Unit 3 test

Unit 3 · Unit 3 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 rate, count and reading in this test is imagined for the question unless the question says otherwise.

Question 1

The energy profile of the reaction penicillin + water → an inactive product, with penicillinase present. Energy in kJ/mol.

Penicillinase, an enzyme made by some bacteria, speeds up the reaction penicillin + water → an inactive product. The figure shows the energy profile of the reaction with penicillinase present. Penicillinase lowers the activation energy of the reaction by 42 kJ/mol.

Which of the following is the activation energy of the reaction with no enzyme present?

A. 27 kJ/mol B. 33 kJ/mol C. 42 kJ/mol D. 69 kJ/mol

Question 2

A bacterium in river mud makes an enzyme that speeds up the breakdown of one kind of resin that washes into the river from a factory. Researchers give the enzyme four other molecules, one at a time.

Which of the following molecules is the enzyme most likely to break down as well?

A. A second kind of resin with a different shape B. A molecule with almost the same shape and charges as the resin C. A molecule that dissolves in water as readily as the resin does D. A molecule with the same charges as the resin but twice its size

Question 3

A student wants to find out whether catalase from celeriac or from guava releases oxygen faster from hydrogen peroxide. She puts a cube of celeriac into a tube of hydrogen peroxide at 20 °C and an equal cube of guava into an equal tube of the same peroxide at 35 °C. A third tube holds peroxide with no tissue. She collects the oxygen from each tube for 3 minutes.

Which of the following changes would make the comparison a fair test?

A. Putting a cube of a third tissue into the tube with no tissue B. Collecting the oxygen from the 35 °C tube for a shorter time than from the 20 °C tube C. Keeping the celeriac tube and the guava tube at the same temperature D. Using a larger cube of guava than of celeriac

Question 4

Students test whether the concentration of cobalt ions changes the rate at which an enzyme from pond bacteria breaks down a sugar.

Which of the following is the null hypothesis for this test?

A. The concentration of cobalt ions has no effect on the rate at which the enzyme breaks down the sugar B. A higher concentration of cobalt ions lowers the rate at which the enzyme breaks down the sugar C. Cobalt ions bind to the enzyme and pull its fold apart D. The enzyme breaks down the sugar at the same rate in every pond

Question 5

Cells lining the gut of a fish make an enzyme that breaks down a toxin in the algae the fish eats. A student proposes that the fish's cells control how fast the toxin is broken down by controlling how much of the enzyme they make.

Which of the following observations would support the student's proposal?

A. The enzyme lowers the activation energy of the reaction that breaks the toxin down B. Fish fed toxin-free food for a month have less of the enzyme and break the toxin down more slowly C. The enzyme breaks down the toxin but leaves the proteins of the algae untouched D. The enzyme breaks the toxin down faster at 25 °C than at 10 °C, with the same amount of enzyme

Question 6

An enzyme from a glacier bacterium is held at 0 °C. Its rate falls from 16 μmol/min to 1.5 μmol/min.

Which of the following results would show that the cold slowed the enzyme rather than denaturing it?

A. A second sample held at 0 °C also gives 1.5 μmol/min B. Held at 0 °C for longer, the sample's rate falls to 0 μmol/min C. Warmed to 70 °C, the sample's rate falls to 0 μmol/min D. Warmed back to 20 °C, the sample's rate returns to 16 μmol/min

Question 7

The fluid inside a plant cell's vacuole holds 100 times the concentration of hydrogen ions (H⁺) of the cell's cytosol. The cytosol is at pH 7.5.

Which of the following is the pH of the fluid inside the vacuole?

A. 0.075 B. 5.5 C. 6.5 D. 9.5

Question 8

A student gives a fixed amount of a purified enzyme its substrate at rising concentrations, with every tube at the enzyme's optimal temperature of 37 °C. Above 22 mmol/L of substrate, the initial rate stays at 64 μmol/min however much more substrate she adds.

Which of the following changes would raise the initial rate above 64 μmol/min?

A. Warming the tube to 60 °C B. Adding some of the product to the tube at the start C. Adding more of the enzyme to the tube D. Raising the substrate concentration to 44 mmol/L

Question 9

The enzyme's rate at 4 μM and at 400 μM of its substrate, with no drug and with drug J, K or M present.

