250+ TOP MCQs on Role of Bioinformatics in Gene Manipulation and Answers

Gene Manipulation Interview Questions and Answers focuses on “Role of Bioinformatics in Gene Manipulation”.

1. When was the first database of protein sequences established?
A. 1940
B. 1950
A. 1960
D. 1970
Answer: C
Clarification: In 1960s, Margaret Dayhoff established the first database of protein sequences, a database that was published annually as a series of volumes entitled Atlas of Protein Sequence and Structure.

2. Which of the following was the first protein to be sequenced?
A. Pectin
B. Insulin
A. Lectin
D. Rhodopsin
Answer: B
Clarification: Bioinformatics was born when the first complete protein sequence was determined. This was bovine insulin sequenced between 1951 and 1955.

3. When was ‘Atlas of Protein Sequence and Structure’ published?
A. 1955
B. 1965
A. 1975
D. 1985
Answer: B
Clarification: By 1965, when the ‘Atlas of Protein Sequence and Structure’ was first published, there were more than 100 sequences in the scientific literature.

4. “Globins” is a family of ____________
A. Datasets
B. Genes
A. Proteins
D. Hosts
Answer: C
Clarification: Most of the sequences contained in ‘Atlas of Protein Sequence and Structure’ were redundant and were used to investigate sequence diversity between homologous proteins in large families such as the globins.

5. When was the first nucleotide sequence determined?
A. 1966
B. 1946
A. 1976
D. 1986
Answer: A
Clarification: The first nucleotide sequence to be determined was that of a yeast transfer-RNA by Madison in the year 1966. Most nucleotide sequences prior to about 1975 were from RNA molecules.

6. When was the first nucleotide sequence database developed?
A. 1942
B. 1972
A. 1982
D. 2002
Answer: C
Clarification: In 1982 there were enough DNA sequences to justify the establishment of the first nucleotide sequence database, GenBank.

7. By the end of 1982, what was the approximate number of sequences in GenBank?
A. 100
B. 200
A. 400
D. 600
Answer: D
Clarification: By the end of 1982, GenBank contained a grand total of 606 sequences. The database grew steadily until about 1994 when the genomics era really kicked in.

8. In 1994, the approximate number of sequences in GenBank rose to ____________
A. 200
B. 2000
A. 20000
D. 200000
Answer: D
Clarification: In 1994 the number of sequences in GenBank was just over 200,000. Two decades later, the figure stands at 30 million and shows no sign of slowing down.

9. Primary sequence databases are repositories for _______________
A. Nucleotide sequence
B. Protein sequence
A. Genome sizes
D. Host range
Answer: A
Clarification: The primary sequence databases are repositories for annotated nucleotide sequence data. These are the most important databases in molecular biology.

10. DDBJ is a _________________
A. Repository
B. Protein bank
A. Nucleotide sequence database
D. Secondary database
Answer: C
Clarification: DDBJ is the abbreviation of the DNA Databank of Japan, and is a primary database. New sequence data can be deposited with this databank.

11. The primary sequence databases are repositories for _______ sequence data.
A. Incomplete
B. Complete
A. Contaminated
D. Raw
Answer: D
Clarification: The primary sequence databases are repositories for raw sequence data derived directly from experiments and sequencing projects.

12. The dbEST is a subsidiary of ___________
A. EMBL
B. GenBank
A. DDBJ
D. SwissProt
Answer: B
Clarification: GenBank has a subsidiary called the dbEST, which is a database of ESTs. It has been instrumental in generating gene maps by in-silico analysis.

13. SWISS-PROT is a repository for ____________
A. Nucleotide sequences
B. Protein sequences
A. Vectors
D. Genome arrays
Answer: B
Clarification: The SWISS-PROT is not just a repository for protein sequences. Rather, it is a collection of confirmed protein sequences annotated with related information.

14. Curated data in a database means _________ data.
A. Actively managed
B. Duplicate
A. Incomplete
D. Raw
Answer: A
Clarification: The quality of data in a database like SWISS-PROT is very high because it is actively managed, that is the data is curated, making all the necessary information accessible.

15. TrEMBL consists of entries in the same format as SWISS-PROT.
A. True
B. False
Answer: A
Clarification: The less robust TrEMBL database consists of entries in the same format as those in SWISS-PROT, derived from the translation of all coding sequences in EMBL.

