Category: Objective Questions

300+ TOP MCQs on Cloning the cDNA and Specialized Libraries – 1 and Answers

Genetic Engineering Multiple Choice Questions on “Cloning the cDNA and Specialized Libraries – 1”.

1. RNaseH method and homopolymer tailing method generates blunt ended cDNA molecules. Which of the following can be used for attaching them to vector?
a) Blunt ended ligation
b) Addition of linkers
c) Using appropriate restriction enzymes
d) All the methods can be used equivalently
Answer: d
Explanation: As the products of RNaseH method and homopolymer tailing method generates dsDNA and blunt ended molecules, there are many methods to attach them to vector. Blunt ended ligation, the addition of linkers and restriction enzymes all can be used.

2. Choose the correct statement for modification of homopolymer tailing method.
a) It includes modification of primers
b) Primers are varied by simply altering their size by randomly adding or removing bases
c) The 5’ end of the first cDNA strand is tailed with C residues
d) A single stranded oilgonucleotide is then used for second strand synthesis
Answer: a
Explanation: The modification of homopolymer tailing method includes modification of primers. Primers can be modified by the introduction of restriction sites in them. The 3’ end of the first cDNA strand is tailed with C residues. Another oligo-dG primer precedes the introduced restriction site and is contained with the double stranded oligonucleotide region. It is used for second strand synthesis.

3. By synthesizing two strands separately then annealing them leads to formation of double stranded oligonucleotide.
a) True
b) False
Answer: a
Explanation: Double stranded oligonucleotides are required for the synthesis of second cDNA strand. This double stranded oligonucleotide is synthesized by separately synthesizing two strands and then annealing them.

4. Choose the incorrect statement for the homopolymer tailing of cDNA strands.
a) The blunt ended double stranded cDNA molecules are treated with terminal transferase and dCTPs
b) Vector is also treated with terminal transferase and dGTPs
c) The vector and cDNA can now anneal with the help of DNA ligase
d) If gaps are created they can be repaired by physiological processes
Answer: c
Explanation: For homopolymer tailing of cDNA strands, the blunt ended double stranded molecules are treated with terminal transferase and dCTPs. This leads to the addition of C residues at 3’ end. Vector is also treated with terminal transferase and dGTPs. This leads to annealing of vector and cDNA molecules and DNA ligase is not required. The gaps created can be repaired by physiological processes once the recombinant molecules enter the host.

5. Choose the correct statement if the RNA is non polydenylated.
a) A collection of chemically synthesized oligonucleotides is used as primers
b) They are usually tetramer
c) Unequal quantities of A, G, T and C are used
d) The primers attach at only specific sequences for first strand synthesis
Answer: a
Explanation: As the RNA is non polyadenylated, oilgo-dT primer can’t be used and in place of it a collection of chemically synthesized oligonucleotides is used as primers. They are usually hexamers and are made by equal quantities of A, G, C and T. And thus all hexameric sequences can be synthesized. The primers can attach to the RNA sequences throughout.

6. In case if molecules smaller than the fragments required for making a full genomic library are used for making a collection. This collection is called as ___________
a) library
b) shelf
c) small library
d) mini library
Answer: b
Explanation: If the molecules smaller than the fragments required for making a full genomic library are used for collection, this collection is called a shelf. It is a subsection of the library.

7. Choose the correct statement for construction of a library subsection.
a) The size of a particular restriction fragment on which the gene is located is not known
b) The size of the restriction fragment can be known by carrying out southern blotting
c) Another digest of the genomic DNA is carried out by a different enzyme
d) DNA fragments of different size are recovered after carrying out gel electrophoresis
Answer: b
Explanation: For the construction of a subsection of the library, some steps are followed. The size of a particular restriction fragment on which gene is located is known at times. The size of the fragment can be known by carrying out southern blotting. After the size is known, another digest of the genomic DNA is carried out and is done by the same enzyme. DNA fragments of the approximately same size are recovered from the gel after carrying out gel electrophoresis. They can be further cloned into a vector.

8. Any cDNA library would represent a fraction of RNA species of an organism.
a) True
b) False
Answer: a
Explanation: Any cDNA library would represent a fraction of RNA species of an organism, the whole organism can’t be represented in a library. It depends on the developmental stage, physiological state and the tissue from which RNA was isolated.

