Cell Size

Unit 2 · Topic 2.2 end-of-topic test

Suggested time: about 46 minutes. Answer everything, then press Submit the test to see the feedback and scoring guides.

Answer every question. For each multiple-choice question, pick one option. When you have answered every question, press Submit the test; the feedback then gives the reasoning for each. For the free-response questions, 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 cell has sides of 20 μm. For a cube of side s, surface area = 6s² and volume = s³.

What are its surface area and its volume?

A. 400 μm² and 8,000 μm³ B. 2,400 μm² and 8,000 μm³ C. 2,400 μm² and 400 μm³ D. 120 μm² and 8,000 μm³

Question 2

A cube-shaped cell has a surface area of 180 μm² and a volume of 150 μm³.

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

A. 1.2 per μm B. 0.83 per μm C. 30 μm D. 1.2 μm²

Question 3

For a cell with a surface area of 60 μm² and a volume of 40 μm³, a student writes: $S A / V = \frac{40}{60} = 0.67 μm$ .

What is wrong with the student's working?

A. Nothing; 0.67 μm is correct B. Upside down; it should be 60 divided by 40 = 1.5 per μm C. It should be 60 − 40 = 20 μm D. Right division; the unit should be μm³

Question 4

An amoeba grows before it divides. Its width increases from 50 μm to 100 μm while its shape stays the same.

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

A. Surface ×2, volume ×2, ratio unchanged B. Surface ×2, volume ×4, ratio halved C. Surface ×4, volume ×8, ratio halved D. Surface ×8, volume ×4, ratio doubled

Question 5

Surface area and volume of the four cells.

Four cells, K, L, M and N, are measured. Their surface areas and volumes are in the table.

Which cell exchanges materials with its surroundings most efficiently, and why?

A. K: it has the most surface B. K: it has the most volume to fill C. N: its surface is bigger than its volume D. M: it has the most surface per unit of volume

Question 6

Cells in a plant tissue are modeled as cubes. A 10 μm cube’s membrane can bring in at most 180 units of oxygen a minute, and the cell uses 100 units a minute. Oxygen intake is proportional to membrane area; oxygen use is proportional to volume.

What percentage of its oxygen need can the membrane supply in a 20 μm cube and in a 30 μm cube?

A. 180% and 180% B. 720% and 1,620% C. 90% and 60% D. 45% and 20%

Question 7

A student writes: "A larger cell exchanges materials with its surroundings more efficiently than a smaller one, because it has more membrane."

Which statement about the larger cell corrects the student?

A. The larger cell has less membrane in total than a smaller cell, so it exchanges less efficiently B. The student is right: more membrane in total means more exchange with the surroundings C. Size makes no difference to efficiency, since the membrane is the same thickness in both cells D. The larger cell has more membrane, but far more interior to serve, so it exchanges less efficiently

Question 8

Two samples of cells come from the lining of a kidney tubule, and the two kinds of cell have the same volume. In one sample the top face of each cell carries about a thousand finger-like folds of its plasma membrane; in the other, the top face is flat. Under the same conditions, the cells with the folded face take up glucose 3.1 times as fast.

Why do the cells with the folded face take up glucose faster?

A. The folds beat and sweep glucose into the cell B. The folds add membrane at the same volume: more surface for each unit of interior C. The folds add volume, so the cell has more room to hold glucose D. The folds make glucose move from low to high concentration into the cell

Question 9

On the underside of a leaf are thousands of tiny pores, each set between a pair of cells. When the pores are open, air reaches the many cell surfaces inside the leaf; when they are shut, the air stays outside.

What are the pores and the paired cells, and how do the cells open the pore?

A. Stomata and guard cells; the cells beat like cilia and fan air in through the pore B. Stomata and guard cells; the cells lose water and shrink, pulling the pore open C. Stomata and guard cells; the cells take in water, swell and bow apart, opening the pore D. Microvilli and gut lining cells; the folds spread apart as the cells fill up with water

Question 10

Root cells of the same plant each grow a single thin outgrowth into the soil. Cells whose outgrowths are 0.2 mm, 0.6 mm and 1.0 mm long take up water at 0.8, 1.9 and 3.1 units a minute. The outgrowths are all about the same width.

What is the outgrowth, and why does a longer one take up water faster?

