Q: Particle size cut — Conventional (cake) filtration typically separates relatively large particles of approximately what minimum size?

Introduction: Filtration modalities are often classified by particle size: conventional filtration for larger particles, microfiltration for submicron to several micrometers, ultrafiltration for macromolecules, and so on. Knowing the approximate cut helps select the right technology for cell removal or clarification steps. Given Data / Assumptions: Gravity or pressure-driven conventional filter media (e.g., depth filters, screens). Rigid particles forming a cake. Qualitative threshold understanding. Concept / Approach: Conventional filtration is typically applied to particles larger than roughly 10 μm, where sieving and cake mechanisms dominate. Smaller particles require microfiltration or finer depth media with controlled pore structures. Step-by-Step Solution: Place conventional filtration at coarser end of the spectrum.Recall that microfiltration spans approximately 0.1 to 10 μm.Therefore, a practical minimum size for classical filtration is on the order of 10 μm. Verification / Alternative check: Vendor cut-sheets for strainers and coarse filters typically list reliable retention above 10 μm, aligning with the rule of thumb. Why Other Options Are Wrong: 5 μm overlaps microfiltration territory. 15–20 μm are possible but unnecessarily restrictive as a general lower bound. 0.5 μm is well within microfiltration. Common Pitfalls: Overlooking particle deformability or compressible cakes; ignoring that depth media can blur the exact cut based on load and flow conditions. Final Answer: dp > 10 μm

Question 1

Ion-exchange basics — An ion-exchange resin is composed of which structural and chemical elements?

Accepted Answer

Introduction: Ion-exchange resins are ubiquitous in water treatment and bioseparations. Their performance relies on a solid matrix bearing charged groups and mobile counter-ions that can exchange with ions from the liquid phase. Recognizing all three components is essential to understand capacity and selectivity. Given Data / Assumptions: Typical styrene–divinylbenzene or acrylic-based matrices. Functional groups tailored as cationic or anionic exchangers. Presence of exchangeable counter-ions (e.g., H+, Na+, Cl−, OH−). Concept / Approach: The polymeric network provides mechanical strength and porosity. Functional groups provide the fixed charge sites. Counter-ions balance charge and participate in exchange equilibria with solutes, enabling separations based on charge and affinity. Step-by-Step Solution: Confirm the need for a solid backbone to hold charged groups and create pores.Identify the role of fixed ionic groups in binding and exchanging ions.Recognize that counter-ions are essential for electroneutrality and exchange processes. Verification / Alternative check: Titration of resins reveals exchange capacity corresponding to functional groups and associated counter-ions; microscopy shows porous polymer matrices supporting transport. Why Other Options Are Wrong: Any single component alone cannot perform ion exchange. An inert polymer without charged groups cannot exchange ions. Common Pitfalls: Ignoring water content and swelling that affect transport; conflating chelating resins with simple ion exchangers but both still require the triad of matrix, groups, and counter-ions. Final Answer: All of these

Question 2

Cost analysis — In fermentor system costing, the annual depreciation is typically what percentage range of the installed capital cost?

Accepted Answer

Introduction: Economic evaluation of bioprocess plants requires estimating annualized capital charges, including depreciation. Thumb rules provide reasonable ranges for early-stage feasibility studies where detailed tax and depreciation schedules are not yet set. Given Data / Assumptions: Installed capital includes vessels, utilities tie-ins, and instrumentation. Straightforward rule-of-thumb budgeting stage. Industry-typical service life and salvage assumptions. Concept / Approach: For preliminary estimates, annual depreciation is often approximated as a percentage of installed capital. A common range used in bioprocess conceptual design is 6–10 percent, capturing typical asset lifetimes and salvage values before full discounted cash flow models are built. Step-by-Step Solution: Identify the commonly cited range from design heuristics.Select the choice that spans expected variability in equipment types and lifetimes: 6–10%.Note that detailed financial models may shift the effective rate. Verification / Alternative check: Texts on process design list annual capital charges (including depreciation and interest) often totaling 15–25 percent; the depreciation subcomponent is frequently taken as 6–10 percent for quick estimates. Why Other Options Are Wrong: Narrower bands like 5–6% or 6–7% may understate typical variability. 8–10% covers the upper slice but omits lower common values. 12–18% is too high for depreciation alone in most bioprocess contexts. Common Pitfalls: Confusing depreciation with total capital charge (which may include interest on investment and maintenance); assuming tax code specifics without jurisdictional detail. Final Answer: 6–10% of capital cost

Question 3

Liquid–liquid extraction — For a useful separation, the distribution selectivity should be...

