Q: Dynamic mass balance for a CSTR: Which statement correctly expresses the unsteady-state (time-dependent) total mass balance for a reacting species in a continuous stirred tank reactor?

Introduction / Context: CSTR modeling starts with a general balance: accumulation = in − out + generation. The question checks recognition of the unsteady (time-dependent) form for a species or conserved quantity in a well-mixed reactor. Given Data / Assumptions: Well-mixed volume, variable with time in principle, though often constant. Inlet and outlet streams with known compositions and flow rates. Homogeneous reaction with a net formation rate inside the reactor. Concept / Approach: The correct statement must include a time derivative on the left-hand side (accumulation term). The right-hand side consists of input, output, and net reaction formation rate terms. Wording differences matter: “time rate of total amount” explicitly captures the derivative d(n)/dt. Step-by-Step Solution: Write general balance: d(Amount in system)/dt = In − Out + Generation.Map to CSTR: accumulation = F_inC_in − F_outC + rV.Recognize that options lacking the time derivative are steady-state forms, not unsteady. Verification / Alternative check: At steady state, the accumulation term goes to zero, recovering the familiar In − Out + Generation = 0, consistent with the same framework. Why Other Options Are Wrong: Option (b) omits time dependence; it is dimensionally inconsistent. Option (c) is close in meaning but is ambiguously phrased and redundant; option (a) states it clearly. “None of the above” is incorrect since (a) is correct. Common Pitfalls: Forgetting the accumulation term when modeling start-up or disturbances. Confusing total-mass versus species balances; the structure is the same but variables differ. Final Answer: Time rate of total amount in the reactor = rate of addition to reactor − rate of removal + rate of formation

Question 1

Fed-batch material balance: a reactor initially contains 2 L of medium with 1 g·L^-1 substrate. It is fed with substrate solution at 1 g·L^-1 at 1 L·h^-1 for 10 h. If after 10 h the reactor substrate concentration is 0.5 g·L^-1, what mass of substra…

Accepted Answer

Introduction: Fed-batch calculations rely on a running mass balance that accounts for initial inventory, cumulative feed addition, withdrawals (if any), and current reactor concentration. This question tests the ability to translate concentration-time information into total substrate consumption by the culture. Given Data / Assumptions: Initial volume V0 = 2 L with C0 = 1 g·L^-1 ⇒ initial substrate mass = 2 g. Feed: C_f = 1 g·L^-1 at F = 1 L·h^-1 for 10 h ⇒ added mass = 10 g; volume added = 10 L. No harvest or bleed (volume strictly increases by feed). Final volume V_f = 2 + 10 = 12 L; final concentration C_f,out = 0.5 g·L^-1 ⇒ final mass = 6 g. Concept / Approach: Total mass consumed = (initial mass + mass in feed) − final mass in reactor. This directly applies the integral species balance over the fed-batch period without needing rates. Step-by-Step Solution: Step 1: Compute initial mass: m0 = 2 L * 1 g·L^-1 = 2 g.Step 2: Compute feed mass: m_in = 1 g·L^-1 * (1 L·h^-1 * 10 h) = 10 g.Step 3: Compute final mass: m_f = 12 L * 0.5 g·L^-1 = 6 g.Step 4: Consumption = (m0 + m_in − m_f) = (2 + 10 − 6) g = 6 g. Verification / Alternative check: A quick reasonableness check: despite substantial feed, the final concentration is lower than the feed, indicating net consumption; the computed 6 g used is consistent with the dilution and uptake. Why Other Options Are Wrong: 2 g or 4 g: Underestimates; ignore the large feed volume. 8 g or 12 g: Overestimates; do not reconcile initial, feed, and final inventory correctly. Common Pitfalls: Forgetting the reactor volume increase, using concentrations instead of masses directly, or overlooking any withdrawals (none here). Final Answer: 6g

Question 2

Continuous reactor—yield from steady-state concentrations: at steady state the reactor contains X = 0.04 g·L^-1 biomass and S_out = 0.02 g·L^-1 phenol; the feed contains S_in = 0.10 g·L^-1 phenol. Assuming biomass in feed is negligible, what is the …

