ Acid Phosphatase The wheat kernel consists of three parts: (1) the embryo or germ that produces the new plant (2) the starchy endosperm which serves as a food source for the embryo and (3) covering layers which protect the grain. The endosperm is the raw material for flour production, while the wheat germ is used in food products as a source of vitamins. Wheat germ is also frequently used in the molecular biology laboratory as a source of plant cell proteins and nucleic acids. In today s laboratory, you will analyze an enzyme called Wheat Germ Acid Phosphatase.
The study of enzymes frequently begins with the extraction of these proteins from tissue. Enzyme extraction procedures usually require a method that destroys the integrity of the cell. The broken cells then release their molecular constituents including enzymes. One method that can be used to lyse (break open) cells involves treating tissue with a detergent that breaks or dissolves cell membranes. In the prodecure described below, you will use an enzyme extraction buffer that contains the detergent NP-40 in order to prepare an enzyme extract from wheat germ containing acid phosphatase. Acid phosphatase catalyzes the removal of phosphate groups from macromolecules and smaller molecules that are stored in the wheat seed. The free phosphate is then used by the growing embryo. In the excercise described below, you will measure the velocity of the reaction catalyzed by purified acid phosphatase and by the acid phosphatase that is found in the wheat germ extract that you prepare. A synthetic phosphatase substrate called nitrophenol phosphate will be used in the experiment. Nitrophenol phosphate is colorless but is broken down to free phosphate and nitrophenol as shown below. Nitrophenol is yellow in alkaline solution so that the appearance of yellow color is indicative of the amount of product formed in the reaction. SUBSTRATE Nitrophenol Phosphate (colorless) Acid Phosphatase PRODUCTS Nitrophenol + Phosphate (yellow) Wheat Germ Acid Phosphatase Enzyme Assay Preparation of enzyme extract 1. Place 0.5 grams of wheat germ into a mortar and add 5 ml of enzyme extraction buffer. (This is supplied as a 100 X solution and will have to be diluted.) 2. Grind the tissue with the pestle until a homogenous suspension is formed. 3. Transfer 1 ml of this solution to a microfuge tube and centrifuge 5 minutes. Carefully transfer the supernatant to a clean tube labeled WGE for wheat germ extract and store the tube on ice for later use. The enzyme assay 1. Obtain a test tube rack with 12 test tubes. Label 6 tubes as A1-A6 and six tubes as B1-B6. Tubes labeled A will be used for our pure acid phosphatase and tubes labeled B will be used for our wheat germ extract. 2. Add 0.5 ml of KOH each to tubes A1-A6 and B1-B6. 3. Obtain 2 new test tubes and label one as A and the other as B.
4. Add 5ml of phosphatase substrate solution each to tubes A and B. 5. Remove 0.5ml of phosphatase substrate solution from tube A and add it to tube A1. Repeat this for tubes B and B1. A1 and B1 now represent our zero time reactions or our starting point (T=0). 6. Concurrently, add 50µl (0.05ml) of pure acid phosphatase solution (about 5 µg) to tube A and 200µl (0.2ml) of wheat germ extract to tube B. Mix the tubes by gently swirling. 7. At the times indicated below, remove 0.5ml of A and B and place it in the tubes listed. Time Tube Number 0 minutes A1 and B1 (already done) 2.5 minutes A2 and B2 5 minutes A3 and B3 10 minutes A4 and B4 15 minutes A5 and B5 20 minutes A6 and B6 *** The KOH in tubes A1-A6 and B1-B6 stops the reaction and causes the product of the reaction (nitrophenol) to turn yellow. *** The intensity of the yellow color is proportion to the amount of product produced by the enzyme. 8. Note the increasing intensity of the yellow color as a function of time elapsed then add 5ml of water to each tube. Mix again by swirling. 9. Observe the instructor as the amount of yellow color (absorbence of light at 410nm wavelength) is determined for a standard set of nitrophenol amounts using a spectrophotometer. Record the standard results in the table below. 10. Measure the absorbence for each of your reactions and record the data in the table on the next page. Standard A410 Tube A410 nmoles Tube A410 nmoles 0 A1 B1 25 nmoles A2 B2 50 nmoles A3 B3 100 nmoles A4 B4 200 nmoles A5 B5
400 nmoles A6 B6 11. Use the graph below to plot the standard curve for the assay then extrapolate your experimental data to determine the nmoles of product produced for each of your reaction times and record that data in the table above. 1.0 0.8 Absorbance at 410nm 0.6 0.4 0.2 0 50 100 200 300 400 Amount of Nitrophenol (nmoles) 12. Use the graph below to plot your experimental results. 400 Nitrophenol Produced (nmoles) 300 200 100 0 2.5 5 10 15 20 Time (minutes)
13. From the above data, calculate the initial velocities for the two reactions. Vo = nmoles at time 2 - nmoles at time 1 time 2 - time 1 Reaction A1-A6 Vo = nmoles nitrophenol/minute Reaction B1-B6 Vo = nmoles nitrophenol/minute Questions 1. Identify the substrate, product and enzyme for the reaction we observed. 2. If tube A contained 5 mg of purified enzyme, how much enzyme was in the 0.2 ml of Wheat Germ extract added to tube B? 3. Which time points in your second graph were useable for calculating the enzyme velocity? 4. Why were the other points not useable?
5. If this reaction is reversible, why didn't the yellow color go away? Estimation of Km and Vmax In a previous experiment, the intial velocity of Acid Phosphatase (V0) was measured at differring concentrations of substrate [S]. The results are tabulated below. V0 (nmoles/min) [S] (µm) 0 0 2 20 10 100 20 200 40 300 60 500 70 1000 Use the graph below to plot these results then estimate the values of Vmax and Km. 80 60 40 20 [S] (µm) 500 1000 References: This lab exercise used information and material provided by the following: Modern Biology Experiment # IND-14 Enzyme Kinetics 1992 by John N. Anderson