BRS Physiology
29. A man presents with hypertension and hypokalemia. Measurement of his arterial blood gases reveals a pH of 7.5 and a calculated HCO3- of 32 mEq/L. His serum cortisol and urinary vanillylmandelic acid (VMA) are normal, his serum aldosterone is increased, and his plasma renin activity is decreased. Which of the following is the most likely cause of his hypertension? (A) Cushing's syndrome (B) Cushing's disease (C) Conn's syndrome (D) Renal artery stenosis (E) Pheochromocytoma
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Which of the following substances has the highest renal clearance? A) Para-aminohippuric acid (PAH) (B) Inulin (C) Glucose (D) Na+ (E) Cl-
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A 45-year-old woman develops severe diar- rhea while on vacation. She has the following arterial blood values: pH = 7.25 PcO2 = 24 mm Hg [HCO3-] = 10 mEq/L Venous blood samples show decreased blood [K+] and a normal anion gap. 3. The correct diagnosis for this patient is (A) metabolic acidosis (B) metabolic alkalosis (C) respiratory acidosis (D) respiratory alkalosis (E) normal acid-base status
The answer is A [IXD1a-c;Tables5-8and5-9].AnacidpH,togetherwithdecreased HCO3- and decreased PCO2, is consistent with metabolic acidosis with respiratory com- pensation (hyperventilation). Diarrhea causes gastrointestinal (GI) loss of HCO3-, creating a metabolic acidosis.
16. Compared with a person who ingests 2 L of distilled water, a person with water depri- vation will have a (A) higher free-water clearance (CH2O) (B) lower plasma osmolarity (C) lower circulating level of antidiuretic hormone (ADH) (D) higher tubular fluid/plasma (TF/P) osmolarity in the proximal tubule (E) higher rate of H2O reabsorption in the collecting ducts
TheanswerisE[VIID;Figures5-14and5-15].Thepersonwithwaterdeprivationwill have a higher plasma osmolarity and higher circulating levels of antidiuretic hormone (ADH). These effects will increase the rate of H2O reabsorption in the collecting ducts and create a negative free-water clearance (-CH2O). Tubular fluid/plasma (TF/P) osmolarity in the proximal tubule is not affected by ADH.
17. Which of the following would cause an increase in both glomerular filtration rate (GFR) and renal plasma flow (RPF)? (A) Hyperproteinemia (B) A ureteral stone (C) Dilation of the afferent arteriole (D) Dilation of the efferent arteriole (E) Constriction of the efferent arteriole
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18. A patient has the following arterial blood values: pH = 7.52 PCO2 = 20 mm Hg [HCO3-] = 16 mEq/L Which of the following statements about this patient is most likely to be correct? (A) He is hypoventilating (B) He has decreased ionized [Ca2+] in blood (C) He has almost complete respiratory compensation (D) He has an acid-base disorder caused by overproduction of fixed acid (E) Appropriate renal compensation would cause his arterial [HCO3-] to increase
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19. Which of the following would best dis- tinguish an otherwise healthy person with severe water deprivation from a person with the syndrome of inappropriate antidiuretic hormone (SIADH)? (A) Free-water clearance (CH2O) (B) Urine osmolarity (C) Plasma osmolarity (D) Circulating levels of antidiuretic hormone (ADH) (E) Corticopapillary osmotic gradient
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20. Which of the following causes a decrease in renal Ca2+ clearance? (A) Hypoparathyroidism (B) Treatment with chlorothiazide (C) Treatment with furosemide (D) Extracellular fluid (ECF) volume expansion (E) Hypermagnesemia
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21. A patient arrives at the emergency room with low arterial pressure, reduced tissue tur- gor, and the following arterial blood values: pH = 7.69 [HCO3-] = 57 mEq/L PCO2 = 48 mm Hg Which of the following responses would also be expected to occur in this patient? (A) Hyperventilation (B) Decreased K+ secretion by the distal tubules (C) Increased ratio of H2PO4- to HPO4-2 in urine (D) Exchange of intracellular H+ for extra- cellular K+
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22. A woman has a plasma osmolarity of 300 mOsm/L and a urine osmolarity of 1200 mOsm/L. The correct diagnosis is (A) syndrome of inappropriate antidiuretic hormone (SIADH) (B) water deprivation (C) central diabetes insipidus (D) nephrogenic diabetes insipidus (E) drinking large volumes of distilled water
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23. A patient is infused with para- aminohippuric acid (PAH) to measure renal blood flow (RBF). She has a urine flow rate of 1 mL/min, a plasma [PAH] of 1 mg/mL, a urine [PAH] of 600 mg/mL, and a hematocrit of 45%. What is her "effective" RBF? A. 600 mL/min B. 660 mL/min C. 1091 mL/min D. 1333 mL/min
