Category: Vector Biology Objective Questions

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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Gene Manipulation Question Bank focuses on “Basic Laboratory Techniques – 6”.

1. In Western blotting, probe is mostly _______________
A. Toxic
B. Labeled
A. Unlabeled
D. Protein itself
Answer: C
Clarification: The probe is mostly unlabeled in the Western blotting technique, the detection or probing is hence usually done by a second molecule.

2. Protein A is obtained from _______________
A. Bacillus
B. Pseudomonas
A. Staphylococcus
D. E.coli
Answer: C
Clarification: The probe is often unlabeled and is itself detected in a sandwich reaction using a second molecule which is labeled, a species specific second antibody, for instance, protein A from Staphylococcus Aureus.

3. Protein A binds to __________________
A. IgG antibodies
B. Genes
A. Proteins
D. Membrane
Answer: A
Clarification: Protein A from Staphylococcus Aureus; which binds to certain subclasses of IgG antibodies. Protein A is actually a probe.

4. Dextran sulfate is a ____________
A. Antibody
B. Protein
A. Vector
D. Rate enhancer
Answer: D
Clarification: Dextran sulfate and other polymers act as volume excluders to increase both the rate and extent of hybridization.

5. Heparin is a _______________
A. Membrane
B. Rate enhancer
A. Antibody
D. Detergent
Answer: D
Clarification: Dried milk, heparin and SDS have been used to depress the non-specific binding of the probe to the membrane. Hence they are detergents as well as blocking agents.

6. The Denhardt’s solution does not use which of the following?
A. Ficoll
B. PVP
A. BSA
D. Heparin
Answer: D
Clarification: Denhardt’s solution, developed in 1966 consists of ficoll, polyvinylpyrrolidone and bovine serum albumin as detergents and blocking agents.

7. Denaturants are used for depressing _________________
A. Melting temperature
B. Freezing temperature
A. Boiling temperature
D. Denaturation
Answer: A
Clarification: Denaturants such as urea and formamide can be used to depress the melting temperature of the hybrid so that reduced temperature of hybridization can be used.

8. Formamide is used as a ___________ in Western blotting.
A. Rate enhancer
B. Denaturant
A. Antibody
D. Membrane
Answer: B
Clarification: Denaturants such as urea and formamide can be used to depress the melting temperature of the hybrid so that reduced temperature of hybridization can be used.

9. Heterologous DNA can increase non-specific binding.
A. True
B. False
Answer: B
Clarification: Heterologous DNA can reduce non-specific binding of probes to non-homologous DNA on the blot. This increases the overall efficiency.

10. Stringency can be regarded as a ________________
A. Toxic treatment
B. Specificity
A. Gel formation
D. Purification
Answer: B
Clarification: Stringency can be regarded as the specificity with which a particular target sequence is detected by hybridization to a probe.

11. Stringency is most commonly controlled by the temperature and ________________
A. Humidity
B. Salt concentration
A. Size of molecule
D. Antibodies
Answer: B
Clarification: Stringency is most commonly controlled by the temperature and salt concentration in post-hybridization washes.

12. The melting temperature TM is the temperature of ___________
A. Probe
B. Hybrid
A. Probe-hybrid
D. Antigens
Answer: C
Clarification: The melting temperature of a probe hybrid system can be calculated to provide a starting point for the determination of correct stringency.

13. What is the usual Tm (melting temperature) of long probes?
A. 30⁰
B. -25⁰
A. 25⁰
D. -30⁰
Answer: B
Clarification: With long probes, the hybridization is usually carried out at -25⁰C of melting temperature. When the probe is used to detect partially matched sequences, the hybridization temperature is reduced by 1⁰ for every 1% sequence divergence between probe and target.

14. A 1⁰ reduction is temperature is done for _____ percent divergence between probe and target.
A. 1
B. 5
A. 10
D. 100
Answer: B
Clarification: When the probe is used to detect partially matched sequences, the hybridization temperature is reduced by 1⁰ for every 1% sequence divergence between probe and target.

15. Which of the following can give a rapid hybridization rate?
A. Oligonucleotides
B. Nucleotides
A. Short probes
D. Long probes
Answer: A
Clarification: Oligonucleotides can give a more rapid hybridization rate than long probes as they can be used at a higher molarity.

