OS II Exam 1 Physiology Questions - Pt.2

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  • 1/27 Questions

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

    • 27. Reabsorption of new HCO3- occurs predominately in the late distal tubule and the collecting duct only when
    • 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.
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About This Quiz


The study of physiology gives us a chance to understand the normal functioning of a living organism’s body. As we continue in the revision process for the OS II exam 1, I have prepared part two of the physiology practice tests. Why don’t you give it a try and see which topics might be problematic for you? All the best!

OS II Exam 1 Physiology Questions - Pt.2 - Quiz

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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 an osmotically active solute that is freely filtered by the nephron but is neither reabsorbed nor secreted. This means that it remains in the tubules and creates an osmotic force that prevents water reabsorption. As a result, more water is excreted in the urine, leading to increased urine output. This is known as osmotic diuresis. Therefore, option a is the correct statement.

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  • 3. 

    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

    Correct Answer
    A. E. Na+ reabsorption by principal cells
    Explanation
    Aldosterone is a hormone that plays a key role in regulating the balance of electrolytes in the body. It acts on the principal cells in the renal tubules, specifically in the distal convoluted tubules and collecting ducts. The correct answer, option e, states that aldosterone stimulates Na+ reabsorption by principal cells. This means that aldosterone promotes the reabsorption of sodium ions from the tubules back into the bloodstream, helping to maintain the body's sodium balance and blood pressure.

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  • 4. 

    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 percentage of water reabsorption 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 percentage (67%) of water reabsorption by the proximal tubule, ensuring that the body maintains proper fluid balance. Therefore, the correct answer is e. Glomerulotubular balance.

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  • 5. 

    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 urea in the tubular fluid. Therefore, in the absence of ADH, the urea concentration at the tip of the longest loop of Henle would be less than the concentration in the proximal tubule, which is 7 mM. Therefore, the correct answer is c. 28 mM, as it is the only option that is less than 100 mM but greater than 7 mM.

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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. During extreme vasoconstriction, the afferent arterioles, which supply blood to the glomerulus in the kidneys, may become constricted. Prostaglandin I2 acts as a local vasodilator in the kidneys, helping to counteract the vasoconstriction and maintain blood flow. This is important for maintaining proper kidney function and preventing damage to the glomerulus. The other options, such as Angiotensin II, ADH, Renin, and Increased sympathetic activity, are more commonly associated with vasoconstriction rather than dilation.

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  • 7. 

    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 commonly used as an osmotic diuretic. When infused into a patient, mannitol increases the osmotic pressure in the tubular fluid. This increased osmotic pressure causes water to be drawn from the surrounding tissues and into the tubular fluid, resulting in increased urine production and diuresis. Therefore, the correct answer is a. Osmotic pressure in the tubular fluid.

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  • 8. 

    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 Cl- ions are transported out of the tubular fluid and into the cells of the proximal tubule in exchange for formate anions. This antiporter mechanism helps to maintain the balance of ions and regulate the concentration of Cl- in the body.

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  • 9. 

    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 is not being reabsorbed in the medullary collecting ducts. This leads to increased urea concentration in the tubular fluid, which in turn increases the osmolality of the medullary interstitium. Therefore, a higher osmolality, such as 500 mOsm/L, could be found in the medullary interstitium when urea recycling is blocked.

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  • 10. 

    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+. This is because the hydrostatic pressure is the force that pushes fluid and solutes out of the glomerular capillaries and into the renal tubules for filtration. When the hydrostatic pressure increases, more Na+ will be pushed out of the capillaries and into the tubules, leading to an increased filtered load of Na+.

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  • 11. 

    A major difference between cortical and juxtamedullary nephrons is that

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

    • B. Only cortical nephrons have peritubular capillaries arising from efferent arterioles

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

    • D. Only juxtamedullary nephrons have collecting ducts

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

    Correct Answer
    A. E. Vasa recta arise from the efferent arterioles of juxtamedullary nephrons but not cortical nephrons.
    Explanation
    The major difference between cortical and juxtamedullary nephrons is that vasa recta arise from the efferent arterioles of juxtamedullary nephrons but not cortical nephrons. Vasa recta are specialized capillaries that run parallel to the loops of Henle and are involved in the reabsorption of water and solutes in the kidney. In cortical nephrons, the vasa recta do not arise from the efferent arterioles, while in juxtamedullary nephrons, the vasa recta arise from the efferent arterioles and play an important role in establishing the concentration gradient in the medulla.

