Multiple choice question for engineering
Set 1
1. It is very important to study lambda biology as lambda phages are used for cloning purposes. Which of the statement is true for lambda phage?
a) It is an example of temperate phage
b) The fate of the phage is decided before it infects the cell
c) The lysis fate is that where the phage inserts it genome into the bacterial genome and the replication goes on
d) The lysogenic fate is that where the phage infects the cell and lysis is carried out
Answer
Answer: a [Reason:] Lambda phage is a temperate phage and its fate is decided as it infects the cell. It can be either be lytic or lysogenic in nature. Lytic phage is that which infects the cell and it is lysed whereas lysogenic phage is that in which phage genome is inserted into the bacterial genome and is replicated along with it.
2. Choose the correct statement for the infectious particle of lambda phage.
a) It is single stranded genome
b) It is circular double stranded genome
c) The ends are blunt with cos (cohesive) sequences
d) The ends are created by cleavage at cos sites during phage packaging
Answer
Answer: d [Reason:] The infectious particle of lambda phage id linear double stranded genome. The ends are staggered with cos (cohesive) sequences and they are created by cleavage at cos sites during phage packaging.
3. Some changes are encountered in the infectious particle as it infects the cell. Choose the incorrect statement.
a) The infectious particle circularizes upon causing infection by annealing at cos sites
b) There are two promoters present, PR and PL
c) Upon infection only PR is activated
d) Promoters give rise to immediate early transcripts
Answer
Answer: c [Reason:] As the infectious particle infects the cell it circularizes by annealing at cos sites. Both the promoters PR and PL are activated upon infection. Promoters give rise to immediate early transcripts which terminate at rightward and leftward terminators.
4. The immediate early transcripts direct synthesis of which genes?
a) N genes
b) cro genes
c) Both N and cro genes
d) PR and PL genes
Answer
Answer: c [Reason:] The immediate early transcripts are transcribed by promoters and they direct the synthesis of gene products of N and cro genes.
5. What is the function of N gene?
a) It promotes the function of terminators
b) It overrides the function of terminators
c) It promotes the production of cro gene
d) It is responsible for making the infectious particle linear
Answer
Answer: b [Reason:] N gene is responsible for overriding the function of terminators. As the terminators are overridden, the transcription is carried in the surrounding regions.
6. Choose the correct statement for cII gene and its related function?
a) The transcription is extended to the region of cII gene and it is termed as late transcription
b) They are responsible for deactivating the promoter for repressor establishment
c) This promoter is responsible to decide whether the life cycle would be lysogenic or lytic
d) cII gene is responsible for other gene which is gene cIII
Answer
Answer: c [Reason:] The cII gene product is formed when the transcription is extended because of the overriding of terminators by the N gene. It is responsible for activation of promoter for repressor establishment. This promoter is responsible for deciding whether it would be in lysogenic or lytic mode. The transcripts formed by extension are called as delayed early transcripts and cIII is also produced along with cII.
7. Which gene is required for the lysogenic phage to be adopted?
a) cI
b) cII
c) cIII
d) All the three genes are interrelated
Answer
Answer: d [Reason:] All the three genes are interrelated. The cII gene is protected from proteases by cIII gene and it further promotes expression of cI gene. cI gene is necessary for silencing of rest of the lambda genome. cII is most important but as it related with other two as well, thus they all are required for lysogenic mode.
8. Choose the incorrect statement for lysogenic mode.
a) PRE forms cI gene product which is responsible for activation of promoters
b) The cII protein is responsible for interfering with cro and Q proteins and they are required for lytic life cyle
c) The cII protein activates the promoter PINT
d) The Int protein with host protein is responsible for site specific recombination
Answer
Answer: a [Reason:] The cI gene product is responsible for inactivation of promoters and thus silencing rest of the lambda genome. The cII gene product is responsible for interfering with cro and Q proteins and they are required for lytic life cycle. The cII gene also activates promoter PINT and the Int protein produced along with host protein is required for site specific recombination.
9. Choose the correct statement for cI gene.
a) It represses only cII gene
b) It activates cII gene and represses cIII gene
c) It represses both cII and cIII gene
d) It activates both cII and cIII gene
Answer
Answer: c [Reason:] cI gene represses both cII and cIII gene and thus expression of cIII is also suppressed.