Accepted Answer

Introduction: In continuous cultures at steady state, yield coefficients are often estimated from the difference between inlet and outlet substrate concentrations and the steady biomass concentration, assuming negligible biomass in the feed. This provides a quick screen of culture efficiency without detailed kinetic modeling. Given Data / Assumptions: Biomass concentration in reactor (and effluent) X = 0.04 g·L^-1. Substrate in feed S_in = 0.10 g·L^-1; in effluent S_out = 0.02 g·L^-1. Negligible product/byproduct carbon sinks; yield approximated from material differences. Concept / Approach: Substrate consumed per liter = S_in − S_out = 0.08 g·L^-1. Biomass present per liter is the result of that consumption at steady state. The apparent yield is Y_x/s ≈ X / (S_in − S_out) under the stated simplifications. Step-by-Step Solution: Step 1: Compute substrate consumed: 0.10 − 0.02 = 0.08 g·L^-1.Step 2: Take the ratio with biomass: Y_x/s = 0.04 / 0.08 = 0.5 g.g^-1.Step 3: Select 0.5 g.g^-1 from the options. Verification / Alternative check: This back-of-the-envelope method aligns with steady-state CSTR balances when maintenance and product formation are modest relative to growth-associated biomass formation. Why Other Options Are Wrong: 0.4 or 0.2 g.g^-1: Miscompute the ratio. 1.0 or 5.0 g.g^-1: Implausibly high given the data and would require different S_out or X. Common Pitfalls: Ignoring non-growth-associated product formation or maintenance terms; for precise yields, full carbon balances are preferred. Final Answer: 0.5 g.g^-1

Question 3

Continuous reactor productivity from outlet composition and flow: the feed contains 40 g·L^-1 maltose; the reactor effluent contains 20 g·L^-1 lactate at a flow rate of 10 L·h^-1. What is the (mass) productivity of lactate leaving the reactor, based…

Accepted Answer

Introduction: Productivity in a continuous process can refer to the mass of product discharged per unit time at the outlet flow. When reactor volume is not provided, one reports the mass flow rate of product based on outlet concentration and volumetric flow. This quick calculation is useful for comparing throughput across operating conditions. Given Data / Assumptions: Outlet (effluent) lactate concentration = 20 g·L^-1. Volumetric flow rate F = 10 L·h^-1. No additional information about reactor volume; thus we compute mass per hour leaving in the effluent stream. Concept / Approach: Mass productivity as mass per unit time is simply the product of outlet concentration and volumetric flow rate: ṁ_product = C_product,out * F. Here it yields a numeric mass per hour. (Note: a volumetric productivity g·L^-1·h^-1 would require reactor volume to compute dilution rate D = F/V.) Step-by-Step Solution: Step 1: Identify C_lactate,out = 20 g·L^-1.Step 2: Identify F = 10 L·h^-1.Step 3: Compute mass flow: ṁ = 20 * 10 = 200 g·h^-1.Step 4: Report the corresponding choice consistent with the numeric value among options. Verification / Alternative check: Dimensional check confirms g·L^-1 * L·h^-1 = g·h^-1. The computed 200 g·h^-1 aligns with the throughput implied by the data. Why Other Options Are Wrong: 50 g entries: Do not match the product C*F. 200 g maltose.l^-1: Refers to a different species and mismatched interpretation. 20 g lactate.h^-1: Off by a factor of 10 from the correct mass flow. Common Pitfalls: Confusing mass flow rate with volumetric productivity (g·L^-1·h^-1). Always check whether reactor volume is known before computing per-volume rates. Final Answer: 200 g lactate.l^-1

Question 4

Chemostat steady state (Monod kinetics): An organism grows on glucose with parameters μm = 0.3 h^-1, Ks = 0.05 g·L^-1, and yield Yxs = 0.3 g·g^-1 (biomass per substrate). It is cultured in a 2 L continuous stirred-tank reactor (chemostat). Feed cont…

Accepted Answer

Introduction / Context: Chemostat problems test understanding of dilution rate, Monod kinetics, washout, and how steady-state biomass concentration depends on the balance between growth and removal. Here, we decide whether the reactor operates at a viable steady state or washes out based on μ (specific growth rate) versus D (dilution rate). Given Data / Assumptions: μm = 0.3 h^-1 (maximum specific growth rate). Ks = 0.05 g·L^-1 (Monod constant). Yxs = 0.3 g·g^-1 (not directly needed if washout occurs). V = 2 L (reactor volume). Feed flow F = 1 L·h^-1. Feed substrate S_in = 10 g·L^-1 (glucose). Ideal, well-mixed chemostat; single-limiting substrate; no cell recycle. Concept / Approach: In a chemostat, the dilution rate is D = F / V. At non-washout steady state, the net growth rate equals the washout rate: μ(S) = D. If the required μ(S) to match D exceeds μm, the culture cannot sustain itself and the reactor washes out (X → 0). Step-by-Step Solution: Compute dilution rate: D = F / V = 1 / 2 = 0.5 h^-1.Compare D with μm: μm = 0.3 h^-1, so D > μm.Growth cannot keep pace with washout because μ(S) ≤ μm for all S.Therefore, steady-state biomass X must go to zero (washout). Verification / Alternative check: Even if S rose to very high levels, μ(S) is bounded by μm. Since D = 0.5 h^-1 > 0.3 h^-1, equality μ = D is impossible; the only steady state is washout (X = 0; S approaches S_in). Why Other Options Are Wrong: 0.4 or 0.5 g·L^-1: imply a viable steady state with μ(S) = D, which cannot happen when D > μm. 10 g·L^-1: that is the feed substrate concentration, not biomass. Biomass cannot exceed limits set by μ and D; here it goes to zero. Common Pitfalls: Forgetting to compute D and compare it with μm. Trying to use yield Yxs when washout already determines X = 0. Final Answer: 0 g·L^-1

Question 5

Bioprocess modes: A culture system in which environmental conditions are kept constant by continuously supplying fresh nutrients and simultaneously removing culture broth (cells, products, and wastes) is called a ________ culture system.

