Category: Objective Questions

Chemical Reaction Engineering Multiple Choice Questions on Equations for Continuous – Flow Reactors”.

1. Find the volume of reactor needed for 70% conversion of A to product. Aqueous feed of A and B (200lit/min, 20mmolA/lit, 50mmolB/lit) is fed to MFR. Reaction kinetics is as follows
A+B → P
-r_A = 200C_AC_B mol/lit-min
A. 12.5
B. 3.6
C. 6.4
D. 1.5

Answer: C
Explanation:
C_A0*X_A = C_B0*X_B
X_B = (frac{20*0.7}{50}) = 0.28
(frac{τ}{CA0} = frac{XA}{-rA})
τ = (frac{0.7*20*10^{-3}}{200*10^{-6}[20(1-0.7)*50(1-0.28)])}) = 0.032 = (frac{V}{v0} )
V = 0.032*200 = 6.4 lit.

2. An existing MFR is to be replaced by the one 4 times larger than it. Find the new conversion if a homogeneous liquid phase reaction with reaction kinetics as follows.
A → R
-r_A = kC_A² with 30% conversion.
A. 0.67
B. 0.45
C. 0.53
D. 0.92

3. Replace the existing MFR with a PFR and calculate the new conversion if a homogeneous liquid phase reaction with reaction kinetics as follows.
A → R
-r_A = kC_A² with 30% conversion
A. 0.65
B. 0.43
C. 0.38
D. 0.92

Answer: C
Explanation:
(frac{τ}{CA0} = frac{XA}{-rA} )
(frac{V}{CA0 V0} = frac{0.3}{k[ CA0(1-0.3)]^2} )
(frac{Vk}{CA0 V0} ) = kτC_A0 = 0.6122
When replaced with a new PFR
(frac{τ}{CA0} = int_0^{XA} frac{dXA}{-rA} )
For 2^nd order on Integration we get,
kτC_A0 = (frac{XA}{1-XA}) = 0.6122
X_A = 0.38.

4. Find the conversion of a 25 litre PFR if a gaseous feed of pure A (2 mol/lit, 150 mol/min)
Is decomposed, kinetics are as follows.
A → 2.5R
-r_A = (15min^-1) C_A
A. 0.89
B. 0.92
C. 0.45
D. 0.68

Answer: B
Explanation:
(frac{τ}{CA0} = frac{V}{FA0})
τ = (25*2)/150 = 0.33 min
For a 1^st order reaction with variable volume on integration we get
kτ = (1+εA.ln((frac{1}{1-XA})) – ε_AX_A
15*0.33 = (1+1.5) ln((frac{1}{1-XA})) – 1.5X_A
X_A = 0.92.

5. For a constant volume system, the size of batch reactor is _______________ PFR.
A. smaller than
B. larger than
C. same as
D. cannot be compared

Answer: C
Explanation:
For PFR, (frac{τ}{CA0} = int_0^{XA}frac{dXA}{-rA} )
For Batch reactor, (frac{t}{CA0} = int_0^{XA}frac{dXA}{-rA} )
From the above equations it is clear that theoretically the element of fluid reacts for same length of time in PFR and Batch reactor. But shutdown time as to be considered for actual design.

Chemical Reaction Engineering Multiple Choice Questions & Answers on “Mechanisms of Catalyst Deactivation”.

1. State true or false.
In parallel deactivation, the reactant produces a side product which deposits on the catalytic surface and deactivates it.
A. True
B. False
Answer: A
Explanation: A parallel deactivation is the one in which reactant A produces the product R and the poison P.
A → R + P↓

2. The catalyst deactivation caused by deposition on surface and pores of catalyst is called ____
A. Adsorption
B. Regeneration
C. Fouling
D. Desorption
Answer: C
Explanation: Fouling is caused by the blockage of pores on the catalyst. The life and the efficiency of the catalyst is reduced by fouling.

