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ICAR AICE-JRF/SRF General Veterinary Microbiology: practice questions

ICAR AICE-JRF/SRF (PhD) · India · NTA for ICAR

Original practice questions written to the published syllabus · not past papers
10worked here, free
33written to this exam in this subject
10syllabus units

About ICAR AICE-JRF/SRF

ICAR AICE-JRF/SRF (PhD) is a PhD, fellowship and lectureship exam run by NTA for ICAR, for candidates in India. Computer-based, 2 hours, 120 questions: Part A 20 (general agriculture, animal husbandry, current events), Part B 50 (core group), Part C 50 (specialised subject). +4 / -1, maximum 480.

Source: icar.org.in.

Checked against the exam body's published information. Formats and fees change; confirm the current edition there before you book.

General Veterinary Microbiology in ICAR AICE-JRF/SRF

Molecular taxonomy (16S rRNA, ANI, MALDI-TOF), pathogenicity islands and secretion systems, lysogenic conversion and plasmid-borne virulence, integrons and transposons in resistance spread, sterilisation kinetics (D, z and F values), MIC, MBC, ECOFF and breakpoints, and quantitative problems in growth and counting.

How bacteria and fungi are built, grown, identified and killed, and how bacteria become resistant to antimicrobials.

Our 33 ICAR AICE-JRF/SRF questions in this subject: 0 easy, 8 medium, 25 hard, across 10 units.

10 practice questions, worked

Choose an answer, then open the working. Each explanation says why the right answer is right and why the others are not.

These 10 were written to ICAR AICE-JRF/SRF and are not shown on any other exam's page.

Question 1 · hard

Gram-negative bacteria treated with lysozyme and EDTA in an isotonic sucrose solution become spherical but keep remnants of the outer membrane. Such forms are called:

  1. Endospores
  2. Mesosomes
  3. Spheroplasts
  4. Protoplasts
Answer and working

C. Spheroplasts

Removing peptidoglycan from Gram-negative cells leaves the outer membrane partly attached, giving spheroplasts. Gram-positive cells treated the same way lose all wall material and become protoplasts. Both lyse in hypotonic solution. Mesosomes are now regarded as fixation artefacts.

Question 2 · hard

In genome-based bacterial taxonomy, two isolates are usually regarded as the same species when their average nucleotide identity (ANI) is at least about:

  1. 99.9%
  2. 70%
  3. 95-96%
  4. 80%
Answer and working

C. 95-96%

An ANI of about 95-96% corresponds to the classical species boundary of about 70% DNA-DNA hybridisation. The 70% figure is the tempting wrong answer: it is the DDH threshold, not the ANI threshold. 99.9% would split most species into many.

Question 3 · hard

A food-poisoning organism with a generation time of 30 minutes is left in warm cooked food with an initial count of 10^2 CFU/g. Assuming no lag and unlimited exponential growth, about how long will it take to reach 10^8 CFU/g?

  1. About 10 hours
  2. About 7 hours
  3. About 3 hours
  4. About 20 hours
Answer and working

A. About 10 hours

n = 3.32 x (8 - 2) = 19.9 generations (exactly log2 10^6 = 19.93). Time = n x g = 19.93 x 30 min = 598 min = about 10.0 h. The common slip is to treat each log rise as one generation (6 x 30 min = 3 h); another is to confuse generations with hours (20 h); using ln instead of log2 (13.8 generations) gives about 7 h.

Question 4 · hard

In Bacillus species, sporulation is typically triggered when:

  1. The culture is exposed to 121 degrees C
  2. Cells enter the lag phase after inoculation into fresh medium rich in amino acids and sugars
  3. Oxygen tension rises above atmospheric levels
  4. Cells leave exponential growth and enter stationary phase as nutrients run out
Answer and working

D. Cells leave exponential growth and enter stationary phase as nutrients run out

Endospore formation is a starvation response that begins as cultures enter stationary phase; it is not triggered by fresh medium (which favours germination) or by heat, which spores resist rather than form in response to.

Question 5 · medium

The genes for botulinum neurotoxin types C and D, and for Shiga toxins of Escherichia coli, are carried on bacteriophages. The acquisition of a new property by a bacterium through a prophage is called:

  1. Phase variation
  2. Transposition
  3. Lysogenic conversion
  4. Transformation (uptake of free DNA)
Answer and working

C. Lysogenic conversion

Lysogenic conversion occurs when a temperate phage integrated as a prophage carries genes that change the host's phenotype, often virulence genes. Losing the prophage can mean losing toxin production.

