MEDICAL PHYSIOLOGY
                         Renal Conference Quiz 7A
                             February 19, 2002

1. How does inhibition of salt reabsorption in the thick ascending limb
inhibit water reabsorption by the collecting tubule?

          Salt reabsorption by the thick ascending limb is the primary
          engine for the countercurrent mechanism which increases the
          osmotic concentration of interstitial fluid in the medulla. It is
          this osmotic force which drives water reabsorption by the
          collecting tubule. Inhibition of that salt reabsorption disrupts
          the countercurrent mechanism.

2. Why did vomiting worsen the hypokalemia in this patient?

          Primarily because the vomited fluid contains potassium at a
          concentration higher than in plasma. In addition dietary K+ is
          not absorbed into the body.

3. Name two mechanisms that stimulated renin release in the patient.

            a. Inhibition of salt reabsorption by the macula densa cells
               resulted in increased renin secretion.
            b. Inhibition of salt and water reabsorption resulted in a
               decrease in blood volume. Thus atrial volume receptors
               reduce their inhibition of the sympathetic nervous system.
               Increased efferent sympathetic nerve activity to the kidney
               stimulates renin release.
            c. A decrease in blood volume leads to a decrease in blood
               pressure sensed by the afferent arteriole leading to
               increased renin release.

4. Briefly explain what was causing the hypokalemia in this patient.

          Hypokalemia is a result of the aldosterone induced increase of K+
          secretion at the collecting tubules and its loss in urine.

5. The following values were obtained from a patient:
          V = 120 mL/hr
          Pcr = 1.0 mg/dl
          Ucr = 50 mg/dl
          PNa = 140 mEq/L
          UNa = 28 mEq/L.
What is the GFR?

          V= 120 ml/hr = 2 ml/min
          GFR = (Ucr × V) / Pcr = (50 mg/dl)(2 ml/min) / 1.0 mg/dl = 100
          ml/min.







                            MEDICAL PHYSIOLOGY
                         Renal Conference Quiz 7B
                             February 21, 2002

1. Name two factors/processes that stimulate K+ secretion into the tubular
fluid.

            a. Increase in salt delivery to the collecting tubule.
            b. Increase in flow rate through the collecting tubule.
            c. Increased circulating levels of renin and aldosterone
            d. Alkalosis.

2. Why was water deprivation and an injection of vasopressin ineffective in
raising urine osmotic concentration in this patient?

          Salt transport by the thick ascending limb was blocked by the
          genetic defect. This disrupted the countercurrent mechanism which
          establishes a high osmotic concentration in the medullary
          interstitial fluid. Without that gradient water deprivation and
          the injection of vasopressin is ineffective in increasing water
          reabsorption in the collecting duct.

3. Why did vomiting contribute to the alkalosis?

          Acid is lost from the body in the vomitus.

4. Briefly describe a mechanism by which aldosterone leads to hypokalemia.

          Aldosterone stimulates secretion of K+ through different
          mechanisms:
            a. Increasing the activity of the Na,K-ATPase, therefore
               increasing the intracellular K+ concentration in the tubular
               cells and favoring the chemical gradient for movement of K+
               into the tubular fluid.
            b. Through its effect on apical Na channels, the apical
               membrane depolarizes reducing the gradient that opposes K+
               secretion.

5. The following values were obtained from a patient:
          V = 120 mL/hr
          Pcr = 1.0 mg/dl
          Ucr = 50 mg/dl
          PNa = 140 mEq/L
          UNa = 28 mEq/L.
What is FENa?

          FENa = (UNa / PNa) / (Ucr / Pcr) X 100
          FENa = (28 mEq/L / 140 mEq/L) / (50 mg/dl / mg/dl) X 100 = 0.4%