Practice questions · Topic 2.2

Unit 2 · Practice for the Topic 2.2 end-of-topic test

You’ve gone through everything in this topic. The summary video below recaps it all, so you’re ready for the questions.

Watch first: Sum up: why a cell can’t just grow bigger

The cube, its ratio, the ratio falling as it grows, and why cells stay small or divide.

Watch next: folds, shapes, and heat per gram

Ways around the surface limit, shapes by the numbers, the leaf’s pores, and why a mouse eats a fifth of its mass a day.

These are practice questions in the shape of the topic test. Work through them before you take the test; every question tells you what it wanted.

These practice questions have the shape of the Topic 2.2 test. Answer every question. For each multiple-choice question, pick one option and press Check; the feedback gives the reasoning. For the first free-response question, you work through one scenario in small steps, and each step offers a hint if you want one. For the second, write one short sentence for each step of your reasoning, each on its own line, and make every link clear (so, because, therefore). That is what the exam’s ‘paragraph form’ means for you: linked sentences, not bullet points. Then open the scoring guide and mark your own work against it. Formulas are given wherever a calculation needs them; use the π key on your calculator and give answers to three significant figures.

Question 1

A cube-shaped plant cell has sides 12 μm long. For a cube of side s, surface area = 6s² and volume = s³.

What are its surface area and its volume?

A. 144 μm² and 1,728 μm³ B. 864 μm² and 1,728 μm³ C. 864 μm² and 144 μm³ D. 576 μm² and 1,728 μm³

Question 2

A cell has a surface area of 220 μm² and a volume of 200 μm³.

What is its surface-area-to-volume ratio?

A. 1.1 per μm B. 0.91 per μm C. 20 μm D. 1.1 μm²

Question 3

A spherical cell grows without changing shape. Its radius increases from 1 μm to 4 μm.

Compared with the small cell, how have its surface area, volume and surface-area-to-volume ratio changed?

A. Surface ×4, volume ×4, ratio unchanged B. Surface ×16, volume ×16, ratio unchanged C. Surface ×16, volume ×64, ratio 25% of what it was D. Surface ×64, volume ×16, ratio 4 times what it was

Question 4

Cells of a kind of bacterium normally divide whenever they reach 2 μm long. A drug stops them dividing but leaves them growing. At 8 μm long the cells take in oxygen too slowly for their needs and stop growing.

Why does the oxygen intake fall behind the cell's needs as the cell grows?

A. The membrane grows thicker as the cell grows, so oxygen crosses it more slowly than before B. A larger cell has less membrane in total than a small one, so less oxygen can cross into it C. The cell has used up all the oxygen dissolved in the liquid around it, so none is left to enter D. Oxygen enters only across the surface, while the need for it grows with the volume, which grows faster

Question 5

A cube-shaped cell with sides 6 μm long can bring in, across its membrane, 120% of the oxygen its interior uses each minute. A cell of the same kind grows to a cube with sides 12 μm long. Oxygen intake is proportional to membrane area; oxygen use is proportional to volume.

What percentage of its oxygen need can the 12 μm cell's membrane supply?

A. 60% B. 120% C. 240% D. 30%

Question 6

The top face of each cell lining a kidney tubule carries thousands of finger-like folds of its plasma membrane, called microvilli. The cell takes useful substances back from the tubule fluid across that face. A cell with microvilli has almost the same volume as a cell with a flat top face.

What do the microvilli do to the cell’s surface-area-to-volume ratio, and why?

A. They lower the ratio: each fold adds interior that the membrane must serve B. They raise the ratio: the folds add membrane area while adding almost no volume C. They leave the ratio unchanged: the cell’s volume is almost the same D. They raise the ratio: the folds add volume for the cell to hold what it takes back

Question 7

A hamster with a mass of 30 g and a dog with a mass of 30 kg pass a cold night in the same barn. Both animals keep their bodies at about 38 °C.

Which animal loses more heat per gram of body each hour, and why?

A. The dog: it has more surface in total, so more heat leaves it B. The dog: a larger body is warmer inside, so heat leaves each gram faster C. The hamster: it has more surface for each gram of body, so each gram loses heat faster D. Both animals lose the same per gram: they are at the same temperature in the same barn

Question 8

A scientist measures how fast a resting bat and a resting cow use energy. The bat has a mass of 10 g and the cow a mass of 600 kg. Per gram of body, the bat uses energy about 20 times as fast as the cow.

