Practice questions · Topic 2.6

Unit 2 · Practice for the Topic 2.6 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: doors for ions and sugars, summed up

A blocked solute keeps its gradient; facilitated diffusion; channels open and closed; aquaporins; what a change to a door does.

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

Root cells of a bean plant take in sucrose (table sugar) from the fluid around them, where it is 12 mmol/L; inside the cells it is 3 mmol/L. The sucrose enters through a carrier protein.

Which of the following describes this crossing?

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

Question 2

Root cells of a barley plant hold K⁺ at 100 mmol/L; the soil water around them holds 1 mmol/L. A toxin locks every potassium channel in their membranes shut.

What is the net movement of K⁺ while the channels are locked?

A. K⁺ leaves the cells B. K⁺ enters the cells C. There is no net movement of K⁺ D. Not enough information to say

Question 3

Two frogs lay eggs. The eggs of frog 1 have many aquaporins in their membranes; the eggs of frog 2 have very few. Both sets of eggs sit in a dilute solution, in which water moves into the eggs. After 30 minutes the eggs of frog 1 have swelled and burst; the eggs of frog 2 have gained about 2% in volume.

What does the comparison show about aquaporins?

A. Aquaporins push water into a cell using the cell's energy B. Aquaporins are the only route by which water can cross a membrane C. Aquaporins reverse the direction in which water crosses a membrane D. Aquaporins let large quantities of water cross a membrane quickly

Question 4

A cell’s energy supply has been cut off. At its membrane, O₂, more concentrated outside than inside, crosses the bilayer inward. An amino acid, 6 mmol/L outside and 2 mmol/L inside, crosses inward, but only through a membrane protein. Na⁺, 145 mmol/L outside and 15 mmol/L inside, stays put while every sodium channel is shut.

Which of the following crossings is facilitated diffusion?

A. The amino acid's crossing B. The O₂'s crossing C. The Na⁺'s crossing D. All three: the O₂, the amino acid and the Na⁺

Question 5

Lactate concentrations at a leg muscle cell and a heart muscle cell during a sprint. Each cell has the same kind of lactate carrier in its membrane.

The figure shows lactate, a small molecule with a full charge that muscle makes during hard exercise, at two cells during a sprint. In a leg muscle cell lactate is 15 mmol/L inside and 3 mmol/L in the blood; in a heart muscle cell it is 3 mmol/L in the blood and 1 mmol/L inside. Both cells carry the same kind of lactate carrier.

What is the net movement of lactate at each cell?

A. Out of both cells B. Into both cells C. Out of the leg muscle cell and into the heart muscle cell D. Into the leg muscle cell and out of the heart muscle cell

Question 6

Sweat flows along a sweat duct. The cells lining the duct take Cl⁻ out of the sweat and into themselves as the sweat flows past: Cl⁻ is 60 mmol/L in the sweat and 15 mmol/L inside the cells, and it enters the cells through chloride channels. In one person those chloride channels are faulty and stay shut.

Predict what happens to the Cl⁻ in this person's sweat.

A. Cl⁻ enters the cells through the phospholipid bilayer B. Cl⁻ stays in the sweat C. Cl⁻ enters the cells through their sodium channels D. The cells pull Cl⁻ in through the shut channels

Question 7

During each heartbeat, the sodium channels of a heart muscle cell open for about one millisecond and then shut again. Na⁺ is 140 mmol/L outside the cell and 10 mmol/L inside.

Which of the following describes the movement of Na⁺ over one heartbeat?

A. Na⁺ flows in only during the millisecond the channels are open B. Na⁺ flows in steadily through the beat, channels open or shut C. Na⁺ flows out during the millisecond the channels are open D. Na⁺ flows in only if the cell uses energy while the channels are open

Question 8

Root cells take up water from soil water through their aquaporins. A gardener over-fertilizes. The soil water becomes far more concentrated in solute than the root cells are. So water now moves out of the root cells into the soil water.

What part do the aquaporins play in this water movement?

A. The aquaporins close, so the water stays inside the cells B. The aquaporins push water back in using the cells' energy C. The aquaporins let solute into the cells along with the water D. The aquaporins let water leave faster, following its own gradient

Question 9

Each of the following describes a substance crossing a cell’s plasma membrane.

Which of the following four crossings is an example of facilitated diffusion?

