Filtration and Reabsorption Amount Filter/d

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1 Renal Physiology 2011 Lisa M. Harrison-Bernard, PhD Contact me at Renal Physiology Lecture 3 Renal Clearance and Glomerular Filtration Filtration and Reabsorption Amount Filter/d Amount Excrete/d % Reabsorb Water (L) K + (meq) Ca 2+ (meq) HCO 3 - (meq) 4, Cl - (meq) 18, Na + (g) Glucose (g) Urea (g) LSU Medical Physiology

2 Tubular Secretion Most important: H + K + Organic anions choline creatinine Foreign chemicals penicillin Handling of Substance L by Kidney Filtered Load GFR X Plasma [L] GFR P L Excretion Rate Urine [L] X Urine flow V U L LSU Medical Physiology

3 Summary 1. Kidney is a very important organ. 2. Juxtaglomerular apparatus is coolest basic renal processes Filtration, reabsorption, secretion 4. Damage to filtration barrier results in glomerular disease Renal Physiology - Lectures Physiology of Body Fluids Structure & Function of the Kidneys 3. Renal Clearance & Glomerular Filtration 4. Regulation of Renal Blood Flow 5. Transport of Sodium & Chloride 6. Transport of Urea, Glucose, Phosphate, Calcium & Organic Solutes 7. Regulation of Potassium Balance 8. Regulation of Water Balance 9. Transport of Acids & Bases 10. Integration of Salt & Water Balance 11. Clinical Correlation Dr. Credo 12. PROBLEM SET REVIEW May 9, EXAM REVIEW May 9, EXAM IV May 12, 2011 LSU Medical Physiology

4 Renal Physiology Lecture 3 Renal Clearance and Glomerular Filtration Chapter 3 Koeppen & Stanton Renal Physiology 1. Starling Forces 2. Control of GFR 3. Concept of Renal Clearance 4. Clearance of Inulin & Creatinine = Estimates of GFR 5. PAH = Estimate of RPF Fluid Movement Out of Glom Cap Into Bowman s Space LSU Medical Physiology

5 Glomerular Ultrafiltration - Eq 3-10 Starling Forces (J v ) = K f [(P GC -P BS ) - σ( GC - BS )] Jv volume flux across the capillary wall Glomerular Ultrafiltration - Eq 3-10 GFR = K f [(P GC -P BS ) - σ( GC - BS )] σ reflection coefficient for protein = 1 protein cannot cross glomerular membrane LSU Medical Physiology

6 K f Ultrafiltration Coefficient GFR = K f [(P GC -P BS ) - ( GC - BS )] K f ml/min/mmhg Intrinsic permeability glom capillary Product of hydraulic conductivity (L p ) & surface area (S f ) L p & S f X > other beds K f Ultrafiltration Coefficient K f = Lp x A A - cm 2 Area - total capillary surface area ~ 1 cm length/glom cap ~ 12 miles total length (2 mil glom) 6,000 cm 2 total t surface area filtration area ~10% (Fenestrae) Boyle et al. Kidney International 1998 LSU Medical Physiology

7 Ultrafiltration Coefficient GFR = K f (P UF ) K f ml/min/mmhg 125 ml/min = 8.3 ml/min/mmhg (15 mmhg) Forces Involved in Glomerular Filtration Fig 3-6 Afferent Arteriole Efferent Arteriole Bowman s space 2 forces favor fluid filtration P GC BS GC P BS 2 forces oppose fluid filtration LSU Medical Physiology

8 Forces Involved in Glomerular Filtration Fig 3-6 Afferent arteriole Efferent arteriole Bowman s space P GC GC Two forces favor fluid filtration. Two forces oppose fluid filtration. BS P BS P GC = Glomerular capillary hydrostatic pressure BS = Bowman s space oncotic pressure P BS = Bowman s space hydrostatic pressure GC = Glomerular capillary oncotic pressure Net Filtration Pressure = P GC P BS GC + BS Forces Involved in Glomerular Filtration Net glomerular filtration pressure = P GC P BS GC + BS LSU Medical Physiology