A. A root hair; a longer one adds membrane across which water can enter B. A root hair; a longer one reaches wetter soil, and its cell wall pumps the water in C. A cilium; a longer one beats harder and drives water into the cell D. A root hair; a longer one holds more cytosol, which soaks up more water

Question 11

A hummingbird has a mass of about 5 g; a swan has a mass of about 10 kg. Both birds keep their bodies at about 40 °C while they rest in the same cool air. Four statements compare the heat the two birds lose to the air.

Which statement is wrong?

A. Each gram of the hummingbird loses heat faster than each gram of the swan B. The hummingbird has more skin for each gram of body than the swan has C. The swan’s total heat loss is the greater, because it has far more skin altogether D. The hummingbird loses more heat altogether, because each gram of it loses heat so fast

Question 12

Mass and heat loss per gram for the three mammals.

Three mammals rest in the same cool room. The table gives each animal’s mass and its heat loss per gram of body per hour.

What explains this pattern?

A. A larger body loses less heat in total, so each gram loses less B. A smaller body has more surface for each gram of mass, so each gram loses heat faster C. A smaller body is warmer inside, so heat leaves it faster D. A larger body has thicker skin, which is what slows heat loss

Question 13

A hummingbird eats about half its body mass in food every day. A horse eats about 2% of its body mass a day. The food an animal eats each day supplies the energy it uses.

What does this show about the metabolic rates of the two animals?

A. The horse has the higher metabolic rate per gram of body B. Both animals have the same metabolic rate per gram; the hummingbird just eats more often C. The hummingbird has the higher metabolic rate per gram of body D. The hummingbird has the higher metabolic rate in total

Question 14

A spherical cell has a radius of 10 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. 1,260 μm², 4,190 μm³, 0.30 per μm B. 1,260 μm², 419 μm³, 3.0 per μm C. 126 μm², 4,190 μm³, 0.030 per μm D. 314 μm², 4,190 μm³, 0.075 per μm

Question 15

A flattened cell is modeled as a rectangular solid 10 μm long, 5 μm wide and 0.5 μm thick. For a rectangular solid, surface area = 2lh + 2lw + 2wh and volume = lwh.

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

A. 57.5 μm², 25 μm³, 2.3 per μm B. 115 μm², 25 μm³, 0.22 per μm C. 100 μm², 25 μm³, 4.0 per μm D. 115 μm², 25 μm³, 4.6 per μm

Question 16

Here are two cells: a spherical cell 7 μm across (radius 3.5 μm) and a cube-shaped cell with sides of 11 μm. Sphere: surface area = 4πr², volume = 4/3 πr³; cube: 6s² and s³.

Which cell exchanges materials with its surroundings more efficiently, and why?

A. The cube: it has the larger surface area in total B. The cube: it has the larger volume, so more interior takes part in exchange C. The sphere: its ratio is 0.86 per μm against the cube’s 0.55 per μm D. The sphere: a sphere always has more surface for its size than a cube has

Question 17

Two cells from a frog’s skin each have a volume of 120 μm³. One is flattened and has 160 μm² of plasma membrane; the other is nearly spherical and has 140 μm² of plasma membrane.

Which cell exchanges materials more efficiently, and what does that say about its shape?

A. The flattened cell, 1.33 against 1.17 per μm: flattening adds surface without adding volume B. The rounder cell, 0.86 against 0.75 per μm: a rounder shape holds more surface C. Neither cell: the two volumes are equal, so the two ratios must be equal too D. The flattened cell, because being thinner it holds less volume for its membrane to serve

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 · 4 points

The model shows three cube-shaped plant cells with sides of 15 μm, 30 μm and 60 μm, with the surface area and volume of the first two; the values for the 60 μm cell are left blank. For a cube of side s, surface area = 6s² and volume = s³. The 15 μm cell’s surface-area-to-volume ratio is 0.4 per μm and the 30 μm cell’s is 0.2 per μm. Everything a cell takes in, and every waste it gets rid of, crosses its surface.

Three cube-shaped model cells drawn to the same scale, with the surface area and volume of the first two.

(a) Describe how the surface area and the volume change each time the side of the cell doubles, using the numbers for the 15 μm and 30 μm cells. (1 point)

A full-credit answer: When the side doubles from 15 μm to 30 μm, the surface area changes from 1,350 μm² to 5,400 μm².
That is a factor of 4.
The volume changes from 3,375 μm³ to 27,000 μm³.
That is a factor of 8.
So the volume grows faster than the surface area.
So the surface-area-to-volume ratio halves, from 0.4 per μm to 0.2 per μm.

