IB Biology Quiz: Apply Membranes And Membrane Transport

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Apply Membranes And Membrane TransportQuestion 1 of 19

A researcher is studying the transport of substance X into a cell. They observe that the rate of transport increases linearly with the concentration of X outside the cell. However, the rate of transport for substance Y increases with its external concentration but eventually reaches a maximum rate (saturates). What can be deduced about the transport mechanisms for X and Y?

X is transported by active transport, and Y is transported by simple diffusion.

X is transported by simple diffusion, and Y is transported by facilitated diffusion.

X is transported by endocytosis, and Y is transported by active transport.

X is transported by facilitated diffusion, and Y is transported by simple diffusion.

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IB Biology Quiz

IB Biology Quiz: Apply Membranes And Membrane Transport

Practice Apply Membranes And Membrane Transport in IB Biology with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

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This quiz focuses on Apply Membranes And Membrane Transport, giving you a quick way to practice the rules, question types, and explanations that matter most for IB Biology.

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Question 1

X is transported by active transport, and Y is transported by simple diffusion.
X is transported by simple diffusion, and Y is transported by facilitated diffusion. (correct answer)
X is transported by endocytosis, and Y is transported by active transport.
X is transported by facilitated diffusion, and Y is transported by simple diffusion.

Explanation: Simple diffusion rate is directly proportional to the concentration gradient, so it increases linearly without saturation. Facilitated diffusion and active transport both rely on a finite number of membrane protein transporters (channels or carriers). Once all transporters are occupied, the rate of transport reaches a maximum (Vmax), a phenomenon known as saturation. Since Y's transport saturates, it must be protein-mediated (facilitated diffusion or active transport). Since X's transport does not saturate, it is most likely simple diffusion.

Question 2

A red blood cell is placed in a solution of 0.3 M urea. Urea is a small, uncharged molecule that can slowly diffuse across the red blood cell membrane. Initially, the total solute concentration inside the cell is also approximately 0.3 M. What is the expected final outcome?

The cell will shrink because the external urea concentration causes initial water loss.
The cell will remain the same size because the solution is isotonic.
The cell will swell and burst because urea enters the cell, increasing internal solute concentration.
The cell will first shrink slightly, then swell and burst as urea diffuses in. (correct answer)

Explanation: This question tests the difference between osmolarity and tonicity. Initially, the solution is isosmotic (same solute concentration). However, because urea is a penetrating solute, it will diffuse down its concentration gradient into the cell. As urea enters, the internal solute concentration rises above 0.3 M, making the cell hypertonic relative to the solution. This causes water to follow urea into the cell via osmosis, leading to swelling and lysis (bursting). The initial slight shrinkage (crenation) can occur as water initially moves out before significant urea has entered, but the dominant final effect is swelling and lysis.

Question 3

A marine amoeba, which is isotonic with seawater, is transferred to a freshwater pond. To survive, it must actively pump water out using its contractile vacuole. A researcher adds a substance that specifically inhibits ATP synthesis in the amoeba. What is the most likely immediate consequence?

The amoeba will shrink as salts diffuse out into the hypotonic environment.
The amoeba will swell and likely burst due to the inability to power its contractile vacuole. (correct answer)
The amoeba will maintain its size as the rate of water influx equals the rate of passive water efflux.
The amoeba will begin synthesizing a rigid cell wall to withstand the osmotic pressure.

Explanation: The amoeba is in a hypotonic environment (freshwater), causing water to constantly enter via osmosis. The contractile vacuole is an organelle that performs active transport to pump this excess water out, a process requiring ATP. Inhibiting ATP synthesis will disable the vacuole, leading to uncontrolled water influx, swelling, and eventual lysis (bursting).

Question 4

The bacterium Vibrio cholerae produces a toxin that causes chloride ion channels in the intestinal epithelial cells to remain permanently open. This leads to a massive loss of chloride ions into the intestinal lumen. Which statement best explains the resulting severe diarrhea?

The loss of Cl⁻ ions causes water to be actively transported out of the cells to maintain charge balance.
The high concentration of Cl⁻ in the lumen lowers its water potential, causing a large net movement of water out of the cells by osmosis. (correct answer)
The open channels allow sodium ions to rush into the cells, causing them to swell and leak their contents into the lumen.
The toxin directly damages the cell membranes, causing them to become fully permeable to water and solutes, resulting in cell death.

Explanation: The movement of chloride ions into the intestinal lumen dramatically increases the solute concentration there, which in turn lowers the water potential. Due to this osmotic gradient, water moves from the epithelial cells and underlying tissues into the lumen via osmosis, leading to watery diarrhea and dehydration. Water is not actively transported; its movement is passive and follows the solute gradient.

