Generic selectors
Exact matches only
Search in title
Search in content
Search in posts
Search in pages
Filter by Categories
nmims post
Objective Type Set
Online MCQ Assignment
Question Solution
Solved Question
Uncategorized

Multiple choice question for engineering

Set 1

1. Which of the following is true regarding taking up of plasmid DNA in the bacterial cells?
a) There are more chances of having two plasmids in a single cell
b) There are more chances of having a single plasmid in one cell
c) Uptake upto two plasmids is possible but not more than that
d) Both are taken up with the same efficiency

View Answer

Answer: b [Reason:] There are more chances of having a single plasmid in one cell. But there are possibilities of having two plasmids in a single cell but the chances are less.

2. If genomic DNA has been inserted instead of the plasmid, what will happen?
a) It would lead to inactivation of lacZ gene
b) The X-gal substrate would be broken down
c) The colonies formed are blue in colour
d) The lacZ gene would be intact

View Answer

Answer: a [Reason:] If genomic DNA is inserted instead of the plasmid the lacZ gene would be inactivated. The X-gal substrate won’t be broken down and thus the colonies formed would be white or off-white.

3. Often PCR can be performed in order to confirm whether an insert is present in the plasmid. Cells are taken directly and PCR is performed, this type of PCR is known as:
a) direct PCR
b) colony PCR
c) quantitative PCR
d) in-situ PCR

View Answer

Answer: b [Reason:] When cells from colonies are taken and directly PCR is performed, it is called as colony PCR. Primers anneal to each side of the cloning site and there is no need of purification in such cases.

4. Generation of recombinants by randomly cloning fragments of total DNA from an organism is called as:
a) genomic library
b) screening
c) recombination
d) shotgun cloning

View Answer

Answer: d [Reason:] Generation of recombinants by randomly cloning the fragments of total DNA from an organism is called as shotgun cloning. Collection of such fragments is known as library.

5. The phenomenon of alpha complementation is:
a) α + β = ω
b) α = β + ω
c) β = α + ω
d) It is either α + β = ω or β = α + ω

View Answer

Answer: a [Reason:] Alpha complementation is the phenomenon of having the alpha and beta subunit combining to give the omega subunit. These are the different genes of the lac operon.

6. An operon is defined as:
a) A related set of genes each having different promoters and are present differently
b) A set of genes which are present together but are controlled by different promoters
c) A set of genes which are present together and are controlled by the same promoter
d) A set of genes which are not present together but controlled by the same promoter

View Answer

Answer: c [Reason:] An operon is defined as the cluster of genes which are present together and are controlled by the same promoter. An example is lac operon. It consists of three subunits, alpha, beta and omega.

7. Why the whole lacZ gene can’t be present in the plasmid at one time?
a) Because the whole lacZ gene can’t be present anywhere
b) The whole lac Z gene is very large in size and the plasmid size is small
c) The whole lacZ gene is never functional
d) Because the plasmid takes is having a restriction site only for taking up a portion of lacZ gene

View Answer

Answer: b [Reason:] The whole lacZ gene is not present in a plasmid because the plasmids are generally small in size and the lacZ gene is large in size. And if the whole gene is present, it makes the overall size very large.

8. Alpha complementation is an indicator of lacZ system. Which of the statement is incorrect for it?
a) One portion of the lacZ gene known as minigene is present in the plasmid
b) Another portion is present in the host itself
c) If they both are allowed to combine in the presence of IPTG, X-gal and ampicillin, blue colonies are observed
d) If insert is also present along with host and plasmid, it results in formation of blue colonies

View Answer

Answer: d [Reason:] Alpha complementation system is used as lacZ indicator system. In this, one portion of the lacZ gene is present in the plasmid and is termed as minigene. The other portion is present in the host itself. The combining of two portions is very important to have an intact lacZ gene and in the case it is intact blue colonies are observed in the presence of IPTG, ampicillin and X-gal. It is so because lacZ breaks down X-gal and it gives blue colour. If the insert is also there, the lacZ gene is interrupted and thus X-gal is not broken down and hence white colonies are obtained.

