Mini 3 Practice Questions
.JPG "Mini 3 Practice Questions - Quiz")
- 1.
An 8-year-old boy presents with a sore throat and fever. He was known to be allergic to penicillin, and had recently returned from travel in Europe. Physical examination reveals an erythematous pharynx, enlarged tonsils and a whitish exudate in the throat. Initial treatment with erythromycin was not successful, and antibiotic susceptibility testing showed that the organism was resistant to quinupristin, clindamycin and clarithromycin. What is the most likely mechanism of resistance being displayed by this isolate?
- A.
A. Point mutation in the 16S rRNA subunit
- B.
B. Acquired methyltransferase
- C.
C. Acquired transpeptidase
- D.
D. Point mutation in the gyrA gene
- E.
E. Altered outer membrane porin expression
Correct Answer
B. B. Acquired methyltransferaseExplanation
Study question: What other mechanisms of resistance could give a similar phenotype?
B. This Streptococcus pyogenes isolate is displaying MLSB resistance, which stands for macrolide, lincosamide and streptogramin B resistance. It is most likely caused by acquisition of an Erm methyltransferase, which methylates a base in the 23S rRNA gene. Interestingly, this type of resistance in this bacterial species is not very common in the US, but is more so in some European countries.Rate this question:
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- 2.
A 23-year-old woman presents with pain upon urination and increased urinary frequency. Urine dipstick analysis shows the presence of WBCs, RBCs, and nitrite. The causative agent is a Gram-negative, oxidase-negative rod that is resistant to the actions of trimethoprim-sulfamethoxazole and vancomycin. What is the most likely mechanism of glycopeptide resistance in this isolate?
- A.
A. D-ser or D-lac in the peptide tail of the peptidoglycan
- B.
B. Point mutation in penicillin-binding proteins (PBPs)
- C.
C. Presence of an outer membrane
- D.
D. Acquisition of an antibiotic acetyltransferase
- E.
E. Expression of a thickened cell wall
Correct Answer
C. C. Presence of an outer membraneExplanation
C. Vancomycin is unable to cross the outer membrane and is thus inactive against Gram-negative bacteria such as Escherichia. This is innate because the outer membrane is a fundamental part of the cell, it does not change it or express it differently in any way to resist the action of Vn.Rate this question:
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- 3.
A 16-year-old boy with cystic fibrosis presents with an exacerbation of his respiratory disease. Sputum cultures reveal the presence of a Gram-negative, oxidase-positive bacillus. Tobramycin treatment is initiated, but antibiotic susceptibility testing returned the next day reveals a tobramycin MIC by Etest of 128 mg/L (S≤4, I = 8, R ≥ 16 mg/L). What is the mechanism of resistance most likely responsible for this result?
- A.
A. Presence of a lipid bilayer membrane
- B.
B. Point mutation in the transglycosylase and/or transpeptidase enzymes
- C.
C. Acquired antibiotic adenyltransferase enzyme
- D.
D. Altered peptidoglycan structure
- E.
E. Point mutation in the DNA gyrase enzyme
Correct Answer
C. C. Acquired antibiotic adenyltransferase enzymeExplanation
Study question: What is the causative agent? How does the Etest work?
C. Tobramycin is an aminoglycoside, and covalent modification with a small molecule (in this case, adenylation) via an acquired transferase enzyme is an important mechanism of resistance to this class of antibiotic. The causative agent here is Pseudomonas aeruginosa—remember its importance to CF prognosis? How does this organism cause disease progression in these patients?Rate this question:
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- 4.
- A 19-year-old man presents with a large abscess on his left thigh. It is painful upon palpation, erythematous, and has pus draining from several openings. After incision and drainage, a sample of the pus is sent to the laboratory for identification and antibiotic susceptibility testing. A Gram-positive, catalase-positive, coagulase-positive coccus is isolated. The following MIC results are obtained by Etest:
- A.
A. Antibiotic efflux and antibiotic modification
- B.
B. Antibiotic modification and target point mutation
- C.
C. Antibiotic degradation and target enzyme replacement
- D.
D. Decreased permeability and target structure replacement
- E.
E. Antibiotic degradation and target structure replacement
Correct Answer
C. C. Antibiotic degradation and target enzyme replacementExplanation
C. S. aureus became resistant to penicillins by acquiring penicillinases, and then acquired the mecA gene to become MRSA. The intermediate Van resistance shown here is actually due to a mechanism distinct from that seen in VRSA. This mechanism is not terribly well understood, but involves a thickening of the cell wall layer and possibly reduced cross-linking. This means that there are more D-ala-D-ala tails present throughout the wall, which bind the drug as it penetrates and prevent it from getting to the cell surface, where the new monomers appear and are incorporated. We didn’t look at this particular mechanism in class, but I hope you realize that VRSA displays higher-level resistance (definitely in the R range, usually up at the higher end).Rate this question:
- 5.
