Microbiology Quiz: Adhesion Invasion And Biofilms
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Adhesion Invasion And BiofilmsQuestion 1 of 20

Helicobacter pylori adheres to gastric epithelium via its BabA adhesin binding to the Lewis b (Leᵇ) blood group antigen. A non-pathogenic E. coli strain, which cannot normally colonize the stomach, is engineered to express BabA on its surface. This engineered strain is introduced into two groups of mice: wild-type mice with Leᵇ-positive gastric epithelia, and knockout mice whose gastric epithelia lack the Leᵇ antigen. What is the most probable outcome?

The engineered E. coli will colonize the stomach of the wild-type mice but will fail to colonize the stomach of the Leᵇ-knockout mice.
The engineered E. coli will colonize both types of mice equally, as it possesses other, non-specific adhesins that can compensate for the lack of Leᵇ.
The engineered E. coli will fail to colonize either mouse type because it lacks other essential H. pylori virulence factors, such as urease for acid survival.
The engineered E. coli will colonize the Leᵇ-knockout mice but not the wild-type mice, because BabA recognizes a cryptic receptor masked by Leᵇ.
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Microbiology Quiz: Adhesion Invasion And Biofilms

Practice Adhesion Invasion And Biofilms in Microbiology 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 Adhesion Invasion And Biofilms, giving you a quick way to practice the rules, question types, and explanations that matter most for Microbiology.

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Question 1

Helicobacter pylori adheres to gastric epithelium via its BabA adhesin binding to the Lewis b (Leᵇ) blood group antigen. A non-pathogenic E. coli strain, which cannot normally colonize the stomach, is engineered to express BabA on its surface. This engineered strain is introduced into two groups of mice: wild-type mice with Leᵇ-positive gastric epithelia, and knockout mice whose gastric epithelia lack the Leᵇ antigen. What is the most probable outcome?

  1. The engineered E. coli will colonize the stomach of the wild-type mice but will fail to colonize the stomach of the Leᵇ-knockout mice. (correct answer)
  2. The engineered E. coli will colonize both types of mice equally, as it possesses other, non-specific adhesins that can compensate for the lack of Leᵇ.
  3. The engineered E. coli will fail to colonize either mouse type because it lacks other essential H. pylori virulence factors, such as urease for acid survival.
  4. The engineered E. coli will colonize the Leᵇ-knockout mice but not the wild-type mice, because BabA recognizes a cryptic receptor masked by Leᵇ.
Explanation: When you encounter questions about bacterial adhesion and colonization, focus on the specific receptor-ligand interactions that enable pathogens to bind to host tissues. This question tests whether you understand that adhesion is often the critical first step in bacterial colonization. The BabA adhesin from H. pylori specifically binds to Lewis b (LebLe^b) blood group antigens on gastric epithelial cells. When you engineer E. coli to express BabA, you're giving it the ability to recognize and bind to Le^b antigens. In wild-type mice with Le^b-positive gastric epithelia, the engineered E. coli can now adhere to the stomach lining through this specific interaction. However, in knockout mice lacking Le^b antigens, there are no receptors for BabA to bind to, preventing colonization. Let's examine why the other options are incorrect: Option B assumes the engineered E. coli has other compensatory adhesins, but the question states it's a non-pathogenic strain that cannot normally colonize the stomach. Option C suggests urease is essential, but while urease helps H. pylori survive acid, the question focuses specifically on adhesion capability—many bacteria can survive gastric conditions temporarily. Option D proposes an inverse relationship that contradicts established BabA-Le^b binding specificity. Remember this principle: bacterial adhesion typically follows a lock-and-key model. When studying pathogenic mechanisms, always consider that specific adhesin-receptor pairs are usually required for successful colonization, and removing either component breaks the interaction.

Question 2

A strain of Staphylococcus aureus can form biofilms on both plastic catheters (abiotic surface) and on host heart valves during endocarditis (biotic surface). A mutant strain is constructed that specifically lacks fibronectin-binding proteins (FnBPs). What is the most likely phenotype of this FnBP mutant strain?

