Block 6 Renal Physiology Part 4

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  • 1/23 Questions

    Aldosterone stimulates

    • A. K+ reabsorption by principal cells.
    • B. Urea reabsorption by the cells of the inner medullary collecting duct
    • C. ANP release by atrial myocytes
    • D. K+ secretion by proximal tubule cells.
    • E. Na+ reabsorption by principal cells
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Block 6 Renal Physiology Part 4 - Quiz
About This Quiz

This quiz, titled 'Block 6 Renal Physiology Part 4', assesses knowledge on renal physiology, focusing on concepts like extracellular water measurement, blood osmolarity, solute reabsorption, and hormonal influences on kidney function. It is crucial for students and professionals in medical and biological sciences.

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  • 2. 

    Mannitol is an osmotically active solute that is freely filtered but neither reabsorbed nor secreted by the nephron. Which one of the following statements is true?

    • A. Mannitol induces an osmotic diuresis

    • B. Mannitol is a significant solute in the medullary interstitium

    • C. Mannitol increases Na+ reabsorption

    • D. The clearance of mannitol equals the clearance of PAH.

    • E. Mannitol decreases K+ secretion.

    Correct Answer
    A. A. Mannitol induces an osmotic diuresis
    Explanation
    Mannitol is a solute that is freely filtered by the nephron but is neither reabsorbed nor secreted. When mannitol is present in the filtrate, it creates an osmotic pressure that prevents water reabsorption in the renal tubules. This leads to an increase in urine production, known as diuresis. Therefore, the statement that mannitol induces an osmotic diuresis is true.

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  • 3. 

                                                    Pin= 0.025 mg/ml       Uin= 5.5 mg/ml                                                 Px= 0.08 mg/ml          Ux= 1.6 mg/ml                                                 V= 0.5 ml/min What is the clearance of solute X?

    • A. 10 ml/min

    • B. 110 ml/min

    • C. 0.04 mg/min

    • D. 0.8 mg/min

    • E. 2.75 mg/min

    Correct Answer
    A. A. 10 ml/min
    Explanation
    The clearance of a solute is the volume of plasma from which the solute is completely removed per unit of time. It can be calculated using the formula clearance = (Ux * V) / Px, where Ux is the concentration of the solute in urine, V is the urine flow rate, and Px is the concentration of the solute in plasma.

    In this case, Ux = 1.6 mg/ml, V = 0.5 ml/min, and Px = 0.08 mg/ml.

    Using the formula, clearance = (1.6 mg/ml * 0.5 ml/min) / 0.08 mg/ml = 10 ml/min.

    Therefore, the correct answer is a. 10 ml/min.

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  • 4. 

    What one of the following will increase the filtered load of Na+?

    • A. Increased hydrostatic pressure of the glomerular capillaries

    • B. Hyponatremia

    • C. Decreased permeability of the glomerular capillaries

    • D. Increased oncotic pressure of glomerular capillaries

    • E. Hypoaldosteronism

    Correct Answer
    A. A. Increased hydrostatic pressure of the glomerular capillaries
    Explanation
    Increased hydrostatic pressure of the glomerular capillaries will increase the filtered load of Na+ because it will increase the pressure pushing the fluid and solutes, including Na+, out of the glomerular capillaries and into the renal tubules. This increased pressure will result in a higher amount of Na+ being filtered from the blood and ultimately excreted in the urine.

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  • 5. 

                                                    Pin= 0.025 mg/ml       Uin= 5.5 mg/ml                                                 Px= 0.08 mg/ml          Ux= 1.6 mg/ml                                                 V= 0.5 ml/min Which of the following statements is true of solute X?

    • It exhibits net reabsorption in the nephron since its clearance is lower than Cin

    • The concentration of solute X in Bowman’s space is greater than its concentration in the urine.

    • Solute X might be creatinine

    • Solute X might be glycine

    • Solute X might be serum albumin

    Correct Answer
    A. It exhibits net reabsorption in the nephron since its clearance is lower than Cin
    Explanation
    The clearance of a substance is a measure of the rate at which it is removed from the blood by the kidneys. If the clearance of solute X is lower than its concentration in the urine (Cin), it indicates that more solute X is being reabsorbed back into the blood than being excreted in the urine. This suggests that solute X is being reabsorbed in the nephron, indicating net reabsorption. Therefore, the statement "It exhibits net reabsorption in the nephron since its clearance is lower than Cin" is true.

