Download applied pharmacokinetics and more Exercises Pharmacology in PDF only on Docsity!
THERAPEUTIC DRUG MONITORING
ANTIEPILEPTIC DRUGS
- Carbamazepine
Pharmacocinétinetics
Elimination: active anticonvulsant metabolite, 10,11-epoxide.
Therapeutic window (Cp): 4 – 12 μg/ml ,(F)=0.8, ( S)=1, (Vd)=1.4 L/kg
(Cl)=0.064 L/h/kg, (T
1/
) =15 h en steady state , (T
1/
)=30-35 h en dose
unique., (𝛼) =0.2 – 0.3, Peak absorption =6h
Problem
- Calculate the daily dose for Mss XUE 36 years old and 60 kg weight, at
the beginning of treatment and at the second week to achieve Cp of
6mcg/ml.
- Under this regimen, after two months of treatment, crises seem not
stabilized and on lab check the Cp measured is 4 μg/ml. Explain the raison
for the low Cp and calculate the new dose to adjust the concentration
to 6 μg/ml.
At the beggining of the treatment , patient will need loading dose
𝑉𝑑∗𝑐𝑝
𝑆∗𝐹
- 4 𝑙
𝑘𝑔
A)
At the second week , patient will need the maintainence dose
- 693
𝑡½
- 693
15ℎ
𝐶𝑙∗Ʈ∗𝐶𝐸
𝑆∗𝐹
- 96 𝑀𝐺
1 ∗ 0. 8
6 𝑚𝑔
𝑚𝑙
6 𝑚𝑔
𝑚𝑙
4 𝑚𝑐𝑔
𝑚𝑙
12 𝑚𝑐𝑔
𝑚𝑙
𝑄𝑜
𝑉𝑑∗𝐾𝑒∗Ʈ
𝑄𝑜
𝐶𝑙∗Ʈ
𝑄𝑜
𝐶𝐸∗Ʈ
- 96 𝑚𝑔
4 𝑚𝑔
𝑙
∗24ℎ
Administration rate = CE ∗ CL = 6mg/l ∗ 5 .76l/h = 34. 56 mg/h
- 84 𝑙
ℎ
6 𝑚𝑔
𝑙
𝑑𝑜𝑠𝑒
𝐶𝐸∗Ʈ
- 96 𝑚𝑔
4 ∗ 24
- 76 𝑙
ℎ
𝑄𝑜
𝑉𝑑
1
1 −𝑒
−𝑘𝑒Ʈ
−𝑘𝑒Ʈ
- 7 𝑙
𝑘𝑔
- 693 𝑉𝑑
𝑡½
𝐶𝑙
𝑉𝑑
40625 𝑙/ℎ
5 𝑙
250 𝑚𝑔
- 5 𝑙
1
1 −𝑒
− 0. 0232 ∗ 12
− 0. 0232 ∗ 12
250 𝑚𝑔
- 5 𝑙
𝑒
− 0. 2784
1 −𝑒
− 0. 2784
The reason of faster decrease in the plasma below therapeutic window
is due to the rapid clearance.
𝑄𝑜
𝑉𝑑∗𝑘𝑒∗Ʈ
𝑄𝑜
𝐶𝑙∗Ʈ
𝑄𝑜
𝐶𝐸∗Ʈ
250 𝑚𝑔
35 𝑚𝑔/𝑙∗12ℎ
3. Phenobarbital
Pharmacotherapeutics
- Barbiturate long acting used in convulsive crises and insomnia ;
250 𝑚𝑔
- 5 𝑙
75699
24301
- Forms (tablets 100 mg ; 50 mg ; injection) ;
- Dosing (2 mg/kg/day per os or IM).
Pharmacokinetics
- Inducer of metabolism ;
- (Cp): 10 – 30 mg/L , (F)= 1 , ( S)=0.9 (sodium salt), (Vd)=0.6 à 0.7 L/kg
- (Cl) adult=4 ml/h/kg, (Cl) child >1 year =2 times adult dose
- (T
1/
) adult= 5 days, (𝛼)=0.5, Cl (hemodialysis)= 3 L/h
Problems
- Mr X aged 39 with 70 kg developed convulsive seizures after a car
accident that triggered cranial trauma. He has been prescribed
Phenobarbital to control seizures. Calculate the loading dose needed to
achieve directly the therapeutic concentration.
- Calculate the maintenance dose needed to keep Cp into the therapeutic
window after the initial loading dose
- Calculate the time required to attain the steady state concentration of
20 mg/l if the loading dose is not administered.
- Another patient aged 25 years and weighing 70 kg presenting chronic renal
failure is taking 60 mg phenobarbital 2 times a day to achieve 20 mg/L at
steady state. Three months late, the renal failure state worsens and he is
prescribed dialysis sessions every 6 hours to be repeated 3 times a week.
Discuss whether the replacement dose is necessary or not and how to
calculate it if necessary.
𝑉𝑑∗𝐶𝑝
𝑆∗𝐹
o 𝐼𝑑𝑒𝑎𝑙 𝑉𝑑 𝑖𝑠 𝑣𝑎𝑟𝑖𝑒𝑠 𝑓𝑟𝑜𝑚 0. 6 𝑡𝑜 𝑜.
7 𝑙
𝑘𝑔
o 𝑇ℎ𝑒 𝑝𝑎𝑡𝑖𝑒𝑛𝑡 𝑟𝑒𝑎𝑙 𝑉𝑑 =
- 6 𝑙
𝑘𝑔
o 𝑇ℎ𝑒 𝐶𝑝 =
30 𝑚𝑔
𝑙
10 𝑚𝑔
𝑙
𝐿𝑜𝑎𝑑𝑖𝑛𝑔 𝑑𝑜𝑠𝑒 𝐷𝑎 𝑎𝑡 min 𝐶𝑙𝑒𝑎𝑟𝑛𝑐𝑒 =
−𝟎.𝟎𝟎𝟓𝟕𝟕𝟓𝒕
𝟐𝟎
𝟏𝟒.𝟕𝟖𝟓
−𝟎.𝟎𝟎𝟓𝟕𝟕𝟓𝒕
o 𝑇ℎ𝑒 𝑔𝑖𝑣𝑒𝑛 , 𝑝𝑎𝑡𝑖𝑒𝑛𝑡 𝑎𝑔𝑒𝑑 25 𝑦𝑒𝑎𝑟𝑠 𝑎𝑛𝑑 𝑤𝑒𝑖𝑔ℎ 70 𝑘𝑔
o 𝑡𝑎𝑘𝑖𝑛𝑔 𝐷𝑜𝑠𝑒 = 60 𝑚𝑔 𝑝ℎ𝑒𝑛𝑜𝑏𝑎𝑟𝑏𝑖𝑡𝑎𝑙 , 𝑡𝑤𝑖𝑐𝑒 𝑑𝑎𝑦 ↔ Ʈ = 12ℎ
o 𝐶𝑝 = 20 𝑚𝑔/𝑙 𝑎𝑡 𝑠𝑡𝑒𝑎𝑑𝑦 𝑠𝑡𝑎𝑡𝑒
1. 𝑡ℎ𝑒 𝑑𝑟𝑢𝑔 ℎ𝑎𝑙𝑓 𝑙𝑖𝑓𝑒 𝑖𝑠 5 𝑑𝑎𝑦 ↔ 120ℎ ≥ 2ℎ, 𝑐𝑜𝑛𝑑𝑖𝑜𝑛 𝑓𝑢𝑙𝑙𝑓𝑖𝑙𝑙𝑒𝑑
𝑑𝑟𝑢𝑔 𝑐𝑙𝑒𝑎𝑟𝑛𝑐𝑒 = 4 𝑚𝑙/ℎ/𝑘𝑔 ∗ 70 𝑘𝑔 = 280 𝑚𝑙/ℎ = 4. 666 𝑚𝑙/𝑚𝑖𝑛
+𝑑𝑖𝑎𝑙𝑦𝑧𝑎𝑏𝑙𝑒 𝑐𝑙𝑒𝑎𝑟𝑎𝑛𝑐𝑒 =
3 𝑙
ℎ
=
