Category: Objective Questions

Chemical Reaction Engineering MCQs focuses on “Choosing of Right Reactor Series or Connections (Maximum Profitability)”.

1. How should the two MFR’s of unequal size be connected in series to maximize production for a first order homogenous reaction?
A. Smaller MFR followed by larger MFR
B. Larger MFR followed by smaller MFR
C. Any arrangement
D. MFR’s should not be connected in series
View Answer

Answer: C
Explanation: The final concentration for both the cases is same and can be calculated using the formula
C_A2 = (frac{CA0}{(τ1K+1)(τ2k+1)} )
Hence, it can be inferred that for first order reactions, irrespective of the arrangements the final conversions are going to be the same.

2. For all positive reaction orders and any particular duty the MFR is always larger than the PFR.
A. True
B. False
View Answer

Answer: A
Explanation: Comparison of performance of single MFR and PFR are made for the nth order and was inferred that the MFR is always larger than a PFR for a given duty.

3. How should a PFR and MFR be connected to maximize production from a first order reaction?
A. MFR followed by PFR
B. PFR followed by MFR
C. Any arrangement
D. MFR and PFR should not be connected
View Answer

Answer: C
Explanation: The final concentration for both the cases is same and can be calculated using the formula
C_A2 = (frac{CA0 e^{-kτp}}{(τmk+1)} )
Therefore, any arrangement does not make any difference for first order reactions.

4. Graphical procedure for finding conversion in MFR’s of different sizes in series is called ______ method.
A. Newton’s
B. Wheeler’s
C. Octaves
D. Jones
View Answer

Answer: D
Explanation: A graphical procedure was presented by Jones in 1951 to estimate the outlet composition of MFR’s in series.

5. Best arrangement of reactors for a reaction whose rate-concentration curve that rises monotonically.
A. Plug, Large MFR, Small MFR
B. Plug, Small MFR, Large MFR
C. Large MFR, Small MFR, Plug
D. Small MFR, Plug, Large MFR
View Answer

Answer: B
Explanation: For this kind of rate-concentration curve, the concentration should be kept as high as possible, therefore this arrangement should be used.

A. 0.805
B. 0.123
C. 0562
D. 0.439
View Answer

Answer: A
Explanation: The design equation of PFR is τ = (int_{CA0}^{CAf}frac{-dCA}{-rA} )
For the 1^st order reaction the equation is simplified as kτ = -ln⁡((frac{CA}{CA0}))
τ = -0.5ln⁡((frac{10}{50})) = 0.805 minutes.

9. Effect of flow type on reactor performance when the conversion is small.
A. no effect
B. cannot be determined
C. very small
D. very large
View Answer

Answer: C
Explanation: The reactor performance is not affected by flow type when the conversion is small. If the conversion is very high, flow type largely affects the reactor performance.

Chemical Reaction Engineering Multiple Choice Questions & Answers on “RTD for Chemical Reactors – Heat Effects”.

1. Increase in temperature for an endothermic reaction ____
A. Decreases equilibrium conversion
B. Increases equilibrium conversion
C. Does not affect equilibrium conversion
D. Initially decreases and then increases equilibrium conversion
Answer: B
Explanation: The rate of endothermic reaction increases as the temperature is increased. With irreversible reactions, highest allowable temperature should be used to achieve maximum conversion.

2. The optimum temperature progression for exothermic reversible reaction is ____
A. The locus of maximum rates
B. Highest allowable temperature
C. The isothermal path
D. Lowest allowable temperature
Answer: A
Explanation: For a reversible exothermic reaction, the rate of forward reaction increases but the maximum attainable conversion decreases. Hence, it starts at high temperature which decreases as conversion rises.