3. Which of the following represents series type deactivation model? (Where R is the product and P is the poison)
A. A → R → P↓
B. A → P↓
C. A → R + P↓
D. A → R, A → P
Answer: A
Explanation: The series deactivation is the mechanism in which the product may decompose to produce the poison or further react to produce the poison. The poison deposits on the catalyst and causes deactivation.

4. The activity of a catalyst pellet is defined as the ratio of ___
A. Rate of adsorption of reactants to the catalyst surface to the rate at which the catalytic pellet converts the reactant
B. Rate at which the catalytic pellet converts the reactant to the rate of adsorption of reactants to the catalyst surface
C. Rate at which the catalytic pellet converts the reactant to the rate of reaction with a fresh pellet
D. The rate of reaction with a fresh pellet to the rate at which the catalytic pellet converts the reactant
Answer: C
Explanation: Activity is a measure of the catalyst performance. It is defined as the ratio of rate at which the catalytic pellet converts the reactant to the rate of reaction with a fresh catalyst pellet.

5. If k_d is the rate constant of deactivation and d is the order of deactivation, then the rate equations representing independent deactivation are ___
A. -r’_A = k’C_Aⁿa
–(frac{da}{dt}) = k_da^d
B. -r’_A = k’a
–(frac{da}{dt}) = k_da^d
C. -r’_A = k’C_Aⁿa
–(frac{da}{dt}) = k_d
D. -r’_A = k’C_Aⁿ
–(frac{da}{dt}) = k_da^d
Answer: A
Explanation: Independent deactivation is the one that is independent of the concentration. Hence, the rate of change of activity is –(frac{da}{dt}) = k_da^d.

6. Which of the following reactor arrangements causes fast deactivation?
A. Mixed flow for fluid
B. Plug flow for fluid
C. Fluidised bed reactor
D. Batch for fluid and solid
Answer: C
Explanation: Fluidisation causes high catalytic deactivation. The catalytic particles are well mixed by the reactant fluid.

7. The rate expression for independent deactivation for batch solids is ____
A. ln((lnfrac{C_A}{C_{A∞}})) = ln (frac{Wk’}{Vk_d}) – k_dt
B. ln (frac{C_A}{C_{A∞}}) = ln (frac{Wk’}{Vk_d}) – k_dt
C. lnln (frac{C_A}{C_{A∞}}) = ln (frac{Wk’}{Vk_d})
D. lnln (frac{C_A}{C_{A∞}}) = (frac{Wk’}{Vk_d}) – k_dt
Answer: A
Explanation: For batch solids – fluid, –(frac{-dC_A}{dt} = frac{W}{V})k’C_Aa and –(frac{da}{dt}) = k_da^d. Integrating and combining the equations, ln((lnfrac{C_A}{C_{A∞}})) = ln (frac{Wk’}{Vk_d}) – k_dt.

8. The slope of the curve of ln((lnfrac{C_A}{C_{A∞}})) and t for batch solid – fluid independent deactivation system is ___
A. ln(lnk_d)
B. ln(k_d)
C. k_d
D. -k_d
Answer: D
Explanation: ln((lnfrac{C_A}{C_{A∞}})) = ln (frac{Wk’}{Vk_d}) – k_dt. The slope of the curve is -k_d and the intercept is ln(frac{Wk’}{Vk_d}. )

9. If the rate of deactivation is given by –(frac{da}{dt}) = 0.0064 day^-1, the expression relating activity and time is ___
A. a = 0.0064t
B. a = 1-0.0064t
C. a = t
D. a = 0.0064-t
Answer: B
Explanation: –(frac{da}{dt}) = 0.0064
–(int_1^A.da = 0.0064(int_0^t)dt
Integrating, a = 1-0.0064t

10. The activity of a catalyst at a time t = 0 is ____
A. Negative
B. Zero
C. Unity
D. ∞
Answer: C
Explanation: When the reaction has not yet started to occur, the catalyst surface is not poisoned. The entire surface on the catalyst is available to promote the reaction. At a time t=0, the activity is the highest and equal to 1. The value of ‘a’ lies between 0 and 1 as ‘a’ is the ratio of rate of reaction achieved by the catalyst at any given time to the maximum rate that can be achieved by a fresh catalyst.

