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Multiple choice question for engineering

Set 1

1. Which stage is preferred for the estimation of the rate of enzymatic reaction?
a) Initial
b) Mid
c) Final
d) Stationary

View Answer

Answer: a [Reason:] Typically, only initial rate data are used. This means that several batch experiments are carried out with different initial substrate concentrations; from each set of data the reaction rate is evaluated at time zero. Initial rates and corresponding initial substrate concentrations are used as (v, s) pairs which can then be plotted in various ways for determination of Vmax and Km. Initial rate data are preferred for enzyme reactions because experimental conditions such as enzyme and substrate concentrations are known most accurately at the start of the reaction.

2. Which of the following plot has the difficulty of extrapolating vmax?
a) Line-weaver burk plot
b) Eadie-Hofstee plot
c) Michaelis-menten plot
d) Langmuir plot

View Answer

Answer: c [Reason:] This simple procedure involves plotting (v, s) values directly Vmax is the rate as s→∞ and Km is the value of s at v = Vmax/2 . The accuracy of this method is usually poor because of the difficulty of extrapolating to Vmax.

3. Which of the following plot is also known as a double reciprocal plot?
a) Line-weaver burk plot
b) Eadie-Hofstee plot
c) Michaelis-menten plot
d) Langmuir plot

View Answer

Answer: a [Reason:] The double-reciprocal equation is obtained by taking the reciprocal of both sides of the Michaelis-Menten equation. The double-reciprocal (also known as the Lineweaver-Burk) plot is created by plotting the inverse initial velocity (1/V0) as a function of the inverse of the substrate concentration (1/[S]). The Vmax can be accurately determined and thus KM can also be determined with accuracy because a straight line is formed. The slope of the resulting line is KM/Vmax, the y-intercept is 1/Vmax, and the x-intercept is -1/KM.

4. Which plot represents the following equation?
bioprocess-engineering-questions-answers-quiz-q4
a) Line-weaver burk plot
b) Eadie-Hofstee plot
c) Michaelis-menten plot
d) Langmuir plot

View Answer

Answer: d [Reason:] A Langmuir plot of s/v versus s should give a straight line with slope l/vmax and intercept Km/ vmax. Linearisation of data for the Langmuir plot minimises distortions in experimental error. Accordingly, its use for evaluation of Vmax and Km is recommended.

5. Which plot represents the following equation?
bioprocess-engineering-questions-answers-determining-enzyme-kinetic-constants-batch-data-q5
a) Line-weaver burk plot
b) Eadie-Hofstee plot
c) Michaelis-menten plot
d) Langmuir plot

View Answer

Answer: c [Reason:] This method uses a linearisation procedure to give a straight- line plot from which Vmax and Km can be determined. So that a plot of 1/v versus 1/s should give a straight line with slope Km/vmax and intercept l/vmax. This double-reciprocal plot is known as the Lineweaver-Burk plot, and is frequently found in the literature on enzyme kinetics. However, the linearisation process used in this method distorts the experimental error in v, so that these errors are amplified at low substrate concentrations.

6. What is the term “Km”?
a) Concentration of the enzyme
b) Concentration of the catalyst
c) Concentration of the product
d) Concentration of the substrate

View Answer

Answer: d [Reason:] For practical purposes, Km is the concentration of substrate which permits the enzyme to achieve half Vmax. An enzyme with a high Km has a low affinity for its substrate, and requires a greater concentration of substrate to achieve Vmax.

7. What is the unit of Vmax?
a) mmol
b) mol/sec
c) mol
d) mol/hr

View Answer

Answer: b [Reason:] Vmax “represents the maximum rate achieved by the system, at maximum (saturating) substrate concentrations”, Unit: μmol/min (or mol/s).

8. Is Vmax affected by enzyme concentration?
a) True
b) False

View Answer

Answer: a [Reason:] The reaction rate still increases with increasing substrate concentration, but levels off at a much lower rate. By increasing the enzyme concentration, the maximum reaction rate greatly increases. However, enzymes become saturated when the substrate concentration is high.

9. Km is directly proportional to the rate of enzyme binding?
a) True
b) False

View Answer

Answer: b [Reason:] Km reflects the enzyme’s dissociation constant. A high Km means weak binding (the enzyme likes to dissociate from its substrate), and a low Km means strong binding (it doesn’t like to dissociate from its substrate, meaning that it has a strong affinity for the substrate).