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  • 12. 

    The Tm for PAH

    • 0.15 mg/ml

    • 1mg/ml

    • 2mg/ml

    • 80mg/min

    • 350mg/min

    Correct Answer
    A. 80mg/min
    Explanation
    The given options are different concentrations of PAH (0.15 mg/ml, 1 mg/ml, 2 mg/ml) and different rates of administration (80 mg/min, 350 mg/min). The correct answer, 80 mg/min, represents the maximum rate at which PAH can be administered. This suggests that administering PAH at a rate higher than 80 mg/min may exceed the maximum safe limit and could potentially cause adverse effects.

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  • 13. 

    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 concentration of glucose in the blood at which the kidneys start to excrete it into the urine. In this case, the correct answer is 2mg/ml, which means that when the concentration of glucose in the blood reaches or exceeds 2mg/ml, the kidneys will begin to eliminate it from the body through urine.

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  • 14. 

    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 there is 1 milligram of glucose dissolved in every milliliter of fluid in Bowman's space.

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  • 15. 

                                                    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 = concentration of inulin in urine (Ux) * urine flow rate (V). In this case, Ux = 1.6 mg/ml and V = 0.5 ml/min. Therefore, the excretion rate of inulin is 1.6 mg/ml * 0.5 ml/min = 0.8 mg/min.

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  • 16. 

                                                    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 a measure of the rate at which the solute is removed from the body. It is calculated by dividing the rate of elimination of the solute by its concentration in a reference fluid. In this case, the rate of elimination of solute X is given by the product of the concentration of solute X in urine (Ux) and the urine flow rate (V), which is 1.6 mg/ml * 0.5 ml/min = 0.8 mg/min. The concentration of solute X in plasma (Px) is given as 0.08 mg/ml. Therefore, the clearance of solute X is 0.8 mg/min / 0.08 mg/ml = 10 ml/min.

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  • 17. 

                                                    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?

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

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

    • C. Solute X might be creatinine

    • D. Solute X might be glycine

    • E. Solute X might be serum albumin

    Correct Answer
    A. A. It exhibits net reabsorption in the nephron since its clearance is lower than Cin
    Explanation
    The clearance of a substance measures the rate at which it is removed from the blood by the kidneys. If the clearance of a substance is lower than its concentration in the urine (Cin), it indicates that the substance is being reabsorbed by the nephron. In this case, since the clearance of solute X is lower than Cin, it suggests that solute X is being reabsorbed in the nephron, leading to net reabsorption. Therefore, option a is the correct answer.

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  • 18. 
    Which patient has hyperaldosteronism? - ProProfs

    Which patient has hyperaldosteronism?

    • A

    • B

    • C

    • D

    • E

    Correct Answer
    A. A
  • 19. 
    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 is the correct answer because hyperaldosteronism is a condition characterized by excessive production of aldosterone, which can lead to fluid retention and ultimately ascites. Ascites refers to the accumulation of fluid in the abdominal cavity, causing abdominal swelling. Therefore, patient D is the most likely to have ascites due to hyperaldosteronism.

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  • 20. 
    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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  • 21. 

    Patient Y drinks 2 L of pure water within 5 minutes. Twenty minutes later which variable will be increased as compared to normal?

    • A. Posm.

    • B. Plasma oncotic pressure.

    • C. Cosm.

    • D. Plasma concentration of angiotensin II.

    • E. CH20.

    Correct Answer
    A. E. CH20.
    Explanation
    When Patient Y drinks 2 L of pure water within 5 minutes, it will lead to an increase in the water content in the body. This excess water will cause a decrease in the osmolality of the blood, leading to a decrease in CH20 (free water clearance). Therefore, twenty minutes later, the variable that will be increased as compared to normal is CH20.

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  • 22. 

