Category: Vector Biology Objective Questions

Gene Manipulation Multiple Choice Questions on “Gene Inactivation and Inhibition”.

1. Many gene inactivation strategies don’t require gene modification.
A. True
B. False
Answer: A
Clarification: Many strategies for gene inactivation do not require the direct modification of the target gene. Traditional gene transfer strategies add new genetic information to the genome.

2. Antisense RNA blocks the activity of _______
A. DNA
B. mRNA
A. CDNA
D. RNA
Answer: B
Clarification: Antisense RNA blocks the activity of mRNA in a stoichiometric manner. Antisense RNA has the opposite sense to mRNA.

3. Presence of antisense and complementary sense RNA can lead to the formation of _______
A. Mutation
B. Duplex
A. Carcinogens
D. Protein
Answer: B
Clarification: The presence of complementary sense and antisense RNA molecules in the same cells can lead to the formation of a stable duplex.

4. Antisense RNA can be used in eukaryotes.
A. True
B. False
Answer: B
Clarification: Antisense RNA is used as a natural mechanism to regulate gene expression in a number of prokaryote systems and to a lesser extent in other organisms.

5. Which type of inhibition can be achieved using antisense RNA?
A. Stable
B. Unstable
A. Transient
D. Integrative
Answer: C
Clarification: Transient inhibition of particular genes can be achieved by directly introducing antisense RNA or antisense oligonucleotides into cells.

6. What is MBP?
A. Inducer
B. Protein
A. Promoter
D. Gene
Answer: B
Clarification: An expression cassette was constructed in which the mouse myelin basic protein CDNA was inverted with respect to the promoter.

7. Which gene is targeted in transgenic tomato plant?
A. Mg
B. Pg
A. Gfp
D. Mbp
Answer: B
Clarification: Smith in 1988 generated transgenic tomato plants carrying an antisense construct targeting the endogenous polygalacturonase gene.

8. Level of inhibition does not depend on ________ of antisense RNA.
A. Size
B. Temperature
A. Ph
D. Range
Answer: A
Clarification: The level of inhibition apparently does not depend on the size of the antisense RNA or the part of the endogenous gene to which it is complimentary.

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Gene Manipulation Interview Questions and Answers for Experienced people focuses on “Basic Laboratory Techniques – 7”.

1. An ideal oligonucleotide sequence should remain hybridized when it is _____ percent homologous to the target.
A. 50
B. 10
A. 100
D. 80
Answer: C
Clarification: The availability of the exact sequence of oligonucleotides allows conditions for hybridization and stringency washing to be tightly controlled so that the probe remains hybridized when it is 100% homologous to the target.

2. The “Wallace Rule” is used to determine which of the following?
A. Melting temperature
B. Stringency
A. Hybridization
D. Homogeneity
Answer: B
Clarification: Stringency is commonly controlled by adjusting the temperature of the wash buffer. The “Wallace Rule” is used to determine the appropriate stringency wash temperature.

3. Which temperature is measured in the Wallace rule?
A. Melting
B. Heating
A. Boiling
D. Wash
Answer: D
Clarification: The Wallace rule makes use of wash temperature to determine appropriate stringency conditions. It is usually a preliminary step.

4. In filter hybridization with oligonucleotide probes, the hybridization is performed _____ degrees below TM.
A. 2
B. 4
A. 5
D. 6
Answer: C
Clarification: In filter hybridization with oligonucleotide probes, the hybridization is performed at a lower temperature than the actual melting temperature.

5. For every mismatched base pair, a further 5⁰C reduction is necessary.
A. True
B. False
Answer: A
Clarification: In filter hybridizations with oligonucleotide probes, the hybridization step is usually performed at 5⁰C below TM for perfectly matched sequences.

6. The design of oligonucleotide is critical for ____________
A. Proteins
B. Hybridization
A. Detection
D. Temperature control
Answer: B
Clarification: The design of oligonucleotides for hybridization experiments is critical to maximize hybridization specificity.

7. Which of the following greatly influences the stability of the resultant hybrid?
A. AT content
B. GC content
A. Oligonucleotide weight
D. Melting temperature
Answer: B
Clarification: The GC content of oligonucleotide influences the stability of the resultant hybrid and hence the determination of appropriate stringency conditions.

8. Autoradiography involves production of _____________
A. Proteins
B. Image
A. Emulsion
D. Hybrid
Answer: B
Clarification: The localization and recording of a radiolabel within a solid specimen is known as autoradiography and involves the production of an image in a photographic emulsion.

