Isothermal titration calorimetry in drug development
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Isothermal titration calorimetry in drug development
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[post_date] => 2025-01-10 06:34:20
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[post_content] => Practice Passage (Question 1-6)
*This passage is the property of Khan Academy and has been reformatted into an AAMC-style interface in their entirety by MedLife Mastery. MedLife Mastery does not endorse and is not an affiliate of Khan Academy.
Isothermal Titration Calorimetry (ITC) is a powerful technique used in pharmaceutical research to identify lead compounds for therapeutic intervention and to study the binding affinity and selectivity of drug candidates. Additionally, ITC can be employed to study enzyme kinetics and confirm mechanisms of action. Unlike earlier assays, ITC is more versatile and can be applied to a wide range of ligands, including proteins, lipids, DNA, and ions, without restrictions on molecular weight.
A typical isothermal titration calorimeter consists of several key components, as shown in Figure 1: the reference cell, sample cell, and an adiabatic shield or jacket. The sample cell, housed within the shield, contains the protein target dissolved in a buffer, while the reference cell holds just the buffer solution. During the experiment, the ligand is gradually titrated into the sample cell, where it interacts with the protein target.
Attribution: Spinal83, CC-BY-SA 3.0.
Figure 1. Schematic diagram of an isothermal titration calorimetry instrument
During ITC, a constant power is supplied to the heater of the reference cell. As the ligand is added and binds to the protein in the sample cell, this binding interaction releases or absorbs heat, causing a temperature difference between the sample and reference cells. To maintain equal temperatures between the two cells, the instrument adjusts the power supplied to the feedback heater of the sample cell. The amount of power needed to restore temperature balance is measured and used to calculate the thermodynamic parameters of the ligand-protein binding interaction.
At the end of the procedure, the following parameters can be attained: affinity constant (Ka), enthalpy changes (ΔH), and binding stoichiometry (n). ∆S and ∆G can be calculated from the measured parameters. The binding energy can also be quantified by an enthalpic and entropic term. The enthalpic term (ΔH) is a measure of the changes in hydrogen and van der Waals bonding, and the entropic term (-TΔS), on the other hand, is an indication of changes in hydrophobic interaction and conformational changes.
A research group used ITC to investigate the role of two small molecule inhibitors, PES and VER-155008, both of which are proposed to inhibit heat shock protein 70 (Hsp70). The molecular chaperones of the Hsp70 family have traditionally been targeted for anti-cancer therapy, and past research has demonstrated that a significant reduction in cancer cell viability necessitates the simultaneous knockdown of both Hsp70 and heat shock cognate 70 (Hsc70), a closely related paralog that also plays a crucial role in cellular stress responses.
Figure 2 shows the ITC results of the interaction between ADP, VER-155008, and PES with full-length human Hsp70. In the case of PES, titration was performed in the presence of ADP. The upper graph shows raw calorimetric data, and the bottom graph shows integration of peaks, with thermodynamic parameters indicated in the center panel.
Attribution: Rainer Schlecht, CC-BY 4.0
Figure 2. ITC results
[post_title] => Isothermal titration calorimetry in drug development
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[question] => In Figure 2, each dip represents an injection of the binding agent, and each subsequent injection produces a smaller energy change due to saturation of the macromolecule. After 45 minutes, with the continued injections in the ADP trial, what enthalpy is most likely observed?
[value] => Array
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[answer] => 3
[description] => Reason for the Correct Answer:
Every dip corresponds to an injection of the binding agent. At first, the energy change is attributable to binding of the binding agent to the macromolecule of interest. However, as time progresses, the macromolecule in the sample cell is getting saturated with the binding agent.
When saturation occurs, the small enthalpy change recorded is no longer attributable to binding. This occurs around 45 minutes for the ADP graph.
There is no vaporization involved in this procedure, so the heat of vaporization is not observed.
Heat of hydration refers to the energy change when a substance dissolves in water, which is not being measured as the binding agent is already dissolved in solution.
After 45 minutes of continued injections in the ADP trial, the most likely enthalpy observed would be the enthalpy of dilution. At this point, the macromolecule is saturated, so the heat change primarily reflects the dilution of the binding agent in the solution, rather than any binding interaction.
