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Consider again a substance X that is excreted in the urine. How do we actually calculate clearance in the proper units The amount excreted exists in some volume of urine, and that amount must have been contained in some volume of plasma. This is depicted in Figure 3 1. The large boxes on the left show the plasma before, during, and after clearance of substance X (dots), while the smaller boxes on the right show the urine gaining X as it is removed from the plasma. The amount cleared per unit time is the product of the volume of plasma cleared per unit time (Cx) and the plasma concentration (Px). Amount cleared 5 Cx Px. The amount appearing in the urine during this time is the product of the urine flow rate (V) and the urine concentration of X (Ux). Amount in urine 5 V Ux. The amount removed from the plasma must equal the amount appearing in the urine. This equality is shown in Equations 3-1 and 3-2. Finally, by rearrangement we solve for clearance (Cx) as shown in Equation 3-3. Thus we have equated amounts per unit time, but by rearrangement we end up with clearance in its proper units volume per time.
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A final complicating factor exists if either the anchor system or the serving system cannot complete the unique-challenge operation. For example:
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