Set 2

1. Choose the incorrect statement for second DNA strand synthesis.
a) The RNA: DNA complex is treated with the enzyme RNaseH and DNA pol
b) The enzyme RNaseH is responsible for nicking the RNA strand and leaving free 3’ hydroxyl ends
c) These RNA fragments can be used as primers
d) There is no portion of RNA left attached to the DNA strand

View Answer

Answer: d [Reason:] For second strand DNA synthesis, the RNA: DNA complex is treated with RNaseH enzyme along with DNA pol. The enzyme is responsible for nicking the RNA strand and leaving 3’ hydroxyl ends. These free ends are used for second strand synthesis. These RNA fragments can be used as primers. After DNA: DNA duplex formation, only a small portion of RNA is left at the 5’ end.

2. What is the final product of the RNaseH method?
a) blunt ended dsDNA
b) staggered dsDNA at both ends
c) staggered dsDNA at 3’ end
d) staggered dsDNA at 5’ end

View Answer

Answer: a [Reason:] The final product of the RNaseH method is blunt ended dsDNA. The RNA piece left at the 5’ end is removed by RNase and thus blunt ended dsDNA is left.

3. What would not happen if the RNA strand is completely removed from RNA: DNA hybrid?
a) There are no chances of the synthesis of second DNA strand
b) Chance complementarity would take place
c) Hairpin structure would be formed
d) Hairpin structure if formed is not the final structure

View Answer

Answer: a [Reason:] If the RNA strand is completely removed from RNA: DNA hybrid, the single stranded DNA remains. Then chance complementarity takes place and the 3’ end complements with any other portion in the single stranded DNA. It leads to the formation of a hairpin structure and the 3’ end extends in the presence of DNA pol and nucleotides.

4. The loop region is single stranded. It can be cleaved by using which enzyme?
a) Exonuclease
b) S1 nuclease
c) RNaseH
d) DNase

View Answer

Answer: b [Reason:] The loop region in the hairpin structure is single stranded in nature. The single stranded portion can be cleaved by S1 nuclease.

5. Choose the correct statement in respect to the self priming method of cDNA synthesis.
a) It is less preferred than RNaseH method
b) A hairpin structure is formed with guarantee
c) The sequence corresponding to the 5’ end is lost
d) Reverse transciptase is not used

View Answer

Answer: c [Reason:] The self priming method is based on the formation of hairpin structure. But there are only chances and no guarantee of formation of hairpin structure. In the first strand synthesis, reverse transciptase is used. The sequence corresponding to the 5’ end is lost and in cDNA large deletions are observed. This method is less preferred than RNaseH method.

6. Choose the incorrect statement for the method homopolymer tailing.
a) The first step is the RNA: DNA hybrid synthesis
b) Terminal transferase is used for addition of nucleotides on 3’ end
c) Terminal transferase adds only at DNA strands
d) The DNA strand is now having known sequence at 3’ end

View Answer

Answer: c [Reason:] Homolpoymer tailing is also used for cDNA synthesis. The first step remains same as the other methods, which is synthesis of RNA: DNA hybrid. It is then treated by terminal transferase, the enzyme is responsible for adding nucleotides on 3’ end of both DNA and RNA. The 3’ end is now having known DNA sequence and it is used as a tail in the reaction.

7. Choose the correct statement for RACE.
a) It stands for Random Amplification of cDNA ends
b) It is for cloning particular cDNA ends
c) It is only of one type, which is 5’ RACE
d) Sequence data is not available in any case

View Answer

Answer: b [Reason:] RACE stands for Rapid Amplification of cDNA ends. Sometimes, we wish to clone specific cDNA portion and are having some sequence data. It is meant for cloning specific cDNA ends and is calssified into two types, 5’ RACE and 3’RACE.

8. The first primer in the case of 3’ RACE is:
a) internal sequence
b) oligo-dT adaptor molecule
c) oligo-dA adaptor molecule
d) adaptor oligo-dT primer

View Answer

Answer: b [Reason:] The first primer in the case of 3’ RACE is oligo-dT adaptor molecule. The second primer is an internal sequence.

