Practice questions · Topic 2.7a

Unit 2 · Practice for the Topic 2.7a 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: osmosis and tonicity, summed up

From the sugar bag in pure water to the three beakers: osmosis, hypotonic, hypertonic and isotonic, and what pure water does to a red cell.

Watch next: keeping water in balance, summed up

The Paramecium’s contractile vacuole, the ATP it uses, and how a river fish and a marine fish keep their water and salt in balance.

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

A hen's egg has its shell dissolved away, leaving the thin membrane beneath, which lets water through but not sugar. A student measures the egg's mass and leaves the egg overnight in corn syrup, a concentrated sugar solution. In the morning the egg has less mass and is shrunken.

What moved, and which way?

A. Water left the egg for the syrup, toward the side with more solute B. Sugar entered the egg from the syrup, toward the side with less sugar C. Water entered the egg from the syrup, toward the side with more water D. Sugar left the egg for the syrup, and water followed it out

Question 2

A student puts sea urchin eggs, each 270 pL (picoliters) in volume, into three solutions of a solute that cannot cross their membranes. After ten minutes: in 0.60 mol/L the eggs were 350 pL; in 1.00 mol/L, 270 pL; in 1.40 mol/L, 210 pL.

Which term describes the 0.60 mol/L solution relative to the eggs, and what did water do?

A. Hypotonic: water entered the eggs B. Hypertonic: water left the eggs C. Isotonic: water crossed both ways at equal rates D. Hypertonic: water entered the eggs

Question 3

A 0.30 mol/L solution of a solute that cannot cross cell membranes is isotonic to human red blood cells. Cells from a crab that lives in seawater, a marine crab, hold about 1.0 mol/L of solute.

Which term describes the same 0.30 mol/L solution relative to the crab cells, and why?

A. Isotonic: a solution's tonicity is fixed by its own concentration B. Hypertonic: 0.30 mol/L is a strong solution for any cell C. Hypotonic: the crab cells hold more solute than the solution D. Isotonic: both kinds of cell are animal cells

Question 4

An animal cell holds 0.30 mol/L of solute that cannot cross its membrane. It sits in 0.30 mol/L glycerol, a small molecule that crosses the membrane freely.

Predict what happens to the cell's volume over the next half hour, and why.

A. The volume stays constant: the total solute concentration is 0.30 mol/L on both sides B. The volume grows: only the cell's own solute now drives osmosis, so water enters C. The volume shrinks: the glycerol inside adds to the cell's solute and draws water out D. The volume stays constant: a solute that crosses the membrane carries water along with it

Question 5

Red blood cells from a fish that lives in seawater, a marine fish, hold about 0.35 mol/L of solute that cannot cross their membranes. A few drops of the fish's blood fall into a freshwater stream, whose water holds almost no solute.

Predict what happens to the red blood cells.

A. The cells shrink as water leaves them for the stream B. The cells hold their volume: the stream is isotonic to them C. The cells swell until their membranes tear: osmotic lysis D. The cells swell a little and then hold, once the stream water matches their inside

Question 6

An Amoeba, a single-celled organism with no cell wall, lives in pond water that holds far less solute than its cytosol. In pond water its contractile vacuole fills and empties every 30 seconds. Moved into water holding 0.08 mol/L of a solute its membrane blocks, it empties its vacuole every 4 minutes.

Explain the change in the emptying rate.

A. The vacuole now keeps water out, so it rarely needs to empty B. Less water enters in the new solution, so the vacuole fills more slowly C. Solute enters the Amoeba, and the vacuole must remove the solute instead D. More water enters in the new solution, and the vacuole stores it

Question 7

Contractile vacuole emptying rate and cell width after the chemical is added.

A single-celled pond organism, a Paramecium, sits in pond water, which holds far less solute than its cytosol. At 0 minutes a student adds a chemical that stops the cell making ATP, and records how often the contractile vacuole empties and how wide the cell is. The results are in the table.

