Practice questions · Topic 2.5

Unit 2 · Practice for the Topic 2.5 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: diffusion and gradients, summed up

The mole and the millimole, the concentration gradient, random motion and diffusion, dynamic equilibrium, passive transport.

Watch next: too big for any door, summed up

Endocytosis, exocytosis, and the five ways across a membrane named by direction, protein, energy and size.

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.

These practice questions have the shape of the Topic 2.5 test. Answer every question. For each multiple-choice question, pick one option and press Check; the feedback gives the reasoning. For the first free-response question, you work through one scenario in small steps, and each step offers a hint if you want one. For the second, 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. Then open the scoring guide and mark your own work against it.

Question 1

Two solutions are prepared, each 0.20 mol/L and each one liter: one of glycine, an amino acid of which one mole has a mass of 75 g, and one of tryptophan, an amino acid of which one mole has a mass of 204 g.

How do the numbers of dissolved amino acid molecules in the two liters compare?

A. The tryptophan liter holds about 2.7 times as many molecules B. The glycine liter holds about 2.7 times as many molecules C. The two liters hold the same number of molecules, about 1.2 × 10²³ each D. The glycine liter holds 75 molecules and the tryptophan liter holds 204

Question 2

In a kidney, fluid inside a tube holds urea at 0.30 mol/L. Blood flowing beside the tube holds urea at 0.005 mol/L. The tube wall between them lets urea through.

Which way is down urea's concentration gradient?

A. From the tube fluid into the blood, from 0.30 mol/L toward 0.005 mol/L B. From the blood into the tube fluid, from 0.005 mol/L toward 0.30 mol/L C. Along the tube, in the direction the fluid flows D. Neither way: urea is spread evenly once both sides hold some

Question 3

A crystal of purple dye dissolves at the bottom of a beaker of still water, and over an hour the color spreads through the whole beaker. A student says: “The dye molecules sense where the water is clear and head for it.”

Which observation shows that the student is wrong?

A. The color spreads until the whole beaker is evenly purple from top to bottom B. The color spreads upward and sideways from the crystal at the same time C. The color spreads through the beaker faster when the water is stirred D. Under a microscope, some dye molecules move back into the purple region

Question 4

In a fish's gill, water flowing past holds oxygen at 8 mg/L, and the blood inside the gill holds oxygen at 3 mg/L. Oxygen molecules cross the gill's thin lining in both directions.

What is the net movement of oxygen at the gill?

A. None: oxygen crosses in both directions, so the two movements cancel B. Into the blood: more oxygen molecules cross from water to blood than from blood to water C. Into the blood: oxygen molecules cross only from the water into the blood D. Into the water: the fish breathes out its oxygen through the gill

Question 5

A membrane separates two urea solutions, both at 0.40 mol/L. Urea can cross the membrane. The number of urea molecules crossing the membrane from left to right each minute is counted for an hour.

How does that number change over the hour?

A. Molecules cross for a few minutes, then crossing stops B. Crossing slows gradually over the hour, toward zero C. Molecules cross at about the same rate all hour D. No molecule crosses at any time in the hour

Question 6

Molecule V is a small nonpolar molecule at 6 mmol/L outside a skin cell and 1 mmol/L inside, and it enters the cell. V enters protein-free bubbles of phospholipid bilayer just as fast, and blocking the cell’s ATP leaves the rate unchanged.

Which of the following is the type of transport shown?

A. Simple diffusion B. Active transport C. Facilitated diffusion D. Endocytosis

Question 7

Inside a leaf cell, sucrose is at 0.30 mol/L; in the watery spaces of the cell wall outside it, sucrose is at 0.02 mol/L. Sucrose is a large polar molecule, and no sucrose carrier in the cell’s membrane is working.

What keeps the sucrose more concentrated inside the cell than outside?

A. The sucrose molecules inside have stopped moving, so none reach the membrane B. The cell uses ATP holding each sucrose molecule in place C. The cell keeps making sucrose faster than sucrose leaks out through the bilayer D. The membrane blocks sucrose, so diffusion cannot even the two sides out

Question 8

A cell lining the stomach pushes hydrogen ions (H⁺) out into the stomach fluid, where H⁺ is at 150 mmol/L. Inside the cell, H⁺ is at 0.0001 mmol/L.