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Tough Vector Biology Questions and Answers focuses on “Special Vectors for Expression in E.Coli – 3”.

1. The lambda P L promoter is a weak promoter.
A. True
B. False
Answer: B
Clarification: The lambda P L is a very strong promoter and is one of the promoters responsible for transcription of the lambda DNA molecule.

2. Lambda vector subverts which important enzyme of E.coli?
A. DNA polymerase
B. Helicase
A. RNA polymerase
D. Nuclease
Answer: C
Clarification: Lambda is a very strong promoter recognized by RNA polymerase which is subverted by lambda into transcribing phage DNA.

3. The lambda promoter is repressed by which of the following gene?
A. T gene
B. cI gene
A. Amp gene
D. Trp gene
Answer: B
Clarification: The lambda promoter is repressed by cI gene. Expression vectors that carry the lambda P L promoter are used with mutant E.coli strain.

4. Expression vectors that carry the lambda P L promoter are used with ________ E.coli strain.
A. Double
B. Hybrid
A. Mutant
D. Wild-type
Answer: C
Clarification: Expression vectors that carry the lambda P L promoter are used with a mutant E.coli host that synthesizes a temperature-sensitive form of the cI protein.

5. Below what temperature can the cI protein repress the lambda promoter?
A. 10 degrees
B. 20 degrees
A. 30 degrees
D. 40 degrees
Answer: C
Clarification: At a low temperature that is less than 30 degrees, the mutant cI protein is able to repress the lambda pL promoter.

6. At higher temperature the cI protein is __________
A. Inactivated
B. Increased
A. Decreased
D. Moderated
Answer: A
Clarification: At a low temperature that is less than 30 degrees, the mutant cI protein is able to repress the lambda pL promoter, but at higher temperatures, the protein is inactivated.

7. The T7 promoter is specific for the __________
A. Mutant strains
B. Bacteriophage
A. RNA polymerase
D. DNA polymerase
Answer: C
Clarification: The T7 promoter is specific for the RNA polymerase coded by the T7 bacteriophage. The phage RNA polymerase is much more active.

8. Gene expressed downstream of T7 promoter will be expressed at _____________
A. Low level
B. High level
A. Moderate level
D. Intermittently
Answer: B
Clarification: The phage RNA polymerase is much more active than the E.coli promoter which means that the gene inserted downstream of the promoter is expressed at a high level.

9. Which of the following is present in the lysogenic special strain of E.coli?
A. Trp promoter
B. Lac promoter
A. Ampicillin
D. IPTG
Answer: B
Clarification: A lysogen contains an inserted copy of the phage DNA in its genome. The phage DNA has been altered by placing a lac promoter.

10. How can the synthesis of T7 RNA polymerase can be switched on in a culture?
A. IPTG addition
B. IPTG removal
A. Ampicillin addition
D. Ampicillin removal
Answer: A
Clarification: Addition of IPTG to the growth medium switches on the synthesis of the T7 RNA polymerase, which in turn leads to gene activation.

11. The cassette in an expression vector comprises of ribosome binding site and ____________
A. Promoter
B. Terminator
A. Repressor
D. Inducer
Answer: B
Clarification: An efficient expression vector, in addition to a strong and regulatable promoter also contains a cassette containing terminator and binding site.

12. Insertion of the foreign gene into the expression vector’s cassette fuses ____ reading frames.
A. 1
B. 2
A. 3
D. 4
Answer: B
Clarification: Insertion of the foreign gene into the restriction site fuses two reading frames that start with the E.coli segment and progress without a break.

13. Which type of protein is produced with cassette vectors?
A. Plant protein
B. Mammalian protein
A. Fusion protein
D. Hybrid protein
Answer: C
Clarification: The product of gene expression is a fusion protein, consisting of short peptide coded by the E.Coli reading frame fused to foreign protein.

14. P-terminus of the fusion protein is fused with a short peptide.
A. True
B. False
Answer: B
Clarification: The product of gene expression is a fusion protein, consisting of short peptide coded by the E.Coli reading frame fused to amino terminus.

15. Which of the following is responsible for the mRNA translation effectively?
A. Ribosome binding site
B. IPTG
A. Ubiquitin
D. Ampicillin
Answer: A
Clarification: Efficient translation of mRNA produced from cloned gene depends on ribosomal binding site and starting nucleotide sequence.

