Big ideas of the unit
- Organelles divide the work. Ribosomes make proteins; the ER, Golgi and vesicles modify and ship them; lysosomes digest; mitochondria and chloroplasts handle energy (Topic 2.1).
- Exchange happens at surfaces, demand comes from volume. As a cell grows, SA/V falls, so cells stay small, flatten or fold their membranes (Topic 2.2).
- The membrane is a fluid mosaic. Heads face water, tails face each other, and proteins sit where their R groups fit (Topic 2.3).
- The hydrophobic interior chooses. Small nonpolar molecules cross freely, water slowly, ions and large polar molecules only through proteins. Walls stop osmotic lysis (Topic 2.4).
- Direction decides cost. Down a gradient is passive; against it needs a protein and ATP; bulk transport needs energy whichever way (Topic 2.5).
- Proteins open passive routes. Channels, carriers and aquaporins speed movement down a gradient but cannot reverse it (Topic 2.6).
- Water follows water potential. Water moves from higher to lower Ψ; solutes lower Ψ, pressure raises it (Topic 2.7).
- Pumps build gradients. The Na⁺/K⁺ pump moves 3 Na⁺ out and 2 K⁺ in per ATP (Topic 2.8).
- Compartments separate and multiply. Internal membranes keep competing reactions apart and add working area (Topic 2.9); mitochondria and chloroplasts began as engulfed prokaryotes (Topic 2.10).
Key relationships and methods
| Idea | Rule or method |
|---|---|
| SA/V | cube 6/s, sphere 3/r; folds add area, not volume |
| Route across a membrane | small nonpolar → bilayer; water → bilayer slowly, aquaporins fast; ions and large polar → proteins |
| Energy | down gradient: none; against gradient: ATP and a protein; vesicles: energy |
| Carrier limit | maximum rate = number of carriers × rate per carrier |
| Water potential | Ψ = Ψp + Ψs; Ψs = −iCRT; water moves to lower Ψ |
| Pump ratio | 3 Na⁺ : 2 K⁺ : 1 ATP, one net positive charge out |
| Origin evidence | two membranes, own circular DNA, bacteria-like ribosomes, division, sequences |
These are original Marlbridge practice questions, not past exam questions. The rubrics are a suggested Marlbridge rubric, not official scoring guidelines. All data sets, drugs and organisms are fictional.
Question 1 (multiple choice · mixed)
A glycoprotein is made on the rough ER, gets its sugar chains in the ER and Golgi complex, and reaches the plasma membrane in a vesicle. On which side of the plasma membrane do its sugar chains end up, and why?
- (A) Outside the cell: the chains are added on the inner (lumen) side of the ER and Golgi, and that side becomes the outside when the vesicle fuses.
- (B) Outside the cell: the chains face the cytosol until fusion, then the protein flips over.
- (C) Facing the cytosol: the lumen side of a vesicle becomes the inside of the cell after fusion.
- (D) Facing the cytosol: free ribosomes add the sugars as the protein is made.
Answer and explanation
Answer: (A). When the vesicle membrane joins the plasma membrane, its inner (lumen) face becomes the outer face of the cell. That is why sugar chains used in cell recognition sit outside.
- (B) is tempting because membrane proteins move, but sideways, not by flipping across the bilayer.
- (C) gets the geometry of fusion backwards.
- (D) confuses where proteins are made with where sugars are added.
Question 2 (multiple choice · mixed)
Cube-shaped cells with sides of 1 µm and 4 µm have the same cytoplasm and aquaporin density. Both are placed in the same hypotonic solution. Which statement is correct?
- (A) The large cell’s volume rises by a larger percentage per second, because it has 16 times the membrane area.
- (B) Both change at the same percentage rate, because aquaporin density is the same.
- (C) The small cell swells faster because it spends less ATP pumping water.
- (D) The small cell’s volume rises by a larger percentage per second at first, because it has 4 times as much membrane per µm³.
