AP Biology Quiz: Cell Size

What this quiz covers

This quiz focuses on Cell Size, giving you a quick way to practice the rules, question types, and explanations that matter most for AP Biology.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

Question 1

1. The larger cell has decreased volume, so fewer nutrients are needed to maintain concentration.
2. The larger cell has increased surface area, so each transporter must work less efficiently.
3. The larger cell has a reduced surface area–to–volume ratio, limiting influx per unit cytoplasm. (correct answer)
4. The larger cell has more transporters total, so nutrient concentration should rise faster inside.
5. The larger cell has a thicker membrane, increasing the diffusion distance across the bilayer.

Explanation: This question evaluates surface area-to-volume ratio reasoning in nutrient transport as cells grow. As the cell's diameter increases, its surface area-to-volume ratio decreases because volume scales with the cube of the radius while surface area scales with the square, reducing the membrane available for transport per unit of internal volume. This limits the influx of nutrients relative to the metabolic demand, which remains constant per unit volume, leading to lower internal concentrations in the larger cell despite unchanged transporter density. Thus, the transport efficiency per cytoplasmic unit drops, causing nutrient scarcity inside. A tempting distractor is choice D, which suggests more total transporters improve efficiency, but this ignores the ratio and falls into the misconception of prioritizing absolute numbers over proportional capacity. A useful strategy is to recall scaling laws—surface with r², volume with r³—to predict limitations in growing cells or organisms.

Question 2

1. Cell Large has higher surface area–to–volume ratio, so solute influx per unit volume is lower in Cell Small.
2. Cell Small has higher surface area–to–volume ratio, increasing solute influx per unit volume early in diffusion. (correct answer)
3. Cell Small has lower total surface area, so solute enters faster because fewer molecules are needed to saturate it.
4. Cell Large has greater volume, which increases concentration gradient and slows diffusion into the smaller cell.
5. Cell Small has more transport proteins, which increases passive diffusion by decreasing membrane selectivity.

Explanation: This question probes surface area-to-volume ratio effects on solute influx rates. Cell Small (radius 2 µm) has a higher surface area-to-volume ratio than Cell Large (6 µm), enabling faster solute entry per unit volume and a quicker rise in internal concentration early on. The smaller cell's geometry supports more efficient diffusion relative to its volume. This explains the differential rates. A tempting distractor is choice C, claiming lower total surface area speeds entry in small cells, but this ignores that efficiency stems from the ratio, not total area, a frequent misconception. For time-dependent diffusion, use surface area-to-volume to compare per-volume changes.

Question 3

1. Cell Z has the highest uptake per unit volume because its total surface area is greatest among the three cells.
2. Cell Y has the highest uptake per unit volume because intermediate size maximizes diffusion across membranes.
3. Cell X has the highest uptake per unit volume because it has the greatest surface area–to–volume ratio. (correct answer)
4. All three cells have equal uptake per unit volume because membrane permeability is the same for glucose.
5. Cell Z has the highest uptake per unit volume because larger volume produces a larger concentration gradient.

Explanation: This question tests the role of surface area-to-volume ratio in glucose uptake efficiency via diffusion. Cell X (radius 4 µm) has the highest surface area-to-volume ratio among the three, providing more membrane area per unit volume for glucose to diffuse inward compared to the larger Cells Y and Z. This greater ratio enhances transport efficiency, allowing higher glucose uptake per unit volume despite identical permeability and consumption rates. Consequently, smaller cells like X maintain better nutrient supply relative to their cytoplasmic needs. A tempting distractor is choice A, which claims Cell Z has the highest uptake due to greatest total surface area, but this overlooks the misconception that total area alone determines efficiency, ignoring the critical ratio to volume. For problems involving cell size and diffusion, prioritize comparing surface area-to-volume ratios to predict per-unit-volume outcomes.