Which statement explains both trends, the falling emptying rate and the growing width?

A. The vacuole was pumping water into the cell; without ATP it stopped, so water now leaks in through the membrane instead B. Osmosis stopped when the cell used up its ATP, so the vacuole had no water to empty, and pond solute swelled the cell C. Without ATP the membrane lets the pond’s solute in, and the entering solute makes the cell swell and drowns the vacuole D. Water still enters by osmosis, but without ATP the vacuole can no longer empty it outside, so the water stays in the cell

Question 8

A seabird drinks seawater, which holds far more solute than its blood. A gland above each eye pumps salt out of the blood into a very salty fluid that drips from the bird's beak, and the bird's blood stays near 0.30 mol/L solute all day.

Which term names what the bird is doing?

A. Osmotic lysis B. Facilitated diffusion C. Plasmolysis D. Osmoregulation

Question 9

A frog's skin lets water through, and pond water holds far less solute than the frog's blood. The frog makes large amounts of very dilute urine, and cells in its skin pump salt in from the pond by active transport.

Which problem is each of these two activities solving?

A. The urine removes water that enters through the skin; the pumps replace salt that leaks out B. The urine removes salt that enters through the skin; the pumps bring in water C. The urine removes water the frog drank; the pumps remove salt that entered with it D. Both activities keep water out of the frog

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 · Analyze Model or Visual Representation · 5 points

Cells scraped from the inside of a person's cheek hold about 0.30 mol/L of solute that cannot cross their membranes. A student puts some of the cells into each of three dishes of a solute that cannot cross the membranes: dish A at 0.10 mol/L, dish B at 0.30 mol/L and dish C at 0.50 mol/L. The drawing shows one cell from each dish after ten minutes, numbered 1 to 3 in no particular order.

One cheek cell from each dish after ten minutes, numbered 1 to 3 in no particular order.

(a) Identify the dish whose solution is hypotonic to the cells, and identify which drawn cell came from that dish. (1 point)

Hint: Hypo- means less. Compare each dish's solute with the 0.30 mol/L inside the cells. Then ask what a cell does in a solution like that, and look for the drawn cell that has done it.

A full-credit answer: Dish A is hypotonic to the cells.
Dish A holds 0.10 mol/L of solute.
The cells hold 0.30 mol/L.
So dish A holds less solute than the cells, and so dish A is hypotonic to the cells.
Water moves toward the side with more solute.
So water entered the cells in dish A, and those cells swelled.
Cell 3 is the swollen cell.
So cell 3 came from dish A.

Check the box for each point your answer earns

dish A (0.10 mol/L), because it holds less solute than the cells, AND cell 3 (the swollen cell) as the cell from dish A.

Do not award the point for dish C (that is the hypertonic dish), for dish A with the reason reversed (more solute), or for dish A matched to cell 1 or cell 2.

Common slip: Picking the dish where the cells shrank, or matching dish A to the shrunken cell. Hypo- means less solute than the cell, and in less solute the cells swell.

(b) Describe what water is doing at the membranes of the cells in dish B. (1 point)

Hint: The cells in dish B kept their volume. Ask whether water molecules have stopped hitting the membrane, and what a steady volume tells you about any water that does cross.

A full-credit answer: In dish B, water is crossing the membranes constantly in both directions.
The 0.30 mol/L solution holds the same solute as the cells, so the solution is isotonic to them.
The two rates of crossing are equal.
So there is no net movement of water.
Therefore the cells keep their volume.

Check the box for each point your answer earns

water is crossing the membranes in both directions at equal rates, so there is no net movement (the solution is isotonic to the cells).

Do not award the point for "water has stopped crossing" or for "nothing is happening".

Common slip: Saying the water has stopped. Water molecules keep crossing; only the net movement is zero.

(c) Identify which drawn cell came from dish C, and explain why its shape changed. (1 point)

Hint: Compare the solute inside the cells with the solute in dish C. Which way does water move across a membrane that lets water through but holds the solute back? Then look for the drawn cell that shows the result.

