Block 6 Renal Physiology BRS W Expl Prt 1
The Block 6 Renal Physiology Brs W Expl Prt 1 below is the first in the revision quizzes that is designed to ensure you understood all the aspects on the renal system that we covered these last few weeks. Give it a try and keep an eye out for part two of the same.
- 1.
Secretion of 1{ ± 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
Correct Answer
D. Spironolactone administrationExplanation
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.Rate this question:
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- 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 (1CF) volume
- B.
Higher positive free-water clearance (CH2O)
- C.
Greater change in plasma osmolarity
- D.
Higher urine osmolarity
- E.
Higher urine flow rate
Correct Answer
D. Higher urine osmolarityExplanation
After drinking distilled water, subject A 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 (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 C H2O than subject A.Rate this question:
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- 3.
A woman with a history of severe diarrhea 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
Correct Answer
A. Metabolic acidosisExplanation
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 HCO 3--, creating a metabolic acidosis.Rate this question:
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- 4.
A woman with a history of severe diarrhea 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 [HCO 3-] is a result of buffering of excess H + by HCO3-.
- C.
The decreased blood [K+] is a result of exchangeof 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).
Correct Answer
D. The decreased blood [K+] is a result of increased circulating levels of aldosterone.Explanation
The decreased arterial [1-1CO 3-] is caused by gastrointestinal (GI) loss of HCO 3- 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.Rate this question:
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- 5.
Use the values below to answer the following question. Glomerular capillary = 47 mm Hg hydrostatic pressure Bowman's space = 10 mm Hg hydrostatic pressure Bowman's space = 0 mm Hg oncotic pressure 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
Correct Answer
C. 37 mm HgExplanation
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).Rate this question:
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- 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
Correct Answer
D. Is inhibited by decreases in arterial PcO2Explanation
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 HCO 3- 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+.Rate this question:
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- 7.
The following information was obtained in a human subject: Plasma Urine [Inulin] = 1 mg/ml [Inulin] = 150 mg/ml [X] = 2 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.
Correct Answer
B. There is net reabsorption of X.Explanation
To answer this question, calculate the glomerular filtration rate (GFR) and C. GFR = 150 mg/ml x 1 ml/min / 1 mg/m1= 150 ml/min. C. = 100 mg/ml x 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.Rate this question:
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- 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±
Correct Answer
A. Fixed acid production plus fixed acid ingestionExplanation
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 NH 4+ to maintain acid-base balance.Rate this question:
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- 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. The subject'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
Correct Answer
C. ECF volume is 10 LExplanation
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.Rate this question:
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- 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 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 will occur in a person who
Correct Answer
D. Excretion rate of glucose increases with increasing plasma glucose concentrationsExplanation
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 T m, 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.Rate this question:
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- 11.
A negative free-water clearance (-CH20)
- 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 1000 mOsm/L
Correct Answer
D. Has an oat cell carcinoma of the lung, and excretes urine with an osmolarity of 1000 mOsm/LExplanation
A person who produces hyperosmotic urine (1000 mOsra/L) will have a negative free-water clearance (-C H20) [ CH2O = V - Cosm]. All of the others will have a positive CH20 because they are producing hyposmofic urine as a result of the suppression of antidiuretic hormone (ADH) by water drinking, central diabetes insipidus, or nephrogenic diabetes insipidus.Rate this question:
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- 12.
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
Correct Answer
A. 1/100 that of A-Explanation
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
2.0 = log A-/HA
100 = A-/HA or HA/A- is 1/100Rate this question:
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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 expansion
- C.
Decreased peritubular capillary protein concentration
- D.
Increased peritubular capillary hydrostatic pressure
- E.
Oxygen deprivation
Correct Answer
A. Increased filtration fractionExplanation
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.Rate this question:
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- 14.
Which of the following substances or combinations of substances could be used to measure interstitial fluid volume?
- A.
Mannitol
- B.
D20 alone
- C.
Evans blue
- D.
Inulin and D20
- E.
Inulin and radioactive albumin
Correct Answer
E. Inulin and radioactive albuminExplanation
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.Rate this question:
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- 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
Correct Answer
D. Concentration in the renal vein is close to zeroExplanation
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.Rate this question:
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- 16.
Compared with a person who ingests 2 L of distilled water, a person with water deprivation will have a
- A.
Higher free-water clearance (CH20)
- 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 H20 reabsorption in the collecting ducts
Correct Answer
E. Higher rate of H20 reabsorption in the collecting ductsExplanation
The person with water deprivation will have a higher plasma osmolarity and higher circulating levels of antidiuretic hormone (ADH). These effects will increase the rate of H20 reabsorption in the collecting ducts and create a negative freewater clearance (- CH20). Tubular fluid/plasma (TF/P) osmolarity in the proximal tubule is not affected by ADH.Rate this question:
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- 17.
Which of the following would cause an increase in glomerular filtration rate (GFR) & 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
Correct Answer
C. Dilation of the afferent arterioleExplanation
Dilation of the afferent arteriole will increase both renal plasma flow (RPF) [because renal vascular resistance is decreased] and glomerular filtration rate (GFR) [because glomerular capillary hydrostatic pressure is increased]. Dilation of the efferent arteriole will increase RPF, but decrease GFR. Constriction of the efferent arteriole will decrease RPF (due to increased renal vascular resistance) and increase GFR. Both hyperproteinemia (inc in the glomerular capillaries) and a ureteral stone (inc hydrostatic pressure in Bowman's space) will oppose filtration and decrease GFR.Rate this question:
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Mar 21, 2023Quiz Edited by
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Mar 01, 2012Quiz Created by
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