Set 2
1. Bacteriophage Mu is also a phage, choose the correct statement for it.
a) It is an example of temperate phage
b) It is packaged into heads containing 50 kbp of DNA
c) When the phage replicates it goes transposition only a few times
d) Mu can’t be used for cloning in vivo
Answer
Answer: a [Reason:] Bacteriophage Mu is also a phage, it is an example of temperate phage. It is packaged into heads containing 39 kbp of DNA. When the phage replicates it goes transposition a few hundred times and inserts itself into many places in the host genome. Mu can be used for cloning in vivo as a ‘mini Mu’ that contains reduced phage genome.
2. Choose the incorrect statement for ‘Mini Mu’?
a) It is produced by cloning of bacteriophage Mu in vivo
b) It contains reduced phage genome
c) It doesn’t contains replication of origin of plasmid
d) A selectable marker is added
Answer
Answer: c [Reason:] Mini Mu is produced by cloning of bacteriophage Mu in vivo. It is having reduced phage genome. A selectable marker and an origin of replication of plasmid are added.
3. Which of the following is not included as a step for cloning mini Mu in vivo?
a) A culture of target strain, whose DNA we want to clone is infected with mini Mu
b) The phage is inserted at multiple sites in the genome
c) Some of the packaged strain is flanked by two mini Mu genes
d) The packaged phage is introduced into a second strain and Mu would replicate as a phage
Answer
Answer: d [Reason:] For cloning in mini Mu, a culture of target strain whose DNA we want to clone is infected with mini Mu. The phage is inserted at multiple sites in the genome. Some of the packages strain is flanked by two mini Mu genes. The packaged phage is introduced into a second strain and the Mu won’t be able to replicate as a phage.
4. If recombination takes place between elements in the second strain, it would replicate by origin of replication of mini Mu. Is the given statement true or false.
a) True
b) False
Answer
Answer: a [Reason:] If recombination takes place between elements in the second strain, it would replicate as a plasmid by origin of replication of mini Mu. Thus a plasmid is formed with a piece of DNA from first strain and stably maintained in the second strain.
5. Which of the following is a restriction for cloning in bacteriophage Mu?
a) It is limited to handling only small number of phage particles
b) It relies on whether the DNA is flanked by two mini Mu genes or not
c) Recombination between the elements in the second strain is not a restriction
d) Fragments of any genomic DNA can be cloned irrespective of the fact whether or not the phage can replicate in it or not
Answer
Answer: b [Reason:] The cloning in bacteriophage Mu is restricted by some factors. These include, it depends whether the DNA is flanked by two mini Mu genes or not. Recombination between the elements in the second strain is also important. Fragments of only those genomic DNA can be cloned, in which Mu can replicate.
6. What is the size of the insert that can be accommodated in the head of bacteriophage PI?
a) 30-40 kbp
b) 80-95 kbp
c) 110-115 kbp
d) 200-300 kbp
Answer
Answer: c [Reason:] Bacteriophage PI is known to accommodate larger size of fragments. The size of the fragments that can be accommodated in the head of this phage lies between 110-115 kbp.
7. What is the function of pac site in the bacteriophage PI?
a) It is responsible for initiation of packaging
b) It is responsible for termination of packaging
c) It is responsible for initiation of replication
d) It is responsible for termination of replication
Answer
Answer: a [Reason:] The pac site in the bacteriophage PI is responsible for initiating the packaging of DNA into the phage heads.
8. What is the function of loxP sites?
a) It is responsible for linearization of DNA in the host bacterium
b) It is responsible for circularization of DNA in the host bacterium
c) It is responsible for conversion of single stranded DNA into double stranded DNA
d) It is responsible for conversion of double stranded DNA into single stranded DNA
Answer
Answer: b [Reason:] The lox P sites are responsible for the circularization of the linear DNA molecules in the host bacterium.
