Multiple choice question for engineering
1. Is substrate only needed for product formation?
Answer: b [Reason:] In some cultures there is no extracellular product formation; for example, biomass itself is the product in manufacture of bakers’ yeast and single-cell protein. In the absence of product formation, we assume that all substrate entering the cell is used for growth and maintenance functions.
2. If the external supply is exhausted will there be exhaustion of maintenance energy?
Answer: b [Reason:] There is substrate uptake as long as there is external substrate available; when the substrate is exhausted maintenance energy is generally supplied by endogenous metabolism.
3. The supply of substrate for production and maintenance is different.
Answer: a [Reason:] Substrate consumed for maintenance does not contribute to growth; it therefore constitutes a separate substrate flow into the cell. In contrast, when production is not linked or only partly linked to energy metabolism, all or some of the substrate required for product synthesis is additional to, and separate from, that needed for growth and maintenance.
4. What is the growth condition for Saccharomyces cerevisiae with glucose as energy source?
c) Tryptophan limited
d) Glucose limited
Answer: b [Reason:] Yeast belongs to the class of Protoascomyceten and more specifically to the physiological class of Saccharomyces Cerevisiae. Saccharomyces Cerevisiae is crabtree positive yeast, which is sensitive to high substrate concentrations that results in the reduction of the oxygen uptake rate. As a result glucose is metabolized to ethanol. This metabolic pathway can be reduced by introduction of a regulated feed procedure adapted to the specific growth rate of the yeast.
5. What is the growth condition for Penicillium chrysogenum with glucose as energy source?
c) Tryptophan limited
d) Glucose limited
Answer: a [Reason:] Penicillin was the first important commercial product produced by an aerobic, submerged fermentation. Penicillin is produced by the fungus Penicillium chrysogenum which requires lactose, other sugars, and a source of nitrogen (in this case a yeast extract) in the medium to grow well.
6. In which phase Penicillin is produced?
a) Lag phase
b) Log phase
c) Exponential phase
d) Stationary phase
Answer: d [Reason:] Like all antibiotics, penicillin is a secondary metabolite, so is only produced in the stationary phase.
7. Which type of fermenter and process does Penicillin production requires?
a) Batch fermenter and fed-batch process
b) Batch fermenter and batch process
c) Continuous fermenter and fed-batch process
d) Continuous fermenter and batch process
Answer: a [Reason:] It requires a batch fermenter, and a fed batch process is normally used to prolong the stationary period and so increase production.
8. A completely mixed continuous stirred-tank reactor for the cultivation of cells is called?
Answer: b [Reason:] A completely mixed continuous stirred-tank reactor for the cultivation of cells are called chemostats. Chemostat controls flow rate and concentration of growth-limiting nutrient of liquid medium entering and exiting a growth chamber (bioreactor).
9. Which of the following equation describes the relationship between μ and residual growth limiting substrate?
a) Eyring equation
b) Van’t Hoff equation
c) Arrhenius equation
d) Monad equation
Answer: d [Reason:] The decrease in growth rate and the cessation of growth due to the depletion of substrate, may be described by the relationship between μ and the residual growth limiting substrate. This relationship is represented by an equation given by Monad in1942 which is known as Monad equation.
10. The monod equation is based on which type of kinetics?
a) Zero order kinetics
b) First-order kinetics
c) Second order kinetics
d) First-zero order kinetics
Answer: d [Reason:] Biochemical reactions involving a single substrate are often assumed to follow Michaelis–Menten kinetics, without regard to the model’s underlying assumptions. The Michaelis-Menten relationship for the effect of substrate concentration on the rate of an enzyme catalysed reaction is given by the equation:
where V = rate of reaction, Vmax = maximal rate of reaction, S = substrate concentration, and Km = substrate concentration at which reaction rate = 0.5 Vmax.
The analogous Monod equation for effect of substrate concentration on the specific growth rate of a microorganism is as follows:
where µ = specific growth rate, µmax = maximum specific growth rate in absence of substrate limitation, S = concentration of growth limiting substrate, Ks = substrate concentration which allows the organism to grow at 0.5 µmax.
