Category: Vector Biology Objective Questions

Vector Biology Multiple Choice Questions on “Vectors for Mammals – 4”.

1. How many routes are available for the biosynthesis of nucleotides in human beings?
A. 1
B. 2
A. 3
D. 4
Answer: B
Clarification: In mammals, nucleotides are produced via two alternate routes, the de novo and the salvage pathway.

2. What are the basic precursors in the de novo pathway of nucleotide synthesis?
A. Carbohydrate, proteins
B. Sugars, amino acids
A. Sugars, vitamins
D. Minerals, carbohydrates
Answer: B
Clarification: In mammals, nucleotides are produced via two alternate routes, the de novo and the salvage pathway. In the de novo pathway, nucleotides are synthesized from basic precursors such as sugars and amino acids.

3. In the salvage pathway, nucleotides are synthesized by the recycling of ___________
A. DNA
B. RNA
A. DNA, RNA
D. Sugars
Answer: C
Clarification: In mammals, nucleotides are produced via two alternate routes, the de novo and the salvage pathway. Salvage pathway recycles nucleotides from DNA and RNA.

4. De novo pathway is exploited for the selection of cells carrying functional HPRT and TK genes.
A. True
B. False
Answer: B
Clarification: If the de novo pathway is blocked, nucleotide synthesis becomes dependent on the salvage pathway, and this can be exploited for the selection of cells carrying functional HPRT and TK genes.

5. The drug aminopterin blocks the _________ of two enzymes.
A. Salvage pathway
B. De novo synthesis
A. Expression
D. Recombination
Answer: B
Clarification: The drug aminopterin blocks the de novo synthesis of inosine monophosphate (IMP) and thymidine monophosphate (TMP).

6. What do you understand by the term “co-transformation”?
A. Integration of 2 transgenes
B. Integration of similar transgenes
A. Integration of a group of transgenes
D. Integration of chromosomal DNA
Answer: A
Clarification: The transfection with two physically unlinked DNAs results in co-transformation that is the integration of both the transgenes into the genome.

7. Southern blot hybridization is done for testing the presence of _________ in a selection of transformants.
A. Non-selected DNA
B. Selected plasmid DNA
A. Genomic DNA
D. Selected genomic DNA
Answer: A
Clarification: To obtain co-transformants, cells were transfected with HSV Tk gene and well defined plasmid DNA. Cells selected on HAT medium are then tested by southern blotting for the presence of non-selected DNA.

8. What is the prerequisite for co-transformation phenomenon to occur?
A. Selectable marker
B. HSV
A. Thymidine kinase
D. PBR322 DNA
Answer: A
Clarification: The phenomenon of co-transformation allows the stable introduction of any foreign DNA sequence into mammalian cells as long as a selectable marker is introduced at the same time.

9. HSV TK gene is a type of __________ marker.
A. Exogenous
B. Endogenous
A. Recombinant
D. Hybrid
Answer: B
Clarification: The HSV Tk gene is a representative of a class of genes known as endogenous markers because they confer a property that is already present in the wild-type cells.

10. The endogenous selectable markers can be used only with mutant cell lines.
A. True
B. False
Answer: A
Clarification: The major disadvantage of the endogenous selectable markers is that they can only be used with mutant cell lines in which the corresponding host gene is non-functional.

11. Dominant selectable markers can be used with __________
A. Any cell type
B. Mutant cells
A. Wild-type cells
D. Recombinant cells
Answer: A
Clarification: Endogenous markers are largely superseded by so-called dominant selectable markers, which confer a phenotype that is entirely novel to the cell and can hence be used in any cell type.

12. What are the dominant selectable markers?
A. Drug-resistance genes
B. Inducing genes
A. Exogenous genes
D. Endogenous genes
Answer: A
Clarification: The dominant selectable markers are usually drug-resistance genes of bacterial origin and transformed cell is selected on a medium that contains the drug at an appropriate concentration.

13. Methotrexate is an analog of __________
A. Aminopterin
B. Kanamycin
A. Folic acid
D. Gentamycin
Answer: C
Clarification: Methotrexate is a folic acid analog, which is a competitive inhibitor of the enzyme dihydrofolate reductase (DHFR).

14. With respect to mammalian cell cloning, salmon sperm DNA can serve as a source of ____________
A. Non-specific carrier
B. Specific carrier
A. Genomic DNA
D. Plasmid DNA
Answer: A
Clarification: Calcium phosphate transfection is mostly used and the specific donor DNA is often bulked with a non-specific carrier such as cleaved Salmon sperm.

