Answers to Integrative Case Studies





Case 1




  • 1a.

    The ECF volume of this man is increased above normal. The presence of edema, distention of the neck veins, and rales (sounds related to fluid in the lungs) is evidence of this increased volume. Additional evidence could be obtained by measuring weight gain because accumulation of each liter of extracellular fluid would increase body weight by 1 kg.


  • 1b.

    The ECV in this man would be decreased from normal. With damage to the myocardium, cardiac output and therefore tissue perfusion would be reduced. This decreased cardiac performance will be sensed by the vascular baroreceptors as a decrease in the ECV.


  • 1c.

    As noted previously, the reduced cardiac performance would be sensed by the vascular baroreceptors in the body as a decreased ECV. This will activate the sympathetic nervous system and the renin-angiotensin-aldosterone system and stimulate AVP secretion. Because ANP and BNP are secreted from the cardiac myocytes in response to stretch and because the heart is dilated (expanded ECF and vascular volume), ANP and BNP secretion will be stimulated.


  • 1d.

    The kidneys would be avidly retaining Na + . With a decrease in ECV, vascular volume sensors, especially in the high-pressure (juxtaglomerular apparatus, aortic arch, and carotid sinus) side of the circulation, would be activated and signals sent to the kidneys to retain Na + .




    • Sympathetic nerves innervating the afferent and efferent arterioles of the glomeruli would cause vasoconstriction. The net result would be to reduce the GFR. This in turn would reduce the filtered load of Na + .



    • Sympathetic innervation of the proximal tubule, thick ascending limb of Henle’s loop, distal tubule, and collecting duct will also increase Na + reabsorption at these sites. This response is mediated by α 1 -adrenoceptors on the cells.



    • Increased sympathetic nerve activity, together with decreased perfusion pressure at the afferent arteriole, will result in the secretion of renin. This activation of the renin-angiotensin-aldosterone system will further stimulate Na + reabsorption because angiotensin II increases both proximal and distal tubule Na + reabsorption and aldosterone increases Na + reabsorption in the ASDN.



    • With the decreased ECV, the GFR decreases and the filtration fraction increases. This in turn decreases the hydrostatic pressure and increases the oncotic pressure in the peritubular capillaries and thereby enhances overall reabsorption of fluid by the proximal tubule.



    • With the increases in the ECF and vascular volumes, the heart dilates and ANP and BNP levels are elevated. However, the effect of these natriuretic peptides (inhibition of renin secretion and natriuresis) appears to be blunted by the effect of the other factors, all of which act to reduce Na + excretion. The net effect of these responses is retention of Na + by the kidneys. As a result of this Na + retention (positive Na + balance), the ECF volume will increase, leading to the formation of edema as seen in the physical examination of this man.



  • 1e.

    The development of hyponatremia indicates that this man is in positive water balance. In this case, the ingestion of water has exceeded the capacity of the kidneys to excrete solute-free water. There are several reasons why solute-free water excretion is impaired in this man.




    • AVP secretion is stimulated because of the decreased ECV. As a consequence, the collecting duct reabsorbs more solute-free water.



    • The decreased ECV results in a reduction in the filtered amount of solute (NaCl) and water and an increase in fractional reabsorption by the proximal tubule (see earlier). As a result, there is decreased delivery of solute and water to the thick ascending limb, the primary site where solute-free water is generated.



  • 1f.

    Hypokalemia in this man is the result of increased renal K + excretion. The major reason for increased K + excretion is related to the elevated levels of aldosterone in this man (secondary to decreased ECV). Aldosterone then acts on the ASDN to stimulate K + secretion. The enhanced secretion of K + by the ASDN will result in increased K + excretion and the development of hypokalemia. Because delivery of Na + to the collecting duct is reduced in this man (see earlier), the hypokalemia will be mild. It is likely that extrarenal factors will also contribute to the development of hypokalemia because aldosterone causes K + to move into cells.


  • 1g.

    To obtain an appropriate response to a loop diuretic, there needs to be adequate delivery of NaCl to the loop of Henle. As outlined before, GFR is reduced in congestive heart failure, which reduces the filtered amount of NaCl. In addition, proximal tubule reabsorption is enhanced. The net result is a significant reduction in the delivery of NaCl to the thick ascending limb and thus a blunted response to the diuretic.


