Multiple choice question for engineering
1. Which of the statement holds for long-range PCR and in its relation?
a) It is the PCR in which longer templates are used
b) DNA polymerases which don’t have proof-reading activity give larger products
c) DNA polymerases’ processivity is not a measure to have larger products
d) It is PCR in which a mixture of enzymes is used to have larger products
Answer: d [Reason:] Long range PCR is the PCR in which a mixture of enzymes is used to have larger products. If DNA polymerases are having proof reading activity then we can obtain larger products because in this case chain terminators are not used. Also, the processivity of the enzymes is also very important.
2. Which of the following conditions don’t contribute to wrong annealing to primer?
a) Chance complementarity
b) Conditions of annealing
c) The original sequence of the primers
d) Both the conditions of annealing and the original sequence don’t play any role
Answer: d [Reason:] There are cases when wrong annealing of the primer takes place. It can happen because of chance complementarity, conditions of annealing such as ionic concentration and temperature. The original sequence of the primers is very important.
3. How can specificity of primer annealing be increased?
a) Use of short primers
b) Raising temperature
c) Adjusting the concentration of sodium ions
d) Using polymerase with proof reading activity
Answer: b [Reason:] The specificity of primers annealing can be increased in various ways such as increasing temperature, using long primers. If long primers are used there are increased chances of having correct matching. Also, if the magnesium ion concentration is adjusted, annealing can be done more effectively. It is so because they stabilize primer-template binding.
4. There are basically two types of contamination, laboratory and external. If a PCR product is found to be contaminated by bacteria. It comes under laboratory contamination. Is the given statement true or false?
Answer: b [Reason:] Laboratory contamination consists of aerosols in pipettes from previously formed PCR products or related DNA sequences. External contamination includes contamination from bacteria, fungi and other human contamination.
5. Which can be used as a precaution in order to minimize contamination?
a) Careful use and design of pipettes
b) Placing the pre-PCR and post-PCR stages in same rooms
c) Extracting the DNA along with surface layers
d) Use of primers carefully is not very important
Answer: a [Reason:] Pipettes are very important in minimizing contamination. Thus they should be designed and used carefully. The pre-PCR and post-PCR stages should be places in separate rooms. While extraction of DNA surface layers should be removed because they might be containing bacteria. Sometimes use of species specific primers is also very important, thus they should be used carefully.
6. During amplification, there are chances of having a product of mixture of different sequences. There are various ways to detect it. Which of the statement is true in regard to it?
a) Direct sequencing can’t be used in the case if the template DNA is heterozygous at the locus
b) Direct sequencing can be used if the template DNA is heterozygous at the locus
c) If cloning is done before sequencing, then it is detected via using only a single clone for sequencing
d) In the case several recombinants are used, it can’t go undetected
Answer: b [Reason:] If the template DNA is heterozygous at the locus, it can be detected via using direct sequencing. It is so because it would give rise to two different signals at same nucleotide position at the sequence output. If the cloning is done before sequencing, a single clone won’t be helpful to detect heterogeneity. It is because single clones are derived from single PCR product. Also, if several recombinants are used, there are chances that they go undetected.
7. If the template DNA belongs to several individual rather than single one, this type of heterogeneity is known as:
b) Product heterogeneity
c) Population heterogeneity
d) Template heterogeneity
Answer: c [Reason:] If the template DNA belongs to several individual then this type of heterogeneity is known as population heterogeneity. It also gives rise to heterogeneity in PCR products.
8. Heterogeneity can also arise if DNA is damaged before amplification. Which of the following doesn’t cause DNA damage?
a) Amination of bases
b) Chemical cross-linking between the strands
c) Chemical cross-linking within the strands
d) Slowing down the polymerase
Answer: a [Reason:] There are various reasons of DNA damage such as chemical cross-linking both between and within the strands. Deamination of bases is also one of the reasons. If polymerase is slowed down there are chances of incorporation of incorrect bases.
9. If cytosine is deaminated, which of the base is formed?
Answer: d [Reason:] If cytosine is deaminated it leads to formation of Uracil. This Uracil is further read as Thymine during DNa synthesis.
