3.5 Cellular Respiration

1. AIt transfers energy from organic molecules such as glucose to ATP, which then powers cell work. Correct
   In cellular respiration, energy from biological macromolecules is used to make ATP. Energy already stored in organic molecules is transferred to ATP (and partly released as heat); ATP then powers cellular work.
2. BIt creates new energy from glucose and stores that energy in ATP, which then powers cell work.
   A student who thinks respiration makes energy picks this. Energy is not created; it is transferred from glucose to ATP and heat.
3. CIt converts the matter of glucose into energy, which is stored in ATP to power cell work.
   A student who thinks food is turned into energy picks this. The atoms of glucose leave as CO₂ and H₂O; only the energy is transferred to ATP.
4. DIt is the exchange of O₂ and CO₂ by breathing, which supplies the cell with energy for its work.
   A student who equates cellular respiration with breathing picks this. Breathing and gas exchange supply O₂ and remove CO₂, but the energy is transferred to ATP by reactions inside the cells.

CED 3.5.A.1 · Read this in Fix

1. ABreaking glucose down in many steps releases more total energy than burning it in one step.
   A student who thinks a multistep enzyme pathway releases more energy picks this. The model shows that both routes start and end at the same energy levels, so the total energy released is the same.
2. BEnergy is released in small amounts through many steps, and part of it is captured in ATP. Correct
   The model shows the same starting and ending energy levels for both routes. In respiration, a series of coordinated enzyme-catalyzed reactions releases the energy in small steps, and part of the energy released at some steps is captured in ATP; burning releases it all at once as heat and light.
3. CEach step breaks bonds in glucose, and breaking each bond releases energy that forms ATP.
   A student who thinks breaking bonds releases energy picks this. Breaking bonds takes in energy; each step releases energy overall because the bonds formed release more than the bonds broken take in.
4. DEach step creates a small amount of new energy, and these amounts add up to form ATP.
   A student who thinks respiration creates energy picks this. No energy is created; each step transfers part of the energy already present in the molecules.

CED 3.5.A.2 · Read this in Fix

1. AIt makes ATP directly, as each carrier adds a phosphate to ADP when an electron passes through it.
   A student who thinks the ETC itself makes ATP picks this. The carriers move H⁺ across the membrane; ATP is made separately, by ATP synthase.
2. BIt passes electrons along a series of redox reactions and uses the energy released to move H⁺ across a membrane. Correct
   The ETC transfers electrons in a series of oxidation-reduction reactions; the energy released moves H⁺ across the inner mitochondrial membrane, establishing an electrochemical gradient that ATP synthase then uses.
3. CIt moves electrons across the inner membrane, forming a gradient of electrons that is used to make ATP.
   A student who thinks the gradient is made of electrons picks this. Electrons pass along the carriers to O₂; the gradient that forms is a gradient of H⁺.
4. DIt combines O₂ with carbon atoms removed from glucose, forming the CO₂ that the cell gives out.
   A student who thinks O₂ becomes CO₂ picks this. O₂ accepts electrons at the end of the ETC and forms H₂O; the CO₂ comes from carbon removed in pyruvate oxidation and the Krebs cycle.

CED 3.5.A.3 · Read this in Fix

1. AGlucose is oxidized by O₂ to pyruvate, forming ATP from ADP and releasing CO₂ and H₂O as well.
   A student who thinks glycolysis needs O₂ picks this. Glycolysis passes electrons to NAD⁺, not O₂, and releases no CO₂.
2. BGlucose is broken down to pyruvate, forming ATP from ADP and NADH from NAD⁺; no O₂ is used. Correct
   Glycolysis releases energy in glucose to form ATP from ADP and inorganic phosphate, NADH from NAD⁺, and pyruvate. It does not use O₂.
3. CGlucose is split into pyruvate and NADH, and all of the ATP is made later, in the mitochondria.
   A student who thinks all ATP is made in mitochondria picks this. Glycolysis forms ATP from ADP and inorganic phosphate in the cytosol.
4. DGlucose is converted to lactic acid and ATP, and this happens only when no O₂ is present in the cell.
   A student who equates glycolysis with fermentation picks this. Glycolysis produces pyruvate whether or not O₂ is present; lactic acid is formed by fermentation.