Accepted Answer

Introduction: In extraction, selectivity quantifies preferential partitioning of a solute between phases, typically defined as a ratio of distribution coefficients or partition coefficients for desired vs undesired species. A favorable selectivity is critical for stage efficiency and solvent economy. Given Data / Assumptions: Standard definition of selectivity S as a ratio exceeding unity for favorable separation. Isothermal operation without chemical reaction. Steady partitioning behavior. Concept / Approach: When S > 1, the target component prefers the extract phase relative to the raffinate or relative to impurities, enabling effective separation with a reasonable number of stages. S ≤ 1 implies poor or no selectivity, demanding impractical solvent flows or excessive stages. Step-by-Step Solution: Recall definition: S = K_A / K_B for components A and B or equivalent selectivity expression.Assess requirement: For enrichment, S must exceed unity.Conclude that S > 1 is desired. Verification / Alternative check: McCabe–Thiele style constructions for extraction show rapidly converging stage counts when S > 1 and divergence when S ≤ 1. Why Other Options Are Wrong: < 1 or = 1 provide little to no driving force. Zero or negative are nonphysical in this context. Common Pitfalls: Confusing distribution ratio D with selectivity S; assuming constant S when complexation or pH effects change partitioning. Final Answer: > 1 (greater than one)

Question 4

Cell disruption — In a bead mill, the overall efficiency of cell breakage depends primarily on which factors?

Accepted Answer

Introduction: Bead milling disrupts cells by mechanical shear and impact between beads and cell suspensions. Performance depends on multiple interacting variables that determine energy dissipation, collision frequency, and stress magnitude on cells. Given Data / Assumptions: Typical horizontal or vertical bead mills. Appropriate bead loading and size for the organism. Operated within manufacturer-recommended speeds. Concept / Approach: Higher cell concentration increases probability of collisions but can raise viscosity. Bead size and loading control kinetic energy and contact events. Agitator design and speed set power density and flow patterns. Optimal disruption maximizes specific energy to cells while avoiding excessive heating. Step-by-Step Solution: Identify key levers: biomass loading, bead properties, and agitator speed.Recognize their roles: collision frequency, stress magnitude, residence time distribution.Select the comprehensive option that includes all three factors. Verification / Alternative check: Empirical optimization curves show disruption efficiency versus bead size and speed, with performance sensitive to cell concentration and viscosity. Why Other Options Are Wrong: Each single-factor option is incomplete. Coolant temperature matters for viability but is not the primary determinant of mechanical breakage efficiency. Common Pitfalls: Overloading beads leading to poor mixing; running at excessive speed causing bead wear and product heating; ignoring screen size that affects bead retention and product quality. Final Answer: All of these

Question 5

Adsorption fundamentals — Conventional adsorption is generally considered a...

Accepted Answer

Introduction: Adsorption involves the accumulation of molecules at a solid or liquid surface. Conventional physisorption is dominated by weak interactions and is characterized by equilibria that can be reversed by changing conditions such as concentration or temperature. Given Data / Assumptions: Physisorption context (van der Waals, hydrogen bonding). Isothermal operation or known isotherm behavior. Absence of strong chemisorption with irreversible bonding. Concept / Approach: Physisorption exhibits equilibrium isotherms (e.g., Freundlich, Langmuir). Because the binding is not covalent, desorption is possible by dilution, heating, or pressure swing, making the process reversible in the engineering sense. Step-by-Step Solution: Identify interaction type: weak, noncovalent for conventional adsorption.Relate to isotherms: surface coverage follows equilibrium relationships.Conclude reversibility under appropriate operating changes. Verification / Alternative check: Temperature-programmed desorption and pressure swing adsorption cycles demonstrate recoverable capacity, confirming reversibility of standard adsorption processes. Why Other Options Are Wrong: Irreversibility is more typical of strong chemisorption or reactive adsorption. Claiming non-thermodynamic or no equilibrium contradicts observed isotherm behavior. Common Pitfalls: Conflating physisorption with chemisorption; ignoring hysteresis that can occur due to pore structure yet does not preclude overall reversibility. Final Answer: reversible process

Question 6

Membrane separations — If a solute passes through a membrane freely, what is the value of the rejection coefficient s?