9. What do we mean by housekeeping genes?
a) Housekeeping genes are those genes which are specific to an organism
b) Housekeeping genes are those genes which are present in all the organisms
c) Housekeeping genes are those genes which are meant for repair and maintenance in a species of organism
d) Housekeeping genes are those genes which required for the replication process
Answer: b
Explanation: Housekeeping genes are those genes which are present in all the organisms. cDNA libraries may therefore contain housekeeping genes and genes specific to that organism.

10. Choose the correct statement for RNA fractionation.
a) The RNA is fractioned by size but before separating on oligo-dT cellulose
b) A sucrose density gradient is used
c) The RNA is applied to the top of a pre-poured gradient and during centrifugation smaller molecules move down the tube faster
d) Different bands are formed according to the density in the sucrose density gradient
Answer: b
Explanation: RNA fractionation is carried out and the basis is the size. It is fractioned by size after carrying out separation on oligo-dT cellulose. A sucrose density is used for size based separation. The RNA is applied to the top of a pre-poured gradient and during centrifugation larger molecules move down the tube faster. There are different bands formed on the basis of size in the sucrose density gradient.

11. What is done after RNA fractionation is carried out?
a) Each band is translated in vivo
b) Translation is carried out in wheat gram or lysate of rabbit reticulocyte cells
c) Translation is carried out with a high background
d) Amino acid is not radioactively labelled
Answer: b
Explanation: RNA fractionation is carried out and it is followed by a translation of each band in vitro. It is carried out in wheat gram or lysate of rabbit reticuloycte cells. Ribosomes, tRNAs are added in order to carry out the translation with low background. An amino acid is radioactively labelled and thus the polypeptide sequence synthesized is labelled.

12. The polypeptides produced after addition of mRNA are analysed with antibodies. Choose the incorrect statement for this analysis.
a) Antibodies are added to each reaction tube and precipitation is simply based on antigen-antibody reaction
b) Along with simple antigen-antibody complex, a substrate is added for easy precipitation
c) Protein A-Sepharose is added
d) Protein A-Sepharose binds to IgG antibodies
Answer: a
Explanation: The analysis of polypeptides after addition of mRNA can be carried out by the addition of antibodies. But it is not simply based on antigen-antibody reaction. A substrate for easy precipitation is also added. For this, protein A-Sepharose is added. Protein A-sepharose binds to IgG antibodies. After carrying out centrifugation, it can be pelleted easily.

13. What is done after the recovery of pellets has been carried out in order to know the amount of polypeptides?
a) Denaturing and gel electrophoresis in SDS- Polyacryamide gel
b) Gel electrophoresis in agarose gel
c) Quantitative PCR
d) Weighing pellets
Answer: a
Explanation: After the recovery of pellets has been carried out by the use of antigen-antibody reaction, denaturation and gel electrophoresis in SDS-Polyacrylamide gel. The amount of radioactivity gives the amount and location of polypeptides.

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300+ TOP MCQs on Purification of Plasmid DNA and Answers

Genetic Engineering Multiple Choice Questions on “Purification of Plasmid DNA”.

1. Isolation of genomic DNA follows the same principles as that of obtaining plasmid from E. coli. Which of the following is not included in it?
a) Cell lysis
b) Removal of proteins
c) Removal of chromosomal DNA
d) Dissolving plasmid in water

Answer: d
Explanation: There are some basic steps which are included in obtaining plasmid DNA from E. coli. Firstly, the cell is lysed, further removal of proteins and chromosomal DNA is done. A plasmid is obtained and collected but not in water. Also, further purification is done if necessary.

2. How many methods are there for obtaining the plasmid DNA from the bacteria?
a) 1
b) 2
c) 3
d) 4

Answer: b
Explanation: There are two methods which are used for obtaining the plasmid DNA from the bacteria. They are named as alkaline lysis method and boiling lysis method. They both are having different working principles.

3. Cell lysis is carried out by which substance?
a) Lysozyme and detergents
b) Water
c) Sugar solution
d) Suphuric Acid

Answer: a
Explanation: Cell lysis is carried out by adding lysozyme and detergents. The cell wall is made up of N-acetyl glucoasamine and N-acetyl muramic acid and they are having cross links. The agents added to break the cross links present between the molecules of the cell wall.

4. Chromosomal or genomic DNA is separated by ____________
a) Sedimentation
b) Dissolution in water
c) Centrifugation
d) Distillation

Answer: c
Explanation: Chromosomal or genomic DNA is comparatively heavier and large in size than that of plasmid DNA. Hence, centrifuging at a high speed leads to settling down of the genomic DNA and thus can be separated easily.

