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Multiple choice question for engineering

Set 1

1. Promoter-probe vectors are used often. Choose the correct statement for these vectors.
a) They are used for identifying sequences that can function as promoter in vivo
b) It contains a reporter gene which contains its promoter along
c) The reporter gene is having a cloning site
d) After insertion of DNA into the cloning site, selection of plasmids is carried out by blue white screening

View Answer

Answer: a [Reason:] Promoter- probe vectors are used for identification of sequences that can function as promoter in vivo. It consists of a reporter gene which is not having its promoter along but other genes such as those for chloramphenicol resistance are present. The reporter gene is preceded by the cloning site and after the insertion of DNA has been done, selection of plasmids is carried out by using chloramphenicol resistance.

2. GFP is one of a marker which is used for screening libraries in hosts other than E. coli. Choose the incorrect statement for GFP.
a) It stands for Green Fluorescent Protein
b) It is obtained from a bio-luminescent jellyfish and produces a protein aequorin which emits blue light
c) The blue light is produced because of binding of sodium ions
d) The absorbed blue light produces green light which can be detected further

View Answer

Answer: c [Reason:] It stands for green fluorescent protein and is obtained from a bio-luminescent jellyfish. The jellyfish produces a protein aequorin which produces a blue light. This blue light is produced because of binding Calcium ions. The absorbed blue light emits green light which is detected further.

3. Luciferase genes are also used at times for detection. Choose the correct statement for them.
a) They are obtained from fire flies only
b) The detection requires provision of substrate which produces light
c) Enzymes such as beta-galactosidase requires substrate X-gluc to produce light
d) Lucifearse genes are preferred over fluorescent proteins

View Answer

Answer: b [Reason:] Luciferase genes are not obtained only from fire flies but also from bio-luminescent bacteria and a sea pansy. The section requires a substrate which produces light. Enzymes such as beta-galactosidase, requires X-gal for the production of blue light. Fluorescent proteins are more preferable over these because they don’t require substrates like luciferase genes.

4. In case of promoter-probe vectors, the same or related species should be as a vector whose DNA is to be screened. Is the given statement true or false?
a) True
b) False

View Answer

Answer: a [Reason:] For promoter-probe vectors, the same species should be used as host for the vector whose DNA is to be screened. It is so because a gene which is promoter in one species may not be promoter in other gene and thus it becomes difficult to detect.

5. Sequences that can function as origins of replication are called as:
a) partial replicating sequences
b) self replicating sequences
c) autonomously replicating sequences
d) modified replicating sequences

View Answer

Answer: c [Reason:] There are some sequences which can function as origins of replication and they are known as autonomously replicating sequences (ARS). At times, selection is carried out for these sequences.

6. Screening is often carried out for a sequence that interacts with a protein for which already a clone is present. It is carried out in which host?
a) Bacterial host
b) Fungal host
c) Parasitic host
d) Yeast host

View Answer

Answer: d [Reason:] Screening is often carried out for a sequence that interacts with a protein for which already a clone is present. This type of screening is often carried out in yeast host.

7. In two hybrid screening system, the activator binds through _______ domain to a sequence upstream of the gene under its control, and ________ domain stimulates transcription.
a) DNA binding, activation
b) Activation, DNA binding
c) Activation, transcription
d) DNA binding, transcription

View Answer

Answer: a [Reason:] The two hybrid-screening system is based on the fact that many transcriptional activators consist of two domains, the DNA binding domain and activation domain. The activator binds through the DNA binding domain to a sequence upstream of the gene under its control. The activation domain stimulates transcription.

8. For two hybrid systems, activation domains can be present in different proteins also rather than being on a single protein. A sequence encodes _______ protein for which we want to find an interacting protein.
a) prey
b) bait
c) binding
d) activation

View Answer

Answer: b [Reason:] Bait protein is the name given to the protein for whom we are looking for an interacting protein. Bait protein is cloned adjacent to the DNA binding protein. The protein interacting with bait protein is called as prey protein and leading to activation of domains.