A. CO₂ leaving a cell straight through the bilayer, down its gradient B. An amino acid entering a cell through a carrier, from 9 mmol/L outside to 3 mmol/L inside, with no energy used C. A sugar entering a cell through a protein, from 1 mmol/L outside to 6 mmol/L inside, stopping when the energy supply is cut D. Water crossing a phospholipid bilayer slowly, on its own

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

Cells lining a kidney tubule pass glucose from the tubule fluid back to the blood, so that the body keeps the glucose rather than losing it in urine. On the side of each cell that faces the blood there are glucose carrier proteins. Between meals, glucose is 8 mmol/L inside these cells and 5 mmol/L in the blood.

(a) Identify the kind of membrane protein glucose uses to cross the membrane facing the blood, and explain why glucose needs a protein at all. (1 point)

Hint: What fills the middle of the membrane, and what kind of molecule is glucose?

A full-credit answer: Glucose crosses through a carrier protein.
Glucose needs a protein because glucose is a large polar molecule.
Water is attracted to glucose.
The hydrocarbon tails in the middle of the membrane carry no charges or partial charges.
So nothing in the tails attracts glucose.
So glucose cannot cross the bilayer on its own.

Check the box for each point your answer earns

a carrier protein (accept: a channel or carrier, or a membrane protein that binds glucose), AND glucose is a large polar molecule that water attracts, and nothing in the hydrophobic interior attracts it, because the hydrocarbon tails carry no charges or partial charges, and it cannot cross the bilayer on its own.

Accept: 'the oily middle of the membrane has nothing that attracts a polar molecule'. Both the protein and the reason are needed for the point.

Common slip: Naming the carrier without saying why glucose needs it, or saying glucose is 'too big'. It is the polarity of glucose against the uncharged interior that keeps it out.

(b) Describe the direction of the net movement of glucose, using the two concentrations, and name this kind of transport. (1 point)

Hint: Which side has the higher glucose concentration, and what is the name for a crossing of this kind?

A full-credit answer: Glucose moves from inside the cell (8 mmol/L) to the blood (5 mmol/L), down its concentration gradient; this kind of transport is facilitated diffusion, a form of passive transport.

Check the box for each point your answer earns

net movement from inside the cell (8 mmol/L) into the blood (5 mmol/L), down glucose's concentration gradient, AND the name facilitated diffusion (a form of passive transport).

Accept: 'from high to low, through the carrier, with no energy used: facilitated diffusion'. Both the direction with the concentrations and the name are needed.

Common slip: Giving the name without the direction, or the direction without the name. The name and the direction are both needed.

(c) State whether the cells use energy on this crossing, and explain why. (1 point)

Hint: Think about what makes glucose molecules move down a gradient in the first place, and what the carrier adds to that.

A full-credit answer: The cells use no energy because glucose is moving down its concentration gradient, and movement down a gradient happens by itself.
Glucose molecules move at random.
More of them leave the crowded inside than arrive from the less crowded blood.
The carrier only provides the route.
So the cells use no energy.

Check the box for each point your answer earns

glucose is moving down its concentration gradient, and movement down a gradient happens by itself, because molecules move at random so more of them leave the more concentrated side than arrive from the less concentrated one; the carrier only provides the route.

Accept: 'movement down a gradient needs no energy; the carrier is just the door'. Do not award the point for 'because it is passive' with no reason.

Common slip: Saying 'because it is passive transport' and stopping. The point wants the reason downhill movement needs no energy.

(d) After a meal, blood glucose rises to 8 mmol/L, equal to the concentration inside the cells. Predict the net movement of glucose through the carriers. (1 point)

Hint: Think about what a carrier does to a glucose molecule that binds to it, from either face of the membrane. Then ask what is different about the two faces once the two concentrations are equal.

A full-credit answer: When the two concentrations are equal, the net movement of glucose is zero, because glucose molecules cross the carriers in both directions in equal numbers.
Molecules still cross the carriers.
As many go one way as the other.
That is a dynamic equilibrium.

Check the box for each point your answer earns

no net movement, because glucose molecules cross the carriers in both directions at equal rates when the concentrations are equal (a dynamic equilibrium).

Accept: 'glucose still crosses but the two flows cancel'. Do not award the point for 'glucose stops moving'.

Common slip: Saying glucose stops moving. It keeps crossing both ways; only the net movement is zero.

(e) A drug halves the number of glucose carriers in these cells. Predict the effect on the movement of glucose between meals, and justify your prediction. (1 point)

Hint: What does the number of carriers control, and what sets the direction of net movement?

A full-credit answer: With half the carriers, glucose still leaves the cells at about half the rate, in the same direction, because each carrier is one route, and halving the routes halves how many molecules cross per second, while the gradient from 8 mmol/L to 5 mmol/L is unchanged and still sets the direction.

