2.2 Cell Size

1. A0.50 μm⁻¹ Correct
   SA = 2lh + 2lw + 2wh = 2(20)(10) + 2(20)(10) + 2(10)(10) = 1000 μm². V = lwh = 20 × 10 × 10 = 2000 μm³. SA/V = 1000/2000 = 0.50 μm⁻¹.
2. B2.00 μm⁻¹
   A student who divides the volume by the surface area picks this: 2000/1000 = 2.00. The ratio of surface area to volume is SA ÷ V = 0.50 μm⁻¹.
3. C0.25 μm⁻¹
   A student who adds only the three faces visible in the drawing picks this: (200 + 200 + 100)/2000 = 0.25. A rectangular solid has six faces, each visible face having a hidden partner, so SA = 1000 μm².
4. D0.10 μm⁻¹
   A student who uses the area of one 20 μm × 10 μm face as the surface area picks this: 200/2000 = 0.10. Surface area is the total area of all six faces, 1000 μm².

Working SA (rectangular solid) = 2lh + 2lw + 2wh = 2(20 μm)(10 μm) + 2(20 μm)(10 μm) + 2(10 μm)(10 μm) = 400 + 400 + 200 = 1000 μm². V = lwh = (20 μm)(10 μm)(10 μm) = 2000 μm³. SA/V = 1000 μm² ÷ 2000 μm³ = 0.50 μm⁻¹. Distractors: V/SA = 2000/1000 = 2.00; three visible faces only, (200 + 200 + 100)/2000 = 0.25; one 20 μm × 10 μm face only, 200/2000 = 0.10.

CED 2.2.A.1 · Read this in Fix

1. ACell B has the higher SA/V, as its larger membrane gives it more surface per unit of volume.
   A student who thinks a larger cell has a higher SA/V because it has more membrane picks this. Cell B's surface area is 16 times cell A's, but its volume is 64 times larger, so its SA/V is a quarter of cell A's.
2. BCell A has the higher SA/V, so it exchanges materials with its surroundings more efficiently. Correct
   For a sphere, SA/V = 3/r, so cell A's SA/V (0.6 μm⁻¹) is four times cell B's (0.15 μm⁻¹). With more membrane per unit volume, the smaller cell exchanges materials with its surroundings more efficiently.
3. CThe two cells have the same SA/V, since both are spheres and so have the same shape.
   A student who thinks cells of the same shape have the same SA/V picks this. SA/V = 3/r for a sphere, so it depends on the radius: 0.6 μm⁻¹ for cell A and 0.15 μm⁻¹ for cell B.
4. DCell A has the higher SA/V, but cell B exchanges more efficiently due to its larger surface area.
   A student who thinks total surface area, not SA/V, sets how well a cell exchanges materials picks this. Cell B has more membrane in total, but much less membrane per unit of the volume it must supply.

CED 2.2.A.2.i · Read this in Fix

1. AGraph 1
   A student who thinks a larger cell has a higher SA/V because it has more membrane picks this rising line. Surface area does increase, but volume increases faster, so SA/V falls (6/s).
2. BGraph 2
   A student who assumes that SA/V falls by the same amount for each 1 μm increase picks this straight line. SA/V = 6/s falls by 3 μm⁻¹ between 1 and 2 μm but by only 0.3 μm⁻¹ between 4 and 5 μm, so the graph is a curve.
3. CGraph 3 Correct
   SA/V of a cube = 6s²/s³ = 6/s, so it falls as s increases: 6 μm⁻¹ at 1 μm, 3 μm⁻¹ at 2 μm, 2 μm⁻¹ at 3 μm and 1.2 μm⁻¹ at 5 μm. The fall is steep at first and then gradual, which gives a curve that flattens as side length increases.
4. DGraph 4
   A student who thinks surface area and volume increase in proportion, so that their ratio stays the same, picks this horizontal line. SA/V = 6/s, which changes with s.