10. Which of the statement is correct for misincorporation?
a) If direct sequencing is carried this misincorporation is a great problem
b) The erroneous molecules give strong signals than genuine molecules in case of misincorporation
c) If the misincorporation in cloned PCR products it is a problem
d) Even if the error is induced at an early stage it is not incorporated in many sequences
Answer: c [Reason:] If the misincorporation takes place and direct sequencing is carried out, it is not a major problem. It is so because only small portion of molecules have it and thus the signals are weaker than that of genuine molecules. If the misincorporation is in cloned PCR product, it is of great problem. It is so because if it is included at an early stage, they are are incorporated in many sequences.
11. Errors can be introduced because of many reasons such as polymerase error or because of heterogeneity. Which of the statement holds true?
a) The error caused because of polymerase is biased for the first position in the codon
b) The error caused because of polymerase is evenly distributed on all the positions in the codon
c) The error caused by sequence heterogeneity is mainly because of first position in the codon
d) The error caused by sequence heterogeneity is evenly distributed on all the codon positions
Answer: b [Reason:] The error caused by polymerase whether due to template change or not, it is always distributed evenly over all the codon positions. If the error is caused by sequence heterogeneity is concentrated on the third codon position. It generally doesn’t leads to amino acid substitution.
12. Which of the statement is incorrect for jumping PCR?
a) It is used in the case when the DNA fragment is degraded
b) In this type of PCR, the molecules are not long enough to span between the two primer sites
c) At the end of first round of synthesis, the extension of the molecule from the primer site to the end of the fragmented molecule takes place
d) It doesn’t leads to the formation of chimeric product
Answer: d [Reason:] Jumping PCR is used in the case when the molecule is not long enough to span between the two primer sites. At times, the whole amplification doesn’t takes place at one go and hence molecule anneals to other fragment having another portion. Thus at times PCR products longer than the template are designed. But the disadvantage is that it leads to formation of chimeric products at times.
1. Primers are generally:
a) 20-30 nucleotides long
b) 40-50 nucleotides long
c) as long as the template is
d) taken according to the amount available
Answer: a [Reason:] Primers are generally short in length. They are 20-30 nucleotides long. It is easier to match short primers with the template in comparison to long primers. But in the case of eukaryotic DNA as template, long primers are preferred.
2. Which end of the primer should be matched properly in order to carry out the amplification?
a) 5’ end
b) 3’ end
c) Both of the ends should be matched properly
d) Any one of the ends should match
Answer: b [Reason:] For carrying out the amplification, it is not necessary that the whole primer should match with the template. But it is necessary that 3’ end should match because if 3’ end is not matched the polymerase won’t be able to carry out elongation.
3. Which of the following nucleotides should be there at 3’ end?
a) Any of A, T, G or C will work out
b) Either A or T
c) Either G or C
d) Specifically G
Answer: c [Reason:] At 3’ end, either G or C should be present. It is so because; these form three hydrogen bonds and thus are stronger. Whereas, A and T form only two hydrogen bonds and thus are comparatively less strong than G or C.
4. Melting temperature is given by:
a) 4(G+C) + 2(A+T)
b) 2(G+C) + 4(A+T)
c) 2(A+G) + 4(C+T)
d) 4(A+G) + 2(C+T)
Answer: a [Reason:] The melting temperature is that temperature at which the primers associate with the template. The melting temperature is decided by the amount of different nucleotides present.
5. Both the primers, the start primer and the end primer should have nearly same melting temperature. Is the given statement true or false?
Answer: a [Reason:] Both the primers should have nearly same melting temperature. It is so because, then they would nearly bind at the same time as the temperature is being lowered for their annealing.
6. Why internal secondary structures are not preferred for primers?
a) Internal structures are very bulky and thus elongation is not preferred
b) Because of it, primer may fold back on itself and won’t be available for template
c) Internal secondary structures require more amount of template
d) If internal structures are present, no proof reading would be observed
Answer: b [Reason:] Internal structures are not preferred because if they are present the primer may fold back on itself. As the primer folds back on it, it is not available for the template. As an intramolecular reaction, self annealing is preferred over intermolecular annealing of primer to the template.
7. Which of the following is favoured for primer design?
a) The melting temperature should be different for both the primers
b) Primers should be long in length
c) Primers should not be complementary to each other
d) Matching should be of whole primer to the template
Answer: c [Reason:] Primers should not be complementary to each other. It is so because if they are complementary, primer dimer formation takes place. If primer dimer is formed, proper elongation won’t be taking place.
