Category: Objective Questions

Bioprocess Engineering Multiple Choice Questions on “By-Products”.

1. What is the term “Gin”?
A. A magical spirit
B. Liquor
C. Non-liquor
D. Gas
Answer: B
Explanation: Gin is liquor which derives its predominant flavour from juniper berries (Juniperus communis). Gin is one of the broadest categories of spirits, all of various origins, styles, and flavour profiles that revolve around juniper as a common ingredient.

2.What do you mean by “Stillage”?
A. Still water
B. Still waste
C. Storage and transport
D. Growth medium
Answer: C
Explanation: A stillage is like a pallet or skid but with a cage or sides or some form of support specifically tailored to the material it is intended to carry. Some are designed to be stackable. A stillage is any device on which a cask of ale is placed for service.

3. The color of the beer is proportional to the roasting of grains.
A. True
B. False
Answer: A
Explanation: Naturally, the roasting process has a considerable impact on the flavor of the beer. The more the grain is roasted, the more complex and rich the brew becomes. In particular, roasting tends to bring out stronger flavor notes, including chocolate and toffee. Darker beers also tend to have a higher alcohol content than light beers. The darker the grain, the darker the beer.

4. Single cell protein cannot be produced from stillage.
A. True
B. False
Answer: B
Explanation: A number of processes to produce SCP from stillage using specific microbes like Geotrichum candidum, Candida utilis or C.tropicalis.

5. Silage is not a feremented fodder.
A. True
B. False
Answer: B
Explanation: Silage is fermented, high-moisture stored fodder which can be fed to cattle, sheep and other such ruminants (cud-chewing animals) or used as a biofuel feedstock for anaerobic digesters. It is fermented and stored in a process called ensilage, ensiling or silaging, and is usually made from grass crops, including maize, sorghum or other cereals, using the entire green plant (not just the grain). Silage can be made from many field crops, and special terms may be used depending on type: oatlage for oats, haylage for alfalfa.

6. What are Hops?
A. A flower
B. A fruit
C. A vegetable
D. A horse shoe
Answer: A
Explanation: Hops are the flowers (also called seed cones or strobiles) of the hop plant Humulus lupulus They are used primarily as a flavoring and stability agent in beer, to which they impart bitter, zesty, or citric flavours; though they are also used for various purposes in other beverages and herbal medicine.

7. “The yeast may also be used directly as a source of vitamin”. Is the statement True or False?
A. True
B. False
Answer: A
Explanation: The yeast may also be used directly as a source of vitamins. If it is to be used as a human food it must be debittered to remove the hop bitter substances absorbed on to the yeast cells.

8. Amino acid waste cannot be used as a fertilizer.
A. True
B. False
Answer: B
Explanation: The main wastes from glutamic acid or lysine fermentations are cells, a liquor with a high amino-acid content which can be used as an animal-feed supplement, and the salts removed from the liquor by crystallization, which is a good fertilizer.

9. The baker’s yeast are now produced as a brewery by-product.
A. True
B. False
Answer: B
Explanation: Although baker’s yeast was originally obtained as a brewery by-product, this market has diminished considerably. Most baker’s yeast is now produced directly by a distinct production process.

10. What is the sludge thickener?
A. Increasing the sludge volume
B. Reducing the sludge volume
C. Constant sludge volume
D. The second step of concentration of sludge
Answer: B
Explanation: Thickening is the first step for reducing the sludge volume by removal of free sludge water. Thin sludge is concentrated to thick sludge. Thick sludge has a higher viscosity, but must still be pumpable. Sludge settles in gravity thickeners and is compressed by the weight of its own solids.

11. What can sludge from a sewage treatment plant be used?
A. Non-organic solid formation
B. Non-biosolid formation
C. Waste formation
D. Biosolid formation
Answer: D
Explanation: Biosolids is a term often used in conjunction with reuse of sewage solids after sewage sludge treatment. Biosolids can be defined as organic wastewater solids that can be reused after stabilization processes such as anaerobic digestion and composting.

