Lecture 11

Lecture 11 - Lecture 11: Clearance Contd and Glomerular...

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Lecture 11: Clearance Cont’d and Glomerular Filtration Topics to be Covered : o review of clearance equation o low vs high clearance of circulating products o clearance of PAH as an estimate of RPF o clearance of inulin, creatinine, and glomerular filtration Conservation of Mass in Kidney o Input vs Output of Substance X input = only 1 route to kidney release (output) = couple routes o law of conservation of mass : the 2 routes, the sum of them is equal to input of kidney Arterial Input = Venous output + Urine Output o egress: the action of going out—exiting Equations of Conservation of Mass o P Xa = renal arterial concentration of X RPF a = renal arterial plasma flow rate P Xv = renal venous plasma concentration RPF v = renal venous plasma flow rate U x (x) V = rate of excretion renal arterial of conc. X (x) renal arterial plasma flow is equal to rates of release of kidneys o Excretion : elimination of urine excretion = urinary concentration of X (x) rate of urine flow o Clearance equation : allows us to calculate rates of clearance Both equations are dependent upon same hypothesis: all of substance X in urine, comes from a single virtual volume of renal plasma arterial blood single volume of plasma can yield all of X in the urine 1) C x = Clearance of X flow rate There is no venous egress of X therefore 0 o meaning, there’s no X in venous plasma U x = urine 2) C x = U x (x) V/Pxa the renal clearance equation ---memorize this! C = UV/P ** o C = clearance o u = concentration of x in urine Slide 1 Slide 2 Slide 3
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o v = rate of urine flow o p = concentration of x in systemic arterial plasma blood Renal Clearance Continued o Range of clearance: 0 (no clearance) 600-700 mL/min (which is volume of renal plasma flow) o Glucose: under normal circumstances doesn’t get excreted in urine bc it gets reabsorbed in tubules egress of glucose is 0 clearance for glucose is 0 no clearance of glucose no virtual or actual vol of blood is cleared of glucose o PAH : not naturally occurring in body—has to be administered if it’s added to body and when it passes through kidney, in one circulation, ALL of it gets cleared—gets filtered out into urine clearance of PAH is 600-700 mL/min the rate of renal plasma flow o Inulin : not insulin! naturally occurring compound in nature but not in body Inulin clearance = rate of glomerular filtration Inulin can measure the rate at which the glomerular capillaries filter plasma Glucose and Clearance o focus on dark grey column shows range of circulating concentration of glucose that are physiological—what’s in us now o 3 curves: Green: excretion of urine Gold: filtration Red: reabsorption by nephron tubules o grey = physiological range below or in physiological range: everything that is filtered in this range is reabsorbed so there is 0 glucose excreting
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This note was uploaded on 03/07/2012 for the course 830 201 taught by Professor Leyton during the Fall '08 term at Rutgers.

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Lecture 11 - Lecture 11: Clearance Contd and Glomerular...

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