301, Biology Exercise #2 Enzyme Kinetics of -Galactosidase
Introduction Enzymes as Catalysts. Enzymes are a subgroup of proteins which catalyze biochemical reactions. Without enzymes most necessary chemical reactions in our bodies would be unable to proceed. An enzyme can increase the rate of a reaction but it cannot change the equilibrium of the reaction. For example, if substrate S is converted to product P and the equilibrium constant, Keq, for this reaction is equal to 100, then the concentration of P at equilibrium is 100 times greater than that of S. However, it may take a long time to reach this equilibrium in the absence of an enzyme. An enzyme increases the rate of the reaction by decreasing the activation energy (Ea). Without an enzyme, a graph of free energy versus progress of the reaction is as shown in Figure A (see below). A large input of energy is necessary to make the reaction proceed. However, when an enzyme is present, it lowers the energy of activation so that the reaction can proceed with a much smaller amount of energy (Figure B).
Enzyme Reaction. The simplest reaction that can be catalyzed by an enzyme is one in which a single substrate is converted to a single product. In this reaction, the substrate binds to the active site of the enzyme, a relatively small three dimensional area that is specific for the substrate, to form an enzyme-substrate complex. The "binding" is usually by weak bonds such as hydrogen bonds, electrostatic or hydrophobic interactions, or van der Waals forces. The enzyme then converts the substrate to product, which is released to give free enzyme again. This reaction is described by the following equation in which E, S and P represent enzyme, substrate and product, respectively, and the terms k 1, k2, and k3 are the rate constants for the indicated reactions:
k1 E + S k2 ES
k3 E + P
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Enzyme Kinetics. The rate of catalysis or velocity (v), increases as the substrate concentration is increased. However, the velocity eventually reaches a maximum (V max) when all the available active sites are filled, and the enzyme is said to be saturated. If more enzyme is added, a larger number of active sites are available for catalysis and the Vmax increases in proportion to the enzyme added. Km is a measure of the strength of binding of the substrate to the active site and can be defined in two ways. First, it is the substrate concentration at which half of the active sites are filled. Since V max occurs when all the sites are filled, Km is the substrate concentration ([S]) at which V is equal to one half of Vmax (see below).
The second way to define Km is by the rate constants of the individual steps: Km = (k2 + k3) / k1. One application of this occurs when k2 is much larger than k3. This circumstance essentially reduces the equation to k2 / k1 which is the dissociation constant of the ES complex. Thus, when k2>>k3, Km is an indication of the strength of enzyme-substrate binding. If Km is high, there is weak binding; whereas, if Km is low, strong binding occurs. The variables v, Vmax, S and Km can be related to each other in the Michaelis-Menten equation (see below) where v is velocity at a particular substrate concentration and [S] is the substrate concentration. Vmax [S] v = ------------[S] + Km
Michaelis-Menten Eq.
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The Km of an enzyme for a specific substrate is a fixed property of the enzyme. Evaluating the Km is difficult if the data are plotted on a Michaelis-Menten plot of v vs [S] (see above) because of the hyperbolic shape of the curve. However, the Michaelis-Menten equation can be transformed into a linear form (i.e. y = mx + b) by taking the reciprocals of both sides of the equation and rearranging the terms: 1 --v Km ------Vmax 1 1 --- + -----[S] Vmax
=
If the reciprocals are plotted as 1/v vs 1/[S], which is called a Lineweaver-Burk plot, straight lines are generated from which both Km and Vmax can be readily evaluated (see below). The y intercept is equal to 1/Vmax and the x intercept is equal to -1/Km. The slope of the line is Km/Vmax.. Thus, by plotting the data in this form, one can easily determine the Vmax as the Y intercept, then use the slope to determine the Km.