12. What is activated sludge?
A. Separation of liquid and solid phase
B. Product of wastewater treatment
C. Type of wastewater treatment
D. Suspended solids
Answer: C
Explanation: The activated sludge process is a type of wastewater treatment process for treating sewage or industrial wastewaters using aeration and a biological floc composed of bacteria and protozoa.

13. What is dewatered sludge?
A. Separation of liquid and solid phase
B. Product of wastewater treatment
C. Type of wastewater treatment
D. Suspended solids
Answer: A
Explanation: Sludge dewatering is the separation of a liquid and solid phase whereby, generally, the least possible residual moisture is required in the solid phase and the lowest possible solid particle residues are required in the separated liquid phase (“the centrate”).

14. What is sewage sludge?
A. Separation of liquid and solid phase
B. Product of wastewater treatment
C. Type of wastewater treatment
D. Suspended solids
Answer: B
Explanation: Sewage sludge is a product of wastewater treatment. Wastewater and stormwater enter the sewage system and flow into wastewater treatment facilities, where the solid wastes are separated from the liquid wastes through settling. At this point, they are processed and “digested”, or decomposed by bacteria.

15. What is a primary sludge?
A. Separation of liquid and solid phase
B. Capture of suspended solids
C. Type of wastewater treatment
D. Suspended solids
Answer: B
Explanation: Primary sludge is a result of the capture of suspended solids and organics in the primary treatment process through gravitational sedimentation, typically by a primary clarifier. The secondary treatment process uses microorganisms to consume the organic matter in the wastewater.

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Chemical Reaction Engineering Multiple Choice Questions & Answers on “Reaction Kinetics Elements – Modelling the Rate Coefficient – 1”.

1. The value of rate constant for a zero order reaction is obtained by the plot between ____
A. Concentration vs time
B. Concentration vs rate
C. Rate vs time
D. Concentration vs rate
View Answer

Answer: A
Explanation: For a zero order reaction, (frac{-dC_A}{dt}) = k
Integrating the above equation between initial concentration C_A0 and final concentration C_A, (C_A0 – C_A)=kt
The slope of the graph of concentration vs time gives the rate constant for a zero order reaction.

2. Which of the following theories does not propose the temperature dependence of rate constant?
A. Arrhenius theory
B. Collision theory
C. Transition state theory
D. Bohr’s theory
View Answer

Answer: D
Explanation: Bohr’s theory proposes the model of an atom. Arrhenius theory, Collision theory, Transition state theory propose the variation of rate constant with temperature.

4. For a first order reaction, the rate constant as a function of half life is given as ____
A. (frac{0.6931}{k} )
B. 0.6931×k
C. 0.6931
D. (frac{k}{0.6931} )
View Answer

Answer: A
Explanation: Half life period is the time taken for one-half of the reactant to be consumed. For a first order reaction, ((frac{-dC}{dt})) = kCⁿ. Integrating between initial concentration C_A0 and final concentration (frac{C_{A0}}{2}), ln(2) = kt_0.5
t_0.5 = (frac{0.6931}{k}.)

5. The rate constant is a ____
A. Linear function of frequency factor
B. Exponential function of frequency factor
C. Logarithmic function of frequency factor
D. Sinusoidal function of frequency factor
View Answer

Answer: A
Explanation: By Arrhenius equation, k = (Ae^{frac{-E_a}{RT}})
K varies directly with respect to the frequency factor, A.

6. For any order n, the rate constant is expressed as ____
A. (-r_A)
B. (-r_A) × C_Aⁿ
C. (-r_A) × C_A
D. (frac{-r_A}{C_A^n} )
View Answer

Answer: D
Explanation: For any order n, (-r_A) = kC_Aⁿ. Hence, k = (frac{-r_A}{C_A^n}. )

7. For a reaction with negative order n, which one of the following is true about reaction rate constant?
A. k = (-r_A)
B. k = 1
C. k = (frac{-r_A}{C_A^n} )
D. k = (-r_A) × C_Aⁿ
View Answer

Answer: D
Explanation: For n<0, (-r_A) = kC_Aⁿ
Hence, k = (-r_A) × C_Aⁿ

8. Which among the following is true for liquid phase reactions of order n?
A. ln((frac{-dC}{dt})) = lnk + n(lnC.
B. ln((frac{-dC}{dt})) = lnk + lnC
C. ln((frac{-dC}{dt})) = nlnk + lnC
D. ln((frac{-dC}{dt})) = lnk + (frac{lnC}{n} )
View Answer

Answer: A
Explanation: ((frac{-dC}{dt})) = kCⁿ
Taking natural logarithm on both sides, ln((frac{-dC}{dt})) = lnk + n(lnC.