What does this show?

A. The bat's metabolic rate is 20 times the cow's in total B. Both animals have the same metabolic rate per gram; the bat is simply more active C. The cow's cells use energy faster than the bat's D. Metabolic rate per gram of body is higher in the smaller organism

Question 9

A spherical yeast cell has a radius of 1 μm. For a sphere, surface area = 4πr² and volume = 4/3 πr³.

What are its surface area, volume and surface-area-to-volume ratio?

A. 12.6 μm², 4.19 μm³, 3.0 per μm B. 12.6 μm², 12.6 μm³, 1.0 per μm C. 3.14 μm², 4.19 μm³, 0.75 per μm D. 12.6 μm², 4.19 μm³, 0.33 per μm

Question 10

Two single-celled fungi take up nutrients across their surfaces. Cell J has a surface area of 80 μm² and a volume of 40 μm³. Cell K has a surface area of 120 μm² and a volume of 100 μm³.

Which cell exchanges nutrients with its surroundings more efficiently, for its size, and why?

A. K: it has more surface in total, 120 μm² against J's 80 μm² B. J: its ratio is 2.0 per μm against K's 1.2 per μm, more surface per μm³ C. K: it has more volume, 100 μm³ against 40 μm³, so it holds more nutrients D. Neither: both are wrapped in the same kind of membrane

How to tackle the free-response questions. Read the verb first: describe asks what you see or know; explain asks why or how, so name the mechanism; predict asks what will happen and why; justify asks for the evidence that supports a claim. Each point is earned by one idea, stated in a sentence that names the thing and the mechanism. Extra words earn nothing; a wrong extra can lose the point. If there is a figure or table, use what it shows. When you finish, check the box for each point your answer earns and compare your sentences with the full-credit answer.

Free response 1 · Analyze Model or Visual Representation · 5 points

The model shows two cube-shaped cells from an early embryo. Cell A has sides 5 μm long; its surface area is 150 μm², its volume 125 μm³ and its surface-area-to-volume ratio 1.2 per μm. Cell B has sides 10 μm long; its values are left blank. For a cube of side s, surface area = 6s² and volume = s³. Everything a cell takes in, and every waste it gets rid of, crosses its surface.

Two cube-shaped cells from an early embryo, drawn to the same scale; cell B's values are left blank.

(a) Calculate the surface area and the volume of cell B, with units. (1 point)

Hint: Six faces for the surface; side × side × side for the volume.

A full-credit answer: Surface area of B = 6(10)² = 600 μm².
Volume of B = (10)³ = 1,000 μm³.

s = 10 μm

surface area = 6 s^{2}

volume = s^{3}

surface area = 6 {(10)}^{2}

volume = {(10)}^{3}

surface area = 600 {μm}^{2}

volume = 1, 000 {μm}^{3}

Check the box for each point your answer earns

surface area 600 μm² and volume 1,000 μm³, both with units.

Accept: 600 μm² and 1000 μm³. Do not award the point for 100 μm² (one face) or for volume given in μm².

Common slip: Giving 100 μm² for the surface. That is one face; a cube has six, and all six are surface.

(b) Calculate the surface-area-to-volume ratio of cell B, with its unit. (1 point)

Hint: Which quantity goes on top and which underneath? Then ask what unit is left when μm² is divided by μm³.

A full-credit answer: Ratio of B = 600 divided by 1,000 = 0.6, with the unit per μm.
The ratio is surface area divided by volume, and dividing μm² by μm³ leaves per μm.

surface area = 600 μm²

volume = 1,000 μm³

S A / V = \frac{surface area}{volume}

S A / V = \frac{(600)}{(1, 000)}

S A / V = 0.6 per μm

Check the box for each point your answer earns

600 divided by 1,000 = 0.6 per μm (surface area divided by volume, with the unit).

Accept: 0.6 μm⁻¹. Do not award the point for the inverted ratio, 1.67.

Common slip: Dividing the volume by the surface area to get 1.67. The ratio is surface over volume: it says how much surface serves each μm³.

(c) Describe how the surface area, the volume and the ratio changed when the side doubled from 5 μm to 10 μm, using the numbers in the model. (1 point)

Hint: Divide each of B's values by A's to find the factor; then compare the two factors.