CED 2.2.A.2.i · Read this in Fix

1. AQ's microvilli beat back and forth, sweeping the nutrients from the gut into the cell.
   A student who thinks microvilli beat like cilia picks this. Microvilli are folds of the membrane that add surface area; they do not beat or sweep material.
2. BQ's microvilli add extra volume, so the cell has more room to take nutrients in.
   A student who thinks more volume lets a cell absorb faster picks this. The model states that the two cells have equal volumes; it is the extra membrane area, not extra room, that speeds absorption.
3. CQ grew microvilli because it needed to take in more nutrients than P did.
   A student who thinks cells develop structures because they need them picks this. A need does not produce microvilli, and the statement does not explain how microvilli speed absorption: they add membrane surface area.
4. DQ has far more membrane in contact with the gut contents, yet its volume is similar to P's. Correct
   The microvilli are folds of the plasma membrane, so in the model cell Q has much more membrane facing the gut contents than cell P while enclosing about the same volume. More membrane area means a higher SA/V and more surface across which nutrients can be absorbed at once.

CED 2.2.A.2.iii · Read this in Fix

1. AThe largest model, which has the most surface area, cooled the fastest of the three.
   A student who thinks a larger body loses heat faster because it has more total surface area picks this. The 2.7 kg model's curve is the least steep: it has more surface in total but much more mass to cool.
2. BAll three models cooled at the same rate, since all of them started at 40 °C.
   A student who thinks objects starting at the same temperature cool at the same rate picks this. The three curves start together at 40 °C but separate at once; the 0.1 kg model cools fastest.
3. CThe smaller the model, the faster its temperature fell toward the bath temperature. Correct
   After 10 minutes the 0.1 kg model had cooled to about 25.7 °C, the 0.8 kg model to about 30.7 °C and the 2.7 kg model to about 33.2 °C; the smaller the model, the more steeply its curve falls. Smaller masses have a higher SA/V and exchange proportionally more heat with their surroundings.
4. DThe 0.1 kg model lost more thermal energy than either of the two larger models.
   A student who treats temperature as the same thing as thermal energy picks this. The 0.1 kg model's temperature fell the most, but it has far less mass: by the end, the 2.7 kg model had lost about 22 times as much thermal energy.

CED 2.2.A.2.iv · Read this in Fix

1. ASpecies A, the smallest, used more O₂ in total each hour than species D did.
   A student who reads a rate per gram as the whole animal's rate picks this. Species A uses 20 × 1.70 = 34 mL O₂ per hour in total; species D uses 20,000 × 0.30 = 6,000 mL per hour.
2. BO₂ consumption per gram decreased as body mass rose from species A to D. Correct
   The rate per gram falls from 1.70 mL O₂/g per h in the 20 g species to 0.30 in the 20,000 g species, consistent with the pattern that the smaller a multicellular organism, the higher its metabolic rate per unit body mass.
3. CO₂ consumption per gram was about the same in all four of the species.
   A student who thinks total metabolic rate is proportional to body mass, so that each gram uses O₂ at the same rate, picks this. The rate per gram in species A (1.70) is more than five times that in species D (0.30).
4. DO₂ consumption per gram was highest in the species with the greatest body mass.
   A student who thinks larger bodies lose heat faster because of their greater surface area, and so need more energy per gram, picks this. The table shows the reverse: species D, the largest, has the lowest rate per gram.

CED 2.2.A.2.v · Read this in Fix

1. AIts SA/V will be twice the usual value, so exchange will keep pace with its needs more easily.
   A student who thinks a larger cell has a higher SA/V because it has more membrane picks this. Surface area has increased fourfold but volume eightfold, so SA/V has halved.
2. BIts SA/V will stay the same, so its membrane will meet its needs just as well as before.
   A student who thinks cells of the same shape have the same SA/V whatever their size picks this. Surface area and volume do not increase in proportion; SA/V has halved.
3. CIt will form membrane folds to restore its SA/V, as it needs more surface to supply itself.
   A student who thinks cells develop structures because they need them picks this. Nothing in the scenario suggests that the cell will change its membrane; a need does not cause a structure to form.
4. DIts SA/V will halve, so it will be harder for exchange to keep up with its needs. Correct
   Eight times the volume with the same shape means twice the linear size, four times the surface area and so half the SA/V. The cytoplasm's demand for resources has increased eightfold, but the membrane that supplies it has increased only fourfold, so exchange may not keep up with demand.