Accepted Answer

Introduction: The rejection coefficient s (also called sigma) quantifies solute retention by a membrane. It is defined such that s = 1 indicates complete rejection and s = 0 indicates no rejection (free passage). Understanding s is essential for designing ultrafiltration, nanofiltration, and reverse osmosis steps. Given Data / Assumptions: Definition: s = 1 − (C_p / C_f), where C_p is permeate concentration and C_f is feed or bulk concentration at the membrane surface. Steady operating conditions; negligible concentration polarization for the qualitative conclusion. Solute passes freely. Concept / Approach: If the solute passes freely, then C_p ≈ C_f, giving s ≈ 0. As retention increases, C_p decreases relative to C_f and s increases toward 1. Values outside 0 to 1 are nonphysical for ideal definitions and data without artifacts. Step-by-Step Solution: Start from s = 1 − (C_p / C_f).For free passage: C_p / C_f ≈ 1.Compute s ≈ 1 − 1 = 0. Verification / Alternative check: Membrane characterization tables list s near zero for fully permeating tracers and near unity for fully retained macromolecules, matching the calculation. Why Other Options Are Wrong: 1 implies total rejection, not free passage. > 1 or negative contradict the standard definition. < 1 but greater than 0 implies partial rejection, not free passage. Common Pitfalls: Using bulk feed concentration instead of membrane-surface concentration when polarization is significant; misinterpreting observed s due to analytical noise at very low C_p. Final Answer: 0

Question 7

Particle size cut — Conventional (cake) filtration typically separates relatively large particles of approximately what minimum size?

Accepted Answer

Introduction: Filtration modalities are often classified by particle size: conventional filtration for larger particles, microfiltration for submicron to several micrometers, ultrafiltration for macromolecules, and so on. Knowing the approximate cut helps select the right technology for cell removal or clarification steps. Given Data / Assumptions: Gravity or pressure-driven conventional filter media (e.g., depth filters, screens). Rigid particles forming a cake. Qualitative threshold understanding. Concept / Approach: Conventional filtration is typically applied to particles larger than roughly 10 μm, where sieving and cake mechanisms dominate. Smaller particles require microfiltration or finer depth media with controlled pore structures. Step-by-Step Solution: Place conventional filtration at coarser end of the spectrum.Recall that microfiltration spans approximately 0.1 to 10 μm.Therefore, a practical minimum size for classical filtration is on the order of 10 μm. Verification / Alternative check: Vendor cut-sheets for strainers and coarse filters typically list reliable retention above 10 μm, aligning with the rule of thumb. Why Other Options Are Wrong: 5 μm overlaps microfiltration territory. 15–20 μm are possible but unnecessarily restrictive as a general lower bound. 0.5 μm is well within microfiltration. Common Pitfalls: Overlooking particle deformability or compressible cakes; ignoring that depth media can blur the exact cut based on load and flow conditions. Final Answer: dp > 10 μm

Question 8

Microfiltration range — Microfiltration for removing suspended material such as bacteria generally uses membranes with pore sizes in which range?

Accepted Answer

Introduction: Microfiltration membranes are commonly used to remove bacteria, cell debris, and other suspended solids from liquids. Correctly identifying the pore size range is important for selecting between microfiltration, ultrafiltration, and nanofiltration technologies in downstream processing. Given Data / Assumptions: Bacterial cell diameters typically range from roughly 0.5 to several micrometers. Membranes are characterized by nominal pore size. Pressure-driven, crossflow or dead-end operation. Concept / Approach: Microfiltration is widely defined as spanning approximately 0.1 to 10 μm. Pores smaller than roughly 0.1 μm belong to ultrafiltration or tighter classifications; pores much larger than a few micrometers drift toward coarse screening rather than membrane microfiltration. Step-by-Step Solution: Relate target (bacteria) to pore size needed for retention.Recall standard membrane taxonomy: microfiltration 0.1–10 μm, ultrafiltration ~1–100 nm, nanofiltration ~0.5–2 nm by effective MWCO behavior.Select the range that matches microfiltration: 0.1 to 10 μm. Verification / Alternative check: Supplier catalogs list microfiltration cartridges at 0.1, 0.2, 0.45, and 1.0 μm nominal ratings for bioburden control, consistent with this range. Why Other Options Are Wrong: Ångström ranges (options b and d) are ultrafiltration or nanofiltration scales. 20–30 μm are coarse screens or prefilters, not typical microfiltration membranes. 0.005–0.05 μm (5–50 nm) is ultrafiltration, not microfiltration. Common Pitfalls: Confusing nominal pore size with absolute retention rating; forgetting that real retention depends on particle shape and operating conditions. Final Answer: 0.1 to 10 μm

Question 9

Bioprocessing — Pretreatment Steps for Biological Feed Streams: Which of the following unit operations are considered valid pretreatments to condition feeds before downstream processing (e.g., clarification, filtration, or adsorption)?