10. What is the unit of catalytic efficiency kcat/Km?
a) conc-1 time-1
b) conc time
c) conc-2time-2
d) conc time-1

View Answer

Answer: a [Reason:] When calculating Kcat, the concentration units cancel out, so Kcat is expressed in units of inverse time. It is the turnover number- the number of substrate molecule each enzyme site converts to product per unit time. The units of Km are those of concentration i.e. mM, mM or Km is the concentration of substrate at which half maximal velocity is observed.

11. Which of the following state represents the condition – “Initial mixing of E+S, while [ES] builds up”.
a) Enzyme is saturated with substrate
b) Pre- Steady state
c) Steady – state
d) Steady-state kinetics

View Answer

Answer: b [Reason:] In the first moment after an enzyme is mixed with substrate, no product has been formed and no intermediates exist. The study of the next few milliseconds of the reaction is called pre-steady-state kinetics.

12. What competes with substrate for active site?
a) Competitive inhibition
b) Uncompetitive inhibition
c) Non-competitive inhibition
d) Pure Non-competitive inhibition

View Answer

Answer: a [Reason:] When a fake substrate binds to the active site of an enzyme, it can’t be processed in the same way and it won’t turn into a product. A fake substrate is called a competitive inhibitor. Competitive inhibitors bind the active site of an enzyme, preventing a real substrate from binding and a product from being formed.

13. The concept of “induced fit” refers to the fact that _______
a) When a substrate binds to an enzyme, the enzyme induces a loss of water (desolvation) from the substrate.
b) Substrate binding may induce a conformational change in the enzyme, which then brings catalytic groups into proper orientation.
c) Enzyme-substrate binding induces an increase in the reaction entropy, thereby catalyzing the reaction.
d) Enzyme specificity is induced by enzyme-substrate binding.

View Answer

Answer: b [Reason:] Induced-fit model A proposed mechanism of interaction between an enzyme and a substrate. It postulates that exposure of an enzyme to a substrate causes the active site of the enzyme to change shape in order to allow the enzyme and substrate to bind.

14. A given substrate may be acted upon by a number of different enzymes, each of which uses the same substrate (s) and produces the same product (s). The individual members of a set of enzymes sharing such characteristics are known as-
a) Group specific enzymes
b) Isoenzymes
c) Allosteric enzymes
d) Substrate specific enzymes

View Answer

Answer: b [Reason:] Isozymes (also known as isoenzymes or more generally as multiple forms of enzymes) are enzymes that differ in amino acid sequence but catalyze the same chemical reaction. These enzymes usually display different kinetic parameters (e.g. different KM values), or different regulatory properties. Allozymes represent enzymes from different alleles of the same gene, and isozymes represent enzymes from different genes that process or catalyse the same reaction, the two words are usually used interchangeably.

15. 1.5 μg of enzyme gives a Vmax of 3 μmol product produced per minute. What is the turnover number for this enzyme?
a)100 sec-1
b) 200 sec-1
c) 1000 sec-1
d) 2000 sec-1

View Answer

Answer: c bioprocess-engineering-questions-answers-determining-enzyme-kinetic-constants-batch-data-q15

Set 2

1. Single cell protein (SCP) is the production of ?
a) Extracellular proteins
b) Fermentation of waste products
c) Intracellular proteins extraction
d) Metabolites

View Answer

Answer: b [Reason:] Single-cell proteins develop when microbes ferment waste materials (including wood, straw, cannery, and food-processing wastes, residues from alcohol production, hydrocarbons, or human and animal excreta). Single-cell protein (SCP) refers to edible unicellular microorganisms. The biomass or protein extract from pure or mixed cultures of algae, yeasts, fungi or bacteria may be used as an ingredient or a substitute for protein-rich foods, and is suitable for human consumption or as animal feeds.

2. What do you mean by “Trophophase”?
a) Production of waste materials
b) Production of topical products
c) Production of primary metabolites
d) Production of secondary metabolites

View Answer

Answer: c [Reason:] Trophophase: The phase in the active growth of a culture in which primary metabolites are formed.