9. The first cDNA strand in 5’ RACE is tailed with oligo-dA tail. Is the given statement true or false?
a) True
b) False

View Answer

Answer: a [Reason:] The first cDNA strand in the case of 5’ RACE is tailed with an oligo-dA tail. The synthesis of first cDNA strand is by an internal primer.

10. What is the second primer in the case of 5’ RACE?
a) Internal primer
b) Oligo-dA sequence
c) Adaptor-oligo-dT primer
d) Oligo-dT adaptor molecule

View Answer

Answer: c [Reason:] The second primer ie the primer used for synthesis of second cDNA strand is adaptor-oligo-dT primer. Subsequent PCR is carried out by using internal primer for coding sequence.

Set 3

1. If it is desirable to express two or more proteins simultaneously, then it is known as:
a) hybrid of proteins
b) fusion of proteins
c) co-expression of proteins
d) combination of proteins

View Answer

Answer: c [Reason:] If it is desirable to express two or more proteins simultaneously, then it is called as co-expression of proteins. There are varied uses of such proteins.

2. For co-expression of proteins, it is possible to have several independent sets of promoters and other signals in a single vector. Is the given statement true or false?
a) True
b) False

View Answer

Answer: a [Reason:] For co-expression of proteins, there are two strategies which are used. The first is to use several different expression vectors at a single time. The second method is to use several independent sets of promoters and signals in a single vector.

3. A considerable amount of modification can take place in the ______ region of RNA in _______ before translation. It affects the yield of proteins.
a) coding, prokaryotes
b) coding, eukaryotes
c) non-coding, prokaryotes
d) non-coding, eukaryotes

View Answer

Answer: b [Reason:] A considerable amount of modification in the coding region of the RNA can take place in eukaryotes before translation. This can have an effect on the yield of the protein.

4. Strong secondary structure in a message may affect ________
a) translation
b) transcription
c) both translation and transcription
d) none of translation and transcription

View Answer

Answer: a [Reason:] It is not necessary that if transcription and translation initiate well, they continue well. Factors such as secondary structure, affect translation. There are other factors such as codon usage and codon meaning which affect translation.

5. What is the function of T7 lysozyme?
a) Attacking the peptidoglycan in bacterial cell walls
b) Inhibitor of T7 polymerase
c) Activator of T7 polymerase
d) Both attacking the peptidogylcan in bacterial cell walls and inhibitor of T7 polymerase

View Answer

Answer: d [Reason:] T7 lysozyme is used for reducing the level of expression from T7 promoters. It is phage encoded and serves a dual purpose. It attacks the peptidoglycan in bacterial cell walls and is inhibitor of T7 polymerase.

6. As the activity of T7 polymerase is reduced by lysosome, what is effect on the rate of synthesis after induction?
a) Increase
b) Decrease
c) May increase or decrease
d) No effect

View Answer

Answer: b [Reason:] As the activity of T7 polymerase is reduced by the lysosome, resultantly the rate of synthesis after induction is also reduced.

7. Proteins at times are not soluble in the cell and form aggregates known as:
a) coagulations
b) aggregated mass
c) inclusion bodies
d) insoluble mass

View Answer

Answer: c [Reason:] Proteins are not always soluble and at times they form insoluble aggregates. These are called as inclusion bodies. It is to separate them out by centrifugation.

8. It is not possible to solubilise proteins from aggregated mass of proteins. Is the given statement true or false?
a) True
b) False

View Answer

Answer: b [Reason:] At times, it is possible to solubilise proteins from the aggregated mass of proteins. And this further helps in correct folding.

9. Lon protease is used for degradation of foreign proteins in E.coli cells. It is active against which type of proteins?
a) Partially denatured proteins
b) Completely denatured proteins
c) Both partially and completely denatured proteins
d) Intact proteins

View Answer

Answer: c [Reason:] Lon protease is used for degradation of foreign proteins. It is active against both partially and completely denatured proteins.

