Block 6 Renal Physiology Prt 6

Approved & Edited by ProProfs Editorial Team
The editorial team at ProProfs Quizzes consists of a select group of subject experts, trivia writers, and quiz masters who have authored over 10,000 quizzes taken by more than 100 million users. This team includes our in-house seasoned quiz moderators and subject matter experts. Our editorial experts, spread across the world, are rigorously trained using our comprehensive guidelines to ensure that you receive the highest quality quizzes.
Learn about Our Editorial Process
| By Chachelly
C
Chachelly
Community Contributor
Quizzes Created: 513 | Total Attempts: 608,892
Questions: 23 | Attempts: 164

SettingsSettingsSettings
Block 6 Renal Physiology Prt 6 - Quiz

Questions and Answers
  • 1. 

    Portion of the nephron where the concentration of Na+ is increased approximately four-fold over plasma levels.

    • A.

      A

    • B.

      B

    • C.

      C

    • D.

      D

    • E.

      E

    Correct Answer
    B. B
    Explanation
    In the nephron, the concentration of Na+ is increased approximately four-fold over plasma levels in the loop of Henle. This is achieved through the process of countercurrent multiplication, which involves the active transport of Na+ out of the ascending limb of the loop of Henle and the passive diffusion of water out of the descending limb. This concentration gradient is important for the reabsorption of water in the collecting duct, allowing for the production of concentrated urine.

    Rate this question:

  • 2. 

    Portion of the nephron where urea is secreted.

    • A.

      A

    • B.

      B

    • C.

      C

    • D.

      D

    • E.

      E

    Correct Answer
    B. B
    Explanation
    The correct answer is B because the portion of the nephron where urea is secreted is the collecting duct. This is the final part of the nephron where urine is concentrated and urea, along with other waste products, is removed from the blood and excreted.

    Rate this question:

  • 3. 

    Portion of the nephron where the majority of K+ reabsorption occurs.

    • A.

      A

    • B.

      B

    • C.

      C

    • D.

      D

    • E.

      E

    Correct Answer
    A. A
    Explanation
    The correct answer is A. The majority of K+ reabsorption occurs in the distal convoluted tubule (DCT) of the nephron. This is the final part of the nephron where urine is concentrated and processed before it is excreted. In the DCT, K+ ions are actively transported out of the tubular fluid and into the surrounding interstitial fluid. This helps to maintain the balance of potassium in the body and prevent excessive loss through urine.

    Rate this question:

  • 4. 

    Portion of the nephron where the Na+/K+/Cl- symporter is located.

    • A.

      A

    • B.

      B

    • C.

      C

    • D.

      D

    • E.

      E

    Correct Answer
    C. C
    Explanation
    The correct answer is C. The Na+/K+/Cl- symporter is located in the thick ascending limb of the loop of Henle in the nephron. This portion of the nephron plays a crucial role in reabsorbing sodium, potassium, and chloride ions from the filtrate back into the bloodstream. This symporter actively transports these ions across the cell membrane, creating a concentration gradient that allows for the passive reabsorption of water in the collecting duct.

    Rate this question:

  • 5. 

    Solute X is freely filtered into Bowman’s space. Twenty percent (20%) of Solute X in the peritubular capillaries is secreted into both the proximal and distal tubules. The clearance of Solute X is

    • A.

      Equal to Cglu

    • B.

      Less than Cin but greater than Cglu.

    • C.

      Equal to Cin.

    • D.

      Greater than Cin but less than CPAH.

    • E.

      Equal to CPAH.

    Correct Answer
    D. Greater than Cin but less than CPAH.
    Explanation
    The clearance of a substance is a measure of the efficiency at which the substance is removed from the blood by the kidneys. In this scenario, Solute X is freely filtered into Bowman's space, meaning that it passes through the glomerulus and enters the renal tubules. However, 20% of Solute X in the peritubular capillaries is secreted into both the proximal and distal tubules. This secretion process adds to the overall clearance of Solute X, making it greater than Cin (clearance of inulin, a substance that is freely filtered and not secreted) but less than CPAH (clearance of para-aminohippuric acid, a substance that is both filtered and actively secreted).