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[answers] => Array
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[0] => Array
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[each_answer] => A. Binding energy
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[1] => Array
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[each_answer] => B. Heat of vaporization
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[2] => Array
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[each_answer] => C. Enthalpy of dilution
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[3] => Array
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[each_answer] => D. Heat of hydration
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[1] => Array
(
[quiz_unique_key] => 3873426850
[question] => Under physiological conditions, what change in free binding energy (∆G) in kcal/mol is needed to change the affinity constant (Kₐ) 100-fold (R = 1.98 cal/mol∙K; 2.303⋅logx = lnx)?
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[description] => Reason for the Correct Answer:
An increase in the affinity constant will be associated with a decrease in Gibbs free energy of binding, as the process is becoming more favorable.
ΔG = – RTlnKₐ is the relationship between the free binding energy (∆G) and the affinity constant (Kₐ).
If we change the affinity constant 100-fold, then the change in Gibbs free energy equals – RTln(100Kₐ) – (-RTln(Kₐ)), or – [RTln(100Kₐ) – RTln(Kₐ)].
This can be simplified to – RTln(100Ka/Kₐ) = – RTln(100).
Substitute ln(100) with 2.303⋅log(100), which is given in the question stem. Since log(100) = 2, ΔG = – RT(2.303)(2).
To calculate RT, be sure to convert a body temperature of 37°C to an absolute temperature of 310 K to plug in for T and add in the value for R, so ΔG = – (1.98)(310)(2.303)(2) kcal/mol.
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[answers] => Array
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[0] => Array
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[each_answer] => A. – (1.98)(310)(2.303)(2) cal/mol
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[1] => Array
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[each_answer] => B. +(1.98)(273)(2.303)(2) cal/mol
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[2] => Array
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[each_answer] => C. – (1.98)(273)(2.303)(1/2) J/mol
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[each_answer] => D. – (1.98)(310)(2.303)(1/2) cal/mol
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[quiz_unique_key] => 83407773
[question] => Which of the following conditions is specified for standard state conditions?
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[answer] => 1
[description] => Reason for the Correct Answer:
Values of thermodynamic quantities are commonly expressed for standard state conditions – here, a superscript circle is used to denote a thermodynamic quantity that is under standard state conditions: ΔH°, ΔS°, ΔG°.
STP is short for Standard Temperature and Pressure, which is defined to be 273 K (0° Celsius) and 1 atm pressure. STP is often used for measuring gas density and volume using the Ideal Gas Law, where 1 mole of ideal gas occupies 22.4 L.
The standard state temperature is 25°C (298 K). Note that temperature is not specified for standard state conditions, but most tables are compiled for this temperature.
All gases are at 1 atmosphere of pressure, and all liquids and gases are pure.
The energy of formation of an element in its normal state is defined as zero. All elements are in their elemental state, not always gaseous.
For standard conditions, all solutions are at 1 molar concentration, and not 1 mole of each reactant.
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[answers] => Array
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[0] => Array
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[each_answer] => A. 1 molar concentration of all solutions
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[1] => Array
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[each_answer] => B. 1 mole of each reactant
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[2] => Array
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[each_answer] => C. Elements in their gaseous state
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[3] => Array
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[each_answer] => D. Ambient temperature of 273 K
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[3] => Array
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[quiz_unique_key] => 2377279144
[question] => The apparatus uses an adiabatic shield to enclose the sample and reference cells. Which of the following expressions holds true for the system?
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[answer] => 3
[description] => Reason for the Correct Answer:
The first law of Thermodynamics states that ∆U = Q – W, where Q is heat absorbed or released by the system and W is the work done by the system.
The other permutation of the first law states that ∆U = Q + W, where W is work done on the system.
When the heat flow is equal to the work done (Q = W), the process must be isothermal. It can be deduced that ∆U = 0 or ∆T = 0.
When the change in internal energy is equal to the heat flow of the system (∆U = Q), then there can be no work done by or on the system. Since work is the PV-work kind (W=0), there can be no change in volume (∆V = 0), and the process must be isochoric.