  • 1h.

    As noted, the diuretic is given to prevent NaCl retention by the kidneys and thus reduce his edema. The additional NaCl lost from the ECF by the action of the diuretic will further reduce his ECV. As a result, the sympathetic nervous system will be further activated, as will the renin-angiotensin-aldosterone system. AVP secretion will also be stimulated.


  • 1i.

    Loop diuretics increase the excretion of K + and thus can lead to the development of hypokalemia. Two important effects contribute to this response. First, the thick ascending limb of Henle’s loop reabsorbs approximately 20% of the filtered amount of K + . Loop diuretics inhibit this process. K + is reabsorbed by the Na + -K + -2Cl symporter in the apical membrane and by the paracellular pathway driven by the lumen-positive transepithelial voltage (inhibition of the symporter results in a reduction in the lumen-positive voltage). Second, the loop diuretic will cause increased delivery of Na + and fluid to the ASDN and thereby stimulate K + secretion. Because aldosterone levels are also elevated (secondary to decreased ECV), K + secretion is further stimulated. Together these effects will enhance K + secretion and thus renal K + excretion, leading to a worsening of the hypokalemia.


  • 1j.

    Creatinine is excreted from the body primarily by glomerular filtration. Therefore the amount of creatinine excreted is determined primarily by its filtered amount. With a reduction in the ECV, the glomerular filtration rate is reduced. The reduced filtration rate will decrease the filtered amount of creatinine and thus its excretion. As a result, the serum [creatinine] will increase. With the added decrement in the ECV caused by the loop diuretic, the glomerular filtration rate will fall further and thereby cause the serum [creatinine] to increase even more.





Case 2




  • 2a.

    This woman’s symptoms and the electrolyte disturbances are most characteristic of decreased levels of adrenal cortical steroids and especially the mineralocorticoid hormone aldosterone. This is a patient with Addison’s disease. The presence of hyperpigmentation suggests that the primary problem is at the level of the adrenal gland (i.e., nonresponsive to ACTH). ACTH levels are elevated in response to the decreased circulating levels of adrenal cortical steroids. (ACTH is synthesized as pre-proopiomelanocortin (POMC), and when this molecule is processed to ACTH, several of the cleavage products have melanocyte-stimulating properties). These cleavage products act on the epidermal melanocytes, leading to the hyperpigmentation of the gums and skin.


  • 2b.

    Normally the kidneys would respond to the decreased ECF volume present in this woman by dramatically reducing the excretion of Na + . The urine [Na + ] is unexpectedly high in this woman because of the reduced ability of her kidneys to reabsorb Na + in the thick ascending limb and especially the ASDN. This “Na + wasting” is a result of the low levels of aldosterone. Although the angiotensin II levels would be elevated in this woman, her adrenal glands are not secreting aldosterone in response to the angiotensin II (or her hyperkalemia). The hypotension is a result of decreased vascular volume secondary to the negative Na + balance present in this woman, which in turn reflects the decreased circulating levels of aldosterone.


  • 2c.

    Hyponatremia indicates a problem in water balance. Thus the ability of this woman’s kidneys to excrete solute-free water is impaired, and she is in positive water balance (solute-free water ingestion > solute-free water excretion). There are several reasons for the decreased ability of this woman’s kidneys to excrete solute-free water. Because of her decreased ECF and vascular volumes, her vascular baroreceptors are activated. As a result, the sympathetic nervous system and the renin-angiotensin-aldosterone system are activated. Nonosmotic release of AVP also occurs because of a fall in the ECF volume. As a consequence of the elevated AVP levels, solute-free water is reabsorbed by the late portion of the distal tubule and collecting duct. In addition, the decreased ECF volume results in a decrease in the filtered amount of solute (NaCl) and water and an increase in fractional reabsorption by the proximal tubule. As a result, there is decreased delivery of solute and water to the thick ascending limb, the primary site where solute-free water is generated.


  • 2d.

    Urinary K + excretion is determined in large part by the amount of K + secreted into tubular fluid by the ASDN. K + secretion at this site is reduced by the low levels of plasma aldosterone in this woman. In addition, aldosterone causes the uptake of K + into cells (e.g., skeletal muscle). In the absence of aldosterone, there will be less cellular uptake of K + . This will also contribute to the development of hyperkalemia.