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Molecular Biology Multiple Choice Questions on “Ribosomes Read the Genetic Code”.

1. With respect to polymerization of amino acids which of the following statements is wrong?
A. The polymerization machinery consists of fifty proteins and two RNA molecules
B. Overall molecular mass 2.5 megadalton
C. Occurs at a slow pace of only 2 – 20 amino acids per second
D. In eukaryotes the process occurs in the cytoplasm

Answer: A
Explanation: Though the polymerizing machinery consists of 50 proteins, it atleast should consist of 3 RNA molecules. The molecules that take part in translation are mRNA, tRNA and rRNA. Messenger or mRNA is the transcript, transfer or tRNA is the carrier of amino acid and the ribosome contains rRNA or ribosomal RNA which decodes the transcript codon.

2. Ribosome begins its translation at the 3’ end of the mRNA in prokaryotes.
A. True
B. False

Answer: B
Explanation: In prokaryotes the transcription and translation occurs simultaneously. Thus, the end of mRNA which emerges from transcription is translated first. As the RNA polymerase reads the frame in 5’ to 3’ direction thus the 5’ end of mRNA is synthesized first thus, it is translated first.

3. Check which one of the following mechanisms is wrongly paired with its location of occurrence?
A. Transcription – nucleus
B. Post transcriptional mechanism – ER
C. Translation – cytoplasm
D. Post translational mechanism – ER

Answer: B
Explanation: Post transcriptional mechanisms are splicing, capping and tailing. These three mechanisms occurs in the nucleus as if the nascent mRNA (hnRNA. leaves the nucleus without the modification ribosome would not be able to read it and also it would not be able to read it and also it would be hydrolysed by the enzyme RNase present in the cell cytoplasm.

4. With respect to the subunits of ribosome which of the following is wrongly paired?
A. Ribosome = rRNA + protein
B. Large subunit = decoding center
C. Small subunit = decoding center
D. Subunit sedimentation unit = Svedberg

Answer: B
Explanation: The large subunit contains the three sites A, P and E. These are the centers where the amino acyl tRNA brings the amino acid peptide linkage is formed and the tRNA exits respectively. Thus the subunit is known as peptidyl transferase center.

5. An mRNA bearing multiple ribosomes is known as______________
A. Small subunit-mRNA-initiator tRNA complex
B. mRNA ribosome complex
C. Polyamine-ribosome complex
D. Polysome

Answer: D
Explanation: A single ribosome contacts around so nucleotides on mRNA whereas the due to the large density of ribosome it allows one ribosome per 80 nucleotides. Thus a typical mRNA could be very long (example: 1000 nucleotides). Therefore more than one ribosome can attach to the mRNA forming a structure known as polyribosome or polysome.

6. What is the actual substrate for each round of amino acid addition?
A. Ribosome
B. Amino acid
C. mRNA
D. tRNA

Answer: D
Explanation: The actual substrate for the addition of amino acid to the growing polypeptide chain in each round are two charged species of t RNAs. They are an aminoacyl tRNA which is the carrier of amino acid to the ribosome-mRNA complex and the peptidyl tRNA which forms the peptide linkages between the amino acids.

7. The reaction to form a new peptide bond is known as the peptidyl transferase reaction.
A. True
B. False

Answer: A
Explanation: The mechanism for the formation of new peptide bond first requires the synthesis of amino terminus before carboxyl terminus. The second step of the mechanism is the transfer of growing polypeptide chain from the peptidyl tRNA to the amino acyl tRNA. Thus reaction is called peptidyl transferase reaction.

