Cell Size

Unit 2 · Topic 2.2 end-of-topic test

Answer every question. For each multiple-choice question, pick one option and press Check; the feedback tells you what a wrong choice assumed. For the two free-response questions, write your answer in full sentences, then open the scoring guide and mark your own work against it. Formulas are given wherever a calculation needs them; use π = 3.14.

Question 1

A model cell is a cube with sides of 3 μm. For a cube of side s, surface area = 6s² and volume = s³.

What are its surface area and its volume?

A. 9 μm² and 27 μm³ B. 54 μm² and 27 μm³ C. 54 μm² and 9 μm³ D. 18 μm² and 27 μm³

Question 2

A cube-shaped cell has a surface area of 150 μm² and a volume of 125 μm³.

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

A. 1.2 per μm B. 0.83 per μm C. 25 μm D. 1.2 μm²

Question 3

For the 2 μm cube (surface area 24 μm², volume 8 μm³), a student writes: SA/V = 8 ÷ 24 = 0.33 μm.

What is wrong with the student's working?

A. Nothing; 0.33 μm is correct B. Upside down; it should be 24 ÷ 8 = 3 per μm C. It should be 24 − 8 = 16 μm D. Right division; the unit should be μm³

Question 4

A yeast cell grows before it divides. Its diameter increases from 4 μm to 8 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

Four cube-shaped model cells: Model K has a surface area of 150 μm² and a volume of 125 μm³; Model L, 54 μm² and 27 μm³; Model M, 24 μm² and 8 μm³; Model N, 96 μm² and 64 μm³.

Which model would exchange 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 and volume are closest in size 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 120 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. 120% and 120% B. 240% and 360% C. 60% and 40% D. 30% and 13%

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

Question 8

Two samples of cells from the lining of the small intestine are compared. The cells in one sample have microvilli, folds of the plasma membrane, and have three times the membrane area of the cells in the other sample. Both kinds of cell have the same volume. Under the same conditions, the cells with microvilli take up nutrients 2.7 times as fast.

Why do the cells with microvilli take up nutrients faster?

A. The folds beat and sweep nutrients 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 nutrients D. The folds make nutrients 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 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 mouse weighs about 20 g; an elephant weighs about 4,000 kg. Both keep their bodies at about 37 °C in cool air.

Which animal loses more heat each hour in total, and which loses more heat each hour per gram of body?

A. Mouse in total; mouse per gram B. Elephant in total; elephant per gram C. Same per gram; the elephant more in total D. Elephant in total; mouse per gram

Question 12

Three mammals rest in the same cool room. A shrew (mass 5 g) loses 30 units of heat per gram per hour; a rabbit (2,000 g) loses 4 units per gram per hour; a cow (500,000 g) loses 0.5 units per gram 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 mouse eats about a fifth of its body mass in food every day. An elephant eats about 4% 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 elephant has the higher metabolic rate per gram of body B. Both have the same metabolic rate per gram; the mouse just eats more often C. The mouse has the higher metabolic rate per gram of body D. The mouse has the higher metabolic rate in total

Question 14

A spherical cell has a radius of 3 μm. For a sphere, surface area = 4πr² and volume = 4/3 πr³. Use π = 3.14.

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

A. 113 μm², 113 μm³, 1.0 per μm B. 113 μm², 38 μm³, 3.0 per μm C. 38 μm², 113 μm³, 0.33 per μm D. 28 μm², 113 μm³, 0.25 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

Two cells are compared: a spherical cell 5 μm across (radius 2.5 μm) and a cube-shaped cell with sides of 7 μm. Sphere: surface area = 4πr², volume = 4/3 πr³; cube: 6s² and s³. Use π = 3.14.

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

A. The cube: 294 μm² of surface against the sphere's 78.5 μm² B. The cube: 343 μm³ of volume against the sphere's 65 μm³ C. The sphere: 1.2 per μm against the cube's 0.86 per μm D. Neither: both are wrapped in the same kind of membrane

Question 17

Two blood cells each have a volume of 90 μm³. One is flattened and has 120 μm² of plasma membrane; the other is nearly spherical and has 100 μm² of plasma membrane.