15. One application in which the use of plasmid vectors is critical, in the case of mammals is ____________
A. Stable transformation
B. Transient transformation
A. Transfection
D. Transduction
Answer: B
Clarification: One application in which the use of plasmid vectors is critical, in the case of mammals is a transient transformation. Here the goal is to exploit the short term persistence of extrachromosomal DNA.

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Vector Biology Interview Questions and Answers focuses on “Cloning Vectors for E.Coli – 2”.

1. What is an additional feature of M13mp7?
A. 2 antibiotic resistance genes
B. Bigger size
A. Multiple cloning sites
D. Smaller size
Answer: C
Clarification: The M13mp7 contains multiple cloning sites all clustered into the polylinker and this polylinker is artificially synthesized and then inserted in the intergenic sequence.

2. What is a phagemid?
A. A hybrid vector
B. Phage vector
A. Plasmid vector
D. Viral vector
Answer: A
Clarification: Phagemid is a hybrid of M13 phage and Pbr322 plasmid. Pembl8 is an example of phagemid which was created by transferring into a pUC8 a 1300 bp fragment of M13 genome.

3. What does the M13 fragment in a phagemid contain?
A. BamHI restriction site
B. Signal sequences
A. Origin of replication
D. Promoter sequence
Answer: B
Clarification: These signal sequences are recognized by the enzymes that convert the normal double-stranded M13 molecule into single-stranded DNA before secretion of new phage particles.

4. Why is a helper phage needed when cloning experiments which a hybrid vector such as Pembl8 is done?
A. Efficient insertion
B. Attachment to host
A. To provide replicative enzymes
D. Stable transformation
Answer: C
Clarification: The hosts for Pembl8 cloning experiments are subsequently infected with a normal M13 to act as a helper phage, providing necessary replicative enzymes and phage coat proteins.

5. How can the recombinants of phagemid vector Pembl8 be identified?
A. Agar plating
B. Agar + Antibiotic
A. Agar + X-gal
D. Minimal media plating
Answer: C
Clarification: Pembl8, being derived from Puc8, has the polylinker cloning sites within the lacZ’ gene. So recombinants can be identified by the lac selection system.

6. What is the size of fragments that can be obtained by using a phagemid vector?
A. 1 kb
B. 10 kb
A. 1500 bp
D. 50 kb
Answer: B
Clarification: With a phagemid vector such as Pembl8, single-stranded versions of cloned DNA fragments up to 10 kb can be obtained, greatly extending the range of M13 cloning system.

7. What is the size limit for in vitro packaging of an unmodified lambda vector?
A. 10 kb
B. 52 kb
A. 100 kb
D. 10 bp
Answer: B
Clarification: The lambda DNA molecule can be increased in size by only 5%, representing the addition of only 3 kb of new DNA. If total size of the molecule id more than 52 kb, then it cannot be packaged.

8. Which non-essential region of the lambda phage can be deleted without impairing viability?
A. Protein coding
B. Promoter region
A. Integration and excision region
D. Terminator region
Answer: C
Clarification: The removal of non- essential region between 2 to 35 positions on the restriction map decreases the size of the vector by 15 kb and increasing its efficiency of carrying foreign DNA.

9. What is the basic difference between a modified (non-essential regions removeD. and an unmodified lambda vector?
A. Gene expression increases
B. Stable infection
A. Non- lysogenic cycle
D. Star activity
Answer: C
Clarification: Since the non-essential region of the lambda genome which is responsible for integration into the host genome is removed, it no longer follows the lysogenic cycle hence goes infecting the cell and eventually losing them.

10. Why is natural selection used to isolate modified lambda that lacks certain restriction sites?
A. Strains that lack sites are known
B. Easier than in vitro mutagenesis
A. There are no restriction sites in lambda
D. Natural selection is less time consuming
Answer: A
Clarification: There are multiple recognition sites in a lambda phage and only mutant strains that contain not many of these sites can be used as cloning vectors. To get hold of such mutant strains a host E.Coli strain that produces EcoR1 is used.