A Gram-negative, oxidase-negative, lactose-fermenting, sorbitol non-fermenting bacillus is submitted for susceptibility testing. It is reported to be resistant to erythromycin, gentamycin and tetracycline. If only one resistance mechanism/event had occured in this strain, explaining all of the resistance phenotypes seen, which of the following could it have been (choose all that apply)?
- A.
A. Antibiotic destruction
- B.
B. Decreased permeability
- C.
C. Mutation of DNA gyrase
- D.
D. Increased efflux
- E.
E. Replacement of penicillin-binding proteins
- F.
F. Antibiotic modification
Correct Answer(s)
B. B. Decreased permeability
D. D. Increased effluxExplanation
B and D. These are both possible—E. coli is Gram-negative, so can modify outer membrane proteins to decrease permeability, and also expresses efflux pumps which can be quite non-specific in terms of what they pump out. The other mechanisms would likely be limited to a single or related classes of antibiotic, and/or don’t fit with these antibiotics.Rate this question:
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- 6.
A 6-year-old girl presents with a severe sore throat and low fever. Her history is remarkable for a complete lack of immunization. Physical examination reveals cervical lymphadenopathy and an adherent membrane in the pharynx. Antitoxin is administered, and susceptibility testing reveals that the isolate is resistant to erythromycin. Which resistance mechanisms could have led to this finding?
- A.
A. Acquired efflux pump / ribosome mutation
- B.
B. Transpeptidase mutation / peptidyltransferase mutation
- C.
C. Acquired efflux pump / outer membrane porin mutation
- D.
D. Altered peptidoglycan structure / 16S rRNA mutation
- E.
E. Topoisomerase mutation / acquisition of erm gene
Correct Answer
A. A. Acquired efflux pump / ribosome mutationExplanation
Study question: How does the exotoxin produced by this organism work? How is this organism cultured? What are its characteristics?
A. The other choices all have a component referring to a different target or require a structure not present in this organism.Rate this question:
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- 7.
In a monomicrobial infection, successive isolates can be used to monitor the progress of therapy. Occasionally, antibiotic susceptibility levels can be seen to change over time in these series of isolates, particularly in response to antibiotic treatment. Why might this happen?
- A.
A. The patient becomes desensitized to the drug, thereby allowing the bacteria to survive
- B.
B. The antibiotic becomes inactivated over time, allowing the bacteria to survive
- C.
C. Bacterial cells with random mutations providing resistance are selected for by treatment
- D.
D. The bacteria acquire resistance genes over time, allowing them to survive
- E.
E. Bacteria become less tolerant of the antibiotic, and start to develop resistance mechanisms
Correct Answer
C. C. Bacterial cells with random mutations providing resistance are selected for by treatmentExplanation
C. The important concept here is selective pressure. This could mean selecting for mutations or acquired elements; however, in an infection caused by a single organism, there is no source of acquired DNA.Rate this question:
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- 8.
Many efforts are being made to reduce antibiotic resistance rates. One approach is to reduce selective pressure. Which of the following methods makes use of this approach? i)Antibiotic cycling ii)Use of probiotics iii)Reducing agricultural use iv)Appropriate prescribing v)Targeting non-essential bacterial processes vi)Ensuring patient compliance
- A.
A. i, iii, v
- B.
B. i, ii, iv, vi
- C.
C. ii, iv, v, vi
- D.
D. ii, iii, iv, v, vi
- E.
E. i, iii, iv, v, vi
- F.
F. All of the above
Correct Answer
F. F. All of the aboveExplanation
F. At the heart of it, the thing driving resistance is bacterial survival, and use of antibiotics is the pressure that selects for resistance. Anything that reduces antibiotic use (either totally, by preventing infections, or of specific drugs) or allows bacterial survival without disease (effectively making a pathogen a commensal) will reduce this pressure.Rate this question:
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- 9.
A 74-year-old woman living in a long-term care facility is being admitted to hospital for a surgery. Active surveillance screening cultures show that she is colonized with a Gram-positive, catalase-negative coccus that is resistant to the action of bile and hydrolyzes esculin to esculetin, producing a black precipitate in the presence of ferric citrate. The isolate is also resistant to vancomycin. What is the most likely mechanism of vancomycin resistance in this isolate?
- A.
A. Presence of a lipid bilayer membrane
- B.
B. Point mutation in the transglycosylase and/or transpeptidase enzymes
- C.
