Answers to Self-Study Problems





Chapter 1




  • 1.



































    Molarity (mmol/L) Osmolality (mOsm/kg H 2 O)
    9 g NaCl 154 308
    72 g Glucose 400 400
    22.2 g CaCl 2 200 600
    3 g Urea 50 50
    8.4 g NaHCO 3 100 200


  • 2.

    The cell will swell when placed in the solution because the solute is only a partially effective osmole (i.e., this is a hypotonic solution). Because the reflection coefficient (σ) is 0.5, the effective osmolality of the solution is only 150 mOsm/kg H 2 O. The solution would have to contain 600 mmol/L of this solute to be isotonic.


  • 3.

    Na + , with its anions Cl and , constitutes most particles in the ECF and is therefore the major determinant of plasma osmolality. Consequently, plasma osmolality can be estimated by simply doubling the plasma [Na + ]. Thus the estimated plasma osmolality in this individual is as follows:


<SPAN role=presentation tabIndex=0 id=MathJax-Element-1-Frame class=MathJax style="POSITION: relative" data-mathml='Posm=2(130)=260mOsm/kgH2O’>Posm=2(130)=260mOsm/kgH2OPosm=2(130)=260mOsm/kgH2O
Posm=2(130)=260mOsm/kgH2O
Posm=2(130)=260mOsm/kgH2O
Posm=2(130)=260mOsm/kgH2O



  • This value is well below the normal range of 285 to 295 mOsm/kg H 2 O and will result in movement of water from the extracellular fluid (ECF) into the intracellular fluid (ICF). Because ions move freely across the capillary wall, the [Na + ] (and osmolality) of the plasma and interstitial fluid will be the same. Therefore water movement across the capillary endothelium will not be affected.


  • 4.

    The increase in venous pressure causes increased movement of fluid out of the capillary. As a result, fluid accumulates in the interstitial space. Some of this fluid will be taken up by the lymphatics, and lymphatic flow will increase. However, when the capacity of the lymphatics to remove this fluid is exceeded, the volume of the interstitial space increases and edema forms (see also Chapter 6 ).


  • 5.

    This is an isotonic solution and will therefore be confined initially to the ECF. The ECF volume will increase by 1 L, and ICF volume will be unchanged (initial volumes being 10 L [ECF] and 20 L [ICF], respectively). Plasma [Na + ] will decrease because of the addition of 1 L of Na + -free solution to the ECF. The new plasma [Na + ] is calculated as follows:

















Initial ECF Na + content = 145 mEq/L × 10 L
= 1450 mEq
New [Na + ] = 1450 mEq/11 L
= 132 mEq/L



  • Therefore the immediate effect is:

















ECF volume 11 L
ICF volume 20 L
Plasma [Na + ] 132 mEq/L



  • In the long term, the dextrose will be metabolized to CO 2 and H 2 O. Thus infusion of a dextrose solution is equivalent to infusion of solute-free water. After metabolism of dextrose (several hours) and equilibration, the 1 L of infused H 2 O will distribute into both the ICF and ECF in proportion to the ratio of their volumes (two-thirds into the ICF and one-third into the ECF):

















ICF volume = 20 L + 0.67 L = 20.67 L
ECF volume = 10 L + 0.33 L = 10.33 L
New [Na + ] = 1450 mEq/10.33 L = 141 mEq/L



  • Therefore the long-term effect is:














ECF volume 10.3 L
ICF volume 20.7 L
Plasma [Na + ] 141 mEq/L



  • As noted previously, a dextrose solution is equivalent to solute-free water. Therefore these solutions would be used when the patient had lost solute-free water and body fluid osmolality is elevated (i.e., hypernatremia).


  • 6.

