Practice questions · Topic 2.5

Unit 2 · Practice for the Topic 2.5 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.

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 tells you what a wrong choice assumed. 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 your answer in full sentences as you would in the test, then open the scoring guide and mark your own work against it.

Question 1

A lab prepares two solutions, each 0.20 mol/L and each one liter: one of glycine, an amino acid whose molecules weigh 75 g per mole, and one of tryptophan, an amino acid whose molecules weigh 204 g per mole.

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, since more grams went in B. The glycine liter holds about 2.7 times as many molecules, since its molecules are lighter C. The two liters hold the same number of molecules, about 1.2 × 10²³ each D. The glycine liter holds 75 molecules per liter and the tryptophan liter 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 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 drop of perfume evaporates on a desk in a still room with no drafts. A minute later a student at the far side of the room smells it.

Why does the scent reach the far side of the room?

A. Each perfume molecule is pushed along by the molecules evaporating behind it B. The perfume molecules head for the regions of the room that hold none C. Warm air from the desk carries the scent across the room in a stream D. Random motion: more molecules leave the crowded region than return to it

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 thin gill wall 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 fish tank is divided by a fine mesh that dissolved salt ions can cross. Both halves have held salt at 0.5 g/L for a day. A student says: "The salt ions have stopped crossing the mesh, because the two sides are equal."

Which statement corrects the student?

A. The salt ions have stopped moving altogether, and that is why the two sides stay equal B. The salt ions still cross, but only from the half that was filled first C. The salt ions keep crossing in both directions at equal rates, so there is no net movement D. The salt ions now move together in one direction, circling through the mesh and back

Question 6

Molecule V is a small nonpolar molecule. It is at 6 units per liter outside a skin cell and 1 unit per liter inside, and it enters the cell straight through the bilayer while the cell's supply of ATP is blocked.

How should this crossing be classified?

A. Passive transport by simple diffusion: down its gradient, at no cost B. Active transport: any substance entering a cell costs the cell energy C. Facilitated diffusion: a molecule this small needs a carrier to get in D. Endocytosis: the membrane wraps around each molecule of V

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 the cell's sucrose carriers are all shut.

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 spends ATP holding each sucrose molecule in place C. Sucrose is heavier than water, so it settles inside the cell 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 (thousandths of a mol/L). Inside the cell, H⁺ is at 0.0001 mmol/L.

What must this crossing involve?

A. A channel protein only, since H⁺ is an ion B. A pump protein and energy from ATP, since H⁺ moves against its gradient C. No protein, since H⁺ is the smallest ion there is D. A carrier working at no cost, since H⁺ moves down its gradient

Question 9

A cell lining a newborn's gut takes in whole antibody proteins from milk. Each antibody is far too large for any channel or carrier. Under the microscope the cell's membrane is seen folding around a cluster of antibodies and closing behind them.

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 costs the cell energy 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 costs 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 crossings of the plasma membrane of a cell lining the small intestine are described, an hour after a meal. Crossing 1: fructose, a sugar, enters from the gut fluid, where it is at 12 mmol/L (thousandths of a mol/L), into the cytosol, where it is at 1 mmol/L, through a membrane protein; when the cell's ATP supply is blocked, fructose keeps entering at the same rate. Crossing 2: amino acid L enters from the gut fluid at 0.5 mmol/L into the cytosol, where it is already at 8 mmol/L, through a different membrane protein; when the cell's ATP supply is blocked, this uptake stops. Crossing 3: oxygen enters from the blood, where it is more concentrated than in the cytosol, straight through the bilayer with no protein. Crossing 4: the cell releases mucus, a very large polysaccharide, from membrane sacs that fuse with the plasma membrane and open outward.

(a) Identify the direction of crossing 1 relative to fructose's 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 crossing 2 show about how amino acid 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 the uptake stops when the cell's ATP is blocked. Together these show that the cell spends 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 spending 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 paid for with ATP.

