Category: Vector Biology Objective Questions

Vector Biology Puzzles focuses on “Vectors for Plants – 2”.

1. A two vector system contains a large normal vector and a small vector containing solely T-DNA fragment.
A. True
B. False
Answer: A
Clarification: A two vector system is basically constructed to surmount the problem of the big size of Ti plasmid. The basic principle is that no physical attachment of the plasmid with T-DNA is necessary.

2. The virulence region is contained in __________ of the binary vector system.
A. Large vector
B. Large vector along with specificity regions
A. Large vector along with restriction site
D. Small vector
Answer: B
Clarification: The virulence region and the host specificity region are contained in the larger plasmid and the T-DNA region along with a restriction site are contained in the smaller plasmid.

3. The unique restriction sites in a “binary vector system” are located in _________
A. T-DNA
B. Telomere sequence
A. Large plasmid
D. Small plasmid
Answer: D
Clarification: The unique restriction sites are located within the T-DNA region on the smaller plasmid which is about 20 kb in size. The sites are required for manipulations.

4. Which vector does the co-integration strategy use?
A. E.coli vector
B. Bacteriophage
A. Bacterial plasmid
D. Ti plasmid
Answer: A
Clarification: The co-integration strategy is adapted to surmount the problem of the huge size of the Ti plasmid. In this strategy, E.coli vector is used that contains a part of T-DNA.

5. In which region does the infection of wound happen in crown gall disease?
A. Leaf
B. Root
A. Stem
D. Leaf node
Answer: C
Clarification: The infection of the wound happens in the stem, the bacteria attack the plant at the wounded stem portion and the process of transfer proceeds.

6. What does disarming the vector mean?
A. Removing cancerous genes
B. Removing flanks of lacZ’
A. Mutating DNA
D. Removing T-DNA
Answer: A
Clarification: A mature plant regenerated from transformed cells will contain the cloned gene in every cell and will pass the cloned gene to its offspring.

7. Which parts of the T-DNA are involved in infection?
A. Interstitial sequence
B. 25bp sequence
A. Origin of replication
D. Opine genes
Answer: B
Clarification: The only parts of the T-DNA involved in infection are two 25 bp repeat sequences found at the left and right borders of the region integrated into the plant DNA.

8. What is Pbin19?
A. Binary vector
B. Hybrid vector
A. Plasmid vector
D. Ti plasmid vector
Answer: A
Clarification: PBIN19 is a disarmed cloning vector which is a binary vector. The left and right borders of this vector flank copy of lacZ’ gene.

9. What is the basic difference between Ri plasmid and Ti plasmid of Agrobacterium?
A. Disease caused
B. Parent molecule
A. Types of plants prone
D. Mode of infection
Answer: A
Clarification: Both Ri and Ti plasmid are present in the parent molecule of Agrobacterium Tumefaciens. The basic difference lies in the type of disease which the plasmid causes; due to its inherent capability of transferring a part of its own plasmid DNA to the genome of the plant.

10. Which disease is caused by Ri plasmid of Agrobacterium tumefaciens?
A. Crown gall disease
B. Hairy root disease
A. Carcinoma
D. Fungal infection
Answer: B
Clarification: The Ri plasmid of the soil bacterium species Agrobacterium Tumefaciens causes the hairy root disease in the attacked dicot plants. The underlying mechanism of this infection is the transferal of T-DNA of plasmid to the plant host genome.

11. Which of the following is a symptom of hairy root disease in plants?
A. Massive proliferation
B. Decreased proliferation
A. Cancerous outgrowth
D. Dwarfness in plants
Answer: A
Clarification: The Ri plasmid of Agrobacterium Rhizogenes causes a hairy root disease, typified by the massive proliferation of a highly branched root system.

12. Cloning genes in dicots are more difficult than cloning genes in monocots.
A. True
B. False
Answer: B
Clarification: Several factors have made it much more difficult to clone genes in monocots than in dicots such as tomato, tobacco, potato, peas.

13. Which of the following is an example of monocot?
A. Tobacco
B. Potato
A. Wheat
D. Peas
Answer: C
Clarification: Monocots include wheat, barley, rice, and maize, which are the most important crop plants and hence the most desirable targets for genetic engineering projects.