Accepted Answer

Introduction / Context: Understanding operational modes—batch, fed-batch, and continuous—is fundamental in bioprocess engineering. The question targets the definition of a system with constant operating conditions achieved by steady inflow and outflow, which is the hallmark of continuous culture. Given Data / Assumptions: Fresh medium is continuously added. An equal volume is continuously withdrawn. Environmental variables (e.g., volume, temperature, pH, and—ideally—substrate level) are maintained approximately constant. Concept / Approach: A continuous culture (e.g., chemostat) operates at steady state when dilution rate and growth rate balance. This distinguishes it from batch (closed, no inflow/outflow) and fed-batch (inflow without outflow, volume increases, conditions change over time). Step-by-Step Solution: Identify the presence of both inflow and outflow → not batch or fed-batch.Note constant conditions and fixed volume → defines a chemostat-type operation.Conclude the correct term: continuous culture system. Verification / Alternative check: In a chemostat at steady state, μ = D and state variables (X, S, P) are time-invariant, consistent with “constant environmental conditions.” Why Other Options Are Wrong: Batch: closed system; no inflow/outflow; conditions evolve with time. Fed-batch: inflow without outflow; volume and conditions change. Semi-continuous: intermittent withdrawals/additions; conditions are not strictly constant. Common Pitfalls: Confusing fed-batch (which can approximate constant substrate) with true continuous culture. Assuming “continuous monitoring” implies continuous culture; the key is continuous flow. Final Answer: continuous

Question 6

Vinegar (acetic acid) production: Why are fed-batch reactors preferred for acetobacter processes producing vinegar?

Accepted Answer

Introduction / Context: Acetic acid bacteria oxidize ethanol to acetic acid. However, high ethanol levels are toxic and can cause over-oxidation and stress. Fed-batch strategies are widely used to control ethanol in the optimal range while allowing acetic acid to accumulate toward the desired product strength. Given Data / Assumptions: Organism: acetic acid bacteria (e.g., Acetobacter, Komagataeibacter). Limitation: ethanol inhibition at high substrate levels. Goal: produce vinegar (acetic acid) efficiently and safely. Concept / Approach: In fed-batch, ethanol feed is metered so its concentration stays low, keeping cells in their optimal oxidation window. This avoids substrate inhibition and reduces by-products from over-oxidation, while acetic acid is allowed to rise (within tolerance). Continuous or batch with high initial ethanol would be less controlled or inhibitory. Step-by-Step Solution: Identify inhibition: excessive ethanol harms acetic acid bacteria.Use controlled feeding to keep ethanol near a safe setpoint.Permit acetic acid accumulation toward target strength.Conclude: the main advantage is maintaining low ethanol concentrations. Verification / Alternative check: Process profiles show higher productivity and stability when ethanol is fed to maintain low, near-constant concentrations compared with one-shot charges that spike ethanol and reduce activity. Why Other Options Are Wrong: Maintaining low acetic acid: not the objective; product concentration increases during the run. “Best at high ethanol”: contradicts known ethanol inhibition. All of the above: cannot be true because (b) and (c) are false. Common Pitfalls: Confusing fed-batch control of ethanol with controlling product acidity; the latter is a separate limitation. Feeding ethanol too quickly, which defeats the purpose by causing inhibitory spikes. Final Answer: They can maintain low ethanol concentrations to avoid substrate inhibition and over-oxidation risks

Question 7

Fluidized-bed bioreactors: What is the primary function of the disengagement (freeboard) zone at the top of a fluidized bed?

Accepted Answer

Introduction / Context: Fluidized beds suspend solid particles in an upward fluid stream. The top zone is engineered to minimize particle carryover so that the bed remains stable and material is conserved. Understanding this function is important for reactor design and scale-up. Given Data / Assumptions: Three regions: distributor, fluidized bed, disengagement/freeboard. Concern: entrainment of fine or partially fluidized particles into the outlet. Goal: retain catalyst/biocatalyst beads or carriers inside the reactor. Concept / Approach: The disengagement zone provides a larger cross-sectional area and lower superficial velocity, allowing entrained particles to decelerate, settle back, and disengage from the gas/liquid. This reduces particle loss and stabilizes the inventory. It may incidentally reduce splashing/foaming, but that is not its principal design objective. Step-by-Step Solution: Recognize entrainment as the key risk at the reactor outlet.Expand area (or provide internals) to reduce local velocities.Allow particles to disengage and return to the bed by gravity.Outcome: reduced particle loss and steadier bed height. Verification / Alternative check: Material balance around the reactor shows lower solids loss and less fines capture in downstream equipment when an adequate disengagement zone is provided. Why Other Options Are Wrong: Reducing particle velocity alone (b) is a means, not the primary function; the objective is solids retention. Foaming control (c) can be a secondary benefit but is not the main purpose. All of the above (d) overstates secondary effects; the key function is solids disengagement. Common Pitfalls: Undersizing the freeboard, leading to chronic carryover. Ignoring particle size distribution changes that affect entrainment. Final Answer: Reduce loss of particles from the reactor by allowing entrained solids to disengage and fall back

