Practice questions · Topic 3.5

Unit 3 · Practice for the Topic 3.5 end-of-topic test

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

Watch first: Topic 3.5 summary: cellular respiration

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Glycolysis, pyruvate oxidation and the Krebs cycle send carbons out as carbon dioxide and load the carriers; the chain pumps protons and ATP synthase makes most of the ATP; the energy not caught as ATP leaves as heat.

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

Answer every question. For each multiple-choice question, pick one option and press Check; the feedback gives the reasoning. For the free-response questions, write one short sentence per step of your reasoning, in order, and show any calculation. In the exam, write the steps as a paragraph: a bulleted list is not scored. The first free-response question walks you through the reasoning one part at a time, and you can open a hint for each part; the second and third are at the level of the test. When you finish a question, open the scoring guide and mark your own work against it. Error bars in these questions are ±2SE.

Question 1

A student wants to show that slices of carrot root, a plant organ with no chloroplasts, respire. The student seals the slices in a clear jar with an oxygen sensor and a carbon dioxide sensor and leaves the jar for two hours.

Which of the following results would show that the root cells are respiring?

A. The oxygen rises and the carbon dioxide falls B. The oxygen falls and the carbon dioxide rises C. The oxygen and the carbon dioxide both fall D. The oxygen stays the same and the carbon dioxide rises

Question 2

In the cytosol, an enzyme of glycolysis removes two electrons (with hydrogen) from a three-carbon molecule, and NAD⁺ picks them up, becoming NADH. The enzyme comes out of the reaction unchanged.

Which of the following describes NAD⁺?

A. An enzyme: a protein that speeds the reaction up without being changed by it B. A protein of the electron transport chain, set in the inner membrane of the mitochondrion C. A product of the reaction: the three-carbon molecule the enzyme has changed D. A coenzyme: a non-protein molecule that works alongside the enzyme and takes part in the reaction

Question 3

In a muscle fiber short of oxygen, NADH hands its two electrons (traveling with hydrogen) to pyruvate, which becomes lactate.

Which molecule is oxidized?

A. NADH B. Pyruvate C. Lactate D. NAD⁺

Question 4

A researcher breaks yeast cells open and removes every mitochondrion, leaving only the cytosol fluid. The researcher seals this fluid in a tube with glucose, ADP, Pi and NAD⁺ under pure nitrogen. Within minutes the tube holds pyruvate, ATP and NADH.

What does this result show about glycolysis?

A. Glycolysis needs oxygen, so some must have dissolved in before the tube was sealed B. Glycolysis happens in the matrix, so a few mitochondria must have stayed in the fluid C. Glycolysis happens entirely in the cytosol and at no step uses oxygen D. Glycolysis releases carbon dioxide, which is what filled the gas space above the fluid

Question 5

Three batches of artificial vesicles hold only ATP synthase, ADP and phosphate: no fuel and no electron transport chain. Inside every vesicle the pH is 7.5. A researcher keeps the outside of one batch at pH 7.5, of a second batch at pH 6.5 and of a third batch at pH 5.5.

Which batch forms ATP fastest, and why?

A. The batch with pH 5.5 outside, because the proton gradient across its membrane is the steepest, so protons flow fastest through ATP synthase B. The batch with pH 7.5 outside, because with the same pH on both sides nothing disturbs ATP synthase C. The batch with pH 6.5 outside, because a moderate proton gradient suits ATP synthase best and too many protons outside block it D. All three batches at the same rate, because ATP synthase alone makes ATP whatever the pH outside

Question 6

A mitochondrion in cross-section, inside a cell. Four positions are numbered: 1 is pointed to from the right, 2 is pointed to from below, 3 sits at the left, and 4 is pointed to from inside.

The figure shows a mitochondrion in cross-section, inside a cell. Four positions are numbered.

Which number marks the fluid in which the Krebs cycle happens?

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

Question 7

A bacterium from pond water, which has no mitochondria, is growing on glucose and taking up oxygen. The water just outside its plasma membrane is at pH 6.5 and its cytosol is at pH 7.6. A researcher adds a poison that blocks the bacterium's electron transport chain. Within a minute the water outside is at pH 7.4 and the cytosol is at pH 7.4.

Which of the following explains the change in the two pH values?