11. Logistic growth model describes which type of growth?
a) Product- based growth
c) Substrate-based growth
d) Maintenance-based growth
Answer: c [Reason:] Logistic growth model describes the Substrate-based growth. It is used from beginning of exponential phase through end of max population (or stationary phase).
Modification of unlimited growth eqn:
dX/dt = µm X[1 – X/Xf]
Where, Xf = max population size or concentration
Term [1 – X/Xf] describes reduction of specific growth rate as exponential growth phase gives way to retardation phase.
12. What do you mean by the low Ks value?
a) Low affinity for the limiting substrate
b) Medium affinity for the limiting substrate
c) High affinity for the limiting substrate
d) No affinity for the limiting substrate
Answer: c [Reason:] If the organism has a very high affinity for the limiting substrate (a low Ks value) the growth rate will not be affected until the substrate concentration has declined to a very low level. Thus, the deceleration phase for such a culture would be short.
13. The biomass concentration is at the highest level in which phase?
a) Lag phase
b) Log phase
c) Exponential phase
d) Stationary phase
Answer: c [Reason:] The biomass concentration at the end of the exponential phase is at its highest level. Therefore the decline in substrate concentration will be very rapid so that the time period during which the substrate concentration is close to Ks is very short.
14. Which of the following phase is known as the “Maximum population phase”?
a) Lag phase
b) Log phase
c) Exponential phase
d) Stationary phase
Answer: d [Reason:] The stationary phase in batch culture is that point where the growth rate has declined to zero. This phase is also known as the maximum population phase.
15. Monod and Michealis-Menten kinetics are same.
Answer: b [Reason:] Difference between the two types of kinetics (Monod and M ichaelis-Menten):
– Michaelis-Menten – quantity of reactive material (enzyme) is constant
– Monod – quality of reactive material (cells) is increasing.
1. What do you mean by “kLa”?
a) Volumetric mass transfer coefficient
b) Henry’s law coefficient
c) Volumetric oxygen transfer coefficient
d) Volumetric Solute transfer coefficient
Answer: c [Reason:] The value of the specific exchange surface (a) is difficult to determine for small bubbles found in a bioreactor. So, the entire term “kLa” is often called the volumetric oxygen transfer coefficient.
2. For which type of mass transfer does Oxygen-Balance method is used?
a) Gas- Gas
Answer: c [Reason:] This technique is based on the equation for gas-liquid mass transfer. The oxygen content of gas streams flowing to and from the fermenter is measured.
3. What is the proper concentration unit of kLa?
Answer: a [Reason:] In this relation, the volumetric oxygen transfer coefficient, KLa, has the units of mmol, of O2/ml. h. unit concentration gradient. Using the proper concentration units, KLa has the unit of reciprocal of time (i.e., time-1).
4. Which of the following is the efficient value for oxygen concentration?
a) Equal to Ccrit
b) Below Ccrit
c) Above Ccrit
Answer: c [Reason:] It is important that the oxygen concentration remains above Ccrit so that the rate of oxygen uptake by the cells is independent of oxygen level.
5. A 20-1 stirred fermenter containing a Bacillus thuringiensis culture at 30°C is used for production of microbial insecticide, kLa is determined using the dynamic method. Air flow is shut off for a few minutes and the dissolved-oxygen level drops; the air supply is then re-connected. When steady state is established, the dissolved-oxygen tension is 78% air saturation. The following results are obtained.
a) 0.080 s-1
b) 0.083 s-1
c) 0.085 s-1
d) 0.081 s-1
Explanation: (C_AL ) ̅ = 78% air saturation. Let us define t1 = 5s, CAL1 = 50%, t2 = 15 s and CAL2 = 66%. From the equation:
6. Refer to Q5 and, calculate: An error is made determining the steady-state oxygen level which, instead of 78%, is taken as 70%. What is the percentage error in kLa resulting from this 10% error in C ̅AL?
Answer: d [Reason:] If C ̅AL is taken to be 70% air saturation:
The error in kLa is almost 100%.
7. “The kLa value will depend upon the design and operating conditions of the fermenter”, is this statement true or false?