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25. A woman runs a marathon in 90°F weather and replaces all volume lost in sweat by drinking distilled water. After the marathon, she will have (A) decreased total body water (TBW) (B) decreased hematocrit (C) decreased intracellular fluid (ICF) volume (D) decreased plasma osmolarity (E) increased intracellular osmolarity
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26. Which of the following causes hyper- kalemia? (A) Exercise (B) Alkalosis (C) Insulin injection (D) Decreased serum osmolarity (E) Treatment with β-agonists
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27. Which of the following is a cause of metabolic alkalosis? (A) Diarrhea (B) Chronic renal failure (C) Ethylene glycol ingestion (D) Treatment with acetazolamide (E) Hyperaldosteronism (F) Salicylate poisoning
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28. Which of the following is an action of parathyroid hormone (PTH) on the renal tubule? (A) Stimulation of adenylate cyclase (B) Inhibition of distal tubule K+ secretion (C) Inhibition of distal tubule Ca2+ reab- sorption (D) Stimulation of proximal tubular phos- phate reabsorption (E) Inhibition of production of 1,25-dihydroxycholecalciferol
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13. Which of the following would produce an increase in the reabsorption of isosmotic fluid in the proximal tubule? (A) Increased filtration fraction (B) Extracellular fluid (ECF) volume expan- sion (C) Decreased peritubular capillary protein concentration (D) Increased peritubular capillary hydro- static pressure (E) Oxygen deprivation
The answer is A[IIC3;IVC1d(2)]. Increasing filtration fraction means that a larger portion of the renal plasma flow (RPF) is filtered across the glomerular capillaries. This increased flow causes an increase in the protein concentration and oncotic pressure of the blood leaving the glomerular capillaries. This blood becomes the peritubular capillary blood supply. The increased oncotic pressure in the peritubular capillary blood is a driving force favoring reabsorption in the proximal tubule. Extracellular fluid (ECF) volume expan- sion, decreased peritubular capillary protein concentration, and increased peritubular capillary hydrostatic pressure all inhibit proximal reabsorption. Oxygen deprivation would also inhibit reabsorption by stopping the Na+-K+ pump in the basolateral membranes.
2. A buffer pair (HA/A-) has a pK of 5.4. At a blood pH of 7.4, the concentration of HA is (A) 1/100 that of A- (B) 1/10 that of A- (C) equal to that of A- (D) 10 times that of A- (E) 100 times that of A-
The answer is A[IXB3].TheHenderson-Hasselbalch equation can be used to calculate the ratio of HA/A-: pH=pK+logA− HA 7.4=5.4+logA− HA 2.0=logA− HA 100=A− HAorHAA−is1100
5. Use the values below to answer the fol- lowing question. Glomerular capillary hydrostatic pressure = 47mmHg Bowman's space hydrostatic pressure = 10mmHg Bowman's space oncotic pressure = 0 mm Hg At what value of glomerular capillary oncotic pressure would glomerular filtration stop? (A) 57 mm Hg (B) 47 mm Hg (C) 37 mm Hg (D) 10 mm Hg (E) 0 mm Hg
The answer is C [IIC4,5].Glomerular filtration will stop when the net ultrafiltration pressure across the glomerular capillary is zero; that is, when the force that favors filtration (47 mm Hg) exactly equals the forces that oppose filtration (10 mm Hg + 37 mm Hg).
A 45-year-old woman develops severe diar- rhea while on vacation. She has the following arterial blood values: pH = 7.25 PcO2 = 24 mm Hg [HCO3-] = 10 mEq/L Venous blood samples show decreased blood [K+] and a normal anion gap. Which of the following statements about this patient is correct? (A) She is hypoventilating (B) The decreased arterial [HCO3-] is a result of buffering of excess H+ by HCO3- (C) The decreased blood [K+] is a result of exchange of intracellular H+ for extra- cellular K+ (D) The decreased blood [K+] is a result of increased circulating levels of aldosterone (E) The decreased blood [K+] is a result of decreased circulating levels of antidiuretic hormone (ADH)
The answer is D [IXD1a-c;Tables5-8and5-9].The decreased arterial [HCO3-] is caused by gastrointestinal (GI) loss of HCO3- from diarrhea, not by buffering of excess H+ by HCO3-. The woman is hyperventilating as respiratory compensation for metabolic acido- sis. Her hypokalemia cannot be the result of the exchange of intracellular H+ for extracel- lular K+, because she has an increase in extracellular H+, which would drive the exchange in the other direction. Her circulating levels of aldosterone would be increased as a result of extracellular fluid (ECF) volume contraction, which leads to increased K+ secretion by the distal tubule and hypokalemia.