A researcher measures a liver enzyme's rate at a low and a high concentration of its substrate, with no drug and with each of three drugs, J, K and M. The amount of enzyme, the temperature and the pH were the same in every test. The rates, at 4 μM and at 400 μM of substrate, are in the table.

Which of the drugs is a competitive inhibitor?

A. Drug J only B. Drug K only C. Drugs J and K D. Drugs J, K and M

Question 10

The optimal temperature and the optimal pH of four protein-digesting enzymes.

The table gives the optimal temperature and the optimal pH of four enzymes that digest protein. Human stomach fluid is at 37 °C and about pH 2.

Which of the four enzymes could digest protein in human stomach fluid?

A. Enzyme 1 B. Enzyme 2 C. Enzyme 3 D. Enzyme 4

Question 11

Twenty-five tubes of an enzyme from a cave bacterium, all at 30 °C, released product at a mean rate of 20.0 μmol/min. The standard deviation of the 25 readings is 1.5 μmol/min.

Which of the following are the two ends of the ±2SE error bar on this mean?

A. 19.7 to 20.3 μmol/min B. 19.4 to 20.6 μmol/min C. 18.5 to 21.5 μmol/min D. 17.0 to 23.0 μmol/min

Question 12

The mean rate of the enzyme at three concentrations of chromium ions, five tubes each, same temperature and pH. Error bars represent ±2SE. Gridlines every 1 μmol/min.

Students measured the rate of an enzyme from a bacterium that breaks down fuel oil at three concentrations of chromium ions, five tubes at each, with temperature and pH the same in every tube. The graph shows the three means with error bars representing ±2SE.

For which of the following comparisons is the null hypothesis of no difference in rate rejected?

A. 0 and 10 mg/L only B. 0 and 20 mg/L only C. 0 and 10 mg/L, and 0 and 20 mg/L D. All three pairs, 0 and 10, 0 and 20, and 10 and 20 mg/L

Question 13

A class measures an enzyme's rate in five tubes at each of two temperatures. The two ±2SE error bars overlap. The class wants to know whether the two rates differ.

Which of the following changes would make the standard error of each mean smaller, if the spread of the readings stays the same?

A. Measuring twenty tubes at each temperature instead of five B. Measuring each tube's rate for twice as long C. Reading each tube's rate to one more decimal place D. Repeating the calculation with the same five readings

Question 14

A working muscle cell holds 11 mmol of ATP per liter of cytosol and hydrolyzes 3.5 mmol of ATP per liter of cytosol every second. Its ATP level stays steady.

How much ATP does the cell remake from ADP and Pi each minute, per liter of cytosol?

A. 3.5 mmol B. 199 mmol C. 210 mmol D. 221 mmol

Question 15

The energy each animal transferred from its food in a day and the energy it captured in ATP, in kJ.

The table gives, for three animals, the energy transferred from their food in a day and the energy captured in ATP.

Which animal captured in ATP the largest share of the energy it transferred from its food?

A. All three animals captured the same share B. The pony C. The wallaby D. The leopard

Question 16

An adult pony has kept the same body mass for a year. A student says: “Its mass is steady, so the pony has no need to take in energy.” The student is wrong.

Which of the following is the reason the student is wrong?

A. At every energy transfer, some of the pony's energy leaves as heat that can do no more work B. An animal that has stopped growing needs food only for the matter it is built from C. The pony creates the energy for its work inside its own cells D. A pony standing still in its field uses no energy

Question 17

A branched pathway in a cell. Each arrow is one reaction, catalyzed by the enzyme written beside it.

The figure shows a branched pathway in a cell. Y is the reactant of two enzymes: E3 makes Z from Y, and E4 makes V from Y. W keeps arriving, and E4 is far from saturated. A drug stops E3 working.

Which of the following happens to the amounts of Y, Z and V?

A. Y falls, Z falls and V rises B. Y rises, Z falls and V rises C. Y rises, Z is unchanged and V falls D. Y rises, Z falls and V is unchanged

Question 18

An aquarium plant is sealed in a jar with an oxygen sensor. In bright light the jar's oxygen reading rises by 22 mL an hour. In that light the plant's true rate of photosynthesis is 35 mL of oxygen an hour. Respiration uses oxygen at the same rate in light and dark.

Which of the following is the plant's rate of respiration?

A. 13 mL of oxygen an hour B. 22 mL of oxygen an hour C. 35 mL of oxygen an hour D. 57 mL of oxygen an hour

Question 19

Geologists find fossils generally taken to be cyanobacteria in a rock layer 3.1 billion years old. Oxygen from photosynthesis began to build up in the air and sea about 2.4 billion years ago.

