Practice questions · Topic 2.6

Unit 2 · Practice for the Topic 2.6 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. 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 (thousandths of a mol/L); inside the cells it is 3 mmol/L. The sucrose enters through a carrier protein, and the uptake continues at the same rate when the cells' energy supply is cut off.

How should this crossing be classified?

A. Simple diffusion: sucrose slips between the hydrocarbon tails on its own B. Facilitated diffusion: down its gradient, through a carrier, at no cost to the cell C. A crossing that costs the cell energy, because a carrier protein is doing the work D. A crossing that costs the cell energy, because sucrose is moving against its gradient

Question 2

Guard cells on a leaf hold K⁺ at 300 mmol/L; the fluid around them holds 10 mmol/L. A toxin locks every potassium channel in their membranes shut. The cells spend no energy on K⁺.

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

A. K⁺ leaves through the bilayer, down its steep gradient B. K⁺ enters, because the toxin has reversed the gradient C. Almost none: an ion crosses only through an open channel D. K⁺ leaves through the cells' sugar carriers instead

Question 3

Frog eggs have very few aquaporins in their membranes. Researchers make one batch of eggs insert many aquaporins and leave a second batch as it is. Both batches are placed in a dilute solution, one in which water moves into the eggs. After 30 minutes the eggs with added aquaporins have swelled and burst; the untreated eggs 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

Three substances were tracked at the membrane of a cell whose energy supply had been cut. O₂, more concentrated outside than inside, crossed the bilayer inward. An amino acid, 6 mmol/L outside and 2 mmol/L inside, crossed inward, but only through a membrane protein. Na⁺, 145 mmol/L outside and 15 mmol/L inside, stayed put while every sodium channel was shut.

Which crossing is facilitated diffusion?

A. The amino acid's crossing B. The O₂'s crossing C. The Na⁺'s crossing D. All three crossings

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 polar molecule 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, and the cells spend no energy on lactate.

What is the net movement of lactate at each cell?

A. Out of both cells: the carrier moves lactate in one direction only B. Into both cells: carriers bring substances into 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

Cells lining a sweat duct take Cl⁻ back out of the sweat as it passes: 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. The cells spend no energy on Cl⁻.

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

A. Cl⁻ crosses the bilayer into the cells instead, more slowly B. Cl⁻ stays in the sweat: with no open channel it has no route in C. Cl⁻ enters the cells through their sodium channels D. The cells spend energy to 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. The cell spends no energy on this crossing.

Describe 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 spends 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, and water now moves out of the root cells into the soil water. The aquaporins stay open.

What do the aquaporins do for water in the over-fertilized soil?

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

Four crossings are observed at cell membranes.

Which crossing is 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 to 3 mmol/L, at no cost C. A sugar entering a cell through a protein, from 1 to 6 mmol/L, stopping when the energy supply is cut D. Water crossing a bare 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 glucose is kept rather than lost 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, and glucose leaves the cells into the blood. The cells spend no energy on this step.

(a) Identify the kind of membrane protein glucose uses to leave the cell, 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 leaves through a carrier protein. It needs a protein because it is a large polar molecule that water holds on to; the hydrocarbon tails in the middle of the membrane carry no charges or partial charges, so they have nothing to hold it, and it 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 holds on to, so it has nothing to hold it in the hydrophobic interior, whose 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 to hold 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, at 8 mmol/L, into the blood, at 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, for free: 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. Both are needed.

(c) Explain why the cell spends no energy on this crossing. (1 point)

Hint: Which way is glucose moving relative to its concentration gradient, and what does a crossing in that direction cost a cell?

A full-credit answer: The cell spends nothing because glucose is moving down its concentration gradient, and movement down a gradient happens by itself: glucose molecules move at random, so more of them leave the crowded inside than arrive from the less crowded blood. The carrier only provides the route.

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: 'downhill movement is free; 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: With the two concentrations equal, how does the number of glucose molecules crossing one way compare with the number crossing the other?

A full-credit answer: When the two concentrations are equal there is no net movement of glucose. Molecules still cross the carriers, but as many go one way as the other, 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 into the blood, at about half the rate and in the same direction, because each carrier is one route: halving the routes halves how many molecules cross per second, while the gradient from 8 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

Researchers study how a labeled amino acid enters cultured mouse cells. Dish 1 holds untreated cells. Dish 2 holds cells 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. The concentration inside the cells is measured at 0, 5, 10 and 20 minutes; the table shows the results. The cells spend no energy on this amino acid.

Concentration of the labeled 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, at no cost to the cell. This is facilitated diffusion, a form of passive transport.

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, for free'. 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) Justify the researchers' choice to put the same number of cells in each dish. (1 point)

A full-credit answer: With the same number of cells in each dish, any difference in how much amino acid ends up inside the cells can be put down to the blocked carriers, and not to one dish simply having more cells, more membrane and more carriers taking the amino acid up.

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

A full-credit answer: About 10 mmol/L, the same as the solution around the cells; the rise levels off there.

Check the box for each point your answer earns

about 10 mmol/L, the same as the solution outside (accept 9.5 to 10 mmol/L).

Do not award the point for a value above 10 mmol/L, or for a fall back toward 0 mmol/L.

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

(d) Justify your prediction. (1 point)

A full-credit answer: 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 left: molecules cross in both directions at equal rates, a dynamic equilibrium, so there is no net movement. Because the cells spend no energy on this amino acid, the carriers cannot push the inside any higher.

Check the box for each point your answer earns

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, molecules cross in both directions at equal rates (a dynamic equilibrium) and there is no net movement, and with no energy spent 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 run out of room'.

Common slip: Saying the carriers 'get full' or the cell 'runs out of room'. 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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