14. What is the technique of biolistics used for?
A. Regeneration of plants
B. Regeneration of plasmid
A. Introduction of DNA
D. Infection by Agrobacterium
Answer: C
Clarification: Biolistics are the techniques of bombardment with microprojectiles to introduce plasmid DNA directly into plant embryos.

15. Using the biolistics approach, which step is skipped in the process of cloning plants?
A. Use of Agrobacterium
B. Use of plasmids
A. Use of foreign gene
D. Use of embryos
Answer: A
Clarification: Biolistics circumvents the need to use Agrobacterium (Tumefaciens or Rhizogenes)as the means of transferring DNA into the plant cells.

.

Vector Biology Question Bank focuses on “Bacteriophages – 2”.

1. What is the function of a tail of the head-and-tail lambda phage?
A. Stabilization of phage DNA
B. Attachment of phage
A. Release of replicative enzymes
D. Helps in motility
Answer: B
Clarification: The DNA is contained in the polyhedral head structure and the tail serves to attach the phage to the bacterial surface and to inject the DNA into the cell.

2. How are genes arranged in a lysogenic phage (for example a lambda phage)?
A. Alternatively
B. In clusters
A. On opposite ends
D. Similar genes at distant positions
Answer: B
Clarification: Genes related in terms of function are clustered together in the genome. Clustering is profoundly important for controlling the expression of the lambda genome as it allows genes to be switched on and off rather than individually.

3. Circularization of the injected linear phage DNA molecule is facilitated by which of the following?
A. Polyhedral head
B. Cos sites
A. Phage tail
D. Capsid proteins
Answer: B
Clarification: The lambda cohesive ends are called the cos sites and they play different roles during the infection cycle. They allow the linear DNA molecule that is injected into the cell to be circularized, which is a necessary prerequisite for insertion into the bacterial genome.

4. By which mechanism does the replication of new lambda DNA molecules are produced in the host bacterium?
A. Rolling Circle Mechanism
B. DNA Polymerase Binding
A. The same mechanism as that of host
D. Replication fork propagation
Answer: A
Clarification: A large number of new lambda DNA molecules are produced by the rolling circle mechanism of replication, in which a continuous DNA strand is rolled off the template molecule.

5. M13 filamentous phage DNA molecule is smaller than the head-and-tail lambda phage.
A. True
B. False
Answer: A
Clarification: M13 DNA molecule is much smaller than the lambda DNA molecule, being only 6407 nucleotides in length. It is circular and consists entirely of single stranded DNA.

6. Which of these is not a feature associated with M13 phage?
A. Doesn’t need genes for insertion into host genome
B. Capsid constructed of multiple copies of 3 genes
A. Simpler infection cycle than lambda
D. Capsid constructed of 15 different proteins
Answer: D
Clarification: The small size of M13 DNA molecule means that it has a room for fewer genes and this is possible because the M13 capsid is constructed from multiple copies of just 3 proteins (3 genes) whereas synthesis of lambda phage capsid involves 15 different proteins.

7. How does an injection of a filamentous phage (example- M13) DNA into the host bacterium occur?
A. Via pilus
B. Surface attachment
A. Transformation
D. Transfection
Answer: A
Clarification: Injection of an M13 DNA molecule into an E.coli occurs via pilus, the structure that connects two cells during sexual conjugation. Once inside the cell, single stranded molecule acts as the template for synthesis of the complementary strand.

8. Replicative form (RF) refers to what part of the M13 bacteriophage?
A. M13 genome
B. Restriction sites
A. Endonuclease activity
D. Replication phase
Answer: A
Clarification: The double-stranded form of M13 genome is known as the Replicative form, which behaves much like a plasmid; this is what makes it an attractive cloning vector.

9. Phage display technique makes use of which of the following vectors?
A. M13
B. Lambda
A. 2 micron circle
D. BAC
Answer: A
Clarification: M13 vectors are used for phage display, a technique for identifying genes whose protein products interact with one another. M13 is also used in DNA sequencing and in vitro mutagenesis.