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Vector Biology Multiple Choice Questions on “Vectors for Insects”.

1. Recombinant proteins are expressed in _____________
A. E.coli
B. Yeast
A. Animals
D. Fungi
Answer: C
Clarification: Vector systems are needed in biotechnology for cloning genes in animal cells, for the synthesis of recombinant proteins from genes that are not correctly expressed in E.coli or yeast.

2. What is gene therapy?
A. Technique of curing a disease
B. Introduction of a disease
A. Introduction of a gene in E.coli
D. Introduction of a gene in yeast
Answer: A
Clarification: Methods for cloning in humans are being sought by clinical molecular biologists attempting to devise techniques for gene therapy, in which a disease is treated by the introduction of cloned gene into the patient.

3. What is Drosophila Melanogaster?
A. Fruit fly
B. House fly
A. Damsel fly
D. Dragon fly
Answer: A
Clarification: Drosophila melanogaster is a popular model organism which is basically a fruit fly. It came into laboratory recognition with Morgan’s experiments.

4. When were the experiments with Drosophila Melanogaster first attempted?
A. 1910
B. 1810
A. 1710
D. 1610
Answer: A
Clarification: The potential of Drosophila Melanogaster was first recognized by geneticist Thomas Hunt Morgan, who is 1910 started to carry out genetic crosses between the flies.

5. Several genes constituting the genome of Drosophila are closely related to equivalent genes in mammals.
A. True
B. False
Answer: A
Clarification: The homeotic selector genes of Drosophila- the genes that control the overall body plan of the fly, are closely related to equivalent genes in mammals.

6. What is the P element?
A. Popular element
B. Transposon
A. Homeotic gene
D. Pressure inducing gene
Answer: B
Clarification: No plasmids are known in Drosophila and no viruses that infect the organism are used as vectors for it. Instead, cloning in Drosophila makes use of a transposon called the P element.

7. What is the approximate length of the P element?
A. Less than 1 kb
B. Less than 10 kb
A. Less than 20 kb
D. Less than 20 mb
Answer: B
Clarification: P element is a transposon found in the fruit fly Drosophila Melanogaster. Transposons are common in all organism, they are short pieces of DNA, usually less than 10 kb in length.

8. Transposons can move from one position to another in the ________ of a cell.
A. Nucleus
B. Chromosome
A. Cell membrane
D. Mitochondria
Answer: B
Clarification: Transposons, also known as jumping genes can change their positions from one location to the other in a chromosome. The transposon named P element found in Drosophila is 2.9 kb in size.

9. How many genes are contained in a P element?
A. 1
B. 2
A. 3
D. 4
Answer: C
Clarification: P elements found in the fruit fly Drosophila Melanogaster are actually the transposons residing in the chromosome of the organism. It contains three genes flanked by two unique regions on both the ends.

10. Which sequences flank the P element of the Drosophila?
A. Repeated
B. Interstitial
A. Inverted repeats
D. Random
Answer: C
Clarification: P elements, which are one of the several types of transposons found in Drosophila, are 2.9 kb in length and contain three genes flanked by short inverted repeat sequences.

11. What do the genes of the P element code for?
A. Amylase
B. Lactase
A. Transposase
D. Transcriptase
Answer: C
Clarification: P elements are used as cloning vectors in Drosophila, these
are 2.9 kb in length and contain three genes flanked by short inverted repeat sequences, the genes code for the enzyme transposase which carries out the transposition activity.

12. What is the function of flanking inverted repeats in a P element?
A. Maintaining homeostasis
B. Recognition sequence
A. Repression/induction activity
D. Operators
Answer: B
Clarification: The three genes present in the P element code for transposase enzyme and the repeat sequence on two ends act as recognition sequences that enable the enzyme to identify.

13. A P element can change position/ jump between a ________ and one of the fly’s chromosome.
A. Plasmid
B. Bacterial chromosome
A. Yeast chromosome
D. Another fly’s chromosome
Answer: A
Clarification: As well as moving from one site to another within a chromosome, P elements can also jump between chromosomes, or between a plasmid carrying a P element and one of the fly’s chromosomes.

