P01-a · 2 actual pages
Card A · From a cell to its boundary
A typical cell has a plasma membrane separating its inside from its surroundings. The membrane is not the same as a cell wall. In a typical plant-cell model a wall lies outside the membrane. Typical prokaryotic and eukaryotic models both have membrane, cytoplasm, DNA and ribosomes; DNA location and internal organisation differ. Specialised mature cells may not fit every typical inventory.
An organelle has a function within a cell. A tissue is a group of cells and associated material organised for a function; multiple tissues contribute to an organ. A membrane's transport properties help a cell exchange materials, but one membrane drawing does not explain an entire organ or body system. If a sketch leaves a structure unlabelled, say “not shown” unless its stated inventory permits a claim of absence.
Bridge case: the complete comparison inventory of Model Leaf L states membrane, wall, cytoplasm, nucleus, mitochondria, chloroplasts and ribosomes. Model Bacterium B states membrane, wall, cytoplasm, DNA in a nucleoid region and ribosomes, with no membrane-bound nucleus. These are limited comparison inventories, not exhaustive descriptions of real specimens.
Slot 11 · Place the boundary
Read Cards A and B. Name three features shared by Leaf L and Bacterium B and one difference in DNA organisation. Describe where a plant cell wall sits relative to its membrane. On a two-line membrane sketch, place heads and tails and explain one placement. Finish: “This model shows …; it does not tell me …”.
P01-b · 2 actual pages
Card B · A fluid mosaic
A simplified typical animal plasma membrane model has a phospholipid bilayer. Each phospholipid has a water-interacting head and water-avoiding tails. Heads face the watery spaces on both sides; tails face inward toward other tails. Many lipids and proteins can move laterally, although not every component moves freely. “Fluid” describes molecular mobility, not a hole or a flowing river.
Proteins have several roles. A channel can provide a selective hydrophilic route across the membrane. A carrier binds and changes shape to move a substance; its mechanism may be passive or energy coupled. A glycoprotein has an attached carbohydrate chain and can contribute to recognition, adhesion or signalling. Not every protein is a channel. Cholesterol lies among phospholipids in this animal model and helps regulate fluidity and permeability; its effects depend on conditions. Do not claim that all cells have the same cholesterol content or that cholesterol always makes a membrane more fluid.
Our word map, from outside to inside: watery outside; outward-facing carbohydrate on a glycoprotein; heads; inward tails with cholesterol; heads; watery cytoplasm. A channel spans the bilayer. This is a simplified model, with no scale and no full inventory of membrane components.
Slot 12 · Give a component a job
Use Card B. Explain how a channel, a glycoprotein and cholesterol contribute to this animal membrane model. Add a carrier's possible role. Correct this caption: “Every protein is a pump, and cholesterol always increases fluidity.” Keep a specific limit beside your correction.
P02-a · 2 actual pages
Card C · Net movement and mechanisms
Particles move randomly in both directions. “Net movement” compares those opposing movements. For a permitted neutral substance, simple diffusion has net movement down its concentration gradient through the bilayer, without cellular energy supplied to that transport. Small nonpolar molecules such as oxygen can take this route.
Facilitated diffusion uses a suitable channel or carrier and moves down the relevant gradient, without cellular energy supplied to that transport. A route being a protein does not make it active. For charged substances the relevant electrochemical gradient combines concentration and electrical effects. Do not settle their net passive direction from concentration alone.
Protein-mediated active transport is coupled to an energy source and can maintain movement against a gradient. Energy can come directly from ATP or indirectly from a stored ion gradient. Our simplified pump cases explicitly use ATP coupling; this does not mean all active transport uses ATP directly. An energy-coupled pump is not just an open channel.
At dynamic equilibrium, opposing molecular movements continue but there is no net movement for that substance under the stated stable conditions. A labelled difference without membrane permeability information is insufficient to predict crossing.
