Unit 2 test

Unit 2 · Unit 2 end-of-unit test (also the test-out)

Suggested time: about 80 minutes. Answer everything, then press Submit the test to see the feedback and scoring guides.

Answer every question, then press Submit the test. Feedback and the scoring guides appear after you submit. For the three 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. After you submit, mark your own free-response work against each scoring guide. Suggested time: 80 minutes. Every case that opens with ‘Suppose’ is imagined for the question. R = 0.0831 L·bar/(mol·K); temperatures in kelvin are °C + 273; M is short for mol/L.

Question 1

Suppose a gland cell makes two proteins at the same time. One protein leaves the cell and works in the gut. The other stays inside the cell, in the fluid around the organelles, and helps break glucose down there.

Where are the ribosomes that make the glucose-breaking protein?

A. On the outer surface of the rough ER B. Inside the interior of the rough ER C. Free in the cytosol, attached to no membrane D. Inside the Golgi complex

Question 2

Suppose a biologist examines a newly found single-celled organism, 1.5 μm long, under an electron microscope. It has a cell wall. Its DNA lies in one region of the cell with no membrane around it. Its cytosol is full of ribosomes.

Which of the observations decides that the cell is prokaryotic?

A. The DNA lies in a region with no membrane around it B. The cell has a cell wall C. The cytosol is full of ribosomes D. The cell is 1.5 μm long

Question 3

Suppose a gland cell in a fish makes two proteins at the same time on its rough ER. Both proteins pass through the Golgi complex. One protein then goes into a lysosome. The other is released from the cell into a duct.

What sends each protein to its own destination?

A. The ribosome that made each protein B. The size of each protein C. A signal sent from the nucleus to each vesicle D. The carbohydrate chain the Golgi complex attaches to each protein

Question 4

Rough ER membrane area, mitochondria and lysosomes in four cells from one fish.

Suppose researchers measure four kinds of cell from one fish, as the table shows: the area of rough ER membrane in each cell, and the number of mitochondria and of lysosomes.

Which cell makes protein for release from the cell?

A. Tail-muscle cell B. Poison-gland cell, from a fin C. Spleen cell that breaks down worn-out blood cells D. Fat-storing skin cell

Question 5

In the thymus, an organ in the chest, most of the immune cells that form there die within days and are cleared away.

Which observation would show that these cells die by apoptosis rather than by injury?

A. The cells break into pieces that neighboring cells take in and digest B. The cells swell and burst, spilling their contents C. The cells are killed by bacteria that enter the thymus D. The cells stop dividing and stay where they are

Question 6

A muscle cell is supplied with glucose and oxygen.

Which substances leave the cell’s mitochondria as the products of aerobic cellular respiration?

A. Oxygen and glucose B. Glucose and carbon dioxide C. Carbon dioxide and water D. Oxygen and carbon dioxide

Question 7

The endomembrane system is the set of membranes that work together to modify, package and transport proteins, lipids and polysaccharides within a cell. The mitochondrion has two membranes, yet biologists leave it out of the system.

Why is the mitochondrion left out of the endomembrane system?

A. The mitochondrion has an outer and an inner membrane rather than one B. Vesicles move between the other members of the system but never to or from the mitochondrion C. The mitochondrion makes ATP rather than proteins D. The mitochondrion sits too far from the nucleus

Question 8

In a eukaryotic cell the nucleus is enclosed by the nuclear envelope, a double membrane.

Which of the following is directly joined to the nuclear envelope, so that their membranes are continuous?

A. The plasma membrane B. The Golgi complex C. The outer membrane of a mitochondrion D. The endoplasmic reticulum

Question 9

Suppose a cell from the root of a daikon, a plant, is cut open. One fluid-filled sac fills most of the cell and presses the cytosol into a thin layer against the cell wall.

What does this sac do for the cell?

A. The sac holds digestive enzymes and breaks down worn-out parts B. The sac makes the proteins the cell needs C. The sac captures light energy and makes sugar D. The sac stores water and dissolved nutrients for the cell

Question 10

Suppose a round cell is modeled as a sphere of radius 2.2 μm. For a sphere of radius r, surface area = 4πr² and volume = 4/3 πr³.

Which of the following is its surface-area-to-volume ratio?

A. 0.68 per μm B. 0.73 per μm C. 1.36 per μm D. 1.82 per μm

Question 11

The air sacs of a lung are lined with cells flattened into thin plates. Oxygen crosses these cells to reach the blood.