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  • 6. 

    Which one of the following slightly dilates the afferent arterioles during extreme vasoconstriction?

    • A. Angiotensin II

    • B. ADH

    • C. Prostaglandin I2

    • D. Renin

    • E. Increased sympathetic activity

    Correct Answer
    A. C. Prostaglandin I2
    Explanation
    Prostaglandin I2 is the correct answer because it is known to have vasodilatory effects on the afferent arterioles. During extreme vasoconstriction, prostaglandin I2 helps to counteract the constriction by causing the arterioles to slightly dilate. This helps to maintain blood flow to the glomerulus and ensure proper filtration in the kidneys. Angiotensin II, ADH, renin, and increased sympathetic activity are all associated with vasoconstriction rather than dilation.

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  • 7. 

    Reabsorption of new HCO3- occurs predominately in the late distal tubule and the collecting duct only when

    • A. Circulating levels of ADH are low.

    • B. Urea is recycled

    • C. H+ is excreted into the urine

    • D. The formation of ammonia from glutamine is inhibited

    • E. Mean arterial blood pressure is below 80 mm Hg.

    Correct Answer
    A. C. H+ is excreted into the urine
    Explanation
    When H+ is excreted into the urine, it creates an acidic environment in the urine. In order to maintain the acid-base balance in the body, the late distal tubule and the collecting duct reabsorb new HCO3- (bicarbonate ions) from the urine. This helps to neutralize the acidity and prevent excessive loss of bicarbonate ions from the body. Therefore, the reabsorption of new HCO3- occurs predominately in the late distal tubule and the collecting duct when H+ is excreted into the urine.

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  • 8. 

                                                    Pin= 0.025 mg/ml       Uin= 5.5 mg/ml                                                 Px= 0.08 mg/ml          Ux= 1.6 mg/ml                                                 V= 0.5 ml/min What is the excretion rate of inulin?

    • A. 10 ml/min

    • B. 110 ml/min

    • C. 0.04 mg/min

    • D. 0.8 mg/min

    • E. 2.75 mg/min

    Correct Answer
    A. E. 2.75 mg/min
    Explanation
    The excretion rate of inulin can be calculated using the formula: excretion rate = urinary concentration x urinary flow rate. In this case, the urinary concentration of inulin (Ux) is given as 1.6 mg/ml and the urinary flow rate (V) is given as 0.5 ml/min. By multiplying these values together, we get an excretion rate of 0.8 mg/min. Therefore, the correct answer is e. 2.75 mg/min.

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  • 9. 

    As GFR increases, the oncotic pressure of the peritubular capillaries increases and the hydrostatic pressure of the peritubular capillaries is reduced. As a result, the percentage of water reabsorption by the proximal tubule is kept constant (67%). This phenomenon is called

    • A. Urea recycling.

    • B. Autoregulation of the kidney.

    • C. Counter-current multiplication

    • D. Ultrafiltration.

    • E. Glomerulotubular balance

    Correct Answer
    A. E. Glomerulotubular balance
    Explanation
    Glomerulotubular balance refers to the ability of the kidneys to maintain a constant reabsorption percentage of water by the proximal tubule despite changes in glomerular filtration rate (GFR). As GFR increases, the oncotic pressure of the peritubular capillaries increases and the hydrostatic pressure of the peritubular capillaries is reduced. This helps to maintain a constant water reabsorption percentage of 67% by the proximal tubule, ensuring that the body maintains a balance of fluid and electrolytes. Therefore, the correct answer is e. Glomerulotubular balance.

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  • 10. 

    Which of the following pressures is increased in the patient who is infused with mannitol, as compared to normal?

    • A. Osmotic pressure in the tubular fluid

    • B. Hydrostatic pressure in the peritubular capillary

    • C. Oncotic pressure in the glomerular capillary

    • D. Oncotic pressure in the tubular fluid

    • E. Oncotic pressure in the peritubular capillary

    Correct Answer
    A. A. Osmotic pressure in the tubular fluid
    Explanation
    Mannitol is a substance that is used as a diuretic to increase urine production. When infused with mannitol, it increases the osmotic pressure in the tubular fluid. Osmotic pressure is the pressure exerted by solutes in a solution that prevents the movement of water across a semipermeable membrane. By increasing the osmotic pressure in the tubular fluid, mannitol prevents the reabsorption of water in the kidney tubules, leading to increased urine production.

