250+ TOP MCQs on Reducing the Size of Kinetic Models and Answers

Chemical Reaction Engineering Questions and Answers for Campus interviews focuses on “Reducing the Size of Kinetic Models”.

1. For the parallel reaction A → B and A → C, of rate constants k1 and k2 respectively, both reactions of order 1, the rate expression is given as ____
A. (-rA) = k1CA + k2CA
B. (-rA) = k1CA – k2CA
C. (-rA) = k2CA
D. (-rA) = k1CA

Answer: A
Explanation: For a zero order reaction, (frac{-dC_A}{dt}) = k1CA + k2CA
A forms two products: B and C. The concentration of A decreases as the reaction progresses.

2. The rate determining step of a series of reactions is the one ____
A. That is fastest
B. That is slowest
C. That does not contribute to the reaction
D. That does not occur

Answer: B
Explanation: Rate determining step is the slowest of all the reaction steps. The rate determining step is important in deriving the rate equation of a chemical reaction.

3. An assumption of Steady State Approximation is ____
A. Reaction occurs without the formation of intermediates
B. Catalyst does not accelerate reaction rate
C. One of the intermediates in the reaction is consumed as quickly as it is generated
D. Equilibrium state is attained at the end of a reaction

Answer: C
Explanation: The steady-state approximation is a method used to derive the rate expression. It is based on the assumption that one intermediate in the reaction mechanism is consumed as fast as it is generated.

4. The reaction involving oxidation of nitric oxide to nitrous oxide follows the mechanism ____
A. (frac{-dO_2}{dt}) = k[NO2]2[O2]
B. (frac{-dO_2}{dt}) = k[NO2][O2]
C. (frac{-dO_2}{dt}) = k[NO2]2[O2]2
D. (frac{-dO_2}{dt}) = k[NO2][O2]2

Answer: A
Explanation: In the first step, 2 molecules of NO form N2O2. Further oxidation of N2O2 is the slowest step and it is the rate determining step.

5. The order of the reaction involving the conversion of ozone to oxygen is ____
A. Zero order
B. First order
C. Second order
D. Third order

Answer: B
Explanation: The rate of decomposition of ozone is:
-rO3= k[O3]2[O2]-1
Order = 2-1 = 1

6. At very low concentrations of azomethane, its decomposition follows which order?
A. First order
B. Zero order
C. Third order
D. Second order

Answer: D
Explanation: The rate expression for decomposition of azomethane is:
rN2 = (frac{k_1 k_3 C_{Azo}^2}{k_3+k_2 C_{Azo}} )
Where, k1, k2 and k3 are rate constants of the 3 intermediate reactions involved in azomethane formation.
At low concentration, k2 CAzo << k3
Hence, rN2=k1CAzo2.

7. State true or false.
Gas phase decomposition of N2O follows first order mechanism for low concentrations of N2O and second order mechanism for high concentrations of N2O.
A. True
B. False

Answer: A
Explanation: The rate expression is, rN2 = (frac{k_1 NO_2^2}{1+kNO_2} )
At low concentrations of N2O, kNO2 << 1 and the reactions follows second order. At high concentrations of N2O, kNO2 >> 1 and the reaction follows first order.

8. State true or false.
Decomposition of nitrogen pentoxide follows second order mechanism, assuming that the kinetics is controlled by a single step mechanism.
A. True
B. False

Answer: B
Explanation: The rate of N2O5 decomposition is given by, r = k[N2O5].
Hence, the reaction follows first order.

9. Formation of hydrogen iodide from the corresponding elements is a ____
A. Single step mechanism
B. Two step mechanism
C. Three step mechanism
D. Four step mechanism

Answer: B
Explanation: In the slow step, iodine decomposes to give 2[I]. In the consequent step, hydrogen reacts with 2[I] to give hydrogen iodide.

10. In the formation of phosgene from carbon monoxide and chlorine, which of the following is rate determining?
A. Decomposition of chlorine
B. Formation of COCl.
C. Reaction of COCl. with Cl2
D. Decomposition of CO2

Answer: C

250+ TOP MCQs on Stoichiometry – Constant – Volume Batch Reaction Systems and Answers

Chemical Reaction Engineering Assessment Questions and Answers focuses on “Stoichiometry – Constant – Volume Batch Reaction Systems”.

