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Three basic renal processes

Three basic renal processes. Introduction • Formation of urine by the kidney involves three main processes: 1. Glomerular filtration (180ltrs/day,20% of plasma filtered) 2. Tubular reabsorption (178.5ltrs) 3. Tubular secretion. Glomerular Filtration Rate (GFR)

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Three basic renal processes

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  1. Three basic renal processes • Introduction • • Formation of urine by the kidney involves three main processes: • 1. Glomerular filtration(180ltrs/day,20% of plasma filtered) • 2. Tubular reabsorption(178.5ltrs) • 3. Tubular secretion

  2. Glomerular Filtration Rate (GFR) • • The total amount of filtrate formed by all the renal corpuscles in both kidneys per minute is called the glomerular filtration rate, or GFR. • In normal kidneys, the 17 millimeters of mercury of net filtration pressure produces approximately 125 milliliters of filtrate per minute. This translates to about 180 liters in 24 hours !!! • Fortunately, 99% of this volume will be reabsorbed as the filtrate passes through the tubules. • 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.

  3. The Filtration Membrane and Glomerular Filtration • Passage through the filtration membrane is limited not only on the basis of size but also by charge. • Glomerular filtration is a process of bulk flow driven by the hydrostatic pressure of the blood. • Small molecules pass rapidly through the filtration membrane, while large proteins and blood cells are kept out of the capsular space.

  4. Neutral molecule Anion Anion proteinuria Shape Filtration barrier • nephrotic syndrome • Glomerulonephritis: loss of negative charge on glomerular membranes

  5. no protein Glomerular filtration Afferent arteriole Efferent arteriole Glomerular capillaries Bowman’s space • all other molecules and ions at ~same concentration as plasma

  6. Determinants of GFR The rate of filtration in any of the body's capillaries, including the glomeruli, is determined by the hydraulic permeability of the capillaries, their surface area, and the net filtration pressure (NFP) acting across them. Rate of filtration = hydraulic permeability x surface area x NFP Because it is difficult to estimate the area of a capillary bed, a parameter called the filtration coefficient (Kf) is used to denote the product of the hydraulic permeability and the area.

  7. Net Filtration Pressure • There are three forces affecting net filtration at the glomerulus: 1. Hydrostatic Pressure within the Glomerular Capillaries 2. Back Pressure from Fluid Within the Capsule 3. Osmotic Pressure within the Glomerular Capillaries NFP = (PGC – ЛGC) – (PBC – ЛBC) GFR = Kf (PGC – PBC – ЛGC). Net Filtration Pressure = 60 mm Hg - (15 mm Hg + 28 mm Hg) = 17 mm Hg “kidney shut-down” or acute renal failure.

  8. GFR Measurement of GFR is the best single means of assessing kidney function quantitatively Sum of filtration rates of all the nephrons = total glomerular filtration rate = total GFR = GFR If total number of functioning nephrons is reduced (chronic renal failure)

  9. Normally, renal blood flow doesn’t change much Flow = Δ Pressure Resistance Renal blood flow GFR 1500 150 ml/min 1000 100 Autoregulation 500 50 100 150 200 50 Mean arterial pressure (mmHg)

  10. Afferent arteriole Efferent arteriole Arterial pressure Mechanism? vasoconstriction During autoregulation, as blood pressure increases, so does vascular resistance PGC ↓ • myogenic • tubuloglomerular feedback

  11. afferent arteriole A1 receptor adenosine ATP Na+ 2Cl- K+ NaCl delivery macula densa cell Tubuloglomerular feedback Vasoconstriction macula densa

  12. A drug is noted to cause a decrease in GFR. Identify 4 possible actions of the drug that might decrease GFR. A person is given a drug that dilates the afferent arteriole and constricts the efferent arteriole by the same amounts. Assuming no other actions of the drug, what happens to this person's GFR A clamp around the renal artery is partially tightened to reduce renal arterial pressure from a mean of 120 to 90 mm Hg. do you predict RBF will change?

  13. 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: Plasma concentration = Px Rate of excretion = Ux.V = ml/min liters/day gallons/second

  14. 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

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

  16. 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

  17. 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)

  18. 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)

  19. 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 ?

  20. 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)

  21. 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) %

  22. 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

  23. 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:

  24. 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? If CPAH = 650ml/min and the patient has a hematocrit of 45%, what is the rate of renal blood flow?

  25. 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)

  26. 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

  27. Where Does Na+ Reabsorption Occur? FE = 10% [Na+]=145 FE = 3% FE = 35% FE = 0.1 - 2% [Na+] units = mmole/L FE= Fractional excretion

  28. The hospital lab reports that your patient's creatinine clearance is 120 g/day. This value is A. Normal B. Significantly below normal C. Not an interpretable number as presented List, in order of decreasing renal clearance, the following substances: glucose, urea, sodium, inulin, creatinine, PAH.

  29. The following test results were obtained during a clearance experiment: UIn = 50 mg/L PIn = 1 mg/L V = 2 mL/min UNa = 75 mmol/L PNa = 150 mmol/L What is the fractional excretion of sodium?

  30. Micturition • Once urine enters the renal pelvis, it flows through the ureters and enters the bladder, where urine is stored. • Micturition is the process of emptying the urinary bladder. • Two processes are involved: • The bladder fills progressively until the tension in its wall is raised above a threshold level, and then • A nervous reflex called the micturition reflex occurs that empties the bladder. • The micturition reflex is an automatic spinal cord reflex; however, it can be inhibited or facilitated by centers in the brainstem and cerebral cortex.

  31. stretch receptors Urine Micturition

  32. 1) APs generated by stretch receptors • 2) reflex arc generates APs that • 3) stimulate smooth muscle lining bladder • 4) relax internal urethral sphincter (IUS) • 5) stretch receptors also send APs to Pons • 6) if it is o.k. to urinate • APs from Pons excite smooth muscle of bladder and relax IUS • relax external urethral sphincter • 7) if not o.k. • APs from Pons keep • EUS contracted stretch receptors

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