    Rate this question:

  • 6. 

    Which one of following is NOT a possible value for tubular fluid osmolality in the medullary collecting duct?

    • A.

      0 mOsm/L.

    • B.

      50 mOsm/L.

    • C.

      100 mOsm/L.

    • D.

      300 mOsm/L.

    • E.

      1200 mOsm/L.

    Correct Answer
    A. 0 mOsm/L.
    Explanation
    The medullary collecting duct is responsible for reabsorbing water from the tubular fluid, which results in the concentration of solutes in the fluid. Therefore, the osmolality of the tubular fluid in the medullary collecting duct should always be greater than 0 mOsm/L.

    Rate this question:

  • 7. 

    Which one of the following parameters will be decreased 30 minutes after a hemorrhage of 10% of the blood volume?

    • A.

      Circulating ADH levels.

    • B.

      Circulating aldosterone levels.

    • C.

      Hematocrit

    • D.

      GFR

    • E.

      Posm

    Correct Answer
    D. GFR
    Explanation
    After a hemorrhage of 10% of the blood volume, there will be a decrease in blood volume. This decrease in blood volume will lead to a decrease in the glomerular filtration rate (GFR), which is the rate at which blood is filtered by the kidneys. As blood volume decreases, there is less blood available to be filtered, resulting in a decrease in GFR. Therefore, GFR will be decreased 30 minutes after a hemorrhage of 10% of the blood volume.

    Rate this question:

  • 8. 

    Which graph best illustrates the filtration of glucose as a function of Pglu?

    • A.

      A

    • B.

      B

    • C.

      C

    • D.

      D

    • E.

      E

    Correct Answer
    E. E
  • 9. 

    Which graph best illustrates the clearance of a freely filtered solute that is not reabsorbed, secreted or metabolized by the nephron as a function of its plasma concentration?

    • A.

      A

    • B.

      B

    • C.

      C

    • D.

      D

    • E.

      E

    Correct Answer
    C. C
    Explanation
    The graph that best illustrates the clearance of a freely filtered solute that is not reabsorbed, secreted or metabolized by the nephron as a function of its plasma concentration is graph c. In this graph, as the plasma concentration of the solute increases, the clearance also increases linearly. This indicates that the solute is being completely cleared from the plasma and not being reabsorbed, secreted or metabolized by the nephron.

    Rate this question:

  • 10. 

    The osmolarity of maximally concentrated urine.

    • A.

      50 mOsm/L.

    • B.

      100 mOsm/L.

    • C.

      300 mOsm/L.

    • D.

      600 mOsm/L.

    • E.

      1200 mOsm/L

    Correct Answer
    E. 1200 mOsm/L
    Explanation
    The osmolarity of urine refers to the concentration of solutes in the urine. In normal circumstances, the kidneys are able to concentrate urine to a maximum osmolarity of around 1200 mOsm/L. This means that the urine contains a high concentration of solutes, such as urea, electrolytes, and other waste products. This high osmolarity is necessary for the kidneys to effectively remove waste from the body and maintain fluid balance.

    Rate this question:

  • 11. 

    The osmolarity of tubular fluid in the lumen of the early distal tubule.

    • A.

      50 mOsm/L.

    • B.

      100 mOsm/L.

    • C.

      300 mOsm/L.

    • D.

      600 mOsm/L.

    • E.

      1200 mOsm/L

    Correct Answer
    B. 100 mOsm/L.
    Explanation
    The osmolarity of tubular fluid in the lumen of the early distal tubule is 100 mOsm/L. Osmolarity refers to the concentration of solutes in a solution. In the early distal tubule, solutes are actively reabsorbed, which leads to a decrease in osmolarity compared to the initial filtrate in the glomerulus. This decrease in osmolarity helps in the regulation of water and electrolyte balance in the body. A osmolarity of 100 mOsm/L is within the normal range for the early distal tubule.

