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Transcript
W.L. Backes
CSRB 4D3
Principles of Pharmacology
Drug Metabolism
A.
Characteristics
1.
2.
3.
B.
chemical change in a drug
forms more water soluble metabolites
usually terminates drug action
Types of non-synthetic (phase 1) reactions
1.
oxidation reactions – Mostly by cytochrome P450
NADPH
RH
O2
NADPH
cyt. P450
reductase
Fe3+
FAD
FMN
P450
NADP+
a.
types of microsomal oxidations
1)
aromatic hydroxylation
OH
CH2
CH 3
2)
3)
CH 2
aliphatic hydroxylation
epoxidation
CH2OH
H 2O
ROH
Principles of Pharmacology
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O
4)
N-, O-, and S-dealkylations
HO
CH 3O
+
HCHO
O
O
NCH 3
NCH 3
HO
HO
Morphine
Codeine
b.
a.
b.
5)
N-hydroxylation (not P450)
6)
N-oxidation (not P450)
7)
oxidative deamination
8)
sulfoxide formation
9)
desulfuration
broad selectivity of cytochrome P-450
1)
multiple forms
2)
mechanistic broad selectivity
classification based on sequence similarity
1) 40% families
2) 59% subfamilies
regulation of P450 expression
1)
enzyme induction
a. transcriptional activation
b. also stabilization of mRNA, protein, and increased protein
synthesis
2)
polymorphisms
a. are frequently silent
b. expression polymorphisms
c. polymorphisms in structural gene
Principles of Pharmacology
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c.
d.
f.
g.
h.
CYP1A
1) inducible by polycyclic hydrocarbons – large compounds that look
like cholesterol
2) found in cigarette smoke
3) 1A1 – inducible
4) 1A2 constitutively expressed - little expression in extrahepatic tissues
5) CYP1A1 polymorphisms
a) Ah receptor polymorphism
b) polymorphisms in structural gene
6) CYP1A2 polymorphisms
a) detected by caffeine metabolism
b) CYP1A2 is expressed only in liver, but appears to be
associated with cancer at distal sites
CYP2A
1) hormonally regulated forms
2) 2A1 - increased in female and young male rats
3) 2A2 - not found in female rats
4) 2A6 in humans
5) CYP2A6 polymorphism
a) significant ethnic variability
b) variant 2A6 alleles
c) coumarin 7-hydroxylation
CYP2B - induced by phenobarbital
1)
2B1 inducible in liver but constitutive in lung
2)
2B2 constitutive and inducible
3)
2B6 human P450 isozyme
CYP2C – constitutive hormonally regulated isozymes
1)
gender specific expression in rodents
2)
regulated by growth hormone levels
3)
CYP2C19 polymorphisms
a) there are 7 different 2C isozymes in human
b) mephenytoin metabolism is marker activity
(1) 3 – 5 % of Caucasians
(2) 20 – 25% of Asians
c) CYP2C19 metabolizes many drugs of pharmacologic
significance
(1) omeprazole, imipramine, propranolol, hexobarbital,
diazepam, etc.
(2) taxol
CYP2D6 – human P450 isozyme – exhibits polymorphisms
1)
debrisoquine
polymorphism in 4-hydroxylation
autosomal recessive trait
related to increased cancer risk – extensive metabolizers were
more susceptible to cancer
P450 2D appears to have an abnormal substrate binding site
Principles of Pharmacology
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2.
2)
dextromethorphan
3)
codeine
i.
CYP2E1
1) ethanol-inducible isozyme
2) production of reactive oxygen species - toxicity
3) polymorphisms
4) role in cancer
5) role in alcohol-related liver dysfunction
j.
CYP3A
1)
hormonally regulated P450 isozymes
2)
CYP3A4 major isozyme found in humans
3)
no discernable polymorphisms
4)
10% of patients show poor response to immunosuppressant
therapy with the 3A4 substrate cyclosporine
k.
non-microsomal oxidations
1)
alcohol dehydrogenase
2)
aldehyde dehydrogenase
3)
xanthine oxidase
4)
tyrosine hydroxylase
5)
monoamine oxidase
reduction
a.
location of reductases
1)
microsomes
2)
cytosol
3)
anaerobic microorganisms in the gut
b.
