Question

Determine the mg maintenance dose (to the nearest tenth) of amikacin for a 42-year-old female patient weighing 218 lb. and measuring 5 ft. 1 inches in height with a serum creatinine of 1.6 mg/dL. The physician requests a maintenance dose of 5 mg/kg of ideal body weight (adjusted for renal impairment - based on the calculated value of creatinine clearance) to be administered continually at intervals of 8 hours.

Please show all work.

Answer 1

Body weight of the female patient : 218lbs = 98.8kgs

Height of the patient = 5ft 1 inch = 154.9 cm

Serum clearance level = 1.6mg/dl

Maintenance dose instructed by physician = 5mg/kg of ideal body weight ,Q8H (8th hourly)

Step 1: Ideal BW in a women =  45.5 + (0.91 × [height in centimeters − 152.4])

= 45.5 + (0.91 x [154.9 - 152.4])

= 45.5 + (0.9 x 2.5)

= 45.5 + 2.3 = 47.8 kg.

So the ideal BW of this client is 47.8 kg.

[Another formula to calculate ideal BW = 45.5+2.3∗(height over 60inches)]

Step 2: Dosing Weight = Ideal BW + 0.4 x (Actual BW−Ideal BW)

= 47.8 + 0.4 x (98.8 - 47.8)

= 47.8 + (0.4 x 51)

= 47.8 + 20.4 = 68.2

Dosing weight = 68.2 kg.

Step 3:

Maintenance Dose(mg)= dose for ideal BW x calculated dosing weight

= 15 mg/kg x 68.2kg =1000mg

Peak = Dose / Vd = Dose / Volume of distribution.

    So peak = 1000mg / (68.2kg x 0.3L/kg) = 48.9mcg/mL