Answer: a [Reason:] The value of kLa is unique to both the size and configuration of a reactor vessel. Although some empirically derived expressions have been published for predicting kLa values in non-Newtonian fluids, there is no agreed upon set of equations that account for all of the variables that can affect the results. Accordingly, predicting kLa is not possible, and kLa studies need to be performed for bioreactors individually. The kLa value will depend upon the design and operating conditions of the fermenter and will be affected by the variables such as
agitation rate and
8. Which of the following technique does not require the measurement of dissolved oxygen concentrations?
a) Dynamic gassing out method
b) Static gassing out method
c) Oxygen-Balance method
d) Sulphite oxidation method
Answer: d [Reason:] The oxygen-transfer rate is determined by the oxidation of sodium sulphite solution. This technique does not require the measurement of dissolved oxygen concentrations.
As oxygen enters solution it is immediately consumed in the oxidation of sulphite, so that the sulphite oxidation rate is equivalent to the oxygen-transfer rate. Since the dissolved oxygen concentration, is zero then the kLa may then be calculated from the equation:
kLa = OTR/ C*
where OTR is the oxygen transfer rate.
9. Speed is the factor affecting the value of kLa.
Answer: a [Reason:] Because kLa measurements involve monitoring levels of DO following a system perturbation, the results can be influenced by the response time (or “speed”) of a sensor making those determinations. Sensors with response times (τr) on the order of the first-order time constant of the mass transfer (1/ kLa) require special treatment of their data to correct for the time lag in readings introduced by the oxygen sensor.
10. Which of the following is not a chemical method to measure kLa?
a) Dynamic gassing out method
b) Sodium sulfite oxidation method
c) Carbon dioxide absorption method
d) Glucose oxidase method
Answer: a [Reason:] This method hinges on the measurement of the dissolved oxygen concentration that is altered by absorption or desorption, facilitated by flushing with inert gases like nitrogen. The instantaneous dissolved oxygen concentration can be measured by using electrodes and kLa is hence estimated from the slope of the resulting plot.
11. Which of the following does not affect KLa value?
a) Air flow rate
b) Presence of enzymes
c) Presence of antifoam agents
d) Degree of agitation
Answer: b [Reason:] The mass transfer coefficient is strongly affected by agitation speed and air flow rate. The mass transfer coefficient increases with agitation speed and air flow rate. Since fermentation is usually conducted at constant temperature and pressure so thermodynamically antifoam agents make the foam unstable causing ∆G < va ∆A (where ΔG is the free energy change, va is the surface tension and ∆A is the change in area).
12. A 10,000 liter (of liquid) bioreactor contains 5 g / L of growing cells qO2 = 20 mmoles O2 / (g cells hr) DT = 2 m, Di = 1 m, (6 – blade turbine agitator) x 3 blades and CL = 1 mg O2/L. Calculate OUR.
a) 200 mmoles O2 / (g cells hr)
b) 250 mmoles O2 / (g cells hr)
c) 100 mmoles O2 / (g cells hr)
d) 150 mmoles O2 / (g cells hr)
Answer: c [Reason:] OUR = X qO2 = (5 g / L) (20 mmoles O2 / (g cells hr)) = 100 moles O2 / (g cells hr).
13. Refer to Q12 and, calculate OTR. (Given: kLa = 169 mmol O2/ 1 hr atm, P* = 0.0263 atm and PO2 = 0.21 atm).
a) 30.05 mmoles O2/ liter hr
b) 31.05 mmoles O2/ liter hr
c) 20.05 mmoles O2/ liter hr
d) 21.05 mmoles O2/ liter hr
Answer: b [Reason:] OTR = kLa(PO2 – P*)
= 169 mmol O2/ 1 hr atm(0.21- 0.0263) atm
= 31.05 mmoles O2/ liter hr.
14. From Q12 and Q13, which of the following condition is relevant?
Answer: b [Reason:] Since OUR > OTR, we must modify the bioreactor operation in order to bring them into balance:
• increase N
• use pure O2 rather than air.
15. KLa is measured in the absence of microorganisms?
Answer: a [Reason:] The methods to measuring the kLa in a microbial bioprocess can be classified into the absence of microorganisms or with dead cells and in the presence of biomass that consumes oxygen at the time of measurement.
1. The equation of aerobic fermentation is a qualitative term –
Carbon and energy source + Nitrogen source + O2 + other requirements → Biomass + products + CO2 + H2O + heat.
