Which of the following is not a tool in recombinant DNA technology MHT CET 2022?

Latest Recombinant DNA Technology MCQ Objective Questions

Recombinant DNA Technology Question 1:

What initiation and termination factors are involved in transcription in Eukaryotes?

1. β and γ, respectively
2. α and σ, respectively
3. σ and ρ, respectively
4. α and β, respectively

Answer (Detailed Solution Below)

Option 3 : σ and ρ, respectively

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The correct answer is option 3.

Solution

Concept:

• The process of the formation of mRNA from DNA is called transcription.
• The process of transcription occurs in the nucleus.

Explanation:

There are three major steps in the transcription:

• Initiation
  • First, the DNA double helix is uncoiled by the enzyme RNA polymerase leading to the formation of a template strand and a coding strand.
  • The strand with polarity 3' to 5' acts as a template strand and mRNA is synthesized complementary to it. In the place of thymine, Uracil is added.
  • The template strand has a promoter, a structural gene, and a terminator.
  • The RNA polymerase further binds to the promoter and starts the initiation.
  • For the initiation to begin, it couples with an initiation factor called the sigma factor (σ).
  • The RNA polymerase is capable of only catalyzing elongation.
  • Hence, it requires an initiation factor and termination factor for initiation and termination, respectively
• Elongation
  • The RNA polymerase uses nucleoside triphosphate and polymerizes in a template-dependent manner following the complementary rule.
  • The elongation continues until the RNA polymerase reaches the terminator.
• Termination
  • Once RNA polymerase encounters the terminator, the nascent RNA falls off.  
  • The RNA polymerase combines with the termination factor (ρ) to terminate the process of transcription. 
• RNA Polymerase I, RNA Polymerase II, and RNA Polymerase III play an important role in eukaryotes in synthesizing different types of RNA.

So, the correct answer is option 3.

Recombinant DNA Technology Question 2:

Which of the following is not used as a vector in recombinant DNA technology?

1. Plasmid
2. Phagemid
3. YAC
4. EcoR1

Answer (Detailed Solution Below)

Option 4 : EcoR1

Concept:

• Cloning vectors are defined as DNA molecules that can carry a foreign DNA fragment when inserted into it.
• Cloning vectors are an essential tool for recombinant DNA technology.
• Criteria for an ideal vector:
  • Small and easy to isolate.
  • Origin of Replication (ori) - It is the DNA sequence from where replication starts and any DNA fragment linked to it can be made to replicate within the host cell.
  • Cloning sites - There should be one or more unique restriction sites into which DNA fragment can be inserted.
  • Selectable Markers - They should have a selectable marker that helps in recognition of transformants. E.g.- Antibiotic resistance genes.

• EcoRI is a restriction enzyme.
• Restriction enzymes or restriction endonucleases are endonucleases that have the ability to cut at a specific site within DNA.
• Restriction enzymes were first isolated in 1963 as an enzyme that restricted growth of bacteriophage in E. coli.
• Restriction enzymes have a specific recognition sequence which is a palindromic nucleotide sequence.
• These sequences are base pair sequences that read the same on both the strands of DNA when orientation of reading is same, i.e., either 5' to 3' or 3' to 5'.
• The sites at which they cut are called restriction sites.
• The action of restriction endonucleases on DNA is known as restriction digestion as it cleaves the DNA.
• Restriction endonucleases cut at the same points on both the strands.
• Example -
  • The recognition sequence for EcoRI is 5'- GAATTC -3'.
  • The restriction site is between G and A on both the strands.
  • EcoRI produces sticky-ends or cohesive ends of DNA as it cuts away from the center of the recognition sequence.

Therefore, EcoRI is NOT a vector in recombinant DNA technology.