Question 8

Chemostat calculation: A chemostat of liquid volume 2 L is fed at 4 L·h^-1. What is the dilution rate D for this reactor?

Accepted Answer

Introduction / Context: Dilution rate (D) is the defining parameter of a chemostat, connecting feed flow and reactor volume. This item checks correct unit usage and the simple formula for D. Given Data / Assumptions: Reactor volume V = 2 L. Feed flow F = 4 L·h^-1. Well-mixed, constant volume operation. Concept / Approach: The dilution rate is defined as D = F / V with units of time^-1. It represents the fraction of reactor volume replaced per unit time. Step-by-Step Solution: Compute D = F / V = 4 / 2 = 2 h^-1.Check units: (L·h^-1)/L = h^-1.Therefore, D = 2 h^-1. Verification / Alternative check: Residence time τ = 1 / D = 0.5 h, consistent with a high-throughput chemostat replacing its content twice per hour. Why Other Options Are Wrong: “2 litres” and “4 litres per hour” are not rates normalized by volume; they are volume or volumetric flow, not D. “2 litres per hour” has wrong units; D must be in h^-1. Common Pitfalls: Forgetting to normalize by reactor volume. Reporting D with incorrect units. Final Answer: 2 h^-1

Question 9

CSTF (continuous stirred tank fermenter): Cellular productivity typically rises with dilution rate up to a maximum. What happens to productivity if the dilution rate is increased beyond the point of maximum productivity?

Accepted Answer

Introduction / Context: In a CSTF, volumetric productivity often shows a peak versus dilution rate because biomass concentration falls as D increases, while specific rate terms may rise. Recognizing the post-peak behavior is key to safe operation and avoiding loss of culture. Given Data / Assumptions: Steady-state operation with growth-limited substrate. Productivity refers to biomass (or product tied to growth) per reactor volume per time. Maximum occurs below washout (D < μm). Concept / Approach: As D increases, X decreases because μ(S) = D and S rises, but near μm the system approaches washout: X → 0. Hence the volumetric productivity (proportional to D * X for growth-associated products) declines steeply beyond the optimum and can collapse as D → μm. Step-by-Step Solution: Note productivity trend: rises with D at first due to increased throughput.Approaching peak: falling X offsets gains in D.Beyond the maximum: X drops quickly; productivity decreases abruptly.Near washout (D ≈ μm): productivity approaches zero. Verification / Alternative check: Plots from chemostat experiments show a dome-shaped curve of productivity versus D, with a sharp decline on the high-D side as cultures destabilize. Why Other Options Are Wrong: “Increases” or “increases drastically”: contradict empirical and theoretical trends. “Becomes zero immediately”: only becomes zero at washout; beyond the maximum it declines toward zero, not instantly zero. Common Pitfalls: Operating too close to μm; small disturbances can cause washout. Ignoring that product formation linked to growth mirrors biomass behavior. Final Answer: Decreases abruptly (approaching washout as D nears μm)

Question 10

Dynamic mass balance for a CSTR: Which statement correctly expresses the unsteady-state (time-dependent) total mass balance for a reacting species in a continuous stirred tank reactor?

Accepted Answer

Introduction / Context: CSTR modeling starts with a general balance: accumulation = in − out + generation. The question checks recognition of the unsteady (time-dependent) form for a species or conserved quantity in a well-mixed reactor. Given Data / Assumptions: Well-mixed volume, variable with time in principle, though often constant. Inlet and outlet streams with known compositions and flow rates. Homogeneous reaction with a net formation rate inside the reactor. Concept / Approach: The correct statement must include a time derivative on the left-hand side (accumulation term). The right-hand side consists of input, output, and net reaction formation rate terms. Wording differences matter: “time rate of total amount” explicitly captures the derivative d(n)/dt. Step-by-Step Solution: Write general balance: d(Amount in system)/dt = In − Out + Generation.Map to CSTR: accumulation = F_inC_in − F_outC + rV.Recognize that options lacking the time derivative are steady-state forms, not unsteady. Verification / Alternative check: At steady state, the accumulation term goes to zero, recovering the familiar In − Out + Generation = 0, consistent with the same framework. Why Other Options Are Wrong: Option (b) omits time dependence; it is dimensionally inconsistent. Option (c) is close in meaning but is ambiguously phrased and redundant; option (a) states it clearly. “None of the above” is incorrect since (a) is correct. Common Pitfalls: Forgetting the accumulation term when modeling start-up or disturbances. Confusing total-mass versus species balances; the structure is the same but variables differ. Final Answer: Time rate of total amount in the reactor = rate of addition to reactor − rate of removal + rate of formation

Question 11

Overflow metabolism and yield in Escherichia coli: Under which operating scenario would the biomass yield (g cells per g glucose) be the lowest for E. coli?