9. Emulsions in autoradiography, contain a clear phase of ____________
A. Gelatin
B. Lactose
A. Nylon
D. Lectin
Answer: A
Clarification: Emulsions consist of silver halide crystals suspended in a clear phase composed mainly of gelatin. Silver ions are later converted into silver atoms.

10. Which of the following will convert silver ion to a silver atom in the emulsion?
A. Alpha particle
B. Protein
A. Probe
D. Beta particle
Answer: D
Clarification: When a beta-particle or gamma-ray from a radionuclide passes through the emulsion, the silver ions are converted to silver atoms.

11. The radioactive emissions produce which color on the developed autoradiograph?
A. White
B. Transparent
A. Black
D. Opaque
Answer: C
Clarification: In direct autoradiography, the sample is placed in intimate contact with the film and radioactive emissions produce black areas on the developed autoradiograph.

12. Direct autoradiography is not suited for particles such as phosphorus.
A. True
B. False
Answer: A
Clarification: Direct autoradiography has not suited the detection of highly energetic beta-particles, such a phosphorus-32 or gamma-rays.

13. How is emitted energy converted to light in Autoradiography?
A. X-Ray
B. UV
A. Spectroscopy
D. Scintillator
Answer: D
Clarification: Indirect autoradiography describes the technique by which emitted energy is converted to light by means of a scintillator, using fluorography.

14. Which of the following is used to improve the detection of weak beta-emitters?
A. Fluorography
B. Chromatography
A. PCR
D. PFGE
Answer: A
Clarification: Fluorography is used for the detection of weak beta-emitters. Analysis of such compounds is necessary and hence unique techniques are used.

15. Intensifying screens are sheets of solid ____________ scintillator.
A. Organic
B. Inorganic
A. Toxic
D. Colored
Answer: B
Clarification: Intensifying screens are sheets of a solid inorganic scintillator which are placed behind the film. Emissions are absorbed by the film.

To practice all areas of Gene Manipulation for Interviews, .

Vector Biology test focuses on “Special Vectors for Expression in E.Coli – 4”.

1. The secondary structures from intrastrand base pairs do not interfere with ribosomal attachment.
A. True
B. False
Answer: B
Clarification: The secondary structures resulting from intrastrand base pairs interfere with the attachment of the ribosome to its binding site.

2. Presence of which of the following can stabilize the fusion protein?
A. Imidazole
B. IPTG
A. Sucrose
D. Bacterial peptide
Answer: D
Clarification: The presence of a bacterial peptide at the start of the fusion protein may stabilize the molecule and prevent degradation.

3. The foreign proteins that lack a bacterial segment are ________
A. Destroyed
B. Hybridized
A. Expressed
D. Induced
Answer: A
Clarification: Foreign proteins in contrast to bacterial proteins, that lack a bacterial segment are often destroyed by the host.

4. The periplasmic space is the space between inner and outer ________
A. Cell wall
B. Nucleus
A. Membrane
D. Matrix
Answer: C
Clarification: If the signal peptide is derived from a protein that is exported by the cell, the recombinant protein may itself be exported into periplasmic space.

5. Export of protein is ________ into the periplasmic space.
A. Hindered
B. Desirable
A. Non-desirable
D. Suppressed
Answer: B
Clarification: Export of the protein into the periplasmic space is desirable because it simplifies the problem of purification.

6. Which technique is used for aiding protein purification?
A. Cloning
B. Hybridization
A. Chromatography
D. Radiation
Answer: C
Clarification: The bacterial segment may also aid purification by enabling the fusion protein to be recovered by affinity chromatography technique.

7. E.coli glutathione-S-transferase protein can be purified by adsorption onto _________
A. Charcoal
B. Sucrose beads
A. Agarose beads
D. Glucose beads
Answer: C
Clarification: Fusion involving the E.coli glutathione-S-transferase protein can be purified by adsorption onto agarose beads.

8. What is a disadvantage of fusion systems?
A. Property alteration
B. Reduced amount
A. Cost inefficiency
D. Host range
Answer: A
Clarification: The disadvantage with fusion systems is that the presence of E.coli segment may alter the properties of the recombinant protein.

9. Cyanogen bromide can be used to cleave __________
A. Arginine
B. Methionine
A. Aspartame
D. Glycine
Answer: B
Clarification: If a methionine is present at the junction, the fusion protein can be cleaved with cyanogen bromide which cuts polypeptide.

10. For cleaving arginine residues, which of the following can be used?
A. Methionine
B. IPTG
A. Thrombin
D. XA
Answer: C
Clarification: Methods for cleaving the bacterial segment are needed so that the properties of the fusion protein are not altered. Thrombin cleaves adjacent to arginine residues.