CED 3.5.B.1 · Read this in Fix

1. AGlucose is oxidized to CO₂ inside the mitochondria, while the cytosol fraction stores pyruvate.
   A student who thinks mitochondria take up and break down glucose picks this. Mitochondria given glucose (tube 2) showed no change.
2. BPyruvate is oxidized to CO₂ in the cytosol, while the mitochondria take up the O₂ this requires.
   A student who thinks the Krebs cycle takes place in the cytosol picks this. The cytosol given pyruvate (tube 3) released no CO₂.
3. CPyruvate forms from glucose because O₂ is present to react with glucose in the cytosol.
   A student who thinks glycolysis needs O₂ picks this. Every tube contained O₂, so the data cannot show that O₂ is needed, and glycolysis does not use O₂.
4. DGlucose is converted to pyruvate in the cytosol, and pyruvate is oxidized to CO₂ in the mitochondria. Correct
   Pyruvate formed only when glucose was given to the cytosol (tube 1), and CO₂ was released only when pyruvate was given to mitochondria (tube 4). Glycolysis occurs in the cytosol; pyruvate is transported into the mitochondrion, where it is oxidized.

CED 3.5.B.2 · Read this in Fix

1. ARegion W Correct
   The Krebs cycle takes place in the mitochondrial matrix, the compartment enclosed by the folded inner membrane.
2. BRegion X
   A student who thinks the Krebs cycle takes place in the cytosol, with glycolysis, picks this region, which is outside the mitochondrion.
3. CRegion Y
   A student who thinks the intermembrane space is the inner compartment picks this region. It lies between the outer and inner membranes; the matrix is enclosed by the inner membrane.
4. DRegion Z
   A student who thinks every stage of respiration takes place on the inner membrane picks this. The inner membrane holds the ETC and ATP synthase; the Krebs cycle is in the matrix.

CED 3.5.B.3 · Read this in Fix

1. ATube 3, which has NADH but has the matrix fraction instead
   A student who thinks a control keeps the tested factor and changes something else picks this. Tube 3 differs in the fraction, so it tests where the ETC is, not whether O₂ use depends on NADH.
2. BTube 2, which has the inner-membrane fraction but no NADH Correct
   Tube 2 differs from tube 1 only in the absence of NADH, so a difference in O₂ consumption between them can be attributed to NADH delivering electrons to the ETC in the inner membrane.
3. CTube 4, which has NADH but no mitochondrial fraction at all
   A student who thinks the control is always the tube without biological material picks this. Tube 4 lacks the inner-membrane fraction, not NADH, so it shows whether NADH alone uses O₂, not whether tube 1's O₂ use depends on NADH.
4. DTube 1 again, since a repeat of a tube serves as its control
   A student who thinks replication is a control picks this. Repeating tube 1 shows how much its result varies, but not what happens without NADH.

CED 3.5.B.4 · Read this in Fix

1. AThe matrix has the lower pH, as the ETC moves H⁺ into the matrix from the intermembrane space.
   A student who thinks the ETC pumps H⁺ into the matrix picks this. H⁺ is moved out of the matrix, so the matrix has the lower H⁺ concentration and the higher pH.
2. BThe matrix has the lower pH, as the ETC moves H⁺ out of the matrix into the intermembrane space.
   A student who thinks a higher H⁺ concentration means a higher pH picks this. Moving H⁺ out of the matrix lowers its H⁺ concentration, which raises its pH.
3. CThe matrix has the higher pH, as the ETC moves H⁺ out of the matrix into the intermembrane space. Correct
   As electrons pass along the ETC, H⁺ is moved from the matrix into the intermembrane space. The intermembrane space therefore has the higher H⁺ concentration and the lower pH, and the matrix the higher pH.
4. DThe two have the same pH, as H⁺ crosses the lipid bilayer freely until it is spread evenly.
   A student who thinks ions cross the lipid bilayer freely picks this. The bilayer is a barrier to H⁺, so the ETC can keep a higher H⁺ concentration in the intermembrane space.

CED 3.5.B.5 · Read this in Fix

1. AGlycolysis soon stops in the mutant but not the wild type, as NAD⁺ is not regenerated from NADH. Correct
   Glycolysis needs NAD⁺ to accept electrons. Without O₂, fermentation regenerates NAD⁺ from NADH as pyruvate is converted to ethanol and CO₂. The mutant cannot do this, so its NAD⁺ runs out and glycolysis stops; the wild type continues.
2. BGlycolysis continues in the mutant, but it makes less ATP, as the fermentation step makes the ATP.
   A student who thinks fermentation reactions make ATP picks this. The ATP comes from glycolysis, which stops in the mutant once NAD⁺ runs out.
3. CGlycolysis continues in the mutant, as NADH passes electrons through the ETC to CO₂ instead.
   A student who thinks cells find another way to make ATP because they need to picks this. Without O₂ the mutant has no terminal electron acceptor for its ETC, and CO₂ cannot take O₂'s place, so its NADH is not reoxidized and glycolysis stops once NAD⁺ runs out.
4. DGlycolysis stops in both strains, as glycolysis needs O₂ to oxidize glucose to pyruvate.
   A student who thinks glycolysis needs O₂ picks this. Glycolysis passes electrons to NAD⁺, not to O₂, so it continues in the wild type, whose fermentation regenerates NAD⁺.

