# Choose the correct answer with solution if possible: 1.Consider a woman BD who has liver disease and cannot metabolize the drug as effectively so that the half-life in her is doubled. The half-life of the drug is typically 4 hours in an individual with normal liver function. If BD stopped taking the drug after 48 hours of administration, estimate the total time that the drug was in her body. (Options: 48 hours, 52 hours, 62 hours, 70 hours, 88 hours). 2.Consider BD again and that the drug dose was not adjusted based on her liver condition. She is given 100mg of the drug every 4 hours, as would typically be done for a normal individual with her weight and age. What would be steady state concentration of the drug in BD’s plasma in that case? (The concentration normal individual at steady state is typically 3mg/ml.) (Options: 1.5 mg/mL, 3 mg/mL, 4.5 mg/mL, 6 mg/mL, 12 mg/mL) 3.Glipizide is an antidiabetic drug that has a pKa of 5.9. Approximately what percentage of the drug would be in the lipid soluble form in the duodenal lumen which has a pH of 7? (Options: 1, 7, 14, 28, 90) 4.Passive diffusion through the cellular membranes of gastric cells can lead to very high concentrations of the drug inside gastric cells if the drug has which of the following characteristics?  Basic: larger pKb weaker base (Options: It is a base with pKa of 7, It is an acid with pKa of 3.5, It is a base with pKa of 5.2, It is an acid with pKa of 8.5, It is non-ionizable).

1. The total time that the drug was in BD’s body can be estimated by considering the half-life of the drug. In this case, the half-life is doubled for BD due to her liver disease. The half-life of the drug is typically 4 hours in an individual with normal liver function. This means that it takes 4 hours for the concentration of the drug in the body to decrease by half.

If BD stopped taking the drug after 48 hours of administration, we can calculate the number of half-lives that occurred in that time period. Since the drug’s half-life is now doubled for BD, each half-life would be 8 hours. Therefore, the number of half-lives that occurred in 48 hours would be 48/8 = 6.

To estimate the total time that the drug was in BD’s body, we can multiply the number of half-lives by the original half-life of the drug (4 hours). In this case, the total time would be 6 x 4 = 24 hours. However, since the half-life is now doubled for BD, the total time would be 24 x 2 = 48 hours.

Therefore, the correct answer is 48 hours.

2. To calculate the steady state concentration of the drug in BD’s plasma, we need to consider the dosing regimen and the drug’s pharmacokinetics. In this case, BD is given 100mg of the drug every 4 hours, which is the typical dosing for a normal individual with her weight and age.

The steady state concentration of a drug is achieved when the rate of drug administration is equal to the rate of drug elimination. In this case, we can assume that the drug follows linear pharmacokinetics and that it has a one-compartment model.

The steady-state concentration can be calculated using the equation:

C_ss = (Dose / Clearance) x F x (1 – e^(-k x tau))

Where C_ss is the steady state concentration, Dose is the dose administered, Clearance is the drug clearance, F is the bioavailability, k is the elimination rate constant, and tau is the dosing interval.

Since the drug dose is 100mg, the clearance is unknown, the bioavailability is assumed to be 100% (F=1), and the dosing interval is 4 hours (tau = 4), we need to find the elimination rate constant (k).

The elimination rate constant can be determined using the half-life of the drug. Since the half-life of the drug is not provided, we can’t calculate the elimination rate constant and therefore, we can’t determine the steady state concentration.

Unfortunately, without the necessary information, we cannot provide a solution for this question.

3. To determine the percentage of the drug in the lipid soluble form in the duodenal lumen, we need to consider the pKa of the drug and the pH of the duodenal lumen. The pKa is a measure of the drug’s acidity or basicity, and it tells us at what pH the drug will be mostly ionized or mostly unionized.

In this case, the pKa of Glipizide is 5.9, and the pH of the duodenal lumen is 7. To estimate the percentage of the drug that will be in the lipid soluble form, we can use the Henderson-Hasselbalch equation:

% unionized = 100 / (1 + 10^(pH – pKa))

Plugging in the values, we get:

% unionized = 100 / (1 + 10^(7 – 5.9))
% unionized = 100 / (1 + 10^1.1)
% unionized = 100 / (1 + 12.59)
% unionized = 100 / 13.59
% unionized ≈ 7.35%

Therefore, approximately 7.35% of the drug would be in the lipid soluble form in the duodenal lumen.

4. Passive diffusion through the cellular membranes of gastric cells can lead to very high concentrations of the drug inside gastric cells if the drug has characteristics such as being a base with a larger pKb or a weaker base.

The pKb is a measure of the basicity of the drug, and it tells us at what pH the drug will be mostly ionized or mostly unionized. A higher pKb indicates a stronger base.

In the given options, the characteristics that would lead to high concentrations of the drug inside gastric cells are:
– It is a base with pKa of 7 (higher pKa indicates a stronger acid, not base)
– It is a base with pKa of 5.2 (higher pKa indicates a stronger acid, not base)
– It is an acid with pKa of 3.5 (not a base)
– It is an acid with pKa of 8.5 (not a base)
– It is non-ionizable (ionizability doesn’t affect passive diffusion)

Therefore, the correct option is “It is a base with pKa of 5.2” as this indicates a weaker base, which can lead to high concentrations of the drug inside gastric cells through passive diffusion.