Question 5

The secretion of insulin from pancreatic β-cells involves the packaging of insulin into vesicles which then fuse with the plasma membrane to release their contents. Which combination of processes is required for this to occur?

Endocytosis for packaging, followed by simple diffusion for release.
Exocytosis for release, which requires ATP for vesicle movement and fusion. (correct answer)
Active transport of insulin across the plasma membrane via carrier proteins.
Exocytosis for release, a passive process driven by insulin gradients.

Explanation: The release of large molecules like insulin is accomplished by exocytosis. This process involves vesicles, formed by the Golgi apparatus, moving to the plasma membrane and fusing with it. Both the movement of vesicles along the cytoskeleton and the fusion of the membranes are active processes that require energy in the form of ATP. It is not passive and does not involve carrier proteins for the insulin itself.

Question 6

A dialysis bag, permeable to water but not to sucrose, contains a 0.5 M sucrose solution. It is placed in a beaker containing a 0.2 M sucrose solution. What will be the net movement of water and the change in the bag's mass?

Water will move out of the bag, and its mass will decrease.
Water will move into the bag, and its mass will increase. (correct answer)
There will be no net movement of water, and its mass will remain constant.
Water will move into the bag, and its mass will decrease as sucrose moves out.

Explanation: The solution inside the bag (0.5 M sucrose) is hypertonic to the solution outside (0.2 M sucrose). This means the water potential inside the bag is lower (more negative) than outside. By osmosis, there will be a net movement of water from the area of higher water potential (the beaker) to the area of lower water potential (the bag). This influx of water will cause the bag to swell and its mass to increase.

Question 7

A scientist creates artificial vesicles (liposomes) with a high internal concentration of K⁺ ions and a low internal concentration of Na⁺ ions, similar to a real cell. These vesicles are placed in a solution with high Na⁺ and low K⁺. If a drug is added that creates pores in the membrane permeable only to K⁺, what will occur?

K⁺ ions will flow out of the vesicle, making the interior of the vesicle negative relative to the exterior. (correct answer)
Na⁺ ions will flow into the vesicle, creating a negative membrane potential.
K⁺ ions will flow out of the vesicle, making the interior of the vesicle positive relative to the exterior.
Both Na⁺ and K⁺ ions will flow until their concentrations are equal inside and outside the vesicle.

Explanation: The vesicles have a high internal K⁺ concentration and are placed in a low K⁺ solution. This creates a strong concentration gradient for K⁺ to move out. Since the added pores are only permeable to K⁺, these positive ions will diffuse out of the vesicle, down their concentration gradient. This efflux of positive charge leaves behind unbalanced negative charges inside the vesicle, making the interior negative relative to the exterior, thus establishing a membrane potential.

Question 8

A cell requires the uptake of molecule Z. The process is observed to be highly specific for Z, requires ATP, and is not affected by changes in the membrane potential. Which transport mechanism is most likely responsible?

Simple diffusion through the lipid bilayer.
Secondary active transport using a proton gradient.
Facilitated diffusion via a channel protein.
Receptor-mediated endocytosis. (correct answer)

Explanation: Receptor-mediated endocytosis is highly specific because the substance (Z) must first bind to a specific receptor protein on the cell surface. The process of forming a vesicle to internalize the receptor-ligand complex requires ATP. Unlike ion pumps or cotransporters, this bulk transport mechanism is generally not directly affected by the membrane's electrical potential. The other options are incorrect: simple diffusion is not specific and uses no ATP; facilitated diffusion uses no ATP; secondary active transport would be affected by membrane potential if an ion gradient is used.

Question 9

The fusion of a vesicle with the plasma membrane during exocytosis and the fusion of a phagosome with a lysosome both depend on the ability of membranes to change shape, break, and reconnect. What property of the plasma membrane is essential for these events?

The presence of a rigid cell wall outside the membrane.
The high concentration of aquaporins for water movement.
The fluidity of the phospholipid bilayer. (correct answer)
The impermeability of the membrane to polar molecules.

Explanation: The fluid mosaic model describes the plasma membrane as a fluid structure where phospholipids and proteins can move relative to one another. This fluidity allows the membrane to be dynamic: it can curve to form vesicles, break, and fuse with other membranes without losing its integrity. Rigidity (A) would prevent this, while aquaporins (B) and impermeability (D) are properties of the membrane but not the key feature that allows for fusion and fission.

Question 10

[HL] The resting potential of a neuron is primarily maintained by the Na⁺/K⁺ pump and K⁺ leak channels. The pump creates steep gradients for Na⁺ (high outside) and K⁺ (high inside). Why does the opening of K⁺ leak channels cause the membrane potential to become negative on the inside?