Question 4

1. Doubling diameter increases membrane thickness, which decreases permeability to nonpolar molecules.
2. Doubling diameter increases surface area faster than volume, lowering diffusion time for equilibration.
3. Doubling diameter decreases surface area–to–volume ratio, reducing exchange capacity per unit volume. (correct answer)
4. Doubling diameter decreases total surface area, reducing the number of molecules that can cross per second.
5. Doubling diameter increases external concentration gradient, slowing net diffusion into the cell.

Explanation: This question tests understanding of how surface area-to-volume ratio affects cellular transport efficiency. Doubling the diameter halves the surface area-to-volume ratio, reducing diffusion capacity per unit volume and slowing equilibration with external concentrations. As use occurs per volume, the larger cell requires more time to match influx to demand through concentration gradients. The transport efficiency logic shows size increases hinder rapid exchange due to volume outpacing area. A tempting distractor is choice B, which incorrectly states SA increases faster than V, a reversal of the actual geometric relationship. To approach similar problems, recall SA/V formulas and predict effects on diffusion timescales.

Question 5

1. Cell Y has fewer transport proteins because larger cells reduce membrane protein density.
2. Cell Y has a lower surface area–to–volume ratio, limiting glucose entry relative to demand. (correct answer)
3. Cell Y has a higher surface area, so glucose leaves the cell faster than it enters.
4. Cell Y has a smaller cytoplasmic volume, so glucose concentration drops more quickly.
5. Cell Y has a higher metabolic rate because larger cells always metabolize faster per unit volume.

Explanation: This question tests understanding of how surface area-to-volume ratio affects cellular processes like glucose import. The correct answer is B because cell Y, being larger, has a lower surface area-to-volume ratio, which limits the membrane area available for glucose transport relative to the cytoplasmic volume that metabolizes it. Consequently, glucose entry via facilitated diffusion cannot match the demand in the larger volume, causing a faster drop in concentration. This transport efficiency logic explains why larger cells face challenges in sustaining nutrient levels through membrane-based uptake. A tempting distractor is C, which wrongly suggests that higher surface area increases glucose loss, based on the misconception that total area directly correlates with outward diffusion rates without considering the ratio. To approach similar problems, always compare surface area-to-volume ratios to evaluate how size impacts supply relative to demand.

Question 6

1. Cell NN has a higher surface area–to–volume ratio, allowing more water movement per unit volume. (correct answer)
2. Cell NN has a lower surface area–to–volume ratio, allowing more water movement per unit volume.
3. Cell NN has less total surface area, which increases water efflux per unit volume by concentrating aquaporins.
4. Cell NN has a larger total volume, which increases osmotic driving force and speeds water loss.
5. Cell NN has a thicker membrane, which increases permeability and accelerates water loss per unit volume.

Explanation: This question tests understanding of how surface area-to-volume ratio affects cellular transport efficiency. Cell NN's smaller size gives it a higher surface area-to-volume ratio, enabling more water efflux per unit volume under the same osmotic gradient. This results in a greater fractional volume loss as water moves faster relative to initial volume. The transport efficiency logic applies to osmosis, where SA scales flux and V determines relative change. A tempting distractor is choice C, suggesting less total SA increases rate, but smaller cells have less total SA, confusing absolute vs. relative measures. To approach similar problems, analyze fractional changes by linking SA/V to osmotic rates per volume.

Question 7

1. Total channel number decreases as cell size increases, reducing total nutrient influx below demand.
2. Surface area increases with the cube of radius, so influx per unit volume decreases as size grows.
3. Volume increases faster than surface area, decreasing transport capacity relative to cellular demand. (correct answer)
4. Nutrient demand decreases with increasing volume, so larger cells should accumulate extra nutrients.
5. Channel flux increases automatically in larger membranes, offsetting any surface area limitations.