    A 0.9% NaCl solution is isotonic saline. Therefore the entire infused volume will remain in the ECF. In this example, the ECF will increase by 1 L and the ICF will not change. One approach to this problem is to calculate the amount of Na + in the infused volume and then determine the effect on the plasma [Na + ]. For example:


























0.9% NaCl = 154 mEq/L
Amount of infused Na + = 1 L × 154 mEq/L
= 154 mEq
New ECF Na + content = 1450 mEq + 154 mEq
= 1604 mEq
New plasma [Na + ] = 1604 mEq/11 L
= 146 mEq/L



  • Therefore the immediate effect is:














ECF volume 11 L
ICF volume 20 L
Plasma [Na + ] 146 mEq/L



  • The long-term effect of this infusion is identical to that seen immediately. Thus the long-term effect is:














ECF volume 11 L
ICF volume 20 L
Plasma [Na + ] 146 mEq/L



  • From this it is apparent that infusion of isotonic saline into an individual with a normal serum [Na + ] will result in an increase in the volume of the ECF, equal to the entire infused volume, with no appreciable change in the serum [Na + ].


  • 7.

    The initial volumes of the body fluid compartments and the osmoles in these compartments are calculated as follows (osmolality is estimated as 2 × [Na + ] = 280 mOsm/kg H 2 O):























Initial total body water = 0.6(60 kg) = 36 L
Initial ICF volume = 0.4(60 kg) = 24 L
Initial ECF volume = 0.2(60 kg) = 12 L
Initial total body osmoles = (total body water)(body fluid osmolality) = (36 L)(280 mOsm/kg H 2 O) = 10,080 mOsm
Initial ICF osmoles = (ICF volume)(body fluid osmolality) = (24 L)(280 mOsm/kg H 2 O) = 6720 mOsm
Initial ECF osmoles = Total body osmoles − ICF osmoles = 10,080 mOsm − 6720 mOsm = 3360 mOsm



  • Four kilograms of body weight are lost. It is assumed that this entire weight reduction reflects fluids lost through vomiting and diarrhea. Thus 4 L of fluid are lost. Because the plasma [Na + ] is unchanged, a proportional amount of solute was also lost (isotonic loss of fluid). No fluid shifts occur between the ECF and ICF because of the absence of an osmotic gradient between these compartments. Therefore the ECF loses 4 L of volume and 4 × 280 = 1120 mOsm of solute.























New total body water = 36 L − 4 L = 32 L
New ICF volume = 24 L (unchanged)
New ECF volume = 12 L − 4 L = 8 L
New total body osmoles = 10,080 mOsm − 1120 mOsm = 8960 mOsm
New ICF osmoles = 6720 mOsm (unchanged)
New ECF osmoles = 3360 mOsm − 1120 mOsm = 2240 mOsm


  • 8.

    The initial volumes of the body fluid compartments and the osmoles in these compartments are calculated as in problem 7:























Initial total body water = 0.6(50 kg) = 30 L
Initial ICF volume = 0.4(50 kg) = 20 L
Initial ECF volume = 0.2(50 kg) = 10 L
Initial total body osmoles = (total body water) × (body fluid osmolality) = (30 L)(290 mOsm/kg H 2 O) = 8700 mOsm
Initial ICF osmoles = (ICF volume)(body fluid osmolality) = (20 L)(290 mOsm/kg H 2 O) = 5800 mOsm
Initial ECF osmoles = Total body osmoles − ICF osmoles = 8700 mOsm − 5800 mOsm = 2900 mOsm



  • The total amount of mannitol added to the ECF must be calculated to determine its effect on body fluids. At 5 g/kg, a total of 250 g was added to the ECF (1.374 moles of mannitol). Because mannitol is a single particle in solution, this adds 1374 mOsm to the ECF. The mannitol will raise ECF osmolality and result in the shift of fluid from the ICF into the ECF.


























