- A freeze-fracture electron micrograph of a bacterial cytoplasmic membrane reveals a surface covered in bumps and corresponding pits. The bumps are observed to be more numerous on the P-face (protoplasmic face) than the E-face (exoplasmic face). These bumps represent:
Answer: integral membrane proteins that remain with one leaflet after the bilayer is split.
Correct. The freeze-fracture technique splits the membrane bilayer along its weakest plane: the hydrophobic interior. This separates the inner (protoplasmic) leaflet from the outer (exoplasmic) leaflet. Integral membrane proteins, which are embedded within the bilayer, are fractured along with the leaflets. They appear as bumps (or particles) on the face of the leaflet they remain with and leave corresponding pits on the complementary face. They are typically more abundant on the P-face.
A is incorrect because the fracture occurs between the tails, not at the polar heads.
B is incorrect because the bumps are proteins, not individual hopanoid molecules.
C is incorrect because peripheral proteins are associated with the surface and would be removed during the process or would not appear as embedded bumps.
- A novel prokaryotic organism is isolated from a hydrothermal vent. Analysis of its cytoplasmic membrane reveals lipids with branched, 20-carbon chains linked to a glycerol backbone. This structure is most characteristic of which of the following?
Answer: Isoprenoid units linked to glycerol by ether bonds.
Correct. The membranes of Archaea are chemically distinct from those of Bacteria and Eukarya. They contain branched chains built from isoprenoid units, not fatty acids. These chains are linked to glycerol by ether bonds, which are more stable to hydrolysis at high temperatures than the ester bonds found in bacterial and eukaryotic lipids. The hydrothermal vent environment strongly suggests the organism is an extremophile, likely an archaeon.
A describes the lipids of Bacteria and Eukarya.
C describes a component (sterols) primarily of eukaryotic membranes, though some bacteria have analogous hopanoids.
D describes peptidoglycan, the main component of the bacterial cell wall, not the cell membrane lipids.
- Some hyperthermophilic archaea that thrive in boiling water have membranes composed of a tetraether lipid monolayer. Compared to a typical bacterial diether lipid bilayer, the primary advantage of this monolayer structure is that it:
Answer: is inherently more resistant to thermal separation of the membrane leaflets.
Correct. A lipid bilayer is composed of two independent leaflets held together by non-covalent hydrophobic forces. At very high temperatures, thermal energy can be sufficient to cause these leaflets to pull apart, destroying the membrane. A tetraether monolayer consists of single molecules that span the entire width of the membrane. This covalent linkage from one side to the other prevents the leaflets from separating, providing superior thermal stability.
A is incorrect because a monolayer is generally more rigid, not more flexible, than a bilayer.
C is incorrect as protein density is not a direct function of the lipid structure (monolayer vs. bilayer).
D is incorrect because the type of lipid linkage (ether vs. ester) does not directly contribute to the magnitude of the proton motive force, which is generated by proton pumps.
- A bacterial transport protein moves substance S across the cytoplasmic membrane. The process exhibits Michaelis-Menten kinetics and is highly specific for S. Experiments show that the intracellular concentration of S can reach levels 100 times greater than the extracellular concentration. This transport process is best described as:
Answer: active transport, because the substance is moved against its concentration gradient.
Correct. The defining characteristic of active transport is the movement of a substance against its concentration gradient, which requires the expenditure of energy. The observation that the intracellular concentration of S is 100 times the extracellular concentration is conclusive evidence for active transport.
A is incorrect. Although facilitated diffusion uses specific proteins and shows saturation, it is a passive process that cannot move a substance against its concentration gradient. It can only equilibrate concentrations.
B is incorrect because simple diffusion does not involve a protein, does not show saturation kinetics, and cannot concentrate a substance.
D is incorrect. While group translocation is a form of active transport, there is no information to suggest that S is chemically modified. Active transport is the broader, more certain conclusion.