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250+ TOP MCQs on Vectors and Cloning in Gram – Positive Bacteria – 2 and Answers

Vector Biology MCQs focuses on “Vectors and Cloning in Gram – Positive Bacteria – 2”.

1. All B. subtilis vectors replicate by _______________ mechanism.
A. Translation
B. Meiosis
A. Mitosis
D. Rolling circle
Answer: D
Clarification: All the B. subtlis vectors replicate by the rolling circle mechanism. Nearly every step in the process digresses from its usual function, thus effecting rearrangements.

2. ____________ DNA is generated during rolling circle mechanism.
A. Double stranded
B. Mutated
A. Single stranded
D. Wild type
Answer: C
Clarification: Single-stranded DNA is generated during rolling circle mechanism. It is known to be a reactive intermediate in every recombination process.

3. Vectors replicating by ____________ mechanism are more stable than those replicating by rolling circle mechanism.
A. Alpha
B. Theta
A. Beta
D. Gamma
Answer: B
Clarification: If structural instability is a consequence of rolling circle mechanism, then vectors which replicate by the alternative theta mechanism could be more stable.

4. Plasmid Pamb1 is derived from _____________
A. B. subtilis
B. Streptococcus
A. Yeast
D. Bacillus
Answer: B
Clarification: Janniere et al. (1990) have studied two potentially useful plasmids, Pamb1 which is a large natural plasmid derived from Streptococcus.

5. Replication of Pamb1 and Ptb19 does not lead to accumulation of _______________
A. Single-stranded DNA
B. Toxins
A. Genes
D. Nucleases
Answer: A
Clarification: Replication of Pamb1 and Ptb19 does not lead to accumulation of detectable amounts of single-stranded DNA, whereas the rolling-circle mode of replication does.

6. Classical E.coli vectors replicate via theta-like structures.
A. True
B. False
Answer: A
Clarification: The classical E. coli vectors, which are derived from plasmid ColE1, all replication via theta-like structures and not the rolling-circle mechanism.

7. All the series of cloning vectors developed from PAMB1 carry repE and copF which is a __________
A. Promoter
B. Regulator
A. Gene
D. Protein
Answer: B
Clarification: All the vectors developed from PAMB1 carry a gene essential for replication, repE, and its regulator copF. These are two important constituents.

8. CopF of PAMBI-derived vectors can be deactivated by inserting linker into ____________
A. Promoter
B. Inducer
A. KpnI site
D. Tol site
Answer: C
Clarification: The copF gene of PAMB1-derived vectors can be deactivated by inserting a linker into a unique Kpn1 site. Deactivation of copF leads to increase in the plasmid copy number per cell.

9. Deactivation of copF gene leads to _________
A. Increase in cell size
B. Increase in plasmid copy number
A. Decrease in cell size
D. Decrease in plasmid copy number
Answer: B
Clarification: The copF gene of PAMB1-derived vectors can be deactivated by inserting a linker into a unique Kpn1 site. Deactivation of copF leads to increase in the plasmid copy number per cell.

10. The original low copy number state of the PAMB1-derived vectors can be restored by _________ of the linker.
A. Induction
B. Deletion
A. Deactivation
D. Removal
Answer: D
Clarification: The copF gene of PAMB1-derived vectors can be deactivated by inserting a linker into unique Kpn1 site. Deactivation of copF leads to increase in the plasmid copy number per cell.

11. Poyart and Trie-Cuot constructed a shuttle vector based on PAMB1 for construction of ____________
A. Transcriptional fusions
B. Translational fusions
A. Biotherapeutics
D. Fusions
Answer: A
Clarification: Poyart and Trie-Cuot constructed a shuttle vector based on PAMB1 for construction of transcriptional fusions, it can be conjugally transferred between E. coli and a wide range of gram-positive bacteria.

12. Compared with E.coli, B. subtilis has additional requirements, it can _________ the expression of genes from Gram-negative organisms in ones that are Gram-positive.
A. Induce
B. Increase
A. Prevent
D. Decrease
Answer: C
Clarification: Compared with E.coli, B. subtilis has additional requirements for efficient transcription and translational and this can prevent the expression of genes from Gram-negative organisms in ones that are Gram-positive.

13. Which of the following is the principal sigma factor?
A. Sigma A
B. Sigma B
A. Sigma C
D. Sigma D
Answer: A
Clarification: The number of sigma factors is different in each of the various genera but the principal sigma factor is sigma A. The composition of the core RNA polymerase is the same in most general.

