Consider the very simple chemical reaction shown below, where a, b, c, and d ..... However, this problem can be solved by using a competitive binding assay, ...
Exp: Isothermal Titration Calorimetry for Lysozyme/Inhibitors Introduction Isothermal titration calorimetry (ITC) is a relatively new technique that is used to determine thermodynamic parameters for a reaction. Modern, ultrasensitive nano-ITC (< 1 mL scale using M and nM solutions) has found many applications in Biochemistry and Pharmaceutical research: proteinprotein, protein-drug, protein-DNA, protein-ligand and metal-ligand interactions are just a few examples of intermolecular interactions that can be studied. A schematic diagram of the ITC apparatus is shown in Figure 1. Briefly, a reference cell and a sample cell are submerged in an adiabatic chamber. One reagent is contained in the sample cell, and the other reagent is loaded into the injecting syringe. Both reagents are in an identical buffer solution to minimize mixing/dilution enthalpy. In a standard ITC experiment, the automatic injector introduces a known amount of solution into the sample cell using a programmable series of small titrations. A sensitive thermocouple on each cell activates feedback electronics which produce an electrical current to maintain the exact same temperature difference between the two cells. A graph of current vs. time produces an ITC isotherm. The heat enthalpy of the reaction is obtained by integrating the area under the curve, allowing for both exothermic and endothermic reactions to be monitored. Each titration peak is integrated separately, and these values are used to create the sigmoid-shaped curve of the integrated isotherm. The equilibrium constant (Ka) can be determined by the severity of the sigmoid curve, and the stoichiometry (n) can be determined using the inflection point of the sigmoid curve. The experimentally determined enthalpy change (ΔHo) in the standard state and Ka from the ITC analysis can be used to calculate the standard Gibbs free energy difference or the Gibbs free energy of the reaction (ΔGo) and the entropy change (ΔSo) based on the well-known thermodynamic relationships at a constant temperature (T): ΔGo = - RT ln(Ka) = ΔHo - TΔSo. However, it should be noted that Gibbs free energy (G) is a function of the extent of reaction (ξ) such that G is minimized (i.e. dG(ξ)/dξ = 0) for a closed system that reaches equilibrium at constant T and P. Thus, the relationship between Go and Ka originated from “bookkeeping” rather than physics.1
Figure 1: ITC schematic2
Consider the very simple chemical reaction shown below, where a, b, c, and d are stoichiometry variables for the reaction and nA,0, nB,0, nC,0, and nD,0 are the initial numbers of moles for species A, B, C, and D, respectively. aA + bB
cC + dD
The extent of reaction (ξ) is defined in such a way that the final numbers of moles for A, B, C, and D are: nA = nA,0 – aξ, nB = nB,0 – bξ, nC = nC,0 + cξ and nD = nD,0 + dξ assuming the reaction proceeds from left to right. Differentiation of the above equations leads to: dnA = - adξ, dnB = - bdξ, dnC = cdξ, and dnD = ddξ Since G = H - TS dG = dH – TdS –SdT = dE + PdV + VdP – TdS – SdT , since H = E + PV = qrev + rev + PdV+ VdP –TdS –SdT, since E = qrev + rev (i.e. reversible net heat transferred to the system and net work done on the system) = TdS + rev + PdV + VdP –TdS –SdT , since qrev = TdS = rev + PdV+ VdP –SdT = (rev - PdV) + PdV+ VdP –SdT, where rev is defined as non P-V work.