3.
types of reduction reactions
1)
azoreduction
2)
nitroreduction
3)
ketoreduction
hydrolysis
a.
esterase
1) succinylcholine apnea
a) due to an esterase deficiency
b) enzyme appears to have an altered affinity for esters
c) affects about 1 in 3000
b.
amidase
c.
epoxide hydrolase
Principles of Pharmacology
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B.
synthetic (phase II) reactions
1.
conjugation reactions
OSO3H
OH
O-glucuronide
a.
b.
d.
glucuronidation
1)
found in endoplasmic reticulum
2)
reacts with hydroxyl, amine or carboxylic acid functions
sulfate conjugation
1)
found in the cytosol
2)
reacts with alcohols, phenols, and aromatic amines
3)
saturable
acetylation
Principles of Pharmacology
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CH3
C O
O
N
NH2
C
NH
NH
Acetyl CoA
N
C
NH
O
1)
2)
3)
d.
e.
found in the cytosol
reacts with primary amines
Acetylation polymorphism
a) differences not due to differences in quantity of enzyme
b) in U.S. mostly slow acetylators (0.72)
c) Japanese have a higher percentage of rapid acetylators,
whereas peoples of Middle Eastern descent, Scandanavians
and Finns have more slow acetylators
d) slow acetylators more likely to exhibit toxic effects
e) specific acetylation disorders
(1) Isoniazid-induced neurotoxicity – slow acetylators
(2) Isoniazid-induced hepatitis – rapid acetylators
(3) Drug-induced lupus erythematosus – slow acetylators
(4) Arylamine induced bladder cancer – higher incidence
in slow acetylators
(5) colorectal cancer – higher incidence in rapid
acetylators particularly when associated with rapid
CYP1A2 phenotype
methylation
1)
found in both the cytosol and the endoplasmic reticulum
2)
S-adenosyl methionine is the methyl donor
3)
reacts with hydroxyl, sulfhydryl, and amine functions
glutathione conjugation
1)
superfamily of genes found on 11 chromosomes
2)
found in the cytosol and mitochondria (kappa) of liver and
numerous other tissues
3)
exists as a functional dimer
4)
mu, pi, sigma (tyrosine) and theta (serine)
5)
drug binds to glutathione (glu - - cys-gly)
6)
can ultimately form mercapturic acids
7)
reacts with:
a)
aromatic hydrocarbons (epoxides)
b)
arylamines
c)
organic halides
a)
phenols
8)
Principles of Pharmacology
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b)
some have ligand-binding capabilities (ligandins)
alpha class GSTs are inducible by ingestion of high levels of
brussels spouts
H
S
C
H2
O
H
OH
H
C
C
H2
H
N
H2
C COOH
NH
C
CH2
CH2
CHNH2
COOH
9)
f.
polymorphisms
a) four classes – GSTM1 to GSTM4
b) GSTM1 is polymorphic (*0, *a, and *b)
1) null gene, increased risk of cancer
2) essentially absent in 20 – 50% of all individuals
c) GSTP1 (four allelic variants)
d) ethnic component
(1) higher percentage of null gene in Asians than
Caucasians
(2) lower percentage in blacks
e) links between GSTM1 null phenotype and pituitary
adenomas, head and neck cancer, malignant melanoma,
colorectal cancer and bladder carcinoma
f) association of lack of GSTM1 with lung cancer in heavy
smokers
g) appears to be linked to a combined GSTM1 null and
CYP1A1 Msp I.
glycine conjugation
1)
found in both the cytosol and mitochondria
2)
reacts with aliphatic acids and aromatic carboxylic acids
3)
can be saturated
4)
developmentally induced
Factors that Affect Drug Metabolism
I.
Age
A.
B.
Developmental changes
Changes with senescence
Principles of Pharmacology
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II.
Induction and Inhibition by Foreign Compounds
A.
B.
III.
Dietary factors
A.
B.
C.
D.
IV.
Administration of drugs and chemicals can induce P450 and other drug
metabolizing enzymes
1.
phenobarbital and other drugs
2.
polycyclic hydrocarbons and carcinogenesis
3.
ethanol
Inhibitors of drug metabolism
1.
ethanol
2.
cimetidine
charcoal broiled beef
vitamin deficiencies
brussel sprouts
grapefruit
Species differences
A.
B.
quantitative differences
qualitative differences
Principles of Pharmacology
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Clinical Pharmacokinetics
I.
Compartmental Modeling
A.