How does the flattened shape help the exchange of oxygen?

A. The flattened shape gives each cell a larger volume for the same surface B. The flattened shape gives each cell more surface for its volume than a rounded shape would C. The flattened shape makes the plasma membrane thinner, so oxygen crosses it faster D. The flattened shape lets each cell hold more mitochondria near its surface

Question 12

Suppose two mammals, E and G, rest quietly in the same warm room at the same body temperature. Per gram of body, E uses oxygen five times as fast as G.

Which of the following is most likely true?

A. E has the smaller body mass B. E has the larger body mass C. E has less surface for each gram of body than G D. E loses heat more slowly per gram of body than G

Question 13

A membrane protein’s chain of amino acids has five stretches in order: a stretch with charged R groups, a stretch of 20 nonpolar R groups, a stretch with charged R groups, a second stretch of 20 nonpolar R groups, and a final stretch with charged R groups. The chain’s first end lies in the cytosol.

Where does the chain’s other end lie?

A. In the extracellular fluid B. In the hydrophobic interior of the membrane C. Among the phospholipid heads of the outer layer D. In the cytosol

Question 14

A section of a plasma membrane: the extracellular fluid (shaded gray) above, the cytosol (white) below, and four lettered pieces.

The model shows a section of a plasma membrane, with the extracellular fluid above and the cytosol below. Four pieces are lettered G, J, L and M. One of the four pieces is drawn out of place.

Which piece is drawn out of place?

A. G B. J C. L D. M

Question 15

Suppose researchers add a new drug to the solution around protein-free bubbles of phospholipid bilayer. Within two minutes the drug is at the same concentration inside the bubbles as outside: it crosses as fast as oxygen crosses the same bubbles, and far faster than water does.

Which of the following describes the drug’s molecules?

A. Small nonpolar molecules B. Small polar molecules with no charge C. Molecules with a full charge D. Large polar molecules

Question 16

Suppose four samples of a rod-shaped bacterium are prepared. In two samples an enzyme has removed the cell wall; in the other two the cell wall is intact. One sample of each kind sits in pure water and one in a solution that matches the cytosol.

Which cells burst?

A. The cells with a cell wall, in pure water B. The cells with no cell wall, in pure water C. The cells with a cell wall, in the matching solution D. The cells with no cell wall, in the matching solution

Question 17

Suppose a membrane separates two solutions of acetone, a small polar molecule, both at the same concentration. Acetone crosses this membrane. Each minute about 90 million acetone molecules cross from the left side to the right side.

About how many acetone molecules cross from the right side to the left side each minute?

A. 0 B. 45 million C. 90 million D. 180 million

Question 18

Suppose the cells of a sphagnum moss take up a dissolved mineral from the water around them through a membrane protein.

Which further observation would show that the uptake is active transport?

A. The mineral is an ion B. The uptake is faster when the water holds more of the mineral C. The mineral enters through a membrane protein rather than through the bilayer D. The cells take up the mineral when it is already more concentrated inside them than in the water

Question 19

Suppose a gland cell makes a protein for release from the cell. The protein leaves the ribosomes on the rough ER with no carbohydrate chain. Inside the Golgi complex a carbohydrate chain is attached to it, on the side facing the interior of the Golgi complex. The protein is then carried to the plasma membrane in a vesicle and released by exocytosis. Some copies of the protein stay set in the plasma membrane.

For a copy that stays in the plasma membrane, where does its carbohydrate chain face?

A. The cytosol B. The extracellular fluid C. The hydrophobic interior of the membrane D. The interior of the Golgi complex, where the chain stays behind

Question 20

Suppose glucose is at 5.5 mmol/L inside a cell and 5.5 mmol/L in the fluid outside it. Glucose crosses the membrane through carrier proteins. A signal makes the cell move twice as many glucose carriers into its membrane.

What happens to the net movement of glucose after the signal?

A. The net movement doubles, into the cell B. The net movement stays at zero C. A net movement into the cell begins D. A net movement out of the cell begins

Question 21

Solute concentration of the water each organism lives in and of its cytosol.

Suppose four single-celled organisms with no cell wall each live in a different water. Each organism has a contractile vacuole. The table gives the solute concentration of each organism’s water and of its cytosol.

Which organism’s contractile vacuole empties most often?

