OS II Exam 1 Physiology Questions - Pt.2

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

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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OS II Exam 1 Physiology Questions - Pt.2 - 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... see morewhich topics might be problematic for you? All the best! see less

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

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

Explanation

Reabsorption of new HCO3- occurs predominantly in the late distal tubule and the collecting duct only when H+ is excreted into the urine. This is because the excretion of H+ leads to the reabsorption of HCO3-, which helps to maintain acid-base balance in the body. When H+ is excreted into the urine, it combines with HCO3- to form H2CO3, which then dissociates into CO2 and H2O. The CO2 is reabsorbed into the tubular cells and converted back into HCO3-, which can be reabsorbed into the bloodstream. Therefore, the reabsorption of new HCO3- depends on the excretion of H+ in the urine.

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4. Aldosterone stimulates

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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5. The renal threshold for glucose

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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6. Blockage of the ureter by a kidney stone will increase

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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7. The release of ANP is stimulated by

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

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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9. 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?

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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10. 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

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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11. 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

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

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

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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14. In the proximal tubule, Cl- reabsorption

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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15. 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?

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

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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17. A major difference between cortical and juxtamedullary nephrons is that

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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18. The Tm for PAH

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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19. Which patient has hyperaldosteronism?

Explanation

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20. The concentration of glucose in Bowman's space

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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21.                                                 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?

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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22.                                                 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?

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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23.                                                 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?

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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24. Which patient has ascites? Which patient has hyperaldosteronism?

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

Explanation

Patient C suffers from hyperaldosteronism.

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

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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27. Radioactive Cl- is a marker for

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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Aquaporins are inhibited by mercury. In a patient with mercury...
The clearance of Na+ will be increased most by inhibition of the
Reabsorption of new HCO3- occurs predominately in the late distal...
Aldosterone stimulates
The renal threshold for glucose
Blockage of the ureter by a kidney stone will increase
The release of ANP is stimulated by
In congestive heart failure, edema occurs as a direct result of
Mannitol is an osmotically active solute that is freely filtered but...
As GFR increases, the oncotic pressure of the peritubular capillaries...
In the presence of maximal circulating levels of ADH, the...
Which one of the following slightly dilates the afferent arterioles...
Which of the following pressures is increased in the patient who is...
In the proximal tubule, Cl- reabsorption
Drug U blocks urea recycling. Assume enough time has passed for drug U...
What one of the following will increase the filtered load of Na+?
A major difference between cortical and juxtamedullary nephrons is...
The Tm for PAH
Which patient has hyperaldosteronism?
The concentration of glucose in Bowman's space
                                               ...
                                               ...
                                               ...
Which patient has ascites?...
Which patient suffers from nephrogenic diabetes insipidus?...
Patient Y drinks 2 L of pure water within 5 minutes. Twenty minutes...
Radioactive Cl- is a marker for
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