• Plasmid -
  • They are closed, circular, extra-chromosomal, self-replicating double-stranded DNA molecules present in bacterial cells.
  • They are most suitable as cloning vectors and widely used in recombinant DNA technology.
  • They can carry an insert of up to 15 kb.
• Phagemid - 
  • These are vectors with both plasmid and bacteriophage properties.
  • Bacteriophage is a virus that can infect a bacteria and is commonly called phage.
  • It contains the F1 origin of replication of the F1 phage, and can be replicated in a cell like a plasmid.
• Yeast Artificial Chromosome (YAC) - 
  • These are designed to replicate as plasmids in bacteria when no foreign DNA is present.
  • Once a DNA fragment is inserted, YACs are transferred to cells, and they then replicate as eukaryotic chromosomes.
  • They can carry up to 500 kb of DNA.

Recombinant DNA Technology Question 3:

During DNA replication, Okazaki fragments are used to elongate

1. The lagging strand away from the replication fork
2. The leading strand towards replication fork
3. The lagging strand towards replication fork
4. The leading strand away from replication fork

Answer (Detailed Solution Below)

Option 1 : The lagging strand away from the replication fork

Correct Answer: 1)

Solution:

Concept: 

• At the time of DNA replication, two DNA polymerase molecules work simultaneously at the DNA fork, one on the lagging strand and the other on the leading strand.

• At lagging strand in 5' → 3' direction, Okazaki fragments are synthesized by DNA polymerase. As the replication fork opens further, new Okazaki fragments appear.

• Because of the opposite direction of the lagging strand, discontinuous synthesis takes place and Okazaki fragments are produced.

• These Okazaki fragments are then joined by a DNA ligase enzyme.

Fig: Lagging and Leading strands at the time of DNA Replication

Explanation:

Option 1: The lagging strand away from the replication fork – CORRECT

• Because of the opposite direction of the lagging strand, discontinuous synthesis takes place and Okazaki fragments are produced.
• Hence, this option is correct.

Option 2: The leading strand towards the replication fork– INCORRECT

• Continuous synthesis is there in the leading strand, so, Okazaki fragments are not produced.

• Hence, this option is incorrect.

Option 3: The lagging strand towards the replication fork– INCORRECT

• The lagging strand is synthesized in opposite direction.

• Hence, this option is incorrect.

Option 4: The leading strand away from replication fork – INCORRECT

• The synthesis of the leading strand is in the direction of replication and not away from it.

• Hence, this option is incorrect.

So, the correct answer is option 1).

Recombinant DNA Technology Question 4:

Under which of the following conditions will there be no change in the reading frame of the following mRNA? 5'AACAGCGGUGCUAUU3'

1. Insertion of G at 5th position
2. Deletion of G from 5th position
3. Insertion of A and G at 4th and 5th positions respectively
4. Deletion of GGU from 7th, 8th and 9th positions

Answer (Detailed Solution Below)

Option 4 : Deletion of GGU from 7th, 8th and 9th positions

The correct answer is option 4.

Concept:

• The genetic code is a set of three-nucleotide combinations called codons.
• Each codon codes for a specific amino acid or stop codon.
• There are 64 trinucleotide codons out of which 61 code for specific amino acids and three are stop codons (UAA, UAG, and UGA).

Explanation:

Frameshift insertion or deletion mutations

• Effects of large deletions, and rearrangement in the DNA sequence results in loss/gain of a gene or function of the gene.
• Ex. change in single base pair in the beta-globin gene results in a change in amino acid sequence from glutamate to valine.
• This condition is called sickle cell anemia.
• Insertion or deletion of one or two bases changes the reading frame of the DNA from the point of insertion or deletion.
• Such mutations are called frameshift insertion or deletion mutations

Insertion or deletion of three or multiples of three bases

• Insertion or deletion of three or multiples of three bases from the DNA does not change the reading frame.
• This is because a set of three nucleotides (codons) codes for a specific amino acid.