A full-credit answer: Cell 2 came from dish C.
Cell 2 is the shrunken, crinkled cell.
The cells hold 0.30 mol/L of solute.
Dish C holds 0.50 mol/L.
Water moves toward the side with more solute.
So water moved from the cells into the solution.
The solute cannot cross the membrane, so only water moved.
Therefore the cells lost volume, and their outlines crinkled.

Check the box for each point your answer earns

cell 2, because dish C holds more solute (0.50 mol/L) than the cells (0.30 mol/L), so water moved out of the cells by osmosis, from the side with less solute toward the side with more, and the cells lost volume; the solute itself cannot cross.

Accept "the solution is hypertonic, so water left the cells" for the explanation. Do not award the point for cell 3 or cell 1, for "solute entered the cells", or for water leaving with no comparison of the two sides.

Common slip: Having the solute move into the cells, or matching dish C to the swollen cell. The membrane holds the solute back; it is water that moves, toward the side with more solute.

(d) Predict what would happen to the cells in a fourth dish of pure water, and explain your prediction. (1 point)

Hint: Pure water holds no solute at all. Compare that with the 0.30 mol/L inside the cells, and ask whether the two sides could ever become equal.

A full-credit answer: In pure water the cells would swell until their membranes tear and the cells burst.
This bursting is osmotic lysis.
Pure water holds no solute.
So pure water can never match the 0.30 mol/L inside the cells.
Water moves toward the side with more solute.
So water keeps moving into the cells by osmosis.
A cheek cell has no cell wall to stop the swelling.
Therefore the membrane stretches until it tears.

Check the box for each point your answer earns

the cells would swell until their membranes tear and they burst (osmotic lysis), because pure water has less solute than the cells, and no solute at all, so water keeps entering by osmosis and the two sides can never match; an animal cell has no cell wall to hold the swelling.

Accept "burst" with the reason that water keeps entering. Do not award the point for "swell and then stop" or for the cells shrinking.

Common slip: Having the cells swell and then hold. With no solute outside the two sides never match, so water keeps entering until the membrane tears.

(e) A fifth dish, D, holds 0.50 mol/L of a small solute that crosses the membranes freely. Make a claim about how the cells in dish D compare with the cells in dish C after an hour, and support your claim. (1 point)

Hint: Dish D's solute can cross the membrane; dish C's cannot. Ask where dish D's solute is after an hour, and what solute is then left unmatched on one side of the membrane.

A full-credit answer: After an hour the cells in dish D are larger than the cells in dish C, swollen past their starting size.
Only solutes that cannot cross the membrane set the tonicity of a solution.
The small solute in dish D crosses into the cells until it is equal on both sides.
So it no longer drives osmosis.
The cells’ own 0.30 mol/L of solute is then unmatched outside, so water moves in and the cells swell.
In dish C the solute stays outside, so those cells stay shrunken.

Check the box for each point your answer earns

the claim that the cells in D are larger (swollen past their starting size, or burst) while the cells in C stay shrunken, supported by evidence AND reasoning: the small solute crosses into the cells until it is equal on both sides, so it no longer drives osmosis; only the cells' own 0.30 mol/L of solute that cannot cross is left counting, with none outside to match it, so water moves in.

Accept "D's cells shrink at first, then swell" with the reason. Do not award the claim alone, or "the same as C, because both dishes hold 0.50 mol/L".

Common slip: Treating dish D like dish C because the totals match. Equal total solute is not isotonic; a solute that crosses stops counting once it is equal on both sides.

Free-response score: 0 of 5

Free response 2 · Conceptual Analysis · 4 points

Brine shrimp live in salt lakes whose water holds about 4.0 mol/L of solute, far more than seawater. A brine shrimp's body fluid stays near 0.40 mol/L. Its body surface is thin and lets water through, and it has gills on its legs whose cells can pump salt.

