Bchm 480

Practice Exam #1                                                                           BCHM 480

1) GPCR30 is a G-Protein coupled receptor whose natural ligand is estrogen and whose Gα-GTP subunit activates Adenylate cylcase.  In a system with 10 nM Adenylate cyclase, 25 nM GPCR and 100 μM Estrogen (total) , the occupancy of Gα-Adenylate Cyclase (Y) is 0.56. and the occupancy of Estrogen –GPCR is 0.95.  What is the Kd of Adenylate Cyclase and Gα, the Kd of Estrogen and GPCR30?

Assume that binding of Estrogen to GPCR and release of Gα is a 1:1 event, meaning [LR] = [Gα]total

LR from GPCR30

[pic 1]    [LR] = 23.75 nM = [Gα]total

Free Gα

[pic 2][LR] = 5.6          [Gα] = 23.75 – 5.6 = 18.15 nM

Kd Adenylate Cyclase and Gα

[pic 3]   Kd = 14.26

Free Estrogen = 100000 – 23.75 = 99976.25

Kd Estrogen and GPCR30

[pic 4]   Kd = 5260

2)  The ionotropic 5-HT3 receptors are ligand gated ion channels found in the digestive tract and the vagus nerve (responsible for vomiting).  Estimated concentrations of serotonin in the synaptic cleft after exocytosis is about 20 nM, if serotonin has a Kd of 1.7nM what is the occupancy of 5-HT3 at this concentration? You can assume the 20 nM is free ligand.

[pic 5]

b) Tropisetron is a competitive inhibitor of the 5-HT3 used to treat nausea associated with chemotherapy and has a KI = 11nM.  If the occupancy of serotonin has to drop below 0.3 to suppress nausea, what is the concentration of tropisetron needed in the synaptic cleft to be effective?

[pic 6]        x = 290 nM

c) If Tropisetron has a volume of distribution of 400L, what dose needs to be taken to reach the concentration determined in part b?  Assume that the concentration in the synaptic cleft is equal to the concentration in the blood plasma.  MW = 284.35 g/mol

Vd = dose / concentration      400L = dose / 290 nM      Dose = 116000 nmol

116000 nmol (284.35 ng/nmol) = 3.298x107 ng or 32.98 mg

d)  Tropisetron has a half life of 5.7 hours. If 50 mg of the drug is given, how long will it take to reach the concentration calculated in part b?

5.6 = 0.7/k    k=0.122

32.98 = 50.00 e-0.122t

t = 3.41 hours

3) The graph below shows the response of a natural ligand to a receptor in the absence of inhibitor and in the presence of 100 nM reversible and 100 nM irreversible inhibitor.  Use the graph to estimate Kd of the ligand, KI of the reversible inhibitor and [IR] for the irreversible inhibitor assuming 1 nM of receptor is present.

[pic 7]

Estimating from the graph the half way point on the no inhibitor curve is roughly 10-7.5 or 3.18x10-8

The apparent Kd on for the reversible inhibitor is 1x10-6

[pic 8]

[IR] for irreversible

Maximum with no inhibitor = 2.25     Maximum irreversible = 0.8

Maximum Y irreversible = 0.8 / 2.25 = 0.356

0.356 = 1 – ([IR] / 1 nM)         0.644 = [IR]/1 nM       [IR] = 0.644 nM

4)  A particular β1-adrenergic receptor has a Kintr = 10.  What is the fa of this receptor?

[pic 9]

b) Drug X under saturating conditions drives the fa of the receptor to 0.545, how would you classify Drug X?  If the Ka for Drug X is 0.001, what is the Ki?

It is a partial agonist

[pic 10]0.0120

Concept Practice  

(These are just some general questions to practice and do not cover every topic that might appear on the exam.  For a complete list of topics, see the end of Lecture 6)

1)   Define the following terms: Rather than retype everything, I am just going to send you to the lecture slide that answers the question

Pharmacodynamics: Lecture 1, Slide 7

Pharmacogenetics:  Lecture 1, Slide 9

Pharmacophore:  Lecture 2, Slide 8

ADME: Lecture 3, Slide 3

OAT:  Lecture  3, Slide 33

Phase 2 drug metabolism: Lecture 4, Slide 8

First Pass effect: Lecture 3, Slide 39

Efficacy:  Lecture 1, Slide 33

Therapeutic effect:  Lecture 2, Slide 4

Non-ionic Diffusion:  Lecture 3, Slide 19

Lipinkski rule of five:  Lecture 3, Slide 18

Ionic Lock:  Lecture 4, Slide 29

Selective Toxicity:  Lecture 1, Slide 14

2)  How does the concept of ionic lock and activation energy play a role in determining the fa of a receptor?

The Ionic lock is the intermolecular forces holding the receptor in the inactive conformation.  These are the interactions that must be broken to move from inactive conformation to active conformation, therefore they represent the activation energy needed to move from one conformation to another.