3. Which of the following correctly relates conversion to the heat of adiabatic reaction?
A. C_p ∆T(-∆H_R) = X_A
B. (frac{C_p}{-H_R}) = X_A
C. (frac{C_p ∆T}{-H_R}) = X_A
D. (frac{-H_R C_p}{∆T}) = X_A
Answer: C
Explanation: Conversion is obtained by the ratio of heat needed to raise the feed stream to the final temperature, T₂ to the heat released by the reaction at T₂.
X_A = (frac{C_p (T_2- T_1)}{-H_R}. )

4. State true or false.
An exothermic reaction approaches isothermal conditions by increasing the addition of inerts to the reaction mixture.
A. True
B. False
Answer: A
Explanation: For an exothermic reaction, H_R < 0. Addition of inerts causes the the curve of conversion and temperature to shift to isothermal conditions.

5. State true or false.
For exothermic reactions with large temperature drop, PFR is used.
A. True
B. False
Answer: B
Explanation: The rate rises from low value to a maximum at some value of conversion. After rising, the curve of conversion versus temperature falls. This is a characteristic of autocatalytic reactions. Hence, CSTR is used for exothermic reactions with large temperature drop.

6. Which of the following reactors is used for large (frac{C_p}{-H_R})?
A. CSTR
B. PFR
C. Photocatalytic reactor
D. Membrane reactor
Answer: B
Explanation: If (frac{C_p}{-H_R}) is high, then the gas either contains too much of inerts or the reaction system is in the liquid state. To achieve optimum temperature progression in such cases, PFR is used.

7. A liquid phase reaction occurring in adiabatic PFR has a heat of reaction -120000 J/molK. Specific heat of the stream is 10000 J / molK. The temperature difference for complete conversion to be achieved is ___
A. 560 K
B. 8.33 K
C. 12 K
D. 83.33K
Answer: C
Explanation: (frac{C_p ∆T}{-H_R}) = X_A
X_A = 1 for complete conversion.
∆T = (frac{120000}{10000}) = 12 K.

8. The conversion achieved for the liquid phase reaction occurring in an adiabatic PFR with a heat of reaction -100 kJ/molK, specific heat of the stream being 100 J / molK and the temperature difference being 800K is ___
A. 0.65
B. 0.75
C. 0.90
D. 0.80
Answer: D
Explanation: X_A = (frac{100 × 800}{100000}) = 0.8.

9. The point of intersection of energy balance and material balance curves for exothermic irreversible reactions, wherein instability is witnessed is stated as ____
A. Pour point
B. Boiling point
C. Flash point
D. Ignition point
Answer: D
Explanation: The point of instability where the energy balance and material balance curves meet is known as ignition point. If temperature is increased beyond ignition point, the reaction is self – sustaining.

10. Decrease in inerts for an endothermic reaction ____
A. Increases conversion
B. Decreases conversion
C. Has no effect on conversion
D. Initially increases and then decreases conversion
Answer: B
Explanation: Upon increasing inerts for an endothermic reaction, the curve shifts away faster from isothermal conditions. The conversion decreases and finally drops to zero as the inert concentration decreases.

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Molecular Biology Multiple Choice Questions on “Various Classes of RNA Have Different Functions – 1”.

1. Which of the following is a character of ORF?
A. Contiguous
B. 3 – nucleotide codons
C. Intron
D. Non – overlapping

Answer: C
Explanation: Exons are the part of nascent RNA that are connected together to form RNA after the removal of introns by the splicosome. The splicosome in the complex formed in the nucleus with nascent RNA being the substrate for the formation of mature RNA after its splicing and other post transcriptional mechanisms.

2. With respect to the open reading frame (ORF) which of the following is wrong?
A. Starts at 5’ end
B. ‘AUG’ is the only start codon for both prokaryotes and eukaryotes
C. The start codon defines the reading frame for all the subsequent codons
D. The stop codon is distinct from the ends of the mRNA

Answer: B
Explanation: ‘AUG’ is the only start codon for eukaryotes. For prokaryotic system ‘AUG’, ‘GUG’ and sometimes even ‘UUG’ act as the start codon.

