Free MCAT-style practice, 3B
MCAT Kidney, Nephron Function and Acid-Base Practice Questions
Eight free MCAT-style practice questions on glomerular filtration, clearance, tubular reabsorption, the loop of Henle, ADH, the renin-angiotensin system and acid-base balance, with calculations and explanations.
Quick concept primer
The nephron filters plasma at the glomerulus, then modifies the filtrate: the proximal tubule reclaims most of the filtered Na+, water, glucose and bicarbonate; the descending loop of Henle is permeable to water; the thick ascending limb pumps out salt but is impermeable to water, building the medullary gradient; and the collecting duct adjusts water permeability under ADH control and Na+/K+ handling under aldosterone control. Clearance is C = (U x V) / P, and the clearance of inulin, which is filtered but neither reabsorbed nor secreted, equals the glomerular filtration rate.
Filtration is driven by Starling forces: glomerular capillary hydrostatic pressure favors filtration, while Bowman's space pressure and capillary oncotic pressure oppose it. The renin-angiotensin-aldosterone system raises blood pressure and volume when renal perfusion falls. The kidney regulates pH by reabsorbing bicarbonate and generating new bicarbonate while excreting acid as titratable acid and ammonium.
Traps: a low urine osmolality can result from lack of ADH or from a kidney that does not respond to ADH, and loop diuretics work by taking away the gradient that concentrates urine.
8 practice questions
Question 1
A patient receives an inulin infusion. Plasma inulin is 0.5 mg/mL, urine inulin is 30 mg/mL and urine flow is 2 mL/min. What is the glomerular filtration rate?
- 120 mL/min
- 240 mL/min
- 12 mL/min
- 60 mL/min
Show answer and explanation
Answer: A. 120 mL/min
Clearance is U x V / P = (30 mg/mL x 2 mL/min) / 0.5 mg/mL = 60 / 0.5 = 120 mL/min. Because inulin is freely filtered and neither reabsorbed nor secreted, its clearance equals GFR. A typical adult value is around 120 mL/min.
Question 2
A person with poorly controlled diabetes has a blood glucose of 400 mg/dL and excretes large volumes of urine containing glucose. Which explanation is correct?
- Glucose is actively secreted by the collecting duct
- The loop of Henle is permeable to glucose
- Glucose is not filtered at the glomerulus
- The filtered glucose load exceeds the maximal transport capacity of the proximal tubule, so glucose remains in the tubular fluid and draws water by osmosis
Show answer and explanation
Answer: D. The filtered glucose load exceeds the maximal transport capacity of the proximal tubule, so glucose remains in the tubular fluid and draws water by osmosis
Glucose is freely filtered and is normally reabsorbed completely in the proximal tubule by SGLT2 and SGLT1 transporters. When the filtered load exceeds the transport maximum, the excess is excreted, and the unreabsorbed glucose raises tubular osmolarity, retaining water and causing an osmotic diuresis (polyuria).
Question 3
A patient is given furosemide, a loop diuretic that inhibits the Na+/K+/2Cl- cotransporter in the thick ascending limb. What is the effect on urine?
- Larger volume of less concentrated urine, because salt reabsorption in the thick ascending limb is blocked and the medullary gradient is lost
- No change in urine volume
- Smaller volume of more concentrated urine
- Increased water reabsorption in the collecting duct
Show answer and explanation
Answer: A. Larger volume of less concentrated urine, because salt reabsorption in the thick ascending limb is blocked and the medullary gradient is lost
The thick ascending limb normally pumps NaCl into the medullary interstitium, helping create the osmotic gradient that allows water reabsorption from the collecting duct. Blocking this transporter delivers more NaCl to distal segments and reduces the gradient, so less water can be reabsorbed. The result is a large volume of dilute urine.
