2. Basic Phk-IV and EV.pptx

3/10/2023 IV BOLUS 1
Pharmacokinetics
IV bolus, Multiple and Infusion

3/10/2023 IV BOLUS 2
Pharmacokinetics of IV bolus
One compartment open model

3/10/2023 IV BOLUS 3
Session objectives
• Determine p’k parameters
• Interpret the parameters

Assumptions
One compartment
A. Rapid Mixing
– drug is mixed instantaneously in
blood or plasma
B. Linear Model (dose-independent)
- First Order Kinetics
• Cp versus Time
3/10/2023 4

3/10/2023 IV BOLUS 5
Plasma Data
• Differential Equation
Developing Equation
• From rate vs Cp

3/10/2023 IV BOLUS 6
Plasma data…
Integrated equation
• Cp Vs time
CP= Cp
Oe-Kt
log Cp= logCp
O - kt/2.3
• DB Vs time
DB = DB
0e–kt
log DB = logDB
0 + - kt/2.3
where
• CP → plasma conc. at time t
• Cp
O → conc. at t=0
• DB → amount drug in the body at time t
• DB
0 → amount at t= 0

3/10/2023 7
Pharmacokinetic Parameters
1. Elimination rate constant, K
– fraction of drug removed per unit of
time
– Overall elimination rate constant
– Terminal elimination rate constant
K from
equation and slope
K from
• Slope of the semilog plot
• 1st order Half life, t1/2
t1/2 = 0.693/k

Parameters…
2. Amount of drug in the body, DB
DB = Cp VD
– Where VD relates [drug] in plasma (Cp) and the
amount of drug in the body (DB)
3. Apparent Volume of distribution, Vd
• VD = the volume in which the extent of drug distributed in the body
• VD for a given drug is constant
VD = DB
0/Cp
0 = dose/Cp
0
3/10/2023 IV BOLUS 8

Plasma data…
Apparent Volume of distribution...
3/10/2023 IV BOLUS 9
20
10
5
1
100mg injected at t=0
VD=100mg = 10L
10mg/L
VD= amount of drug injected
plasma concentration at t=o
Cp0
time

3/10/2023 IV BOLUS 10
Parameters…
4. Clearance, CL
• is a measure of drug elimination from the body
– Fixed volume of fluid (containing the drug) cleared of
drug per unit of time(volume/time, L/hr)
• Cl = elimination rate/CP = K Cp VD/Cp = K VD
5. Half life, t1/2
t1/2 = 0.693/K

Practical examples
1. A 70-kg volunteer is given an iv dose of an antibiotic, and
serum drug conc. were determined at 2 hours and 10
hours after administration. The drug conc. were 100µg/ml
and 10 µg/mL, respectively. What is the biologic half-life
for this drug?
2. If the amount of drug in the body declines from 100% of
the dose (IV bolus injection) to 25% of the dose in 8 hours,
what is the elimination half-life for this drug? (Assume first-
order kinetics.)
3/10/2023 IV BOLUS 11

3. A new drug was given in a single iv dose of 200 mg to an
80-kg adult male patient. After 6 hrs, the plasma drug conc.
of drug was 1.5 mg/100 mL of plasma. Assuming that the
apparent VD is 10% of body weight, compute the total
amount of drug in the body fluids after 6 hours. What is the
half-life of this drug?
4. A drug has an elimination t 1/2 of 6 hours. If a single 200-
mg dose is given to an adult male patient (68 kg) by IV
bolus, what percent of the dose is lost in 24hrs?
3/10/2023 IV BOLUS 12

3/10/2023 IV Multiple dose 13
Multiple IV bolus dosing
One compartment model

Objectives
• Discuss drug accumulation after repeated
dose administration
• Explain loading dose, maintenance dose, and
dosing interval
• Calculate appropriate multiple dose regimen
– to achieve desired Cpmin/Cpmax values