- A mesophilic bacterium is cultured at its optimal temperature of 37°C. The culture is then abruptly shifted to a lower temperature of 15°C. To maintain optimal membrane fluidity, which of the following compositional changes is most likely to be observed in the bacterial membrane over the next several generations?
Answer: An increase in the proportion of unsaturated fatty acids and a decrease in the average fatty acid chain length.
Correct. At lower temperatures, membranes tend to become more rigid. To counteract this and maintain fluidity, bacteria incorporate more unsaturated fatty acids, which have kinks that prevent tight packing, and shorter fatty acid chains, which have fewer van der Waals interactions. Both of these changes increase membrane fluidity.
A is incorrect because increasing saturated fatty acids would decrease fluidity, exacerbating the effect of the cold. Hopanoids buffer fluidity, so a change would be complex, but this combination is incorrect.
C is incorrect because these changes (decreasing unsaturation, increasing chain length) are adaptations to higher temperatures to decrease excessive fluidity.
D is incorrect because the primary adaptation to temperature involves lipid composition, not protein concentration, and increasing saturated fatty acids would be counterproductive.
- The ABC (ATP-Binding Cassette) transport system for maltose in E. coli uses a substrate-binding protein, a transmembrane channel, and a cytoplasmic nucleotide-binding domain (NBD). A mutation in the NBD that prevents it from hydrolyzing ATP, but still allows it to bind ATP, would most likely result in:
Answer: the transporter being locked in a conformation where maltose is bound but not translocated into the cell.
Correct. The transport cycle of an ABC importer involves the substrate-binding protein delivering the substrate to the transmembrane domain, followed by ATP binding to the NBDs. ATP binding causes a conformational change that exposes the substrate to the cytoplasm. ATP hydrolysis is then required to reset the transporter to its original conformation so it can bind another substrate molecule. If hydrolysis is blocked, the transporter becomes trapped in the post-transport, pre-hydrolysis state, unable to complete the cycle.
B is incorrect because inactivating the energy-providing step would not cause the transporter to work in reverse.
C is incorrect because ATP hydrolysis is required for the cycle to continue; without it, transport will stop after a single turnover.
D is incorrect because the initial binding of maltose to the periplasmic protein is independent of the events at the NBD.
- Gram-negative bacteria utilize siderophores to acquire iron. The transport of the iron-siderophore complex across the outer membrane is an energy-dependent process that directly involves:
Answer: a specific outer membrane receptor protein that uses energy transduced from the cytoplasmic membrane by the TonB system.
When you encounter questions about iron transport in Gram-negative bacteria, focus on the unique challenge these organisms face: their double membrane structure requires specialized transport mechanisms that can't rely solely on ATP hydrolysis at the outer membrane.
Iron-siderophore complexes are too large and polar to cross the outer membrane through normal porins or diffusion. Instead, they require specific outer membrane receptor proteins (like FhuA or FepA) that recognize and bind particular siderophore types. The critical insight is that the outer membrane lacks ATP, so energy must come from elsewhere. The TonB system solves this by transducing energy from the proton-motive force at the cytoplasmic membrane. TonB proteins span the periplasm and physically interact with outer membrane receptors, providing the energy needed for conformational changes that allow siderophore transport.
Answer D correctly identifies this TonB-dependent mechanism. Answer A is wrong because ABC transporters operate at the cytoplasmic membrane, not the outer membrane, and the outer membrane lacks ATP. Answer B fails because the outer membrane's lipid bilayer is impermeable to large, charged siderophore complexes regardless of concentration gradients. Answer C incorrectly suggests porins can become "energized" - porins are passive channels that allow small molecules through, but siderophore complexes require active transport through specialized receptors.
Remember: Gram-negative bacteria have compartmentalized their transport systems - the outer membrane uses TonB-dependent transport for large substrates, while ABC transporters function at the cytoplasmic membrane where ATP is available.