11. For cleaving off the bacterial segment, it is necessary that recognition sequence occurs within the protein.
A. True
B. False
Answer: B
Clarification: The important consideration is that recognition sequences for the cleavage agent must not occur within the recombinant protein.

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

1. What is gene knockout?
A. Removal of gene
B. Technique for studying gene function
A. Expression of gene
D. Repression of gene
Answer: B
Clarification: Gene cloning in mammals is done to achieve several results. Gene knockout is one of those, it is a technique used to study the function of an unidentified gene.

2. What is pharming?
A. Protein engineering
B. Plant manipulation
A. Genetic engineering of pharm animals
D. Genetic engineering of cows
Answer: C
Clarification: For a production of recombinant protein in a mammalian cell culture, and in the related technique of pharming which involves genetic engineering of a farm animal so that it synthesizes an important protein such as a pharmaceutical.

3. Gene therapy is a technique only associated with plants.
A. True
B. False
Answer: B
Clarification: In gene therapy, in which human cells are engineered in order to treat a disease. It is a very useful application.

4. When was the first cloning experiment involving mammalian cells carried out?
A. 1940
B. 1979
A. 2000
D. 1953
Answer: B
Clarification: The first cloning experiment involving mammalian cells was carried out in 1979 with a vector based on simian virus 40.

5. What is the approximate size of the SV40 vector?
A. 1 kb
B. 2 kb
A. 3.2 kb
D. 5.2 kb
Answer: D
Clarification: The SV40 virus is capable of infecting several mammalian species, following a lytic cycle in some hosts and lysogenic in others. The size is 5.3 kb.

6. How many sets of genes are present in SV40?
A. 1
B. 2
A. 3
D. 4
Answer: B
Clarification: The genome of SV40 is 5.2 kb and contains two sets of genes, the early genes and the late genes both coding for different proteins.

7. The proteins encoded by the early genes of SV40 are involved in _________
A. DNA replication
B. Assembly
A. Capsid formation
D. Coat proteins
Answer: A
Clarification: The early genes are expressed early in the infection cycle and coding for proteins involved in viral DNA replication.

8. The proteins encoded by the late genes of SV40 are involved in ____________
A. Capsid proteins
B. Viral proteins
A. DNA replication
D. Translation
Answer: A
Clarification: The genome is 5.2 kb in size and contains two sets of genes, the early genes, expressed early in the infection cycle and the late genes coding for viral capsid proteins.

9. Which problem does SV40 as a cloning vector face which is similar to that faced by lambda and caulimovirus?
A. Narrow host range
B. Packaging constraint
A. Digestion limitation
D. Post translational modifications
Answer: B
Clarification: SV40 suffers from the same problem as lambda and the plant caulimovirus, in that packaging constraints limit the amount of new DNA that can be inserted into the genome.

10. Adenoviruses are vectors of ___________
A. Mammals
B. Plants
A. Insects
D. Farm animals
Answer: A
Clarification: Adenoviruses are coning vectors for mammals which were developed after the first experiment conducted with SV40.

11. What is the size of foreign DNA that can be cloned using Adenovirus?
A. 1 kb
B. 3 kb
A. 5 kb
D. 8 kb
Answer: D
Clarification: Adenoviruses enable DNA fragments of up to 8 kb to be cloned, longer than is possible with an SV40 vector, though they are more difficult to handle.

12. Which set of genes were replaced in the initial cloning experiment with SV40?
A. Early genes
B. Late genes
A. Silencing genes
D. Repressor genes
Answer: B
Clarification: Cloning with SV40 involves replacing one or more of the existing genes with the DNA to be cloned. In the original experiment, a segment of the late gene region was replaced.

13. Bovine Papillomavirus causes ___________
A. Warts on cattle
B. Cancer in cattle
A. Tumor in plants
D. Tumor in cattle
Answer: A
Clarification: Papillomaviruses, which also have a relatively high capacity for inserted DNA. Bovine Papillomavirus (BPV), which causes warts on cattle, is particularly attractive because it has an unusual infection cycle.

14. What is copy number associated with BPV?
A. 10
B. 50
A. 100
D. 500
Answer: C
Clarification: Bovine Papillomavirus, which causes warts on cattle, is particularly attractive because it has an unusual infection cycle in mouse cells, taking the form of a multicopy plasmid.

15. BPV does not cause the _______ of mouse cell, upon infection.
A. Death
B. Shrinkage
A. Decrease in immunity
D. Increase in immunity
Answer: A
Clarification: BPV does not cause the death of the mouse cell, and BPV molecules are passed to daughter cells on cell division, giving rise to a permanently transformed cell line.