Answer: b [Reason:] This equation should be expressed in quantitative terms, which is important in the economical design of media if component wastage is to be minimal. Thus, it should be possible to calculate the minimal quantities of nutrients which will be needed to produce a specific amount of biomass.
2. Biomass and product formation are related.
Answer: a [Reason:] Knowing that a certain amount of biomass is necessary to produce a defined amount of product, it should be possible to calculate substrate concentrations necessary to produce required product yields. There may be medium components which are needed for product formation which are not required for biomass production.
3. For bacteria with a Y for glucose of 0.5, which is 0.5 g cells g-1 glucose, the concentration of glucose needed to obtain 30 g dm-3 cells will be ____________
a) 20 g dm-3 glucose
b) 40 g dm-3 glucose
c) 50 g dm-3 glucose
d) 60 g dm-3glucose
4. For bacteria with a Y for glucose of 0.5, which is 0.5 g cells g-1 glucose, the concentration of glucose needed to obtain 60 g dm-3 cells will be ______________
a) 50 g dm-3 glucose
b) 60 g dm-3glucose
c) 100 g dm-3glucose
d) 120 g dm-3glucose
5. For bacteria with a Y for glucose of 0.2, which is 0.2 g cells g-1 glucose, the concentration of glucose needed to obtain 60 g dm-3 cells will be __________________
a) 60 g dm-3 glucose
b) 30 g dm-3glucose
c) 300 g dm-3glucose
d) 600 g dm-3 glucose
6. The culture medium should not ________________
a) Be sterilized
b) Be cheap and readily available
c) Contain desired products
d) Allow high yield of undesired products
Answer: d [Reason:] There should not be any undesirable changes in the consistency of the medium during preparation of media and after sterilization. There should be minimum production of unwanted products.
7. Medium containing pure compounds is suitable for large scale.
Answer: b [Reason:] It is easy to devise a medium containing pure compounds on a small scale but this medium may be unsuitable for use in a large scale fermentation processes.
8. Which of the following is not an antifoaming agent?
a) Sodium lauryl sulfate
d) Alkyl polyacrylates
Answer: a [Reason:] Sodium laureth sulfate, or sodium lauryl ether sulfate (SLES), is a detergent and surfactant found in many personal care products (soaps, shampoos, toothpastes, etc.). It is an inexpensive and effective foamer. Sodium lauryl sulfate (also known as sodium dodecyl sulfate or SDS) and ammonium lauryl sulfate (ALS) are commonly used alternatives to SLES in consumer products.
9. Chelating agents prevent formation of insoluble of _________________________
a) Calcium precipitates
b) Enzyme precipitates
c) Metal precipitates
d) Product precipitates
Answer: c [Reason:] A chelate is a chemical compound composed of a metal ion and a chelating agent. A chelating agent is a substance whose molecules can form several bonds to a single metal ion. In other words, a chelating agent is a multidentate ligand. An example of a simple chelating agent is ethylenediamine.
10. Which of the following is an upstream process?
a) Product recovery
b) Product purification
c) Media formulation
d) Cell lysis
Answer: c [Reason:] Upstream processing includes formulation of the fermentation medium, sterilisation of air, fermentation medium and the fermenter, inoculum preparation and inoculation of the medium.
11. Suppression of regulatory mechanism is necessary in mutation for strain improvement.
Answer: a [Reason:] Microorganisms usually have regulatory mechanisms that control the amount of metabolites synthesized, therefore suppression of regulatory mechanisms is necessary to develop the strains for higher yields.
1. What does X in “Xn” represents in the large number of experiments?
a) Number of variables
b) Number of levels
c) Representation of any element
d) Number of elements
Answer: b [Reason:] To make a full factorial search which would examine each possible combination of independent variable at appropriate levels could require a large number of experiments, xn, where x is the number of levels and n is the number of variables. This may be quite appropriate for three nutrients at two concentrations but not for six nutrients at three concentrations.
