Glomerular filtration US Medical PG Practice Questions and MCQs
Practice US Medical PG questions for Glomerular filtration. These multiple choice questions (MCQs) cover important concepts and help you prepare for your exams.
Glomerular filtration US Medical PG Question 1: Which region of the nephron reabsorbs the highest percentage of filtered bicarbonate?
- A. Collecting duct
- B. Thick ascending limb
- C. Distal tubule
- D. Proximal tubule (Correct Answer)
Glomerular filtration Explanation: ***Proximal tubule***
- The **proximal convoluted tubule (PCT)** reabsorbs approximately 80-90% of the **filtered bicarbonate** through a process involving **carbonic anhydrase** and the **Na+/H+ exchanger**.
- This vital function ensures that the majority of bicarbonate, a key buffer, is returned to the blood to maintain **acid-base balance**.
*Collecting duct*
- While the collecting duct does have the ability to reabsorb and secrete bicarbonate, its contribution is minor compared to the PCT, primarily for fine-tuning acid-base balance.
- Cells in the collecting duct, particularly **Type A intercalated cells**, are important for secreting acid (H+) in acidosis and therefore reabsorbing bicarbonate, but not the bulk of it.
*Thick ascending limb*
- The primary role of the **thick ascending limb** is the reabsorption of **sodium**, **potassium**, and **chloride** to create a concentrated interstitium, not significant bicarbonate reabsorption.
- It is largely impermeable to water and is relatively impermeable to bicarbonate.
*Distal tubule*
- The **distal convoluted tubule (DCT)** reabsorbs a small percentage of filtered bicarbonate, but its main role is regulated reabsorption of **sodium** and **calcium**, and secretion of **potassium** and **hydrogen ions**.
- Its contribution to bicarbonate reabsorption is much less significant than that of the proximal tubule.
Glomerular filtration US Medical PG Question 2: A healthy 30-year-old female has a measured creatinine clearance of 100 mL/min. She has a filtration fraction (FF) of 25%. Serum analysis reveals a creatinine level of 0.9 mg/dL and an elevated hematocrit of 0.6. Which of the following is the best estimate of this patient’s renal blood flow?
- A. 1.2 L/min
- B. 600 mL/min
- C. 800 mL/min
- D. 400 mL/min
- E. 1.0 L/min (Correct Answer)
Glomerular filtration Explanation: ***1.0 L/min***
- The **renal plasma flow (RPF)** can be calculated by dividing the **creatinine clearance (which approximates GFR)** by the **filtration fraction (FF)**: RPF = GFR / FF = 100 mL/min / 0.25 = 400 mL/min.
- To find the **renal blood flow (RBF)**, we use the formula RBF = RPF / (1 - Hematocrit). Given RPF = 400 mL/min and Hematocrit = 0.6, RBF = 400 mL/min / (1 - 0.6) = 400 mL/min / 0.4 = 1000 mL/min, or **1.0 L/min**.
*1.2 L/min*
- This value would result if the hematocrit were lower (e.g., 0.5) or if the GFR or FF were different, leading to an incorrect RPF or RBF calculation.
- It does not align with the provided values when applying the standard physiological formulas relating GFR, FF, RPF, and hematocrit.
*600 mL/min*
- This value might be obtained if the hematocrit was significantly underestimated or if the RPF calculation was incorrect in determining the RBF.
- It arises from using an incorrect formula or misinterpreting the relationship between plasma flow and blood flow.
*800 mL/min*
- This result would occur if the calculation for RPF or the subsequent RBF was erroneous, possibly by using an incorrect denominator in the RBF formula.
- For example, if RPF was incorrectly assumed to be 320 mL/min and divided by 0.4 (1-Hematocrit).
*400 mL/min*
- This value represents the calculated **renal plasma flow (RPF)**, not the **renal blood flow (RBF)**.
- RBF is always higher than RPF because it includes both plasma and cellular components of blood.
