MCAT practice · Biological and Biochemical Foundations · 2A Cells and membranes · 1A Proteins and enzymes
Three routes into an epithelial cell
An original practice passage written for this site, not an AAMC item. Exam pace is about 8 minutes for a passage like this.
Epithelial cells take up nutrients and drugs by several routes, and the way uptake rate changes with concentration is often the first clue to which route a molecule uses. Researchers compared three radiolabeled, uncharged test solutes, designated X, Y, and Z, in a cultured human epithelial cell line. Cell monolayers were incubated at 37 °C in a buffer containing 140 mM Na⁺ and one labeled solute, and cell-associated radioactivity was measured after 10 s, an interval over which uptake rose linearly with time. Initial uptake rates were normalized to cell protein (Figure 1).
In a second experiment, uptake of each solute at 1 mM was measured under altered conditions (Table 1). To block the Na⁺-K⁺ ATPase, cells were preincubated for 30 min with 100 µM ouabain. For Na⁺-free buffer, NaCl was replaced mole for mole with N-methyl-D-glucamine chloride; N-methyl-D-glucamine is a large organic cation that does not cross the membrane appreciably. In a third condition, 20 mM unlabeled Y was added together with the labeled solute.
Two further observations were recorded. First, when ouabain was added at the same moment as the labeled solute, instead of 30 min beforehand, uptake of Z over the 10-s assay was 94% of control. Second, after 60 min of incubation with 0.1 mM external solute, the ratio of intracellular to extracellular solute concentration, estimated from measured cell water volume, was 1.0 for X, 1.0 for Y, and 14 for Z. None of the treatments changed cell protein content or the cells' exclusion of a membrane-impermeant dye, indicating that the membranes remained intact.
The researchers proposed that each solute crosses the plasma membrane by a different mechanism, and that the three mechanisms can be told apart by combining concentration dependence, competition, ion dependence, and the final distribution of solute across the membrane.
- X
- Y
- Z
| Condition | X (% of control) | Y (% of control) | Z (% of control) |
|---|---|---|---|
| Control | 100 | 100 | 100 |
| Ouabain, 100 µM, 30-min preincubation | 98 | 97 | 22 |
| Na⁺ replaced by N-methyl-D-glucamine | 101 | 96 | 7 |
| Plus 20 mM unlabeled Y | 99 | 10 | 95 |
Data are illustrative, constructed for practice.
1. Which comparison of the carriers for Y and Z is best supported by Figure 1?
- A. Wrong: Y's rates are larger only at high concentration, which reflects its higher maximal rate (about 30 vs about 12 pmol/s/mg); Y reaches half of its maximum near 1 mM, Z near 0.2 mM, so Z's carrier has the higher apparent affinity.
- B. Right: Z's uptake is half-maximal (about 6 of 12) near 0.2 mM and Y's (about 15 of 30) near 1 mM, so Z's apparent Km is lower and its affinity higher, while Y's plateau, and so its maximal rate, is higher.
- C. Wrong: Z does have the higher apparent affinity (half-maximal near 0.2 mM), but its uptake levels off near 12 pmol/s/mg while Y approaches about 30, so Z's maximal rate is lower.
- D. Wrong: this reverses both relationships. A larger plateau means a larger maximal rate (Y), and reaching half of the plateau at a lower concentration means a higher apparent affinity (Z).
2. Taken together, the results indicate that solute Y most likely enters the cells by:
- A. Wrong: simple diffusion gives a rate proportional to concentration, as for X. Y's uptake saturates and is cut to 10% by excess unlabeled Y, which shows a limited number of binding sites.
- B. Wrong: saturation shows a carrier but not an energy source. Removing Na⁺ or blocking the pump left Y uptake near 96–97% of control, and Y never accumulated above the outside concentration, as Z did.
- C. Right: Y's uptake saturates and is competed by unlabeled Y, so a carrier is involved; it is unaffected by ouabain or Na⁺ removal, and Y equilibrates at an inside-to-outside ratio of 1.0, so the carrier moves Y only down its gradient.
- D. Wrong: an ATP-driven pump would concentrate Y inside the cell, but Y's steady-state inside-to-outside ratio is 1.0, so Y moves only down its gradient.
3. Which explanation best accounts for the effect of ouabain on the uptake of Z?
- A. Wrong: if the pump itself moved Z, blocking it should cut Z uptake without a long preincubation, and Z would not need external Na⁺ to enter. The delay, and the dependence on external Na⁺, point instead to a separate carrier powered by the pump's gradient.
- B. Wrong: a competitive inhibitor of the carrier would act as soon as it was added, but ouabain added with the solute left Z uptake at 94% of control; the effect requires time, which points to an indirect action.
- C. Wrong: Z uptake saturates, depends on Na⁺, and accumulates against its gradient, none of which fits simple diffusion; the intact membranes also argue against a general permeability change.
- D. Right: Z needs external Na⁺ (7% of control without it) and accumulates 14-fold, the signature of Na⁺-coupled secondary active transport. Ouabain acts only after a 30-min preincubation, the time needed for the Na⁺ gradient to run down once the pump stops.
4. Why did the researchers replace NaCl with N-methyl-D-glucamine chloride, rather than simply leaving NaCl out of the Na⁺-free buffer?
- A. Wrong: N-methyl-D-glucamine is not a pump inhibitor, and the design uses ouabain separately for that purpose; a substitute that also blocked the pump would confound the two treatments.
- B. Wrong: removing external Na⁺ abolishes the inward Na⁺ gradient rather than raising it; the purpose of the condition is to test whether Z uptake needs that gradient.
- C. Wrong: the substitute was chosen because it is not transported; if it could replace Na⁺, Z uptake would not have fallen to 7% of control.
- D. Right: omitting 140 mM NaCl would leave the buffer strongly hypotonic and low in Cl⁻, so cells would swell and any change could reflect osmotic stress. An impermeant cation substitute keeps everything but Na⁺ the same.
5. Per ATP hydrolyzed, the enzyme that ouabain inhibits in this study moves which ions across the plasma membrane?
- A. Wrong: this reverses the counts; the pump exports three Na⁺ and imports two K⁺.
- B. Right: the Na⁺-K⁺ ATPase exports 3 Na⁺ and imports 2 K⁺ per ATP, building the inward Na⁺ gradient that Na⁺-coupled carriers such as Z's spend.
- C. Wrong: this reverses the directions; the pump keeps Na⁺ low and K⁺ high inside the cell, so it moves Na⁺ out and K⁺ in.
- D. Wrong: a 3-for-3 exchange would be electroneutral; the real 3-for-2 stoichiometry moves one net positive charge out per cycle, making the pump electrogenic.
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