Transcript Tutorial answers x

Tutorial
Exercise D
Angela Arsova
1.
Make a literature in PubMed or ChEMBL and see
if you can find the determined EC50/Ki/Kd-values
for fenoterol, propranolol, xamoterol and
acebutolol, remember to note the method by
which the EC50/Ki/Kd values have been
determined.
Jozwiak et al. 2011
Fenoterol EC50 1-600 nM
Acebutolol EC50 1μM
Propranolol IC50 1-50 nM
ChEMBL search
Type in compound name
and click at ‘compounds’
Click the ChEMBL ID
Scroll down and find a pie chart with the
bioactivity summary (e.g click at EC50)
2.
Discuss any discrepancies between your
calculated EC50 values and the Ki/Kd values in the
literature.
Litterature:
Fenoterol EC50 1-600 nM
Acebutolol EC50 1μM
Propranolol IC50 1-50 nM
EC50 and IC50
EC50: The EC50 is defined quite simply as the concentration of agonist that provokes a
response halfway between the baseline (Bottom) and maximum response (Top). It is
impossible to define the EC50 until you first define the baseline and maximum
response.
IC50: In many experiments, you vary the concentration of an inhibitor. With more
inhibitor, the response decreases, so the dose-response curve goes downhill. With
such experiments, the midpoint is often called the IC50 ("I" for inhibition) rather than
the EC50 ("E" for effective). This is purely a difference in which abbreviation is used,
with no fundamental difference.
Kd and Ki
Kd: Equillibration dissociation constant for labeled compound
([R]*[L])/ [RL] = Kd
The Kd has a meaning that is easy to understand. Set [Ligand] equal to Kd in the
equation above. The Kd terms cancel out, and you will see that [Receptor]/
[Ligand×Receptor]=1, so [Receptor] equals [Ligand×Receptor]. Since all the receptors
are either free or bound to ligand, this means that half the receptors are free and half
are bound to ligand. In other words, when the concentration of ligand equals the Kd,
half the receptors will be occupied at equilibrium. If the receptors have a high affinity
for the ligand, the Kd will be low, as it will take a low concentration of ligand to bind
half the receptors.
Ki: Equillibration dissociation constant for the unlabeled compound
3.
Explain why the concentration response curves
for fenoterol, propranolol, xamoterol and
acebutolol looks different when IBMX is present.
Is there any difference in the max response and
EC50 values for each drug when IBMX is present?
Concentration-response curves (AUC)
26
Fenoterol
Fenoterol + IBMX
Propranolol
Propranolol + IBMX
Xamoterol
Xamoterol + IBMX
Acebutolol
Acebutolol + IBMX
AUC
24
22
20
18
-10
-5
log(M) Fenoterol
EC50
Fenoterol
7.533e-010
Fenoterol + IBMX
5.048e-011
Propranolol
~ 2.215e-006
Xamoterol
0.0001086
Xamoterol + IBMX
1.490e-007
Acebutolol
7.518e-006
Propranolol + IBMX
1.166e-008
Acebutolol + IBMX
1.748e-007
Concentration-response curves (12 min)
1.0
Fenoterol
Fenoterol + IBMX
Propranolol
Propranolol + IBMX
Xamoterol
AUC
0.9
0.8
Xamoterol + IBMX
0.7
Acebutolol
Acebutolol + IBMX
0.6
-10
-5
log(M) Fenoterol
EC50
Fenoterol
7.067e-010
Fenoterol + IBMX
4.264e-011
Xamoterol
~ 3.686e+009
Propranolol
~ 5.872e-007
Xamoterol + IBMX
1.364e-007
Acebutolol
1.699e-006
Propranolol + IBMX
6.076e-008
Acebutolol + IBMX
2.033e-007
4.
Explain what happens to the fenoterol curves
when you add increasing concentrations of
propranolol. Which type of ligand is
propranolol?