8. Where does ribosome gets the energy for the formation of peptide bond?
A. Breaking of the keto bond
B. Breaking of acyl bond
C. Simultaneous hydrolysis of ATP
D. Energy released during peptidyl transferase reaction

Answer: B
Explanation: The energy provided for the formation of peptide bond is provided by the breaking of high energy acyl bonds. This is because the charging reaction of the tRNA involves the hydrolysis of a molecule of ATP. Thus, the energy of peptide bond formation originates from the molecule of ATP that was hydrolysed during tRNA charging reaction but not by simultaneous hydrolysis.

9. With respect to the mRNA – ribosome complex for translation mechanism which of the following pair is wrong?
A. Small subunit – decoding center
B. A site – aminoacylated tRNA binding site
C. P site – peptidyl tRNA binding site
D. E site – polypeptide exit site

Answer: D
Explanation: E site is the binding site for the tRNA that is released after the growing polypeptide chain has been transferred to the aminoacyl tRNA. Thus E site is the peptidyl tRNA exit site.

10. Only a single stranded RNA can pass through the ribosome for decoding.
A. True
B. False

Answer: A
Explanation: The mRNA enters the decoding center through two narrow channels in the small subunit. The entry channel is only wide enough for unpaired RNA to pass through. These features ensures that mRNA is in a single stranded from as it enters the decoding center by removing any intermolecular base pairing interactions that might have been formed in the mRNA.

11. How many channels are present in the ribosome?
A. 2
B. 3
C. 4
D. 5

Answer: B

Molecular Biology Multiple Choice Questions on “Fundamentals of PCR”.

1. PCR is a DNA amplifying in vivo method.
A. True
B. False

Answer: B
Explanation: A powerful method for amplifying a particular DNA segment is the polymerase chain reaction. This procedure is carried out entirely biochemically, that is invitro.

2. Which of the following is a mismatch?
A. Polymerase – Taq polymerase
B. Template – double stranded DNA
C. Primer – oligonucleotide
D. Synthesis – 5’ to 3’ direction

Answer: B
Explanation: PCR uses the enzyme DNA polymerase that directs the synthesis of DNA from deoxynucleotide substrates on one stranded DNA template. DNA polymerase synthesizes DNA in a 5’→3’ direction and can add nucleotides in the 3’ end of a custom – designed oligonucleotide.

3. Annealing of primer is facilitated by complementary region.
A. True
B. False

Answer: A
Explanation: A synthetic oligonucleotide is annealed to a single stranded DNA template that contains a complementary region. DNA polymerase thus uses this annealed primer to elongate the new strand.

4. Primer used for the process of polymerase chain reaction are ______________
A. Single stranded DNA oligonucleotide
B. Double stranded DNA oligonucleotide
C. Single stranded RNA oligonucleotide
D. Double stranded RNA oligonucleotide

Answer: A
Explanation: Two single stranded, oligonucleotide is synthesized and used as a primer. The primers thus produces binds to the denatured DNA and polymerase starts its synthesis in the 5’ to 3’ direction.

5. Polymerase used for PCR is extracted from _____________
A. Escherichia coli
B. Homo sapiens
C. Thermus aquaticus
D. Saccharomyces cerevisiae

Answer: C
Explanation: Amplification is usually carried out by the DNA polymerase I enzyme from Thermus aquaticus. As this organism lives in hot springs, Taq polymerases are thermostable thus they are resistant to high temperature.

6. How many DNA duplex is obtained from one DNA duplex after 4 cycles of PCR?
A. 4
B. 8
C. 16
D. 32

Answer: C
Explanation: After each cycle the number of duplexes doubles itself thus after the first cycle there are 2 DNA duplexes. A duplex has 2 DNA strands thus after second cycle there will be 4 Duplexes, after third cycle there will be 8 DNA duplex. Lastly, after the 4th cycle there will be 16 duplexes.

7. At what temperature do denaturation of DNA double helix takes place?
A. 60˚
B. 54˚
C. 74˚
D. 94˚

Answer: D
Explanation: The polymerase chain reaction is started by heating the mixture at 94˚C. At this temperature the hydrogen bonding between the nucleotide bases melts thus separating the two DNA strands.

