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Multiple choice question for engineering

Set 1

1. Bisulphite ions are used to deaminate ________ residues in _______ DNA.
a) C, double stranded
b) C, single stranded
c) U, double stranded
d) U, single stranded

View Answer

Answer: b [Reason:] Bisulphite ions are used to deaminate C residues in single stranded DNA. These C residues on deamination give U residues.

2. For mutagenesis without PCR, which of the following can be used as a template?
a) Single stranded DNA
b) Double stranded DNA
c) Circular DNA
d) Both single and double stranded DNA

View Answer

Answer: d [Reason:] If mutagenesis is carried without PCR, both single and double stranded DNA can be used as a template. The single stranded DNA is obtained from filamentous phages and double stranded DNA is obtained from conventional vectors.

3. An oligonucleotide is synthesized which contains the mutation and the rest is _______ to the template DNA.
a) complementary
b) non-complementary
c) can either be complementary or non-complementary
d) not related

View Answer

Answer: a [Reason:] The oligonucleotide is synthesized which contains the mutation and the rest of the sequence is complementary to the template DNA. And the oligonucleotide is allowed to anneal to the template DNA.

4. If the template is double stranded, they need to be separated before annealing of oligonucleotide. Is the given statement true or false?
a) True
b) False

View Answer

Answer: a [Reason:] If the template is double stranded, then the two strands are separated before annealing of the oligonucleotide. This separation is done by heating or alkali denaturation.

5. Which of the following statement is incorrect for synthesis of second strand?
a) The oligonucleotide is acting as a primer for the synthesis of second strand
b) DNA polymerase and dNTPs are added for synthesis
c) The polymerase should have 5’-3’ exonuclease activity
d) A polymerase having 5’-3’ exonuclease activity would degrade the primer that carries the mutant sequence

View Answer

Answer: c [Reason:] The oligonucleotide is acting as a primer for synthesis of second strand. DNA polymerase and dNTPs are added for synthesis and the polymerase should not have 5’-3’ exonuclease activity. If the polymerase is having exonuclease activity it would degrade the primer that carries the mutant sequence and the mutated portion is replaced by template DNA.

6. Once the double stranded molecule with mutation is introduced into E. coli for replication, how many types of molecules are produced?
a) 1
b) 2
c) 3
d) 4

View Answer

Answer: b [Reason:] Once the double stranded molecule with mutation is introduced into E. coli for replication, there are basically two types of molecules produced after replication. One is the wild type molecule and the other is having mutated sequence.

7. How many sites can be mutated at a time?
a) 1
b) 2
c) 3
d) Many

View Answer

Answer: d [Reason:] Many sites can be mutated at a time. For inducing more than one mutation, more than one mismatch are there in oligonucleotide to the target sequence.

8. Several different mutations can be induced at one site. Is the given statement true or false?
a) True
b) False

View Answer

Answer: a [Reason:] Several mutations can be induced at one site. This is done by using a mixture of oligonucleotides which are having different nucleotides at the same position.

9. If mixed oligonucleotides are used, it is regarded as:
a) mixed mutagenesis
b) multiple mutagenesis
c) cassette mutagenesis
d) polymutagenesis

View Answer

Answer: c [Reason:] Mixed oligomucleotides is the collection of those oligonucleotides which are having different nucleotides at the same position. This type of mutagenesis is called as cassette mutagenesis. It is a fast method for inducing multiple mutations.

10. It is easier to subclone a restriction fragment if it belongs to?
a) small gene
b) large gene
c) prokaryotic organism
d) eukaryotic organism

View Answer

Answer: b [Reason:] If the target sequence belongs to a large gene, it is easier to sub-clone a restriction fragment from it and mutate the fragment. Chances of having unwanted mutations are also reduced.

Set 2

1. The process of finding a particular member of library which is having some defined properties is called as:
a) searching
b) screening
c) locating
d) narrowing

View Answer

Answer: b [Reason:] The process of finding a particular member of library which is having some defined properties is called as screening. Mostly screening is carried out for a particular coding sequence.

2. If for a particular organism sequence data is available and we have to simply search in the data through computer, then this method is called as:
a) annotation
b) database search
c) in silico
d) electronic search

View Answer

Answer: c [Reason:] The easiest way to find for a particular sequence is to exploit the genomic analyses. If the database is available for a particular sequence and the task is to simply search the data through the computer then the procedure is called as in silico.