K⁺ leak channels actively pump positive K⁺ ions into the cell, making the outside negative.
The movement of K⁺ through leak channels triggers the Na⁺/K⁺ pump to expel three positive charges for every two it brings in.
The negative charge of the channel protein itself is transferred to the inside of the cell as K⁺ passes through.
The diffusion of positive K⁺ ions out of the cell, down their concentration gradient, leaves a net negative charge inside. (correct answer)

Explanation: The Na⁺/K⁺ pump establishes a high concentration of K⁺ inside the cell. The K⁺ leak channels are always open, allowing K⁺ ions to diffuse passively out of the cell, moving down their steep concentration gradient. As these positively charged ions leave the cell, they leave behind negatively charged proteins and anions that cannot cross the membrane, resulting in a net negative charge on the interior of the plasma membrane and establishing the negative resting potential.

Question 11

[HL] The absorption of glucose from the gut into an epithelial cell is achieved by a Na⁺-glucose cotransporter. This protein moves one Na⁺ ion down its concentration gradient and one glucose molecule against its concentration gradient into the cell. How is this process best classified?

Primary active transport, as it moves glucose against its gradient.
Facilitated diffusion, as it utilizes a protein carrier and the movement of Na⁺.
Secondary active transport, as it uses the electrochemical potential of Na⁺ established by a separate pump. (correct answer)
Simple diffusion, as the overall process is driven by the strong Na⁺ gradient.

Explanation: This is a classic example of secondary active transport (or cotransport). The transport of glucose against its gradient does not directly use ATP. Instead, it uses the kinetic energy stored in the Na⁺ concentration gradient. This Na⁺ gradient is maintained by the Na⁺/K⁺-ATPase pump (a primary active transporter) located elsewhere on the cell membrane, which does use ATP. Therefore, the glucose transport is indirectly powered by ATP.

Question 12

Freshwater fish face the constant problem of water entering their bodies by osmosis. They do not drink water and excrete large volumes of dilute urine. What does this indicate about the cells of the fish relative to the freshwater environment?

The cells are isotonic with the freshwater, so water movement is minimal.
The cells are hypotonic to the freshwater, causing water to diffuse out of the fish.
The cells are hypertonic to the freshwater, causing a constant influx of water. (correct answer)
The cells use active transport to pump water in, which must then be excreted.

Explanation: Freshwater is a hypotonic environment, meaning it has a very low solute concentration and high water potential. The cells and body fluids of a fish have a higher solute concentration, making them hypertonic relative to the freshwater. This osmotic gradient causes water to constantly move from the environment into the fish's body via osmosis, particularly across the gills. The physiological adaptations (not drinking, producing dilute urine) are to cope with this constant water influx.

Question 13

Cystic fibrosis is a genetic disorder caused by a mutation in the CFTR gene, which codes for a chloride ion channel protein. This faulty protein is often degraded by the cell before it reaches the plasma membrane. In lung epithelial cells, this prevents the normal movement of chloride ions out of the cell.

Based on the passage, what is the direct effect on water movement across the apical membrane of lung epithelial cells in a person with cystic fibrosis?

Water moves into the cells via osmosis, making the mucus layer outside the cells abnormally thick and sticky. (correct answer)
Water moves out of the cells via osmosis, diluting the mucus layer and making it too thin.
Water is actively transported out of the cells, attempting to compensate for the lack of chloride ion movement.
Water movement ceases entirely because the faulty CFTR protein also functions as an aquaporin.

Explanation: Normally, chloride ions are transported out of the cell, making the fluid outside (the mucus layer) hypertonic and drawing water out by osmosis. In cystic fibrosis, the inability to move Cl⁻ out of the cell means the mucus layer is less concentrated with solutes (less negative water potential) compared to the cell cytoplasm. Consequently, water does not move out sufficiently, and may even move in, resulting in a dehydrated, thick, and sticky mucus layer.

Question 14

Some cancer cells overexpress a type of protein called a multidrug resistance (MDR) transporter. This protein is an ATP-powered pump in the plasma membrane that expels a wide variety of chemotherapy drugs from the cell. How does this mechanism allow cancer cells to survive treatment?

It prevents the drugs from entering the cell by blocking membrane channels.
It uses exocytosis to expel large quantities of the drug at once.
It uses primary active transport to pump the drug out, keeping its intracellular concentration low. (correct answer)
It alters the membrane fluidity so that the drugs cannot diffuse across it.

Explanation: The MDR transporter is described as an ATP-powered pump. This indicates it is a primary active transporter. It actively moves chemotherapy drugs, which have already entered the cell (likely by diffusion), out of the cell against a concentration gradient. By continuously pumping the drug out, the transporter maintains a low intracellular concentration, preventing the drug from reaching its target and killing the cell.

