Problem 1: Consider two connected compartments, aqueous (20 L of water) and lipid (5 L of fat). 330 μmol of drug with logP of 1.6 gets absorbed and distributed between the two compartments according to its octanol-water partition coefficient. Estimate the aqueous concentration of the drug.
Answers:

• ~132 nM
• ~13.2 μM
• ~ 8μM
• ~1.5 μM
• ~30 μM

Problem 2: In a solution at equilibrium, concentrations of unbound protein and unbound drug are 2.5 μM and 15 μM, respectively. Given the dissociation constant Kd of 5 μM, estimate the drug-bound-fraction of total protein (the binding reaction has 1:1 stoichiometry)
Answers:

• 90%
• 75%
• 50%
• 100%
• 80%
• 10%

Problem 3: Given a drug with logP = 3.5, estimate the molar ΔG of its transition from octanol to water.
Answers:

• 4.83 kcal/mol
• 0.042 kcal/mol
• -0.042 kcal/mol
• -3.5 kcal/mol
• 3.5 kcal/mol
• -4.83 kcal/mol

Problem 4: In an equilibrium mixture of two chiral isomers of a particular compound, about the less favorable S-isomer is found in 20% of the molecules, and the remaining 80% are the R-isomers. Compare the molar Gibbs free energies of formation of the two isomers, G^o (R) and G^o (S) at 29 °C . (Hint: use the ΔG = -RT ln K formula for the S → R equilibrium )
Answers:

• G^o (R) - G^o (S) = -11.5 kcal/mol
• G^o (R) - G^o (S) = -830 cal/mol
• G^o (R) - G^o (S) = 11.5 kcal/mol
• G^o (R) - G^o (S) = 830 cal/mol
• G^o (R) - G^o (S) = -960 cal/mol

Problem 5: Calorimetric data on binding of several anti-HIV drugs to their common target, HIV protease, is given in the table (all values in kcal/mol reasured at 25°C). One of the drugs has the affinity constant (a.k.a. the binding constant, K_b ) of ~9.5x10⁸ M^-1. Which drug is it? (Hint: ΔG = -RT ln K )
Answers:

• Amprenavir
• Indinavir
• Saquinavir
• Ritonavir
• Tipranavir
• Atazanavir
• Darunavir
• Nelfinavir

Problem 6: Adding the X = 0.04 molar fraction of solute to pure water at room temperature leads to the following outcome:
Answers:

• decreased chemical potential of water by ln(0.04)
• the chemical potential of water will not change, but the chemical potential of the solute will decrease.
• increase of the chemical potential of water proportionally to RTX
• decrease of the chemical potential of water by about 100 J/mol
• increase of the chemical potential of water by about 0.01 J/mol

Problem 7: 5g of a protein drug (MW = 50 kDa) was injected into patient's body. The drug circulated in the blood plasma (total volume 4L) but could not spread to the interstitial space. What is the increase in the osmotic pressure between the blood capillaries and the interstitial fluid? Is it likely to cause edema, or the opposite effect?
Answers:

• 1.2 Pa and no change fluid volumes
• 1.2 atm and increased edema
• 12.4 mmHg and reduced edema
• 1.2 mmHg and increased edema
• 62 Pa and reduced edema

Problem 8: Half a mole of glucose is dissolved in 1 kg of water. Which answer is closer to the extra osmotic pressure exerted on small lipid vesicles surrounded by glucose-inpermeable membranes? The glucose molecules are not ionized or changed in solution.
Answers:

• 0.01 atm
• 12 Pa
• can't tell without MM of glucose
• 12.7 atm
• 0.8 atm
• 83 N/m2

Problem 9: Osmolarity of a fluid preparation equals to 300% the osmolarity of a plasma-isotonic solution. The consequences of injecting this preparation intravenously include:
Answers:

• shrinking of the blood cells
• inflating and potentially bursting of the blood cells
• decrease of the external osmotic pressure onto the blood cell membranes
• higher freezing temperature for this preparation than for a plasma isotonic solution

Problem 10: A patient suffers from a mild edema. After some treatment patient's albumin concentration in blood plasma started growing. Which statement is correct? (hint: albumin stays in plasma and does not distribute to the interstitial fluid).
Answers:

• the edema will be increased
• the edema will not change significantly
• the excess albumin will bind Na cations and the balance will not change
• the additional albumin molecules will distribute equally in plasma and interstitial fluid
• the edema will be reduced

Problem 11: The freezing point of blood serum is about -0.56 °C. What happens if sea water (T_f = -1.8 °C) is added to red blood cells culture?
Answers:

• blood cells swell up and birst
• blood cells shrink
• impossible to say without knowing molecular masses of the constituents of blood and sea water.
• blood cells preserve their volume

Problem 12: 0.5 mol KCl, 0.3 mol Na₂ SO₄ , and 1.5 mol glucose are dissolved in three separate 10L containers of pure water to produce solutions A, B, and C, respectively. Ionic compounds dissociated completely. Rank the solutions by their freezing points from highest to lowest.
Answers:

• T_A > T_C > T_B
• all three boiling points are equal
• T_C > T_A > T_B
• T_C > T_B > T_A
• T_B > T_A > T_C

Problem 13: Oxygen ( O₂ ) is absorbed from the inhaled air in alveoli to the oxygen low capillaries of pulmonary arteries. The oxygen diffuses/moves from the alveoli into the blood until the following condition is met:
Answers:

• the chemical potentials of O₂ in blood and air become equal
• the concentrations of O₂ in air and blood become equal
• the temperatures of O₂ in blood and air become equal
• partial pressures of O₂ in blood and air become equal