- Thermodynamic and kinetic relationships. Convince yourself (showing your work!) that:
- A 1.4 kcal/mol change in DG¢0 corresponds to a ~10-fold change in the reaction equilibrium constant Keq¢.
- A 1.4 kcal/mol change in DG‡ corresponds to a ~10-fold change the reaction rate constant k.
- Data fitting to obtain Kd. Hormone receptors are a class of proteins that recognize and tightly bind small molecule hormones; many of these proteins also act as transcription factors that will activate or suppress the transcription of target genes in response to ligand binding, allowing for the hormone-dependent regulation of gene expression.
In PS4_data.xls, there is binding data for three different hormone receptors (HRA, HRB, and HRC) binding their cognate ligands L. In each experiment, the concentration of the hormone receptor is 0.001nM and the fraction bound (FB) was measured at ligand concentrations ranging from 0.1 – 25 nM.
For each set of binding data, plot and fit the data to the ligand binding equation(s) we derived in class (and are listed in Ch 5) to determine the Kd for each hormone receptor HRA, HRB, and HRC. You can use whatever program you like to fit the data, or the simple and free online curve-fitting tool mycurvefit.com. Using MyCurveFit, you can (1) clear the sample X and Y data at the bottom of the window, and directly paste both L and FB data into the table from Excel; then (2) select Fit Method > User Defined and enter the ligand binding equation you’d like to use to fit data (note, use single letter variables in this equation, so K instead of Kd, to keep mycurvefit happy); (3) clicking Apply will carry out the nonlinear least squares fit of the data and show the plotted data points and fit at the top of the window; (4) just below the plot you will see the fitted values for the equation you used. You can click the gray arrow next to these values to see an expanded table with the error on each fitted value. Note: if you sign up for a free account at mycurvefit, you get 50 free fits per month, which will be more than enough for this assignment as long as you paste the L and FB data into the table, rather than entering each individual value by hand.
- For each hormone receptor HRA, HRB, and HRC, what is the Kd?
- Do any of these receptor-ligand interactions exhibit cooperative binding? If so, what is the Hill coefficient, and what does it tell us about the cooperativity and number of binding sites in that receptor?
- Negative Cooperativity. Which of the following situations would produce a Hill plot with Hill coefficient of nH < 1.0? Explain your reasoning.
- The protein has multiple subunits, each with a single ligand-binding site. Binding of ligand to one site decreases the binding affinity of other sites for the ligand.
- The protein is a single polypeptide with two ligand-binding sites, each having a different affinity for the ligand.
- The protein is a single polypeptide with a single ligand-binding site. As purified, the protein preparation is heterogeneous, containing some protein molecules that are partially denatured and thus have a lower binding affinity for the ligand.
- Reversible (but Tight) Binding to an Antibody An antibody binds to an antigen with a Kd of 5 x 10–8 At what concentration of antigen will the fraction bound be (a) 0.2, (b) 0.5, (c) 0.6, (d) 0.8?
- Cooperativity in Hemoglobin. Under appropriate conditions, hemoglobin dissociates into its four subunits. The isolated a subunit binds oxygen, but the O2-saturation curve is hyperbolic rather than sigmoidal. In addition, the binding of oxygen to the isolated a subunit is not affected by the presence of H+, CO2 or BPG. What do these observations indicate about the source of the cooperativity in hemoglobin?
- Enzyme kinetics. Bacterial mRNA molecules have a triphosphate group at their 5’ end that can be cleaved by the enzyme RppH. You’re studying two homologs of this enzyme, EcRppH and BsRppH, and want to know which homolog has greater activity on nucleotide triphosphate substrates. You’ve carried out some enzymology experiments and measured the initial velocity (V0) at nucleotide triphosphate substrate concentrations ranging from 1 to 20 mM and a total enzyme concentration of 2 m The data for each enzyme, EcRppH and BsRppH is given in PS4_data.xls.
- Fit the data to obtain Vmax, Km, and kcat for both EcRppH and BsRppH.
- For each enzyme, what will be the initial rate V0 at 8 mM substrate? What will be the kcat for each enzyme at 8 mM substrate?
- Which enzyme homolog is more efficient?
- Competitive Inhibition. For a simple enzyme catalyzed reaction in the presence of a competitive inhibitor I, we have the following equilibria:
The Michaelis-Menten equation in the presence of the competitive inhibitor I becomes:
- Derive the Michaelis-Menten equation above for competitive inhibition, clearly noting any key assumptions you make. [hints: start with the d[P]/dt, d[ES]/dt, and [E]tot expressions as we did in class to derive the MM equation, but think carefully about which of these expressions will change in the case of a competitive inhibitor; once these equations are set up, you can follow a similar procedure as the MM derivation we did in class, using the expressions for KI and a listed above along the way, to arrive at the Michaelis-Menten equation for competitive inhibition]
- How do Vmax or KM change in the presence of a competitive inhibitor?
- The ‘aKM’ term is often referred to as the ‘apparent KM’. How does the apparent KM change in the presence of a competitive inhibitor?
- What does the KI tell you about the enzyme inhibitor interaction?
- Describe (qualitatively) how you could use enzyme kinetics experiments to measure the KI of a competitive inhibitor.
- Free energy of binding and catalysis. Draw a free energy diagram (G versus reaction coordinate) of (a) a nonenzymatic reaction and the corresponding enzyme-catalyzed reactions in which (b) S binds loosely to the enzyme and (c) S binds very tightly to the enzyme. Compare DG‡ for each case. Why is tight binding of S not advantageous?
- The pH optimum of lysozyme. The active site of lysozyme contains two acidic amino acid residues essential for catalysis: Glu35 and Asp52.
- Describe and/or sketch the roles of Glu35 and Asp52 in the lysozyme catalyzed cleavage of peptidoglycan, assuming the SN2 mechanism of catalysis (see Fig 6-29).
- The pKa values of the carboxyl side chains of these residues are 5.9 for Glu35 and 4.5 for Asp52. What is the ionization state (protonated or unprotonated) of each residue at pH 5.2, the pH optimum of lysozyme? How can the ionization states of these residues explain the pH-activity profile of lysozyme shown below?