  • 2e.

    This woman has a metabolic acidosis, as evidenced by her low serum [ <SPAN role=presentation tabIndex=0 id=MathJax-Element-1-Frame class=MathJax style="POSITION: relative" data-mathml='HCO3−’>HCO3HCO3−
    HCO 3 −
    ]. This is a result of the inability of her kidneys to excrete sufficient net acid (RNAE) to balance the nonvolatile acids produced each day through metabolism (NEAP). The defect in her kidneys is reduced H + secretion by α-intercalated cells present in the ASDN. H + secretion by these cells is dependent on aldosterone. This is both a direct and an indirect effect. Aldosterone directly stimulates α-intercalated cells to secrete H + . It is also responsible for stimulating Na + reabsorption by principal cells in this segment. This reabsorption of Na + generates a lumen-negative transepithelial voltage that facilitates H + secretion by intercalated cells. Therefore, in the absence of aldosterone, H + secretion is impaired. This will reduce <SPAN role=presentation tabIndex=0 id=MathJax-Element-2-Frame class=MathJax style="POSITION: relative" data-mathml='HCO3−’>HCO3HCO3−
    HCO 3 −
    reabsorption, the excretion of H + with urinary buffers, and the excretion of NH 4 + . Thus RNAE is reduced.





Case 3




  • 3a.

    Na + excretion is determined by diet and ECF volume status. In a euvolemic individual, the amount of Na + excreted in the urine reflects the amount ingested in the diet. Because the man appears to be euvolemic, his daily Na + intake would be approximately 80 mEq/day. The cause of this man’s SIAD mostly likely is due to the unregulated secretion of AVP by the lung cancer cells (small cell carcinoma cells in the lung secrete AVP). The hyponatremia results because the unregulated secretion of AVP prevents his kidneys from excreting solute-free water. As a result, his water intake exceeds the ability of his kidneys to excrete solute-free water, and he is in positive water balance.


  • 3b.

    Infusion of 1 L of isotonic saline will add 300 mOsm of NaCl (150 mmol of NaCl = 300 mOsm) to his ECF together with 1 L of water. This will transiently increase the ECF volume and raise the plasma osmolality and [Na + ]. However, over time the entire amount of NaCl will be excreted because he is in steady-state Na + balance (excretion = intake). However, because of the inappropriate secretion of AVP, he will not be able to excrete the water. With a U osm of 600 mOsm/kg H 2 O, he will excrete the infused NaCl (300 mOsm) in 0.5 L of urine. Thus 0.5 L of solute-free water will remain in the body, and the plasma [Na + ] will decrease. For this man, the addition of 0.5 L of solute-free water to his body fluid will reduce the plasma [Na + ] to 119 mEq/L:


<SPAN role=presentation tabIndex=0 id=MathJax-Element-3-Frame class=MathJax style="POSITION: relative" data-mathml='Totalbodyosmoles=240mOsm/L×42L(unchanged)=10,080mOsmNewtotalbodywater=42L+0.5L=42.5LNewplasmaosmolality=10,080mOsm/42.5L=237mOsm/kgH2ONewplasma[Na+]=237mOsm/kgH2O/2=119mEq/L’>Totalbodyosmoles=240mOsm/L×42L(unchanged)=10,080mOsmNewtotalbodywater=42L+0.5L=42.5LNewplasmaosmolality=10,080mOsm/42.5L=237mOsm/kgH2ONewplasma[Na+]=237mOsm/kgH2O/2=119mEq/LTotalbodyosmoles=240mOsm/L×42L(unchanged)=10,080mOsmNewtotalbodywater=42L+0.5L=42.5LNewplasmaosmolality=10,080mOsm/42.5L=237mOsm/kgH2ONewplasma[Na+]=237mOsm/kgH2O/2=119mEq/L
Totalbodyosmoles=240mOsm/L×42L(unchanged)=10,080mOsmNewtotalbodywater=42L+0.5L=42.5LNewplasmaosmolality=10,080mOsm/42.5L=237mOsm/kgH2ONewplasma[Na+]=237mOsm/kgH2O/2=119mEq/L

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Oct 10, 2019 | Posted by in NEPHROLOGY | Comments Off on Answers to Integrative Case Studies
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