Unit 2 · Topic 2.5 end-of-topic test
Suggested time: about 46 minutes. Answer everything, then press Submit the test to see the feedback and scoring guides.
A 0.25 mol sample of ethanol, with a mass of 11.5 g, is dissolved in one liter of water. A 0.25 mol sample of glycerol, with a mass of 23.0 g, is dissolved in another liter of water.
How do the numbers of dissolved molecules in the two liters compare?
A bag made of thin membrane holds a dye solution at 0.10 mol/L. It is lowered into a beaker of the same dye at 0.60 mol/L. The membrane lets the dye through.
Which way is down the dye's concentration gradient, and what does 0.60 mol/L tell you about the beaker solution?
A spoonful of sugar dissolves at the bottom of a glass of still, cold tea. An hour later the tea tastes equally sweet at the top and at the bottom. A student explains: "The sugar molecules sense where the tea is unsweetened and swim toward it."
Which statement corrects the student?
In a placenta, the mother’s blood and the baby’s blood flow past each other with only thin cell layers between them, and the two bloods never mix. Oxygen crosses these cell layers freely. Dissolved oxygen was measured at three points along one placental vessel; the results are in the table.
What is the net movement of oxygen at each point?
Two chambers are separated by a membrane that a dye can cross. Both chambers hold the dye at 0.40 mol/L. Labeled dye molecules are followed for one minute: 85 cross from left to right and 85 cross from right to left. The concentrations stay at 0.40 mol/L throughout.
What do these counts show?
Cells lining a fish’s gut hold amino acid Y at 3 mmol/L. Just after a meal the gut fluid holds Y at 12 mmol/L. Between meals the gut fluid holds Y at 0.5 mmol/L.
Which way is down the concentration gradient of Y at each time?
A leaf cell in bright light takes in carbon dioxide and gives off oxygen. Dissolved carbon dioxide is at 0.6 mg/L in the water film outside the cell and 0.2 mg/L inside, and it enters. Dissolved oxygen is at 12 mg/L inside the cell and 9 mg/L outside, and it leaves.
How should the movement of carbon dioxide and the movement of oxygen be classified?
Substance A, a small nonpolar molecule, gets into protein-free bubbles of phospholipid bilayer within minutes; fructose, a sugar, never does. Living cells with fructose carriers take in both: A from 3 mmol/L outside to 1 mmol/L inside, and fructose from 6 mmol/L outside to 2 mmol/L inside. The cells’ ATP use stays the same while they do.
How do A and fructose enter the living cells?
A bag of thin membrane is filled with a cloudy starch solution and lowered into a beaker of amber iodine solution. Iodine turns blue-black when it mixes with starch. After twenty minutes the inside of the bag is blue-black and the beaker is still amber.
Which statement explains the two colors?
A muscle cell holds calcium ions (Ca²⁺) at 0.0001 mmol/L inside against 1.2 mmol/L outside. For a short interval every calcium channel in its membrane is shut, and no protein is moving Ca²⁺.
What happens to the Ca²⁺ gradient during the interval, and why?
Cells of a kelp, a large seaweed, hold iodide ions (I⁻) at 30 mmol/L, while the seawater around them holds I⁻ at 0.0005 mmol/L, and the cells keep taking more in. When a drug lowers the cells’ ATP supply, iodide uptake falls by 90%.
How is iodide entering the kelp cells?
A nerve cell releases a chemical messenger at its tip by exocytosis, thousands of vesicles a minute. Measurements show that while this goes on, the area of the plasma membrane at the tip grows.
Why does the membrane area grow?
The figure shows a single-celled pond organism touching a food particle far too large for any channel or carrier, and the same cell a little later. When the organism’s ATP supply is blocked, particles stay stuck to the outside of the cell.
What is the process shown, and what does the ATP result show?