Explanation: This question assesses surface area-to-volume ratio in nutrient supply limitations. With constant channel density and flux, volume increases faster (r³) than surface area (r²), so transport capacity grows slower than demand, causing a nutrient deficit in the larger spherical cell. This mismatch means influx per unit volume decreases, failing to meet the fixed requirement per volume. Hence, larger cells face supply constraints despite unchanged per-channel properties. A tempting distractor is choice A, which claims channel numbers decrease with size, but actually total channels increase with area; the misconception lies in not recognizing relative insufficiency. For broader application, use proportional scaling to forecast metabolic constraints in enlarging cells or microbes.

Question 8

1. Larger cells have more total surface area, so toxin entry per minute is lower and concentration rises slowly.
2. Larger cells have lower surface area–to–volume ratio, so toxin entry per unit volume is reduced. (correct answer)
3. Larger cells have higher surface area–to–volume ratio, so toxin entry per unit volume is reduced.
4. Larger cells have thicker membranes, which always decreases permeability to all toxins in the bilayer.
5. Larger cells have lower external toxin concentration because they absorb toxins faster from the environment.

Explanation: This question tests understanding of how surface area-to-volume ratio affects cellular transport efficiency. Larger cells have a lower surface area-to-volume ratio, so even with identical permeability, toxin entry per unit volume is reduced. This results in a slower concentration increase as influx doesn't scale with the larger volume. The transport efficiency logic shows concentration changes depend on SA for entry and V for dilution. A tempting distractor is choice A, which incorrectly links more total SA to lower entry, confusing overall flux with per-volume effects. To approach similar problems, focus on rates of concentration change by considering SA/V impacts on accumulation.

Question 9

1. The larger cell has a lower surface area–to–volume ratio, reducing O$_2$ entry per unit volume. (correct answer)
2. The smaller cell has less membrane, so O$_2$ molecules accumulate outside and diffuse inward faster.
3. The larger cell has more total surface area, so O$_2$ diffusion must outpace consumption.
4. The smaller cell consumes less total O$_2$, so its membrane becomes more permeable to oxygen.
5. The larger cell has higher cytosolic viscosity, which directly decreases membrane permeability to O$_2$.

Explanation: This question probes surface area-to-volume ratio concepts in oxygen diffusion and consumption. The larger cell's lower surface area-to-volume ratio means less membrane surface per unit volume for O₂ entry, which cannot keep pace with the consumption rate that is uniform per volume, resulting in lower internal O₂ levels. In the smaller cell, the higher ratio provides more efficient diffusion relative to its smaller volume, maintaining higher internal concentrations under the same conditions. This disparity arises because O₂ supply is surface-limited while demand is volume-dependent. A tempting distractor is choice C, which highlights total surface area but overlooks the ratio, representing the misconception that larger absolute area compensates for increased volume without considering efficiency per unit. To generalize, apply ratio calculations to assess gas exchange constraints in cells or tissues of varying sizes.

Question 10

1. The larger cell has a lower surface area–to–volume ratio, reducing diffusion supply relative to cytoplasmic demand. (correct answer)
2. The larger cell has a higher surface area–to–volume ratio, reducing diffusion supply relative to cytoplasmic demand.
3. The larger cell has a greater total surface area, so diffusion supply should exceed demand and raise concentration.
4. The larger cell has a smaller cytoplasmic volume, so consumption is lower and concentration becomes lower.
5. The larger cell has fewer collisions with molecules outside, so diffusion slows due to reduced Brownian motion.

Explanation: This question tests understanding of how surface area-to-volume ratio affects cellular transport efficiency. The larger Cell KK has a lower surface area-to-volume ratio, reducing diffusion supply of the molecule per unit volume relative to consumption. This leads to a lower steady-state internal concentration to balance influx and use. The transport efficiency logic illustrates that without active transport, equilibrium depends on SA-limited passive flux matching volumetric demand. A tempting distractor is choice C, suggesting greater SA raises concentration, but lower ratio actually constrains it per volume. To approach similar problems, use steady-state concepts and SA/V to predict concentration differences.