(c) Explain why crossing 1 costs the cell nothing while crossing 2 must cost it energy. (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 is moving down its gradient. Particles move constantly and at random, so more leave the crowded gut fluid than return, and the net movement into the cell happens by itself; the protein only gives fructose a route through the membrane it cannot cross alone. This is facilitated diffusion, a form of passive transport, and it costs nothing. L is moving against its gradient, from 0.5 to 8 mmol/L. Random movement alone gives a net flow the other way, so the cell must spend 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 spend energy to drive it uphill.

Accept: "downhill is free; uphill has to be paid for" with the mechanism (random movement gives a net flow downhill only) stated.

Common slip: Saying crossing 1 is free because fructose is a sugar the cell needs. What makes it free is the direction: down the gradient, which random movement supplies on its own.

(d) Identify the mechanism at crossing 3 and at crossing 4, 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: Crossing 3 is simple diffusion: oxygen moves down its gradient straight through the bilayer, with no protein, which is the route open to a small nonpolar molecule. Crossing 4 is exocytosis: mucus is far too large for any protein door, so it leaves inside a vesicle that fuses with the plasma membrane and opens outward.

Check the box for each point your answer earns

crossing 3 is simple diffusion (oxygen, a small nonpolar molecule, passes straight through the bilayer with no protein, down its gradient); crossing 4 is exocytosis (a very large molecule leaves inside a membrane sac, a vesicle, that fuses with the plasma membrane).

Accept: 'passive transport through the bilayer' for crossing 3. Both mechanisms with their clues are needed for the point.

Common slip: Calling crossing 4 active transport because it costs energy. Exocytosis does cost energy, but it moves a whole vesicle's worth of material, not one molecule through a pump.

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

Hint: When a molecule is at the same concentration on both sides of the membrane, what happens to the net traffic through its protein?

A full-credit answer: When fructose is at 12 mmol/L on both sides, the net movement into the cell stops. Fructose molecules keep crossing through the protein, but as many leave the cell as enter it, so the concentration inside stops rising: a dynamic equilibrium. The molecules still move; 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 (thousandths of a mol/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, with temperature and solution volumes the same in every trial. Trial 1, untreated cells: labeled Na⁺ appears inside the cells, but the total Na⁺ inside stays at 10 mmol/L. Trial 2, cells treated with a drug that blocks ATP production: labeled Na⁺ appears inside, and the total Na⁺ inside rises to 150 mmol/L by the end of the hour. Trial 3, cells treated with a compound that blocks one channel protein, with ATP production normal: almost no labeled Na⁺ appears inside, and the total inside stays at 10 mmol/L.

(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 it moves down its gradient, or a pump when it 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, so water holds the ion and nothing in the interior can take it.

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 an ion, which water holds on to, has nothing to hold it there.

Accept: "the oily interior has nothing to hold a charge, so ions need a protein".

Common slip: Saying Na⁺ is too big to cross. It is one of the smallest particles present; its charge, held by water, is what keeps it 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, and their Na⁺ climbed from 10 toward 200 mmol/L. So the low Na⁺ in trial 1 was being maintained by work: a pump spending ATP moves Na⁺ out against its gradient as fast as it enters. With no ATP the pump stops, and 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) Propose 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) Justify the claim that these cells keep their Na⁺ low by combining a channel and a pump, using the results of all three trials. (1 point)

A full-credit answer: Trial 3 shows the channel is the way in: with it blocked, almost no labeled Na⁺ reaches the cytosol. Trial 2 shows the pump is the way out and that it runs on ATP: with ATP blocked, the Na⁺ that leaks in through the channel is no longer removed, and the inside climbs to 150 mmol/L. Trial 1 shows the two working together: labeled Na⁺ enters through the channel, yet the total stays at 10 mmol/L because the pump, spending ATP, sends Na⁺ back out against its gradient as fast as it comes in.

Check the box for each point your answer earns

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 point for a justification that uses only one trial.

Common slip: Justifying 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 evidence.

Free-response score: 0 of 4

Multiple choice checked: 0 of 10 correct.

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