Category: Objective Questions

Bioinformatics Question Bank focuses on “Secondary Structure Prediction for Transmembrane Proteins”.

1. Which of the following is untrue regarding the transmembrane proteins?
A. Constitute up to 30% of all cellular proteins
B. They are responsible for performing a wide variety of important functions in a cell, such as signal transduction, cross-membrane transport, and energy conversion
C. The membrane proteins are also of tremendous biomedical importance
D. They are not drug targets or receptors

Answer: D
Explanation: The membrane proteins are also of tremendous biomedical importance, as they often serve as drug targets for pharmaceutical development. There are two types of integral membrane proteins: α-helical type and β-barrel type. Most transmembrane proteins contain solely α-helices, which are found in the cytoplasmic membrane. A few membrane proteins consist of β-strands forming a β- barrel topology, a cylindrical structure composed of antiparallel β-sheets.

2. The structures of this group of proteins, however, are comparatively a lot difficult to resolve either by x-ray crystallography or nuclear magnetic resonance (NMR) spectroscopy.
A. True
B. False

Answer: A
Explanation: For this group of proteins, prediction of the transmembrane secondary structural elements and their organization is particularly important. Fortunately, the prediction process is somewhat easier because of the hydrophobic environment of the lipid bilayers, which restricts the transmembrane segments to be hydrophobic as well.

3. Which of the following is untrue regarding Prediction of Helical Membrane Proteins?
A. For membrane proteins consisting of transmembrane α–helices, these transmembrane helices are predominantly hydrophobic with a specific distribution of positively charged residues
B. The α-helices generally run perpendicular to the membrane plane
C. The α-helices generally run parallel to the membrane plane
D. The α-helices have an average length between seventeen and twenty-five residues

Answer: c
Explanation: The hydrophobic helices are normally separated by hydrophilic loops with average lengths of fewer than sixty residues. The residues bordering the transmembrane spans are more positively charged. Another feature indicative of the presence of transmembrane segments is that residues at the cytosolic side near the hydrophobic anchor are more positively charged than those at the lumenal or periplasmic side. This is known as the positive-inside rule.

4. The early algorithms based their prediction on hydrophobicity scales.
A. True
B. False

Answer: A
Explanation: A number of algorithms for identifying transmembrane helices have been developed where the early algorithms based their prediction on hydrophobicity scales. They typically scan a window of seventeen to twenty-five residues and assign membrane spans based on hydrophobicity scores. Some are also able to determine the orientation of the membrane helices based on the positive-inside rule.

5. Predictions solely based on hydrophobicity profiles have lowest error rates.
A. True
B. False

Answer: B
Explanation: Predictions solely based on hydrophobicity profiles have high error rates. As with the third-generation predictions for globular proteins, applying evolutionary information with the help of neural networks or HMMs can improve the prediction accuracy significantly.

6. The presence of ______ signal peptides can significantly compromise the prediction _______ because the programs tend to confuse hydrophobic signal peptides with membrane helices.
A. hydrophobic, accuracy
B. hydrophobic, error
C. hydrophilic, accuracy
D. hydrophilic, error

Answer: A
Explanation: Predicting transmembrane helices is relatively easy. The accuracy of Some of the best predicting programs, such as TMHMM or HMMTOP, can exceed 70%. To minimize errors, the presence of signal peptides can be detected using a number of specialized programs and then manually excluded.

7. Which of the following is untrue regarding TMHMM?
A. It is a web-based program based on an HMM algorithm
B. It is trained to recognize transmembrane helical patterns
C. It is not trained to recognize transmembrane helical patterns
D. When a query sequence is scanned, the probability of having an α-helical domain is given

Answer: c
Explanation: It is trained to recognize transmembrane helical patterns based on a training set of 160 well-characterized helical membrane proteins. The orientation of the α-helices is predicted based on the positive-inside rule. The prediction output returns the number of transmembrane helices, the boundaries of the helices, and a graphical representation of the helices. This program can also be used to simply distinguish between globular proteins and membrane proteins.