When there is a decrease in internal energy, it can be attributed to work done by the system on the surroundings. Implicit is that there is no heat exchange or Q = 0, which is called adiabatic.
An adiabatic process is one in which there is no heat exchange (Q = 0) between the system and surroundings. If W is the work done by the system and no heat is exchanged, then the equation ∆U = Q – W would simplify to ∆U = -W.
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[answers] => Array
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[0] => Array
(
[each_answer] => A. Any decrease in internal energy is offset by a compression that increases the temperature.
)
[1] => Array
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[each_answer] => B. The heat flow and work done exactly cancel each other out.
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[2] => Array
(
[each_answer] => C. Any increase in internal energy is equal to the negative of work done by the system.
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[3] => Array
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[each_answer] => D. The change in internal energy is equal to the heat flow of the system.
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[4] => Array
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[quiz_unique_key] => 2261298308
[question] => ΔG is a measure of overall binding affinity. Based on the information in the passage, which of the following configurations would correspond to a high binding affinity?
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[answer] => 1
[description] => Reason for the Correct Answer:
The binding free energy can be quantified by an enthalpic and entropic term. The more negative ΔG is, the higher the binding affinity.
The enthalpic term (ΔH) is a measure of the energy content of the bonds broken and created, and enthalpic interactions are driven by H-bonding and van der Waals interactions.
Good H-bonding would be more exothermic than van der Waals bonding, so good H-bonding would correspond to a high binding affinity.
The entropic term (-TΔS), on the other hand, is an indication of changes in hydrophobic interaction and conformational changes.
Favorable entropic binding affinity is derived from hydrophobic interactions. The release of tightly bound water upon ligand binding, i.e. in hydrophobic binding pockets, contribute significantly to the entropic term.
Unfavorable entropy results from an overall decrease in the degrees of freedom of the interacting species. It means that the protein has undergone a conformational change or that the ligand is very flexible.
Good H-bonding creates a favorable enthalpic term, and hydrophobic interaction and little to no conformational change creates a favorable entropic term. Both together result in a higher binding affinity.
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[answers] => Array
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[0] => Array
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[each_answer] => A. Strong hydrogen-bonding and hydrophobic interactions
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[1] => Array
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[each_answer] => B. Strong van der Waals bonding, and possibly “rigid body”
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[2] => Array
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[each_answer] => C. Strong hydrogen-bonding coupled to conformational change
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[each_answer] => D. Strong hydrogen-bonding coupled to conformational change
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[quiz_unique_key] => 955613802
[question] => Which statement most accurately describes the data in Figure 2?
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[answer] => 1
[description] => Reason for the Correct Answer:
The interaction of Hsp70 with ADP and VER-155008 are both exothermic due to the negative values on the y-axis of the graphs.
Knowing that ΔG = ΔH – TΔS and knowing that ΔG (for a spontaneous process) and ΔH (in the graph) are both negative, ΔS could be positive or negative.
The orange line on the graph shows that 1/Kꭰ (which equals Kₐ) can be determined by the slope of the graph.
If we look at the sigmoidal curves for ADP and VER-155008, it can determined that the curve for ADP has a higher slope than VER-155008. Since dissociation constant is the inverse of this slope as indicated in Figure 2, then VER-155008 has a greater dissociation constant.
The ADP graph has the greatest slope and therefore the highest affinity for Hsp70, and PES does not bind to Hsp70 which can be determined by the lack of any sigmoidal curve in the results.
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[answers] => Array
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[0] => Array
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[each_answer] => A. ADP has the highest affinity for Hsp70, while PES does not bind at all in any specific way to Hsp70.
)
[1] => Array
(
[each_answer] => B. The dissociation constant (Kꭰ) value is greater for ADP than VER-155008 due to the greater slope of the sigmoid curve.
)
[2] => Array
(
[each_answer] => C. The interaction of Hsp70 with VER-155008 is slightly more endothermic than the interaction with ADP.
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[3] => Array
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[each_answer] => D. Knowing that Kꭰ is positive and that ΔG and ΔH are negative for VER-155008, it can be determined that the change in entropy has to be negative.
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