PHCL 510
Prof. Hisham
Theophylline Problem Set
with answers
1. JB is a 50-year-old, 60-kg (5 ft 7 in) male with heart failure started on a 50 mg/hr
aminophylline infusion after being administered an intravenous loading dose. The
theophylline concentration was 15.6 μg/mL at 1000 H and 18.3 μg/mL at 1400 H.
What aminophylline infusion rate is needed to achieve Css = 15 μg/mL?
Solution:
Apply Chiou method:
CL 
2SR o
2 Vd (C1  C2 )

C1  C2 (C1  C2 )(t 2  t1 )
2  0.8  50 mg.hr
2  (0.5 L.kg  60 kg)(15.6 mg.L  18.3 mg.L )

15.6 mg.L  18.3 mg.L
(15.6 mg.L  18.3 mg.L )(4 hr )
 1.17 L/hr
1
CL 
1
1
1
1
1
1
1
Css  CL (15 mg/L  1.17 L.hr )
=
=22 mg/hr
S
0.8
(Round this figure to 20 mg/hr aminophylline)
1
Ro =
2. LK is a 50-year-old, 75-kg (5 ft 10 in) male with chronic bronchitis who requires
therapy with oral theophylline. He currently smokes 2 packs of cigarettes daily, and
has normal liver and cardiac function. Suggest an initial theophylline dosage
regimen (t.i.d) designed to achieve a steady-state theophylline concentration equal
to 8 μg/mL. Assume that oral sustained-release theophylline tablets will be
prescribed to this patient (F=1, S=1) and the half-life of the drug is 8 hr.
Solution:
0.693 0.693
kd 

 0.0866 hr 1
t 12
8 hr
Vd  0.5 L.kg  75 kg=38 L 
1
CL  kd Vd  0.0866 hr  38 L  1.6  effect of smoking   5.27 L/hr
  8 hr
C  CL  
D  ss
SF
8 mg.L1  5.27 L.hr 1   8 hr 


 337.28 mg
11
(Round it to 300 mg every 8 hr)
1
PHCL 510
Prof. Hisham
3. A.M. is a 80-kg, 5 foot 11 inch, 50-year-old male who is to be treated with
theophylline for an asthmatic attack. Theophylline’s clearance in A.M. is estimated
to be 1.64 L/h based on his demographic features: severe obstructive pulmonary
disease; congestive heart failure; smoking a pack of cigarettes a day. The population
average volume of distribution of theophylline is 0.5 L/kg.
a) Advise on a suitable infusion rate of aminophylline (S= 0.8) to achieve a
plasma concentration of 15 mg/L theophylline.
b) Determine a suitable intravenous loading dose of aminophylline.
c) An infusion of 30.75 mg/h is used. A steady-state plasma sample is obtained
and reveals a theophylline Cp value of 25 mg/L (A.M.’s clearance was only
estimated). Recommend a more suitable infusion rate based on this
information.
d) Assume that the new infusion rate is used and that steady state is achieved. The
infusion is terminated. How long would it take plasma concentrations to fall to
3.5 mg/L?
e) Comment on how steady-state plasma concentrations of theophylline may be
affected by:
1. Compounds such as cimetidine and the quinolone antibiotics that
inhibit theophylline’s intrinsic clearance
2. Compounds that alter hepatic blood flow
3. Phenytoin, which induces theophylline’s metabolism
Solution:
Css  CL 15 mg.L1  1.64 L.hr 1

 30.75 mg/hr
S
0.8
Vd  Css 0.5 L.kg 1  15 mg.L1

 30.75 mg/hr
b) D 
0.8
0.8
R
30.75 mg.hr 1
 1.23 L/hr
c) CL  o 
Css
25 mg.L1
a) R o 
Css  CL 15 mg.L1  1.23 L.hr 1

 23.06 mg/hr
S
0.8
Or a more direct approach since the pharmacokinetics of theophylline are linear:
Ro 
desired Css old R o 15 mg.L1 30.75 mg.hr 1



 23.06 mg/hr
observed Css
S
25 mg.L1
0.8
d) C  3.5 mg/L
Co  15 mg/L
New R o 
kd 
CL
1.23 L.hr 1