1. State true or false.
Constant volume system means that the volume of reactor is constant.
A. True
B. False
Answer: B
Explanation: The volume of the reaction mixture remains constant for a constant volume system. The system is incompressible.

2. Which of the following represents constant volume system?
A. C + O2 → CO2
B. CO + H2O → CO2 + H2
C. N2 + 3H2 → 2NH3
D. SO2 + 0.5O2 → SO3
Answer: B
Explanation: Gas phase reactions in which the number of moles of the product is equal to the number of moles of reactant is a constant volume system. In CO + H2O → CO2 + H2, 2 moles of reactants form 2 moles of product.

3. What is the partial pressure of the product related to the total pressure for a constant volume system? (Where pR is the partial pressure of the product, pR0 is the initial partial pressure of the product, P is the total pressure, P0 is the initial pressure of reaction mixture, r is the stoichiometry of the product and ∆n is the difference between the number of moles of the product and the number of moles of the reactant)
A. pR = (frac{r}{∆n}) (P – P0)
B. pR = pR0 – (frac{r}{∆n}) (P – P0)
C. pR = pR0 + (P – P0)
D. pR = pR0 + (frac{r}{∆n}) (P – P0)
Answer: D
Explanation: For the reaction, aA + bB → rR + sS, pR = CRRT. CR is the product concentration. CRRT = pR0 + (frac{r}{∆n}) (P – P0).

4. The desired relationship between the partial pressure of reactant and total pressure for the reaction N2O4 → 2NO2 is _____
A. PA = PA0 – 2(P – P0)
B. PA = 2PA0 – (P – P0)
C. PA = PA0 – (P – P0)
D. PA = PA0 + (P – P0)
Answer: C
Explanation: ∆n = 2-1 =1. pA = pA0 – (frac{a}{∆n}) (P – P0), a is the stoichiometry of the reactant. Hence, PA = PA0 – (P – P0).

5. If the total pressure for reaction is initially is 2 atm and the pressure is changed to 3 atm for the reaction to occur, then the value of (PA – PA0) for the reaction N2O → N2 + 0.5O2 is ____
A. -2
B. 2
C. -1
D. 1
Answer: A
Explanation: ∆n = 0.5
pA = pA0 – (frac{a}{∆n}) (P – P0)
(PA = PA0) = (frac{-1}{0.5})(3-2)
(PA = PA0) = -2.

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250+ TOP MCQs on Non Ideal Flow Basics – RTD for Combination of Reactors and Answers

1. State true or false.
A corner peak in the E(t) indicates bypassing of reactants.
A. True
B. False
View Answer

Answer: A
Explanation: Corner peak is an identification of reactor bypassing. It is because the reactants bypassed appear on the reactor outlet as soon as they are fed in.

3. If the residence time obtained for a reactor by the RTD experiments is less than that expected, it indicates the presence of ____
A. CSTR
B. PFR
C. Bubbling bed reactor
D. Dead volume
View Answer

Answer: D
Explanation: The dead volume of a reactor does not contribute to the reactor function. The entire volume of a reactor is the sum of the active volume and dead volume.

4. If the entire volume of a PFR is 10 m3 and dead volume is 2 m3, then the active volume of the reactor is ____
A. 20
B. 5
C. 8
D. 12
View Answer

Answer: C
Explanation: Volume of PFR, V = Vp + Vd
Vp is the active volume of PFR and Vd is the dead volume.
10 = Vp + 2
Vp = 2.

6. State true or false.
The exit age distribution for the series combination of a PFR followed by a CSTR is same as that for CSTR followed by a PFR.
A. True
B. False
View Answer

Answer: A
Explanation: There is no axial mixing in a PFR. The output of the PFR is delayed no matter it is arranged in front of CSTR or following CSTR.

7. For two PFRs P1 and P2 in parallel, the exit age distribution gives ____
A. A single peak at t = tp1
B. A single peak at t = tp2
C. Two peaks, one at t = tp1 and the other at t = tp2
D. A corner peak
View Answer

Answer: C
Explanation: Two peaks are obtained. One is at tp1 = (frac{V_{p1}}{v_1}) another is at tp2 = (frac{V_{p2}}{v_2}. )
v1 and v2 are the volumetric flowrates in P1 and P2, respectively.