At constant T and P, the above equation becomes to dG = rev = AdnA + BdnB + CdnC +μDdnD, where A, B, C and D are chemical potentials. = - aAdξ - bBdξ +cCdξ + cDdξ = (cC + dD - aA- bB) dξ The equation is similar to G = dG/dξ (or ∂G/∂ξ) = (cC + dD - aA - bB) Since biochemical reactions are in solution, then = + RTln rA[A], where r is the activity coefficient and [A] is the concentration of species A (note that molarity is replaced with molality), and assuming this is an ideal solution (i.e. dilute solution) then = + RT ln [A]. Thus G = 0 leads to Go = cC + dD aA bB = - RT lnQ, where Q = Keq (or Ka) = ([C]c[D]d)/([A]a[B]b), where Q is the quotient. If we consider a simple biochemical reaction with a 1:1 stoichiometry binding: A + B C, where the reaction starts with 1 mole of A and B at 25oC and 1 atm (standard state) and proceeds to completion to yield 1 mole of the complex, C, the extent of reaction ξ is ranges from 0 to 1. The free energy at ξ = 0 is Go for reactants (i.e. Gos) and that at ξ = 1 is Go for products (i.e. Gop). Panel A of the diagram shown below is for a favorable reaction (i.e. Go < 0) in the standard state, which is indicated by the symbol “naught” and G is minimized at equilibrium at constant P and T. Panels B and C show two
extreme cases, in which G is minimized closer to that of reactants or product, which defines the Ka value to be very small or infinite.
Thermodynamic binding parameters provide profound insights into the forces stabilizing ligandbound complex structures and can be applied in the investigation of protein folding and binding. A polypeptide backbone only has a limited number of chemical groups, such as terminal carboxyl and amino groups and side chain amino acids, which can form a set of non-covalent weak interactions including van der Waals, hydrogen bonds, ionic interactions, and dipole-dipole interactions, as well as solvation. The changes of these weak interactions in the unfolded and folded states (in protein folding) and un-ligand and ligand receptors can be detected by ultrasensitive ITC.
In this experiment, you will determine the binding between HEW lysozyme (E.C. 3.2.1.17) and its inhibitors. Lysozyme is a small, stable protein that is found in a variety of species. Animals, insects, and plants use lysozyme as a primary agent to kill invading Gram-positive bacteria. Specifically, lysozyme possesses neuraminidase activity which cleaves glycosidic bonds between two carbohydrate moieties, such as N-acetylglucosamine and N-acetylmuramic acids. This action results in the degradation of the bacterial cell walls. Lysozyme binds to and hydrolyzes N-acetylglucosamine oligosaccharides (NAGn). For example, it converts a NAG hexamer (NAG6) to NAG4 and NAG2 as the major products, and to two NAG3 molecules as the minor product. NAG3 strongly binds to lysozyme, and thus can inhibit the enzymatic activity. A representative set of raw data for the lysozyme/NAG3 experiment conducted at 25oC and pH 5 is illustrated in Figure 2A. The reaction is exothermic as indicated by the negative sign of heat evolved. The heat generated per injection (peak integral) starts to decline and reaches a minimum at a molar ratio (NAG3/lysozyme) of approximately 2.5 (Figure 2B). The heat evolved in the later injections is mainly due to the heat of titrant dilution and mechanical work from syringe injections.
Though the basic principle of ITC is simple, the experimental approach and the associated data fitting is not straight forwarded. In general, an ITC measurement uses a total-fill nature (in contrast to partial-fill), in which the working volume of the sample (Vcell) is filled with the receptor solution before the titration. Each subsequent injection acts to drive an equal amount of liquid out to an inactive overfill tube on the top of sample cell (in this region, the reaction heat is not detected). Therefore, each peak in Figure 2A corresponds to the heat evolved on addition of an aliquot of ligand (L) to receptor (M) and its integration yields apparent heat change, qi,app, between the (i-1)th and ith injections. qi,app = qi – qi-1
eq. 1
Given thatqi,app is proportional to the change of bound ligand for the ith injection (i.e. [Li]bound = [Li]bound -[Li-1]bound), Vcell, and the apparent molar enthalpy of association (Happ), the following equation can be applied. qi,app = qi – qi-1 = [Li]bound Vcell Happ
eq. 2
Since Vcell and Happ are kept constant in the measured condition, one has to evaluate[Li]bound. Consider the following binding example with one set of n identical binding sites: L + M ML
eq. 3
where ML is the ligand/receptor complex. We could quickly obtain the following equations (refer to your Biochemistry Textbook for the derivation):
Ka= 1/Kd = Θi/((1-Θi)[Li])
eq. 4
where Θi is the fraction of receptor binding sites (ranging from 0 to 1) occupied by the ligand after the ith injection, [Li] is the concentration of free ligand after the ith injection, and Kd is the dissociation binding constant for the ligand/receptor complex.