C. Acquired antibiotic acetyltransferase enzyme
- D.
D. Altered peptidoglycan structure
- E.
E. Decreased uptake of the antibiotic
- F.
F. Increased antibiotic efflux
Correct Answer
D. D. Altered peptidoglycan structureExplanation
D. The organism being described is Enterococcus, specifically vancomycin-resistant Enterococcus (VRE). The resistance mechanism in this case is acquisition of a van element, which encodes proteins resulting in a change in the peptidoglycan monomer structure. This altered structure is not efficiently bound by vancomycin.Rate this question:
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- 10.
In the case above, how was this resistance mechanism most likely acquired?
- A.
A. Point mutation
- B.
B. Specialized transduction
- C.
C. Generalized transduction
- D.
D. Transformation
- E.
E. Conjugation
Correct Answer
E. E. ConjugationExplanation
E. The van element is on a transposon, which is thought to have been present on a conjugative plasmid and acquired via conjugation. Most acquired antibiotic resistance is thought to have been via conjugation, although transformation is a significant route in naturally competent organisms. Make sure you can explain these processes!Rate this question:
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- 11.
A 22-year-old man presents to the emergency department with a high fever and disorientation. He has a large furuncle (skin abscess) on his buttock. Blood cultures are positive for a beta-hemolytic coccus that may produce a number of exotoxins, including a superantigen and a pore-forming toxin called alpha-toxin (alpha-hemolysin). Empiric treatment with nafcillin is unsuccessful, and molecular testing reveals the presence of the mecA gene. What does the product of this gene do?
- A.
A. Degrades a broad range of beta-lactamase antibiotics
- B.
B. Degrades penicillins
- C.
C. Creates and connects strands of peptidoglycan
- D.
D. Acetylates nafcillin
- E.
E. Incorporates D-serine into the peptidoglycan monomer
Correct Answer
C. C. Creates and connects strands of peptidoglycanExplanation
C. The mecA gene encodes PBP2’ (PBP2a), which is a transglycosylase/transpeptidase enzyme that is insensitive to penicillins and cephalosporins.Rate this question:
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- 12.
A patient is treated with doxycycline. After 48 hours, no improvement is seen, and culture and susceptibility results come back showing resistance to this antibiotic. What are characteristics of the most likely mechanism of resistance?
- A.
A. Transmembrane protein requiring energy
- B.
B. Outer membrane protein forming a passive pore
- C.
C. Hydrolytic enzyme
- D.
D. Enzyme forming a covalent bond between the antibiotic and a small molecule
- E.
E. Enzyme forming a covalent bond between the target and a small molecule
Correct Answer
A. A. Transmembrane protein requiring energyExplanation
A. The most common mechanism of tetracycline resistance (in the absence of any contravening evidence) is efflux. Efflux pumps are transmembrane proteins that require either energy from the proton motive force or ATP to move the antibiotic from a volume of low concentration (inside the cell) to high concentration (outside the cell). Other mechanisms definitely exist, such as that described in E or B, point mutations in ribosomal components, as well as a concept called ‘ribosome protection’ which we did not discuss in class but is, in fact, the 2nd most common mechanism. This involves proteins which bind the ribosome and prevent tetracycline binding.Rate this question:
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- 13.
A 38-year-old homeless man is being treated in the emergency department for a wound. The attending physician notes that the patient has a cough, which the patient claims to have had for “ages”. Chest xray shows a region of caseation, and acid-fast staining of sputum reveals red bacilli. Antibiotic susceptibility testing shows that the causative agent has a reduced susceptibility to isoniazid. How might this resistance have come about?
- A.
A. Acquisition of an isoniazid acetyltransferase enzyme
- B.
B. Point mutation in inhA or katG genes
- C.
C. Acquisition of a mex efflux pump
- D.
D. Production of mycolic acids
- E.
E. Acquisition of SCCmec transposon
Correct Answer
B. B. Point mutation in inhA or katG genesExplanation
B. All resistance in Mycobacterium tuberculosis seen to date is due to point mutation. In this case, it can occur either in the gene for the antibiotic target (inhA), or in the gene for a catalase that activates isoniazid to its bactericidal form (katG). What other class of pathogen is also restricted to point mutation to develop resistance?Rate this question:
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- 14.
What is a good mnemonic device to help you remember the various bacteria causing many problems in terms of resistant infections?
- A.
A. ESKAPET
- B.
B. HACEK
- C.
C. SMASHED
- D.
D. MRSA
- E.
E. VRE
Correct Answer
A. A. ESKAPETExplanation
A. Make sure you know what the organisms are, and realize that this doesn’t cover all of them
(far from it).
