Category: Objective Questions

Genetic Engineering Multiple Choice Questions on “BAC Vector, M13 and its Derivatives – 1”.

1. F (fertility) factor is used in many bacterial systems for different purposes. Choose the incorrect statement with respect to the F factor.
a) It was identified in E. coli for catalysing genetic exchange between cells
b) Cells having F factor are called as female
c) The F factor is transferred from one cell to another via factors present outside
d) They are not related to BAC systems
Answer: a
Explanation: BAC vectors are based on the F (fertility) factor. They are known for transferring genetic material between cells and they do it by a proteinaceous filament called sex pilus. The cells having F factor are called as male cells.

2. F factor integration into the host chromosome is called as __________
a) F’ plasmid
b) F factor and host chromosome recombination
c) Hfr (High frequency of recombination)
d) Recombinant host chromosome
Answer: c
Explanation: F factor integration can be done into the host chromosome and this is called as Hfr (High frequency of recombination) strain.

3. Excision of the F factor can also be done from the host chromosome. Choose the incorrect statement in respect to it.
a) This excision is possible in vivo
b) Along with the F factor, host chromosome’s DNA sequences are also excited
c) The sequences of host chromosome which can be accommodated in the F plasmid are small in size
d) These plasmids are termed as F’ plasmids
Answer: c
Explanation: It is possible to carry out the excision of the F factor from the host chromosome. The excision is possible in vivo. Along with the F factor, the DNA sequences of host chromosome are also excised. The F plasmid is able to accommodate very large size f host chromosome fragments and these are known as F’ plasmids.

4. The low copy number of F plasmid is essential for the stability of these plasmids.
a) True
b) False
Answer: a
Explanation: Usually, the F plasmids are present in a low copy number in cells. These low copy numbers are very essential for the stability because then only a few molecules can act as substrates for recombination mediated deletion.

5. Choose the correct statement for BAC vector system.
a) BAC vector system stands for bacteria and chromosome
b) It usually accepts insert of size approximately 1000kbp
c) The repE and oriS sequences are required for controlling the copy number and par A-C sequences are required for replication
d) A selectable marker is there for chloramphenicol resistance
Answer: d
Explanation: BAC vector system stands for Bacterial Artificial Chromosome. It accepts an insert of size 100kbp-300kbp. The repE and oriS sequences are required for controlling the replication and the par A-C sequences are required for controlling the copy number. A selectable marker is there for chloramphenicol resistance. A lacZ gene is also present and recombination of BAC vectors is done into E. coli by electropoartion.

6. The bacteriophage M13 vectors belong to a group of phages called as __________
a) skinny or filamentous phage
b) M phage group
c) single stranded phage
d) double stranded phage
Answer: a
Explanation: Bacteriophage M13 vectors belong to a group of vectors known as skinny or filamentous phage. They have a dimension of 850 nm * 6 nm * 6nm. They are very useful as they convert the double stranded DNA into single stranded DNA.

7. Choose the correct statement for the infection process of M13.
a) The infectious particle is double stranded
b) It is contained in a protein coat which is made up of products of gene III or gene VIII
c) The phage infects only female cells
d) The phage enters the bacterial cells by the filament which is meant for movement
Answer: b
Explanation: M13 infectious particles are single stranded in nature. It is contained in a protein coat which is made up of products of gene III or gene VIII. The phage infects the bacterial cells by sex pilus and thus it infects only male cells.

8. Once the bacteriophage enters the bacterial cells, some changes are carried out in it. Which of the following is not included?
a) The single stranded molecule entering the cell is single stranded and is called as a + strand
b) The single stranded molecule is converted into double stranded molecule which is called as replicative form (RF)
c) It is carried out in a process similar to normal replication process of E. coli
d) The positive strand is used for the transcription of viral proteins
Answer: d
Explanation: Once the bacteriophage enters the bacterial cells it is converted into double stranded form from the single stranded form. The double stranded form is known as replicative form (RF). The single strand enteri g is called as a + strand. It is carried out in a process similar to that the normal replication process. A specific origin of synthesis is used for transcription of the complementary negative strand. The minus strand can be used to synthesize viral proteins.

9. Replication can also be carried out by rolling circle replication. Choose the correct statement for this type of replication?
a) It can be used for the formation of RF
b) The product of gene III binds on a specific site of double stranded genome
c) It creates a nick in the – strand generating a free 3’ hydroxyl
d) This strand is extended by the polymerase and displaces the original – strand
Answer: a
Explanation: Rolling circle replication is used for the formation of RF. The product of gene II binds on a specific site of double stranded genome. It creates a nick in the + strand generating a free 3’ hydroxyl. This strand is extended by the polymerase and it displaces the original + strand. This strand is removed from the newly synthesized strand because of another nick by gene II. Thus a separate + strand is created and it is circularized by gene II product.