9. The transcription domain is ________ if some of the bait and prey proteins are non-specific in nature?
a) deactivated
b) activated
c) destroyed
d) may be activated or not

View Answer

Answer: d [Reason:] If the bait and prey proteins are non-specific in nature, some of them might lead to activation of transcription domain without the presence of another domain.

10. _____ should enter the cell in the case of activation of reporter gene for two hybrid system.
a) Bait protein
b) Prey protein
c) Both bait and prey protein
d) Either one of them

View Answer

Answer: c [Reason:] Both prey and bait proteins should enter inside the cell for activation of reporter gene for two hybrid system. It is so because activation is based on the interaction of these two proteins.

11. Protein-protein interactions such as in electron transport lead to activation of reporter gene. Is the given statement true or false?
a) True
b) False

View Answer

Answer: b [Reason:] Protein-protein interactions are weak interactions and thus they are not sufficient for the activation of reporter gene.

Set 2

1. The term ‘endonuclease’ refers to cutting the DNA sequence from:
a) only within the polynucleotide chain, not at the ends
b) the ends of the chain
c) anywhere in the chain
d) exactly in the middle of the chain

View Answer

Answer: a [Reason:] The cleavage is done within the polynucleotide chain and not at the ends. The enzyme which cuts the sequence at the ends is known as exonuclease.

2. The restriction endonuclease is having a defence mechanism in bacterial system against foreign DNA such as viruses. But how it is able to protect its own DNA?
a) By methylation of bacterial DNA by restriction enzyme
b) By methylation of foreign DNA by restriction enzyme
c) By phosphorylation of bacterial DNA by restriction enzyme
d) By phosphorylation of foreign DNA by restriction enzyme

View Answer

Answer: a [Reason:] The bacterial DNA is methylated by restriction enzyme and thus now it is not recognized by the restriction endonuclease. Thus methylation prevents the restriction endonuclease from cutting its own DNA.

3. Even after replication, how the modified DNA remains protected?
a) It remains protected because of conservative mode of replication
b) It remains protected because of semi-conservative mode of replication
c) The mode of replication has no role to play in protection
d) It is again modified after replication

View Answer

Answer: b [Reason:] Because of the semi-conservative mode of replication, one of the DNA strands remain methylated even after replication. One methylated strand is sufficient for providing protection against cleavage by restriction endonuclease.

4. How many classes of restriction enzymes are there?
a) 2
b) 1
c) 3
d) 4

View Answer

Answer: c [Reason:] Three classes of restriction enzymes are there, I,II and III. These classes are having different characteristics such as the site of cleavage on the basis of recognition sequence.

5. Type II cuts the sequence in the following way:
a) Within the recognition sequence
b) At 100-1000 nucleotides away from the recognition sequence
c) At 27-30 nucleotides away from the recognition sequence
d) It cuts randomly

View Answer

Answer: a [Reason:] Recognition sequence is the set of nucleotides which are identified by the enzyme and then it cleaves. In class II, the cleavage is done within the recognition sequence.

6. After cleaving the sequence, the nature of the ends created by the type II endonuclease is:
a) The ends created are always single stranded
b) The ends created are always double stranded
c) Either the ends are single stranded or they are double stranded
d) One end is single stranded and one end is double stranded

View Answer

Answer: c [Reason:] Either the ends are both double stranded or are both single stranded. The double stranded ends are blunt ends whereas the single stranded ends are sticky ends.

7. A sequence is having two ends, 5’ and 3’. Which of the following statements is correct regarding the nature of the ends?
a) The 5’ end is having hydroxyl group
b) The 5’ end is having phosphate group
c) The 3’ end is having phosphate group
d) Any group can be present at any end

View Answer

Answer: b [Reason:] The 5’ end is having a phosphate group. As in a DNA sequence, the 5’ end is characterized by phosphate group and the 3’ end is characterized by a hydroxyl group.

8. Blunt ends created by the restriction endonuclease can be joined. True or false?
a) True
b) False

View Answer

Answer: a [Reason:] The blunt ends created can be joined by the enzyme responsible for ligation such as DNA ligase. It is not necessary to have sticky or single stranded ends.