Therefore, the total ligand concentration in Vcell (i.e. [Li]tot) is comprised both of the free and bound ligands after the ith injection and is equal to [Li]tot = [Li] + [Li]bound and [Li] = [Li]tot - n Θi [M]tot
eq. 5
where [Li]bound = n Θi [M]tot. Combining eq. 4 and 5 gives Θi2 - Θi (1 + [Li]tot/n[Mi]tot + 1/nKa[Mi]tot) + [Li]tot/n[Mi]tot = 0
eq. 6
th
where [Mi]tot is the total concentration of M after the i injection. Based on eq. 2, eq. 6 can be extended to qi,app = n ([Mi]totΘi - [Mi-1]totΘi-1) Vcell Happ
eq. 7
Because ITC analysis software generally presents qi,app versus the molar ratio of ligand and receptor (Xi = [Li]tot/[Mi]tot), eq. 6 can be presented by the following if we define ri = 1/Ka[Mi]tot Θi2 - Θi (1 + Xi/n + r i/n) + Xi/n = 0
eq. 8
Note that the reciprocal of r happens to be c (= Ka[Mi]tot), a “window” value used to evaluate optimal conditions for an ITC experiment (see Appendix 1 for some interesting situations). Also note that qi,app is the experimental value obtained from the integration of each peak from ITC and Vcell is constant such that a non-linear regression method that fulfills both eq. 7 and 8 should yield Ka, Happ, and n. Here n is treated as a float number (variable), which can provide valuable information: an n value that deviates significantly from the actual stoichiometry indicates problems in determining the accurate sample concentrations. The concentrations of ligand and receptor remaining in the sample cell after the ith injection (i.e. [Li]tot and [Mi]tot) have to be calculated according to liquid replacement. A simple presentation using mass conservation gives the concentration after the ith injection: [M]tot Vcell = [Mi]tot Vcell + ½ ([M]tot + [Mi]tot) Vi
eq. 9
where [M]tot is the total concentration of M before the 1st injection and Vi is the sum of the volume of each injection after the ith injection. The second term in eq. 9 expresses that the concentration of receptor in Vi is the mean of the beginning and present concentrations. And therefore [Mi]tot = [M]tot (1- Vi/2Vcell) / (1 + Vi/2Vcell)
eq. 10
Similarly, the actual ligand concentration in the sample cell after the ith injection, [Li]tot, is derived from the hypothetical bulk concentration [L] itot , in which assumes that all injected ligand remain in the sample cell (i.e. [L] itot = concentration of ligand in the syringe x Vi / Vcell). [L] itot Vcell = [Li]totVcell + (½ ([Li])tot Vi)
eq. 11
[Li]tot = [L] itot (1/ (1 + Vi/2Vcell))
eq. 12
For a given situation in which the heat evolved in a fast-binding reaction also contributes to the ITC signal before it passes out of the working volume, further correction on the individual heat evolved is required. Here the H determined from the fitting is the apparent enthalpy difference (Happ) because it includes other enthalpy sources, such as dilution (Hd) and buffer protonation/ionization (Hi). The correction for Hd is made with an identical titration without receptor, in which the heat changes due to dilution for each injection (qi,d) are subtracted from qi,app. The correction for Hi will require performing an identical experiment in different buffers with the known values of Hi. Thus, the actual binding enthalpy (Hb) can be determined from the following equation: Hb = Happ - Hd - nHHi
eq. 13
where nH is the number of protons released or absorbed in the complex formation. In the case that there is very little protonation in the lysozyme/NAG3 complex at pH 5, and Hd is corrected by control subtraction, the determined Happ becomes Hb. Note that this correction is only applied to a twocomponent system. For a three-component system (lysozyme, NAG3, NAG) described below, the determined Happ value is still an apparent enthalpy difference, attributed from both lysozyme/NAG3 and lysozyme/NAG. Data analysis is required to obtain the Hb value for each complex. In short, an ITC experiment starts with a full Vcell, and each injection replaces an identical injected volume in the sample cell that minimally decreases ligand and receptor concentrations, leading to less detected heat than what is needed to be considered. In order to fit ITC data, a program generally involves: 1) initial guess for the n, Ka, and Happ values. 2) calculation of qi for each injection and comparison with the values determined from ITC. 3) improvement of the parameters using standard Marquardt methods. 4) repetition of steps 1-3 until no further significant improvement.