5. Proteins can be removed via treatment by?
a) Phenol and chloroform treatment
b) Treatment with sodium hydroxide
c) Chloroform treatment alone
d) Centrifuging

Answer: a
Explanation: Proteins can be removed via treatment with phenol and chloroform treatment. Chloroform is alone not sufficient. The phenol added helps in the destruction of proteins and chloroform helps in its dissolution under acidic conditions.

6. The nucleic acid remaining in the solution can be precipitated by addition of sodium or ammonium acetate and ethanol.
a) False
b) True

Answer: b
Explanation: The nucleic acid is present in the solution and is precipitated by the addition of sodium or ammonium acetate and ethanol. It is because; nucleic acid is polar in nature and thus easily dissolves in water. Hence, to avoid this sodium acetate and ethanol is added. Sodium acetate is shields the charge present on the sugar phosphate backbone and further bonds are easily formed between ethanol and phosphate. It leads to separating out of nucleic acids.

7. Nucleic acid precipitated constitutes of ____________
a) plasmid DNA
b) plasmid DNA, along with RNA and chromosomal DNA
c) rna alone
d) chromosomal DNA only

Answer: b
Explanation: Nucleic acid precipitated contains the plasmid DNA and along with it RNA and remnants of chromosomal DNA are also present. RNA can be removed via adding RNase.

8. Treatment with exonuclease leads to removal of ____________
a) remnants of chromosomal DNA
b) RNase
c) plasmid DNA which is circularized
d) proteins

Answer: a
Explanation: Exonuclease leads to removal of remnants of chromosomal DNA because they are usually having linear ends. The circularized ends of plasmid are protected from the action of exonuclease because they don’t have any free ends for their action.

9. Adsorption onto a solid phase support followed by elution is used as an alternative for separation of which component?
a) chromosomal DNA
b) plasmid DNA
c) RNA alone
d) other impurities

Answer: b
Explanation: This method is used for separation of plasmid DNA. It is advantageous because it avoids the use of phenol and also removes RNA at times along with plasmid DNA.

10. Which of the following components bind to the solid column made of silica, under high salt concentration?
a) Proteins
b) Polysaccharides
c) Both proteins and polysaccharides
d) Plasmid DNA

Answer: d
Explanation: Plasmid DNA binds to a solid support which is made of silica and under high salt concentrations. A high salt concentration doesn’t allow less polar molecules to bind such as polysaccharides and proteins. The binded DNA molecule is further eluted by using a low salt concentration.

11. Purification of DNA by using silica derivatized groups by DEAE is termed as ____________
a) ion exchange resin based method.
b) silica based purification
c) atom based resin exchange method
d) packed bed purification

Answer: a
Explanation: Silica derivatized groups by DEAE are used for purification of DNA. These groups are positively charged and the DNA gets attached to it, along with other species such as RNA which are negatively charged. Further, DNA can be obtained by varying the ionic concentrations.

12. Which of the following is correct with respect to caesium chloride centrifugation?
a) Caesium is light in weight
b) The dissolution of caesium and nucleic acids leads to the formation of gradients
c) According to the amount of supercoiling and A+T content, the DNA settles
d) Nicked DNA settles below than supercoiled DNA

Answer:b
Explanation: Caesium chloride is heavy and when the nucleic acid is dissolved with it, density gradients are formed. According to the extent of supercoiling and G+C content, settling of DNA takes place. The nicked DNA settles above than the nicked DNA.

13. Which of the following components settles at the bottom?
a) RNA
b) Proteins
c) Nicked DNA
d) Supercoiled DNA

Answer: a
Explanation: The component settling at the bottom is RNA. And the proteins float on the free surface. The nicked DNA forms a band above the supercoiled form.

14. The location of plasmid DNA can be visualized by addition of:
a) bromophenol blue
b) ethidium bromide
c) ortho xylene
d) texas red

Answer: b
Explanation: Ethidium bromide is added before centrifugation. It is an orange-red coloured stain which gives rosy coloured bands when placed under UV light. It acts upon by intercalating between the bases.

300+ TOP MCQs on Cultured Cells in Mammals and Restricted Areas in Intact Organisms – 1 and Answers

Genetic Engineering Multiple Choice Questions on “Cultured Cells in Mammals and Restricted Areas in Intact Organisms – 1”.