A. The organism from a rain puddle B. The organism from a lake C. The organism from the sea D. The organism from a very salty inland lake

Question 22

A cell from a plant stem has a rigid cell wall. Its pressure potential is +9.1 bar and its solute potential is −11.8 bar. It sits in an open beaker of 0.19 M sucrose at 25 °C. Take R as 0.0831 L·bar/(mol·K).

Which of the following describes the net movement of water?

A. Water enters the cell B. Water crosses both ways at equal rates C. Water leaves the cell D. Water enters at first and then leaves

Question 23

A 0.31 M NaCl solution sits in an open beaker at 29 °C. Dissolved NaCl is fully split into sodium ions (Na⁺) and chloride ions (Cl⁻). Take R as 0.0831 L·bar/(mol·K).

Which of the following is the solution’s solute potential?

A. −15.6 bar B. −7.78 bar C. −1.49 bar D. +15.6 bar

Question 24

Four solutions at the same temperature.

Four open beakers at the same temperature hold the solutions in the table. NaCl splits into two ions when it dissolves and CaCl₂ into three; glucose and sucrose stay as whole molecules.

Which solution has the lowest water potential?

A. The glucose solution B. The CaCl₂ solution C. The NaCl solution D. The sucrose solution

Question 25

Chloride ion concentration outside and inside the cell.

The figure shows the concentration of chloride ions (Cl⁻) on the two sides of a nerve cell’s plasma membrane. The inside of the cell is slightly negative relative to the outside. A chloride channel opens.

How do the two pulls on Cl⁻ compare?

A. Both favor entering B. Both favor leaving C. The concentration difference favors leaving; the charge across the membrane favors entering D. The concentration difference favors entering; the charge across the membrane favors leaving

Question 26

Four pumps: the ions each moves in one cycle.

Suppose four pumps are described in the table: the ions each moves in one cycle, and the direction. Each sodium ion (Na⁺), potassium ion (K⁺) and hydrogen ion (H⁺) carries one positive charge; each calcium ion (Ca²⁺) carries two.

Considered on its own, which pump’s cycling leaves the inside of its cell more negative?

A. The pump in a yeast cell’s membrane B. The pump in a muscle cell’s membrane C. The pump in a bacterium’s membrane D. The pump in a kidney cell’s membrane

Question 27

Suppose that in the muscle cells of a patient, a digestive enzyme that is normally locked inside lysosomes is found scattered through the cytosol.

Which feature have these cells most likely lost?

A. The enzyme’s ability to work in acid B. The ribosomes that make the enzyme C. An intact membrane around their lysosomes D. The folds of their mitochondria’s inner membranes

Question 28

Suppose biologists compare a 25-base stretch of DNA from a mitochondrion with the same stretch from two kinds of bacteria. Bacterium T’s DNA matches the mitochondrion’s at 22 of the 25 bases; bacterium Z’s matches at 10.

Which of the following does the comparison show?

A. Bacterium T lives closer to the cell than bacterium Z does B. Bacterium T’s ancestors separated from the mitochondrion’s more recently C. Bacterium T holds more DNA in total than bacterium Z does D. Bacterium T’s stretch of DNA is the same length as the mitochondrion’s

Question 29

Suppose a drug stops the mitochondria of a single-celled eukaryote from splitting in two, but the cell keeps dividing normally.

What happens to the number of mitochondria in each cell over several cell divisions?

A. The number rises B. The number stays the same C. The number falls to zero at the first division D. The number falls by about half at each division

Question 30

Four associations between a bacterium and a larger organism.

Suppose biologists describe four associations between a bacterium and a larger organism, as the table shows.

Which association most resembles the one biologists propose for the origin of mitochondria?

A. Bacterium living and dividing inside a cell B. Bacterium attached to the outside of a cell C. Bacterium taken into a cell and digested D. Bacterium in an animal’s gut

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 · Interpreting and Evaluating Experimental Results · 9 points

Suppose researchers study the excretory tubes of silverfish, small insects. The cells lining a tube move potassium ions (K⁺) from the insect’s blood into the fluid inside the tube; the fluid then leaves the body. In Experiment 1 the researchers bathe isolated tubes in a solution like the insect’s blood and measure the mass of fluid each tube produces in one hour, with and without a poison that stops the cells making ATP (Table 1). In Experiment 2 they measure the K⁺ concentration of the bath and of the fluid produced, with and without the poison (Table 2).

Table 1: fluid produced by isolated tubes in one hour. Table 2: K⁺ concentration of the bath and of the tube fluid.

