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Unformatted text preview: 7.05 Problem Set 2 Answer Key 1. From Question 1, Exam 2, 2005 2. From Question 1, Exam 2, 2005 3. From Question 2, Exam 2, 2004 4. From Question 5, Exam 1, 2007 5. You have identified a novel protease that cleaves on the carboxyl side of both hydrophobic (i.e., phenylalanine) and basic (i.e., arginine) residues. A. You are interested by its unique specificity. Predict, based upon your knowledge of protease specificity pockets, a residue (or residues) that might be within your novel protease’s specificity pocket (assume that like other proteases, your protease has only one specificity pocket). And account for the unique specificity of your protease. How does its specificity pocket accommodate both hydrophobic and basic residues? Because we know that basic residues are readily recognized by the protease, the specificity pocket of the protease should contain stabilizing, recognition residues that are acidic. Therefore, aspartic acid or glutamic acid are two residues that could be anticipated to lie within the protease’s specificity pocket. We also know that the protease uniquely recognizes hydrophobic residues. Therefore, we could also expect to find stabilizing hydrophobic residues within the protease’s specificity pocket. The unique specificity of the protease could be a result of conformational changes of these amino acid residues within the specificity pocket. For example, if glutamic acid is the acidic residue within the specificity pocket, it could orient itself towards the pocket and form stabilizing interactions (like salt bridges) with a basic residue. Alternatively, the glutamic acid residue could orient itself towards to outside of the protein, away from the specificity pocket, and interact with peripheral solvent molecules, while allowing the remaining hydrophobic residues within the specificity pocket to interact favorably with a hydrophobic residue. This sort of conformational flexibility of specificity pockets nicely allows for the dual recognition of distinct amino acid residues. B. You suspect that your protease is a serine protease, and you are determined to solve its crystal structure. A post‐doc in your lab is quick to inform you that it would be wise to consider crystallizing your protease with inhibitors. Why is it wise to crystallize your protein with an inhibitor(s)? What are the requirements of the inhibitor(s) with which you will crystallize your protein? Crystallization of the protein with an inhibitor would create an environment in the active site of the protein that is similar to that of a protease caught in the act of proteolysis, with its catalytic machinery poised for chemistry. You would not be able to catch the protease acting on its native substrates by x‐ray crystallography. The inhibitor, therefore, is a powerful tool to use. An ideal inhibitor would be one that covalently attaches to your protease and renders it enzymatically dead by virtue of competitive, covalent inhibition. The inhibitor selected should also mimic the predicted transition state of the enzyme‐substrate complex. In the case of crystallization and structural determination of the protease, you could solve the structures of your protease bound to one inhibitor, which resembles a basic residue substrate, and another inhibitor, which resembles a hydrophobic residue substrate. The former structure would provide insight into the specificity pocket of your protease when interacting with a basic residue substrate. The latter structure would provide insight into the conformation of your protease upon interaction with a hydrophobic residue substrate. C. You successfully solved the crystal structure of your protease. You find that it was indeed a serine protease, and you are now eager to biochemically characterize it. You purify the enzyme and assay its activity by utilizing a substrate molecule that generates a colored product when cleaved by your protease. Your data is summarized in the table below. Construct a Lineweaver‐Burk plot and calculate Vmax and Km. [S] (M) 2.0 x 10‐6 5.0 x 10‐6 6.0 x 10‐5 1.5 x 10‐4 First find 1/[S] and 1/Vo. [S] (M) 2.0 x 10‐6 5.0 x 10‐6 6.0 x 10‐5 1.5 x 10‐4 Then plot 1/Vo vs. 1/[S]. Velocity (μmol/min) 50 100 200 238.1 1/[S] (L/mol) 5.0 x 105 2.0 x 105 1.7 x 105 6.7 x 103 1/V (min/μmol) 0.02 0.01 0.005 0.004 Velocity (μmol/min) 50 100 200 238.1 y‐int. = 1/Vmax ≈ 0.004 Vmax ≈ 250 μmol/min x‐int. = ‐1/Km ≈ ‐1.3 x 105 Km ≈ 7.5 x 10‐6 M ≈ 7.5 μM D. Under what conditions would you expect the velocity to be independent of substrate concentration? At high concentrations of substrate, when [S] >> Km, vo ≈ k2[E]T. This implies that the reaction under these conditions is zero order with respect to [S]. E. You find that addition of 100 nM of a compound known as “X” alters the rate of product formation. Values for the velocities of the reaction at the same concentrations of substrate are given below. Calculate the Vmax and Km in the presence of “X”. Is “X” a competitive or a non‐competitive inhibitor? [S] (M) Velocity (μmol/min) (in the presence of “X”) ‐6 2.0 x 10 9.60 5.0 x 10‐6 22.73 6.0 x 10‐5 136.4 ‐4 1.5 x 10 187.5 First, find 1/Vo. 1/[S] will be the same as derived from 5C. [S] (M) Velocity (μmol/min) (in 1/[S] (L/mol) 1/V (min/μmol) the presence of “X”) 2.0 x 10‐6 9.60 5.0 x 105 0.104 ‐6 5.0 x 10 22.73 2.0 x 105 0.044 ‐5 5 6.0 x 10 136.4 1.7 x 10 0.0073 1.5 x 10‐4 187.5 6.7 x 103 0.0053 Data in presence of “X” can be plotted as follows: y‐int. = 1/Vmax ≈ 0.004 Vmax ≈ 250 μmol/min x‐int. = ‐1/Km ≈ ‐2.0 x 104 Km ≈ 5.0 x 10‐5 M ≈ 50 μM Compound “X” is a competitive inhibitor. We see that Vmax remains unchanged in the presence of “X”, but that Km has increased almost seven‐fold. This implies a less stable enzyme‐substrate complex because the inhibitor is competing for the same substrate binding site and therefore affecting substrate binding to enzyme. ...
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- Spring '09