10. Which of the following is a positive regulatory gene on the lambda phage genome?
A. N
B. O
A. P
D. R
Answer: A
Clarification: Except for the two positive regulatory genes N and Q, all functionally related genes are clustered together on the phage lambda genome map.

To practice Vector Biology Question Bank, .

Gene Manipulation MCQs focuses on “Gene Manipulation in Xenopus”.

1. Gene transfer to Xenopus results in ________ possibilities.
A. 1
B. 2
A. 3
D. 4
Answer: B
Clarification: Gene transfer to Xenopus can result in transient expression or germline transformation. Gurdo in 1971 first showed these possibilities.

2. Xenopus were first shown to express which protein as a result of gene transfer?
A. Insulin
B. Globin
A. Pectin
D. Chitin
Answer: B
Clarification: Gurdon in 1971 first showed that Xenopus oocytes synthesized large amounts of globin after they had been microinjected with rabbit globin mRNA.

3. Which expression system of Xenopus is used to express a wide range of proteins?
A. Repair
B. Replication
A. Oocyte
D. Sperm
Answer: C
Clarification: Xenopus oocyte expression system has been a valuable tool for expressing a very wide range of proteins from plants and animals.

4. X. laevis is a __________
A. Insect
B. Fish
A. Snake
D. Frog
Answer: D
Clarification: X. laevis is an African clawed frog; oocytes can be obtained in large numbers by removal of the ovary of an adult female. Each fully grown oocyte is large cell arrested at first meiotic prophase.

5. The germinal vesicle is a ___________
A. Cell
B. Membrane
A. Nucleus
D. Cytosol
Answer: C
Clarification: The large oocyte has a correspondingly large nucleus, called the germinal vesicle which is located in the darkly pigmented hemisphere of the oocyte.

6. MRNA for microinjection into the Xenopus can be synthesized by in vitro ___________
A. Transcription
B. Translation
A. Replication
D. Fusion
Answer: A
Clarification: Due to the large size of the oocytes, mRNA synthesized by transcription in vitro using phage T7-RNA polymerase can be used for microinjection.

7. The finely drawn glass capillary to be used for gene transfer is the ____________
A. Needle
B. Temperature probe
A. pH probe
D. separator
Answer: A
Clarification: The process of microinjection is achieved by using a finely drawn glass capillary as the injection needle, held in a simple micromanipulator.

8. The oocyte nucleus of Xenopus contains ______ RNA polymerases.
A. 1
B. 2
A. 3
D. 4
Answer: C
Clarification: The oocyte nucleus contains a store for the three RNA polymerases, enough to furnish the needs of the developing embryos.

9. Heat-shock promoter can be used to express cDNAs in Xenopus.
A. True
B. False
Answer: A
Clarification: It is possible to express complementary DNA (cDNA. linked to a heat shock promoter or to a mammalian virus promoter for transcription of injected exogenous DNA.

10. Vaccinia virus vectors can be used for gene expression in ________
A. Nucleus
B. Cytosol
A. Cytoplasm
D. Membrane
Answer: C
Clarification: Various vectors can be used with Xenopus species to lead to gene expression in various locations of the cell. Vaccinia virus vectors are used for expression in cytoplasm.

To practice MCQs on all areas of Gene Manipulation, .

Gene Manipulation online quiz focuses on “Basic Laboratory Techniques – 3”.

1. What is used to transfer nucleic acid from gels to membranes for further analysis?
A. Gel electrophoresis
B. PFGE
A. Blotting
D. PCR
Answer: C
Clarification: Blotting is used to transfer nucleic acids from gels to membranes for further analysis. Nucleic acid labeling and hybridization have formed the basis for a range of experimental techniques.

2. Blotting describes the __________ of nucleic acids.
A. Monitoring
B. Immobilization
A. Racing
D. Comparison
Answer: B
Clarification: Blotting describes the immobilization of sample nucleic acids on to a solid support, generally on nylon or nitrocellulose membrane.

3. Which membrane is used in blotting?
A. Agarose
B. Sucrose
A. Polythene
D. Nylon
Answer: D
Clarification: The main aim of blotting is immobilization of the nucleic acid samples on a membrane. The membrane must promote binding.