New total body water = 30 L (unchanged)
New total body osmoles = 8700 mOsm + 1374 mOsm = 10,074 mOsm
New ICF osmoles = 5800 mOsm (unchanged)
New ECF osmoles = 2900 mOsm + 1374 mOsm = 4274 mOsm
New plasma osmolality = New total osmoles/Total body water = 10,074 mOsm/30L= 336 mOsm/kg H 2 O
New ICF volume = ICF osmoles/New P osm = 5800 mOsm/336 mOsm/kg H 2 O = 17.3 L
New ECF volume = Total body water − ICF volume = 30 L − 17.3 L = 12.7 L



  • Because mannitol increases the osmolality of the ECF, 2.7 L of fluid shifts from the ICF into the ECF. To calculate the new plasma [Na + ], assume that the amount of Na + in the ECF is unchanged after mannitol infusion. Originally there were 2900 mOsm attributable to Na + (2 × [Na + ] = ECF volume) in the ECF. Because the Na + osmoles are unchanged but are now present in a larger volume, the new plasma [Na + ] is calculated as follows:











New plasma Na + osmoles = 2900 mOsm from Na + /12.7L = 228 mOsm/L
New plasma [Na + ] = Na + Osm/2= 114 mEq/L


  • 9.

    Both individuals start out with the same total body water (36 L) and total body osmoles (10,440 mOsm) assuming a plasma [Na + ] of 145 mEq/L and a plasma osmolality of 290 mOsm/kg H 2 O. Subject A loses 1 L of total body water and 1000 osmoles of total body solute, resulting in a new plasma osmolality of


<SPAN role=presentation tabIndex=0 id=MathJax-Element-2-Frame class=MathJax style="POSITION: relative" data-mathml='Posm=(10,440mOsm−1000mOsm)/35L=270mOsm/kgH2O’>Posm=(10,440mOsm1000mOsm)/35L=270mOsm/kgH2OPosm=(10,440mOsm1000mOsm)/35L=270mOsm/kgH2O
Posm=(10,440mOsm−1000mOsm)/35L=270mOsm/kgH2O
Posm=(10,440mOsm1000mOsm)/35L=270mOsm/kgH2O
Posm=(10,440mOsm−1000mOsm)/35L=270mOsm/kgH2O



  • Subject B loses 4 L of total body water and 1600 mOsm of total body solute, resulting in a new plasma osmolality of


<SPAN role=presentation tabIndex=0 id=MathJax-Element-3-Frame class=MathJax style="POSITION: relative" data-mathml='Posm=(10,440mOsm−1600mOsm)/32L=276mOsm/kgH2O’>Posm=(10,440mOsm1600mOsm)/32L=276mOsm/kgH2OPosm=(10,440mOsm1600mOsm)/32L=276mOsm/kgH2O
Posm=(10,440mOsm−1600mOsm)/32L=276mOsm/kgH2O
Posm=(10,440mOsm1600mOsm)/32L=276mOsm/kgH2O
Posm=(10,440mOsm−1600mOsm)/32L=276mOsm/kgH2O




Chapter 2




  • 1.

    The gross anatomic features of the kidney include the cortex, medulla, nephrons, blood vessels, lymphatics, nerves, renal pyramids, papilla, minor calyx, major calyces, and pelvis.


  • 2.

    The nephron consists of a renal corpuscle, proximal tubule, loop of Henle, distal tubule, and collecting duct system.


  • 3.

    The renal artery branches progressively to form the interlobar artery, the arcuate artery, the interlobular artery, and the afferent arteriole, which leads into the glomerular capillaries (i.e., glomerulus). The glomerular capillaries come together to form the efferent arteriole, which leads into a second capillary network, the peritubular capillaries, which supply blood to the nephron. The vessels of the venous system run parallel to the arterial vessels and progressively form the interlobular vein, arcuate vein, interlobar vein, and renal vein, which courses beside the ureter.


  • 4.

    The renal corpuscle is the first part of the nephron and is composed of glomerular capillaries and Bowman’s capsule.


  • 5.

    The glomerular capillary endothelium, basement membrane, and foot processes of podocytes form the so-called filtration barrier. Proteins are filtered primarily based on size.


  • 6.

    Structures that compose the juxtaglomerular apparatus include the macula densa of the thick ascending limb, extraglomerular mesangial cells, and the renin-producing granular cells of the afferent arterioles.


  • 7.

    The juxtaglomerular apparatus is one component of a feedback mechanism (i.e., tubuloglomerular feedback) that regulates renal blood flow and glomerular filtration rate. It also regulates renin secretion by the granular cells of the afferent arteriole.


  • 8.