Transport file T: neutral molecule Q is 8 mmol/L outside and 2 mmol/L inside. Q cannot cross the bare bilayer but has a suitable passive carrier. Neutral molecule R is 1 mmol/L outside and 5 mmol/L inside; an ATP-coupled pump is explicitly carrying R inward. Oxygen is 6 mmol/L outside and 2 mmol/L inside and can cross the bilayer. These concentrations are invented teaching values, not physiological records.
Slot 13 · Draw a justified net arrow
Use Card C's Transport File T. For Q, R and oxygen, name the route, net direction and relevant energy statement. Explain why an arrow describes net movement. A separate ion is 10 mmol/L outside and 2 mmol/L inside; a suitable channel is open but no electrical conditions are given. Can concentration alone settle its net direction? Explain the information you still need.
P02-b · 2 actual pages
Card D · Osmosis with stated conditions
Osmosis is net water movement across a selectively permeable membrane down a water-potential gradient. For the simple cases here, one neutral solute cannot cross, water can cross, temperature is equal, and initial hydrostatic pressure is equal. Under those conditions, the side with more dissolved solute has lower water potential, so initial net water movement is toward that side. Solute itself is not the substance moving by osmosis.
File O: a flexible model pouch contains 0.30 mol/L of impermeant neutral solute. Its surrounding solution contains 0.10 mol/L of the same solute. At the stated equal temperature and equal initial pressure, initial net water movement is into the pouch. We have not supplied enough information for final volume, time or how a real cell survives.
If pressure differs or is unknown, solute concentration alone may not determine net water direction. A cell wall can contribute to pressure. Water molecules still move both ways when net movement is zero. These are initial predictions about supplied model conditions, not instructions to prepare solutions or change a person's hydration.
Slot 14 · Which material crosses?
Use File O in Card D. State what can cross, the temperature and pressure conditions, and the initial net water direction. Give a reason without saying that solute moves by osmosis. Use Card E to distinguish V and X, explaining one feature of each sequence. Name one thing these descriptions cannot tell you about a real cell.
P02-c · 1 actual pages
Card E · Crossing by vesicles
Endocytosis encloses material in a membrane-bound vesicle formed from the cell boundary. Exocytosis releases vesicle contents when a vesicle fuses with that boundary. These are energy-requiring cellular processes, different from a single molecule passing through a channel or bilayer. A large particle enclosed by membrane is not simply “diffusing through a large hole.”
File V has three described frames: 1, a large external particle near the plasma membrane; 2, the membrane wraps around the particle; 3, a closed vesicle lies inside. This represents endocytosis. File X has an internal vesicle approaching the membrane, fusion, then contents outside: exocytosis. The frames are original conceptual descriptions, not filmed events. Neither frame set identifies the specific organism, cargo chemistry or measured rate.
Slot 14 · Which material crosses?
Use File O in Card D. State what can cross, the temperature and pressure conditions, and the initial net water direction. Give a reason without saying that solute moves by osmosis. Use Card E to distinguish V and X, explaining one feature of each sequence. Name one thing these descriptions cannot tell you about a real cell.
P03-a · 1 actual pages
Card F · Shape, size and units
For a cube with edge a, surface area = 6a² and volume = a³. Surface-area-to-volume ratio = 6/a. Keep area and volume units: dividing µm² by µm³ gives µm⁻¹. A ratio can be written as a surface area : volume pair, but “6 : 1” without dimensions can conceal a size change. A centre is a/2 from the nearest face.
| Model cube | Edge | Surface area | Volume | SA/V | Face-to-centre distance |
|---|---|---|---|---|---|
| P | 10 µm | 600 µm² | 1,000 µm³ | 0.60 µm⁻¹ | 5 µm |
| Q | 20 µm | 2,400 µm² | 8,000 µm³ | 0.30 µm⁻¹ | 10 µm |
| R | 40 µm | 9,600 µm² | 64,000 µm³ | 0.15 µm⁻¹ | 20 µm |
Larger similar shapes have more total surface but less surface per unit volume and a longer route to the centre. These features can constrain exchange relative to demand and help explain cell-size limits. They do not alone prove an exact maximum size for all cells: permeability, shape, internal transport and metabolic demand also matter. Folds can add exchange surface; a thin shape can shorten diffusion distance. This table is geometry, not a measured biological experiment.