9. For a first order reaction, the rate constant is expressed in terms of initial concentration C_Ao and final concentration C_A as ____
A. C_A = C_Ao × k
B. C_A = C_Ao × e^-kt
C. C_A = C_Ao × kt
D. C_A = C_Ao × e^kt
View Answer

Answer: B
Explanation: ((frac{-dC}{dt})) = kC_A
ln((frac{CAo}{CA})) = kt
C_A = C_Ao × e^-kt

10. If rate of a zero order reaction is 10 (frac{mol}{m^3×s}) and reaction time is 5s, then the value of rate constant is ____
A. 1 s^-1
B. 2 s^-1
C. 5 s^-1
D. 10 s^-1
View Answer

Answer: B
Explanation: (-r_A) = kC_Aⁿ
10 = k (5)
Hence, k = 2 s^-1.

Chemical Reaction Engineering Multiple Choice Questions & Answers on “Non – Elementary Rate Laws”.

1. Which of the following reactions follows elementary rate law?
A. Formation of hydrogen bromide
B. Vapor phase decomposition of ethanal
C. Cis-trans isomerization
D. Reversible catalytic decomposition of isopropylbenzene
Answer: C
Explanation: Cis-trans isomerization follows elementary rate laws. Hydrogen bromide formation reaction as well as the reversible catalytic decomposition of isopropylbenzene are non-elementary in nature. Rate of vapor phase decomposition of ethanal is proportional to C_ethanal^3/2.

2. Langmuir Hinshelwood kinetics is followed by the reaction:
A. Formation of hydrogen bromide
B. Vapor phase decomposition of ethanal
C. Cis-trans isomerization
D. Reversible catalytic decomposition of isopropylbenzene
Answer: D
Explanation: C₆H₅CH(CH₃)₂ ←→ C₆H₆ + C₃H₆
Let’s represent this reaction symbolically as C ←→ B + P
the reaction follows the rate law -r’_C = [k(P_C – P_BP_P/K_P)] / [1 + K_CP_C + K_BP_B] which is the Langmuir Hinshelwood model.

3. Which of the following is true for fluidized catalytic beds?
A. They cannot be used for multi-phase chemical reactions
B. The bulk density is a function of the flow rate through the bed
C. They come under the category of batch reactors
D. There is no pressure drop
Answer: B
Explanation: Fluidized catalytic beds are mostly used for multi-phase reactions, for example, cumene decomposition. They come under flow reactors. The inclusion of fluidized bed offers more resistance to flow and hence results in a pressure drop.

4. The reaction H₂ + Br₂ → 2HBr proceeds via which mechanism?
A. Free-radical
B. Ionic substitution
C. Elimination
D. Pericyclic
Answer: A
Explanation: Bromine forms a free-radical and reacts with H₂. Thus, free-radical mechanism is followed. There is no elimination and no cyclic structure is formed as product or intermediate state.

5. Which of the following does not hold true for gas-solid catalyzed reactions?
A. Rate law is preferably written in terms of partial pressures
B. Rate law is preferably written in terms of concentrations
C. Rate law can be written only in terms of partial pressures
D. Rate law can be written only in terms of concentrations
Answer: A
Explanation: In most cases, for gas-solid catalyzed reactions we write the rate law in terms of partial pressures as it makes the analysis simpler. Also, it is preferable because measuring pressure is easier than measuring concentration. Partial pressure and concentration are directly related and the rate law can very easily be represented in either form.