What must this crossing involve?

A. A channel protein only B. A pump protein and energy from ATP C. No protein at all D. A carrier protein that uses no energy

Question 9

A cell lining a newborn’s gut takes in whole antibody proteins from milk. Under the microscope, the cell’s membrane where a cluster of antibodies touches it is seen to dent inward, and then to close over the cluster.

How do the antibodies get into the cell, and where are they once inside?

A. By simple diffusion through the bilayer; loose in the cytosol B. Through a channel protein; loose in the cytosol C. By exocytosis; enclosed in a vesicle of membrane D. By endocytosis; enclosed in a vesicle of membrane

Question 10

A cell in a mammary gland holds milk proteins inside vesicles. The vesicles move to the plasma membrane and fuse with it, and the proteins appear in the milk outside. When a drug blocks the cell's ATP production, the vesicles stay inside and protein stops appearing in the milk.

What do these observations show?

A. The proteins leave by exocytosis, and fusing a vesicle with the membrane uses energy from the cell B. The proteins cross the bilayer on their own, using ATP to squeeze between the tails C. The proteins leave through a carrier protein that ATP powers D. The vesicles are taken in from the milk, which uses ATP

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 · 5 points

Four substances cross the plasma membrane of a cell lining the small intestine an hour after a meal. Fructose, a sugar, enters through a membrane protein from the gut fluid, at 12 mmol/L, into the cytosol, at 1 mmol/L; blocking the cell’s ATP supply leaves its entry unchanged. Amino acid L enters through a different membrane protein from the gut fluid, at 0.5 mmol/L, into the cytosol, already at 8 mmol/L; blocking ATP stops this uptake. Oxygen enters from the blood, at 0.15 mmol/L, into the cytosol, at 0.05 mmol/L, and crosses protein-free bubbles of phospholipid bilayer just as readily. The cell releases mucus, a very large polysaccharide: small sacs of membrane move to the plasma membrane, and the mucus appears outside just after each sac reaches it.

(a) Identify the direction fructose moves relative to its concentration gradient, and state the two concentrations that tell you. (1 point)

Hint: Compare the concentration where fructose starts with the concentration where it ends up.

A full-credit answer: Fructose moves down its concentration gradient, from 12 mmol/L in the gut fluid to 1 mmol/L in the cytosol: from where it is more concentrated to where it is less.

Check the box for each point your answer earns

fructose moves down its concentration gradient, from 12 mmol/L in the gut fluid to 1 mmol/L in the cytosol.

Accept: "from high to low" with the two values quoted.

Common slip: Reading 'through a protein' as a sign of uphill movement. The direction is set by the two concentrations, and 12 to 1 mmol/L is downhill.

(b) Describe what the two observations about amino acid L show about how L enters. (1 point)

Hint: Taking the two concentrations and the ATP result together, what does each one tell you about how L gets in?

A full-credit answer: L moves from 0.5 mmol/L in the gut fluid to 8 mmol/L in the cytosol, which is against its concentration gradient; and when ATP is blocked the uptake stops, which shows that the cell uses energy from ATP to move L uphill through a pump: active transport.

Check the box for each point your answer earns

L moves against its gradient (0.5 mmol/L outside up to 8 mmol/L inside), and the uptake stops when ATP is blocked, so the cell is using energy to move it: active transport.

Accept: 'uphill' for against the gradient; both observations are needed for the point.

Common slip: Naming active transport from the protein alone. Channels and carriers are proteins too; what marks active transport is movement against the gradient using ATP.

(c) Explain what decides whether a crossing through a membrane protein uses energy from the cell, using fructose and L as your examples. (1 point)

Hint: Think about which way random movement alone would carry each substance, and which way each one is actually going.

A full-credit answer: Fructose moves down its gradient, from 12 mmol/L to 1 mmol/L.
Fructose molecules move constantly and at random, so more leave the crowded gut fluid than return to it.
So the net movement into the cell happens by itself; the protein only gives fructose a route: facilitated diffusion.
L moves against its gradient, from 0.5 mmol/L to 8 mmol/L.
Random movement alone gives a net flow the other way, so the cell must use energy from ATP to push L uphill.

