Practice questions · Topic 2.9

Unit 2 · Practice for the Topic 2.9 end-of-topic test

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Hold conditions, keep reactions apart, add surface: what a cell’s internal membranes do, and how to predict what fails when a compartment breaks.

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 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. 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 is at the level of the test. When you finish a question, open the scoring guide and mark your own work against it.

Question 1

In the storage cells of a sprouting bean seed, digestive enzymes sit inside small membrane-bound bodies together with stored protein. The interior of each body is acidic; the cytosol around it is close to neutral. The enzymes cut stored protein quickly in acid and barely at all at the cytosol's acidity.

What lets the enzymes work inside the bodies and hardly at all in the cytosol around them?

A. The cytosol holds a substance that destroys the digestive enzymes B. The membrane of each body speeds the digestive enzymes up C. The digestive enzymes only cut protein that is stored inside a body D. The membrane of each body keeps its interior acidic while the cytosol is not

Question 2

A scientist builds an artificial cell: a single outer membrane filled with fluid. The scientist considers four ways of arranging its contents.

Which arrangement gives the artificial cell compartmentalization?

A. Its proteins gathered into a dense cluster at the center, with no membrane around them B. Several membrane-bound sacs inside it, each holding fluid with its own conditions C. A rigid cell wall added around the outside of the outer membrane D. The fluid inside stirred so that every substance is spread evenly throughout

Question 3

Cells of a mustard leaf make a bitter compound that damages a plant's protein-building machinery when the compound reaches it. The cells store the compound inside their central vacuoles, while ribosomes in the cytosol build new proteins all day.

Why does the cell benefit from storing the compound in the vacuole?

A. The compound never reaches the ribosomes, so protein building goes on undisturbed B. The vacuole is where the compound is made, so it has to stay there C. The compound would leak out of the cell if it were loose in the cytosol D. The vacuole's membrane makes the compound more bitter to insects

Question 4

Cells of a mustard leaf store a bitter compound inside their central vacuoles. The compound damages ribosomes when it reaches them, and the ribosomes in the cytosol of these cells build new proteins all day. The vacuole membrane becomes leaky, so the bitter compound spreads into the cytosol.

Predict the effect on protein building in the cytosol, and explain why.

A. Protein building speeds up: the compound now reaches the ribosomes and helps them B. Protein building is unchanged: the ribosomes are still in the cytosol where they were C. Protein building slows: the compound now reaches the ribosomes it damages D. Protein building stops only inside the vacuole, where the ribosomes have moved

Question 5

The flight muscle cells of a hummingbird hold mitochondria whose inner membranes are folded far more than those in a chicken's breast muscle cells. The reactions that make ATP happen on the inner membrane. The two kinds of mitochondria are about the same size.

What do the extra folds do for the hummingbird's muscle cells?

A. The folds store fuel for the flight muscle to burn later B. The folds keep the inside of each mitochondrion warmer, so the reactions speed up C. The folds give more membrane surface, so more ATP-making reactions happen at once D. The folds let each mitochondrion hold more fluid than a chicken's

Question 6

Cells lining the gut that take in fat from food have far more smooth ER membrane than their neighbors. The reactions that rebuild absorbed fat into fats the body can use happen on the membranes of the smooth ER.

What does the extra ER membrane do for these cells?

A. The extra membrane gives more surface on which the fat-building reactions can happen at once B. The extra membrane stores the absorbed fat until the cell needs it C. The extra membrane lets the cell get by with fewer mitochondria than its neighbors D. The extra membrane thickens the plasma membrane so that fat can cross it more easily

Question 7

In many plant cells the central vacuole is acidic while the cytosol is close to neutral, and the reactions that happen in the cytosol work best close to neutral. The vacuole’s membrane becomes leaky to acid.

Predict the effect on the reactions in the cytosol, and explain why.

A. The reactions are unaffected: the acid stays in the vacuole B. The reactions slow: the acid spreads into the cytosol, which is no longer close to neutral C. The reactions speed up: acid speeds up every reaction D. The reactions move into the vacuole, where the conditions are now the same as outside

Question 8

In a damaged cell, the membrane of one lysosome breaks. The lysosome’s acid and its digestive enzymes reach the mitochondria next to it. The reactions that make ATP happen on proteins in each mitochondrion’s inner membrane, and those reactions work well only in conditions close to neutral.

Predict the effect on how fast the cell makes ATP, and why.

A. Faster: the acid speeds up the ATP-making reactions on the inner membrane B. Unchanged: the inner membrane and its ATP-making proteins are still in place C. Faster: the digestive enzymes break down fuel right beside the inner membrane D. Slower: the acid and the digestive enzymes damage the inner membrane’s proteins

Question 9

One compartment inside a cell, before and after a change to its membrane, with the concentration of substance Y inside the compartment and in the cytosol.

