1 / 19

Chapter 17 Physiology of the Kidneys

Chapter 17 Physiology of the Kidneys. Dr. David Washington. Glomerular Filtration. A person with a cardiac output of 5,600 ml/min would have a flow rate through the kidneys of about 1,200 ml/min, or 21% of the C.O. (renal fraction = = 21%)

sawyer
Download Presentation

Chapter 17 Physiology of the Kidneys

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. Chapter 17 Physiology of the Kidneys Dr. David Washington

  2. Glomerular Filtration A person with a cardiac output of 5,600 ml/min would have a flow rate through the kidneys of about 1,200 ml/min, or 21% of the C.O. (renal fraction = = 21%) The glomeruli filter particles up to 80 A, therefore; proteins and blood cells are not filterable. 1,200 5,600 o

  3. Uv p GFR = Measurement of GFR(Glomerular Filtration Rate) U = concentration of substance in urine P = concentration of substance in plasma V = volume of urine example (inulin) U = 125 mg/ml P = 1 mg/ml V = 1ml/min therefore,

  4. Measurement of GFR(Glomerular Filtration Rate) U = 125 mg/ml P = 1 mg/ml V = 1ml/min therefore, That is; 125 ml of plasma had to be filtered to get 125 mg of inulin in the urine. 125 mg/ml X 1 ml/min 1 mg/ml GFR = GFR = 125 ml/min

  5. Cortex Outer Medula Inner Medulla Counter-Current Mechanism(Urine Formation) K+ Na+ K+ Na+ H2O H2O Glucose Amino acid 300 400 600 320 400 600 200 400 H2O Na+ 800 1000 600 800 800 1000 Na+ Na+ Diffusion passive active 1200 Highly permeable to Na Impermeable to H2O

  6. Cortex Outer Medula Inner Medulla Counter-Current Mechanism(Urine Formation) Net action in loop of Henle - H2O reabsorption Net action at distal tubule: a. ADH - increases reabsorption b. Aldosterone - increases Na reabsorption

  7. Countercurrent Multiplier Mechanism Proximal Tubule Tubular lumen Peritubular fluid Na+ K + H2O Glucose Amino acids Glucose 70%

  8. Countercurrent Multiplier Mechanism Loop of Henle (Descending limb) H2O Na+ (Ascending limb) Na+ H2O Distal Tubule Na+ H2O 15% 14% x membrane impermeable K+ Diabetes insipidus

  9. Bowman’s capsule Glomerulus Juxtaglomerular cells Macula densa cells Afferent arteriole Efferent arteriole Autoregulation of Glomerular Filtration Rate (J-G apparatus) Distal tubule

  10. Renin-Angiotensin-Aldosterone Pathway 14 Angiotensinogen Asp-Arg-Val-Tyr-Ile-His-Pro-Phe-His-Leu-Leu-Val-Tyr-Ser Renin Angiotensin I Asp-Arg-Val-Tyr-Ile-His-Pro-Phe-His-Leu Ang.I Converting enzyme (in lungs) Angiotensin II Asp-Arg-Val-Tyr-Ile-His-Pro-Phe Vasoconstriction Aldosterone 10 8

  11. Low plasma Na+ concentration Hypothalamus Posterior pituitary ADH Water reabsorption in collecting ducts Urine Vol Blood Vol Na+ reabsorption in distal tubules Aldosterone Adrenal cortex Angiotensin II Renin Juxtaglomerular apparratus Compensation for Low Sodium Low Na+ Intake Na+ retention in blood Sympathetic nerve activity

  12. H H O H H O H H O Acids, Bases and Buffers H+ OH- H+ OH- H - O - H H+ + OH- hydroxide H2O

  13. H H O H H O H H O Acids, Bases and Buffers HCL CL- H+ H+ CL- H+ CL- HCL H+ + CL- Acid in H2O

  14. H H O H H O H H O Acids, Bases and Buffers NaOH H+ H+ OH- Na+ OH- NAOH Na+ + OH- Base in H2O

  15. Bicarbonate Buffer System Contains mixture for converting strong acids + bases to weak acids and bases. Example: (H2CO3 and NaHCO3) a) HCL + NaHCO3 H2Co3 + NaCL sodium bi- carbonic carbonate acid b) NaOH + H2Co3 NaHCO3 + H20 pk - the ph at which the concentration of the two portions of the buffer are equal. Buffer power - ability to maintain a ph following the addition of acid or base

  16. Bicarbonate Buffer System Buffer Curve: 0 25 50 75 100 75 50 25 Acid added H2CO3 NaHCO3 Base added pk 6.1 4 5 6 7 8 ph

  17. Henderson-Hasselbalch Equarion (pH caused by the concentration of CO2 in blood) HA H+ + A- 1) H2 CO3 H++ HCO3- carbonic bicarbonate ion acid Extracellular Fluid pH

  18. 2) KA = 3) KA = 4) log KA = log (H+) + log 5) - log(H+) = -logKA + log (HCO3-) (CO2) (HCO3-) (CO2) (H+)x(HCO3-) (H2CO3) (H+)x(HCO3-) (CO3) Extracellular Fluid pH Since [H2CO3] in extra-cellular fluids is an express of [CO2], we write --- equation 3.

  19. 6) pH = pKA + 7) pH =p KA + log pKA= 6.1; [HCO3-] = 24mM/L in arterial blood pH = 6.1 + log = 6.1 + log 20 = 6.1 + 1.3 = 7.4 24 0.03 x 40 (HCO3-) (CO2) (HCO3-) 0.03 x Pco2 Extracellular Fluid pH (solubility coefficient in mM/L)

More Related