Concentration-response curves (AUC)
28
Fenoterol + buffer
Fenoterol + propranolol (-9)
Fenoterol + propranolol (-8.5)
Fenoterol + propranolol (-8)
Fenoterol + propranolol (-7.5)
Fenoterol + propranolol (-7)
Forskolin
AUC
26
24
22
20
18
-13 -12 -11 -10
-9
-8
-7
-6
-5
-4
log(M) drug
EC50
Fenoterol + buffer
1.043e-009
Fenoterol + propranolol (-8)
6.215e-008
Fenoterol + propranolol (-9)
1.923e-009
Fenoterol + propranolol (-7.5)
2.058e-007
Fenoterol + propranolol (-8.5)
5.859e-009
Fenoterol + propranolol (-7)
8.950e-007
Forskolin
1.801e-006
Concentration-response curves (12 min)
1.0
Fenoterol + buffer
Fenoterol + propranolol (-9)
Fenoterol + propranolol (-8.5)
Fenoterol + propranolol (-8)
Fenoterol + propranolol (-7.5)
Fenoterol + propranolol (-7)
Forskolin
12 min
0.9
0.8
0.7
0.6
-13 -12 -11 -10
-9
-8
-7
-6
-5
-4
log(M) drug
EC50
Fenoterol + buffer
1.004e-009
Fenoterol + propranolol (-8)
6.953e-008
Fenoterol + propranolol (-9)
2.152e-009
Fenoterol + propranolol (-7.5)
2.280e-007
Fenoterol + propranolol (-8.5)
6.228e-009
Fenoterol + propranolol (-7)
1.022e-006
Forskolin
1.923e-006
5.
What does the Schild plot tell you about
propranolol?
Schild plot
Log ((A'/A)-1)
3
AUC
12 min
2
1
0
-1
-10
-9
-8
-7
-6
log(M) propranolol
AUC
Best-fit values
Slope
Y-intercept when X=0.0
X-intercept when Y=0.0
1/slope
95% Confidence Intervals
Slope
Y-intercept when X=0.0
X-intercept when Y=0.0
12 min
1.376 ± 0.08988
11.61 ± 0.7219
-8.439
0.7266
1.383 ± 0.09489
11.73 ± 0.7621
-8.481
0.7229
1.090 to 1.662
9.317 to 13.91
-8.636 to -8.281
1.081 to 1.685
9.308 to 14.16
-8.698 to -8.312
GraphPad Prism Help File
SchildSlope quantifies how well the shifts correspond to the prediction of
competitive interaction. If the competitor is competitive, the SchildSlope will
equal 1.0. You should consider constraining SchildSlope to a constant value of
1.0. antagonist term, [B], is now raised to the power S, where S denotes the
Schild slope factor. If the shift to the right is greater than predicted by
competitive interactions, S will be greater than 1. If the rightward shift is less
than predicted by competitive interaction, then S will be less than 1.
The slope of a Schild plot should equal 1 if all of the assumptions underlying
the method of analysis are fulfilled. A slope which is significantly greater than
1 may indicate positive cooperativity in the binding of the antagonist,
depletion of a potent antagonist from the medium by receptor binding or
non-specific binding (e.g. to glassware or partitioning into lipid), or lack of
antagonist equilibrium.
6.
Are you familiar with other types of cAMP
assays? Which and what are the pros/cons
compared to the real-time cAMP biosensor
(EPAC)?
FlashPlate®
cAMP dynamic 2 assay
radiolabeled cAMP and a AB against cAMP
AB competion based assay based on FRET
EPAC vs other cAMP assays
Pros
• Real-time
• Cheap!
Cons
• Time-delay (due to
handeling)
• Occupy the machine in at
least 30 min
7.
If you were to characterize a drug from a
pharmaceutical company targeting a GPCR
signaling through the cAMP assay how would
you then do it? Can you use the EPAC cells for
that?
Real-time cAMP assay
HEK293
Excitatio
n
Wild type human
β2-AR
Donor
EPAC149 cAMP biosensor
FRET
Acceptor
Emission
2013.09.17
Slide no 22
Real-time cAMP assay
cAMP
cAMP
cAMP
cAMP
cAMP
cAMP
cAMP
cAMP
cAMP
cAMP
cAMP
( d o n o r /a c c e p to r )
cAMP
cAMP
c A M P re s p o n s e
0 .9
0 .8
0 .7
0 .6
0 .5
0
30
60
90
120
150
T im e ( m in )
FRET
Acceptor signal
2013.09.17
Donor signal
 cAMP
=  FRET
=  donor/acceptor
Slide no 23
Receptor reserve/spare receptors
• Full agonists need to activate only a small
fraction of the receptors in the tissue to elicit
a response (the rest of the receptors are
‘spare’)
• Spare receptors make tissues more sensitive
to an agonist
Relationship between EC50 and Kd
• Full agonist: Kd agonist concentration is not an
approximation of EC50 agonist concentration
because maximal response of the full agonist
does not correlate with 100% receptor occupancy
• For example the full agonist maybe needs to bind
only 20% of the receptors to elicit a maximal
response and this does not correlate with 100%
binding of the receptors from which you can
determine the Kd
Relationship between EC50 and Kd
• Partial agonist: Kd agonist concentration is an
approximation of EC50 agonist concentration
because the concentration producing maximal
response approximates the saturation binding
concentration of the agonist
• The partial agonist will need to bind 100% of
the receptors to reach a maximal response