Microbiology - MCQ Practice Questions

Understanding:

We need to identify which of the given viruses has a negative-sense (antisense) single-stranded RNA genome.

Step 1: Classify each virus by genome type.

Poliovirus belongs to the family Picornaviridae and has a positive-sense single-stranded RNA genome (+ssRNA). It can directly serve as mRNA.

Step 2: Evaluate Influenza virus.

Influenza virus belongs to the family Orthomyxoviridae and possesses a negative-sense single-stranded RNA genome (−ssRNA). Its genome is complementary to mRNA and requires an RNA-dependent RNA polymerase (RdRp) carried within the virion to first transcribe it into positive-sense mRNA before translation can occur.

Step 3: Evaluate the remaining options.

Hepatitis A virus is a Picornavirus with +ssRNA. Dengue virus belongs to the family Flaviviridae and also has a +ssRNA genome.

Answer:

Influenza virus is the negative-sense single-stranded RNA virus among the given options.

Quick Tip:

A useful mnemonic — negative-sense RNA viruses (e.g., Orthomyxoviruses, Paramyxoviruses, Rhabdoviruses, Bunyaviruses) must carry their own RNA polymerase into the host cell since the host ribosomes cannot directly translate negative-sense RNA.

Understanding:

We need to identify the basis of Baltimore's classification of viruses.

Step 1: Recall the principle behind Baltimore classification.

David Baltimore proposed this classification in 1971. It groups all viruses into seven classes (Groups I–VII) based on the nature of their nucleic acid genome (DNA or RNA, single-stranded or double-stranded, positive or negative sense) and the pathway used to generate messenger RNA (mRNA) from that genome.

Step 2: Evaluate why other options are incorrect.

Host range/tissue tropism and mode of transmission are epidemiological criteria, not used in Baltimore classification. Capsid morphology is the basis of a different classification approach (structural classification). Pathogenicity is a clinical criterion.

Step 3: Confirm the correct answer.

The central idea is that mRNA production is the common requirement for all viruses since ribosomes only translate mRNA. Baltimore defined each class by how the virus achieves this, making it a universally applicable system.

Answer:

Baltimore classification is based on the nature of the viral genome and the strategy used to produce mRNA.

Quick Tip:

Remember that Baltimore Group IV (+ssRNA) viruses use their genome directly as mRNA, whereas Group V (−ssRNA) must first transcribe it — this distinction is a favourite exam point.

Understanding:

We need to identify what characterises the eclipse period in viral replication.

Step 1: Define the one-step growth curve phases.

When a synchronised viral infection is studied using a one-step growth experiment, three phases are observed: the latent period, the rise period, and the plateau. The latent period is further divided into the eclipse period and a post-eclipse accumulation phase.

Step 2: Characterise the eclipse period specifically.

The eclipse period begins immediately after viral entry and uncoating. During this time, the original infecting virion has been disassembled, new virions have not yet been assembled, and consequently no infectious virus can be detected either inside or outside the cell. The virus exists only as disassembled components (nucleic acid and proteins being synthesised separately).

Step 3: Distinguish from the latent period.

The latent period includes both the eclipse period and the time when newly assembled (but not yet released) virions are accumulating inside the cell. Infectious virus can be detected inside the cell in the late latent period but not during the eclipse.

Answer:

During the eclipse period, infectious virus cannot be detected inside or outside the host cell because the infecting virion has been uncoated and new virions have not yet assembled.

Quick Tip:

Eclipse period ⊂ Latent period. After the eclipse period ends, intracellular infectious virus can be detected, but extracellular virus remains undetectable until lysis or budding begins.

Understanding:

We need to identify the correct mechanism by which Type I interferons protect uninfected cells.

Step 1: Recall the nature of interferon action.

Type I interferons (IFN-α and IFN-β) are produced by virus-infected cells. They are secreted and bind to IFN receptors (IFNAR) on the surface of neighbouring uninfected cells. This binding triggers the JAK-STAT signalling pathway, leading to the upregulation of hundreds of Interferon-Stimulated Genes (ISGs).

Step 2: Identify the key antiviral proteins induced.

Two classic antiviral effectors induced by Type I IFN are:

These mechanisms create an antiviral state in uninfected cells before the virus reaches them.