Which of the following would the geologists expect to find among the iron minerals of the 3.1-billion-year-old layer?

A. Red, rusted iron minerals B. Red, rusted iron minerals, but fewer than in layers 2.0 billion years old C. Dark iron minerals in the lower part of the layer and red, rusted ones in the upper part D. Dark iron minerals of the kind that form where there is almost no oxygen

Question 20

Researchers measure the oxygen release and the sugar output of isolated chloroplasts under four treatments. In one treatment, in the first minute, the chloroplasts went on releasing oxygen at their normal rate, but their sugar output fell.

Which of the following treatments gives this result?

A. Bright light with a herbicide that stops electron transfer along the electron transport chain in the thylakoid membrane B. Dim light with carbon dioxide plentiful C. Bright light with the carbon dioxide removed D. Darkness with carbon dioxide plentiful

Question 21

Four candidate rows, each naming one space for the mitochondrion, the chloroplast and the bacterium.

A mitochondrion, a chloroplast and an aerobic bacterium each pump protons across a membrane with an electron transport chain. Each row of the table names one space for each of the three.

Which row names the spaces the protons are pumped into?

A. Row 1 B. Row 2 C. Row 3 D. Row 4

Question 22

Biologists claim that the chloroplast descends from a cyanobacterium taken inside a eukaryotic cell long ago.

Which of the following findings would count against this claim?

A. The chloroplast holds its own closed loop of DNA, just like a bacterium B. The chloroplast's ATP synthase is built to the same plan as a cyanobacterium's C. Cyanobacteria photosynthesize with no chloroplast at all inside them D. The chloroplast's DNA matches the plant's own nucleus more closely than any cyanobacterium's

Question 23

Electrons are passing along the electron transport chain of a respiring mitochondrion.

Compared with the matrix, which of the following describes the intermembrane space?

A. A higher concentration of protons and a positive charge B. A lower concentration of protons and a positive charge C. A lower concentration of protons and a negative charge D. A higher concentration of protons and a negative charge

Question 24

NADH hands its electrons to the first protein of the electron transport chain. FADH₂ hands its electrons to a protein further along the chain.

Compared with the electrons from one NADH, what do the electrons from one FADH₂ do?

A. Pump protons the other way across the inner membrane B. Pump more protons across the inner membrane C. Pump fewer protons across the inner membrane D. Pump the same number of protons across the inner membrane

Question 25

In a muscle fiber short of oxygen, an enzyme turns pyruvate into lactate. A drug blocks that enzyme. Within a minute, glycolysis in the fiber stops.

Why does glycolysis stop?

A. Pyruvate builds up and blocks the first enzyme of glycolysis B. The fiber has no glucose left C. The lactate-making step made the ATP that starts glycolysis D. The fiber has no NAD⁺ left to accept electrons

Question 26

A yeast culture with plenty of oxygen makes 84 mmol of ATP an hour and uses 2.8 mmol of glucose an hour. Its oxygen is cut off, and the culture goes on making 84 mmol of ATP an hour by fermenting.

About how much glucose does the fermenting culture use an hour?

A. 2.8 mmol B. 5.6 mmol C. 42 mmol D. 84 mmol

Question 27

Yeast cells in a sealed flask with no oxygen are fed glucose in which every carbon atom is the heavy form ¹³C. The yeast ferment the glucose.

What share of the ¹³C ends up in the carbon dioxide the yeast give off?

A. None of it B. About 33% C. About 67% D. All of it

Question 28

The mean oxygen released in 20 minutes at five light intensities, five tubes each.

Students measure the oxygen an aquarium plant releases in 20 minutes at five light intensities, five tubes at each, and calculate the standard error of each mean. The means are in the table.

Which of the following is the correct way to draw the graph?

A. Points, each with a ±2SE bar, joined one to the next B. Points, each with a ±2SE bar, with one straight best-fit line through them C. Bars, one per light intensity, with no error bars D. Points joined one to the next, with no error bars

Question 29

The ATP made per minute in the two tubes.

Researchers set up two tubes, each with plenty of NADH, ADP, Pi and dissolved oxygen. In tube 1 the proteins of the electron transport chain and ATP synthase are set in the membrane of sealed vesicles. In tube 2 the same proteins float free in the solution, with no membrane. The table gives the ATP each tube makes.

