Question: 2-The following table gives plasma drug concentrations obtained following an IV bolus administration of a 250 mg/L dose of a drug that exhibited the characteristics of a one-compartment model and was eliminated exclusively by urinary excretion. t (hr) C_p (\u03bcg/mL) 0.5 68.0 1.0 54.0 2.0 30.0 3.0 18.5 5.0 6.0 7.0 1.8 \u2022 Plot the data and determine the following: A. t 0.5 , K, C0 and Vd B. The drug plasma concentration at 75 min following the administration of a 2.5 mg/kg dose to a subject weighing 70 kg. C. The time at which the plasma concentration of the drug will fall below 20 \u03bcg/mL, following the administration of a 275 mg dose.

1. **Problem Statement:** We have plasma drug concentration data after an IV bolus dose of 250 mg/L for a drug following a one-compartment model eliminated by urinary excretion. We need to find: A. Half-life ($t_{0.5}$), elimination rate constant ($K$), initial concentration ($C_0$), and volume of distribution ($V_d$). B. Drug plasma concentration at 75 minutes after a 2.5 mg/kg dose for a 70 kg subject. C. Time when plasma concentration falls below 20 $\mu g/mL$ after a 275 mg dose. --- 2. **Formulas and Important Rules:** - The drug concentration in a one-compartment model after IV bolus follows: $$C = C_0 e^{-Kt}$$ - Half-life formula: $$t_{0.5} = \frac{\ln 2}{K}$$ - Volume of distribution: $$V_d = \frac{X_0}{C_0}$$ where $X_0$ is the initial dose. - To find concentration at time $t$: $$\log C = \log C_0 - \frac{K t}{2.303}$$ --- 3. **Part A: Determine $t_{0.5}$, $K$, $C_0$, and $V_d$ from the graph/data** - From the graph or data, given: $$C_0 = 90 \text{ mg/L}$$ $$t_{0.5} = 1.275 \text{ hr}$$ - Calculate elimination rate constant $K$: $$K = \frac{\ln 2}{t_{0.5}} = \frac{0.693}{1.275} = 0.543 \text{ hr}^{-1}$$ - Calculate volume of distribution $V_d$: $$V_d = \frac{X_0}{C_0} = \frac{250}{90} = 2.77 \text{ L}$$ --- 4. **Part B: Calculate plasma concentration at 75 min (1.25 hr) after 2.5 mg/kg dose for 70 kg subject** - Calculate dose $X_0$: $$X_0 = 2.5 \times 70 = 175 \text{ mg}$$ - Calculate initial concentration $C_0$: $$C_0 = \frac{X_0}{V_d} = \frac{175}{2.77} = 63.176 \text{ mg/L}$$ - Calculate concentration at $t=1.25$ hr: $$\log C = \log 63.176 - \frac{0.543 \times 1.25}{2.303}$$ $$\log C = 1.800 - 0.295 = 1.505$$ $$C = 10^{1.505} = 31.956 \text{ mg/L}$$ --- 5. **Part C: Calculate time when plasma concentration falls below 20 $\mu g/mL$ after 275 mg dose** - Calculate initial concentration $C_0$: $$C_0 = \frac{275}{2.77} = 99.277 \text{ mg/L}$$ - Use formula to find time $t$ when $C = 20$: $$\log 20 = \log 99.277 - \frac{0.543 t}{2.303}$$ $$1.301 = 1.997 - 0.236 t$$ - Solve for $t$: $$0.236 t = 1.997 - 1.301 = 0.696$$ $$t = \frac{0.696}{0.236} = 2.94 \text{ hr}$$ --- **Final answers:** - A: $t_{0.5} = 1.275$ hr, $K = 0.543$ hr$^{-1}$, $C_0 = 90$ mg/L, $V_d = 2.77$ L - B: Plasma concentration at 75 min = 31.956 mg/L - C: Time to fall below 20 $\mu g/mL$ = 2.94 hr