3. If we are having sequence data for a particular organism, but screening is carried out for homologues the program used is BLAST. Is the given statement true or false?
a) True
b) False

View Answer

Answer: a [Reason:] Homologues are group of organisms which contain similar sequences. Screening for such organisms can be carried out by using programs such as BLAST (Basic Local Alignment Search Tool). Search can be carried out within or for nucleotide and protein sequences.

4. If screening is carried out on the basis of sequences which are related to the desired sequence, then the process is called as:
a) in-silico
b) homologue search
c) annotation
d) partial search

View Answer

Answer: c [Reason:] If the search is carried out on the basis of related sequence rather than looking for the actual sequence then the process is termed as annotation.

5. How many techniques are there for carrying out the screening of sequences encoding for RNAs?
a) 1
b) 2
c) 3
d) 4

View Answer

Answer: b [Reason:] There are basically two techniques used for carrying out the screening of sequences encoding for RNAs. The first technique is based on nucleic acid hybridization and the second is based on coding function in vivo.

6. Choose the incorrect statement for colony or plaque lift.
a) It is the base for screening based on nucleic acid hybridization
b) It is also known as Grunstein-Hogness technique
c) It is based on the fact that which members of the library have the same sequence as the DNA probe
d) The library may be plaques on a bacterial lawn and in that case it is known as plaque lift

View Answer

Answer: c [Reason:] Colony lift or plaque lift is the basis for nucleic acid hybridization. It is also known as Grunstein-Hogness technique named after the discoverer. It is based on the fact that which members of the library are complementary to the sequence of DNA probe can be known. The library may be plaques on a bacterial lawn and in that case it is known as plaque lift. There may be colonies on a plate and in that case it is known as colony lift.

7. The screening of libraries us carried out by nucleic acid hybridisation and constitutes of following steps:
i) Peeling of membranes carrying away bacterial cells with it
ii) Cells are lysed and denaturing of DNA is being carried out
iii) Hybridization with labelled DNA
iv) Placing the membrane onto plate containing recombinant cells
Choose the correct sequence in which the steps are carried out (starting to ending).
a) iv)-i)-ii)-iii)
b) iv)-ii)-i)-iii)
c) i)-ii)-iii)-iv)
d) iii)-ii)-i)-iv)

View Answer

Answer: a [Reason:] Firstly, the membrane is placed onto the plate containing recombinant cells. Then the membrane is peeled off and the bacterial cells are carried away with it. After this, cells are lysed and denaturing of DNA is done. It is followed by hybridization with labelled DNA and then visualization is done.

8. Choose the incorrect statement with respect to the membranes used for adhering of bacterial cells onto them.
a) Nitrocellulose membranes were preferred earlier
b) They bind DNA very efficiently
c) They can be handled easily without breakage
d) They are inflammable

View Answer

Answer: c [Reason:] Nitrocellulose membranes were preferred earlier. They bind DNA very efficiently and are brittle thus can’t be handled efficiently without breaking. They are inflammable.

9. Nitrocellulose membranes are less sensitive than nylon membranes. Is the given statement true or false?
a) True
b) False

View Answer

Answer: b [Reason:] Nylon membranes are less sensitive than nitrocellulose membranes. Though nylon membranes are used more often today but nitrocellulose membranes are more sensitive. To make nylon membranes more sensitive, they are derivatized.

10. What is used for lysing of bacterial cells and denaturation of DNA?
a) Exonuclease
b) Sulphuric Acid
c) Sodium Hydroxide
d) Heat

View Answer

Answer: c [Reason:] The bacterial cells are lysed by the use of sodium hydroxide. The phage proteins and DNA are also denatured by the use of sodium hydroxide.

Set 3

1. Transgenic Drosophila can be created by microinjection of DNA into the embryos. These embryos are at which stage of development?
a) One-cell stage
b) Pre-blastoderm stage
c) Blastoderm
d) Morula

View Answer

Answer: b [Reason:] Transgenic Drosophila can be created by microinjection of DNA into the embryos. The embryo should be at pre-blastodermal stage. Other developmental stages such as blastoderm and morula follow the pre-blastodermal stage.