9. The recognition sequence for BamHI is 5’ G|GATCC 3’. The ‘|’ represents the cutting site. What can be inferred about the ends from it?
a) The ends created are double stranded
b) The single stranded end is 5’ in nature
c) The single stranded end is 3’ in nature
d) To decide about the nature of the ends more information is needed

View Answer

Answer: b [Reason:] The other strand is just complementary to it and can be written in the following way: 5’ G|GATCC 3’ 3’ CCTAG|G 5’ After cleavage, the sequences are represented as: 5’ G 3’ 5’ GATCC 3’ 3’ CCTAG 5’ 3’ G 5’ Thus, we can see that the ends generated are single stranded at 5’ end.

10. The recognition sequence of Sau3A is 5’ |GATC 3’ and that for DpnI is 5’ GA|TC 3’. Which of the statements is true?
a) The ends created by both the enzymes are compatible
b) The ends created by both the enzymes are not-compatible
c) The ends created by DpnI are single stranded
d) The ends created by Sau3A are single stranded

View Answer

Answer: b [Reason:] Though, the recognition sequence is same, the ends are not compatible. It is so because both the enzymes leave double stranded ends and thus can’t be ligated.

11. The recognition sequence is at times palindromic in nature. Which of the following statements is correct in respect to it?
a) The molecules which are cut by the same enzyme, anneal only if the sequence is palindromic in nature
b) When the molecules are cleaved by the same enzyme and the recognition sequence is palindromic in nature, there is no effect on annealing
c) There are increased chances of annealing if the recognition sequence is palindromic in nature
d) The term ‘palindromic’ can be used whether the sequence is read from 5’ to 3’ or 3’ to 5’

View Answer

Answer: c [Reason:] The term palindromic can be used only when sequence is read along the same polarity ie either 5’ to 3’ or 3’ to 5’. When the recognition sequence is palindromic in nature, there are increased chances of annealing because now there are increased orientations. If the sequence is non-palindromic in nature, then also annealing would take place but in fewer orientations.

12. If all the nucleotides are present with equal frequencies and at random, what are the chances of having a particular four nucleotide long motif?
a) 1/256
b) 1/64
c) 1/16
d) 1/8

View Answer

Answer: a [Reason:] There are four nucleotide bases present in a DNA sequence A, T, C and G. If the bases are present with equal frequency and at random the chances of having a particular 4 nucleotide long motif is 1/ (4*4)= 1/256.

Set 3

1. At times screening is done for the protein product of DNA of interest rather than the sequence itself. How many methods are there to carry out this?
a) 1
b) 2
c) 3
d) 4

View Answer

Answer: b [Reason:] The two methods for this procedure are direct selection for insert function and the second is ligand binding by the expressed protein.

2. Choose the correct statement if the screening is carried out by screening by expression in vivo.
a) The proportion of recombinants carrying gene of interest is small
b) The recombinants carrying gene of interest don’t complement the host mutation
c) Mutation should be affecting many genes
d) Most of the products selected would be result of complementation

View Answer

Answer: a [Reason:] The proportion of recombinants carrying the gene of interest is small. The host is mutant towards the gene of interest and thus the recombinants complement the host mutation. Mutation used should affect only one gene. The most of the products selected would be result of reversion of only mutation and not complementation.

3. Choose the incorrect statement in respect to host used.
a) The host must carry mutation for the gene of interest
b) The host should not contain restriction enzymes
c) The host if having restriction enzymes, then the incoming DNA should not be methylated
d) The host should be deficient in recombination

View Answer

Answer: c [Reason:] The host used must carry the mutation for the gene of interest. The host must not contain restriction enzymes. If the host is having restriction enzymes then the incoming DNA must be methylated. If the incoming DNA is not methylated, then it would be cleaved by restriction enzymes. The host should also be deficient in recombination otherwise it would be difficult to get it.

