Antibodies Ab are proteins capable of selective binding with specific antigen Ag species (usually pr — Biochemistry Chemistry Question
Analysis of complex formation
Antibodies Ab are proteins capable of selective binding with specific antigen Ag species (usually protein or polysaccharide), thus forming the so-called immune complex Ab*Ag. The binding constant of the process Kb is very high (around 10^9), however, binding is reversible.
Ab + Ag <-> Ab*Ag
Despite of seeming complexity of biological objects, their functional features can often be analyzed by simply treating Ag and Ab as a ligand and complexing agent, respectively, in a common reaction of the Ab*Ag complex formation. Moreover, specific binding of proteins with other ligands (enzyme inhibitors, lipids, metal ions, etc.) can be analyzed by using the same approach.
Express Kb as a function of equilibrium concentrations [Ab], [Ag], [Ab*Ag] (consider that 1 : 1 Ab*Ag complex is formed).
Model Answer
Kb = [Ab*Ag] / ([Ab][Ag])
Parameter r (often represented as r or n) is defined as the average number of Ag molecules bound to one Ab molecule. In the case of only one binding site in Ab, r = [Ab*Ag] / C_Ab. Express r as a function of Kb and equilibrium concentration of the unbound ligand [Ag] for this simplest case of a single binding site in Ab molecule. Assume that Kb remains unchanged in course of the binding process. Draw schematically the r vs [Ag] plot (“titration” curve of Ab with Ag).
Model Answer
r = (Kb * [Ag]) / (Kb * [Ag] + 1). The titration curve is a hyperbolic binding curve asymptotically approaching r = 1.0 as [Ag] increases. Plots for higher Kb values are steeper, reaching saturation at lower free ligand concentrations. [VISUAL]
a) Plot the Experimental data A (see the table below) as [Ab*Ag] / [Ag] vs [Ab*Ag].
b) Express [Ab*Ag] / [Ag] as a function of [Ab*Ag].
c) One of the data points in the Experimental data A set has been determined incorrectly. Encircle this outlier in the plot.
d) Suggest a way for Kb determination from the plot analysis.
e) In the same plot, draw schematically a curve for ADP binding with another ligand, if the latter is characterized by a 10 times higher Kb value (as compared to that for ADP*Mg2+ complex formation).
Experimental data set A
ADP protein binds with Mg2+ in 1:1 complex (single binding site, one Mg2+ per site). Kb is not dependent on r. ADP total concentration is kept constant at 80 µM.
Table of Experimental data set A:
Mg2+ total concentration, (× 10^-6 mol dm-3) | Bound Mg2+ concentration, (× 10^-6 mol dm-3)
20.0 | 11.6
50.0 | 26.0
100 | 42.7
150 | 52.8
200 | 59.0
300 | 61.1
400 | 69.5
[VISUAL]
Model Answer
a) Plotting [Ab*Ag] / [Ag] vs [Ab*Ag] (a Scatchard plot) yields a straight line with a negative slope.
b) [Ab*Ag] / [Ag] = Kb * (C_Ab - [Ab*Ag])
c) Point #6 (Mg2+ total = 300 µM, Bound Mg2+ = 61.1 µM) is determined incorrectly and lies below the linear trend.
d) Kb is determined as the negative of the slope of the Scatchard plot (Kb = 2·10^4 M^-1).
e) With a 10 times higher Kb, the plot is a straight line with a slope and a y-intercept that are both 10 times larger than the original curve (slope = -2·10^5, intercept = 16.4). [VISUAL]
Some antibodies can only bind a single antigen molecule, whereas others bind two (or even more) antigen molecules. Maximal number of Ag molecules that can be bound to a single Ab is referred to as the Ab valence.
a) Derive an expression to be used for determination of the Ab valence from the plot analysis in coordinates [Ab*Ag] / [Ag] vs [Ab*Ag].
b) Plot the Experimental data B using the above coordinates. Determine the enzyme valence.
Experimental data set B
An enzyme binds with its inhibitor I, the binding to different sites is independent, and Kb is the same. Enzyme total concentration is kept constant at 11 ⋅ 10^-6 mol dm-3.
Table of Experimental data set B:
I total concentration, (× 10^-6 mol dm-3) | Free (unbound) I concentration, (× 10^-6 mol dm-3)
5.2 | 2.3
10.4 | 4.95
15.6 | 7.95
20.8 | 11.3
31.2 | 18.9
41.6 | 27.4
62.4 | 45.8
[VISUAL]
Model Answer
a) [Ab*Ag] / [Ag] = Kb * (N * C_Ab - [Ab*Ag]), where N is the antibody valence.
b) Plotting the data in Scatchard coordinates yields a straight line with a slope of -0.0660 (Kb = 6.6·10^4 M^-1) and a y-intercept of 1.46. Using the intercept value: 1.46 = Kb * N * C_Enzyme = 6.6·10^4 * N * 11·10^-6 ⇒ N = 2. The enzyme valence is 2. [VISUAL]
Ab specimen often contains admixtures of other proteins not capable of binding with Ag. Thus, a “known” total Ab concentration includes both functionally active antibodies and unreactive proteins.
a) Suggest a way for determination of the actual Ab concentration from the data analysis in coordinates [Ab*Ag] / [Ag] vs [Ab*Ag].
b) Does the ADP specimen contain any unreactive admixtures (Experimental data A)?
c) Why is it impossible to conclude unambiguously about the presence of unreactive admixtures in the enzyme specimen (Experimental data B)? What helpful (to determine the admixtures concentration) data is missing?
Model Answer
a) Since the x-intercept of the Scatchard plot is equal to N * C_Ab, if the valence N is known, the active concentration C_Ab can be directly calculated from the intercept divided by the binding constant Kb (or found as the x-intercept divided by N).
b) For Set A (N = 1, Kb = 2·10^4), the Scatchard plot has a y-intercept of 1.64. C_Ab = 1.64 / (2·10^4) = 82 µmol L^-1. This is in close agreement with the specified constant concentration of 80 µM, indicating that the ADP protein specimen contains no unreactive admixtures or functionally inactive proteins.
c) It is impossible to determine purity for Set B because the true structural or biochemical valence (N) of the enzyme is not known independently. The valence of N = 2 was calculated under the assumption of 100% enzyme purity. To verify purity, the true valence (the number of binding sites per enzyme molecule) must be known from independent structural or biochemical data.