2. In the Plackett-Burman method, the evaluation of x-1 variables by x experiments, where x must be the multiple of?
Answer: b [Reason:] The Plackett-Burman method allows for the evaluation of x-1 variables by x experiments, x must be a multiple of 4, e.g. 8, 12, 16, 20, 24, etc. Normally one determines how many experimental variables need to be included in an investigation and then selects the Plackett-Burman design which meets that requirement most closely in multiples of 4.
3. Experimental error and dummy variables in Placket-Burman method are linked.
Answer: a [Reason:] Any factors not assigned to a variable can be designated as dummy variables. Alternatively, factors known to not have any effect may be included and designated as dummy variables. The incorporation of dummy variables into an experiment makes it possible to estimate the variance of an effect (experimental error).
4. The response surface optimization technique is also given by Plackett- Burman, which is another stage in medium optimization.
Answer: b [Reason:] The response surface optimization techniques were introduced by Box and Wilson in 1951.
5. The contour plot is __________
d) Any dimension
Answer: b [Reason:] The axes of the contour plot are the experimental variables and the area within the axes is termed the response surface. To construct a contour plot, the results of a series of experiments employing different combinations of variables are inserted on the surface of the plot at the points delineated by the experimental conditions. Points giving the same results are then joined together to make a contour line. In its simplest form two variables are examined and the plot is two dimensional.
6. Is there any serum named as “Chicken serum”?
Answer: a [Reason:] Chicken serum is collected from healthy birds slaughtered for human consumption. Chicken serum is tested specifically for cytotoxicity. Test samples are examined for the ability to support the attachment and proliferation of three adherent cell lines. The test and reference cultures are monitored for evidence of nutritional deficiency, abnormal morphology, or cytotoxicity.
7. Is serum-free media beneficial?
Answer: a [Reason:] The advantages of removing serum from media include:
i) More consistent and definable medium composition to reduce batch variation.
ii) Reduction in potential contamination to make sterility easier to achieve.
iii) Potential cost savings because of cheaper replacement components.
iv) Simplifying downstream processing because the total protein content of the medium has been reduced.
8. What do you mean by the term “Trace elements”?
a) Needed in small amount
b) Needed at medium amount
c) Needed at very high amount
d) The elements which are discovered
Answer: a [Reason:] A chemical element required only in minute amounts by living organisms for normal growth.
9. Hybridoma technique is not required for antibody production.
Answer: b [Reason:] Hybridoma technology is a method for producing large numbers of identical antibodies (also called monoclonal antibodies).The myeloma cell line that is used in this process is selected for its ability to grow in tissue culture and for an absence of antibody synthesis.
10. What is Hepes?
a) An animal culture
b) A buffer
c) A type of non-ionic element
d) A horse serum
Answer: b [Reason:] HEPES (4-(2-hydroxyethyl)-1-piperazineethanesulfonic acid) is a zwitterionic organic chemical buffering agent; one of the twenty Good’s buffers. HEPES is widely used in cell culture, largely because it is better at maintaining physiological pH despite changes in carbon dioxide concentration (produced by cellular respiration) when compared to bicarbonate buffers, which are also commonly used in cell culture.
11. Which of the following is not a foaming agent?
a) Sodium lauryl sulfate
b) Ammonium lauryl sulfate
c) Titanium hydride
d) Pluronic f-68
Answer: d [Reason:] Pluronic F68 is used in cell culture as a stabilizer of cell membranes protecting from membrane shearing and additionally acts as an anti-foaming agent.
12. Casein is converted to yoghurt because of _____________
a) Lactic acid
b) Salicylic acid
c) Benzoic acid
d) Picric acid
Answer: a [Reason:] Compared to other milk products such as cheese, ice cream, or butter, yogurt contains all milk proteins and salts whereas the milk sugar (lactose) is mostly converted into lactic acid which gives yogurt a pleasant acidic flavor.
13. A trickling filter is used for –
a) Antibiotic production
b) Beer production
c) Waste water treatment
d) Citric acid manufacturing
Answer: c [Reason:] A trickling filter , also called trickling biofilter, biofilter, biological filter and biological trickling filter , is a fixed-bed, biological reactor that operates under (mostly) aerobic conditions. Pre-settled wastewater is continuously ‘trickled’ or sprayed over the filter.
