Q. A patient is given 500 mg of a new antibiotic orally. The...

Q. A patient is given 500 mg of a new antibiotic orally. The bioavailability is found to be 80% and the mass of drug recovered in the urine in unchanged form is 100 mg. The volume of distribution is 69 L. the elimination half-life is 0.5h and the fraction of drug unbound in plasma is 0.01. What is the renal clearance of the antibiotic? - 23.9 L/h - 95.6 L/h - 71.7 L/h - 48.4 L/h

Transcript

00:01 Hello student, here the question is given and the question is about renal clearance of antibiotic and here these are the given total dose administered of antibiotic is 500 mg and the bioavailability of these is 80 % while unchanged drug excreted in urine is 100 mg and the half life of these drug elimination is of 0 .5 hour while the volume of distribution is 69 liter.

00:25 So here we have to calculate this total clearance.

00:30 So here the first we are this fraction of drug we have to calculate which are going to be absorbed.

00:41 So this is the fraction and this is the drug which is going to be absorbed.

00:45 So this is can be calculated by total dose which is administered and the bioavailability.

00:52 So here this is the given and here this is the 500 and the 80 % is going to be bioavailability.

01:00 So from here we are getting only this 400 mg of this fraction of these drug which are going to be absorbed and the similarly if we are calculating this rate constant and this is going to be k and lnk so this is the 2 divided by half life of the drug.

01:21 So ln2 the value of 6 .693 and the half life is given which is the 0 .5 hours.

01:29 So from here we are getting the value of this 1 .386 and this is going to be our inverse and if we are calculating the excretion rate constant so this is going to be ke and divided by this rate constant which we are calculated and this is going to equal to this d alpha u and the fraction of this drug which is going to be absorbed.

02:05 So this ke divided by k and here this is d alpha u and divided by this f d0...

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