8. What will happen if the amino acid sequence is used directly for primer designing?
a) There would be certainty because the genetic code is unique for each amino acid
b) There would be uncertainty as the genetic code is degenerate and none of the amino acid is having a unique code
c) There would be uncertainty as the genetic code is degenerate but some of the amino acids such as methionine are having a unique codon
d) The amount of uncertainty or certainty is a matter of chance
Answer: c [Reason:] If the amino acid sequence is used directly, there would be uncertainty. It is so because, the genetic code is degenerate. It means that, one codon can correspond to many amino acids. There are some exceptions such as methionine which are having a unique codon.
9. In the case of uncertainty, if more than one nucleotide is included at a position it is called as:
a) mixed site
b) polynucleotide site
c) unique site
d) degenerate site
Answer: a [Reason:] If uncertainty is there at times, more than one nucleotide can be included at one position. And in this condition it is called as mixed site.
10. What is the property of inosine?
a) Having narrow range of pairing capabilities
b) Having a broad range of pairing capabilities
c) The pairing capability is same as the normal nucleotides
d) It is abbreviated as A
Answer: b [Reason:] It is having a broad range of pairing capabilities and thus can be used in order to have less amount of uncertainty.
1. Some vectors carry a mutant form of promoter known as lacUV5 promoter. It carries ______ in the promoter region and _____ the efficiency.
a) point mutations, decreases
b) point mutations, increases
c) frameshift mutations, increases
d) frameshift mutations, increases
Answer: b [Reason:] In some vectors, lacUV5 promoter is present. It is a mutant form and carries point mutations which increase the efficiency.
2. Lambda PL promoter is used in which vectors?
a) Cloning vectors
b) Expression vectors
c) Both cloning and expression vectors
d) Bacteriophage Mu
Answer: b [Reason:] Lambda PL promoters are the promoters for the left region in bacteriophage lambda. It is widely used in expression vectors.
3. The promoter can be controlled by a repressor which is temperature sensitive. Is the given statement true or false?
Answer: a [Reason:] The promoter is activated at a temperature higher than 30 degrees because at this temperature repressor is inactivated.
4. Choose the correct statement for rifampcin.
a) It inactivates both E. coli polymerase and T7 polymerase
b) It activates both E. coli polymerase and T7 polymerase
c) It inhibits T7 polymerase but doesn’t inhibits E. coli polymerase
d) It inhibits E. coli polymerase but doesn’t inhibits T7 polymerase
Answer: d [Reason:] Rifampcin is added in order to reduce the transcription of other genes. It inhibits the E. coli RNA polymerase but doesn’t inhibit the T7 polymerase.
5. The ______ operon encodes proteins involved in arabinose metabolism.
Answer: c [Reason:] The araBAD operon encodes proteins involved in arabinose metabolism. It is controlled by AraC transcriptional regulator.
6. tac promoter is an example of which type of promoter?
a) hybrid promoter
b) fusion promoter
c) lacZ promoter
d) araBAD promoter
Answer: a [Reason:] Hybrid promoters are those promoters which are produced by two promoters from different sources. Tac promoter is a hybrid promoter produced from trp promoter and lacUV5 promoter.
7. The tac promoter is made by _____ region of trp promoter and _____ region of lacUV5 promoter.
a) 10, 35
b) 35, 10
c) 10, 10
d) 35, 35
Answer: b [Reason:] It is a hybrid promoter made by 35 region of trp promoter and 10 region of lacUV5 promoter.
8. The tac promoter includes lac operator and is regulated by repressor. Is the given statement true or false?
Answer: a [Reason:] The tac promoter includes lac operator and is regulated by the repressor. The repressor is to be supplied by the host.
9. Expression of T7 promoter- lac operator hybrid requires:
a) T7 RNA polymerase
b) An inducer such as IPTG
c) Both T7 RNA polymerase and inducer such as IPTG
d) T7 DNA polymerase
Answer: c [Reason:] Expression of T7 promoter- lac operator hybrid requires both T7 RNA polymerase and inducer such as IPTG. If inducer is absent levels of expression are very low.
10. When lacUV5 control system is used, addition of IPTG _______ the expression of _________
a) activates, T7 RNA polymerase
b) inactivates, T7 RNA polymerase
c) activates, T7 DNA polymerase
d) inactivates, T7 DNA polymerase
Answer: a [Reason:] When lacUV5 control system is used, addition of IPTG activates the expression of T7 RNA polymerase. Thus transcription of sequences under the control of T7 promoter is controlled.