Now, let us look at each option:

• Insertion of G at 5th position:
  • As it is a single-point mutation it will alter the frame.
  • It will result in the frame as shown below:
    • 5' AAC AGG CGG UGC UAU U 3'
• ​Deletion of G from 5th position:
  • It is also a single-point mutation it will alter the frame.
  • It will result in the frame as shown below:
    • ​5' AAC ACG GUG CUA UU 3'
• Insertion of A and G at 4th and 5th positions respectively:
  • This will result in the insertion of two bases and will alter the frame.
  • The resultant frame will be:
    • 5' AAC AAG GCG GUG CUA UU 3'
• Deletion of GGU from 7th, 8th, and 9th positions:
  • Since it is the deletion of three nucleotides (a codon) it will not have any effect on the reading frame.
  • The resultant frame will be:
    • 5'AAC AGC GCU AUU 3'

So, the correct answer is option 4.

Recombinant DNA Technology Question 5:

Expressed Sequence Tags (ESTs) refer to :

1. Genes expressed as RNA
2. Polypeptide expression
3. DNA polymorphism
4. Novel DNA sequences

Answer (Detailed Solution Below)

Option 1 : Genes expressed as RNA

Correct Answer: 1)

Solution:

Concept:

• Expressed Sequence Tags (ESTs) are the sequence of DNA (genes) that are expressed as mRNA for protein synthesis.
• These are used in human Genome Project because analysis of ESTs constitutes a useful approach in the identification of genes.
• Expressed Sequence Tags (ESTs) are derived through single sequencing reactions which are performed on randomly selected clones of cDNA libraries.

Explanation:

Option 1: Genes expressed as RNA – CORRECT

• ESTs are the sequence of DNA (genes) that are expressed as mRNA.
• And are useful for the identification of genes.
• Hence, this option is correct.

Option 2: Polypeptide expression – INCORRECT

• A polypeptide is a chain of amino acids.
• The polypeptide is the final product of a gene expression.
• While ESTs are the sequences of mRNA.
• Hence, this option is incorrect.

Option 3: DNA polymorphism – INCORRECT

• DNA polymorphism is the presence of variants of specific DNA sequences in different populations or individuals.
• And as we already know that the ESTs are the sequences of mRNA.
• Hence, this option is incorrect.

Option 4: Novel DNA sequences – INCORRECT

• Novel DNA sequences are present in the genome of an individual but are not there in the human reference assembly of the Human Genome Project.
• We can say novel DNA sequences are individual and population specific.
• Hence, this option is also incorrect.

So, the correct answer is option 1).

First restriction endonuclease enzyme was-

1. Hind-II
2. EcoR I
3. Hae III
4. Bam-I

Answer (Detailed Solution Below)

Option 1 : Hind-II

Concept-

• Restriction enzymes belong to a larger class of enzymes called nucleases.
• Restriction enzymes are used to break DNA molecules.
• Restriction enzymes are of three types- Exonuclease, endonuclease, and restriction endonuclease.
  • Exonuclease- They cut off nucleotides from 5' or 3' ends of DNA molecule.
  • Endonuclease- They break DNA duplex at any point except the end.
  • Restriction endonuclease- They cleave DNA duplex at specific points.

Explanation-

• Restriction endonuclease was found by Arber.
• They act as "molecular scissors".
• They recognize the base sequence at palindrome sites in the DNA duplex and cut its strands.
• Hind II, whose functioning depended on a specific DNA nucleotide sequence was isolated.
• It was found that Hind II always cut DNA molecules at a particular point by recognizing a specific sequence of six base pairs.
• This specific base sequence is known as the recognition sequence for Hind II.

The first restriction endonuclease-Hind II.

Name	Recognition sequence	Source
EcoRI		Escherichia coli-containing drug resistance plasmid RI.
Bam I		Bacillus amyloliquefaciens
Hae III		Haemophilus aegyptius
Hind III		Haemophilus influenzae

Technique for transferring foreign DNA into a host organisms DNA is known as

1. Blotting technique
2. Recombinant DNA technology
3. Gene cloning technique
4. PCR technique

Answer (Detailed Solution Below)

Option 2 : Recombinant DNA technology

The correct answer is Recombinant DNA technology.