A. The blocked chain pumps protons the other way, from the water outside into the cytosol B. The bacterium switches to fermentation, and the lactate it releases neutralizes the water outside C. ATP synthase in the plasma membrane starts pumping protons into the cell to make up for the chain D. The chain in the plasma membrane has stopped pumping protons out of the cell, so the protons outside flow back in

Question 8

Which of the following animals warms itself from within, using brown fat?

A. A shark in cold water B. A newborn seal pup on the ice C. A tortoise on a sunny rock D. A toad in a cold ditch

Question 9

A bacterium from waterlogged rice-paddy soil is grown on glucose with no oxygen. With dissolved iron(III) ions added, it makes 9 ATP per glucose and the iron(III) becomes iron(II). With the iron left out, it makes 2 ATP per glucose and releases lactate. The same bacterium can also use oxygen as the terminal electron acceptor of its electron transport chain.

Which of the following is the best prediction for its ATP per glucose with oxygen, and why?

A. About 30, because oxygen is a stronger acceptor than iron(III), so the electrons release more energy B. About 9, because the chain makes the same ATP whatever molecule accepts the electrons at its end C. About 11, because oxygen adds its own 2 ATP to the 9 that the chain to iron(III) gives D. About 2, because oxygen stops the fermentation and the bacterium then keeps only glycolysis's ATP

Question 10

A class feeds yeast four flours, wheat, rye, corn and rice, and measures the carbon dioxide each culture makes in 20 minutes, with a mean and a standard error for each flour. A student plots the four means as points in the order wheat, rye, corn, rice, joins the points with a line, and adds a ±2SE error bar to each point.

Which of the following is the mistake in the student's graph?

A. The error bars should be ±1SE, because one standard error is what the class calculated B. The four flours are separate categories, so the means should be bars, one per flour, and never joined C. The flours should be plotted in alphabetical order, so that the line rises steadily D. A straight best-fit line should be drawn through the four points instead of joining them

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 · Scientific Investigation · 6 points

Students measured the oxygen used by mealworms (beetle larvae) with respirometers: sealed tubes in which a chemical absorbs all the carbon dioxide given off, so the gas volume falls only as oxygen is used. At each of 15, 25 and 35 °C, six respirometers each held 5 g of live mealworms; six more held 5 g of glass beads at 25 °C. Each sat for 10 minutes. The table gives the mean oxygen used and the standard error (SE) of each set of six.

Oxygen used in 10 minutes by the contents of each respirometer; six respirometers per condition.

(a) Identify the set of respirometers that serves as the control, and state what it lets the students rule out. (1 point)

Hint: What is a control for in an experiment, and which set of respirometers here does that job? What would a change in that set's gas volume have to be caused by?

A full-credit answer: The control is the set of respirometers holding glass beads.
It lets the students rule out any fall in gas volume that does not come from a living organism.
The bead respirometers hold everything the other tubes hold except the mealworms.
So any fall in their volume (0.02 mL) comes from the apparatus: the chemical, the seal, or a change in temperature.
So the far larger fall in the mealworm tubes came from the mealworms.

Check the box for each point your answer earns

the glass-bead respirometers are the control, and they let the students rule out a change in gas volume caused by anything other than the mealworms (temperature or pressure changes in the tube, the absorbing chemical itself), so the fall in the mealworm tubes can be put down to the mealworms.

Accept "the beads show the volume change with no organism" as the ruled-out cause. Do not award the point for naming the 15 °C tubes as the control, or for the beads with no statement of what they rule out.

Common slip: Calling the 15 °C tubes 'the control'. The 15 °C tubes are a second treatment. A control has the organism removed and everything else kept the same.

(b) Calculate the upper and lower ends of the ±2SE error bar on the 25 °C mean. (1 point)

Hint: Which formula gives each end of a ±2SE bar from the mean and its standard error?

A full-credit answer: The 25 °C bar runs from 0.93 mL to 1.17 mL.
Two standard errors is 2 × 0.06 = 0.12 mL.
The bar reaches 0.12 mL below the mean of 1.05 mL and 0.12 mL above it.

mean at 25 °C = 1.05 mL

SE = 0.06 mL

bar ends = 𝑥̄ \pm 2 S E

bar ends = 𝑥̄ \pm 2 S E

lower end = 1.05 - (2 \times 0.06) = 1.05 - 0.12 = 0.93 mL

upper end = 1.05 + (2 \times 0.06) = 1.05 + 0.12 = 1.17 mL

Check the box for each point your answer earns

0.93 mL to 1.17 mL (the mean of 1.05 mL, minus and plus 2 × 0.06 mL).

