Category: Vector Biology Objective Questions

Vector Biology Quiz focuses on “Special Vectors for Expression in E.Coli – 2”.

1. What is the most critical component of an expression vector?
A. Terminator
B. Suppressor
A. Inducer
D. Promoter
Answer: D
Clarification: The promoter is the most important component of an expression vector because it controls the very first stage of gene expression.

2. The rate at which messenger RNA is synthesized is determined by ______________
A. Promoter
B. Gene
A. Host
D. Nucleus
Answer: A
Clarification: The promoter controls the attachment of RNA polymerase enzyme to the DNA and determines the rate at which messenger RNA is synthesized.

3. The amount of recombinant protein obtained depends on the nature of nature of host.
A. True
B. False
Answer: B
Clarification: The amount of recombinant protein obtained depends on the nature of nature of the promoter. The promoter controls the attachment of RNA polymerase enzyme to the DNA and determines the rate at which messenger RNA is synthesized.

4. A small variation in the consensus sequence of a promoter will lead to change in the variation of ______________
A. Efficiency of transcription
B. Translational efficiency
A. Cloning efficiency
D. Host range
Answer: A
Clarification: Although most E.coli promoters do not differ much from the consensus sequences, a small variation may have a major effect on the efficiency with which the promoter can direct transcription.

5. For genes whose products are needed in small amounts, which types of promoters are required?
A. Strong promoters
B. Consensus promoters
A. Weak promoters
D. Any promoter
Answer: C
Clarification: Weak promoters are inefficient, direct transcription of genes whose products are needed in only small amounts.

6. Strong promoters sustain a _______ rate of transcription.
A. Medium rate
B. Low rate
A. High rate
D. Average rate
Answer: C
Clarification: Strong promoters are those that can sustain a high rate of transcription; strong promoters usually control genes whose translational products are required in large amounts by the cell.

7. Induction and repression are types of ___________
A. Regulation
B. Expression
A. Signals
D. Processes
Answer: A
Clarification: An important factor to be considered during construction of an expression vector is that whether it will be possible to regulate the promoter.

8. An inducible gene is the one whose __________ can be switched on by the addition of a chemical to the medium.
A. Translocation
B. Transcription
A. Replication
D. Translation
Answer: B
Clarification: An inducible gene is the one whose transcription can be switched on by the addition of a chemical to the medium, which is a substrate.

9. A repressible gene is the one that can be _________
A. Switched off
B. Switched on
A. Expressed
D. Killed
Answer: A
Clarification: A repressible gene is switched off by the addition of the regulatory chemical. This chemical is one of the substrates for the enzymes.

10. High level of transcription may affect the ability of the recombinant plasmid to _____________
A. Express
B. Sustain
A. Replicate
D. Take up DNA
Answer: C
Clarification: Regulation of the cloned gene is desirable, as a continuously high level of transcription may affect the ability of the recombinant plasmid to replicate, leading to its eventual loss from the culture.

11. The “tac promoter” is induced by _______
A. Lactose
B. Amylose
A. Lactose
D. IPTG
Answer: D
Clarification: The “tac promoter” is a hybrid between the trp and lac promoters. It is stronger than either but still induced by IPTG.

12. The “lac promoter” is the sequence that controls transcription of _____________
A. LacZ’ gene
B. Trp gene
A. Lambda DNA
D. Amylose
Answer: A
Clarification: The “lac promoter” is the sequence that controls transcription of LacZ’ gene, coding for beta-galactosidase.

13. Tryptophan synthesis is controlled by _________
A. Lac promoter
B. Trp promoter
A. Tac promoter
D. Lambda promoter
Answer: B
Clarification: The trp promoter is normally upstream of the cluster of genes coding for several of the enzymes involved in the biosynthesis of the amino acid tryptophan.

14. The lac promoter is induced by _____
A. IPTG
B. Lactose
A. Tryptophan
D. Amylose
Answer: A
Clarification: The lac promoter is induced by isopropylthiogalactosidase (IPTG), so an addition of a gene inserted downstream of the lac promoter.

15. Lambda P L promoter is responsible for the transcription of ____________
A. Lambda DNA
B. Lambda RNA
A. Tryptophan
D. IPTG
Answer: A
Clarification: Lambda P L promoter is responsible for the transcription of lambda DNA. It is a very strong promoter recognized by a polymerase.

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Vector Biology Interview Questions and Answers for Experienced people focuses on “Vectors for Plants – 3”.

