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Multicompartment Models Prof. Dr. Henny Lucida, Apt These models were developed to explain the observation that, after a rapid IV injection, the Cp vs t curve does not decline linearly as a single, first order process. The drug distributes at various rates into different tissue groups. After drug equilibration within the peripheral tissues, the Cp vs t curve reflects first-order elimination of the drug from the body General Groupin of Tissues according to Blood Supply Blood Supply Tissue Group % BW Highly perfused 9 Slowly perfused Heart, brain, hepatic-portal system, kidney and endocrine glands Skin and muscle Adipose (fat) tissue & marrow Bone, ligaments, tendons, cartilage, teeth and hair 50 19 22 Two-Compartment open model • The plasma drug concentration declines biexponentially as the sum of two firstorder processes– distribution and elimination. • The drug distributes into 2 comp: the central comp and the tissue (peripheral) comps. Drug transfer between the two comps is assumed to take place by firstorder process. Basic Equations C p Ae at Be bt D0 a k 21 A V p a b D0 k 21 b B V p a b a b k12 k 21 k a b k 21 k Residual Method Example: 100 mg of a drug was administered by rapid IV injection to a 70-kg, healthy adult male. Blood samples were taken periodically after the administration of drug, and the plasma fraction of each sample was assayed for drug. The following data were obtained: Time (hr) 0.25 Cp (mg/mL) Cp (mg/mL) 43.00 Time (hr) 4.0 0.5 32.00 8.0 2.80 1.0 20.00 12.0 1.20 1.5 14.00 16.0 0.52 2.0 11.00 6.50 Step1. Transform data to log Cp Time (hr) 0.25 Log Cp Log Cp 1.6335 Time (hr) 4.0 0.5 1.5051 8.0 0.4472 1.0 1.3010 12.0 0.0792 1.5 1.1461 16.0 -0.2840 2.0 1.0414 0.8129 Plot log Cp vs time a b Step 2. Determine the linear regression of line b (after 4 hr), these data represent only the elimination process thus the slopeof line is equal to K C p Be bt bt logC p logB 2.303 The linear regression: Y1=1.1785 – 0.0915 X1 (r=-0.9999) Slope = -0.0915; b = 0.2107/hr B=antilog 1.1785= 15.0834 mg/mL Equation for b : Cp = 15.08 e-0.21t Step 3. The ka is determined residually by subtracting the data point at < 4 hr with the fitted data using linear regression of b. At these data point, the absorption process is predominant, then the slope would be the ka. This step would be easily solved by using a semi-logarithmic paper. Then the linear regression of the subtracted data is determined: Y2 = 1.6727 – 0.8055 X2 The slope = -0.8055, then a or ka = 1.8551/hr The intercept = 1.6727, then A or the C0 = antilog 1.6727 = 47.0652 mg/mL. Equation for a : Cp = 47.06 e-1.85t Thus the profile of plasma conc based on two compartment model is Cp = 47.06 e-1.85t + 15.08 e-0.21t Then: k, k12 and k21 can be calculated. Problems A drug was administered by rapid iv injection into a 70-kg adult male. Blood samples were withdrawn over a 7-hour period and assayed for intact drug. The results are tabulated below. Using the method of residuals, calculate the values for intercepts A and B and slopes a, b, k, k12 and k21. Time (hr) Cp (mg/mL) Time (hr) Cp (mg/mL) 0.00 70.0 2.5 14.3 0.25 53.8 3.0 12.6 0.50 43.3 4.0 10.5 0.75 35.0 5.0 9.0 1.00 29.1 6.0 8.0 1.50 21.2 7.0 7.0 2.00 17.0