14. In B. subtilis many promoters contain essential ________ motif.
A. AGAG
B. TGTG
A. GGGG
D. GCGC
Answer: B
Clarification: In B. subtilis many promoters contain an essential TGTG motif upstream of the -10 region. Mutations of this region reduce promoter strength.

15. The TGTG motif is located at ______ region.
A. -10
B. -16
A. -20
D. -25
Answer: B
Clarification: In B. subtilis many promoters contain an essential TGTG motif upstream of the -10 region. Mutations of this region reduce promoter strength.

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250+ TOP MCQs onVectors for Mammals – 5 and Answers

Vector Biology Questions & Answers for Exams focuses on “Vectors for Mammals – 5”.

1. What is reporter genes used for?
A. To confirm transformation
B. To confirm transfection
A. To induce expression
D. To cease expression
Answer: A
Clarification: When controlled by a strong promoter, reporter genes are often used as markers to confirm stable or transient transformation.

2. What is “CAT”?
A. Repressor gene
B. Promoter
A. Reporter gene
D. Hybrid vector
Answer: C
Clarification: The first reporter gene to be used in animal cells, derived from E.coli transposons Tn9; it has also been used to a certain extent in plants.

3. The CAT gene confers antibiotic resistance.
A. True
B. False
Answer: A
Clarification: CAT gene encodes the enzyme chloramphenicol acetyltransferase, which confers resistance to antibiotic chloramphenicol.

4. Simian Virus 40 (SV40) is an example of ____________
A. Caulimovirus
B. Polyomavirus
A. Plant virus
D. Retrovirus
Answer: B
Clarification: Transient transformation can be achieved using replicon vectors that contain origins of replication derived from certain viruses of the polyomavirus family such as SV40.

5. Polyomavirus vectors such as SV40 are used especially for ____________
A. Gene expression
B. Gene manipulation
A. Recombinant proteins
D. Hybrid vector production
Answer: C
Clarification: These viruses cause lytic infections, during the infection cycle viral gene products accumulate at high levels. Hence this strategy is used to produce recombinant proteins.

6. Which is the first virus to be developed as an animal vector?
A. Papillomavirus
B. SV40
A. Adenovirus
D. HIV
Answer: B
Clarification: SV40 was the first animal virus to be characterized in detail at the molecular level and for this reason, it was also the first to be developed as a vector.

7. What is the approximate size of the SV40 vector?
A. 1 kb
B. 3 kb
A. 5 kb
D. 7 kb
Answer: C
Clarification: Simian Virus 40 (SV40) has a small icosahedral capsid and a circular double-stranded DNA genome of approximately 5 kb. It contains the early genes and the late genes.

8. Transcription in SV40 is controlled by ___________
A. Regulatory element
B. Late genes
A. Early genes
D. Promoter
Answer: A
Clarification: Transcription in SV40 is controlled by a complex regulatory element located between the early and late regions, and this includes early and late promoters, an enhancer and the origin of replication.

9. How many proteins are produced during the first stage of SV40 infection cycle?
A. 1
B. 2
A. 3
D. 4
Answer: B
Clarification: During the first stage of SV40 infection cycle, the early transcript produces two proteins, known as large T and small t tumor antigens.

10. What is the function of the T-antigen of SV40 viral vectors?
A. Genome replication
B. Translation
A. Transcription
D. Conjugation
Answer: A
Clarification: The function of T-antigen is particularly important as this protein binds to the viral origin of replication and is absolutely required for genome replication.

11. What function does an “oncoprotein” serve?
A. Controlled cell proliferation
B. Ceased cell proliferation
A. Uncontrolled cell proliferation
D. Initiates cell proliferation
Answer: C
Clarification: The T-antigen also acts as an oncoprotein, working with the host cell’s cycle machinery and causing uncontrolled cell proliferation.

12. Which type of infection is caused by the human BK virus?
A. Viral
B. Lytic
A. Latent
D. Carcinogenic
Answer: C
Clarification: Human BK virus causes latent infections in which the viral genome is maintained as a low to moderate copy-number replicon that does not interfere with host cell growth.

13. An epstein-barr virus causes a latent infection.
A. True
B. False
Answer: A
Clarification: Epstein-Barr virus causes latent infections in which the viral genome is maintained as a low to moderate copy-number replicon that does not interfere with host cell growth.