A full-credit answer: When the side doubled, the surface area went from 150 μm² to 600 μm².
That is a factor of 4.
The volume went from 125 μm³ to 1,000 μm³.
That is a factor of 8.
So the volume grew faster than the surface area.
So the ratio halved, from 1.2 per μm to 0.6 per μm.

cell A: surface area = 150 μm², volume = 125 μm³, SA/V = 1.2 per μm

cell B: surface area = 600 μm², volume = 1,000 μm³, SA/V = 0.6 per μm

factor = \frac{value for cell B}{value for cell A}

surface area factor = \frac{(600)}{(150)}

volume factor = \frac{(1, 000)}{(125)}

ratio factor = \frac{(0.6)}{(1.2)}

surface area factor = 4

volume factor = 8

ratio factor = 0.5

Check the box for each point your answer earns

surface area ×4 (150 → 600 μm²), volume ×8 (125 → 1,000 μm³), so the volume grew faster than the surface and the ratio halved (1.2 → 0.6 per μm).

Accept: the factors 4 and 8 with the ratio described as falling or halving. Do not award the point for "both got bigger" with no factors.

Common slip: Saying both got bigger. The point needs the two factors, 4 and 8, or the ratio halving, to show that volume outran surface.

(d) Explain why the change in the ratio you described in (c) limits how large a cell can grow. (1 point)

Hint: What does each quantity in the ratio stand for in a living cell? Which one is the route in and out, and which one is the amount of cell that has to be supplied?

A full-credit answer: A cell takes in what it needs only across its surface.
The cell gets rid of waste only across its surface.
The amount it needs grows with its volume.
As the cell grows, its volume grows faster than its surface.
So less and less surface serves each μm³ of interior.
Past a certain size the surface can no longer bring in enough for the interior, and the cell cannot keep growing.

Check the box for each point your answer earns

a cell takes in what it needs and gets rid of waste only across its surface, while the amount it needs grows with its volume; as the cell grows its volume outruns its surface, so past a certain size the surface cannot supply the interior fast enough.

Accept: "supply grows with surface, demand grows with volume, and volume grows faster".

Common slip: Saying 'bigger cells need more' without linking surface to supply and volume to demand. The point is that the two grow at different rates.

(e) Predict what the embryo's cells do as they grow toward the size of cell B, and justify your prediction using the ratio. (1 point)

Hint: Think about what a growing cell can do so that its surface keeps up with its interior, and which of those an embryo's cells do.

A full-credit answer: As the cells approach B’s size, they divide rather than growing without limit.
A 10 μm cell has only 0.6 μm² of surface for each μm³ of interior.
That is half what a 5 μm cell has.
Dividing a 10 μm cell into smaller cells brings the ratio back up to 1.2 per μm.
So each new cell’s surface can again keep up with its interior.

Check the box for each point your answer earns

the cells divide (or stop growing) rather than growing without limit, because dividing a large cell into smaller ones raises the surface-area-to-volume ratio again (each 5 μm cell has 1.2 per μm against the 10 μm cell's 0.6), so the surface can keep up with the interior.

Accept: "they divide, restoring more surface per μm³". Do not award the point for "they grow bigger to get more surface".

Common slip: Predicting that the cells grow larger to gain surface. They do gain surface in total, but they gain volume faster; dividing is what restores surface per μm³.

Free-response score: 0 of 5

Free response 2 · Conceptual Analysis · 4 points

Two kinds of bacterium live in the same soil water and take up dissolved nutrients across their surfaces. Species R is rod-shaped and is modeled as a cylinder of radius 0.5 μm and length 2 μm. Species S is round and is modeled as a sphere of radius 0.75 μm. Formulas: cylinder, surface area = 2πrh + 2πr², volume = πr²h; sphere, surface area = 4πr², volume = 4/3 πr³.

(a) Calculate the surface-area-to-volume ratio of each cell. (1 point)

A full-credit answer: R: SA/V = (2π(0.5)(2) + 2π(0.5)²) ÷ (π(0.5)²(2)) = 5.00 per μm.
S: SA/V = 4π(0.75)² ÷ (4/3 π(0.75)³) = 4.00 per μm.