One compartment model
Total Body Volume
A
B.
absorption
elimination
A
A
Two compartment model
Central compartment
A
absorption
elimination
A
distribution
A
Peripheral compartment
A
Principles of Pharmacology
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[drug] in plasma
g/100 ml
Intravenous Bolus (One compartment model)
30
20
50
5
10
2
4
6
8
10
12
0.5
0
14
4
Time (h.)
8
12
Time (h.)
1. Elimination rate constant
2. Half-life
3. Relationship between elimination rate constant and half-life
t1/2 = 0.693/kel
4. Zero order kinetics
zero order
30
20
[drug] in plasma
g/100 ml
0
0
[drug] in plasma
g/100 ml
[drug] in plasma
g/100 ml
C.
zero order
50
5
1st order
0.5
0
4
8
12
Time (h.)
10
first order
0
0
2
4
6
8
Time (h.)
10
12
14
Principles of Pharmacology
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Single Extravascular Administration
30
100
[drug] in plasma
g/100 ml
[drug] in plasma
g/100 ml
C.
20
10
1
0
10
10 20 30 40 50
Time (h.)
0
0
10
20
30
40
50
Time (h.)
1.
2.
3.
Elimination rate constant is calculated the same way as for an i.v.
bolus
Absorption rate constant
Effect of absorption rate
[drug] in plasma
g/100 ml
Effect of absorption rate constant
ka=1.39
ka=0.693
20
ka=0.277
ka=0.069
10
0
0
10
20
30
40
50
Ti me (h.)
Intravenous Bolus (Two compartment model)
75
[drug] in plasma
g/100 ml
100
[drug] in plasma
g/100 ml
D.
50
1
0
25
0
0
10
4
8 12 16 20 24
Time (h.)
2
4
6
8
Time (h.)
10
12
14
Principles of Pharmacology
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1.
2.
E.
calculation and significance of the disposition rate constant
distribution rate constant
Single Extravascular Dose (Two compartment model)
50
40
[drug] in plasma
g/Liter
[drug] in plasma
g/100 ml
100
30
20
10
1
0
10
20
30
40
Time (h.)
10
0
0
4
8
12
16
20
24
Time (h.)
Volume of Distribution
A.
B.
Apparent volume into which a drug distributes in the body
How it is calculated
1.
Inject a known quantity of drug as an
i.v. bolus
2.
Take blood samples at various times
after injection
3.
Plot data on a semilogarithmic plot and
extrapolate back to t = 0 to calculate the
concentration prior to elimination
4.
Take values and plug into the following
formula:
VD = __amount of drug___
plasma concentration
C.
III.
Factors affecting volume of distribution
Clearance
A.
Total clearance
1.
ClT = kel * VD
2.
ClT = X0/Area under the curve
= X0/Cp
[drug] in plasma
g/100 ml
II.
50
Co p
5
0.5
0
4
8
Time (h.)
12
Principles of Pharmacology
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B.
C.
Bioavailability
A.
B.
Intravenous Infusion
A.
Conditions
1.
constant rate of drug administration (zero order)
2.
proportional elimination of drug (first order)
[drug] in plasma
g/100 ml
V.
area under the curve
time to peak drug concentration
30
20
10
t1/2 = 4 h
0
0
2
4
6
8 10 12 14 16 18 20 22 24
Time (h.)
B.
C.
D.
E.
F.
Css = infusion
rate/ClT
Adjustments to
dosage – the Css is
directly proportional
to the infusion rate
(dose)
Time required to
achieve Css
Relationship
between i.v. infusion
and i.v. bolus
Effect of alteration in
infusion rate
[drug] in plasma
g/100 ml
IV.
3.
ClT = Clliver + Clkidney + Clother
Hepatic clearance
1.
Extraction ratio
2.
Relationship between Cl
Renal clearance
30
infusion
20
10
bolus
0
0
2
4
6
8 10 12 14 16 18 20 22 24
Time (h.)
Principles of Pharmacology
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VI.
Repeated intravenous injections
A.
Relationship between frequency of administration and Css
10
[drug] in plasma
[drug] in plasma
20
5
0
0
50
100
150
10
0
0
200
Time (h)
10
20
30
40
Time (h)
Left panel – A drug with a t1/2 = 5 h is administered after virtually all of the drug is
eliminated. Right panel – A drug with a t1/2 = 5 is administered about once every half-life.