14. The basic vector used for cloning in Drosophila is __________
A. Plasmid
B. Baculovirus
A. Culimovirus
D. YAC
Answer: A
Clarification: The vector is a plasmid that carries two P elements, one of which contains the insertion site for the DNA that will be cloned. The second P element has its inverted repeat sequences removed.

15. Baculovirus are vectors used for cloning in _____________
A. Mammals
B. Insects
A. Reptiles
D. Plants
Answer: B
Clarification: Baculovirus is important vectors that play an important role in gene cloning in insects other than Drosophila. Their main use is a production of recombinant proteins.

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Basic Vector Biology Questions and Answers focuses on “Restriction Endonucleases – 2”.

1. EcoR1 has which of the following hexanucleotide recognition sequence?
A. GAATTC
B. GGATCC
C. AGATCT
D. GCAGCA
Answer: A
Clarification: E.coli isolated from the gram-negative bacterium Escherichia coli. A recognition sequence is the one where an enzyme cleaves.

2. Which of the following is an example of flush end cutter?
A. PvuI
B. PvuII
C. EcoR1
D. SfiI
Answer: B
Clarification: Many restriction endonucleases like PvuII, AluI, HaeIII simply cleave the double stranded DNA producing bunt or flush ends.

3. Which of the following does not produce cohesive ends upon cleavage on their recognition sequences?
A. EcoRI
B. PvuII
C. BamHI
D. HindIII
Answer: B
Clarification: PvuII produces blunt ends whereas all others produce sticky ends. Some endonucleases cleave the DNA strand in a staggered way usually by two or four nucleotides, so that the resulting DNA fragments have short single-stranded overhangs at each end.

4. What is dithiothreitol?
A. Reducing agent
B. Restriction endonuclease
C. Ligase
D. Exonuclease
Answer: A
Clarification: Most endonuclease function at pH 7.4, but different enzymes vary in their requirements for ionic strength and concentration. It is advisable to add a reducing agent like DTT which stabilizes the enzyme and prevents its deactivation. Before adding the enzyme, solution containing DNA must be adjusted to provide the correct conditions for maximum enzyme activity.

5. Restriction digests with which enzyme needs to be kept at a higher temperature than the usual 37 degree temperature?
A. TaqI
B. EcoRI
C. BamHI
D. HindIII
Answer: A
Clarification: Most endonucleases work best at 37 degrees, TaqI enzyme from Thermus Aquaticus has a high working temperature like Taq DNA polymerase.

6. How can it be determined that a DNA molecule is cleaved by the action of endonuclease?
A. Checking the viscosity of the solution
B. Temperature check
C. Color indication
D. pH check
Answer: A
Clarification: Whether or not a DNA molecule is cut can be determined by viscosity examination. Larger DNA molecules result in more viscous solutions, DNA cleavage is associated with a decrease in viscosity.

7. How can the sizes of individual cleavage products be determined?
A. Temperature check
B. Molecular weight check
C. Chromatography
D. Gel electrophoresis
Answer: D
Clarification: Gel electrophoresis is a technique by which the sizes of fragmented DNA molecules can be measured against a molecular weight ladder of standard weights.

8. The rate of migration of a DNA molecule in an electrophoresis depends on which two factors?
A. Temperature, length
B. Length, source
C. Shape, charge-to-mass ratio
D. Shape, viscosity
Answer: C
Clarification: Shape and charge-to-mass ratio of nearly all DNA molecules is same and therefore to separate them gel electrophoresis is done in place of conventional electrophoresis.

9. How many basepairs long DNA molecules can be distinguished by the polyacrylamide (40%) gel electrophoresis?
A. 1-100 kbp
B. 1-300 bp
C. 100-200 kbp
D. 300-600 bp
Answer: B
Clarification: A very thin (0.3mm) polyacrylamide gel with extremely small pores can be used to distinguish small DNA molecules, can distinguish between molecules differing in length by just a single nucleotide.

10. What are double digestions?
A. Cutting with an enzyme twice on the same site
B. Cutting with two different enzymes on the same site
C. Cutting with an enzyme twice on different sites
D. Cutting with two different enzymes on the different sites
Answer: D
Clarification: Cutting with two different enzymes and at two different sites is done to create a restriction map of the genome of an organism.