  1. It will be unable to form biofilms on either the catheter or the heart valve, as FnBPs are essential for all initial attachment.
  2. It will form robust biofilms on both surfaces, but the biofilm on the heart valve will lack the normal structure provided by cross-linked fibronectin.
  3. It will be able to form a robust biofilm on the heart valve by using alternative protein adhesins but will be unable to attach to the catheter.
  4. It will form a biofilm on the catheter similar to wild-type but will be significantly attenuated in its ability to form a biofilm on the heart valve. (correct answer)
Explanation: When analyzing bacterial biofilm formation, you need to distinguish between attachment mechanisms used on abiotic (non-living) versus biotic (living) surfaces. Staphylococcus aureus uses different strategies for these environments. Fibronectin-binding proteins (FnBPs) are crucial for attachment to host tissues because fibronectin is abundant in the extracellular matrix of tissues like heart valves. These proteins specifically recognize and bind to fibronectin, facilitating initial adhesion to biotic surfaces. However, plastic catheters lack fibronectin and other host proteins, so S. aureus relies on different adhesins for catheter colonization, such as proteins that bind directly to plastic polymers or form hydrophobic interactions. Answer D correctly identifies that the FnBP mutant will retain normal biofilm formation on catheters (since FnBPs aren't needed for plastic attachment) but will be significantly impaired on heart valves where fibronectin binding is critical for initial adhesion. Answer A incorrectly assumes FnBPs are essential for all biofilm formation, ignoring the different mechanisms used on abiotic surfaces. Answer B wrongly suggests the mutant can still form robust biofilms on heart valves—without FnBPs, initial attachment itself would be compromised, not just biofilm structure. Answer C reverses the actual phenotype, incorrectly stating the strain would fail on catheters but succeed on heart valves. Remember: bacterial pathogens typically use surface-specific adhesins. When evaluating adhesin mutants, consider what type of surface each adhesin targets and whether alternative attachment mechanisms exist for different environments.

Question 3

A mature biofilm of Staphylococcus epidermidis grown on a catheter is treated with different enzymes. Treatment with Dispersin B, an enzyme that specifically hydrolyzes poly-N-acetylglucosamine (PNAG/PIA), causes massive biofilm detachment. Treatment with a broad-spectrum protease causes partial detachment, but less than Dispersin B. Treatment with DNase I has a minimal effect on the biofilm's integrity. Based on these results, what is the primary structural component of the S. epidermidis biofilm in this context?

  1. Extracellular DNA (eDNA), which forms a mesh-like structure that is essential for trapping the bacterial cells and providing structural integrity.
  2. Surface-associated proteins, such as Bap, which are the main factors mediating both cell-cell and cell-surface interactions in the mature biofilm.
  3. A polysaccharide adhesin, specifically PNAG, which serves as the key intermolecular glue linking cells to each other and to the abiotic surface. (correct answer)
  4. A complex co-polymer of eDNA and proteins, as no single enzyme treatment was capable of completely eradicating the entire biofilm structure.
Explanation: The experiment is a functional dissection of the biofilm matrix. The profound effect of Dispersin B, which specifically targets PNAG, indicates that this polysaccharide is the dominant and essential structural scaffold of the biofilm. The partial effect of protease suggests that proteins play a role, but a secondary one compared to PNAG. The minimal effect of DNase I shows that eDNA is not a major structural component in this specific biofilm. Therefore, PNAG is the primary structural element.

Question 4

The invasion of epithelial cells by Listeria monocytogenes is a classic example of the zipper mechanism, mediated by the binding of bacterial Internalin A (InlA) to host E-cadherin. Which of the following experimental interventions would most specifically block this invasion pathway without directly killing the bacteria or host cells?

  1. Treating the culture with an inhibitor of Type III secretion system needle formation.
  2. Depleting host cell membrane cholesterol using methyl-β-cyclodextrin to disrupt lipid rafts.
  3. Adding a high concentration of soluble, recombinant E-cadherin extracellular domain to the medium. (correct answer)
  4. Pre-treating the host cells with a broad-spectrum inhibitor of matrix metalloproteinases (MMPs).
Explanation: The zipper mechanism relies on a high-affinity interaction between a bacterial surface protein (InlA) and a host cell receptor (E-cadherin). Adding a soluble form of the receptor's extracellular domain acts as a competitive inhibitor or decoy. The InlA on the bacterial surface will bind to the soluble E-cadherin in the medium instead of the E-cadherin on the host cell surface, thus specifically blocking the interaction required for invasion. T3SS inhibitors (A) are irrelevant as Listeria doesn't use this system. Cholesterol depletion (B) and MMP inhibition (D) target other cellular processes that are not the primary, specific interaction of this invasion mechanism.

Question 5

A researcher is testing two antibiotics against a mature Pseudomonas aeruginosa biofilm. Antibiotic X is a small, hydrophilic molecule that targets DNA gyrase. Antibiotic Y is a large, amphipathic molecule that disrupts the cell membrane. The minimum inhibitory concentration (MIC) for both antibiotics against planktonic bacteria is identical. However, against the biofilm, Antibiotic X shows significantly reduced efficacy compared to Antibiotic Y, though both are less effective than on planktonic cells. What is the most likely explanation for the differential efficacy observed?