11. Why is natural selection used to isolate modified lambda that lacks certain restriction sites?
A. Strains that lack sites are known
B. Easier than in vitro mutagenesis
A. There are no restriction sites in lambda
D. Natural selection is less time consuming
Answer: A
Clarification: There are multiple recognition sites in a lambda phage and only mutant strains that contain not many of these sites can be used as cloning vectors. To get hold of such mutant strains a host E.Coli strain that produces EcoR1 is used.

12. Insertion and replacement vectors are modified vectors of which of the following?
A. Plasmid
B. Lambda phage
A. M13 phage
D. Yeast artificial chromosome
Answer: B
Clarification: Once the problems associated with lambda vector; packaging constraints and multiple restriction sites had been solved, development of different types of modified lambda vectors is done.

13. What type of vector is the lambda-gt10?
A. Insertion vector
B. Replacement vector
A. Hybrid vector
D. Unmodified lambda vector
Answer: A
Clarification: It is a vector which can carry up to 8 kb of new DNA, inserted into unique EcoR1 site located in the Ci gene. Recombinants are distinguished as clear plaques.

14. Which property is not associated with a lambda insertion vector?
A. Non-essential region removed
B. Two vector arms ligated together
A. At least one restriction site is present
D. Expression of the gene can be obtained
Answer: D
Clarification: Insertional lambda vectors are cloning vectors and not expression vectors because they do not contain promoter sequences and ribosomal binding sites and hence the expression of the inserted gene cannot be obtained.

15. How can identification of recombinants of lambda-gt10 vector be done?
A. Ampicillin resistance
B. Lac selection
A. cI gene insertional inactivation
D. Agar and X-gal plating
Answer: C
Clarification: The foreign DNA in this vector is cloned within the cI gene containing the EcoR1 restriction site. The recombinants hence are distinguished as clear plaques from the non-recombinant turbid plaques.

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Gene Manipulation Multiple Choice Questions on “Applications of Gene Manipulation”.

1. How many routes are available for the biosynthesis of nucleotides in human beings?
A. 1
B. 2
A. 3
D. 4
Answer: B
Clarification: In mammals, nucleotides are produced via two alternate routes, the de novo and the salvage pathway.

2. What are the basic precursors in the de novo pathway of nucleotide synthesis?
A. Carbohydrate, proteins
B. Sugars, amino acids
A. Sugars, vitamins
D. Minerals, carbohydrates
Answer: B
Clarification: In mammals, nucleotides are produced via two alternate routes, the de novo and the salvage pathway. In the de novo pathway, nucleotides are synthesized from basic precursors such as sugars and amino acids.

3. In the salvage pathway, nucleotides are synthesized by the recycling of __________
A. DNA
B. RNA
A. DNA, RNA
D. Sugars
Answer: C
Clarification: In mammals, nucleotides are produced via two alternate routes, the de novo and the salvage pathway. Salvage pathway recycles nucleotides from DNA and RNA.

4. De novo pathway is exploited for the selection of cells carrying functional HPRT and TK genes.
A. True
B. False
Answer: B
Clarification: If the de novo pathway is blocked, nucleotide synthesis becomes dependent on the salvage pathway, and this can be can be exploited for the selection of cells carrying functional HPRT and TK genes.

5. The drug aminopterin blocks the _________ of two enzymes.
A. Salvage pathway
B. De novo synthesis
A. Expression
D. Recombination
Answer: B
Clarification: The drug aminopterin blocks the de novo synthesis of inosine monophosphate (IMP) and thymidine monophosphate (TMP).

6. What do you understand by the term “co-transformation”?
A. Integration of 2 transgenes
B. Integration of similar transgenes
A. Integration of a group of transgenes
D. Integration of chromosomal DNA
Answer: A
Clarification: The transfection with two physically unlinked DNAs results in co-transformation that is the integration of both the transgenes into the genome.

7. Southern blot hybridization is done for testing the presence of _________ in the selection of transformants.
A. Non-selected DNA
B. Selected plasmid DNA
A. Genomic DNA
D. Selected genomic DNA
Answer: A
Clarification: To obtain co-transformants, cells were transfected with HSV Tk gene and well-defined plasmid DNA. Cells selected on HAT medium are then tested by southern blotting for the presence of non-selected DNA.

8. Incorporation of ribozymes into antisense RNA leads to their __________
A. Activity
B. Temperature change
A. Cleavage
D. Regeneration
Answer: C
Clarification: The incorporation of ribozyme catalytic centers into antisense RNA allows the ribozyme to be particularly targeted and then cleaved and degraded.