9. Which of the following is not included as a step in cloning in bacteriophage PI?
a) The growth of vector is done as a plasmid by using pBR322 origin of replication
b) Cleaving with BamH1 is done and then products cleaved with Sau3A are ligated
c) Incubation with mutant lysogen is done which doesn’t cleaves the recombinants at pac sites
d) Incubation is done with mutant lysogen which doesn’t contains head and tail proteins
Answer
Answer: c [Reason:] Bacteriophage PI is a complex vector. The growth of vector is done as a plasmid by using pBR322 origin of replication. Cleaving with Bam H1 is done and it is followed by ligation with Sau3A partial digestion products. It is followed by incubation with mutant lysogen which cleaves the recombinants at pac sites but there are no head and tail proteins. It is followed by incubation in lysogen in which cleavage is not carried out at pac sites but there are head and tail proteins. It leads to packaging into heads.
10. What happens after packaging is carried out in the phage head?
a) A sequence dependent cleavage of DNA is carried out
b) A sequence independent cleavage of DNA is carried out
c) They are linearized before they are infected into the bacterial cells
d) They are circularized before they are infected into the bacterial cells
Answer
Answer: b [Reason:] After the packaging is carried out in the phage head, a sequence independent cleavage of DNA is carried out. Molecules with a very small size insert can’t fill into the whole phage head before the next pac site is reached.
11. Circularization must take place in the bacterial cells before the replication starts. The given statement is true or false?
a) True
b) False
Answer
Answer: a [Reason:] Circularization must take place in the bacterial cells before the replication is initiated. The cells are introduced into the bacterial cells after the packaging has been carried out.
12. How many origin of replication are there once the cyclization is carried out by loxP sites?
a) 1
b) 2
c) 3
d) 4
Answer
Answer: b [Reason:] There are two origin of replication once the cyclization is carried out by loxP sites. One origin of replication is for low copy number and the other can be activated for carrying out a high copy number replication.
13. Choose the incorrect statement for PAC vectors.
a) In these vectors, the phage packaging signals are removed but the two P1 origin of replication persist
b) They resemble BAC vectors
c) They accept an insert of upto 300 kbp
d) Instability is observed
Answer
Answer: d [Reason:] PAC vectors are those vectors, in which the phage packaging signals are removed but the two P1 origin of replication persist. These resemble BAC vectors. They accept upto an insert of upto 300 kbp. Maintenance at low copy numbers is helpful in avoiding instability.
14. Choose the correct statement in respect to sacB gene?
a) The cloning site for PAC lies on the side of sacB gene
b) It is responsible for producing an enzyme which is responsible for catalysing sucrose into glucose and fructose
c) Expression of sacB in the presence of sucrose is beneficial
d) Levan is non-toxic for E. coli
Answer
Answer: b [Reason:] sacB gene along with E. coli promoter flanks the cloning site in PAC vector. It is responsible for producing an enzyme which is responsible for catalysing sucrose into glucose and fructose. The fructose is ploymerized to levan and is toxic to E. coli. Thus expression of sacB in the presence of sucrose is lethal.
15. Selection of molecules by lack of inserts on the basis of sacB gene is known as:
a) positive selection
b) negative selection
c) counter selection
d) sacB selection
Answer
Answer: c [Reason:] The sacB gene is abolished if an insert is there in the cloning site and thus can survive in the presence of sucrose. It is known as positive selection if it is directly looked for an insert. If instead of insert, absence of sacB gene is looked upon, it is called as counter selection.
Set 3
1. The process of amplification of specific DNA sequences by an enzymatic process is termed as:
a) amplification
b) polymerase chain reaction (PCR)
c) translation
d) microarrays
Answer
Answer: a [Reason:] The process of amplification of specific DNA sequences by an enzymatic process is termed as Polymerase Chain Reaction (PCR). For the PCR to take place there should be small sequences at each end which should be known.
2. What are primers?
a) Primers are the short sequences at the end of the nucleotide sequences which are used for amplification
b) Primers are the short sequences which are complementary to the nucleotides at the end of the sequence which is to be amplified
c) Primers are the short sequences present anywhere in the nucleotide sequence to be amplified
d) Primers are the short sequences which are complementary to the nucleotides anywhere in the sequence to be amplified
Answer
Answer: b [Reason:] Primers are short nucleotide sequences which are complementary to the stretches at the ends of the DNA sequence to be amplified.
