Samenvatting
Summary Advanced Pharmacology (XM_0112)
- Instelling
- Vrije Universiteit Amsterdam (VU)
This document consists of a summary of the course Advanced Pharmacology, corresponding to the Master Drug Discovery Sciences.
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Voorbeeld van de inhoud
Advanced Pharmacology (XM_0112)Green equations need to be learned by heart!
Lecture 1. Ligand-receptor binding
Equilibrium binding [A] + [R] →[AR]
If a ligand (A) has
an affinity to the Association rate = 𝐾1 × [𝑅] × [𝐴]
receptor (R), it will Dissociation rate = 𝐾2 × [𝐴𝑅]
bind to the
receptor (forming At equilibrium
AR). Over time the Association rate = dissociation rate
specific binding 𝑲𝟏 × [𝑹] × [𝑨] = 𝑲𝟐 × [𝑨𝑹]
increases, but at [𝑹]×[𝑨] 𝒌 𝒌𝟐
some point the [𝑨𝑹]
= 𝒌𝟐 𝒌𝟏
= 𝒌𝒆𝒒
𝟏
association rate is equal to the dissociation rate, resulting in a
flatting binding curve and equilibrium binding. The law of mass action means that the velocity of a
chemical reaction is proportional to the product of concentrations (or mass) of reactants. If the
concentration of a ligand is increased, the association rate is higher and the specific binding curve will
be higher. On the other hand, when the concentration of a ligand is decreased, the association rate is
lower and the specific binding curve will be lower.
Saturation binding
The receptor number is limited in biological/experimental
sample (𝑅𝑡𝑜𝑡𝑎𝑙 or 𝐵𝑚𝑎𝑥 ). If the concentration of ligands (A)
increase, the concentration of the ligand bound to the
receptor (AR) also increases, until 𝑅𝑡𝑜𝑡𝑎𝑙 or 𝐵𝑚𝑎𝑥 is reached.
This is a saturable relationship between [A] and [AR].
At any [A], the receptors are either [R] or [AR], leading to the
fact that [𝑅] + [𝐴𝑅] = 𝑅𝑡𝑜𝑡𝑎𝑙 or 𝐵𝑚𝑎𝑥 .
[𝑹]
The proportion of free receptors (𝑷𝑹 )=
𝑹𝒕𝒐𝒕𝒂𝒍
[𝑨𝑹]
The proportion of occupied receptors (𝑷𝑨𝑹 )= 𝑹
𝒕𝒐𝒕𝒂𝒍
𝑃𝑅 + 𝑃𝐴𝑅 = 1
Mathematical relationship between 𝑃𝐴𝑅 and [A]
[𝑅]×[𝐴] 𝑘 [𝐴]×[𝑅]
[𝐴𝑅]
= 𝑘2 = 𝑘𝑒𝑞 [𝐴𝑅] = 𝑘𝑒𝑞
1
Vermenigvuldigen
Invullen Delen door
[𝐴]×[𝑅] [𝐴] met 𝒌𝒆𝒒
( ) [R] ( )
[𝐴𝑅] [𝐴𝑅] 𝑘𝑒𝑞 𝑘𝑒𝑞 [𝐀]
𝑃𝐴𝑅 = 𝑅 = [𝑅]+[𝐴𝑅] → 𝑃𝐴𝑅 = [𝐴]×[𝑅]
→ 𝑃𝐴𝑅 = [𝐴]
→ 𝑷𝑨𝑹 = 𝒌
𝑡𝑜𝑡𝑎𝑙 [𝑅]+( ) 1+( ) 𝒆𝒒 +[𝑨]
𝑘𝑒𝑞 𝑘𝑒𝑞
1
,Advanced Pharmacology (XM_0112)
Green equations need to be learned by heart!
For a curve plotting 𝑃𝐴𝑅 and [A], the formula
[A]
𝑃𝐴𝑅 = 𝑘 is used. When you want a
𝑒𝑞 +[𝐴]
curve plotting [AR] and [A], the formula
[𝐴]×𝐵𝑚𝑎𝑥
needs to be rewritten to [𝐴𝑅] = 𝑘𝑒𝑞 +[𝐴]
.
This both results in a non-linear curve fitting [A] [𝐴𝑅]
one-site saturation binding and is the occupancy when the 𝑃𝐴𝑅 = 𝑘 𝑃𝐴𝑅 = 𝑅
𝑒𝑞 +[𝐴] 𝑡𝑜𝑡𝑎𝑙
omschrijven
reaction is at equilibrium. [A] [𝐴𝑅] [𝑨]×𝑩𝒎𝒂𝒙
𝑘𝑒𝑞 +[𝐴]
=𝑅 → [𝑨𝑹] = 𝒌𝒆𝒒 +[𝑨]
𝑡𝑜𝑡𝑎𝑙
Scatchard plot
It is hard to obtain information from a non-
linear curve, so ideally a linear curve is formed.
