- Some single-celled freshwater organisms possess contractile vacuoles that periodically expel water. What is the primary function of this organelle in this context?
Answer: To maintain water balance by removing excess water that enters the cell via osmosis.
Freshwater is strongly hypotonic to the cytoplasm of single-celled organisms. Consequently, water constantly enters the cell via osmosis, threatening to swell and lyse it. The contractile vacuole is an adaptation that actively collects this excess water from the cytoplasm and expels it from the cell, thus maintaining osmotic balance (osmoregulation). A is incorrect because the goal is to get rid of water, not take in more salt which would worsen the problem. C is incorrect as the organism is in a hypotonic, not hypertonic, environment. D is incorrect as the primary function is osmoregulation, not thermoregulation.
- The sodium-glucose cotransporter (SGLT1) in the small intestine moves glucose into epithelial cells against its concentration gradient. It does this by coupling the import of one glucose molecule to the import of two sodium ions, which are moving down their electrochemical gradient. How should this transport mechanism be classified?
Answer: Secondary active transport, as it uses the energy stored in an ion gradient established by a separate pump.
This process is a classic example of secondary active transport. It is 'active' because glucose is moved against its concentration gradient. It is 'secondary' because the energy for this movement does not come directly from ATP hydrolysis at the transporter itself. Instead, it uses the potential energy stored in the Na+ electrochemical gradient, which is maintained by the Na+/K+ pump (a primary active transporter) located elsewhere on the cell membrane. A is incorrect because the SGLT1 protein does not directly use ATP. C is incorrect because glucose is moved against its gradient, which is not diffusion. D is incorrect as it is a protein-mediated process, not simple diffusion.
- A red blood cell is placed in a solution of 0.3 M urea. The cell initially shrinks but then swells and returns to its original size before lysing. Urea is a small, uncharged molecule that can slowly cross the cell membrane. What is the best explanation for this sequence of events?
Answer: The solution is initially hypertonic, causing water to leave the cell. As urea diffuses into the cell, the intracellular solute concentration increases, causing water to re-enter by osmosis.
Initially, the total solute concentration outside (0.3 M urea) is higher than inside the red blood cell (approx. 0.3 Osm), making the solution hypertonic. This causes water to exit the cell via osmosis, leading to shrinkage (crenation). However, because urea is a penetrating solute, it slowly diffuses down its concentration gradient into the cell. As urea enters, the solute concentration inside the cell rises, eventually exceeding that of the external solution. This reverses the osmotic gradient, causing water to enter the cell, leading to swelling and eventual lysis. A is incorrect as urea is not actively pumped in. C is incorrect as the solution is not initially hypotonic. D is incorrect as the solution is hypertonic and urea diffuses into, not out of, the cell.
- Why does the process of exocytosis require energy in the form of ATP, even though it often involves substances moving down their concentration gradient out of the cell?
Answer: ATP hydrolysis powers vesicle transport along cytoskeleton and facilitates membrane fusion processes.
Exocytosis is an active process. While the final release of contents might follow a concentration gradient, the complex machinery involved requires energy. ATP is needed to power motor proteins (like kinesins and dyneins) that transport the vesicles from their site of origin (e.g., Golgi apparatus) to the plasma membrane along microtubule tracks. Furthermore, the process of fusing two lipid bilayers is energetically unfavorable and requires ATP and specific protein complexes (like SNAREs) to occur. A is part of general cell metabolism, not the immediate energy cost of exocytosis. C and D are incorrect descriptions of the process.
- A transmembrane protein has a single alpha-helical domain that spans the phospholipid bilayer. Which characteristic is most likely for the amino acids within this domain?
Answer: They are predominantly hydrophobic and have nonpolar R-groups.
The region of a transmembrane protein that passes through the lipid bilayer is in direct contact with the hydrophobic fatty acid tails of the phospholipids. To be thermodynamically stable in this nonpolar environment, the amino acid R-groups (side chains) in this domain must also be nonpolar and therefore hydrophobic. A and D are incorrect because hydrophilic or polar R-groups would be repelled by the lipid tails. C is incorrect because the stem specifies an alpha-helical domain, not a beta-pleated sheet, and the R-groups would need to be uniformly nonpolar to embed in the membrane core.
- The secretion of neurotransmitters at a synapse involves the fusion of vesicles with the presynaptic membrane. Which process is most directly coupled with this event?
Answer: Exocytosis, which increases the surface area of the presynaptic membrane.
The release of substances packaged in vesicles by fusing the vesicle membrane with the plasma membrane is the definition of exocytosis. This process adds the vesicle's membrane to the plasma membrane, thereby temporarily increasing its surface area. A (endocytosis) is involved in recycling the membrane later, but it is not the release mechanism itself. C is incorrect as phagocytosis is the engulfment of large particles, not the release of small molecules. D is incorrect because the entire vesicle fuses; the neurotransmitters are not released through channels in the vesicle membrane during this process.
- Which of the following molecules would be expected to have the lowest rate of simple diffusion across a pure phospholipid bilayer, assuming no transport proteins are present?
Answer: A chloride ion (Cl⁻), a small charged particle.