14. Blood agar is an example of which type of media?
Answer: b [Reason:] Blood agar is a differential medium that distinguishes bacterial species by their ability to break down the red blood cells included in the media. Blood agar is often used to distinguish between the different species of pathogenic Streptococcus bacteria.
15. Sodium carboxy methyl cellulose is not a type of non-nutritional media supplements.
Answer: b [Reason:] Sodium carboxy methyl cellulose may be added to media at 0.1% to help to minimize mechanical damage caused by the shear force generated by the stirrer impeller.
1. Thiele modulus is used for the solid catalyst.
Answer: b [Reason:] The Thiele Modulus was developed to describe the relationship between diffusion and reaction rate in porous catalyst pellets with no mass transfer limitations. This value is generally used in determining the effectiveness factor for catalyst pellets.
2. Reducing the reaction rate rA,obs improves the effectiveness of mass transfer aimed at increasing the reac tion rate.
Answer: a [Reason:] When the catalyst is very active with a high demand for substrate, mass transfer is likely to be slow relative to reaction so that steep concentration gradients are produced. However, limiting the reaction rate by operating at sub-optimum conditions or using an organism or enzyme with low intrinsic activity does not achieve the overall goal of higher conversion rates. Because rA,obs is the reaction rate per volume of catalyst, another way of reducing rA,obs is to reduce the cell or enzyme loading in the solid. This reduces the demand for substrate per particle so that mass transfer has a better chance of supplying it at a sufficient rate. Therefore, if the same mass of cells or enzyme is distributed between more particles, the rate of conversion will increase.
3. Thiele modulus will increase with the decrease of the size of the catalyst.
Answer: b [Reason:] Thiele modulus (ф) is proportional to the square of catalyst size (R2 for spheres or b2 for flat plates), reducing the catalyst size has a more dramatic effect on ф than changes in any other variable. It is therefore a good way to improve the reaction rate. In principle, mass-transfer limitations can be completely overcome if the particle size is decreased sufficiently.
4. “Increasing the bulk concentration of substrate CAb”, is included in which parameter?
a) Internal mass-transfer
b) External mass-transfer
c) Internal-External mass- transfer
d) Heat transfer
Answer: b [Reason:] External mass transfer is more rapid at high bulk substrate concentrations; the higher the concentration, the greater is the driving force for mass transfer across the boundary layer.
5. The boundary layer is necessary for the analysis of data.
Answer: b [Reason:] In large-scale reactors, external mass-transfer problems may be unavoidable if sufficiently high liquid velocities cannot be achieved. However, when evaluating biocatalyst kinetics in the laboratory, it is advisable to eliminate fluid boundary layers to simplify analysis of the data. Several laboratory reactor configurations allow almost complete elimination of interparticle and interphase concentration gradients. Operation with high liquid velocity through the bed reduces boundary-layer effects.
6. Increase in pump speed does not change the overall reaction rate.
Answer: a [Reason:] Increase in pump speed do not change the overall reaction rate. Therefore, if we can identify a liquid velocity u*L at which reaction rate becomes independent of liquid velocity, operation at uL > u*L will ensure that ηC = 1.
7. Accessibility of the surface area of the catalyst is typically non-limiting.
Answer: b [Reason:] Rate of catalytic surface reactions is proportional to the catalyst surface area. However, accessibility of that surface area is typically limiting, as most of it is inner surface area inside the porous catalyst structure. Essentially all the surface area of most typical technical catalysts is internal surface area. Hence, a catalyst can only be used at its maximum potential, if diffusion inside the pore structure is not limiting.
8. Rate of reaction over a finely crushed catalyst of radius of 0.5 mm was measured as 10.0 mole/sm3 catalyst. Temperature is 400 K and pressure is 105 Pa and mole fraction of reactant in the gas is 0.1. Find the rate for a catalyst of pellet radius of 3mm. (Assume ηC = 1 for small catalyst).
a) 7.01 mole/ m3s
b) 7.03 mole/ m3s
c) 7.05 mole/ m3s
d) 7.07 mole/ m3s
9. The internal effectiveness factor is symbolized by ________
Answer: c [Reason:] The internal effectiveness factor (based on CAs
10. The overall effectiveness factor is symbolized by __________
Answer: d [Reason:] The overall effectiveness factor (based on CAb
) is defined as: