Microbiology - MCQ Practice Questions

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).