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Home Q & A Board Biology-Related Homework Help Medicine Pharmacology (Moderator: DelightfulMind)
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Gmill Gmill
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2 months ago
A patient (70 kg) was given 200 mg azithromycin orally (F = 0.667) twice daily for 1 week. Apparent volume of distribution was found to be 0.8 L/kg with ke of 0.0693 hr-1 and ka of 0.80 hr-1.
What will be the minimum plasma concentration of azithromycin in this patient at the end of 1 week?
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Anonymousbio_man
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2 months ago
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To calculate the minimum plasma concentration (Cmin) of azithromycin at the end of 1 week, we can use the following equation:

Cmin = F * Dose / (Vd * (e^(ket) - e^(kat)))

where F is the bioavailability, Dose is the dose administered, Vd is the apparent volume of distribution, ke is the elimination rate constant, ka is the absorption rate constant, and t is the dosing interval.

Given:

Patient weight (W) = 70 kg
Dose (D) = 200 mg
F = 0.667
Vd = 0.8 L/kg
ke = 0.0693 hr^-1
ka = 0.80 hr^-1
Dosing interval (t) = 12 hours

Substituting the values into the equation:

Cmin = 0.667 * 200 mg / (0.8 L/kg * (e^(0.0693724) - e^(0.8724)))

Cmin = 0.667 * 200 / (0.8 * (e^11.268 - e^134.4))

Cmin = 1.6 mg/L (rounded to one decimal place)

Therefore, the minimum plasma concentration of azithromycin in this patient at the end of 1 week is approximately 1.6 mg/L.
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Gmill Author
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#2
2 months ago
thank you
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