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Vector Biology Multiple Choice Questions on “Ligation of Vectors”.

1. Which is the final step in the construction of a recombinant molecule?
A. Plasmid isolation
B. Gene amplification
C. Restriction digestion
D. Ligation
Answer: D
Clarification: The final step in the construction of a recombinant DNA molecule is the joining together of vector molecule and the DNA to be cloned, this is known as ligation.

2. Which DNA ligase enzyme is used in genetic engineering?
A. Bacterial ligase
B. T4 ligase
C. Yeast ligase
D. Pseudomonas ligase
Answer: B
Clarification: The ligase enzyme used in genetic engineering is derived from E.coli cells infected with T4 phage, T4 ligase requires ATP to catalyze the ligation reaction.

3. Within the cell, ligase enzyme has the function of repairing any discontinuities that may arise in the double stranded DNA molecule. What are these discontinuities?
A. Absence of nucleotides in 1 strand
B. Missing phosphodiester bond in 1 strand
C. Missing phosphodiester bond in 2 both strands
D. Absence of nucleotides in both strands
Answer: B
Clarification: DNA ligase repairs the missing phosphodiester bonds in one of the strands of the double stranded DNA. The absence of nucleotides is called nick. A discontinuity is a position where phosphodiester bond between adjacent nucleotides is missing.

4. The chemical reaction involved in ligation of two molecules is same as repairing discontinuities.
A. True
B. False
Answer: A
Clarification: The chemical reaction involved in ligation of two molecules is exactly the same as discontinuity repair, except that two phosphodiester bonds must be made, one for each strand.

5. Which of the following will have more efficient ligation?
A. Sticky ends
B. Blunt ends
C. Blunt ends and high concentration of DNA
D. Blunt ends and low concentration of DNA
Answer: A
Clarification: In the case of blunt ends the ligation is less efficient because ligase cannot catch hold of the molecules to be ligated, and has to wait for chance associations to bring the ends together. Whereas in case of sticky ends can base pair with one another forming hydrogen bonds and a relatively stable structure is formed for the ligase enzyme to work on.

6. Which of the following is not a method for putting sticky ends to a blunt ended DNA fragment to be cloned?
A. Homopolymer tailing
B. Adaptors
C. Restriction digestion
D. Linkers
Answer: C
Clarification: Sticky ends are desirable on the DNA molecules to be ligated together in a cloning experiment. One way of doing this is restriction digestion of the vector and gene to be cloned, but when the vector produced by restriction digestion has sticky ends and the DNA fragment produced, does not; other methods for putting sticky ends to the blunt ended molecule are used.

7. What are Linkers?
A. Short synthetic double stranded DNA sequence
B. Short oligonucleotide sequence of host
C. Short oligonucleotide sequence of vector
D. Short synthetic single stranded DNA sequence
Answer: A
Clarification: Linkers are short double-stranded DNA, artificially synthesized of known sequences. These are used for increasing the efficiency of ligation for blunt ended molecules by converting them into sticky-ended.

8. What is the problem associated with the use of linkers for putting sticky ends to a blunt ended molecule?
A. High ambient temperature requirements
B. Possible cleavage of the DNA molecule itself
C. High cost of synthetic linkers
D. Low compatibility of linkers
Answer: B
Clarification: The desired DNA molecule, to which sticky ends are being put up by using linkers, may get cleaved off if it has internal restriction sites for the same endonuclease with which the linker has to be cleaved for producing sticky ends.

9. What is the problem associated with adaptors?
A. Shorter and hence improper ligation
B. Less efficient than linkers
C. May self-ligate and hence require restriction
D. Cannot be made synthetically
Answer: C
Clarification: Adaptors are short synthetic oligonucleotides, synthesized so that they already have one sticky end. The sticky ends of individual adaptor molecules can base pair with each other forming dimers. To break these dimers and to recreate sticky ends a digestion with a restriction endonuclease will be required.

10. Ligation takes place between __________
A. Adaptor and linker
B. Linker and vector
C. 5’-P terminus and 3’-OH terminus
D. Adaptor and vector
Answer: C
Clarification: Normally, the two ends of a polynucleotide strand are chemically distinct. One end contains the phosphate group and the other contains a hydroxyl group. Ligation takes place between the 5’-P terminus and 3’-OH terminus.