3. A reaction mixture for PCR consists of:
a) heat unstable polymerase
b) primers in limited amount
c) deoxynucleoside triphosphate (dNTPs)
d) a region complementary to the sequence to be amplified
Answer
Answer: c [Reason:] A reaction mixture for PCR consists of the region of the DNA sequence to be amplified, primers in large molar excess, heat stable polymerase and dNTPs. Heat stable polymerase which is used commonly is Taq polymerase.
4. Which of the following is a characteristic of Taq polymerase?
a) It is a RNA polymerase
b) It is heat stable
c) It is obtained from thermophilic bacterium and can be grown in laboratory below a temperature of 75 degrees
d) It is used in cellular synthesis processes and the optimum temperature is atleast 90 degrees
Answer
Answer: b [Reason:] Taq polymerase is a DNA polymerase obtained from thermophilic bacterium Thermus aquatics. It is heat stable and can be grown in laboratory at a temperature above 75 degrees. It is used in cellular DNA synthesis processes and the optimum temperature is atleast 80 degrees. It is not readily denatured by repeated cooling and heating cycles and thus is used in amplification processes.
5. These are steps taken in carrying out the PCR reaction:
i) Attaching of primers by cooling
ii) Denaturation of strands
iii) DNA synthesis
iv) Heating
Which is the correct order?
(Mentioned from starting to ending the reaction)
a) i)-ii)-iii)-iv)
b) ii)-i)-iii)-iv)
c) iv)-iii)-ii)-i)
d) iv)-ii)-i)-iii)
Answer
Answer: d [Reason:] PCR consists of series of steps. Firstly, the reaction mixture is heated so that the strands are separated i.e. their denaturation takes place. Then it is again cooled, so that the primers are able to attach. Once the primers are attached, synthesis of DNA is allowed. This whole process is repeated.
6. Which of the following is not a condition for PCR?
a) Initial melting carried out for 5 minutes at 94 degrees
b) Initial melting followed by 30 cycles each consisting of melting for 1 minute at 94 degrees
c) Renaturation for 5 minutes at 60 degrees
d) DNA synthesis at 72 degrees for 1.5 minutes
Answer
Answer: c [Reason:] The melting temperature is 94 degrees and the initial melting is carried for 5 minutes at this temperature. It is followed by 30 cycles each consisting of melting for 1 minute at 94 degrees. The renaturation is carried for 1 minute at 60 degrees. The DNA synthesis is carried out at 72 degrees for 1.5 minutes. After the 30 cycles, a final round of extension is carried out for 10 minutes.
7. Primers and polymerases are added again during the reaction, because they get consumed as the reaction proceeds. Is the given statement true or false?
a) True
b) False
Answer
Answer: b [Reason:] Primers and polymerases are not added more as the reaction proceeds. It is so because, the polymerase is heat stable and is not destroyed during the reaction. Primers on the other hand are added in excess at beginning of the reaction.
8. All the molecules generated during PCR will not be full length. Some will also be of intermediate length. Which of the statements is correct?
a) After first cycle, majority of the molecules will be full length and only some will be of intermediate length
b) In the next cycle, each intermediate molecule will generate one intermediate molecule and one target molecule
c) The number of full length molecules increase as number of cycles proceed
d) The number of intermediate molecules increase geometrically and the number of target molecules increase arithmetically
Answer
Answer: b [Reason:] Intermediate molecules are those which would be having primer at one end and the other end won’t be defined. After the first cycle, half of the molecules would be full length and half would be of intermediate length. In the next cycle, each intermediate molecule generates one intermediate molecule and one target molecule. Target molecule is that which is defined by primers at both the ends. The number of full length molecules remains constant. But, target molecules increase geometrically and the intermediate molecules increase arithmetically.
9. Which of the following activity is not present in Taq polymerase?
a) 5’-3’ polymerase
b) 5’-3’ exonuclease
c) 3’-5’ exonuclease
d) Both 5’-3’ polymerase and 5’-3’ exonuclease
Answer
Answer: c [Reason:] Taq polymerase is not having 3’-5’ exonuclease activity. It is a proof reading activity and it is very important to have a check on the mutations if they are encountered in the PCR products.