Question 11

1. Cube B has more surface area, so dye entry must be faster into each region of cytoplasm.
2. Cube A has a greater surface area–to–volume ratio, increasing exchange capacity per unit volume. (correct answer)
3. Cube A has a smaller surface area, so dye molecules collide less and move inward faster.
4. Cube B has a larger volume, which increases the external dye concentration gradient at its surface.
5. Cube B contains more cytoplasm, which increases the rate of membrane transport proteins.

Explanation: This question assesses the role of surface area-to-volume ratio in modeling diffusion efficiency in cells. Cube A, with a side length of 1 unit, has a surface area-to-volume ratio of 6/1 = 6, while Cube B's ratio is 6/3 = 2, meaning Cube A has proportionally more surface area per unit volume for dye entry. This higher ratio enables faster dye penetration throughout Cube A's smaller volume, resulting in uniform coloring sooner as the dye doesn't have as far to diffuse relative to the exchange surface. In contrast, Cube B's larger volume demands more dye influx to achieve the same concentration change, but its lower ratio limits the efficiency of surface-mediated transport. A tempting distractor is choice A, which emphasizes total surface area over the ratio, embodying the misconception that bigger absolute area always means faster overall uptake without considering volume demands. For transferable strategy, remember to compare ratios rather than absolutes when analyzing how size affects diffusion-limited processes in biology.

Question 12

1. Cell B has greater membrane surface area, so passive ion entry should be greater per unit volume.
2. Cell A has a smaller surface area, which increases the net diffusion rate across its membrane.
3. Cell B has a lower surface area–to–volume ratio, reducing influx relative to cytoplasmic volume. (correct answer)
4. Cell B has a larger volume, which increases the concentration gradient across the membrane over time.
5. Cell A has less cytoplasm, so ions do not diffuse once they cross the membrane.

Explanation: This question explores surface area-to-volume ratio effects on passive ion influx. Cell B, being larger, has a lower ratio, meaning less surface for ion entry relative to its greater cytoplasmic volume, resulting in smaller concentration changes per unit volume compared to Cell A. Despite identical permeability, the volume in Cell B dilutes the impact of influx more significantly, leading to the observed difference. This demonstrates transport efficiency's dependence on size scaling. A tempting distractor is choice A, which suggests greater total area increases influx per volume, but this confuses absolute with relative measures, embodying the misconception of ignoring volume's cubic scaling. Strategically, scale linear dimensions to ratios when comparing exchange in geometrically similar structures.

Question 13

1. Reducing radius increases surface area–to–volume ratio, increasing efflux capacity per unit volume. (correct answer)
2. Reducing radius decreases surface area–to–volume ratio, increasing efflux capacity per unit volume.
3. Reducing radius decreases total surface area, increasing efflux rate because fewer molecules collide with the membrane.
4. Reducing radius increases volume faster than surface area, lowering internal concentration by dilution effects.
5. Reducing radius changes membrane permeability by compressing phospholipids and increasing diffusion across the bilayer.

Explanation: This question tests understanding of how surface area-to-volume ratio affects cellular transport efficiency. Reducing the radius increases the surface area-to-volume ratio, enhancing efflux capacity per unit volume for the produced molecule. This allows better removal relative to production, lowering internal concentration. The transport efficiency logic demonstrates smaller sizes improve exchange efficiency by boosting SA relative to V. A tempting distractor is choice C, claiming decreased total SA increases rate, but smaller radius decreases total SA, misleading on absolute terms. To approach similar problems, simulate size changes and evaluate SA/V impacts on concentration equilibria.

Question 14

1. Larger cells have less membrane surface area, so diffusion stops entirely.
2. Larger cells have a lower surface area–to–volume ratio, reducing exchange per volume. (correct answer)
3. Larger cells have higher surface area–to–volume ratio, reducing exchange per volume.
4. Larger cells increase gradient magnitude, so diffusion cannot replenish cytoplasm fast enough.
5. Larger cells reduce cytoplasmic volume relative to membrane area, increasing demand.