The rate of simple diffusion across a lipid bilayer is determined by size, polarity, and charge. Small, nonpolar molecules like CO₂ cross most easily. Small, polar molecules like ethanol and glycerol can cross, but more slowly. Charged particles (ions) like Cl⁻ are extremely impermeable to the hydrophobic core of the membrane, regardless of their small size, because of the strong hydration shell of water molecules surrounding them. Therefore, a chloride ion would have the lowest rate of simple diffusion.
- Which statement correctly distinguishes between ligand-gated and voltage-gated ion channels?
Answer: Ligand-gated channels open in response to the binding of a specific molecule, while voltage-gated channels open in response to a change in membrane potential.
The key distinction between these two types of gated channels is their gating mechanism. Ligand-gated channels are controlled by the binding of a chemical signal (a ligand), such as a neurotransmitter. Voltage-gated channels are controlled by changes in the electrical potential difference across the membrane. A is incorrect as both are forms of facilitated diffusion (passive transport). C is incorrect; leakage channels (which are different from gated channels) are responsible for the resting potential, and both types of gated channels are typically closed at rest. D is incorrect as selectivity varies for both types; some are highly selective for one ion, while others can be less selective.
- A scientist places a marine alga, which is isotonic to seawater, into a freshwater pond. Which of the following is the most likely immediate outcome for the algal cells?
Answer: The cells will take up water, and the turgor pressure will increase significantly against the cell wall.
Seawater is hypertonic compared to freshwater. Therefore, the cytoplasm of the marine alga is hypertonic to the pond water. When placed in the freshwater (a hypotonic solution), water will move down its water potential gradient and enter the algal cells via osmosis. Because the alga has a cell wall, it will not lyse. Instead, it will swell, and the internal hydrostatic pressure (turgor pressure) will build up against the rigid cell wall. A describes the opposite effect. B is incorrect because the cell wall does not prevent water movement, only lysis. D describes a long-term adaptation, not an immediate osmotic outcome, and it is more likely the cell would have mechanisms to expel water (like a contractile vacuole in some species).
- In an experiment, identical potato cylinders are placed in sucrose solutions of varying concentrations. After one hour, the change in mass is recorded. Which of the following describes the point at which the line of best fit for a graph of percentage mass change versus sucrose concentration crosses the x-axis?
Answer: The point where the water potential of the sucrose solution is equal to the initial water potential of the potato cells.
The x-axis crossing point represents zero percentage mass change. This indicates that there was no net movement of water between the potato cells and the surrounding solution. No net movement of water occurs when the water potential inside the cells is equal to the water potential of the external solution. This is the isotonic point. A describes a point of mass gain. C describes a point of maximum mass loss. D is incorrect as sucrose is not typically actively transported into potato tuber cells in this context, and the mass change is primarily due to water movement (osmosis).
- The rate of transport of substance X into a cell is measured at various external concentrations. The rate increases linearly with concentration and does not plateau. The rate of transport of substance Y increases with concentration but reaches a maximum rate (Vmax). Neither process requires ATP. What are the most likely modes of transport for X and Y?
Answer: X: simple diffusion; Y: facilitated diffusion
A transport rate that is linearly dependent on concentration and does not saturate is characteristic of simple diffusion across the membrane. The rate is limited only by the concentration gradient. A transport rate that shows saturation kinetics (reaches a Vmax) indicates the involvement of a limited number of membrane proteins (channels or carriers). Since the transport of Y does not require ATP, it must be facilitated diffusion, not active transport. Therefore, X is transported by simple diffusion and Y by facilitated diffusion.
- Aquaporins are channel proteins that facilitate the rapid movement of water across membranes. A mutation that inactivates aquaporins in the epithelial cells of the kidney's collecting ducts would most likely result in what condition?
Answer: The production of a large volume of dilute urine because water reabsorption is impaired.
The collecting ducts of the kidney are where final water reabsorption occurs, under the control of the hormone ADH which promotes the insertion of aquaporins into the membrane. If these aquaporins are non-functional, water cannot be effectively reabsorbed from the filtrate back into the blood. This leads to the excretion of a large volume of dilute urine, a condition known as diabetes insipidus. B is the opposite of what would happen. C is incorrect as filtration in the glomerulus is a separate process. D is incorrect because glucose reabsorption occurs earlier in the nephron and is not directly dependent on aquaporins in the collecting duct.
- A researcher uses ouabain, a specific inhibitor of the Na+/K+ pump, to treat a culture of neurons. Which of the following would be an immediate and direct consequence of the pump's inhibition?
Answer: A gradual decrease in the magnitude of the membrane potential, moving it closer to zero.
The Na+/K+ pump is electrogenic; it pumps 3 Na+ ions out for every 2 K+ ions it pumps in, contributing to the negative resting membrane potential. Inhibiting the pump stops this net outward movement of positive charge. Additionally, the ion gradients (high extracellular Na+, high intracellular K+) will slowly dissipate due to leakage through channels. Both effects cause the membrane potential to become less negative (depolarize), moving closer to zero. A is incorrect because the pump's inhibition leads to depolarization, not hyperpolarization. B is incorrect as the pump's ATP consumption is a small fraction of the cell's total, so a significant increase in ATP is unlikely and not the most direct consequence on the membrane. D is incorrect as the primary glucose transporter in neurons (GLUT) uses facilitated diffusion, which is independent of the Na+/K+ pump's ion gradients.