2. Syncytium is a layer of ______ that have not been separated into individual cells.
a) nuclei
b) mitochondria
c) cytoplasm
d) nuclei and cytoplasm

View Answer

Answer: a [Reason:] Syncytium is a layer of nuclei which is not divided into separate individual cells. This is found at pre-blastodermal stage of development.

3. In Drosophila, only nuclei cells give rise to germline cells. Is the given statement true or false?
a) True
b) False

View Answer

Answer: a [Reason:] In Drosophila, only nuclei cells give rise to germline cells. These germline cells are located at one end of the embryo and acquire DNA stably.

4. ______ integration systems are used for transfer of DNA in Drosophila and it is composed of ______
a) Artificial, P elements
b) Artificial, S elements
c) Natural, P elements
d) Natural, S elements

View Answer

Answer: c [Reason:] In Drosophila, the natural integration systems are used for transfer of DNA and it is composed of transposable elements. These transposable elements are known as P elements.

5. Only the _____ part of Drosophila is transgenic and the rest is not. This is known as ______
a) germline cell, mosaic
b) nurse cells, mosaic
c) germline cells, hybrid
d) nurse cells, hybrid

View Answer

Answer: a [Reason:] Only the germline cells portion (but not necesaarily all germline cells) is modified and the rest is non-transgenic. This is collectively known as mosaic.

6. rosy gene is used a selectable marker for transformation in Drosophila, it produces an enzyme required for the synthesis of:
a) wing pigment
b) eye pigment
c) both eye and wing pigment
d) thorax pigment

View Answer

Answer: b [Reason:] rosy gene is used as a selectable marker for transformation in Drosophila, it produces enzyme xanthine dehydrogenase and it is required for pigments of eye.

7. Wild type flies have crimson red eyes whereas rosy mutant have brick-red eyes. Is the given statement true or false?
a) True
b) False

View Answer

Answer: b [Reason:] Wild type genes have brick-red eyes whereas rosy mutant have crimson red eyes. Thus, rosy genes can be used as selectable marker.

8. Wild-type Drosophila flies are _____ to ethanol supplied in food.
a) resistant
b) non-resistant
c) resistant at low concentration and non-resistant at higher concentration
d) resistant at higher concentrations and non-resistant at lower concentration

View Answer

Answer: a [Reason:] The wild type flies are resistant to ethanol supplied in the food. This property can be exploited as a selectable marker and which is in the form of alcohol dehydrogenase gene.

9. Cis-acting sites should be present in the vector for ________
a) replication
b) selecting recombinants by acting as a marker
c) transposition
d) providing high copy number

View Answer

Answer: c [Reason:] Apart from selectable markers, cis acting sites should also be present. These are required for transposition.

10. The host should ____ P elements, these elements lead to _______
a) not have, instability
b) have, stability
c) not have, increase the time taken for integration
d) have, reduces the time taken for integration

View Answer

Answer: a [Reason:] The host should not have P elements, these elements lead to instability. It is so because they are responsible for transposition.

11. Where do P elements integrate in the genome?
a) At specifically defined sites
b) Randomly
c) Only at the ends of the genome
d) Integration depends on reaction conditions

View Answer

Answer: b [Reason:] P elements integrate in the genome randomly. This makes it non-feasible to be used directly for gene disruption.

12. Transient gene silencing can be carried out by microinjecting ______
a) single stranded RNA
b) double stranded DNA
c) double stranded RNA
d) either double stranded DNA or RNA

View Answer

Answer: c [Reason:] Transient gene silencing is carried out by microinjecting double stranded RNA. Long term gene silencing can be carried out by synthesizing inverted copy of the target gene.

Set 4

1. Enzyme commonly used for carrying out ligation reaction is:
a) Transferase
b) Reverse transciptase
c) Ligase
d) DNase

View Answer

Answer: c [Reason:] Ligase is the enzyme which is commonly used for carrying out the ligation reaction. They can either be obtained from E. coli or from cells that have been infected by virus.

2. Which of the following statements is correct with respect to T4 DNA ligase?
a) It can carry out only blunt ended ligations
b) It doesn’t requires ATP
c) It requires a phosphate group at 3’ end and a hydroxyl group at 5’ end for the molecule to be joined
d) It is obtained from T4 bacteriophage upon infection by E. coli

View Answer

Answer: d [Reason:] T4 DNA ligase is obtained from T4 bacteriophage upon infection by E. coli. It carries out ligation both in blunt ended and sticky ended molecules and requires ATP. It requires a phosphate group at 5’ end and a hydroxyl group at 3’ end.