10. Gene V blocks the production of which gene?
a) Gene III
b) Gene II
c) Gene VIII
d) Gene IV
Answer: b
Explanation: Gene V blocks the production of gene II. As the gene V blocks the production of gene II, conversion of single strands into double strands is also blocked.

Genetic Engineering Multiple Choice Questions on “Synthesis of RNA”.

1. Cloning vectors designed for the purpose of synthesis of RNA and proteins are known as ___________
a) Cloning vectors
b) RNA vectors
c) Bacteriophage vectors
d) Expression vectors
Answer: d
Explanation: Expression vectors are those vectors which are used for the synthesis of RNA and protein i.e. cloned DNA sequences are used for expression.

2. For studying processes such as splicing and cleavage, RNA is required. Choose the correct statement for this.
a) A mixture of different types of RNA is required
b) Two types of RNA are required
c) A few contaminating proteins are required
d) Sodium hydroxide is required
Answer: c
Explanation: For studying RNA processing events such as splicing and cleavage, RNA is required. Here, a single type of RNA is required and along with it, a few contaminating proteins are also used.

3. RNA can be synthesized by using vector. A vector with _______ is used and further through ________ RNA is isolated.
a) origin of replication, translation
b) promoter, transcription
c) promoter, translation
d) origin of replication, transcription
Answer: b
Explanation: Vectors can be used at times for synthesis of RNA. A vector with the promoter is used and transcription is carried out in a bacterial cell and afterward RNA is isolated.

4. Isolation of RNA can be carried out easily from bacterial cells.
a) True
b) False
Answer: b
Explanation: Isolation of RNA can’t be carried out easily from bacterial cells. Hence, the production of RNA is carried out by the transcription of cloned DNA.

5. Promoters are generally used after isolation from ___________
a) bacteriophage T7
b) bacteriophage SP6
c) baceriophage Mu
d) both bacteriophage T7 and SP6
Answer: d
Explanation: Promoters used for the synthesis of RNA are generally obtained from bacteriophage T7 and SP6. They can also be obtained rarely from bacteriophage T3.

6. By selecting the appropriate polymerase to activate the promoter ________ can be carried out _______
a) transcription, regardless of orientation
b) transcription, only in one orientation
c) translation, regardless of orientation
d) translation, only in one orientation
Answer: a
Explanation: By selecting the appropriate polymerase to activate the promoter, transcription can be carried out but it is regardless of the orientation.

7. Guanylyl transferase and GTP are used for?
a) Transcription
b) Translation
c) Capping of the message
d) Packaging
Answer: c
Explanation: At times, capping of the message is very necessary. Capping can be achieved by incubating the transcripts with enzymes such as guanylyl transferase and GTP.

8. Capping can be introduced by the use of cap analogue. Which of the statement is true?
a) Cap analogue can be introduced at the end of the transcript
b) Cap analogue can be introduced at the start of the transcript
c) Cap analogue can be introduced anywhere in the transcript
d) Cap analogue can be introduced both at the end and starting of the transcript
Answer: b
Explanation: For achieving capping, cap analogue should be introduced. Cap analogue should be introduced at the starting of the transcript. It is so because it requires 5’ end.

9. If transcription should not be carried out beyond the insert in the vector, then it should be linearized.
a) True
b) False
Answer: a
Explanation: If the transcription should not be carried out beyond the insert in the vector, then it should be linearized. It can be done by digesting with a restriction enzyme.

10. If the cells containing plasmids are infected with helper phage, which type of DNA can be produced, packaged and secreted into the medium?
a) Single stranded DNA
b) Double stranded DNA
c) Both single and double stranded DNA
d) Circular DNA
Answer: c
Explanation: If the cells containing plasmids are infected with helper phage, then single stranded DNA can be produced, packaged and secreted into the medium.

Bioinformatics Multiple Choice Questions on “Position – Specific Scoring Matrices”.

1. Analysis of s for conserved blocks of sequence leads to production of the position-specific scoring matrix.
A. True
B. False

Answer: A
Explanation: The analysis of MSAs (Multiple Sequence Alignment) for conserved blocks of sequence leads to production of the position-specific scoring matrix or PSSM. The PSSM may be used to search a sequence to obtain the most probable location or locations of the motif represented by the PSSM. Alternatively, the PSSM may be used to search an entire database to identify additional sequences that also have the same motif.