    Radioactive Cl- is a marker for

    • A. PV

    • B. ISF

    • C. ICW

    • D. ECW

    • E. TBW

    Correct Answer
    A. D. ECW
    Explanation
    Radioactive Cl- is a marker for extracellular water (ECW). This is because Cl- ions are primarily found outside of cells in the extracellular fluid. By using radioactive Cl- as a marker, researchers can track the movement and distribution of water in the extracellular compartment of the body. This can provide valuable information about fluid balance and hydration status.

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  • 23. 

    Blockage of the ureter by a kidney stone will increase

    • A. GFR

    • B. Urine flow

    • C. The hydrostatic pressure of tubular fluid

    • D. The oncotic pressure of tubular fluid

    • E. The oncotic pressure in glomerular capillaries

    Correct Answer
    A. C. The hydrostatic pressure of tubular fluid
    Explanation
    When the ureter is blocked by a kidney stone, it prevents the flow of urine from the kidney to the bladder. As a result, the urine backs up in the tubular system, increasing the hydrostatic pressure of the tubular fluid. This increased pressure can lead to dilation of the renal pelvis and calyces, causing pain and discomfort. It can also impair the filtration process in the kidney, leading to a decrease in glomerular filtration rate (GFR). Therefore, the correct answer is c. The hydrostatic pressure of tubular fluid.

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  • 24. 

    Aquaporins are inhibited by mercury. In a patient with mercury poisoning, which clearance is LEAST changed?

    • A. Water

    • B. Na+.

    • C. HCO3-

    • D. K+.

    • E. Glucose.

    Correct Answer
    A. E. Glucose.
    Explanation
    Aquaporins are membrane proteins that facilitate the transport of water across cell membranes. Inhibition of aquaporins by mercury would lead to a decrease in water clearance. However, the clearance of other solutes such as Na+, HCO3-, K+, and glucose would not be directly affected by the inhibition of aquaporins. Therefore, the clearance of glucose would be least changed in a patient with mercury poisoning.

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  • 25. 

    The clearance of Na+ will be increased most by inhibition of the

    • A. Na+/glucose symporter.

    • B. Na+/H+ antiporter

    • C. Na+/K+ ATPase.

    • D. Na+/K+/Cl- symporter.

    • E. Aldosterone-sensitive Na+ channels

    Correct Answer
    A. C. Na+/K+ ATPase.
    Explanation
    Inhibition of the Na+/K+ ATPase would result in an increase in the clearance of Na+. The Na+/K+ ATPase is responsible for maintaining the concentration gradients of Na+ and K+ across the cell membrane. By actively pumping Na+ out of the cell and K+ into the cell, it helps to maintain the low intracellular Na+ concentration. Inhibition of this pump would disrupt the normal balance of Na+ and K+ and lead to an increase in the clearance of Na+.

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  • 26. 

    The release of ANP is stimulated by

    • A. Angiotensin II.

    • B. Hypovolemia.

    • C. Stretch of the macula densa cells

    • D. Increased Posm.

    • E. Increased blood volume.

    Correct Answer
    A. E. Increased blood volume.
    Explanation
    The release of ANP (atrial natriuretic peptide) is stimulated by increased blood volume. ANP is a hormone released by the atria of the heart in response to stretching of the atrial walls caused by increased blood volume. It acts to decrease blood volume and blood pressure by promoting sodium and water excretion in the kidneys, leading to diuresis and vasodilation. Therefore, when blood volume increases, ANP is released as a regulatory mechanism to maintain fluid balance in the body.

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  • 27. 

    In congestive heart failure, edema occurs as a direct result of

    • A. Renal failure.

    • B. Increased hydrostatic pressure in systemic capillaries.

    • C. Decreased ADH

    • D. Increased plasma oncotic pressure

    • E. Increased urine output.

    Correct Answer
    A. B. Increased hydrostatic pressure in systemic capillaries.
    Explanation
    In congestive heart failure, the heart is unable to pump blood effectively, leading to a buildup of fluid in the body. This fluid accumulation causes increased hydrostatic pressure in the systemic capillaries, leading to the development of edema. This occurs because the increased pressure forces fluid out of the capillaries and into the surrounding tissues. Renal failure, decreased ADH, increased plasma oncotic pressure, and increased urine output are not direct causes of edema in congestive heart failure.

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  • Aug 31, 2023
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  • Mar 31, 2012
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