1. Lesch-Nyhan syndrome is caused by the deficiency of the enzyme hypoxanthine-guanine phosphoribosyl transferase (HGPRT). Cells deficient in HGPRT die in a medium containing which of the following?
a) Hypoxanthine and thymidine
b) Thymidine
c) Aminopterin and thymidine
d) Hypoxanthine, thymidine and aminopterin (HAT medium)
Answer: d
Explanation: HGPRT deficient cells in the medium containing all three hypoxanthine, thymidine and aminopterin. It is so because aminopterin blocks the endogenous synthesis of purines needed for the synthesis of nucleic acid.

2. Presence of wild-type DNA onto the HGPRT– in the presence of _______ led to DNA uptake and stable transformation.
a) lithium acetate
b) calcium phosphate
c) sodium chloride
d) aluminum sulphate
Answer: b
Explanation: Presence of wild type in HGPRT—in the presence of calcium phosphate led to DNA uptake and stable transformation.

3. Cells deficient in thymidine kinase (TK) are also killed in HAT medium.
a) True
b) False
Answer: a
Explanation: Cells deficient in thymidine kinase are also killed in HAT medium. It is so because pyrimidine synthesis is also blocked in the presence of aminopterin and utilization of thymidine in the HAT medium requires a functional TK.

4. Hygromycin is used as a selectable marker in mammalian cultured cells. It is used for ____________
a) initiating protein synthesis
b) inhibiting protein synthesis
c) initiating DNA binding process
d) inhibiting DNA binding process
Answer: b
Explanation: Hygromycin is used as a protein synthesis inhibitor. It is conferred by a bacterial hph gene which encodes hygromycin phosphotransferase. Resistance to hygromycin is used as a selectable marker.

5. Puromycin is a protein synthesis inhibitor. It is conferred by _______ gene.
a) streptococcal
b) bacilovirus
c) streptomyces
d) both streptococcal and streptomyces
Answer: c
Explanation: Puromycin is a protein synthesis inhibitor and is conferred by streptomyces gene. It does so by encoding puromycin-N-acetyltransferase.

6. Resistance to bleomycin (zeocin) is used as a selectable marker for mammalian cultured cells and its function is ____________
a) DNA damaging agent
b) DNA synthesis promoter
c) Inhibiting RNA synthesis
d) Activating RNA synthesis
Answer: a
Explanation: Resistance to bleomycin (zeocin) is used as a selectable marker for mammalian cultured cells and it is a DNA damaging agent. It does so by expression of a binding protein.

7. Resistance to methotrexate, which inhibits the enzyme dihydrofolate reductase (DHFR) is used as a selectable marker. This enzyme is involved in the synthesis of _____ carbon units and is required for _____ biosynthesis.
a) two, nucleoside
b) two, nucleotide
c) one, nucleotide
d) one, nucleoside
Answer: d
Explanation: Resistance to methotrexate, which is used as a selectable marker inhibits DHFR enzyme. This enzyme is involved in the synthesis of one carbon units and is required for nucleoside synthesis.

8. Histidinol dehydrogenase allows synthesis of histidine from exogenous histanol.
a) True
b) False
Answer: a
Explanation: Histidinol dehydrogenase is also used as a selectable marker. It allows synthesis of histdidine from exogenous histanol and protects from toxic effects of histidinol.

9. Many mammalian cells contain Thymidine Kinase, the mammalian enzyme uses the analogue _____ than does the viral enzyme.
a) more efficiently
b) less efficiently
c) with same efficiency
d) either with same or more efficiency
Answer: b
Explanation: Many mammalian cells contain thymidine kinase, the mammalian enzyme uses the analogue less efficiently than the viral enzyme. If cells lack viral enzyme they are resistant to analogue and are able to grow in a mammalian enzyme.

10. A wide range of host cell lines are available and commonly used human cell lines are obtained from ___________
a) kidney
b) liver
c) lymphoblast from leukaemia patient
d) both kidney and lymphoblast from a leukaemia patient
Answer: d
Explanation: There is a wide range of host cell lines available. Commonly used human cell lines are HeLa 293T, obtained from kidney and Jurkat which is obtained from lymphoblast from a leukaemia patient.

11. DEAE-dextran is used for introduction of DNA. It is a modified _____ and is ______
a) polysaccharide, negatively
b) polysaccharide, positively
c) monosaccharide, positively
d) monosaccharide, negatively
Answer: b
Explanation: There are various methods for the uptake of transformation DNA such as electroporation and DEAE-dextran method. It is a modified polysaccharide and is positively charged. It forms a complex with negatively charged DNA and is taken into cells by endocytosis.