In a pancreatic cell, vesicles filled with insulin, a protein hormone, wait near the plasma membrane. After a signal, the vesicles move to the membrane, fuse with it, and the insulin appears in the blood. A student says: "That cannot be exocytosis. Exocytosis is how a cell gets rid of waste."
Which response is correct?
Liver cells hold glycerol, a small polar molecule, at 0.5 mmol/L. Cells are placed in solutions with glycerol at 0.1, 0.5 and 2.0 mmol/L, and the net movement of glycerol is recorded after one minute. Untreated cells: out, none, in. Cells treated with a drug that blocks one channel protein: almost none in every solution. The cells’ ATP use is the same in every trial.
By what mechanism does glycerol cross these membranes, and what in the results shows it?
A single-celled organism living in a pond keeps its inside far saltier than the pond water. Its membrane has channels through which salt ions constantly leak out, down their gradient.
Which mechanism lets the cell keep its inside salty, and why?
A dye that cannot cross any membrane is dissolved in the fluid around a white blood cell. The cell then takes in a bacterium by endocytosis.
Where is the dye found inside the cell afterward, and why?
(a) Identify the mechanism by which oxygen enters the cell and the mechanism by which glucose enters the cell, and justify each identification with observations from the table. (1 point)
A full-credit answer: Oxygen enters by simple diffusion, straight through the bilayer: it is at 0.20 mmol/L outside and 0.05 mmol/L inside, so it moves down its gradient, and it crosses protein-free bubbles just as readily, so it needs no protein.
Glucose enters by facilitated diffusion, through a carrier: it is at 8 mmol/L outside and 2 mmol/L inside, so it moves down its gradient; it does not enter protein-free bubbles, so it needs a protein; and blocking ATP leaves its rate unchanged, so no energy is used.
Check the box for each point your answer earns
Accept: "passive transport through the bilayer" for oxygen and "passive transport through a carrier (or a membrane protein)" for glucose. Do not award the point if either substance is said to need ATP, or if glucose is said to cross the bilayer itself.
Common slip: Calling glucose's entry simple diffusion because glucose moves down its gradient, or calling it active transport because a protein is involved. The bubble result decides the route; the ATP result decides whether the cell uses energy.
(b) Identify the mechanism by which Cl⁻ leaves the cell, and justify your identification with two observations from the table. (1 point)
A full-credit answer: Cl⁻ leaves by active transport, through a pump, because Cl⁻ is at 40 mmol/L inside and 550 mmol/L outside and the cell moves Cl⁻ out, toward the side where Cl⁻ is already more concentrated, so Cl⁻ is moved against its concentration gradient, which a substance left to itself never does; and because blocking the cell’s ATP supply stops the movement, so the cell is using energy from ATP to move Cl⁻.
Check the box for each point your answer earns
Accept: "the cell pumps Cl⁻ out" as the name. One observation alone does not earn the point; the direction against the gradient and the dependence on ATP are both needed.
Common slip: Naming active transport from the direction alone, or from the ATP result alone. A test item asks for both observations because either one on its own could be read another way.
(c) Identify the process that takes the bacterium into the cell and the process that releases the mucus, and describe what surrounds the bacterium once it is inside the cell. (1 point)
A full-credit answer: The bacterium is taken in by endocytosis: the membrane dents inward around the bacterium and closes over it.
The mucus is released by exocytosis: each sac of membrane is a vesicle that fuses with the plasma membrane and opens outward, releasing its mucus into the seawater.
Inside the cell the bacterium is surrounded by a closed vesicle of membrane, pinched off from the plasma membrane.
Check the box for each point your answer earns
Accept: "phagocytosis" for the bacterium. Do not award the point if the bacterium is said to be loose in the cytosol, or if the two process names are swapped.
Common slip: Leaving the bacterium loose in the cytosol. The pocket that closed over it becomes a sealed vesicle, and the bacterium stays inside that vesicle.
(d) Explain why passive crossings alone would erase the difference between the concentration of Cl⁻ inside the cell and in the seawater, and how the cell keeps the difference. (1 point)
A full-credit answer: Passive crossings can only run down a gradient.
