Renal resorption and excretion: calculated from the clearance

If we know the level of glomerular filtration and the concentration of a substance in the urine, we can tell whether the substance is reabsorbed or excreted more. For example, if a substance's excretion flow (Us x V) is less than its filtration flow (GFR x Ps), then that substance is inevitably reabsorbed at a certain segment of the renal tubule.

Conversely, if the discharge of a substance is greater than the filtration flow of that substance, then the excretion flow will be the sum of the filtration and excretion flow of the renal tubules.

The following example demonstrates tubular reabsorption. Suppose we have the following test parameters for a patient:

Urine flow = 1 ml / min.

Urinary sodium concentration = 70 mEq / L = 70µEq / ml.

Plasma Ntri concentration = 140 mEq / L = 140 µEq / ml.

GFR (inulin clearance) = 100 ml / min.

In this example, the filtration flow rate of sodium is GFR x PNa or 100 ml / min x 140 µEq / ml = 14,000 µEq / min. The excretion rate for sodium (UNa × V) is 70 µEq / min. Hence, the sodium reabsorption is the difference of the filtration and excretion flow, or 14,000 µEq / min - 70 µEq / min = 13,930 µEq / min.

Compare Inulin clearance and other substances. The following summary will compare the clearance of substances and inulin clearance (a method of calculating glomerular filtration rate):

(1) If the clearance of a substance equals that of Inulin, that substance is only filtered without being reabsorbed or excreted; (2) If the clearance of a substance is less than the Inulin clearance, that substance is inevitably reabsorbed; (3) If the clearance of a substance is greater than that of inulin, that substance is certainly excreted by the renal tubules. Below are the clearance of some substances processed in the kidneys (approximate calculation).