12. Cells whose contents have been removed and replaced, by swelling and shrinking in solutions of suitable osmotic strength are called as ____________
a) protoplast
b) ghosts
c) shrunken cells
d) vacuole
Answer: b
Explanation: Ghosts are those cells whose contents have been removed and replaced by swelling and shrinking in solutions of suitable osmotic strength. Red blood ghosts are also used for the introduction of DNA into mammalian cells.

13. In ___________ cells, virus replication doesn’t take place and viral DNA ___________
a) non-permissive, is also not expressed
b) permissive, is also not expressed
c) non-permissive, can be expressed
d) non-permissive, is always expressed
Answer: c
Explanation: In non-permissive cells, virus replication doesn’t take place and viral DNA can be expressed though.

14. SV40 is a virus and it produces how many transcripts?
a) 1
b) 2
c) 3
d) 4
Answer: b
Explanation: SV40 is a virus and it produces two transcripts. These transcripts are produced by early and late transcripts.

15. Splicing of the transcripts is necessary for efficient expression.
a) True
b) False
Answer: a
Explanation: Splicing of both the transcripts is necessary and it is for efficient expression. Sequences that have not been through the splicing process are not expressed efficiently even if introns are removed.

300+ TOP MCQs on Applications of Genetic Engineering and Answers

Genetic Engineering Multiple Choice Questions on “Applications of Genetic Engineering”.

1. Amplification of specific region can be done by using primers for specific regions. If the PCR product is ______ and is in sufficient quantity, then sequence can be determined ________
a) non-specific, directly
b) non-specific, indirectly or directly
c) specific, directly
d) specific, indirectly
Answer: c
Explanation: Amplification of specific region can be done by using primers for specific regions. If the PCR product is specific, it means that only a single band is obtained in gel electrophoresis and is insufficient quantity then the sequence can be determined directly.

2. Which of the following is not suitable if the PCR product is non-specific?
a) Adjusting the concentration of magnesium ions
b) Increasing annealing temperature
c) Using touchdown PCR
d) Using inverse PCR
Answer: d
Explanation: In case, products are non-specific then sequence can’t be known directly. Optimization of PCR should be carried out and various other strategies should be used such as adjusting the concentration of magnesium ions, increasing annealing temperature or using touch-down PCR.

3. The disadvantage in the approach based on using PCR is that there is no permanent record until some of the bacterial cells are preserved.
a) True
b) False
Answer: a
Explanation: The disadvantage of this approach is that there is no permanent record until some of the bacterial cells are preserved. If enough bacteria are there initially then genomic library can be constructed.

4. How many approaches are there in order to clone the complete genome?
a) 1
b) 2
c) 3
d) 4
Answer: b
Explanation: There are basically two approaches in order to clone the complete genome. In the first approach, systematic cloning is cloned is carried out. The second approach is based on cloning overlapping fragments at random.

5. If a full proteomic analysis of growth medium is carried out and is combined with ____________ genome sequence, genes for other _____________ proteins are also obtained.
a) partial, defensive
b) partial, secreted
c) complete, defensive
d) complete, secreted
Answer: d
Explanation: If a full proteomic analysis of growth medium is carried out and is combined with complete genome sequence, genes for other secreted proteins can also be obtained.

6. If a putative protein sequence is cloned in an expression vector and the expressed protein is not showing protease activity, then which of the following is not correct?
a) The protein is not protease
b) The protein can be incorrectly folded which can block the protease activity
c) There might be some other cofactor required for protease activity
d) The most commonly used expression system is E.coli
Answer: a
Explanation: If the putative protein sequence is cloned and the expressed protein is not showing protease activity, it is not necessary that it is not protease. Suppression of protease activity can be because of incorrect folding or that some cofactor is required for protease activity. The most commonly used expression system is E.coli.

7. For getting a large amount of proteins to crystallize, which of the following should be used as an expression system?
a) Bacterial system
b) Yeast systems
c) Eukaryotic systems
d) Both eukaryotic and bacterial systems can be used
Answer: d
Explanation: The system to be used for getting large amounts of proteins to crystallize can be either bacterial or eukaryotic. It depends on the source of the gene and whether post-translational modification is necessary or not.

8. If a mutation perturbs the structure, then stability and folding are not affected.
a) True
b) False
Answer: b
Explanation: At times mutated proteins are not expressed well. Mutation perturbs a structure at times and it affects the stability and folding of the protein.