Question 6 · hard

The Sterne strain used in live anthrax spore vaccines is toxigenic but non-capsulated. Genetically this is because it:

  1. Has a chromosomal mutation that inactivates lethal factor but leaves oedema factor and the capsule
  2. Lacks both pXO1 and pXO2
  3. Carries plasmid pXO1 (toxin genes) but lacks plasmid pXO2 (capsule genes)
  4. Carries pXO2 (capsule genes) but lacks pXO1 (toxin genes)
Answer and working

C. Carries plasmid pXO1 (toxin genes) but lacks plasmid pXO2 (capsule genes)

Full virulence of Bacillus anthracis needs both plasmids: pXO1 encodes protective antigen, lethal factor and oedema factor; pXO2 encodes the poly-D-glutamic acid capsule. The Sterne strain lacks pXO2, so it still induces antibody to protective antigen but is attenuated.

Question 7 · hard

Class 1 integrons, common in Gram-negative bacteria from animals, are important in resistance spread because they:

  1. Use an integrase to capture gene cassettes that are then expressed from a shared promoter
  2. Encode a sex pilus for cell-to-cell transfer
  3. Replicate independently of the chromosome as small circular DNA that carries its own origin
  4. Are bacteriophages that inject resistance genes into every cell they infect in the gut
Answer and working

A. Use an integrase to capture gene cassettes that are then expressed from a shared promoter

An integron has an integrase gene (intI), a recombination site (attI) and a promoter; resistance gene cassettes are captured at attI and expressed together. Integrons are not mobile by themselves; they move on transposons and plasmids. Class 1 integrons often carry sul1 in the 3' conserved segment.

Question 8 · medium

In a heat-resistance test at 121 degrees C, the number of viable spores falls from 10^7 to 10^3 in 8 minutes. The D-value is:

  1. 0.5 minutes
  2. 2 minutes
  3. 4 minutes
  4. 8 minutes
Answer and working

B. 2 minutes

D = time for a 1-log (90%) reduction = t / (log N0 - log N) = 8 / (7 - 3) = 8/4 = 2 min. 0.5 min comes from inverting the ratio (4/8); 8 min is the total time, not per log.

Question 9 · hard

In a broth MBC test the starting inoculum is 5 x 10^5 CFU/mL. The MBC is defined as the lowest concentration that kills at least 99.9% of this inoculum. What is the highest surviving count allowed at the MBC?

  1. 5,000 CFU/mL
  2. 50,000 CFU/mL
  3. 500 CFU/mL
  4. 50 CFU/mL
Answer and working

C. 500 CFU/mL

99.9% kill leaves 0.1% = 0.001 of the inoculum: 5 x 10^5 x 0.001 = 500 CFU/mL (a 3-log reduction). 5,000 would be 99% kill, 50,000 only 90%, and 50 is 99.99%.

Question 10 · hard

What is the main difference between an epidemiological cut-off value (ECOFF) and a clinical breakpoint?

  1. An ECOFF is always higher than the clinical breakpoint, because it includes an extra safety margin for dosing
  2. An ECOFF is used only for fungi
  3. An ECOFF applies to disk diffusion and a breakpoint only to MIC tests
  4. An ECOFF separates wild-type from acquired-resistance isolates; a breakpoint predicts treatment outcome
Answer and working

D. An ECOFF separates wild-type from acquired-resistance isolates; a breakpoint predicts treatment outcome

ECOFFs are drawn from MIC distributions and are used in resistance surveillance. Clinical breakpoints also include pharmacokinetic and pharmacodynamic data, dose and outcome evidence, so they can be above, equal to or below the ECOFF.

A study plan for General Veterinary Microbiology in ICAR AICE-JRF/SRF

This plan follows where our questions actually fall, so the time goes where the marks are likely to be. It is our reading of the syllabus, not the exam body's weighting.

  1. First, the heavy units (70% of the 33 questions): Mechanisms of antimicrobial resistance (9, 27%); Bacterial classification (5, 15%); Sterilisation (5, 15%); Bacterial genetics (4, 12%).

    Read the manual chapters for these first and do every worked question you can find on them.

  2. Then the middle band: Bacterial growth (3, 9%); Bacterial identification (2, 6%); General mycology (2, 6%); Bacterial structure (1, 3%).

    Aim for one sitting per unit, then mixed questions across them.

  3. Then the lighter units: Culture methods (1, 3%); Disinfection (1, 3%).

    Learn the definitions, classifications and standard numbers; these units are more often recalled than worked.

  4. Leave time for the hard ones: Bacterial growth (3 hard of 3), Bacterial identification (2 hard of 2), Bacterial structure (1 hard of 1).

    Come back to these after a first pass, untimed, and read every explanation, including for the options you rejected.

  5. Finish timed. Sit the subject as a timed set without notes, mark it, and send each wrong answer back to its unit above.

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