In this exercise, you will use NanoAnalyze software which comes with the ITC instrument to perform the fitting (see the manual from TA instruments). An example of the integrated and dilution corrected heat for each injection versus molar ratio is shown in Figure 2B.
Determination of Extremely Weak Binding by ITC Measuring extremely low binding constants by a direct titration possesses difficulties, such as the large amounts of ligand and receptor molecules required to achieve a suitable detection signal and ligandbinding saturation. However, this problem can be solved by using a competitive binding assay, in which a strong-binding ligand is titrated to the receptor in the presence of a weak-binding ligand. A decrease in apparent binding affinity is observed since the strong-binding ligand has to compete for the same binding site occupied by the weak-binding ligand. In addition to NAG3 binding, lysozyme also binds NAG and NAG2, albeit with a much weaker interaction. In this exercise, you will conduct a competitive binding experiment by titrating a NAG3 solution into a lysozyme solution complexed with NAG so that the Ka value of lysozyme/NAG (i.e. KNAG) can be determined by the following equation.
KNAG3 = Kapp (1 + KNAG [NAG])
eq. 14
where KNAG3 and KNAG are Kas for NAG3 and NAG to lysozyme, respectively, and Kapp is the apparent binding constant determined in the presence of a specific concentration of NAG. K NAG can be determined more precisely with a set of data measured at different NAG concentrations as shown in eq. 14.
K NAG3 K app
= 1 + KNAG [NAG]
eq. 15
By plotting KNAG3/Kobs versus [NAG], a linear correlation is established with a slope representing KNAG. The mathematical derivation of eq. 14 and 15 can be found in Appendix 2. The competitive binding assay does not only provide Ka for the weaker binding, but also provides information about the weak interactions between the enzyme and substrate (inhibitor). For example, the weak interactions in the initial weaker enzyme/inhibitor complex are interrupted during the competitive binding and subsequently reformed in the strong enzyme/inhibitor complex. Thus, if the weak binding interactions in the strong and weak enzyme/inhibitor complex are different, the heat evolved in a competitive binding experiment can be detected by ITC. Thermodynamic parameters G, H, and S are state functions such that Go, Ho, and So for NAG3 binding to lysozyme (i.e. GNAG3, HNAG3, and SNAG3) can be expressed by the following: GNAG3 = Gocomplex - Golysozyme - GoNAG3 = HNAG3 - TSNAG3
eq. 16
HNAG3 = Hocomplex - (Holysozyme + HoNAG3)
eq. 17
SNAG3 = Socomplex - (Solysozyme + SoNAG3)
eq. 18
where Gocomplex, Golysozyme, GoNAG3, Hocomplex, Holysozyme, HoNAG3, Socomplex, Solysozyme, and SoNAG3 represent binding free energies of the complex, lysozyme, and NAG3; conformational (potential) energies of the complex, lysozyme, and NAG3; and the entropies of the complex, lysozyme, and NAG3, respectively in the solution.