3. What do you mean by “Idiophase”?
a) Production of waste materials
b) Production of topical products
c) Production of primary metabolites
d) Production of secondary metabolites

View Answer

Answer: d [Reason:] Idiophase: The phase in the growth of a culture during which secondary metabolites are produced.

4. Which of the following does not have the property of production of secondary metabolites?
a) Filamentous fungi
b) Filamentous bacteria
c) Sporing bacteria
d) Enterobacteria

View Answer

Answer: d [Reason:] Not all microorganisms undergo secondary metabolism –it is common amongst the filamentous bacteria and fungi and the sporing bacteria but it is not found, for example, in the Enterobacteriaceae. Thus, the taxonomic distribution of secondary metabolism is quite different from that of primary metabolism.

5. Is secondary metabolite useful?
a) True
b) False

View Answer

Answer: a [Reason:] Many secondary metabolites have antimicrobial activity, others are specific enzyme inhibitors, some are growth promoters and many have pharmacological properties. Thus, the products of secondary metabolism have formed the basis of a number of fermentation processes.

6. Microbial process is advantageous than chemical process.
a) True
b) False

View Answer

Answer: a [Reason:] Microbial processes have the additional advantage over chemical reagents of operating at relatively low temperatures and pressures without the requirement for potentially polluting heavy-metal catalysts.

7. Which of the following is an upstream process?
a) Product recovery
b) Product purification
c) Media formulation
d) Cell lysis

View Answer

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.

8. Which of the following is a downstream process?
a) Product recovery
b) Screening
c) Media formulation
d) Sterilization of media

View Answer

Answer: a [Reason:] Downstream Processing includes the recovery of the products in a pure state and the effluent treatment. Product recovery is carried out through a series of operations including cell separation by settling, centrifugation or filtration; product recovery by disruption of cells (if the product is produced intracellularly); extraction and purification of the product. Finally, the effluents are treated by chemical, physical or biological methods.

9. Which of the following is not a product of fermentation?
a) Oxygen
b) Carbon dioxide
c) Ethanol
d) Lactate

View Answer

Answer: a [Reason:] Fermentation is a metabolic process that consumes sugar in the absence of oxygen. The products are organic acids, gases, or alcohol. It occurs in yeast and bacteria, and also in oxygen-starved muscle cells, as in the case of lactic acid fermentation.

10. Alcoholic fermentation is carried by yeast known as ___________
a) Lactobacillus
b) Bacillus
c) Saccharomyces cerevisiae
d) Escherichia coli

View Answer

Answer: c [Reason:] Saccharomyces cerevisia is a species of yeast. It has been instrumental to winemaking, baking, and brewing since ancient times. It is believed to have been originally isolated from the skin of grapes (one can see the yeast as a component of the thin white film on the skins of some dark-colored fruits such as plums; it exists among the waxes of the cuticle).

11. Which of the following is not a probiotic?
a) Fungi
b) Saccharomyces cerevisiae
c) Escherichia coli
d) Lactobacillus

View Answer

Answer: a [Reason:] Probiotics are live bacteria and yeasts that are good for your health, especially your digestive system. Example – Lactobacillus species (L. acidophilus, L. casei, L. reuteri, L. bulgaricus, L. plantarum, L. johnsonii, and L. lactis), Bifidobacterium species (B. bifidum, B. longum, B. breve, B. infantis, B. lactis, and B. adolescentis), and others microbes like Bacillus cereus, Non pathogenic Escherichia coli, Saccharomyces cerevisiae, Enterococcus faecalis, and Streptococcus thermophilus.

12. Biofuels are products of fermentation.
a) True
b) False

View Answer

Answer: a [Reason:] Study and development of cell factories for production of biofuels (bioethanol, biobutanol, biodiesel) and biochemicals (3-hydroxypropionic acid and ethylene). A common challenge for these cell factories is the requirement of high yield and productivity to make the potential production cost effective and competitive with petroleum based production. An efficient cell factory requires many rounds of metabolic engineering as well as carefully designed and optimized fermentation process. Saccharomyces cerevisiae, is commonly used.