3000 𝑚𝑙
60 𝑚𝑖𝑛
= 50 𝑚𝑙/𝑚𝑖𝑛
2. 𝑡ℎ𝑒 𝑡𝑜𝑡𝑎𝑙 𝑐𝑙𝑒𝑎𝑟𝑎𝑛𝑐𝑒 𝐶𝑙 = 50 𝑚𝑙/𝑚𝑖𝑛 + 4. 666 𝑚𝑙/𝑚𝑖𝑛 = 54. 666 𝑚𝑙/𝑚𝑖𝑛 ≤
500 𝑚𝑙/𝑚𝑖𝑛 𝑖. 𝑒 𝑐𝑜𝑛𝑑𝑖𝑡𝑖𝑜𝑛 𝑜𝑓 𝐷 𝑟𝑒𝑝𝑙𝑎𝑐𝑒𝑚𝑒𝑛𝑡 𝑖𝑠 𝑎𝑙𝑠𝑜 𝑓𝑢𝑙𝑙𝑓𝑖𝑙𝑙𝑒𝑑
𝑇ℎ𝑒 𝑑𝑖𝑎𝑙𝑖𝑠𝑎𝑏𝑙𝑒 𝑣𝑜𝑙𝑢𝑚𝑒
( 𝑉𝑑
′
)
𝑉𝑑
𝛼
=
42 𝑙
- 5
= 84 𝑙
3. 𝑉𝑑
′
≤ 250 𝑙 𝑡ℎ𝑢𝑠 3 𝑟𝑑 𝑐𝑜𝑛𝑑𝑖𝑡𝑖𝑜𝑛 𝑖𝑠 𝑓𝑢𝑙𝑙𝑓𝑖𝑙𝑙𝑒𝑑
𝑐𝑜𝑛𝑐𝑙𝑢𝑠𝑖𝑜𝑛 𝑖𝑠 𝑡ℎ𝑎𝑡 𝑡ℎ𝑒 𝑟𝑒𝑝𝑙𝑎𝑐𝑒𝑚𝑒𝑛𝑡 𝑑𝑜𝑠𝑒 𝑛𝑒𝑒𝑑𝑒𝑑
( 𝑟𝑒𝑞𝑢𝑖𝑟𝑒𝑑
) 𝑓𝑜𝑟 𝑡ℎ𝑒 𝑝𝑎𝑡𝑖𝑒𝑛𝑡
−𝑪𝒍 𝒕𝒐𝒕𝒂𝒍∗𝑻𝒅
𝑽𝒅
)
𝑻𝒉𝒖𝒔
𝑪𝒍𝒆𝒂𝒓𝒏𝒄𝒆 𝒕𝒐𝒕𝒂𝒍∗𝑻𝒅
𝑽𝒅
=
𝟑.𝟐𝟖𝒍
𝒉
∗𝟔𝒉
𝟒𝟐𝒍
= 𝟎. 𝟒𝟔𝟖𝟓𝟕
𝒍𝒏𝟎.𝟗𝟎𝟏𝟏𝟖
𝟎.𝟎𝟎𝟓𝟕𝟕𝟓
𝟐𝟎𝒎𝒈
𝒍
−𝟎.𝟒𝟔𝟖𝟓𝟕
𝟐𝟎𝒎𝒈
𝒍
−𝟎.𝟒𝟎𝟏𝟔𝟑
𝟑𝟏𝟒.𝟐𝟒+𝟑𝟐𝟓.𝟏𝟓𝒎𝒈
𝟐
𝒓𝒂𝒏𝒈𝒆
𝑷𝒂𝒓𝒂𝒎𝒆𝒕𝒆𝒓
𝒊𝒅𝒆𝒂𝒍 𝑽𝒅 𝒑𝒂𝒕𝒊𝒆𝒏𝒕 𝑽𝒅
𝑽𝒅
′
=
𝒗𝒅
𝜶
𝑫𝒓
𝟕. 𝟗𝟐𝒅𝒂𝒚
𝑫𝒓𝒆𝒑𝒍
𝒅𝒂𝒚
𝒎𝒊𝒏𝒊𝒎𝒖𝒎 𝟎. 𝟔𝒍/𝒌𝒈 𝟒𝟐𝒍 𝟖𝟒𝒍 𝟑𝟏𝟒. 𝟐𝟒𝒎𝒈 𝟑𝟗. 𝟔𝟕𝒎𝒈
𝒎𝒆𝒂𝒏 𝟎. 𝟔𝟓𝒍/𝒌𝒈 𝟒𝟓. 𝟓𝒍 𝟗𝟏𝒍 𝟑𝟏𝟗. 𝟔𝟗𝒎𝒈 𝟒𝟎. 𝟑𝟔𝒎𝒈
𝒎𝒂𝒙𝒊𝒎𝒖𝒎
𝟎. 𝟕𝒍/𝒌𝒈 𝟒𝟗𝒍 𝟗𝟖𝒍
𝟑𝟐𝟓. 𝟏𝟓𝒎𝒈 𝟒𝟏. 𝟎𝟓𝒎𝒈
ANTIARRYTHMIC DRUGS
4. Lidocaine
Pharmacotherapeutics :
- Antiarrythmic indicated in severe ventricular arrhythmias.
- Local Anesthetic.
Pharmacokinetics
- Important first pass effect.