1. Which of the following entities of a plant contain their own genomes, distinct from the nucleus?
A. Golgi apparatus
B. Ribosomes
A. Chloroplast
D. Ribozymes
Answer: C
Clarification: If biolistics is used to integrate DNA in a plant embryo, then some particles may penetrate one or more of the chloroplasts present in the cells.

2. Chloroplast genomes are smaller than the nuclear genomes present in the plants.
A. True
B. False
Answer: A
Clarification: Chloroplasts contain their own genomes, distinct from and much shorter than the DNA molecules in the nucleus and DNA can become integrated into this chloroplasts genome.

3. By which mechanism does the transfer of genes into the chloroplast genome take place?
A. Replication
B. Homologous recombination
A. Restriction digestion
D. Apoptosis
Answer: B
Clarification: The DNA to be cloned must be flanked by sequences similar to the region of the chloroplast genome into which the DNA is to be inserted so that insertion can take place through homologous recombination.

4. What is the advantage of inserting a gene of interest into the chloroplast?
A. Increased expression
B. Decreased expression
A. Ease of replication
D. Lower capital costs
Answer: A
Clarification: As a plant cell contains tens of chloroplasts whereas a nucleus is present only as a single copy, a gene inserted into the chloroplast genome is likely to be expressed at a higher level.

5. What is the drawback of using plant viruses as cloning vectors?
A. Small size
B. Large size
A. RNA as genetic material
D. DNA as genetic material
Answer: C
Clarification: The main problem of using plant viruses as cloning vectors is that the vast majority of plant viruses have genomes not of DNA but RNA, and manipulations with RNA are more difficult.

6. Which of the following is a plant virus?
A. Geminivirus
B. M13
A. HIV
D. RP4
Answer: A
Clarification: Only two classes of DNA virus are known to infect higher plants, the caulimovirus and the geminivirus, and neither is ideally suited for gene cloning.

7. Which was the first plant used for cloning experiment?
A. Turnip
B. Potato
A. Tomato
D. Brinjal
Answer: A
Clarification: One of the first successful plant genetic experiments happened in 1984, used a caulimovirus vector to clone a new gene into turnip plants.

8. Which problem is associated with a caulimovirus?
A. Size
B. In vivo packaging
A. Restriction issues
D. Poor ligation
Answer: A
Clarification: Caulimovirus vector was first used to clone genes in a turnip plant. The basic issue with this vector is that the total size of the genome is constrained by the need to package it into its protein coat.

9. Which basic strategy is used in the generation of cloning vector CaMV?
A. Helper phage
B. Co-integration
A. Hybrid vector
D. Phagemid
Answer: A
Clarification: The Cauliflower mosaic virus (CaMV) genome lacks several essential genes, which means that it can carry a large DNA insert but cannot by itself direct infection.

10. What is provided by the viral genome in cloning experiments using CaMV?
A. Virus proteins
B. Endonucleases
A. Ligases
D. Origin
Answer: A
Clarification: Plants are inoculated with the vector DNA along with a normal CaMV genome. The normal viral genome provides the genes needed for the cloning vector to be packaged into virus proteins.

11. Caulimoviruses have a narrow host range.
A. True
B. False
Answer: A
Clarification: The host range of caulimovirus is extremely narrow. This restricts cloning experiments to just a few pants, mainly brassicas such as turnips, cabbages.

12. Which of the following exceptional quality is related to caulimovirus?
A. Active promoters
B. Active repressors
A. Small size
D. Host range
Answer: A
Clarification: Caulimoviruses have been important in genetic engineering as the source of highly active promoters that work in all plants and used to obtain expression of genes introduced by Ti plasmid cloning vector.

13. Natural hosts of geminivirus include _________
A. Wheat
B. Ginger
A. Potato
D. Tomato
Answer: A
Clarification: Geminiviruses are particularly interesting because their natural hosts include monocots such as wheat and maize and therefore they are potential vectors for these.

14. What are problems associated with geminiviruses?
A. Deletions and rearrangements
B. Size irregularity
A. Infect only monocots
D. Low gene expression
Answer: A
Clarification: The biggest problem associated with monocot viruses Geminivirus is that they undergo rearrangements and deletions, which would scramble up any additional DNA that has been inserted.