Answer and explanation
Answer: (D). Small cell: SA = 6 µm², V = 1 µm³, SA/V = 6 µm⁻¹. Large cell: SA = 96 µm², V = 64 µm³, SA/V = 1.5 µm⁻¹. Each µm³ of the small cell is served by 4 times as much membrane.
- (A) is tempting: the large cell does take in more water in total, but it has 64 times the volume to fill.
- (B) ignores how much volume the membrane must serve.
- (C) is wrong because osmosis through aquaporins is passive.
Question 3 (multiple choice · mixed)
Fictional animal cells have cytoplasm equivalent to 0.30 M of solutes that cannot cross the membrane. Glycerol is small, polar and uncharged: it crosses the bilayer slowly and is not pumped. The cells are placed in a large volume of 0.30 M glycerol. Which prediction is best?
- (A) No net water movement ever, because both sides have the same molarity.
- (B) Little change at first; then the cells swell and may burst as glycerol diffuses in and water follows.
- (C) The cells shrink and stay small, because the glycerol outside draws water out of them.
- (D) The cells swell at once, because they pump glycerol in using ATP.
Answer and explanation
Answer: (B). At first both sides hold 0.30 M of solute, so little water moves. But glycerol crosses, and none is inside, so it diffuses in down its gradient. The inside now holds its own solutes plus glycerol, so its water potential falls below the outside and water enters by osmosis. With no wall, the cell keeps swelling.
- (A) is tempting, but tonicity depends on solutes that cannot cross; glycerol can.
- (C) would need a solute outside that cannot enter.
- (D) contradicts the stem: entry is passive.
Question 4 (constructed response · mixed)
A fictional cyanobacterium is a sphere of radius 1.5 µm. Inside it are thylakoid membranes, where light is captured, with a total area 6.0 times the area of its plasma membrane.
(a) Calculate the cell’s surface area, volume and SA/V, and its total membrane area per µm³ of volume. (b) Explain how the thylakoid membranes benefit the cell. (c) A student says: “This cell has internal membranes, so it must be a eukaryote.” Evaluate the claim. (d) Give two pieces of evidence that chloroplasts descend from cyanobacteria-like cells, and one way a chloroplast differs from a free-living cyanobacterium.
Worked solution and suggested Marlbridge rubric
(a) SA = 4π × 1.5² = 28.3 µm². V = (4/3)π × 1.5³ = 14.1 µm³. SA/V = 3 ÷ 1.5 = 2.0 µm⁻¹. Thylakoids: 6.0 × 28.27 = 169.6 µm²; total membrane = 197.9 µm², so 14.0 µm² per µm³, 7 times the plasma membrane alone.
(b) Light is captured on membranes, so internal membranes give far more working area in the same volume.
(c) Not supported. Prokaryotes typically lack internal membranes but can have specialized regions. This cell still has no nucleus (its DNA is in a nucleoid) and no membrane-bound organelles such as mitochondria.
(d) Evidence (any two): two membranes; own circular DNA; bacteria-like ribosomes; division by splitting; DNA sequences closest to cyanobacteria. Difference (any one): a chloroplast cannot live outside the cell; most of its proteins are coded by nuclear genes.
| Point | What earns it |
|---|---|
| 1 | SA, V and SA/V correct |
| 1 | 14.0 µm² per µm³ (or 197.9 µm² total with the division shown) |
| 1 | Membranes as the site of the reactions, with more area in the same volume |
| 1 | Claim rejected: no nucleus or membrane-bound organelles |
| 1 | Two pieces of evidence |
| 1 | One valid difference |
Total: 6 points.
Question 5 (constructed response · mixed)
A fictional salt-marsh plant grows in soil water equivalent to 0.20 M NaCl at 20 °C. Its root cells use ATP to pump Na⁺ from the cytosol into the central vacuole, whose sap is equivalent to 0.40 M NaCl and sets the cell’s solute potential. Pressure potential is +3.0 bar.
