Chemistry Questions Revised

CHEMISTRY QUESTIONS
By Name
Course
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Date
1 E) 1. Plasma Clearance
Renal Clearance + Hepatic clearance + other clearance = 120+90+24 = 234mg/ml
2C) Renal Clearance + Hepatic clearance + other clearance= 212+200+20 = 432 L/Hr
3 C) Volume of Distribution
Amount of drug in body (dose)/ [drug] in plasma = 180/144 = 1.25 L
3 D) Plasma Clearance
Renal Clearance + Hepatic clearance + other clearance (Loo, 2011, 920).
189+36+40 = 265L/Hr
4 A)Plasma half-life (T1/2) [20%]
Kel = In (3.9/150)/144 = 0.02 (hr-1)
Half-life = 0.693/-0.02= 34.6
4 B)Overall elimination rate constant (Kel) [20%]
Kel = In (3.7/140)/144 = 0.04 (hr-1)
4 C)Absorption rate constant (Kab) [30%]
180/150 = 1.2 L
4 D)The bioavailability (F) of the oral formulation using: [30%]
F = In (2.96/14.4)/20 = -0.06 (hr-1)
5 A)What is the plasma half-life of the drug? [10%]
Half-life = 0.692/-0.06 = 8.54(hr-1)
5 B)what would the plasma half-life be if the metabolism of the drug was completely blocked? [10%]
In (2.94/14.4)/25 = -0.07 (hr-1)
5 C)What would the plasma half-life be if the patient suffered from a liver disease such that biliary flow was completely blocked? [10%]
In (2.96/14.4)/30 = -0.09 (hr-1)
References
Loo, J.C.K. and Riegelman, S., 2011. New method for calculating the intrinsic absorption rate of drugs. Journal of pharmaceutical sciences, 57(6), pp.918-928.