Accept 0.93–1.17 mL with the working shown as 1.05 − 0.12 and 1.05 + 0.12. Do not award the point for 0.99–1.11 mL (±1SE) or for "1.05 ± 0.06" written as the bar.

Common slip: Using ±1SE (0.99 to 1.11 mL). The convention is two standard errors each side, so the bar is twice that long.

(c) Describe the trend in oxygen use across the three temperatures, and state how the plotted means should be connected. (1 point)

Hint: Compare the size of the rise between 15 and 25 °C with the size of the rise between 25 and 35 °C. Do the three means follow one straight trend? What does that decide about how they are connected?

A full-credit answer: Oxygen use rises steeply from 15 °C to 25 °C and only a little from 25 °C to 35 °C, so the means should be joined one to the next.
The rise from 15 °C to 25 °C is 0.45 mL.
The rise from 25 °C to 35 °C is 0.07 mL: the trend levels off.
The means do not follow one straight trend.
So they are joined one to the next, not fitted with one straight best-fit line.

Check the box for each point your answer earns

the trend (oxygen use rises with temperature: steeply from 15 to 25 °C and much less from 25 to 35 °C, or "rises and then levels off") AND the connection: the means are joined one to the next, because they do not follow one straight trend.

Accept a smooth curve through the means for the connection. Do not award the point for one straight best-fit line through the three means, or for a trend with no connection stated.

Common slip: Drawing one straight best-fit line through the three means. A best-fit line is for means that follow one straight trend. These means rise steeply and then level off, so they are joined one to the next.

(d) Determine, using the error bars, whether the data show a difference in oxygen use between 15 °C and 25 °C. (1 point)

Hint: Do the two bars overlap? What does a gap between two bars, or an overlap, let you claim about the two means?

A full-credit answer: The 15 °C bar runs from 0.50 to 0.70 mL.
The 25 °C bar runs from 0.93 to 1.17 mL.
There is a clear gap between the two bars, so they do not overlap.
So the difference is unlikely to be chance.
Therefore the data show that the mealworms used more oxygen at 25 °C than at 15 °C.

Check the box for each point your answer earns

the decision (the data do show a difference: the mealworms used more oxygen at 25 °C than at 15 °C) AND the observation it rests on: the 15 °C bar (0.50–0.70 mL) and the 25 °C bar (0.93–1.17 mL) do not overlap, so the difference is unlikely to be chance.

Determine needs the decision and the observation it rests on. Accept "the null hypothesis of no difference between the temperatures is rejected". Do not award the point for "25 °C is higher" with no reference to the bars, or for a claim that the bars overlap.

Common slip: Comparing the two means alone. Two means always differ a little. Only a gap between the ±2SE bars lets you claim the difference is real.

(e) Explain why the gas volume in a respirometer of live mealworms falls. (1 point)

Hint: Where in cellular respiration is oxygen actually used, and what happens to the carbon dioxide inside this apparatus?

A full-credit answer: The volume falls because the mealworms’ cells take up oxygen, which is used to accept the electrons at the end of the electron transport chain, while the carbon dioxide they release is absorbed by the chemical.
The cells are respiring: they break food down to make ATP.
Oxygen takes the electrons off the end of the chain, with hydrogen ions, and becomes water.
The chemical absorbs the carbon dioxide, so the only gas change left is the oxygen taken up.

Check the box for each point your answer earns

the mealworms' cells are respiring (breaking food down to make ATP) and take up oxygen, the terminal electron acceptor at the end of the electron transport chain (it takes the electrons and hydrogen ions and becomes water), while the carbon dioxide they release is absorbed by the chemical, so the volume falls by the oxygen used.

Accept "oxygen accepts the electrons at the end of the chain" for oxygen's role. Do not award the point for "the mealworms breathe the air" with no role for oxygen, for oxygen turned into carbon dioxide, or for an answer that leaves out the absorbed carbon dioxide.

Common slip: Saying the mealworms 'use up the air'. The point needs oxygen's job, taking the electrons at the end of the chain, and a word about where the carbon dioxide went.