1. Introduction
• Most drugs are given as MDR
–MDR → prolonged therapeutic activity
• In designing a MDR,
–the target Cp must be related to the therapeutic response
–CP must be maintained within the TW
–p’k parameters are first obtained from single-dose studies
• Two main parameters - adjusted in MDR
1. Dose (Do)
2. Dosing interval (T)

1. Introduction…
• Superimposition is assumed
►early doses don’t affect the p’k of subsequent doses
►CP after the 2nd , 3rd or nth dose will overlay/superimpose
the CP attained after the (n–1)th dose
►Allows prediction of CP after multiple doses from a dose
Accumulation
• Subsequent doses given before drug from
previous dose(s) eliminated

Accumulation...
• Example: Rx: 100 mg dose qid (4mg/ml every 6h)
- (Literature data t1/2 = 6 hr and V = 25 L)

2. Accumulation...
• As a table
Start End Cp lost
4 --> 2 mg/L 2 mg/L
6 --> 3 mg/L 3 mg/L
7 --> 3.5 mg/L 3.5 mg/L
7.5 --> 3.75mg/L 3.75mg/L
... --> ... ...
8 --> 4mg/L 4mg/L

Drug accumulation...
Development of a General Equation
• At start of the first dose
Cp1= Cp1
0
• At the end of the first dosing interval, 
Cp1
= Cp1
0*e-k*
• At the start of the second interval
Cp2
0 = Cp1
0 + Cp1
= Cp1
0 + Cp1
0*e-k*
• At the end of the 2nd dosing interval, 
Cp2
= Cp1
0*e-k* +Cp1
0*(e-k*)2
• At the start of the 3rd interval
Cp3
= Cp1
0 + Cp1
0*e-k* +Cp1
0*(e-k*)2

Drug accumulation
Geometric Series
– Start of Interval, n
• Cpn
0 = Cp1
0 + Cp1
0 • R + Cp1
0 •R2 +...+ Cp1
0 •Rn-1
– End of Interval, n
» Cpn
T = Cp1
0 • R + Cp1
0 •R2 + Cp1
0 • R 3+ ...+ Cp1
0 •Rn
Sum of Geometric Series
At Start of Interval n (Cmax)
– Cpn
0 = Cp1
0[(1-Rn)/(1-R)] = D/VD[(1-e-nkT)/(1-e-kT)]
– At the End of Interval n (Cmin)
– Cpn
T = Cp1
0[(1-Rn)/(1-R)].R = D/VD[(1-e-nkT)/(1-e-kT)]. e-kT

Steady state/plateau
• Peak or max. conc. at steady state
Cmax,ss = Cp
o
* 1 = Do x 1
(1-e-k) Vd (1-e-k)
• Minimum or trough conc. at steady state
Cmin,ss = Cp
o
* e-k = Do x e-k
(1-e-k) Vd (1-e-k)
• Cmax,ss & Cmax,ss must be with in the TW
• 4 to 5 t1/2 to reach steady state

Drug accumulation
• Example
– Rx: 100 mg q6h; Assume the drug has t1/2 = 4 hr;
k=0.17h-1; V = 10L; R = e-0.17x6 = 0.35
• Question: What are Cmax and Cmin
– Cmax = Cp1
0 /(1-R) = Dose/v.(1-R)= 100/10.(1-0.35)=15.5mg/L
and
– CpT
∞ = Cmin=Dose.R/v.(1-R)=15.5mg/L x 0.35=5mg/L

Loading and maintenance doses
• Loading dose/priming
–The required Cp is quickly achieved
–DL = DM/(1-e-k)
 DL→ loading dose
 Dm→ maintenance dose
• Example Approach to Cmax
– Loading Dose = Cmax •V = 15.5 x 10 = 155mg
– Dosing Regimen: 155 mg IV loading dose
followed by 100 mg q6h

