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**a. Define Cp _{min}/Cp_{max}.** From information on the drug with reference to
the patient's clinical requirements. For example the normal upper limit for
an aminoglycoside peak concentrations might be 6 mg/L, however in case of life-threatening
infection higher levels may be approached. Initial calculation might be based
on a peak of 6 mg/L and a trough below 1 mg/L. (Use 1 mg/L as the trough and
extend the interval when making the adjustment in τ.

**b. Determine CL _{Cr}.** Probably from serum creatinine levels using the
Cockcroft-Gault equation.

**c. Determine kel.** Using the equation kel = km + b • CL_{Cr} with km and b
values from the literature

**d. Calculate Tau.** Since
and we know Cp_{min}, Cp_{max}, and kel we can calculate tau, τ. Typically this
will be some uneven time value.

**e. Round Tau.** A more usual dosing interval should now be chosen. For
example a tau of 7.8 or 6.7 hour could be rounded to 8 hours, thus dosing three
times a day.

**f. Recalculate R.** A new value of tau results in a new value of R.

**g. Calculate Maintenance Dose.** The maintenance dose can be calculated
from the minimum or the maximum plasma concentration. Thus

Maintenance dose = Cp_{max} • V • (1 - R) **OR**
= Cp_{min} • V • (1 - R)/R

**h. Calculate Loading DOSE.** The loading dose can be calculated directly
(for an IV bolus) by equating Cp^{0} and the Cp_{max} value. Thus,

Loading dose = Cp_{max} • V

**Example:**

A 75 kg, 65 year old male patient, serum creatinine concentration of 2.3 mg/100
ml, is to be given an aminoglycoside IV to achieve a peak plasma concentration of 6
mg/L and trough concentration **below** 1 mg/L. The apparent volume of
distribution is reported to be 0.28 L/kg. From Table 3-9 (Wagner, 1975 p161), km and b values are 0.02 and 0.0028, respectively.

a. Cp_{max} = 6 mg/L and Cp_{min} = 1 mg/L

b. CL_{Cr} =

c. kel = km + b • CL_{Cr} = 0.02 + 0.0028 x 34 = 0.115 hr^{-1}

d. R =

ln (0.1667) = -1.792 = -0.115 x τ

τ = 15.6 hour

e. Since a longer dosing interval is needed to keep the trough level
**below** 1 mg/L use a tau value of 18 hours.

f. New R value. R = e^{-0.115 x 18} = 0.1262

g. Calculate maintenance dose using Cp_{max} = 6 mg/L as reference point. Thus

Maintenance dose = Cp_{max} • V • (1 - R)
= 6 x 75 x 0.28 x (1 - 0.1262)
= 110 mg

Thus use 100 mg iv every 18 hours

Cp_{min} = Cp_{max} • R = 5.45 x 0.1262 = 0.69 mg/L

h. The loading dose can be calculated as:-

Loading dose = Cp_{max} • V = 6 x 75 x 0.28 = 126 mg. Using 125 mg would give a
Cp_{max} =
= 5.95 mg/L.

Thus a loading dose of 125 mg followed by 100 mg every 18 hours should be satisfactory.

Comparison with PDR recommendation.

Usual dose for 75 kg patient is 75 mg q8h. With serum creatinine 2.3 mg/100 ml
give 40 percent of 75 mg q8h. That is 30 mg q8h (R = 0.399) giving 2.38 and
0.95 for Cp_{max} and Cp_{min}, respectively.

Practice calculating a patient's elimination rate constant from creatinine clearance

- Wagner, J.G. 1975
**Fundamentals of Clinical Pharmacokinetics**, Drug Intelligence Publications, Inc., Hamilton, IL

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