Check the box for each point your answer earns

glucose still leaves the cells into the blood, in the same direction, but at about half the rate, because each carrier is one route and halving the routes halves how many molecules cross per second, while the gradient (8 to 5 mmol/L) is unchanged and still sets the direction.

Accept: 'slower, same direction, because fewer doors'. Do not award the point for a change of direction or for 'the cell now needs energy'.

Common slip: Changing the direction, or saying the cell now needs energy. Fewer doors slow a crossing; only the gradient sets its direction.

Free-response score: 0 of 5

Free response 2 · Scientific Investigation · 4 points

A scientist studies how an amino acid enters mouse cells. She sets up two dishes. Dish 1 holds untreated cells. Dish 2 holds cells she has treated with a substance that blocks their amino acid carriers. Each dish holds 50,000 cells in a solution with the amino acid at 10 mmol/L, and inside the cells it starts at 0 mmol/L. She measures the concentration inside the cells at 0, 5, 10 and 20 minutes; the table shows the results.

Concentration of the amino acid inside the cells over 20 minutes. The solution around the cells holds 10 mmol/L throughout.

(a) Describe how the amino acid enters the cells in dish 1, and name the kind of transport. (1 point)

A full-credit answer: In dish 1 the amino acid enters through carrier proteins, moving down its concentration gradient from 10 mmol/L outside toward 0 mmol/L inside, using no energy from the cell; this is facilitated diffusion, a form of passive transport.
Movement down a gradient happens by itself, so the cell uses no energy on it.

Check the box for each point your answer earns

the amino acid enters through carrier proteins, moving down its concentration gradient from 10 mmol/L outside toward 0 mmol/L inside, with no energy from the cell: facilitated diffusion (a form of passive transport).

Accept: 'through the carriers, from high to low, with no energy used'. The route and the name are both needed.

Common slip: Saying the amino acid 'diffuses in' with no mention of the carriers. Dish 2 shows that with the carriers blocked almost none gets in.

(b) Explain why the investigation uses the same number of cells in each dish. (1 point)

A full-credit answer: More cells would give more membrane and more carriers.
So more cells take up more amino acid.
With the same number of cells in each dish, that cause is ruled out.
So any difference between the dishes can be put down to the blocked carriers.

Check the box for each point your answer earns

with equal numbers of cells, a difference in the amino acid taken up between the dishes can be put down to the blocked carriers rather than to one dish having more cells, and so more membrane and more carriers, taking up more.

Do not award the point for 'it was a control' or 'to make it fair' with no statement of what an unequal count would have muddled.

Common slip: Writing 'it was a control' or 'to keep it fair' and stopping. Say what an unequal count would have muddled: more cells take up more, whatever their carriers are doing.

(c) Determine the concentration of the amino acid inside the dish-1 cells at 60 minutes. (1 point)

A full-credit answer: The concentration inside is about 10 mmol/L, the same as the solution around the cells.
The amino acid moves down its gradient through the carriers.
Once the inside reaches 10 mmol/L there is no gradient left.
So the rise levels off there.

Check the box for each point your answer earns

the decision that the concentration is about 10 mmol/L, the same as the solution outside (accept 9.5 to 10 mmol/L), AND the reasoning it rests on: the rise levels off once the inside matches the outside, because there is then no gradient.

Do not award the point for a value above 10 mmol/L, for a fall back toward 0 mmol/L, or for the value alone with no reasoning.

Common slip: Continuing the rise past 10 mmol/L because it rose steadily before. The carriers cannot take the inside above the outside.

(d) Support the claim you made in (c), using what a carrier does for the amino acid. (1 point)

A full-credit answer: The carriers only let the amino acid run down its gradient.
Once the inside reaches 10 mmol/L, the outside’s value, there is no gradient left.
So molecules cross in both directions at equal rates.
That is a dynamic equilibrium, so there is no net movement.
The cells use no energy on this amino acid.
Therefore the carriers cannot push the inside concentration any higher.

Check the box for each point your answer earns

the evidence AND the reasoning: the carriers only let the amino acid run down its gradient; once the inside reaches the outside's 10 mmol/L there is no gradient, so molecules cross in both directions at equal rates (a dynamic equilibrium) and there is no net movement; the cells use no energy on this crossing, so the carriers cannot push the inside any higher.

Accept: 'at 10 mmol/L the gradient is gone, so the net movement is zero; carriers give a route, not a push'. Do not award the point for 'the carriers get full' or 'the cells have no room left'.

Common slip: Saying the carriers 'get full' or the cell 'has no room left'. The rise stops because the gradient is gone, not because anything is full.

Free-response score: 0 of 4

Multiple choice checked: 0 of 9 correct.

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