15. What is the viral induced gene silencing technique (VIGS) that uses geminivirus?
A. Silencing technique
B. Gene silencing technique
A. Expression technique
D. Overexpression technique
Answer: B
Clarification: VIGS is a technique that incorporates plant virus Geminivirus for investigating the functions of individual plant genes.

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Vector Biology Multiple Choice Questions on “Restriction Endonucleases – 1”.

1. Cutting and joining of the DNA are which techniques?
A. DNA degradation
B. DNA replication
A. DNA manipulation
D. DNA synthesis
Answer: C
Clarification: Cutting and joining are examples of manipulative techniques most often used in the construction of recombinant DNA molecule. To produce this molecule, the vector and the DNA to be cloned must be cut at specific points and then joined together in a controlled manner.

2. What type of DNA enzymes is made use of in most of the DNA manipulative techniques?
A. Partially degraded
B. Purified
A. Degraded or denatured
D. Enclosed in a parent cell
Answer: B
Clarification: Although the enzymatic reactions such as DNA replication and transcription, recombination between different DNA molecules, are often straightforward but are impossible to perform by standard chemical methods. Purified enzymes are therefore crucial to genetic engineering.

3. Enzymes that remove nucleotides one at a time from the end of a DNA molecule are called ____________
A. Ligases
B. Exonucleases
A. Endonucleases
D. Modifying enzymes
Answer: B
Clarification: Nucleases degrade the DNA molecules by breaking the phosphodiester bonds that link one nucleotide to another in a DNA strand. There are two different kinds of nucleases.

4. The enzyme Bal31 purified from the bacterium Alteromonas Espejiana is an example of which enzyme?
A. Exonuclease
B. Endonuclease
A. Ligase
D. Phosphatase
Answer: A
Clarification: Bal31 removes nucleotides from both ends of a double stranded molecule. The greater the length of time that Bal31 is allowed to act on a group of DNA molecules, the shorter the resulting fragments will be endonuclease.

5. Which endonuclease cleaves both single and double stranded DNA molecules, in a non-specific manner?
A. S1
B. Bal31
A. DNase I
D. BamHI
Answer: C
Clarification: DNase I, prepared from cow pancreas cuts both single and double stranded molecules. DNase I is a non-specific in that it attacks at any internal phosphodiester bond, so the end result of prolonged DNase I action is a mixture of mononucleotides and very short oligonucleotides.

6. Klenow fragment is the modified enzyme of which of the parent DNA polymerase?
A. DNA polymerase I
B. DNA polymerase II
A. DNA polymerase III
D. DNA polymerase IV
Answer: A
Clarification: DNA pol I attach to a short single-stranded region in a double-stranded DNA molecule and then synthesize a completely new strand, degrading the existing strand as it moves forward. Hence it possesses both nuclease and polymerase activity. Removal of the segment controlling nuclease activity renders the enzyme modified and it is then called Klenow fragment.

7. The Taq DNA polymerase is DNA polymerase _________ enzyme from the bacterium Thermus aquaticus.
A. I
B. II
A. III
D. Klenow fragment
Answer: A
Clarification: Taq DNA polymerase is used in polymerase chain reaction and is DNA polymerase I enzyme. It is highly thermostable and is hence used in PCR.

8. Which of the following statement is not true in case of DNA Polymerase- Reverse transcriptase?
A. Involved in the replication of bacteriophage
B. Uses RNA as a template
A. Used in complementary DNA cloning
D. Synthesizes DNA from RNA
Answer: A
Clarification: Reverse transcriptase is involved in the replication of several kinds of viruses. Reverse transcriptase is unique in that it uses as a template not DNA but RNA.

9. Which of the following is not a source of alkaline phosphatase enzyme?
A. E.coli
B. Calf intestinal tissue
A. Arctic shrimp
D. Calf thymus tissue
Answer: D
Clarification: Alkaline phosphatase which removes the phosphate group present at the 5’ terminus of a DNA molecule is found in E.coli, calf intestinal tissue, arctic shrimp. Calf thymus tissue whereas is the source of another DNA modifying enzyme- Terminal deoxynucleotidyl transferase.

10. The DNA to be cloned must be cleaved along with the vector and with the same restriction enzymes.
A. True
B. False
Answer: A
Clarification: Large DNA molecules have to be broken down to produce fragments small enough to be carried by the vector. Most of the cloning vectors are very inefficient in carrying DNA more than 8kb in length.