    The mesangium consists of mesangial cells and the mesangial matrix. Mesangial cells, which possess many properties of smooth muscles cells, surround the glomerular capillaries, provide structural support for the glomerular capillaries, secrete the extracellular matrix, exhibit phagocytic activity that removes macromolecules from the mesangium, and secrete prostaglandins and proinflammatory cytokines. Because they also contract and are adjacent to glomerular capillaries, mesangial cells may influence the glomerular filtration rate (GFR) by regulating blood flow through the glomerular capillaries or by altering the capillary surface area. Mesangial cells located outside the glomerulus (between the afferent and efferent arterioles) are called extraglomerular mesangial cells.


  • 9.

    Renal nerves regulate renal blood flow, glomerular filtration rate, and salt and water reabsorption by the nephron. The nerve supply to the kidneys consists of sympathetic nerve fibers that originate in the celiac plexus. There is no parasympathetic innervation. Adrenergic fibers that innervate the kidneys release norepinephrine. The adrenergic fibers lie adjacent to the smooth muscle cells of the major branches of the renal artery (interlobar, arcuate, and interlobular arteries) and the afferent and efferent arterioles. Moreover, sympathetic nerves innervate the renin-producing granular cells of the afferent arterioles. Renin secretion is stimulated by increased sympathetic activity. Nerve fibers also innervate the proximal tubule, loop of Henle, distal tubule, and collecting duct; activation of these nerves enhances Na + reabsorption by these nephron segments.





Chapter 3




  • 1.

    Before phlorhizin administration:























Serum [inulin] 1 mg/mL
Serum [glucose] 1 mg/mL
Inulin excretion rate 100 mg/min
Glucose excretion rate 0 mg/min
Inulin clearance 100 mL/min
Glucose clearance 0 mL/min



  • After phlorhizin administration:























Serum [inulin] 1 mg/mL
Serum [glucose] 1 mg/mL
Inulin excretion rate 100 mg/min
Glucose excretion rate 100 mg/min
Inulin clearance 100 mL/min
Glucose clearance 100 mL/min



  • Before treatment with phlorhizin, the filtered amount of glucose (glomerular filtration rate [GFR] × serum [glucose]) is 100 mg/min (GFR calculated from inulin clearance). With this filtered amount of glucose all the glucose is reabsorbed and none is excreted. Thus the clearance of glucose is zero. After phlorhizin the filtered amount is unchanged, but there is no glucose reabsorption. Therefore all the glucose that is filtered is excreted, and the clearance of glucose equals that of inulin.


  • 2.



    • a.

      Although the appearance of red cells in the urine can result from damage to the glomerular filtration barrier, red cells can also appear in the urine for other reasons. For example, they can appear because of bleeding in any part of the lower urinary tract. Such bleeding is seen with kidney stones and occasionally during a bacterial infection of the lower urinary tract, which causes bleeding. Thus the appearance of blood in the urine does not necessarily mean the glomerular filtration barrier is damaged.


    • b.

      Because glucose is filtered and completely reabsorbed by the proximal tubule, it is not normally found in the urine. Its presence in the urine indicates an elevated serum glucose level such that the filtered amount (i.e., GFR × serum [glucose]) is greater than the ability of the proximal tubule to reabsorb glucose. Because glucose is freely filtered by the normal glomerulus, damage to the ultrafiltration barrier would not increase its filtration.


    • c.

      In healthy individuals, Na + normally appears in the urine. Like glucose, Na + is freely filtered by the normal glomerulus. Therefore damage to the filtration barrier does not increase the rate of Na + excretion.


    • d.

      This is the correct answer. Normally the urine contains essentially no protein. The glomerulus prevents the filtration of plasma proteins. However, when the glomerulus is damaged, large amounts of plasma proteins are filtered. If the amount filtered overwhelms the resorptive capacity of the proximal tubule, protein appears in the urine (proteinuria).



  • 3.

    The equation for blood flow through an organ is Q = ΔP/R. Sympathetic agonists, angiotensin II, and prostaglandins change blood flow by altering the resistance (R). Whereas sympathetic agonists and angiotensin II increase R and thereby decrease renal blood flow (RBF), prostaglandins decrease R and thereby increase RBF.