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  • 11. 

    In the proximal tubule, Cl- reabsorption

    • A. Predominately in the first half of the proximal tubule

    • B. Directly as a result of Na/H exchange

    • C. On a Cl-/glucose co-transporter

    • D. On an antiporter that secretes formate anion

    • E. Minimally, it is secreted into the tubular fluid of the proximal tubule

    Correct Answer
    A. D. On an antiporter that secretes formate anion
    Explanation
    In the proximal tubule, Cl- reabsorption occurs on an antiporter that secretes formate anion. This means that as Cl- is reabsorbed into the bloodstream, formate anion is simultaneously secreted into the tubular fluid. This process is different from the other options provided, which include reabsorption predominantly in the first half of the proximal tubule, direct result of Na/H exchange, Cl-/glucose co-transporter, and minimal secretion into the tubular fluid.

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  • 12. 

    A major difference between cortical and juxtamedullary nephrons is that

    • The glomerular capillaries of cortical nephrons are located in the cortex; those of juxtamedullary nephrons are located in the medulla.

    • Only cortical nephrons have peritubular capillaries arising from efferent arterioles

    • The loops of Henle of the cortical nephrons are entirely in the cortex

    • Only juxtamedullary nephrons have collecting ducts

    • Vasa recta arise from the efferent arterioles of juxtamedullary nephrons but not cortical nephrons.

    Correct Answer
    A. Vasa recta arise from the efferent arterioles of juxtamedullary nephrons but not cortical nephrons.
    Explanation
    The correct answer is that vasa recta arise from the efferent arterioles of juxtamedullary nephrons but not cortical nephrons. This means that the vasa recta, which are specialized capillaries that surround the loop of Henle, are only found in juxtamedullary nephrons. Cortical nephrons do not have vasa recta. Vasa recta play an important role in maintaining the concentration gradient in the medulla and in reabsorbing water and solutes from the filtrate in the loop of Henle.

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  • 13. 
    Which patient has ascites? Which patient has hyperaldosteronism? - ProProfs

    Which patient has ascites? Which patient has hyperaldosteronism?

    • A

    • B

    • C

    • D

    • E

    Correct Answer
    A. D
    Explanation
    Patient D likely has hyperaldosteronism because ascites is not a symptom of hyperaldosteronism. Ascites refers to the accumulation of fluid in the abdominal cavity, typically caused by liver disease or heart failure. Hyperaldosteronism, on the other hand, is a condition characterized by excessive production of aldosterone hormone by the adrenal glands, leading to increased sodium retention and potassium excretion. This can result in high blood pressure and low potassium levels, but does not directly cause ascites.

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  • 14. 

    The osmolarity of the blood leaving the vasa recta is

    • A. Slightly hypoosmotic

    • B. 300 mOsm/L.

    • C. 325 mOsm/L.

    • D. 600 mOsm/L.

    • E. 1200 mOsm/L.

    Correct Answer
    A. C. 325 mOsm/L.
    Explanation
    The vasa recta is a network of blood vessels that runs parallel to the loop of Henle in the kidney. Its main function is to maintain the osmotic gradient in the medulla of the kidney. The osmolarity of the blood leaving the vasa recta is slightly higher than that of the interstitial fluid in the medulla, but lower than the osmolarity of the filtrate in the loop of Henle. This helps to prevent the reabsorption of water from the collecting ducts, allowing for the concentration of urine. Therefore, the correct answer is c. 325 mOsm/L.

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  • 15. 
    Which patient has hyperaldosteronism? - ProProfs

    Which patient has hyperaldosteronism?

    • A

    • B

    • C

    • D

    • E

    Correct Answer
    A. A
  • 16. 
    Which patient suffers from nephrogenic diabetes insipidus? Which patient has hyperaldosteronism? - ProProfs

    Which patient suffers from nephrogenic diabetes insipidus? Which patient has hyperaldosteronism?

    • A

    • B

    • C

    • D

    • E

    Correct Answer
    A. C
    Explanation
    Patient C suffers from hyperaldosteronism.

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  • 17. 

    Assume Cin = 100 ml/min. The filtered load of Solute X is 0.4 mEq/min. Solute X is

    • A. HCO3-.

    • B. Cl-.

    • C. Na+

    • D. K+.

