Unit 2 · Topic 2.7a end-of-topic test
Suggested time: about 48 minutes. Answer everything, then press Submit the test to see the feedback and scoring guides.
A student fills a bag made of a thin membrane that lets water through but not sucrose with 0.25 mol/L sucrose (table sugar) solution, ties it, measures its mass and lowers it into a beaker of 0.05 mol/L sucrose solution. After an hour the bag has more mass. The student explains: 'Sucrose spread out of the bag into the beaker until it was even on both sides, and that is why the bag has more mass.'
What is wrong with the student's explanation?
A tank is divided into two chambers by a membrane that lets water through but not glucose. Chamber A holds 0.05 mol/L glucose and chamber B holds 0.25 mol/L glucose, filled to the same level, as in the drawing.
What is the net movement across the membrane over the next hour?
A cook slices strawberries, sprinkles sugar over them and leaves the bowl for an hour. The slices are now sitting in a pool of sweet liquid.
Where did the liquid come from?
A student puts cells from a dog's kidney, each 300 μm³ in volume, into three sucrose solutions. Sucrose cannot cross their membranes. After ten minutes their volumes were: 0.20 mol/L sucrose, 400 μm³; 0.30 mol/L sucrose, 300 μm³; 0.50 mol/L sucrose, 215 μm³.
Which term describes the 0.50 mol/L solution relative to the cells, and what did water do?
A saline solution for rinsing contact lenses is isotonic to the cells on the surface of the eye. After an hour in the saline, the volume of those cells is unchanged.
What is water doing at the membranes of those cells during that hour?
A student puts cells from a mouse's liver into three solutions of a solute that cannot cross their membranes. The table gives each solution's concentration and the cells' volume at the start and after ten minutes.
Which solution is isotonic to the cells, and what is water doing at their membranes in it?
A 0.15 mol/L solution of a solute that cannot cross cell membranes is hypotonic to mouse cells, which hold about 0.30 mol/L of solute. Cells from a freshwater mussel hold about 0.05 mol/L of solute.
Which term describes the same 0.15 mol/L solution relative to the mussel cells, and why?
A student puts sheep red blood cells, each 30 μm³ in volume, into 0.40 mol/L urea and, separately, into 0.40 mol/L sucrose. Urea is a small molecule that crosses the cell membrane; sucrose cannot. The cells’ volumes are in the table.
Why did the cells in urea shrink and then swell?
Animal cells hold 0.30 mol/L of solute that cannot cross their membranes. Solution X holds 0.30 mol/L sucrose, which cannot cross. Solution Y holds 0.15 mol/L sucrose together with 0.15 mol/L urea, a small molecule that crosses the membrane freely. Both solutions total 0.30 mol/L of solute.
Predict the cells' volume after an hour in each solution.
Animal cells hold 0.25 mol/L of solute that cannot cross their membranes. Some of these cells are to be kept at their starting volume for a two-hour experiment.
Which solution keeps the cells at their starting volume for the whole two hours?
A cell from the lining of a fish's gut holds 0.35 mol/L of solute that cannot cross its membrane. A student puts it into a solution of 0.15 mol/L of the same kind of solute. After a few minutes the student replaces that solution with one of 0.50 mol/L. The cell stays intact throughout.
How does the cell's volume change, in order?
A few drops of blood fall into a tube of distilled water. Within a minute the mixture, which was cloudy red, turns a clear red that light passes straight through.
What happened to the red blood cells?
Animal cells hold 0.30 mol/L of solute that cannot cross their membranes. A student puts two samples of these cells into solutions of the same solute: one sample into 0.20 mol/L, the other into 0.05 mol/L.
Compare the two samples after five minutes.
A crayfish lives in a river. The river water holds far less solute than the crayfish’s blood. All day, cells in the crayfish’s gills pump salt in from the river by active transport.
Which problem does the salt pumping solve?
A single-celled freshwater organism with no cell wall is moved from pond water into a solution whose solute concentration matches that of its cytosol.
Predict what its contractile vacuole does in the new solution, and why.
A single-celled organism from a freshwater stream has no cell wall, and the solute used in this experiment cannot cross its membrane. The graph gives how often its contractile vacuole empties in four solutions of that solute.
The organism is moved from the 0.12 mol/L solution back into the 0.00 mol/L solution. Predict the change in its vacuole, and why.
A sea turtle lives in seawater that has far more solute than its blood. It drinks seawater, and a gland beside each eye uses active transport to push salt out of its blood into very salty tears.
What problem is the turtle solving by drinking and pushing out salt?
A salmon hatches in a river and later swims out to sea. In the river its blood holds far more solute than the water around it; in the sea it holds far less. In the river the salmon makes large amounts of dilute urine and pumps salt in through its gills.
Which set of changes must happen when the salmon reaches the sea?
After drinking a liter of water, a healthy person's blood becomes slightly more dilute. Over the next two hours the kidneys make a large volume of very dilute urine, and the blood's solute concentration returns to where it was.
What is the body doing over those two hours?
(a) Identify the tonicity of the 0.40 mol/L solution relative to the cells, and predict what the cells in that dish looked like after ten minutes. (1 point)
A full-credit answer: The 0.40 mol/L solution holds more solute than the 0.22 mol/L inside the cells.
So the 0.40 mol/L solution is hypertonic to the cells.
Water moves toward the side with more solute.
So water left the cells.
After ten minutes the cells were smaller than the cell drawn from the 0.22 mol/L dish, and their outlines were crinkled.
Check the box for each point your answer earns
Accept "more concentrated than the cells" for hypertonic if the word is not used, and "shrivelled" or "crenated" for the shape. Do not award the point for hypotonic, or for shrinking with the tonicity unnamed.