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Gene Manipulation Multiple Choice Questions on “Gene Sequencing”.

1. Chain-termination is a type of ______________
A. Sequencing
B. Vector generation
A. Antibiotic production
D. Gene manipulation
Answer: A
Clarification: One of the commonest methods of sequencing is the Sanger sequencing which is also known as chain-termination.

2. The first significant DNA sequence to be obtained was that of ________
A. Lambda
B. Plasmid
A. Lactose
D. Mammals
Answer: A
Clarification: The first significant DNA to be obtained was that of cohesive ends of lambda which were 12 bases long; in the year 1971.

3. Plus and minus sequencing is the other name for Sanger sequencing.
A. True
B. False
Answer: A
Clarification: Plus and minus sequencing was used by Sanger in 1977 to sequence the 5386 base pairs long PhiX174 genome. This method was later superseded by Maxam Gilbert method.

4. Which type of DNA cleavage is done in the Maxam Gilbert method?
A. Edge
B. Interstitial
A. Base-specific
D. Gene-specific
Answer: C
Clarification: The Maxam Gilbert method of sequencing was devised in 1977. It uses a variety of chemical reagents to bring about base-specific cleavage of DNA.

5. Sequence of which of the following cannot be determined using the Maxam Gilbert method?
A. Bacteria
B. Plants
A. Bacteriophage T7
D. Plasmid
Answer: C
Clarification: Although Maxam Gilbert is a popular technique of gene sequencing and it superseded the Sanger sequencing method; it has not been used to sequence the genome of T7 phage.

6. What is the main enzyme component of Sanger sequencing?
A. Helicase
B. Polymerase
A. Nuclease
D. Gyrase
Answer: B
Clarification: The chain-termination or dideoxy method of DNA sequencing capitalizes on two unique properties of DNA polymerase enzyme.

7. Which of the following is used by DNA polymerase as a substrate?
A. Sucrose
B. Lactose
A. Nucleotide
D. Nucleoside
Answer: C
Clarification: The DNA polymerase enzymes used in the chain-termination method of sequencing can synthesize a complimentary copy of single-stranded DNA and can use nucleotides as substrates.

8. Which of the following act as chain terminator?
A. Exogenous
B. DNA
A. Deoxynucleotides
D. Dideoxynucleotides
Answer: D
Clarification: A complementary strand is synthesized by the DNA polymerase and whenever an analog which is basically a dideoxynucleotide is incorporated in the growing chain, the chain is terminated.

9. The Klenow fragment is basically a _______________
A. DNA hybrid
B. DNA polymerase
A. RNA polymerase
D. Promoter
Answer: B
Clarification: Klenow fragment is a DNA polymerase which is used in the Sanger sequencing method. It lacks the exonuclease activity, associated with intact enzyme.

10. ____________ is a chemically synthesized oligonucleotide.
A. Klenow fragment
B. DNA
A. Primer
D. RNA
Answer: C
Clarification: Initiation of DNA synthesis requires a primer and usually this is a chemically synthesized oligonucleotide which is annealed close to the sequence being analyzed.

11. How many types of deoxynucleoside triphosphates are used in Sanger sequencing?
A. 1
B. 2
A. 3
D. 4
Answer: D
Clarification: Four different types of deoxynucleoside triphosphates are used, one or more of which is labeled with phosphorus-32.

12. Prior to getting electrophoresed in the sequencing gel, DNA is ____________
A. Purified
B. Denatured
A. Synthesized
D. Fragmented
Answer: B
Clarification: After a suitable incubation period for the DNA mix that contains deoxynucleoside triphosphates, dideoxynucleotides and sample DNA, the DNA is denatured before electrophoresis.

13. A sequencing gel is a ________________ gel.
A. Toxic
B. Highly-polymerized
A. High resolution
D. Low resolution
Answer: C
Clarification: The sequencing gel is a high resolution gel designed to fractionate single-stranded DNA fragments on the basis of their size.

14. What is the molarity of urea used in sequencing gels?
A. 1 M
B. 3 M
A. 5 M
D. 7 M
Answer: D
Clarification: The high-resolution sequencing gels can resolve fragments differing by just one base. These contain 6-20% acrylamide and 7 M urea.

15. The function of urea in the sequencing gels is to promote adherence.
A. True
B. False
Answer: B
Clarification: The sequencing gels contain 6-20% acrylamide and 7 M urea. The function of urea is to minimize DNA secondary structure which affects electrophoretic mobility.