4. Which of the following is used as targets in the blotting techniques?
A. Nucleic acids
B. Agarose
A. Proteins
D. E. coli
Answer: A
Clarification: The blotted nucleic acids are then used as targets in subsequent hybridization experiments. There are a few blotting procedures.

5. Which of the following is used for the analysis of compositional properties of DNA?
A. Southern blotting
B. Northern blotting
A. PCR
D. CHEF
Answer: A
Clarification: Southern blotting is the method used to transfer DNA from agarose gels to membranes so that the compositional properties of DNA can be analyzed.

6. When was the original method of southern blotting developed?
A. 1975
B. 1964
A. 1954
D. 1944
Answer: A
Clarification: The original method for southern blotting was developed in 1975, for detecting fragments in an agarose gel that are complementary to a given RNA or DNA.

7. Southern blotting cannot be used for RNA molecules.
A. True
B. False
Answer: B
Clarification: Southern method was developed as a variation to the Northern method of DNA analysis. Southern blotting is solely used for RNA.

8. In southern blotting ___________ is present in the reservoir.
A. DNA
B. Buffer
A. Gel
D. Agarose
Answer: B
Clarification: In the southern blotting technique, the agarose gel is mounted on filter paper wick which dips into a reservoir containing a transfer buffer.

9. The DNA molecules are immobilized on the _____________ in southern blotting technique.
A. Reservoir
B. Gel
A. Tray
D. Membrane
Answer: D
Clarification: The DNA molecules are carried out of the gel by the buffer flow and immobilized on the membrane. Initially, nitrocellulose was used as the membrane.

10. Which membranes have greater binding capacity than nitrocellulose membranes?
A. Sucrose
B. Agarose
A. Nylon
D. Teflon
Answer: C
Clarification: The main drawback of nitrocellulose membranes is their fragile nature. Supporting nylon membranes have a greater binding capacity and high tensile strength.

11. Larger DNA fragments require a ___________ transfer time.
A. Longer
B. Shorter
A. Medium
D. Very high
Answer: A
Clarification: Large DNA fragments, greater than 10 kb require a longer transfer time than short fragments. Uniform transfer hence must be allowed.

12. The depurination treatment in blotting, involves the use of HCl and ____________
A. Alkali
B. Acid
A. Proteins
D. Nucleic acids
Answer: A
Clarification: To allow uniform transfer of a wide range of DNA fragment sizes, the electrophoresed DNA is exposed to short depurination treatment with HCl followed by alkali.

13. The depurination treatment __________ the DNA fragments.
A. Increases
B. Shortens
A. Inactivates
D. Fluoresces
Answer: B
Clarification: The depurination treatment shortens the DNA fragments by alkaline hydrolysis at depurinated sites. It also denatures the fragments.

14. Depurination denatures the DNA fragments.
A. True
B. False
Answer: A
Clarification: Depurination denatures the fragments prior to transfer, ensuring that they are in the single stranded state and accessible for probing.

15. Gel is _____________ in neutralizing solution prior to blotting.
A. Inactivated
B. Boiled
A. Polymerized
D. Equilibrated
Answer: D
Clarification: The gel is equilibrated in neutralizing solution prior to blotting. This is the final step in the southern blotting technique.

To practice all areas of Gene Manipulation for online Quizzes, .

Tricky Vector Biology Questions and Answers focuses on “Vectors for Mammals – 13”.

1. What is the reason for a large genome of poxyvirus?
A. Encodes for all human genes
B. Encodes for replication machinery
A. Encodes all promoters
D. Contains a large portion of junk DNA
Answer: B
Clarification: The large genome and structural complexity of the virus is because it must encode and package all its own DNA replication and transcription machinery.

2. Why are recombinant genomes of poxyvirus non-infectious?
A. Virus packages its own DNA
B. Host packages viral DNA
A. DNA remains unpacked
D. Virus packages unpacked
Answer: A
Clarification: Since the virus normally packages its own replication and transcription enzymes, recombinant genomes introduced into cells by transfection are generally non-infectious.

3. 5-bromodeoxyuridine is an analog of ___________
A. Pyrimidine
B. Alanine
A. Valine
D. Thymidine
Answer: D
Clarification: In one selection regime strategy, the transgene is inserted into the viral TK gene and negative selection using the thymidine analog 5-bromodeoxyuridine.