1. Isolation of genomic DNA follows same principles as that of obtaining plasmid from E. coli. Which of the following is not included in it?
a) Cell lysis
b) Removal of proteins
c) Removal of chromosomal DNA
d) Dissolving plasmid in water
Answer: d [Reason:] There are some basic steps which are included in obtaining plasmid DNA from E. coli.
Firstly, the cell is lysed, further removal of proteins and chromosomal DNA is done. Plasmid is obtained and collected but not in water. Also, further purification is done if necessary.
2. How many methods are there for obtaining the plasmid DNA from the bacteria?
Answer: b [Reason:] There are two methods which are used for obtaining the plasmid DNA from the bacteria. They are named as alkaline lysis method and boiling lysis method. They both are having different working principles.
3. Cell lysis is carried out by which substance?
a) Lysozyme and detergents
c) Sugar solution
d) Suphuric Acid
Answer: a [Reason:] Cell lysis is carried out by adding lysozyme and detergents. The cell wall is made up of N-acetyl glucoasamine and N-acetyl muramic acid and they are having cross links. The agents added break the cross links present between the molecules of the cell wall.
4. Chromosomal or genomic DNA is separated by:
b) Dissolution in water
Answer: c [Reason:] Chromosomal or genomic DNA is comparatively heavier and large in size than that of plasmid DNA. Hence, centrifuging at a high speed leads to settling down of the genomic DNA and thus can be separated easily.
5. Proteins can be removed via treatment by?
a) Phenol and chloroform treatment
b) Treatment with sodium hydroxide
c) Chloroform treatment alone
Answer: a [Reason:] Proteins can be removed via treatment with phenol and chloroform treatment. Chloroform is alone not sufficient. The phenol added helps in destruction of proteins and chloroform helps in its dissolution under acidic conditions.
6. The nucleic acid remaining in the solution can be precipitated by addition of sodium or ammonium acetate and ethanol. Is the statement true or false?
Answer: b [Reason:] The nucleic acid is present in the solution and is precipitated by addition of sodium or ammonium acetate and ethanol. It is because; nucleic acid is polar in nature and thus easily dissolves in water. Hence, to avoid this sodium acetate and ethanol is added. Sodium acetate is shields the charge present on the sugar phosphate backbone and further bonds are easily form between ethanol and phosphate. It leads to separating out of nucleic acids.
7. Nucleic acid precipitated constitutes of:
a) plasmid DNA
b) plasmid DNA, along with RNA and chromosomal DNA
c) RNA alone
d) chromosomal DNA only
Answer: b [Reason:] Nucleic acid precipitated contains of the plasmid DNA and along with it RNA and remnants of chromosomal DNA are also present. RNA can be removed via adding RNase.
8. Treatment with exonuclease leads to removal of:
a) remnants of chromosomal DNA
c) plasmid DNA which is circularized
Answer: a [Reason:] Exonuclease leads to removal of remnants of chromosomal DNA because they are usually having linear ends. The circularized ends of plasmid are protected from the action of exonuclease because they don’t have any free ends for their action.
9. Adsorption onto a solid phase support followed by elution is used as an alternative for separation of which component:
a) chromosomal DNA
b) plasmid DNA
c) RNA alone
d) other impurities
Answer: b [Reason:] This method is used for separation of plasmid DNA. It is advantageous because it avoids the use of phenol and also removes RNA at times along with plasmid DNA.
10. Which of the following components bind to the solid column made of silica, under high salt concentration?
c) Both proteins and polysaccharides
d) Plasmid DNA
Answer: d [Reason:] Plasmid DNA binds to solid support which is made of silica and under high salt concentrations. A high salt concentration doesn’t allow less polar molecules to bind such as polysaccharides and proteins. The binded DNA molecule is further eluted by using a low salt concentration.
11. Purification of DNA by using silica derivatized groups by DEAE is termed as:
a) ion exchange resin based method.
b) silica based purification
c) atom based resin exchange method
d) packed bed purification
Answer: a [Reason:] Silica derivatized groups by DEAE are used for purification of DNA. These groups are positively charged and the DNA gets attached to it, along with other species such as RNA which are negatively charged. Further, DNA can be obtained by varying the ionic concentrations.