3. If blunt ended ligations are to be carried out. Which of the following enzymes can be used?
a) E. coli DNA ligase
b) T4 DNA ligase
c) Both of these enzymes act equally in carrying out blunt ended ligations
d) None of them is able to carry out blunt ended ligations

View Answer

Answer: b [Reason:] If blunt ended ligations are to be carried out, T4 DNA ligase should be used. E. coli DNA ligase is unable to carry out blunt ended ligation.

4. E. coli DNA ligase doesn’t requires a cofactor. The statement is true or false?
a) True
b) False

View Answer

Answer: b [Reason:] The E. coli DNA ligase requires a cofactor nicotamine adenine dinucleotide (NAD), along with 5’ phosphate group and 3’ hydroxyl group.

5. Mechanism of ligation for both T4 DNA ligase and E. coli DNA ligase makes use of Adenosine Monophosphate (AMP). Which of the steps is involved in the ligation mechanism?
a) AMP is added to the 5’ phosphate of one of the DNA molecule
b) It leads to liberation of pyrophosphate from NAD and nicotinamide mononucleotide from ATP
c) The AMP is further displaced by an electrophilic attack
d) The AMP is further displaced by nucleophilic attack by 3’ hydroxyl of the same DNA molecule

View Answer

Answer: a [Reason:] AMP is added to the 5’ phosphate of one of the DNA molecule. It leads to liberation of pyrophosphate from ATP and nicotanamide mononucleotide from NAD. It is further displaced by nucleophilic attack by 3’ hydroxyl of the other DNA molecule.

6. Topoisomerase is also an enzyme which is used for carrying out ligation. The correct statement for topoisomerase is?
a) They act only on double stranded molecules
b) They alter the degree of supercoiling of DNA molecules
c) They are less effective than conventional DNA ligase
d) There are three types of topoisomerases

View Answer

Answer: b [Reason:] Topoisomerases alter the degree of supercoiling of both single and double stranded DNA molecules. There are two types of topoisomerases, I and II. I is responsible for altering single stranded molecules and type II for altering double stranded molecules. They are more effective than conventional DNA ligase.

7. Mobile genetic elements can be transferred from one DNA portion to another. The enzyme carrying out this is:
a) Ligase
b) Transciptase
c) Transposase
d) Endonuclease

View Answer

Answer: c [Reason:] Transposase is used for transferring mobile genetic elements from one portion to another. They are useful in inserting origin of replication or anti-biotic resistance genes.

8. Phage based recombination systems are used for:
a) cleaving the molecules at specific sites
b) adding the molecules at specific sites
c) breakage and rejoining the molecules at specific sites
d) breakage at random sites

View Answer

Answer: c [Reason:] Phage based recombination systems are used for breaking and rejoining the moleculesat specific sites and not at random sites.

9. Bacteriophage lambda is having a phage recombination system. Following are the characteristics of this system:
a) It is used for inserting phage genome into the bacterium
b) It is used for inserting bacterial genome into the phage
c) The specific site in bacteria is attB and that in phage is attP
d) The specific sites in both of them are called as attP

View Answer

Answer: a [Reason:] The phage genome is inserted into the bacterium. The site for insertion at the phage genome is attP and that in the case of bacterium is attB.

10. Ligation enzymes are used for ligating newly synthesized okazaki fragments. What holds true for okazaki fragments?
a) Okazaki fragments are short fragments of DNA formed on leading strand
b) Okazaki fragments are large fragments of DNA formed on lagging strand
c) Okazaki fragments are short fragments of DNA formed on lagging strand
d) Okazaki fragments are large fragments of DNA formed on leading strand

View Answer

Answer: c [Reason:] Okazaki fragments are short DNA fragments on the lagging template strand during replication. They are complementary to the lagging strand.

Set 5

1. Which of the following statements is correct with respect to exonuclease?
a) They only act on single stranded DNA molecules
b) They only act on double stranded DNA molecules
c) They remove a single nucleotide base at a time
d) They remove nucleotide bases from the middle of polynucleotide chain

View Answer

Answer: c [Reason:] Exonuclease is responsible for removal of a single nucleotide base at a time, from the end of the polynucleotide sequence. They can act on both single and double stranded DNA molecules.