8. Which of the following is untrue regarding Phobius?
A. It is a web-based program designed to overcome false positives caused by the presence of signal peptides
B. The program incorporates distinct HMM models for signal peptides only
C. The program incorporates distinct HMM models for signal peptides as well as transmembrane helices
D. After distinguishing the putative signal peptides from the rest of the query sequence, prediction is made on the remainder of the sequence

Answer: B
Explanation: In addition to the given data, it has been shown that the prediction accuracy can be significantly improved compared to TMHMM (94% by Phobius compared to 70% by TMHMM). In addition to the normal prediction mode, the user can also define certain sequence regions as signal peptides or other nonmembrane sequences based on external knowledge.

9. Which of the following is true regarding Prediction of β-Barrel Membrane Proteins?
A. For membrane proteins with β-strands only, the β-strands forming the transmembrane segment are amphipathic in nature
B. For membrane proteins with β-strands only, the β-strands forming the transmembrane segment are only hydrophilic in nature
C. For membrane proteins with β-strands only, the β-strands forming the transmembrane segment are only hydrophobic in nature
D. They contain six to nine residues

Answer: A
Explanation: As stated, for membrane proteins with β-strands only, the β-strands forming the transmembrane segment are amphipathic in nature. They contain ten to twenty-two residues with every second residue being hydrophobic and facing the lipid bilayers whereas the other residues facing the pore of the β-barrel are more hydrophilic.

10. Scanning a sequence by hydrophobicity does not reveal transmembrane β-strands.
A. True
B. False

Answer: A
Explanation: These programs for predicting transmembrane α-helices are not applicable for this unique type of membrane proteins. To predict the β-barrel type of membrane proteins, a small number of algorithms have been made available based on neural networks and related techniques.

Genetic Engineering Multiple Choice Questions on “Restriction Endonuclease & Phosphatases – 1”.

1. The term ‘endonuclease’ refers to cutting the DNA sequence from ______________
a) only within the polynucleotide chain, not at the ends
b) the ends of the chain
c) anywhere in the chain
d) exactly in the middle of the chain
Answer: a
Explanation: The cleavage is done within the polynucleotide chain and not at the ends. The enzyme which cuts the sequence at the ends is known as exonuclease.

2. The restriction endonuclease is having a defence mechanism in the bacterial system against foreign DNA such as viruses. But how it is able to protect its own DNA?
a) By methylation of bacterial DNA by restriction enzyme
b) By methylation of foreign DNA by restriction enzyme
c) By phosphorylation of bacterial DNA by restriction enzyme
d) By phosphorylation of foreign DNA by restriction enzyme
Answer: a
Explanation: The bacterial DNA is methylated by restriction enzyme and thus now it is not recognized by the restriction endonuclease. Thus methylation prevents the restriction endonuclease from cutting its own DNA.

3. Even after replication, how the modified DNA remains protected?
a) It remains protected because of conservative mode of replication
b) It remains protected because of semi-conservative mode of replication
c) The mode of replication has no role to play in the protection
d) It is again modified after replication
Answer: b
Explanation: Because of the semi-conservative mode of replication, one of the DNA strands remain methylated even after replication. One methylated strand is sufficient for providing protection against cleavage by a restriction endonuclease.

4. How many classes of restriction enzymes are there?
a) 2
b) 1
c) 3
d) 4
Answer: c
Explanation: Three classes of restriction enzymes are there, I, II and III. These classes are having different characteristics such as the site of cleavage on the basis of the recognition sequence.

5. Type II cuts the sequence in the following way __________
a) Within the recognition sequence
b) At 100-1000 nucleotides away from the recognition sequence
c) At 27-30 nucleotides away from the recognition sequence
d) It cuts randomly
Answer: a
Explanation: Recognition sequence is the set of nucleotides which are identified by the enzyme and then it cleaves. In class II, the cleavage is done within the recognition sequence.

6. After cleaving the sequence, the nature of the ends created by the type II endonuclease is __________
a) The ends created are always single stranded
b) The ends created are always double stranded
c) Either the ends are single stranded or they are double stranded
d) One end is single stranded and one end is double stranded
Answer: c
Explanation: Either the ends are both double stranded or are both single stranded. The double stranded ends are blunt ends whereas the single stranded ends are sticky ends.