Accepted Answer

Introduction / Context: Biomass yield in E. coli depends strongly on glucose availability and the onset of overflow metabolism (acetate formation). High glucose pulses drive rapid glycolytic flux beyond respiratory capacity, lowering ATP per substrate and reducing biomass yield. Given Data / Assumptions: Aerobic conditions for all cases. Yield is defined on glucose (g cells per g glucose consumed). Overflow metabolism (acetate) is triggered at high glucose uptake rates. Concept / Approach: When glucose is initially high in a batch, E. coli often secretes acetate (overflow), which reduces carbon efficiency and yield. In contrast, low-glucose strategies (fed-batch or well-controlled continuous) limit substrate uptake to respiratory capacity, minimizing by-products and improving yield. Step-by-Step Solution: High initial glucose → high uptake → overflow → acetate formation.Carbon lost as acetate lowers biomass yield on substrate.Maintaining low glucose (fed-batch, continuous) controls uptake and improves yield.Therefore, the lowest yield occurs in the high-glucose batch case. Verification / Alternative check: Typical yield trends: batch (high glucose) < batch (low glucose) < continuous/fed-batch (controlled low glucose), consistent with measured acetate profiles and oxygen uptake limits. Why Other Options Are Wrong: Low-glucose batch: lower risk of overflow, higher yield. Fed-batch with low glucose: designed to maximize yield by avoiding overflow. Continuous at low glucose: also maintains high yield by matching uptake to respiration. Common Pitfalls: Confusing growth rate with yield; high initial growth can coincide with low yield. Ignoring oxygen transfer limits that exacerbate overflow metabolism. Final Answer: An aerated batch culture with an initially high glucose concentration

Question 12

Bioprocess rationale — Why are antibiotic production processes commonly run in fed-batch reactors rather than simple batch or continuous modes? (Consider precursor toxicity, growth-phase effects, and yield optimization.)

Accepted Answer

Introduction: Antibiotic fermentations are classic examples where the bioreactor operating mode strongly influences productivity. This question examines why fed-batch culture is preferred for many secondary metabolites such as polyketides and beta-lactams, focusing on toxicity management, growth-phase control, and yield optimization. Given Data / Assumptions: Antibiotic synthesis is typically a secondary metabolism phenomenon. Some pathway precursors or intermediates can inhibit growth at high concentrations. Feed composition and rate can be manipulated during the run. Goal is to maximize volumetric and specific product formation. Concept / Approach: Fed-batch combines the simplicity of batch with the controllability of continuous feeding. By metering carbon, nitrogen, or specific precursors, we can keep substrate levels non-inhibitory, steer cultures into stationary or quasi-stationary states that favor secondary metabolism, and avoid catabolite repression. The operating strategy therefore tackles toxicity, timing of pathway expression, and by-product control simultaneously. Step-by-Step Solution: Identify toxicity: many antibiotic precursors (e.g., organic acids, side-chain donors) inhibit at high levels; feeding avoids spikes.Link to growth phase: secondary metabolism is upregulated as growth slows; feed limitation can drive the culture toward stationary state.Optimize yield: controlled carbon limitation reduces overflow metabolism and enhances product yields.Conclude: all three listed reasons independently support fed-batch; together they make a compelling case. Verification / Alternative check: Pilot data frequently show higher titers in fed-batch than in simple batch for antibiotic producers; time-varying feeds tune the specific production rate while preserving viability, unlike chemostats that hold cells in exponential growth where secondary metabolism may be repressed. Why Other Options Are Wrong: (a), (b), (c) are each true but incomplete; the most comprehensive and correct choice is all of the above. (e) Incorrect; steady-state chemostats often underperform for secondary metabolites that need non-growth or stationary-phase conditions. Common Pitfalls: Confusing feed for “more food” rather than “precise limitation”; ignoring precursor toxicity thresholds; assuming continuous mode is universally superior. Final Answer: All of the above.

Question 13

Metabolic coupling in yeast — When a fermenting Saccharomyces culture becomes metabolically uncoupled, ethanol production becomes...