Slot 16 · Keep the units
Use Card F's rules. A new cube S has edge 30 µm. Calculate its surface area, volume, SA/V and face-to-centre distance. Compare it with P's 10 µm edge: does S have more total surface? More surface per unit volume? Explain both answers.
P03-b · 1 actual pages
Slot 17 · Same volume, different shape
A cube is 2 × 2 × 2 units. A rectangular model is 1 × 2 × 4 units. For a rectangular box, area = 2(lw + lh + wh), volume = lwh. Calculate each area, volume and area/volume. Find each shortest centre-to-face distance. Explain how the thinner model might help exchange and name one reason geometry alone cannot predict an actual cell's exchange rate.
P04-a · 2 actual pages
Card G · Two measures that answer different questions
Three hypothetical cube models have edges 2, 4 and 8 mm. At 4 minutes a supplied idealised penetrated layer is 0.5 mm deep from every face. Same material, temperature and surrounding conditions are assumed. 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. The unpenetrated central cube has edge max(a − 2d, 0). Fraction penetrated = 1 − [max(a − 2d, 0)/a]³. This is a geometrical fraction, with no time unit. A layer is counted as penetrated; it is not a direct measure of molecule number or equal concentration throughout that layer.
| Edge a | Depth d at 4 min | Central edge | Fraction penetrated |
|---|---|---|---|
| 2 mm | 0.5 mm | 1 mm | 0.875 = 87.5% |
| 4 mm | 0.5 mm | 3 mm | 0.578125 = 57.8125% |
| 8 mm | 0.5 mm | 7 mm | 0.330078125 = 33.0078125% |
The average front speed over this interval is (0.5 − 0) mm ÷ (4 − 0) min = 0.125 mm/min for all three supplied models. Smaller cubes have a larger fraction penetrated at this time, despite the same stated depth and average front speed. Do not conclude that molecules intrinsically diffuse faster in the smaller cube. This imposed constant-depth example is not the physical claim that diffusion depth always increases linearly with time.
Slot 18 · How much has been reached?
Use Card G's geometric rule. A model cube has edge 6 mm and a layer 1 mm deep from every face at 5 minutes. 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. Find the central edge and fraction penetrated. Separately calculate the interval-average change in depth divided by elapsed time in mm/min. Explain why those two results answer different questions. Do not extend this one-time value into a constant speed for every later time.
P04-b · 2 actual pages
Card H · Evidence worth asking for
Supplied plan H compares models with different edge sizes. It proposes measuring time to a defined centre endpoint while holding material, external concentration, temperature, stirring and measurement rule constant. Independent model samples would be repeated at each size; the plan does not say they have been tested. Holding conditions constant helps make a size comparison interpretable. Repeats can reveal variation but do not remove a consistent measurement bias.
Supplied invented repeat set: model A's endpoint times are 100, 104 and 108 seconds; model B's are 160, 164 and 168 seconds. Means are 104 s and 164 s; ranges are 8 s each. A's listed times are shorter, with no overlap in these small supplied sets. That supports a bounded comparison of the supplied numbers. It does not prove statistical significance, population behaviour, an exact cell-size law or that an actual investigation happened.
When evaluating a claim, name the measure, give the relevant numbers, state the conditions, and say what would improve the evidence. Avoid treating an unfamiliar value as an error merely because it disagrees with a prediction. Check the method and retain the value unless there are documented grounds for exclusion.
Original educational explanations and supplied models © NeuroForgeIO Pty Ltd 2026, CC BY 4.0. Scientific source scopes and simplifications are recorded in SOURCE-SCOPE.md.
Slot 19 · Improve a plan
Read Card H. Name the proposed independent variable, the dependent measure, and three conditions to keep constant. Calculate both supplied means and ranges. Give a conclusion about those numbers, one limitation, and a practical improvement to the proposed measurement or repeats. State whether a learner investigation has actually occurred.