10. Choose the correct statement.
a) Taq polymerase is having high processivity
b) Processivity is defined in this case as synthesis of DNA by polymerase
c) It requires a 5’ end for the elongation to take place
d) The maximum size of the molecules which can be synthesized is 10kbp
Answer
Answer: b [Reason:] Taq polymerase is having low processivity. It means that it falls off from the template before it has synthesized a large piece of DNA. It requires a 3’ OH for carrying out the elongation. The maximum size of the molecules which can be synthesized is 2-4 kbp.
11. What is the half life cycle for Taq polymerase?
a) 40 minutes
b) 80 minutes
c) 10 minutes
d) 50 minutes
Answer
Answer: a [Reason:] The half life cycle for Taq polymerase is 40 minutes at 94 degrees. Thus, at the end of the 30 cycles, a significant loss of activity takes place.
12. Taq polymerase incorporates which residue at 3’ end?
a) G
b) T
c) A
d) C
Answer
Answer: c [Reason:] It incorporates A residue at the 3’ end. This overhang is often useful in carrying out the cloning of these PCR products.
13. Polymerases are also available from other Thermus species. Which of the following is correct?
a) Thermus flavus gives Tfl enzyme
b) Thermus thermopilus gives Tfl enzyme
c) They are having proof reading activity
d) Thermus flavus gives Tth enzyme
Answer
Answer: a [Reason:] Other Thermus species also provide polymerases. Thermus flavus gives Tfl enzyme and Thermus thermopilus gives Tth enzyme. They are also 3’-5’ proof-reading activity.
14. Polymerases are available with proof reading activity. Which of the following are the characteristics of these types of polymerases?
a) They add a A residue at 3’ end
b) They are obtained from Thermococcus litoralis
c) They can’t be obtained from archaebacteria
d) The marine bacteria from which they are obtained grow at temperatures lower than that of Thermus aquatics
Answer
Answer: b [Reason:] As these polymerases are having a proof-reading activity, they generally don’t add any residue at the end. They are obtained from bacteria such as Thermococcus litoralis and grow at temperatures above than that of Thermus aquatics. They can also be obtained from archaebacteria.
15. Thermococcus litoralis grows at a temperature upto 98 degrees. The given statement is true or false?
a) True
b) False
Answer
Answer: a [Reason:] Thermococcus litoralis grows at a temperature upto 98 degrees. The half life is also high, 90% of its activity is retained after 1 hour of incubation at 95 degrees.
Set 4
1. Ligation is defined as:
a) Alignment of only double stranded DNA molecules at the ends and the formation of phosphodiester bonds between both the strands
b) Alignment of either of the double or single stranded DNA molecules and formation of glycosidic bonds between both the strands
c) Alignment of either of the double or single stranded DNA molecules and then formation of phosphodiester bonds. The bond can be between either one or both the strands
d) Alignment of single stranded DNA molecules and formation of glycosidic bonds between these strands
Answer
Answer: c [Reason:] Ligation is basically the formation of bonds between the ends of two DNA strands. The strands can be single or double. The nature of the bond is phosphodiester, which means that bond is formed between sugar and phosphate. The phosphodiester bond can either be formed between one strand or both the strands.
2. If only one bond is broken in the sugar-phosphate backbone, it is called as:
a) gap
b) nick
c) break
d) leakage
Answer
Answer: b [Reason:] In the case if only one bond is broken in the sugar phosphate backbone, it is termed as nick. If a number of nucleotides are missing, it is termed as gap. Nick can be sealed by ligation reactions but gap can’t be sealed by ligation reaction.
3. How many categories of ligation reaction are there on the basis of ends created?
a) 1
b) 2
c) 3
d) 4
Answer
Answer: b [Reason:] There are two categories of ligation reaction and are named as blunt-ended and sticky ended reaction. They are named because of the nature of the ends of the molecules to be ligated.
4. In the case of blunt-end ligation, blunt ends can be generated by:
a) simply the action of restriction endonuclease which gives straight ends
b) the polishing of staggered ends
c) both the action of restriction endonuclease which give straight ends and polishing of staggered ends
d) by the action of restriction endonuclease which gives staggered ends
Answer
Answer: c [Reason:] Blunt ends are those which don’t have single stranded ends as overhangs. The ends are double stranded. And these can be generated by polishing the staggered ends or the action of restriction endonuclease which generate straight ends.