(a) Describe the property of the plasma membrane that stops K⁺ crossing the phospholipid bilayer on its own. (1 point)

A full-credit answer: The hydrocarbon tails in the hydrophobic interior of the membrane carry no charges or partial charges.
Water attracts a K⁺ ion, and nothing in the interior attracts it.
So K⁺ cannot cross the bilayer on its own.

Check the box for each point your answer earns

the hydrophobic interior (the hydrocarbon tails) carries no charge, so a charged ion has nothing to attract it there and it stays in the water.

Accept: ‘the interior is nonpolar, so ions cannot pass through it’.

Common slip: Saying the ion is too big. A K⁺ ion is smaller than an oxygen molecule; its charge, not its size, keeps it out.

(b)(i) Identify the dependent variable in Experiment 1. (1 point)

A full-credit answer: The mass of fluid each tube produced in one hour.

Check the box for each point your answer earns

the mass (amount) of fluid produced (in one hour).

(b)(ii) Justify the researchers’ inclusion of untreated tubes in Experiment 1. (1 point)

A full-credit answer: The untreated tubes show how much fluid a tube produces with its ATP supply intact.
So a fall in the poisoned tubes can be attributed to the loss of ATP, not to the tube being isolated in a bath.

Check the box for each point your answer earns

the untreated tubes give the fluid production with ATP available, a comparison that lets a change in the poisoned tubes be attributed to the poison.

Scoring note: ‘it is the control’ with no purpose stated does not earn the point.

(b)(iii) Based on Table 1, describe the effect of the poison on fluid production. (1 point)

A full-credit answer: The poison cut fluid production from 4.0 mg to 0.7 mg in one hour.

Check the box for each point your answer earns

fluid production fell (from 4.0 mg to 0.7 mg per hour).

Scoring note: a direction is required; ‘the poison changed fluid production’ does not earn the point.

(c)(i) Identify the independent variable in Experiment 2. (1 point)

A full-credit answer: Whether the poison was present.

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the presence or absence of the poison (the treatment).

(c)(ii) Based on Table 2, identify the treatment in which the tube fluid held K⁺ at the higher concentration. (1 point)

A full-credit answer: The untreated tubes: 128 mmol/L in the fluid against 21 mmol/L with the poison.

Check the box for each point your answer earns

the untreated tubes (128 mmol/L against 21 mmol/L).

(c)(iii) Based on Table 1, calculate the percent change in fluid production caused by the poison, giving a fall as a negative value. (1 point)

Write down the values in the question:

fluid with the poison (new) = 0.7 mg

fluid untreated (old) = 4.0 mg

Write down the equation:

percent change = \frac{new - old}{old} \times 100

Substitute the values into the equation:

percent change = \frac{0.7 - 4.0}{4.0} \times 100

Calculate:

percent change = −82.5%

A full-credit answer: (0.7 − 4.0) ÷ 4.0 × 100 = −82.5%.

(d)(i) The researchers claim that the tube cells move K⁺ into the tube fluid by active transport. Using data from the tables, support the claim. (1 point)

A full-credit answer: In untreated tubes K⁺ reached 128 mmol/L in the fluid from a bath at 18 mmol/L, so it moved against its concentration gradient.
With the poison, which stops the cells making ATP, the fluid’s K⁺ stayed near the bath’s, at 21 mmol/L, and fluid production fell.
Movement against the concentration gradient that stops without ATP is active transport.

Check the box for each point your answer earns

K⁺ was moved from 18 mmol/L to 128 mmol/L (against its concentration gradient) AND the movement stopped when the poison removed the cells’ ATP.

Scoring note: either half alone does not earn the point.

(d)(ii) The researchers also claim that water enters the tube fluid by osmosis and that the cells use energy only to move K⁺. Justify the claim. (1 point)

A full-credit answer: The pumped K⁺ raises the solute concentration of the tube fluid above the blood’s.
Water moves by osmosis toward the side with more solute, down its own concentration gradient.
Movement down a concentration gradient uses no energy from the cell.
The cells use energy to move K⁺; water follows by osmosis with no further energy.

Check the box for each point your answer earns

pumping K⁺ raises the solute concentration of the tube fluid, so water moves into it by osmosis (down its own gradient), which needs no energy from the cell.

Accept: water follows the solute / moves toward the more concentrated fluid.

Free-response score: 0 of 9

Free response 2 · Analyze Model or Visual Representation · 4 points

Suppose researchers break open cells from the digestive gland of a cuttlefish and spin the mixture so that its parts separate by size into four fractions, Q, R, X and Z. They test each fraction for the substances and structures in Table 1; a plus sign marks each one found.

