4 questions, one for each idea where we can. Answer them, then see which ideas to fix.
Question 1 of 4
In a hypothetical bacterium, the concentration of K⁺ is much higher in the cytoplasm than in the surroundings. A toxin inserts large pores into the bacterium's plasma membrane; the pores stay open and let ions and small polar molecules pass freely in either direction. Which prediction about the K⁺ concentration gradient is best supported?
Answer and reasoning
AK⁺ will move out through the pores, and the gradient will shrink as the concentrations even out.Correct The gradient could form only because the membrane restricted the movement of K⁺. With open pores, K⁺ moves by net diffusion from the higher concentration inside to the lower concentration outside, so the difference between the two sides shrinks.
BK⁺ will move in through the pores, making the gradient across the membrane even steeper than before. A student who thinks molecules move from lower to higher concentration picks this. Net movement through the open pores is from the high K⁺ concentration inside to the low concentration outside.
CThe gradient will stay as it is, because the bacterium keeps all of the K⁺ that it needs inside it. A student who thinks a cell holds on to substances because it needs them picks this. Need does not stop K⁺ from moving through open pores down its concentration gradient.
DK⁺ will leak out through the open pores until no K⁺ at all is left anywhere inside the bacterium. A student who thinks diffusion empties the starting region picks this. Net movement continues only until the concentrations inside and outside are equal.
Which statement correctly describes passive transport across a plasma membrane?
Answer and reasoning
ANet movement from lower to higher concentration, needing no direct input of metabolic energy A student who thinks passive transport moves molecules toward higher concentration picks this. Passive transport moves molecules down the gradient, from higher to lower concentration.
BNet movement from higher to lower concentration, requiring the direct input of metabolic energy A student who thinks passive transport needs energy from the cell picks this. The random motion of the molecules produces passive transport; no metabolic energy is supplied.
CNet movement from higher to lower concentration, with no direct input of metabolic energyCorrect Passive transport is the net movement of molecules down their concentration gradient, from higher to lower concentration, without the direct input of metabolic energy.
DNet movement from lower to higher concentration, requiring a direct input of metabolic energy A student who has swapped the definitions of passive and active transport picks this. Movement that requires a direct input of energy and can move molecules against a gradient is active transport.
Cells of a hypothetical alga were placed in a solution containing 4.0 mM of ion X; the outside concentration stayed at 4.0 mM throughout. The graph shows the concentration of X inside the cells over 40 minutes. What was the mean rate of increase of the concentration of X inside the cells during the first 10 minutes?
Answer and reasoning
A1.6 mM/min A student who divides the concentration at 10 minutes by the time, without subtracting the starting value, picks this: 16.0 ÷ 10 = 1.6. The cells already contained 4.0 mM at the start.
B1.2 mM/minCorrect Rate = change in concentration ÷ time = (16.0 − 4.0) mM ÷ (10 − 0) min = 1.2 mM/min. The inside concentration rose well above the outside concentration of 4.0 mM, so X was moved against its concentration gradient.
C0.6 mM/min A student who assumes the rate was constant and uses the whole experiment picks this: (28.0 − 4.0) ÷ 40 = 0.6. The rate was much higher in the first 10 minutes than later, when the curve levels off.
D0.8 mM/min A student who divides the time by the change picks this: 10 ÷ 12.0 = 0.8. A rate is the change per unit time, so the change is divided by the time.
Working Read from the graph: 4.0 mM at 0 min and 16.0 mM at 10 min. Rate = dY/dt = (16.0 mM − 4.0 mM)/(10 min − 0 min) = 12.0 mM/10 min = 1.2 mM/min. Distractors: 16.0/10 = 1.6 (starting value not subtracted); (28.0 − 4.0)/40 = 0.6 (average over the whole 40 min); 10/12.0 = 0.83, shown as 0.8 (time divided by change).
Which statement correctly describes endocytosis and exocytosis?
Answer and reasoning
ABoth are passive, since vesicles form and fuse with membranes without any energy from the cell itself. A student who thinks vesicle transport is passive picks this. Forming vesicles and fusing them with the plasma membrane require energy.
BBoth require energy and move large substances, or large amounts of material, across the membrane.Correct Endocytosis and exocytosis use vesicles to move large substances, or large amounts of substances, into and out of the cell, and both require energy from the cell.
CIn endocytosis material leaves the cell; in exocytosis material is brought into the cell. A student who confuses the two terms picks this. Endocytosis brings material into the cell; exocytosis releases material from it.
DBoth move small molecules, such as O₂ and CO₂, into and out of cells inside vesicles. A student who thinks all substances cross the membrane in vesicles picks this. O₂ and CO₂ are small nonpolar molecules that cross the phospholipid bilayer directly.
In preparation: 0 of 4 sections compiled and reviewed. The rest show key terms and common mistakes from our question bank until they are.
2.5.A.1 Concentration gradient Fix
Concentration gradient
A difference in the concentration of a substance between two regions, such as the two sides of a membrane.
Selective permeability and gradients
Because a membrane restricts the movement of some solutes, differences in their concentration can be formed and kept across it; a membrane that let every solute through freely could not hold such gradients.