7. A sequence is having two ends, 5’ and 3’. Which of the following statements is correct regarding the nature of the ends?
a) The 5’ end is having hydroxyl group
b) The 5’ end is having phosphate group
c) The 3’ end is having phosphate group
d) Any group can be present at any end
Answer: b
Explanation: The 5’ end is having a phosphate group. As in a DNA sequence, the 5’ end is characterized by phosphate group and the 3’ end is characterized by a hydroxyl group.

8. Blunt ends created by the restriction endonuclease can be joined.
a) True
b) False
Answer: a
Explanation: The blunt ends created can be joined by the enzyme responsible for ligation such as DNA ligase. It is not necessary to have sticky or single stranded ends.

9. The recognition sequence for BamHI is 5’ G|GATCC 3’. The ‘|’ represents the cutting site. What can be inferred about the ends from it?
a) The ends created are double stranded
b) The single stranded end is 5’ in nature
c) The single stranded end is 3’ in nature
d) To decide about the nature of the ends more information is needed
Answer: b
Explanation: The other strand is just complementary to it and can be written in the following way:
5’ G|GATCC 3’
3’ CCTAG|G 5’
After cleavage, the sequences are represented as:
5’ G 3’ 5’ GATCC 3’
3’ CCTAG 5’ 3’ G 5’
Thus, we can see that the ends generated are single stranded at 5’ end.

10. The recognition sequence of Sau3A is 5’ |GATC 3’ and that for DpnI is 5’ GA|TC 3’. Which of the statements is true?
a) The ends created by both the enzymes are compatible
b) The ends created by both the enzymes are not-compatible
c) The ends created by DpnI are single stranded
d) The ends created by Sau3A are single stranded
Answer: b
Explanation: Though the recognition sequence is the same, the ends are not compatible. It is so because both the enzymes leave double stranded ends and thus can’t be ligated.

11. The recognition sequence is at times palindromic in nature. Which of the following statements is correct with respect to it?
a) The molecules which are cut by the same enzyme, anneal only if the sequence is palindromic in nature
b) When the molecules are cleaved by the same enzyme and the recognition sequence is palindromic in nature, there is no effect on annealing
c) There are increased chances of annealing if the recognition sequence is palindromic in nature
d) The term ‘palindromic’ can be used whether the sequence is read from 5’ to 3’ or 3’ to 5’
Answer: c
Explanation: The term palindromic can be used only when a sequence is read along the same polarity ie either 5’ to 3’ or 3’ to 5’. When the recognition sequence is palindromic in nature, there are increased chances of annealing because now there are increased orientations. If the sequence is non-palindromic in nature, then also annealing would take place but in fewer orientations.

12. If all the nucleotides are present with equal frequencies and at random, what are the chances of having a particular four nucleotide long motif?
a) 1/256
b) 1/64
c) 1/16
d) 1/8
Answer: a
Explanation: There are four nucleotide bases present in a DNA sequence A, T, C and G. If the bases are present with equal frequency and at random the chances of having a particular 4 nucleotide long motif is 1/ (4*4)= 1/256.

Genetic Engineering Multiple Choice Questions on “Expression of Coding Section In-Vitro”.

1. If an expression of the coding sequence is to be carried in the vitro, then which of the statement holds true?
a) The extract in which the host DNA constructs are incubated is capable of translation only
b) The extract in which the host DNA constructs are incubated is capable of transcription only
c) The extract is made up of lysate of E. coli cells, containing RNA polymerase, ribosomes, tRNAs etc
d) Non radiolabelled amino acids are used
Answer: c
Explanation: If an expression of the coding sequence is to be carried in vitro, then the extract in which host DNA constructs are incubated is capable of both transcription and translation. The extract is made up of lysate of E. coli cells, containing RNA polymerase, ribosomes, tRNAs etc. Radiolabelled amino acids are used and thus detection becomes easier.