Step 3: Eliminate incorrect options.

Interferons do not directly neutralise virions (that is the role of antibodies), do not immediately stimulate antibody production (that is adaptive immunity, which takes days), and do not activate complement directly.

Answer:

Type I interferons protect neighbouring cells by inducing antiviral proteins such as 2′-5′ oligoadenylate synthetase and PKR through the JAK-STAT pathway.

Quick Tip:

Interferons do not protect the cell that produces them — they warn neighbouring cells. This paracrine signalling is a key concept frequently tested in virology.

Understanding:

We need to identify the Baltimore Group VI virus that uses reverse transcriptase.

Step 1: Review the relevant Baltimore groups.

Baltimore Group VI: ssRNA-RT viruses — these have a positive-sense ssRNA genome but replicate through a DNA intermediate using reverse transcriptase. Example: Retroviruses, including HIV.

Baltimore Group VII: dsDNA-RT viruses — these have a dsDNA genome and also use reverse transcriptase, but the template is DNA. Example: Hepadnaviruses (Hepatitis B virus) and Cauliflower Mosaic Virus.

Step 2: Classify each option.

Step 3: Confirm answer.

HIV is specifically a Group VI (ssRNA-RT) virus that uses reverse transcriptase to convert its RNA genome into dsDNA, which then integrates into the host chromosome as a provirus.

Answer:

HIV is the Baltimore Group VI virus that uses reverse transcriptase in its replication cycle.

Quick Tip:

The key distinction: Group VI (Retroviruses) starts as RNA and makes DNA; Group VII (Hepadnaviruses) starts as DNA and makes RNA via reverse transcriptase. Both use reverse transcriptase, but in opposite directions relative to their replication cycle.

Understanding:

We need to identify the correct function of the haemagglutinin (HA) protein of Influenza virus.

Step 1: Recall the structure of Influenza surface proteins.

Influenza virus has two major surface glycoproteins: Haemagglutinin (HA) and Neuraminidase (NA). They have distinct and complementary roles.

Step 2: Define the role of HA.

HA performs two sequential functions in viral entry:

1. Receptor binding: HA binds to sialic acid (N-acetylneuraminic acid) residues present on glycoproteins and glycolipids on the surface of host respiratory epithelial cells.

2. Membrane fusion: After endocytosis, the acidic pH in the endosome triggers a conformational change in HA, exposing the fusion peptide. This mediates fusion of the viral envelope with the endosomal membrane, releasing the viral RNA segments into the cytoplasm.

Step 3: Identify the role described in each wrong option.

Answer:

Haemagglutinin binds sialic acid on host cell surfaces and mediates viral envelope–endosomal membrane fusion to release the viral genome.

Quick Tip:

Remember: HA = entry (binds in, fuses in); NA = exit (cleaves sialic acid so new virions can escape). Oseltamivir (Tamiflu) inhibits NA, preventing viral release.

Understanding:

We need to distinguish antigenic shift from antigenic drift and identify the mechanism of antigenic shift.

Step 1: Define antigenic drift.

Antigenic drift refers to the gradual accumulation of point mutations in the HA and NA genes due to the error-prone nature of viral RNA polymerase. This causes minor antigenic changes and is responsible for seasonal influenza epidemics. This corresponds to option A, which is incorrect for antigenic shift.

Step 2: Define antigenic shift.

Antigenic shift is an abrupt, major change in the HA and/or NA antigens of Influenza A virus. It occurs when two different strains of Influenza A (e.g., a human strain and an avian or swine strain) simultaneously infect the same host cell. Because the Influenza A genome is segmented (8 segments of −ssRNA), the genome segments from both strains can be packaged together in new virions in random combinations — a process called reassortment. This can produce a novel virus with a completely new HA or NA subtype to which the human population has little or no pre-existing immunity, potentially causing a pandemic.

Step 3: Evaluate the other options.

Influenza virus does not integrate into the host chromosome (that is a feature of retroviruses). Reassortment between Influenza A and Influenza B does not occur naturally due to incompatibility of their internal proteins.

Answer:

Antigenic shift results from the reassortment of genome segments between two different Influenza A strains co-infecting the same cell.