6. For the reaction A + 2B ←→ C, what is the value of K_C if the rate law is given by
–r_A = k(C_A^2/3/C_B – K_CC_C)? Given that initial concentration of A is equal to that of B = 5M and the equilibrium conversion of B is found to be 0.3. k = 100.
A. Kc = 0
B. Kc = 0.5
C. Kc = 1
D. Kc = 1.5
Answer: C
Explanation: X_A = C_BoaX_B/C_Aob = 0.15
C_A = C_Ao(1 – X_A)
C_B = C_Ao(1 – 2X_A)
C_C = C_AoX_A
At equilibrium, take –r_A equal to 0 and solve for K_C.

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Chemical Reaction Engineering Multiple Choice Questions & Answers on “ Non Ideal Flow Basics – Properties of C, F and E Curves – 1”.

1. If M is the total amount of tracer injected and v is the volumetric flow rate of the effluent, then what is the Relationship between cumulative and exit age distribution?
A. E = (frac{C}{(frac{M}{v})} )
B. E = (frac{M}{(v)} )
C. E = (frac{C}{(Mv)} )
D. E = (frac{MC}{(v)} )
View Answer

Answer: A
Explanation: The distribution of the time spent by each fluid element leaving the reactor is called exit age distribution.
Edt = (frac{C}{(frac{M}{v})} )dt
E = (frac{C}{(frac{M}{v})}. )

2. State true or false.
The time spent by different fluid elements in the reactor is measured by a technique called the stimulus–response technique.
A. False
B. True
View Answer

Answer: B
Explanation: The stimulus response technique determines the time spent by each fluid element inside a reactor. This is done by measuring the response of the effluent stream to changes in the concentration of inert species in the feed stream.

3. State true or false.
The average residence time (t) of a fluid element is given by t = ʃ tF(t)dt, integrated between time t = 0 and t = ∞.
A. True
B. False
View Answer

Answer: A
Explanation: F(t + dt) represents the volume fraction of the fluid having a residence time of less than t + dt, and F(t) represents that having a residence time of less than t. The differential of F(t), will be the volume fraction of the effluent stream having a residence time between t and t + dt. From the principles of probability, the average residence time (t) of a fluid element is given by t = ʃ tF(t)dt.

4. Exit age distribution and F(t) are related as ____
A. (frac{dF}{dt}) = E
B. (frac{dE}{dt}) = F
C. F × t = E
D. E × V = F
View Answer

Answer: A
Explanation: F = ʃ E.dt, integrated between time t = 0 and t = t. t is the time spent by the fluid element in the reactor. (int_t^α) E dt = 1-(int_0^t)E dt. On differentiating with respect to time, (frac{dF}{dt}) = E.

7. The average residence time obtained from cumulative age distribution is ____
A. τ = (frac{ʃtC(t)dt}{ʃ C(t)dt} )
B. τ = ʃ tC(t)dt
C. τ = (frac{ʃ tC(t)dt}{ʃ dt} )
D. τ = (frac{ʃ C(t)dt}{ʃ tC(t)dt} )
View Answer

Answer: A
Explanation: The average residence time is obtained as the mean of the C curve. It is obtained as the ratio of volume to volumetric flowrate.

8. The average residence time obtained from exit age distribution is ____
A. τ = t²ʃ tC(t)dt
B. τ = (int_0^∞)E(t)dt
C. τ = (frac{ʃ tE(t)dt}{ʃ E(t)dt})
D. τ = (frac{ʃ tC(t)dt}{ʃ dt})
View Answer

Answer: B
Explanation: τ = ʃ tF(t)dt. But, (frac{dF}{dt}) = E
Hence, τ=(int_0^∞)E(t)dt.

9. The area under the curve of exit age distribution integrated between time, t = 0 and t = ∞ is ____
A. 0
B. 2
C. Unity
D. Infinite
View Answer

Answer: C
Explanation: Fraction of all the components that exit the reactor after residing inside the reactor between time t = 0 and t = α is unity. This is based on the assumption that no tracer can remain indefinitely inside the reactor.

Molecular Biology Interview Questions on “Genome Organization – 3”.