Enzyme Inhibition. One of the major control mechanisms of enzymes is inhibition. There are two general categories of inhibitory molecules, irreversible and reversible. Irreversible inhibitors covalently bind to an enzyme and cannot dissociate from it. An example of this type of inhibitor is nerve gas. Reversible inhibitors, on the other hand, do not bind covalently and can dissociate from the enzyme. Two types of reversible inhibitors are competitive and non-competitive. Competitive inhibitors bind to the active site, and thus compete with the substrate for access to the enzyme. In theory, the effects of the inhibitor can be negated if enough substrate is added; therefore, a competitive inhibitor should increase the Km of an enzyme but not alter the Vmax. A non-competitive inhibitor binds to the enzyme at a place other than the active site (i.e. the substrate binding site). The binding of the inhibitor alters the enzyme so that the catalysis of substrate to product is inhibited and, as a result, the Vmax is decreased. A true non-competitive inhibitor does not change the Km; however, many inhibitors that decrease the Vmax also increase the Km.
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Experiment In this lab you will be studying the enzyme -glactosidase. This is the enzyme that makes up one of the three proteins encoded by the lactose operon in bacteria. This is an enzyme which has been studied extensively for decades. In addition to its role as a model for regulation of gene activity, it is also used widely as a marker for molecular biology experiments, since it is a bacterial enzyme, not found in mammals, and it can convert a non-colored substrate to a colored product, making detection easy. The basic reaction catalyzed by this enzyme is:
You will be using a colorimetric assay to measure -galactosidase activity activity. In this assay, a colorless substrate (ONPG or o-nitrophenyl--D-galactopyranoside) is converted to a colored product (o-nitrophenol) which absorbs at 420 nm and can be measured using a spectrophotometer. The molar absorptivity of o-nitrophenol = .0045 OD/n mol. You can use this information to calculate the amount of product formed in your reaction. For the initial part of the lab (Part a), you will measure the rate of product formation using a series of different substrate concentrations and a fixed enzyme concentration. This will allow you to determine the Km and the Vmax for the enzyme. You will then look at the effect of changing the enzyme concentration (Part b), adding an inhibitor, D-galactose (Part c), and heating the enzyme on enzyme activity (Part d). For Part (e), you will set up an experiment to test the requirement for -mercaptoethanol in this system. mercaptoethanol is a disulfide reducing agent. The enzyme -galactosidase is a tetramer, which reportedly requires disulfides in its quaternary structure. Note: You may find it helpful to label your tubes differently for each section, that is Part (a) 1-5, Part (b) 6-10, etc. This may make things easier to follow when you write your lab report.
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Methods 1. Turn on the spectrophotometer, set the wavelength to 420 nm, and allow it to warm up for about 15 min. With no tube in place and the lid closed, set the indicator on the spectrophotometer to read 0% transmittance using the left front knob. Fill a cuvette with water and insert it into the holder. Close the lid and set the indicator on the spectrophotometer to read 0 absorbance (ABS) units using the right front knob. Before each reading, add the reaction mix to the tube without enzyme, put the tube in place, close the lid and zero again. 2. Dilutions: (a) Substrate: To set up the individual reactions, you will need to make a series of dilutions of the substrate. The stock solution of ONPG substrate is 0.015M (15mM). In each reaction, you will then use 0.5 mL of a specific dilution. Calculate the final concentration of each substrate in the reaction mix and enter it in the chart below: Tube A B C D E (b) Enzyme You must also make a dilution of the enzyme. Take 100 L of the stock enzyme and add to 9.9 mL of enzyme diluent (100x dilution). It is important to keep this diluted enzyme on ice throughout the rest of the experiment. 3. Experiments: (a): Velocity vs. substrate concentration. Set up your tubes with everything listed below. Wear gloves. -mercaptoethanol is a carcinogen. Use a pipet for the assay buffer, the mercaptoethanol and the water. Use a P1000 set at 500 for the ONPG. Be sure to use a different tip for each concentration. ONPG 6 mL stock 2 mL (A) 3 mL (B) 2 mL (B) 3 mL (D) Substrate diluent 0 8 mL 3 mL 8 mL 3 mL Final ONPG