The Scatchard plot makes use of the formula:
[𝐴𝑅] 𝐵𝑚𝑎𝑥 [𝐴𝑅]
[𝐴]
= 𝑘𝑒𝑞
− 𝑘𝑒𝑞
.
[𝐴]×𝐵𝑚𝑎𝑥
[𝐴𝑅] = → [𝐴𝑅] × (𝑘𝑒𝑞 × [𝐴]) = [𝐴] × 𝐵𝑚𝑎𝑥 → ([𝐴𝑅] × 𝑘𝑒𝑞 ) + ([𝐴𝑅] × [𝐴]) =
𝑘𝑒𝑞 ×[𝐴]
([𝐴𝑅]×𝑘𝑒𝑞 ) ([𝐴𝑅]×[𝐴]) [𝐴]×𝐵𝑚𝑎𝑥 ([𝐴𝑅]×𝑘𝑒𝑞 )
[𝐴] × 𝐵𝑚𝑎𝑥 → + = → + [𝐴𝑅] = 𝐵𝑚𝑎𝑥 →
[𝐴] [𝐴] [𝐴] [𝐴]
([𝐴𝑅]×𝑘𝑒𝑞 ) [𝑨𝑹] 𝑩𝒎𝒂𝒙 [𝑨𝑹]
[𝐴]
= 𝐵𝑚𝑎𝑥 − [𝐴𝑅] → [𝑨]
= 𝒌𝒆𝒒
− 𝒌𝒆𝒒
(Scatchard plot)
The Scatchard plot should not be used for
parameter estimations because that will lead to
violation of linear regression, but the Scatchard
plot is useful for visualization of tendencies. The
Scatchard plot is useful in cases in which the
saturation curve seems normal, but a non-linear scatchard plot is formed. This is caused by the fact that
the receptor has 2 independent binding sites, resulting in 2 linear components, together forming a non-
[A]×Bmax ,1 [A]×Bmax,2
linear scatchard plot. In this case: [AR] = + . This can then be solved by blocking
keq,1 ×[A] keq,2 ×[A]
one of the binding sites on the receptor using a ligand that is specific for that binding site. Another
reason can be that the reaction has not reached equilibrium yet.
How to (experimentally) quantify binding?
The quantification of the binding can be done using labeled
ligands. These labels can be either radioactive labels or
fluorescent labels. Radioactive labels are more tolerated for
ligands because the label is very small and does not cause
steric hindrance of the binding. Fluorescent labels on the
other hand are very big and do Normal ligand Ligand with Ligand with fluorescent
sometimes cause steric radioactive label label
hindrance of the binding. For
this reason, when using a
fluorescent label, it is very
important to investigate
whether the label does or
doesn’t cause steric hindrance.
2
,Advanced Pharmacology (XM_0112)
Green equations need to be learned by heart!
Radioligands
When using radioactive labels on ligands, the ligands with
label need to be incubated with the receptors until
equilibrium is reached. Then the free ligands need to be
separated from the bound ligands using centrifugation or
filtration to quantify the bound radioactivity. In this way
the total binding can be measured, which consists of
specific binding and non-specific binding. This
specific binding forms a saturation curve
[𝐴]×𝐵𝑚𝑎𝑥
(𝑌 = 𝑘𝑒𝑞 ×[𝐴]
), while the non-specific binding forms
a linear curve (𝑌 = 𝑁𝑆 × [𝐴]). The total binding can
[𝐴]×𝐵𝑚𝑎𝑥
then be calculated as (𝑌 = 𝑘𝑒𝑞 ×[𝐴]
+ 𝑁𝑆 × [𝐴]).
To quantify the non-specific binding, a saturating concentration of unlabeled ligand [B] can be used.
Competition binding
To perform a competition binding assay, one labeled
[A*] is used with increasing concentrations of
unlabeled [B]. The higher the concentration of
unlabeled [B], less [A*] is able to bind to the receptor,
causing a decrease in fluorescence/radioactivity.
