Today our model is that enzymes promote chemical reactions by lowering the activation energy necessary for reaction by forming an enzyme-substrate complex:
The activity of the enzyme has to do with the properties of its active site. This location in the protein's tertiary structure binds with a specific substrate, stresses the bonds in that molecule in a particular way, and thereby accelerates a specific reaction, resulting in a unique product. This specific binding of particular substrates to form unique products will be tested today.
Since the enzyme is not used up in the reaction, only a small amount of enzyme is necessary to bring about a large amount of reaction product. In fact, the amount of a single typical enzyme in a cell would be below the detection limits of even the most sensitive protein assays.
The amount of enzyme present is therefore estimated by measuring the amount of reaction it promotes. Under what circumstances would this be an invalid approach? The RATE of reaction is the change in concentration of a substrate or a product over a period of time. In most cases measurement of the formation of product is preferred over measurement of the disappearance of a substrate. Why?
Today, you will investigate a battery of compounds that differ structurally from a known substrate to determine what makes a proper substrate molecule, what structure renders a molecule a competitive inhibitor, and what, if anything, makes a molecule simply innocuous. In this way you should learn what the relationship is between potential substrates and the active site of the enzyme.
THE ENZYME REACTION
To test our model, the enzyme POLYPHENOLOXIDASE will be examined. The enzyme is found in plants (responsible for browning of freshly peeled fruit or vegetables). We will use potatoes as a source of the enzyme and will use pyrocatechol (catechol) as the substrate:
The reaction converts catechol (a POLYPHENOL) to an OXIDized form with the aid of the enzyme (-ASE); hence the name of the enzyme is polyphenoloxidase. The ortho-quinone is rapidly converted to red products that can be measured in a spectrophotometer at 540 nm. Since these subsequent reactions are relatively very rapid, the measurement of the red products at time intervals gives an immediate indication of the rate of the enzyme-catalyzed reaction. Since you have already used this enzyme in Principles of Biology, the instructions here are greatly abbreviated.
A. PREPARATION OF THE ENZYME
The instructor will homogenize a peeled potato with an equal weight of distilled water. The homogenate will be filtered through a nylon fabric and centrifuged for about five minutes. The supernatant will be used as our source of enzyme. You must keep it on ice. Why?
B. THE BASIC REACTION
You will now calibrate the enzyme assay so that the reactions will complete in a reasonable amount of time. The basic reaction mixture is: 3 mL distilled water, 1 mL of 0.1 M potassium phosphate pH 6 buffer, 1 mL of 0.006 M Catechol. At t=0 0.5 mL of homogenate is added and the tube immediately and thoroughly mixed. Absorbance at 540 nm is measured at 30-sec intervals. You should run this reaction to reacquaint yourself with the basic reaction and data collection techniques. Once this basic process is completed, you are ready to proceed to some investigative science.
C. DETERMINING MICHAELIS-MENTEN PARAMETERS
Now that you have been through the usual reaction process, it is time to find out some properties of the enzyme. A stock solution of 0.024 M Catechol is available. Make a dilution series to produce 0.012 M, 0.006 M, 0.003 M, 0.0015 M and 0.00075 M solutions. In the end you need to have 1 mL in each of the six (right?) tubes. Are the levels in each tube the same? Add 3 mL of distilled water and 1 mL of 0.01 M potassium phosphate pH 6 buffer to each tube. Are the levels in each tube the same? Is each tube about half-full? This ballpark checking helps avoid disasters. At t=0 for each tube, add 0.5 mL of potato homogenate, thoroughly mix, and begin absorbance measurements at 30-sec intervals. Obviously you want to stagger the starting times!
On your own time, plot absorbance vs time and determine the initial reaction rate (absorbance min-1). In general only the first three or four points are used. Why? Cricket Graphª can do this quite easily. Then plot reaction rate vs concentration of substrate to make a Michaelis-Menten plot. Be sure to use the ACTUAL (in the mixture) final concentration of the compound!
Finally, determine the Km and Vmax for our enzyme. This can be done in one of two ways. If available, use non-linear regression analysis to determine the best values statistically; this is the best way to proceed. If that method is not available, then calculate values for (rate)-1 and (substrate concentration)-1. Plot these inverted values to make a Lineweaver-Burk plot.
D. TESTING STRUCTURAL ANALOGS OF CATECHOL
Your group will select (or be assigned) a family of three compounds to test. Each family shows some slight deviations from catechol (1,2-benzenediol). First one might repeat the process in B above to determine whether each compound functions as a substrate or not. Observations of colored products might indicate that 540 nm is not a reasonable choice for further measurement.
1. The compound is a substrate. You should vary the concentration of the compound used in the reaction mixture. Serial dilutions come to mind. A stock solution of 0.024 M could be diluted to make 0.012 M, 0.006 M, 0.003 M, 0.0015 M, and 0.00075 M. All six (are you sure?) concentrations can be used in the reaction mixture in place of the catechol concentrations in part C above. Absorbance should be measured as before at 30-sec intervals after adding the homogenate and thorough mixing.On your own time, determine the Km and Vmax for this enzyme-substrate combination, and compare them to those with catechol from part C above. Does the enzyme have greater or less affinity for this substrate than it does for catechol? Does the enzyme catalyze the reaction more quickly with this substrate than with catechol? What does this tell you about the active site of the enzyme? How does the structure of the catechol analog possibly make these figures different than those for catechol?
2. The compound is not a substrate. The structure of the compound may be similar enough to the substrate, catechol, to bind at the active site, but may structurally different enough to avoid producing colored products. Recall the enzyme-substrate complex model:
enzyme + substrate <---> enzyme-substrate complex <---> enzyme + product(s) In the case of a structural analog of the substrate we might predict the following:
enzyme + analog <---> enzyme-analog complex <---> NO REACTION If this model holds, then the analog may compete with catechol for the active site on the enzyme. We would then call the compound a competitive inhibitor. A competitive inhibitor alters the Km but not the Vmax for the catechol-polyphenoloxidase reaction system. You would need to know the normal Vmax and Km for comparison (part C above). Then you would repeat part C but include 1 mL of a 0.006 M solution of the potential inhibitor to replace 1 mL of the usual amount of water. Have the instructor verify your tube recipes. Add the homogenate at t=0 and measure absorbance as before after thorough mixing.
On your own time, plot absorbance vs time for each tube. Calculate the initial rate for each tube. Compare the Km and Vmax for the system both in the presence and absence of the analog. How does the analog alter these two parameters? Is the compound truly a competitive inhibitor? What does this fact tell you about the active site of the enzyme? What is the change in chemical structure that causes this alteration in results?
3. The compound is innocuous. The compound may not be a substrate and may not be a competitive inhibitor either. We are trying some compounds selected because they have similar functional groups, we do not know in advance what each compound can do. It is possible that the chemical structure is different enough that it is neither substrate nor competitive inhibitor. Don't despair, that result in itself is very useful in showing essential components! Remember, the rejected hypothesis is valuable! What does this compound's chemical structure have that renders it both unable to bind at the active site and also to fail to react? What does this tell you about the active site of the enzyme? What does it tell you about the mechanism of the normal reaction?
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