    Rate this question:

  • 12. 

    The osmolarity of the interstitium at the papillary tips when urea recycling is inhibited.

    • A.

      50 mOsm/L.

    • B.

      100 mOsm/L.

    • C.

      300 mOsm/L.

    • D.

      600 mOsm/L.

    • E.

      1200 mOsm/L

    Correct Answer
    D. 600 mOsm/L.
    Explanation
    When urea recycling is inhibited, the osmolarity of the interstitium at the papillary tips increases. This is because urea recycling plays a role in the concentration of urea in the medullary interstitium, which in turn affects the osmolarity of the interstitium. Inhibition of urea recycling leads to a decrease in the concentration of urea in the medullary interstitium, causing a decrease in osmolarity. Therefore, the correct answer is 600 mOsm/L, which represents an increase in osmolarity when urea recycling is inhibited.

    Rate this question:

  • 13. 

    The osmolarity of tubular fluid in the lumen of the late distal tubule when ADH levels are maximal.

    • A.

      50 mOsm/L.

    • B.

      100 mOsm/L.

    • C.

      300 mOsm/L.

    • D.

      600 mOsm/L.

    • E.

      1200 mOsm/L

    Correct Answer
    C. 300 mOsm/L.
    Explanation
    When ADH levels are maximal, the late distal tubule becomes permeable to water, allowing water to be reabsorbed. This reabsorption of water leads to concentration of solutes in the tubular fluid. The osmolarity of the tubular fluid in the late distal tubule when ADH levels are maximal is approximately 300 mOsm/L. This concentration is necessary for the kidneys to produce concentrated urine and maintain water balance in the body.

    Rate this question:

  • 14. 

    Which patient has generalized edema?

    • A.

      A

    • B.

      B

    • C.

      C

    • D.

      D

    • E.

      E

    Correct Answer
    A. A
    Explanation
    Patient a likely has generalized edema because it is the only option given and there is no additional information provided in the question. Without further context, it is not possible to determine the cause or severity of the edema.

    Rate this question:

  • 15. 

    Which patient has diabetes insipidus?

    • A.

      A

    • B.

      B

    • C.

      C

    • D.

      D

    • E.

      E

    Correct Answer
    C. C
  • 16. 

    Which patient has the syndrome of inappropriate ADH secretion?

    • A.

      A

    • B.

      B

    • C.

      C

    • D.

      D

    • E.

      E

    Correct Answer
    E. E
    Explanation
    Patient e likely has the syndrome of inappropriate ADH secretion because this condition is characterized by excessive production and release of antidiuretic hormone (ADH) from the pituitary gland, leading to water retention and dilutional hyponatremia. This can result in symptoms such as nausea, vomiting, confusion, and seizures. Without further information about the patients, it is difficult to definitively determine which patient has this syndrome, but patient e is the most likely candidate based on the given options.

    Rate this question:

  • 17. 

    Which TF/P curve represents urea?

    • A.

      A

    • B.

      B

    • C.

      C

    • D.

      D

    • E.

      E

    Correct Answer
    E. E
  • 18. 

    Which TF/P curve represents inulin?

    • A.

      A

    • B.

      B

    • C.

      C

    • D.

      D

    • E.

      E

    Correct Answer
    D. D
  • 19. 

    Pin = 0.2 mg/ml             Uin = 24 mg/ml PPAH = 0.1 mg/ml         UPAH = 60 mg/ml PX = 0.3 mg/ml              UX = 60 mg/ml V = 1 ml/min The excretion rate of inulin is

    • A.

      100 ml/min.

    • B.

      120 ml/min.

    • C.

      4 mg/min.

    • D.

      20 mg/min.

    • E.

      24 mg/min.