- Polymyxins are antibiotics effective against Gram-negative bacteria. Their mechanism involves binding to Lipid A and disrupting both the outer and inner membranes. This physical disruption primarily compromises the membrane's function in:
Answer: maintaining selective permeability, leading to leakage of cytoplasmic contents.
When analyzing antibiotic mechanisms, you need to distinguish between antibiotics that target specific biochemical processes versus those that cause direct physical damage to cellular structures.
Polymyxins work by binding to Lipid A, a component of lipopolysaccharide (LPS) in the outer membrane of Gram-negative bacteria. This binding disrupts both the outer and inner membranes physically, creating holes and compromising membrane integrity. When membranes lose their structural integrity, their primary function—maintaining selective permeability—is destroyed. This leads to uncontrolled leakage of essential cytoplasmic contents like ions, ATP, and small molecules, ultimately killing the bacterial cell. Answer A correctly identifies this mechanism.
Answer B is incorrect because while polymyxins do disrupt membranes, they don't specifically target ribosomes or protein synthesis machinery. The cell death results from membrane damage, not from blocking translation.
Answer C confuses polymyxins with quinolone antibiotics like ciprofloxacin, which specifically inhibit DNA gyrase. Polymyxins don't target DNA replication enzymes—they physically damage membranes.
Answer D describes the mechanism of β-lactam antibiotics (penicillins, cephalosporins) that inhibit peptidoglycan synthesis by blocking transpeptidation. This is completely unrelated to polymyxin's membrane-disrupting action.
Remember this pattern: when you see "membrane disruption" or "physical damage" in antibiotic questions, think about loss of selective permeability as the primary consequence. Polymyxins are membrane-active agents, not enzyme inhibitors, so their effects stem from structural damage rather than blocking specific biochemical pathways.
- Lipopolysaccharide (LPS) is a major component of the outer membrane of Gram-negative bacteria and consists of three parts: Lipid A, a core polysaccharide, and an O-antigen. A mutation that completely prevents the synthesis of Lipid A would be lethal to the cell primarily because Lipid A is essential for:
Answer: anchoring the entire LPS molecule into the outer membrane, which is critical for membrane integrity.
Correct. Lipid A is the hydrophobic anchor of the LPS molecule. Its fatty acid chains are embedded in the outer leaflet of the outer membrane, securing the entire structure in place. It is the most conserved part of LPS and is absolutely essential for the assembly and structural integrity of the outer membrane. Its absence leads to a non-viable cell.
A is incorrect because the endotoxic response is a consequence of Lipid A's presence during infection, not a requirement for the bacterium's own viability.
B is incorrect as other structures can serve as phage receptors, and this is not an essential function for viability.
D is incorrect because antigenic variability is provided by the O-antigen, not Lipid A. Furthermore, this function relates to virulence, not fundamental viability.
- A mutant strain of Escherichia coli is generated that lacks the gene for OmpF, a major non-specific porin protein. This mutant is then grown in a minimal medium containing a low concentration of glucose as the sole carbon source and a high concentration of a small, hydrophilic antibiotic. Which of the following outcomes is most probable compared to the wild-type strain?
Answer: The mutant will exhibit a slower growth rate than the wild-type and may show increased resistance to the antibiotic.
Correct. Porins like OmpF form channels in the outer membrane of Gram-negative bacteria, allowing passive diffusion of small, hydrophilic molecules like glucose and many antibiotics. Deleting a major porin will significantly slow down the influx of both the necessary nutrient (glucose) and the harmful substance (antibiotic). The reduced nutrient uptake will lead to a slower growth rate, while the reduced antibiotic uptake will confer a degree of resistance.
A is incorrect because the negative effect of reduced nutrient uptake will outweigh any benefit from reduced antibiotic entry, resulting in slower, not faster, growth.
B is incorrect because 'unable to grow' is too absolute; other, less efficient porins may exist, or very slow diffusion might occur, permitting slow growth.