  1. The dense, negatively charged exopolysaccharide (EPS) matrix more effectively hinders the diffusion of the small, hydrophilic Antibiotic X than the larger, amphipathic Antibiotic Y. (correct answer)
  2. Efflux pumps are highly upregulated in the biofilm and exhibit a greater binding affinity for Antibiotic X, leading to its rapid expulsion from the cell.
  3. Persister cells within the biofilm are metabolically dormant and thus inherently resistant to the bactericidal action of Antibiotic X, but not the membrane-disrupting action of Y.
  4. Quorum sensing signals within the mature biofilm specifically induce the expression of enzymes that catalytically degrade Antibiotic X but not Antibiotic Y.
Explanation: The biofilm's exopolysaccharide (EPS) matrix acts as a diffusion barrier. Its dense, viscous, and often charged nature can significantly impede the penetration of molecules. Small, hydrophilic molecules like Antibiotic X can be slowed by interactions with the hydrated, negatively charged polysaccharide components. In contrast, a larger amphipathic molecule like Antibiotic Y may be able to penetrate more effectively by partitioning into hydrophobic microdomains or interacting differently with the matrix. While other resistance mechanisms exist, the physical properties of the EPS provide the most direct explanation for differential penetration and efficacy based on antibiotic structure.

Question 6

A team investigates the invasion of an unknown bacterial pathogen into non-phagocytic epithelial cells. They observe that invasion is completely blocked by treating the host cells with cytochalasin D, an actin polymerization inhibitor. In a separate experiment, treating the host cells with a potent PI3-kinase inhibitor significantly enhances the rate of bacterial uptake. Based on these findings, which invasion mechanism is this pathogen most likely utilizing?

  1. A zipper mechanism, as it requires localized actin rearrangement directly at the site of bacterial attachment which is often mediated by PI3-kinase signaling.
  2. A trigger mechanism, as it involves massive, global cytoskeletal ruffling that is actin-dependent but can be negatively regulated by pathways like PI3K that promote membrane stability. (correct answer)
  3. A toxin-mediated pore formation mechanism that allows direct entry into the cytoplasm, bypassing the need for host cell signaling pathways like PI3K and actin.
  4. A mechanism dependent on microtubule rearrangement for trafficking, which is functionally independent of the actin cytoskeleton and PI3K signaling.
Explanation: The absolute requirement for actin polymerization (blocked by cytochalasin D) is common to both zipper and trigger mechanisms. The key differentiator here is the effect of PI3K inhibition. Zipper mechanisms (e.g., Listeria) often co-opt PI3K signaling to promote localized membrane engulfment. In contrast, trigger mechanisms (e.g., Salmonella) involve the injection of effectors that induce widespread membrane ruffling. Some host pathways, including PI3K signaling, can act to stabilize the membrane and counteract this ruffling. Inhibiting such a negative regulatory pathway would thus enhance uptake, as observed.

Question 7

Pathogen A can reversibly switch expression of its single pilin gene on and off via inversion of a DNA promoter element. Pathogen B maintains a single pilin expression locus but possesses multiple, distinct, silent pilin gene cassettes elsewhere in its chromosome. It can recombine different silent cassettes into the expression locus, producing pili with novel amino acid sequences. Which statement best describes these strategies?

  1. Both Pathogen A and Pathogen B are utilizing phase variation to control the presentation of their adhesins to the host.
  2. Pathogen A is utilizing antigenic variation by altering its pilin promoter, while Pathogen B is utilizing phase variation.
  3. Pathogen A is utilizing phase variation (an ON/OFF switch), while Pathogen B is utilizing antigenic variation (changing the protein's structure). (correct answer)
  4. Pathogen A's strategy is for adapting to different nutrient environments, while Pathogen B's strategy is purely for immune evasion.
Explanation: This question tests the precise definitions of two key immune evasion mechanisms related to adhesins. Phase variation is the all-or-nothing (ON/OFF) expression of a gene, as described for Pathogen A. Antigenic variation is the modification of the protein product itself, usually by expressing one of several different versions of the gene, as described for Pathogen B. This allows the pathogen to present a new antigenic face to a host that has developed an immune response to the previous version. While both can have roles in niche adaptation, their primary function in this context is evading host immunity.