14. What are polyadenylation signals?
A. Inducers
B. Terminators
A. Promoters
D. Replicons
Answer: B
Clarification: Polyadenylation signals are terminators required in eukaryotic genes to generate a defined 3’ end to the mRNA.

15. Which of the following is a source of Poly(A. sites, incorporated into mammalian vectors?
A. Mouse beta-globin
B. Bovine serum albumin
A. Maltose binding protein
D. Agrobacterium
Answer: A
Clarification: Poly-adenyl sites from the Simian 40 Virus early transcription unit or mouse beta-globin gene are often incorporated into mammalian expression vectors.

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250+ TOP MCQs onCloning Vectors for E.Coli – 3 and Answers

Vector Biology Multiple Choice Questions on “Cloning Vectors for E.Coli – 3”.

1. Lambda-ZAPII can carry what size of the foreign DNA?
A. 1 kb
B. 5 kb
A. 10 kb
D. 15 kb
Answer: C
Clarification: Lambda-ZAPII is a lambda insertion vector that can take up to 10 kb of foreign DNA into any one of 6 restriction sites within a polylinker, incorporated within the lacZ’ gene.

2. How can lambda-ZAPII recombinants be distinguished?
A. Agar + ampicillin plating
B. Agar + tetracycline plating
A. Plaque morphology analysis
D. Agar + X-gal plating
Answer: D
Clarification: Insertion of foreign DNA in the polylinker containing restriction sites leads to inactivation of the lacZ’ gene. Hence recombinants can be distinguished as white colonies from non-Recombinant blue colonies.

3. What is a “stuffer fragment” in cloning jargon?
A. A replaceable fragment
B. Promoter region
A. Ribosomal binding site
D. Clustered with restriction sites
Answer: A
Clarification: Stuffer fragment is a replaceable fragment flanked by two recognition sites for the restriction endonuclease. Often this fragment carries additional sites that can be used to cut it up into small pieces so that its own re-insertion is unlikely.

4. What is not true in regard to a lambda replacement vector?
A. It contains two/more restriction sites
B. Contains stuffer fragment
A. Recombinant selection by size
D. Carries smaller DNA than insertion vectors
Answer: D
Clarification: Replacement lambda vectors are designed to carry larger pieces of DNA than the insertion vectors and upto 20 kb is the capacity whereas that of an insertion vector is 10 kb.

5. How can recombinant selection for lambda replacement vectors be done?
A. In vitro packaging
B. cI gene inactivation
A. Lac selection
D. Antibiotic resistance
Answer: A
Clarification: The replacement vectors are created so as to carry larger pieces of foreign DNA ad hence size constraints are put for in vitro packaging to occur. Small DNA chunks hence will not pack into the phage head.

6. Lambda-EMBL4 is an example of _________
A. Insertion vector
B. Replacement vector
A. Hybrid vector
D. Mammalian vector
Answer: B
Clarification: It is a lambda replacement vector and can carry upto 20 kb of foreign DNA by replacing a segment flanked by EcoRI, BamHI and SalI restriction endonucleases sites.

7. Recombinant selection of lambda-EMBL4 can be done on the basis of Spi phenotype.
A. True
B. False
Answer: A
Clarification: Spi phenotype can be used to distinguish recombinants of the replacement vector from the non-recombinants. Sensitive to P2 phage infection: for phages that do not infect hosts if it is already infected by P2 phage.

8. What is a pre-requisite for cloning experiments with lambda insertion or replacement vectors?
A. Circular form of the vector
B. Small size
A. Helper phage integration
D. Two-phage system
Answer: A
Clarification: For a replacement or insertion vector to stably infect the host, it is required that the vector be in its circular form with cos sites hydrogen bonded to each other.

9. What are cosmid vectors?
A. Hybrid between phage and plasmid
B. Hybrid between M13 and lambda phage
A. Modified lambda vector
D. Modified M13 vector
Answer: A
Clarification: Cosmids are hybrid between lambda phage and bacterial plasmid. The construction of cosmids centers on the fact that only cos sites are required for packaging the DNA into a phage protein coat.

10. What is the basis of cosmid construction?
A. Origin of replication is vital
B. Only cos sites are required for packaging
A. Promoter sequences for expression
D. Different restriction sites
Answer: B
Clarification: The design of a cosmid vector exploits the fact that only the cos sites are responsible for in vitro packaging of foreign DNA into phage protein coat.