Table 1: what each fraction was found to hold.

(a) Determine which fraction holds the mitochondria, and state what in the table your decision rests on. (1 point)

A full-credit answer: Fraction R.
It holds the enzymes of aerobic cellular respiration, which take place inside mitochondria.

Check the box for each point your answer earns

fraction R, because it holds the enzymes of aerobic cellular respiration (the reactions of a mitochondrion).

(b) Explain the advantage to the cell of keeping the enzymes found in fraction X inside their own compartment. (1 point)

A full-credit answer: The enzymes of fraction X break large molecules down and work only in acid.
A membrane around them keeps their compartment acidic while the cytosol is not, so they work inside it and hardly at all outside it.
So the cell can digest what it takes in without digesting its own proteins.

Check the box for each point your answer earns

the membrane keeps the acid (and the enzymes) inside the compartment, so the enzymes digest material there and do not break down the cell’s own contents.

Accept: ‘keeps the digestive reactions apart from the rest of the cell’.

(c) The researchers feed living cells labeled amino acids, wait 15 minutes, then break the cells open and separate them the same way. Predict the fraction in which the labeled protein that the cell makes for release will be found. (1 point)

A full-credit answer: Fraction Z.

Check the box for each point your answer earns

fraction Z.

(d) Justify your prediction in part (c). (1 point)

A full-credit answer: A protein for release is made by ribosomes on the rough ER and passed into the ER’s interior.
It then moves in a vesicle to the Golgi complex, the stack of flattened sacs.
At 15 minutes it is inside the ER or the Golgi complex, and both are in fraction Z.

Check the box for each point your answer earns

a protein for release is made on the rough ER and passes through the Golgi complex (in vesicles), and fraction Z holds the ribosome-studded sacs and the stacks of flattened sacs (the ER and the Golgi complex).

Scoring note: naming the route with no link to what fraction Z holds does not earn the point.

Free-response score: 0 of 4

Free response 3 · Analyze Data · 4 points

Suppose a class cuts cores of the same size from a rutabaga and from a kohlrabi, two vegetables, measures the mass of each core, and leaves each core for an hour in a sucrose solution of one of five concentrations at room temperature. The class then measures each core’s mass again and calculates its percent change in mass. Figure 1 shows the results.

Figure 1. Percent change in mass of rutabaga cores (dark bars) and kohlrabi cores (pale bars) after one hour in sucrose solutions.

(a) Based on Figure 1, identify the sucrose concentration at which the mass of the kohlrabi cores barely changed. (1 point)

A full-credit answer: 0.46 M.

Check the box for each point your answer earns

0.46 M (the kohlrabi bar closest to zero).

(b) Based on Figure 1, identify the two sucrose concentrations between which the rutabaga cores changed from gaining mass to losing mass. (1 point)

A full-credit answer: Between 0.23 M and 0.46 M.

Check the box for each point your answer earns

between 0.23 M and 0.46 M.

(c) A student hypothesizes that kohlrabi cells hold a higher concentration of solute than rutabaga cells. Use the data to evaluate the hypothesis. (1 point)

A full-credit answer: The hypothesis is supported.
The kohlrabi cores stopped gaining mass at about 0.46 M sucrose; the rutabaga cores stopped between 0.23 M and 0.46 M.
The solution that is isotonic to a tissue holds as much solute as its cells.
The isotonic solution for kohlrabi is more concentrated, so kohlrabi cells hold more solute.

Check the box for each point your answer earns

supported, because the kohlrabi cores reached zero change at a higher sucrose concentration (about 0.46 M) than the rutabaga cores (between 0.23 M and 0.46 M), so the solution isotonic to kohlrabi tissue is more concentrated and its cells hold more solute.

Accept: supported, because at every concentration the kohlrabi cores gained more or lost less mass than the rutabaga cores.

Scoring note: a verdict with no comparison of the two isotonic points does not earn the point.

(d) Explain why both kinds of core lost mass in the 0.92 M solution. (1 point)

A full-credit answer: 0.92 M sucrose holds more solute than the cells of either vegetable.
Water moves by osmosis toward the side with more solute, so water left the cells.
The cores held less water, so their mass fell.

Check the box for each point your answer earns

the solution held more solute than the cells (a lower water potential), so water left the cells by osmosis and the cores lost mass.

Free-response score: 0 of 4

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Multiple choice checked: 0 of 30 correct.