Explanation: This question requires understanding how surface area-to-volume ratio affects concentration maintenance. As a cell's radius increases, its surface area grows proportionally to r² while its volume grows proportionally to r³, causing the surface area-to-volume ratio to decrease. With a constant concentration gradient driving diffusion, the rate of molecule entry (proportional to surface area) increases more slowly than the cell's internal requirement (proportional to volume). This means less molecule enters per unit volume of cytoplasm in larger cells, making it harder to maintain the same internal concentration. Choice C incorrectly claims larger cells have higher surface area-to-volume ratios, which contradicts basic geometry. The strategy is to recognize that maintaining concentrations becomes progressively harder as cells grow because exchange capacity per unit volume always decreases.

Question 15

1. Cell M has higher entry per requirement because larger cells have more total membrane area.
2. Cell N has higher entry per requirement because smaller cells have higher surface area–to–volume ratio. (correct answer)
3. Both cells have equal entry per requirement because permeability is identical.
4. Cell M has higher entry per requirement because volume increases slower than surface area.
5. Cell N has lower entry per requirement because its membrane area is too small to matter.

Explanation: This question requires comparing surface area-to-volume ratios between cells of different sizes. If Cell M has 4× the volume of Cell N and they have the same shape, then Cell M's linear dimensions are ∛4 ≈ 1.59× those of Cell N. This means Cell M's surface area is only (∛4)² ≈ 2.52× that of Cell N. Therefore, Cell M has a surface area-to-volume ratio of 2.52/4 = 0.63× that of Cell N. Since solute entry depends on surface area and requirement depends on volume, Cell N has higher entry per unit requirement due to its higher surface area-to-volume ratio. Choice A incorrectly focuses on total membrane area rather than the ratio to volume. The strategy is to recognize that when comparing cells of different sizes, smaller cells always have advantages in exchange efficiency per unit volume.

Question 16

1. Cell T has more total surface area, so toxin influx per unit volume is lower than in Cell S.
2. Cell S has higher surface area–to–volume ratio, increasing toxin influx per unit cytoplasm. (correct answer)
3. Cell T has greater volume, so toxin molecules move faster once inside the cytoplasm.
4. Cell S has less volume, so the membrane becomes more permeable to toxins over time.
5. Cell T has a larger size, so the external toxin concentration near its membrane is always lower.

Explanation: This question examines how surface area-to-volume ratio affects toxin accumulation in cells. Cell S (3 μm sides) has a surface area-to-volume ratio of 2 μm⁻¹, while Cell T (9 μm sides) has a ratio of 0.67 μm⁻¹. Since the toxin enters by simple diffusion through the membrane and isn't broken down, the rate of concentration increase depends on influx per unit volume. Cell S, with its threefold higher surface area-to-volume ratio, experiences three times more toxin influx per unit of cytoplasm, causing it to reach a higher internal concentration over the same time period. Choice A incorrectly focuses on total surface area without considering the even larger difference in volume—while Cell T has more total surface area, it has proportionally much more volume to fill. The strategy is to calculate surface area-to-volume ratios to predict relative rates of concentration change.

Question 17

1. Larger cells have reduced surface area–to–volume ratio, limiting oxygen entry per unit cytoplasmic volume. (correct answer)
2. Larger cells have increased total surface area, so oxygen entry per unit volume is always higher than in smaller cells.
3. Larger cells maintain higher external oxygen concentration near the membrane, increasing diffusion into the cell interior.
4. Larger cells have more mitochondria, so oxygen diffuses into the cell faster to match increased demand.
5. Larger cells have lower cytoplasmic viscosity, which increases oxygen movement across the plasma membrane.