4. If the gene to be screened belongs to prokaryotic species then there are chances that it is not expressed. Is the given statement true or false?
a) True
b) False

View Answer

Answer: a [Reason:] In the case if gene which is to be screened belongs to prokaryotic species than there are chances that it is not expressed. There are chances that promoter is not present in the insert region.

5. If the library to be screened and the host belong to different species, then it is called as:
a) homologous selection
b) heterologous selection
c) intraspecies selection
d) mixed selection

View Answer

Answer: b [Reason:] If the library to be screened and the host belong to different species, then it is called as heterologous selection. In this case, there are chances that protein is expressed but it is non functional.

6. The direct selection allows _______ number of recombinants to be screened and _______
a) less, slowly
b) less, quickly
c) more, quickly
d) more, slowly

View Answer

Answer: c [Reason:] If the direct selection is carried out on a plate, then the process results in screening of more recombinants and it is done quickly and easily.

7. Xenopus oocyte cells are also used at time to carry out screening. Choose the incorrect statement for this procedure.
a) DNA from collections is transcribed within in vitro and the transcription products are microinjected into oocytes
b) The RNA is translated within oocytes
c) Screening is carried out on the basis of DNA
d) The screening is helpful for the proteins whose function can be easily screened such as transport of ions

View Answer

Answer: c [Reason:] DNA from collections is transcribed within in vitro and the transcription products are microinjected into the oocytes. The RNA is translated within the oocytes and screening is carried out on the basis of RNA. The screening is helpful for those proteins whose functions can be easily screened such as transport of ions.

8. In the case of ligand binding by the expressed protein, the library can be screened by:
a) immunochemically, by the use of antibodies
b) ligands if the sequence we are looking for encodes a protein specific to a ligand
c) using a specific DNA sequence which can bind to the protein encoded by the sequence of interest
d) using all the above methods given above

View Answer

Answer: d [Reason:] There are various ways to screen such a library. It can be done immunochemically i.e. by the use of antibodies. Ligands can also be used which bind to the specific proteins encoded by the sequence. There are also cases in which specific DNA sequences bind to the protein encoded by the sequence of interest.

9. In the case of immunochemical screening, the position of ________ antibody is detected by________ antibody.
a) secondary, primary
b) primary, secondary
c) primary, tertiary
d) secondary, tertiary

View Answer

Answer: b [Reason:] In the case of immunochemical screening, the position of primary antibody is detected with secondary antibody. Secondary antibody is common for all the primary antibodies and thus detection becomes easier.

10. The ligand should bind to a single protein only for the screening process. Is the given statement true or false?
a) True
b) False

View Answer

Answer: a [Reason:] The ligand should bind to a single protein only for the screening process, because if binds to a heterodimer than both the proteins should be present and then it would become difficult.

11. Choose the incorrect statement for panning.
a) It is a variant of ligand binding approach.
b) The ligand is immobilized on the solid support and the cells are passed over it
c) If the ligand-binding domain is exposed on surface of the any recombinant molecules then they will bind to ligand
d) They are not useful for scanning libraries in cultured mammalian cells

View Answer

Answer: d [Reason:] Panning is a variant of ligand binding approach. In this, the ligand is immobilized on the solid support and the cells are passed over. If the ligand binding domain is exposed on the cells they would bind to the ligand and would be recovered afterwards. They are useful in screening libraries in cultured mammalian cells because the cell wall is not there and thus access is easy.

12. If screening is carried out by using a combination of nascent peptide and mRNA, then it is called as:
a) nascent peptide display
b) mRNA display
c) ribosome display
d) ribozyme display

View Answer

Answer: c [Reason:] In the case of ribosome display, a pool of translating ribosomes is used. In this, nascent peptide and mRNA is there. It is passed over a solid support which is having ligand attached to it. Ribosomes which encode a protein that bind to the ligand are attached.

Set 4

1. A megaprimer method is a ____ stage approach and uses _____ oligonucleotide primers.
a) two, two
b) two, three
c) one, two
d) one, three

View Answer

Answer: b [Reason:] A megaprimer method is an approach of carrying out mutagenesis by using PCR. It is a two stage approach and uses three oligonucleotide primers.