4. When transgene is inserted into the viral hemagglutinin locus, wild-type plaques turn ________
A. Red
B. White
A. Blue
D. Clear
Answer: A
Clarification: When transgene is inserted into the viral hemagglutinin locus, if chicken erythrocytes are added to the plate of infected cells, wild-type plaques turn red and recombinants turn clear.

5. NEO is a type of ____________
A. Selectable marker
B. Screenable marker
A. Promoter
D. Inducer
Answer: A
Clarification: Since vaccinia vectors have a high capacity for foreign DNA, selectable markers such as NEO can be co-introduced with the experimental transgene to identify recombinants.

6. LacZ’ is a screenable marker used with Vaccinia vectors.
A. True
B. False
Answer: A
Clarification: Since vaccinia vectors have a high capacity for foreign DNA, screenable markers such as LacZ’ can be co-introduced with the experimental transgene to identify recombinants.

7. Transgene expression in vaccinia vectors depends on ___________
A. Exogenous promoter
B. Endogenous promoter
A. Endogenous inducer
D. Exogenous inducer
Answer: B
Clarification: Transgene expression usually needs to be driven by an endogenous vaccinia promoter, since transcription relies on proteins supplied by the virus.

8. Highest expression levels in Vaccinia vectors are produced by ______ promoters.
A. P11
B. P7.5
A. 4b
D. P10
Answer: A
Clarification: The highest expression levels are provided by late promoters such as P11, allowing the production of up to 1 microgram of protein.

9. Vaccinia vectors cannot be used to express genes with __________
A. Exons
B. Junk DNA
A. Viral components
D. Introns
Answer: D
Clarification: Since the cytoplasm lacks not only host transcription factors but also the nuclear splicing apparatus, vaccinia vectors cannot be used to express genes with introns.

10. Which of the following sequence must be removed from DNA expressed in Vaccinia vectors?
A. TTTTTNT
B. TTTTTTT
A. NNNNNNN
D. TNTNTNT
Answer: A
Clarification: The sequence TTTTTNT must be removed from all foreign DNA sequences expressed in Vaccinia vectors since the virus uses this as a transcriptional terminator.

11. A useful binary expression system contains Vaccinia virus and bacteriophage ________
A. T7
B. T5
A. T8
D. T9
Answer: A
Clarification: A useful binary expression system has been developed, in which the transgene is driven by the bacteriophage T7 promoter and the T7 polymerase.

12. The vaccinia virus cannot be used to express antigens from other infectious agents.
A. True
B. False
Answer: B
Clarification: The vaccinia virus cannot be used to express antigens from other infectious agents by replacing the Vaccinia Tk locus with transgene encoding hepatitis B surface antigen.

13. Which cells were transfected in an early demonstration that vaccinia virus could be used to express antigens from other infectious agents?
A. Human cells
B. Insect cells
A. Plant cells
D. Monkey cells
Answer: D
Clarification: The transgene is cloned in a plasmid and this plasmid is then transfected into vaccinia-infected monkey cells, and recombinant vectors are then selected.

14. Vaccinia viruses expressing the influenza hemagglutinin gene were used to immunize ____________
A. Hamsters
B. Humans
A. Rabbits
D. Cow
Answer: A
Clarification: Vaccinia viruses expressing the influenza hemagglutinin gene are used to immunize hamsters, and induce resistance to influenza.

15. Resistance to SIV and HIV-2 was shown by infecting _______ with Vaccinia vectors.
A. Cow
B. Plants
A. Drosophila
D. Mouse
Answer: D
Clarification: Monkeys infected with recombinant vaccinia and canarypox vectors have shown resistance SIV and HIV-2.

To practice Tricky questions and answers on all areas of Vector Biology, .

Vector Biology Multiple Choice Questions on “Vectors for Plants – 1”.

1. When were the cloning vectors for plants developed?
A. 1900
B. 1980
A. 2000
D. 1910
Answer: B
Clarification: The cloning vectors for plants were first developed in the 1980s and their use has led to the genetically modified crops.