Recombinant DNA technology:

• This technique is mainly used to change the phenotype of an organism (host) when a genetically altered vector is introduced and integrated into the genome of the organism.
• So, this process involves the introduction of a foreign piece of DNA structure into the genome which contains our gene of interest. This gene that is introduced is the recombinant gene. So, option 2 is correct.

​

• Blotting technique: Blots are techniques for transferring DNA, RNA, and proteins onto a carrier so they can be separated, and often follows the use of gel electrophoresis. The Southern blot is used for transferring DNA, the Northern blot for RNA, and the western blot for PROTEIN.
• Gene cloning is a traditional method or technique used in molecular genetics to make DNA copies.
• Polymerase chain reaction, or PCR, is a technique to make many copies of a specific DNA region in vitro (in a test tube rather than an organism).

The first Human hormone product by Recombinant DNA technology is

1. Insulin
2. Thyroxine
3. Estrogen
4. Progesterone

Answer (Detailed Solution Below)

Option 1 : Insulin

Concept-

• In genetic engineering, we can break the DNA molecule at two desired places with the help of endonuclease.
• And then insert it in another DNA molecule at the desired place.
• The new DNA molecule is called recombinant DNA.
• Genetic engineering can be used to improve the quality of human life.

Explanation-

• The first genetically engineered insulin obtained by recombinant DNA technique with the help of E-Coli was developed by the American firm, Eli-Lilly on 5 July 1983.
• Insulin is used for diabetes.
• Insulin consists of two short polypeptide chains: chain A and chain B, that are linked together by disulfide bridges.
• Eli Lilly an American company prepared two DNA sequences corresponding to A and B, chains of human insulin, and introduced them in plasmids of E. coli to produce insulin chains.
• Chains A and B were produced separately, extracted, and combined by creating disulfide bonds to form human insulin. 

Therefore the first Human hormone product by Recombinant DNA technology is insulin.

• Estrogens produce wide-ranging actions such as stimulation of growth and activities of female secondary sex organs, development of growing ovarian follicles, the appearance of female secondary sex characters (e.g., the high pitch of voice, etc.), mammary gland development.
• Estrogens also regulate female sexual behavior.

Progesterone

• Progesterone supports a pregnancy.
• Progesterone also acts on the mammary glands and stimulates the formation of alveoli (sac-like structures which store milk) and milk secretion. 

Thyroxine

• Thyroxine is the hormone secreted by the thyroid gland.

Which of the following feature(s) facilitate cloning into a vector?

1. Origin of replication

2. Selectable marker

3. Cloning sites

1. 1 only
2. 2 and 3
3. 1 and 3
4. 1, 2 and 3

Answer (Detailed Solution Below)

Option 4 : 1, 2 and 3

The correct answer is Option 4, i.e. 1, 2 and 3.

• Vectors used at present, are engineered in such a way that they help easy linking of foreign DNA and selection of recombinants from non-recombinants.
• The following are the features that are required to facilitate cloning into a vector.

1. Origin of replication (ori):
   • This is a sequence from where replication starts and any piece of DNA when linked to this sequence can be made to replicate within the host cells.
   • This sequence is also responsible for controlling the copy number of the linked DNA.
   • So, if one wants to recover many copies of the target DNA it should be cloned in a vector whose origin support high copy number.
2. Selectable marker:
   • In addition to ‘ori’, the vector requires a selectable marker, which helps in identifying and eliminating non-transformants and selectively permitting the growth of the transformants.
3. Cloning sites:
   • In order to link the alien DNA, the vector needs to have very few, preferably single, recognition sites for the commonly used restriction enzymes.
   • The ligation of alien DNA is carried out at a restriction site present in one of the two antibiotic resistance genes.

Explanation:

• For example, the above figure is of plasmid pBR322, which is a cloning vector.
• It contains an origin of replication (ori).
• It has several cloning sites like PstI, PvuI, EcoRI, BamHI and SalI.
• The selectable markers here are ampicillin resistance (ampR) and tetracycline resistance (tetR).