9 Net Glomerular Filtration Pressure- Glomerular Capillary = P GC P BS GC + BS P GC = 50, P BS = 10, GC = 25, BS = 0 mmhg mmhg = 15 mmhg P GC only force favors filtration 2X > most capillaries Glomerular Ultrafiltration - Eq 3-10 GFR = K f [(P GC -P BS ) - ( GC - BS )] Rate of glomerular ultrafiltration = P UF ultrafiltration coefficient (K f ) X net Starling forces (P UF ) LSU Medical Physiology

10 Starling Forces Change Along the Length of Capillaries: Skeletal Muscle (Non-Renal Capillary) Arteriole Venule Net Filtration Net Absorption Equilibration Point Overall: Filtration ~ Absorption Starling Forces Along Glomerular Capillaries Afferent Distance along glom Efferent arteriole cap network arteriole GC increases along glom capillary NEVER Absorptive flux LSU Medical Physiology

11 Forces Involved in ABSORPTION by Peritubular Capillaries Net peritubular capillary pressure = P PC P O PC + O P PC P O O PC Net Glomerular Filtration Pressure- Peritubular Capillary = P PC P O PC + O P PC = 20, P O = 8, PC = 35, O = 6 mmhg mmhg = minus 17 mmhg Negative filtration absorption Forces favor reabsorption of fluid LSU Medical Physiology

12 Renal Physiology Lecture 3 Starling Forces 2. Control of GFR 3. Concept of Renal Clearance 4. Clearance of Inulin & Creatinine 5. Clearance of PAH What would happen to GFR if RAP? LSU Medical Physiology

13 Control of GFR P UF = P GC P BS ( GC BS ) GFR = K f (P UF ) Renal Artery Pressure (RAP) = P GC = GFR AA resistance P GC = GFR EA GC AA Control of GFR P UF = P GC P BS ( GC BS ) GFR = K f (P UF) EA AA resistance P GC = GFR EA resistance P GC = GFR GC AA LSU Medical Physiology

14 P UF = P GC P BS ( GC BS ) GFR = K f (P UF ) f Control of GFR UF GC - GFR protein metabolism - malnutrition, hepatic disease, GI losses protein excretion - kidney disease = proteinuria P UF = P GC P BS ( GC BS ) GFR = K f (P UF ) Control of GFR P BS - GFR acute obstruction stone, enlarged prostate BS - GFR filter protein - proteinuria LSU Medical Physiology

15 Control of GFR P UF = P GC P BS ( GC BS ) GFR = K f (P UF ) K f = L p x S f K f = GFR reduce surface area or # filtering glomeruli hypertension diabetes glomerulosclerosis Decreases in GFR - Disease GFR = K f [(P GC -P BS ) - ( GC - BS )] Acute Renal Failure RAP, R A, R E, P G, GFR LSU Medical Physiology

16 Renal Physiology Lecture 3 Starling Forces Control of GFR 3. Concept of Renal Clearance 4. Clearance of Inulin & Creatinine 5. Clearance of PAH Renal Plasma Clearance Renal CLEARANCE of any substance volume of plasma from which a substance is completely removed (cleared) by kidneys per unit time Units = Volume plasma per time ml/min QUANTITATIVE evaluation: how kidney handles specific substance LSU Medical Physiology

17 Clearance S = Mass S excreted / time Plasma [S] Cl S P S = U S V Cl S = U S V P S What is the renal clearance of glucose? LSU Medical Physiology

18 Substance CLEARANCE (ml/min) Glucose 0 Na K + 12 Inulin 125 Creatinine 140 PAH 560 Renal Physiology Lecture 3 Starling Forces Control of GFR Concept of Renal Clearance 4. Clearance of Inulin & Creatinine 5. Clearance of PAH LSU Medical Physiology