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

A. The flattened cell, 1.3 against 1.1 per μm: flattening adds surface without adding volume B. The rounder cell, 0.9 against 0.75 per μm: a rounder shape holds more surface C. Neither: equal volume means equal ratio D. The flattened cell, because being thinner it holds less volume

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 cells with sides of 2 μm, 4 μm and 8 μm. For a cube of side s, surface area = 6s² and volume = s³. The 2 μm cell has a surface area of 24 μm², a volume of 8 μm³ and a surface-area-to-volume ratio of 3 per μm. The 4 μm cell has 96 μm², 64 μm³ and 1.5 per μm. The values for the 8 μm cell are left blank. 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 2 μm and 4 μm cells. (1 point)

A full-credit answer: When the side doubles from 2 to 4 μm, the surface area is multiplied by 4, from 24 to 96 μm², while the volume is multiplied by 8, from 8 to 64 μm³. The volume grows faster than the surface area, so the surface-area-to-volume ratio halves, from 3 to 1.5 per μm.

Write down the values in the question:

2 μm cell: surface area = 24 μm², volume = 8 μm³
4 μm cell: surface area = 96 μm², volume = 64 μm³

Write down the equation:

          value for the 4 μm cell
factor = ─────────────────────────
          value for the 2 μm cell

Substitute in the values, and calculate:

surface area factor = 96 ÷ 24
surface area factor = 4
volume factor = 64 ÷ 8
volume factor = 8

Check the box for each point your answer earns

the surface area is multiplied by 4 (24 → 96 μm²) while the volume is multiplied by 8 (8 → 64 μm³), so the volume grows faster than the surface area (or: the ratio halves, 3 → 1.5 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) Explain why a falling surface-area-to-volume ratio limits how large a cell can grow. (1 point)

A full-credit answer: A cell takes in what it needs and gets rid of wastes only across its surface, while the amount it needs and the waste it makes grow with its volume. As the cell grows its surface cannot keep up with its interior, so past a certain size the surface cannot supply the volume.

Check the box for each point your answer earns

a cell takes in what it needs and gets rid of wastes 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.

(c) Complete the model for the 8 μm cell: give its surface area, its volume and its surface-area-to-volume ratio, each with its unit. (1 point)

A full-credit answer: The 8 μm cell has a surface area of 384 μm², a volume of 512 μm³ and a surface-area-to-volume ratio of 0.75 per μm.

Write down the values in the question:

s = 8 μm

Write down the equations:

surface area = 6s²
volume = s³
        surface area
SA/V = ──────────────
           volume

Substitute in the values, and calculate:

surface area = 6 × 8²
surface area = 6 × 64
surface area = 384 μm²
volume = 8³
volume = 512 μm³
SA/V = 384 ÷ 512
SA/V = 0.75 per μm

Check the box for each point your answer earns

surface area 384 μm², volume 512 μm³, ratio 0.75 per μm (all three, with units).

Accept: 0.75 μm⁻¹ or "3 : 4" for the ratio. Do not award the point if the ratio is inverted (1.33) or if the units are missing from all three values.

Common slip: Inverting the ratio to 1.33, or leaving the units off. The ratio is surface area divided by volume, in per μm.

(d) Explain how the pattern in the model relates to the way living cells are built: either to the fact that cells stay small or divide, or to the folded membranes of cells specialized for exchange, such as the microvilli on the cells lining the gut. (1 point)

A full-credit answer: 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 the interior it has to supply. A cell built for exchange goes further: folding its membrane into microvilli adds surface while adding almost no volume, raising the ratio so that 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 0.5 μm and length 4 μm. Both take up dissolved nutrients across their surfaces. Formulas: sphere, surface area = 4πr², volume = 4/3 πr³; cylinder, surface area = 2πrh + 2πr², volume = πr²h. Use π = 3.14.