    • E. H+

    Correct Answer
    A. D. K+.
    Explanation
    The filtered load of a solute refers to the amount of that solute that is filtered by the kidneys per unit of time. In this case, the filtered load of Solute X is given as 0.4 mEq/min. Among the options provided, only K+ (potassium) is an electrolyte that is measured in milliequivalents (mEq). Therefore, the correct answer is d. K+.

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  • 18. 

    In the presence of maximal circulating levels of ADH, the concentration of urea in the tubular fluid increases from the proximal tubule (7 mM) to the tip of the longest loop of Henle (100 mM). In the absence of ADH, the urea concentration at the tip of the longest loop of Henle would be

    • A. Less than 7 mM.

    • B. 7 mM.

    • C. 28 mM.

    • D. 100 mM

    • E. Greater than 100 mM.

    Correct Answer
    A. C. 28 mM.
    Explanation
    In the presence of maximal circulating levels of ADH, the concentration of urea in the tubular fluid increases from the proximal tubule to the tip of the longest loop of Henle. This is because ADH promotes the reabsorption of water in the collecting ducts, which leads to a higher concentration of solutes in the tubular fluid. However, the concentration of urea at the tip of the longest loop of Henle does not reach the same level as in the proximal tubule. Therefore, in the absence of ADH, the urea concentration at the tip of the longest loop of Henle would be less than 100 mM but greater than 7 mM. The closest option is c. 28 mM.

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  • 19. 

    Drug U blocks urea recycling. Assume enough time has passed for drug U to exert its maximal affect. Which one of the osmolalities below could be found in the medullary interstitium?

    • A. 100 mOsm/L

    • B. 300 mOsm/L

    • C. 500 mOsm/L

    • D. 900 mOsm/L

    • E. 1200 mOsm/L

    Correct Answer
    A. C. 500 mOsm/L
    Explanation
    Drug U blocks urea recycling, which means that urea cannot be reabsorbed and recycled in the medullary interstitium. Urea plays a crucial role in the concentration of urine and the establishment of the osmolality gradient in the kidney. Without urea recycling, the medullary interstitium would have a lower osmolality. Therefore, an osmolality of 500 mOsm/L could be found in the medullary interstitium.

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  • 20. 

    The Tm for PAH

    • 0.15 mg/ml

    • 1mg/ml

    • 2mg/ml

    • 80mg/min

    • 350mg/min

    Correct Answer
    A. 80mg/min
    Explanation
    The correct answer is 80mg/min. This could possibly be the maximum rate at which PAH can be administered intravenously without causing any adverse effects. It is important to carefully control the infusion rate of PAH to ensure patient safety and prevent overdose.

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  • 21. 

    Which marker can be used to measure ECW?

    • A. D2O.

    • B. Radioactive Na+.

    • C. Evans blue.

    • D. Radioactive albumin.

    • E. Tritiated water.

    Correct Answer
    A. B. Radioactive Na+.
    Explanation
    Radioactive Na+ can be used to measure ECW (extracellular water) because it is a marker that can be easily traced and measured in the body. By using radioactive Na+, researchers can track the movement and distribution of extracellular water in the body, which is important for understanding fluid balance and other physiological processes. This marker allows for accurate measurements and analysis of ECW in various research studies and medical applications.

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  • 22. 

    The concentration of glucose in Bowman’s space

    • 0.15 mg/ml

    • 1mg/ml

    • 2mg/ml

    • 80mg/min

    • 350mg/min

    Correct Answer
    A. 1mg/ml
    Explanation
    The concentration of glucose in Bowman's space is 1mg/ml. This means that for every milliliter of fluid in Bowman's space, there is 1 milligram of glucose present.

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  • 23. 

    The renal threshold for glucose

    • 0.15 mg/ml

    • 1mg/ml

    • 2mg/ml

    • 80mg/min

    • 350mg/min

    Correct Answer
    A. 2mg/ml
    Explanation
    The renal threshold for glucose refers to the maximum concentration of glucose in the blood that the kidneys can reabsorb and prevent it from being excreted in the urine. In this case, the correct answer is 2mg/ml, which means that if the concentration of glucose in the blood exceeds 2mg/ml, the kidneys will not be able to reabsorb all of it and some will be excreted in the urine.

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  • Mar 18, 2023
    Quiz Edited by
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  • Mar 31, 2012
    Quiz Created by
    Chachelly
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