Accepted Answer

Introduction: Yeast can channel carbon to ethanol either as part of growth-associated catabolism or as a non-growth-associated overflow, depending on metabolic coupling between energy generation and biosynthesis. This question probes what happens when a fermenting culture becomes metabolically uncoupled. Given Data / Assumptions: Fermentative metabolism is active (anaerobic or Crabtree effect under high glucose). Uncoupling means ATP generation no longer drives proportional biomass synthesis. Carbon flux to ethanol can persist independently of growth. Concept / Approach: In coupled metabolism, catabolic ATP and reducing power support macromolecular synthesis; product formation is often growth associated. Under uncoupling, ATP dissipation and redox balancing can proceed without parallel biomass accumulation. Ethanol, serving as an electron sink and ATP-yielding product via substrate-level phosphorylation, may continue to form even when net growth stalls, thus appearing non-growth-associated. Step-by-Step Solution: Recognize the definition: growth-associated product formation scales with dX/dt; non-growth-associated scales with X (maintenance).Uncoupling reduces biomass yield Yx/s but does not necessarily stop glycolytic flux.Therefore ethanol formation can persist at roughly constant or maintenance-linked rates even with little biomass increase. Verification / Alternative check: Fitting Luedeking–Piret parameters often shows a drop in the growth-associated term (alpha) and a rise in the non-growth term (beta) after uncoupling stress (e.g., inhibitors, ionophores, heat shock), consistent with continued ethanol production at low growth. Why Other Options Are Wrong: (a) Ethanol remains a primary catabolic product, not a secondary metabolite. (b) Opposite trend; uncoupling weakens growth association. (d) Ethanol is not a growth nutrient for yeast under these conditions. (e) Ethanol formation is often used to restore redox balance rather than being suppressed. Common Pitfalls: Equating “uncoupled” with “no metabolism”; assuming ethanol is only formed during rapid growth. Final Answer: non-growth-associated (continues even without net biomass increase).

Question 14

Chemostat principle — For organisms growing at steady state in a chemostat, how is the specific growth rate determined?

Accepted Answer

Introduction: A chemostat maintains a microbial culture at a defined physiological state by continuous feeding and withdrawal. The central steady-state relationship links the organism’s specific growth rate to the hydraulic operating condition. This question tests the core identity mu = D. Given Data / Assumptions: Perfect mixing; constant volume V and flow F. Single limiting nutrient; stable steady state (no washout). Biomass concentration and growth rate are time-invariant at steady state. Concept / Approach: At steady state, biomass accumulation is zero. The mass balance for cells is dX/dt = (mu − D) * X = 0, giving mu = D provided X > 0. Thus the specific growth rate is set by, and equals, the dilution rate D = F/V. It does not generally equal mu_max unless D is chosen near that value without causing washout. Step-by-Step Solution: Write biomass balance: dX/dt = mu * X − D * X.Impose steady state (dX/dt = 0) with nonzero X.Solve: mu = D.Interpretation: measuring or setting D determines mu at steady state. Verification / Alternative check: Independent estimates of mu from substrate uptake or CO2 evolution agree with D under steady conditions; when D exceeds mu_max the culture washes out, confirming the relationship’s limit. Why Other Options Are Wrong: (a) False; mu is directly controlled by D. (c) False; mu equals mu_max only at the brink of washout. (d) Incorrect given (b) is true. (e) Not a general identity; mu depends on D and kinetics, not simply S_feed / tau. Common Pitfalls: Confusing instantaneous mu with mu_max; overlooking transient behavior before steady state; neglecting cell recycle systems where the simple balance changes. Final Answer: It equals the dilution rate and can be determined from it.

Question 15

Mixing patterns — In which reactor configuration are mixing profiles closest to ideal plug flow (minimal back-mixing)?

Accepted Answer

Introduction: Reactor performance depends on hydrodynamics. Ideal plug flow has negligible axial mixing and a narrow residence-time distribution, while stirred tanks approach complete back-mixing. This question identifies which configuration best approximates plug flow in practice. Given Data / Assumptions: Steady operation with constant flow. Non-channeling, well-designed packed bed with uniform packing. Comparable scales across options. Concept / Approach: In a packed bed, fluid moves through tortuous interstitial pathways with limited axial dispersion, approaching plug-flow behavior at suitable Reynolds and Peclet numbers. Fluidized beds suspend particles and increase mixing; stirred tanks with or without recycle are highly back-mixed. Therefore, packed beds most closely approach plug flow among the listed choices. Step-by-Step Solution: Compare dispersion: packed bed has low axial dispersion when designed properly.Stirred tank (with recycle) intentionally promotes mixing, broadening residence times.Fluidized bed adds particle motion, enhancing axial mixing relative to packed bed.Conclude packed bed is closest to plug flow. Verification / Alternative check: Tracer tests yield residence-time distributions: packed beds show narrow E(t) functions; CSTRs show exponential decay; fluidized beds lie in between. Modeling with dispersion coefficients confirms smallest axial dispersion for packed beds under non-channeling conditions. Why Other Options Are Wrong: (b) and (e) are mixing-dominant systems; recycle further increases back-mixing. (c) More mixed than packed beds due to solids motion. (d) Not equal; designs differ fundamentally. Common Pitfalls: Assuming all packed beds behave ideally; ignoring channeling or maldistribution that can degrade plug-flow performance. Final Answer: Continuous packed-bed reactor

Question 16

Toxic by-products and yields — Accumulated toxins in a bioreactor often lower biomass yield because...