R: r = 0.5 μm, h = 2 μm

S: r = 0.75 μm

cylinder: S A / V = \frac{2 π r h + 2 π r^{2}}{π r^{2} h}

sphere: S A / V = \frac{4 π r^{2}}{\frac{4}{3} π r^{3}}

R: S A / V = \frac{2 π (0.5) (2) + 2 π {(0.5)}^{2}}{π {(0.5)}^{2} (2)}

S: S A / V = \frac{4 π {(0.75)}^{2}}{\frac{4}{3} π {(0.75)}^{3}}

R: S A / V = 5.00 per μm

S: S A / V = 4.00 per μm

Check the box for each point your answer earns

R 5.0 per μm and S 4.0 per μm (surface area divided by volume; accept R 4.8–5.2 and S 3.8–4.2 per μm).

Accept: the ratios alone, with or without the surface area and volume written. Do not award the point for inverted ratios or for one cell only.

Common slip: Forgetting the two end circles of the cylinder (2πr²), or dividing volume by surface. Surface over volume, for both cells, with every face counted.

(b) Determine which species takes up nutrients more efficiently for its size, using the two ratios from (a). (1 point)

A full-credit answer: Species R takes up nutrients more efficiently for its size.
R’s ratio is 5.0 per μm and S’s is 4.0 per μm.
So each μm³ of R has more membrane serving it.
Nutrients enter a cell only across its membrane.
The need for nutrients grows with the cell’s volume.
Therefore more surface for each μm³ of volume gives a faster supply.

Check the box for each point your answer earns

the decision that species R takes up nutrients more efficiently AND the reasoning it rests on: R's surface-area-to-volume ratio is higher (5.0 against 4.0 per μm), so R has more membrane for each μm³ of interior; nutrients enter only across the membrane while the need for them grows with the volume.

Accept: "R: more surface per unit of volume, so faster exchange for the same interior". Do not award the decision alone.

Common slip: Choosing S because its volume is larger. More volume is more interior to supply; what matters for efficiency is surface for each unit of volume.

(c) Predict what happens to the ratio of species S, and to how well its surface can supply its interior, if a cell doubles its radius to 1.5 μm before dividing. (1 point)

A full-credit answer: Doubling the radius multiplies the surface area by 4, to 28.3 μm².
It multiplies the volume by 8, to 14.1 μm³.
So the ratio halves, to 2.0 per μm.
The surface supplies the interior less well, because each μm³ now has half as much surface to be supplied across.
So exchange cannot keep pace with the larger volume.

r = 1.5 μm

surface area = 4 π r^{2}

volume = \frac{4}{3} π r^{3}

S A / V = \frac{4 π r^{2}}{\frac{4}{3} π r^{3}}

surface area = 4 π {(1.5)}^{2}

volume = \frac{4}{3} π {(1.5)}^{3}

S A / V = \frac{4 π {(1.5)}^{2}}{\frac{4}{3} π {(1.5)}^{3}}

surface area = 28.3 {μm}^{2}

volume = 14.1 {μm}^{3}

S A / V = 2.00 per μm

Check the box for each point your answer earns

the ratio halves to 2.0 per μm (surface ×4 to 28.3 μm², volume ×8 to 14.1 μm³), so the surface supplies the interior less well; exchange cannot keep pace with the larger volume.

Accept: "the ratio falls (halves) and exchange becomes less efficient" with the direction of both changes stated.

Common slip: Saying the ratio rises because the cell has more surface. It has more surface in total but far more volume, so less surface for each μm³.

(d) The cells lining a mammal's gut have surfaces folded into thousands of tiny projections. Support the claim that these folds use the same principle as the shapes of the two bacteria. (1 point)

A full-credit answer: The folds are microvilli.
Microvilli add membrane surface to the cell while adding almost no volume.
So they raise the cell’s surface-area-to-volume ratio.
That is the same quantity that makes the rod-shaped R more efficient than the round S.
Therefore, in both cases, more surface for each unit of volume lets the membrane take up nutrients faster for the interior it serves.

Check the box for each point your answer earns

the evidence AND the reasoning: the folds (microvilli) add membrane surface at almost the same volume (evidence), so they raise the cell's surface-area-to-volume ratio, just as the rod shape gives R more surface per μm³ than the sphere; in both cases more surface for each unit of volume means faster exchange across the membrane (reasoning).

Accept: "folds add surface without adding volume; higher ratio; faster exchange", with the link to the bacteria's shapes stated. Do not award the point for "folds make the cell bigger", or for the principle stated with no link to the folds.

Common slip: Saying the folds make the cell larger. The volume barely changes; the folds add surface, and surface per unit of volume is what speeds exchange.

Free-response score: 0 of 4

Multiple choice checked: 0 of 10 correct.

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