8. The variance of a PFR is ____
A. 1
B. 0
C. ∞
D. -1
View Answer

Answer: B
Explanation: There is no mixing inside a PFR. There is no variation in the concentration in the PFR.

9. The variance of a CSTR is ____
A. 1
B. 0
C. ∞
D. -1
View Answer

Answer: C
Explanation: There is complete mixing inside a CSTR. The reactants are uniformly mixed. Hence, there is a variation in CSTR.

250+ TOP MCQs on Palindromes, Inverted Repeats and Stem and Loop Structures and Answers

Molecular Biology Multiple Choice Questions on “Palindromes, Inverted Repeats and Stem and Loop Structures”.

1. Double-stranded structure of nucleic acid is the basic requirement for palindromic sequences.
A. True
B. False

Answer: A
Explanation: Palindromic sequence is a nucleic acid sequence present in double-stranded nucleic acid. This is because 5’ to 3’ reading frame on one strand matches the 5’ to 3’ reading frame of its complementary strand, which is requisite for any sequence to be called a palindrome.

2. Which of the following words represents a palindrome?
A. DUCK
B. MAMA
C. BOOM
D. MADAM

Answer: D
Explanation: A palindrome is a word which can be read as the same from both forward and backward. Thus, MADAM is the same when spelt in either way.

3. Which of the following will form a palindromic sequence?
A. ATTGCAAT
B. AGTCCTGA
C. GTTCCAAG
D. GTTGGAAC

Answer: A
Explanation: The complementary sequence of ATTGCAAT is TAACGTTA. Thus, when the first is read from left to right and the later read from right to left the sequence of the bases is exactly the same. This is the criteria for a sequence to be palindromic.

4. Which of the following is affected by palindromic sequences?
A. Acetylation site
B. Phosphorylation site
C. Methylation site
D. Promoter site

Answer: C
Explanation: Palindromic sites are frequently methylated in many organisms. These are the sites recognized by methylases where a methyl group is attached to inactivate a gene or mark a restriction site for an endonuclease.

5. Palindromic sequences play a very important role in gene manipulation.
A. True
B. False

Answer: A
Explanation: Many restriction endonucleases recognize and cut the palindromic sequence, for example, EcoR1. This is important for the genetic manipulation to meet the purpose of insertion and excision of gene or genome fragment from certain genetic material (eg. plasmid).

6. Which of the following palindromes is not a restriction site?
A. GAATTC
B. TACGTA
C. CCTAGG
D. AGCT

Answer: B
Explanation: GAATTC serves as the restriction site for the EcoR1 endonuclease. CCTAGG serves as the restriction site for the endonuclease BamH1. Again AGCT acts as the restriction site for the endonuclease Alu1. Only TACGTA does not serve as a restriction site or else it is not yet known.

7. Molecular cutters do not recognize the palindromic sequence.
A. True
B. False

Answer: B
Explanation: Molecular cutters or endonucleases always recognize a palindromic sequence. Type I endonucleases generally cuts the DNA 1000 bp away from the 5’ end of its recognition sequence. Similarly, Type II cuts within the sequence and Type III cuts 25-27 bp away.

8. Which of the following is an example of an inverted repeat?
A. ATTCGCGAAT
B. ATCGNNNNCGAT
C. TACCNNNNGGAT
D. AGCCNNNCCGA

Answer: B
Explanation: Inverted repeat is a single stranded sequence of nucleotides followed downstream by its reverse complement separated by a few or more nucleotides. When the intervening length in between the repeats is zero, the composite sequence is a palindromic sequence.

9. Inverted repeat have a number of biological functions. Which of the following is a biological function of an inverted repeat?
A. Diseases
B. Central dogma
C. Cellular metabolism
D. Genetic stability

Answer: A
Explanation: The Inverted repeat defines the region of self complementation thus can be identified by many proteins, enzymes and transposons. They also serve as sites for many mutations. Thus all of these factors may lead to alteration in the basic genome sequence. If any such alteration occurs in any vital gene it could cause genetic disorders and diseases. Example of certain disease is osteogenesis imperfecta.