2. The quality and quantity of information provided by the PSSM also varies for ________ in the motif.
A. each row
B. each column
C. rows and columns
D. neither the rows nor the columns

Answer: B
Explanation: The quality and quantity of information provided by the PSSM also varies for each column in the motif, and this variation profoundly influences the matches found with sequences. This situation can be accurately described by information theory, and the results can be displayed by a colored graph called a sequence logo.

3. Two considerations arise in trying to tune the PSSM so that it adequately represents the training sequences. Which of the following is not their description?
A. If a given column in 20 sequences has only isoleucine, it is not very likely that different amino acid will be found in other sequences with that motif because the residue is probably important for function
B. If a given column in 20 sequences has only isoleucine, it is very likely that different amino acid will be found in other sequences with that motif because the residue is probably important for function
C. If the number of sequences with the found motif is large and reasonably diverse, the sequences represent a good statistical sampling of all sequences that are ever likely to be found with that same motif
D. Another column in the motif from the 20 sequences may have several amino acids, and some amino acids may not be represented at all

Answer: B
Explanation: The PSSM is constructed by a simple logarithmic transformation of a matrix giving the frequency of each amino acid in the motif. Even more variation may be expected at that position in other sequences, although the more abundant amino acids already found in that column would probably be favored.

4. If a good sampling of sequences is _______ the number of sequences is _________ and the motif structure is ________ it should, in principle, be possible to obtain frequencies highly representative of the same motif in other sequences also.
A. available, sufficiently large, not too complex
B. unavailable, sufficiently large, not too complex
C. unavailable, sufficiently small, not too complex
D. available, sufficiently large, too complex

Answer: A
Explanation: The more abundant amino acids already found in that column would probably be favored. Thus, if a good sampling of sequences is available, the number of sequences is sufficiently large, and the motif structure is not too complex, it should, in principle, be possible to obtain frequencies highly representative of the same motif in other sequences also (Henikoff and Henikoff 1996).

5. If the data set is _______ then unless the motif has __________ amino acids in each column, the column frequencies in the motif may not be highly representative of all other occurrences of the motif.
A. small, distinct
B. small, almost identical
C. large, almost identical
D. large, distinct

Answer: B
Explanation: The number of sequences for producing the motif may be small, highly diverse, or complex, giving rise to a second level of consideration. If the data set is small, then unless the motif has almost identical amino acids in each column, the column frequencies in the motif may not be highly representative of all other occurrences of the motif. In such cases, it is desirable to improve the estimates of the amino acid frequencies by adding extra amino acid counts, called pseudocounts, to obtain a more reasonable distribution of amino acid frequencies in the column.

6. Even if many pseudocounts are added in comparison to real sequence counts, the amino acid frequencies will not have any effect or influence.
A. True
B. False

Answer: B
Explanation: Knowing how many counts to add is a difficult but fortunately solvable problem. On the one hand, if too many pseudocounts are added in comparison to real sequence counts, the pseudocounts will become the dominant influence in the amino acid frequencies and searches using the motif will not work. On the other hand, if there are relatively few real counts, many amino acid variations may not be present because of the small sample of sequences.

7. Which of the following is not a feature of editors and formatters?
A. provision for displaying the sequence on a color monitor with residue colors to aid in a clear visual representation of the alignment
B. recognition of the multiple sequence format that was output by the MSA (Multiple Sequence Alignment) program
C. maintenance of the alignment in a suitable format when the editing is completed
D. disallowing shading conserved residues in the alignment

Answer: D
Explanation: In addition to this, provision of a suitable windows interface, allowing use of the mouse to add, delete, or move sequence followed by an updated display of the alignment, is a feature. In addition, there are other types of editing that are commonly performed on MSAs (Multiple Sequence Alignment) program such as, for example, shading conserved residues in the alignment.

8. GDE (Genetic Data Environment) provides a general interface on UNIX machines for sequence analysis, sequence alignment editing, and display.
A. True
B. False

Answer: A
Explanation: It is available from several anonymous FTP sites. This interface requires communication with a host UNIX machine running the Genetics Computer Group software. Interface with MS-DOS or Macintosh is possible if the computer is equipped with the appropriate X-Windows client software.

9. MACAW is a local multiple sequence alignment program only.
A. True
B. False

Answer: B
Explanation: MACAW is both a local multiple sequence alignment program and a sequence editing tool. Given a set of sequences, the program finds ungapped blocks in the sequences and gives their statistical significance. Later versions of the program find blocks by one of three user-chosen methods.

