MCAT Review: HIV-1 Protease Inhibitors PDF

MCAT Review 1. Human immunodeficiency virus 1 (HIV-1) protease inhibitors have been successful at lowering HIV-1 levels and blocking the onset of acquired immune deficiency syndrome (AIDS). However, because HIV- 1 mutants are developing resistance to current protease inhibit...

MCAT Review 1. Human immunodeficiency virus 1 (HIV-1) protease inhibitors have been successful at lowering HIV-1 levels and blocking the onset of acquired immune deficiency syndrome (AIDS). However, because HIV- 1 mutants are developing resistance to current protease inhibitors, new inhibitors such as Compound 1 (molar mass: 483.5 g mol−1) are being tested. Compound 1 Compound 1 has been shown to inhibit HIV-1 protease with Ki = 60.3 μM (Table 1). Ki is the dissociation constant for the enzyme-bound inhibitor, which is either EI or ESI, depending on the type of inhibitor. The peptide substrate for HIV-1 protease is IRKILFLDG. HIV-1 protease hydrolyzes only the peptide bond after leucine and before phenylalanine. Table 1 Kinetic Parameters of HIV-1 Protease with and without Compound 1 Compound 1 KM kcat kcat/KM Ki (μM) (mM) (s−1) (mM−1 s−1) (μM) 0 1.678 0.250 00.149 — 60.0 0.832 0.126 0.151 60.3 120 0.430 0.064 0.149 60.3 Note: HIV-1 protease follows Michaelis−Menten kinetics. A Lineweaver–Burk plot of Vo−1 versus [S]−1, both without Compound 1 and with different concentrations of Compound 1, generates a set of parallel straight lines with positive slopes.  Adapted from T. Sperka, J. Pitlik, P. Bagossi, and J. Tösnér, Bioorganic & Medicinal Chemistry Letters. ©2005 Elsevier Inc. What quantity of Compound 1 must be provided to prepare 100.00 mL of solution with a concentration equal to Ki? A. 48.4 mg B. 24.2 mg C. 5.64 mg D. 2.92 mg Solution: The correct answer is D. This quantity corresponds approximately to the amount of Compound 1 in 100 mL of a 1000 μM solution, which is not the Ki of Compound 1. This quantity corresponds approximately to the amount of Compound 1 in 100 mL of a 500 μM solution, which is not the Ki of Compound 1. This quantity corresponds approximately to the amount of Compound 1 in 100 mL of a 117 μM solution, which is not the Ki of Compound 1. In 100.00 mL solution, 60.3 μM Compound 1 contains 6.03 μmol, which when converted to mol and multiplied by the molar mass, yields 0.00292 g or 2.92 mg. What quantity of Compound 1 must be provided to prepare 100.00 mL of solution with a concentration equal to Ki? Rephrase the question: The question is basically asking you how much of compound 1 is needed to prepare 100 ml of solution with a concentration of Ki You can use the Molarity equation to help solve: Molar concentration = mol of solute/ L M= 60.3 μmol /100 ml The Ki= 60.3 ( assumingly in 1 Liter of solution if using the above equation.. ) Lets convert 100 ml to L ( there are 1000 ml in a liter, because were converting from a small unit to a bigger unit you must divide 100 ml/1000 L= 0.1 L 60.3 μmol / 0.1L