2. If translation and transcription are taking place together, the reaction is called as ______ and if they are carried out separately, it is called as _______
a) coupled, linked
b) linked, coupled
c) grouped, separate
d) grouped, individual
Answer: a
Explanation: If translation and transcription are carried out together, it is known as a coupled reaction. If they are carried out separately then they are known as linked reactions.

3. If translation and transcription are carried out separately, then what is added to carry out translation?
a) DNA polymerase
b) RNA polymerase
c) Taq polymerase
d) Tfl polymerase
Answer: b
Explanation: The added RNA polymerase carries out the translation in a separate reaction by the use of lysate of wheat gram cells.

4. If the screening of the DNA constructs is carried out in pooled batches then it takes less time.
a) True
b) False
Answer: a
Explanation: If the screening of DNA constructs is carried out in pooled batches then less time is taken. Once the pooled batch is identified with recombinants then each member is screened.

5. Choose the incorrect statement for the open reading frame.
a) It is helpful in deciding whether the clone is correct or not
b) It consists of an initiation codon
c) It consists of a termination codon
d) The presence of introns doesn’t affect the open reading frame
Answer: d
Explanation: Open reading frame is very important in deciding whether the clone is correct or not. It consists of an initiation codon, followed by a stretch of DNA and termination codon. The presence of introns the open reading would be affected and makes detection difficult.

6. Introns are ________ in nature and the sequence is _______ than exons.
a) coding, less random
b) non-coding, less random
c) non-coding, more random
d) coding, more random
Answer: c
Explanation: Introns are non-coding in nature and thus the sequence is having a more random base composition than exons. Exons are coding sequences.

7. If the genome size is small, then sequencing of the entire genome is ___________
a) easy
b) difficult
c) independent of the size of genome
d) not possible
Answer: a
Explanation: If the genome size is small, then the sequencing of entire genome is carried out easily. Hence, direct sequencing is carried out if the genome size is small.

8. Choose the correct statement with respect to the bacterial genome.
a) They are easy to sequence
b) They are difficult to sequence
c) Methods such as transcription and translation are satisfactory
d) Only transcription can be carried out
Answer: b
Explanation: Bacterial genome is of large size and thus it is difficult to carry out sequencing. Methods such as transcription and translation are unsatisfactory.

9. For identification of genes that are expressed under special conditions, _______ libraries are used.
a) cDNA
b) genomic
c) subtractive
d) shelf
Answer: c
Explanation: If genes are expressed under special conditions then specialized libraries should be used. These specialized libraries are subtractive. In this, driver and tracers are used.

10. In ______ organisms, firstly the gene is looked into model organisms.
a) eukaryotic
b) prokaryotic
c) both eukaryotic and prokaryotic
d) large sized
Answer: a
Explanation: In eukaryotic organisms, firstly the gene is looked into the model organism. In the model organism, the genome size is small. Eukaryotic organisms generally have a varying size of the genome.

Bioinformatics Interview Questions and Answers focuses on “Motif and Domain Databases Using Regular Expressions”.

1. While scanning for similarities in motifs, how regular expressions’ techniques work?
A. It represents a sequence family by a string of characters and further compares them
B. An algorithm similar to dynamic programming is used
C. Dot matrix analysis is used in this type of sequence analysis
D. Matrix analysis methods are used in this type
Answer: A
Explanation: In regular expressions’ techniques Pattern matching is defined as true or false in answer or outcome. In other words, if the pattern described in regex is found in a string of letters, the answer is true.

2. Which of the following best defines regular expressions?
A. They are made up of terms, operators and modifiers
B. They describe string or set of strings to find matching patterns
C. They are strictly restricted to alignment and corresponding score
D. They consist of set of rules for the connotations of various amino acid residues
Answer: B
Explanation: Regular expressions are powerful notable algebra that describe string or set of strings to find matching patterns. Pattern matching is defined as true or false in answer or outcome. And it is true that they are made up of terms, operators and modifiers but they are terminologies further used in matching process.