(f) Predict what happens to the oxygen used in 10 minutes if the students give the mealworms at 25 °C a substance that lets protons leak straight back across the inner mitochondrial membrane, bypassing ATP synthase, and justify your prediction. (1 point)

Hint: Which stage of respiration uses the oxygen, and how is that stage linked to the proton gradient? What does the leak change for it?

A full-credit answer: Oxygen use would rise, because the leak lets protons return to the matrix without passing through ATP synthase, so the electron transport chain pumps against a smaller proton gradient and passes electrons faster.
A faster chain passes more electrons to oxygen each minute.
So more oxygen is used, and the volume falls faster than 1.05 mL in 10 minutes.

Check the box for each point your answer earns

oxygen use rises (the volume falls faster), because the leak drains the proton gradient, so the chain pumps against a smaller proton gradient and passes electrons faster, passing more electrons to oxygen; the proton gradient's energy leaves as heat instead of ATP.

Accept a prediction that ATP output falls or that the mealworms warm up, alongside the oxygen prediction. Do not award the point for "oxygen use falls because ATP is no longer made", or for a prediction with no link between the proton gradient and the chain's speed.

Common slip: Predicting that oxygen use falls because 'less ATP is made'. ATP output does fall. But the chain, freed from a steep proton gradient, passes electrons faster and uses more oxygen, as it does in brown fat.

Free-response score: 0 of 6

Free response 2 · Conceptual Analysis · 6 points

A researcher grows yeast cells in a stirred flask with plenty of glucose and dissolved oxygen, then adds a poison that stops electron transfer at a protein near the end of the electron transport chain; glucose and oxygen stay plentiful. In the hour before the poison the culture used 5 mmol of glucose and made 140 mmol of ATP, and the flask's oxygen sensor showed a steady fall. In the hour after, the culture used 40 mmol of glucose and made 80 mmol of ATP, the oxygen reading stopped falling, and ethanol began to build up in the flask.

(a) Describe how the untreated yeast cells made ATP by oxidative phosphorylation, from the electrons delivered by NADH to the ATP. (1 point)

A full-credit answer: NADH delivers its electrons to the electron transport chain.
The electrons pass from protein to protein toward oxygen.
Oxygen, the terminal electron acceptor, takes the electrons off the end and becomes water.
The energy the electrons release pumps protons from the matrix into the intermembrane space: a proton gradient.
Protons flow back into the matrix down their proton gradient, almost only through ATP synthase.
That flow drives the formation of ATP from ADP and inorganic phosphate.

Check the box for each point your answer earns

electrons delivered by NADH pass down the electron transport chain from protein to protein to oxygen, the terminal electron acceptor, which becomes water; the energy the electrons release pumps protons out of the matrix into the intermembrane space; the protons flow back down their proton gradient into the matrix almost only through ATP synthase, and that flow drives the formation of ATP from ADP and inorganic phosphate (chemiosmosis).

Accept "a difference in proton concentration and charge across the membrane" for the proton gradient. Do not award the point for ATP synthase pumping the protons, for oxygen supplying the energy, or for the chain making the ATP directly.

Common slip: Giving ATP synthase the pumping job, or the chain the ATP-making job. The chain pumps protons out; ATP synthase is the way back, and that is where the ATP is made.

(b) Calculate the ATP made per glucose before and after the poison, and explain why the yeast used more glucose after it. (2 points)

A full-credit answer: Before the poison the yeast made 140 ÷ 5 = 28 ATP per glucose.
After it, they made 80 ÷ 40 = 2 ATP per glucose.
Each glucose now gives 2 ATP instead of 28.
So the yeast must break down far more glucose for the ATP they make.

before the poison: glucose used = 5 mmol, ATP made = 140 mmol

after the poison: glucose used = 40 mmol, ATP made = 80 mmol

ATP per glucose = \frac{ATP made}{glucose used}

ATP per glucose = \frac{ATP made}{glucose used}

before the poison: ATP per glucose = \frac{140}{5} = 28

after the poison: ATP per glucose = \frac{80}{40} = 2

Check the box for each point your answer earns

140 ÷ 5 = 28 ATP per glucose with oxygen, and 80 ÷ 40 = 2 ATP per glucose without. without oxygen only glycolysis makes ATP, about 2 per glucose instead of about 30, so the yeast must split far more glucose for the same ATP.

Common slip: Subtracting instead of dividing, or giving the two values with no reason for the extra glucose. The yield is ATP made divided by glucose used, for each hour separately; a smaller yield per glucose means more glucose for the same ATP.