Question 8

A scientist characterizes a novel adhesin in an pathogenic bacterium. Adhesion to host cells is specifically blocked by soluble heparin. Genetic analysis reveals the adhesin is secreted via a Type V (autotransporter) secretion system. High-resolution electron microscopy does not reveal any pilus-like appendages on the bacterial surface, even under conditions of high adhesin expression. These characteristics are most consistent with the adhesin being which of the following?

  1. A Type IV pilus, which is assembled from pilin subunits and often binds to a variety of host cell receptors.
  2. An afimbrial adhesin, which is a single, large protein that remains associated with the bacterial surface after secretion and binds to host receptors. (correct answer)
  3. A capsular polysaccharide that mediates non-specific, charge-based interactions with the host cell surface.
  4. The needle-complex component of a Type III secretion system that mediates the initial contact with the host cell.
Explanation: The combination of observations points strongly to an afimbrial adhesin. The lack of visible pili rules out fimbrial structures like Type IV pili. The involvement of an autotransporter is a classic secretion mechanism for large, single-protein adhesins like filamentous hemagglutinin (FHA) of Bordetella. The specific inhibition by heparin (a sulfated glycosaminoglycan) indicates a specific protein-carbohydrate interaction, not a non-specific polysaccharide capsule. While a T3SS needle makes contact, it is not typically described with these secretion and binding characteristics.

Question 9

Uropathogenic E. coli (UPEC) uses Type 1 pili to bind mannose residues on bladder epithelial cells. Pili expression is subject to phase variation, controlled by the inversion of a promoter-containing DNA element. A researcher compares two strains in a mouse model of UTI: Strain A is 'phase-locked ON' (constitutively expresses pili), and Strain B is 'phase-locked OFF' (cannot express pili). Which outcome is most likely?

  1. Strain B will be more virulent than Strain A because the absence of pili allows it to evade recognition by the host immune system from the outset.
  2. Both strains will be equally avirulent, as the dynamic switching of pili expression is the only way to establish a productive infection.
  3. Strain A will establish an initial infection effectively but will be cleared more rapidly by the host immune system than a wild-type strain, while Strain B will fail to establish an infection. (correct answer)
  4. Strain A will be hypervirulent because its constant piliation will maximize its ability to form intracellular bacterial communities (IBCs) shielded from immunity.
Explanation: This question requires integrating the roles of adhesion and immune evasion. Strain B cannot express pili, so it cannot perform the initial, critical step of adhering to the bladder epithelium; it will fail to infect. Strain A can adhere effectively and initiate infection. However, Type 1 pili are highly antigenic. A wild-type strain can switch pili expression off (phase vary) to evade the host immune response (e.g., neutrophils). Since Strain A cannot turn off its pili, it remains a constant target for the immune system and will be cleared more efficiently than a wild-type strain that can 'hide'.

Question 10

Pathogen A can reversibly switch expression of its single pilin gene on and off via inversion of a DNA promoter element. Pathogen B maintains a single pilin expression locus but possesses multiple, distinct, silent pilin gene cassettes elsewhere in its chromosome. It can recombine different silent cassettes into the expression locus, producing pili with novel amino acid sequences. Which statement best describes these strategies?

  1. Both Pathogen A and Pathogen B are utilizing phase variation to control the presentation of their adhesins to the host.
  2. Pathogen A is utilizing antigenic variation by altering its pilin promoter, while Pathogen B is utilizing phase variation.
  3. Pathogen A is utilizing phase variation (an ON/OFF switch), while Pathogen B is utilizing antigenic variation (changing the protein's structure). (correct answer)
  4. Pathogen A's strategy is for adapting to different nutrient environments, while Pathogen B's strategy is purely for immune evasion.
Explanation: This question tests the precise definitions of two key immune evasion mechanisms related to adhesins. Phase variation is the all-or-nothing (ON/OFF) expression of a gene, as described for Pathogen A. Antigenic variation is the modification of the protein product itself, usually by expressing one of several different versions of the gene, as described for Pathogen B. This allows the pathogen to present a new antigenic face to a host that has developed an immune response to the previous version. While both can have roles in niche adaptation, their primary function in this context is evading host immunity.

Question 11

Extracellular DNA (eDNA) is now recognized as a key component of the matrix of many bacterial biofilms, a shift from the older view that it was merely debris from dead cells. Which statement most accurately reflects the origin and function of eDNA in a biofilm?