3. In regular expressions, which of the following pair of pattern is wrongly matched with its significance?
A. [ ] – Or
B. { } – Not
C. ( ) – Repeats
D. Z – Any
Answer: D
Explanation: Regular Expression Symbols have their own significances in regular expressions system means [GA] .g.e rFo ‘G or A’, {V,P} means not P or V, x(4) means (xxxx). Likewise, X denotes any character.

4. In terminologies related to regular expressions which of the following is false about terms and operators?
A. Terms are strings or substrings
B. Operators combine terms and expressions
C. Operators do not have precedence
D. Operators have precedence like arithmetic operators
Answer: C
Explanation: For harmonious, efficient and error-free functioning of the matching preocess, operators have precedence in order to set the priority of the operations to be carried out during the alignment.

5. In regular expressions, which of the following pair of pattern is wrongly matched with its significance?
A. ‘-’ – separator
B. < – N-terminal
C. > – C-terminal
D. ‘>>’ – end
Answer: D
Explanation: Regular Expression Symbols have their own significances in regular expressions’ system. For e.g. x(2,3) means x-x or x-x-x. Similarly, ‘.’ means end.

6. Emotif uses which databases for alignment of sequences?
A. BLOCKS and PRINTS databases
B. PROSITE
C. BLOCKS
D. PRINTS
Answer: A
Explanation: Emotif is a motif database that uses multiple sequence alignments from both the BLOCKS and PRINTS databases with an alignment collection much larger than PROSITE. It identifies patterns by allowing fuzzy matching of regular expressions. Therefore, it produces fewer false negatives than PROSITE.

7. While analysing motif sequences, what is the major disadvantageous feature of PROSITE?
A. The database constructs profiles to complement some of the sequence patterns
B. The functional information of these patterns is primarily based on published literature
C. Some of the sequence patterns are too short to be specific
D. Lack of specificity about probability and variation and relation between them
Answer: C
Explanation: The major pitfall with the PROSITE patterns is that some of the sequence patterns are too short to be specific. Rest of the options are advantages. The problem with these short sequence patterns is that the resulting match is very likely to be a result of random events. Overall, PROSITE has a greater than 20% error rate. Thus, either a match or non-match in PROSITE should be treated with caution.

8. Which of the following is not a characteristic of Fuzzy or approximate matches in regular expression?
A. This method is able to include more variant forms of a motif with a conserved function
B. the rule of matching is based on observations, not actual assumptions
C. with the more relaxed matching, there is increase of the noise level and false positives
D. the rule of matching is based on assumptions not actual observations
Answer: B
Explanation: The rule of matching is based on assumptions not actual observations in Fuzzy or approximate matches in regular expression. This provides more permissive matching by allowing more flexible matching of residues of similar biochemical properties. For example, if an original alignment only contains phenylalanine at a particular position, fuzzy matching allows other aromatic residues (including unobserved tyrosine and tryptophan) in a sequence to match with the expression.

9. Which of the following is not a characteristic of exact matches in regular expression?
A. There must be a strict match of sequence patterns
B. Any variations in the query sequence from the predefined patterns are not allowed
C. Provide more permissive matching by allowing more flexible matching of residues of similar biochemical properties
D. Searching a motif database using this approach results in either a match or non-match
Answer: C
Explanation: In this type of matching, there has to be a strict match of sequence patterns. This way of searching has a good chance of missing truly relevant motifs that have slight variations, thus generating false-negative results. As new sequences of motif are being accumulated, the rigid regular expression tends to become obsolete if not updated regularly to reflect the changes.

10. What does this representation mean- R.L.[EQD]?
A. An arginine- Amino acid- Leucine- Amino acid- Either Apartic acid, glutamic acid or glutamine
B. An arginine- Leucine- Either Apartic acid, glutamic acid or glutamine
C. An arginine- Leucine- Amino acid- Either Apartic acid, glutamic acid or glutamine
D. An arginine- Leucine- Apartic acid and glutamic acid and glutamine
Answer: A
Explanation: This is an example of pexel motif. Here, the ‘.’ represents the ‘end’ i.e. the amino acid as mentioned in the answer and the [ ] means ‘or’ i.e. either of the mentioned residue is present in the given position.