CED 3.5.B.6 · Read this in Fix

1. AThey respire only because they are in the dark; in light, photosynthesis would replace it.
   A student who thinks plants respire only in the dark picks this. Plant cells respire in the light and in the dark; photosynthesis does not take the place of respiration.
2. BTheir stored starch combines directly with O₂ in one reaction, as it does when starch burns.
   A student who thinks O₂ reacts directly with food molecules in one step, as in burning, picks this. In cells the starch is broken down to glucose and oxidized in a series of enzyme-catalyzed reactions; O₂ accepts electrons only at the end of the ETC.
3. CTheir cells turn the stored starch into energy, and this is why the amount of starch falls.
   A student who thinks matter is converted into energy picks this. The starch falls because its atoms leave the seed as CO₂ and H₂O; the energy released is transferred to ATP and heat.
4. DTheir cells carry out cellular respiration, using energy from stored starch to make ATP. Correct
   Germinating seeds are living cells that break down their stored starch (a macromolecule) to glucose and oxidize it in cellular respiration, using O₂ and releasing CO₂. The energy transferred to ATP powers germination.

CED 3.5.A.1 · Read this in Fix

1. A2.7 mL/h
   A student who does not use the control's readings picks this: 0.90/20 × 60 = 2.7 mL/h. The 0.10 mL change in the glass-bead respirometer is not caused by respiration and must be subtracted.
2. B2.3 mL/h
   A student who thinks dry seeds are not alive, and so treats them as the non-respiring control, picks this: (0.90 − 0.13)/20 × 60 ≈ 2.3 mL/h. Dry seeds are alive and respire slowly, so their change includes some respiration.
3. C2.4 mL/h Correct
   Subtract the change caused by temperature and pressure, shown by the glass beads: 0.90 − 0.10 = 0.80 mL in 20 min, or 0.040 mL/min, which is 2.4 mL/h.
4. D0.8 mL/h
   A student who reports the volume consumed as the rate picks this. 0.80 mL is the amount consumed in 20 min; the rate is 0.80 mL/20 min = 2.4 mL/h.

Working The glass beads do not respire, so the 0.10 mL change in their respirometer is caused by temperature and pressure changes that affect every respirometer. Corrected volume consumed by the germinating seeds in 20 min = 0.90 − 0.10 = 0.80 mL. Rate = 0.80 mL/20 min = 0.040 mL/min = 0.040 × 60 = 2.4 mL/h. Distractors: ignoring the control gives 0.90/20 × 60 = 2.7 mL/h; subtracting the dry-seed reading, as if dry seeds did not respire, gives (0.90 − 0.13)/20 × 60 = 2.31 ≈ 2.3 mL/h; reporting the corrected volume, 0.80 mL, as the rate gives 0.8 mL/h.

CED 3.5.A.1 · Read this in Fix

1. AThe rate is highest at 35 °C, as expected if heat above that temperature kills the living enzymes.
   A student who thinks enzymes are living things picks this. Enzymes are molecules, not living things; heat lowers their activity by disrupting their shape, not by killing them. The rate is still 91% of the highest at 40 °C, and it is the steep fall at higher temperatures, not the position of the peak, that points to protein catalysts.
2. BThe rate is lowest at 5 °C, as expected if cold temperatures denature the enzymes that catalyze the reactions.
   A student who thinks cold denatures enzymes picks this. The low rate at 5 °C is explained by slower molecular motion and fewer collisions, which slow any chemical reaction, so it does not single out enzymes.
3. CThe rate peaks between 35 and 40 °C, as expected because every enzyme works best near 37 °C.
   A student who thinks every enzyme has an optimum near 37 °C picks this. Optima differ between enzymes and organisms, so the position of the peak alone is not evidence that the reactions are enzyme-catalyzed.
4. DThe rate falls steeply above 40 °C, as expected if the proteins that catalyze the reactions lose their shape. Correct
   Rates of all chemical reactions rise with temperature, but a steep fall above an optimum is characteristic of enzyme-catalyzed reactions: high temperatures disrupt the structure of the protein catalysts and denature them. This feature of the data therefore supports the claim.