  1. eDNA is primarily derived from lysed host immune cells and serves mainly as a source of carbon and nitrogen for the growing bacteria.
  2. eDNA is released by a subpopulation of bacteria through mechanisms like programmed cell lysis and acts as an electrostatic scaffold, promoting cell-cell adhesion. (correct answer)
  3. The presence of eDNA is largely an in vitro artifact and it does not play a significant structural role in the majority of biofilms formed in vivo.
  4. The primary role of eDNA is to facilitate horizontal gene transfer within the biofilm community; its structural contribution is a minor, secondary effect.
Explanation: Modern biofilm research has shown that eDNA is an active and crucial component of the matrix. It is typically of bacterial origin, released through regulated processes, not just random cell death. Its negatively charged phosphate backbone allows it to act as an electrostatic 'glue', cross-linking cells and other matrix components like proteins and polysaccharides, thereby contributing significantly to the biofilm's structural integrity, especially during the early stages of development. While it can also be a source of nutrients or DNA for HGT, its structural role is well-established and primary in many species.

Question 12

A microbiologist compares biofilm formation in wild-type Staphylococcus aureus, a mutant unable to produce autoinducing peptides (AIPs), and a mutant with a constitutively active AIP receptor (AgrC). The Agr quorum sensing system in S. aureus is known to promote expression of dispersal-related proteases at high cell density. Which of the following outcomes after 48 hours of growth is most consistent with this regulatory role?

  1. The AIP-deficient mutant will form the thickest biofilm, while the constitutively active mutant will be unable to initiate biofilm formation.
  2. The AIP-deficient mutant will form a thick, unstructured biofilm, while the constitutively active mutant will show robust initial attachment but then premature dispersal, resulting in a thin biofilm. (correct answer)
  3. Both the wild-type and the constitutively active mutant will form hyper-robust biofilms, while the AIP-deficient mutant will be unable to form a biofilm.
  4. The constitutively active mutant will form the thickest, most structured biofilm due to constant upregulation of adhesion factors controlled by the Agr system.
Explanation: The Agr system has a complex, biphasic role. While it can contribute to initial stages, its activation at high cell density (high AIP concentration) triggers the expression of proteases and other factors that actively break down the biofilm matrix and promote dispersal. Therefore, the AIP-deficient mutant, which never activates this dispersal program, becomes 'stuck' in the accumulation phase, forming a thick biofilm. Conversely, the constitutively active mutant will prematurely and continuously express dispersal factors, leading to detachment and a resulting thin biofilm.

Question 13

A mature bacterial biofilm is treated with a high concentration of a bactericidal antibiotic for 24 hours, killing >99.9% of the cells. The antibiotic is then removed, and the few surviving cells are transferred to fresh, antibiotic-free medium. After these cells grow to a high density, their minimum inhibitory concentration (MIC) for the same antibiotic is determined. What is the most likely result of this MIC test?

  1. The MIC will be significantly higher than that of the original strain, as the surviving cells were genetically resistant mutants selected by the antibiotic.
  2. The MIC will be slightly elevated due to stable epigenetic modifications induced by antibiotic stress, which are passed on to all subsequent progeny.
  3. The MIC cannot be determined because the surviving cells will have entered a viable but non-culturable (VBNC) state and will not grow in the fresh medium.
  4. The MIC will be identical to that of the original, antibiotic-naive strain, because the survivors were phenotypically tolerant persister cells, not resistant mutants. (correct answer)
Explanation: When you encounter questions about antibiotic treatment of biofilms followed by regrowth, you're dealing with the critical distinction between antibiotic resistance and tolerance. This distinction is fundamental to understanding bacterial survival strategies. The correct answer is D because the scenario describes classic persister cell behavior. Persister cells are a small subpopulation within biofilms that enter a dormant, metabolically inactive state, making them phenotypically tolerant to antibiotics without acquiring genetic resistance mutations. When the antibiotic is removed, these cells "wake up" and resume normal growth and division, producing offspring genetically identical to the original population. Since no genetic changes occurred, the MIC remains unchanged. Option A is incorrect because true genetic resistance would require specific mutations in target genes, efflux pumps, or enzymatic pathways. The >99.9% kill rate followed by normal regrowth patterns suggests tolerance, not resistance. Option B misrepresents bacterial epigenetics – while bacteria do have some epigenetic mechanisms, stable inheritance of antibiotic resistance through epigenetic modifications is not well-established or common. Option C confuses VBNC states with persister cells. VBNC cells have lost culturability under standard conditions, but the question states the survivors successfully grew to high density in fresh medium. Remember this key distinction: resistant bacteria can grow in the presence of antibiotics due to genetic changes, while tolerant persister cells survive antibiotic exposure through dormancy but remain genetically susceptible. Look for survival percentages and regrowth patterns to distinguish between these mechanisms on exams.

