Kidney and Body Fluids 3

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Physiology 1B (Kidney and Body Fluids) Mapa Mental sobre Kidney and Body Fluids 3, criado por Daniel Elandix G em 01-11-2013.
Daniel Elandix G
Mapa Mental por Daniel Elandix G, atualizado more than 1 year ago
Daniel Elandix G
Criado por Daniel Elandix G mais de 10 anos atrás
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Resumo de Recurso

Kidney and Body Fluids 3
  1. Measurement of GFR

    Anotações:

    • Need a substance that is freely filtered at the glomerulus and neither reabsorbed or secreted along the remainder of tubules. Other minor criteria includes: Not protein bound, not metabolised, not stored in kidneys, not toxic, no effect in filtration rate, relatively easy to measure in plasma and urine.
    1. Calculations

      Anotações:

      • for a substance (y) total amount filtered will be excreted in the urine. Amount filtered: GFR X Arterial plasma concentration of y Amount excreted: Uy X V, urinary concentration of y multiplied by urinary flow rate. GFR = Uy X V/Py
      1. Substance Used

        Anotações:

        • Starch like polymer of fructose... GFR substances used is inulin or creatinine clearance that provides a reasonable approximate.                                  
      2. Measurement of RPF

        Anotações:

        • Renal plasma flow: Calculated via the Fick principle, amount of substance entering the kidney is equal to the amount of substance leaving the kidney assuming if the kidney does not metabolise/make substance.
        1. Using Substance that is completely cleared

          Anotações:

          • RPF= Ux(ruinary concentration) X V (urine flow rate) / Pax (venous plasma concentration)
          1. ERPF

            Anotações:

            • Effective Renal Plasma flow. For example, the para-amino hippuric acid (PAH)  it appears in the renal vein. ERPF = 90% of true RPF.
          2. Renal Clearance

            Anotações:

            • Renal Clearance is a measure of the success of how much plasma cleared per unit time. Renal clearance of a substance is the minimal volume of plasma which could have supplied the amount of the substance which is excreted in the urine per unit time. Clearance A = Ua (urinary concentration of A) X V (urine flow rate) / Pa (
            1. Significance

              Anotações:

              • Clearance of inulin and para-amino hippuric acid are special examples of clearance formula. If we know the clearance for other substances in addition to inulin we can tell how the kidney handles the substance. if Cs less then GFR it must be reabsorbed.
            2. Calculations

              Anotações:

              • Normally plasma concentration of x X GFR. Excretion rate of x = amount of x excreted in urine per unit time or urinary concentration of x X urine flow rate. If less x: It may have undergone net reabsorption:  GFR X px - Ux X V If more X: The opposite formula is true. Ux X V - GFR X Px.
              1. Tubular Modifications
                1. Mechanisms
                  1. Reabsorption

                    Anotações:

                    • Few kinds: passive: water and urea Active: Primary or secondary active Tubular maximum limited reabsorption (carrier can be saturated) such as glucose, phosphate, bit c amino acids. Active gradient-time limited mechanisms, gradient established across the tubular wall within interval of time that the fluid is in contact with epithelium such as sodium.
                    1. Secretions

                      Anotações:

                      • Similar to reabsorption Passive: NH3 Active: Tubular maximum limited. Gradient time limited: H+

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