5. The ligation reaction is more efficient in which case?
a) Blunt end ligation
b) Sticky end ligation
c) Both have same efficiency
d) Depends on the reaction conditions
Answer
Answer: b [Reason:] Sticky end ligation is generally more efficient than blunt end ligation. It is so because sticky end ligation is carried out because of complementary base pairing.
6. The sticky ends are held together by which type of bonds?
a) Hydrogen bond
b) Covalent bond
c) Ionic bond
d) Van-der-waal forces
Answer
Answer: a [Reason:] The sticky ends are held together by the hydrogen bonds. Hydrogen bonds are weak in nature and can easily be broken.
7. Why sticky ended ligations are carried out at temperatures lower than room temperature?
a) It is so because the vibrational and kinetic energy of the molecules at room temperature is lower than that of the energy required to break the bonds holding the ends
b) The energy required to break the bonds holding the ends is very less than that of the kinetic and vibrational energy at room temperature
c) The enzyme carrying out ligation is unstable at low temperature
d) The sticky ends created, don’t just religate at low temperature
Answer
Answer: a [Reason:] The sticky ends are held together by the hydrogen bonds. These hydrogen bonds are weak in nature and the energy required to break them is very less than the kinetic and vibrational energy of the molecules at room temperature. Hence the reaction is carried at 4 degrees.
8. Ligation reaction can be both intramolecular and intermolecular in nature. True or false?
a) True
b) False
Answer
Answer: a [Reason:] Ligation reaction can be both intramolecular and intermolecular in nature. The intermolecular reaction is that in which ends ligated are of two different molecules. The intramolecular ligation is that in which ends ligated belong to the same molecule.
9. If a ligation reaction is being carried out and recircularization is observed, which type of reaction is being carried out?
a) Intramolecular
b) Intermolecular
c) Both observe recircularization equally
d) Recircurlization is not possible in any of the case
Answer
Answer: a [Reason:] Recircularization is the phenomenon of joining the ends of the same molecule. It happens in the case of intramolecular ligation reaction.
10. What is the kinetics of the intramolecular and intermolecular ligation reactions?
a) Second order kinetics for intramolecular and first order for intermolecular
b) First order kinetics for intramolecular and second order for intermolecular
c) Both are first order
d) Both are second order
Answer
Answer: b [Reason:] As the intramolecular reaction requires only type of reaction species, the reaction is first order.Whereas, on the other hand intermolecular reaction requires collision between two different species and hence it is second order reaction.
11. What are the effects of increasing concentration of reaction components?
a) It increases chances of ligation in both intramolecular and intermolecular reactions
b) It increases chances of ligation only in intermolecular and no effect on intramolecular
c) It decreases chances of ligation in intramolecular and increase in that of intermolecular
d) It decreases chances of ligation in both types of reaction
Answer
Answer: b [Reason:] As the concentration of reaction components is increased, there are increased chances of ligation in intermolecular reaction because frequency of collision of two different molecules increases. The intramolecular reaction is unaffected because the probability of meeting the ends of a molecule remains same.
Set 5
1. Choose the incorrect statement for cDNA libraries.
a) They constitute of DNA copies produced from the RNA sequences and usually mRNA
b) They represent expressed sequences
c) Introns are not represented
d) Comparison of cDNA sequences with genomic sequences leads to determination of polyadenylation sites
Answer
Answer: c [Reason:] cDNA libraries constitute of DNA copies produced from RNA sequences and usually mRNA. They are not only the collection of expressed sequences but post transcriptional changes are also recorded. Introns can also be represented. Comparison of cDNA sequences with genomic sequences can lead to determination of positions of introns, polyadenylation sites etc.
2. A times partial sequencing of cloned cDNAs is carried out. These cDNA are known as:
a) expressed RNA sequences
b) expressed sequence tags (ESTs)
c) expressed cDNA sequences
d) library
Answer
Answer: b [Reason:] The partial sequencing of cloned cDNA is the first step of genome characterization project. These cloned cDNA sequences are known as expressed sequence tags (ESTs).