1. Which of the following facilitates nucleosome positioning?
A. Nucleosome remodeling complex
B. Topoisomerase II
C. SMC protein
D. DNA – binding protein

Answer: D
Explanation: Nucleosome positioning can be directed by DNA – binding protein or particular DNA sequences. Just as many proteins cannot bind to DNA within the nucleosome, prior binding of a protein to a site on DNA can prevent association of the core histone with that stretch of DNA. If two such DNA – binding proteins are bound to sites closer than the minimal nucleosomal DNA requirement (≈150 bp) the DNA stretch between the two proteins will remain nucleosome free.

2. Which of the following regions promote histone – DNA association?
A. A, T
B. A, G
C. G, C
D. C, T

Answer: A
Explanation: A:T rich DNA has an intrinsic tendency to bend toward the minor grove. Thus A:T rich DNA is favored in positions in which the minor grove faces the histone octamer. G:C rich DNA has the opposite tendency thus, is favored when the major grove faces away from the histone octamer.

3. With respect to end terminal modification of the histone tail which of the following options is true?
A. Lysine – phosphate modification
B. Serine – methylation
C. Acetylation – transcriptionally active
D. Modifications – involvement in gene expression

Answer: D
Explanation: Acetylation marks transcriptionally active region whereas methylation marks both transcriptionally active and repressed regions. Finally, phosphorylation histone H3 is commonly seen in highly condensed chromatin. Thus, these modifications result in a code that can be read by the proteins involved in gene regulation and expression.

4. Which of the following is a function of a modified histone tail?
A. Association with adjacent nucleosome
B. Interaction with DNA backbone
C. Interaction with chromodomains
D. Formation of 40-fold DNA structure

Answer: C
Explanation: Modification of histone tail generates binding sites for proteins. Specific protein domains called bromodomains and chromodomains mediate these interactions. Chromodomains containing proteins interact with methylated histone tails and are generally associated with tail specific methylating enzymes.

5. Bromodomains of proteins are associated with phosphorylated histone tails.
A. True
B. False

Answer: B
Explanation: Bromodomains of proteins interacts with the acetylated histone tails. These proteins are generally associated with histone tail – specific acetyl transferases. Such complexes facilitate the maintenance of acetylated chromatin by further modifying the already acetylated regions.

6. Which of the following enzymes of transcription commonly contains a bromodomain?
A. TFIID
B. CDK
C. Pre – RC
D. DNA polymerase

Answer: A
Explanation: Bromodomain containing proteins are involved in regulating transcription or formation of heterochromatin. Only TFIID of the above options is involved in transcription of DNA whereas the others are involved in DNA replication.

7. Which of the following enzymes is not correctly paired with its function?
A. Acetyl transferase – adds acetyl group to lysine
B. Methyl transferase – adds methyl group to serine
C. Topoisomerase – associated with nuclear scaffold
D. Deacetylases – removes acetyl groups from serine

Answer: B
Explanation: Methylation is an N – terminal modification of histone tail induced by the enzyme methyl transferase. But methylation is induced in the amino acid lysine. Serine is generally involved with phosphorylation.

8. Which of the following is not promoted by histone tail modification?
A. Formation of repressive structures
B. Gene expression
C. Nucleosome remodeling
D. Nucleosome sliding

Answer: D

Molecular Biology Multiple Choice Questions on “DNA Replication Occurs at the Replication Fork”.

1. Semiconservative nature of replication of eukaryotic chromosome was first demonstrated by _______________
A. Walter Flemming on root tip cells of Vicia faba
B. J. Herbert Taylor, Philip Wood and Walter Hughes on root tip cells of Vicia faba
C. Walter Flemming on root tip cells of Phaseolus vulgaris
D. J. Herbert Taylor, Philip Wood and Walter Hughes on root tip cells of Phaseolus vulgaris

Answer: B
Explanation: Walter Flemming discovered cell division not semi conservative nature of replication. Semiconservative nature of replication was first demonstrated by J. Herbert Taylor, Philip Wood and Walter Hughes on root tip cells of broad bean which is the common name for Vicia faba.