Check the box for each point your answer earns

fructose is moving down its gradient, and particles moving at random give a net movement down a gradient by themselves, so the protein only provides a route (facilitated diffusion, a form of passive transport); L is moving against its gradient, which random movement never does on its own, so the cell must use energy to drive it uphill.

Accept: 'downhill uses no energy; uphill needs energy', with each substance's direction relative to its gradient named; the random-movement mechanism is welcome but not required.

Common slip: Saying fructose's entry uses no energy because fructose is a sugar the cell needs. What makes it use no energy is the direction: down the gradient, which random movement supplies on its own.

(d) Identify the mechanism by which oxygen enters and the mechanism by which the mucus leaves, giving the clue in the description that decides each. (1 point)

Hint: For each one, ask: protein or no protein, and how big is the thing that moves?

A full-credit answer: Oxygen enters by simple diffusion: it moves from 0.15 mmol/L to 0.05 mmol/L, down its gradient, and it crosses protein-free bubbles just as readily, so it needs no protein.
Oxygen is a small nonpolar molecule, and the bilayer itself is its route.
The mucus leaves by exocytosis: mucus is far too large for any channel or carrier, and it appears outside just as each sac of membrane reaches the plasma membrane.
So the sacs fuse with the plasma membrane and open outward.

Check the box for each point your answer earns

oxygen enters by simple diffusion (oxygen moves down its gradient, 0.15 to 0.05 mmol/L, and crosses protein-free bilayer, so it needs no protein); the mucus leaves by exocytosis (mucus is far too large for any channel or carrier, and it appears outside as each membrane sac reaches the plasma membrane, so the sacs fuse with the membrane and release it).

Accept: 'passive transport through the bilayer' for oxygen. Both mechanisms with their clues are needed for the point.

Common slip: Calling the release of mucus active transport because it uses energy. Exocytosis does use energy, but it moves a whole vesicle's worth of material, not one molecule through a pump.

(e) Predict what happens to the entry of fructose if fructose builds up in the cytosol until it is also at 12 mmol/L, and justify your prediction. (1 point)

Hint: Once fructose is at 12 mmol/L on both sides, compare the number of fructose molecules crossing into the cell each second with the number crossing out.

A full-credit answer: When the two concentrations are equal, fructose still crosses through the protein in both directions, but the net movement of fructose into the cell stops; this is because as many fructose molecules leave the cell as enter it.
So the concentration inside stops rising.
That is a dynamic equilibrium: the molecules still move, and only the net movement is zero.

Check the box for each point your answer earns

net movement of fructose into the cell stops; fructose molecules keep crossing through the protein in both directions at equal rates (a dynamic equilibrium), so the concentration inside no longer rises.

Accept: "no net movement; equal traffic both ways". Do not award the point for "fructose stops moving" or "the protein closes".

Common slip: Saying the fructose molecules stop moving, or the carrier shuts. They keep crossing both ways; equal concentrations give equal traffic, so no net change.

Free-response score: 0 of 5

Free response 2 · Scientific Investigation · 4 points

Root cells of a salt-marsh plant keep sodium ions (Na⁺) at 10 mmol/L inside while the soil water around them holds Na⁺ at 200 mmol/L. Researchers add labeled Na⁺ to the soil water and follow it for one hour in three trials, keeping temperature and solution volumes the same. The results are in the table.

Labeled Na⁺ and total Na⁺ inside the root cells after one hour in each trial.

(a) Describe how an ion such as Na⁺ can cross a plasma membrane, and what stops it crossing the bilayer on its own. (1 point)

A full-credit answer: An ion such as Na⁺ crosses the membrane only through a membrane protein: a channel when the ion moves down its gradient, or a pump when the ion is moved against it.
It cannot cross the bilayer on its own because the hydrocarbon tails in the middle of the membrane carry no charges or partial charges.
Water attracts the ion.
Nothing in the interior attracts the ion.
So Na⁺ stays in the water.

