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Renal clearance

Clearance of x (C x ) = U x .V. P x. Renal clearance. The renal clearance of a substance is an expression of the degree to which the substance is removed from the blood plasma and excreted into the urine. e. g. Substance x. Rate of excretion of x = U x .V .

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Renal clearance

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  1. Clearance of x (Cx) = Ux.V Px Renal clearance The renal clearance of a substance is an expression of the degree to which the substance is removed from the blood plasma and excreted into the urine e. g. Substance x Rate of excretion of x = Ux.V Need to relate the amount excreted to the plasma concentration (= Px)

  2. Clearance of x (Cx) = Ux.V Px Renal clearance Definition- The renal clearance of a substance is: • “The VOLUME of plasma rendered free of a given • substance in one minute” Substance x: Rate of excretion = Ux.V Plasma concentration = Px = ml/min liters/day gallons/second

  3. Questions: • How much urea is excreted per min? • What is the clearance of urea? Purea = 0.2 mg/ml Uurea = 6 mg/ml V = 2ml/min Answers: • Urea excretion = 6 mg/ml x 2 ml/min = 12 mg/min • Urea clearance = 12 mg/min = 60 ml/min i.e. 60 ml of plasma loses its urea into the urine in one minute 0.2 mg/ml

  4. Filtration Reabsorption Secretion Excretion Representative Nephron Glomerular capillaries Afferent arteriole Efferent arteriole Renal artery Peritubular capillaries Renal vein

  5. 1. Freely filtered 2. Not reabsorbed 3. Not secreted H2O Amount of a excreted per minute = amount of a filtered per minute Uax V = GFa xGFR GFR = Ua x V C1 x V1 = C2 x V2 Using Clearance to estimate/measure GFR Need a substance (a) with special properties: (concentration in Bowman’s space = concentration in plasma) Urine conc. of a x urine flow rate = glomerular filtrate conc. of a x GFR (GFa = Pa) GFa

  6. Very few substances meet the criteria required to measure GFR Best known: Inulin • polysaccharide of fructose • molecular weight ~5000 daltons • molecular diameter ~3 nm Typical value for GFR ~120 ml/min ~180 liters/day! • Inulin clearance = GFR Inulin is cleared from plasma by filtration alone. Since none of filtered inulin returns to the plasma, volume of plasma cleared of inulin per min must equal the volume of plasma filtered per min (= GFR)

  7. Useful alternative: • Freely filtered  • Not reabsorbed  • Not secreted ( true Ccr normally overestimates GFR by ~10-20%) Problems with measuring Cinclinically: • Must be infused intravenously • Continual blood sampling required • Chemical analysis cumbersome Creatinine (notcreatine) • Endogenous (by-product of muscle metabolism) • Released into blood at relatively constant rate (plasma concentration fairly stable; therefore only need one blood sample)

  8. Patient (Jim) with suspected renal failure • 24 hour urine volume = 600 mL • Urine [creatinine] = 240 mg/dL • (40-300 mg/dLin males and 37-250 mg/dL in females) • Plasma [creatinine] = 10 mg/dL • (0.5 to 1.0 mg/dL for women and 0.7 to 1.2 mg/dL for men) • BUN = 80 mg/dL(normal 10-20 mg/dL) Calculate his creatinine clearance • BUN:creatinine ratio ?

  9. Ccr = 240 mg/dL x 600 mL/day (0.42 ml/min) 10 mg/dL Low GFR means renal insufficiency Jim’s creatinine clearance (Ccr): Ccr = Ucr x V / Pcr Ccr = 10ml/min = GFR for Jim. (normal = 90-140 mL/min)

  10. GFR can be estimated from Pcr Jim GFR is lower in young children and declines in old age. The Gault-Cockroft formula is used clinically to estimate GFR in the steady state from a measurement of plasma creatinine: % GFR(ml /min) = [(140 - age)XWeight(kg)] (Pcr(mg / dL)x72)

  11. Renal clearance of selected substances provides special information Para-amino hippurate (PAH) • organic anion • freely filtered • avidly secreted by peritubular capillaries into proximal tubule • at low plasma concentrations, the combination of filtration and secretion means that all PAH arriving in the renal plasma is excreted in the urine • Clearance of inulin = GFR • Clearance of creatinine = ~ GFR • Clearance of para-amino hippurate = renal plasma flow

  12. Volume of plasma cleared of PAH (= CPAH) = volume of plasma entering kidneys = renal plasma flow ~ 1/5 filtered enters in renal artery remaining 4/5 secreted ~ none in the renal vein all excreted in the urine PAH handling by the kidney At low plasma concentrations:

  13. RPF = CPAH = 65 mg/ml x 1 ml/min = 650 ml/min 0.1 mg/ml Urine flow rate = 1 ml/min PPAH = 0.1 mg/ml UPAH = 65 mg/ml Renal plasma flow? Filtration Fraction (FF) using a value of GFR of 125 ml/min and ERPF = 700 ml/min, the fraction of plasma that is filtered is 125/700 = 18%. FF is typically in the range 15-25%. That is 15-25% of the plasma volume becomes filtrate in the nephrons of the kidney.

  14. Quantifying tubular function Filtration = GFR x PZ • (“filtered load”) Excretion = UZ x V • (“excretion rate”) Excretion = filtration – reabsorption + secretion Useful data about the transport of a substance by the renal tubule can be obtained by applying a few simple calculations: Reabsorpn = (GFR x Pz) – (Uz x V) Secretn = (Uz x V) - (GFR x Pz)

  15. Fractional excretion Fractional excretion of Z (FEZ) = rate of excretion of Z rate of filtration of Z FEZ = UZ x V = CZ / GFR GFR x PZ Proportion of the filtered load was excreted = fractional excretion i.e. excretion of Z as a percentage of the filtered load

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