Each Restriction enzyme cleaves a DNA molecule only at :

1. the ends of genes
2. methyl groups
3. specific nucleotide sequence
4. center of DNA molecule

Answer (Detailed Solution Below)

Option 3 : specific nucleotide sequence

Concept:

• A restriction enzyme is a protein produced by bacteria that cleaves DNA at specific sites along with the molecules.
• It is also known as restriction endonuclease and "molecular scissors".
• The first restriction endonuclease is Hind II.
• In the bacterial cell, restriction enzymes cleave foreign DNA, thus eliminating infecting organisms.
• This enzyme can be isolated from the bacterial cells and used in the laboratory to manipulate fragments of DNA.
• Restriction enzymes were characterized by molecular biologist Werner Arber.
• Work of Restriction enzyme:
  • A bacterium uses a restriction enzyme to defend against bacterial viruses called bacteriophages.
  • When a phage infects a bacterium, it inserts its DNA into the bacterial cell so that it might be replicated.
  • It prevents replication of the phage DNA by cutting it into many pieces.
  • Restriction enzymes were named for their ability to restrict.

Explanation:

•  A restriction enzyme is used to cut DNA molecules.
• It is of 3 types: Exonucleases, Endonucleases, and Restriction endonucleases.
  • Exonucleases-  They cut off nucleotides from 5" or 3" ends of DNA molecules.
  • Endonucleases-  They break DNA duplexes at any point except ends.
  • Restriction endonucleases- They cleave DNA duplexes at a  specific point.
• Each restriction endonucleases functions by inspecting the length of a DNA sequence.
• And each restriction endonucleases cleave DNA molecules at a specific nucleotide sequence. So, it recognizes a specific palindromic nucleotide sequence in the DNA.

​Therefore, the correct answer is a specific nucleotide sequence.

• DNA (Deoxyribonucleic aid) is a nucleic acid made up of nucleotides.
• It is a long polymer of deoxyribonucleotides.
• DNA was first identified by Fredrich Meischer in 1869. He named it Nuclein.
• DNA acts as the genetic material in most organisms.
• It is a double-helix structure of protein and nucleic acid which carry all the genetic information in the form of genes.
• A nucleotide has 3 components: a nitrogenous base ( A, T, G, C), a pentose sugar, and a phosphate group.
• The length of the DNA is usually defined as the number of nucleotides in it.

Steps of Recombinant DNA technology are given below:

A. Identification and isolation of the genetic material.

B. Fragmentation of DNA.

C. Obtaining the foreign gene product

D. Downstream processing.

E. Ligation of DNA fragmentation into the vector.

F. Isolation of desired DNA fragments.

G. Amplification of gene of interest.

H. Transfer of Recombinant DNA into the host cell/organism.

The correct sequence of steps is

1. A → D → C → B → E → G → F → H
2. A → B → F → G → E → H → C → D
3. H → F → G → E → A → D → B → C
4. C → A → B → D → F → E → G → H

Answer (Detailed Solution Below)

Option 2 : A → B → F → G → E → H → C → D

The correct answer is A → B → F → G → E → H → C → D.

• The complete process of recombinant DNA technology includes multiple steps, maintained in a specific sequence to generate the desired product
  • Identification and isolation of the genetic material: The first and the initial step in Recombinant DNA technology is to isolate the desired DNA in its pure form i.e. free from other macromolecules.
  • Fragmentation of DNA: The restriction enzymes play a major role in determining the location at which the desired gene is inserted into the vector genome. These reactions are called ‘restriction enzyme digestions’.
  • Isolation of desired DNA fragments.
  • Amplification of gene of interest: It is a process to amplify a single copy of DNA into thousands to millions of copies once the proper gene of interest has been cut using the restriction enzymes.
  • Ligation of DNA fragmentation into the vector: In this step of Ligation, joining of the two pieces – a cut fragment of DNA and the vector together with the help of the enzyme DNA ligase.
  • Transfer of Recombinant DNA into the host cell/organism: The recombinant DNA is introduced into a recipient host cell. This process is termed Transformation. Once after the insertion of the recombinant DNA into the host cell, it gets multiplied and is expressed in the form of the manufactured protein under optimal conditions.
  • Obtaining the foreign gene product:Multiplication/expression of the introduced gene in the host.
  • Downstream processing: Selection of the transformed host cells and identification of the clone containing the desired gene/DNA fragment.