15 μm cell: surface area = 1,350 μm², volume = 3,375 μm³

30 μm cell: surface area = 5,400 μm², volume = 27,000 μm³

factor = \frac{value for the 30 μm cell}{value for the 15 μm cell}

surface area factor = \frac{(5, 400)}{(1, 350)}

volume factor = \frac{(27, 000)}{(3, 375)}

surface area factor = 4

volume factor = 8

Check the box for each point your answer earns

the surface area is multiplied by 4 (1,350 → 5,400 μm²) while the volume is multiplied by 8 (3,375 → 27,000 μm³), so the volume grows faster than the surface area (or: the ratio halves, 0.4 → 0.2 per μm).

Accept: "surface ×4, volume ×8" with the numbers from the model cited. Do not award the point for "both get bigger" or "the volume gets bigger" with no factors or numbers.

Common slip: Saying both get bigger. The point needs the factors, 4 for surface and 8 for volume, or the ratio halving.

(b) Calculate, for the 60 μm cell, its surface area, its volume and its surface-area-to-volume ratio, each with its unit. (1 point)

A full-credit answer: The 60 μm cell has a surface area of 21,600 μm², a volume of 216,000 μm³ and a surface-area-to-volume ratio of 0.1 per μm.

s = 60 μm

surface area = 6 s^{2}

volume = s^{3}

S A / V = \frac{6 s^{2}}{s^{3}}

surface area = 6 {(60)}^{2}

volume = {(60)}^{3}

S A / V = \frac{6 {(60)}^{2}}{{(60)}^{3}}

surface area = 21, 600 {μm}^{2}

volume = 216, 000 {μm}^{3}

S A / V = 0.1 per μm

Check the box for each point your answer earns

surface area 21,600 μm², volume 216,000 μm³, ratio 0.1 per μm (all three, with units).

Accept: 0.1 μm⁻¹ or "1 : 10" for the ratio. Do not award the point if the ratio is inverted (10) or if the units are missing from all three values.

Common slip: Inverting the ratio to 10, or leaving the units off. The ratio is surface area divided by volume, in per μm.

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

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, and the waste it makes, grow with its volume.
As the cell grows, its volume grows faster than its surface.
So past a certain size the surface cannot supply the volume, 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 (or the waste it makes) grows with its volume; so as the cell grows its surface cannot keep up with (cannot supply) its interior.

Accept: "there is not enough membrane to supply the volume" or "exchange across the surface cannot keep pace with the demand of the interior". Do not award the point for "the cell gets too big" with no reference to both surface and volume.

Common slip: Saying the cell gets too big. The point needs both halves: exchange happens across the surface, and need grows with the volume.

(d) Explain how the pattern in the model relates to the way living cells are built. (1 point)

A full-credit answer: The ratio falls as a cell grows.
So a cell stays small, or divides, rather than growing without limit.
A small cell has enough surface to supply its interior.
A cell built for exchange goes further.
The gut-lining cell folds its membrane into microvilli.
The folds add surface while adding almost no volume.
So the folds raise the cell’s ratio, and exchange keeps pace with the interior.

Check the box for each point your answer earns

EITHER: because the ratio falls as a cell grows, a cell stays small or divides rather than growing without limit, so its surface keeps up with its interior; OR: folding the membrane (microvilli, root hairs, a flattened shape) adds surface while adding almost no volume, raising the surface-area-to-volume ratio so that exchange keeps pace with the interior.

Accept: either link made in full (the pattern in the model tied to the reason). Do not award the point for naming microvilli or division without saying what they do to surface relative to volume.

Common slip: Naming microvilli or division without saying what they do to surface relative to volume. Tie the structure to the ratio.

Free-response score: 0 of 4

Free response 2 · Conceptual Analysis · 4 points

In pond water live a single-celled alga, modeled as a sphere of radius 2 μm, and a rod-shaped bacterium, modeled as a cylinder of radius 0.4 μm and length 4 μm. Both cells take up dissolved nutrients across their surfaces. Formulas: sphere, surface area = 4πr², volume = 4/3 πr³; cylinder, surface area = 2πrh + 2πr², volume = πr²h.