(c) Explain why ethanol appeared in the flask after the poison. (1 point)

A full-credit answer: With the chain blocked, NADH has nowhere to unload its electrons.
So the cell’s NAD⁺ is soon all tied up as NADH.
Glycolysis needs NAD⁺ to accept electrons, so glycolysis would stop.
Fermentation rescues it.
Pyruvate loses a carbon as carbon dioxide.
NADH hands its electrons to the two-carbon fragment that is left, making ethanol.
That regenerates NAD⁺, so glycolysis keeps going and keeps making its two ATP per glucose.

Check the box for each point your answer earns

with the chain blocked, NADH can no longer unload its electrons to the chain, so the cell's NAD⁺ would all become NADH and glycolysis would stop; fermentation passes NADH's electrons to the two-carbon fragment left when pyruvate loses a carbon as carbon dioxide, making ethanol and regenerating the NAD⁺ that glycolysis needs.

Accept "NADH is oxidized back to NAD⁺ by making ethanol" with the blocked chain as the reason. Accept "the yeast ferment, making 2 ATP per glucose" provided NAD⁺ regeneration or the blocked chain is given as the reason for the ethanol. Do not award the point for "the yeast switch to fermentation to make ATP" with no mention of NAD⁺, for the ethanol-forming step itself producing the ATP, or for ethanol as a leftover of the blocked chain.

Common slip: Saying that the ethanol-forming step itself makes the ATP. Glycolysis makes the two ATP per glucose. Fermentation regenerates NAD⁺, so glycolysis can keep making them.

(d) The researcher gives a second flask of the same yeast, with plenty of glucose and oxygen, a drug that blocks only the channel through ATP synthase. Predict what happens to the pH of the intermembrane space in the first minute after the poison in the first flask, and in the first minute after the drug in the second flask. Justify each prediction. (2 points)

A full-credit answer: In the first flask, the pH of the intermembrane space rises toward the matrix pH.
The poison blocks transfer near the end of the chain, so the whole chain backs up and pumping stops.
Protons already outside keep flowing back into the matrix through ATP synthase.
So the concentration of protons in the intermembrane space falls, and its pH rises.
In the second flask, the pH of the intermembrane space falls.
The chain is not blocked, so it keeps pumping protons out of the matrix into the intermembrane space.
The blocked ATP synthase lets none flow back into the matrix.
So protons pile up in the intermembrane space, and its pH falls.

Check the box for each point your answer earns

the first flask: the pH rises (toward the matrix pH), because the blocked chain stops pumping protons into the intermembrane space while protons keep flowing back into the matrix through ATP synthase, so the concentration of protons in the intermembrane space falls. the second flask: the pH falls (the intermembrane space becomes more acidic), because the chain is not blocked and keeps pumping protons into the intermembrane space, while the blocked ATP synthase no longer lets them flow back into the matrix, so protons pile up there.

Accept "the proton gradient collapses" for the first flask and "the proton gradient grows" for the second, each with the direction of the pH change stated. Do not award a point for a prediction with no cause, or for the two directions swapped.

Common slip: Swapping the two flasks. A blocked chain stops the pump, so the protons outside drain back through ATP synthase and the pH rises. A blocked ATP synthase shuts only the way back while the pump keeps working, so protons accumulate outside and the pH falls.

Free-response score: 0 of 6

Free response 3 · Conceptual Analysis · 5 points

A hummingbird beats its wings about 50 times a second. Its flight-muscle cells are packed with mitochondria, and their inner membranes are folded far more densely than the inner membranes of the mitochondria in the bird’s liver cells. A researcher feeds a hummingbird sugar water in which every carbon atom of the sugar is the heavy form ¹³C. Within minutes, the carbon dioxide the bird breathes out carries ¹³C. The table shows what one glucose gives a flight-muscle cell, stage by stage, by the end of the Krebs cycle.

What one glucose gives a flight-muscle cell at each stage, by the end of the Krebs cycle.

(a) Describe what the mitochondria do for a flight-muscle cell, and what happens to each pyruvate from glycolysis when it reaches a mitochondrion. (2 points)

A full-credit answer: Most of the steps of respiration happen inside the mitochondria.
There the cell breaks food down with oxygen and makes most of its ATP.
Pyruvate is transported from the cytosol into the mitochondrial matrix.
There it is oxidized: one of its three carbons leaves as carbon dioxide.
The electrons it gives up load NAD⁺, making NADH.
The two-carbon fragment left enters the Krebs cycle.