10. Most commonly known hairpin structures are found in _____________
A. DNA
B. mRNA
C. tRNA
D. rRNA

Answer: C
Explanation: Hairpin structure is a type of stem-loop structure involving intra molecular base pairing. This is generally found in single-stranded DNA and RNA. Most commonly known hairpin structures are observed in the cloverleaf model of tRNA namely the anticodon loop, D loop and the ΨU loop.

11. Which of the following does not promote stability in stem-loop structure?
A. Length of the stem
B. Size of the loop
C. A : U base pairing
D. π orbital of aromatic ring

Answer: C
Explanation: A : U base pairing has only two hydrogen bonds to hold them together whereas G : C has three hydrogen bonds. Thus a G : C rich stem promotes stem-loop structure more by providing more stability to the stem.

12. What is the minimum number of bases required for loop stability?
A. 2
B. 3
C. 4
D. 5

Answer: B
Explanation: Loops less than 3 bases long are sterically impossible to form. The optimal loop length is 4 – 8 bases which perfectly stabilize the loop of the stem-loop structure.

13. Which of the following does not contain a stem-loop structure?
A. tRNA
B. Pseudoknot
C. Ribozyme
D. rRNA

Answer: D

250+ TOP MCQs on Properties of DNA Polymerase – 2 and Answers

Molecular Biology MCQs on “Properties of DNA Polymerase – 2”.

1. How many active sites are present in the DNA polymerase to catalyze the addition of the four dNTPs?
A. 1
B. 2
C. 3
D. 4

Answer: A
Explanation: Only 1 active site is dedicated for the addition of all the four types of dNTPs in the DNA polymerase. This is due to the identical geometry of all the four types of dNTPs.

2. How does the polymerase recognize the correct dNTP for an addition?
A. Structure of dNTP
B. Molecular weight of dNTP
C. Purine and pyrimidine orientation
D. Ability of hydrogen bond formation

Answer: D
Explanation: Only 1 active site is dedicated for the addition of all the four types of dNTPs in the DNA polymerase. This is due to the identical geometry of all the four types of dNTPs. The polymerase recognizes the correct dNTP for addition is by the ability of hydrogen bond formation of the new bases with the existing bases on the template strand.

3. DNA polymerase can distinguish between dNTPs and rNTPS due to ___________
A. Structural differentiation
B. Steric exclusion
C. Steric hindrance
D. Enzyme substrate mismatch

Answer: B
Explanation: DNA polymerase shows an impressive ability of distinguishing between dNTPs and rNTPS. This is facilitated by the steric exclusion of the rNTPS from the DNA polymerase active site which is too small to accommodate the 2’-OH of rNTPs.

4. When we compare the structure of DNA polymerase to the structure of a body part, it resembles to _________________
A. Right hand
B. Left hand
C. Right foot
D. Left foot

Answer: A
Explanation: From the studies of the atomic structures of the various DNA polymerases bound to the primer:template junction it reveals that the structure resembles partially to that of a closed right hand. Based on the analogy to a hand the three domains of polymerase are called the thumb, fingers and palm.

5. With respect to the palm domain of the DNA polymerase which of the following is not its property?
A. Contains primary elements of the catalytic site
B. Binds to 2 divalent ions
C. Composed of α helix
D. Brings about the environmental changes around 3’-OH of dNTP

Answer: C
Explanation: The palm domain of the DNA polymerase is composed of β sheet and contains the primary elements of the catalytic site. In particular, this region of the polymerase binds to 2 divalent metal ions that bring about the chemical environmental changes around 3’-OH of dNTP for its polymerization.

6. The two divalent metal ions of the active sites of DNA polymerase are major catalytic elements that bring about changes required for the joining of the dNTPs.
A. True
B. False

Answer: A
Explanation: One of the metal ions reduces the affinity of the 3’-OH for its hydrogen. This generates a 3’-O- that is primed for nucleophilic attack of the α – phosphate of the incoming dNTP. The second metal ion coordinates the negative charges of the β – and γ – phosphates of the dNTPs and stabilizes the pyrophosphate produced by the joining the primer and the incoming nucleotide.