(a) Calculate the water potential of the soil water. (b) Calculate the water potential of a root cell, and predict the direction of water movement. (c) Explain what produces the pressure potential and why it does not burst the cell. (d) Give two advantages of storing Na⁺ in the vacuole rather than the cytosol. (e) A drug stops ATP production in the root cells. Predict the effect on vacuole Na⁺ and on water uptake.
Worked solution and suggested Marlbridge rubric
(a) T = 293 K. Ψ = Ψs = −2 × 0.20 × 0.0831 × 293 = −9.74 bar (Ψp = 0 in open soil water).
(b) Ψs = −2 × 0.40 × 0.0831 × 293 = −19.48 bar. Ψ = +3.0 + (−19.48) = −16.5 bar. This is lower than −9.74 bar, so water moves from the soil into the cell.
(c) Water entering the vacuole pushes the cell against its wall, which resists expansion, so pressure builds. The wall prevents osmotic lysis.
(d) (1) High Na⁺ would interfere with cytosolic enzymes; a separate compartment keeps it away from them. (2) The stored salt lowers the vacuole’s water potential, so the cell still draws water from salty soil.
(e) Pumping Na⁺ against its gradient is active transport, so it stops. Na⁺ leaks back into the cytosol, so vacuole Ψs rises (less negative). As the cell’s Ψ nears −9.74 bar, water uptake slows and may reverse.
| Point | What earns it |
|---|---|
| 1 | −9.74 bar with i = 2 and T = 293 K |
| 1 | −16.5 bar |
| 1 | Water enters, because the cell’s Ψ is lower |
| 1 | Pressure from the wall resisting expansion; wall prevents lysis |
| 1 | Protects cytosolic enzymes or reactions by separation |
| 1 | Vacuole solutes lower Ψ so water still enters |
| 1 | Pumping stops, Na⁺ leaks out, Ψ rises and water uptake falls |
Total: 7 points.
Question 6 (constructed response · mixed)
Eggs of a fictional fish have very few water channels. A researcher injects eggs with mRNA coding for a membrane protein, W, or with water (control). Days later she moves them into a hypotonic solution and measures the initial rate of volume increase (% per minute). A third group with W is first treated with compound T, which binds to W.
| Group | Egg 1 | Egg 2 | Egg 3 | Egg 4 | Egg 5 |
|---|---|---|---|---|---|
| Control | 0.8 | 1.1 | 0.9 | 1.2 | 1.0 |
| W | 7.9 | 9.4 | 8.6 | 8.1 | 9.0 |
| W + compound T | 1.4 | 1.1 | 1.6 | 1.2 | 1.2 |
(a) Calculate the mean and mean ± 2 SE for the control and W groups. (b) State a null hypothesis for these two groups and decide whether to reject it. (c) W is made on ribosomes from the mRNA. Describe its route to the plasma membrane, and where its nonpolar R groups sit once there. (d) Explain why control eggs swell slowly and W eggs swell fast. (e) Explain what the compound T group adds, and propose an experiment to test whether water movement through W needs ATP.
Worked solution and suggested Marlbridge rubric
(a) Control: mean 1.00, s = 0.158, SE = 0.158 ÷ √5 = 0.071, interval 0.86 to 1.14. W: mean 8.60, s = 0.620, SE = 0.277, interval 8.05 to 9.15.
(b) Null hypothesis: there is no difference in the initial rate of swelling between control eggs and eggs with W. The intervals are far apart, so reject it. W eggs swell about 8.6 times as fast.
(c) Ribosomes on the rough ER make W and insert it into the ER membrane; vesicles carry it to the Golgi, then to the plasma membrane, where they fuse. Its nonpolar R groups face the fatty acid tails.