• E.g., Rx: a physician wants to achieve plasma drug level of
Cmax=35 mg/L; Cmin=10mg/L after adm of an antibiotics as
MRD; (assume the Vd and k of drug are 25 L; and k=
0.15h-1, respectively).
– Question: Calculate Dosage Regimen
• Cmax/Cmin = 1/R
• R=10/35 = 0.2857= e-k,
• =8.35
– Using a dosing interval of 8h, R = e-k= e-8´0.15 = 0.3012
– Cmax = Do
V• (1-R)
– Dose = Cmax •V• (1-R) = 35 ´ 25 ´ (1- 0.3012) = 611 mg

• The Dosing Regimen
– Loading Dose = Cmax • V = 35 x 25 = 875 mg
– Loading Dose: 875
– Maintenance Dose: 600 mg q8h
– Rx = 600mg TID, first adm a loading dose of
875mg

Average steady state blood levels
• Where
– Do = maintenance dose,
– F = fraction of dose absorbed,
–  = dosing interval
– t1/2 = half life
– 1.44 = reciprocal of 0.693
10 March 2023 MDR 26

Designing dosage regimen
• Example 1: The elimination half-life of an
antibiotic is 3 hours and the apparent volume
of distribution is 20% of body weight. The
therapeutic window for this drug is from 2 to
10 µg/mL. Adverse toxicity is often observed
at drug concentrations above 15 µg/mL. The
drug will be given by multiple IV bolus
injections.
– a. Calculate the dose for an adult male patient
(68 years old, 82 kg) with normal renal function to
be given every 8 hours.
– b. Calculate the C ∞
av value.

• A
• B

• 2. Gentamicin has an average elimination
half-life of approximately 2 hours and an
apparent volume of distribution of 20% of
body weight. It is necessary to give
gentamicin, 1 mg/kg every 8 hours by
multiple IV injections, to a 50-kg woman
with normal renal function.
• Calculate (a) Cmax, and (b) C ∞
av.

b.

• Example 3: A patient receives 1000 mg every 6 hours
by repetitive IV injection of an antibiotic with an
elimination half-life of 3 hours. Assume the drug is
distributed according to a one-compartment model
and the volume of distribution is 20 L.
– Find the average amount of drug in the body.

• Example 4. Patient C.S. is a 35-year-old
male weighing 76.6 kg. The patient is to
be given multiple IV bolus injections of an
antibiotic every 6 hrs. The effective conc.
of this drug is 15 µg/mL. After the patient
is given a single IV dose, the elimination
half-life for the drug is determined to be
3.0 hr and the VD is 196 mL/kg. Determine
a multiple IV dose regimen for this drug.

IV INFUSION
ONE COMPARTMENT OPEN MODEL
3/10/2023 IV INFUSION 34

Introduction
• Given slowly at a constant or
zero-order rate
• maintains an effective constant
plasma [drug]
– For drugs with a narrow TW
• allows precise control of plasma
[drug] to fit the individual needs
of the patient
• [Drug] is constant after a plateau
or steady-state

CP - t relationship
At steady state:
• Rate at which the drug is leaving the body is equal to
the rate entering the body
• Infusion rate = elimination rate
• Rate of drug input = rate of drug output
• zero-order input and first-order output
• No net change in the Cp
dCp/dt = 0

CP - t relationship….
• Change in the amount of drug in the body at any time =
– rate of input minus the rate of output
– (dDB/dt) = R - kDB …………1
R = infusion rate→ Dose/duration of infusion
DB = amount of drug in the body
K = 1st order elimination rate constant
• Integration above eq. and substitution of DB =C pVD gives
– Cp = R (1-e-kt)………………2
kVD

Steady state Concentration (CSS)
• no net change in the amount of
drug in the body
Cp = (R/kVD)(1-e-kt)
• Theoretical Css is at t = ∞
• At t = ∞, e–kt approaches zero
Css= R/kVD = R/Cl
• CSS depends on R, k & VD

Time to reach CSS
• Drug elimination is exponential (1st order)
– CP becomes asymptotic to the theoretical CSS
• Mathematically, the time to reach true steady-state drug
concentration, CSS, would take an infinite time
• The time for the CP to reach more than 95% of the CSS is
often taken as time for CSS
• About 4 to 5 t1/2
• = Time to reach desired therapeutic conc.

