Multiple choice question for engineering
Set 1
1. How many types of reporter gene constructs are there?
a) 1
b) 2
c) 3
d) 4
Answer
Answer: b [Reason:] There are mainly two constructs for the reporter gene. In the first construct, reporter gene completely replaces the target gene. In the second construct, there is a fusion between the target gene and the reporter gene.
2. Desired changes are made in the structure of the promoter and is ligated onto reporter gene. The alteration can be known by measuring the activity. Is the given statement true or false?
a) True
b) False
Answer
Answer: a [Reason:] Desired changes are made in the structure of the promoter in order to know about the region of expression.
3. ______ fluorescent protein should be used for determination of _______ of subcellular components.
a) One, age
b) One, location
c) More than one, location
d) More than one, age
Answer
Answer: c [Reason:] Fluorescent protein is used for location of subcellular components. More than one fluorescent protein should be used for exact determination of the location of the subcellular components.
4. Sequences can be cloned flanking the reporter genes using the transposon tagging approach. The use of reporter genes to identify sequences expressed in this way is called as:
a) trapping
b) tagging
c) identification
d) blotting
Answer
Answer: a [Reason:] Sequences can be cloned flanking the reporter genes using the transposon tagging approach and if sequences are identified in this way, then it is called as trapping.
5. If detection of enhancers is done it is called as ______ and for detection of transcribed sequences it is called as ______
a) enhancer trapping, promoter trapping
b) enhancer trapping, gene trapping
c) promoter trapping, gene trapping
d) promoter trapping, enhancer trapping
Answer
Answer: b [Reason:] Trapping can be classified into different types by using different types of constructs. If enhancer is detected it is called as enhancer trapping. For promoter it is called as promoter trapping and for transcribed sequences it is called as gene trapping.
6. The enhancer trap system will _____ in the ______
a) not replicate independently, host
b) replicate independently, host
c) not replicate independently, vector
d) replicate independently, vector
Answer
Answer: a [Reason:] The enhancer trap system will not replicate independently in the host. But the detection can be done by using a selectable marker.
7. For gene trap vector system, reporter gene and _____ are present. Reporter gene is _____
a) selectable marker, having promoter
b) selectable marker, promoterless
c) helper phage, promoterless
d) helper phage, having promoter
Answer
Answer: b [Reason:] For gene trap vector system, reporter gene and a selectable marker are present. The reporter gene is promoterless.
8. The reporter gene in enhancer trap system is preceded by ______
a) intron
b) exon
c) promoter
d) origin of replication
Answer
Answer: a [Reason:] The reporter gene in the enhancer trap system is promoterless and it is preceded by introns.
9. Insertion of vector into intron gives rise to ________. It contains a part coming from intron and a part from ______
a) hybrid intron, exon
b) hybrid vector, exon
c) hybrid intron, target DNA sequence
d) hybrid vector, target DNA sequence
Answer
Answer: c [Reason:] Insertion of vector into intron gives rise to hybrid intron. It contains a part coming from intron and a part from target DNA sequence.
10. The hybrid intron is spliced from the transcript using the splice acceptor site in the vector. Is the given statement true or false?
a) True
b) False
Answer
Answer: a [Reason:] The hybrid intron is spiced from the transcript using the splice acceptor site in the vector and then the reporter sequence is fused next to the target exon.
Set 2
1. If the embryo is at one-cell stage then it is found in:
a) ovary
b) oviduct
c) uterus
d) either ovary or uterus
Answer
Answer: b [Reason:] When embryo is at one-cell stage it is found in oviduct. This is important for generation of whole organisms that are transgenic. Isolation of one-cell embryo is done and then it is micro-injected into the pro-nucleus.