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Tube #
1 2 3 4 5
Assay Buffer (mL) 1.0 1.0 1.0 1.0 1.0
1M Water mercaptoethanol (mL) (mL) 0.3 0.3 1.1 0.3 0.3 0.3 1.1 1.1 1.1 1.1
ONPG (L) 500 (A) 500 (B) 500 (C) 500 (D) 500 (E)
Add the contents of Tube #1 to the Spec 20 tube. Insert into the instrument and set the absorbance to zero using the right knob. Remove the tube, add 100 L of diluted enzyme, put a piece of parafilm on the top of the tube and mix quickly. Then insert the tube into the instrument and take your 0' reading as fast as possible. Take readings at 30 seconds, 1minute and 2 minutes. 4. Remove the cuvette and rinse it several times with distilled water. Repeat the procedure used for Tube #1 for each of the remaining reactions (2-5). Determine the average change in absorbance per minute (the velocity) for reactions 1-5. For your report, divide the change in OD per minute by 0.0045 to determine the change in product per minute (v) in nmoles. (b) Effect of lowering enzyme concentration Repeat the experiment in Part (a) but use a 1in 3 dilution of the enzyme solution used above (i.e. add 1 mL of diluted enzyme to 2 mL of enzyme dilution buffer final enzyme concentration is 1/3 of concentration used in part (a)) c. Effect of an Inhibitor Test the effect of adding D-galactose to the reaction. The galactose is made up as a 0.15M solution. Add 500 L to each reaction (add to your notebook what the final concentration of inhibitor is). Note: It is essential to keep the final volume of the reaction constant. When you make up your assay tubes, change the volume of water used to compensate for the volume of substrate added. d. Effect of heating the enzyme. Take 1 mL of your standard enzyme dilution and heat to 95 C for 2 minutes before using in the reaction. e. The effect of omission of -mercaptoethanol Repeat Part (a) omitting the -mercaptoethanol.
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Lab Notebook Include the following in your notebook (point values are indicated): 1. Goal of the lab. This should be one to two sentences to describe the purpose of the exercise. (5 points) 2. Here you will need to put in your notebook a chart for the dilutions (with the final substrate concentrations calculated) and charts for each of the individual parts, a-e. You should use the model shown for Part (a) on the lab sheet, but put into your notebook what you will do for each of the other parts. IT IS ESSENTIAL THAT YOU DO THIS BEFORE YOU COME TO THE LAB. IF YOU DO NOT, YOU WILL NOT HAVE TIME TO COMPLETE ALL THE STEPS INDICATED. THE TAS WILL NOT STAY OVER TIME TO ALLOW PEOPLE TO COMPLETE THE WORK IF THEY HAVE NOT COME PREPARED. (15 points) 3. Record any observations about the lab or any changes from the hand out. Include your observations of how the tubes looked when you examine them by eye. What color is the substrate solution? the product? (5 pts). 4. Record results. Here you need to make a table in your lab notebook in which you indicate for each of the five parts above the reaction number and the spectrophotometer readings (5 pts).
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Lab Report #2 For this lab, you will write a FULL LAB REPORT. The report should flow like an essay and should be written for an unschooled audience. However, be succinct! Many journal have page charges, so it is advisable to be concise in your writing. See the "Instructions for Formal Lab Reports", available in the Lab Materials section of Blackboard. If you need more information about how to write a report, check the book A Short Guide to Writing about Biology, by Jan A. Pechenik, on reserve in the Chemistry Biology Library. NUMBER YOUR PAGES! You will need the following elements: 1. Cover Page with Title (use your own, not the one on the lab sheet). 1. Introduction (5 points). Write a paragraph or two explaining the point of the experiment. What can one learn about an enzyme or an enzyme inhibitor by doing this type of experiment? Key words to include: the enzyme used, the substrate, method of detection of activity, Michaelis-Menten, Lineweaver-Burk, Km, Vmax, pH, different classes of inhibitors. 2. Materials and Methods (5 points). Describe in your own words how the experiment was done. Do not just refer to the lab handout. Instead write a series of steps that would be clear for anyone reading them to follow. Write out what you did. Use either active or passive voice as you prefer, but be consistent throughout. 