Accepted Answer

Introduction: Biomass yield Yx/s reflects how efficiently substrate carbon and energy are converted into new cell material. When toxins accumulate, cells spend more resources on survival rather than growth. This question targets the energy-allocation mechanism behind reduced yields. Given Data / Assumptions: Toxins may include organics, oxygen radicals, acids, or solvents. Cells possess maintenance energy requirements for homeostasis (repair, ion gradients). Total ATP generation from metabolism is finite at given conditions. Concept / Approach: Maintenance models partition ATP usage into growth-associated (anabolism) and non-growth-associated (homeostasis). Toxins raise maintenance demands (e.g., efflux pumps, repair, chaperone activity), diverting ATP from biosynthesis. Consequently, less biomass is produced per unit substrate, reducing Yx/s despite similar or higher substrate consumption. Step-by-Step Solution: Recognize that toxic stress increases maintenance energy m_ATP.ATP budget shifts from polymerization and macromolecule synthesis to repair and protection.Thus, ATP diversion from anabolism to homeostasis lowers biomass yield. Verification / Alternative check: Chemostat data fitted with Pirt-type relationships (q_s = m + mu/Yx/s) show increased maintenance term under stress, matching reduced yields. Transcriptomics often reveals upregulation of stress-response genes with higher ATP demand. Why Other Options Are Wrong: (a) and (c) misstate the role of NAD+ (a redox carrier) relative to ATP budgeting. (b) Suggests the opposite direction of diversion. (e) Ribosome allocation is important but not the primary, general explanation for yield drop due to toxins. Common Pitfalls: Blaming only substrate limitation; ignoring proton leak, efflux, or repair costs; conflating redox balance with ATP availability. Final Answer: ATP is diverted away from anabolism to homeostasis (maintenance).

Question 17

Industrial practice — Why are continuous cultures not widely used for many biotechnological products?

Accepted Answer

Introduction: Continuous culture (chemostat or perfusion) offers steady-state operation and constant product quality in theory, yet many industrial fermentations remain batch or fed-batch. This question asks which limitation most credibly explains the limited industrial adoption across diverse products. Given Data / Assumptions: Long campaigns increase exposure to contamination risk. Microbial evolution and mutation can occur over extended times. Regulatory frameworks exist for continuous bioprocessing, though implementation is case-specific. Concept / Approach: The principal practical barrier is operational risk: contamination or genetic drift during long continuous runs can collapse productivity or alter product quality. While some secondary metabolites are less amenable to chemostats, the statement that they are universally unsuitable is too strong. Likewise, regulators have approved continuous steps (e.g., perfusion cell culture) under quality systems, so a blanket ban is inaccurate. Step-by-Step Solution: Evaluate (a): Some secondary metabolite processes can be adapted (e.g., cell recycle, two-stage systems); not categorically impossible.Evaluate (c): Regulations do not prohibit continuous per se; compliance and control are required.Identify (b) as a widely recognized, overarching risk that constrains adoption. Verification / Alternative check: Industry case studies show strict asepsis and strain stability are critical; failures due to contamination/mutation cause costly downtime, explaining preference for finite-duration fed-batch campaigns. Why Other Options Are Wrong: (a) Overgeneralization; some continuous processes make secondary products. (c) Factually incorrect; licensing is possible with proper control. (d) Not all statements are true, so “all of the above” is wrong. (e) Many organisms reach steady state; that is the basis of chemostats. Common Pitfalls: Assuming technical feasibility equals operational desirability; ignoring cell-line instability over long times at constant selective pressure. Final Answer: Contamination or mutation can have a disastrous effect on a long-running operation.

Question 18

Maximizing biomass productivity — Saccharomyces cerevisiae shows high biomass yields at low glucose and high dissolved oxygen. Which reactor strategy best supports maximum biomass productivity?