2. How many approaches are there which can be used for exonucleolytic activity in double stranded DNA molecules?
a) 1
b) 2
c) 3
d) 4

View Answer

Answer: b [Reason:] There are 2 approaches which can be used to carry out exonucleolytic activity in double stranded DNA. The two approaches which are used are separation of both the strands together and separation of both the strands separately.

3. What is the mode of action of exonuclease III?
a) Exonuclease III acts on single stranded DNA in 3’-5’ direction
b) Exonuclease III acts on double stranded DNA in 5’-3’ direction
c) Exonuclease III acts on single stranded DNA in 5’-3’ direction
d) Exonuclease III acts on double stranded DNA in 3’-5’ direction

View Answer

Answer: d [Reason:] It is an enzyme which is having exonuclease activity in 3’-5’ direction and only on double stranded DNA. It doesn’t acts on single stranded molecules.

4. Which of the following statements is correct regarding S1 nuclease?
a) It acts on double stranded DNA
b) It acts on single stranded DNA
c) It acts on both types of strands
d) It is obtained from E. coli

View Answer

Answer: b [Reason:] S1 nuclease is obtained from Aspergillus orzyae and it acts on single stranded DNA molecules. It is having exonucleolytic activity.

5. What happens if a DNA molecule is treated by first Exonuclease III and then followed by treatment with S1 nuclease?
a) The molecule is shortened only from 3’ en.
b) The molecule is shortened only from 5’ end
c) The molecule is shortened from both the ends
d) Only Exonuclease acts and S1 doesn’t acts

View Answer

Answer: c [Reason:] The molecule is shortened from both the ends. It is so because as firstly Exonuclease III acts, single stranded ends are produced on both the sides. These are further acted upon by S1 nuclease and thus the molecule is shortened from both the sides.

6. How can one end be protected from the action of Exonuclease III, so that the molecule is not shortened from both the ends?
a) By using Phosphorothioate nucleotide analogue
b) By making both the ends double stranded in nature
c) By labelling one end with a radioactive compound
d) By increasing the time of exposure of the DNA molecule to the enzyme

View Answer

Answer: a [Reason:] Phosphorothioate nucleotide analogues are used to replace some of the nucleotides at the ends with it. It doesn’t allows the action of the enzyme on it and thus the molecule won’t be shortened from that end on which replacement is done.

7. Bal31 is also an enzyme which is used. Which of the following statements hold true in its context?
a) It is having only 3’-5’ exonuclease activity and no endonuclease activity
b) It is having only an endonuclease activity
c) It is having 5’-3’ exonuclease activity
d) It leads to shortening from both the ends

View Answer

Answer: d [Reason:] Bal31 is having an exonuclease activity in 3’-5’ direction and also an endonuclease activity. Thus, after having 3’-5’ exonuclease activity, the endonuclease activity takes place and the molecule is shortened from both the ends.

8. The extent of deletions can be manipulated by controlling which of them?
a) Time of incubation only
b) Amount of nuclease added
c) They both have a role to play in the extent of deletion
d) The amount of deletion which can be carried out for a particular amount of DNA is fixed regardless of the amount of nuclease and incubation time

View Answer

Answer: c [Reason:] The extent of deletion can be varied by varying the incubation time and the amount of nuclease added. Thus, they both play a role in controlling the extent of deletion.

9. What is the function of methylase?
a) Addition of methyl groups to DNA
b) Removal of methyl groups from DNA
c) Both in removal and addition of methyl groups from DNA
d) It is used in production of methane gas

View Answer

Answer: a [Reason:] Methylase is used for addition of methyl groups on DNA. It is done via placing the methyl groups which are taken from S-adenosyl methionine.

10. Methylase is useful in cloning experiments. True or false?
a) True
b) False

View Answer

Answer: a [Reason:] It is useful in cloning experiments because as it methylates DNA, that DNA is protected from that group of restriction enzyme. It means that if we carry out methylation by EcoRI methylase, the DNA won’t be cleaved by the restriction enzyme EcoRI. In cloning experiments, sometimes it is necessary to protect the DNA from cleavage by a particular enzyme.