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Samenvatting + oefenvragen en antwoorden- Medicinal Chemistry and Biophysics (WBFA056-5)
Samenvatting + oefenvragen en antwoorden- Medicinal Chemistry and Biophysics (WBFA056-5).
[Meer zien]Voorbeeld 4 van de 46 pagina's
Voorbeeld 4 van de 46 pagina's
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Medicinal Chemistry and BiophysicsLectures 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.
4
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