The figure shows one compartment inside a cell, before and after a change. A reaction that happens inside the compartment needs substance Y at 10 mmol/L or more. Before the change Y is at 30 mmol/L inside the compartment and 0.3 mmol/L in the cytosol; afterward it is at 1.2 mmol/L in both.

What most likely changed, and what follows for the reaction?

A. Most of the Y in the cell was destroyed, so the reaction stops for lack of Y B. Y was pumped out of the compartment, so the reaction now happens in the cytosol instead C. The membrane became leaky to Y, so Y spread out by diffusion and the reaction stops D. Y was concentrated in the cytosol, so the reaction speeds up there

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 · Conceptual Analysis · 4 points

A company wants yeast cells to make a useful compound, Q, in two steps. Enzyme 1 turns a starting substance into an intermediate, J, and works best in acidic conditions; J damages ribosomes if it reaches them. Enzyme 2 turns J into Q and works best in conditions close to neutral, like the cytosol. Design A places both enzymes in the cytosol: after a day the cells hold 6 units of Q, and they grow slowly. Design B places enzyme 1 inside the yeast's vacuole, whose interior is acidic, and enzyme 2 in the cytosol; the vacuole's membrane lets J pass slowly out into the cytosol. After a day the cells hold 48 units of Q and grow normally.

A sketch of the two designs: the first move in a question like this is to draw the cell, the vacuole inside it, and where each enzyme sits. Dots are ribosomes.

(a) Explain why enzyme 1 makes more J in design B than in design A. (1 point)

Hint: In which conditions does enzyme 1 work best, and which conditions does it sit in, in each design?

A full-credit answer: Enzyme 1 works best in acidic conditions.
In design B, enzyme 1 sits inside the vacuole.
The vacuole’s membrane keeps the interior acidic, so enzyme 1 works at its best.
In design A, enzyme 1 sits in the cytosol.
The cytosol is close to neutral, so enzyme 1 works slowly.
So enzyme 1 makes more J in design B.

Check the box for each point your answer earns

enzyme 1 works best in acid; in design B it sits inside the acidic vacuole, so it works at its best, whereas in design A it sits in the cytosol, close to neutral, where it works slowly.

Accept: 'the vacuole's membrane gives enzyme 1 the acid it needs; the cytosol does not'.

Common slip: Saying the vacuole 'protects' enzyme 1 with no mention of acidity. It is the acidic conditions the membrane holds that matter.

(b) Explain why the cells grow normally in design B but slowly in design A. (1 point)

Hint: In each design, where does J form, and what sits nearby that J could affect?

A full-credit answer: In design A, J forms in the cytosol.
The ribosomes are in the cytosol, so J reaches them and damages them.
Protein building slows, so the cells grow slowly.
In design B, J forms inside the vacuole, away from the ribosomes.
The J that leaks out slowly is turned into Q by enzyme 2 in the cytosol.
So little J ever reaches the ribosomes, and the cells grow normally.

Check the box for each point your answer earns

in design A, J forms in the cytosol, where the ribosomes are, and damages them, so protein building and growth slow; in design B, J forms inside the vacuole, away from the ribosomes, and the J that leaks out slowly is turned into Q by enzyme 2, so little of it reaches the ribosomes.

Accept: 'the vacuole membrane keeps J apart from the ribosomes'. Do not award the point for 'the cells have more Q' with no mention of J and the ribosomes.

Common slip: Explaining the growth from the amount of Q. Growth depends on the ribosomes, and the question is whether J reaches them.

(c) Predict the yield of Q from a design C: enzyme 1 in the vacuole and enzyme 2 in the cytosol, as in design B, but with a vacuole membrane that blocks J completely. (1 point)

Hint: In design C, where is J made and where is enzyme 2? What lies between them?

A full-credit answer: Design C would give almost no Q, less even than design A’s 6 units.
The J that enzyme 1 makes stays trapped in the vacuole.
Enzyme 2 is in the cytosol, so it never receives any J.

Check the box for each point your answer earns

the prediction that design C gives very little Q, close to zero, lower than design A's 6 units. No reasoning is required for this point.

Do not award a prediction near design B's 48 units, or one equal to design A.

Common slip: Predicting a yield near 48 units because the acidity is right. Right conditions do nothing for enzyme 2 if J never reaches it.

(d) Justify your prediction in (c), using what the vacuole's membrane must hold in and what it must let through for this process to work. (1 point)

Hint: What must stay inside the vacuole for enzyme 1 to do its job, and what must be able to leave it for enzyme 2 to do its job? A quick sketch of the vacuole inside the cell, with each enzyme in its place, makes this easier to see.

A full-credit answer: For the process to work, the membrane must do two things.
It must hold in the acid, so that enzyme 1 works.
It must let J through to the cytosol, where enzyme 2 turns it into Q.
Design C’s membrane holds the acid but traps J.
So enzyme 2 receives no J, and no Q forms, however well enzyme 1 works.
Design B gave 48 units only because J could pass out of the vacuole.