13. Biorefinery is not an example of fermentation.
a) True
b) False

View Answer

Answer: b [Reason:] An important motivation for fermentation processes is that they may become a key component in a biobased economy. Major focus is on using sustainable raw materials for the fermentation processes are to develop processes using plant cell wall material from the agricultural and forestry sectors including waste and side streams. Using these streams, leads to a number of challenges with respect to its complex composition, in particular inhibitory compounds present in the streams. Work on understanding and improving robustness of yeast strains to make them more suited to ferment efficiently in plant cell wall material derived streams is being done.

Set 3

1. The size of eddy is proportional to the streamline flow.
a) True
b) False

View Answer

Answer: b [Reason:] Turbulence in fluids produces bulk mixing on a scale equal to the smallest eddy size. Within the smallest eddies, flow is largely streamline so that further mixing must occur by diffusion of fluid components. Mixing on a molecular scale therefore relies on diffusion as the final step in the mixing process.

2. Diffusion maintains the concentration difference between high and low concentrations.
a) True
b) False

View Answer

Answer: b [Reason:] Molecular diffusion is the movement of component molecules in a mixture under the influence of a concentration difference in the system. It occurs in the direction required to destroy the concentration difference. If the gradient is maintained by constantly supplying material to the region of high concentration and removing it from region of low concentration, diffusion will continue.

3. The total rate of oxygen uptake is proportional to the concentration of __________
a) Medium
b) Liquid
c) Number of cells
d) Gas

View Answer

Answer: c [Reason:] The concentration of cell increases during course of batch culture and the total rate of oxygen uptake is proportional to the no. of cell present. The rate of oxygen consumption per cell also known as oxygen uptake rate (OUR): Qo = qox Where: Qo = oxygen uptake rate per volume broth (gL-1 s-1) qo = specific oxygen transfer rate (gg-1 s-1).

4. Which one of the following is not the type of transfer in molecular diffusion?
a) Density
b) Mass
c) Momentum
d) Energy

View Answer

Answer: a [Reason:] The molecular transfer equations of Newton’s law for fluid momentum, Fourier’s law for heat, and Fick’s law for mass are very similar. One can convert from one transfer coefficient to another in order to compare all three different transport phenomena.

5. If the rain drop drags and enter into the atmosphere, which type of diffusion will it experience?
a) Energy
b) Osmosis
c) Momentum
d) Mass

View Answer

Answer: c [Reason:] The drag experienced by a rain drop as it falls in the atmosphere is an example of momentum diffusion (the rain drop loses momentum to the surrounding air through viscous stresses and decelerates).

6. The S.I. unit of thermal diffusivity is ___________
a) m/s
b) m2/s
c) m/s2
d) m2/s2

View Answer

Answer: b [Reason:] The units of thermal diffusivity are length2/time and a common set of SI units is m2/sec (the use of cm or mm as the length scale is frequent as it allows reporting of values closer to the value one).

7. In the below equation, what does “D” represents?

bioprocess-engineering-questions-answers-role-diffusion-bioprocessing-q7

a) Length of the membrane
b) Distance
c) Diffusion
d) Thickness of the membrane

View Answer

Answer: d [Reason:] It states that the rate of diffusion of a gas across a membrane is: ▪ Constant for a given gas at a given temperature by an experimentally determined factor, K. ▪ Proportional to the surface area over which diffusion is taking place, A. ▪ Proportional to the difference in partial pressures of the gas across the membrane, P2 – P1. ▪ Inversely proportional to the distance over which diffusion must take place, or in other words the thickness of the membrane, D.

8. A bioreactor of working volume 50 m3 produces a metabolite (X) in batch culture under given operation conditions form a substrate (S). The final concentration of metabolite at the end of each run was 1.1kg m-3. The bioreactor was operated to complete 60 runs in each year. What will be the annual output of the system (production of metabolite (X)) in Kg per year?
a) 70
b) 66
c) 77
d) 60

View Answer

Answer: b [Reason:] Volume = 50 m3 Total output = no of runs * product = 60 * 1.1 = 66.

9. Example of steady-state diffusion is _______________
a) Hydrogen purification by palladium sheets
b) Doping semi-conductors
c) Corrosion resistance of aluminium
d) Decarburization

View Answer

Answer: a [Reason:] A practical example of steady-state diffusion – the purification of hydrogen gas. One side of a thin sheet of palladium metal is exposed to the impure gas composed of hydrogen and other gaseous species such as nitrogen, oxygen, and water vapor. The hydrogen selectively diffuses through the sheet to the opposite side, which is maintained at a constant and lower hydrogen pressure.