CED 3.5.A.2 · Read this in Fix

1. AModel 1 Correct
   Electrons from NADH pass from carrier to carrier in redox reactions to O₂, the terminal electron acceptor, and the energy released moves H⁺ from the matrix into the intermembrane space. This builds the H⁺ gradient across the inner membrane.
2. BModel 2
   A student who thinks the ETC pumps H⁺ into the matrix picks this model. The ETC moves H⁺ out of the matrix, so H⁺ becomes more concentrated in the intermembrane space.
3. CModel 3
   A student who thinks the ETC builds a gradient of electrons picks this model. Electrons pass along the carriers to O₂; it is H⁺ that is moved across the membrane.
4. DModel 4
   A student who thinks O₂ is the source of the cell's energy picks this model, in which electrons flow from O₂ to NADH. Electrons flow from NADH, which carries them from organic molecules, to O₂, the terminal acceptor.

CED 3.5.A.3 · Read this in Fix

1. AOnly the last carrier of the ETC stops, so the carriers before it keep pumping H⁺ as before.
   A student who thinks a block affects only one step picks this. A carrier can pass electrons on only if the next one accepts them, so a block at the end stops flow along the whole chain.
2. BThe mitochondria switch to passing electrons to CO₂ instead, so electron transport continues.
   A student who thinks organelles change their reactions because they need energy picks this. Mitochondria use O₂ as their terminal electron acceptor and cannot switch to CO₂ because O₂ has run out.
3. CThe ETC carriers stay reduced, so H⁺ pumping stops and NADH builds up in the matrix. Correct
   O₂ is the terminal electron acceptor. Without it, the last carrier cannot pass on its electrons, so every carrier before it stays reduced, electron flow and H⁺ pumping stop, and NADH can no longer be oxidized, so it accumulates.
4. DPyruvate oxidation in the matrix continues as before, since these reactions do not use O₂.
   A student who thinks the matrix reactions are independent of the ETC picks this. They need NAD⁺, which is regenerated only when NADH passes electrons to the ETC, so they slow and stop.

CED 3.5.A.3.i · Read this in Fix

1. AIt is fermentation, since any process that makes ATP without O₂ is a form of fermentation.
   A student who thinks respiration always needs O₂ picks this. Fermentation uses no ETC; a process with an ETC and a terminal electron acceptor is respiration, whatever the acceptor.
2. BIt is cellular respiration, with sulfate replacing O₂ as the terminal electron acceptor. Correct
   Electrons from organic molecules pass along an ETC to a terminal electron acceptor, and the resulting H⁺ gradient drives ATP synthesis: this is cellular respiration. Anaerobic prokaryotes use molecules other than O₂, here sulfate, as the terminal acceptor.
3. CIt is a way of obtaining energy by breaking down sulfate in place of organic molecules.
   A student who thinks the terminal electron acceptor is the fuel picks this. The organic molecules are oxidized and supply the electrons; sulfate accepts them at the end of the chain.
4. DIt is a process that turns sulfate into the CO₂ released, as aerobic cells turn O₂ into CO₂.
   A student who thinks O₂ becomes CO₂ in respiration picks this. In both cases the CO₂ comes from the carbon of organic molecules; the terminal acceptor is reduced (O₂ to H₂O).

CED 3.5.A.3.i · Read this in Fix

1. AWithout O₂, the bacterium cannot break down glucose to make ATP at all unless nitrate is present.
   A student who thinks glucose cannot be broken down without O₂ picks this. In glucose without O₂ or nitrate, the density rose from 0.1 to 1.2 units, so the cells made some ATP, for example by glycolysis.
2. BNitrate supplies the energy for growth, since the bacterium grows whenever nitrate is in the medium.
   A student who thinks the terminal electron acceptor is the fuel picks this. With nitrate but no glucose, the density stayed at 0.1 units, so nitrate alone does not supply energy; glucose is the fuel.
3. CWithout O₂, nitrate greatly increases growth on glucose, as nitrate serves as a terminal electron acceptor. Correct
   Without O₂, growth on glucose reached 6.1 units with nitrate but only 1.2 units without it, and nitrate alone supported no growth. This fits nitrate acting as a terminal electron acceptor for electrons from glucose, as anaerobic prokaryotes use molecules other than O₂.
4. DGrowth is greatest with O₂ because O₂ combines directly with glucose and releases its energy.
   A student who thinks O₂ reacts directly with glucose picks this. Growth is greatest with O₂, but O₂ accepts electrons only at the end of the ETC, after glucose has been broken down in many steps.