7. Mismatched DNA does not affect the rate of activity of DNA polymerase.
A. True
B. False

Answer: B
Explanation: Te accuracy of the base – pairing is monitored by the palm domain of the polymerase. The palm makes extensive hydrogen bond contacts with the base pairs in the minor groove of the newly synthesized strand. These contacts are not base pair specific but only bond if the base – pairing is correct. Thus mismatched DNA dramatically slows the rate of activity of DNA polymerase.

8. Which is the rate limiting step of DNA replication?
A. Formation of the RNA primer
B. Binding of primer to the DNA template
C. Binding of DNA polymerase to the primer:template junction
D. Binding of the first dNTP to the primer

Answer: C

250+ TOP MCQs on Coding and Non-Coding RNA and Answers

Molecular Biology Multiple Choice Questions on “Coding and Non-Coding RNA”.

1. Non – coding sequence in mRNA is known as __________
A. Template
B. Non – template
C. Intron
D. Exon

Answer: C
Explanation: Long stretches of non – coding DNA are found within most eukaryotic genes. These sequences are known as introns which are removed from the pre-mature RNA to produce mature RNAs.

2. Which of the following organisms have monocistronic DNA?
A. Virus
B. Bacteria
C. Fungi
D. Yeast

Answer: B
Explanation: The bacterial genome is monocistronic. This means that the genes are not disrupted by the presence of introns.

3. Which of the following organisms have overlapping genes?
A. Virus
B. Bacteria
C. Fungi
D. Yeast

Answer: A
Explanation: The viruses have overlapping genes to accommodate the genome in such a small cell. The bacterial cell is monocistronic whereas all eukaryotic cells are polycistronic.

4. Introns were first discovered in __________
A. 1958
B. 1955
C. 1966
D. 1977

Answer: D
Explanation: Introns were first discovered in 1977 during the studies of replication of adenovirus in cultured human cells. It was discovered independently in the laboratories of Phillip Sharp and Richard Roberts.

5. Adenovirus is a useful model of studying gene expression because __________
A. High infection rate
B. Assured transformation
C. High protein production
D. Short duplication time

Answer: C
Explanation: Adenovirus is a useful model of studying gene expression because of two reasons:
i) The size of the viral genome is small that is only 3.5*104 base pairs long.
ii) The adenovirus mRNAs are produced at high levels in infected cells thus yielding a high amount of protein.

6. Mouse β – globin gene has __________ introns.
A. 1
B. 2
C. 3
D. 4

Answer: B
Explanation: Mouse β – globin gene has 2 introns. This gene encodes the β – subunit of haemoglobin. This is found by the observations produced by the electron microscopic analysis of RNA – DNA hybrids and subsequent nucleotide sequencing of cloned genomic DNA and cDNAs.

7. The amount of exon in the eukaryotic genome is higher than that of the introns.
A. True
B. False

Answer: B
Explanation: The intron – exon structure of many eukaryotic genes is quite complicated, and the amount of DNA in the intron sequences is often greater than that in the exons. For example, an average human gene contains about 10 exons, interrupted by introns distributed over approximately 56 kb of genomic DNA. The exons generally total only about 4.1 kb including both the 5’ and 3’ ends of the mRNA that are not translated into proteins.

8. The mRNA of which eukaryotic protein lacks introns?
A. Haemoglobin
B. Myoglobin
C. Histone
D. Polymerase

Answer: C
Explanation: Introns are present in almost all genes of complex eukaryotes, although they are not universal. Almost all histone genes for example, lacks introns, so introns are not clearly not required for gene function in eukaryotes.

9. Introns are nothing more than genetic load in the eukaryotic genome.
A. True
B. False

Answer: B
Explanation: Although most introns do not specify the synthesis of protein product, they have other cellular activities. Many introns encode functional RNAs, including the small nucleolar RNAs that function in ribosomal RNA processing and the microRNAs, which are major regulators of gene expression in eukaryotic cells. Introns also contain regulatory sequences that control transcription and mRNA processing.

10. Presence of introns facilitates the formation of several different mRNAs, thus increasing protein yield of different types but from same gene.
A. True
B. False

Answer: A