10. Fick’s second law is applicable to steady –state diffusion?
a) True
b) False

View Answer

Answer: b [Reason:] Non- steady state diffusion – Concentration of solute atoms at any point in metal changes with time in this case. And the Fick’s second law states that rate of compositional change is equal to diffusivity times the rate of change of concentration gradient.

11. The most influencing factor of diffusivity of molecules is _______________
a) Lattice structure
b) Temperature
c) Presence of defects
d) Diffusing species

View Answer

Answer: b [Reason:] Temperature. As temperature increases the average kinetic energy of particles increases. The diffusion coefficient is a physical constant dependent on molecule size and other properties of the diffusing substance as well as on temperature and pressure.

Set 4

1. What do you mean by “Axenic culture”?
a) Containing single type of organism
b) Containing two types of organism
c) Containing multiple types of organism
d) Not containing any type of organism

View Answer

Answer: a [Reason:] In biology, axenic describes the state of a culture in which only a single species, variety, or strain of organism is present and entirely free of all other contaminating organisms. Axenic culture is also an important tool for the study of symbiotic and parasitic organisms in a controlled manner.

2. Cell death in solids is more effective than liquid.
a) True
b) False

View Answer

Answer: b [Reason:] If the liquid contains contaminant particles in the form of flocs or pellets, temperature gradients may develop. Because heat transfer within solid particles is slower than in liquid, the temperature at the centre of the solid will be lower than that in the liquid for some proportion of the sterilising time. As a result, cell death inside the particles is not as effective as in the liquid. Longer holding times are required to treat solid-phase substrates and media containing particles.

3. Hold-up of large volumes of medium for longer periods of time is efficient.
a) True
b) False

View Answer

Answer: a [Reason:] Sustained elevated temperatures during heating and cooling are damaging to vitamins, proteins and sugars in nutrient solutions and reduce the quality of the medium. Because it is necessary to hold large volumes of medium for longer periods of time. Continuous sterilisation, particularly a high-temperature, short-exposure-time process, can significantly reduce damage to medium ingredients while achieving high levels of cell destruction.

4. Plug flow contains _______________
a) There is mixing
b) There is variation
c) There is no mixing but variation
d) There is neither mixing nor variation

View Answer

Answer: d [Reason:] The type of flow in pipes where there is neither mixing nor variation in fluid velocity is called Plugflow. Plug flow is an ideal flow pattern; in reality, fluid elements in pipes have a range of different velocities.

5. For an ideal plug flow, Reynolds number should be ____________
a) Low
b) Variably low
c) High
d) Not very high

View Answer

Answer: c [Reason:] Plug flow is approached in pipes at turbulent Reynolds numbers above about 2 × 104; operation at high Reynolds numbers minimises fluid mixing and velocity variation.

6. Deviation from plug flow behavior is ____________
a) Linear dipersion
b) Axial dispersion
c) Circular dispersion
d) Non-dispersion

View Answer

Answer: b [Reason:] Deviation from plug-flow behaviour is characterised by the degree of axial dispersion in the system, i.e. the degree to which mixing occurs along the length or axis of the pipe. Axial dispersion is a critical factor affecting design of continuous sterilisers.

7. Peclet number has no units.
a) True
b) False

View Answer

Answer: a [Reason:] The Peclet number is a dimensionless number used in calculations involving convective heat transfer. It is the ratio of the thermal energy convected to the fluid to the thermal energy conducted within the fluid.

8. Performance of sterilizer is directly proportional to the Peclet number.
a) True
b) False

View Answer

Answer: a [Reason:] At any given sterilisation temperature defining the value of kd and Da, performance of the steriliser declines significantly as the Peclet number decreases.

9. The pores of filter for filtration should be between in diameter is _________
a) 0.1 – 0.55 μm
b) 0.2 – 0.45 μm
c) 0.3 – 0.45 μm
d) 0.3 – 0.50 μm

View Answer

Answer: b [Reason:] Membranes used for filter sterilisation of liquids are made of cellulose esters or other polymers and have pores between 0.2 and 0.45 μm in diameter. The membranes themselves must be sterilised before use, usually by steam.