Proportional occupancy 𝑝𝐴𝑅,0 in the absence of [B]
[𝐴 ∗] + [𝑅] ↔ [𝐴𝑅]
[𝐴𝑅]
𝑝𝐴𝑅,0 = 𝑅𝑡𝑜𝑡𝑎𝑎𝑙 = [𝑅] + [𝐴𝑅]
𝑅𝑡𝑜𝑡𝑎𝑎𝑙
[𝑨𝑹]
𝒑𝑨𝑹,𝟎 = [𝑹]+[𝑨𝑹]
Proportional occupancy 𝑝𝐴𝑅,0 in the absence of [B] in cases of an equilibrium
[𝐴∗]×[𝑅]
[𝐴𝑅] = 𝑘𝑒𝑞
[𝐴] Vermenigvuldigen
( ) met 𝒌𝒆𝒒
𝑘𝑒𝑞 [𝐀]
𝑃𝐴𝑅,0 = [𝐴]
→ 𝑷𝑨𝑹,𝟎 =
1+( ) 𝒌𝒆𝒒 +[𝑨]
𝑘𝑒𝑞
Proportional occupancy 𝑝𝐴𝑅,0 in the presence of [B]
[𝐴𝑅]
𝑝𝐴𝑅,𝐵 = 𝑅 𝑅𝑡𝑜𝑡𝑎𝑎𝑙 = [𝑅] + [𝐴𝑅] + [𝐵𝑅]
𝑡𝑜𝑡𝑎𝑎𝑙
[𝑨𝑹]
𝒑𝑨𝑹,𝑩 = [𝑹]+[𝑨𝑹]+[𝑩𝑹]
Proportional occupancy 𝑝𝐴𝑅,0 in the presence of [B] in cases of an equilibrium
[𝐴∗]×[𝑅] [𝐵]×[𝑅]
[𝐴𝑅] = [𝐵𝑅] =
𝑘𝑒𝑞 𝑘𝑒𝑞,𝐵
[𝐴]×[𝑅] [𝑨]
( ) Delen door [R] ( )
𝑘𝑒𝑞 𝒌𝒆𝒒
𝑃𝐴𝑅,𝐵 = → 𝑷𝑨𝑹,𝑩 =
[𝐴]×[𝑅] [𝐵]×[𝑅] [𝑨] [𝑩]
[𝑅]+( )+( ) 𝟏+( )+( )
𝑘𝑒𝑞 𝑘𝑒𝑞,𝐵 𝒌𝒆𝒒 𝒌𝒆𝒒,𝑩
[𝑨]
( ) ×𝑩𝒎𝒂𝒙
𝒌𝒆𝒒
[𝐴𝑅]𝐵 =
[𝑨] [𝑩]
𝟏+( )+( )
𝒌𝒆𝒒 𝒌𝒆𝒒,𝑩
3
, Advanced Pharmacology (XM_0112)
Green equations need to be learned by heart!
In a competition binding assay using one labeled [A*] with increasing
concentrations of unlabeled [B], a curve is formed, plotting 𝑃𝐴𝑅,𝐵 and log[B]. In
this curve the 𝐼𝐶50 is equal to [B] giving 𝑃𝐴𝑅,𝐵 is 0.5 × 𝑃𝐴𝑅,0 .
The 𝐼𝐶50 measures the ability of [B] to compete for [A*] binding. The 𝐼𝐶50 is a
variable and changes depending on the
concentration of [A*]. The 𝐼𝐶50 is thus
not equal to the affinity of the
competitor [B]. If you compensate for
the difference in the concentration of
[A] by the concentration of [B], you get
𝑘𝑒𝑞,𝐵 .
𝑃
𝑃𝐴𝑅,𝐵 = 0.5 × 𝑃𝐴𝑅,0 0.5 = 𝑃𝐴𝑅,𝐵
𝐴𝑅,0
[𝐴] [𝐴]
( ) 𝑰𝑪𝟓𝟎 invullen voor [B] ( )
𝑘𝑒𝑞 𝑘𝑒𝑞
• 𝑃𝐴𝑅,𝐵 = → 𝑃𝐴𝑅,𝐵 =
[𝐴] [𝐵] [𝐴] 𝐼𝐶50
1+( )+( ) 1+( )+( )
𝑘𝑒𝑞 𝑘𝑒𝑞,𝐵 𝑘𝑒𝑞 𝑘𝑒𝑞,𝐵
[𝐴]
( )
𝑘𝑒𝑞
• 𝑃𝐴𝑅,0 = [𝐴]
1+(
𝑘𝑒𝑞
) Invullen in
𝑷
𝟎. 𝟓 = 𝑷 𝑨𝑹,𝑩
𝑨𝑹,𝟎
[𝐴]
( )
𝑘𝑒𝑞
[𝐴] 𝐼𝐶 [𝐴]
1+( )+( 50 ) Omschrijven
1+(
𝑘𝑒𝑞,𝐴
)
𝑘𝑒𝑞 𝑘𝑒𝑞,𝐵
0.5 = [𝐴]
→ 0.5 =
[𝐴] 𝐼𝐶
(
𝑘𝑒𝑞
) 1+(
𝑘𝑒𝑞,𝐴
)+(𝑘 50 )
𝑒𝑞,𝐵
[𝐴]
1+( )
𝑘𝑒𝑞
This leads to the Cheng-Prusoff equation, which is used to calculate the 𝑘𝑒𝑞,𝐵 from the 𝐼𝐶50 :
𝑰𝑪𝟓𝟎
𝒌𝒆𝒒,𝑩 =
[𝑨]
𝟏+( )
𝒌𝒆𝒒,𝑨
The reversed Cheng-Prusoff equation can be used to calculate the 𝐼𝐶50 from the 𝑘𝑒𝑞,𝐵 :
𝒌𝒆𝒒,𝑩 ×[𝑨∗]
𝑰𝑪𝟓𝟎 = 𝒌𝒆𝒒,𝑩 + 𝒌𝒆𝒒,𝑨
4
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