Membrane Transport, Quantified

You’ll be able to

Contrast the rate-versus-concentration behavior of simple diffusion, facilitated diffusion, and active transport, explaining why carrier-mediated transport saturates
Calculate water potential using Ψ = Ψp + Ψs and the solute-potential relationship Ψs = −iCRT to predict both the direction and the magnitude (in MPa) of water movement
Compare the responses of walled plant cells and wall-less animal cells to hypertonic, hypotonic, and isotonic solutions, and identify when bulk transport is required

Three transport modes, three rate signatures

Plot transport rate against solute concentration and the three mechanisms look different. Simple diffusion climbs in a straight line: the more solute, the steeper the gradient, and there is no limit — no protein is involved, so the rate never levels off. Facilitated diffusion rises, then plateaus: because it works through a finite number of channel or carrier proteins, once every protein is occupied (saturated) the rate hits a maximum (a Vmax), and adding more solute cannot speed it up. Active transport shows the same saturating curve — it too runs on a limited set of carrier proteins — but it moves solute up the gradient and therefore consumes ATP. Saturation is the tell-tale sign that a protein is mediating transport.

Water potential, made quantitative

Water moves from higher Ψ to lower Ψ, and Ψ has two parts: Ψ = Ψp + Ψs. The solute potential is given by Ψs = −iCRT, where i is the ionization constant (1 for glucose or sucrose; 2 for NaCl, which splits into Na⁺ and Cl⁻), C is molar concentration, R = 0.00831 L·MPa·mol⁻¹·K⁻¹, and T is temperature in kelvin (°C + 273). Adding solute makes Ψs — and therefore Ψ — more negative. The pressure potential Ψp is 0 in an open beaker but becomes positive when a plant cell wall pushes back (turgor). Comparing the total Ψ of a cell and its surroundings tells you exactly which way, and how strongly, water will flow.

Tonicity: walls change everything (and when proteins are not enough)

The same external solution treats plant and animal cells differently because a wall lets Ψp build up. In a hypotonic bath, water enters both cell types; a wall-less animal cell keeps swelling and lyses (bursts), but a plant cell stops when rising Ψp (turgor) raises its Ψ to match the outside — it becomes firm and turgid, the healthy state. In a hypertonic bath, both lose water: the animal cell crenates (shrivels) and the plant cell plasmolyzes as its membrane pulls from the wall. When cargo is simply too large for any channel or pump — a whole particle, a droplet, a macromolecule — the cell abandons protein transport for bulk transport: endocytosis (phagocytosis of solids, pinocytosis of fluids, receptor-mediated uptake) and exocytosis, both of which package cargo in vesicles and both of which require energy.

Water potential

Ψ = Ψp + Ψs

Ψ is water potential, Ψp is pressure potential, Ψs is solute potential (all in MPa). Water always moves toward lower (more negative) Ψ. Pure water at atmospheric pressure has Ψ = 0.

Solute potential

Ψs = −iCRT

i = ionization constant (1 for sucrose/glucose, 2 for NaCl); C = molarity; R = 0.00831 L·MPa·mol⁻¹·K⁻¹; T = temperature in kelvin (°C + 273). Ψs is always 0 or negative — solutes lower water potential.

Worked example

A 0.1 M NaCl solution sits in an open beaker at 22 °C. A plant cell whose water potential is Ψ = −0.75 MPa is dropped in. Calculate the solution’s water potential and predict which way water moves and what happens to the cell. (R = 0.00831 L·MPa·mol⁻¹·K⁻¹.)

1.Convert temperature to kelvin: T = 22 + 273 = 295 K. NaCl dissociates into Na⁺ and Cl⁻, so i = 2.
2.Solute potential: Ψs = −iCRT = −(2)(0.1)(0.00831)(295). Work it in order: 2 × 0.1 = 0.2; 0.2 × 0.00831 = 0.001662; 0.001662 × 295 = 0.49. So Ψs = −0.49 MPa.
3.The beaker is open to the atmosphere, so its pressure potential Ψp = 0. Then Ψ(solution) = Ψp + Ψs = 0 + (−0.49) = −0.49 MPa.
4.Compare potentials: the solution is at −0.49 MPa and the cell is at −0.75 MPa. Since −0.49 is higher (less negative) than −0.75, the solution has the higher Ψ, so water moves from the solution into the cell.

Answer: The solution’s water potential is Ψ = −0.49 MPa. Because water flows from higher Ψ (−0.49, the solution) to lower Ψ (−0.75, the cell), water moves into the cell; it takes up water, and as Ψp rises it moves toward turgor.

Checkpoint

As solute concentration increases, the rate of facilitated diffusion rises and then levels off at a plateau, while simple diffusion keeps rising in a straight line. What best explains the plateau?

Watch out

Two easy point-losers in water-potential math: always convert temperature to kelvin (°C + 273), and set i = 2 for NaCl (it dissociates) but i = 1 for sucrose or glucose (it does not). Forgetting either one changes Ψs by a large factor.

Checkpoint

A 0.3 M glucose solution sits in an open beaker at 27 °C. Using Ψs = −iCRT with R = 0.00831 L·MPa·mol⁻¹·K⁻¹, what is the solution’s water potential?

Checkpoint

A plant cell sitting in pure water (Ψ = 0) has reached equilibrium — no net water movement — and its solute potential is Ψs = −0.6 MPa. What must its pressure potential Ψp be?

On the exam

For a water-potential free-response, show the pipeline explicitly: convert to kelvin, compute Ψs = −iCRT, set Ψp = 0 for an open solution (or solve for it at equilibrium), add to get Ψ = Ψp + Ψs, then state the rule — water moves from higher Ψ to lower Ψ. Carry the MPa units and the negative signs through every line; graders award the reasoning, not just the final number.

Answer the 3 checkpoints as you read.