Question 14

A researcher is testing two antibiotics against a mature Pseudomonas aeruginosa biofilm. Antibiotic X is a small, hydrophilic molecule that targets DNA gyrase. Antibiotic Y is a large, amphipathic molecule that disrupts the cell membrane. The minimum inhibitory concentration (MIC) for both antibiotics against planktonic bacteria is identical. However, against the biofilm, Antibiotic X shows significantly reduced efficacy compared to Antibiotic Y, though both are less effective than on planktonic cells. What is the most likely explanation for the differential efficacy observed?

  1. The dense, negatively charged exopolysaccharide (EPS) matrix more effectively hinders the diffusion of the small, hydrophilic Antibiotic X than the larger, amphipathic Antibiotic Y. (correct answer)
  2. Efflux pumps are highly upregulated in the biofilm and exhibit a greater binding affinity for Antibiotic X, leading to its rapid expulsion from the cell.
  3. Persister cells within the biofilm are metabolically dormant and thus inherently resistant to the bactericidal action of Antibiotic X, but not the membrane-disrupting action of Y.
  4. Quorum sensing signals within the mature biofilm specifically induce the expression of enzymes that catalytically degrade Antibiotic X but not Antibiotic Y.
Explanation: The biofilm's exopolysaccharide (EPS) matrix acts as a diffusion barrier. Its dense, viscous, and often charged nature can significantly impede the penetration of molecules. Small, hydrophilic molecules like Antibiotic X can be slowed by interactions with the hydrated, negatively charged polysaccharide components. In contrast, a larger amphipathic molecule like Antibiotic Y may be able to penetrate more effectively by partitioning into hydrophobic microdomains or interacting differently with the matrix. While other resistance mechanisms exist, the physical properties of the EPS provide the most direct explanation for differential penetration and efficacy based on antibiotic structure.

Question 15

In Pseudomonas aeruginosa, the intracellular concentration of the second messenger cyclic-di-GMP (c-di-GMP) is a critical regulator of the switch between motile and sessile lifestyles. High c-di-GMP levels promote biofilm formation, while low levels promote motility and dispersal. A mutant strain is created with a deletion in a gene encoding a key phosphodiesterase (PDE), an enzyme that degrades c-di-GMP. How would this mutation most likely affect the biofilm life cycle?

  1. The mutant would be unable to initiate biofilm formation due to a defect in the initial surface sensing and attachment signaling cascade.
  2. The mutant would form a hyper-adherent, robust biofilm that is unable to undergo dispersal, even under nutrient-limiting conditions. (correct answer)
  3. The mutant would exhibit a hyper-motile phenotype, be unable to form microcolonies, and only form a thin, unstructured layer of cells.
  4. The mutant would form a morphologically normal biofilm, but the cells that eventually disperse would lack flagella and be non-motile.
Explanation: This is a two-step logic problem. Step 1: A phosphodiesterase (PDE) degrades c-di-GMP. Therefore, a PDE deletion mutant will be unable to break down c-di-GMP, leading to constitutively high intracellular levels of this second messenger. Step 2: High levels of c-di-GMP are the signal that promotes biofilm formation (e.g., by stimulating exopolysaccharide production) and represses motility (e.g., by inhibiting flagellar synthesis and function). Combining these, the mutant will be 'locked' in a pro-biofilm state, leading to a hyper-adherent biofilm that cannot execute the dispersal program, which requires low c-di-GMP.

Question 16

Helicobacter pylori adheres to gastric epithelium via its BabA adhesin binding to the Lewis b (Leᵇ) blood group antigen. A non-pathogenic E. coli strain, which cannot normally colonize the stomach, is engineered to express BabA on its surface. This engineered strain is introduced into two groups of mice: wild-type mice with Leᵇ-positive gastric epithelia, and knockout mice whose gastric epithelia lack the Leᵇ antigen. What is the most probable outcome?