Check the box for each point your answer earns

the evidence AND the reasoning: the membrane must hold in the acid, so that enzyme 1 works, AND let J through to the cytosol, where enzyme 2 turns it into Q; design C's membrane holds the acid but traps J, so enzyme 2 receives no J and no Q forms, however well enzyme 1 works.

Accept: reasoning that cites design B's yield as depending on J passing out of the vacuole. Do not award the point for support from acidity alone.

Common slip: Justifying from acidity alone. A compartment helps only if what the next step needs can still get through.

Free-response score: 0 of 4

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

The model shows part of a cell from a mustard leaf. The central vacuole is shaded; its shading marks an acidic interior that holds a bitter defensive compound at 80 mmol/L, while the cytosol holds the compound at 0.2 mmol/L. The compound stops ribosomes from working if it reaches them. The mitochondrion is drawn with its outer membrane and its folded inner membrane, where the reactions that make ATP happen. The dots in the cytosol are ribosomes, building new proteins.

Part of a mustard leaf cell: the central vacuole (shaded: acidic, holding the defensive compound at 80 mmol/L); a mitochondrion with a folded inner membrane; ribosomes in the cytosol. The cytosol holds the compound at 0.2 mmol/L.

(a) Describe what the vacuole's membrane does for the cell, in terms of the conditions it holds inside and what it keeps apart. (1 point)

A full-credit answer: The vacuole’s membrane keeps the inside acidic.
It holds the compound at 80 mmol/L inside while the cytosol has only 0.2 mmol/L.
It also keeps the compound apart from the ribosomes in the cytosol.
The compound would stop those ribosomes if it reached them.

Check the box for each point your answer earns

the membrane lets the vacuole's interior hold conditions different from the cytosol (acidic, and the compound at 80 mmol/L against 0.2 mmol/L outside), AND it keeps the compound away from the ribosomes it would stop.

Both halves, the conditions held and what is kept apart, are needed for the point.

Common slip: Saying the membrane 'holds the contents in place'. Name the conditions it holds and the thing it keeps the compound away from.

(b) Explain how the membrane makes possible a concentration of 80 mmol/L inside the vacuole beside 0.2 mmol/L in the cytosol. (1 point)

A full-credit answer: Dissolved particles move at random and spread out.
Without a membrane, the compound would spread until it was evenly mixed.
The compound stays 400 times more concentrated inside the vacuole.
So the membrane must not let the compound diffuse freely across it.
The membrane is what makes the difference in concentration possible.

Check the box for each point your answer earns

dissolved particles move at random and spread, so the compound could stay 400 times more concentrated in the vacuole only because the membrane does not let it diffuse freely across; without the membrane, diffusion would even the two sides out.

Accept: 'a gradient can exist across a membrane only for a substance the membrane blocks'. Do not award the point for 'the vacuole is where it is made' with no reference to the membrane blocking diffusion.

Common slip: Saying the compound is concentrated because it is made in the vacuole. Whatever makes it, only a membrane that blocks it keeps it from spreading.

(c) Predict the effect on the ribosomes if the vacuole’s membrane is dissolved, and justify your prediction. (1 point)

A full-credit answer: The ribosomes slow or stop building proteins.
With the membrane gone, the compound spreads out of the vacuole by diffusion into the cytosol.
It reaches the ribosomes and stops them.
The acid spreads out too, so the cytosol becomes acidic as well.

Check the box for each point your answer earns

the ribosomes slow or stop building proteins, because the compound spreads out of the vacuole by diffusion into the cytosol, reaches the ribosomes and stops them (the cytosol may also turn acidic).

Accept either the compound or the acid as the agent, provided it is said to reach the ribosomes. Do not award the point for 'nothing changes' or for a prediction with no justification.

Common slip: Predicting no change because the ribosomes themselves are untouched. Their surroundings change: what the membrane held in now reaches them.

(d) Support the claim that the mitochondrion in the model shows a third benefit of internal membranes, different from the two in part (a). (1 point)

A full-credit answer: The mitochondrion’s inner membrane is folded.
Folding fits far more membrane into the same space.
The ATP-making reactions happen on that membrane, so more of them happen at the same time.
So the mitochondrion shows a third benefit of internal membranes: they add surface for reactions, as well as holding conditions and keeping things apart.

Check the box for each point your answer earns

the evidence AND the reasoning: the mitochondrion’s inner membrane is folded; folding fits more membrane into the same space, so more of the ATP-making reactions happen at the same time; that is a third benefit (more surface for reactions), different from holding conditions and keeping things apart.

Accept: 'more surface area for membrane-bound reactions' with the link to more reactions at once or faster ATP-making, named as a benefit different from the two in (a).

Common slip: Saying the folds store energy or let fuel in. They add surface for the reactions that make ATP.

Free-response score: 0 of 4

Multiple choice checked: 0 of 9 correct.

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