Summary + Practice Questions and Answers - Medicinal Chemistry and Biophysics (WBFA056-5)

Medicinal Chemistry and Biophysics

Lectures Groves: Biophysics

Lecture 1: Kinetics and Thermodynamics

80% of all new medicines are small molecules.

Biophysics  use of light, sound, or particle emission (waves)to study a (bio) sample.
- Most of biophysics relies on building simplistic models and measuring differences in energy.

The 3D structure determines biological activity of the drugs by altering  membrane passage,
binding to targets, metabolism, and pharmacokinetics.

Membranes
- LogP can be used to estimate/predict the passive membrane passage of a drug.
- Active transport relies on molecular recognition (shape) by transport proteins.
- Most drugs rely on passive membrane passage.

Log P
- pH of solute chosen to generate neutral molecules.
- Measurement of lipophilicity.
- Example: if Log P = 4.49 than is the drug around 30.000 (10^4.49) times more soluble in
membranes than in water.

Lipinski’s rule of five
Does the drug ‘obey’ Lipinski?
1. Molecular mass less than 500.
2. Log P less then 5.
3. Less then 10 hydrogen bonds acceptors (-O-, -N-, etc).
4. Less than 5 hydrogen bond donors (NH, OH etc).
Good absorption requires good solubility in both water and membranes.

Rule 3 and 4 mean that if there are more acceptors and doners there is more charge which makes
the drug more soluble and a lower Log P which is unwanted.

Thermodynamics and kinetics
- Thermodynamics describe the equilibrium state  K (big k)
- Kinetics describe the rate (speed) of the process  k (little k)

The more stable the product, the more there is present at equilibrium.

Types of interactions
- Covalent
• Dissociation does NOT occur (one way).
• No equilibrium, only a kinetic rate.
- Non-covalent
• Goes both ways  association and
dissociation.
• Keq is the associating constant also called Ka.
• Ka = 1/ Kd

1

,Problem
At equilibrium, 90% of an equimolar mixture of a drug and its target is in the bound state. What is the
KD of binding in this case?




How about when this is 99.9999%?




Gibbs Free Energy
- Each new target interaction provides a change in Gibb’s Free Energy (ΔG).
- The energy required to create a state from nothing (vacuum).
- Important descriptor of the changes that occur over a binding or reaction process

Which occur naturally?
 a+ b

Which occurs the fastest
a

Which have the lowest KD
b


Thermodynamics and kinetics
Enthalpy and Entropy are the driving forces of a reaction.
Drug design is about making ΔG as small as possible (-500).
- Enthalpy (ΔH)
• ΔH > 0  (heat )energy is absorbed (endothermic).
• ΔH < 0  (heat) energy is released (exothermic)
- Entropy (ΔS)
• Measure of the ordering of the system.
• Reduced ligand flexibility  lower entropy.
• Removal of solvation shell around both binding partners  increased entropy.

Kinetics
- The speed of the reaction depends on the activation energy.
- Equilibrium populations only on energy level changes.
- ΔG = RT Ln KD  ΔG = -RT Ln KA

Non-covalent binding is achieves by many simultaneous interactions between the ligand and the
macromolecule.

Other types of interactions
- Electrostatic interactions (ion-ion)  opposite charges attract and equal charges repel.
- Ion-ion dipole interactions (hydrogen bonds)  contributes to enthalpy, geometry
important.
- Hydrophobic interactions  entropy is the driving force, geometry less important.


2

,Weaker ionic interactions (ΔH) can be compensated for by improves hydrophobics (ΔS) and vice
versa.




Exercise 1
Calculate the ΔG of an interaction with
KD = 10-3 M
KD = 10-7 M
KA = 106 M-1
At room temperature, at zero degrees and at bod temperature.
t = 37 degrees  T = 310 K.
R = 8.314 JK-1mol-1

Answer
ΔG = R x T x Ln (KD) = 8.314 x 293 x Ln(10-3) = -16800
ΔG = R x T x Ln (KD) = 8.314 x 273 x Ln(10-3) = -15700
ΔG = R x T x Ln (KD) = 8.314 x 310 x Ln(10-3) = -17800

ΔG = R x T x Ln (KD) = 8.314 x 293 x Ln10-7 = -39300
ΔG = R x T x Ln (KD) = 8.314 x 273 x Ln10-7 = -36600
ΔG = R x T x Ln (KD) = 8.314 x 310 x Ln10-7 = -41500

ΔG = -R x T x Ln (KA) = -8.314 x 293 x Ln106 = -33700
ΔG = -R x T x Ln (KA) = -8.314 x 273 x Ln106 = -31400
ΔG = -R x T x Ln (KA) = -8.314 x 310 x Ln106 = -35600


Exercise 2
a. Explain why solubility in water as well as in a membrane is necessary for absorption of a drug?
b. How can you explain answer a using the Lipinski’s Rule of Five?

Answer
a. Passage of multiple cell mambranes passievlt requires solubility in both milleus.
b. Log P < 5 and Log P = log (max concentration in otc/max concentration in water unionized).




3

, Test Exam Question
The ΔG of the isolated protein and drug is 32000 J/mol. The ΔG of an equimolar mixture of the
protein and drug is 8000 J/mol. The Enthalpy change upon binding of the drug to a protein is -6000
J/mol at room temperature (298 K, R = 8.134 J K -1 mol-1).
a. Will this drug naturally bind to the target? Express your answer in terms of thermodynamic
parameters.
b. What will be the KD n the scenario describes above?
c. Is this reaction driven by hydrophobic or electrostatic interactions?

Answer
a.
b. ΔG = RT ln KD  KD = exp (ΔG/RT) = = exp (-24000/298 x 8.134) =
exp -10.9 = 0.0186 = 18.6 uM.
c. -TΔS much more negative than ΔH, hydrophobic interactions
correlate with -TΔS, so hydrophobic interactions contribute more.




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