10. Heat sterilization is more effective than filtration for liquid medium.
a) True
b) False

View Answer

Answer: a [Reason:] Liquid filtration is generally not as effective or reliable as heat sterilisation. Viruses and mycoplasma are able to pass through membrane filters; care must also be taken to prevent holes or tears in the membrane. Usually, filter-sterilised medium is incubated for a period of time before use to test its sterility.

11. Membrane filters and depth filters are same.
a) True
b) False

View Answer

Answer: b [Reason:] A Depth Filter is a filter consisting of either multiple layers or a single layer of a medium having depth, which captures contaminants within its structure, as opposed to on the surface. Depth filters typically have nominal pore size ratings, Whereas, Membrane filters or “membranes” are microporous plastic films with specific pore size ratings. Also known as screen, sieve or microporous filters, membranes retain particles or microorganisms larger than their pore size primarily by surface capture.

12. Membrane filters are more efficient than depth filters.
a) True
b) False

View Answer

Answer: a [Reason:] Depth filters do not perform well if there are large fluctuations in flow rate or if the air is wet; liquid condensing in the filter increases the pressure drop, causes channelling of the gas flow, and provides a pathway for organisms to grow through the bed, Whereas, Membrane filter cartridges typically contain a pleated, hydrophobic filter with small and uniformly-sized pores 0.45 pm or less in diameter. The hydrophobic nature of the surface minimises problems with filter wetting while the pleated configuration allows a high filtration area to be packed into a small cartridge volume.

13. Deep bed filters are used for filtration of?
a) Liquid
b) Solid
c) Gas
d) Non-fluid

View Answer

Answer: a [Reason:] Deep bed filtration is a rapid and efficient method for removing small particles from liquids. Such dispersions of particles in liquids are common in a wide range of industries.

14. 20-micron filter is more efficient than 5-micron filter.
a) True
b) False

View Answer

Answer: b [Reason:] The average size of the openings between pieces of the filter media are represented in microns. For example, a 20-micron filter has larger openings than a 5-micron filter. Consequently, the 20-micron filter element will let larger particles pass through the filter than the 5-micron media would.

15. What is an MPR rating on air filters?
a) Magnitude performance rating
b) Micro-particle performance rating
c) Macro-particle performance rating
d) Moles per rate

View Answer

Answer: b [Reason:] MPR (Micro-Particle Performance Rating) MPR Rating is a rating system developed by 3M. It rates the manufacturer’s filters and their ability to capture airborne particles smaller than 1 micron.

Set 5

1. Which order represents Michelis-Menten kinetics?
a) First-Second order
b) Zero-First order
c) Zero-Second order
d) Second order

View Answer

Answer: b [Reason:] The kinetics of many biological reactions are either zero-order, first-order or a combination of these called Michaelis-Menten kinetics. The reaction order depends on the relative size of the two terms in the denominator. At low substrate concentration [S] <<KM, under these conditions the reaction rate varies linearly with substrate concentration [S] – First order kinetics. However at higher [S] with [S] >> KM, the reaction becomes independent of [S] – Zero order kinetics.

2. The relationship between reaction rate and reactant concentration is of which order from the following relation?
rA = k1 CA
a) Zero order
b) First order
c) Zero-First order
d) Second order

View Answer

Answer: b [Reason:] If a reaction obeys first-order kinetics, the relationship between reaction rate and reactant concentration is as follows: rA = k1 CA where rA is the volumetric rate of reaction and k1 is the first- order rate constant with dimensions T-1.

3. The following enzyme reaction is applicable for which order?
bioprocess-engineering-questions-answers-test-q3
a) Zero order
b) Zero-First order
c) Zero-Second order
d) First order

View Answer

Answer: b [Reason:] The kinetics of most enzyme reactions are reasonably well represented by the Michaelis-Menten equation, where rA is the volumetric rate of reaction, CA is the concentration of reactant A, Vmax is the maximum rate of reaction at infinite reactant concentration, and Km is the Michaelis constant for reactant A. Vmax has the same dimensions as rA; Km has the same dimensions as CA.

4. Refer to Q3, and estimate the unit of Km?
a) kg mol m-2
b) kg mol
c) kg mol m-1
d) kg mol m-3

View Answer

Answer: d [Reason:] Km is the Michaelis constant for reactant A. Km has the same dimensions as CA. Typical units for Vmax are kgmol mm-3 s-I; typical units for Km are kgmolmm-3.