11. Host controlled restriction is a phenomenon related to ________
A. Bacteria
B. Virus
A. Plasmid
D. Gene of interest
Answer: A
Clarification: The initial observation that led to the discovery of restriction endonucleases was made in early 1950s when it was shown that some strains of bacteria are immune to bacteriophage infection.

12. Why does the restriction phenomenon in bacteria naturally occur?
A. For efficient cloning
B. Bacteria produce an enzyme
A. Destruction of bacterium’s own DNA
D. For survival
Answer: B
Clarification: Restriction occurs because the bacterium produces an enzyme that degrades phage DNA before it has time to replicate and direct synthesis of new phage particles.

13. Which type of restriction endonucleases is used most in genetic engineering?
A. Type I
B. Type II
A. Type III
D. Type IV
Answer: B
Clarification: Type I and Type III are complex and have only a limited role in genetic engineering. Type II restriction endonucleases are used mostly as the cutting enzymes in gene cloning.

14. The restriction endonuclease PvuI (isolated from Proteus Vulgaris) cuts DNA at which position?
A. Hexanucleotide CGATCG
B. Random position
A. Towards the end
D. Hexanucleotide CAGCTG
Answer: A
Clarification: A particular enzymes cleaves DNA at the recognition sequence and nowhere else. PvuI cuts DNA at hexanucleotide sequence CGATCG.

15. The restriction endonuclease AluI is isolated from which microbe?
A. Proteus Vulgaris
B. Staphylococcus Aureus
A. Arthrobacter Luteus
D. Haemophilus Influenzae
Answer: C
Clarification: AluI restriction endonuclease derived from Arthrobacter Luteus, cleaves a DNA molecule at nucleotide long site- AGCT. There are other endonucleases that cleave at bigger sites.

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Gene Manipulation test focuses on “Gene Manipulation in Fruitflies”.

1. Gene transfer to fruit flies involves microinjection of DNA into the nucleus.
A. True
B. False
Answer: B
Clarification: Gene transfer to fruit flies involves the microinjection of DNA into the pole plasma. P elements are used to introduce exogenous DNA.

2. P elements introduce DNA into the Drosophila _________
A. Somatic cells
B. Cytosol
A. Germline
D. Totipotent cells
Answer: C
Clarification: P elements, also known as transposons and jumping genes are used to introduce DNA into the Drosophila germline. These are transposable DNA elements.

3. P elements are highly __________
A. Toxic
B. Mobile
A. Erroneous
D. Acidic
Answer: B
Clarification: P elements are transposable DNA elements that under certain circumstances can be highly mobile in the germline of D. melanogaster.

4. The P-M hybrid dysgenesis is a _________
A. Technique
B. Syndrome
A. Apparatus
D. Strain
Answer: B
Clarification: P elements cause a syndrome of related genetic phenomena called P-M hybrid dysgenesis. It occurs when males of a P strain are mated with females of an M strain.

5. The P-M hybrid dysgenesis syndrome results in abnormal _____________
A. Offspring
B. Strain
A. Gene
D. Protein
Answer: A
Clarification: The syndrome predominantly affects the germline and induces a high rate of mutation and frequent chromosomal aberrations, resulting in an abnormal offspring.

6. Eggs permissive for P-element transposition are called __________
A. P-cytotype
B. M-cytotype
A. L-cytotype
D. K-cytotype
Answer: A
Clarification: P-elements are mobilized in the eggs of M-strain females. Eggs that are permissive for P-element transposition are described as M-cytotype.

7. P-element encodes a repressor of its own.
A. True
B. False
Answer: A
Clarification: The P-elements cause dysgenesis in crosses within P strains, because they are not mobilized in P-cytotype eggs. P-element encodes a repressor for itself which prevents transposition.

8. The P-elements have ______ base pairs inverted terminal repeats.
A. 11
B. 21
A. 31
D. 41
Answer: D
Clarification: The elements are characterized by perfect 31 base pairs inverted terminal repeats which are recognized by the transposase.

9. A truncated version of which enzyme can act as repressor?
A. Transposase
B. Polymerase
A. Nuclease
D. Ligase
Answer: A
Clarification: The prototype element contains a single gene, comprising four exons, encoding the transposase; a truncated version of transposase may act as a repressor.

10. Primary transcript of which enzyme is differentially spliced in germ cells and somatic cells?
A. Polymerase
B. Helicase
A. Ligase
D. Transposase
Answer: D
Clarification: The transposase primary transcript is differentially spliced in germ cells and somatic cells, such that functional transposase is produced only in germ cells.