(a) Describe the direction of net water movement across the brine shrimp's body surface in the lake, and explain why water moves that way. (1 point)

A full-credit answer: Water moves out of the shrimp across its body surface into the lake.
The lake water holds about 4.0 mol/L of solute.
The shrimp’s body fluid holds 0.40 mol/L.
So the lake water holds far more solute than the body fluid: the lake is hypertonic to the shrimp.
Water moves by osmosis toward the side with more solute.
Therefore water leaves the shrimp.

Check the box for each point your answer earns

water moves out of the shrimp into the lake, because the lake water holds far more solute than the body fluid (it is hypertonic to the shrimp) and water moves by osmosis toward the side with more solute.

Accept "the shrimp loses water" with the solute comparison. Do not award the point for water entering, or for the direction with no comparison of the two sides.

Common slip: Sending water into the shrimp because it lives in water. The lake is far saltier than the shrimp, so water leaves it.

(b) Describe two things the brine shrimp must do to keep its body fluid near 0.40 mol/L. (1 point)

A full-credit answer: The shrimp drinks the lake water.
Drinking replaces the water that leaves across its body surface.
The lake water the shrimp drinks brings in salt.
So the shrimp’s gill cells pump that salt back out into the lake by active transport.
The shrimp also makes very little urine, so it keeps the water it has.

Check the box for each point your answer earns

two of: drink the lake water to replace the water it loses; pump the salt that comes in (with the drink and from the lake) out through its gills by active transport; make very little urine (keep the water it has).

Accept "pump salt out" without naming active transport. Do not award the point for one action only, or for making large amounts of dilute urine.

Common slip: Describing a freshwater animal: dilute urine and salt pumped in would lose water and gain salt, the opposite of what a brine shrimp needs.

(c) A single-celled organism with no cell wall is moved from a freshwater pond into the salt lake. Predict what happens to its volume and to how often its contractile vacuole empties, and justify your prediction. (1 point)

A full-credit answer: In the pond, the cell’s cytosol held more solute than the water around it.
So water entered the cell constantly, and the contractile vacuole filled and emptied many times a minute.
In the lake, the outside holds far more solute than the cytosol.
So water leaves the cell by osmosis, and the cell shrinks.
The contractile vacuole only ever removes water that has entered.
Now no water enters.
So the vacuole stops emptying.

Check the box for each point your answer earns

the cell shrinks (loses water), because the lake holds far more solute than its cytosol and water moves out by osmosis, AND its contractile vacuole stops emptying (or empties very rarely), because the vacuole only removes water that has entered the cell and now little or none does.

Accept "the vacuole has nothing to remove". Do not award the point for the vacuole emptying more often, or for the cell swelling.

Common slip: Having the vacuole speed up to fight the salt. It removes incoming water; when water leaves instead, it has nothing to do.

(d) Evaluate the claim that keeping its body fluid at 0.40 mol/L uses energy for as long as the brine shrimp lives in the lake. (1 point)

A full-credit answer: The claim is correct.
Salt enters the shrimp all the time: with every drink of lake water, and from the lake down its gradient.
The only way to get that salt out is to pump it against its gradient by active transport.
Active transport uses ATP.
The water loss and the salt gain never stop, so the pumping never stops either.
Therefore holding the body fluid at 0.40 mol/L uses energy for as long as the shrimp lives in the lake.

Check the box for each point your answer earns

the judgement that the claim is correct AND the ground for it: salt enters the shrimp constantly (with every drink and from the lake, down its gradient) and must be pumped out against its gradient by active transport, which uses ATP; the leaks never stop, so the pumping never stops.

Accept "salt keeps coming in, so it keeps having to be pumped out, and pumping uses ATP". Do not award the judgement alone, or "water is pumped out" (cells pump solutes, not water), or a one-off use of energy.

Common slip: Having the shrimp pump water out. Cells pump solutes; water follows by osmosis. The energy goes on moving salt against its gradient.

Free-response score: 0 of 4

Multiple choice checked: 0 of 9 correct.

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