Highly repetitive DNA is seen in:-

(a) Telomere

(b) Microsatellite DNA

(c) Centromere

(d) Cloning of DNA

1. (a), (b), (c), (d)
2. (a), (b)
3. (c), (d)
4. (a), (b), (c)

Answer (Detailed Solution Below)

Option 4 : (a), (b), (c)

• Prokaryotes have DNA with unique, non-repeated base-pair sequences, meaning that each linear sequence constituting a gene is present only once in the genome.
• On the other hand, eukaryotic chromosomes are much more complex as they have repetitive sequences of DNA along with the unique sequences.
• Repetitive DNA refers to such DNA containing repetitive sequences that constitute a major part of eukaryotic genome.
• Repetitive DNA can be:
  • Tandem repetitive sequences that are located in one locus in cluster form. Example - Satellite DNA.
  • Interspersed repeats that are scattered throughout the genome along with other sequences. Example - Long Interspersed Nucleotide Element (LINE).
• In humans, about 40% of DNA is repetitive.

• Telomere -
  • It refers to the terminal end of eukaryotic chromosome.
  • It is made of repetitive sequences of non-coding DNA.
  • A characteristic repeat found at the telomeres is GGGGATT.
  • It is responsible for protection against chromosome degradation and maintaining genomic integrity.
  • Each time a cell divides, the telomere becomes shorter.
  • The progressive shortening of the telomere leads to aging.
• Microsatellite DNA -
  • These are repetitive DNA sequences of non-coding DNA.
  • The repeating units consist of 2-6 base-pairs of DNA.
  • They are widely used as genetic markers in paternity tests and population genetics studies.
• Centromere -
  • It is a dense, constricted region of the chromosome where the kinetochore is assembled during cell division.
  • They are composed of highly repetitive DNA elements.

• Cloning of DNA refers to the process of making multiple copies of DNA sequences.
• This is done to amplify a desired DNA segment.
• DNA cloning is done with any DNA segment with the gene of interest.
• This does not contain repetitive DNA, but unique sequence of DNA.

Which one of the following is not required for DNA-cloning?

1. DNA ligase
2. A vector
3. Methylases
4. Restriction endonucleases

Answer (Detailed Solution Below)

Option 3 : Methylases

Concept-

• DNA is a long polymer of deoxyribonucleotides.
• The length of DNA is usually defined as a number of nucleotides (or a pair of nucleotides referred to as base pairs) present in it.
• This also is the characteristic of an organism.
• An alien piece of DNA has become part of a chromosome, which has the ability to replicate.
• In a chromosome, there is a specific DNA sequence called the origin of replication, which is responsible for initiating replication.
• Therefore, for the multiplication of any alien piece of DNA in an organism, it needs to be a part of a chromosome(s) that has a specific sequence known as ‘origin of replication.
• Thus, alien DNA is linked with the origin of replication, so that, this alien piece of DNA can replicate and multiply itself in the host organism.
• This can also be called cloning or making multiple identical copies of any template DNA.

Methylases are not required for DNA cloning.

• Methylases add methyl groups (—CH3) to adenine or cytosine bases within the recognition sequence, which is thus modified and protected from the endonuclease.

DNA ligase

• DNA ligase is also called molecular glue.
• This enzyme joins DNA fragments together by forming phosphodiester bonds between nucleotides.

A vector

• Plasmid (autonomously replicating circular extra-chromosomal DNA).
• These plasmid DNA act as vectors to transfer the piece of DNA attached to it. 