Accepted Answer

Introduction: Pretreatment prepares biological feed streams for efficient downstream processing. The goal is to protect product quality while improving separations such as filtration, centrifugation, and chromatography. This question checks recognition of common, legitimate pretreatment strategies used in bioprocessing. Given Data / Assumptions: The feed contains biomolecules (cells, proteins, polysaccharides, lipids) typical of fermentation broths or extracts. Operations occur before primary clarification or capture steps. Product stability limits (temperature, pH, shear) are respected. Concept / Approach: Pretreatment aims to modify physical/chemical properties to enhance unit operations. Heat can selectively denature unwanted enzymes or contaminants, reducing viscosity or fouling and improving stability. Filter aids increase cake porosity and permeability, enabling higher flux and avoiding compressible cakes. Electrolytes alter ionic strength and zeta potential, promoting flocculation/precipitation of impurities or stabilizing target proteins depending on the objective. Together, these tools tune rheology, particle size, and interfacial interactions for better separations. Step-by-Step Solution: Identify typical pain points: high viscosity, fine colloids, proteolysis, fouling.Map to pretreatments: heat to inactivate proteases; filter aids to form a porous precoat; salts to aggregate fines or adjust binding behavior.Evaluate each option as a legitimate, widely used pretreatment under proper controls.Conclude that all listed actions qualify as pretreatments. Verification / Alternative check: Process development routinely screens temperature holds, precoat types/doses, and conductivity adjustments to maximize clarification efficiency and capture yields without harming the critical quality attributes of the product. Why Other Options Are Wrong: E: Contradicted by standard bioprocess texts and practice; all listed methods are valid when appropriately applied. Common Pitfalls: Overheating can denature the product; excessive filter-aid dosing can entrap product; too much salt can cause irreversible precipitation. Always optimize dose and conditions via small-scale studies. Final Answer: All of the above

Question 10

Protein Adsorption Isotherms — Identify the appropriate correlation: Given Ymax (maximum adsorbed mass per mass of adsorbent), X (solute mass fraction in the diluent phase on a solute-free basis), KL (constant), and Y (equilibrium adsorbed mass per …

Accepted Answer

Introduction: Adsorption of proteins onto solids is often described using a Langmuir-type isotherm that assumes a finite number of equivalent binding sites and monolayer coverage. Recognizing the correct algebraic form is essential for parameter estimation and scale-up of capture steps. Given Data / Assumptions: Y is the mass of solute adsorbed per mass of adsorbent at equilibrium. Ymax is the saturation capacity (monolayer limit). X is the solute mass fraction in the liquid phase (solute-free basis). KL is an empirical constant linked to affinity. Concept / Approach: The classic Langmuir model in concentration terms is q = qmax * C / (KL + C). Replacing concentration with a proportional composition variable X yields Y = Ymax * X / (KL + X). This captures initial linear uptake at low X and asymptotic saturation as X increases, consistent with site-limited adsorption behavior often observed with proteins on hydrophobic or ion-exchange surfaces under noncompetitive conditions. Step-by-Step Solution: Match variables: Y ↔ adsorbed loading, X ↔ liquid-phase composition variable.Recall Langmuir structure: numerator includes the driving variable (X), denominator adds the affinity term KL.Select expression that shows saturation to Ymax as X → ∞.Confirm linear Henry-like behavior for small X (Y ≈ (Ymax/KL) * X). Verification / Alternative check: Plotting 1/Y versus 1/X (Lineweaver–Burk linearization) yields a straight line if Langmuir holds, enabling estimates of Ymax and KL. Why Other Options Are Wrong: B omits X in the numerator and does not approach Ymax correctly; C lacks saturation behavior; D has incorrect subtraction form; E is not a standard isotherm representation and fails limiting checks. Common Pitfalls: Misusing X units or mixing mass fraction with concentration without appropriate conversion; ensure consistency when fitting data. Final Answer: Y = Ymax * X / (KL + X)

Downstream Processing Questions