  • 4.

    Normally, renal prostaglandin production is low and nonsteroidal antiinflammatory drugs (NSAIDs) do not have an appreciable effect on prostaglandin production. However, during reductions in GFR and RBF, elevated prostaglandin levels cause vasodilation of the afferent and efferent arterioles. This effect prevents excessive decreases in RBF and GFR. Administration of NSAIDs to patients with low GFR and RBF inhibits prostaglandin production and further reduces GFR and RBF.





Chapter 4




  • 1.

    The glomeruli filter 25,200 mEq of Na + and 18,000 mEq of Cl each day (assuming glomerular filtration rate [GFR] = 180 L/day), and 99% is reabsorbed by the nephrons, with less than 1% appearing in the urine. Although Na + and Cl uptake into cells across the apical membrane and NaCl reabsorption across the paracellular pathway are passive processes (i.e., they do not require the direct input of adenosine triphosphate [ATP]), they ultimately depend on the operation of the Na + -K + –adenosine triphosphatase (ATPase). Accordingly, reabsorption of NaCl requires a considerable quantity of ATP, the synthesis by kidney cells of which requires large amounts of oxygen and hence high blood flow.


  • 2.

    “Normal” or “average” urine composition does not actually exist because of the variability in the volume of urine excreted, as well as variations in the intake of solutes in the diet ( Table D.1 ). Urine was collected on three different days from a participant who ate a consistent diet but ingested different amounts of water each day. Although the amount of each solute excreted was similar each day ( Table D.2 ), the concentration of each solute in the urine was different because the volume of urine varied each day. This question demonstrates that the amount (or rate) of a solute excreted, not the concentration of the solute in the urine, is important in the clinical evaluation of urine.



    TABLE D.1

    Urine Flow Rate

    Modified from Valtin HV: Renal function , ed 2, Boston, 1983, Little, Brown.





































































    0.5 L/day 1 L/day 2 L/day
    Na + (mEq/L) 300 150 75
    K + (mEq/L) 200 100 50
    Cl (mEq/L) 300 150 75
    <SPAN role=presentation tabIndex=0 id=MathJax-Element-4-Frame class=MathJax style="POSITION: relative" data-mathml='HCO3−’>HCO3HCO3
    HCO3−
    HCO3
    HCO 3 −
    (mEq/L)
    4 2 1
    Ca ++ (mg/dL) 40 20 10
    <SPAN role=presentation tabIndex=0 id=MathJax-Element-5-Frame class=MathJax style="POSITION: relative" data-mathml='NH4+’>NH+4NH+4
    NH4+
    NH+4
    NH 4 +
    (mEq/L)
    100 50 25
    Creatinine (mg/L) 2000 1000 500
    Glucose (mmol/L) 1.0 0.5 0.25
    Urea (mmol/L) 600 300 150
    Urea (mg/L) 14000 7000 3500
    pH 5.0 to 7.0
    Osmolality (mOsm/kg H 2 O) 1600 800 400


    TABLE D.2








































    Solute Solute Excretion/Day
    Na + (mEq) 150
    K + (mEq) 100
    Cl (mEq) 150
    <SPAN role=presentation tabIndex=0 id=MathJax-Element-6-Frame class=MathJax style="POSITION: relative" data-mathml='HCO3−’>HCO3HCO3
    HCO3−
    HCO3
    HCO 3 −
    (mEq)
    2
    Ca ++ (mg) 20
    <SPAN role=presentation tabIndex=0 id=MathJax-Element-7-Frame class=MathJax style="POSITION: relative" data-mathml='NH4+’>NH+4NH+4
    NH4+
    NH+4
    NH 4 +
    (mEq)
    50
    Creatinine (mg) 1000
    Glucose (mmol) 0.5
    Urea (mmol) 300
    Urea (mg/L) 7000
    Osmolytes (mOsm) 800

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Oct 10, 2019 | Posted by in NEPHROLOGY | Comments Off on Answers to Self-Study Problems
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