2. Vectors based on naturally occurring ____________ of Agrobacterium are used in plants.
A. Plasmids
B. Phages
A. Cos sites
D. Chromosome
Answer: A
Clarification: Three types of vector systems with varying degrees of success are used in plants. One of them is vectors based on naturally occurring plasmids of Agrobacterium.

3. What is Agrobacterium tumefaciens?
A. Plant species
B. Plant virus
A. Soil microorganism
D. Fertilizer
Answer: C
Clarification: Agrobacterium tumefaciens is a soil micro-organism that causes crown gall disease in many species of dicotyledonous plants.

4. Which disease is caused by Agrobacterium tumefaciens?
A. Crown gall
B. Carcinoma
A. Angiogenesis
D. Fungal infection
Answer: A
Clarification: Crown gall occurs when a wound on the stem allows A. Tumefaciens bacteria to invade the plant. After the infection bacteria causes a cancerous proliferation of the stem tissue in the region of the crown.

5. The ability to cause crown gall disease is associated with the presence of __________ within the bacterial cell.
A. Plasmid
B. Origin
A. Replication sites
D. Polymerase
Answer: A
Clarification: The tumor inducing plasmid present in the bacterial cell of agrobacterium tumefaciens is responsible for causing the crown gall disease.

6. What is the size of the Ti plasmid?
A. 10 kb
B. 100 kb
A. 20 kb
D. 200 kb
Answer: D
Clarification: The Ti plasmid is a large plasmid of 200 kb size or greater. It carries numerous genes involved in the infective process.

7. A remarkable feature of the Ti plasmid is that after infection a part of the molecule is integrated into plant chromosomal DNA.
A. True
B. False
Answer: A
Clarification: The part of the Ti plasmid that gets integrated into the plant chromosome after infection is called the T-DNA or the transfer DNA.

8. What is the size range of T-DNA?
A. 15 and 30 kb
B. 5 and 10 kb
A. 50 and 100 kb
D. 1 and 5 kb
Answer: A
Clarification: The size range of T-DNA is between 15 and 30 kb depending upon the strain of bacteria. The whole of this region is integrated into a host chromosome.

9. The T-DNA is maintained in ______ form in the plant.
A. Stable
B. Unstable
A. Integrated
D. Loosely bound
Answer: A
Clarification: The T-DNA after infection and transfer to the host chromosome remains in the plant in a stable form and is passed to the daughter cells after each cell division.

10. What is the special feature of T-DNA?
A. Expression of unusual genes
B. Repression of unusual genes
A. Cell proliferation
D. Cell death
Answer: A
Clarification: The T-DNA contains eight or so genes that are expressed in the plant cell and are responsible for cancerous properties of the transformed cells.

11. What are the plant host cells that contain T-DNA called?
A. Infected
B. Attacked
A. Transformed
D. Transfected
Answer: C
Clarification: When the T-DNA from Ti plasmid present inside the bacterial cell Agrobacterium tumefaciens is transferred to the chromosome of the plant host, the host cells are called transformed.

12. What are opines?
A. Bacterial nutrients
B. Plant nutrients
A. Telomere sites
D. T-DNA ends
Answer: A
Clarification: The genes present on the T-DNA fragment direct the synthesis of unusual compounds called opines, that the bacteria use as nutrients.

13. What is the binary vector strategy used for?
A. Vector insertion into host
B. Insertion of new DNA
A. Manipulation of vector
D. Manipulation of host
Answer: B
Clarification: Due to the big size of Ti plasmid it is extremely difficult to find a unique restriction site. Novel strategies are hence used for the insertion of foreign DNA in the plasmid.

14. What is the basis of “binary vector” strategy?
A. No physical attachment
B. Big size
A. Strain dependence
D. Lysogenic/lytic cycle
Answer: A
Clarification: The binary vector strategy is based on the observation that the T-DNA does not need to be physically present/attached to the rest of the Ti plasmid.

15. In the binary vector strategy, the sizes of the two vectors used are _____ for bigger plasmid and ______ for smaller T-DNA plasmid.
A. 200, 20
B. 170, 20
A. 200, 50
D. 170, 50
Answer: B
Clarification: There are two vectors in the binary vector system. The bigger one is 170 kb in size and a smaller plasmid containing only the T-DNA region is 20 kb in size.

.