Glomerular filtration US Medical PG Question 3: A patient is receiving daily administrations of Compound X. Compound X is freely filtered in the glomeruli and undergoes net secretion in the renal tubules. The majority of this tubular secretion occurs in the proximal tubule. Additional information regarding this patient's renal function and the renal processing of Compound X is included below:
Inulin clearance: 120 mL/min
Plasma concentration of Inulin: 1 mg/mL
PAH clearance: 600 mL/min
Plasma concentration of PAH: 0.2 mg/mL
Total Tubular Secretion of Compound X: 60 mg/min
Net Renal Excretion of Compound X: 300 mg/min
Which of the following is the best estimate of the plasma concentration of Compound X in this patient?
- A. 2 mg/mL (Correct Answer)
- B. 3 mg/mL
- C. There is insufficient information available to estimate the plasma concentration of Compound X
- D. 1 mg/mL
- E. 0.5 mg/mL
Glomerular filtration Explanation: ***2 mg/mL***
* The **net renal excretion of Compound X (300 mg/min)** is the sum of the filtered load and the net tubular secretion.
* Given that Compound X is **freely filtered** and undergoes **net secretion (60 mg/min)**, we can calculate the filtered load and subsequently its plasma concentration.
* **Net excretion = Filtered load + Net tubular secretion**
* **300 mg/min = Filtered load + 60 mg/min**
* **Filtered load = 300 mg/min - 60 mg/min = 240 mg/min**
* Since **Filtered load = Glomerular Filtration Rate (GFR) * Plasma concentration (P_X)**, and GFR is estimated by **inulin clearance (120 mL/min)**:
* **240 mg/min = 120 mL/min * P_X**
* **P_X = 240 mg/min / 120 mL/min = 2 mg/mL**.
*3 mg/mL*
* This value would imply a significantly higher filtered load or a different contribution from tubular secretion.
* Calculations using this plasma concentration would not align with the provided excretion and secretion rates.
*There is insufficient information available to estimate the plasma concentration of Compound X*
* The problem provides all necessary values: **Inulin clearance (GFR)**, **net tubular secretion of Compound X**, and **net renal excretion of Compound X**.
* These parameters are sufficient to determine the filtered load and thus the plasma concentration of Compound X.
*1 mg/mL*
* A plasma concentration of 1 mg/mL would result in a lower filtered load than calculated and would not account for the observed net renal excretion.
* **Filtered load = 120 mL/min * 1 mg/mL = 120 mg/min**. Total excretion would then be 120 mg/min + 60 mg/min = 180 mg/min, which contradicts the given 300 mg/min.
*0.5 mg/mL*
* This plasma concentration would lead to an even lower filtered load, making it impossible to achieve the *net renal excretion of Compound X* given the tubular secretion.
* **Filtered load = 120 mL/min * 0.5 mg/mL = 60 mg/min**. Total excretion would be 60 mg/min + 60 mg/min = 120 mg/min, which is much lower than the given 300 mg/min.
Glomerular filtration US Medical PG Question 4: On cardiology service rounds, your team sees a patient admitted with an acute congestive heart failure exacerbation. In congestive heart failure, decreased cardiac function leads to decreased renal perfusion, which eventually leads to excess volume retention. To test your knowledge of physiology, your attending asks you which segment of the nephron is responsible for the majority of water absorption. Which of the following is a correct pairing of the segment of the nephron that reabsorbs the majority of all filtered water with the means by which that segment absorbs water?
- A. Distal convoluted tubule via passive diffusion following ion reabsorption
- B. Distal convoluted tubule via aquaporin channels
- C. Thick ascending loop of Henle via passive diffusion following ion reabsorption
- D. Proximal convoluted tubule via passive diffusion following ion reabsorption (Correct Answer)
- E. Collecting duct via aquaporin channels
Glomerular filtration Explanation: ***Proximal convoluted tubule via passive diffusion following ion reabsorption***
- The **proximal convoluted tubule (PCT)** is responsible for reabsorbing approximately **65-70% of filtered water**, making it the primary site of water reabsorption in the nephron.