(d) The solution has a higher water potential, so water enters by osmosis in both groups. Without channels, water crosses the bilayer only in small amounts. W behaves as an aquaporin: its pore lets many water molecules through, still down the gradient.
(e) Compound T cuts the rate by about 85% (mean 1.30, close to the control), linking the fast swelling to W itself. Test: treat W eggs with an inhibitor of ATP production, keeping temperature, solution and egg batch the same, and measure the initial rate. If the route is passive, it stays near 8.6.
| Point | What earns it |
|---|---|
| 1 | Both means and intervals |
| 1 | Null hypothesis stated and rejected, using non-overlapping intervals |
| 1 | Route: rough ER → Golgi → vesicle fusing with the plasma membrane |
| 1 | Nonpolar R groups against the fatty acid tails |
| 1 | Slow bilayer crossing versus fast passage through a water channel, both down the gradient |
| 1 | Compound T as evidence that W causes the effect |
| 1 | ATP-inhibitor test with a controlled variable and the predicted result |
Total: 7 points.
Question 7 (constructed response · mixed)
A fictional spherical bacterium takes up a sugar only by facilitated diffusion. Its membrane has 150 carriers per µm², and each carrier moves at most 400 sugar molecules per second. Each µm³ of cytoplasm uses 1.5 × 10⁵ sugar molecules per second.
(a) For cells of radius 1.0 µm and 2.0 µm, calculate the maximum supply and the demand. State which cell can meet its demand. (b) Calculate the carrier density the 2.0 µm cell would need to meet its demand. (c) A student says the 2.0 µm cell could meet demand by spending more ATP on its carriers. Evaluate this, and suggest two other changes that would work. (d) Explain why real uptake is usually below the maximum supply, and why the carriers can never make the inside more concentrated than the outside.
Worked solution and suggested Marlbridge rubric
(a) Maximum supply per µm² = 150 × 400 = 6.0 × 10⁴ molecules s⁻¹.
- r = 1.0 µm: SA = 12.57 µm², V = 4.19 µm³. Supply = 7.54 × 10⁵; demand = 6.28 × 10⁵. Supply is larger: meets demand.
- r = 2.0 µm: SA = 50.27 µm², V = 33.51 µm³. Supply = 3.02 × 10⁶; demand = 5.03 × 10⁶. Cannot meet demand.
(b) Needed supply per µm² = demand ÷ SA = 5.03 × 10⁶ ÷ 50.27 = 1.0 × 10⁵ molecules s⁻¹, so carriers = 1.0 × 10⁵ ÷ 400 = 250 per µm², two-thirds more than now.
(c) Wrong. Facilitated diffusion is passive: the carriers use no ATP. Changes (any two): a smaller cell (r ≤ 1.2 µm gives the needed SA/V of 2.5 µm⁻¹); a thin rod shape; membrane folds.
(d) The maximum assumes every carrier is always busy, which needs a high outside concentration. Carriers only give the sugar a path, so net movement stops when inside equals outside; going higher needs active transport.
| Point | What earns it |
|---|---|
| 1 | Supply and demand for r = 1.0 µm with the correct decision |
| 1 | Supply and demand for r = 2.0 µm with the correct decision |
| 1 | 250 carriers per µm² |
| 1 | ATP rejected: facilitated diffusion is passive |
| 1 | Two valid changes linked to supply or SA/V |
| 1 | Saturation and the equal-concentration limit explained |
Total: 6 points.
How did you do?
Add up your points (3 multiple choice, 26 constructed response). Compare only with your own next attempt; the total does not predict an exam score.
- SA/V (Questions 2, 4, 7): rework Topic 2.2.
- Water potential and tonicity (Questions 3, 5, 6): check i, kelvin, signs and which solutes can cross in Topic 2.7.
- Passive or active (Questions 5(e), 6(e), 7(c)): reread Topic 2.5 and Topic 2.8.
- Quick recheck: retake the unit diagnostic.
Then tick off each topic checklist: