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Block: URIN 313 Physiology of THE URINARY SYSTEM Lecture 7

Block: URIN 313 Physiology of THE URINARY SYSTEM Lecture 7. Dr. Amel Eassawi Dr. Shaikh Mujeeb Ahmed. RENAL BLOOD FLOW AND CLEARANCE. Objectives. Describe the concept of renal plasma clearance. Use the formula for measuring renal clearance.

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Block: URIN 313 Physiology of THE URINARY SYSTEM Lecture 7

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  1. Block: URIN 313Physiology of THE URINARY SYSTEM Lecture 7 • Dr. AmelEassawi • Dr. ShaikhMujeeb Ahmed

  2. RENAL BLOOD FLOW AND CLEARANCE

  3. Objectives • Describe the concept of renal plasma clearance. • Use the formula for measuring renal clearance. • Use clearance principles for inulin, creatinine etc. for determination of GFR. • Use PAH clearance for measuring renal blood flow. • Outlines the factors affecting the renal blood flow.

  4. RENAL BLOOD SUPPLY • Although kidneys constitute less than 0.5% of total body mass, they receive 20-25% of resting cardiac output. • Left and right renal artery enters kidney. • Branches into segmental, interlobar, arcuate, interlobular arteries. • Each nephron receives one afferent arteriole. • Divides into glomerulus – capillary ball. • Reunite to form efferent arteriole (unique). • Divide to form peritubular capillaries or some have vasa recta. • Peritubularvenule, interlobar vein and renal vein exits kidney.

  5. RENAL BLOOD SUPPLY

  6. Renal blood flow and Oxygen consumption • Oxygen consumption per 100 g of renal tissue is 5 ml/min. which is next highest to the heart (8ml/min). • Total renal blood flow is ≈ 1200/min. • RBF per 100 g tissue is ≈ 400 mL/min which is disproportionately high as compared to heart which is only 80 mL/min. • A-V O2 difference is 1.5ml O2/100ml of blood flow. In heart it is ≈ 10 ml O2/100ml • 90 % of RBF goes to the renal cortex, 9% goes to outer medulla and 1% to the inner medulla

  7. REGULATION OF RENAL BLOOD FLOW RBF (Q):is directly proportional to the pressure gradient (ΔP) between the renal artery and the renal vein Is inversely proportional to the resistance(R) of the renal vasculature (Q) = Δ P R The major mechanism of changing Renal blood flow is by changing Afferent or Efferent Arteriolar resistance.

  8. 1)SYMPATHETIC NERVES AND CIRCULATING CATACHOLAMINES Both afferent and efferent arterioles are innervated by sympathetic nerves that act via α1 receptors to cause vasoconstriction. However ,since far more α1 receptors are present on afferent arterioles, increased sympathetic stimulation will cause a decrease in both RBF & GFR.

  9. 2) ANGIOTENSIN II This is a potent vasoconstrictor. However Efferent arteriole is more sensitive to Angiotensin II. Hence low levels of Angiotensin II causes increase in GFR while high levels of Angiotensin II will decrease GFR. RBF is decreased. 3) PROSTAGLANDINS PGE 2, PGI 2 are produced locally in the kidneys – cause vasodilation of both afferent & efferent arterioles. This effect is protective for renal blood flow , it modulates the vasoconstriction produce by sympathetic & angiotensin-II 4) DOPAMINE At low levels Dopamine dilates Cerebral, Cardiac, Splanchnic & Renal arterioles and constricts Skeletal Muscle and Cutaneous arterioles. Hence low dose Dopamine can be used in the treatment of hemorrhage .

  10. AUTOREGULATION OF RENAL BLOOD FLOW • Myogenic theory 2. Tubuloglomerular feedback by Juxta Glomerular Apparatus (JGA)

  11. Assessing Kidney Function • Albumin excretion (microalbuminuria) • Plasma concentration of waste products (e.g. Blood Urea Nitrogen (BUN), creatinine) • Urine specific gravity, urine concentrating ability • Imaging methods (e.g. MRI, PET, arteriograms, IVpyelography, ultrasound etc) • Isotope renal scans • Biopsy • Clearance methods (e.g. 24-hr creatinine clearance)

  12. Clearance • Clearance is a general concept that describes the • rate at which substances are removed (cleared) • from the plasma. • Renal clearance of a substance is the volumeof • plasma completely cleared of a substance • per min by the kidneys.