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Vector Biology Question Paper focuses on “Single Stranded DNA Vectors Cloning – 2”.

1. Single-stranded DNA was used in __________
A. Dideoxy method
B. Replication
A. Translational
D. Hybridization
Answer: A
Clarification: Sequencing by the original dideoxy method required single stranded DNA, as do techniques for oligonucleotide directed mutagenesis.

2. Single stranded form combines cloning, amplification and __________________ of and originally double-stranded DNA fragment.
A. Strand separation
B. Replication
A. Transcription
D. Translation
Answer: A
Clarification: Single-stranded form is an attractive means of combining the cloning, amplification and strand separation of an originally double-stranded DNA fragment.

3. RF acts just like a ___________
A. Phage
B. Host
A. Plasmid
D. Hybrid
Answer: C
Clarification: The phage DNA is replicated via a double stranded circular intermediate. This replicative form can be purified and manipulated in vitro just like a plasmid.

4. RF and single-stranded DNA __________ the competent E.coli cells.
A. Infect
B. Transfect
A. Kill
D. Translate
Answer: B
Clarification: Both RF and single stranded DNA will transfect competent E.coli cells to yield either plaques or infected colonies, depending on the assay method.

5. There are no packaging constraints in single-stranded vectors.
A. True
B. False
Answer: A
Clarification: The size of the phage particle is governed by the size of the viral DNA and therefore there are no packaging constraints associated.

6. Agarose gel electrophoresis is used to determine the _________ of the insert.
A. Activity
B. Size
A. Orientation
D. Replication
Answer: C
Clarification: If two clones carry the insert in opposite directions, the single stranded DNA from them will hybridize and this can be detected by agarose gel electrophoresis.

7. Filamentous phages do not possess the advantages of plasmids.
A. True
B. False
Answer: B
Clarification: As vectors, filamentous phages possess all the advantages of plasmids while producing particles containing single stranded DNA in an easily obtainable form.

8. Unlike lambda, filamentous phages do not have ____________
A. Codons
B. Nucleases
A. Cloning sites
D. Transcription factors
Answer: C
Clarification: Unlike lambda, the filamentous coliphages do not have any non-essential genes which can be used as cloning sites, but intergenic sequences are present.

9. The M13 intergenic sequence contains ________
A. Origins
B. Termination sequence
A. Replication site
D. Nucleosome
Answer: A
Clarification: In M13 there is a 507 bp intergenic region, from position 5498 to 6005 of the DNA sequence, which contains the origins of DNA replication for both the viral and the complementary strands.

10. The first use of M13 as a cloning vector made use of _______ sites.
A. 4
B. 8
A. 10
D. 12
Answer: C
Clarification: The first example of M13 cloning made use of one of 10 BsuI sites in the genome, two of which are in the intergenic sequence region.

11. Insertion of fragments into lac region destroys __________
A. Blue color
B. White color
A. Transparency
D. Opaqueness
Answer: A
Clarification: Insertion of DNA fragments into the lac region of M13 mp1 destroys its ability to form the blue plaques, making detection easy.

12. The lac region does not contain any restriction site for which of the following?
A. AvaII
B. PvuII
A. EcoR1
D. BglII
Answer: C
Clarification: The lac region only contains unique sites for AvaII, BglII, and PvuI and three sites for PvuII and there are no sites anywhere on the complete genome.

13. Which process was used to create an EcoR1 site in the lac region of coliphage vectors?
A. Hybridization
B. Mutagenesis
A. Replication
D. Restriction
Answer: B
Clarification: Groneborn and Messing (1978) used in vitro mutagenesis to change a single base pair, thereby creating a unique EcoR1 site within.

14. The M13 mp2 contained ________
A. Cloning site
B. Nuclease
A. Codon
D. EcoR1 site
Answer: D
Clarification: Groneborn and Messing (1978) used in vitro mutagenesis to change a single base pair, thereby creating a unique EcoR1 site within the lac fragment, called M13 mp2.

15. The derivatives of M13 vector are exact counterparts of _________
A. Pbr322
B. PUC
A. Lambda
D. Adenovirus
Answer: B
Clarification: The derivatives mp7-11, mp18, mp19 are the exact M13 counterparts of the PUC plasmids, containing many common cloning sites.

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Vector Biology Problems focuses on “Vectors for Mammals – 3”.