CED 3.5.A.3.i · Read this in Fix

1. AA can make ATP faster, as its larger inner membrane area holds more ETCs and ATP synthases. Correct
   The folding of the inner membrane increases its surface area. The ETC and ATP synthase are in the inner membrane, so the more folded inner membrane of A holds more of them and allows more ATP to be synthesized.
2. BA can make ATP faster, as its folded membrane lets glucose diffuse into the matrix faster.
   A student who thinks glucose enters mitochondria picks this. Glucose is broken down in the cytosol; the inner membrane's folds increase the area for the ETC and ATP synthase.
3. CB can make ATP faster, as its larger matrix holds more Krebs cycle enzymes, which make most ATP.
   A student who thinks the Krebs cycle makes most of the ATP picks this. Most ATP is made by ATP synthase in the inner membrane, which is more extensive in A.
4. DBoth make ATP at the same rate, as the ATP made depends only on the glucose that is supplied.
   A student who thinks ATP output is set only by the fuel supply picks this. The amount of inner membrane, which holds the ETCs and ATP synthases, also limits how fast ATP can be made.

CED 3.5.A.3.ii · Read this in Fix

1. AIt forms across the inner membrane of its mitochondria, with more H⁺ in the intermembrane space.
   A student who thinks bacteria have mitochondria picks this. Bacteria have no mitochondria; their ETC and ATP synthase are in the plasma membrane.
2. BIt forms across the plasma membrane, with more H⁺ in the cytoplasm than outside it.
   A student who thinks H⁺ is pumped into the compartment where ATP is made picks this. The ETC moves H⁺ out of the cytoplasm, and H⁺ flows back in through ATP synthase.
3. CIt forms across the cell wall, with more H⁺ outside the wall than between the wall and the membrane.
   A student who thinks the cell wall is a selectively permeable barrier picks this. The wall is permeable to ions; the gradient forms across the plasma membrane.
4. DIt forms across the plasma membrane, with more H⁺ outside the membrane than in the cytoplasm. Correct
   Prokaryotes have no mitochondria; in aerobic prokaryotes the passage of electrons along the ETC moves protons out across the plasma membrane, so H⁺ is more concentrated outside the membrane, and H⁺ flowing back in through ATP synthase drives ATP synthesis.

CED 3.5.A.3.ii · Read this in Fix

1. ACells with more inner mitochondrial membrane per cell will have a higher rate of ATP synthesis.
   A student who thinks the null hypothesis is the expected result picks this. This is the alternative hypothesis that the student expects to support.
2. BCells with more inner mitochondrial membrane per cell will have a reduced rate of ATP synthesis.
   A student who thinks the null hypothesis is the opposite of the prediction picks this. A negative relationship is another alternative hypothesis; the null hypothesis states no relationship.
3. CThe rate of ATP synthesis is not related to the area of inner mitochondrial membrane per cell. Correct
   A null hypothesis states that there is no relationship between the variables being studied. The data will be used to decide whether this can be rejected.
4. DThe muscle cells will not carry out ATP synthesis at any time during the investigation.
   A student who reads 'null' as 'nothing happens' picks this. The null hypothesis concerns the relationship between the variables, not whether ATP synthesis occurs.

CED 3.5.A.3.ii · Read this in Fix

1. AH⁺ moving from side 1 to side 2 through ATP synthase, down the H⁺ gradient Correct
   The carriers have moved H⁺ to side 1, where it is more concentrated. H⁺ flows back to side 2 through ATP synthase by chemiosmosis, and this flow drives the formation of ATP from ADP and inorganic phosphate on side 2.
2. BATP synthase moving H⁺ from side 2 to side 1, against its concentration gradient
   A student who thinks ATP synthase pumps H⁺ in the same direction as the carriers picks this. Moving H⁺ against its gradient would take energy; ATP synthase uses the energy of H⁺ flowing down its gradient.
3. CElectrons passing from the last carrier of the chain directly into ATP synthase
   A student who thinks electrons flow through ATP synthase picks this. The model shows the electrons going to O₂, which is reduced to H₂O; ATP synthase is powered by H⁺.
4. DO₂ binding to ATP synthase and adding an oxygen atom to each ADP molecule
   A student who reads 'oxidative phosphorylation' as adding oxygen to ADP picks this. O₂ is the terminal electron acceptor and forms H₂O; ATP forms when inorganic phosphate is added to ADP.