10. Two commonly encountered examples are the Genetics Computer Group’s MSF format and the CLUSTALW ALN format.
A. True
B. False

Answer: A
Explanation: This is because these formats follow a precise outline, one may be readily converted to another by computer programs. READSEQ by D.G.Gilbert at Indiana University at Bloomington is one such program.

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Genetic Engineering Questions and Answers for Entrance exams focuses on “Choosing the Right Mutations, Database of Mutant Lines and Gene Disruption”.

1. The minor change in amino acid sequence can lead to _______ effect on three dimensional structure and there _____ in the primary sequence.
a) huge, may be no change
b) no, may be huge change
c) very less, is very less change
d) huge, would be huge change
Answer: a
Explanation: The minor change in the amino acid sequence can lead to a huge effect on the three dimensional structure and might also abolish its function. In the primary sequence, there might be no change.

2. It is often useful to inactivate endogenous genes in an organism. It might be helpful in finding out _________ role of the wild type gene.
a) biological
b) chemical
c) physiological
d) anatomical
Answer: c
Explanation: Inactivation of endogenous genes in an organism is very important at times. It is helpful in finding out the physiological role of the wild type gene.

3. The inactivation of endogenous genes may also be helpful in directing the expression of the mutated gene in the absence of background expression of wild type gene.
a) True
b) False
Answer: a
Explanation: Inactivation of endogenous genes is very useful. It can be used to direct the expression of the mutated gene in the absence of background expression of wild type gene.

4. How can mutant strains be produced?
a) In systematic mutagenesis programmes
b) In individual organisms
c) Both by individual organisms and systematic mutagenesis programmes
d) Apart from these two, other methods are also used
Answer: c
Explanation: For popular model organisms, there are databases for mutant strains. These mutant strains can either be produced by systematic mutagenesis programmes or by individual organisms.

5. The principle of gene disruption is ________ to replace the endogenous chromosomal copy of a gene with __________
a) homologous recombination, inactivated gene
b) reciprocal translocation, inactivated gene
c) homologous recombination, activated gene
d) reciprocal translocation, activated gene
Answer: a
Explanation: For gene disruption, the basic principle is homologous recombination to replace the endogenous chromosomal copy of a gene with the inactivated gene.

6. The gene to be disrupted is cloned and a selectable marker is inserted. What should be the effect of selectable marker?
a) It should have no effect on target gene
b) It should make the target gene non-functional
c) There is no restriction; it can be either functional or non-functional
d) It should improvise the chances of survival of the target gene
Answer: b
Explanation: In the target gene, a selectable marker should be inserted. It can be either in the form of ampicillin resistance or a nutritional marker. The selectable marker should render the target gene non-functional.

7. The disrupted gene is excised from the vector and is inserted into the target organism. The excised gene should be in which form?
a) Circular
b) Supercoiled
c) Either supercoiled or circular
d) Linear
Answer: d
Explanation: The disrupted gene is excised from the vector and is inserted into the target organism. The excised gene should be linear in form.

8. Stable acquisition of the marker can take place only if a double crossover over the flanking sequence and their chromosomal counterparts causes the marker’s integration into the chromosome.
a) True
b) False
Answer: a
Explanation: The incoming molecule having the marker will not replicate stably because it is linear. For the stable replication, double crossover of the flanking sequence should take place, replacing the endogenous gene with the disrupted gene.

9. If ______ gene is there, the double crossover may leave ________ in the chromosome.
a) linear, functional copy
b) circular, functional copy
c) linear, a non-functional copy
d) circular, a non-functional copy
Answer: b
Explanation: If the disrupted gene is in a circular form, there is a possibility that the double crossover may still leave a functional copy in the chromosome.

10. If the target organism contains more than one copy of the gene, what is the effect on these copies?
a) Only copy is disrupted
b) All the copies are disrupted
c) It is difficult to ensure that all the copies are disrupted
d) Only a specified number of copies are disrupted
Answer: c
Explanation: If more than one copy of the gene is there in the target organism, it is difficult to ensure that all the copies are disrupted. Multiple copies of the gene can be present, if the organism is not haploid.

To practice all areas of Genetic Engineering for Entrance exams,

Genetic Engineering Multiple Choice Questions & Answers focuses on “cDNA Libraries, Polyadenylated RNA and cDNA Synthesis – 2”.

