462aH Homework Assignment 5
Homework Due 9/27/06
Homework (10 points). For numerical questions, please show your work.
(1)An experiment was performed to measure the equilibrium constants for dissociation of the HIV protease to two inhibitors. The chemical equation describing the dissociation of the protease from an inhibitor, for simple equilibria, is:
PI P + I
where P is free protease, I is free inhibitor, and PI is protease-bound inhibitor. We define the fraction of protein with bound inhibitor as a, thus a = [PI] / [Pt], where [Pt] is the total concentration of protein, or [P] + [PI].
The following data were generated ([Pt] = 1 x 10-13 M):
[I] x 10-12 M / a (Inhibitor 1) / a (Inhibitor 2)0.050 / 0.010 / 0.014
0.071 / 0.014 / 0.018
0.10 / 0.020 / 0.023
0.30 / 0.057 / 0.067
0.71 / 0.12 / 0.14
1.0 / 0.17 / 0.19
3.0 / 0.38 / 0.42
8.0 / 0.62 / 0.69
25 / 0.83 / 0.91
50 / 0.91 / 0.95
100 / 0.95 / 0.98
200 / 0.98 / 0.99
(a) Using the chemical equation shown above, write the mathematical equation for the dissociation constant Kd for this system and show that when Y = 0.5, Kd = [I].
The model for the chemical dissociation is: PI ↔ P + I
The equilibrium constant for this reaction written as a dissociation is:
Kd = [products]/[reactants] = [P][I] / [PI]
a is defined to be [PI] / [Pt] = [PI] / [P] + [PI]
When Y = 0.5,
0.5 = [PI] / [P] + [PI] or [P] = [PI].
Thus, when Y = 0.5, [P] = [PI] and Kd = [P][I] / [PI} = [P][I] / [P], since [P] / [P] =1,
Kd = [I] at the special point of Y = 0.5
(b) Plot (on graph paper) Y vs. [I] for inhibitor 1, and draw the curve which best fits the data. What is the value for Kd?
(c) The expression for Kd can be rearranged to give 1/Y = Kd/[I] + 1, which is in the form of y = mx + b and linear in 1/Y and 1/[I]. Straight-line plots are useful in that deviations from the expected curve are more easily spotted by eye than is the case for hyperbolic curves. Plot 1/Y vs. 1/[I] for both inhibitors. Do both inhibitors obey simple equilibria? Please explain your answer (recall that the equation was derived assuming simple equilibria). If you can determine Kd for each inhibitor from your plots, do so. If you cannot, explain why.
No, both inhibitors do not follow simple equilibria. If simple equilibria was obeyed, the plot would yield straight lines. Only inhibitor 1 (red) plots as a straight line, the line for inhibitor 2 (blue) shows curvature at low concentrations. Therefore Kd cannot be determined for inhibitor 2.
(d) Biochemists normally define equilibrium constants for ligand dissociation, as we have done here. However, chemists normally define such constants for association, which are often symbolized as Ka. Write the chemical and mathematical equations for association of P and I.
The model for the chemical association is P + I ↔ PI
The equilibrium constant for this reaction written as an association is:
Ka = [products]/[reactants] = [PI] / [P][I]
(e) If an inhibitor had Ka = 1 x 109 M-1 and a solution contained [Pt] = 1 x 10-11 M, what concentration of the inhibitor would be needed to ensure that 50% of the protein was inhibited? (Hint: see part (a)).
1 x 10-9 M
(2) A scientist was interested in the details of substrate binding to an enzyme and generated data for substrate dissociation from the protein as a function of pH, shown below. Each value listed for Kd was obtained from a plot of Y as a function of [S] at the given pH, and all such plots were hyperbolic.
pH Kd (x 10-6 M) pH Kd (x 10-6 M)
4.5 0.20 7.0 7.6
5.5 0.99 7.5 9.1
6.0 2.4 8.5 9.9
6.5 5.1
(a) Is binding better at low or high pH?
Kd is smaller at low pH; thus binding is "better" (tighter) at low pH.
(b) Plot the data shown. Is there an ionizable group involved in binding? If so, what is the pKa of this group?
The curve relating the pH effect on Kd is sigmoidal indicating that there is an ionizable group in binding. The apparent pKa for the transition from weaker to stronger binding is 6.5, the midpoint in the sigmoidal curve.
(3) Suggest two possible explanations for the data in question (2) (hint: think about possible ionizable groups in both the protein and the substrate).
A possible explanation for this is that an ionizable group on the protein with a pKa of 6.5 is involved in binding, and that only the protonated form contributes to this binding. The amino acid histidine has an appropriate pKa for this role. A second possibility is that an ionizable group on the ligand has a pKa of 6.5 and must be in the protonated state to contribute to binding.
(4) The scientist that measured the above data thought a particular histidine in the protein might be responsible for the variability in Kd. He therefore prepared a mutant protein in which the histidine was changed to a valine. If the scientist's hypothesis were correct, what would the graph of Kd as a function of pH look like for the mutant protein?
If correct, the Kd would no longer vary with pH and an analogous graph for the mutant would be a horizontal line. The information available is not sufficient to predict the Kd, only that it is invariant with pH.