To practice all areas of Bioinformatics for Interviews, About Us.

Privacy Policy

Bioprocess Engineering Multiple Choice Questions on “Material Balances with Recycle, By – Pass and Purge Systems”.

1. Which of the following represents the figure corresponding to the stream?
A. Recycle system
B. By-pass system
C. Purge system
D. Recover stream

Answer: B
Explanation: A by-pass is one where a portion of the inlet to a process unit is split from the feed and instead of entering the process is combined with the outlet from that process. This practice is far less common than recycle, but may be used if your ultimate goal is a material with properties” in-between” the untreated reactant and the process outlet product.

2. Consider the following labeled flowchart for a simple chemical process based on reaction A -> B and predict the overall and single-pass conversion of the process?
A. 100%, 70%
B. 100%, 50%
C. 100%, 75%
D. 100%, 55%

3. What is the combined feed ratio from the following?
A. Quantity of mixed feed stream to the quantity of fresh feed stream
B. Quantity of recycle stream to the quantity of fresh feed stream
C. Quantity of fresh feed stream to the quantity of recycle stream
D. Quantity of fresh feed stream to the quantity of mixed feed stream

Answer: A
Explanation: Combined feed ratio is the ratio of the quantity of mixed feed stream to the quantity of fresh feed stream.
Combined feed ratio = M/F.

4. A single effect evaporator is fed with 10000 kg / h of weak liquor containing 15 % caustic by weight and is concentrated to get thick liquor containing 40 % by weight caustic. Calculate:
(i) kg / h of water evaporated and
(ii) kg / h of thick liquor
A. 4560 kg/h, 3720 kg/h
B. 3460 kg/h, 7680 kg/h
C. 4350 kg/h, 6732 kg/h
D. 6250 kg/h, 3750 kg/h

5. A solution of potassium dichromate in water contains 15% w/w Kr₂Cr₂O₇. Calculate the amount of Kr₂Cr₂O₇ that can be produced from 1500 kg of solution if 700 kg of water is evaporated and remaining solution is cooled to 293K.
Data: Solubility of Kr₂Cr₂O₇ at 293 K is 115 kg per 1000 kg of water.
A. 158.875 kg
B. 156.678 kg
C. 145.478 kg
D. 148.875 kg

Answer: A
Explanation: 1500 kg of solution.
Kr₂Cr₂O₇ content of solution = 0 15×1500 = 225 kg
Water in 15 % solution = 1500 – 225 = 1275 kg
M.B of water:
Water in final solution = 1275 – 700 = 575 kg
Solubility of Kr₂Cr₂O₇ at 293 K = 115 kg / 1000 kg water
Amount of Kr₂Cr₂O₇ at 293 K = 115/1000 × 575 = 66.125 kg
M.B of Kr₂Cr₂O₇:
[Kr₂Cr₂O₇ in feed solution] = [Kr₂Cr₂O₇ in solution at 293k] + [Kr₂Cr₂O₇ produced as crystals]
Amount of Kr₂Cr₂O₇ crystals produced = 225 – 66.125 = 158.875 kg.

6. Which of the following is not the application of recycle system?
A. Increased reactant conversion
B. Decreased reactant conversion
C. Continuous catalyst regeneration
D. Circulation of the working fluid

Answer: B
Explanation: The most common application of recycle for systems involving chemical reaction is the recycle of reactants, an application that is used to increase the overall conversion in a reactor and not the decreased reactant conversion.

7. The fresh feed to an ammonia synthesis reactor contains nitrogen, hydrogen and 2.0 mole per cent inerts. The molar ratio of H₂:N₂ is 3:1. The product stream consists of pure ammonia. Since conversion in the reactor is only 15%, a recycle stream is used and in order to avoid build-up of inerts, a purge stream is withdrawn. The rate of purge stream is adjusted to keep inert concentration in the recycle stream at 8 mole per cent. For a fresh feed rate of 100 moles/hr. Note that recycle stream contains only nitrogen, hydrogen and inerts. The N₂:H₂ ratio of 1:3 is maintained in every process stream. Calculate the moles of ammonia produced.