2. Embryonic stem cells are also used for generation of transgenic organisms. They are obtained from _______ of a developing ________
a) trophoectoderm, gastrula
b) trophoectoderm, blastula
c) inner cell mass, blastula
d) inner cell mass, gastrula
Answer
Answer: c [Reason:] Embryonic stem cells are also used for generation of transgenic organisms. They are obtained from inner cell mass of a developing blastula. A developing blastula composes of inner cell mass and it is surrounded by trophoectoderm.
3. Embryonic stem cells (ES) are isolated and are injected again into the blastocoel of a developing embryo. The embryo which develops is entirely made up of these cells only. Is the given statement true or false?
a) True
b) False
Answer
Answer: b [Reason:] ES cells are isolated and are injected into the blastocoel of the developing embryo. But the embryo is not composed wholly of ES cells and thus it is said to be chimeric.
4. ES cells are used in order to ensure that insertion is done at required chromosomal location and it is called as:
a) gene targeting
b) knocking out
c) knocking in
d) gene disruption
Answer
Answer: a [Reason:] ES cells are used in order to ensure that insertion is done at required chromosomal location and the process is termed as gene targeting. In some cases it is not necessary to ensure that integration is taking place at a normal chromosomal location.
5. If a gene is inactivated by gene targeting then it is called as:
a) knock-in gene
b) knock-out gene
c) gene disruption
d) insertional inactivation
Answer
Answer: b [Reason:] If a gene is inactivated by gene targeting then it is called as knock-out gene. If a gene is replaced by some other gene then it is called as knocking in.
6. Integration events may be insertional involving _____ crossover or replacement involving _____ crossovers.
a) single, single
b) double, double
c) double, single
d) single, double
Answer
Answer: d [Reason:] Integration can be done by either insertion or replacement. Insertion is carried out by using single crossover and replacement is carried out by double crossover.
7. It is necessary to analyse the genome of transgenic cells to confirm that the site of integration is correct. This is carried out by southern blotting. Is the given statement true or false?
a) True
b) False
Answer
Answer: a [Reason:] It is necessary to analyse the genome of transgenic cells to confirm that the site of integration is correct and it is carried out by southern blotting. DNA from cultures derived from individual transformed cells is used for screening.
8. The gene targeting approach produces individuals which are ______ for inactivation of gene.
a) homozygous
b) heterozygous
c) either only homozygous or only heterozygous
d) both heterozygous and homozygous
Answer
Answer: b [Reason:] The gene targeting approach produces individuals which are heterozygous for inactivation of gene. But it is necessary to generate homozygotes and they are produced by crossing heterozygous individuals and then screening is carried out.
9. If controlled inactivation of gene is carried out and some of the consequences when inactivation of a target gene is deleterious are avoided. It is referred as:
a) specialized gene targeting
b) controlled gene targeting
c) conditional gene targeting
d) specific gene targeting
Answer
Answer: c [Reason:] Conditional gene targeting is that where controlled inactivation is carried out. If inactivation leads to deleterious effects then they are avoided.
10. For carrying out gene manipulation, use of cultured cells is _____ transgenic organisms.
a) less reliable
b) more reliable
c) may be less or more reliable
d) is same reliable as
Answer
Answer: a [Reason:] For carrying out gene manipulation, use of cultured cells is less reliable than that of transgenic organisms. Thus, these transgenic organisms are used greatly.
Set 3
1. Libraries can broadly be classified into how many types?
a) 1
b) 2
c) 3
d) 4
Answer
Answer: b [Reason:] Libraries can be classified basically into two types- genomic libraries and cDNA libraries. Genomic libraries are constructed from the whole genome of the organism. cDNA libraries are constructed from the DNA copies of RNA sequences.
2. Choose the correct statement for genomic libraries.
a) Genomic libraries include the representation of the whole genome of the organism
b) Sequences such as telomeres are also represented
c) Telomeres can be readily cloned
d) The larger the size of insert of genomic DNA in recombinants, the more is the number of recombinants required to represent the genome in the library
Answer
Answer: a [Reason:] Genomic libraries include the representation of whole organism. Sequences such as telomeres can’t be represented in the genomic libraries because they can’t be readily cloned. The larger size of insert of genomic DNA in recombinants, less number of recombinants is required to represent the genome in the library.