D is incorrect because porins are transport channels, not the primary structural elements providing osmotic protection. That role belongs to the peptidoglycan cell wall.
- Valinomycin is an antibiotic that acts as a potassium-specific ionophore, inserting into membranes and allowing K⁺ to move freely across. Bacteria typically maintain a high internal K⁺ concentration and a negative-internal membrane potential. Treatment of a bacterial culture with valinomycin would directly cause:
Answer: an efflux of K⁺ down its concentration gradient, leading to membrane depolarization.
When you encounter questions about ionophores and membrane potential, focus on the direction of ion movement and how it affects the electrical gradient across the membrane.
Valinomycin creates pores that allow K⁺ to move freely across the membrane. Since bacteria maintain high internal K⁺ concentrations, K⁺ will flow down its concentration gradient from inside to outside the cell. Because K⁺ carries positive charge, this efflux removes positive charges from the cell interior, making it less negative (or more positive) relative to the outside - this is depolarization. Answer D correctly describes this process.
Answer A is backwards - positive charges are leaving the cell, not entering, and this causes depolarization, not hyperpolarization (which would make the interior more negative). Answer B confuses osmotic effects with immediate electrical effects. While K⁺ loss might eventually affect osmotic balance, the direct effect of valinomycin is electrical, not osmotic lysis. Answer C misunderstands how ATP synthesis works - the F₁F₀-ATPase is driven by proton gradients (H⁺), not potassium gradients. Additionally, dissipating ion gradients would decrease, not increase, the driving force for ATP synthesis.
Remember that ionophores disrupt normal ion gradients by making membranes permeable to specific ions. Always consider: which direction will the ion flow (down its gradient), what charge does it carry, and how will this movement affect membrane potential? This approach works for any ionophore question, whether dealing with K⁺, Na⁺, or other ions.
- The lac permease of E. coli is a lactose-H⁺ symporter that uses the proton motive force to actively transport lactose into the cell. If a chemical agent that acts as a protonophore (e.g., CCCP), which allows protons to freely diffuse across the membrane, is added to a culture, what is the immediate effect on lactose transport?
Answer: Lactose transport against a concentration gradient will halt, and accumulated lactose may exit the cell.
Correct. The lac permease relies on the proton motive force (PMF), which is an electrochemical gradient of protons across the membrane. A protonophore dissipates this gradient by creating a channel for protons, effectively short-circuiting the energy source. Without the PMF, the symporter cannot transport lactose against its concentration gradient. Furthermore, if the intracellular lactose concentration is high, the permease may facilitate the diffusion of lactose back out of the cell down its now-unopposed concentration gradient.
A is incorrect because dissipating the proton gradient eliminates the energy source, it does not provide more energy.
C is incorrect because transport proteins are specific to their energy source; a symporter cannot switch to using ATP.
D is incorrect because protonophores disrupt the energy gradient, not the protein structure itself. If the agent were removed and the cell could re-establish the PMF, transport could resume.
- A bacterial cell is suspended in a solution containing four different molecules at equal molar concentrations: glycerol (MW 92, polar), ethanol (MW 46, polar), O₂ (MW 32, nonpolar), and tryptophan (MW 204, polar). Assuming no specific transporters are involved, which molecule will have the highest initial rate of diffusion across the cytoplasmic membrane?
Answer: O₂, because it is small and has high lipid solubility.
Correct. The rate of simple diffusion across a lipid bilayer is determined primarily by two factors: size and polarity (lipid solubility). Small, nonpolar (hydrophobic) molecules pass through most easily. Oxygen is both very small and nonpolar, allowing it to rapidly diffuse across the membrane down its concentration gradient.
A is incorrect. While glycerol is a metabolite, its polarity hinders its passage through the hydrophobic membrane core. Its metabolic importance does not affect its rate of simple diffusion.