10. Lon mutation can lead to ______ and _______
a) mucoidy, UV sensitivity
b) mucoidy, UV insensitivity
c) overproduction of fats, UV sensitivity
d) overproduction of fats, UV insensitivity

View Answer

Answer: a [Reason:] Lon mutation can lead to problems such as mucoidy and UV sensitivity. Mucoidy is the phenomenon overproduction of polysaccharide capsules in the cell and thus manipulation becomes difficult.

11. Mutation in gale gene or cpsA-E gene cluster is used for:
a) suppression of mucoidy
b) activation of mucoidy
c) suppression of UV sensitivity
d) both suppression of mucoidy and UV sensitivity

View Answer

Answer: a [Reason:] Mucoidy is undesirable for physical manipulation. It can be suppressed by mutation in gale gene or cspA-E gene cluster.

Set 4

1. Little quantities of radiolabelled proteins are required for which of the following?
a) co or post translational targeting
b) modification of proteins
c) both co or post translational targeting and modification of proteins
d) crystallization for structural studies

View Answer

Answer: c [Reason:] Protein synthesis is very important and they are having varied functions. Small quantities of radiolabelled proteins are required for co or post translational targeting and modification of proteins.

2. __________ quantities of _______ protein are required for determination of properties in biochemical and biological assays.
a) Small, non-radiolablled
b) Small, radiolabelled
c) Large, radiolablelled
d) Large, non-radiolabelled

View Answer

Answer: d [Reason:] At times, large and non-radiolabelled proteins are required for determination of properties in biochemical and biological assays. It is also required for carrying out the crystallization for structural studies.

3. Small quantities of radiolabelled RNA can be produced by translation in vitro which is done by transcription in vitro. Is the given statement true or false?
a) True
b) False

View Answer

Answer: a [Reason:] If small quantities of radiolabelled RNA are required, it can be produced by translation in vitro which is accomplished by transcription in vitro.

4. For production of unlabelled and huge amount of proteins, which of the following is true?
a) Transcription is carried out in vivo and translation in vitro
b) Transcription and translation both are carried out in vivo
c) Transcription and translation both are carried out in vitro
d) Transcription is carried out in vitro and translation in vivo

View Answer

Answer: b [Reason:] For having, large and unlabelled proteins, transcription and translation both should be carried out in vivo.

5. How many methods are there, which are used for protein synthesis in vitro?
a) 1
b) 2
c) 3
d) 4

View Answer

Answer: c [Reason:] There are basically three methods which are used for protein synthesis in vitro. They are based on a lysate of reticulocytes.

6. Reticulocytes are immature red cells, which are obtained from:
a) rabbit
b) pig
c) human
d) cow

View Answer

Answer: a [Reason:] Reticulocytes are immature red blood cells which are obtained from rabbits. The lysate of this is used for transcription in vitro.

7. What is the use of adding micrococcal nuclease in the reticulocyte cells?
a) It degrades DNA
b) It degrades mRNA
c) It degrades proteins
d) It degrades RNA and untranslated DNA

View Answer

Answer: b [Reason:] The micrococcal nuclease is added in the reticulocyte cells in order to degrade the mRNA. This mRNA produces high amount of background in translational products.

8. Treatment of reticulocyte cells is done with EGTA. It chelates the calcium ions which are required for functioning of micrococcal nuclease. Is the given statement true or false?
a) True
b) False

View Answer

Answer: a [Reason:] The treatment of reticulyocte cells is done with EGTA. It chelates the calcium ions and they are required for functioning of micrococcal nuclease. It is required for cleavage of mRNA.

9. In ________ transcription and translation is coupled.
a) prokaryotes
b) eukaryotes
c) both prokaryotes and eukaryotes
d) both prokaryotes and eukaryotes, but favourable in eukaryotes

View Answer

Answer: a [Reason:] In prokaryotes, transcription and translation are coupled. It means that as the transcription is initiated, translation also starts.