3. Choose the incorrect statement for preparation of genomic libraries.
a) The first step is isolation of genomic DNA
b) Physical damage to the DNA should be avoided
c) If a nuclear DNA library is to be constructed, organelle DNA is to be removed
d) For the construction of organelle library, organelle DNA is purified from the nuclear DNA
Answer
Answer: c [Reason:] For the preparation of genomic libraries, the first step is isolation of genomic DNA. Physical damage to the DNA should be avoided and DNA of high molecular weight should be obtained. For the construction of nuclear DNA library, organelle DNA (DNA which is included in mitochondria and chloroplast), is not removed. It is so because they are very less in amount in comparison to nuclear DNA. For the construction of organelle library, organelle DNA is purified from the nuclear DNA.
4. The various steps for construction of libraries are:
i) Fragmentation of DNA
ii) Isolation of genomic DNA
iii) Amplification
iv) Ligation and introduction to the host
v) Vector preparation
The correct order of construction of libraries is (In the order of starting to ending) :
a) i)-ii)-iii)-iv)-v)
b) ii)-i)-v)-iv)-iii)
c) ii)-v)-i)-iv)-iii)
d) v)-ii)-i)-iii)-iv)
Answer
Answer: b [Reason:] Firstly, the isolation of genomic DNA is carried out. It is followed by the fragmentation of DNA. Then the vector preparation is done, which is followed by the ligation and introduction to the host. The final step is the amplification of the recombinants.
5. Complete digestion is preferred over partial digestion for the fragmentation of DNA. The given statement is true or false?
a) True
b) False
Answer
Answer: b [Reason:] Partial digestion is preferred over complete digestion. It is so because if complete digestion is carried out, it is not necessary that all the fragments are represented. Thus partial digestion is preferred over complete digestion.
6. To avoid ligation of separate DNA fragments, which of the enzyme is used?
a) phosphatase
b) kinase
c) ligase
d) endonuclease
Answer
Answer: a [Reason:] To avoid ligation of separate DNA fragments, phosphatase is used. By the use of phosphatase, terminal phosphate groups are removed and thus ligation is not done. Joining of separate fragments is avoided because it would lead to formation of non-contiguous DNA.
7. Vector and insert are mixed, ligated and packaged and introduced into the host by:
a) transformation
b) transduction
c) infection
d) transformation and infection both
Answer
Answer: d [Reason:] Vector and insert are mixed, ligated and packaged into the host by transformation and infection both.
8. Libraries using phage cloning vectors are often kept as:
a) unpackaged phage
b) packaged phage
c) both packaged and unpackaged phage
d) both packaged and unpackaged phage are used but packaged is favoured
Answer
Answer: b [Reason:] Libraries using phage cloning vectors are often kept as packaged phage. This can be plated out on appropriate hosts when needed.
9. Libraries constructed in plasmid vectors can be kept as:
a) plasmid containing cells
b) naked DNA
c) both plasmid containing cells and naked DNA
d) naked DNA is preferred over plasmid containing cells
Answer
Answer: c [Reason:] Libraries constructed in plasmid vectors can be kept as both plasmid containing cells and naked DNA. The naked DNA can be transformed into the host as required.
10. During storage there are chances of degradation of DNA. Larger fragments are having more chances of being got lost. The given statement is true or false?
a) True
b) False
Answer
Answer: a [Reason:] During storage there are chances of degradation of DNA and the larger fragments are at greater risk of being lost. And thus average insert size will gradually fall.
Set 4
1. Choose the correct statement for lambda ZAP vector.
a) It is not based on bacteriophage lambda
b) It contains a region that can be excised in vivo
c) The excision leads to the formation of bluescript plasmids and it contains an initiator region only
d) The multiple cloning site is not flanked by the initiator and the terminator region
Answer
Answer: b [Reason:] Lambda ZAP vector is based on the bacteriophage lambda. It contains a region that can be excised in vivo and leads to the formation of bluescript plasmids. This consists of multiple cloning site which is flanked by the initiator and the terminator region.