2. Choose the correct statement.
a) There are two flanking primers and they are having mutations
b) There is one primer which anneals the target sequence and is having mutation
c) Either of the flanking primers or the primer is annealing to the target sequence is having mutation
d) One of the flanking primer and the primer annealing to the target sequence is having mutation

View Answer

Answer: b [Reason:] In the megaprimer approach three primers are used. There are two primers which flank the target sequence and a third primer anneals to the target sequence. The mutation is there in the third primer.

3. PCR using the mutagenic primer and one of the flanking primers is used to carry out amplification and generates a product corresponding to the part of the gene. It is called as megaprimer. Is the given statement true or false?
a) True
b) False

View Answer

Answer: a [Reason:] Megaprimer is produced when amplification is carried out by using mutagenic primer and one of the flanking primers. It corresponds to the part of the gene.

4. Sometimes mutagenesis is carried out with the help of primers. Choose the correct statement with respect to it.
a) Double stranded circular molecule is used as a template
b) Mutation is introduced into one of flanking primers
c) Single stranded circular molecule is used as a template
d) After amplification, mutation is introduced into one strand

View Answer

Answer: a [Reason:] Double stranded circular molecules are used as a template. Mutation is introduced into both of the primers and then amplification is carried out. After amplification is carried out, mutation can be introduced either in one strand or both the strands.

5. If PCR is used to introduce random mutations rather than specific mutations, it is called as:
a) mutagenic PCR
b) error-prone PCR
c) random PCR
d) general PCR

View Answer

Answer: b [Reason:] At times, PCR is used to introduce random mutations, rather than specific mutations and this type of PCR is called as error-prone PCR. There can be either one or many mutations.

6. For the selection of the molecules having mutated sequence, which of the statement is true?
a) It is suitable for methods which are PCR based
b) It is suitable for methods which are not PCR based
c) It is suitable for both PCR and not PCR based
d) Selection of molecules with mutant sequence is not possible

View Answer

Answer: b [Reason:] Selection is done either of molecules having mutant sequence or degradation of wild type molecules is carried out. For selection of molecules having mutant sequence, it can be used for methods which are not PCR based.

7. Choose the incorrect statement for the methodology of selection of molecules with mutant sequences.
a) A vector is used which is having antibiotic resistance gene
b) Apart from antibiotic resistance gene, a second antibiotic resistance gene is also present
c) There are two mutagenic primers which are used
d) The second strand synthesis is carried out by only one primer

View Answer

Answer: d [Reason:] A vector is used which is having conventional antibiotic resistance and along with a second antibiotic resistance gene. The second gene is inactivated because of point mutation. Also, there are two mutagenic primers used. The first primer is used for directing the incorporation of mutation. The second primer makes the inactive gene inactive. Both the primers are used for second strand synthesis.

8. Once second strand synthesis is carried out, it is introduced into host. Host is having which mutation?
a) mutS mutation
b) mutD mutation
c) mutE mutation
d) mutG mutation

View Answer

Answer: a [Reason:] The host is having mutS mutation. And because of this mutation, no mismatch repair takes place.

9. Replication by first strand leads to the formation of mutated molecules and functional antibiotic resistant gene. Is the given statement true or false?
a) True
b) False

View Answer

Answer: b [Reason:] Replication can be either by first strand or second strand. First strand replication leads to generation of wild type molecules and the antibiotic resistance is inactive. If the replication is carried out by second strand, mutated sequence is formed and the antibiotic resistance gene becomes active.

10. In a phosphothiorate nucleotide, oxygen atom is replaced by with atom?
a) Magnesium
b) Calcium
c) Sodium
d) Sulphur

View Answer

Answer: d [Reason:] In a phosphothiorate nucleotide, oxygen atom is replaced by a sulphur atom. Because of this replacement the DNA molecule becomes resistant to attack by nucleases.

11. What is the function of ung gene?
a) It is responsible for deamination of cytosine
b) It is responsible for deamination of uracil
c) It is responsible for removal of uracil
d) It is responsible for removal of cytosine

View Answer

Answer: c [Reason:] ung gene is responsible for removal of uracil. It does it by the enzyme Uracil-N-glycosylase and apyrimidinic site is created.