Quick Tip:

Antigenic DRIFT = small, gradual (like a boat drifting slowly) → epidemics. Antigenic SHIFT = sudden, large change (like a gear shift) → pandemics. Segmented genome = prerequisite for reassortment.

Understanding:

We need to correctly describe the sequential roles of CD4 and CCR5 in HIV-1 entry.

Step 1: Describe the initial attachment.

The HIV-1 envelope glycoprotein complex consists of gp120 (surface unit) and gp41 (transmembrane unit). The first step of entry is the binding of gp120 to the CD4 receptor, which is expressed primarily on T helper lymphocytes, macrophages, and dendritic cells. This interaction is the primary, high-affinity attachment event.

Step 2: Describe co-receptor engagement.

Binding of gp120 to CD4 causes a conformational change in gp120, exposing the V3 loop region. This newly exposed region then binds to a chemokine co-receptor — either CCR5 (on macrophages and memory T cells) or CXCR4 (on naïve T cells). Engagement of the co-receptor causes a further conformational change.

Step 3: Describe membrane fusion.

The conformational changes triggered by co-receptor binding expose the hydrophobic fusion peptide at the N-terminus of gp41. This fusion peptide inserts into the host cell membrane, and gp41 undergoes a hairpin fold that brings the viral and cellular membranes together, resulting in fusion and release of the viral core into the cytoplasm.

Step 4: Confirm the correct answer.

CD4 is the primary receptor for gp120; CCR5 is the co-receptor whose engagement triggers gp41-mediated fusion. Individuals homozygous for the CCR5-Δ32 mutation are highly resistant to R5-tropic HIV infection.

Answer:

gp120 binds CD4 first, then CCR5 acts as co-receptor, triggering gp41-mediated membrane fusion.

Quick Tip:

Maraviroc, an antiretroviral drug, works by blocking CCR5, preventing HIV entry. This is only effective against R5-tropic (CCR5-using) strains, not X4-tropic strains.

Understanding:

We need to calculate the titre of the original virus stock from plaque assay data.

Formula:

The titre of the original stock is calculated as:

Step 1: Substitute the known values.

Step 2: Verify units.

Dilution is dimensionless; volume is in mL; so the result is in PFU/mL. The calculation is consistent.

Answer:

The titre of the original virus stock is $4.5\times 10^8$ PFU/mL.

Quick Tip:

A common error is forgetting to divide by the inoculation volume. Always account for the volume factor: inoculating 0.1 mL instead of 1 mL means the plaques observed represent only $\frac{1}{10}$th of what would be seen per mL, so the titre is 10× higher than if you only used the dilution factor alone.

Understanding:

We need to correctly define the concept of viral tropism.

Step 1: Define viral tropism.

Viral tropism refers to the specificity of a virus for particular cell types, tissues, or host species. A virus is said to be tropic for the cell type it preferentially infects. This specificity is primarily governed by molecular interactions between viral surface proteins (e.g., attachment proteins such as haemagglutinin, gp120, or fibre proteins) and specific receptor molecules expressed on the target cell surface.

Step 2: Additional determinants of tropism.

While receptor compatibility is the primary determinant, other factors also influence tropism:

Step 3: Distinguish from the other options.

Answer:

Viral tropism describes the preference of a virus for infecting specific cell types or tissues, primarily determined by receptor compatibility.

Quick Tip:

Classic examples of tropism: HIV is lymphotropic and macrophage-tropic (CD4+ cells); Rabies virus is neurotropic (neurons); Hepatitis B virus is hepatotropic (hepatocytes). Receptor expression is the gatekeeper of tropism.

Understanding:

This question asks about the most critical parameter governing oxygen supply to aerobic microorganisms in submerged fermentation.

Step 1: Role of oxygen in SmF

In submerged fermentation, aerobic microorganisms require a continuous supply of dissolved oxygen. Since oxygen has low solubility in aqueous media, its transfer from the gas phase to the liquid phase is often the rate-limiting step.