Why is no ATP made in tube 2?

A. ATP synthase must touch the chain proteins to receive the protons they pump B. With no membrane there is no side for the pumped protons to pile up on C. Without a membrane the oxygen cannot reach the end of the chain D. Free-floating chain proteins cannot take electrons from NADH

Question 30

A researcher gives isolated mitochondria the two-carbon fragment that pyruvate oxidation leaves, with NAD⁺, FAD, ADP and Pi and no oxygen. The mitochondria release carbon dioxide, load NAD⁺ to NADH and FAD to FADH₂, and make a little ATP.

Which of the following stages is taking place in the mitochondria?

A. The electron transport chain B. Glycolysis C. Pyruvate oxidation D. The Krebs cycle

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

Cut persimmon flesh turns brown within an hour. An enzyme in the flesh, polyphenol oxidase, speeds up a reaction in which colorless molecules and oxygen form a brown pigment. Its optimal pH is 7. Scientists propose that a dilute acid on the cut surface slows the browning by moving the pH away from the enzyme's optimal pH. They make an extract of the flesh at pH 7 and measure the rate of pigment formation in three groups of five tubes: the untreated group; the acid group, extract with the acid added, at pH 4; and the neutralized group, extract with the same acid added, then at once brought back to pH 7. Figure 1 shows the mean rates with error bars representing ±2SE. A folding test on the acid group's extract after ten minutes at pH 4 found 89 of every 100 enzyme molecules folded normally.

Figure 1. The mean rate of brown pigment formation in each group, five tubes per group. Error bars represent ±2SE. Gridlines every 1 μmol/min.

(a)(i) Describe how the enzyme's active site holds its substrate. (1 point)

A full-credit answer: The substrate binds in the active site only if its shape fits the pocket.
Its charges must also match the charges on the R groups lining the pocket.

Check the box for each point your answer earns

the substrate is held only if its shape fits the active site, or only if its charges match the R groups lining the site. Accept either the shape or the charge; a full description names both.

Does not earn: 'the substrate binds to the enzyme' with no mention of fit or charge.

Common slip: Saying the substrate binds 'anywhere on the enzyme': only the pocket holds it.

(a)(ii) Explain why a very small amount of the enzyme can turn the whole cut surface brown. (1 point)

A full-credit answer: The enzyme is a catalyst, so the reaction leaves it unchanged.
So each enzyme molecule acts again and again on new substrate molecules.
A small amount of enzyme therefore turns a great deal of substrate into pigment.

Check the box for each point your answer earns

the enzyme is not used up (comes out of each reaction unchanged), so one molecule acts again and again.

Accept: 'a catalyst is not consumed' with the consequence that it keeps working.

Common slip: Saying the enzyme 'is used up slowly': a catalyst is not used up at all.

(b)(i) Identify the control group for the scientists' experiment. (1 point)

A full-credit answer: The untreated group: the extract at pH 7 with nothing added.

Check the box for each point your answer earns

the untreated group (the extract with nothing added, at pH 7).

Common slip: Naming a control variable, such as the temperature, instead of the control group.

(b)(ii) Justify why the scientists included the neutralized group, the extract with the acid added and then brought back to pH 7. (1 point)

A full-credit answer: The neutralized group has the acid present but not the low pH.
If its rate matches the untreated group's, the acid itself does nothing to the enzyme.
So any fall in the acid group can be credited to the low pH, not to something else the acid adds to the extract.

Check the box for each point your answer earns

ONE of: (1) the neutralized group separates the two possible causes: it shows whether the acid slows the enzyme by lowering the pH or by something else the acid adds to the extract; (2) it shows that with the pH brought back to 7 the rate is normal, so the acid lowers the rate only while the pH is low.

Accept: 'to show that the acid on its own, at pH 7, does not change the rate'; 'to show that the acid's effect depends on the pH, not on the acid having been added'.

Does not earn: 'it is a control' or 'to have something to compare with' without naming what the neutralized group rules out.

Common slip: Saying only 'it is a second control': the point is what the neutralized group rules out.

(b)(iii) Identify the independent variable in the scientists' experiment. (1 point)

A full-credit answer: The treatment of the extract: whether the acid is added and the pH the extract is left at.

Check the box for each point your answer earns

the treatment of the extract (pH; or presence of the acid and the pH).

Accept: 'the pH of the extract' or 'whether the acid was added and whether it was neutralized'.