- The myocardium is located in the central compartment (two compartment model)
- (CE)= 1 – 5 mg/L , (F)=0.3 , S)= 1
- patient condition Normal CHF Cirrhosis
V 1 0.5 L/kg 0.3 L/kg 0.6 L/kg
Vd 1.3 L/kg 0.88L/kg 2.3 L/kg
- Clearance (Cl) 10ml/min/kg 6ml/min/kg 6ml/min/kg
- (T 1/
) 𝛼 8 min 8 min 8 min
) β 100 min 100 min 300 min
Problems
𝑄𝑜
𝐾𝑒∗𝑉𝑑∗Ʈ
𝐶𝐸∗𝐶𝑙∗Ʈ
𝑆∗𝐹
2 𝑚𝑔
𝑙
- 420 𝑙
𝑚𝑖𝑛
5. Procainamide
Pharmacotherapeutics :
- Antiarrythmic drug for ventricular tachycardia ;
- Usual doses (250 mg - 500 mg every 3 to 4 hours in IV, IM or peroral.
- Standard loading dose is 15 mg/kg.
Pharmacokinetics
- Elimination via kidney and liver ;
- The metabolite N-acetylprocainamide is active.
- (CE)=4 – 8 mg/L , (F)=0.85 , ( S)=0.87 (HCl) , (Vd)=2 L/kg
- Renal Clearance (Cl) =3x Cl
creatinine
, Clearance (Cl) acetylation =0.13 L/h/kg
- Clearance (Cl) other =0.1 L/h/kg , (T 1/
) 𝛼 = 5 min, (T 1/
) β = 3 hrs
Problems
- A male patient aged 62 years old and 70 kg weight presents premature
ventricular contractions not responding to lidocaine treatment. He has been
found having myocardium infarction and renal failure with SCrss of 1.3 mg/dL
and CLcr as 50 ml/min. Calculate the IV loading dose to achieve the Cp of 8
mg/L.
- For the same patient find the maintenance dose to achieve 6 mg/L Cpss.
𝐀𝐍𝐒𝐖𝐄𝐑 𝟏𝟏.
𝑰𝒇 𝑸𝒐 ⇎ 𝑫𝒎 ↔ 𝑫𝒎 =
𝑸𝒐
𝑺𝑭
=
𝟏𝟓. 𝟔𝟎𝟕𝟐𝒎𝒈
𝟏 ∗ 𝟎. 𝟑
= 𝟓𝟐. 𝟎𝟐𝟒𝒎𝒈 𝒐𝒇 𝑳𝒊𝒅𝒐𝒄𝒂𝒊𝒏𝒆
8 𝑚𝑔
𝑙
2 𝑙
𝑘𝑔
𝑄𝑜
𝑉𝑑∗𝐾𝑒∗Ʈ
𝐶𝐸∗𝐶𝑙 𝑇𝑜𝑡𝑎𝑙∗Ʈ
𝑆∗𝐹
𝑤𝑒𝑖𝑔ℎ𝑡 ( 140 −𝑎𝑔𝑒)
72 ∗𝑆𝑟𝐶𝑟𝑠𝑠
70 ( 140 − 62 )
72 ∗ 1. 3
min 𝑓𝑟𝑜𝑚 𝑡ℎ𝑒 𝑞𝑢𝑒𝑠𝑡𝑜𝑛 𝑤𝑒 ℎ𝑎𝑣𝑒 𝐶𝑙𝐶𝑟𝑠𝑠 = 50 𝑚𝑙/
min 𝑙𝑒𝑡 𝑡𝑎𝑘𝑒 𝑖𝑡 𝑖𝑛 𝑢𝑠𝑒.
𝑉𝑑∗𝐶𝑝
𝑆∗𝐹
140 𝑙∗
8 𝑚𝑔
𝑙
- 87 ∗ 1
= 1287. 3 𝑚𝑔 ↔ IV(F=1)
140 𝑙∗ 8 𝑚𝑔/𝑙
- 87 ∗ 0. 85
- A male patient 70 kg aged 50 years had heart attack with auricular fibrillation. He is
treated with 300 mg quinidine every 6 hours. Calculate Cpmax and Cpmin reached
after 3 doses are given.
- Discuss if it is relevant to take a blood sample during the first 24 hours to measure
the Cpss.