B is incorrect. Although ethanol is smaller than glycerol, its polarity still limits its diffusion rate compared to a nonpolar molecule like O₂.
D is incorrect. Tryptophan is large and polar, making its simple diffusion rate extremely low. Large molecules do not create pores; they are generally excluded by the membrane.
- In a fluid bacterial membrane, the rate of lateral diffusion of a phospholipid is many orders of magnitude faster than the rate of its transverse diffusion ('flip-flop'). This significant difference is primarily due to the:
Answer: high energy barrier required for the hydrophilic head group to traverse the hydrophobic membrane core.
Correct. Lateral diffusion involves a phospholipid moving within the plane of its own leaflet, which is an energetically favorable process. Transverse diffusion, or flip-flop, requires the polar, hydrophilic head group of the phospholipid to leave its aqueous environment and pass through the nonpolar, hydrophobic fatty acid core of the bilayer. This is an extremely energetically unfavorable event, creating a large activation energy barrier that makes the process very rare and slow without enzymatic assistance (from flippases/floppases).
B is incorrect because flippases catalyze transverse diffusion, not lateral diffusion. Lateral diffusion is a spontaneous physical process.
C is incorrect because phospholipids within a leaflet are held by non-covalent forces, not covalent bonds.
D is incorrect because the primary barrier is the hydrophobic nature of the membrane's interior, not the viscosity of the surrounding aqueous compartments.
- Hopanoids in bacterial membranes function analogously to sterols in eukaryotic membranes. A key function of these molecules is to act as membrane fluidity buffers. This means they:
Answer: restrain lipid movement at high temperatures and prevent tight packing at low temperatures.
Correct. Hopanoids and sterols act as fluidity buffers by having a dual effect depending on temperature. At high temperatures, their rigid ring structure intercalates between phospholipids and restrains their movement, preventing the membrane from becoming too fluid. At low temperatures, they prevent the fatty acid tails from packing too closely together and crystallizing, thus maintaining fluidity.
A describes the exact opposite of the buffering function.
C is incorrect. While lipid rafts or domains exist, the primary role of hopanoids is not to form large crystalline structures but to modulate the fluidity of the entire membrane.
D is incorrect because hopanoids do not form covalent bonds with phospholipids; they associate via non-covalent hydrophobic and van der Waals interactions.
- In the bacterial phosphotransferase system (PTS) for glucose transport, a cascade of proteins transfers a phosphate group to the incoming sugar. The ultimate source of the high-energy phosphate for this process is:
Answer: Phosphoenolpyruvate (PEP), which is converted to pyruvate during the initial phosphotransfer step.
Correct. The phosphotransferase system is a form of group translocation where the transported substance is chemically modified. The energy for this process is derived from the high-energy phosphate bond in phosphoenolpyruvate (PEP), a key intermediate in glycolysis. The phosphate from PEP is transferred to Enzyme I of the PTS, initiating a phosphorelay cascade that culminates in the phosphorylation of the incoming sugar.
A is incorrect because ATP is the energy source for ABC transporters, not the PTS.
B is incorrect because the PMF is used to power secondary active transport systems, such as symporters and antiporters.
D is incorrect because GTP is primarily used as an energy source in protein synthesis and signal transduction, not the PTS.
- A protoplast, created by enzymatically removing the cell wall from a Gram-positive bacterium, is transferred from an isotonic buffer to a hypotonic solution (distilled water). What is the most likely immediate outcome?
Answer: The protoplast will swell and lyse due to the unopposed influx of water.
Correct. The cytoplasm has a higher solute concentration than distilled water. Due to osmosis, water will move rapidly from the hypotonic solution into the protoplast. In a normal bacterium, the rigid cell wall would push back against the resulting turgor pressure, preventing the cell from expanding indefinitely. Without the cell wall, the flexible cytoplasmic membrane cannot withstand this pressure, and the cell will swell until it bursts (lyses).
