Membrane transport and osmosis for the MCAT

MCAT content library / Biology & biochemistry / Membrane transport and osmosis

Use permeability, electrochemical gradients, and conserved solute to explain what moves across a membrane—and how the cell changes volume.

About 18 minutes, including the worked examples and checks.

From a membrane to a neural signal · Lesson 2 of 6 · Continue in this path

Useful foundations

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Direction depends on a route and a driving force
Channels, carriers, and energy coupling
Worked application: distinguish saturation from uphill transport
Osmolarity counts particles; tonicity asks about volume
A rule for changed conditions
Check your reasoning
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Direction depends on a route and a driving force

A membrane can maintain very different solutions on its two sides because not every substance crosses it equally well. To predict net movement, identify the substance, an available route, and the energetic conditions. A concentration difference cannot produce substantial transport through an effectively impermeable barrier. Conversely, opening a route does not guarantee inward movement: the direction depends on the driving force. These relationships connect the AAMC topics of permeability, transport energetics, channels, and osmosis.¹

For an uncharged solute under otherwise comparable conditions, passive net movement is down its concentration gradient. For an ion, both concentration and voltage matter. A positive ion may move toward the electrically negative side even when that direction opposes its concentration gradient, provided the combined electrochemical change is favorable. Molecules can still cross in both directions when net flux is zero; equilibrium does not mean molecular motion has stopped.⁵

Channels, carriers, and energy coupling

Simple diffusion crosses the lipid phase. Facilitated diffusion uses a protein while remaining energetically downhill. Channels provide selective aqueous routes, often controlled by gates. Carriers bind a solute and alternate access between membrane sides. Limited carrier number and cycling speed can produce a transport-rate plateau as substrate concentration rises. Specificity and saturation support a carrier mechanism, but a plateau alone does not establish active transport: passive carriers also saturate.²

Primary active transport couples transport directly to an energy source; the sodium–potassium ATPase uses ATP hydrolysis to move three Na⁺ out and two K⁺ in per cycle. Secondary active transport couples an uphill solute movement to another solute moving downhill. A symporter moves its coupled substrates in the same direction; an antiporter moves them in opposite directions. The direction labels describe the transport arrangement, while the energy balance determines whether a particular cycle is favorable.³

For example, if inward glucose movement costs +4 kJ/mol and the coupled inward Na⁺ movement releases 6 kJ/mol, a one-to-one cycle has ΔG = +4 − 6 = −2 kJ/mol of glucose. This addition describes the actual conditions, not standard-state values. The favorable ion movement must be mechanistically coupled to glucose transport. An ion gradient elsewhere in the membrane cannot power an unrelated carrier merely by existing.⁷

Worked application: distinguish saturation from uphill transport

The following measurements are original illustrative data. A preparation has a fixed number of Na⁺-coupled glucose carriers. Intracellular glucose is held at 20 mM, membrane voltage is clamped at zero, and brief measurements maintain the stated concentrations. One condition has an inward Na⁺ concentration gradient; the other has equal Na⁺ concentrations on both sides. No other energy source is available. Positive flux means net glucose entry; negative flux means net exit.

External glucose (mM)	Inward Na⁺ drive: flux (relative units)	No Na⁺ drive: flux (relative units)
1	3.0	−1.00
2	5.0	−0.80
4	7.0	−0.50
8	8.0	−0.20
16	8.4	−0.05

Net glucose flux at different external glucose concentrations

Original illustrative flux measurements. Positive values indicate entry; negative values indicate exit. Lines connect measurements without implying a fitted equation.

With the Na⁺ drive, increasing external glucose from 1 to 16 mM increases inward flux, with progressively smaller gains. That pattern is consistent with limited carrier capacity. More decisively, glucose enters even though every external concentration is below the maintained internal 20 mM. Because glucose is uncharged, voltage cannot supply its uphill energy; the supplied Na⁺ drive can.

Removing that drive changes the sign of net glucose transport. The remaining outward glucose gradient can favor reverse cycling. The experiment supports gradient-dependent coupling, but it does not determine how many sodium ions accompany each glucose molecule. Nor does loss of a gradient imply destruction of the carrier. In a cell, ATP depletion may eventually reduce secondary transport by allowing ion gradients to dissipate, even when the cotransporter does not hydrolyze ATP itself.³⁷

Osmolarity counts particles; tonicity asks about volume

Osmosis is net water movement across a selectively permeable membrane. For dilute ideal solutions, osmotic pressure is Π = iMRT: M is formula-unit molarity, i accounts for the number of dissolved particles per formula unit, R is the gas constant, and T is absolute temperature. Equivalently, use Π = RT times total particle molarity. Thus ideal 0.10 M NaCl contributes about 0.20 Osm/L if fully dissociated. Do not multiply by i again when a concentration is already given in osmoles per liter. A hydrostatic pressure difference can oppose osmosis.⁴

Osmolarity counts all dissolved particles. Tonicity describes a solution’s effect on cell volume and depends on which particles remain effectively nonpenetrating over the relevant interval. Two isoosmotic solutions can have different long-term volume effects if their solutes cross differently. Specify the membrane and observation time before calling a solution isotonic, hypotonic, or hypertonic.⁶

Worked volume calculation with a permeating solute

An ideal cell initially has volume V₀, 300 mOsm/L of trapped particles, and no urea. It is placed in a large bath containing 200 mOsm/L of nonpenetrating particles plus 100 mOsm/L urea. Treat urea as effectively nonpenetrating during the initial water response, but allow it to equilibrate later. Water moves freely. Ignore hydrostatic pressure, solute metabolism, membrane rupture, and volume constraints; bath concentrations remain fixed.