Check the box for each point your answer earns

the mitochondria's job: they make most of the cell's ATP by breaking food down with oxygen (most of the steps of respiration happen inside them). pyruvate: it is moved from the cytosol into the mitochondrial matrix and oxidized there: one of its three carbons leaves as carbon dioxide, its electrons load NAD⁺ to make NADH, and the two-carbon fragment left enters the Krebs cycle.

Accept "the mitochondria carry out cellular respiration, which makes the cell's ATP" for the job, and an answer without the name "pyruvate oxidation" for pyruvate. Do not award the job point for the mitochondria storing energy or storing glucose. Do not award the pyruvate point for pyruvate entering the Krebs cycle whole, or for all three carbons leaving as carbon dioxide at this step.

Common slip: Saying the mitochondria store the cell's energy, or sending the whole pyruvate into the Krebs cycle. A mitochondrion stores nothing: it breaks food down and makes ATP as the cell needs it. One carbon of pyruvate leaves first, as carbon dioxide, and the electrons go to NAD⁺; only the two-carbon fragment enters the cycle.

(b) Identify, using the table, the molecules that carry most of the glucose’s energy by the end of the Krebs cycle. (1 point)

A full-credit answer: The ATP column adds to 4 ATP made directly.
The carrier columns add to 10 NADH and 2 FADH₂.
Each loaded carrier holds two electrons taken from the glucose.
So most of the glucose’s energy is on the 10 NADH and 2 FADH₂, not in the 4 ATP.

Check the box for each point your answer earns

the loaded electron carriers, 10 NADH and 2 FADH₂ (the ATP column adds to only 4).

Accept "NADH and FADH₂" without the totals. Do not award the point for ATP, or for carbon dioxide.

Common slip: Answering 'ATP'. Only 4 ATP have been made directly. The electrons taken from the glucose, on NADH and FADH₂, carry most of the energy, and they have not reached the chain yet.

(c) Explain how the ¹³C in the bird’s breath shows that the carbon of the carbon dioxide came from the sugar. (1 point)

A full-credit answer: Enzymes treat ¹³C like ordinary carbon, so the labeled sugar is broken down in the usual way.
An instrument can tell ¹³C from ordinary carbon by its mass.
Only the sugar was labeled, so wherever ¹³C turns up is where the sugar’s carbon went.
The ¹³C turned up in the carbon dioxide within minutes.
So the carbon of the carbon dioxide came from the sugar: two carbons per glucose at pyruvate oxidation and four in the Krebs cycle.

Check the box for each point your answer earns

enzymes treat ¹³C like ordinary carbon (so the labeled sugar is broken down in the usual way) but an instrument tells ¹³C apart by its mass; only the sugar was labeled, so the ¹³C in the carbon dioxide can only have come from the sugar's carbon.

Accept an answer that names where the carbons leave (two at pyruvate oxidation, four in the Krebs cycle). Do not award the point for "the ¹³C makes the bird respire faster", or for the carbon dioxide's carbon coming from the oxygen the bird breathed in.

Common slip: Saying the carbon dioxide's carbon came from the oxygen the bird breathed in. Oxygen gas becomes water at the end of the chain. The carbon dioxide's carbon is the sugar's carbon, which is why it carries the sugar's label.

(d) Explain why the densely folded inner membranes let the flight-muscle mitochondria make ATP faster than the liver mitochondria. (1 point)

A full-credit answer: The electron transport chain and ATP synthase are set in the inner membrane.
Folding the membrane increases its surface area.
More surface fits more chain proteins and more ATP synthase.
So the flight-muscle mitochondria make ATP faster.

Check the box for each point your answer earns

the electron transport chain and ATP synthase are set in the inner membrane; folding gives the membrane more surface area, so more chain proteins and more ATP synthase fit in it, so more ATP is made each second.

Accept "more surface area, so more ATP synthase". Do not award the point for the folds holding more oxygen or more glucose, or for each ATP synthase turning faster.

Common slip: Saying the folds hold more oxygen or more fuel. Folds hold nothing. They give the membrane more surface, and the surface is where the chain and ATP synthase sit.

Free-response score: 0 of 5

Multiple choice checked: 0 of 10 correct.

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