19 Inulin Cl IN = GFR Inulin MW 5,000 Da P IN X GFR = U IN X V. Freely filterable NOT reabsorbed NOT secreted NOT metabolized, synthesized, stored NOT alter GFR NONtoxic Infusion required P IN,U IN - analytic method LSU Medical Physiology

20 Inulin Measurement of GFR Amount Filtered = Amount Excreted GFR P IN = U IN V GFR = U IN V = Cl IN P IN Creatinine Cl Cr ~ GFR LSU Medical Physiology

21 Creatinine Rate of Production = Rate of Excretion 1 g/day = 1 g/day Index of GFR GFR (Clearance Cr ) = U Cr V P Cr Cl Cr is inversely related to P Cr P Cr X GFR U Cr =. X V Creatinine ~Fig 3-2 Metabolism of creatine phosphate - muscle Produced continuously Freely filtered NOT reabsorbed Small amount secreted NO infusion required Stable P[Cr] P[Cr] & U[Cr] colorimetric method LSU Medical Physiology

22 Plasma Cr Inversely Related to GFR ~Fig 3-3 P Cr = 6mg/dl GFR 1/6 P Cr =2mg/dl GFR ½ Normal P Cr =1mg/dl Normal GFR Plasma Creatinine Concentrations P Cr = mg/dl (1.0mg/dl) normal range for adult Plasma Cr inversely related to GFR GFR ml/min P Cr mg/dl LSU Medical Physiology

23 Normal Values GFR GFR corrected to body surface area 1.73 m 2 Newborns 20 ml/min 120 ml/min 130 ml/min Declines after age 40 ~ 1 ml/min/yr Current Recommendations for Routine Estimation of GFR in Clinical Setting MDRD Study Equation: GFR (ml/min/1.73 m 2 )= 175 X (S Cr ) X (Age) X (0.742 if female) X (1.212 if African American) Cockcroft-Gault Equation: Clearance Cr (ml/min) = (140 age) X (Wt in kg) X (0.85 if female) / (72 X S Cr ) MDRD = Modification of Diet in Renal Disease; SCr = serum creatinine, mg/dl GFR Calculators Are Available Online: National Kidney Disease Education Program (NKDEP) of NIH Kidney Disease Outcomes Quality Initiative (KDOQI) of National Kidney Foundation LSU Medical Physiology

24 Renal Physiology Lecture 3 Starling Forces Control of GFR Concept of Clearance Clearance of Inulin & Creatinine 5. Clearance of PAH PAH Para-amino hippuric acid Cl PAH ~ RPF LSU Medical Physiology

25 Para-amino hippuric acid (PAH) Organic anion Freely filtered Vigorously secreted PT 90% removed single circuit ~ 10% RV NOT produced Infusion required Para-amino hippuric acid (PAH) PAH Clearance ~ Renal Plasma Flow Amount Entering Kidney ~ Excretion Rate RPF P PAH = U PAH V Rearrange RPF = U PAH V P PAH LSU Medical Physiology

26 Summary 1. Glomerular filtrate formation -filtration barrier & Starling forces 2. Clearance of certain substances index of renal function 3. Plasma creatinine - tool for diagnosing and following renal function THE END LSU Medical Physiology

27 Calculate Renal Plasma Flow P PAH = 0.05 mg/ml; U PAH = 30 mg/ml; V = 1.0 ml/min RPF = U PAH V P PAH R Plasma F = RPF = 600 ml/min 0.50 Hct R Blood F = RBF = 1,200 ml/min Review Clearance Principles Cl X Cl X < GFR net reabsorption X > GFR net secretion X Cl X < Cl IN reabsorbed - glucose Cl X > Cl IN secreted - PAH Cl X = Cl IN only filtered - creatinine filtered load > rate of excretion = reabsorption X LSU Medical Physiology

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