3. With respect to polycistronic mRNAs which of the following is wrong?
A. Multiple ORFs
B. Found in eukaryotes
C. Encodes proteins with related functions
D. Multiple polypeptide chain

Answer: B
Explanation: Eukaryotes only contain 1 ORF per mRNA and are thus monocistronic. Polycistronic mRNAs are generally found in prokaryotes with 2 or more ORFs.

4. What was the name of the ribosome binding site?
A. ORF
B. RBS
C. A site
D. Shine – Dalgarno sequence

Answer: C
Explanation: Upstream the ORF a 3 – 9 base pair sequence on the 5’ side of the sequence is identified as the ribosome binding site (RBS). This element is referred to as the Shine – Dalgarno sequence, named after the scientists who discovered it by comparing the sequences of multiple mRNAs.

5. Which part of the ribosome identifies the Shine – Dalgarno sequence?
A. Protein
B. 16S rRNA
C. 23S rRNA
D. 5S rRNA

Answer: B
Explanation: The Shine – Dalgarno sequence is identified by the 16S rRNA. The core of the 16S rRNA has the sequence of 5’…..CCUCCU…..3’ and is located near the 3’ end of the rRNA. Not surprisingly the prokaryotic RBS are most often the subset of sequence 5’…..AGGAGG…..3’. Thus, 16S rRNA is the one that aligns the ribosome with the mRNA.

6. Eukaryotic mRNAs recruit ribosomes using the Shine – Dalgarno sequence.
A. True
B. False

Answer: B
Explanation: Eukaryotic mRNAs recruit ribosomes using specific chemical modifications called 5’ cap. The 5’ end of the mRNA is capped with methylated Guanine nucleotide to the mRNA via an unusual 5’ to 5’ linkage. To this methylated Guanine three phosphates are added. This cap binds to the ribosome which then slides along the mRNA length to find the ‘AUG’ for start of translation.

7. Eukaryotic mRNA is read in the 3’ to 5’ direction.
A. True
B. False

Answer: B
Explanation: The eukaryotic mRNA is read in the 5’ to 3’ direction. This is because capping of mRNA occurs at the 5’ end, thus the ribosome attaches to the 5’ end and starts translating in the 5’ to 3’ direction.

8. Which one of the following is known as the Kozak sequence?
A. 5’…..CNNAUCG…..3’
B. 5’…..GNNAUGG…..3’
C. 5’…..TNNAUGG…..3’
D. 5’…..GNNAUGC…..3’

Answer: B
Explanation: The Kozak sequence is named after its identification by Marilyn Kozak. This sequence contains purines (A/G) three bases upstream of ‘AUG’ and a ‘G’ immediately following it. Thus the sequence stands up to 5’…..A/GNNAUGG…..3’. this sequence is known to increase translation efficiency.

9. With respect to the composition of the ribosome which of the following is correct?
A. Ribosome is composed of 60S and 30S subunit
B. Eukaryotic ribosome small subunit contains only one 16S rRNA
C. 60S subunit consists of 5S rRNA and 23S rRNA
D. 60S and 40S makes up the 80S ribosome

Answer: D
Explanation: There is a discrepancy in the sedimentation velocity of the subunits separately and as a whole. This is because of the fact that the sedimentation velocity is determined both by shape and size and hence, it is not an exact measure of mass. Prokaryotic ribosome consists of 50S and 30S subunit and the 30S subunit contains only one 16S rRNA. The 50S subunit consists of 5S and 23S rRNA.

10. Which element of the ribosome plays a key role in mRNA translation?
A. rRNA of the large subunit
B. Proteins of the large subunit
C. rRNA of the small subunit
D. Proteins of the small subunit

Answer: C

Molecular Biology Multiple Choice Questions on “Inverse PCR & Randomly Amplified Polymorphic DNA(RAPD)”.