9. The RNA level ___________ the steady-state level of the corresponding protein directly and the post-translational modification of the protein ____________
a) reflects, can be determined
b) reflects, can’t be determined
c) doesn’t necessarily reflects, can be determined
d) doesn’t necessarily reflects, can’t be determined
Answer: d
Explanation: The RNA level doesn’t necessarily correspond to the steady level of corresponding protein directly and the post-translational modification or location of the protein can’t be determined.

10. For a convenient transformation system, _____ can be used for gene silencing.
a) antisense RNA
b) transposon insertion
c) either antisense RNA or transposon insertion
d) transposon insertion followed by antisense RNA
Answer: a
Explanation: If a convenient transformation system is used, antisense RNA can be used for gene silencing. If the transformation system is not convenient then transposon insertion is used for silencing.

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300+ TOP MCQs on Organelle Transformation and Answers

Genetic Engineering Multiple Choice Questions on “Organelle Transformation”.

1. Mitochondrial genome encodes tRNAs ___________ and polypeptides involved in _______
a) mRNAs, oxidative phosphorylation
b) rRNAs, oxidative phosphorylation
c) rRNAs, reductive phosphorylation
d) mRNAs, reductive phosphorylation
Answer: b
Explanation: Mitochondrial genome encodes tRNAs, rRNAs and polypeptides which are required for oxidative phosphorylation. tRNAs and rRNAs are also encoded by chloroplast genome.

2. Atrazine is a herbicide and it acts on _________
a) reaction centre in photosystemI
b) reaction centre in photosytemII
c) reaction centre in both the photosystems
d) neither of the photosystems
Answer: b
Explanation: Atrazine is a herbicide and it acts on a reaction centre present in photosystem II. This herbicide is a product of chloroplast psbA gene.

3. Allotopic gene expression is the case when the presence of a normal gene in an organelle is not a problem in expression.
a) True
b) False
Answer: a
Explanation: At times, the presence of a normal gene in the organelle doesn’t create a problem and thus inserting modified organelle genes into the nucleus may be expressed well. Such an expression is called a allotropic gene expression.

4. atpB encodes _______ subunit of ATP synthase, an enzyme used for generation of ______
a) beta, ADP
b) alpha, ATP
c) beta, ATP
d) alpha, ADP
Answer: c
Explanation: atpB encodes a beta subunit of ATP synthase, it is a multisubunit complex used for generation ATP and it is done in the presence of light reaction.

5. Bacterial aadA gene is responsible for conferring resistance to _________
a) spectinomycin
b) streptomycin
c) ampicillin
d) spectomycin and streptomycin
Answer: d
Explanation: Bacterial aadA gene is responsible for conferring resistance to both streptomycin and spectomycin. It is used a selectable marker for chloroplast transformation.

6. How many types of the chloroplast are there in Chlamydomonas?
a) 1
b) 2
c) 3
d) 4
Answer: a
Explanation: Chlamydomonas is having only one type of chloroplast and this property makes it easier to use it for transformation.

7. For transformation of the chloroplast of higher plants, a vector is used which ______ in the chloroplast.
a) doesn’t replicates
b) replicates
c) may or may not replicate
d) replicates under certain specified conditions
Answer: a
Explanation: For transformation of the chloroplast of higher plants, a vector is used which doesn’t replicates in the chloroplast. There is a selectable marker which is present and the gene of interest is flanked by chloroplast DNA.

8. Chloroplast can be transferred through pollen in all crops.
a) True
b) False
Answer: b
Explanation: Cholorplast can’t be transferred through pollen for all the crops. Thus incorporation of transgenes in chloroplast may offer more biological containment then that incorporation into nucleus may offer.

9. Mutant strains of Saccharomyces cervevisiae in which endogenous DNA are deleted are called as _________
a) rho0
b) synthetic rho
c) rho+
d) rho-
Answer: a
Explanation: Mutant strains of Saccharomyces cervevisiae in which endogenous DNA are deleted are called as rho0. Synthetic rho- strains are produced when DNA is introduced into mitochondria and concatamers are produced.

10. COX3 gene is a selectable marker. Choose the correct statement with respect to it.
a) It confers the ability to grow by anaerobic respiration
b) It confers the ability to grow by aerobic respiration
c) It confers the ability to grow in absence of uracil
d) It confers the ability to grow in lithium acetate medium
Answer: b
Explanation: COX3 gene is used as a selectable marker. It confers the ability to grow by aerobic respiration in the mutant cells for mitochondrial COX3 gene.