Common slip: Swapping the prefixes. Hyper- means more: a solution with more solute than the cell is hypertonic, and the cell in it shrinks.
(b) Explain what water is doing at the membranes of the cells in the 0.22 mol/L dish, where the cells kept their volume. (1 point)
A full-credit answer: The 0.22 mol/L solution holds the same solute as the cells.
So the solution is isotonic to the cells.
Water molecules keep crossing the membrane in both directions.
The two rates are equal.
So there is no net movement of water.
Therefore the cells’ volume stays constant.
Check the box for each point your answer earns
Accept "as much water enters as leaves". Do not award the point for "water stopped crossing" or for "the solution is isotonic" with no statement about the two flows being equal.
Common slip: Saying the water stopped crossing. In an isotonic solution water crosses constantly in both directions; only the net movement is zero.
(c) Explain why only torn fragments of membrane were found in the dish of pure water. (1 point)
A full-credit answer: Pure water has no solute at all.
So the pure water can never match the 0.22 mol/L inside the cells.
Water moves toward the side with more solute.
So water keeps moving into the cells by osmosis.
A frog red blood cell has no cell wall to stop the swelling.
So each cell swells until its membrane tears, and its contents spill out.
This is osmotic lysis.
Only the torn membranes are left.
Check the box for each point your answer earns
Accept "the cells burst" with the reason that water kept entering because the outside has less solute (or no solute). Do not award the point for "the cells dissolved" or for water entering with no reason given.
Common slip: Saying the cells dissolved. Cells burst because water keeps entering; the membrane tears when it can stretch no further.
(d) A fifth dish holds 0.40 mol/L of a small solute that crosses the membranes freely. Make a claim about what the cells look like after an hour, and support your claim. (1 point)
A full-credit answer: After an hour the cells have swollen past their starting size, and some may have burst.
At first the outside holds more solute than the cells, so water leaves and the cells shrink.
The small solute crosses the membrane and enters the cells until it is equal on both sides.
So it then drives no osmosis.
the cells’ own 0.22 mol/L of solute is still there, and nothing outside matches it.
Therefore water moves into the cells, and the cells swell.
Check the box for each point your answer earns
Accept an answer that adds an early shrinking before the swelling. Do not award the claim alone, or "the cells stay shrunken, like the 0.40 mol/L dish".
Common slip: Treating the fifth dish like the 0.40 mol/L dish because the totals match. Equal total solute is not isotonic; a solute that crosses stops counting once it is equal on both sides.
(a) Describe the direction of net water movement across the fish's gills at low tide and at high tide. (1 point)
A full-credit answer: At low tide the creek water holds far less solute than the fish’s blood.
So the creek is hypotonic to the fish.
Water moves toward the side with more solute.
So at low tide water moves into the fish across its gills.
At high tide the seawater holds far more solute than the blood.
So the seawater is hypertonic to the fish.
Therefore at high tide water moves out of the fish across its gills.
Check the box for each point your answer earns
Accept the directions with the solute comparison stated in either form (less solute outside; more water outside). Do not award the point for one tide only, or for the directions reversed.
Common slip: Giving one direction for the whole day. The creek changes from hypotonic to hypertonic with the tide, so the net movement of water reverses.
(b) Explain how the fish keeps its blood near 0.35 mol/L at low tide, including what it does with water and what it does with salt. (1 point)
A full-credit answer: At low tide water enters the fish across its gills all the time.
Salt leaks out of the fish into the dilute creek all the time.
The fish removes the water by making large amounts of very dilute urine.
The fish replaces the lost salt by pumping salt in through its gills by active transport.
Active transport uses energy from the fish.
Together the urine and the pumping hold the blood near 0.35 mol/L.
Check the box for each point your answer earns
Accept "pumps salt in at the gills" without naming active transport. Do not award the point for urine alone or salt alone, or for the fish drinking water at low tide.
Common slip: Handling only the water. Salt leaks out down its gradient all the time, so it has to be pumped back in as constantly as the water is removed.
(c) Predict two changes in what the fish does when the tide comes in and the creek fills with seawater, and justify one of them. (1 point)
A full-credit answer: When the tide comes in, the fish starts drinking seawater.
The fish also makes very little urine.
Its gills pump salt out instead of in.
The fish drinks because water now leaves across its gills toward the saltier sea, so the fish must replace the water it loses.
Dilute urine would waste more water, so the fish makes little urine.
Salt comes in with every drink and from the sea, so the fish has to pump that salt back out.
Check the box for each point your answer earns
Accept any two of the three changes. Do not award the point for the changes with no justification, or for a justification in which water enters the fish at high tide.
Common slip: Keeping the low-tide habits. Dilute urine and salt pumped in would dehydrate and over-salt a fish in seawater.
(d) Predict how often the single-celled organism's contractile vacuole empties at high tide compared with low tide, and explain the difference. (1 point)
A full-credit answer: At low tide the creek is hypotonic to the cell, so water enters constantly by osmosis, and the contractile vacuole fills and empties many times a minute to throw it back out.
At high tide the seawater holds far more solute than the cell, so little or no net water enters.
The contractile vacuole only ever removes water that has already entered.
So at high tide the vacuole has almost nothing to collect and empties rarely, or stops.
Check the box for each point your answer earns
Accept "stops emptying" or "empties rarely" for high tide. Do not award the point for "empties more often to pump water in" or for the vacuole keeping water out.
Common slip: Having the vacuole work harder in seawater. It removes water that has entered; when water stops entering, it has nothing to do.