Free MCAT-style practice, 2A
MCAT Membrane Transport, Osmosis and Tonicity Practice Questions
Eight free MCAT-style practice questions on diffusion, facilitated transport, osmosis and tonicity, the Na+/K+ ATPase, secondary active transport, membrane fluidity and cell junctions, with explanations.
Quick concept primer
Transport questions sort into three bins. Passive transport moves solutes down their electrochemical gradient without energy: simple diffusion (rate roughly proportional to the gradient) and facilitated diffusion through channels or carriers (saturable, specific). Primary active transport uses ATP directly, as in the Na+/K+ ATPase, which pumps 3 Na+ out and 2 K+ in per ATP. Secondary active transport uses the stored energy of an ion gradient, such as the Na+ gradient, to move another solute uphill.
Water follows solute. Tonicity is the effect of a solution on cell volume, set by the concentration of solutes that cannot cross the membrane. A solution that is hypertonic to the cell pulls water out and shrinks it; a hypotonic one swells it. Osmolarity counts particles after dissociation, so CaCl2 gives three particles per formula unit.
Traps: urea and other membrane-permeant solutes raise osmolarity but not tonicity, and secondary active transport still depends indirectly on ATP through the pump that maintains the gradient.
8 practice questions
Question 1
Red blood cells with an internal solute concentration equivalent to 0.3 osmolar (nonpenetrating solutes) are placed in 0.1 M glucose that cannot enter the cell. Which outcome is expected?
- Cells shrink as water leaves
- No change in volume
- Cells shrink because glucose enters
- Cells swell and may lyse as water enters
Show answer and explanation
Answer: D. Cells swell and may lyse as water enters
A 0.1 M glucose solution is 0.1 osmolar, far lower than the 0.3 osmolar of the cell interior, so the external solution is hypotonic. Water moves toward the higher solute concentration, into the cells, which swell and may burst. If the solution were hypertonic, cells would crenate.
Question 2
A cell has an internal nonpenetrating solute concentration of 300 mOsm. Which solution is hypertonic to this cell? (Assume complete dissociation of salts and that urea freely crosses the membrane.)
- 0.3 M urea
- 0.1 M CaCl2
- 0.2 M glucose
- 0.2 M NaCl
Show answer and explanation
Answer: D. 0.2 M NaCl
Count particles. 0.1 M CaCl2 gives 0.3 Osm (isotonic). 0.2 M glucose is 0.2 Osm (hypotonic). 0.2 M NaCl gives 0.4 Osm (hypertonic), so water leaves the cell. 0.3 M urea has an osmolarity equal to the cell, but because urea crosses the membrane it equilibrates and exerts no sustained osmotic pull, so the solution is effectively hypotonic and cells placed in it swell.
Question 3
Uptake of a labeled solute into cells was measured at different external concentrations, with and without a metabolic poison that depletes ATP.
| External concentration (mM) | Uptake rate, no poison | Uptake rate, with poison |
|---|---|---|
| 1 | 10 | 10 |
| 2 | 18 | 18 |
| 4 | 30 | 30 |
| 8 | 42 | 42 |
| 16 | 48 | 48 |
Which mechanism best explains the data?
- Primary active transport
- Secondary active transport coupled to Na+
- Facilitated diffusion through a carrier
- Simple diffusion
Show answer and explanation
Answer: C. Facilitated diffusion through a carrier
The rate saturates at high concentration, which indicates a limited number of carrier proteins, so this is not simple diffusion (whose rate would rise in proportion to the gradient). The unchanged rate with ATP depletion shows no direct ATP requirement, which rules out primary active transport and, because the Na+ gradient would collapse without ATP, secondary active transport too. A saturable, energy-independent process is facilitated diffusion.
Question 4
Ouabain inhibits the Na+/K+ ATPase. After exposure of a cardiac myocyte to ouabain, which change in intracellular ions is expected over several minutes?
