Question #1. B.C., a 65-year old, 45-kg man with a serum creatinine concentration of 2.2 mg/dL, is being treated for a presumed hospital-acquired, MRSA infection. Design a dosing regimen that will produce peak concentration less than 40 to 50 mg/L and through concentrations of 5 to 15 mg/L.

Target Plasma Concentration

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Clearance and Volume of Distribution

The first step in calculating an appropriate dosing regimen for B.C. is to estimate his pharmacokinetic parameters (i.e., volume of distribution, clearance, elimination rate constant, and half-life).

The volume of distribution for B.C. can be calculated by using Equation 13.1.

V (L) = 0.17 (age in years) + 0.22 (TBW in kg) + 15

So, B.C.'s expected volume of distribution would be: V (L) = 0.17 (65 yrs) + 0.22 (45 kg) + 15 = 36.0 L [Equation 13.1]

Using Equation 13.2 and Equation 13.4 to calculated B.C.'s expected creatinine clearance and vancomycin clearance.

Clcr for males (mL/min) = (140 – Age)(Weight in kg) / [(72)(SCrss)] [Equation 13.2]

Vancomycin Cl ≈ Clcr [Equation 13.4]

For B.C. the vancomycin Cl ≈ (140 – 65 yrs)(45 kg) / [(72)(2.2 mg/dL)] = 21.3 mL/min = 1.28 L/hr

The calculated vancomycin clearance of 1.28 L/hr and the volume of distribution of 36.0 L then can be used to estimate the elimination rate constant of 0.036 hr-1. And the corresponding vancomycin half-life can be calculated, which equals (0.693)(V) / Cl = 19.5 hr.

Loading Dose

In clinical practice, loading doses of vancomycin are seldom administered. This is probably because most clinicians prescribe about 15 mg/kg as their maintenance dose.

C0 = (S)(F)(Loading Dose) / V = (1)(1)(15 mg/kg x 45 kg) / 36 L = 18.8 mg/L ≈ 20 mg/L (Equation 13.8)

If you want to administer a loading dose, the loading dose = (V)(C) / [(S)(F)] = (36.0 L)(30 mg/L) / [(1)(1)] = 1080 mg or ≈ 1000 mg.

Steady-State

During the steady-state, Css max = Css min + [(S)(F)(Dose) / V] (Equation 13.5). This equation is based on several conditions including: 1) Steady state has been achieved; 2) the measured plasma concentration is a trough concentration; and 3) the bolus dose is an acceptable model (infusion time <1/6 half-life).

In the clinical setting, trough concentrations are often obtained slightly before the true trough. Because vancomycin has a realtive long half-life, most plasma concentrations obtained within 1 hour of the true trough can be assumed to have met condition 2 above.

Since vancomycin follows a multicompartmental model, it is difficult to avoid the distribution phase when obtaining peak plamsa concentrations. If peak levels are to be measured, samples should be obtained at least 1 or possibly 2 hours after the end of the infusion period. It is difficult to evaluate the appropriateness of a dosing regimen that is based on plasma samples obtained before steady state. Additional plasma concentrations are required to more accurately estimate a paient's apparent clearance and half-life, and to ensure that any dosing adjustments based on a non-steady-state trough concentration actually achieve the targeted steady-state concentrations.

Maintenance Dose

The maintenance dose can be calculated by a number of methods. One approach might be to first approximate the hourly infusion rate required to maintain the desired average concentration. Then, the hourly infusion rate can be multiplied by an appropriate dosing interval to calculate a reasonable dose to be given on an intermittent basis. For example, if an average concentraion of 20 mg/L is selected (approximately halfway between the desired peak concentration of ≈ 30 mg/L and trough concentration of ≈ 10 mg/L), the hourly administration rate would be 25.6 mg/hr.

Maintenance Dose = (Cl)(Css ave)(tau) / [(S)(F)] 

For this patient the 24 hour dose should be (1.28 L/hr)(20 mg/L)(24 hr) / [(1)(1)] = 614 mg ≈ 600 mg

– or –

Maintenance delivery rate = Dose/tau = (Cl)(Css ave) / [(S)(F)]

For this patient the maintenance deliver rate = (1.28 L/hr)(20 mg/L) / [(1)(1)] = 25.6 mg/hr

The second approach that can be used to calculate the maintenance dose is to select a desired peak and trough concentration that is consistent with the therapeutic range and B.C.'s vancomcin half-life. For example, it steady-state peak concentrations of 30 mg/L are desired, it would take approximately two half-lives for that peak level to fall to 7.5 mg/L. Since the vancomycin half-life in B.C. is approximately 1 day, the dosing interval would be 48 hours. The dose to be administered every 48 hours can be calculated as follows using Equation 13.5:

Dose = (V)(Css max – Css min) / [(S)(F)] = (36.0 L)(30 mg/L – 7.5 mg/L) / [(1)(1)] = 810 mg ≈ 800 mg

The peak and trough concentrations that are expected using this dosing regimen can be calcualted by using Equations 13.12 and 13.14, respectively.

Css max = (S)(F)(Dose) / {V x [1- e(-k*tau)]} = 27.0 mg/L (Equation 13.12)

Note that although 27 mg/L is an acceptable peak, the actual clinical peak would normally be obtained approximately 1 hour after the end of a 1-hour infusion, or 2 hours after this calculated peak concentration, and would be about 25 mg/L, as calculated by Equation 13.13.

C2 = C1[e(-k*t)] = 25.1 mg/L (Equation 13.13)

The calculated trough concentration would be about 5 mg/L.

Css min = (S)(F)(Dose / V)[e(-k*tau)] / [1 – e(-k*tau)] = (Css max)[e(-k*tau)] = 4.8 mg/L (Equation 13.14 and 13.15)

This process of checking the expected peak and trough concentrations is most appropriate when the dose or the dosing interval has been changed from a calculated value (e.g., twice the half-life) to a practical value (e.g., 8, 12, 18, 24, 36, or 48 hours). Many institutions generally prefer not to use dosing intervals of 18 or 36 hours because the time of day whent the next dose is to be given changes, potentially resulting in dosing errors. If different plasma vancomycin concentrations are desired, Equations 13.12 and 13.14 can be used target specific vancomycin concentrations by adjusting the dose and/or the dosing interval.

A third alternative is to rearrange Equation 13.14, such that the dose can be calculated:

Dose = (Css min)(V)[1 – e(-k*tau)] / {(S)(F)[e(-k*tau)]} (Equation 13.16)