Question

Practice Chapter 5: Renal Excretion 1. After an intravenous injection of 1000 mg of a new drug to a patient, the following urinary and plasmatic information was obtained: Collection Interval (h) Urine Volume Concentration (mg/mL) in Urine Cp(tmid) (mg/L) 0-2 125 15.9 2-4 150 1.776 0.933 0.504 10.0 4-6 175 6.3 6-8 138 0.403 3.97 8-10 2.51 163 313 0.215 0.112 10-14 1.24 Calculate: K, half-life, Kr, Total Clearance, Renal Clearance, Fraction of Urinary excretion (Fr), and Vd.

could two is the volume that you have to mutlply to the cu to find the Au or amount in MG

its not organic chemsitry its pharmacokinetics

Answer 1

The total body clearance (ClT) is the sum of the renal clearance (see the record on Renal drug excretion.

total clearance=renal clearance+plasma clearance=49+6=55mg/min

renal elimination rate constant( k1) is thecalculated by

k1=( inc1-inc2)/(t2-t1)

here cocentration at 2hr(120 min) is1.776

concentration at 4hr(240min) is .933

so k=(1.766-.933)/(240-120)= .006

renal clearance(CLr)

k * V = .006 * 1000 = 6 mg/min

Fraction excreted unchanged

Fr = CLr/CL = 6/55 = .109 mg/min
total elimination constant

k =( k1 + k2)/2 = (.006 + .049)/2 = .027

t1/2 calculation

The t1/2 represents the time needed to eliminate 50% of the drug in the body. After 3.3 half-lives, 90% of the drug will be eliminated from the body.

t1/2 (= 0.693/k)

= .693/.027 =25.6 min