- There is no change in ion concentrations
- Intracellular Na+ falls and intracellular K+ rises
- Intracellular Ca2+ falls
- Intracellular Na+ rises, which reduces the driving force for Ca2+ extrusion by the Na+/Ca2+ exchanger, so intracellular Ca2+ rises
Show answer and explanation
Answer: D. Intracellular Na+ rises, which reduces the driving force for Ca2+ extrusion by the Na+/Ca2+ exchanger, so intracellular Ca2+ rises
Without the pump, Na+ leaks in and accumulates inside the cell. The Na+/Ca2+ exchanger uses the inward Na+ gradient to push Ca2+ out, so a smaller gradient means less Ca2+ extrusion and a rise in cytosolic Ca2+. This is how digoxin, a related cardiac glycoside, strengthens contraction. K+ would fall, not rise, and membrane depolarization would follow.
Question 5
Intestinal epithelial cells absorb glucose from the lumen even when the luminal concentration is lower than the intracellular concentration, using the SGLT1 transporter. What is the immediate energy source for this uphill movement?
- The inward electrochemical gradient of Na+
- The concentration gradient of glucose itself
- Direct hydrolysis of ATP by SGLT1
- The proton gradient across the membrane
Show answer and explanation
Answer: A. The inward electrochemical gradient of Na+
SGLT1 is a symporter that couples glucose entry to the downhill movement of Na+. The Na+ gradient is maintained by the basolateral Na+/K+ ATPase, which is why this is called secondary active transport: the pump uses ATP, and the symporter uses the resulting gradient. SGLT1 itself does not hydrolyze ATP, and moving glucose uphill cannot be powered by its own gradient.
Question 6
Fish moved from warm water to cold water adapt by changing the composition of their membranes to maintain fluidity. Which change is most likely?
- Loss of all membrane proteins
- Increased proportion of saturated fatty acid chains
- Increased proportion of unsaturated fatty acid chains
- Complete replacement of phospholipids with cholesterol
Show answer and explanation
Answer: C. Increased proportion of unsaturated fatty acid chains
At low temperature membranes tend to pack tightly and become rigid. Unsaturated fatty acids contain cis double bonds that introduce kinks, preventing tight packing and keeping the membrane fluid. Saturated chains pack tightly and would make the membrane more rigid. Cholesterol also buffers fluidity, but cannot replace phospholipids.
Question 7
Which statement correctly pairs a cell junction with its function?
- Hemidesmosomes connect two neighboring cells
- Gap junctions form channels that allow ions and small molecules to pass between the cytoplasm of adjacent cells, such as in cardiac muscle
- Tight junctions allow rapid passage of ions between adjacent cells
- Desmosomes seal the space between epithelial cells to prevent leakage
Show answer and explanation
Answer: B. Gap junctions form channels that allow ions and small molecules to pass between the cytoplasm of adjacent cells, such as in cardiac muscle
Gap junctions are made of connexons and let ions and small molecules pass directly, which allows electrical coupling of cardiac myocytes. Tight junctions form a seal that prevents paracellular leakage and separates apical from basolateral membrane domains. Desmosomes provide mechanical strength by attaching adjacent cells through intermediate filaments, and hemidesmosomes anchor cells to the basement membrane.
Question 8
Passive diffusion across an epithelial sheet is proportional to the concentration difference and inversely proportional to thickness (Fick's law). If the epithelium thickens to twice its original thickness and the concentration difference across it falls to one half, by what factor does the diffusion rate change?
- It falls to one eighth
- It falls to one half
- It is unchanged
- It falls to one quarter
Show answer and explanation
Answer: D. It falls to one quarter
Flux is proportional to (concentration difference) divided by thickness. Doubling the thickness halves the rate, and halving the concentration difference halves it again, giving 1/2 x 1/2 = 1/4 of the original rate. This is the same logic behind thickened alveolar walls in disease reducing gas exchange.