Note the greater accumulation of drug and decreased fluctuations.
B.
Relationship between dose and Css
C.
Relationship between t1/2 and Css
[drug] in plasma
75
50
t1/2 = 50 h
25
t1/2 = 5 h
0
0
10
20
30
40
50
60
70
Time (h)
The above graph depicts two drugs that are administered with the same frequency, but have
different half-lives. The drug with the shorter half-life reaches its steady state concentration
more rapidly. The drug with the longer half-life exhibits a much greater accumulation in the
blood and shows smaller fluctuations (on a percentage basis).
Principles of Pharmacology
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D.
Loading dose
20
[drug] in plasma
administered an initial loading dose
10
no loading dose
0
0
10
20
30
Time (h)
VII. Repeated extravascular administration
1
5
[drug]inblod
1
0
5
0
0
5
1
0 1
5 2
0 2
5 3
0 3
5
T
i
m
e
(
h
r
s
)
Principles of Pharmacology
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Principles of Pharmacology
Pharmacokinetics Problem Set
January 2009
1.
A patient was administered an analgesic drug by a single 12 mg intravenous injection.
Blood samples were then taken at various times and the following results were
obtained.
[drug] in plasma
g/Liter
100
10
1
0.1
0
4
8
12
Time (h.)
What type of pharmacokinetic model is followed by this data? Calculate the t1/2, the
elimination rate constant and the apparent volume of distribution for the drug. What is
the total body clearance for this drug?
2.
You have decided to administer the drug described in question #1 by intravenous
infusion at a rate of 3.7 mg/min. How long will it take before steady state is attained?
What is the Css after intravenous infusion? The drug was administered to the patient for
10 days, after which administration was discontinued. What would the drug
concentration be 9 hrs. after administration was stopped?
You want to give the drug orally in four separate doses, and you know that 20% of the
drug will be inactivated by first pass metabolism. What dose of this drug would you
prescribe if you wanted to achieve a steady state plasma concentration of 0.29 mg/L.
A new drug used in the treatment of asthma was administered by i.v. infusion at a rate
of 1 g/min. A graph of the infusion is shown.
20
[drug] (g/liter)
3.
15
10
5
0
0
10
20
30
40
50
60
Time (hours)
70
80
90
100
Principles of Pharmacology
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a. What is the half-life of this drug?
b. Also calculate the elimination rate constant, the volume of distribution, and total body
clearance for this drug.
c. The drug is eliminated 20% by renal and 80% by hepatic mechanisms. If there is a 50%
hepatic failure, then what is the new steady state concentration?
d. If the drug was administered by oral administration how many hours would be required to
attain steady state?
e. If the drug (no hepatic failure) when taken orally has only a 40% bioavailability when
compared to the intravenous route, then what oral dose would be required to attain a
steady state concentration of 2 g/ml (the drug is to be administered every 12 h).
f. Provide a possible explanation for the low bioavailability of this drug.
4.
Blood ethanol data were obtained from a patient on admission and at various times
afterward. The following graph of the data was obtained.
[Ethanol] (mg/100 ml)
400
350
300
250
200
150
100
50
0
0
1
2
3
4
5
6
7
8
Time (hours)
What is the pharmacokinetic model? Calculate the rate of disappearance of the drug.
Answers
1.
One compartment Model
t1/2 = 3 hrs
kel = 0.231/ hr
Vd = 375 liters
ClT = 86.6 Liter/h or 1444 ml/min
2.
Css = 2.56 µg/ml
12 – 15 hrs to attain steady state
C9 hrs = 0.32 µg/ml
Principles of Pharmacology
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Dose = 31.4 mg/hr = 753.6 mg/day (or 4 x 188 mg)
3.
t1/2 = 10 hrs (from graph)
kel = 0.0693/hr
ClT = 50 ml/min
Vd = 43.3 liters
ClR = 10 ml/min
ClH = 40 ml/min
with 50% hepatic failure, ClH is decreased to 20 ml/min
therefore ClTmodified = 30 ml/min
Css = 33µg/liter
Recalculate time to achieve steady state:
t1/2 = 16.7 hrs
time to steady state = 66 - 83 hrs
Calculation of oral dose
4.
180 mg every 12 hrs
zero order decay
20 mg (100 ml)-1(hr)-1
for zero order reactions v = kel, therefore kel = 20 mg/100ml/hr.