Step 2: Significance of kLa

The volumetric oxygen transfer coefficient, kLa, quantifies the rate at which oxygen is transferred from gas bubbles to the liquid medium. A higher kLa means more efficient oxygen delivery. It is influenced by agitation speed, aeration rate, impeller design, and broth viscosity. Oxygen transfer rate (OTR) is expressed as:

OTR = kLa × (C* − CL)

where C* is the saturation concentration of dissolved oxygen and CL is the actual dissolved oxygen concentration in the broth.

Step 3: Why other options are less critical for oxygen transfer

While pH, osmotic pressure, and redox potential influence microbial metabolism, none of them directly governs the physical transfer of oxygen from gas to liquid phase. kLa is the engineering parameter specifically designed to quantify and optimize this transfer.

Answer:

The volumetric oxygen transfer coefficient (kLa) is the most critical parameter for ensuring adequate oxygen transfer to aerobic microorganisms in submerged fermentation, as it directly determines the rate of oxygen movement from the gas phase into the liquid medium.

Quick Tip:

In aerobic fermentation questions, whenever oxygen supply is the subject, kLa is the key engineering parameter — a common exam distinction from biological parameters like pH or redox potential.

Understanding:

This question asks about the specific enzyme that converts pyruvate to acetaldehyde during alcoholic fermentation.

Step 1: Pathway of alcoholic fermentation

In Saccharomyces cerevisiae under anaerobic conditions, glucose is first converted to pyruvate via glycolysis. Pyruvate then undergoes two sequential reactions to produce ethanol:

Step 2: Reaction 1 — Decarboxylation

Pyruvate decarboxylase (PDC) catalyses the irreversible decarboxylation of pyruvate to acetaldehyde, releasing CO₂. This enzyme requires the cofactor thiamine pyrophosphate (TPP):

Pyruvate → Acetaldehyde + CO₂ (catalysed by Pyruvate decarboxylase)

Step 3: Reaction 2 — Reduction

Alcohol dehydrogenase (ADH) then reduces acetaldehyde to ethanol using NADH, regenerating NAD⁺ to sustain glycolysis:

Acetaldehyde + NADH + H⁺ → Ethanol + NAD⁺

Step 4: Eliminating distractors

Pyruvate dehydrogenase converts pyruvate to acetyl-CoA (aerobic pathway, not fermentation). Lactate dehydrogenase converts pyruvate to lactate in lactic acid fermentation, not in alcoholic fermentation.

Answer:

Pyruvate decarboxylase is the enzyme responsible for decarboxylating pyruvate to acetaldehyde in alcoholic fermentation by Saccharomyces cerevisiae.

Quick Tip:

A key distinction: pyruvate decarboxylase is unique to organisms performing alcoholic fermentation (e.g., yeast and some bacteria); mammals lack this enzyme and instead use pyruvate dehydrogenase under aerobic conditions.

Understanding:

This question asks about the fed-batch strategy used to prevent catabolite repression while achieving high cell density.

Step 1: What is catabolite repression?

Catabolite repression occurs when a rapidly metabolisable carbon source (e.g., glucose) is present in excess. High intracellular concentrations of catabolite metabolites repress the expression of genes needed for secondary metabolite production or utilisation of alternative substrates. In yeast, excess glucose also causes the Crabtree effect, leading to ethanol production even under aerobic conditions.

Step 2: The fed-batch solution

In fed-batch fermentation, the carbon source is added continuously or intermittently at a carefully controlled, growth-limiting rate. This keeps the concentration of the carbon source low in the broth, avoiding catabolite repression, while still supplying enough substrate to support high biomass accumulation over time.

Step 3: Why other options fail

Adding all carbon source at the start leads to initial excess and triggers catabolite repression. Switching to anaerobic conditions reduces productivity for aerobic processes. Adding nitrogen cannot compensate for excess carbon — catabolite repression is specifically a carbon-source regulatory phenomenon.

Answer:

Continuously feeding the carbon source at a controlled, limiting rate is the fed-batch strategy that prevents catabolite repression while allowing high cell density to be achieved.

Quick Tip:

Fed-batch is the industry standard for producing insulin, antibiotics, and many enzymes — the controlled feed rate is set to match the maximum specific growth rate (μmax) without causing substrate accumulation.

Understanding:

We must calculate the specific growth rate using the Monod equation.