Question 4
Two urine samples were collected from the same healthy person under different conditions.
| Sample | Urine volume (mL/h) | Urine osmolality (mOsm/kg) |
|---|---|---|
| 1 | 20 | 1000 |
| 2 | 400 | 80 |
Which sample was collected during the period of highest circulating ADH?
- Neither, because ADH does not affect urine
- Sample 1, after water deprivation
- Both, because ADH is constant
- Sample 2, after drinking a large volume of water
Show answer and explanation
Answer: B. Sample 1, after water deprivation
ADH acts on V2 receptors in the collecting duct to insert aquaporin-2 channels, allowing water to leave the tubule down the medullary osmotic gradient. High ADH therefore gives a small volume of concentrated urine. Sample 2 is a large volume of dilute urine, which occurs when ADH is suppressed after a water load.
Question 5
A patient with hypertension takes an ACE inhibitor. Which combination of effects is expected?
- Decreased angiotensin II, decreased aldosterone, vasodilation and a tendency toward higher plasma K+
- Increased renin only, with no change in blood pressure
- Decreased angiotensin I and increased angiotensin II
- Increased angiotensin II, increased aldosterone and vasoconstriction
Show answer and explanation
Answer: A. Decreased angiotensin II, decreased aldosterone, vasodilation and a tendency toward higher plasma K+
ACE converts angiotensin I to angiotensin II, so blocking it lowers angiotensin II, which reduces vasoconstriction and aldosterone release. Less aldosterone means less K+ secretion, so plasma K+ may rise. Renin rises through loss of feedback, but angiotensin II does not.
Question 6
A patient has a tumor producing excess aldosterone. Which laboratory pattern is expected?
- Hypertension, hyperkalemia and metabolic acidosis
- Hypertension, hypokalemia and metabolic alkalosis
- Hypotension, hyperkalemia and metabolic acidosis
- Hypotension, hypokalemia and metabolic acidosis
Show answer and explanation
Answer: B. Hypertension, hypokalemia and metabolic alkalosis
Aldosterone promotes Na+ and water retention, raising blood volume and pressure. Its stimulation of Na+/K+ ATPase and ENaC in the collecting duct also drives K+ secretion and H+ secretion by intercalated cells, causing hypokalemia and a metabolic alkalosis.
Question 7
A patient has a plasma bicarbonate of 12 mM and a PCO2 of 40 mmHg. Using pH = 6.1 + log([HCO3-] / (0.03 x PCO2)), what is the approximate pH, and what compensation is expected?
- pH 7.10; hyperventilation to lower PCO2
- pH 7.10; hypoventilation to raise PCO2
- pH 7.40; no compensation
- pH 7.70; renal bicarbonate loss
Show answer and explanation
Answer: A. pH 7.10; hyperventilation to lower PCO2
Dissolved CO2 is 0.03 x 40 = 1.2 mM, so the ratio is 12 / 1.2 = 10 and log(10) = 1. pH = 6.1 + 1 = 7.10, an acidemia. Because the primary problem is low bicarbonate (metabolic acidosis), the respiratory system compensates by raising ventilation, which lowers PCO2 and moves the ratio back toward normal. For comparison, normal values of 24 mM and 40 mmHg give a ratio of 20 and a pH of 7.40.
Question 8
In a glomerular capillary, the hydrostatic pressure is 50 mmHg, the hydrostatic pressure in Bowman's space is 10 mmHg and the capillary oncotic pressure is 25 mmHg (neglect oncotic pressure of the filtrate). What is the net filtration pressure?
- 85 mmHg
- 65 mmHg
- 35 mmHg
- 15 mmHg
Show answer and explanation
Answer: D. 15 mmHg
Net filtration pressure = glomerular hydrostatic pressure minus Bowman's space pressure minus capillary oncotic pressure = 50 - 10 - 25 = 15 mmHg. The hydrostatic pressure in the glomerulus pushes fluid out, while the other two forces oppose filtration. A fall in glomerular hydrostatic pressure, such as in severe hypotension, can reduce GFR.