Time to reach CSS……
Number of t 1/2 to reach a fraction of C SS
Percent of CSS reached Number of t1/2
90 3.32
95 4.32
99 6.65

Time to reach CSS……
Example: no of t1/2 to reach 99% CSS
• Css= R/kVD
Cp = (R/kVD)(1-e-kt)
0.99 Css = Css (1-e-kt)
0.01 = e-kt
t99%SS= ln(0.01)/-k = 4.61/k = (4.61/0.693) t1/2= 6.65t1/2
• not dependent on the rate of infusion (an increase
infusion rate will not shorten the time to reach steady
state
• Depends on the elimination rate constant or t1/2

Cp after infusion period
• If infusion stops at steady state or before steady, CP
declines according to first-order kinetics
• Stopped after steady state
Ct =Csse-kt
t → time after the infusion
• Similar to IV bolus, Co= Css

Loading Dose (DL) Plus IV Infusion
• initial bolus dose used to obtain desired conc. as rapidly as
possible
– [drug] after an IV bolus dose C1 = Co. e-kt =(DL/VD)e-kt
– [drug] by infusion, C2 = (R/VDK) (1-e-kt)
• If DL & infusion is started at the same, Cp at t hours = C1 + C2
– Cp =(DL/VD) e-kt + (R/VD K) (1-e-kt)
– Cp = (DL/VD) e-kt + (R/VD K - R/VD K e-kt)
– Cp = R/VD K + (DL/VDe-kt - R/VD K e-kt)
• Let DL equal the amount of drug at SS,
DL = Css VD = R/K

DL Plus IV Infusion…
• Cp= R/VD K + (DL/VDe-kt - R/VD K e-kt)
• Substituting D L = R/k in above eq. makes the
expression in parentheses cancel out.
• Equation, therefore, is the same expression for
CSS or steady-state plasma concentration:
• Cp= R/VD K
• Css = R/VD K

Examples
1. A physician wants to administer an anesthetic agent at a
rate of 2 mg/hr by IV infusion. The elimination rate
constant is 0.1 hr– 1, and the volume of distribution is 10 L.
What loading dose should be recommended if the doctor
wants the drug level to reach 2µg/mL immediately?
2. What is the concentration of a drug 6 hours after
administration of a loading dose of 10 mg and
simultaneous infusion at 2 mg/hr (the drug has a t 1/2 of 3
hr and a volume of distribution of 10 L)?

3. . A female patient (35 years old, 65 kg) with normal renal function is to be given a
drug by IV infusion. According to the literature, the elimination half-life of this drug is
7 hours and the apparent V D is 23.1% of body weight. The pharmacokinetics of this
drug assumes a first-order process. The desired steady-state plasma level for this
antibiotic is 10 µg/mL.
• a. Assuming no loading dose, how long after the start of the IV infusion would it
take to reach 95% of the C SS?
• b. What is the proper loading dose for this antibiotic?
• c. What is the proper infusion rate for this drug?
• d. What is the total body clearance?
• e. If the total body clearance declined 50% due to partial renal failure, what new
infusion rate would you recommend to maintain the desired steady-state plasma
level of 10 µg/mL?
•
3/10/2023 IV BOLUS 46

3/10/2023 p'k of extravascular dose 48
Pharmacokinetics of
extravascular dose
One compartment model

1. Introduction
Extravascular
• Not IV
• Involve an Absorption Step
including Oral, IM, SC, etc.
• Absorption Defined by
Rate and Extent

Introduction...
Oral route (oral doses)- Most commonly used route of admn.
• The rate of change in the amount of drug in the body, dDB/dt,
is the function of absorption and elimination rate.
• dDB/dt = dDGI/dt - dDE/dt