Common slip: Naming the rate of pigment formation, which is the dependent variable.

(c) Based on Figure 1, describe the effect on the rate of pigment formation of the acid at pH 4 compared with the acid brought back to pH 7 (the neutralized group). (1 point)

A full-credit answer: The acid at pH 4 lowers the rate: the acid group's mean is about 2.3 μmol/min against about 7.7 μmol/min in the neutralized group.
The two error bars do not overlap, so the difference is shown.
The neutralized group's bar overlaps the untreated group's, so the acid at pH 7 is not shown to change the rate.

Check the box for each point your answer earns

the acid at pH 4 lowers the rate (the acid group lower than the neutralized group), and the bars do not overlap, so the difference is not chance.

Accept: the direction with values read from the figure; 'the neutralized group's bar overlaps the untreated group's, so no difference is shown there' as the comparison.

Does not earn: 'the rates are different' with no direction.

Common slip: Reporting only the means without saying whether the bars overlap.

(d)(i) Predict the rate of pigment formation when a tube from the acid group is brought back to pH 7 after ten minutes at pH 4. (1 point)

A full-credit answer: The rate rises back to close to the untreated group's rate, about 8 μmol/min.

Check the box for each point your answer earns

the rate rises back toward the untreated group's rate (about 8 μmol/min).

Accept: 'returns to normal' or 'rises to about 7 to 9 μmol/min'.

Common slip: Predicting that the rate stays low: the fold was never lost.

(d)(ii) Justify your prediction in part (d)(i). (1 point)

A full-credit answer: At pH 4 the folding test found 89 of every 100 enzyme molecules still folded.
So the low pH only changed the charges on the R groups lining the active site; the fold held.
Back at pH 7 the R groups regain their charges, so the substrate binds well again and the rate returns.

Check the box for each point your answer earns

the fold was not lost (89 of 100 folded), so only the R-group charges changed; at pH 7 the charges return and the substrate binds again.

Does not earn: 'the enzyme was not denatured' with no link to the charges returning at pH 7.

Common slip: Saying the enzyme 'refolds' at pH 7: it never unfolded.

(d)(iii) Explain why a cut persimmon kept in a refrigerator browns more slowly than one left on a counter. (1 point)

A full-credit answer: In the refrigerator the molecules of enzyme and substrate move more slowly.
So they collide less often and with less energy.
So fewer collisions succeed each second, and the pigment forms more slowly.

Check the box for each point your answer earns

the cold slows the molecules, so enzyme and substrate collide less often (and with less energy), so the rate falls.

Does not earn: 'the cold denatures the enzyme' (cold slows; it does not denature).

Common slip: Saying the cold denatures the enzyme: cooling slows it, and the rate returns on warming.

Free-response score: 0 of 9

Free response 2 · Scientific Investigation · 4 points

Compound U is made by some soil bacteria. In a mitochondrion, compound U sets itself in the inner membrane and lets protons cross the membrane without passing through ATP synthase. Table 1 shows its effect at two concentrations. Researchers grew a strain of bacteria that makes compound U and a strain that does not, in the same nutrient broth, then removed the bacteria and kept the broth. They left half of each broth at its high concentration and diluted the other half to a low concentration. They added the four broths to dishes of mammalian cells growing in culture, as Table 2 shows, and measured each dish's oxygen use and ATP production.

Table 1. The effect of compound U on the inner mitochondrial membrane.

(a) Describe the advantage to a cell of making ATP by oxidative phosphorylation rather than by glycolysis alone. (1 point)

A full-credit answer: Oxidative phosphorylation gives about 30 ATP per glucose.
Glycolysis alone gives only 2 ATP per glucose.
So the cell gets far more ATP from each glucose it breaks down.

Check the box for each point your answer earns

far more ATP per glucose (about 30 against 2) when the chain and ATP synthase are used.

Accept: 'about fifteen times as much ATP from each glucose'.

Common slip: Saying oxidative phosphorylation 'is faster': the advantage is the yield per glucose.

(b) Identify the two treatment groups that served as controls in the researchers' experiment. (1 point)

Table 2. The four treatment groups of mammalian cells.

A full-credit answer: Groups 3 and 4: the broths from the bacteria that do not make compound U, at the high and the low concentration.

Check the box for each point your answer earns

groups 3 and 4 (both broths from the strain that makes no compound U). Both groups are needed for the point.