12. Which of the following is correct with respect to caesium chloride centrifugation?
a) Caesium is light in weight
b) The dissolution of caesium and nucleic acids leads to the formation of gradients
c) According to amount of supercoiling and A+T content, the DNA settles
d) Nicked DNA settles below than supercoiled DNA
Answer:b [Reason:] Caesium chloride is heavy and when nucleic acid is dissolved with it, density gradients are formed. According to extent of supercoiling and G+C content, settling of DNA takes place. The nicked DNA settles above than the nicked DNA.
13. Which of the following components settles at the bottom?
c) Nicked DNA
d) Supercoiled DNA
Answer: a [Reason:] The component settling at the bottom is RNA. And the proteins float on the free surface. The nicked DNA forms a band above the supercoiled form.
14. The location of plasmid DNA can be visualized by addition of:
a) bromophenol blue
b) ethidium bromide
c) ortho xylene
d) texas red
Answer: b [Reason:] Ethidium bromide is added before centrifugation. It is an orange-red coloured stain which gives rosy coloured bands when placed under UV light. It acts upon by intercalating between the bases.
1. Which of the promoter requires low level of cI gene for its activation?
Answer: d [Reason:] The promoter requiring low level of cI gene for its activation is PRM (Promoter for repressor maintenance). It is inactivated by high levels of cI gene.
2. Choose the correct statement with respect to lysogen and prophage.
a) The integrated phage is called as lysogen
b) The infected bacterial cell is called as prophage
c) If there is only growth of few lysogens, the E. coli lawn would be turbid
d) If lysogens won’t be formed, the plaques won’t be clear
Answer: c [Reason:] The integrated phage is called as prophage and the infected bacterial cell is known as lysogen. Lysogens are immune to further infection by the same phage. If only growth of few lysogens would take place, the lawn would be turbid. In the acse if lysogens aren’t formed, the plaques would be clear.
3. Choose the incorrect statement for cro and Q proteins.
a) They are required for the lytic life cycle to continue
b) Cro protein inactivates the PRM promoter
c) The Q protein is responsible for expression of late genes
d) cII protein is required for the lytic life cycle to continue
Answer: d [Reason:] The cro and Q proteins are required for the lytic life cycle to continue and cII protein is not required for it. Cro protein inactivates the PRM promoter and the Q protein is responsible for expression of late genes. These late genes are responsible for producing coat proteins, allowing assembly of functional phage and cell lysis.
4. Induction of lysogen takes place because of:
a) Low levels of cII gene
b) Low level of cI gene
c) Low levels of both cI and cII gene
d) High levels of both cI and cII gene
Answer: b [Reason:] Induction of lysogen takes place because of low levels of cI gene. It is caused by specific proteolysis under the action of host recA protein.
5. Which of the phage genes are responsible for phage excision?
a) xis gene
b) int gene
c) recA protein
d) both xis and int gene
Answer: d [Reason:] As the level of cI genes fall, the repression of PR and PL genes is lifted up. Now phage genes are expressed and xis and int genes are expressed. These genes are responsible for causing phage excision.
6. How many phases are there for replication of lambda DNA to take place?
Answer: b [Reason:] There are two phases for the lambda DNA replication to take place. They include the theta mode and rolling circle mode.
7. The phase generating additional circular DNA molecules in the first phase does it by:
a) bidirectional theta mode
b) rolling circle mode
c) separation of the two strands
d) copying only one strand
Answer: a [Reason:] The first phase includes circularization of the molecule by annealing of the cos sites and they replicate in a bidirectional theta mode. This leads to generation of an additional circular DNA molecule.
8. Choose the incorrect statement with respect to rolling circle replication.
a) It is for a single replication fork and yields concatemeric molecule
b) Staggered cleavage of DNA takes place at the cos sites
c) The cleavage generates 10 nucleotide long overhangs
d) Without the cos sites, packaging won’t be possible
Answer: c [Reason:] The second phase constitutes of rolling circle mode. It is done with a single replication fork and yields concatemeric molecules. They are required for assembly into mature phage particles. The packaging region should be flanked by the cos sites. Phage DNA is inserted onto the phage head and the cos sites are brought adjacent to each other. It is followed by cleavage at cos sites and it leaves 12 nucleotide long staggered ends. The cos sites are required for packaging to take place.