For HNAG3, the enthalpy contribution includes ionic, electrostatic (Coulomb), van der Waals, polarization of the interacting groups, and solvation (including hydrogen bonds) between the unbound and bound ligand receptors. In fact, the understanding of the correlation between structural components to thermodynamics is currently beyond theoretical and experimental approaches because one has to include the formation and breaking of all non-covalent bonds, such as 1) the loss of protein-solvent hydrogen bonds and van der Waals, 2) the loss of ligand-solvent interactions, 3) the formation of protein-ligand bonds and van der Waals, and 4) the reorganization of bonds within the protein and hydrogen-bonding network surrounding the protein and ligand (solvation), etc. Here, we assume that the enthalpy change is purely from hydrogen bond formation in the binding interface as seen in the crystal structure. It is important to note that the breakdown or formation of hydrogen bonds determined in ITC is not necessarily located in the interface between the ligand and receptor. In a situation in which there is no net change of the number of hydrogen bonds in the binding interface, the receptor (or ligand) still can undergo a conformational change in the complex which would lead to either more or less hydrogen bond formation. Furthermore, not all hydrogen bond energies are equal.
HAZARDS The buffer used in this laboratory was prepared by the TA in order to avoid buffer mismatch for the ITC titration. However, if you are instructed to make such a buffer, proceed with caution given the solid sodium hydroxide (NaOH), concentrated phosphoric acid, and acetic acid used in buffer preparation are corrosive and can cause severe burns. Lysozyme may cause skin, eye, or digestion tract irritation. NAG and NAG3 are not hazardous according to OSHA criteria. The needle of the sample loading syringe is sharp and should be used with caution. All students must wear safety eye protection and gloves.
General procedure HEW lysozyme and NAG were purchased from Sigma Aldrich (St. Louis, MO) and NAG3 was purchased from Seikagaku Biobusiness Corporation (Tokyo, Japan). You will use a Nano-ITC2G (TA Instruments, New Castle, DE) with a Hastelloy sample cell. To prepare samples for ITC, stock solutions of 1.4 mM HEW lysozyme, 40 mM NAG, and 12 mM NAG3 will be prepared in 0.1 M sodium acetate, pH 5.0 by your TA or instructor. It is crucial to use identical buffers and stock solutions for sample preparation, to avoid any heat evolved by mismatched buffers. All samples need to be degassed before loading. The solutions of lysozyme and NAG3 will be loaded into the sample cell and the titration syringe, respectively. The titration is performed by injecting 10 L 2.4 mM NAG3 solution into 0.14 mM lysozyme every 5 minutes for a total of 25 injections at 25oC. The competitive experiments will be conducted in identical conditions, except that both lysozyme and NAG3 solutions contain NAG (5, 10, or 20 mM). The raw data obtained from ITC measurements is required to be corrected for the heat evolved from titrant dilution, which was obtained by injecting NAG3 to a buffer containing no protein. In principle, two separate ITC experiments are required to generate a binding isotherm. However, in this exercise, you will make the correction of heat dilution by subtracting your raw data with a file generated
from the control titration previously completed by your TA. Alternatively, we could estimate the heat of titrant dilution from the final injections during the fitting process, in which the binding is already complete. The error associated with this approach is within 3%. However, the first approach is universally recommended.
Procedure (1) You will receive stock solutions of NAG3 (12 mM), NAG (40 mM), and lysozyme (1.4 mM) from your TA. All stock solutions were prepared in an identical buffer, 0.1 M sodium acetate, pH 5.0. Calculate the volumes of the stock solutions that are needed to make 1.5 mL 0.14 mM lysozyme and 500 L 2.4 mM NAG3. In the case that you will be working on a competitive binding experiment, you will be informed by your TA or instructor to make appropriate changes. An example to make 0.14 mM lysozyme sample without NAG is shown below.
