Practice questions · Topic 3.5

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

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 tells you what a wrong choice assumed. For the free-response questions, write your answer in full sentences and show any calculation. The first free-response question is broken into small steps, each with a sentence starter and a hint you can open if you need it; the second is at the level of the test. When you finish each one, open the scoring guide and mark your own work against it. Error bars in these questions are ±2SE.

Question 1

Slices of carrot root, a plant organ with no chloroplasts, are sealed in a clear jar under a bright lamp for two hours. The oxygen in the jar falls from 20.9% to 20.3%, and the carbon dioxide rises from 0.04% to 0.55%.

Which process explains both changes?

A. Photosynthesis in the root cells, which takes in oxygen and gives out carbon dioxide when the light is bright B. Cellular respiration in the root cells, which runs in bright light just as it does in the dark C. Fermentation in the root cells, which they switch to as soon as they are cut from the plant D. Breathing by the root, which pulls air in through its surface when the lamp warms it

Question 2

Two fine pH probes are pushed into a respiring mitochondrion, each coming to rest in a different fluid-filled region of the organelle. One probe reads pH 6.9; the other reads pH 7.9.

Which region is the probe reading pH 6.9 resting in?

A. The matrix B. The cytosol C. The inner membrane's folds D. The intermembrane space

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, and which is reduced?

A. NADH is oxidized and pyruvate is reduced B. NADH is reduced and pyruvate is oxidized C. Both are oxidized, because both molecules change D. Both are reduced, because electrons pass between them

Question 4

Yeast cells are broken open and spun so that every mitochondrion is removed, leaving only the cytosol fluid. This fluid is sealed in a tube with glucose, ADP, Pi and NAD⁺, and the gas above it is 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 into the fluid from the air before the tube was sealed B. Glycolysis runs in the matrix of mitochondria, so a few mitochondria must have stayed behind in the fluid C. Glycolysis runs in the cytosol and needs no oxygen, splitting glucose to pyruvate and making ATP and NADH D. Glycolysis releases the carbon of glucose as carbon dioxide, which is what filled the gas space above the fluid

Question 5

Isolated mitochondria are given pyruvate, plenty of NAD⁺ and FAD, ADP, Pi and oxygen. For every pyruvate they take in, they release three molecules of carbon dioxide.

Where do the three carbon dioxide molecules come from?

A. One from the oxidation of pyruvate as it enters the matrix, and two from the Krebs cycle B. All three from the Krebs cycle, which takes the whole pyruvate apart C. All three from the end of the electron transport chain, where oxygen combines with carbon D. Two from the oxidation of pyruvate, and one from glycolysis running inside the mitochondrion

Question 6

Dissolved oxygen in a sealed chamber of isolated mitochondria. Pyruvate was added at 2 minutes and a poison that blocks a protein near the end of the electron transport chain at 6 minutes.

Isolated mitochondria sit in a sealed chamber fitted with a dissolved-oxygen sensor. Pyruvate is added at 2 minutes, and a poison that blocks a protein near the end of the electron transport chain is added at 6 minutes. The graph shows the dissolved oxygen over 8 minutes.

What was the rate of oxygen use between 2 and 6 minutes?

A. 20 μmol/L per minute B. 30 μmol/L per minute C. 60 μmol/L per minute D. 120 μmol/L per minute

Question 7

A soil bacterium, which has no mitochondria, is growing on glucose and taking up oxygen from the water around it. A pH microelectrode held just outside its plasma membrane reads pH 6.5; a second, inside its cytosol, reads pH 7.6.

What do the readings show?

A. The bacterium is fermenting and releasing lactic acid into the soil water around it B. The bacterium's ATP synthase in its plasma membrane pumps protons out of the cell as it makes ATP C. The bacterium's mitochondria lie just inside the plasma membrane and leak protons outward D. The bacterium's electron transport chain sits in its plasma membrane and pumps protons out of the cell

Question 8

In the 1930s a drug was sold for weight loss. It lets protons pass freely across the inner mitochondrial membrane of every cell in the body. People who took it lost fat, felt hot and sweated, and their oxygen use rose by a third.