1. The type of DNA amplification where the region of DNA amplified lies on either side of a known segment __________
A. RT – PCR
B. Anchored – PCR
C. Inverse – PCR
D. Nested – PCR

Answer: C
Explanation: PCR can be used to amplify the sequences flanking on either sides of a known DNA segment. The process uses primers complementary to the 5’ ends of the interested segment. This is a reverse process of the general PCR mechanism thus is known as the inverse PCR.

2. Primer complementary to regions used in inverse PCR is _____________
A. 3’ end of unknown region
B. 5’ end of unknown region
C. 3’ end of known region
D. 5’ end of known region

Answer: C
Explanation: Inverse PCR uses primers complementary to the 3’ ends of the known segment of DNA. this thus, amplifies the unknown regions on the either sides of the known region and thus is known as inverse PCR.

3. With respect to target DNA used in inverted PCR which of the following is not true?
A. Restricted segment
B. Blunt ended segment
C. Intact known segment
D. Flanked unknown segment flanked on either side

Answer: B
Explanation: The target DNA is cut with a restriction enzyme that produces sticky ends and that does not cut within the region of which the sequence is known. The cutter instead cuts at unknown sites on the either sides of the known sequence. This lets the DNA circularize later, which is a very important step for achieving our desired results.

4. The DNA concentration in inverse PCR is kept low.
A. True
B. False

Answer: a
Explanation: The cut made by the restriction enzymes lie beyond the unknown region of our interest. As the restricted fragment thus produced has sticky ends it is left to circularize. The DNA taken for this process is in small quantity or else all the restricted fragments would bind among themselves and we will not get the desired circularized DNA.

5. The linear fragment used for inverted PCR has known ends _____________
A. True
B. False

Answer: A
Explanation: The molecular cutters used are rare and the cut producing the fragment involves the two regions of interest flanked to either sides of a known region. This fragment is left to circularize. This circular DNA is then restricted within the known DNA segment to produce a linear fragment having known ends on either sides.

6. With respect to RAPD which of the following is false?
A. 10 bases long
B. G/C rich
C. Has inverted repeats
D. Radom sequences are used

Answer: C
Explanation: RAPDs are generated by using random sequences of ordinarily 10 bases long oligonucleotides. These oligonucleotides are generally G + C rich and do not contain any repeated sequences.

7. Polymorphism in RAPD is observed because ______________
A. DNA used is from different chromosomes of same species
B. DNA used is from same chromosomes of same species
C. DNA used is from different chromosomes of different species
D. DNA used is from complementary chromosomes of same species

Answer: C
Explanation: In RAPD, primers for PCR amplification of genomic DNAs are from different species/strains. Thus polymorphism is produced due to the complementary sequence for the primer used being present in one strain and absent in another.

8. RAPDs cannot be used for PCR amplification.
A. True
B. False

Answer: B
Explanation: As the RAPDs behave as dominant markers, they are easily recognized by the primers for amplification. Thus, RAPDs turn out to be great templates for the primers for PCR amplification.

9. RAPDs are much more convenient markers than RFLPs.
A. True
B. False

Answer: A
Explanation: RAPDs have similar applications as RFLPs, but they are considerably faster and more convenient, particularly with such species where little previous work has been done. In some cases, RAPDs may generate such an amount of information in 4 weeks, which would have taken about 2 years to obtain using RFLPs.

10. The inheritance pattern of RAPD is _______________
A. Dominant
B. Recessive
C. Codominant
D. Random

Answer: A
Explanation: The inheritance pattern of RAPD is the dominant type. The inheritance pattern of RFLP is Codominant type.

11. RAPDs can be used to detect multiple alleles of a marker.
A. True
B. False

Answer: B

Molecular Biology online test on “Manufacturing the Message – 2”.

1. How many nucleotides of the growing RNA chain remain base paired with the template DNA?
A. 10
B. 12
C. 8
D. 16

Answer: A
Explanation: Only 8 or 9 nucleotides of the growing RNA chain remain base paired with the template DNA. The remainder of mRNA chain is peeled off and directed out of the enzyme through the RNA – exit channel.