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Gene Manipulation Interview Questions and Answers for freshers focuses on “Basic Laboratory Techniques – 4”.

1. Which nylon membranes should be used in southern blotting to reduce the need for gel pretreatment?
A. Positive
B. Negative
A. Neutral
D. Heavy
Answer: A
Clarification: An alternative method is to use positively charged nylon membranes, which removes the need for extended gel pretreatment.

2. What is the native form of DNA?
A. Denatured
B. Non-denatured
A. Small size
D. Composite form
Answer: B
Clarification: The native form of the DNA is the one in which all inter and intra bonds are intact in the helical structure of the DNA.

3. Nitrocellulose is flammable in nature.
A. True
B. False
Answer: A
Clarification: Due to the flammable nature of nitrocellulose and also the nitrocellulose membrane, the baking of the nitrocellulose is done in a vacuum oven.

4. Thymine residues in the DNA and positively charged amino groups on the surface of nylon membranes ___________
A. Are of same size
B. Are of same charge
A. Interact
D. Cross-link
Answer: D
Clarification: A fixation method is based on ultraviolet cross-linking. It is based on the formation of cross-links between thymine groups of DNA and positive amino groups on nitrocellulose membrane.

5. How can the fixation period be determined initially?
A. Weighing
B. Calibration experiment
A. Heating
D. UV analysis
Answer: B
Clarification: A calibration experiment must be performed to determine the optimal fixation period. This fixation period can then be utilized in subsequent experiments.

6. Following the fixation step in blotting, the membrane is placed in RNA or DNA sequence which is _____________ to blot-transferred DNA.
A. Toxic
B. Complementary
A. Larger in size
D. Similar
Answer: B
Clarification: Following the fixation step, the membrane is placed in a solution of labeled RNA or single-stranded DNA or oligodeoxynucleotide which is complementary in sequence to the blot-transferred DNA.

7. The labeled nucleic acid used for detection is called _________
A. Probe
B. Gene
A. Analyte
D. Sample
Answer: A
Clarification: Conditions are chosen so that the labeled nucleic acid hybridizes with the DNA on the membrane. Since the labeled nucleic acid is used to detect and locate the complementary sequence, it is called the probe.

8. The labeled nucleic acid _________ with DNA sample to be analyzed.
A. Interferes
B. Solidifies
A. Crystallizes
D. Hybridizes
Answer: D
Clarification: The labeled nucleic acid acts as a probe. The main function of this nucleic acid is to hybridize with the given DNA or RNA sample.

9. Which film is used in southern blotting?
A. UV film
B. Radiography film
A. X-ray film
D. Cellulose film
Answer: C
Clarification: After the hybridization reaction has been carried out, the membrane is washed to remove unbound radioactivity and regions of hybridization are detected autoradiographically by placing the membrane in direct contact with X-ray film.

10. Lower ionic strength is an example of increased stringency in southern blotting.
A. True
B. False
Answer: A
Clarification: The hybridization can be carried out in conditions of relatively low stringency which permit a high rate of hybridization, followed by a series of high stringencies such as high temperature and low ionic strength.

11. Which of the following will reveal any of the imperfectly hybridized DNA samples?
A. Membrane separation
B. Autoradiography
A. UV radiation
D. Infrared radiation
Answer: B
Clarification: Autoradiography following each washing stage will reveal any DNA bands that are related to, but not perfectly complementary, the probe and will also permit an estimate of the degree of mismatching to be made.

12. The Southern Blotting methodology is extremely _____________
A. Expensive
B. Difficult
A. Easy
D. Sensitive
Answer: D
Clarification: The southern blotting can be extremely sensitive. It can be applied to mapping restriction sites around a single-copy gene sequence.

13. Mini-satellite probes in Southern blotting can be used in ______________
A. DNA forensics
B. Agriculture
A. Cloning
D. Purification
Answer: A
Clarification: When mini-satellite probe is used in Southern blotting, it can be applied forensically to minute amounts of DNA.

14. Northern blotting is used for _______
A. Proteins
B. Hosts
A. RNA
D. DNA
Answer: C
Clarification: Northern blotting technique is a variant of southern blotting technique and is used exclusively for RNA analysis.

15. Which of the following techniques can be used for RNA analysis?
A. Chromatography
B. Dialysis
A. Southern blotting
D. Northern blotting
Answer: D
Clarification: Southern’s technique is of enormous value but cannot be used for RNA analysis because RNA was found not to bind with nitrocellulose.