EV single dose equation
A. zero-order absorption
Cp = ko/(kVD)(1-e-kt)
B. 1st- order absorption
CP = FKaDo[e-kt - e-kat]
VD(Ka-k)
F
Dose Dose and DF parameters
Ka
Kel Drug and Patient parameters
V

Pharmacokinetic Parameters
1. tmax ,Time for maximum CP
• tmax = 1 ln Ka
Ka-K K
• independent of Do
• depend on ka and k
• Cmax is CP at tmax
K
Ka
K)
(Ka
.
t log
303
2
max 


2. determination of K
• Determined from the elimination phase of the Cp–time curve
CP = FKaDo[e-kt - e-kat]
VD(Ka-k)
• Assuming k a >> k in above Equation, the value for the
second exponential will become insignificantly small with
time (ie, e –k
a
t ≈ 0)
CP = FKaDo e-kt
VD(Ka-k)
CP = Ae-kt; A = fDKa/[VD(Ka-K)]

determination of K...
• CP = Ae-kt;
• This equation represents
first-order drug elimination,
• Plot logCp vs t
– slope = -k/2.3
– Intercept A
– Similar to IV bolus dose

3. Determination of Ka- Method of residuals
• Back Extrapolate the
elimination phase curve to t=0
– Curve AC is obtained
• Take ≥3 points on the upper
part of AC
– eg, x'1, x'2, x'3, ...
• Drop vertically to obtain
corresponding points on the
curve
– eg, x1, x2, x3, . . .

Ka-Method of residuals…..
• Plot the values of the d/ces at
the corresponding time points
– x'1-x1, x'2-x2, x'3-x3, ...
• A straight line will be obtained
slope =-ka/2.3
• Equation of the residual line
C’ = [fDKa/(VD(ka-K))]e-kat
C’ = x‘i-xi

4. Lag time-Method of residuals
• Time b/n administration and start of absorption
– Enteric coating tablets
– Gastric emptying
– Dissolution of capsule shell
The lag time t0 is subtracted from each time point
CP = FKaDo [e-k(t-to) - e-ka(t-to)]
VD(Ka-k)
5. Elimination half life
t1/2= 0.693/k

58
Effect of ka on Cmax, tmax, & AUC
• Increasing the
absorption rate
constant (Ka) results
in:
– Shorter tmax
– Higher Cmax
– Unchanged AUC
Changing Ka ( K unchanged)

59
Effect of K on tmax, Cmax, and
AUC
• Increasing the
elimination rate
constant (K) results
in:
– Shorter tmax
– Lower Cmax
– Lower AUC
Changing K ( Ka unchanged)

60
Effect F on tmax, Cmax, and AUC
0
50
100
150
200
250
300
0 20 40 60 80 100
time
Concentration
F = 1
F = 0.5
F = 0.25
 Increasing the bioavailability
results in:
 Unchanged tmax
 Higher Cmax
 Higher AUC

Example
A single oral dose (600 mg) of an antibiotic (90%
bioavailable) was given to an adult male patient (43 years,
72 kg). From the literature, the drug has ka = 1.0 hr-1, kel =
0.15 hr-1, and V = 30 liter.
calculate t max, C max, and t 1/2 for the drug in this patient..

MDR-extravascular

Multiple-Oral-Dose Regimen
• Cp at any time during an oral or EV MDR, assuming a
one-compartment model and constant doses and dose
interval, can be determined as follows:
– where
• n = number of doses,
• T= dosage interval,
• F = fraction of dose absorbed, and
• t = time after administration of n doses

Cp-time

C∞
max, C∞
min, C∞
av
tp –time for peak conc. at steady state
• C∞
av = FDo/VdK

Loading dose, DL
• To reduce the onset time of the
the time it takes to achieve the MEC
• DL/DM = 1/(1-e-kaT)(1-e-kT)
• Practically, for most drugs Ka>>>K
• DL/DM = 1/(1-e-kT)
DL = DM/(1-e-kT)