Enterococcus VRE
Staphylococcus- MRSA, MRSE, b-lactamases
Klebsiella - ESBL, carbapenems
Acinetobacter - ESBL, efflux, OMPs, etc.
Pseudomonas - ESBL, efflux, OMPs, etc.
ESBL - G- rods
Tuberculosis - MDR and XDR
Streptococcus pneumoniae and Neisseria gonorrhoeae are also problematic (see notes)Rate this question:
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- 15.
Which method of antibiotic susceptibility testing involves measuring diameters of zones of inhibition and interpreting them via standardized tables?
- A.
A. Etest
- B.
B. Broth microdilution
- C.
C. Agar microdilution
- D.
D. Kirby Bauer
- E.
E. MIC testing
Correct Answer
D. D. Kirby BauerExplanation
D. Kirby Bauer is also known as the disc diffusion method. Make sure you know what a zone of inhibition is, why it is caused, and how to interpret the measurement from this test. Etest uses zones of inhibition, but does not require measurement of diameters (the MIC is right on the strip).Rate this question:
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- 16.
What element is NOT important in determination of susceptibility breakpoints by the Clinical and Laboratory Standards Institute?
- A.
A. Pharmacokinetics
- B.
B. Clinical outcomes
- C.
C. MIC distributions
- D.
D. Type of infection
- E.
E. Minimum bactericidal concentrations
Correct Answer
E. E. Minimum bactericidal concentrationsExplanation
E. Breakpoints for antibiotic susceptibility testing are based on MIC, not MBC values. All of the other choices are used to help make in vitro breakpoints as predictive as possible for clinical use.Rate this question:
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- 17.
What is shown in the image?
- A.
A. MIC interpretation table
- B.
B. Susceptibility table
- C.
C. Antibiogram
- D.
D. Treatment table
- E.
E. Research result
Correct Answer
C. C. AntibiogramExplanation
C. Antibiograms compile AST data to allow local resistance trends to be seen and increase the odds of picking an effective empiric (i.e. immediate, before getting specific AST results) treatment.Rate this question:
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- 18.
A 54-year-old man presents with endocarditis. The causative agent is a Gram positive coccus that is catalase-negative and alpha-hemolytic. Based only on this image, why would a competent physician be unlikely to prescribe a macrolide for treatment?
- A.
A. 63% of similar isolates are resistant to a member of this antibiotic class
- B.
B. Only 84% of similar isolates are susceptible to a member of this antibiotic class
- C.
C. 9% of similar isolates are resistant to a member of this antibiotic class
- D.
D. 37% of similar isolates are resistant to a member of this antibiotic class
- E.
E. Only 87% of similar isolates are susceptible to a member of this antibiotic class
Correct Answer
D. D. 37% of similar isolates are resistant to a member of this antibiotic classExplanation
D. The antibiogram tells us that 63% of viridans streptococci are susceptible to erythromycin. This is a highly integrated question, so good if you got it on the first try!Rate this question:
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- 19.
What would be the change in in vitro growth characteristics of an organism that became newly resistant to a bacteriostatic antibiotic?
- A.
Blue to Red
- B.
Blue to Green
- C.
Green to Red
- D.
Red to Green
- E.
Green to Blue
- F.
Red to Blue
Correct Answer
C. Green to RedExplanation
Bactericidal drugs kill bacteria, giving the blue line.
Bacteriostatic antibiotics stop growth without killing, giving the green line.
Resistance allows growth to continue in the presence of antibiotic, giving the red line.
Tolerance would cause a bactericidal drug to act in a bacteriostatic way, giving the green line.Rate this question:
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- 20.
What is the most likely way in which an organism would rapidly become resistant to multiple antibiotic classes with different targets?
- A.
Acquisition of plasmid by transformation
- B.
Point mutations
- C.
Acquisition of plasmid by conjugation
- D.
Acquisition of chromosomal genes by conjugation
- E.
Acquisition of genes by transduction
Correct Answer
C. Acquisition of plasmid by conjugationExplanation
Conjugation is the most common mechanism of resistance due to acquired means. HFr strains (i.e. able to conjugate chromosomal genes) are much less common than F+ strains. Transformation is limited to naturally competent species, so this could be a good answer if, for example, Streptococcus or Neisseria is implied. Transduction is not a common mechanism for resistance gene transfer. Point mutation is a very common mechanism of resistance, but is a rare event and thus unlikely to occur in multiple, unrelated genes at the same time.Rate this question:
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- 21.