5. What do you mean by the “Turn- over number”?
a) Occurrence of saturation
b) Conversion of amount of substrate into product
c) Conversion of amount of product into substrate
d) Enzyme unbound with substrate

View Answer

Answer: b [Reason:] When all enzyme is bound to substrate kcat, the turnover number, is the maximum number of substrate molecules converted to product per enzyme molecule per second. Further addition of substrate does not increase the rate which is said to be saturated.

6. High substrate concentration follows which order?
a) Zero order
b) First order
c) Zero- Second order
d) Second order

View Answer

Answer: a [Reason:] v ≈ vmax Therefore, at high substrate concentrations, the reaction rate approaches a constant value independent of substrate concentration; in this concentration range, the reaction is essentially zero order with respect to the substrate.

7. Convert 35℃ into K.
a) 300.08 K
b) 305.18 K
c) 308.15 K
d) 315.18 K

View Answer

Answer: c [Reason:] Convert temperatures to degrees Kelvin (K) using this equation: T(K) = T(℃) + 273.15 T = 35℃ = 308.15 K.

8. Convert 35 K into ℃.
a) 230.18
b) 238.15
c) -230.18
d) -238.15

View Answer

Answer: d [Reason:] Convert degree Kelvin (K) into temperatures using this equation: T(℃) = T(K) – 273.15 = -238.15℃.

9. Catalytic efficiency is defined as__________
a) kcat/KM
b) KM/ kcat
c) Km/ k0
d) Km/k1

View Answer

Answer: a [Reason:] The constant kcat/KM (catalytic efficiency) is a measure of how efficiently an enzyme converts a substrate into product.

10. Unbinding of Enzyme- Substrate complex increases the reaction rate.
a) True
b) False

View Answer

Answer: b [Reason:] The rate of an enzymatic reaction will increase as substrate concentration increases, and that increased unbinding of enzyme-substrate complexes will decrease the reaction rate.

11. Linear inhibition is sometimes called as ____________
a) Complete inhibition
b) Partial inhibition
c) Incomplete inhibition
d) Mixed inhibition

View Answer

Answer: a [Reason:] Linear inhibition is sometimes called complete inhibition, and the contrasting term partial inhibition is sometimes used for a type of non-linear inhibition in which saturation with inhibitor does not decrease the rate to zero. These latter terms are discouraged because they can be misleading, implying, for example, that the rate may indeed be decreased to zero in ‘complete inhibition’ at non-saturating concentrations of inhibitor.

12. The rate limiting step of Michaelis Menten kinetics is _________
a) Complex formation step
b) Non-complex formation step
c) Complex dissociation step
d) Non- complex dissociation step

View Answer

Answer: c [Reason:] The rate limiting step in the enzyme catalyzed transformation of substrate S into product P is the breakdown of the ES complex. The rate-limiting step is usually the product formation step.

13. For a given enzyme catalyzed reaction, the Michaelis constant is 0.5 mM and the substrate concentration is 3.0 mM. What is the fractional saturation of the enzyme under these conditions?
a) 80.57%
b) 85.5%
c) 85.7%
d) 80.75%

View Answer

Answer: c bioprocess-engineering-questions-answers-general-reaction-kinetics-biological-systems -q13

14. In a particular enzyme-catalyzed reaction, Vmax = 0.2 mol/sec and Km = 5 mM. Assume the enzyme shows standard Michaelis-Menten kinetics. What is the rate of the reaction when [S] = 10 mM?
a) 0.133
b) 0.331
c) 0.233
d) 0.332

View Answer

Answer: a [Reason:] v= Vmax[S]/ (Km + [S]) = 0.2×10/ (5+10) = 0.133.

15. In a particular enzyme-catalyzed reaction, Vmax = 0.2 mol/sec and Km = 6 mM. Assume the enzyme shows standard Michaelis-Menten kinetics. What is the rate of the reaction when [S] = 20 mM?
a) 0.155
b) 0.156
c) 0.153
d) 0.152

View Answer

Answer: c [Reason:] v= Vmax[S]/ (Km + [S]) = 0.2×20/ (6+20) = 0.153.