2. The transcription bubble formed for facilitating RNA synthesis is about _____________ nucleotides long.
A. 10
B. 17
C. 14
D. 20

Answer: B
Explanation: The transcription bubble formed for facilitating RNA synthesis is about 17 nucleotides long. This is the region of unwound DNA where the two strands are separated. The bubble seems to move along with the RNA polymerase as the RNA grows in length.

3. The nascent RNA forms a double helix with the non – coding strand.
A. True
B. False

Answer: A
Explanation: The nascent RNA forms a double stranded hybrid helix with the antisense strand which is also known as the non-coding strand. The helix formed by the RNA and DNA is about 12 base pairs long.

4. How many proof-reading activities does RNA polymerase have?
A. 1
B. 2
C. 3
D. 0

Answer: B
Explanation: The RNA polymerase has 2 proof-reading activities. They are pyrophospholytic editing and hydrolytic editing.

5. The RNA polymerase can remove only the incorrect bases from the nascent RNA.
A. True
B. False

Answer: B
Explanation: The RNA polymerase is able to remove both correct and incorrect bases from the nascent RNA. As it spends longer hovering over the mismatched bases than the matched ones thus it removes the mismatched bases more frequently.

6. The Gre factor enhances the hydrolytic editing function of the polymerase.
A. True
B. False

Answer: A
Explanation: The hydrolytic editing is enhanced by the Gre factor and also serves as the elongation stimulating factor. That is, they ensure that the polymerase elongates efficiently and helps overcome “arrest” at sequences that are difficult to transcribe.

7. E. coli polymerase adds upto _____________ nucleotides per second.
A. 20
B. 40
C. 60
D. 80

Answer: B
Explanation: The E. coli polymerase adds upto an average of 40 nucleotides per second. This rate varies depending upon the local DNA sequence.

8. The Gre factor is homologous to the eukaryotic factor _____________
A. TFIIB
B. TFIIH
C. DPE
D. TFIIS

Answer: D

Molecular Biology Multiple Choice Questions on “Proteins are Chains of Amino Acids”.

1. Which of the following statements are true about aminoacyl tRNA?

i. Recognition and attachment of correct amino acid depends on aminoacyl tRNA synthetase.
ii. Transfer of amino acid to the polypeptide chain.
iii. Recognition of specific codon
iv. Recognition of specific anticodon

A. i and ii
B. iii and iv
C. i, ii and iii
D. i, ii, iii and iv

Answer: D
Explanation: Recognition and attachment of correct amino acid is the primary role of aminoacyl tRNA synthetase and requires the identification of anticodon on the tRNA to attach the correct amino acid. This aminoacyl tRNA now transfers the amino acid attached to it to the elongating polypeptide chain by recognizing the codon on the mRNA during translation.

2. For the charging of tRNA molecules, the acyl linkage occurs between the carboxyl group of the amino acid to the ____________
A. 2’ hydroxyl group of A
B. 3’ hydroxyl group of T
C. 2’ hydroxyl group of G
D. 3’ hydroxyl group of C

Answer: A
Explanation: The tRNA molecules to which amino acids are attached are said to be charged. Charging requires an acyl linkage between the carboxyl group of the amino acid and 2’ or 3’ hydroxyl group of the adenosine nucleotide that protrude from the acceptor stem.

3. The formation of the acyl linkage is significant for protein synthesis even though it is a high energy bond.
A. True
B. False

Answer: A
Explanation: Acyl linkage is formed due to hydrolysis which results in the change in free energy. This is significant for protein synthesis: the energy released when the bond is broken helps drive the formation of the peptide bonds that links amino acids to each other in polypeptide chain.