8. At what temperature do annealing of DNA and primer takes place?
A. 42˚
B. 54˚
C. 74˚
D. 96˚

Answer: D
Explanation: After the denaturation of the two strands the temperature is decreased to 50 – 60˚C. At this temperature the primers anneal to their complementary segments. Thus as 54˚ lies within this range, 54˚ is the correct option.

9. The temperature at which DNA synthesis takes place is 74˚C.
A. True
B. False

Answer: A

Molecular Biology Multiple Choice Questions on “Manufacturing the Message – 1”.

1. Without using primers for RNA synthesis the polymerase produces small stretches of RNA of _____________ nucleotides.
A. 8
B. 10
C. 12
D. 14

Answer: B
Explanation: Without using primers for RNA synthesis the polymerase produces small stretches of RNA of 10 nucleotides. These transcripts are released from the polymerase which without dissociating from the template begins the synthesis of RNA again.

2. To facilitate binding of σ⁷⁰ the melting of the template and non – template strands occurs between the positions ____________
A. -11 to +3
B. -9 to +5
C. -10 to +1
D. -10 to +2

Answer: A
Explanation: The transition from the closed to open complex involves structural changes in the enzyme and opening of the DNA duplex to reveal the template and the non-template strands. This melting occurs between positions -11 to +3, in relation to the transcription start site.

3. The isomerisation of σ⁷⁰ of bacterial polymerase is an energy dependent process.
A. True
B. False

Answer: B
Explanation: The isomerisation of σ⁷⁰ of bacterial polymerase is not an energy dependent process. Instead isomerisation is the result of a spontaneous conformational change in the DNA-enzyme complex to a more energetically favorable form.

4. With respect to the formation of a complex between polymerase and DNA template which of the following is correct?
A. Open: reversible; Closed: irreversible
B. Open: irreversible; Closed: reversible
C. Open and closed: reversible
D. Open and closed: irreversible

Answer: B
Explanation: Isomerisation is essentially irreversible thus open complexes are irreversible and once forms transcription will be initiated. But the closed complex is readily reversible and the polymerase can easily dissociate from the promoter and is not bound to transcribe the DNA.

5. How many channels are found in the RNA polymerase clamp?
A. 3
B. 4
C. 5
D. 6

Answer: C
Explanation: The RNA polymerase clamp has 5 channels. They are NTP uptake channel, NT channel, T channel, RNA-exit channel and downstream DNA channel.

6. The channel that allows the exit of the coding strand is known as ____________
A. NTP channel
B. NT channel
C. T channel
D. Downstream DNA channel

Answer: B
Explanation: The channel that allows the exit of the coding strand is known as the NT channel. NT channel is the non-template exit channel. Non-template strand is also known as the coding strand.

7. Region 1.1 of the σ factor mimics the DNA and is negatively charged.
A. True
B. False

Answer: A
Explanation: Yes, the region 1.1 of the σ factor mimics the DNA and is negatively charged. The space in the active center cleft which is either occupied by the region 1.1 or by DNA, is highly positively charged.

8. RNA polymerase adds the nucleotides de novo. The first nucleotide added by the polymerase is ____________
A. ATP
B. TTP
C. GTP
D. CTP

Answer: A
Explanation: The first nucleotide added by the polymerase is always a purine. In this case, adenine is more common than guanine thus ATP is the first nucleotide added.

9. The ternary complex is formed by the polymerase, DNA duplex and the nascent RNA.
A. True
B. False

Answer: B
Explanation: The ternary complex is formed by the polymerase, single stranded DNA template and the nascent RNA. This is formed after the polymerization of atleast 10 nucleotides which stabilizes the structure.

10. After the binding of polymerase to the DNA template the σ factor is ejected.
A. True
B. False

Answer: B

Molecular Biology Questions on “Overview of Protein Synthesis – 2”.

