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Transcript
INTRAVASCULAR DOSING
PHARMACOKINETICS
ALLIE PUNKE
[email protected]
CONTINUOUS IV INFUSION
 Continuous IV infusion (CIVI) releases the drug at a
constant rate.
 In contrast to IV Bolus, where the maximum
concentration occurs immediately/initial
concentration, the maximum concentration for
continuous IV infusion occurs at the end of infusion.
 Basic equation:


IV Bolus equation: Ct=Co*e^-kt
CIVI equation: Ro/CL*(1-e^-ktinf)(e^-ktpi)
Continuous IV infusion
 A drug is being given by continuous IV infusion at a
rate of 10 µg/hour to reach the steady state
concentration of 46 mg/L. What should the infusion
rate be increased to in order to reach a steady state
concentration of 120 mg/L? 26
10 ug/hour x
------------- ---------46 mg/L 120 mg/L
Ln concentration
Concentration Time Profiles
Infusion
is stopped
During
infusion, at
steady state
During
infusion, before
steady state
Post infusion
Time
 General Expression for plasma concentration:
Ct= R0 * (1-e-k * Tinf) (e-k * ( t - Tinf))
Only important before
Only important after
CL
steady state has been
achieved
the infusion is ended
Ln concentration
During infusion, before steady state
Infusion
is stopped
During
infusion, before
steady state
Reasoning:
Since time and time
of infusion are equal,
t = Tinf, this term
becomes e0 = 1.
Time
 General Expression for plasma concentration:
Ct= R0 * (1-e-k * Tinf) (e-k * ( t - Tinf))
CL
Only important before
steady state has been
achieved
Only important after
the infusion is ended
Ln concentration
During infusion, at steady state
Infusion
is stopped
During
infusion, at
steady state
Reasoning:
Since time and time of
infusion are equal, t = Tinf,
this term becomes e0 = 1.
When Tinf > Tss = 5*T1/2,
then we assume e→∞ = 0.
[1 ─ 0 = 1] making this
term 1.
Time
 General Expression for plasma concentration:
Ct= R0 * (1-e-k * Tinf) (e-k * ( t - Tinf))
CL
Only important before
steady state has been
achieved
Only important after
the infusion is ended
Ln concentration
After infusion, before steady state
Infusion
is stopped
Reasoning:
Both terms are
necessary for
finding Ct.
After infusion,
before steady
state
Time
 General Expression for plasma concentration:
Ct= R0 * (1-e-k * Tinf) (e-k * ( t - Tinf))
CL
Only important before
steady state has been
achieved
Only important after
the infusion is ended
After infusion that ended at steady state
Ln concentration
Infusion
is stopped
Post infusion,
after steady state
was reached
Reasoning:
When Tinf > Tss = 5*T1/2,
then we assume e→∞ = 0. [1
─ 0 = 1] making this term
1.
Time
 General Expression for plasma concentration:
Ct= R0 * (1-e-k * Tinf) (e-k * ( t - Tinf))
CL
Only important before
steady state has been
achieved
Only important after
the infusion is ended
Continuous IV infusion
 A patient is receiving a drug by CIVI at a rate of 800 mg/h to reach a
steady state concentration level of 60 mg/L. A blood level drawn at 7
hours results in a concentration of 40 mg/L.










What is the half-life?
Ct=Css (1-e^-ktinf)
40 mg/L=60 mg/L (1-e^-k*7)
K=0.158, Half-life: 4.4 hours
What is the CL?
Css=Ro/CL
60 mg/L=800 mg/h/CL
CL=13.3 L/h
How long to reach steady state?
22 hours
Continuous IV Infusion
 Maximum concentration after 40 hours?
 Cmax=60 mg/L(1-e^-0.158*40)
 Cmax=59.89
 Suppose the team decides that a level of 47 mg/L is desired.
What should the new rate be? 627 mg/h
Continuous IV Infusion
 A patient is started on a CIVI at a rate of 150 mg/h to reach Css
of 6 mg/L. A blood level of 4 mg/L is drawn at 16 hours.