Students often think Cells take in, hold on to or release substances because they need them, and they find another way when the usual route is blocked. In fact No. Movement across a membrane depends on the concentration gradient, the permeability of the membrane, the routes available and, for active transport and bulk transport, a supply of energy. A cell's need for a substance does not by itself move it or open new routes.
Students often think In diffusion, molecules keep moving across until all, or nearly all, of them have left the region where they started. In fact No. Net movement continues only until the concentrations on the two sides are equal; molecules are then spread evenly, not all gathered on one side.
2.5.A.2 Passive transport Fix
Passive transport
The net movement of molecules from a region of higher concentration to a region of lower concentration, without the direct input of metabolic energy.
Net movement
The overall movement that results when movement in one direction exceeds movement in the other. When concentrations on the two sides are equal, molecules keep crossing in both directions, but there is no net movement.
Students often think In passive transport, molecules move from the region of lower concentration to the region of higher concentration. In fact No. In passive transport the net movement is from the region of higher concentration to the region of lower concentration, down the concentration gradient.
Students often think When the concentrations become equal, the molecules stop moving across the membrane. In fact No. Molecules keep moving and crossing in both directions, but at equal rates, so there is no net movement and the concentrations stay equal (dynamic equilibrium).
2.5.A.3 Active transport Fix
Active transport
Movement of molecules across a membrane that requires the direct input of energy. In some cases, it moves molecules from a region of lower concentration to a region of higher concentration.
Students often think The rate of increase over an interval is the value at the end of the interval divided by the time, without subtracting the starting value. In fact No. A rate of change is the change in the quantity divided by the time taken, dY/dt. The starting value has to be subtracted before dividing.
Students often think A rate is calculated by dividing the time taken by the change in the quantity. In fact No. A rate is the change in the quantity per unit time, so the change is divided by the time: (16.0 − 4.0) mM ÷ 10 min = 1.2 mM/min.
2.5.B.1 Endocytosis Fix
Endocytosis
A process in which a cell takes in large molecules or particulate matter by folding its plasma membrane in on itself and forming new, small vesicles that engulf material from outside the cell. It requires energy.
Exocytosis
A process in which internal vesicles fuse with the plasma membrane and release their contents, such as large molecules, from the cell. It requires energy.
Vesicle
A small sac enclosed by a membrane that carries material within a cell; vesicles form from, and fuse with, other membranes, including the plasma membrane.
Bulk transport
A general name for endocytosis and exocytosis, the energy-requiring processes that move large substances or large amounts of substances into and out of cells in vesicles.
Students often think Endocytosis and exocytosis are passive: vesicles form and fuse with membranes on their own, without energy from the cell. In fact No. Endocytosis and exocytosis require energy from the cell to move large substances, or large amounts of substances, into and out of the cell.
Students often think Endocytosis releases material from the cell, and exocytosis takes material into the cell. In fact No. In endocytosis the cell takes material in by folding the plasma membrane inward to form vesicles; in exocytosis internal vesicles fuse with the plasma membrane and release material from the cell.
6 more questions. Every wrong answer here is a real mistake students make, and you see why it is wrong as soon as you answer.
Question 1 of 6
Vesicles made only of phospholipids were placed in a solution containing 10 mM of substance S, a small nonpolar molecule. The concentration of S outside the vesicles stayed at 10 mM. The graph shows the concentration of S inside the vesicles over 40 minutes. Which statement is supported by the data?
Answer and reasoning
AAfter about 30 minutes, molecules of S stopped crossing the membrane in either direction. A student who thinks molecules stop moving when concentrations become equal picks this. The graph shows that the net change stopped; molecules of S keep crossing in both directions at equal rates.
BThe inside concentration kept on rising until it was higher than the outside concentration. A student who thinks diffusion continues until the molecules have left the starting region picks this. The inside concentration leveled off at 10 mM, equal to the outside concentration.
CThe inside concentration rose by the same amount in each of the 5-minute intervals. A student who assumes a constant rate picks this. The rise was 5.1 mM in the first 5 minutes but 0.2 mM in total between 25 and 35 minutes.
DThe inside concentration rose quickly at first, then more slowly, and leveled off at the outside value.Correct The inside concentration rose by 5.1 mM in the first 5 minutes but much less in later intervals, and it reached 10.0 mM, the outside concentration. Net movement of S, down its concentration gradient, slowed as the gradient became smaller and stopped when the concentrations were equal.
Cells of a hypothetical yeast that contained none of ion Y were placed in a solution containing 5 mM Y. Half of the cells had first been treated with an inhibitor that stops the production of ATP. The graph shows the concentration of Y inside the cells in each group. A student claims that the cells take up Y by active transport. Which statement best explains how the data support the claim?
Answer and reasoning
AY entered the cells in both groups, and the uptake of any substance by a cell shows active transport. A student who thinks any uptake shows active transport picks this. Y could enter the inhibited cells passively, down its gradient, until the inside and outside concentrations were equal.