2. The initiator is recognized by which gene?
a) Gene I
b) Gene I and II
c) Only gene II
d) Gene III
Answer
Answer: c [Reason:] The f1 initiator and terminator region flanks the multiple cloning site and the initiator region is recognized by gene II.
3. Choose the incorrect statement for the replication process.
a) The initiator site is nicked and replication of one strand is started
b) Replication takes place in both the directions
c) Replication continues through the bluescript
d) It is stopped at the terminator and then again a nick is made
Answer
Answer: b [Reason:] The initiator site is nicked and the replication of one strand is initiated. Replication takes place in one direction only and continues through the bluescript. It is stopped at the terminator and then a nick is made again there.
4. Which of the following doesn’t takes place after replication?
a) The single stranded sequence is generated
b) It is circularized to form closed single stranded molecule
c) It may circularize or remain linear
d) The double stranded molecule can be synthesized by cellular DNA synthesis
Answer
Answer: c [Reason:] As the replication takes place, a single stranded sequence is generated and it circularizes in vivo in order to form single stranded circular molecule. The double stranded molecule can be synthesized by cellular DNA synthesis.
5. Choose the correct statement for cosmids.
a) It can be regarded as lambda substitution vector
b) Less amount of phage DNA is deleted
c) Only cos packaging sites are left
d) It doesn’t contains a origin of replication
Answer
Answer: c [Reason:] Cosmids can be regarded as lambda replacement vectors. More amount of phage DNA is deleted and only cos packaging sites are left. There are no coat protein genes left. It also contains an origin of replication.
6. Once cosmids are inside the E.coli cells, they don’t generate phage but are propagated as plasmids. Is the given statement true or false.
a) True
b) False
Answer
Answer: a [Reason:] Once cosmids are inside the E. coli cells, they don’t generate more phage but are propagated as plasmids. It is so because no more coat protein genes are present and thus it can’t be packaged. They can’t give rise to plaques.
7. Which of the following is the correct method to check whether the DNA has entered into the cell or not in the case of cosmid?
a) If transformation has taken place turbid plaques are formed
b) If transformation has taken place clear plaques are formed
c) If transformation has taken place, it can be confirmed via ampicillin resistance
d) If transformation has taken place, it can be confirmed if forms plaques and is ampicillin resistant also
Answer
Answer: c [Reason:] In the case of cosmid, transformation can be confirmed via counting the colonies containing ampicillin. As the plaques are not formed by cosmids, it can’t be used as method to detect transformation.
8. Which size of insert is accepted by the cosmids?
a) 10-20 kbp
b) 35-45 kbp
c) 50-60 kbp
d) 100-120 kbp
Answer
Answer: b [Reason:] There is a minimum size limit on the insert which is accepted by the cosmids and it lies in the range of 35-45 kbp. They generally accept large size insert and it is beneficial for construction of genomic libraries.
9. What happens once the cosmid enters the E. coli cells?
a) There is strict size selection inside E. coli cells
b) Partial deletion may take place
c) The tendency of deletion may is increased by using low copy number
d) The tendency of deletion can’t be altered
Answer
Answer: b [Reason:] There is no size selection once the cosmid enters the E. coli cells. Because of it partial deletion may take place. The tendency of deletion can be reduced by using low copy number.
10. If colEI derived origin of replication is replaced by origin of replication of F plasmid, it is called as:
a) phagemid
b) F cosmid
c) plasmid
d) fosmid
Answer
Answer: d [Reason:] If colE1 derived origin of replication is replaced by origin of replication of F plasmid it is known as fosmid. F plasmids are present in a low copy number.