A. 38.90 moles/hr
B. 28.90 moles/hr
C. 37.50 moles/hr
D. 27.50 moles/hr

8. The fresh feed to an ammonia synthesis reactor contains nitrogen, hydrogen and 2.0 mole per cent inerts. The molar ratio of H₂:N₂ is 3:1. The product stream consists of pure ammonia. Since conversion in the reactor is only 15%, a recycle stream is used and in order to avoid build-up of inerts, a purge stream is withdrawn. The rate of purge stream is adjusted to keep inert concentration in the recycle stream at 8 mole per cent. For a fresh feed rate of 100 moles/hr. Note that recycle stream contains only nitrogen, hydrogen and inerts. The N₂:H₂ ratio of 1:3 is maintained in every process stream, and calculate the moles of nitrogen entering the reactor and in the recycle stream?

A. 125 moles/hr, 100.50 moles
B. 135 moles/hr, 50 moles
C. 125 moles/hr, 50 moles
D. 185 moles/hr, 100.50 moles

Answer: A
Explanation: Moles of nitrogen entering the reactor = (37.50/2)/0.15 = 125 moles/hr
Therefore there are 125 – 24.50 = 100.50 moles of nitrogen in the recycle stream.

9. The fresh feed to an ammonia synthesis reactor contains nitrogen, hydrogen and 2.0 mole per cent inerts. The molar ratio of H2:N2 is 3:1. The product stream consists of pure ammonia. Since conversion in the reactor is only 15%, a recycle stream is used and in order to avoid build-up of inerts, a purge stream is withdrawn. The rate of purge stream is adjusted to keep inert concentration in the recycle stream at 8 mole per cent. For a fresh feed rate of 100 moles/hr. Note that recycle stream contains only nitrogen, hydrogen and inerts. The N2:H2 ratio of 1:3 is maintained in every process stream, and calculate the number of moles, moles of inerts and moles of hydrogen in the recycle stream?

A. 437 moles/hr, 35 moles/hr, 301.5 moles/hr
B. 237 moles/hr, 30 moles/hr, 200 moles/hr
C. 567 moles/hr, 35 moles/hr, 205 moles/hr
D. 347 moles/hr, 30 moles/hr, 500 moles/hr

Answer: A
Explanation: Total number of moles in the recycle stream = 100.50×4/0.92 = 437 moles/hr
Moles of inerts in the recycle stream = 437 – 100.50×4=35 moles/hr
Moles of hydrogen in the recycle stream = 100.50×3 = 301.50 moles/hr.

10. What do you mean by the splitting point?
A. The two streams split with different composition
B. The two streams split with equal composition
C. Assuming it’s not a reactor and there’s only 2 streams
D. Assuming it’s not a reactor and there’s only 1 stream

Answer: B

Bioprocess Engineering Assessment Questions and Answers focuses on “Role of Shear in Stirred Fermenters”.

1. Which type of forces stretch and distort the bubbles?
A. Shear forces
B. Strain forces
C. Surface tension
D. Frictional forces

Answer: A
Explanation: Shear forces in turbulent eddies stretch and distort the bubbles and break them into smaller sizes; at the same time, surface tension at the gas-liquid interface tends to restore the bubbles to their spherical shape. In the case of solid material such as cell flocs or aggregates, shear forces in turbulent flow are resisted by the mechanical strength of the particles.

2. Viscosity is inversely proportional to the size of eddies?
A. True
B. False

Answer: B
Explanation: If the viscosity of the fluid is increased, the size of the smallest eddies also increases. Increasing the fluid viscosity should, therefore, reduce shear damage in bioreactors.

3. Larger foam is produced in bioreactors with increased headspace and small workspace?
A. True
B. False

Answer: A
Explanation: A bioreactor is divided in a working volume and a headspace volume. The working volume is the fraction of the total volume taken up by the medium, microbes, and gas bubbles. The remaining volume is called the headspace. Typically, the working volume will be 70-80% of the total fermenter volume. This value will however depend on the rate of foam formation during the reactor. If the medium or the fermentation has a tendency to foam, then a larger headspace and smaller working volume will need to be used.