0.1 M sodium acetate, pH 5.0 (L) 1350
1.4 mM lysozyme in 0.1 M sodium acetate, pH 5.0 (L) 150
40 mM NAG in 0.1 M sodium acetate, pH 5.0 (L) 0
Total volume (L) 1500
(2) Place all samples in a vacuum chamber and start degassing for at least 20 min. (3) During the degassing process, clean up the sample cell by rinsing with water and then buffer. In the end, you will need to remove all remaining solution using the same vacuum pump. (4) Load lysozyme into the sample cell and NAG3 to the syringe. Your instructor will show you the skills needed for appropriately loading both samples. Be aware that the instrument and accessories are very expensive; extra care should be applied. It is also very important to practice using water or a BSA solution before working with your real samples. (5) Launch the ITC run program from the Desktop. You will choose the following settings (underlined) Stirring Rate (rpm): 200 Syringe size (l): 250 (100 or 250) Experimetnal type: Incremental Titration (default) Data interval (s): 1 (default) Check on Save Experiments in Data Protection, and type 1 min in Interval. Experiment detail: type the concentrations of the syringe and sample cell in the appropriate boxes. For the lysozyme/NAG3 experiment, these are 2.4 mM and 0.14 mM for the sample cell and syringe, respectively. You could include any additional information in the Comments area. Note that for now, Stirrer On should NOT be checked. To program the injection, click “Insert”, then add the following: Inject Intervals (s): 300 Inject Volume (l): 10
Number of Injections: 25 Note: The total volume of injections cannot exceed 250 L in our case. (7) Change the following settings to Start Delay (s): 3600 Initial Baseline (s): 300 Final Baseline (s): 300 (8) Check Stirrer On and click on the GO symbol (green arrow) and provide a filename. The machine will automatically complete each programmed measurement. (9) Once the experiment is done, check the following before you clean up your experiment. i. The machine is stopped (the RED button should be highlighted). ii. Stirrer On should be UNCHECKED. iii. Press buret up by clicking the blue arrow symbol. (10) Clean up the sample cell by rinsing with 100 mL dilute detergent solution followed by 1 L denionized water.
Data Analysis (1) Load your lysozyme data into NanoAnalyze and make sure all parameters are correct (for example, the concentrations of lysozyme and NAG3, and the volume of individual injections). Follow the procedure in the notes from the company (NanoAnalyze Software.pdf in BlackBoard), fit the data and obtain the following values: n, Ka, and Ho. Note the heat of dilution is corrected by subtracting a blank file, a control experiment done by the TA, from your raw data. (2) Calculate Go and So (R = 8.314 J/mol.K). If you performed a competitive binding, then determine the binding constant for lysozyme/NAG (KNAG) once you obtain all data from each group. Perform the analysis using eq. 15. (3) Determine the heat enthalpy difference (HNAG) for the lysozyme/NAG complex. You can use a direct titration approach just as you did for lysozyme/NAG3. However, one could also determine HNAG using Hess’s Law. For example, the enthalpy difference due to completely replacing the weak inhibitor by a stronger inhibitor (Hrep) can be obtained by titrating NAG3 to a solution containing lysozyme and a very large excess of NAG to ensure that all lysozyme molecules initially are occupied with NAG. We could then present the experiment with the following two chemical reactions: Lysozyme + NAG3 lysozyme/NAG3
HNAG3
eq. 19
Lysozyme/NAG + NAG3 lysozyme/NAG3 + NAG
Hrep
eq. 20
If we change the direction of the second chemical reaction (i.e. eq. 20) and then combine with eq. 19, we obtain the following equation. Lysosome + NAG lysozyme/NAG
HNAG = HNAG3 – Hrep
eq. 21
Unfortunately in our case, the weak inhibitor has an extremely weak binding affinity, which makes it impossible to determine Hrep because it requires greater than 1 M NAG to reach 98 % saturation (the value is estimated from the Ka value of lysozyme/NAG) for the competitive binding experiment. But we could use an alternative approach. The Ka for lysozyme/NAG was determined to be ~50 M-1 (your data may differ slightly), which determines the fraction of lysozyme initially occupied with the NAG monomer (flysozyme/NAG) before titration (see Appendix 3 for the calculation). Thus, the Ho measured in the presence of 20 mM NAG is attributed to (1-f) x 100% from the binding of unbound lysozyme to the titrant NAG3 and f x 100% from the replacement of lysozyme bound-NAG with NAG3. The relationship can be presented as in eq. 22. H20mM = (1-f) x HNAG3 + f x (HNAG3 – HNAG) = HNAG3 – f x HNAG
eq. 22
where H20mM is the heat enthalpy change value determined from the experiment containing 20 mM NAG.