Escherichia coli strains X and Y are both resistant to ampicillin. Ampicillin resistance is stable in strain X when it is grown for multiple generations in the absence of the antibiotic. However, strain Y loses ampicillin resistance when it is grown in media without the antibiotic. Which of the following best explains the acquisition of ampicillin susceptibility in strain Y?
- A.
Downregulation of the resistance gene
- B.
Insertion of a transposon into the resistance gene
- C.
Loss of a plasmid carrying the resistance gene
- D.
Point mutations in the resistance gene
- E.
Recombination with a defective copy of the resistance gene
Correct Answer
C. Loss of a plasmid carrying the resistance geneExplanation
All of these are possible to some extent. Downregulation could occur via a mutation in a promoter region gene for a regulatory protein, which would be in a similar range of ‘likelihood’ as mutations within the resistance gene. Transposons moving into a resistance gene would reverse resistance, but again, is unlikely. Recombination implies a source of a defective resistance gene, and recombination is also not likely. The MOST likely option is loss of plasmid. Since plasmids do not partition equally upon cell division, daughter cells with no plasmid do arise. In the absence of selective pressure, these plasmid-free cells will replicate better than those carrying plasmid (less DNA and proteins to make, more energy to devote to growth), and will become the predominant/only cell type in the population over successive generations.Rate this question:
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- 22.
A 53-year-old man develops a cough and fever. The causative agent is able to reduce the uptake of various antibiotics. Of the following, what is the genus of the most likely causative agent?
- A.
Staphylococcus
- B.
Listeria
- C.
Pseudomonas
- D.
Streptococcus
- E.
Mycoplasma
- F.
Pneumocystis
Correct Answer
C. PseudomonasExplanation
List two organisms that have relatively impermeable outer membranes. This is one, Acinetobacter is the other one.
Reduced uptake requires an outer membrane (where the porins live). The only organism with an outer membrane on this list is Pseudomonas. It and Acinetobacter have many porins to choose from, which allows them to make their outer membranes more selective than many other bacteria.Rate this question:
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- 23.
Gram-positive cocci are seen in purulent fluid drained from a skin abscess. A rapid agglutination test using latex beads coated with fibrinogen shows clumping. To which of the following antibiotics is this organism most likely to display resistance?
- A.
Mupirocin
- B.
Vancomycin
- C.
Nafcillin
- D.
Linezolid
- E.
Daptomycin
Correct Answer
C. NafcillinExplanation
The organism here is Staphylococcus. Most staphylococci are resistant to penicillins via penicillinase. MRSA/MRSE are also resistant to nafcillin/oxacillin/methicillin, and comprise the majority of clinical staphylococcal isolates in the US. Relatively few staphylococcal isolates are resistant to the other antibiotics in this list (at the moment…).Rate this question:
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- 24.
What is the general resistance mechanism used in the previous example?
- A.
Penicillinase
- B.
ESBL
- C.
Ribosome methylation
- D.
Production of altered PG monomer
- E.
Point mutation in the PBP
- F.
Acquisition of alternate PBP
Correct Answer
F. Acquisition of alternate PBPExplanation
MRSA has acquired the PBP2a gene, which encodes a transglycosylase/transpeptidase enzyme that is insensitive the meth/naf/oxa. Most staphylococci do express penicillinases, but these will not degrade the penicillinase-resistant penicillins. Point mutation in the PBP could give this type of resistance, but it is not commonly seen. ESBLs (Extended Spectrum Beta Lactamases) can also degrade these drugs, but are not found in staphylococci. Altered PG monomer is the mechanism for vancomycin resistance (see further down in the lecture), and ribosome methylation would not provide beta-lactam resistance (wrong target).Rate this question:
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- 25.
Which of the following scenarios is most likely due to gene acquisition?
- A.
Acyclovir-resistant HSV-1
- B.
Vn-resistant E. coli
- C.
Isoniazid-resistant TB
- D.
Aminoglycoside-resistant Enterococcus
Correct Answer
D. Aminoglycoside-resistant EnterococcusExplanation
TB and viral resistance are always due to point mutations. Escherichia, being a Gram-negative, is intrinsically resistant to vancomycin (the drug is large and hydrophobic, and is unable to cross the outer membrane). Aminoglycoside resistance is due to a transferase enzyme (of some sort), which must be acquired from somewhere.Rate this question:
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.JPG "A 54-year-old man presents with endocarditis. The causative agent is a Gram positive coccus that is catalase-negative and alpha-hemolytic. Based only on this image, why would a competent physician be unlikely to prescribe a macrolide for treatment?")
.JPG "What would be the change in in vitro growth characteristics of an organism that became newly resistant to a bacteriostatic antibiotic?")
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Mar 20, 2023Quiz Edited by
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Aug 03, 2012Quiz Created by
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