Renal and Acid-Base Physiology Flashcards
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 [V B 4 b].
Distal K+ secretion is decreased by factors that decrease the driving force for passive diffusion of K+ across the luminal membrane. Because spironolactone 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+ secretion. Thiazide diuretics increase flow through the distal tubule and dilute the luminal [K+] so that the driving force for K+ secretion is increased.
Jared and Adam both weigh 70 kg. Jared drinks 2 L of distilled water, and Adam drinks 2 L of isotonic NaCl. As a result of these ingestions, Adam will have a
(A) greater change in intracellular fluid (ICF) volume
(B) higher positive free-water clearance (CHO2)
(C) greater change in plasma osmolarity
(D) higher urine osmolarity
(E) higher urine flow rate
The answer is D [I C 2 a; VII C; Figure 5.15; Table 5.6].
After drinking distilled water, Jared will have an increase in intracellular fluid (ICF) and extracellular fluid (ECF) volumes, a decrease in plasma osmolarity, a suppression of antidiuretic hormone (ADH) secretion, and a positive free-water clearance (CHO2), and will produce dilute urine with a high flow rate. Adam, after drinking the same volume of isotonic NaCl, will have an increase in ECF volume only and no change in plasma osmolarity. Because Adam’s ADH will not be suppressed, he will have a higher urine osmolarity, a lower urine flow rate, and a lower CH O 2 than Jared.
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.
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 [IX D 1 a–c; Tables 5.8 and 5.9].
An acid pH, together with decreased HCO3- and decreased Pco2, is consistent with metabolic acidosis with respiratory compensation (hyperventilation). Diarrhea causes gastrointestinal (GI) loss of HCO3 -, creating a metabolic acidosis.
A 45-year-old woman develops severe diarrhea 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 extracellular 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 [IX D 1 a–c; Tables 5.8 and 5.9].
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 acidosis. Her hypokalemia cannot be the result of the exchange of intracellular H+ for extracellular 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.
Use the values below to answer the following question.
Glomerular capillary hydrostatic pressure = 47 mmHg
Bowman space hydrostatic pressure = 10 mmHg
Bowman space oncotic pressure = 0 mmHg
At what value of glomerular capillary oncotic pressure would glomerular filtration stop?
(A) 57 mmHg (B) 47 mmHg (C) 37 mm Hg (D) 10 mm Hg (E) 0 mm Hg
The answer is C [II C 4, 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 mmHg) exactly equals the forces that oppose filtration (10 mmHg + 37 mmHg).
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
The answer is D [IX C 1 a, b].
Decreases in arterial Pco2 cause a decrease in the reabsorption 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 normally. This process causes little acidification of the urine and is not linked to net excretion of H+ as titratable acid or NH4+.
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 secretion 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
The answer is B [II C 1].
To answer this question, calculate the glomerular filtration rate (GFR) and CX. GFR = 150 mg/mL × 1 mL/min ÷ 1 mg/mL = 150 mL/min. CX = 100 mg/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.
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+
The answer is A [IX C 2].
Total daily production of fixed H+ from catabolism of proteins 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.
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. The woman’s
(A) extracellular fluid (ECF) volume is 1 L
(B) intracellular fluid (ICF) volume is 1 L
(C) ECF volume is 10 L
(D) ICF volume is 10 L
(E) interstitial volume is 12.5 L
The answer is C [I B 1 a].
Mannitol is a marker substance for the extracellular fluid (ECF) volume. ECF volume = amount of mannitol/concentration of mannitol = 1 g – 0.2 g/(0.08 g/L) = 10 L.
A 58-year-old man is given a glucose tolerance test. In the test, the plasma glucose concentration is increased and glucose reabsorption and excretion are measured. When the plasma glucose concentration is higher than occurs at transport maximum (Tm), the
(A) clearance of glucose is zero
(B) excretion rate of glucose equals the filtration rate of glucose
(C) reabsorption rate of glucose equals the filtration rate of glucose
(D) excretion rate of glucose increases with increasing plasma glucose concentrations
(E) renal vein glucose concentration equals the renal artery glucose concentration
The answer is D [III B; Figure 5.5].
At concentrations greater than at the transport maximum (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 concentrations greater than Tm.
A negative free-water clearance (-CH2O) will occur in a person who
(A) drinks 2 L of distilled water in 30 minutes
(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 1,000 mOsm/L
The answer is D [VII D; Table 5.6].
A person who produces hyperosmotic urine (1,000 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.
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/l00 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 [IX B 3].
The Henderson-Hasselbalch equation can be used to calculate the ratio of HA/A-:
pH = pK + log A-/HA 7.4 = 5.4 + log A-/HA 7.4 - 5.4 = log A-/HA 2 = log A-/HA 100 = A-/HA or HA/A- is 1/100
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 expansion
(C) Decreased peritubular capillary protein concentration
(D) Increased peritubular capillary hydrostatic pressure
(E) Oxygen deprivation
The answer is A [II C 3; IV C 1 d (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 expansion, 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.
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
The answer is E [I B 2 b–d].
Interstitial fluid volume is measured indirectly by determining 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 volume. Radioactive albumin is a marker for plasma volume.
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
The answer is D [III C; Figure 5.6].
At plasma concentrations that are lower than at the transport maximum (Tm) for para-aminohippuric acid (PAH) secretion, PAH concentration 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.
