Unit 2 · Topic 2.5 end-of-topic test
A student weighs out one mole of glucose and one mole of table sugar (sucrose) and dissolves each in its own liter of water. The glucose weighed 180 g; the sucrose weighed 342 g.
How do the numbers of dissolved sugar 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 drop of blue ink falls into a glass of still water. An hour later the whole glass is evenly blue. A student explains: "The ink molecules sense where the clear water is and swim toward it."
Which statement corrects the student?
In a lung, a tiny air sac sits against a blood vessel, with only two thin cell layers between them. Oxygen was measured at three points along one sac. Point 1: air 14 units, blood 6 units. Point 2: air 11 units, blood 9 units. Point 3: air 10 units, blood 10 units. Oxygen crosses these cell layers freely.
What is the net movement of oxygen at each point?
Two chambers are separated by a membrane that a dye can cross. Both hold the dye at 0.40 mol/L. A tracker follows labeled dye molecules 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?
A red blood cell sits in a solution whose glucose concentration, 5 mmol/L (thousandths of a mol/L), equals the concentration inside the cell. The cell's membrane has glucose carriers. A labeled glucose molecule is seen entering the cell.
What does the entering molecule tell you about glucose movement at this cell?
A working muscle cell uses oxygen and makes carbon dioxide. Oxygen is at 40 units outside the cell and 10 inside, and it enters. Carbon dioxide is at 50 units inside and 40 outside, and it leaves.
How should the two crossings be classified?
Protein-free bubbles of bilayer are placed in a solution of substance A, a small nonpolar molecule, and glucose. Within minutes A is found inside the bubbles; glucose never is. Living cells with glucose carriers are then placed in the same solution, with their ATP supply blocked. Both A and glucose enter the cells, each from a higher concentration outside to a lower one inside.
How do A and glucose 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. The membrane lets the small iodine molecules through but not the large starch molecules. Iodine turns blue-black when it meets starch. After twenty minutes the inside of the bag is blue-black and the beaker is still amber.
Why is the starch still more concentrated inside the bag while the iodine has evened out?
A nerve cell holds Na⁺ at 15 mmol/L inside against 145 mmol/L outside. For a short interval every sodium channel in its membrane is shut, and no protein is moving Na⁺.
What happens to the Na⁺ gradient during the interval, and why?
Soil water around a plant root holds nitrate at 0.30 mmol/L; the root cells hold it at 5.4 mmol/L, and they keep taking more in. When a treatment lowers the cells' ATP supply, nitrate uptake falls by 90%.
How is nitrate entering the root 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 meeting a food particle far too large for any channel or carrier. In the second panel its membrane has folded around the particle and pinched off, and the particle is inside the cell, wrapped in membrane. When the organism's ATP supply is blocked, particles stay stuck to the outside.
What is the process, 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?
Kidney cells hold a charged solute X at 4.0 mmol/L. Cells are placed in solutions with X at 2.0, 4.0 and 8.0 mmol/L, and the net movement of X 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 X cross these membranes?
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 liver cell takes in a large protein particle from the blood by endocytosis.
Where is the particle afterward, and what surrounds it?
(a) Name the mechanism at crossing 1 and at crossing 2, and state what the two crossings have in common. (1 point)
A full-credit answer: Crossing 1 is simple diffusion: O₂ passes straight through the bilayer. Crossing 2 is facilitated diffusion: glucose passes through a carrier protein. Both are passive transport: the substance moves down its concentration gradient with no energy spent by the cell.
Check the box for each point your answer earns
Accept: "both move from higher to lower concentration for free" for the shared feature. Both names and the shared feature are needed for the point.
Common slip: Naming the two crossings without saying what they share. Both are downhill and free: passive transport.
(b) Explain why crossing 3 must cost the cell energy while crossing 2 is free. (1 point)
A full-credit answer: At 3, Cl⁻ is moved from 40 to 550 mmol/L, against its concentration gradient, and only energy spent by the cell, from ATP through a pump, can move a substance uphill. At 2, glucose moves from 8 to 2 mmol/L, down its gradient, which happens on its own because more glucose molecules leave the crowded side than return to it.
Check the box for each point your answer earns
Accept: "against the gradient needs ATP; down the gradient is free" provided both directions are read correctly from the numbers. Do not award the point for "because 3 uses a protein": 2 uses a protein too.
Common slip: Saying crossing 3 costs energy because it uses a protein. Crossing 2 uses a protein too; the direction relative to the gradient is what decides.
(c) Name the processes at 4 and 5. Then draw, on paper, what crossing 4 looks like a minute later, labeling the particle and the membrane around it. (1 point)
A full-credit answer: 4 is endocytosis and 5 is exocytosis. A minute later the particle at 4 is inside the cell, enclosed in a closed vesicle of membrane pinched off from the plasma membrane, and the plasma membrane is continuous again above it.
Check the box for each point your answer earns
Accept: a labeled sketch with the particle inside a membrane circle in the cytosol. Do not award the point if the particle is drawn loose in the cytosol or still open to the outside.
Common slip: Drawing the particle loose in the cytosol or still open to the outside. What endocytosis takes in arrives wrapped in a vesicle.
(d) Explain why a cell needs the crossings that cost energy (3, 4 and 5) as well as the free ones (1 and 2) if it is to keep the solute concentrations inside it different from those outside. (1 point)
A full-credit answer: Passive crossings only run down a gradient, so on their own they would even a substance out and erase differences between inside and outside. By spending energy the cell can move substances against their gradients (3) and take in or release material in bulk (4, 5), so it can hold gradients such as low Cl⁻ inside and keep its inside different from its surroundings.
Check the box for each point your answer earns
Accept: "passive transport only evens things out; active transport lets the cell keep gradients" with the link to keeping the inside different. Do not award the point for restating which crossings cost energy without saying what that lets the cell do.
Common slip: Restating which crossings cost energy without saying what that lets the cell do. The point is that energy is what holds the differences.
(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 only through a membrane protein, a channel or a carrier. It cannot cross the bilayer on its own because the hydrocarbon tails in the middle of the membrane carry no charges or partial charges, so a polar molecule, which water holds on to, has nothing to hold it in the hydrophobic interior.
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. Its polarity, held by water, is what keeps it out of the oily middle.
(b) Identify the variable the researchers changed between trials, the variable they 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) Propose 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, reversing the gradient. Prediction: Z leaves the cells through the protein until the two sides are equal, showing 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). 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) Justify the claim that Z enters the cells by facilitated diffusion, using the results of all three trials. (1 point)
A full-credit answer: Z moves down its gradient, from 8 to 2 mmol/L, so no energy is needed, and trial 2 confirms it: blocking ATP changes nothing, so the cell is not paying. Trial 3 shows a membrane protein is required, so this is not simple diffusion. Down the gradient, through a protein, at no cost to the cell: facilitated diffusion, a form of passive transport.
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 point if the justification uses only one trial.
Common slip: Using only one trial. Each of the three pieces, downhill, no ATP needed and protein needed, comes from its own trial.