11. The ARG8m gene which produces an enzyme for arginine biosynthesis is located in _______ and is of ______ origin.
a) mitochondrial, nuclear
b) nuclear, mitochondrial
c) nuclear, nuclear
d) mitochondrial, mitochondrial
Answer: a
Explanation: The ARG8m gene which produces an enzyme for arginine biosynthesis is located in the mitochondria but is of nuclear origin. It is designed for expression in the mitochondrion and will confer the ability to grow in the absence of arginine.

12. Barstar is _________
a) RNAse
b) RNAse inhibitor
c) DNAse
d) DNAse inhibitor
Answer: b
Explanation: Barstar is RNAse inhibitor. And Barsar is RNAse and inhibitor is used as a selectable marker. Barsar is used for degrading the mitochondrial RNA. Barstar added helps in suppressing the function of Barsar and thus restoration of mitochondrial function takes place.

13. Caenorhabditis elegans is a model organism of great importance in biological systems. It is a/an _________
a) algae
b) parasite
c) fungi
d) nematode
Answer: d
Explanation: Caenorhabditis elegans is a model organism of great importance in biological systems and it is a nematode. The genetic manipulation of the organism is quite complex.

14. DNA can be injected into Caenorhabditis elegans by biolistic transformation. The injected DNA forms arrays of extrochromosomal copies which are stable in nature.
a) True
b) False
Answer: b
Explanation: DNA can be injected into Caenorhabditis elegans by biolistic transformation. The injected DNA forms extrachromosomal copies but these are not stable in nature. This can be avoided by the incorporation of poison sequences.

300+ TOP MCQs on The Basic Technique and Answers

Genetic Engineering Multiple Choice Questions on “The Basic Technique”.

1. The process of amplification of specific DNA sequences by an enzymatic process is termed as ____________
a) amplification
b) polymerase chain reaction (PCR)
c) translation
d) microarrays
Answer: a
Explanation: The process of amplification of specific DNA sequences by an enzymatic process is termed as Polymerase Chain Reaction (PCR). For the PCR to take place there should be small sequences at each end which should be known.

2. What are primers?
a) Primers are the short sequences at the end of the nucleotide sequences which are used for amplification
b) Primers are the short sequences which are complementary to the nucleotides at the end of the sequence which is to be amplified
c) Primers are the short sequences present anywhere in the nucleotide sequence to be amplified
d) Primers are the short sequences which are complementary to the nucleotides anywhere in the sequence to be amplified
Answer: b
Explanation: Primers are short nucleotide sequences which are complementary to the stretches at the ends of the DNA sequence to be amplified.

3. A reaction mixture for PCR consists of ____________
a) heat unstable polymerase
b) primers in a limited amount
c) deoxynucleoside triphosphate (dNTPs)
d) a region complementary to the sequence to be amplified
Answer: c
Explanation: A reaction mixture for PCR consists of the region of the DNA sequence to be amplified, primers in large molar excess, heat stable polymerase and dNTPs. Heat stable polymerase which is used commonly is Taq polymerase.

4. Which of the following is a characteristic of Taq polymerase?
a) It is an RNA polymerase
b) It is heat stable
c) It is obtained from thermophilic bacterium and can be grown in the laboratory below a temperature of 75 degrees
d) It is used in cellular synthesis processes and the optimum temperature is at least 90 degrees
Answer: b
Explanation: Taq polymerase is a DNA polymerase obtained from thermophilic bacterium Thermus aquatics. It is heat stable and can be grown in the laboratory at a temperature above 75 degrees. It is used in cellular DNA synthesis processes and the optimum temperature is at least 80 degrees. It is not readily denatured by repeated cooling and heating cycles and thus is used in amplification processes.

5. These are steps taken in carrying out the PCR reaction:
i) Attaching of primers by cooling
ii) Denaturation of strands
iii) DNA synthesis
iv) Heating
Which is the correct order?
(Mentioned from starting to ending the reaction)
a) i)-ii)-iii)-iv)
b) ii)-i)-iii)-iv)
c) iv)-iii)-ii)-i)
d) iv)-ii)-i)-iii)
Answer: d
Explanation: PCR consists of a series of steps. Firstly, the reaction mixture is heated so that the strands are separated i.e. their denaturation takes place. Then it is again cooled so that the primers are able to attach. Once the primers are attached, the synthesis of DNA is allowed. This whole process is repeated.