A more general equation can be expressed as: HNAG = (HNAG3- H20mM)/f
eq. 23
(4) Now assume that only new hydrogen bond formation seen in the crystal structure contributes to H. Calculate the hydrogen bond energy (in kJ/mol) for the lysozyme/NAG3 and lysozyme/NAG complexes. Recall that the number of hydrogen bonds can be visualized by the modeling experiment in PyMol. (5) You will need to correlate your ITC data and the visualization of lysozyme/NAG3 and lysozyme/NAG done with PyMol. As addressed in introduction, we assume that the H difference is purely from the change of the number of hydrogen bonds formed between the ligand and receptor. Explain qualitatively why NAG3 binds tighter than NAG to lysozyme. Explain the correlation between Ho values and hydrogen bond formation. The reported Ho for
(i)
lysozyme/NAG was -24 kJ/mol, which may be different than your determined value. Give a possible explanation for such a discrepancy. (ii)
If you determined the enthalpy difference (Happ) for the competitive binding experiment at the
saturated concentration of NAG (i.e. > 1 M in our case) and the enthalpy difference for the formation of lysozyme/NAG3 (i.e. HNAG3), show your work to determine HNAG for lysozyme/NAG formation.
References (1) Tellinghuisen, J. Achieving Chemical Equilibrium: The Role of Imposed Conditions in the Ammonia Formation Reaction. J. Chem. Educ. 2006, 83, 1090-1093. (2) O’Brien, R.; Ladbury, J. E.; B.Z., C. In Protein-Ligand interactions: hydrodynamics and calorimetry Harding, S. E., Chowdry, B., Eds.; Oxford University Pres: 2000.
Appendix 1. C value Recall eq. 7 and 8 from above, where we define ri = 1/Ka[Mi]tot qi,app = n ([Mi]totΘi - [Mi-1]totΘi-1) Vcell Happ
eq. 7
Θi2 - Θi (1 + Xi/n + r i/n) + Xi/n = 0
eq. 8
There are two very interesting cases: (1) For a very strong binding, c = ∞ (r = 0) and assuming n = 1, then we have Θi2 - Θi(1+ Xi+ ri) + Xi = 0 Θi =
(1 Xi ri ) - (1 Xi ri ) 2 - 4Xi 2
Here, two situations can be emphasized: (i) Xi = 0 (i.e. [Li]tot = 0) Θi =
(1) - (1) 2 0 2
This represents the state before titration where the fraction of ligand-bound protein is zero.
(ii) Xi = 1 (i.e. [Li]tot = [Mi]tot) Θi =
(1 1) - (1 1) 2 - 4 1 2
This represents the state when all protein molecules are in the bound form where the final concentrations of ligand and receptor are equal, consistent with a very tight binding. The ITC thermogram will have a sharp transition, which cannot be deconvoluted to obtain Ka. However, there is no such problem in determining H. (2) For a very weak binding, c = 0 (r = ∞) and assuming n = 1, then we have Θi2 - Θi (1+ Xi + ri) + Xi = 0 Θi =
(ri Xi ) - (ri Xi ) 2 - 4Xi 0 2
In such a situation, the fraction of ligand-bound protein is close to zero initially and increases insignificantly for the sequential injections, which is challenging to be detected in ITC. Also, saturation occurs only when the concentration of ligand (i.e. Xi) is compatible to ri.