Accepted Answer

Introduction: Saccharomyces cerevisiae displays high biomass yields when glucose is low (to avoid the Crabtree effect) and oxygen is abundant. The objective is to choose a reactor mode that sustains these conditions continuously for maximal productivity (g biomass per L per h). Given Data / Assumptions: Crabtree effect: high glucose drives fermentative overflow, lowering Yx/s. High dissolved oxygen supports fully respiratory metabolism with higher ATP yield per glucose. Productivity depends on both yield and growth rate at steady state. Concept / Approach: A chemostat allows tight control of residual glucose at low levels by setting dilution rate and feed concentration, maintaining respiratory metabolism and high biomass yield. Batch with high initial glucose transiently experiences high substrate, inviting ethanol formation and yield loss. Fed-batch can work if residual glucose is kept very low, but without control it often produces spikes; continuous offers the most stable low-glucose environment for maximum sustained biomass productivity. Step-by-Step Solution: Identify desired state: low residual glucose, high DO, steady respiration.Map states to modes: chemostat maintains constant limitation and DO control.Reject modes that cause glucose spikes (batch high initial, uncontrolled fed-batch).Conclude that continuous low-glucose aerobic operation maximizes biomass productivity. Verification / Alternative check: Chemostat studies show higher Yx/s and stable high qO2 under glucose limitation; ethanol is minimized, and productivity mu*X can be optimized by selecting D below washout threshold. Why Other Options Are Wrong: (a) High initial glucose triggers overflow and reduces yield. (c) Without residual glucose control, fed-batch often experiences transient high glucose. (d) Modes are not equivalent. (e) Anaerobic operation produces ethanol with low biomass yield. Common Pitfalls: Confusing titer with productivity; neglecting oxygen transfer limits at high cell densities; ignoring washout limits when setting D. Final Answer: Continuous fermenter operated with low glucose (substrate limitation) and high dissolved oxygen.

Question 19

Stationary-phase antibiotic — A Bacillus species makes an antibiotic only in stationary phase in batch culture. If the same strain is grown in a 5 L continuous culture at steady state, the antibiotic productivity will...

Accepted Answer

Introduction: Some antibiotics are strictly stationary-phase products. Continuous chemostat operation maintains cells in exponential growth at steady state. This question asks you to infer the resulting steady-state productivity for a product that requires stationary-phase physiology. Given Data / Assumptions: The strain produces antibiotic only during stationary phase. Chemostat steady state maintains exponential growth (mu = D > 0). No special two-stage or cell-recycle configuration is used. Concept / Approach: At steady state in a single-stage chemostat, cells neither enter nor remain in stationary phase; they divide at mu = D continuously. If the biosynthetic pathway is strictly turned on only in stationary phase, the specific production rate q_p is zero in exponential phase, so volumetric productivity P_v = q_p * X is also zero under steady conditions. Step-by-Step Solution: Recognize physiological requirement: stationary phase is essential for expression.Recall chemostat state: exponential growth persists; stationary phase is absent.Conclude q_p = 0 at steady state, hence productivity is zero. Verification / Alternative check: Two-stage chemostats (growth then production) or cell-recycle/perfusion can create high-density, low-growth zones to enable production; a single-stage chemostat at steady state does not, validating the zero-productivity conclusion. Why Other Options Are Wrong: (a) and (b) incorrectly suggest production during steady exponential growth. (d) Violates mass balance; productivity is finite and here zero. (e) Oscillation is not intrinsic to steady chemostat operation without feedback instabilities. Common Pitfalls: Confusing steady state with high cell density; assuming any continuous operation supports stationary-phase pathways without special design. Final Answer: be zero at steady state.

Question 20

Mixing intensity per volume — In which bioreactor is mixing per unit volume generally the poorest?

Accepted Answer

Introduction: Mixing quality affects mass and heat transfer, gradients, and overall performance. Different reactor types provide different hydrodynamic environments. This question asks which configuration delivers the weakest mixing per unit volume under typical operating conditions. Given Data / Assumptions: Packed bed with fixed particles and interstitial flow. Fluidized bed with particle motion and enhanced mixing. Airlift with gas-induced circulation providing moderate mixing. Concept / Approach: In a packed bed, fluid flows through narrow channels with limited transverse mixing and low turbulence levels; axial dispersion is minimized (approaching plug flow), but overall bulk mixing is inherently poor. Fluidized beds improve mixing through particle motion and bubbling, and air-lift reactors provide recirculation loops that enhance mixing relative to packed beds. Mechanically stirred tanks provide the strongest mixing among listed configurations. Step-by-Step Solution: Compare hydrodynamics: fixed packing restricts eddy formation and cross-mixing.Recognize that fluidization introduces vigorous mixing via particle–fluid interactions.Airlift circulation produces bulk loops, better than packed-bed interstitial flow.Therefore, packed bed has the poorest mixing per volume. Verification / Alternative check: Tracer studies show the smallest degree of back-mixing and the most pronounced concentration gradients in packed beds under laminar/transition flow, confirming limited mixing. Why Other Options Are Wrong: (b) Fluidized beds are designed to enhance mixing. (c) Air-lift reactors generate circulation that improves mixing over packed beds. (d) Not applicable because option (a) is correct. (e) Agitated tanks have the highest mixing intensity among these. Common Pitfalls: Equating “less back-mixing” with “better mixing”; they are different metrics tied to different performance objectives. Final Answer: Continuous packed-bed reactor

Fermentation Reactors Questions