6. Which of the following is not a condition for PCR?
a) Initial melting carried out for 5 minutes at 94 degrees
b) Initial melting followed by 30 cycles each consisting of melting for 1 minute at 94 degrees
c) Renaturation for 5 minutes at 60 degrees
d) DNA synthesis at 72 degrees for 1.5 minutes
Answer: c
Explanation: The melting temperature is 94 degrees and the initial melting is carried for 5 minutes at this temperature. It is followed by 30 cycles each consisting of melting for 1 minute at 94 degrees. The renaturation is carried for 1 minute at 60 degrees. The DNA synthesis is carried out at 72 degrees for 1.5 minutes. After the 30 cycles, a final round of extension is carried out for 10 minutes.

7. Primers and polymerases are added again during the reaction because they get consumed as the reaction proceeds.
a) True
b) False
Answer: b
Explanation: Primers and polymerases are not added more as the reaction proceeds. It is so because the polymerase is heat stable and is not destroyed during the reaction. Primers on the other hand are added in excess at the beginning of the reaction.

8. All the molecules generated during PCR will not be full length. Some will also be of intermediate length. Which of the statements is correct?
a) After first cycle, majority of the molecules will be full length and only some will be of intermediate length
b) In the next cycle, each intermediate molecule will generate one intermediate molecule and one target molecule
c) The number of full length molecules increase as number of cycles proceed
d) The number of intermediate molecules increase geometrically and the number of target molecules increase arithmetically
Answer: b
Explanation: Intermediate molecules are those which would be having primer at one end and the other end won’t be defined. After the first cycle, half of the molecules would be full length and half would be of intermediate length. In the next cycle, each intermediate molecule generates one intermediate molecule and one target molecule. Target molecule is that which is defined by primers at both the ends. The number of full length molecules remains constant. But, target molecules increase geometrically and the intermediate molecules increase arithmetically.

9. Which of the following activity is not present in Taq polymerase?
a) 5’-3’ polymerase
b) 5’-3’ exonuclease
c) 3’-5’ exonuclease
d) Both 5’-3’ polymerase and 5’-3’ exonuclease
Answer: c
Explanation: Taq polymerase is not having 3’-5’ exonuclease activity. It is a proof reading activity and it is very important to have a check on the mutations if they are encountered in the PCR products.

10. Choose the correct statement.
a) Taq polymerase is having high processivity
b) Processivity is defined in this case as a synthesis of DNA by polymerase
c) It requires a 5’ end for the elongation to take place
d) The maximum size of the molecules which can be synthesized is 10kbp
Answer: b
Explanation: Taq polymerase is having low processivity. It means that it falls off from the template before it has synthesized a large piece of DNA. It requires a 3’ OH for carrying out the elongation. The maximum size of the molecules which can be synthesized is 2-4 kbp.

11. What is the half life cycle for Taq polymerase?
a) 40 minutes
b) 80 minutes
c) 10 minutes
d) 50 minutes
Answer: a
Explanation: The half life cycle for Taq polymerase is 40 minutes at 94 degrees. Thus, at the end of the 30 cycles, a significant loss of activity takes place.

12. Taq polymerase incorporates which residue at 3’ end?
a) G
b) T
c) A
d) C
Answer: c
Explanation: It incorporates A residue at the 3’ end. This overhang is often useful in carrying out the cloning of these PCR products.

13. Polymerases are also available from other Thermus species. Which of the following is correct?
a) Thermus flavus gives Tfl enzyme
b) Thermus thermopilus gives Tfl enzyme
c) They are having proof reading activity
d) Thermus flavus gives Tth enzyme
Answer: a
Explanation: Other Thermus species also provide polymerases. Thermus flavus gives Tfl enzyme and Thermus thermopilus gives Tth enzyme. They are also 3’-5’ proof-reading activity.

14. Polymerases are available with proof reading activity. Which of the following are the characteristics of these types of polymerases?
a) They add A residue at 3’ end
b) They are obtained from Thermococcus litoralis
c) They can’t be obtained from archaebacteria
d) The marine bacteria from which they are obtained grow at temperatures lower than that of Thermus aquatics
Answer: b
Explanation: As these polymerases are having a proof-reading activity, they generally don’t add any residue at the end. They are obtained from bacteria such as Thermococcus litoralis and grow at temperatures above than that of Thermus aquatics. They can also be obtained from archaebacteria.

15. Thermococcus litoralis grows at a temperature upto 98 degrees.
a) True
b) False
Answer: a
Explanation: Thermococcus litoralis grows at a temperature upto 98 degrees. The half life is also high, 90% of its activity is retained after 1 hour of incubation at 95 degrees.