Question 15

Ouabain is a cardiac glycoside that specifically inhibits the Na⁺/K⁺-ATPase pump. If ouabain is applied to a neuron, which of the following effects would be observed over time?

The intracellular concentration of Na⁺ will decrease and the intracellular concentration of K⁺ will increase.
The membrane will immediately hyperpolarize due to the blockage of positive ion movement.
The intracellular concentration of Na⁺ will increase and the intracellular concentration of K⁺ will decrease. (correct answer)
The transport of glucose into the cell via Na⁺-glucose cotransporters will increase to compensate.

Explanation: The Na⁺/K⁺ pump actively transports 3 Na⁺ ions out of the cell and 2 K⁺ ions into the cell. Inhibiting this pump will stop this process. As a result, Na⁺ that leaks in and K⁺ that leaks out will not be pumped back against their gradients. This will cause the intracellular Na⁺ concentration to gradually increase and the intracellular K⁺ concentration to gradually decrease, disrupting the resting membrane potential.

Question 16

Placing a plant tissue in a highly concentrated salt solution causes the cells to undergo plasmolysis. What is the correct sequence of events during this process?

Water leaves the cytoplasm and vacuole by osmosis, causing the plasma membrane to pull away from the cell wall. (correct answer)
The cell wall shrinks, pulling the plasma membrane away from it, and the cell becomes turgid.
Salt enters the cell by active transport, causing water to follow and the cell to become flaccid.
The plasma membrane ruptures, allowing cytoplasm to leak out and the cell wall to collapse.

Explanation: The highly concentrated salt solution is hypertonic to the cell's cytoplasm. This causes a net movement of water out of the cell's large central vacuole and cytoplasm via osmosis. As the cell loses water, its volume decreases, and the flexible plasma membrane pulls away from the rigid cell wall. This process is called plasmolysis, and the cell is described as flaccid or plasmolyzed.

Question 17

[HL] Acetylcholine is a neurotransmitter that binds to receptors on muscle cells. These receptors are ligand-gated ion channels. What happens when acetylcholine binds to its receptor?

The channel protein changes conformation, opening a pore that allows specific ions to diffuse across the membrane. (correct answer)
The receptor acts as an enzyme, using ATP to actively transport sodium ions into the cell.
The binding triggers endocytosis, engulfing the acetylcholine molecule into the muscle cell.
The receptor signals the Na⁺/K⁺ pump to increase its rate of activity, hyperpolarizing the membrane.

Explanation: Ligand-gated channels are a type of membrane protein that opens or closes in response to the binding of a specific chemical messenger (the ligand), in this case, acetylcholine. The binding causes a conformational change in the protein structure, which opens the channel's pore. This allows ions (typically Na⁺ in this case) to flow across the membrane down their electrochemical gradient, a form of facilitated diffusion.

Question 18

Some organisms living in extremely cold environments adapt by altering their plasma membrane composition. Which adaptation would be most effective at maintaining membrane fluidity at low temperatures?

Increasing the proportion of saturated fatty acids and decreasing the cholesterol content.
Increasing the proportion of unsaturated fatty acids and increasing the cholesterol content. (correct answer)
Decreasing the proportion of both unsaturated fatty acids and cholesterol.
Increasing the proportion of saturated fatty acids and increasing the cholesterol content.

Explanation: At low temperatures, membranes risk becoming too rigid. Unsaturated fatty acids have kinks in their tails, preventing tight packing and thus increasing fluidity. Cholesterol acts as a 'fluidity buffer'; at low temperatures, it prevents phospholipids from packing too closely together, thereby increasing fluidity. Therefore, increasing both unsaturated fatty acids and cholesterol is the most effective adaptation.

Question 19

Why does the process of phagocytosis, such as a white blood cell engulfing a bacterium, require a significant amount of cellular energy (ATP)?

To power the synthesis of new membrane lipids needed to form the phagocytic vesicle.
To create a strong osmotic gradient that draws the bacterium towards the cell surface.
To actively transport the bacterium across the plasma membrane into the cytoplasm.
To fuel the movement of the cytoskeleton to extend pseudopods and enclose the bacterium. (correct answer)

Explanation: Phagocytosis is a form of endocytosis that requires major reorganization of the cell's plasma membrane and underlying cytoskeleton. ATP is primarily required to power motor proteins that cause the actin cytoskeleton to polymerize and depolymerize, allowing the cell to extend pseudopods (cellular projections) that surround and engulf the large particle. This is a dynamic mechanical process, not a simple transport or synthesis event.