BY entered the untreated cells fastest at first, and fast uptake is a sign of active transport. A student who thinks active transport is recognized by its speed picks this. Passive uptake is also fastest at first, when the gradient is steepest; the inhibited cells show the same pattern.
CThe untreated cells stopped taking in Y once they held as much of it as they needed. A student who thinks cells take in substances until their needs are met picks this. The data do not show what the cells need, and the statement does not connect the uptake to energy or to the gradient.
DUntreated cells built up Y far above the outside level, and blocking ATP production prevented this.Correct Untreated cells accumulated Y to 24 mM, almost five times the outside concentration, so Y was moved against its gradient. Inhibited cells reached only 5 mM, the outside level, which passive entry could explain. Movement against the gradient that depends on the ATP supply is active transport.
Cells of a hypothetical freshwater alga contain nitrate at a much higher concentration than the pond water around them. Which new investigation would best test the hypothesis that the algal cells take up nitrate by active transport?
Answer and reasoning
ACompare nitrate uptake by cells with and without an inhibitor of ATP production, to see if uptake needs ATP.Correct Active transport requires a direct input of energy, such as from ATP. If blocking ATP production stops the cells from accumulating nitrate, while everything else is kept the same, the result supports active transport; if uptake continues against the gradient, the hypothesis is not supported.
BMeasure how fast nitrate enters the cells, to see whether it enters faster than other ions do. A student who thinks active transport is recognized by its speed picks this. The rate of entry alone cannot show whether energy is used; passive transport can be fast and active transport slow.
CMeasure the nitrate concentration in the cells again, using more precise equipment, to confirm it. A student who thinks repeating an observation tests the explanation for it picks this. A more precise measurement confirms that the gradient exists but cannot show how it is produced.
DPlace cells in a nitrate solution and check whether any nitrate enters, as uptake shows active transport. A student who thinks any uptake of a substance shows active transport picks this. Nitrate could enter by passive transport; the test must show that uptake against the gradient depends on the cells' energy supply.
The model shows three stages of a process at the plasma membrane of a cell. Which statement best describes what the model shows?
Answer and reasoning
AOutside material passes through a pore in the membrane and is then wrapped in a vesicle. A student who thinks large particles enter through pores in the membrane picks this. The model shows no pore; the membrane folds around the particle.
BOutside material breaks through the membrane, which tears and then seals behind it. A student who thinks the membrane tears during endocytosis picks this. In stage 2 the membrane is continuous; it folds and pinches off without tearing.
COutside material is enclosed as the membrane folds inward and pinches off a vesicle.Correct In stage 2 the membrane folds inward around the particle, and in stage 3 the fold has closed off to form a vesicle containing the particle inside the cell. This is endocytosis: the cell takes in material by folding its plasma membrane in on itself and forming a new vesicle.
DOutside material sinks into the membrane, which folds around it with no energy from the cell. A student who thinks endocytosis is passive picks this. The model shows the membrane folding inward, but endocytosis requires energy from the cell.
In each diagram, the extracellular fluid is above the plasma membrane and the cytoplasm is below it. The left drawing shows the start of a process and the right drawing shows its end. Which diagram correctly represents exocytosis?
Answer and reasoning
ADiagram 1Correct In exocytosis a vesicle inside the cell moves to the plasma membrane and fuses with it. Its membrane becomes part of the plasma membrane, and its contents are released outside the cell, as this diagram shows.
BDiagram 2 A student who thinks the vesicle leaves the cell whole picks this diagram. In exocytosis the vesicle's membrane fuses with the plasma membrane; only its contents leave.
CDiagram 3 A student who confuses exocytosis with endocytosis picks this diagram, which shows material from outside being taken into the cell in a vesicle.
DDiagram 4 A student who thinks large molecules can leave the cell by crossing the membrane on their own picks this diagram. The vesicle's contents are released by fusion of the vesicle with the plasma membrane, not by leaking across it.
Cells of a hypothetical gland make a digestive enzyme, a large protein, and package it into vesicles inside the cell. A chemical is added that blocks exocytosis in these cells but does not affect how the enzyme is made or packaged. Which prediction about the cells is best supported?
Answer and reasoning
AThe enzyme will still leave the cells by diffusing directly across the plasma membrane. A student who thinks large molecules can diffuse across the membrane picks this. A large protein cannot cross the hydrophobic interior of the bilayer; it leaves by exocytosis.
BThe cells will stop taking in material from outside, but secretion of the enzyme will continue. A student who confuses exocytosis with endocytosis picks this. Fusion of vesicles with the plasma membrane is the step that releases material, so secretion is what stops.
CThe enzyme will build up in vesicles inside the cells, and little of it will be secreted.Correct The enzyme is released by exocytosis, in which vesicles fuse with the plasma membrane. With exocytosis blocked, vesicles full of enzyme accumulate in the cytoplasm and little enzyme leaves the cells.
DThe cells will release the enzyme through transport proteins instead, as they need to secrete it. A student who thinks cells find another route when they need to release a substance picks this. The need to secrete does not open a new route for a large protein.
Compiled from the AP Biology Course and Exam Description (effective Fall 2025) and our question bank · Specialist review in progress. How these pages are made · Free, no account