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

A full-credit answer: Alga: SA/V = 4π(2)² ÷ (4/3 π(2)³) = 1.50 per μm.
Bacterium: SA/V = (2π(0.4)(4) + 2π(0.4)²) ÷ (π(0.4)²(4)) = 5.50 per μm.

alga: r = 2 μm

bacterium: r = 0.4 μm, h = 4 μm

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

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

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

bacterium: S A / V = \frac{2 π (0.4) (4) + 2 π {(0.4)}^{2}}{π {(0.4)}^{2} (4)}

alga: S A / V = 1.50 per μm

bacterium: S A / V = 5.50 per μm

Check the box for each point your answer earns

alga 1.5 per μm and bacterium 5.5 per μm (surface area divided by volume; accept alga 1.4–1.6 and bacterium 5.3–5.7 per μm).

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

Common slip: Dividing volume by surface area, or working out one cell only. Surface over volume, for both cells.

(b) Determine which cell exchanges nutrients with the water more efficiently, for its size, using the ratios from (a). (1 point)

A full-credit answer: The bacterium exchanges nutrients more efficiently for its size.
Everything a cell takes in crosses its surface.
The amount it needs grows with its volume.
The bacterium has 5.5 μm² of surface for each μm³ of volume, and the alga has 1.5 μm².
So each μm³ of the bacterium is supplied across almost four times as much membrane.
Therefore the bacterium is the more efficient exchanger.

Check the box for each point your answer earns

the decision that the bacterium exchanges more efficiently AND the reasoning it rests on: it has almost four times as much surface for each unit of volume (5.5 against 1.5 per μm), and everything a cell takes in crosses its surface while what it needs grows with its volume.

Accept: "the bacterium; a higher surface-area-to-volume ratio means more membrane per unit of interior to exchange across". Do not award the point for "the bacterium because it is smaller" with no link to surface per unit of volume, or for the decision alone.

Common slip: Saying the bacterium because it is smaller, with no link to surface per unit of volume. Small helps only because it raises the ratio.

(c) Predict what happens to the alga's surface-area-to-volume ratio, and to how well its surface can supply its interior, if it grows to a radius of 6 μm before dividing. (1 point)

A full-credit answer: Tripling the radius multiplies the surface area by 9, to about 452 μm².
It multiplies the volume by 27, to about 905 μm³.
So the ratio falls to a third, 0.50 per μm.
The surface supplies the interior less well, because each μm³ of the alga now has a third as much surface to be supplied across.
So exchange cannot keep pace with the larger volume.

r = 6 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 {(6)}^{2}

volume = \frac{4}{3} π {(6)}^{3}

S A / V = \frac{4 π {(6)}^{2}}{\frac{4}{3} π {(6)}^{3}}

surface area = 452 {μm}^{2}

volume = 905 {μm}^{3}

S A / V = 0.500 per μm

Check the box for each point your answer earns

the ratio falls to a third, 0.50 per μm (surface ×9 to about 452 μm², volume ×27 to about 905 μm³), so the surface supplies the interior less well; exchange cannot keep pace with the larger volume.

Accept: "the ratio falls (to a third) and exchange becomes less efficient" with the direction of both changes stated. Do not award the point for "the ratio rises because the cell has more surface".

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) A related alga has the same volume as the 2 μm-radius sphere (33.5 μm³) but grows as a flat disk 0.5 μm thick, with about 150 μm² of surface. Evaluate the claim that this flat shape is an aid to exchange. (1 point)

A full-credit answer: The claim is supported.
The disk has the same volume as the sphere, 33.5 μm³.
But it has about 150 μm² of surface against the sphere’s 50.3 μm².
So the disk has about three times the surface for the same volume: a ratio of about 4.5 per μm against 1.5 per μm.
Nutrients cross into a cell only across its surface.
Therefore the disk takes in nutrients faster for the same interior, and the flat shape aids exchange.

surface area = 150 μm²

volume = 33.5 μm³

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

S A / V = \frac{(150)}{(33.5)}

S A / V = 4.5 per μm

Check the box for each point your answer earns

the judgement that the claim is supported AND the ground for it: at the same volume, the disk has about three times the surface (150 μm² against 50.3 μm²), so its ratio is about 4.5 per μm against 1.5 per μm; more surface per unit of volume means nutrients can be exchanged faster for the same interior.

Accept: the ground stated as "more surface for the same volume, so a higher ratio and faster exchange" without the 4.5 figure. Do not award the judgement alone, or "flat cells are smaller", or a judgement with no comparison of surface to volume.

Common slip: Saying flat cells are smaller. The volume is the same; the shape adds surface, and that is what raises the ratio.

Free-response score: 0 of 4

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Multiple choice checked: 0 of 17 correct.

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