Initial response: before appreciable urea entry, total particle concentrations are 300 mOsm/L on both sides. There is no initial osmotic volume change in this model.
Later response: urea enters and eventually reaches 100 mOsm/L on both sides. Its contribution then cancels when comparing the two solutions.
Conserve trapped particles: their final intracellular concentration must be 200 mOsm/L. Therefore 300V₀ = 200V_final, giving V_final = 1.5V₀.

The bath is initially isoosmotic yet produces swelling over the later interval. The amount of trapped intracellular solute has not increased; water entry has diluted it. This calculation uses conservation of particles together with the supplied permeability model, rather than treating every dissolved particle as a permanent osmotic barrier.

A rule for changed conditions

For transport, compare the free-energy changes of the complete coupled cycle and include its stoichiometry. For volume, identify trapped particles, conserve their amount, and then impose the stated equilibrium condition. Keep a flux measurement separate from an accumulated amount: a negative instantaneous flux predicts a direction of change, while the eventual concentration also depends on time, volume, metabolism, and changing gradients. These distinctions let the same reasoning survive a changed membrane, ion gradient, or observation interval.

Check your reasoning

Choose an answer before opening its feedback. Each option explains the reasoning behind that choice.

Check 1

An ideal cell begins at volume V₀ with 300 mOsm/L trapped particles and no urea. A large bath has 150 mOsm/L nonpenetrating particles plus 150 mOsm/L urea. Water moves freely. Urea is effectively nonpenetrating during the initial water response but later equilibrates across the membrane. Ignore hydrostatic pressure, metabolism, rupture, and volume limits. What are the initial response and final volume?

A. Initial shrinkage; final volume V₀.

Review this choice. The initial total is 300 mOsm/L on each side, so the stated early model does not predict shrinkage. Once urea equilibrates, the lower external nonpenetrating concentration requires swelling.

B. Initial swelling; final volume 2V₀.

Review this choice. The final volume is correct, but the initial totals are equal. Swelling develops as urea enters, rather than during the stipulated initial response before appreciable entry.

C. No initial volume change; final volume V₀.

Review this choice. The initial response is correct. Later, equilibrated urea contributes equally on both sides; trapped particles must dilute from 300 to 150 mOsm/L, doubling volume.

D. No initial volume change; final volume 2V₀.

Correct. Both parts follow from the supplied time scales. Initial totals are equal; later 300V₀ = 150Vfinal, so Vfinal = 2V₀.

The principle to carry forward

Count all particles for the stipulated initial comparison, then use the persistent nonpenetrating particles to calculate the later volume.

Check 2

Artificial vesicles contain a single transport protein. Uncharged solute X is 5 mM inside and 1 mM outside; membrane voltage and every ion gradient are zero. Net X entry occurs only when hydrolyzable ATP is supplied. Replacing ATP with a nonhydrolyzable analog abolishes entry, without changing protein abundance or integrity. No other energy source is present. Which mechanism best fits?

A. ATP-coupled primary active transport.

Correct. X moves uphill, and the supplied controls link that movement to ATP hydrolysis. Direct ATP coupling is the supported energy mechanism.

B. Channel-mediated passive diffusion.

Review this choice. An open passive channel cannot maintain net movement of an uncharged solute from 1 to 5 mM under these conditions.

C. Passive carrier-mediated diffusion.

Review this choice. A passive carrier can be selective and saturable, but it still requires a favorable gradient. The gradient for X favors exit.

D. Sodium-driven secondary active transport.

Review this choice. The stem removes every ion gradient, including a sodium drive. Sodium-coupled transport cannot obtain energy from an absent gradient.

The principle to carry forward

Identify the available energy source and the direction of the transported solute; the presence of a protein alone does not distinguish active from passive transport.

Check 3

A carrier obligatorily moves two Na⁺ inward for each glucose molecule moved inward. Under a new set of concentrations and voltage, each mole of inward Na⁺ transport has ΔG = −4 kJ/mol, while each mole of inward glucose transport has ΔG = +11 kJ/mol. There is no ATP coupling or other energy input. What is ΔG per mole of complete forward cycles?

A. −19 kJ/mol

Review this choice. This treats the uphill glucose term as favorable. Its supplied contribution is +11, not −11.

B. −3 kJ/mol

Review this choice. This is the negative of the forward-cycle result. The defined forward cycle moves both sodium ions and glucose inward.

C. +3 kJ/mol

Correct. Two sodium contributions give 2(−4) = −8; adding +11 gives +3 kJ/mol. The forward cycle is unfavorable under the new conditions.

D. +19 kJ/mol

Review this choice. This changes the sodium contribution from −8 to +8. The stated inward sodium movements are energetically favorable.

The principle to carry forward

Weight each free-energy contribution by the transport stoichiometry. An ordinarily inward cotransporter need not favor inward cycling after its driving conditions change.