# Twenty worked contexts · staff discussion models

These are original conceptual cases. Numerical values are invented or exact geometric results, not collected biological data. Model the reasoning when teaching an example; do not give the solution as the learner's fresh first-response work. Each context includes its whole source, question, solution and limit.

## W01 · A museum comparison card

**Source:** The complete limited Leaf L inventory lists membrane, wall, cytoplasm, nucleus, mitochondria, chloroplasts and ribosomes. Bacterium B lists membrane, wall, cytoplasm, nucleoid DNA and ribosomes, with no membrane-bound nucleus.

**Question:** Give three shared features and the decisive DNA-organisation difference. Does the wall replace the membrane?

**Worked answer:** Membrane, cytoplasm and ribosomes occur in both listed inventories; wall is also shared here. L's DNA is in a membrane-bound nucleus, while B's DNA is in the nucleoid region without that compartment. Both list a membrane as well as a wall, so wall does not replace membrane. This supports comparison of these models, not every specialised mature cell or every bacterium's wall.

## W02 · A reversed display

**Source:** An original animal-membrane sketch labels water on both sides, heads facing the middle and tails pointing into both watery spaces. Its caption says it is a simplified phospholipid bilayer, not a micrograph.

**Question:** Correct the placements and identify one valid limit.

**Worked answer:** Move water-interacting heads toward the water on each side. Point water-avoiding tails inward toward each other. This arrangement explains the bilayer's hydrophobic interior. The model cannot establish exact thickness, component spacing or what a specimen looks like under a microscope. A corrected drawing is still a representation, not an observation.

## W03 · A label-maker's four clues

**Source:** In an invented animal membrane map, component A offers a selective water-compatible passage across the bilayer; B has an external carbohydrate chain attached to protein; C binds a solute and changes shape; D lies between phospholipid tails and helps regulate fluidity/permeability.

**Question:** Identify each component and explain why all four are not pumps.

**Worked answer:** A is a channel; B a glycoprotein; C a carrier; D cholesterol in this animal model. B may contribute to recognition, adhesion or signalling. D is a lipid component, not a protein pump. A passive channel can allow movement down an electrochemical gradient; C's active/passive status needs mechanism and energy evidence. Shape or position alone does not prove that every transport protein is active.

## W04 · Two captions for a mobile membrane

**Source:** Caption 1 says, “Fluid means the membrane is an open leak; cholesterol always increases fluidity.” Caption 2 says, “Many components move laterally; permeability remains selective; cholesterol's effect depends on conditions.” No temperature experiment is supplied.

**Question:** Which caption is better supported by Card B, and what cannot be claimed?

**Worked answer:** Caption 2 retains molecular mobility and selective permeability. Cholesterol can influence fluidity and permeability in condition-dependent ways, so the “always” in Caption 1 overstates the model. The supplied map gives no measured temperature response or cholesterol concentration. We cannot infer an actual rate change or generalise identical composition to all cells.

## W05 · A transport storyboard

**Source:** Neutral Q is 8 mmol/L outside and 2 mmol/L inside, cannot cross the bare bilayer, and has a suitable passive carrier. Neutral R is 1 mmol/L outside and 5 mmol/L inside; an ATP-coupled pump is specified to carry it inward. Oxygen is 6 mmol/L outside and 2 mmol/L inside and can cross the bilayer.

**Question:** Name each route, net direction and energy statement.

**Worked answer:** Q moves net inward by facilitated diffusion down its concentration gradient, with no energy supplied to that transport. R is carried inward by the given active pump, against the stated concentration difference, using the stated ATP coupling. Oxygen moves net inward by simple diffusion through the bilayer, without cellular energy supplied to that transport. Concentrations and mechanisms are supplied model conditions; individual particles can move both ways and not all active transport uses ATP directly.

## W06 · An equal pair and a missing electrical clue

**Source:** Neutral S is 3 mmol/L on both sides, has a suitable passive route and stable equal conditions. A charged ion is 10 mmol/L outside and 2 mmol/L inside with an open selective channel, but membrane potential is unknown.

**Question:** Predict what is justified in each case.

**Worked answer:** S has no net concentration-driven movement under the stated stable conditions, though molecules keep crossing in both directions: dynamic equilibrium. The ion has a chemical concentration difference, but its net passive direction depends on the combined electrical and concentration effects. We need its charge and the relevant electrical conditions to settle that direction. Neither “channel equals pump” nor a concentration-only certain arrow is justified.

## W07 · Three paper pouches

**Source:** Each pouch permits water, not the same neutral solute. Initial temperature and hydrostatic pressure are equal on both sides. Pouch A: inside 0.30 mol/L, outside 0.10 mol/L. B: inside 0.10, outside 0.30. C: both 0.20.

**Question:** Predict initial net water movement and explain the equal case.

**Worked answer:** A has initial net water movement inward; B outward. Under the stated conditions, water moves net toward the side with higher impermeant-solute concentration and lower water potential. C has no net water movement under those equal conditions, though exchange continues in both directions. None gives a final volume or biological survival result; solute is not moving by osmosis in these cases.

