Biological Chemistry Laboratory Biology 3515/Chemistry 3515 Spring 2019 Lecture 13: Data Analysis and Interpretation of the Michaelis-Menten Parameters 19 February 2019 c David P. Goldenberg University of Utah goldenberg@biology.utah.edu
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Analysis of Data from the V versus [S] Experiment Velocity Substrate Concentration We want to fit the experimental data to the Michaelis-Menten Equation: V = [S]V max [S] + K m From the fit, we obtain estimates of K m and V max.
A Classic Method for Analyzing Enzyme Kinetics Data Rearrangement of the Michaelis-Menten Equation: V = [S]V max [S] + K m 1 V = [S] + K m [S]V max = [S] [S]V max + 1 V = 1 [S] K m + 1 V max V max K m [S]V max A plot of 1/V versus 1/[S] should generate a straight line with a slope of K m /V max and an intercept of 1/V max on the 1/V axis.
The Lineweaver-Burk Plot 0 0 If the data are perfect, this plot gives good estimates of K m and V max. But, experimental error in V can lead to strange effects!
The Effects of Errors on a Lineweaver-Burk Plot Errors in the least precise measurements (low V ) can cause large changes in the line fit to the Lineweaver-Burk plot.
Clicker Question #1 Which parameter is likely to be more sensitive to errors in a Lineweaver-Burk plot? A) K m B) V max Provided that values of V approach V max.
Two Ways to Deal with This Problem Use Lineweaver-Burk, but weight data according to uncertainties in 1/V. Fit velocity data directly to the Michaelis-Menten equation using non-linear least-squares method. Velocity Equal errors in V are weighted equally. Substrate Concentration
Interpreting K m E + S E. S E + P Velocity (M/min) K m = k 1 + k cat k 1 [S] V = V max [S] + K m V = V max 2 when [S] = K m [S] (M) When [S] = K m, half of total enzyme has substrate bound. The larger K m is, the more substrate is required to reach V max /2, or any specified fraction of V max.
Clicker Question #2: Data for three substrates with the same enzyme. 100 A B C 0 0 10 Which substrate binds most tightly to the enzyme? No wrong answers, for now.
A Closer Look at Binding and K m : K m versus K d E + S E. S E + P K m is defined in terms of the rate constants: K m = k 1 + k cat k 1 K d is the equilibrium constant for dissociation. K d = [E][S] [E S] = k 1 k 1 A large K d indicates weak binding.
K m versus K d K m = k 1 + k cat k 1 K d = [E][S] [E S] = k 1 k 1 If k cat k 1, i.e., the E S complex is more likely to dissociate than undergo catalysis: K m k 1 k 1 = K d In general, K m K d Strength of equilibrium binding may be greater than indicated by K m.
Energy Profile for an Enzyme-Catalyzed Reaction transition state Free-energy change for binding: G bind = RT ln K d R = Gas constant T = Temperature E + S E. S Free-energy change from E S complex to transition state: ( ) G kb T = RT ln k cat h E + P k b = Boltzmann constant h = Planck constant
The Significance of k cat /K m Free-energy difference between E + S and the transition state: E + S E.S E + P G total = G bind + G ( kb T = RT ln h ) } {{ } Constant If k 1 k cat, K d K m : G total = C RT ln ( kcat K m ( ) Kd +RT ln k cat ) The ratio k cat /K m reflects the free energy difference between E + S and the transition state. (Assuming K d K m ) k cat /K m is commonly interpreted as a measure of enzymatic efficiency. Catalytic efficiency is favored by a large value of k cat and a small value of K m.
Is a Low K m Always Good? Suppose that we could design a mutant enzyme that forms a more stable complex with substrate. wild type E + S E. S mutant E + P This will lower K m and k cat, but leave k cat /K m the same.
Clicker Question #3 At low substrate concentration ([S] K m ), will the velocity for the mutant enzyme be greater or less than that of the original enzyme? A) Greater than the original enzyme wild type B) Less than the original enzyme C) The same as the original enzyme! mutant E + S E. S V = [S][E] T k cat K m + [S] E + P V k cat K m [S][E] T
What About High Substrate Concentrations? Low Substrate Concentration transition state High Substrate Concentration transition state E + S E. S E. S E + S E + P E + P When [S] K m, the enzyme-substrate complex is favored with respect to the free enzyme.
Clicker Question #4 At high substrate concentration, will the velocity for the mutant enzyme be greater or less than that of the original enzyme? E + S wild type transition state A) Greater than the original enzyme B) Less than the original enzyme! C) The same as the original enzyme E. S mutant V = [S][E] T k cat K m + [S] V k cat [E] T = V max E + P
BPTI is an Extreme Example of a Low-K m substrate BPTI (I) + Trypsin (E) Substrate E + S BPTI E + P K d 10 12 M E. S
To Make a Better Enzyme (or Substrate), Stabilize the Transition State! wild type E + S E. S mutant E + P Increased rate at all substrate concentrations. Easier said than done!
Transition State Stabilization in Serine Proteases Enzyme-Substrate Complex Asp102 Ser195 His57 Gly193 Substrate-enzyme model from structure of trypsin-bpti complex (PDB entry 2FTL)
Transition State Stabilization in Serine Proteases Transition State Asp102 Ser195 + - His57 Gly193 Transition state model from structure of trypsin with boronic inhibitor (PDB entry 1BZT)