1. What are somatic cells?
A. Non-germline cells
B. Yeast cells
A. Genetically engineered cells
D. Hybrid vectors
Answer: A
Clarification: Somatic cells are the ones that do not contribute to the germline. Unlike the plant cells, animal somatic cells are restricted in their developmental potential.

2. Transformation of animal germline cells can be done by gene transfer to _____________
A. Totipotent cells
B. Plant cells
A. Yeast cells
D. Bacterial cells
Answer: A
Clarification: Transformation of the animal germline cells required gene transfer to pluripotent or totipotent cells, such as eggs, early embryos, isolated germ cells or gametes.

3. What is the delivery of exogenous genetic material to animal cells, using a vector known as?
A. Transformation
B. Translocation
A. Transduction
D. Conjugation
Answer: C
Clarification: Delivery of exogenous genetic material using a viral vector is known as transduction. The transgene is either added to a complete viral genome or used to replace one or more genes.

4. What is a transgene?
A. Viral vector gene
B. Exogenous genetic material
A. Endogenous genetic material
D. Bacterial gene
Answer: B
Clarification: The process of gene transfer to animal cells can be done by several mechanisms. In all of these the exogenous genetic material is incorporated into the animal cells. This exogenous genetic material is called a transgene.

5. What is Bactofection?
A. Bacterial infection
B. Bacterial hybrid vector
A. Gene transfer using bacteria
D. Recombinant plasmid
Answer: C
Clarification: Bacterial gene delivery is termed Bactofection. The transgene in this mechanism is delivered as a part of the bacterial plasmid.

6. What is “Transfection” in gene delivery sense?
A. Gene delivery using bacterial vectors
B. Gene delivery using viral vectors
A. Non-biological gene delivery
D. Electroporation
Answer: C
Clarification: Some delivery systems are described as non-biological because biological delivery vectors are not required. These methodical approaches are called transfection.

7. In how many stages does the transformation of animal cells occur?
A. 1
B. 2
A. 3
D. 4
Answer: B
Clarification: The transformation of animal cells occurs in two stages, the first involving the introduction of DNA into the cell and the second involving its incorporation into the nucleus.

8. Transfection is more efficient than the integration of foreign DNA.
A. True
B. False
Answer: A
Clarification: Transfection is much more efficient than integration, hence a large proportion of transfected cells never integrate the foreign DNA they contain.

9. What does Transient transfection reflect?
A. Changed properties of host cell
B. Incorporation of origin
A. The short period of replication
D. Incorporation of exogene
Answer: A
Clarification: The DNA is maintained in the nucleus in an extrachromosomal state for just a short time before it is diluted and degraded.

10. In a stable transfection, which type of genetic locus is formed?
A. Transient
B. Stable
A. Unstable
D. Integrative
Answer: B
Clarification: In a small proportion of transfected cells, the DNA will integrate into the genome, forming a new genetic locus that will be inherited by all clonal descendants.

11. Stable transformation results in a __________
A. Transduced cell
B. Transgene
A. Cell line
D. Hybrid vector
Answer: C
Clarification: The formation of a new genetic locus, termed as stable transformation results in the formation of a cell line.

12. How can the transformed cells be isolated from the large background?
A. By manipulation
B. By selection
A. By using viral vectors
D. By conjugation
Answer: B
Clarification: Since integration is an insufficient process, the rare stably transformed cells must be isolated from the large background of non-transformed and transiently transformed cells by selection.

13. Mutant cell lines are required for which type of selectable markers?
A. Endogenous
B. Exogenous
A. Transgene
D. Foreign gene
Answer: A
Clarification: Three types of selectable markers have been developed for animal cells. Endogenous selectable markers are already present in the cellular genome, and mutant cell lines are required when they are used.

14. What is HSV?
A. Human virus
B. Human enzyme
A. Herpes simplex virus
D. Human simplex virus
Answer: C
Clarification: The mouse cells deficient for the enzyme thymidine kinase could be stably transformed to a wild-type phenotype by transfecting them with the herpes simplex virus (HSV) Tk gene.

15. Salvage pathway is for the synthesis of _________
A. Nucleotides
B. Nucleoside
A. Thymidine kinase gene
D. Transgene
Answer: A
Clarification: Cells positive for TK can be selected on HAT medium. This is because both the enzymes are required for nucleotide biosynthesis via the salvage pathway.

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