Restriction endonucleases

• In the year 1963, the two enzymes responsible for restricting the growth of bacteriophage in Escherichia coli were isolated.
• One of these added methyl groups to DNA, while the other cut DNA. The later was called restriction endonuclease.

Which of the following methods is most useful for the enzymatic amplification of specific gene segment of DNA?

1. DNA hybridization
2. Nucleotide sequencing
3. Polymerase chain reaction
4. Reverse transcription

Answer (Detailed Solution Below)

Option 3 : Polymerase chain reaction

The correct answer is the Polymerase chain reaction.

• The polymerase chain reaction is most useful for the enzymatic amplification of specific gene segments of DNA.

• The polymerase chain reaction is a technique to make copies of a specific DNA region in vitro.
• It is used to make enough of the target DNA region so that it can be analyzed or used in some other way. 
• It is based on a thermostable DNA polymerase, Taq polymerase, and requires DNA primers designed specifically for the DNA region of interest.

• DNA hybridization is the process of combining two complementary single-stranded DNA or RNA molecules and enabling them to form a single double-stranded molecule through base pairing.
• Nucleotide sequencing is defined as a process of determining the order of nucleotides.
  • Nucleotides are defined as organic molecules that act as monomers or the subunits of nucleic acids.
  • They are known as the building blocks for nucleic acids.
• Reverse transcription is the enzyme-mediated synthesis of a DNA molecule from an RNA template.
  • The resulting DNA is known as cDNA and can be used as a template for PCR amplification.
  • Reverse transcription followed by PCR is known as RT-PCR (reverse transcription-PCR).

A vector (cloning vector) is a carrier DNA molecule to which the human DNA fragment is attached in DNA transfer the vectors used from smallest to largest is:-

1. Bacteriophage, Plasmids, BAC (Bacterial Artificial Chromosome), Cosmids
2. Cosmids, Plasmids, Bacteriophage, BAC (Bacterial Artificial Chromosome)
3. Plasmids, Bacteriophage, Cosmids, BAC (Bacterial Artificial Chromosome)
4. BAC (Bacterial Artificial Chromosome), Cosmids, Plasmids, Bacteriophage

Answer (Detailed Solution Below)

Option 3 : Plasmids, Bacteriophage, Cosmids, BAC (Bacterial Artificial Chromosome)

• Cloning vectors are DNA molecules that can carry a foreign DNA  fragment when inserted into it.
• Criteria for an ideal vector:
  • Small and easy to isolate.
  • Origin of Replication (ori) - It is the DNA sequence from where replication starts and any DNA fragment linked to it can be made to replicate within the host cell.
  • Cloning sites - There should be one or more unique restriction sites into which DNA fragment can be inserted.
  • Selectable Markers - They should have a selectable marker that helps in recognition of transformants. E.g.- Antibiotic resistance genes.

• Plasmids -
  • They are closed, circular, extra-chromosomal, self-replicating double-stranded DNA molecules present in bacterial cells.
  • They are most suitable as cloning vectors and widely used in recombinant DNA technology.
  • They can carry an insert of up to 15 kb.
• Bacteriophage -
  • It is a virus that can infect a bacteria.
  • Phage vectors are linear DNA molecules that can be modified to carry the desired DNA fragment.
  • It has an insert size of about 20-25 kb.
• Cosmids -
  • It as a plasmid vector containing cos sites of λ-phage.
  • It can accommodate an insert size of about 30-45 kb.
• Bacterial Artificial Chromosome (BAC) -
  • They are engineered DNA molecules that are used to clone bacterial DNA.
  • They allow cloning of up to 300 kb pairs of DNA, which may represent whole chromosomes.
• Thus, the size of vectors from smallest to largest would be:

Plasmids < Bacteriophage < Cosmids < BAC (Bacterial Artificial Chromosome)

What are the 5 tools used in recombinant DNA technology?

What are three 3 tools used in recombinant DNA technology?

Which of the following are biological tool for recombinant DNA technology?

What are the 7 steps in recombinant DNA technology?