Formula:

Step 1: Substitute the values

Step 2: Interpretation

At $S=0.6\text{g/L}$, which is three times $K_s$, the organism grows at 75% of ${\mu }_{\text{max}}$. This is consistent because when $S=K_s$, $\mu =0.5{\mu }_{\text{max}}$, and as $S$ increases further, $\mu$ approaches but never reaches ${\mu }_{\text{max}}$.

Answer:

The specific growth rate at $S=0.6\text{g/L}$ is $0.6{\text{h}}^{-1}$.

Quick Tip:

A quick mental check: when $S=K_s$, $\mu =0.5{\mu }_{\text{max}}$; when $S=3K_s$, $\mu =0.75{\mu }_{\text{max}}$. Here $S=3\times 0.2=0.6$, confirming $\mu =0.75\times 0.8=0.6{\text{h}}^{-1}$.

Understanding:

This question asks which product is a secondary metabolite synthesised during the idiophase (after exponential growth ceases), rather than during active growth.

Step 1: Primary vs. secondary metabolites in fermentation

Primary metabolites (e.g., ethanol, lactic acid, citric acid) are produced during the exponential growth phase (trophophase) as direct products of central metabolic pathways. Their synthesis is tightly coupled to growth.

Step 2: Secondary metabolites and the idiophase

Secondary metabolites are not essential for growth and are produced after the exponential phase ends, during the idiophase. Their synthesis is often triggered by nutrient depletion (e.g., phosphate, nitrogen limitation) and involves specialised biosynthetic gene clusters.

Step 3: Evaluating the options

Ethanol and lactic acid are end-products of primary fermentative metabolism, produced during active growth. Citric acid, while an overflow metabolite, is also primarily a growth-associated (primary) metabolite produced by Aspergillus niger under phosphate/nitrogen limitation but still during active metabolism. Penicillin, produced by Penicillium chrysogenum, is a classic example of a secondary metabolite whose production begins in the idiophase after growth slows, triggered by glucose depletion.

Answer:

Penicillin is the classic secondary metabolite whose production is characteristic of the idiophase, beginning after the exponential growth (trophophase) ends.

Quick Tip:

In industrial fermentation, trophophase = growth phase; idiophase = production phase. For antibiotics like penicillin, the bioreactor is deliberately managed to extend the idiophase by controlled nutrient feeding.

Understanding:

This question asks what the dilution rate equals at steady state in a chemostat.

Formula:

The dilution rate is defined as:

where $F$ is the volumetric flow rate of medium and $V$ is the fermentor volume.

At steady state, the biomass concentration in the chemostat is constant. The mass balance on biomass gives:

Step 1: Apply the steady-state condition

Step 2: Interpretation

At steady state, the specific growth rate of the organism exactly equals the dilution rate. By adjusting $D$ (i.e., changing the feed flow rate), the experimenter can precisely control $\mu$ — a unique advantage of the chemostat. If $D$ exceeds ${\mu }_{\text{max}}$, washout occurs and the culture is lost.

Answer:

At steady state in a chemostat, the dilution rate $D$ is numerically equal to the specific growth rate $\mu$ of the organism.

Quick Tip:

Washout occurs when $D>{\mu }_{\text{max}}$. The critical dilution rate $D_c$ is approximately equal to ${\mu }_{\text{max}}$ — a key value to remember for chemostat design problems.

Understanding:

This question asks about the primary separation technique used to remove fungal mycelia from fermentation broth at an industrial scale.

Step 1: Nature of fungal mycelia

Fungal mycelia are filamentous, forming large, fibrous cell masses. They are significantly different from bacterial cells, which are small and require higher centrifugal forces for separation. Due to their large particle size and fibrous texture, mycelia can be efficiently separated by filtration.

Step 2: Why rotary vacuum drum filtration?

Rotary vacuum drum filtration (RVDF) is the industry-standard first-step separation for mycelial organisms (e.g., Penicillium, Aspergillus, Streptomyces). The rotating drum covered with a filter cloth creates a vacuum that draws liquid through while retaining the mycelial cake on the surface. It is continuous, scalable, and cost-effective for large volumes.

Step 3: Why not the other options?

