Cell Membrane & Transport

Question 1

(a) $\Psi_s = -iCRT = -(1)(0.15)(8.314)(298)$ $= -371.6 \text{ kPa} \approx -372\text{ kPa}$.

• Correct substitution (1); correct numeric value (1); units kPa and negative sign (1).

(b) $\Psi_{cell} = \Psi_s + \Psi_p = -850 + 300 = -550\text{ kPa}$ (2). External solution $\Psi = -372\text{ kPa}$ (turgor of pure solution = 0, so $\Psi_{ext}=\Psi_s$). Since $\Psi_{cell}(-550) < \Psi_{ext}(-372)$, water moves from solution into cell (down the water-potential gradient, high→low $\Psi$) (2).

(c) $\text{CaCl}_2 \to \text{Ca}^{2+} + 2\text{Cl}^-$, so $i = 3$. $\Psi_s = -(3)(0.15)(8.314)(298) = -1114.8 \approx -1115\text{ kPa}$ (3). Explanation: solute potential depends on number of dissolved particles, not identity; CaCl₂ yields 3 particles per formula unit vs 1 for sucrose, tripling the osmotic effect (1). The membrane is selectively permeable — ions and sucrose cannot freely cross the hydrophobic bilayer core, so the colligative osmotic difference is sustained and drives water movement (1).

(d) In a hypotonic/weakly-concentrated medium water enters, the protoplast presses against the wall, $\Psi_p$ is high (positive) → turgid (2). As external solution becomes hypertonic, water leaves; protoplast shrinks, $\Psi_p$ falls toward zero — incipient plasmolysis is the point where the membrane just begins to pull from the wall and $\Psi_p = 0$ (2). Further water loss shrinks the protoplast away from the rigid cellulose wall, membrane detaching at points — full plasmolysis; the wall itself does not collapse (2).

(e) The red blood cell in strongly hypotonic solution gains water osmotically; having no cell wall, no opposing $\Psi_p$ develops, so it swells and bursts — lysis (haemolysis) (3). The plant cell in the same solution swells only until the strong cellulose wall generates a back-pressure ($\Psi_p$) that halts net water entry — it becomes turgid but does not burst; the wall provides mechanical restraint absent in animal cells (3).

Question 2

(a) Primary active transport: directly hydrolyses ATP to pump solute against its gradient — e.g. Na⁺/K⁺-ATPase; energy source = ATP hydrolysis (2.5). Secondary active transport: uses the electrochemical gradient (set up by primary transport) of one ion to drive another solute against its gradient — e.g. Na⁺–glucose co-transporter (SGLT); energy source = the stored ion gradient (indirect ATP) (2.5).

(b) Exporting 3 Na⁺ for 2 K⁺ moves net +1 charge out per cycle → interior becomes negative (electrical gradient) (2); unequal ion counts create chemical concentration gradients (high Na⁺ out, high K⁺ in) (2). The steep inward Na⁺ electrochemical gradient is harnessed by SGLT: Na⁺ flowing down its gradient into the cell drags glucose in against glucose's gradient (co-transport) (3).

(c) $\Delta G = RT\ln(C_{in}/C_{out}) + zF\Delta\psi$ Term 1: $RT\ln(12/145) = 8.314 \times 310 \times \ln(0.08276)$ $\ln(0.08276) = -2.4919$; $8.314\times310 = 2577.3$; product $= -6423 \text{ J mol}^{-1}$ (3). Term 2: $zF\Delta\psi = (+1)(96485)(-0.070) = -6754 \text{ J mol}^{-1}$ (2). $\Delta G = -6423 - 6754 = -13177\text{ J mol}^{-1} \approx -13.2\text{ kJ mol}^{-1}$ (1). $\Delta G < 0$ → inward Na⁺ movement is spontaneous (exergonic) (1). The cell couples this favourable flux to drive uphill transport of glucose/amino acids via secondary active transport (1).

Question 3

(a) (i) Small non-polar steroid: crosses by simple diffusion directly through the hydrophobic phospholipid bilayer, since non-polar molecules dissolve in the lipid core (3). (ii) Large polar peptide: cannot cross the hydrophobic core; enters by endocytosis (bulk transport / receptor-mediated) or via specific carrier — justified because polar/large solutes are excluded by the bilayer's hydrophobic interior (3).

(b) At high temperature cholesterol restrains phospholipid movement, reducing excessive fluidity and stabilising the membrane (2). At low temperature it wedges between phospholipids, preventing tight packing and crystallisation, maintaining fluidity — acting as a "fluidity buffer" (2).

(c) Receptor-mediated endocytosis uses specific membrane receptors that bind particular ligands, clustering in clathrin-coated pits that invaginate (2); pinocytosis is non-specific bulk uptake of extracellular fluid and dissolved solutes (2). Receptor-mediated is selective (only ligands recognised by receptors are taken up) and concentrative (ligands are gathered even at low external concentration), unlike random pinocytosis — e.g. cholesterol/LDL uptake via LDL receptors (2).

[
  {"claim":"Sucrose solute potential ~ -371.6 kPa", "code":"i=1; C=0.15; R=8.314; T=298; Psi=-i*C*R*T; result = abs(Psi - (-371.6)) < 1.0"},
  {"claim":"CaCl2 solute potential ~ -1114.8 kPa (i=3)", "code":"i=3; C=0.15; R=8.314; T=298; Psi=-i*C*R*T; result = abs(Psi - (-1114.8)) < 1.0"},
  {"claim":"Cell water potential = -550 kPa", "code":"Psi=-850+300; result = Psi == -550"},
  {"claim":"Delta G for Na+ influx ~ -13.2 kJ/mol", "code":"R=8.314; T=310; F=96485; z=1; dpsi=-0.070; import sympy as sp; term1=R*T*sp.log(sp.Rational(12,145)); term2=z*F*dpsi; dG=(float(term1)+term2)/1000; result = abs(dG - (-13.2)) < 0.3"}
]