𝑆𝐹𝐷
𝑉𝑑
1 −𝑒
−𝑛𝑘𝑒Ʈ
1 −𝑒
−𝑘𝑒Ʈ
- 8 𝑙
𝑘𝑔
𝐶𝑙
𝑉𝑑
- 693
𝑡½
𝐶𝑝 𝑚𝑎𝑥 =
300 𝑚𝑔 ∗ 0. 73 ∗ 0. 82
126 𝑙
(
1 − 𝑒
− 3 ∗ 0. 1166 ∗ 6
1 − 𝑒
− 0. 1166 ∗ 6
) =
300 𝑚𝑔
126 𝑙
(
1 − 0. 1226
1 − 0. 4967
)
−𝑘𝑒Ʈ
− 0. 1166 ∗ 6
𝑆𝐹𝐷
𝑉𝑑
1 −𝑒
−𝑛𝑘𝑒Ʈ
1 −𝑒
−𝑘𝑒Ʈ
𝐶𝑝 𝑚𝑎𝑥 =
300 𝑚𝑔 ∗ 0. 70 ∗ 0. 62
126 𝑙
(
1 − 𝑒
− 3 ∗ 0. 1166 ∗ 6
1 − 𝑒
− 0. 1166 ∗ 6
) =
300 𝑚𝑔
126 𝑙
(
1 − 0. 1226
1 − 0. 4967
)
−𝑘𝑒Ʈ
− 0. 1166 ∗ 6
𝐶𝑙
𝑉𝑑
CARDIOTONIC DRUGS
7. Digoxin
Pharmacotherapeutics:
- Digitalis chemical indicated in Chronic heart failure and for auricular
fibrillation; Usual Doses (load=1 - 1.5 mg/70kg ; maintenance= 0.
mg/day) per os.
Pharmacokinetics
𝒍𝒏𝟐−𝒍𝒏𝟏
𝟎.𝟎𝟏𝟏𝟗𝟓𝟗
𝟎.𝟔𝟗𝟑
𝒕½
𝟎.𝟔𝟗𝟑
𝟐𝒉
𝒍𝒏𝟐−𝒍𝒏𝟏
𝟎.𝟑𝟒𝟔𝟓
𝑪𝒍∗Ʈ∗𝒄𝒑
𝑺∗𝑭
𝟔.𝟏𝟏𝟏𝒍/𝒉∗𝟐𝒉∗𝟎.𝟎𝟎𝟏𝟓𝒎𝒈/𝒍
𝟏∗𝟎.𝟕𝟑
LES AMINOGLYCOSIDES
8. Gentamicin, tobramycin, amikacin
Pharmacotherapeutics:
- Antibiotics against gram negative infections;
- Doses (infusion 50 to 120 mg in 30 to 60 min every 8 hours for gentamicin
and tobramycin; 200 to 500 mg for amikacin).
Pharmacokinetics
- Poor oral bioavailability ;
- (Cp) Cpmax Cpthrough
Gentamicin et Tobramycin 4 à 8 mg/L < 2 mg/L
Amikacin 20 – 30 mg/L <10 mg/L
- (F) < 0.3, ( S)= 1 , Vd=7.3 L/kg
- Clearance (Cl) in non renal failure=Cl=Cl cr
- Clearance (Cl) moderate renal failure =Cl=0.0043 L/h/kg
- Clearance (Cl) in severe renal failure =Cl=0.0021 L/h/kg
- Clearance hemodialysis=Cl= 1.8 L/h
- (T
1/
) in non renal failure = 2 - 3 hrs
- (T
1/
) in renal failure = 30 - 60 hrs
Problems
- Mss BETY 30 years old and 70 kg carrying infections received a first dose of
gentamicin 100 mg in infusion during 30 min. Her Scrss is 0.9 mg/dL. Calculate
the concentration Cp achieved one hour after starting infusion.
- Considering the infusion as rapid intravenous injection, re-calculate the
concentration achieved one hour after injection taking CL-6.06 L/h, Vd=17.5 L
and ke=0.346 h
.
- Calculate Cpssmax and Cpssmin if the infusion is given every 8 hours.
𝐾
𝑉𝑑∗𝑘𝑒
−𝑘𝑒Ʈ
−𝑘𝑒(𝑡−𝑇)
𝐷𝑂𝑆𝐸
𝐼𝑛𝑓𝑢𝑠𝑖𝑜𝑛 𝑡𝑖𝑚𝑒
100 𝑚𝑔
0 .5ℎ
- 3 𝑙
𝑘𝑔
- 693
𝑡½
- 693
2 .5ℎ
= 0. 2772 /ℎ or
𝐶𝑙
𝑉𝑑
𝑤𝑒𝑖𝑔ℎ
( 140 −𝑎𝑔𝑒
)
72 𝑆𝑟𝐶𝑟𝑠𝑠
70 𝑘𝑔
( 140 − 30
)
72 ∗ 0. 9
⇒ 𝐶𝑙 = 101 𝑚𝑙/ min = 6. 06 𝑙/ℎ
- 06 𝑙/ℎ
511 𝑙
200 𝑚𝑔/ℎ
- 06 𝑙/ℎ
− 0. 011859 ∗ 0. 5
− 0. 011859 ∗ 0. 5
200 𝑚𝑔/ℎ
- 06 𝑙/ℎ
