1 / 30

Cells

Intake : Na + ~ 5-200 mmoles/day K + ~ 50-300 “ H 2 0 ~ 0.5-3 L/day H + ~ 50-100 mEq/day. Air water ~ 3000 mOsm/Kg H 2 0. Extracellular Fluid Volume ~ 1/3 body water; Osmol 290 mOsm/Kg H 2 0 Na + 142 mM; K + 4 mM; H + 40 nM (pH 7.40). Cells. urine. Kidney.

tasha-blackwell
Download Presentation

Cells

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Intake: Na+ ~ 5-200 mmoles/day K+~ 50-300 “ H20 ~ 0.5-3 L/day H+ ~ 50-100 mEq/day Air water ~ 3000 mOsm/Kg H20 Extracellular Fluid Volume ~ 1/3 body water; Osmol 290 mOsm/Kg H20 Na+ 142 mM; K+ 4 mM; H+ 40 nM (pH 7.40) Cells urine

  2. Kidney

  3. Kidney Vasculature

  4. Glomerulus

  5. Nephron Circulation

  6. Blood Flow in Resting Human RPF ~ 600 m/mi = 36L/h = 864L/day

  7. Volume of Urine GFR ~ 200 L/day ~0.5-2 L/day

  8. Glomerular Capillary Wall

  9. Fractional Clearance Across Glomerular Capillary

  10. Glomerular “Sieve”

  11. Electrical Charge Effect on Fractional Clearance Across Glomerular Capillary

  12. Forces Driving Filtration

  13. Filtration & Reabsorption In Systemic Capillary

  14. Forces Determining GFR Systemic capillary: F = Kf x [(Pc-Pi) - (c-I)] Glomerular capillary: GFR = Kf x [(Pgc-Pu) – (gc-u)] Because Pu is unregulated and believed to be constant and U = 0, then GFR = Kf x (P - gc)

  15. Permeability of Glomerular Capillary

  16. Pressures Inside the Renal Circulation

  17. Filtration Pressures in Glomerular Capillary

  18. Filtration Pressures in Glomerular Capillary DP = PGC-PU 40 PUF 30 GFR=Kf x (P - GC) = Kf x PUF mm Hg 20 GC 10

  19. Factors Regulating PGC Control Angiotensin II ANP P P PUF PUF P PUF 30 pGC mm Hg pGC pGC 10 A E A E A E Distance Along Glomerular Capillary

  20. Volume of Urine GFR ~ 50 - 200 L/d Urine, ~ 0.5 – 15 L/d

  21. Nephron Capillaries

  22. Filtration & Reabsorption In Systemic Capillary

  23. Starling’sForces in Peritubular Capillary PR = Kf (C - i) – (PC- Pi) PR = Kf ( - P)

  24. Glomerular and Peritubular Starling Forces Control Angiotensin II DP DP (PGC – PU) PUF Dp Dp PUF 30 pGC pGC mm Hg PR PR 10 DP DP glomerular peritubular peritubular glomerular Distance Along Capillary

  25. Renin-Angiotensin System angiotensinogen renin angiotensin I converting enzyme angiotensin II increases PT reabs vasoconstriction efferent constriction; net PT reabsorption

  26. Angiotensin II Direct Effects on Kidney Function 1 - constricts Efferent Arteriole: PGC and GC ; does not change GFR but peritubular capillary  and, therefore, net Proximal Tubule sodium reabsorption 2- directly Proximal Tubule Na+ transport

  27. Renin-Angiotensin System aldosterone; CD reabsorption angiotensinogen renin angiotensin I converting enzyme angiotensin II efferent constriction; PT reabsorption vasoconstriction

  28. Aldosterone Effect on Na+ and K+ in the Collecting Duct Aldo

  29. Fractional Na+in Urine along the Nephron Fraction in urine (x 100) Angio IIAldosterone

  30. Measurement of GFR If a substance is freely filterable, then its rate of filtration is, Plasma [x] x GFR vol/time The rate of urinary excretion of any substance is, Urine [x] x Urine volume/time If all of x that is filtered appears in the final urine, then we have P[x] x GFR = U[x] x V and, GFR = U[x] x V / P[x]

More Related