  1. The engineered E. coli will colonize the stomach of the wild-type mice but will fail to colonize the stomach of the Leᵇ-knockout mice. (correct answer)
  2. The engineered E. coli will colonize both types of mice equally, as it possesses other, non-specific adhesins that can compensate for the lack of Leᵇ.
  3. The engineered E. coli will fail to colonize either mouse type because it lacks other essential H. pylori virulence factors, such as urease for acid survival.
  4. The engineered E. coli will colonize the Leᵇ-knockout mice but not the wild-type mice, because BabA recognizes a cryptic receptor masked by Leᵇ.
Explanation: When you encounter questions about bacterial adhesion and colonization, focus on the specific receptor-ligand interactions that enable pathogens to bind to host tissues. This question tests whether you understand that adhesion is often the critical first step in bacterial colonization. The BabA adhesin from H. pylori specifically binds to Lewis b (LebLe^b) blood group antigens on gastric epithelial cells. When you engineer E. coli to express BabA, you're giving it the ability to recognize and bind to Le^b antigens. In wild-type mice with Le^b-positive gastric epithelia, the engineered E. coli can now adhere to the stomach lining through this specific interaction. However, in knockout mice lacking Le^b antigens, there are no receptors for BabA to bind to, preventing colonization. Let's examine why the other options are incorrect: Option B assumes the engineered E. coli has other compensatory adhesins, but the question states it's a non-pathogenic strain that cannot normally colonize the stomach. Option C suggests urease is essential, but while urease helps H. pylori survive acid, the question focuses specifically on adhesion capability—many bacteria can survive gastric conditions temporarily. Option D proposes an inverse relationship that contradicts established BabA-Le^b binding specificity. Remember this principle: bacterial adhesion typically follows a lock-and-key model. When studying pathogenic mechanisms, always consider that specific adhesin-receptor pairs are usually required for successful colonization, and removing either component breaks the interaction.

Question 17

A medical device company is developing a new urinary catheter and comparing three materials for their ability to resist biofilm formation by Proteus mirabilis. Material A is standard medical-grade silicone. Material B is silicone coated with a highly hydrophilic, non-degradable polymer. Material C is silicone impregnated with silver nanoparticles. Which ranking of the materials, from LEAST biofilm formation to MOST biofilm formation, is most likely to be observed after 24 hours?

  1. A < B < C
  2. B < A < C
  3. C < B < A (correct answer)
  4. C < A < B
Explanation: This question assesses understanding of anti-biofilm surface strategies. Material C (silver nanoparticles) has an active antimicrobial effect, releasing silver ions that are toxic to bacteria, making it the most effective at preventing biofilm formation. Material B (hydrophilic coating) works by creating a tightly bound layer of water on the surface, which acts as a physical barrier preventing the initial hydrophobic interactions required for bacterial attachment. This makes it less prone to biofilm than standard silicone. Material A (standard silicone) is a relatively hydrophobic and passive surface, making it the most susceptible to biofilm formation among the three choices. Therefore, the order from least to most biofilm is C < B < A.

Question 18

Pseudomonas aeruginosa and Staphylococcus aureus are often co-isolated from chronic wound biofilms. P. aeruginosa produces the quorum sensing molecule N-(3-oxododecanoyl)-L-homoserine lactone (3-O-C12-HSL), which is known to interact with S. aureus regulatory networks. What is a known consequence of this interspecies signaling within a mixed biofilm?

  1. It acts as a universal chemoattractant, promoting the co-aggregation of S. aureus with P. aeruginosa to enhance initial biofilm formation.
  2. It is directly bactericidal to S. aureus, representing a form of chemical warfare that leads to the rapid dominance of P. aeruginosa.
  3. It induces the stringent response in S. aureus, causing the majority of the population to differentiate into dormant persister cells.
  4. It represses the S. aureus accessory gene regulator (Agr) system, leading to altered expression of virulence factors and changes in antibiotic susceptibility. (correct answer)
Explanation: When you encounter questions about interspecies communication in biofilms, focus on how quorum sensing molecules can cross species barriers and interfere with regulatory networks, often disrupting normal bacterial behavior rather than enhancing cooperation. The correct answer is D because 3-O-C12-HSL from P. aeruginosa is well-documented to interfere with S. aureus gene regulation. This molecule represses the Agr (accessory gene regulator) system in S. aureus, which normally controls virulence factor expression and biofilm development. When Agr is suppressed, S. aureus exhibits altered phenotypes including changes in toxin production, biofilm architecture, and antibiotic resistance patterns. This represents a form of regulatory interference rather than cooperation. Option A is incorrect because 3-O-C12-HSL doesn't function as a chemoattractant for initial biofilm formation between these species. While co-aggregation occurs, it's not mediated by this specific signaling pathway. Option B is wrong because 3-O-C12-HSL isn't directly bactericidal to S. aureus - it's a regulatory molecule that modulates gene expression, not a toxic compound. The interaction is more subtle than chemical warfare. Option C misrepresents the mechanism because while 3-O-C12-HSL affects S. aureus metabolism, it doesn't specifically trigger the stringent response or widespread persister cell formation. Remember that interspecies quorum sensing often involves regulatory interference rather than cooperation. When studying mixed biofilms, focus on how signaling molecules from one species can hijack or disrupt the regulatory systems of another, leading to altered virulence and survival strategies.