  13. Cs=Us x V Ps Clearance Technique Renal clearance of a substance is the volumeof plasma completely cleared of a substance per min. Cs x Ps = Us x V Where: Cs = clearance of substance S Ps = plasma conc. of substance S Us = urine conc. of substance S V = urine flow rate

  14. Glucose INULIN PAH Urea

  15. GFR x Pin = Uin x V Uin x V GFR = Pin Use of Clearance to Measure GFR For a substance that is freely filtered, but not reabsorbed or secreted (inulin, 125 I-iothalamate, creatinine), renal clearance is equal to GFR amount filtered = amount excreted

  16. Uin x V GFR = Cinulin = Pin 125 x 1.0 GFR = = 125 ml/min 1.0 Calculate the GFR from the following data: Pinulin= 1.0 mg / 100ml Uinulin= 125 mg/100 ml Urine flow rate = 1.0 ml/min

  17. Clinically it is not convenient to use inulin clearance – to maintain a constant plasma concentration it must be infused continuously throughout measurement. • Creatinine clearance is used as a rough estimate of GFR • Normal plasma serum creatinine level is 1mg/100ml.

  18. Effect of reducing GFR by 50 % on serum creatinine concentration and creatinine excretion rate

  19. Steady-state relationship between GFR and serum creatinine concentration

  20. Use of Clearance to Estimate Renal Plasma Flow Theoretically, if a substance is completely cleared from the plasma, its clearance rate would equal renal plasma flow. Cx = renal plasma flow

  21. ~ 2. amount entered = amount excreted UPAH x V ERPF = PPAH Use of PAH Clearance to Estimate Renal Plasma Flow Paraminohippuric acid (PAH) is freely filtered and secreted and is almost completely cleared from the renal plasma 1. amount enter kidney = RPF x PPAH 3. ERPF x Ppah = UPAH x V ~ 10 % PAH remains ERPF = Clearance PAH

  22. PAH clearance example

  23. UPAHx V ERPF = CPAH= PPAH 5.85 x 1.0 = 585 ml/min ERPF = 0.01 Calculate the RPF from the following data: PPAH = 0.01 mg / 100ml UPAH = 5.85 mg/100 ml Urine flow rate = 1.0 ml/min

  24. APAH - VPAH APAH = 1.0 – 0.1 = 0.9 1.0 normally, EPAH = 0.9 i.e PAH is 90 % extracted ERPF RPF = EPAH To Calculate Actual RPF , One Must Correct for Incomplete Extraction of PAH Extraction ratio of PAH = Percentage of PAH removed from the blood EPAH = VPAH = 0.1

  25. 585 RPF 650 650 RPF = RBF = RBF = RBF = 0.55 1 - 0.45 0.9 1 - Hct = 1182 = 650

  26. Calculation of Tubular Reabsorption Reabsorption = Filtration -Excretion Filtration s = GFR x Ps Excretions= Us x V

  27. Calculation of Tubular Secretion Secretion = Excretion - Filtration Filtration s = GFR x Ps Excretions = Us x V

  28. Clearances of Different Substances Substance Clearance (ml/min inulin 125 PAH 600 glucose 0 sodium 0.9 urea 70 Clearance of inulin (Cin) = GFR if Cx < Cin: indicates reabsorption of x if Cx > Cin: indicates secretion of x Clearance creatinine (Ccreat) ~ 140 (used to estimate GFR) Clearance of PAH (Cpah) ~ effective renal plasma flow

  29. Sample problem In a 24hr period, 1.44 L of urine is collected from a man receiving an infusion of inulin. In his urine, the [inulin] is 150mg/ml, and [Na+] is 200 mEq/L. In his plasma, the [inulin] is 1mg/mL, and the [Na+] is 140mEq/L What is the clearance ratio for Na+, and what is the significance of its value?

  30. References • Human physiology by Lauralee Sherwood, Seventh edition • Text book of physiology by Linda s. Contanzo, Third edition • Text book physiology by Guyton &Hall,11th edition

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