- This water reabsorption primarily occurs **passively**, following the active reabsorption of solutes (especially **sodium ions**), which creates an osmotic gradient.
*Distal convoluted tubule via passive diffusion following ion reabsorption*
- The **distal convoluted tubule (DCT)** reabsorbs a much smaller percentage of filtered water (around 5-10%) and its water reabsorption is largely **regulated by ADH**, not primarily simple passive diffusion following bulk ion reabsorption.
- While some passive water movement occurs, it is not the main mechanism or location for the majority of water reabsorption.
*Distal convoluted tubule via aquaporin channels*
- While aquaporin channels do play a role in water reabsorption in the DCT, particularly under the influence of **ADH**, the DCT is not the segment responsible for the **majority of all filtered water absorption**.
- The bulk of water reabsorption occurs earlier in the nephron, independently of ADH for the most part.
*Thick ascending loop of Henle via passive diffusion following ion reabsorption*
- The **thick ascending loop of Henle** is primarily involved in reabsorbing ions like Na+, K+, and Cl- but is largely **impermeable to water**.
- Its impermeability to water is crucial for creating the **osmotic gradient** in the renal medulla, which is necessary for later water reabsorption.
*Collecting duct via aquaporin channels*
- The **collecting duct** is critically important for **regulated water reabsorption** via **aquaporin-2 channels** under the influence of **ADH**, allowing for fine-tuning of urine concentration.
- However, it reabsorbs only a variable portion (typically 5-19%) of the remaining filtered water, not the **majority of all filtered water**.
Glomerular filtration US Medical PG Question 5: A scientist is studying the excretion of a novel toxin X by the kidney in order to understand the dynamics of this new substance. He discovers that this new toxin X has a clearance that is half that of inulin in a particular patient. This patient's filtration fraction is 20% and his para-aminohippuric acid (PAH) dynamics are as follows:
Urine volume: 100 mL/min
Urine PAH concentration: 30 mg/mL
Plasma PAH concentration: 5 mg/mL
Given these findings, what is the clearance of the novel toxin X?
- A. 1,500 mL/min
- B. 600 mL/min
- C. 300 mL/min
- D. 60 mL/min (Correct Answer)
- E. 120 mL/min
Glomerular filtration Explanation: ***60 ml/min***
- First, calculate the **renal plasma flow (RPF)** using PAH clearance: RPF = (Urine PAH conc. × Urine vol.) / Plasma PAH conc. = (30 mg/mL × 100 mL/min) / 5 mg/mL = 600 mL/min.
- Next, calculate the **glomerular filtration rate (GFR)**, which is the clearance of inulin. GFR = RPF × Filtration Fraction = 600 mL/min × 0.20 = 120 mL/min. Toxin X clearance is half of inulin clearance, so 120 mL/min / 2 = **60 mL/min**.
*1,500 ml/min*
- This value is likely obtained if an incorrect formula or conversion was made, possibly by misinterpreting the units or the relationship between GFR, RPF, and filtration fraction.
- It significantly overestimates the clearance for a substance that is cleared at half the rate of inulin.
*600 ml/min*
- This value represents the **renal plasma flow (RPF)**, calculated using the PAH clearance data.
- It does not account for the filtration fraction or the fact that toxin X clearance is half of inulin clearance (GFR).
*300 ml/min*
- This value would be obtained if the renal plasma flow (RPF) was incorrectly halved, or if an intermediate calculation was misinterpreted as the final answer.
- It does not align with the given filtration fraction and the relationship between toxin X and inulin clearance.
*120 ml/min*
- This value represents the **glomerular filtration rate (GFR)**, which is equal to the clearance of inulin (RPF × Filtration Fraction = 600 mL/min × 0.20 = 120 mL/min).
- The question states that the clearance of toxin X is **half** that of inulin, so this is an intermediate step, not the final answer.
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