3. Polyadenylation of RNA species is an important criterion for the production of cDNA species. Which of the following holds true?
a) Polyadenylation should be at 3’ end
b) Eukaryotic mRNAs are mostly non-polyadenylated
c) Bacterial mRNAs and organelle mRNAs are polyadenylated
d) It is carried out by addition of T residues after synthesis
Answer
Answer: a [Reason:] Polyadenylation of RNA species is an important criterion for the production of cDNA species and polyadeylation should be at the 3’ end. Mostly eukaryotic mRNAs are polyadenylated. It is carried out by addition of A residues after synthesis. Usually, bacterial mRNAs and organelle mRNAs are non-polyadenylated residues.
4. Choose the incorrect statement for oligo-dT cellulose.
a) It is used for separation of polyadenylated mRNA from other mRNA
b) oligo-dT are covalently attached to the solid support via OH bonds
c) A solution containing RNA is passed through the column
d) Poly A tail attaches to the oligo-dT by ionic bonds
Answer
Answer: d [Reason:] oligo-dT cellulose is used for separation of polyadeylated mRNA from other mRNA. These are covalently attached on a solid support via OH bonds. For separation, a solution containing RNA is passed through the column and the molecules having poly A tails are attached to the column via hydrogen bonds.
5. Poly A tail from the column is eluted by using high salt concentration. Is the given statement is true or false.
a) True
b) False
Answer
Answer: b [Reason:] All non-specific mRNA are firstly eluted from the column. The poly A tail is eluted from using low salt concentration, it is so because low salt concentration destabilizes nucleic acids.
6. At times streptavidin is used in place of cellulose. Choose the correct statement for this alternative.
a) oligo-dT streptavidin conjugate can be extracted by magnetic beads
b) The magnetic beads are attached with activin
c) Recovery is not based on the magnetic properties of the beads
d) The interaction between polyA tail and the oligo-dT is reduced
Answer
Answer: a [Reason:] If streptavidin is used in the place of cellulose, oligo-dT streptavidin conjugate can be extracted by magnetic beads. These magnetic beads are having biotin attached to them. Recovery is based on the magnetic properties of the beads. The interaction between the polyA tail and the oilgo-dT still remains the basic principle and it is not affected.
7. Choose the correct with respect to RNA molecules.
a) They are less labile than DNA molecules
b) The 2’ hydroxyl group of ribose group decreases the activity
c) There is no less of activity while boiling
d) Baking of glassware, treatment with UV can be used for protection against degradation
Answer
Answer: d [Reason:] RNA molecules are more labile as compared to the DNA molecules and they are more reactive because of the 2’ hydroxyl group of ribose group. Baking of glassware, treatment with UV can be used as methods for protection against degradation. If precautions are not taken, there is subsequent loss of activity while boiling.
8. What is the basis of RNaseH method?
a) It is based on RNA synthesis by DNA strand
b) It is based on complementary DNA synthesis by RNA strand through reverse transciptase
c) It is based on complementary DNA synthesis by RNA strand through RNaseH enzyme
d) It is based on getting double strand RNA from a single strand
Answer
Answer: b [Reason:] RNase H method is based on the synthesis of the complementary DNA synthesis by RNA strand through reverse transciptase enzyme and it leads to formation of duplex of RNA and DNA.
9. The first step for RNaseH method is to anneal a chemically synthesized oligo-dT primer to the 3’ polyA tail of RNA. Is the given statement true or false?
a) True
b) False
Answer
Answer: a [Reason:] A chemically synthesized oligo-dT primer is annealed to the 3’ polyA tail of RNA. After this primer is attached the RNA DNA duplex is formed with the help of reverse transciptase.
10. For the synthesis of second DNA strand, incorporated oligo-dT is of no use. Why?
a) It is so because it is unstable
b) It is so because it is at the 3’ end of the template molecule
c) It is so because it would be unable to initiate replication
d) It is so because it is detaches itself after first strand synthesis
Answer
Answer: c [Reason:] For the second DNA strand synthesis, oligo-dT is of no use. It is so because it is present at 5’ end of the template and replication could not start from 5’ end.