## W08 · Two delivery animations on paper

**Source:** V: external large particle; membrane wraps around it; closed vesicle inside. X: internal vesicle approaches membrane; membranes fuse; contents released outside. There is no microscopy record or cargo identity.

**Question:** Identify both processes and distinguish them from facilitated diffusion.

**Worked answer:** V models endocytosis and X exocytosis. Both use vesicle/membrane reorganisation and cellular energy, rather than individual cargo simply moving down a gradient through a channel or carrier. The frame order supports the mechanism labels; it does not show a measured transport rate, organism identity or actual event. Not all cellular export follows this simplified route.

## W09 · Doubling a geometric cell

**Source:** Cube P has edge 10 µm; Q has edge 20 µm. Area = 6a²; volume = a³; centre distance = a/2.

**Question:** Calculate both and compare how doubling affects area, volume and ratio.

**Worked answer:** P: area 600 µm², volume 1,000 µm³, SA/V 0.60 µm⁻¹, distance 5 µm. Q: 2,400 µm², 8,000 µm³, 0.30 µm⁻¹, distance 10 µm. Area multiplies by 4; volume by 8; the ratio halves and the path doubles. More absolute area does not mean more area for each unit of volume. Real cells are not necessarily cubes or identical in permeability/demand.

## W10 · A larger cube for an exhibit

**Source:** P is the 10 µm cube above. R has edge 40 µm. A caption says, “R has less surface area than P, so it exchanges less.”

**Question:** Calculate R and replace the caption.

**Worked answer:** R's area is 6 × 40² = 9,600 µm²; volume 40³ = 64,000 µm³; ratio 9,600/64,000 = 0.15 µm⁻¹; nearest centre distance 20 µm. R has 16 times P's total area, 64 times its volume and one quarter its SA/V. Better caption: “R has more total surface, but less surface per unit volume and a longer path to its centre.” Geometry alone cannot quantify actual exchange.

## W11 · Equal-volume shapes

**Source:** Cube C is 2 × 2 × 2 units. Box D is 1 × 2 × 4 units. Box area = 2(lw + lh + wh); volume = lwh. Use arbitrary model length units u.

**Question:** Compare area, volume, SA/V and shortest centre-to-face distance.

**Worked answer:** C area = 2(4 + 4 + 4) = 24 u², volume 8 u³, ratio 3 u⁻¹, shortest distance 1 u. D area = 2(2 + 4 + 8) = 28 u², volume 8 u³, ratio 3.5 u⁻¹, shortest distance 0.5 u. D is thinner with more surface for the same volume. That may assist exchange under suitable conditions; the nearest-face path is not every possible internal path, and actual permeability/gradients remain unknown.

## W12 · A folded exchange patch

**Source:** Two conceptual patches each have projected footprint 4 mm². The smooth patch has actual membrane area 4 mm²; the folded patch 12 mm². Permeability and concentration gradients are assumed otherwise equal for a hypothetical comparison, but no flux is measured.

**Question:** What increases, by what factor, and what claim should remain conditional?

**Worked answer:** Actual membrane area increases from 4 to 12 mm², a factor of 3. Projected footprint remains 4 mm². Extra surface can offer more opportunity for exchange under the stated suitable conditions. The source does not measure flux or show that every substance crosses three times faster. Folds do not remove selectivity, gradient limits or other constraints.

## W13 · A layer around a four-millimetre cube

**Source:** At 4 min an idealised layer reaches 0.5 mm inward from every face of a 4 mm cube. The source imposes this layer; it is not an observed specimen. For this supplied interval, the layer starts at depth d_start = 0 mm when t_start = 0 min. Average front speed = Δdepth/Δtime = (d_end − d_start)/(t_end − t_start). This is the model's interval average, not a constant future speed, intrinsic diffusion coefficient or molecular flux.

**Question:** Calculate the unpenetrated core and penetrated fraction.

**Worked answer:** Core edge = 4 − 2(0.5) = 3 mm. Total volume = 4³ = 64 mm³; core = 3³ = 27 mm³. Penetrated model volume = 64 − 27 = 37 mm³. Fraction = 37/64 = 0.578125 = 57.8125%. Removing a layer from all six faces changes all three dimensions; a depth fraction is not a volume fraction. This proxy does not count molecules or establish a biological diffusion coefficient.

## W14 · Different fractions, equal stated front speed

**Source:** Edges 2 and 8 mm, each supplied with depth 0.5 mm at 4 min under the same idealised conditions. Core edge = max(a − 2d, 0); fraction penetrated = 1 − (core edge/a)³. For this supplied interval, the layer starts at depth d_start = 0 mm when t_start = 0 min. Average front speed = Δdepth/Δtime = (d_end − d_start)/(t_end − t_start). This is the model's interval average, not a constant future speed, intrinsic diffusion coefficient or molecular flux.

**Question:** Compare fractions and interval-average change in depth divided by elapsed time. Does the smaller cube establish faster intrinsic diffusion?