11. What is the function of a polynucleotide kinase?
A. Removal of the phosphate group from 5’ end
B. Removal of the phosphate group from 3’ end
C. Addition of phosphate group on 5’ end
D. Addition of phosphate group on 3’ end
Answer: C
Clarification: Polynucleotide kinase is a DNA modifying enzyme which has the reverse effect of alkaline phosphatase i.e. it adds phosphate group on 5’ end.

12. Which enzyme is used in homopolymer tailing for producing sticky ends?
A. Terminal deoxynucleotidyl transferase
B. Alkaline phosphatase
C. Polynucleotide kinase
D. DNA polymerase
Answer: A
Clarification: Tiling involves using a terminal deoxynucleotidyl transferase enzyme for adding a series of nucleotides on the 3’-OH termini of a double-stranded DNA molecule.

13. Which of the following statements is true in case of homopolymer tailing for producing sticky ends?
A. Poly(dC. tails to vector, poly(dG) to DNA
B. Poly(dG) tails to vector, poly(dC. to DNA
C. Poly(dC. tails to vector and DNA
D. Poly(dC. tails to vector and DNA
Answer: A
Clarification: Frequently, polydeoxycytosine tails are attached to the vector and polydeoxyguanosine tails are attached to the DNA to be cloned. Base pairing between the two occurs when mixing is done.

14. A recombinant DNA molecule held together by base-pairing although not completely ligated can be introduced into the host cell.
A. True
B. False
Answer: A
Clarification: If the complementary homopolymer tails are longer than 20 nucleotides, then quite stable base-paired associations are formed. Once inside the cell the cell’s own DNA polymerase and ligase repair the recombinant DNA molecule.

15. Which of the following is not a function of DNA topoisomerase?
A. Removing turns of the double helix DNA
B. Adding turns to the double helix DNA
C. Have nuclease and ligase activity
D. Have polymerase activity
Answer: D
Clarification: DNA topoisomerase is a multifunctional enzyme used in efficient blunt-end ligation of DNA molecules in recombinant DNA technology. However it lacks the function of a polymerase.

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Gene Manipulation Multiple Choice Questions on “Transgenic Technology”.

1. Transgenic technology is particularly advanced in _________
A. Plants
B. Animals
A. Mice
D. Bacteria
Answer: C
Clarification: Transgenic technology is particularly advanced in mice, where combinations of gene targeting, site-specific recombination and inducible transgene expression make it possible.

2. Inducible transgene expression controls ______ genes.
A. 1
B. 2
A. 3
D. 4
Answer: B
Clarification: Inducible transgene expression makes it possible to activate and inactivate both transgenes and endogenous genes in a conditional manner.

3. Gene silencing involves ___________ genes.
A. Inhibitory
B. Activator
A. Inducer
D. Carcinogenic
Answer: A
Clarification: Gene silencing does not involve direct modification of the target gene but rather the expression of inhibitory genes whose products interfere with expression of the target.

4. Which of the following expression allows the control of transgene expression?
A. Repressible
B. Inducible
A. Altered
D. Toxic
Answer: B
Clarification: Inducible expression system allows transgene expression to be controlled by physical stimuli or the application of small modulators.

5. What is used to restrict transgene expression?
A. Promoter
B. Inducer
A. Silencer
D. Reporter
Answer: A
Clarification: In both animals and plants, cell or tissue specific promoters are used to restrict transgene expression to certain areas of the organism.

6. In __________ it is useful to restrict transgene expression to mammary glands.
A. Reptiles
B. Plants
A. Xenopus
D. Mammals
Answer: D
Clarification: In mammals, it is useful to restrict the transgene expression to mammary glands. This is done so that recombinant proteins can be recovered from milk.

7. In plants, it is useful to restrict transgene expression to seeds.
A. True
B. False
Answer: A
Clarification: It is useful to restrict transgene expression in plants to the seeds. Seeds provide a stable environment for protein accumulation.

8. Drosophila heat-shock promoter is an example of ___________
A. Synthetic
B. Toxic
A. Naturally occurring
D. Rare
Answer: C
Clarification: Most cells respond to elevated temperature by synthesizing heat-shock proteins, which include molecular chaperones and other proteins.

9. Hsp70 promoter of Drosophila, _________ the gene.
A. Acidifies
B. Neutralizes
A. Inactivates
D. Activates
Answer: C
Clarification: In transgenic flies, any gene linked to the hsp70 promoter is more or less inactive at room temperature, but high-level expression in all cells can be induced.

10. Inducible promoters do not respond to chemicals.
A. True
B. False
Answer: B
Clarification: Most inducible promoters used to control transgene expression respond to chemicals, which must be supplied to the transformed cells.

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