What is the half-life of the drug?
10 hours, k=0.068
4 mg/L=6 mg/L (1-e^-k*16)
What is the CL?
Css=Ro/CL
CL=25L/h
When is steady state reached?
50 hours
What is the maximum concentration achieved if the infusion runs for 36
hours?
Ct=6 mg/L(1-e^-0.068*36 hours)=5.5 mg/L
Loading Dose
 What LD would we give this pt initially to reach a Css of
6mg/L?
 LD=V*Ctarget=6 mg/L*367L=2200 mg
 CL=k*V
 25L/h=0.068h^-1*V
 What LD would we give if the patient has a concentration
of 6 mg/L, and we want to achieve a concentration of 14
mg/L?
LD=(Ctarget-Ccurrent)*Vd
LD=(14-6)*367L=2,936 mg
Maintenance Dose
 What would the MD be to achieve 10 mg/L?
 MD=CL*Ctarget
 MD=25L/h*10 mg/L=250 mg/h
Loading and Maintenance Dosing
 If CL is increased, this affects:
 A. Loading dose
 B. Maintenance dose
 C. Both
 D. Neither loading or maintenance dose
CIVI is Stopped
 After CIVI is stopped, at hours 2 and 12, the
concentrations were 12.9 mg/L and 6 mg/L.







What is the concentration at the end of the infusion?
K=ln(12.9/6)/(12-2)=0.077
Ct=Co*e^-0.077t
C0=15 mg/L
What will the concentration be 24 hours after the infusion was
stopped?
Ct=15 mg/L(e^-0.077*24 hours)
Ct= 2.4 mg/L
Short term infusion
 HP is given a 1000mg dose of vancomycin over a 30 min infusion.
Levels are taken at 1 hour and 3 hours and found to be 20mg/L and
14mg/L respectively. The population Vd for vancomycin is 44L.
 What is the concentration at 5 hours?
 K=0.178
 CL=7.85 L/h
 DR=2000 mg/h
 Ct=255 mg/L (1-e^-0.178*0.5)(e^-0.0178*4.5)=9.8mg/L
Short term infusion
 When will the concentration be below 7mg/L?
 7mg/L=255 mg/L(1-e^-0.178*0.5)(e^-0.178*tpi)
 6.36 hours +0.5 hours=6.88 hours
Cumulative Drug in Urine
 If Drug X has an fe=1 and a pt is given a dose of 2
grams IVB what will the ΣDu be at t=∞? 2 grams
 What if fe=0.75?
 Fe*dose=0.75*2 grams=1.5 grams
Calculate kr
 Calculate kr using the following time points
(specifically using Time 4 and 6):
kr=ln(139/115.5)/(5-3)=0.092
Time
Du (mg)
0
400
1
356
2
301
4
278
6
231
8
199
Du/t
Rate
t*
278/2
hours
139
3
115.5
5
Calculate FE
 What is FE if kr is the value found in the previous
problem and you have the following information:

A patient is given 400 mg IVB. At 4 hours, the concentration
was 16 mg/L and at 10 hours, the concentration was 5 mg/L.

Fe=kr/k
Fe=0.092/0.193=0.47

Extravascular Dosing
 You have to take into account bioavailability now!
 Calculate the bioavailability (F):
 72% is not absorbed, 23% does not get first-pass metabolized
by the liver, and 80% is metabolized by the gut

F=0.28*0.23*0.2=0.0128
Extravascular Dosing
 A patient is taking a new drug (500 mg) that has the
population values of:


Half-life: 8 hours, Vd: 50 L, kr: 0.021 h-1, Ka: 1.45-1, 90%
reaches the bloodstream
Solve for:

K: 0.087 h-1

FE:kr/k=0.242

CL (total):4.35L/h
Extravascular Dosing (Continued)

CL (renal): kr*Vd, Cltotal*fe=1.05L/h

Cmax: 7.266

Tmax: ln(1.45/0.087)/(1.45-0.087)=2 hours

AUC: FD/CL (A/k-A/ka)=0.9*500 mg/(4.35)=103.9
Extravascular Dosing
 What is it called when the
absorption phase is the
rate limiting step?
 If you were given time
points with
concentrations and told
to assume the drug
followed flip-flop kinetics,
how could you calculate
ka?
Ka=ln(C2/C1)/(Change in
time)
Summary
 Things to think about when making calculations for
the exam:


1. Is it IVB, CIVI, or oral medications?
2. If it is CIVI, has it been stopped or is it still running?
 Double check your work!
 Think about if your answers actually make sense based on the
information given.
 Good luck!