Cl⁻ is far more concentrated in the seawater than in the cytosol, so on their own they would let Cl⁻ leak in, toward 550 mmol/L, and erase the difference.
By using energy the cell can move Cl⁻ against its gradient, back out into the seawater, so the inside stays at 40 mmol/L.
A difference between inside and outside is kept only by continual use of energy.
Check the box for each point your answer earns
Accept an answer framed in general terms (passive crossings even things out; active transport keeps them uneven) provided it says what the energy is used for: movement against the gradient.
Common slip: Saying the cell needs energy 'to move things across' without saying which way. Passive crossings run down the gradient on their own; the energy is used for movement against the gradient.
(a) Describe how a large polar molecule such as Z can cross a plasma membrane, and what prevents it from crossing the bilayer on its own. (1 point)
A full-credit answer: A large polar molecule such as Z crosses the membrane by passing through a membrane protein: a channel or a carrier.
What keeps it from crossing the bilayer on its own is the hydrophobic interior: the hydrocarbon tails in the middle of the membrane carry no charges or partial charges.
Water attracts a polar molecule.
Nothing in the hydrophobic interior attracts a polar molecule.
So Z stays in the water and cannot cross the bilayer on its own.
Check the box for each point your answer earns
Accept: "it needs a channel or carrier protein because it cannot dissolve into the oily middle", in any wording that names both the protein route and the reason.
Common slip: Saying Z is too big to fit between the phospholipids. Water attracts its polar groups, and that attraction is what keeps Z out of the oily middle.
(b) Identify the variable changed between trials, the variable measured, and the trial that serves as the control. (1 point)
A full-credit answer: The variable changed is the treatment of the cells: none, ATP blocked, or one membrane protein blocked.
The variable measured is how much Z enters, the concentration of Z inside after twenty minutes.
The control is trial 1, the untreated cells.
Check the box for each point your answer earns
All three are needed for the point. Accept "the amount of Z inside" for the measured variable.
Common slip: Swapping the changed and measured variables, or naming trial 2 as the control. The control is the untreated cells the others are compared with.
(c) Describe one further trial that would test whether the blocked protein can move Z against its concentration gradient, and predict its result. (1 point)
A full-credit answer: Load untreated cells with Z at 8 mmol/L and place them in a solution of Z at 2 mmol/L.
That reverses the gradient.
Prediction: Z leaves the cells through the protein until the two sides are equal.
That shows the protein cannot move Z uphill.
With equal Z inside and out, there would be no net movement.
Check the box for each point your answer earns
Accept: any design that makes the gradient run the other way or removes it, with a prediction that net movement follows the gradient (down it, or none when the sides are equal). Also accept: untreated cells kept in the 8 mmol/L solution until the concentration inside stops changing, with the prediction that the inside levels off at 8 mmol/L and never rises above it. Do not award the point for repeating trial 2 or trial 3 unchanged.
Common slip: Repeating trial 2 or trial 3 unchanged. The new trial has to make the gradient run the other way, or remove it.
(d) Evaluate the claim that Z enters the cells by facilitated diffusion, using the results of all three trials. (1 point)
A full-credit answer: The claim is supported.
Z moves from 8 mmol/L to 2 mmol/L, down its gradient, so no energy is needed.
Trial 2 confirms this: blocking ATP changes nothing, so the cell is not using energy.
Trial 3 shows that a membrane protein is required, so the crossing is not simple diffusion.
Down the gradient, through a protein, with no energy used by the cell is facilitated diffusion.
Therefore the claim fits all three trials.
Check the box for each point your answer earns
Accept: the three pieces (downhill, ATP not needed, protein needed) each tied to its trial. Do not award the judgement alone, or a judgement grounded on only one trial.
Common slip: Judging from one trial. Each of the three pieces, downhill, no ATP needed and protein needed, comes from its own trial.