Why did the drug have these effects?

A. The leak stopped the electron transport chain, so the cells burned fat instead of glucose and released its heat B. The leak let ATP out of the mitochondria, and the hydrolysis of that ATP in the cytosol released the heat C. The leak let the gradient's energy escape as heat, so the chain ran faster and burned more fuel to keep up the ATP supply D. The leak made ATP synthase turn faster, and the extra ATP it made was broken down to heat

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.

What are the iron(III) ions doing?

A. Acting as the terminal electron acceptor of an electron transport chain, in place of oxygen B. Acting as a second fuel that the bacterium oxidizes alongside the glucose C. Accepting the electrons from NADH in fermentation, in place of pyruvate D. Supplying the energy for the extra seven ATP as they change from iron(III) to iron(II)

Question 10

A class feeds yeast four flours, wheat, rye, corn and rice, and measures the carbon dioxide each culture makes in 20 minutes: five tubes per flour, giving a mean and a standard error for each flour.

Which graph fits these results?

A. A line graph with the four means joined in order, so the trend from flour to flour can be read B. A bar graph with one bar per flour, each carrying its ±2SE error bar C. A scatter graph with a straight best-fit line drawn through the four means D. A single bar showing the overall mean of all twenty tubes, with its error bar

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 · 5 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. Six respirometers each held 5 g of live mealworms at 15 °C, six held 5 g of live mealworms at 25 °C, and six held 5 g of glass beads at 25 °C. Each ran for 10 minutes. The mean oxygen used was 0.60 mL (standard error 0.05 mL) at 15 °C, 1.05 mL (SE 0.06 mL) at 25 °C, and 0.02 mL (SE 0.01 mL) for the glass beads. The graph shows the three means with error bars of ±2SE.

Mean oxygen used in 10 minutes by the contents of each respirometer (six respirometers per condition). Error bars are ±2SE.

(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 glass-bead respirometers are the control. They hold everything the other tubes hold except a living organism, so any fall in their volume (0.02 mL) comes from the apparatus: the chemical, the seal, or a change in temperature. That lets the students say the far larger fall in the mealworm tubes came from the mealworms themselves.

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'. They 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 bar runs from 0.93 mL to 1.17 mL: two standard errors, 0.12 mL, either side of the mean of 1.05 mL.

Write down the values in the question:

mean at 25 °C = 1.05 mL
SE = 0.06 mL

Write down the equation:

lower end = mean − 2SE
upper end = mean + 2SE

Substitute in the values, and calculate:

lower end = 1.05 − (2 × 0.06) = 1.05 − 0.12 = 0.93 mL
upper end = 1.05 + (2 × 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 what the error bars show about oxygen use at 15 °C and at 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 and the 25 °C bar from 0.93 to 1.17 mL. There is a clear gap between them, so the data support the claim that the mealworms used more oxygen in 10 minutes at 25 °C than at 15 °C; the difference is unlikely to be chance.

Check the box for each point your answer earns

the 15 °C bar (0.50–0.70 mL) and the 25 °C bar (0.93–1.17 mL) do not overlap, so the data show that the mealworms used more oxygen at 25 °C than at 15 °C (the difference is unlikely to be chance).

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.

(d) 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 mealworms' cells are respiring. Oxygen is taken up because it is the terminal electron acceptor: electrons from food pass down the electron transport chain to oxygen, which becomes water. The carbon dioxide the cells release is absorbed by the chemical in the tube, so the only gas change left is the oxygen taken up, and the volume falls by that amount.

Check the box for each point your answer earns

the mealworms' cells are respiring 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.

(e) Predict what happens to the oxygen used in 10 minutes if the mealworms at 25 °C are given 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 rises, so the volume falls faster than 1.05 mL in 10 minutes. The leak lets protons return to the matrix without passing through ATP synthase, so the gradient drains as fast as the chain builds it. Pumping against a small gradient, the chain runs faster and passes more electrons to oxygen, while the gradient's energy leaves as heat instead of ATP.