CED 2.2.A.2.ii · Read this in Fix

1. AIts uptake of water and mineral ions will fall until new root hairs grow from the root. Correct
   Each root hair is tiny, but together many thousands of them add a large area of plasma membrane across which water and mineral ions are absorbed. Losing many of them removes much of this surface, so uptake falls until new root hairs restore the surface area.
2. BIts uptake will hardly change, as root hairs are too small to affect the whole root's surface.
   A student who thinks microscopic structures are too small to matter picks this. Each root hair is tiny, but together the very many root hairs make up a large part of the absorbing surface.
3. CIts water supply will be unaffected, since a plant takes in most of its water through its leaves.
   A student who thinks plants take in their water through their leaves picks this. A land plant absorbs most of its water from the soil through its roots; leaves lose water to the air.
4. DIts uptake will rise, with the remaining root cells absorbing faster to make up for the loss.
   A student who thinks cells change how fast they take in materials because they need to picks this. Absorption depends on the membrane area available, and much of that area has been lost.

CED 2.2.A.2 · Read this in Fix

1. AThey hold a large volume of warm blood, which stores the heat away from the rest of the elephant's body.
   A student who thinks heat is a substance that can be stored in a body part picks this. Thermal energy in the ears' blood is transferred to the cooler air; the ears are thin and hold little volume.
2. BThey add a large surface area but little volume, raising the rate at which heat leaves the body. Correct
   Thin, flat ears add a large area of skin, close to many blood vessels, while adding little volume. This raises the elephant's overall SA/V, so heat is transferred from its blood to the cooler air faster than it would be without them.
3. CThey make the elephant larger overall, which lowers its SA/V and so its rate of heat loss.
   A student who thinks SA/V depends only on overall size picks this. Thin flaps add much more surface than volume, so they raise SA/V rather than lowering it.
4. DThey add to the elephant's mass, and a body with more mass cools faster than a lighter one.
   A student who thinks larger bodies cool faster because they have more surface picks this. A body with more mass has a lower SA/V and cools more slowly; the ears help because their thin shape adds surface, not because they add mass.

CED 2.2.A.1 · Read this in Fix

1. AIt is valid: larger cubes need longer in the acid, so these times make the comparison fair to each of them.
   A student who thinks a fair test gives larger objects more time picks this. Giving larger cubes more time adds a second variable; a fair test keeps time the same for every cube.
2. BIt is flawed only because it has no control cube that is left in water instead of the acid solution.
   A student who thinks an experiment is valid whenever it has a no-treatment control picks this. A cube in water could show that the color change is caused by acid, but it would not remove the problem that time differs between the cubes.
3. CIt is flawed: the cubes spent different times in acid, so the effect of size cannot be separated from that of time. Correct
   Both cube size and time in acid differ between the cubes, so any difference in the percentage that changed color could be caused by either. For a valid test, every cube should spend the same time in the same acid solution, leaving size as the only difference.
4. DIt is valid: the cubes differ in size, so size must be the cause of any difference in the results.
   A student who thinks any difference in results can be put down to the variable deliberately changed picks this. Time in acid also differs between the cubes, so a difference in results cannot be attributed to size alone.

CED 2.2.A.2.ii · Read this in Fix

1. AA shrew's cells are much smaller than a deer's, so each has a higher SA/V and needs more food per gram.
   A student who thinks larger animals are made of larger cells picks this. Cells of a given type are of broadly similar size in shrews and deer; the shrew needs more food per gram because its whole body has a higher SA/V and loses heat proportionally faster, not because its cells are smaller.
2. BA shrew's small body can store only a little heat, so it must eat often to refill that store.
   A student who thinks heat is a substance stored in the body picks this. Thermal energy is not kept in a store; it is transferred to the cooler surroundings, and a small body, with its higher SA/V, transfers it proportionally faster, so the shrew must release more energy per gram.
3. CA shrew uses more energy in total each day than a deer does, because its rate per gram of body mass is higher.
   A student who treats a rate per gram as the whole animal's rate picks this. A deer has thousands of times more mass, so it uses far more energy in total, even though its rate per gram is lower.
4. DA shrew has a higher SA/V, so it loses heat faster per gram and needs more energy per gram to replace it. Correct
   Because the shrew is small, it has a higher SA/V than the deer and loses heat proportionally faster to its surroundings. To keep its body temperature constant it must release more energy per gram, so its metabolic rate per unit body mass, and its food intake per gram, are higher.

CED 2.2.A.2.v · Read this in Fix