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250+ TOP MCQs on Site Directed Mutagenesis and Answers

Gene Manipulation Multiple Choice Questions on “Site Directed Mutagenesis”.

1. Mutagens are physical or _________ agents.
A. Chemical
B. Mechanical
A. Hybrid
D. Exogenous

Answer: A
Clarification: Mutagens aid in the generation of mutants. These are chemical or physical agents that modify the organism’s DNA.

2. Site-directed mutagenesis is changing a given base in the cloned DNA.
A. True
B. False

Answer: A
Clarification: It is possible to change specifically any given base in a cloned DNA sequence. This technique is known as site-directed mutagenesis.

3. Creation of mutant proteins with novel properties is called ____________
A. Cloning
B. Protein engineering
A. Mutagenesis
D. Sequencing

Answer: B
Clarification: The technique of site-directed mutagenesis allows the creation of mutant proteins with novel properties, this is termed protein engineering.

4. When was the first method of site-directed mutagenesis developed?
A. 1940
B. 1970
A. 1980
D. 1950

Answer: C
Clarification: The first method of site-directed mutagenesis to be developed was the single primer method; developed by Gillam in 1980.

5. For single-primer method the DNA must be __________
A. Long
B. Short
A. Double-stranded
D. Single-stranded

Answer: D
Clarification: The method requires that the DNA to be mutated is available in a single stranded form, and cloning the gene in M13 based vectors makes this easy.

6. The synthetic oligonucleotide __________ the DNA synthesis.
A. Primes
B. Shortens
A. Lengthens
D. Degrades

Answer: A
Clarification: The synthetic oligonucleotide primes the DNA synthesis and is then later itself incorporated into the resulting heteroduplex molecule.

7. Clones can be screened using a _____________
A. PCR
B. Suppressor
A. Probe
D. Promoter

Answer: C
Clarification: The frequency with which mutated colonies arise, compared with wild-type colonies is low. In order to pick mutants, the clones can be screened by nucleic acid hybridization with P-32 labeled oligonucleotide as a probe.

8. The use of high-fidelity DNA polymerases has minimized the problem of ____________ mutations.
A. Internal
B. Site-directed
A. Extraneous
D. Point

Answer: C
Clarification: In earlier methods of mutagenesis, care had to be taken to avoid the introduction of adventitious changes. However, with high fidelity DNA, extraneous mutations can be avoided.

9. Contamination in heteroduplex molecules can be removed by __________________
A. Gel electrophoresis
B. PCR
A. Chromatography
D. Distillation

Answer: C
Clarification: The presence of contaminants reduces the proportion of mutant progeny, they can be removed by sucrose gradient centrifugation and gel electrophoresis.

10. The repair system of E.coli is ___________________
A. Lacking
B. Cysteine-directed
A. Methyl-directed
D. Mutated

Answer: C
Clarification: The major reason for low yield of mutant progeny is that the methyl-directed mismatch repair system of E. coli favors the repair of non-methylated DNA.

11. Which DNA are repaired at the site of mismatch?
A. Long
B. Short
A. Degraded
D. Unmethylated

Answer: D
Clarification: Newly synthesized DNA strands that have not yet been methylated are preferentially repaired at the position of mismatch.

12. Which of the following mutations are not used to overcome problems associated with the mismatch repair system?
A. MutL
B. MutS
A. MutH
D. MutE

Answer: D
Clarification: The problems associated with the mismatch repair system can be overcome by using host strains carrying mutL, muD, or mutH mutations.

13. The mutated strains mutL, mutS, and mutH prevent the methyl-directed repair of mismatches.
A. True
B. False

Answer: A
Clarification: The host strains carrying mutations mutL, mutS, and mutH prevent the methyl-directed repair of mismatches. They hence resolve the problems associated with the repair system.

14. All the primer extension methods of mutagenesis require _____________ template.
A. Double-stranded
B. Degraded
A. Single-stranded
D. RNA

Answer: C
Clarification: A disadvantage of all of the primer extension methods of mutagenesis is that they require a single stranded template.

15. Which kind of DNA are easier to prepare for PCR mutagenesis?
A. Linear
B. Circular
A. Single-stranded
D. Double-stranded

Answer: D
Clarification: With PCR-based mutagenesis, the template can be single-stranded or double-stranded, circular or linear. Double-stranded DNAs are easier to prepare.

250+ TOP MCQs onVectors and Cloning in Gram – Positive Bacteria – 1 and Answers

Vector Biology Multiple Choice Questions on “Vectors and Cloning in Gram – Positive Bacteria – 1”.

