The Countercurrent Exchanger System maintains the hyperosmolarity

gradient of the interstitial fluid in the kidney medulla, which is crucial for urine concentration. The countercurrent exchanger system involves the vasa rectae (capillaries), which form a hairpin loop in which blood flows down the descending side of the loop and then back up the ascending side of the loop; hence the term countercurrent. If the kidney were supplied by an ordinary capillary bed (i.e., no hairpin loop), sodium and chloride ions would diffuse into the capillaries and water would diffuse out of the capillaries, thereby destroying the hyperosmolarity gradient. This is exactly what happens as blood flows down the descending side of the loop, but then the process is reversed or exchanged as blood flows up the ascending side of the loop; hence the term exchanger.

IX. GLOMERULAR FILTRATION RATE (GFR). Inulin (a polysaccharide) can be used to calculate GFR because inulin is freely filterable at the glomerulus, it is not reabsorbed from tubular fluid to blood, and it is not secreted from blood to tubular fluid.

A. The calculation of GFR is given by the following equation:

Uj V

" In where, U(n = urine concentration of inulin V = volume of urine per unit time P| = arterial plasma concentration of inulin

B. Although inulin gives an accurate measurement of GFR, inulin is not convenient to use clinically. It is more convenient to use arterial plasma concentration of creatinine (PCr) as an estimate of GFR because PCr and GFR are related by the graph in Figure 19-2.

1. A normal patient will have a PCr of 10 rng/L, indicating a normal GFR of 180 L/day.

2. A patient with kidney disease will have a PCr of 20 mg/L, indicating an abnormally low GFR of 90 L/day.

Figure 19-2. Glomerular filtration rate (GFR) compared with arterial plasma concentration of creatinine (PCr) in a normal patient and in a patient with kidney disease.

Per mg/L

Figure 19-2. Glomerular filtration rate (GFR) compared with arterial plasma concentration of creatinine (PCr) in a normal patient and in a patient with kidney disease.

X. CLEARANCE is the volume of plasma from which a substance is completely cleared by the kidneys per unit time. The clearance of inulin (C|n) is used as a benchmark because inulin is not reabsorbed or secreted by the kidney tubules. Therefore, the Cln = GFR = 180 L/day.

A. If clearance of substance X (C x) is greater than CIn, then kidney tubule secretion has occurred. This is another way of saying that if the excreted mass is greater than the filtered mass, tubular secretion must have occurred.

B. If clearance of substance Y (CY) is less than C|n, then kidney tubule reabsorption has occurred. This is another way of saying that if the excreted mass is less than the filtered mass, tubular reabsorption must have occurred.

^creatinine = ^00 L/day; creatinine is secreted Cinulin= 180 L/day = GFR

Curea = 90 L/day; 50% of filtered urea is reabsorbed ^glucose = 0 L/day; 100% of filtered glucose is reabsorbed

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