(a) Calculate the surface-area-to-volume ratio of each cell, showing the surface area and the volume you used, with units. (1 point)

A full-credit answer: The alga’s surface-area-to-volume ratio is 1.5 per μm, from 50.2 μm² of surface and 33.5 μm³ of volume. The bacterium’s is 4.5 per μm, from 14.1 μm² and 3.14 μm³.

Write down the values in the question:

alga: r = 2 μm, π = 3.14
bacterium: r = 0.5 μm, h = 4 μm, π = 3.14

Write down the equations:

sphere: surface area = 4πr², volume = (4/3)πr³
cylinder: surface area = 2πrh + 2πr², volume = πr²h
        surface area
SA/V = ──────────────
           volume

Substitute in the values, and calculate:

alga surface area = 4 × 3.14 × 2²
alga surface area = 50.2 μm²
alga volume = (4/3) × 3.14 × 2³
alga volume = 33.5 μm³
alga SA/V = 50.2 ÷ 33.5
alga SA/V = 1.5 per μm
bacterium surface area = 2 × 3.14 × 0.5 × 4 + 2 × 3.14 × 0.5²
bacterium surface area = 12.56 + 1.57
bacterium surface area = 14.1 μm²
bacterium volume = 3.14 × 0.5² × 4
bacterium volume = 3.14 μm³
bacterium SA/V = 14.1 ÷ 3.14
bacterium SA/V = 4.5 per μm

Check the box for each point your answer earns

alga 50.2 μm² ÷ 33.5 μm³ = 1.5 per μm; bacterium 14.1 μm² ÷ 3.14 μm³ = 4.5 per μm (both ratios, surface divided by volume, within rounding: alga 1.4–1.6, bacterium 4.3–4.7 per μm).

Accept: ratios given with the working shown, even if one intermediate value is rounded differently. Do not award the point for inverted ratios (volume ÷ 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) Identify which cell exchanges nutrients with the water more efficiently, for its size, and explain why. (1 point)

A full-credit answer: The bacterium. It has three times as much surface for each unit of volume, 4.5 against 1.5 per μm, and everything a cell takes in crosses its surface while what it needs grows with its volume, so each μm³ of the bacterium is supplied across more membrane.

Check the box for each point your answer earns

the bacterium, because it has three times as much surface for each unit of volume (4.5 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.

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 4 μm before dividing. (1 point)

A full-credit answer: Doubling the radius multiplies the surface by 4, to about 201 μm², and the volume by 8, to about 268 μm³, so the ratio halves to 0.75 per μm. The surface supplies the interior less well: exchange cannot keep pace with the larger volume.

Write down the values in the question:

r = 4 μm
π = 3.14

Write down the equations:

surface area = 4πr²
volume = (4/3)πr³
        surface area
SA/V = ──────────────
           volume

Substitute in the values, and calculate:

surface area = 4 × 3.14 × 4²
surface area = 201 μm²
volume = (4/3) × 3.14 × 4³
volume = 268 μm³
SA/V = 201 ÷ 268
SA/V = 0.75 per μm

Check the box for each point your answer earns

the ratio halves to 0.75 per μm (surface ×4 to about 201 μm², volume ×8 to about 268 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. 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. Justify the claim that this flat shape is an aid to exchange. (1 point)

A full-credit answer: At the same volume, 33.5 μm³, the disk has about three times the surface, 150 μm² against 50.2 μm², so its ratio is about 4.5 per μm against 1.5. More surface per unit of volume means nutrients can be exchanged faster for the same interior, so the flat shape aids exchange.

Write down the values in the question:

surface area = 150 μm²
volume = 33.5 μm³

Write down the equation:

        surface area
SA/V = ──────────────
           volume

Substitute in the values, and calculate:

SA/V = 150 ÷ 33.5
SA/V = 4.5 per μm

Check the box for each point your answer earns

at the same volume, the disk has about three times the surface (150 μm² against 50.2 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 comparison stated as "more surface for the same volume, so a higher ratio and faster exchange" without the 4.5 figure. Do not award the point for "flat cells are smaller" or for a claim 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

Multiple choice checked: 0 of 17 correct.

Next: Two kinds of cell