12. DpnI cuts ________ strands.
a) methylated
b) non-methylated
c) phosphorylated
d) non-phosphorylated

View Answer

Answer: a [Reason:] DpnI is the restriction enzyme which is meant for cutting the DNA strands which are methylated.

Set 5

1. Cloning vectors designed for the purpose of synthesis of RNA and proteins are known as:
a) Cloning vectors
b) RNA vectors
c) Bacteriophage vectors
d) Expression vectors

View Answer

Answer: d [Reason:] Expression vectors are those vectors which are used for the synthesis of RNA and protein i.e. cloned DNA sequences are used for expression.

2. For studying processes such as splicing and cleavage, RNA is required. Choose the correct statement for this.
a) A mixture of different types of RNA is required
b) Two types of RNA are required
c) A few contaminating proteins are required
d) Sodium hydroxide is required

View Answer

Answer: c [Reason:] For studying RNA processing events such as splicing and cleavage, RNA is required. Here, a single type of RNA is required and along with it a few contaminating proteins are also used.

3. RNA can be synthesized by using vector. A vector with _______ is used and further through ________ RNA is isolated.
a) origin of replication, translation
b) promoter, transcription
c) promoter, translation
d) origin of replication, transcription

View Answer

Answer: b [Reason:] Vectors can be used at times for synthesis of RNA. A vector with promoter is used and transcription is carried out in a bacterial cell and afterwards RNA is isolated.

4. Isolation of RNA can be carried out easily from bacterial cells. Is the given statement true or false?
a) True
b) False

View Answer

Answer: b [Reason:] Isolation of RNA can’t be carried out easily from bacterial cells. Hence, production of RNA is carried out by transcription of cloned DNA.

5. Promoters are generally used after isolation from:
a) bacteriophage T7
b) bacteriophage SP6
c) baceriophage Mu
d) both bacteriophage T7 and SP6

View Answer

Answer: d [Reason:] Promoters used for synthesis of RNA are generally obtained from bacteriophage T7 and SP6. They can also be obtained rarely from bacteriophage T3.

6. By selecting the appropriate polymerase to activate the promoter, ________ can be carried out _______
a) transcription, regardless of orientation
b) transcription, only in one orientation
c) translation, regardless of orientation
c) translation, only in one orientation

View Answer

Answer: a [Reason:] By selecting the appropriate polymerase to activate the promoter, transcription can be carried out but it is regardless of the orientation.

7. Guanylyl transferase and GTP are used for?
a) Transcription
b) Translation
c) Capping of the message
d) Packaging

View Answer

Answer: c [Reason:] At times, capping of the message is very necessary. Capping can be achieved by incubating the transcripts with enzymes such as guanylyl transferase and GTP.

8. Capping can be introduced by the use of cap analogue. Which of the statement is true?
a) Cap analogue can be introduced at the end of the transcript
b) Cap analogue can be introduced at the start of the transcript
c) Cap analogue can be introduced anywhere in the transcript
d) Cap analogue can be introduced both at the end and starting of the transcript

View Answer

Answer: b [Reason:] For achieving capping, cap analogue should be introduced. Cap analogue should be introduced at the starting of the transcript. It is so because it requires 5’ end.

9. If transcription should not be carried out beyond the insert in the vector, then it should be linearized. Is the given statement true or false?
a) True
b) False

View Answer

Answer: a [Reason:] If the transcription should not be carried out beyond the insert in the vector, then it should be linearized. It can be done by digesting with restriction enzyme.

10. If the cells containing plasmids are infected with helper phage, which type of DNA can be produced, packaged and secreted into the medium?
a) Single stranded DNA
b) Double stranded DNA
c) Both single and double stranded DNA
d) Circular DNA

View Answer

Answer: c [Reason:] If the cells containing plasmids are infected with helper phage, then single stranded DNA can be produced, packaged and secreted into the medium.