10. In which of the following systems, transcription and translation are carried out together?
a) Reticuloycte lysate
b) Wheat gram extract
c) Both in reticulocyte lysate and wheat gram extract
d) S-30 extract

View Answer

Answer: d [Reason:] S-30 extract is the one in which transcription and translation are carried out together. In wheat gram extract and reticulocyte lysate, they are carried out separately.

Set 5

1. The minor change in amino acid sequence can lead to _______ effect on three dimensional structure and there _____ in the primary sequence.
a) huge, may be no change
b) no, may be huge change
c) very less, is very less change
d) huge, would be huge change

View Answer

Answer: a [Reason:] The minor change in the amino acid sequence can lead to huge effect on the three dimensional structure and might also abolish its function. In primary sequence, there might be no change.

2. It is often useful to inactivate endogenous genes in an organism. It might be helpful in finding out _________ role of the wild type gene.
a) biological
b) chemical
c) physiological
d) anatomical

View Answer

Answer: c [Reason:] Inactivation of endogenous genes in an organism is very important at times. It is helpful in finding out the physiological role of the wild type gene.

3. The inactivation of endogenous genes may also be helpful in directing the expression of mutated gene in the absence of background expression of wild type gene. Is the given statement true or false?
a) True
b) False

View Answer

Answer: a [Reason:] Inactivation of endogenous genes is very useful. It can be used to direct the expression of mutated gene in the absence of background expression of wild type gene.

4. How can mutant strains be produced?
a) In systematic mutagenesis programmes
b) In individual organisms
c) Both by individual organisms and systematic mutagenesis programmes
d) Apart from these two, other methods are also used

View Answer

Answer: c [Reason:] For popular model organisms, there are databases for mutant strains. These mutant strains can either be produced by systematic mutagenesis programmes or by individual organisms.

5. The principle of gene disruption is ________ to replace the endogenous chromosomal copy of a gene with __________
a) homologous recombination, inactivated gene
b) reciprocal translocation, inactivated gene
c) homologous recombination, activated gene
d) reciprocal translocation, activated gene

View Answer

Answer: a [Reason:] For gene disruption, the basic principle is homologous recombination to replace the endogenous chromosomal copy of a gene with the inactivated gene.

6. The gene to be disrupted is cloned and a selectable marker is inserted. What should be the effect of selectable marker?
a) It should have no effect on target gene
b) It should make the target gene non-functional
c) There is no restriction; it can be either functional or non-functional
d) It should improvise the chances of survival of target gene

View Answer

Answer: b [Reason:] In the target gene, a selectable marker should be inserted. It can be either in the form of ampicillin resistance or a nutritional marker. The selectable marker should render the target gene non-functional.

7. The disrupted gene is excised from the vector and is inserted into the target organism. The excised gene should be in which form?
a) Circular
b) Supercoiled
c) Either supercoiled or circular
d) Linear

View Answer

Answer: d [Reason:] The disrupted gene is excised from the vector and is inserted into the target organism. The excised gene should be linear in form.

8. Stable acquisition of the marker can take place only if a double crossover over the flanking sequence and their chromosomal counterparts causes the marker’s integration into the chromosome. Is the given statement true or false?
a) True
b) False

View Answer

Answer: a [Reason:] The incoming molecule having the marker will not replicate stably because it is linear. For the stable replication, double crossover of the flanking sequence should take place, replacing the endogenous gene with the disrupted gene.

9. If ______ gene is there, the double crossover may leave ________ in the chromosome.
a) linear, functional copy
b) circular, functional copy
c) linear, a non-functional copy
d) circular, a non-functional copy

View Answer

Answer: b [Reason:] If the disrupted gene is in circular form, there is possibility that the double crossover may still leave a functional copy in the chromosome.

10. If the target organism contains more than one copy of the gene, what is the effect on these copies?
a) Only copy is disrupted
b) All the copies are disrupted
c) It is difficult to ensure that all the copies are disrupted
d) Only a specified number of copies are disrupted

View Answer

Answer: c [Reason:] If more than one copy of the gene is there in the target organism, it is difficult to ensure that all the copies are disrupted. Multiple copies of the gene can be present if the organism is not haploid.