Question

A 192 lb hospitalized patient received intermittent intravenous infusions of an antibiotic for a serious and life threatening infection from a burn wound. The patient received 9.9 mg/kg infused over 45 minutes administered every 6 hours. A blood samples was collected 15 minutes before the third infusion was started and the drug concentration was found to be 16.8 mcg/mL. Another blood sample was collected 60 minutes after the third infusion was completed and the drug concentration was found to be 41.3 mcg/mL. Assume steady state was achieved before these samples were collected.

What dose should the patient receive in order to produce steady state minimum drug concentrations of 4 mcg/mL? (Note: Please provide your answer in units of mg. Do not include the units in the answer you enter.)

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Answer 1

Here, Wt of patient is 192 lb or 87 kg

Therefore, total amount of drug infused in 45 min =( 9.9mg/kg)*87kg= 861.3 mg

Therefore, the infusion rate R_o = 861.3 mg/45min = 19.14 mg/min= 1148.4mg/hr

Also, given here are-

Time between dose infusion (tau) = 6hr

Steady state drug concentration C_ss= 4mcg/ml = 4mg/L

Drug concentration 15 min before third drug infusion C_min = 16.8 mcg/ml = 16.8 mg/L

and drug concentration 60 min after third drug infusion C_max = 41.3mcg/ml = 41.3 mg/L

Now from relationship C_ss = AUC/tau, this gives AUC = C_ss*tau = 4mg/L*6h = 24mg/L/hr

Now, AUC = (F* Dose)/ CL or Dose = AUC*CL/F (where F is bioavailability and CL is clearance volume)

Now solving for CL/F we get, CL/F = R_o/(C_max - C_min) = (1148.4mg/h)/(41.3 -16.8)mg/L = 46.87 L/h

Putting this in aabove equation for dose, we get Dose = 24 * 46.87 = 1125 mg