Question

Q1. A drug is eliminated through glomerular filtration and hepatic metabolism. It does not bind to plasma proteins. Glomerular filtration rate is normal. The drug does not undergo active renal secretion and passive or active reabsorption after renal filtration is observed. The volume of distribution is 50 L. When given as an i.v. bolus, plasma concentration of this drug was 5.2 mg/L after 1 hour, and 2.6 mg/L after 3 hours. The drug undergoes first order elimination a) What is the plasma concentration 8 hours after administration of the drug? (1.5 point) b) What is the half-life of the drug? (1 point) c) What is the dose that was administered? (1 point) d) What is the total clearance of the drug? (1 point) e) What is the hepatic clearance of the drug? (2 points) f) Based on the hepatic clearance, please evaluate if the drug is a low hepatic extraction drug or high hepatic extraction drug. (2 points)

I know a is .458mg/L; b is 1.977hr; c is 368mg; d is 17.35 L/hr. Co= 7.36mg/L and ke=.347 1/hr. I need help with E and F. I'm not sure how to calculate them.

Answer 1

e) Hepatic clearance of the drug

Step 1:

CLhep = CLtot - CLren

Step 2:

CLtot = clearance total

CLtot = Vd X Ke

Vd = Volume of disstribution = 50

Ke = (In 5.2-In 2.6)/ (3-1)

Ke = 0.3471/hr.

Step 3:

CLren = Renal clearance

Renal clearance is calculated based on the GFR. The normal GFR rate is 130mL/min

Renal clearance = 130mL/min = 7.8L/hr

Step 4:

CLhep = CLtot - CLren

= 17.35 - 7.8

= 9.55L/hr.

The hepatic clearance of the drug is 9.55l/hr.

F.Hepatic extraction drug

Hepatic extraction is the difference between the drugs enter into the liver and the amount of extraction. If the extraction is less than 0.3 (E<0.3) it is considered as low extraction drug, if it is greater than 0.7(E>0.7), then it is considered as high extraction drug.

E = CLhep/Qh

CLhep = 9.55L/hr

Qh = 50L

E = 9.55 / 50

E = 0.191.

E <0.3, so it is a low hepatic extraction drug.

Based on the hepatic clearance, the drug is a low hepatic extraction drug.