Check the box for each point your answer earns

Na⁺ crosses only through membrane proteins (a channel, or a pump for uphill movement); it cannot cross the bilayer on its own because the hydrocarbon tails in the interior carry no charges or partial charges, so water attracts an ion and nothing in the interior does.

Accept: "nothing in the oily interior attracts a charge, so ions need a protein".

Common slip: Saying Na⁺ is too big to cross. Na⁺ is one of the smallest particles present; water attracts its charge, and that attraction is what keeps Na⁺ out of the interior.

(b) Explain what trial 2 shows about how the cells normally keep their Na⁺ at 10 mmol/L. (1 point)

A full-credit answer: In trial 2 the cells could make no ATP.
Their Na⁺ climbed from 10 mmol/L toward 200 mmol/L.
So the low concentration of Na⁺ in trial 1 was being maintained by work.
A pump using ATP moves Na⁺ out against its gradient as fast as Na⁺ enters.
With no ATP the pump stops.
So Na⁺ diffuses in down its gradient until the inside approaches the outside.

Check the box for each point your answer earns

with ATP blocked, Na⁺ inside rose toward the outside level, so the cells normally use energy from ATP to pump Na⁺ out against its gradient (active transport) as fast as it leaks in; without the pumping, Na⁺ diffuses in down its gradient (200 to 10 mmol/L) and the two sides move toward equal.

Accept: "a pump using ATP removes Na⁺ as fast as it enters; stop the pump and diffusion evens the sides out".

Common slip: Saying the drug let Na⁺ in. The drug only stopped the ATP supply; Na⁺ was entering all along, and what failed was the pumping out.

(c) Describe one further trial to test whether the blocked channel of trial 3 is the route by which Na⁺ entered in trial 2, and predict its result. (1 point)

A full-credit answer: Treat cells with both the drug that blocks ATP and the compound that blocks the channel, then add labeled Na⁺ for an hour.
If the channel is the route Na⁺ used in trial 2, almost no labeled Na⁺ will appear inside and the total will stay near 10 mmol/L even though the pump is off.
If Na⁺ still rises toward 150 mmol/L, it is entering by some other route.

Check the box for each point your answer earns

treat cells with both the ATP-blocking drug and the channel-blocking compound; predicted result: almost no labeled Na⁺ enters and the total inside stays near 10 mmol/L (if the channel is the entry route), whereas Na⁺ rising as in trial 2 would show another route.

Accept: any design that combines the two treatments (or blocks the channel in ATP-blocked cells) with a prediction that matches the claim being tested.

Common slip: Proposing to repeat trial 3 with more cells. The question is whether the channel is the entry route when the pump is off, which needs both treatments in one trial.

(d) Evaluate the claim that two different membrane proteins, working together, keep the concentration of Na⁺ inside these cells low, using the results of all three trials. (1 point)

A full-credit answer: The claim is supported.
Trial 3: with the channel blocked, almost no labeled Na⁺ reaches the cytosol, so the channel is the way in.
Trial 2: with ATP blocked, Na⁺ is not removed and the inside climbs to 150 mmol/L, so a pump powered by ATP is the way out.
Trial 1: labeled Na⁺ enters, yet the total stays at 10 mmol/L, so the pump removes Na⁺ as fast as it enters.
So a channel and a pump together keep the concentration of Na⁺ low.

Check the box for each point your answer earns

the judgement that the claim is supported AND the ground for it from the trials: trial 3 shows Na⁺ enters through the channel (block it and labeled Na⁺ stops appearing inside); trial 2 shows a pump using ATP removes it (block ATP and Na⁺ builds up inside); trial 1 shows the two working together, labeled Na⁺ entering yet the total holding at 10 mmol/L because the pump removes Na⁺ as fast as the channel lets it in.

Accept: the three trials tied to entry, removal and the steady state in any order. Do not award the judgement alone, or a judgement grounded on only one trial.

Common slip: Judging the claim from trial 2 alone. The claim has two parts, a channel for entry and a pump for removal, and each needs its own trial as the ground.

Free-response score: 0 of 4

Multiple choice checked: 0 of 10 correct.

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