CED 3.5.A.3.iii · Read this in Fix

1. AA low pH around the mitochondria drove ATP synthesis, whether or not H⁺ could cross the inner membrane.
   A student who thinks the amount of H⁺, rather than a difference across the membrane, does the work picks this. Sample 3 was also at pH 6 but made little ATP, because compound P removed the difference.
2. BA difference in H⁺ concentration across the inner membrane drove ATP synthesis without electron transport. Correct
   Only sample 2 made much ATP. There, H⁺ was more concentrated outside the inner membrane (pH 6) than in the matrix (pH 8), and H⁺ flowing back through ATP synthase drove ATP synthesis even though no electrons were being transported. Sample 3 had the same low outside pH but no gradient, because P let H⁺ cross freely, and it made little ATP.
3. CATP synthesis needs O₂ to be added to ADP, so more O₂ must have reached the mitochondria in sample 2.
   A student who thinks oxidative phosphorylation adds oxygen to ADP picks this. ATP forms from ADP and inorganic phosphate, and nothing in the investigation changed the O₂ supply between samples.
4. DATP synthase made ATP in sample 2 by pumping H⁺ out of the matrix into the more acidic solution.
   A student who thinks ATP synthase pumps H⁺ out of the matrix picks this. The H⁺ in sample 2 flowed into the matrix, down its gradient from pH 6 to pH 8, and that flow drove ATP synthesis.

CED 3.5.A.3.iii · Read this in Fix

1. AThe addition of oxygen atoms from O₂ to ADP, which forms ATP in the mitochondrial matrix
   A student who reads 'oxidative' as 'adding oxygen' picks this. O₂ accepts electrons and forms H₂O; ATP forms by adding inorganic phosphate to ADP.
2. BAny ATP synthesis in glycolysis and the Krebs cycle, as carbon atoms of glucose are oxidized
   A student who thinks all ATP made during glucose oxidation is oxidative phosphorylation picks this. ATP made directly by the reactions of glycolysis and the Krebs cycle is not made by ATP synthase and is not oxidative phosphorylation.
3. CThe addition of a phosphate to ADP by each carrier of the ETC as an electron passes through it
   A student who thinks the ETC makes ATP directly picks this. The carriers move H⁺; ATP synthase makes the ATP.
4. DATP synthesis by ATP synthase, powered by H⁺ flowing back across the inner membrane Correct
   Oxidative phosphorylation is the formation of ATP from ADP and inorganic phosphate by ATP synthase, driven by the flow of protons back through ATP synthase by chemiosmosis, using the gradient that electron transport set up.

CED 3.5.A.3.iii · Read this in Fix

1. AU lets H⁺ return to the matrix without passing through ATP synthase, so the energy is released as heat. Correct
   U decouples oxidative phosphorylation from electron transport. Electron transport continues (O₂ consumption even rises) as H⁺ leaks back, but little H⁺ passes through ATP synthase, so little ATP is made and the energy of the gradient is released as heat.
2. BU stops electron transport along the ETC, so less ATP is made and the unused energy is released as heat.
   A student who thinks electron transport and ATP synthesis are one process picks this. If electron transport had stopped, O₂ consumption would have fallen, but it rose to 160%.
3. CU converts ATP molecules into heat energy, so less ATP is left even though more O₂ is being used.
   A student who thinks matter is converted into energy picks this. ATP is not turned into energy; the data show that ATP synthesis itself fell, while the energy from electron transport was released as heat.
4. DU makes the mitochondria need more ATP, so they take in more O₂ in order to make ATP faster.
   A student who explains the change by the mitochondria's needs picks this. ATP synthesis fell to 10%, so the extra O₂ use was not producing more ATP.

CED 3.5.A.3.iv · Read this in Fix

1. AIts body temperature is unaffected, because its fur, not its cells, produces the heat it needs.
   A student who thinks fur produces heat picks this. Fur reduces heat loss; the heat is produced by metabolism, including decoupled oxidative phosphorylation in brown fat.
2. BIts brown fat cells make a replacement protein, because the animal needs heat to survive the cold.
   A student who thinks organisms produce what they need picks this. The mutation eliminates the protein, and need does not cause cells to make a replacement.
3. CIts brown fat cells release more heat, as they make more ATP and breaking ATP's bonds releases heat.
   A student who thinks energy is released by breaking bonds picks this. Without the protein, the energy of the gradient is no longer released directly as heat, so heat production in brown fat falls.
4. DIts brown fat cells release less heat, so it is less able to keep its body temperature up. Correct
   The protein decouples oxidative phosphorylation from electron transport, so the energy of the H⁺ gradient is released as heat, which endotherms use to regulate body temperature. Without the protein, less heat is generated in brown fat.