1. Eukaryotic ribosome is recruited to the mRNA by _____________
A. Randomly
B. Shine – Dalgarno sequence
C. 5’ capping
D. 3’ tailing

Answer: C
Explanation: Prokaryotic ribosome is recruited to the mRNA by the Shine – Dalgarno sequence. In the case of eukaryotes ribosome is recruited to the mRNA by the 5’ cap.

2. After bounding to the mRNA eukaryotic ribosome scans the entire mRNA until it encounters a start codon.
A. True
B. False

Answer: A
Explanation: The eukaryotic ribosome is recruited to the mRNA by the 5’ cap. The guanine residue of the 5’ cap is connected to the 5’ end of the mRNA through three phosphate groups that recognizes and recruits the ribosome. Once bound to the mRNA, the ribosome moves in a 5’→3’ direction until it encounters a 5’…….AUG…….3’ start codon by a process called scanning.

3. The end of all tRNAs is ___________
A. 5’ ACC 3’
B. 5’ CCA 3’
C. 3’ CAC 5’
D. 3’ GAG 5’

Answer: B
Explanation: All tRNAs end at the 3’ terminus with the sequence 5’ CCA 3’. This is the site that is attached to the cognate amino acid by the enzyme aminoacyl tRNA synthetase.

4. How many unusual bases are observed in a tRNA molecule?
A. 2
B. 3
C. 4
D. 5

Answer: D
Explanation: Five unusual bases, produced by enzymatic modification of the usual bases, may be observed in a tRNA molecule. These are pseudouridine, dihyrdouridine, hypoxanthine, thymine and methylguanine. Of these five, unusual bases pseudouridine and dihyrdouridine are most commonly found in the tRNA molecules.

5. In pseudouridine the attachment of Uracil base to ribose through ___________ of uracil.
A. N at position 3
B. C at position 6
C. C at position 5
D. N at position 1

Answer: C
Explanation: Pseudouridine is produced by the enzymatic modification of the regular base uridine through isomerization. In the modification process of uridine to pseudouridine the attachment of base to the ribose sugar is switched from nitrogen at ring position 1 to the carbon at ring position 5.

6. Dihydrouridine is the result of oxidation of the uracil base.
A. True
B. False

Answer: B
Explanation: Dihydrouridine is a derivative of uridine formed by enzymatic reduction. The process of reduction forms the double bond between the carbons at positions 5 and 6.

7. How many loops are seen in the tRNA?
A. 1
B. 2
C. 3
D. 4

Answer: D
Explanation: Four different loops are observable in the tRNA 3˚ motif. They are the ΨU loop, D loop, variable loop and the anticodon loop. The ΨU loop is named because of the presence of the pseudouridine base. The D loop contains the dihyrdouridine base. The anticodon loop contains the anticodon and the variable loop is named so because its length is variable.

8. How many double helix regions are observed in a tRNA molecule?
A. 4
B. 3
C. 2
D. 1

Answer: A
Explanation: The tRNA molecules consist of 4 double helix regions. They are the acceptor stem and the stem of the three loops, the ΨU loop, D loop and the anticodon loop.

9. The variable loop is found between the ___________
A. Acceptor stem and ΨU loop
B. Acceptor stem and D loop
C. Anticodon loop and ΨU loop
D. Anticodon loop and D loop

Answer: C
Explanation: The variable loop sits between the anticodon loop and ΨU loop. Also as the name implies the size of the variable loop varies from 3 to 21 bases.

10. The 3 dimension structure of an adaptor molecule of translation is revealed to be ___________
A. Y shaped
B. L shaped
C. K shaped
D. X shaped

Answer: B
Explanation: The adaptor molecule of translation is the tRNA molecule which acts as the bridge between the mRNA code and the respective amino acid. X-ray crystallography reveals an L-shaped tertiary structure in which the terminus of the acceptor stem is at one end of the molecule and the anticodon loop is about 70 Å away at the other end.

11. Which of the following does not take part in stabilizing the cloverleaf model of the tRNA?
A. Base stacking
B. Base and sugar-phosphate backbone interaction
C. Ionic bond
D. Hydrogen bond

Answer: C