3. Results. For this section, introduce each part of the experiment with a descriptive text that explains the point of the experiment. Provide titles for tables and figures used. You can either include the figures in the body of the text or put them at the end of the document, providing a list of tables first, then a list of figures. Be sure to label axes!!! The following tables and figures should be included: a. A table of your data for Parts (a-e) listing the velocities for Reactions 1-5 for each part (Table1 divide into parts a-e). An example of the table for Part (a) is below. Each part should have its own title indicating what was varied in each. In making these tables, be sure to set them up so that the reader can clearly see what each column represents (center the header). (5 points) Part (a) Change in velocity as a function of substrate concentration Tube # [ONPG] n moles/min 1/s 1/v OD/min
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b. A single plot of the data for Parts (a) and (b), using Michaelis-Menten (v vs. [s]) (Fig. 1). Briefly describe the results shown in the figure. Your descriptive title should include what you are testing. What is varied in each part? Why put them on a common graph? Estimate Km and Vmax. (5 points) c. Lineweaver-Burk (1/V vs. 1/[S]) of the data shown in Figure 1(Fig. 2). Include a sample calculation here. Determine the Vmax and Km for each enzyme concentration.(5 points) d. Plot the data for Parts (a) and (c) on a Lineweaver-Burk plot (Fig. 3), again providing a brief description of what the plot shows below. Again, calculate Km and Vmax for each. (5 points) e. Plot the data from Part (d) and describe the result (5 points) f. Plot the data from part (e) and describe the results. (5 points) For the last two, you may use whichever type of plot you think is most instructive. 4. Discussion. The elements described below should be addressed with a logical, coherent discussion. Also include any difficulties you encountered, or improvements in experimental design. In your introduction, you addressed the question of what one can learn about an enzyme and an inhibitor by doing this type of experiment. Your discussion should include what you have learned. (10 points for overall discussion in addition to point allocations that follow) a. Using your results, discuss the use of Michaelis-Menten vs. Lineweaver-Burk to determine Km and Vmax. (5 points) b. What type of inhibition (competitive or noncompetitive) was observed for the reversible inhibitor according to Fig. 3 and why? (5 points). c. What is the effect of heating the enzyme? Explain this result based on your knowledge of enzyme structure (5 points). d. What was the effect of omission of -mercaptoethanol? What does this indicate about the structure of the enzyme? (5 points) Hints about writing Lab Report #2 Refer to the general comments about writing formal lab reports. In addition, please pay attention to the following hints specific for this report. Logarithms were not involved in the Michaelis-Menten Equation. Excel can get you into trouble because the plot does roughly resemble a log plot. Run your MM plots to origin. Run your LB plots to x-axis (that means you have to extend the x-axis to negative number and use Exel to extend the line back. Km can't be calculated from v and Vmax with the MM equation. Both [S] & Km are unknowns. The word data is plural. Datum is singular. Double-space or space and a half text. Give figures & tables descriptive titles. Use proper units.
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Appendix Kinetics Example
Data obtained: concentration (mM) 1.0000 2.0000 4.0000 8.0000 10.000 Michaelis-Menten Plot: velocity (M/min) 3.0000 4.0000 5.5000 6.3000 6.5000
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velocity (uM/min)
5 4 3 2 1 0 0 2 4 6 8 10 12
[s] mM
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To graph this as Lineweaver Burk, you have to calculate the inverse of each number:
1/concentration 1/velocity* 1.0000 0.33000 0.50000 0.25000 0.25000 0.18000 0.12500 0.15800 0.10000 0.15000 Note here that when you do the calculation, the inverse value for the highest concentration is the lowest number and also has the lowest inverse rate. So this number would be plotted leftmost on the x axis.
Linweaver Burk
0.4 0.35 0.3 0.25 0.2 0.15 0.1 0.05 0 -0.05 0 -0.5 -0.1
y = 0.2023x + 0.1337
1/velocity
Series1 Linear (Series1)
-1
0.5
1
1.5
1/concentration
This line extrapolates to 0.1337, so Vmax = 7.4 (1/0.1337). Km can be obtained from the slope here, knowing Vmax: The slope is determined from the equation described in the Exel file. It is 0.2023. If you then realize that the slope is Km/Vmax, and you know Vmax = 7.4, then Km works out to be 0.2023 x 7.4 or 1.702. If you look at the M-M graph, that is perfectly plausible.
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