CED 3.5.A.3.iv · Read this in Fix

1. AATP production is unchanged, as mitochondria take up the glucose and break it down instead.
   A student who thinks mitochondria take up glucose picks this. Mitochondria take up pyruvate, not glucose, and without O₂ they cannot carry out electron transport.
2. BATP production is unchanged, as the ATP is made by fermentation, which does not need glycolysis.
   A student who thinks fermentation makes the ATP picks this. The ATP comes from glycolysis; fermentation only regenerates NAD⁺ and needs the pyruvate that glycolysis makes.
3. CATP production falls, as glycolysis was the cells' source of ATP in the absence of O₂. Correct
   Without O₂, the yeast make ATP by glycolysis, with fermentation regenerating NAD⁺. Blocking glycolysis stops pyruvate formation, so ATP production falls (and so does the production of ethanol and CO₂).
4. DATP production rises, as the cells sense the energy shortage and speed up their ETCs.
   A student who thinks cells make ATP faster because they need it picks this. Without O₂ there is no terminal electron acceptor for the ETC, and need does not speed up a blocked pathway.

CED 3.5.B.1 · Read this in Fix

1. ASupply glucose and O₂ to mitochondria isolated from each strain, and compare their rates of CO₂ release.
   A student who thinks mitochondria take up glucose picks this. Mitochondria of both strains would release little CO₂ from glucose, because glycolysis takes place in the cytosol, so the test cannot distinguish them.
2. BSupply pyruvate and O₂ to mitochondria isolated from each strain, and compare their rates of CO₂ release. Correct
   Pyruvate is transported from the cytosol into the mitochondrion, where its oxidation releases CO₂. If isolated mutant mitochondria release little CO₂ from pyruvate while wild-type mitochondria release it readily, the result supports the hypothesis; similar rates would argue against it.
3. CSupply pyruvate and O₂ to the cytosol fraction of each strain, and compare their rates of CO₂ release.
   A student who thinks pyruvate is oxidized in the cytosol picks this. Pyruvate oxidation and the Krebs cycle take place in the matrix, so the cytosol fraction does not test mitochondrial uptake.
4. DSupply glucose and O₂ to whole cells of each strain, and compare how fast the two strains grow.
   A student who thinks a whole-organism comparison can test one cellular step picks this. The strains are already known to differ in growth with O₂, and many defects could cause this; the test does not isolate pyruvate uptake.

CED 3.5.B.2 · Read this in Fix

1. A27%
   A student who reports the difference in rate, 60 − 33 = 27 nmol/min, as the percent change picks this. The change must be divided by the original rate.
2. B82%
   A student who divides the change by the new rate picks this: 27/33 × 100 = 82%. Percent change is calculated relative to the original rate, 60 nmol/min.
3. C45% Correct
   Percent decrease = (60 − 33)/60 × 100 = 45%. Blocking a Krebs cycle reaction reduces the release of CO₂ from organic intermediates.
4. D55%
   A student who gives the new rate as a percentage of the original picks this: 33/60 × 100 = 55%. That is the percentage remaining; the decrease is 100% − 55% = 45%.

Working From the table, the rate is 60 nmol/min with 0 μM R and 33 nmol/min with 10 μM R. Percent decrease = (60 − 33)/60 × 100 = 45%. Distractors: reporting the change in rate, 27 nmol/min, as a percentage gives 27%; dividing the change by the new rate gives 27/33 × 100 = 82%; giving the new rate as a percentage of the original gives 33/60 × 100 = 55%.