Set 5
1. One of the commonly used markers is URA3 gene. It encodes for ______ biosynthesis.
a) adenine
b) uracil
c) guanine
d) cytosine
Answer
Answer: b [Reason:] URA3 gene is a commonly used selectable marker. It encodes an enzyme for uracil biosynthesis. The enzyme encoded is orotidine-5’-phosphate decarboxylase.
2. CAN1 gene encodes a permease that causes uptake of _________ analogue.
a) toxic guanidine
b) toxic arginine
c) non-toxic guanidine
d) non-toxic arginine
Answer
Answer: b [Reason:] CAN1 gene encodes a permease that causes uptake of toxic arginine analogue. It causes consequent cell death.
3. Loss of SUP4 abolishes canavanine uptake, it is arginine analogue. Is the given statement true or false?
a) True
b) False
Answer
Answer: a [Reason:] Loss of SUP4 abolishes canavanine uptake. It is arginine analogue. It causes canavanine resistance and cells can be readily selected.
4. Phosphoribosyl amino-imidazole carboxylase is a component of _______ biosynthesis pathway.
a) pyrimidine
b) purine
c) both purine and pyrimidine
d) only adenine
Answer
Answer: b [Reason:] Phosphoribosyl amino-imidazole carboxylase is a component of purine biosynthesis pathway. The mutation for this enzyme is used in selection of SUP4 gene.
5. Cells that have lost SUP4 gene acquire _____ pigment and are visibly distinguishable from others.
a) red
b) blue
c) green
d) pink
Answer
Answer: a [Reason:] Cells that have lost SUP4 gene acquire red colour and this makes them visibly distinguishable from others. This is the basis for selection.
6. Cells mutant in _____ gene can be selected by using 5-fluoro orotic acid, which is turned into ____ products by wild type protein.
a) SUP4, toxic
b) SUP4, non-toxic
c) URA3, toxic
d) URA3, non-toxic
Answer
Answer: c [Reason:] Cells mutant in URA3 gene can be selected by using 5-fluoro orotic acid, which is turned into toxic products by wild type protein. URA3 gene is responsible for uracil biosynthesis.
7. The replicating plasmid vectors in yeast can be divided broadly into how many categories?
a) 1
b) 2
c) 3
d) 4
Answer
Answer: b [Reason:] The replicating plasmid vectors in yeast can be broadly divided into two main categories. These are yeast centromeric plasmid and yeast episomal plasmid vectors.
8. Yeast centromeric plasmid (YCp50) contains ampicillin and tetracycline resistant gene. Is the given statement true or false?
a) True
b) False
Answer
Answer: a [Reason:] Yeast centromeric plasmid contains ampicillin and tetracycline resistance gene along with an origin of replication.
9. CEN4 in YCp50 is an example of _______
a) tetracycline resistant gene
b) ampicillin resistant gene
c) chromosomal centromeric sequence
d) autonomously replicating sequence
Answer
Answer: c [Reason:] CEN4 in YCp50is an example of chromosomal centromeric sequence. It allows partitioning of the plasmid in the same way as endogenous chromosomes are partitioned.
10. Yeast episomal plasmids have the following feature?
a) They contain ARS and CEL both
b) They contain ARS but not CEL
c) They contain CEL but not ARS
d) CEL is necessarily present but ARS may or may not be present
Answer
Answer: b [Reason:] Yeast episomal plasmids are those plasmids which contain ARS sequence but CEL may not be present necessarily.
11. Cis-acting REP 3 sequence is present in Yeast episomal plasmids. Choose the statement which holds true for it?
a) It is present along with chromosomal centromeric sequence
b) It is the site of action of proteins that help in partitioning
c) It gives less stability to that of YCs
d) The copy number is low than that of YCs
Answer
Answer: b [Reason:] Cis acting REP3 sequence is present in Yeast Episomal plasmids. It is a substitution for chromosomal centromeric sequence. It is having a similar stability than that of YCs but a higher copy number than that of it.