Centrifugation is more appropriate for bacterial or yeast cells, not bulky mycelia — it would be inefficient and costly. Ultrafiltration is used for separating macromolecules (proteins, enzymes) and is not a primary step for whole cells. Liquid-liquid extraction is used for recovery of soluble small molecules, not insoluble cell biomass.

Answer:

Rotary vacuum drum filtration is the most commonly used first-step technique for separating fungal mycelia from fermentation broth in industrial-scale processes.

Quick Tip:

Remember the rule: large, filamentous cells (fungi, Streptomyces) → filtration; small, non-filamentous cells (bacteria, yeast) → centrifugation. This determines the primary harvesting strategy.

Understanding:

This question asks which immobilisation method forms a covalent bond between the enzyme and the support, providing the highest operational stability.

Step 1: Overview of immobilisation methods

Four principal methods exist:

Step 2: Stability comparison

Covalent binding forms an irreversible enzyme–support linkage. This prevents enzyme leaching under operational conditions (variations in pH, temperature, ionic strength), making it the method with the highest operational stability for repeated or continuous use.

Step 3: Industrial relevance

Examples include glucose isomerase covalently bound to ion-exchange resins for high-fructose corn syrup production, and penicillin acylase bound to supports for semi-synthetic antibiotic manufacturing.

Answer:

Covalent binding provides the highest operational stability because it forms an irreversible bond between the enzyme and the support matrix, preventing enzyme leaching.

Quick Tip:

Though covalent binding offers the best stability, it may reduce enzyme activity due to conformational changes near the active site. This trade-off between stability and activity is a common exam theme.

Understanding:

This question asks which microorganism converts ethanol to acetic acid in the Orleans (slow/surface) process of vinegar production.

Step 1: The Orleans process

The Orleans process is a traditional surface fermentation method where wine or cider is allowed to ferment slowly in barrels. A surface film of bacteria forms at the air-liquid interface, where they oxidise ethanol to acetic acid using atmospheric oxygen.

Step 2: The biochemical reaction

The oxidation proceeds as:

Ethanol + O₂ → Acetic acid + H₂O

This is an aerobic, incomplete oxidation (not a true fermentation in the strictest sense, but classified under fermentation technology).

Step 3: The responsible organism

Acetobacter aceti is a Gram-negative, obligately aerobic acetic acid bacterium (AAB) that carries out this conversion. It forms a characteristic pellicle (vinegar mother) at the surface of the liquid.

Step 4: Eliminating distractors

Saccharomyces cerevisiae converts sugars to ethanol (yeast fermentation — the preceding step). Lactobacillus aceticus is not the primary organism for vinegar production at industrial scale. Clostridium acetobutylicum is an anaerobic organism used in ABE (acetone-butanol-ethanol) fermentation, not acetic acid production.

Answer:

Acetobacter aceti is the microorganism responsible for oxidising ethanol to acetic acid in the Orleans process of vinegar production.

Quick Tip:

Vinegar production is a two-stage process: (1) yeast converts sugar → ethanol; (2) Acetobacter converts ethanol → acetic acid. Both stages and their organisms are frequently tested together.

Understanding:

This question asks for the correct mathematical definition of the Del factor (nabla, ∇) used in fermentation media sterilisation.

Formula:

The Del factor (also called the criterion of sterilisation or ∇) is defined as:

Step 1: Derivation basis

Microbial death kinetics follows first-order kinetics:

Integrating from $N_0$ to $N$ over time $0$ to $t$:

Step 2: Relationship between expressions

Although $\nabla =k_d\cdot t$ is mathematically equal to $\ln (N_0/N)$, option A alone does not define what ∇ represents in terms of the sterilisation outcome. The defining expression relating ∇ to the probability of contamination is $\nabla =\ln (N_0/N)$. Option C uses ${\log }_{10}$, which is incorrect — the Del factor uses the natural logarithm. Option D is the efficiency of killing, not the Del factor.

Answer:

The Del factor is correctly expressed as the natural logarithm of the ratio of initial to final microbial numbers.

Quick Tip:

A typical industrial sterilisation target is $\nabla =29.4$, which corresponds to reducing the probability of one contaminant surviving from $N_0=10^{13}$ organisms to $N=10^{-3}$ (i.e., one chance in a thousand per batch).