CED 3.5.B.3 · Read this in Fix

1. AP reduced O₂ consumption, but the data do not show a clear effect of Q. Correct
   The ±2 SE bars for P and the control do not overlap, so P likely reduced O₂ consumption. The bars for Q and the control overlap, so the data do not show a clear difference between them.
2. BP and Q both reduced O₂ consumption, as both means are below the control mean.
   A student who treats any difference in means as a real effect picks this. Q's ±2 SE bar overlaps the control's, so the data do not show that Q reduced O₂ consumption.
3. CP reduced O₂ consumption, and the overlap of Q's error bar proves Q has no effect.
   A student who thinks overlapping error bars prove no effect picks this. The overlap means only that no clear effect was shown; a small effect could exist.
4. DP reduced O₂ consumption, which proves that P blocks electron transfer from NADH.
   A student who thinks supporting data prove a hypothesis picks this. Lower O₂ consumption is consistent with P blocking transfer from NADH, but blocking any later step in the chain would give the same result.

Working Reading the graph: control 100 ± 9 (91 to 109); P 24 ± 7 (17 to 31); Q 88 ± 12 (76 to 100). P's bar does not overlap the control's, so P likely reduced O₂ consumption. Q's bar overlaps the control's, so the data do not show a clear effect of Q, although they do not prove that Q has no effect. A fall in O₂ consumption with P is consistent with P blocking electron transfer from NADH but does not prove it, because blocking any later step of the chain would have the same effect.

CED 3.5.B.4 · Read this in Fix

1. AFrom 5 to 8 min, it was higher in the matrix than in the intermembrane space.
   A student who thinks a higher pH means more H⁺ picks this. The matrix had the higher pH, 7.8, and therefore the lower H⁺ concentration.
2. BAfter 11 min, the difference across the inner membrane was as large as it was at 6 min.
   A student who thinks that running out of O₂ stops only the last carrier, so the carriers before it keep moving H⁺, expects the difference to be kept and picks this. After the O₂ was used up, the two pH values moved toward each other (7.3 and 7.2 at 11 min versus 7.8 and 6.9 at 6 min), so the difference became smaller.
3. CThroughout the 12 min, it was the same in the matrix and in the intermembrane space.
   A student who thinks H⁺ crosses the lipid bilayer freely picks this. From about 4 to 10 min, the two pH values differed by up to 0.9 units.
4. DFrom 5 to 8 min, it was higher in the intermembrane space than inside the matrix. Correct
   From 5 to 8 min, the intermembrane space was at pH 6.9 and the matrix at pH 7.8. A lower pH means a higher H⁺ concentration, so H⁺ was more concentrated in the intermembrane space, as expected while the ETC was moving H⁺ out of the matrix.

CED 3.5.B.5 · Read this in Fix

1. AEthanol production began only once the O₂ had fallen to zero, and it then rose steadily.
   A student who thinks fermentation begins only at zero O₂ picks this. Ethanol was already present at 3 h and 4 h, before the O₂ was used up at 5 h.
2. BOnce the O₂ was used up, the yeast cells stopped breaking down glucose altogether.
   A student who thinks glucose cannot be broken down without O₂ picks this. Ethanol kept accumulating after 5 h, so glycolysis and fermentation continued.
3. CEthanol production began while some O₂ remained, and it continued after the O₂ was used up. Correct
   Ethanol first appeared at about 2 to 3 h, when O₂ was still at about 55% to 28% of its starting value, and it kept rising at a steady rate after O₂ reached zero at 5 h. Fermentation allowed glycolysis to continue without O₂.
4. DEthanol rose as the O₂ fell, which shows that the ethanol caused the O₂ concentration to fall.
   A student who takes a correlation as cause and effect picks this. O₂ was already falling before any ethanol appeared, and the O₂ was used by the yeast's aerobic respiration.

CED 3.5.B.6 · Read this in Fix

1. AIt produces the ATP that the cell uses while O₂ is in short supply.
   A student who thinks fermentation reactions make ATP picks this. The model shows ATP formed only in glycolysis; the conversion of pyruvate to lactic acid makes none.
2. BIt releases CO₂ from pyruvate, as every type of fermentation does.
   A student who thinks all fermentation releases CO₂ picks this. The model shows no CO₂; lactic acid fermentation releases none, unlike alcohol fermentation.
3. CIt sends electrons through an ETC, with lactic acid taking the place of O₂.
   A student who thinks fermentation uses an ETC with a different acceptor picks this. The model shows no ETC; NADH passes its electrons to pyruvate, forming lactic acid.
4. DIt uses NADH and regenerates the NAD⁺ that glycolysis needs to keep going. Correct
   The model shows NAD⁺ being reduced to NADH in glycolysis and NADH being oxidized back to NAD⁺ as pyruvate is converted to lactic acid. Fermentation therefore allows glycolysis to proceed in the absence of O₂.

CED 3.5.B.6 · Read this in Fix