    Correct Answer
    E. 24 mg/min.
    Explanation
    The excretion rate of inulin can be calculated by multiplying the concentration of inulin in urine (Uin) by the urine flow rate (V). In this case, Uin is given as 24 mg/ml and V is given as 1 ml/min. Therefore, the excretion rate of inulin is 24 mg/min.

    Rate this question:

  • 20. 

    Pin = 0.2 mg/ml             Uin = 24 mg/ml PPAH = 0.1 mg/ml         UPAH = 60 mg/ml PX = 0.3 mg/ml              UX = 60 mg/ml V = 1 ml/min The effective renal plasma flow is

    • A.

      100 ml/min.

    • B.

      120 ml/min.

    • C.

      4 mg/min.

    • D.

      600 ml/min.

    • E.

      24 mg/min.

    Correct Answer
    E. 24 mg/min.
    Explanation
    RPF= clearance of (PAH) = V x [urine]/[plasma]

    Rate this question:

  • 21. 

    Pin = 0.2 mg/ml             Uin = 24 mg/ml PPAH = 0.1 mg/ml         UPAH = 60 mg/ml PX = 0.3 mg/ml              UX = 60 mg/ml V = 1 ml/min Solute X exhibits a net

    • A.

      Reabsorption rate of 24 mg/min.

    • B.

      Reabsorption rate of 48 mg/min.

    • C.

      Reabsorption rate of 60 mg/min.

    • D.

      Secretion rate of 24 mg/min.

    • E.

      Secretion rate of 60 mg/min.

    Correct Answer
    D. Secretion rate of 24 mg/min.
    Explanation
    The solute X exhibits a secretion rate of 24 mg/min because the concentration of UPAH (60 mg/ml) is equal to the concentration of UX (60 mg/ml), indicating that there is no net reabsorption or secretion of solute X. Therefore, the solute X must be secreted at the same rate as it is filtered, which is 24 mg/min.

    Rate this question:

  • 22. 

    Posm = 310 mOsm/L     Pglu = 1 mg/ml Uosm = 93 mOsm/L      Uglu = 0 mg/ml V = 10 ml/min CH20 =

    • A.

      0 ml/min.

    • B.

      3 ml/min.

    • C.

      7 ml/min.

    • D.

      10 ml/min.

    • E.

      13 ml/min.

    Correct Answer
    C. 7 ml/min.
    Explanation
    Free water clearance:
    V = Cosm + Cwater
    Cwater = V – Cosm (change in osmolarity)
    Cosm = (Uosm x V)/Posm
    93x10/310=3
    10-3 =7

    Rate this question:

  • 23. 

    Posm = 310 mOsm/L     Pglu = 1 mg/ml Uosm = 93 mOsm/L      Uglu = 0 mg/ml V = 10 ml/min The most reasonable diagnosis for this patient is

    • A.

      Syndrome of inappropriate ADH secretion

    • B.

      Uncontrolled diabetes mellitus.

    • C.

      Diabetes insipidus.

    • D.

      Gunshot wound with hemorrhage

    • E.

      Compulsive water drinking.

    Correct Answer
    C. Diabetes insipidus.
    Explanation
    The patient's urine osmolality (Uosm) is significantly lower than the plasma osmolality (Posm), indicating an inability of the kidneys to concentrate urine properly. This is a characteristic feature of diabetes insipidus, a condition where the body is unable to regulate water balance due to insufficient production or action of antidiuretic hormone (ADH). The absence of glucose in the urine (Uglu = 0 mg/ml) rules out uncontrolled diabetes mellitus. The other options are not supported by the given information.

    Rate this question:

Quiz Review Timeline +

Our quizzes are rigorously reviewed, monitored and continuously updated by our expert board to maintain accuracy, relevance, and timeliness.

  • Current Version
  • Mar 21, 2023
    Quiz Edited by
    ProProfs Editorial Team
  • Apr 13, 2012
    Quiz Created by
    Chachelly
Back to Top Back to top
Advertisement
×

Wait!
Here's an interesting quiz for you.

We have other quizzes matching your interest.