Common slip: Naming only group 4, or naming the low-concentration groups 2 and 4.

(c) Based on the information in both tables, predict the treatment group likely to have the greatest rate of oxygen use of the four. (1 point)

A full-credit answer: Group 1, the cells given the broth from bacteria that make compound U at its high concentration.

Check the box for each point your answer earns

group 1.

Common slip: Choosing group 3, the high-concentration broth with no compound U.

(d) The researchers claim that cells in which protons leak back past ATP synthase will show a fall in ATP production. Support the claim. (1 point)

A full-credit answer: The chain pumps protons into the intermembrane space, and ATP synthase makes ATP as protons flow back through it.
With compound U, protons leak back through the membrane instead of through ATP synthase.
So fewer protons flow through ATP synthase, and it makes less ATP.
The proton gradient's energy leaves as heat instead.

Check the box for each point your answer earns

protons return past ATP synthase, so fewer flow through it, so ATP synthase makes less ATP (the proton gradient's energy is released as heat).

Does not earn: 'the cells make less ATP' restated with no link to the flow through ATP synthase.

Common slip: Saying the leak stops the chain: the chain speeds up and uses more oxygen, while ATP falls.

Free-response score: 0 of 4

Free response 3 · Analyze Model or Visual Representation · 4 points

Figure 1 is a model of photosynthesis in a chloroplast. The thylakoid membranes and the stroma are drawn as two boxes. Arrows show what enters and leaves each box, and arrows 1 to 4 show the molecules that pass between the two boxes.

Figure 1. A model of photosynthesis in a chloroplast: the light reactions in the thylakoid membranes, the Calvin cycle in the stroma, and the molecules that pass between them.

(a) Describe a characteristic of chlorophyll that lets the light reactions capture light energy. (1 point)

A full-credit answer: Chlorophyll absorbs blue and red light strongly.
The light it absorbs boosts one of its electrons to a higher energy level, and the photosystem passes that electron on.

Check the box for each point your answer earns

chlorophyll absorbs light (blue and red strongly), or absorbed light boosts an electron in chlorophyll to a higher energy level.

Accept either the absorption or the boosted electron.

Common slip: Saying chlorophyll 'is green' or 'reflects green light': the reflected light is not captured.

(b) A drug stops the enzyme of the Calvin cycle that fixes carbon dioxide. Based on Figure 1, explain what happens to the amounts of ATP and NADPH in the stroma in the first minute, with the light bright and carbon dioxide plentiful. (1 point)

A full-credit answer: With its fixing enzyme stopped, the Calvin cycle stops, so it uses no ATP or NADPH along arrows 1 and 2.
The light reactions go on making ATP and NADPH.
So both pile up in the stroma.

Check the box for each point your answer earns

the Calvin cycle no longer uses ATP and NADPH while the lit light reactions keep making them, so both rise (pile up) in the stroma.

Does not earn: 'ATP and NADPH fall' or 'the light reactions stop' (the light is unchanged).

Common slip: Saying ATP and NADPH fall because sugar output falls: they are used less, so they build up.

(c) Using Figure 1, identify the molecule that carries electrons from the light reactions to the Calvin cycle. (1 point)

A full-credit answer: NADPH, arrow 2.

Check the box for each point your answer earns

NADPH (arrow 2).

Common slip: Naming ATP, which carries energy but not electrons; or NADP⁺, the emptied carrier returning.

(d) Based on Figure 1, explain how a drug that plugs the channel of ATP synthase in the thylakoid membranes stops sugar being made, even in bright light with carbon dioxide plentiful. (1 point)

A full-credit answer: Protons pumped into the thylakoid space can no longer flow back into the stroma through ATP synthase.
So ATP synthase makes no ATP, and arrow 1 stops.
The Calvin cycle needs that ATP to build sugar from carbon dioxide.
So carbon fixation stops and no sugar is made.

Check the box for each point your answer earns

ATP synthase makes no ATP (no protons flow through it), so no ATP reaches the stroma (arrow 1), so the Calvin cycle cannot build sugar.

Accept with or without any statement about NADPH or the proton gradient: the point rests on the ATP link alone.

Does not earn: 'no ATP is made' with no link to the Calvin cycle's need for it.

Common slip: Saying the drug stops light being absorbed: the photosystems still capture light; the ATP step is what fails.

Free-response score: 0 of 4

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Multiple choice checked: 0 of 30 correct.