Example
• An adult male patient (46 yrs old, 81 kg) was given orally
250 mg of TTC HCl TID for 2 weeks. From the literature,
TTC is about 75% bioavailable and has Vd of 1.5 L/kg. k
is about 0.07 hr– 1 & Ka is 0.9 hr– 1.
From this information, calculate
(a) plasma drug concentration C p at 4 hours after the 7th
dose, Here, n=7, t=4, Ꞇ=8

(b) maximum plasma drug concentration at
steady-state C ∞
max,

(c) minimum plasma drug concentration at steady-state C ∞
min, and
(d) average plasma drug concentration at steady-state C ∞
av.

• Example 2. A physician wants to give theophylline to a
young male asthmatic patient (age 29, 80 kg). According
to the literature, the drug has k of 5 hrs & VD equal to 50%
of body weight. The plasma level of theophylline required
to provide adequate airway ventilation is approximately 10
µg/mL.
– a. The physician wants the patient to take medication
every 6 hours around the clock. What dose of
theophylline would you recommend (assume
theophylline is 100% bioavailable)?
– b. If you were to find that theophylline is available to
you only in 225-mg capsules, what dosage regimen
would you recommend?

Dosage regimen
10 March 2023 73
Dosage Regimen – the manner in which the
drug is taken.
– Some are taken as single dose, however, most
drugs are taken on a repetitive bases
– For successful therapy, design of an optimal
multiple dosage regimen (MDR) is necessary.
• MDR:- is the manner in which the drug is
administered in suitable doses by suitable
route, with sufficient frequency that insures
maintenance of Cp within TI for entire period of
therapy.

Approaches to Design of Dosage
Regimen
74
The various approaches employed in
designing a dosage regimen are –
1. Empirical Dosage Regimen
2. Population Averages Dosage Regimen
(a) Fixed model
(b) Adaptive model
3. Individualized Dosage Regimen
» based on the phks of drug in the individual
patient.

1. Empirical Dosage Regimen
75
Designed by the physician based on
– Empirical clinical data,
– Personal experience and
– Clinical observations.
• In many cases, the physician selects a DR for
the patient without using any ph’k variables.
• This approach is, however, not very accurate.

2. Population Averages Dosage
Regimens
76
DR based on Population Averages–
• Is the most often used approach.
• is based on one of the two models
a) The fixed model
– Based on population average ph’k parameters.
– This parameters are used directly to calculate a DR for
the patient.
• Ph’k parameters such as ka, F, VD, & k, are assumed to remain
constant.
• Drug is assumed to follow the phk of a one-compartment
model.
• When a MDR is designed, multiple-dosage equations based on
the principle of superposition are used to evaluate the dose.

b) The adaptive model
77
The adaptive model is Based on
 Population average ph’k parameters of the drug as well as
 Patient variables such as weight, age, sex, body surface
area and known patient pathophysiology such as renal
disease.
• Calculation of the DR takes into consideration
– any changing pathophysiology of the patient &
attempts to adapt or modify the DR according to
the needs of the patient.

Fixed model: Designing MDR
78
Irrespective of the route of adm. & complexity of ph’k
equations, the two major parameters that can be
adjusted in developing a DR are
1. The dose size
2. The dosing frequency
Both parameters govern the amount of drug in the
body at any given time.

Fixed Model: Dose size
79
The magnitude of both therapeutic and toxic
responses depends upon dose size.
• Greater the dose size, greater the fluctuations
between Css,max & Css,min during each dosing
interval & greater the chances of toxicity (Fig.
below).
• For drugs administered chronically, dose size
calculation is based on average Css & is computed
from equation:

80
Fig: influence of dose size on CpVs t after oral adm of a drug at fixed
intervals of time.

Determination of Dose: Examples
81
According to literature, the t1/2 of tobramycin was reported to be 2.15
hrs & the VD was reported to be 33.5% of body weight. What is the
dose for an 80-kg pt if Css level of 2.5 g/mL is desired? Assume that
the drug is given by IV bolus injection TID.
• Solution Assuming the drug is 100% BA as a result of IV injection,
• The dose should be 173 mg every 8 hours.