1. In Gram-positive bacteria, there is a disparity in ___________ in the genomes.
A. AT content
B. GC content
A. Structure
D. Composition
Answer: B
Clarification: In Gram-positive bacteria, the base composition of the different genomes ranges from less than 30 percent to more than 70 percent.

2. There are no universal cloning vectors for use with all Gram-positive bacteria.
A. True
B. False
Answer: A
Clarification: Given the disparity in GC content, the base composition of the different genomes ranges from less than 30 percent to more than 70 percent.

3. Streptomycetes are ______________
A. High GC content
B. Low GC content
A. Gram-negative bacteria
D. Low AT content
Answer: A
Clarification: There are no universal cloning vehicles for use with all Gram-positive bacteria. One set of systems has been developed for high-GC organisms such as Streptomycetes and another for low-GC organisms.

4. Which of the following is not a lactic acid producing bacteria?
A. Streptococcus
B. Lactococcus
A. Lactobacillus
D. Clostridium
Answer: D
Clarification: The latter group comprises bacteria from unrelated genera Bacillus, Clostridium and lactic acid producing bacteria lactococcus, streptococcus, lactobacillus.

5. Bacillus subtilis is a _________ bacteria.
A. Gram-negative
B. Low-GC
A. High-GC
D. High-AT
Answer: B
Clarification: Many of the cloning vectors used with Bacillus Subtilis and other low-GC bacteria are derived from plasmids found in Staphylococcus aureus.

6. Cloning vectors for low-GC bacteria are derived from ________
A. Plasmids
B. Yeasts
A. Plants
D. Mammals
Answer: A
Clarification: Many of the cloning vectors used with Bacillus Subtilis and other low-GC bacteria are derived from plasmids found in Staphylococcus aureus.

7. Plasmids from S. aureus transform into B. subtilis and express _____________ normally.
A. Lactic acid
B. Antibiotic resistance
A. Antibiotics
D. Crown gall
Answer: B
Clarification: Plasmids from S.aureus can be transformed into B.subtilis, where they replicate and express antibiotic resistance normally.

8. S. aureus plasmids carry ______ selectable markers.
A. More than one
B. One
A. No
D. Two
Answer: B
Clarification: None of the S. aureus plasmids carries more than one selectable marker and so improved vectors have been constructed by gene manipulation.

9. Vector PC194 carries ____ gene of PT127.
A. TCR
B. PCR
A. CCR
D. GCR
Answer: A
Clarification: Improved vectors have been constructed by gene manipulation.
For example, pHV11 is PC194 carrying the TCR gene of pT127.

10. Which of the following is greatly reduced in plasmid vectors post insertion of exogenous DNA?
A. Size
B. Stability
A. Infection
D. Efficiency
Answer: B
Clarification: In general, plasmids are stable in Bacillus subtilis, but segregative stability is greatly reduced following insertion of exogenous DNA.

11. Which of the following is a cryptic Bacillus plasmid?
A. Pta1060
B. Pta1100
A. Pta 10
D. BSA
Answer: A
Clarification: Stable host-vector systems in B.subtilis are more likely if endogenous plasmids are used. Bron and colleagues have developed the cryptic Bacillus plasmid.

12. Hybrid vectors can replicate both in ________ and B. subtilis.
A. Mammals
B. Insects
A. Plants
D. E.coli
Answer: D
Clarification: Because of difficulties experienced in direct cloning in B.subtilis, hybrid plasmids were constructed which can replicate in both E.coli and B. subtilis.

13. Which of the following is not a filamentous coliphage vector?
A. M13
B. Lambda
A. F1
D. Fd
Answer: B
Clarification: M13, F1, FD are filamentous coliphages containing a circular single-stranded DNA molecule. These have been developed as vectors because of various advantages.

14. Coliphages are single-stranded vectors.
A. True
B. False
Answer: A
Clarification: M13, F1, FD are filamentous coliphages containing a circular single-stranded DNA molecule. These have been developed as vectors because of various advantages.

15. What is the average size of single-stranded vector?
A. 6400 nucleotides
B. 1200 nucleotides
A. 2500 nucleotides
D. 5500 nucleotides
Answer: A
Clarification: The phage particles have dimensions 900*9 nm and contain a single-stranded circular DNA molecule, which is 6407 (M13) or 6408 (fD).

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