 0.03075 hr 1
1
Vd 0.5 L.kg  80 kg
PHCL 510
Prof. Hisham
We know that:
C  Co  e  kdt
Solve for t
1 C
t   ln  
k d  Co 
This equation can take more than one form:
1  Co 
C 
t  1.44 t 12  ln  o 
t
ln   or
kd  C 
 C
 15 mg.L1 
1
t
ln 
  4.73 hr
0.0375 hr 1  3.5 mg.L1 
e) Will be discussed in class
4. A 70 kg male with a history of asthma has been admitted to the hospital for a severe
acute asthmatic attack and is given 300 mg aminophylline as an intravenous bolus
of the dihydrate every 8 hr (dosing interval, τ) to control the attack. From prior
history in this patient:
 elimination half-life=7 hr
 apparent volume of distribution of theophylline = 30 L.
 Aminophylline dihydrate is the ethylenediamine salt of theophylline
containing two waters of hydration and 79% as active drug.
a) Calculate the maximum steady-state plasma concentration of theophylline,
b) Calculate the “trough”, or minimum, steady state plasma concentration of
theophylline.
c) Calculate the ‘average’ steady-state concentrations of theophylline, for the
dosage regimen above and for a regimen in which one-half the original dose is
administered at 4 hr intervals.
d) What will be the theophylline plasma concentration immediately prior to the
administration of the fourth dose?
e) What will be the theophylline plasma concentration immediately following the
administration of the fourth dose?
f) Determine the drug accumulation ratios (RAcc), in as many ways as you can think
of for the two different dosage regimens employed in (c) of this question.
g) What is the loading-dose needed to produce an immediate steady-state
concentration?
h) How long will it take to reach one half of the steady-state concentration of
theophylline (i.e., fss  0.5 )?
i) A patient weighing 200 lb receives 520 mg theophylline per day as Tedral
tablets. The drug is 100% available for absorption from the tablets, and its
elimination half-life and apparent volume of distribution are 4.5 hr and 0.48
L/kg, respectively. What ‘average’ steady-state plasma concentration would be
obtained?
PHCL 510
Prof. Hisham
Solution:
a) As salt value of aminophylline is 0.79, 300 mg aminophylline gives a dose of
237 mg theophylline.
SD
0.79  300 mg
max
CSS


 14.44 mg/L
1
 kd
Vd 1  e
30 L 1  e 0.099 hr 8 hr




min
max
b) CSS
 CSS
 e  kd  14.44 mg.L1  e 0.099 hr
1
8 hr
 6.54 mg/L
c) CL  k d Vd  0.099 hr 1  30 L  2.97 L/hr
SD
0.79  150 mg
CSS 

 9.975 mg/L
CL   2.97 L  hr 1  4 hr
min
3
d) C

 e
V 1  e 
S  D 1  e 3 k d 

d

 kd 
 kd 

0.79  300 mg 1  e 3  0.099 hr

30 L 1  e 0.099 hr
 C3min 
e) Cmax
4
1
 8 hr
1

 8 hr
e
0.099 hr 1  8 hr
 5.933 mg/L
SD
0.79  300 mg
 5.933 mg.L1 
 13.833 mg/L
Vd
30 L
Or you can calculate directly using the equation for Cmax
:
4
C
max
4



V 1  e 
S  D 1  e 4 k d 
 kd 
d
f) Accumulation ratio for the dosage regimen of 300 mg q8hr:
1
1
R Acc 

 1.264
k d   0.99 hr 1  8 hr
Accumulation ratio for the dosage regimen of 150 mg q4hr:
1
1
R Acc 

 2.525
k d   0.99 hr 1  4 hr
g) It has already been determined that administration of 300 mg aminophylline
(equivalent to 237 mg theophylline) every 8 hr yields a peak steady-state
theophylline plasma concentration of 14.44 mg/mL. The following equation
allows determination of the loading dose (DL) necessary to attain this
theophylline plasma concentration instantaneously:
DL
1

DM 1  e  kd
PHCL 510
Prof. Hisham
1 
1



DL  DM 
 300 mg 
 548.37 mg aminophylline
 kd 
0.099 hr 1  8 hr 
1 e

1 e

Because we need to reach maximum level directly.
h) Time required to attain any fraction of steady state (fss) may be determined by
using the following equation:
fSS  1  e  kd t
 1   0.5
n
Where n=number of elimination half-lives required to reach a given fraction of
steady State, i.e.,
t
n
t 12
To attain 50% of the steady state (i.e., fSS=0.5):
n
fSS  1   0.5
0.5  1   0.5
n
Taking the natural logarithm of both sides:
ln  0.5  n  ln  0.5
n 1
It will take one half-life (i.e., 7 hr) of the drug to reach 50% of the steady state
concentration. (See explanation on page 117 in the GREEN BOOK).
i) Weight of the patient = 200 lb = 90.909 kg (1 kg = 2.2 lb)
FD
CSS 
CL  
FD

k d  Vd  
We can write the previous equation in terms of half-life to avoid the conversion
to kd:
1.44 FD t 12
CSS 
Vd  

1.44  1  520 mg  4.5 hr
 3.22 mg/L
0.48 L.kg 1  90.909 kg  24 hr