In summary, to obtain the Ka value from the ITC thermogram, the c value ideally must be between 1 to 1000, which can be simulated by the above equation.
Appendix 2. Competitive Binding Equation For a ligand (L) and receptor (R), the chemical equation and equilibrium can be presented as
R + L = RL
KL =
[ RL ] [ R][ L]
eq. 24
where KL is the association binding constant for the ligand/receptor complexes, and [R], [L], and [RL] are the concentrations of free receptor, free ligand, and ligand/receptor complex, respectively.
We learn from a biochemistry course that a binding isotherm for such a binding described in eq. 24 can be expressed as
=
[ L] [ L] 1 / K a
eq. 25
where (theta) is the fraction of receptor with ligand bound (i.e. [RL]/[R] T where [R]T is the total concentration of receptor in free and bound forms) and Ka is the association binding constant. Even though in our experiment, both NAG3 and NAG act as inhibitors to lysozyme, one can view NAG3 as a ligand whereas NAG can be viewed as an inhibitor. Since both NAG3 and NAG occupy an identical lysozyme binding site, this becomes a competitive binding study. For an inhibitor (I), the chemical equation is expressed as
R + I RI
KI =
[ RI ] [ R ][ I ]
eq. 26
where KI is the association binding constant for the inhibitor/receptor complexes, and [R], [I], and [RI] are the concentrations of free receptor, free inhibitor, and inhibitor/receptor complex, respectively.
Eq. 25 now can be manipulated to include the inhibitor, in which Ka now becomes the apparent binding constant (Kapp), which can be analyzed with a normal 1:1 binding isotherm. Note that K app does not have a mathematical relationship to the receptor and ligand (or inhibitor) as shown in eq. 24 or 25.
=
[ L] [ RI ] [ RI ] = = [ R]total [ R] [ RL ] [ RI ] [ L] K app
since [R]total = [R] + [RL] + [RI]
eq. 27
Solving eq. 27 gives
[RL][L] + [RL]/Kapp = [L][R] + [L][RL] + [L][RI]
eq. 28
Canceling the term of [RL][L] on both sides and dividing both sides by [RL] gives
1 [ R][ L] [ R][ I ] = + K app [ RL ] [ RI ]
eq. 29
Substituting KL for [RL]/ [L][R] and KI[R][I] for [RI] gives
1 [ L]K I [ R][ I ] 1 1 K [I ] = + = + I K app [ RL ] KL KL KL
eq. 30
Rearrangement of eq. 30 gives
KL = 1 + KI[I] K app
eq. 31
Appendix 3. Calculation of fraction of receptor bound with ligand (or inhibitor) For a 1:1 ligand-receptor binding as shown below: R
+
L
RL
Ka =
[ RL ] [ R][ L]
If the initial concentrations for L and R are [L]i and [R]i, respectively, and the final concentration of the RL complex is x after equilibrium, the following conditions hold. R
+
L
RL
Initial conc.
[R]i
[L]i
0
Final conc.
[R]i – x
[L]i – x
x
Therefore, Ka =
x ([ R]i x)([ L]i x)
Rearrangement of the above equation yields the following equation. x2 – ([R]i + [L]i + 1/Ka )x + [R]i [L]i = 0 Recall that the solution for the quadratic equation ax2+bx+ c is
So
b b 2 4ac . 2a
([R]i [L]i 1/K a ) ([R]i [L]i 1/K a ) 2 4[R]i [L]i x= 2
And the fraction f = x/[R]i