Question 19

A strain of Staphylococcus aureus can form biofilms on both plastic catheters (abiotic surface) and on host heart valves during endocarditis (biotic surface). A mutant strain is constructed that specifically lacks fibronectin-binding proteins (FnBPs). What is the most likely phenotype of this FnBP mutant strain?

  1. It will be unable to form biofilms on either the catheter or the heart valve, as FnBPs are essential for all initial attachment.
  2. It will form robust biofilms on both surfaces, but the biofilm on the heart valve will lack the normal structure provided by cross-linked fibronectin.
  3. It will be able to form a robust biofilm on the heart valve by using alternative protein adhesins but will be unable to attach to the catheter.
  4. It will form a biofilm on the catheter similar to wild-type but will be significantly attenuated in its ability to form a biofilm on the heart valve. (correct answer)
Explanation: When analyzing bacterial biofilm formation, you need to distinguish between attachment mechanisms used on abiotic (non-living) versus biotic (living) surfaces. Staphylococcus aureus uses different strategies for these environments. Fibronectin-binding proteins (FnBPs) are crucial for attachment to host tissues because fibronectin is abundant in the extracellular matrix of tissues like heart valves. These proteins specifically recognize and bind to fibronectin, facilitating initial adhesion to biotic surfaces. However, plastic catheters lack fibronectin and other host proteins, so S. aureus relies on different adhesins for catheter colonization, such as proteins that bind directly to plastic polymers or form hydrophobic interactions. Answer D correctly identifies that the FnBP mutant will retain normal biofilm formation on catheters (since FnBPs aren't needed for plastic attachment) but will be significantly impaired on heart valves where fibronectin binding is critical for initial adhesion. Answer A incorrectly assumes FnBPs are essential for all biofilm formation, ignoring the different mechanisms used on abiotic surfaces. Answer B wrongly suggests the mutant can still form robust biofilms on heart valves—without FnBPs, initial attachment itself would be compromised, not just biofilm structure. Answer C reverses the actual phenotype, incorrectly stating the strain would fail on catheters but succeed on heart valves. Remember: bacterial pathogens typically use surface-specific adhesins. When evaluating adhesin mutants, consider what type of surface each adhesin targets and whether alternative attachment mechanisms exist for different environments.

Question 20

Pseudomonas aeruginosa and Staphylococcus aureus are often co-isolated from chronic wound biofilms. P. aeruginosa produces the quorum sensing molecule N-(3-oxododecanoyl)-L-homoserine lactone (3-O-C12-HSL), which is known to interact with S. aureus regulatory networks. What is a known consequence of this interspecies signaling within a mixed biofilm?

  1. It acts as a universal chemoattractant, promoting the co-aggregation of S. aureus with P. aeruginosa to enhance initial biofilm formation.
  2. It is directly bactericidal to S. aureus, representing a form of chemical warfare that leads to the rapid dominance of P. aeruginosa.
  3. It induces the stringent response in S. aureus, causing the majority of the population to differentiate into dormant persister cells.
  4. It represses the S. aureus accessory gene regulator (Agr) system, leading to altered expression of virulence factors and changes in antibiotic susceptibility. (correct answer)
Explanation: When you encounter questions about interspecies communication in biofilms, focus on how quorum sensing molecules can cross species barriers and interfere with regulatory networks, often disrupting normal bacterial behavior rather than enhancing cooperation. The correct answer is D because 3-O-C12-HSL from P. aeruginosa is well-documented to interfere with S. aureus gene regulation. This molecule represses the Agr (accessory gene regulator) system in S. aureus, which normally controls virulence factor expression and biofilm development. When Agr is suppressed, S. aureus exhibits altered phenotypes including changes in toxin production, biofilm architecture, and antibiotic resistance patterns. This represents a form of regulatory interference rather than cooperation. Option A is incorrect because 3-O-C12-HSL doesn't function as a chemoattractant for initial biofilm formation between these species. While co-aggregation occurs, it's not mediated by this specific signaling pathway. Option B is wrong because 3-O-C12-HSL isn't directly bactericidal to S. aureus - it's a regulatory molecule that modulates gene expression, not a toxic compound. The interaction is more subtle than chemical warfare. Option C misrepresents the mechanism because while 3-O-C12-HSL affects S. aureus metabolism, it doesn't specifically trigger the stringent response or widespread persister cell formation. Remember that interspecies quorum sensing often involves regulatory interference rather than cooperation. When studying mixed biofilms, focus on how signaling molecules from one species can hijack or disrupt the regulatory systems of another, leading to altered virulence and survival strategies.