1. Secretion of K+ by the distal tubule will be decreased by (A) metabolic alkalosis (B) a high-K+ diet (C) hyperaldosteronism (D) spironolactone administration (E) thiazide diuretic administration
The answer is D [VB4b].Distal K+ secretion is decreased by factors that decrease the driving force for passive diffusion of K+ across the luminal membrane. Because spirono- lactone is an aldosterone antagonist, it reduces K+ secretion. Alkalosis, a diet high in K+, and hyperaldosteronism all increase [K+] in the distal cells and thereby increase K+ secre- tion. Thiazide diuretics increase flow through the distal tubule and dilute the luminal [K+] so that the driving force for K+ secretion is increased.
8. To maintain normal H+ balance, total daily excretion of H+ should equal the daily (A) fixed acid production plus fixed acid ingestion (B) HCO3- excretion (C) HCO3- filtered load (D) titratable acid excretion (E) filtered load of H+
TheanswerisA[IXC2].TotaldailyproductionoffixedH+fromcatabolismofproteins and phospholipids (plus any additional fixed H+ that is ingested) must be matched by the sum of excretion of H+ as titratable acid plus NH4+ to maintain acid-base balance.
7. The following information was obtained in a 20-year-old college student who was participating in a research study in the Clinical Research Unit: Plasma [Inulin] = 1 mg/mL [X] = 2 mg/mL Urine [Inulin] = 150 mg/mL [X] = 100 mg/mL Urine flow rate = 1 mL/min Assuming that X is freely filtered, which of the following statements is most correct? (A) There is net secretion of X (B) There is net reabsorption of X (C) There is both reabsorption and secre- tion of X (D) The clearance of X could be used to measure the glomerular filtration rate (GFR) (E) The clearance of X is greater than the clearance of inulin
TheanswerisB[IIC1].Toanswerthisquestion,calculatetheglomerularfiltrationrate (GFR)andCx.GFR=150mg/mL×1mL/min÷1mg/mL=150mL/min.Cx =100mg/mL× 1 mL/min ÷ 2 mg/mL = 50 mL/min. Because the clearance of X is less than the clearance of inulin (or GFR), net reabsorption of X must have occurred. Clearance data alone cannot determine whether there has also been secretion of X. Because GFR cannot be measured with a substance that is reabsorbed, X would not be suitable.
9. One gram of mannitol was injected into a woman. After equilibration, a plasma sample had a mannitol concentration of 0.08 g/L. During the equilibration period, 20% of the injected mannitol was excreted in the urine. (D) (E) The (A) (B) (C) (D) (E) subject's extracellular fluid (ECF) volume is 1 L intracellular fluid (ICF) volume is 1 L ECF volume is 10 L ICF volume is 10 L interstitial volume is 12.5 L
TheanswerisC[IB1a].Mannitolisamarkersubstancefortheextracellular fluid (ECF) volume. ECF volume = amount of mannitol/concentration of mannitol = 1 g - 0.2 g/0.08 g/L = 10 L.
2. Subjects A and B are 70-kg men. Subject A drinks 2 L of distilled water, and subject B drinks 2 L of isotonic NaCl. As a result of these ingestions, subject B will have a (A) greater change in intracellular fluid (ICF) volume (B) higher positive free-water clearance (CH2O) (C) greater change in plasma osmolarity (D) higher urine osmolarity (E) higher urine flow rate
TheanswerisD[IC2a;VIIC;Figure5-15;Table5-6].Afterdrinkingdistilledwater,sub- ject A will have an increase in intracellular fluid (ICF) and extracellular fluid (ECF) vol- umes, a decrease in plasma osmolarity, a suppression of antidiuretic hormone (ADH) secretion, and a positive free-water clearance (CH2O), and will produce dilute urine with a high flow rate. Subject B, after drinking the same volume of isotonic NaCl, will have an increase in ECF volume only and no change in plasma osmolarity. Because subject B's ADH will not be suppressed, he will have a higher urine osmolarity, a lower urine flow rate, and a lower CH2O than subject A.