4. “Instrument air compressor” should be used generally as a compressor?
A. True
B. False

Answer: B
Explanation: Note that it is very important that an “instrument air” compressor is not used. Instrument air is typically generated at higher pressures but is aspirated with oil. Instrument air compressors are used for pneumatic control. Air compressors used for large scale bioreactors typically produce air at 250 kPa. The air should be dry and oil free so as to not block the inlet air filter or contaminate the medium.

5. What is the function of Pluronic f-68?
A. To increase the foaming
B. To decrease the foaming
C. Increase cell attachment
D. To change the composition of the cells

Answer: B
Explanation: Pluronic F68 is used in cell culture as a stabilizer of cell membranes protecting from membrane shearing and additionally acts as an anti-foaming agent. It is a non-ionic surfactant used to control shear forces in suspension cultures. It can also be used to reduce foaming in stirred cultures and reduce cell attachment to glass. It is provided at a concentration of 10% and effective at a working concentration of 0.1%.

6. Microcarrier beads 120 μm in diameter are used to culture recombinant CHO cells for production of growth hormone. It is proposed to use a 6-cm turbine impeller to mix the culture in a 3.5-1itre stirred tank. Air and carbon dioxide are supplied by flow through the reactor headspace. The microcarrier suspension has a density of approximately 1010 kg m^-3 and a viscosity of 1.3 x 10^-3 Pa s. Estimate the kinematic viscosity.
A. 1.30 × 10^-6 m²s^-1
B. 1.29 × 10^-6 m²s^-1
C. 1.50 × 10^-6 m²s^-1
D. 1.49 × 10^-6 m²s^-1

Answer: B
Explanation: Damage due to eddies is avoided if the (K ddot{o}) 1mogorov scale remains greater than 2/3-1/2 the diameter of the beads. Let us determine the stirrer speed required to create eddies with size, λ = 2/3 (120 μm) = 80 μm = 8 x 10^-5 rn. The stirrer power producing eddies of this dimension can be estimated:
Kinematic viscosity, v = (frac{mu}{rho} = frac{1.3×10^{-3} kg ,m^{-1} s^{-1}}{1010 ,kgm^{-3}}) = 1.29 × 10^-6 m²s^-1.

7. Refer to Q6, and calculate the power dissipated per unit mass of fluid.
A. 0.050 m² s^-3
B. 0.042 m² s^-3
C. 0.040 m² s^-3
D. 0.052 m² s^-3

Answer: D
Explanation: (lambda^4 = frac{v^3}{varepsilon})
(varepsilon = frac{v^3}{λ^4})
Therefore,
(varepsilon = frac{(1.29×10^{-6})^3 ,m^6 ,s^{-3}}{(8×10^{-5})^4 m^4}) = 0.052 m² s^-3.

8. Referring to Q6 and Q7 and, calculate the stirrer power.
A. 1.13×10^-2 W
B. 1.10×10^-2 W
C. 1.20×10^-2 W
D. 1.23×10^-2 W

Answer: A
Explanation: The stirrer power P is equal to ε multiplied by ρD_i³:
P = (0.052m²s^-3) (1010kgm^-3) (6x 10^-2m)³
P = 1.13×10^-2kgm²s^-3 = 1.13 x 10^-2W.

9. Referring to Q6, 7 and Q8, and calculate the stirrer speed.(Given: Np’ = 5)
A. 80.5 rpm
B. 85.0 rpm
C. 85.5 rpm
D. 80.0 rpm

10. What is the function of microcarrier beads?
A. To give the cells the shape of beads
B. It provides non-buoyancy condition
C. It helps in the lysis of cells
D. It provides protection and surface area

Answer: D
Explanation: The microcarrier beads are used to increase the number of adherent cells per flask and are either dextran or glass based, they come in a range of densities and sizes. The beads are buoyant and therefore can be used with spinner culture flasks. The surface area available for cell growth on these beads is huge. It is a support matrix allowing for the growth of adherent cells in bioreactors.