Physiology at a Glance by Jeremy P. T. Ward & Roger W. A. Linden

Author:Jeremy P. T. Ward & Roger W. A. Linden [Ward, Jeremy P. T.]
Language: eng
Format: azw3
ISBN: 9781119247319
Publisher: Wiley
Published: 2017-03-07T05:00:00+00:00

Measurement of the glomerular filtration rate and the concept of clearance

If substance X is freely filtered and neither reabsorbed nor secreted in the nephron, the amount appearing in the urine per minute must equal the amount filtered per minute. Thus, if the plasma concentration of X is Cp and the urine concentration is Cu, and the volume of urine passed per minute is V, then Cp × GFR = Cu × V, or GFR = (Cu × V)/Cp.

Creatinine, which is steadily released from skeletal muscle, is often used for clinical measurements of GFR because it is freely filtered and not reabsorbed; there is a little secretion, but this introduces only a small error, except when plasma creatinine or GFR is abnormally low. More accurate measurements are made by infusing the polysaccharide inulin, which is neither reabsorbed nor secreted.

This is known as a clearance method. The term clearance can be confusing, as it does not refer to what actually happens but is merely a way of looking at how the kidney deals with a substance. It is defined as the volume of plasma that would need to be completely cleared of a substance per minute in order to produce the amount found in the urine, or: clearance = (Cu × V)/Cp (i.e. the same equation as above). Thus, the clearance of inulin is equal to GFR. If a substance is reabsorbed in the nephron, its clearance will be less than the GFR and, if it is secreted, it will be greater than the GFR. Some substances that are normally completely reabsorbed have zero clearance until the reabsorption mechanism becomes saturated (e.g. glucose; Chapter 36).

The renal plasma flow (RPF) can be measured in a similar fashion by infusing para-aminohippuric acid (PAH) which at low concentrations is completely removed from renal blood by both filtration and secretion, so that none remains in the venous outflow. The amount appearing in the urine must therefore equal the amount entering the kidney, and thus the clearance of PAH is equal to RPF. The filtration fraction (GFR/RPF; see previously) can therefore be estimated from inulin clearance/PAH clearance. The renal blood flow is equal to RPF/(1 – haematocrit).

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