2. Pick the correct pair with respect to primers used in DNA replication.
A. RNA primer- for prokaryotes only
B. DNA primer-for eukaryotes only
C. DNA primer- for both prokaryotes and eukaryotes
D. RNA primer- for both prokaryotes and eukaryotes

Answer: D
Explanation: Short oligonucleotides of RNA are required by DNA polymerase for the synthesis of both leading and lagging strands of DNA due to the requirement of free 3’ end for DNA synthesis. As formation of oligonucleotides of DNA by DNA polymerase also requires a free 3’ end thus, DNA primers are not applicable for the synthesis of new strands of DNA during replication.

3. Which of the following is correctly matched with its subsequent role?
A. Topoisomerase II- can remove both positive and negative supercoil in the DNA duplex
B. Polymerase I- larger fragment responsible for exonuclease activity
C. DnaA protein- responsible for “melting” of the DNA double helix during replication
D. DnaB protein- attaches to the newly unwounded single strand of DNA to prevent folding of the strand

Answer: C
Explanation: The reasons are
i) Topoisomerase II- converts a positive supercoil to a negative supercoil, also known as gyrase
ii) Polymerase I- larger (klenow) fragment is responsible for polymerase activity
iii) DnaB protein- also known as helicase helps in unwinding of DNA duplex to form the open complex.

4. Replication fork is the junction between the two ___________
A. Unreplicated DNA
B. Newly synthesized DNA
C. Newly separated DNA strands and newly synthesized DNA strands
D. Newly separated DNA strands and the unreplicated DNA

Answer: D
Explanation: As both the strands of DNA occur simultaneously the two template strands undergo separation. The junction between the newly separated DNA strands and the unreplicated DNA is known as the replication fork.

5. Which of the following does not affect DNA replication?
A. Antiparallel nature of DNA
B. End specificity of polymerase
C. SSB protein
D. Helicase

Answer: C
Explanation: The antiparallel nature of DNA and end specificity of polymerase of polymerase leads to two types of strand synthesis, leading and lagging strands. Helicase helps in the unwinding of the DNA helix. SSB protein binds to the single stranded DNA during replication to stabilize it but does not take any part in new strand synthesis.

6. Who was the first person to analyse the process of replication and on which organism?
A. Arthur Kornberg: E. coli
B. John Cairns: E. coli
C. Arthur Kornberg: Bacillus subtilis
D. John Cairns: Bacillus subtilis

Answer: B
Explanation: The person who first analyzed the process of replication was John Cairns in which E. coli were grown in the presence of radioactive Thymidine. This allowed subsequent visualization of newly replicated DNA by autoradiography.

7. In the case of a circular DNA synthesis how many replication forks are observed?
A. 1
B. 2
C. 3
D. 4

Answer: B
Explanation: In case of replication of a complete circular DNA molecule 2 replication forks are observable. This represents the regions of active DNA synthesis and is represented in form of a bubble known as the replication bubble.

8. Primer synthesis short stretches of DNA for replication.
A. True
B. False

Answer: B
Explanation: Primer synthesizes short stretches of RNA for replication. These RNA serves as primers for the synthesis of both the leading and lagging strand during replication.

9. Which of the following is not used for degrading RNA from RNA:DNA hybrid in replication?
A. RNase A
B. RNase H
C. Polymerase I
D. Exonuclease

Answer: A
Explanation: RNase H dissolves RNA in eukaryotes and the last rNTP connected directly to the dNTP is digested by exonuclease. Polymerase I hydrolyses RNA in the 5’ – 3’ direction in prokaryotes. RNase A is not required in replicative purpose.

10. The gaps formed by the hydrolyzing the RNA from RNA:DNA hybrid is filled by DNA polymerase I and the nicks are ligated by T4 ligase.
A. True
B. False

Answer: B
Explanation: The gaps formed by the hydrolyzing the RNA from RNA:DNA hybrid is filled by DNA polymerase δ. The nicks are then filled in by T4 ligase at the cost of an ATP yielding an intact lagging strand.

11. Which of the following is correctly matched with its subsequent role?
A. Topoisomerase II- can remove both positive and negative supercoil in the DNA duplex
B. Polymerase I- larger fragment responsible for exonuclease activity
C. DnaA protein- responsible for “melting” of the DNA double helix during replication
D. DnaB protein- attaches to the newly unwounded single strand of DNA to prevent folding of the strand

Answer: C