10. At plasma concentrations of glucose higher than occur at transport maximum (Tm), the (A) clearance of glucose is zero (B) excretion rate of glucose equals the fil- tration rate of glucose (C) reabsorption rate of glucose equals the filtration rate of glucose (D) excretion rate of glucose increases with increasing plasma glucose concentra- tions (E) renal vein glucose concentration equals the renal artery glucose concentration
TheanswerisD[IIIB;Figure5-5].Atconcentrationsgreaterthanatthetransportmaxi- mum (Tm) for glucose, the carriers are saturated so that the reabsorption rate no longer matches the filtration rate. The difference is excreted in the urine. As the plasma glucose concentration increases, the excretion of glucose increases. When it is greater than the Tm, the renal vein glucose concentration will be less than the renal artery concentration because some glucose is being excreted in urine and therefore is not returned to the blood. The clearance of glucose is zero at concentrations lower than at Tm (or lower than threshold) when all of the filtered glucose is reabsorbed, but is greater than zero at con- centrations greater than Tm.
15. At plasma para-aminohippuric acid (PAH) concentrations below the transport maximum (Tm), PAH (A) reabsorption is not saturated (B) clearance equals inulin clearance (C) secretion rate equals PAH excretion rate (D) concentration in the renal vein is close to zero (E) concentration in the renal vein equals PAH concentration in the renal artery
TheanswerisD[IIIC;Figure5-6].Atplasmaconcentrationsthatarelowerthanatthe transport maximum (Tm) for para-aminohippuric acid (PAH) secretion, PAH concentra- tion in the renal vein is nearly zero because the sum of filtration plus secretion removes virtually all PAH from the renal plasma. Thus, the PAH concentration in the renal vein is less than that in the renal artery because most of the PAH entering the kidney is excreted in urine. PAH clearance is greater than inulin clearance because PAH is filtered and secreted; inulin is only filtered.
6. The reabsorption of filtered HCO3- (A) results in reabsorption of less than 50% of the filtered load when the plasma concentration of HCO3- is 24 mEq/L (B) acidifies tubular fluid to a pH of 4.4 (C) is directly linked to excretion of H+ as NH4+ (D) is inhibited by decreases in arterial PcO2 (E) can proceed normally in the presence of a renal carbonic anhydrase inhibitor
TheanswerisD[IXC1a-b].DecreasesinarterialPCO2causeadecreaseinthereabsorp- tion of filtered HCO3- by diminishing the supply of H+ in the cell for secretion into the lumen. Reabsorption of filtered HCO3- is nearly 100% of the filtered load and requires carbonic anhydrase in the brush border to convert filtered HCO3- to CO2 to proceed nor- mally. This process causes little acidification of the urine and is not linked to net excre- tion of H+ as titratable acid or NH4+.
11. A negative free-water clearance (-CH2O) will occur in a person who (A) drinks 2 L of distilled water in 30 min- utes (B) begins excreting large volumes of urine with an osmolarity of 100 mOsm/L after a severe head injury (C) is receiving lithium treatment for depression, and has polyuria that is unresponsive to the administration of antidiuretic hormone (ADH) (D) has an oat cell carcinoma of the lung, and excretes urine with an osmolarity of 1000 mOsm/L
TheanswerisD[VIID;Table5-6].Apersonwhoproduceshyperosmoticurine(1000 mOsm/L) will have a negative free-water clearance (-CH2O) [CH2O = V - Cosm]. All of the others will have a positive CH2O because they are producing hyposmotic urine as a result of the suppression of antidiuretic hormone (ADH) by water drinking, central diabetes insipidus, or nephrogenic diabetes insipidus.
14. Which of the following substances or combinations of substances could be used to measure interstitial fluid volume? (A) Mannitol (B) D2O alone (C) Evans blue (D) Inulin and D2O (E) Inulin and radioactive albumin
TheanswerisE[IB2b-d].Interstitialfluidvolumeismeasuredindirectlybydetermining the difference between extracellular fluid (ECF) volume and plasma volume. Inulin, a large fructose polymer that is restricted to the extracellular space, is a marker for ECF vol- ume. Radioactive albumin is a marker for plasma volume.