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Vector Biology Multiple Choice Questions on “Special Vectors for Expression in E.Coli – 1”.

1. If a foreign gene is simply ligated into a standard vector, it is likely that a recombinant protein will be produced.
A. True
B. False
Answer: B
Clarification: If a foreign gene is simply ligated into a standard vector and cloned in E.coli, it is very unlikely that a significant amount of recombinant protein will be synthesized.

2. What are the signals, inside the host that surround the gene?
A. Amino acids
B. Nucleotides
A. Proteins
D. Bacterium
Answer: B
Clarification: Expression is dependent on gene being surrounded by a collection of signals that can be recognized by the bacterium. These signals are short sequences of nucleotides.

3. Which of the following is not an important signal for the E.coli genes?
A. Promoter
B. Terminator
A. Inducer
D. Ribosome binding site
Answer: C
Clarification: Promoters, terminators and ribosome binding site are all important signals for E.coli genes. However, the inducer is not an essential signal.

4. Which of the following process is initiated by the promoter?
A. Translation
B. Replication
A. Apoptosis
D. Transcription
Answer: D
Clarification: The promoter, which marks the point at which transcription of the gene should start. In E.coli the promoter is recognized.

5. The promoter is recognized by ______________
A. RNA polymerase
B. DNA polymerase
A. Replicase
D. Helicase
Answer: A
Clarification: The promoter, which marks the point at which transcription of the gene should start. In E.coli the promoter is recognized by the RNA polymerase.

6. The terminator _______ the transcription process.
A. Decreases
B. Increases
A. Activates
D. Ceases
Answer: D
Clarification: The Terminator, marks the point at the end of the gene where transcription should stop. The terminator sequence can base-pair with itself to form a stem-loop structure.

7. A tem-loop structure in an E.coli cell is formed by __________
A. Gene
B. Terminator
A. Inducer
D. Promoter
Answer: B
Clarification: A terminator is usually a nucleotide sequence that can base pair with itself to form a stem loop structure. Terminator marks the end of the transcription process.

8. Ribosome binding site is recognized by __________
A. Polymerase
B. Ribosome
A. Nucleosome
D. Chromosome
Answer: B
Clarification: The ribosome binding site, a short nucleotide sequence recognized by the ribosome as the point at which it should attach to the messenger RNA molecule.

9. The initiation codon of the gene is located ___________ of the ribosome binding site.
A. Upstream
B. Downstream
A. Away
D. Inside
Answer: B
Clarification: The initiation codon of the gene is always a few nucleotides downstream of the ribosome binding site.

10. The ‘TTGACA Box’ signal is found only in ___________
A. Plants
B. Animals
A. Fungi
D. E.coli
Answer: D
Clarification: The genes of higher organisms are also surrounded by expression signals, but their nucleotide sequences are not the same as the E.coli versions.

11. The ‘TATA Box’ in animals contains one more ______ as compared to that present in E.coli.
A. Adenosine
B. Cytosine
A. Thymine
D. Guanine
Answer: A
Clarification: The genes of higher organisms are also surrounded by expression signals, but their nucleotide sequences are not the same as the E.coli versions.

12. The ‘TATA Box’ is present at _____ in animals and at _______ in E.coli.
A. -25, -10
B. -10, -10
A. -10, -25
D. -25, -25
Answer: A
Clarification: The genes of higher organisms are also surrounded by expression signals, but their nucleotide sequences are not the same as the E.coli versions.

13. What could be a possible reason for non-expression of a foreign gene in an E.coli host?
A. Recognition of expression signals
B. Non-recognition of expression signals
A. Indefinite size
D. Inefficient ligation
Answer: B
Clarification: A foreign gene is inactive in E.coli, simply because the bacterium does not recognize its expression signals.

14. Cloning vectors that can be used for recombinant protein production are called _____________
A. Expression vectors
B. Hybrid vectors
A. Hosts
D. Advanced vectors
Answer: A
Clarification: Expression vectors are the cloning vectors that provide expression signals and therefore can be used in the production of recombinant protein.

15. Placing a foreign gene under the control of expression signals will lead to its ____________
A. Activation
B. Expression
A. Inactivation
D. Knockout
Answer: B
Clarification: If a foreign gene is inserted into the vector in such a way that it is placed under the control of a set of E.coli expression signals.

.