Chapter 3 - Chapter 3 Glomerular Filtration and Renal Blood...

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Unformatted text preview: Chapter 3 Glomerular Filtration and Renal Blood Flow • What is urine?- Ultrafiltrate of plasma across the glomerulus. • What is the importance of GFR?- To evaluate the severity and course of kidney disease. - Equal to the sum of the filtration rates of all the functioning neurons. Therefore, GFR is an index of kidney function. ¡ A fall in GFR … disease is progressing ¡ An increase in GFR … suggests recovery • More on GFR later Renal Clearance • The rate at which substances are removed (or cleared) from plasma is known as clearance. • Renal clearance is the volume of plasma completely cleared of a substance by the kidney per unit time. It is then excreted into the urine. • The concept of renal clearance is based on the Fick principle (mass balance or conservation of mass). • The renal artery is the single input source to the kidney. • The renal vein and ureter constitute the two output routes. • Principle of Mass Balance Input via renal artery = output via renal vein and ureter • The equation for renal clearance: C = [U] x x V [P] x C = clearance (ml/min) [U] = urine concentrations (mg/ml) V = urine flow rate per minute (mg/ml) [P] = plasma concentration (mg/ml) x = substance to be cleared • Thus, renal clearance is the ratio of urinary excretion ([U] x V) to plasma concentration [P]. • For a substance - renal clearance increases as the urinary excretion increases. • Again, the units of clearance are volume per unit time (e.g., ml/min; L/hr; L/day), which means what? - The volumes of plasma cleared of the substance per unit time. The concept of clearance is important. ¡ Can be used to measure the GFR and RPF (renal plasma flow) ¡ Can determine whether a substance is reabsorbed or secreted along the nephron. ¡ For RPF, see equation 3-1, K, p. 32 Excretion Rate • Assume a substance is present in the urine at a concentration of 100 mg/ml and the urine flow rate is 1 ml/min. • The excretion rate can be calculated as follows: Excretion rate = U x x V = 100 mg/ml x (1ml/min) = 100 mg/min • A substance present in the plasma at a concentration of 1 mg/ml. Calculate its clearance. C x = [U] x x V = 100 mg/min = 100 ml/min [P] x 1 mg/ml • Means 100 ml of plasma completely cleared of x each minute. Glomerular Filtration Rate Clearance of Inulin • Used to measure GFR • Is a polymer of fructose, MW = 5000 • Not produced by the body (administered intravenously) • Freely filtered across the glomerulus into Bowman’s space • Not reabsorbed, secreted, or metabolized by nephron cells • Accordingly, Amount filtered = Amount excreted GFR x [P in ] = [U in ] x V Where, GFR = [U in ] x V [P in ] P in and U in = plasma and urine concentrations of inulin V = urine flow • The clearance of inulin, therefore, provides a means or determining the GFR....
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This note was uploaded on 01/07/2012 for the course PHCH 6234 taught by Professor Farqui during the Winter '11 term at Palmer Chiropractic.

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Chapter 3 - Chapter 3 Glomerular Filtration and Renal Blood...

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