Fixed Model: Dosing
Frequency
82
• The dose interval (inverse of dosing
frequency) is calculated on the basis of
half-life of the drug.
• If the interval is increased and the dose is
unchanged, Cmax, Cmin and Cav decrease
but the ratio Cmax/Cmin increases.
• Opposite is observed when dosing interval
is reduced or dosing frequency increased.
– It also results in greater drug accumulation in
the body and toxicity (see Fig.).

Fixed Model: Dosing Frequency
83
Fig: influence of dosing frequency on CpVs t profile after
oral adm of a drug at fixed dose of a drug

Fixed Model: Dosing
Frequency
84
• A proper balance b/n both dose size
and dosing frequency is often desired
to attain steady-state conc. with minimum
fluctuations and to ensure therapeutic
efficacy & safety.
– The same cannot be obtained by giving larger
doses less frequently.
– However, administering smaller doses more
frequently results in smaller fluctuations

Fixed Model: Dosing
Frequency…
85
Generally speaking, every subsequent dose should
be administered at an interval equal to half-life of
the drug.
A rule of thumb is that –
• For drugs with wide TI such as penicillin,
– Larger doses may be administered at relatively longer
intervals (˃t1/2 of drug) without any toxicity problem
• For drugs with narrow TI such as digoxin,
– Small doses at frequent intervals (usually ˂ t1/2) is better
to obtain a profile with least fluctuations (similar to
constant rate infusion or controlled-release system).

Determination of Frequency
Example
86
• For example,
– Dm of digoxin is 0.25 mg/day & its t1/2 is 1.7 days.
– Dm of penicillin G is 250 mg QID, while its t1/2 is 0.75
hr.
– Penicillin is given at  = 8x t1/2 , whereas digoxin is
given at  = 0.59x t1/2 0.59.
– The toxic Cp of penicillin G is over 100x ˃ its effective
Cp,
– The toxic Cp of digoxin is only 1.5x ˃ its effective.
• Therefore,
– a drug with a large TI can be given in large doses and
at relatively long dosing intervals.

Determination of Frequency Example
87
The size of a drug dose is often related to the
frequency of drug administration.
• The more frequently a drug is administered, the
smaller the dose must be to obtain the same C∞
av.
• Thus, a dose of 250 mg every 3 hours could be
changed to 500 mg every 6 hours without affecting
the C∞
av of the drug.
– However, as dosing intervals , the size of the dose
required to maintain the C∞
av gets larger.
– When an excessively long dosing interval is chosen,
the large dose may result in peak plasma levels that
are above toxic drug conc, even though C∞
av will
remain the same.

Fixed Model: Determination of Both Dose
and Dosage Interval
88
• Both Do &  should be considered in the DR calculations.
• Ideally, the calculated DR should maintain the serum drug
conc between C∞
max and C∞
min.
• For IV MDR the ratio of C ∞
max/C ∞
min may be expressed by
• Which can be simplified to

Fixed Model: Determination of
Both Dose and Dosage Interval
89
Practice Problem
• The t1/2 of an antibiotic is 3 hrs with Vd of 20% of body Wt. The
usual therapeutic range for this antibiotic is between 5 &
15g/mL. Calculate a DR (multiple IV doses) that will just
maintain the serum drug conc b/n 5 & 15g/mL.
Solution
• From the above Equation, determine the maximum possible
dosage interval .
Take the natural logarithm (ln) on
both size of the equation

Fixed Model: Determination of
Both Dose and Dosage Interval
90
By calculation, the dose of this antibiotic should be 2 mg/kg every
4.76 hours to maintain the serum drug concentration between 5
and 15  g/mL.

2. Basic Phk-IV and EV.pptx

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2. Basic Phk-IV and EV.pptx