11. What will happen if the incubation period, post restriction digestion is shortened?
A. Indigestion
B. Partial digestion
C. Double digestion
D. Complete digestion
Answer: B
Clarification: To perform partial digestion either incubation period is shortened or incubation is done at a low temperature at around 4 degrees which limits the activity of the enzyme.

12. What does a complex pattern of bands in a gel electrophoresis signify?
A. Partial digestion
B. Double digestion
C. Complete digestion
D. Contaminated sample
Answer: A
Clarification: As well as the standard fragments of total digestion, partially digested fragments are also seen as additional bands giving overall complexity.

13. What is the full form of OFAGE?
A. Orthogonal field alternation gel electrophoresis
B. Orthogonal field attraction gel electrophoresis
C. Oligonucleotide field alternation gel electrophoresis
D. Oligonucleotide field attraction gel electrophoresis
Answer: A
Clarification: OFAGE is an advanced gel electrophoresis technique used for the separation of DNA molecules of over 50 kb in length.

14. Which of the following techniques cannot be used to separate the genome of lower eukaryote yeast?
A. CHEF
B. OGAGE
C. FIGE
D. Gel electrophoresis
Answer: D
Clarification: The eukaryotic genome is large and simple techniques like gel electrophoresis cannot distinguish the DNA molecules ranging in several thousand basepairs.

15. FIGE is the modification of which technique?
A. OFAGE
B. Agarose gel electrophoresis
C. Electrophoresis
D. Acrylamide gel electrophoresis
Answer: A
Clarification: Field Inversion Gel Electrophoresis is a modification of OFAGE in which parallel alternating pairs of electrodes are placed.

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Gene Manipulation Assessment Questions and Answers focuses on “Gene Targeting and Transfer in Plants”.

1. When was the first stable transmission of DNA into plants reported?
A. 1971
B. 1981
A. 1940
D. 1950
Answer: B
Clarification: The introduction of foreign DNA into a plant followed by stable transmission through the germline was first demonstrated in 1981.

2. Which was the first transgenic plant generated?
A. Cumin
B. Tobacco
A. Basil
D. Tomato
Answer: B
Clarification: The first transgenic plant generated was that of tobacco. Soil bacterium was used as a vector for transferring exogenous DNA into the plant.

3. Plant viruses are _______ vectors.
A. Integrative
B. Replacement
A. Episomal
D. Artificial
Answer: C
Clarification: Plant viruses have been developed as versatile episomal vectors, allowing high level transient gene expression.

4. Differentiated plant tissue shows high degree of _________
A. Inactivity
B. Toxicity
A. Growth
D. Plasticity
Answer: D
Clarification: A fundamental difference between animals and plants is that organized, differentiated plant tissue shows a high degree of developmental plasticity.

5. Callus tissue is derived from ________
A. Trunk
B. Root
A. Leaf
D. Seed
Answer: D
Clarification: Seed derived callus tissue may be able to regenerate an entirely new plant under appropriate culture conditions. Depending on different species, different cultures can be obtained.

6. Tissue culture step is necessary for producing transgenic plants.
A. True
B. False
Answer: A
Clarification: For most plant species, some form of tissue culture step is therefore necessary for the successful production of transgenic plants.

7. Need of tissue culture is minimized or eliminated if ___________ is used.
A. Leaf
B. Whole plant
A. Root
D. Seed
Answer: B
Clarification: There is an increasing interest in the use of whole plant transformation strategies, in which the need for tissue culture is minimized or eliminated.

8. In which cultures, cells are maintained in an undifferentiated state?
A. Leaf
B. Root
A. Stem
D. Callus
Answer: D
Clarification: Callus cultures are established under conditions that maintain cells in an undifferentiated state. Plant tissue culture is required for most procedures.

9. Explants are plant ________________
A. Living tissue
B. Stems
A. Roots
D. Seeds
Answer: A
Clarification: Tissue culture is the process whereby small pieces of living tissue (explants) are isolated from an organism and grown aseptically.

10. Undetermined cells are capable of _____________
A. Growth
B. Proliferation
A. Toxication
D. Protein production
Answer: B
Clarification: For successful plant tissue culture it is best to start with an explant rich in undetermined cells because such cells are capable of rapid proliferation.

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