**Worked answer:** For 2 mm, core edge 1 mm, fraction = 1 − 1/8 = 87.5%. For 8 mm, core edge 7 mm, fraction = 1 − 343/512 = 169/512 = 33.0078125%. With d_start = 0 mm at t_start = 0 min, both interval-average front speeds = Δdepth/Δtime = (d_end − d_start)/(t_end − t_start) = (0.5 − 0) mm / (4 − 0) min = 0.125 mm/min. The small cube has a greater fraction penetrated at that time, not a greater supplied front speed. One supplied change-in-depth/elapsed-time pair cannot support a universal linear depth-versus-time rule.

## W15 · A fairer proposed comparison

**Source:** A proposed test compares a 2 mm model at 20°C with an 8 mm model at 30°C, then attributes every time difference to size. No test has been run.

**Question:** Identify the confound and improve the plan.

**Worked answer:** Size and temperature both change, so a difference could not be attributed to size alone. Use the same specified temperature, material, external concentration, stirring and endpoint rule for all size groups; measure a defined dependent quantity, such as seconds to a stated centre endpoint. Use independent samples and record uncertainty. Planning these changes is not evidence that a real experiment has occurred or that the expected result is guaranteed.

## W16 · Three supplied repeats

**Source:** Invented endpoint times A: 100, 104, 108 s. B: 160, 164, 168 s. A consistent endpoint rule and comparable conditions are stated for this arithmetic example.

**Question:** Calculate means, ranges and mean difference; give a bounded conclusion.

**Worked answer:** A mean = (100 + 104 + 108)/3 = 104 s; range 108 − 100 = 8 s. B mean = (160 + 164 + 168)/3 = 164 s; range 168 − 160 = 8 s. B's mean is 60 s longer. A's listed times are all shorter, with no overlap in these small supplied sets. This is not statistical significance, a population estimate, or real practical evidence. More independent data and method detail would be needed for broader claims.

## W17 · Overlap in a second pair

**Source:** Supplied invented times C: 90, 100, 110 s; D: 100, 110, 120 s. Comparable model conditions are stated; only three values per group are given.

**Question:** Compare means/ranges and evaluate “Every D result will exceed every C result.”

**Worked answer:** C mean 100 s, range 20 s. D mean 110 s, range 20 s. The observed intervals overlap from 100 to 110 s; a D value of 100 equals a C value and is below C's 110. The universal claim fails even in the supplied list. D's mean is higher by 10 s here; larger independent samples and uncertainty information are needed for a reliable broader comparison. Do not discard overlap as a mistake.

## W18 · A clock that adds five seconds

**Source:** A conceptual timing display always reports actual interval + 5 s. Three independent readings displayed are 105, 109 and 113 s. The offset is given as known for this model.

**Question:** Find corrected values, mean and range. Would more repeats alone fix this bias?

**Worked answer:** Corrected values are 100, 104 and 108 s. Displayed mean is 109 s; corrected mean 104 s. Both ranges are 8 s because subtracting the same 5 s does not change spread. More repeats with the uncorrected display would not remove the systematic offset. Real calibration would need documented evidence, not an assumed correction because a result is inconvenient.

## W19 · A root-surface explanation

**Source:** A conceptual plant cell has a long projection increasing its membrane area. A specified neutral solute U is 1 mmol/L outside and 4 mmol/L inside; an ATP-coupled protein pump carries U inward. No real plant is tested; real mineral ions have additional electrical considerations.

**Question:** Explain the roles of shape and mechanism without reducing one to the other.

**Worked answer:** The projection offers more membrane surface for potential exchange. The supplied inward U transport is active because its explicitly energy-coupled mechanism carries it against the stated concentration difference. More area alone does not supply energy or make a route permeable. This neutral-solute model illustrates a principle; it is not a literal account of every root ion pathway or a measured absorption rate.

## W20 · Choose a better diffusion caption

**Source:** The Card G 2 mm and 8 mm models have identical supplied 0.5 mm depths at 4 min. For this supplied interval, the layer starts at depth d_start = 0 mm when t_start = 0 min. Average front speed = Δdepth/Δtime = (d_end − d_start)/(t_end − t_start). This is the model's interval average, not a constant future speed, intrinsic diffusion coefficient or molecular flux. Caption A says, “Smaller cubes make molecules move faster.” Caption B says, “At this supplied time, the smaller cube has a larger fraction penetrated; the stated average front speed is the same.”

**Question:** Choose, justify and add one evidence limitation.

**Worked answer:** B distinguishes the 87.5% and 33.0078125% fractions from the common Δdepth/Δtime = (d_end − d_start)/(t_end − t_start) = (0.5 − 0) mm / (4 − 0) min = 0.125 mm/min interval average, using d_start = 0 mm and t_start = 0 min. A asserts an intrinsic speed difference not shown by the source. Add: “These idealised supplied layers are a geometric teaching model, not actual investigation measurements.” An actual inquiry would need a defined measure, comparable conditions, independent repeats and recorded observations.