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 gradient and runs faster, passing more electrons to oxygen; the 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 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 gradient, runs faster and uses more oxygen, as it does in brown fat.

Free-response score: 0 of 5

Free response 2 · Conceptual Analysis · 4 points

Yeast cells are grown in a stirred flask with plenty of glucose and dissolved oxygen. A poison is then added 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 the role of oxygen in the electron transport chain. (1 point)

A full-credit answer: Oxygen is the terminal electron acceptor. Electrons delivered by NADH and FADH₂ pass down the chain from protein to protein, and at the end oxygen takes them, together with hydrogen ions, and becomes water. Taking the electrons off the end is what keeps the chain flowing.

Check the box for each point your answer earns

oxygen is the terminal electron acceptor: it takes the electrons from the end of the chain, with hydrogen ions, and becomes water, which keeps electrons flowing down the chain.

Accept "final electron acceptor". Do not award the point for oxygen supplying the energy for ATP, or for oxygen combining with carbon to make carbon dioxide.

Common slip: Giving oxygen the job of supplying energy or of becoming carbon dioxide. The energy is in the electrons taken from food; oxygen only takes them off the end, and it becomes water.

(b) Calculate the ATP made per glucose before and after the poison. (1 point)

A full-credit answer: Before the poison the yeast made 28 ATP per glucose; after it, 2 ATP per glucose.

Write down the values in the question:

before the poison: glucose used = 5 mmol, ATP made = 140 mmol
after the poison: glucose used = 40 mmol, ATP made = 80 mmol

Write down the equation:

                     ATP made
ATP per glucose = ──────────────
                   glucose used

Substitute in the values, and calculate:

before: 140 ÷ 5 = 28 ATP per glucose
after: 80 ÷ 40 = 2 ATP per glucose

Check the box for each point your answer earns

140 ÷ 5 = 28 ATP per glucose before the poison and 80 ÷ 40 = 2 ATP per glucose after it.

Accept 28 and 2 with the unit written as "ATP per glucose" or "mmol ATP per mmol glucose". Do not award the point for 140 − 5 = 135, or for the two totals restated.

Common slip: Subtracting instead of dividing, or restating the totals. The yield is ATP made divided by glucose used, for each hour separately.

(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, and glycolysis, which needs NAD⁺ to accept electrons, would stop. Fermentation rescues it: pyruvate loses a carbon as carbon dioxide, and NADH hands its electrons to the two-carbon fragment left, making ethanol and regenerating NAD⁺. Glycolysis can then keep making its two ATP per glucose, which is why glucose use rose eightfold.

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. Do not award the point for "the yeast switch to fermentation to make ATP" with no mention of NAD⁺, or for ethanol as a leftover of the blocked chain.

Common slip: Saying fermentation makes the ATP. Fermentation makes none; its job is to regenerate NAD⁺ so that glycolysis can keep making its two per glucose.

(d) Predict what happens to the pH of the intermembrane space in the first minute after the poison, and justify your prediction. (1 point)

A full-credit answer: The pH of the intermembrane space rises toward the matrix pH. The chain pumps protons out of the matrix only while electrons pass along it; with transfer blocked near the end, the whole chain backs up and pumping stops, while protons already outside keep flowing back into the matrix through ATP synthase. The gradient runs down, so the intermembrane space is soon no more acidic than the matrix.

Check the box for each point your answer earns

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 gradient runs down.

Accept "the gradient collapses" with the direction of the pH change stated. Do not award the point for the pH falling (the gradient growing, which is what blocking ATP synthase does), or for a prediction with no cause.

Common slip: Predicting that the pH falls. That is what happens when ATP synthase is blocked and the chain keeps pumping; here it is the pump itself that has stopped.

Free-response score: 0 of 4

Multiple choice checked: 0 of 10 correct.

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