- Familial hypercholesterolemia is a genetic disorder characterized by high levels of LDL cholesterol in the blood. The disorder is often caused by a mutation in the gene coding for the LDL receptor protein. How does this mutation lead to the observed symptoms?
Answer: The defective receptors are unable to bind to LDL particles in the bloodstream, preventing their uptake by receptor-mediated endocytosis.
LDL particles are taken up by cells from the blood via receptor-mediated endocytosis. This process requires the LDL particle to bind to a specific LDL receptor on the cell surface. In familial hypercholesterolemia, a mutation in the receptor protein prevents this binding. As a result, cells cannot efficiently remove LDL from the blood, leading to its accumulation and the associated cardiovascular risks. A describes a different type of disorder, a lysosomal storage disease. C is the opposite of what happens. D is incorrect; the problem is with uptake (endocytosis), not export (exocytosis).
- The fluid mosaic model describes the plasma membrane as a dynamic structure. Which statement provides the strongest evidence for the lateral movement of membrane proteins?
Answer: The fusion of a human cell and a mouse cell shows that their distinct membrane proteins become intermixed over time.
The classic experiment by Frye and Edidin involved fusing human and mouse cells, each with their membrane proteins labeled with different fluorescent dyes. Initially, the proteins were segregated to their respective halves of the fused cell membrane. Over time, the proteins were observed to intermix, providing direct evidence that they are free to move laterally within the plane of the membrane. A provides evidence for the 'mosaic' part (embedded proteins) but not the 'fluid' part. C provides evidence for how proteins are integrated into the membrane but not their movement. D is a general observation but does not demonstrate fluidity.
- The bacterium Psychrobacter arcticus thrives in polar sea ice. Which adaptation would be most advantageous for maintaining appropriate membrane fluidity in its extremely cold environment?
Answer: A high proportion of short-chain, unsaturated fatty acid tails in its phospholipids to reduce packing density.
To maintain fluidity at low temperatures, membranes need to prevent phospholipids from packing too closely together. Unsaturated fatty acids have kinks in their tails that increase space between molecules, and shorter tails have fewer van der Waals interactions. This combination reduces the temperature at which the membrane solidifies. A is incorrect because cholesterol is not found in bacterial membranes; it is characteristic of animal cells. B describes an adaptation for high temperatures, as saturated fats pack tightly and increase viscosity. D is incorrect as peripheral proteins are not primarily involved in regulating membrane fluidity in this manner.
- A plant cell with a solute potential (ψs) of -0.7 MPa is placed in an open beaker of pure water (ψ = 0 MPa). The cell's membrane is fully permeable to water but not to the solutes. What will happen?
Answer: Water will enter the cell until the cell's pressure potential (ψp) reaches +0.7 MPa, at which point net water movement will cease.
Water moves from a region of higher water potential to a region of lower water potential. The pure water has a ψ of 0 MPa. The cell initially has a water potential (ψw) of -0.7 MPa (since ψp is initially 0). Water will therefore move into the cell. As water enters, the cell swells and the rigid cell wall exerts a positive pressure potential (turgor pressure, ψp) back on the cell. Net water movement stops when the cell's water potential equals the surrounding water's potential (0 MPa). This occurs when ψw = ψs + ψp = 0, so -0.7 MPa + ψp = 0. Thus, equilibrium is reached when ψp = +0.7 MPa. B, C, and D all describe scenarios involving water loss, which is incorrect.
- A researcher is studying a carrier protein that actively transports calcium ions (Ca²⁺) out of the cell. They observe that the transport rate reaches a maximum even when both the Ca²⁺ gradient and the supply of ATP are increased. What is the most likely limiting factor?
Answer: The number of available carrier proteins in the membrane.
Like enzymes, carrier proteins involved in both facilitated diffusion and active transport can become saturated. When all the carrier proteins are occupied and working at their maximum conformational change rate, increasing the substrate (Ca²⁺) concentration or energy supply (ATP) will not increase the overall transport rate. The process is limited by the finite number of transporters embedded in the membrane. The stem explicitly states that the Ca²⁺ gradient and ATP supply are not limiting, eliminating A and B. D is irrelevant to the maximum rate of the active transport pump itself.
- What is the primary role of glycoproteins and glycolipids on the outer surface of an animal cell membrane?
Answer: To serve as cell-surface receptors and in cell-to-cell recognition and adhesion.
The carbohydrate portions of glycoproteins and glycolipids form the glycocalyx on the exterior of the cell membrane. This layer is crucial for cell identity, allowing cells to recognize each other (e.g., in tissue formation and immune responses) and for binding signaling molecules as receptors. A is incorrect; this is a function of the cell wall in plants, not the glycocalyx in animals. B is incorrect as this role is performed by specific transmembrane proteins, not the carbohydrate chains. D is incorrect as the primary hydrophobic barrier is the lipid bilayer itself; the glycocalyx is hydrophilic.