Explanation: This question tests understanding of how surface area-to-volume ratio affects oxygen availability in cells of different sizes. Population L cells, with twice the diameter of Population S cells, have half the surface area-to-volume ratio (for spheres, SA/V = 3/r). This means each unit volume of cytoplasm in the larger cells has access to proportionally less membrane area for oxygen entry via diffusion. Since oxygen consumption per unit volume is similar in both populations, the larger cells cannot replenish oxygen as efficiently as smaller cells, making them more prone to developing low oxygen levels. Answer B incorrectly claims that larger total surface area leads to higher oxygen per unit volume, failing to account for the even greater increase in volume. The critical insight is that as cells grow larger, their volume increases faster than their surface area, creating transport limitations.

Question 18

1. Cell Y has lower surface area–to–volume ratio, reducing diffusion across membrane relative to cytoplasmic volume to fill. (correct answer)
2. Cell Y has higher total surface area, so the molecule enters faster per unit volume than in Cell X.
3. Cell Y has a larger radius, which increases membrane permeability to lipid-soluble molecules compared with Cell X.
4. Cell Y has more intracellular targets, which increases the diffusion rate across the membrane into the cytoplasm.
5. Cell Y has a smaller volume, so fewer molecules are required to reach the same intracellular concentration as Cell X.

Explanation: This question examines how surface area-to-volume ratio affects the time to reach equilibrium for diffusing molecules. Cell Y, with three times the radius of Cell X, has one-third the surface area-to-volume ratio (SA/V = 3/r for spheres). This means that for each unit volume of cytoplasm that needs to be filled with the signaling molecule, Cell Y has proportionally less membrane area available for diffusion. Since diffusion rate depends on membrane area, Cell Y fills more slowly per unit volume, taking longer to reach the same intracellular concentration as Cell X. Answer B incorrectly claims that higher total surface area leads to faster entry per unit volume, failing to recognize that volume increases with r³ while surface area only increases with r². When comparing equilibration times, always consider how much membrane area is available per unit of volume to be filled.

Question 19

1. Cell 1, because a single membrane reduces resistance and increases diffusion rate.
2. Cell 2, because greater total membrane surface area increases exchange capacity. (correct answer)
3. Cell 1, because larger spheres have higher surface area–to–volume ratios.
4. Cell 2, because diffusion depends on volume more than surface area.
5. Both cells, because total volume alone determines the maximum diffusion rate.

Explanation: This question tests understanding of how membrane compartmentalization affects surface area-to-volume ratio. Cell 2, divided into many smaller compartments, has much greater total membrane surface area than Cell 1 despite having the same total volume. Since glucose uptake rate depends on membrane surface area while glucose demand depends on volume, Cell 2 has a higher uptake capacity per unit cytoplasm. This is because dividing a large sphere into many small spheres dramatically increases the total surface area while keeping volume constant. Choice A incorrectly assumes that having a single membrane somehow increases diffusion rate, ignoring that total surface area is what matters for total uptake. The key principle is that subdividing cells or creating internal membranes is a biological strategy to overcome surface area-to-volume limitations.

Question 20

1. The larger cell has a higher surface area–to–volume ratio, slowing diffusion out per unit volume.
2. The larger cell has a lower surface area–to–volume ratio, limiting CO$_2$ efflux per unit volume. (correct answer)
3. The larger cell has more membrane area, which decreases the concentration gradient for CO$_2$ diffusion.
4. The larger cell produces less CO$_2$ per unit volume, causing CO$_2$ to accumulate inside.
5. The larger cell has increased cytoplasmic volume, which increases diffusion rate across the membrane.

Explanation: This question tests surface area-to-volume ratio in waste efflux like CO₂ diffusion. The larger cell's reduced ratio means less membrane surface per unit volume for CO₂ to exit, causing buildup inside as production per volume is similar but efflux efficiency drops. In smaller cells, the higher ratio facilitates faster clearance relative to the smaller cytoplasmic volume, preventing accumulation. This highlights how size limits passive diffusion out of cells. A tempting distractor is choice C, which claims more membrane area decreases the gradient, but this misapplies diffusion principles and ignores the volume's faster growth, a common ratio misconception. A transferable approach is to use ratio trends to predict accumulation risks in larger cells or organisms without specialized transport systems.