4. The joining of the amino acid to the tRNA requires ___________ steps.
A. 1
B. 2
C. 3
D. 4

Answer: B
Explanation: All aminoacyl tRNA synthetases attach amino acids to the tRNA in two enzymatic steps. Step 1: adenylation in which the amino acid reacts with ATP to become adenylated. Step 2 is tRNA charging in which the adenylated amino acid reacts with the tRNA while bound to the tRNA synthetases.

5. The principle driving force for adenylation reaction during the formation of the aminoacyl tRNA is carried out by ___________
A. Isomerase
B. Synthetase
C. Pyrophosphatase
D. Phosphokinase

Answer: C
Explanation: The principle driving force for adenylation reaction is the subsequent hydrolysis of pyrophosphate by pyrophosphatases. As a result of adenylation, the amino acid is attached to adenylic acid via a high energy ester bond in which the carbonyl group of the amino acid is joined to the phosphoryl group of the AMP.

6. How many types of tRNA synthetases are found?
A. 4
B. 3
C. 2
D. 1

Answer: C
Explanation: There are two classes of tRNA synthetases. Class I enzymes attach the amino acid to the 2’ – OH of the tRNA and are generally monomeric. Class II enzymes attach the amino acid to the 3’ – OH of the tRNA and are generally dimeric or tetrameric.

7. How many tRNA synthetases are found in a cell?
A. 64
B. 32
C. 10
D. 20

Answer: D
Explanation: Each 20 amino acids are attached to the appropriate tRNA by a single, dedicated tRNA synthetase. Because most amino acid are identified by more than one codon, it is not uncommon for one synthetase to identify and charge more than one tRNA. Thus, only one tRNA synthetase is responsible for charging all tRNAs for a particular amino acid.

8. The quaternary structure of which of the aminoacyl tRNA synthetase is the odd one among the following?
A. Glycine
B. Alanine
C. Proline
D. Serine

Answer: A
Explanation: The quaternary structure of the aminoacyl tRNA synthetase of Alanine, Proline and Serine are monomeric. Glycine has a dimeric aminoacyl tRNA synthetase molecule.

9. Which of the following pair is an example for using only one type of tRNA synthetase in bacteria?
A. Glutamine and cystine
B. Glutamic acid and asparagine
C. Cystine and Valine
D. Glutamine and glutamic acid

Answer: D
Explanation: Some bacteria lack a synthetase for charging the tRNA for glutamine (tRNAGln) with its cognate amino acid. Instead, a single species of aminoacyl tRNA synthetases charges tRNAGln (glutamic aciD. as well as tRNAGlu.

10. Which of the following parts of the mRNA determines the specificity of the amino acid attached?
A. Acceptor stem
B. Variable loop
C. ΨU loop
D. D loop

Answer: A
Explanation: Genetic, biochemical and X-ray crystallographic evidence indicate that the specificity determinants are clustered at the two distant sites on the molecule. They are the acceptor stem and the anticodon loop.

11. Any mutation in the sequence of the acceptor stem does not pose a threat to the amino acid incorporation.
A. True
B. False

Answer: B
Explanation: The acceptor stem is an especially important determinant for the specifically of tRNA synthetase recognition. In some cases changing a single base pair in the acceptor stem is sufficient to convert the recognition specificity of a tRNA from one synthetase to another.

12. For the addition of amino acids to the tRNA molecules the tRNA synthetases rely on the guidance provided by the anticodon sequence.
A. True
B. False

Answer: B
Explanation: For the addition of amino acids to the tRNA molecules by the tRNA synthetases, guidance provided by the anticodon sequence is not enough for the specificity as amino acids are usually specified by more than one codon. Thus to recognize the specific tRNAs, the synthetases for that amino acid must rely on determinants that lie outside of the anticodon.

13. The set of tRNA determinants that enable synthetases to discriminate among tRNAs are called ___________
A. Primary genetic code
B. First genetic code
C. Secondary genetic code
D. Second genetic code

Answer: D