A describes the outcome in a hypertonic solution.
B is incorrect because the cytoplasmic membrane is not strong enough to resist significant osmotic pressure.
D is incorrect because cell wall synthesis is a complex, relatively slow process that could not occur fast enough to prevent immediate osmotic lysis.
- A scientist performs a FRAP (Fluorescence Recovery After Photobleaching) experiment on two bacterial strains, A and B, grown at 37°C. Membrane lipids are fluorescently labeled. After bleaching a small spot, Strain A shows 95% recovery of fluorescence within 30 seconds, while Strain B shows only 50% recovery in the same time. What is the most valid conclusion regarding the membrane of Strain B compared to Strain A?
Answer: A significant fraction of the lipids in the membrane of Strain B are part of an immobile fraction.
When you encounter FRAP experiments in microbiology, you're examining membrane fluidity and lipid mobility. FRAP measures how quickly fluorescently-labeled molecules can move back into a bleached area—this recovery directly reflects molecular movement within the membrane.
Strain A's 95% fluorescence recovery indicates that nearly all lipid molecules can move freely and redistribute into the bleached spot. However, Strain B's limited 50% recovery suggests that a significant portion of its membrane lipids cannot move into the bleached area, creating what we call an "immobile fraction." This makes option A correct—roughly half of Strain B's lipids are essentially stuck in place, unable to participate in lateral diffusion.
Option B is backwards: short-chain, unsaturated fatty acids actually increase membrane fluidity and would promote faster recovery, not slower. Option C contradicts the experimental setup—both strains were grown at 37°C, and higher growth temperatures typically lead to more saturated fatty acids that would decrease fluidity anyway. Option D misses the mark because integral membrane proteins weren't being measured in this experiment; the fluorescent labels were specifically attached to lipids, not proteins.
The key insight is that FRAP recovery percentage directly correlates with the mobile fraction of labeled molecules. When recovery is incomplete, it indicates physical constraints preventing molecular movement—often due to lipid-protein interactions, membrane domains, or cytoskeletal attachments.
Remember: In FRAP experiments, incomplete fluorescence recovery always points to an immobile fraction, while recovery speed indicates how fast the mobile fraction moves.
- A researcher attempting to characterize a membrane-associated protein, ProtX, finds that it can be dissociated from the bacterial membrane by washing the cells with a high-salt buffer (1 M KCl). However, it cannot be removed by washing with a buffer of high pH. Based on these results, ProtX is most likely a(n):
Answer: peripheral protein, associated with the membrane via electrostatic interactions.
When you encounter questions about membrane protein characterization, focus on how different types of membrane associations respond to various extraction methods. The key insight here is that different proteins interact with membranes through distinct mechanisms, which determine how they can be removed.
The correct answer is D because peripheral proteins associate with membranes primarily through electrostatic interactions (ionic bonds) with membrane lipids or other membrane proteins. High-salt solutions like 1 M KCl contain abundant ions that compete for these electrostatic interactions, effectively disrupting the protein-membrane association and releasing the protein. However, high pH alone doesn't provide competing ions, so it cannot disrupt these electrostatic bonds, explaining why ProtX remains membrane-bound under high pH conditions.
Answer A is incorrect because integral proteins are embedded within the lipid bilayer through hydrophobic interactions, which are not disrupted by high-salt solutions. These proteins typically require detergents for removal. Answer B is wrong for the same reason – transmembrane proteins with alpha-helical domains are integral proteins that span the membrane and cannot be removed by salt washes alone. Answer C is incorrect because lipid-anchored proteins are covalently attached to membrane lipids. Covalent bonds are much stronger than electrostatic interactions and cannot be broken by salt solutions or pH changes alone.
Remember this pattern: if a membrane protein can be removed by high salt but not by pH changes alone, think peripheral protein with electrostatic associations. Salt competes for ionic interactions, while detergents are needed for hydrophobic interactions.