1. Choose the incorrect statement for second DNA strand synthesis.
a) The RNA: DNA complex is treated with the enzyme RNaseH and DNA pol
b) The enzyme RNaseH is responsible for nicking the RNA strand and leaving free 3’ hydroxyl ends
c) These RNA fragments can be used as primers
d) There is no portion of RNA left attached to the DNA strand
Answer: d
Explanation: For second strand DNA synthesis, the RNA: DNA complex is treated with RNaseH enzyme along with DNA pol. The enzyme is responsible for nicking the RNA strand and leaving 3’ hydroxyl ends. These free ends are used for second strand synthesis. These RNA fragments can be used as primers. After DNA: DNA duplex formation, only a small portion of RNA is left at the 5’ end.

2. What is the final product of the RNaseH method?
a) blunt ended dsDNA
b) staggered dsDNA at both ends
c) staggered dsDNA at 3’ end
d) staggered dsDNA at 5’ end
Answer: a
Explanation: The final product of the RNaseH method is blunt ended dsDNA. The RNA piece left at the 5’ end is removed by RNase and thus blunt ended dsDNA is left.

3. What would not happen if the RNA strand is completely removed from RNA: DNA hybrid?
a) There are no chances of the synthesis of the second DNA strand
b) Chance complementarity would take place
c) Hairpin structure would be formed
d) Hairpin structure is formed is not the final structure
Answer: a
Explanation: If the RNA strand is completely removed from RNA: DNA hybrid, the single stranded DNA remains. Then chance complementarity takes place and the 3’ end complements with any other portion in the single stranded DNA. It leads to the formation of a hairpin structure and the 3’ end extends in the presence of DNA pol and nucleotides.

4. The loop region is single stranded. It can be cleaved by using which enzyme?
a) Exonuclease
b) S1 nuclease
c) RNaseH
d) DNase
Answer: b
Explanation: The loop region in the hairpin structure is single stranded in nature. The single stranded portion can be cleaved by S1 nuclease.

5. Choose the correct statement with respect to the self priming method of cDNA synthesis.
a) It is less preferred than RNaseH method
b) A hairpin structure is formed with guarantee
c) The sequence corresponding to the 5’ end is lost
d) Reverse transcriptase is not used
Answer: c
Explanation: The self priming method is based on the formation of a hairpin structure. But there are only chances and no guarantee of formation of hairpin structure. In the first strand synthesis, reverse transcriptase is used. The sequence corresponding to the 5’ end is lost and in cDNA large deletions are observed. This method is less preferred than RNaseH method.

6. Choose the incorrect statement for the method homopolymer tailing.
a) The first step is the RNA: DNA hybrid synthesis
b) Terminal transferase is used for the addition of nucleotides on 3’ end
c) Terminal transferase adds only at DNA strands
d) The DNA strand is now having known sequence at 3’ end
Answer: c
Explanation: Homolpoymer tailing is also used for cDNA synthesis. The first step remains same as the other methods, which is the synthesis of RNA: DNA hybrid. It is then treated by terminal transferase, the enzyme is responsible for adding nucleotides on 3’ end of both DNA and RNA. The 3’ end is now having a known DNA sequence and it is used as a tail in the reaction.

7. Choose the correct statement for RACE.
a) It stands for Random Amplification of cDNA ends
b) It is for cloning particular cDNA ends
c) It is only of one type, which is 5’ RACE
d) Sequence data is not available in any case
Answer: b
Explanation: RACE stands for Rapid Amplification of cDNA ends. Sometimes, we wish to clone specific cDNA portion and are having some sequence data. It is meant for cloning specific cDNA ends and is classified into two types, 5’ RACE and 3’RACE.

8. The first primer in the case of 3’ RACE is ___________
a) internal sequence
b) oligo-dT adaptor molecule
c) oligo-dA adaptor molecule
d) adaptor oligo-dT primer
Answer: b
Explanation: The first primer in the case of 3’ RACE is oligo-dT adaptor molecule. The second primer is an internal sequence.

9. The first cDNA strand in 5’ RACE is tailed with oligo-dA tail.
a) True
b) False
Answer: a
Explanation: The first cDNA strand in the case of 5’ RACE is tailed with an oligo-dA tail. The synthesis of the first cDNA strand is by an internal primer.

10. What is the second primer in the case of 5’ RACE?
a) Internal primer
b) Oligo-dA sequence
c) Adaptor-oligo-dT primer
d) Oligo-dT adaptor molecule
Answer: c
Explanation: The second primer ie the primer used for the synthesis of the second cDNA strand is adaptor-oligo-dT primer. Subsequent PCR is carried out by using internal primer for coding sequence.

To practice all areas of Genetic Engineering,