Vector Biology test focuses on “Special Vectors for Expression in E.Coli – 4”.
1. The secondary structures from intrastrand base pairs do not interfere with ribosomal attachment.
A. True
B. False
Answer: B
Clarification: The secondary structures resulting from intrastrand base pairs interfere with the attachment of the ribosome to its binding site.
2. Presence of which of the following can stabilize the fusion protein?
A. Imidazole
B. IPTG
A. Sucrose
D. Bacterial peptide
Answer: D
Clarification: The presence of a bacterial peptide at the start of the fusion protein may stabilize the molecule and prevent degradation.
3. The foreign proteins that lack a bacterial segment are ________
A. Destroyed
B. Hybridized
A. Expressed
D. Induced
Answer: A
Clarification: Foreign proteins in contrast to bacterial proteins, that lack a bacterial segment are often destroyed by the host.
4. The periplasmic space is the space between inner and outer ________
A. Cell wall
B. Nucleus
A. Membrane
D. Matrix
Answer: C
Clarification: If the signal peptide is derived from a protein that is exported by the cell, the recombinant protein may itself be exported into periplasmic space.
5. Export of protein is ________ into the periplasmic space.
A. Hindered
B. Desirable
A. Non-desirable
D. Suppressed
Answer: B
Clarification: Export of the protein into the periplasmic space is desirable because it simplifies the problem of purification.
6. Which technique is used for aiding protein purification?
A. Cloning
B. Hybridization
A. Chromatography
D. Radiation
Answer: C
Clarification: The bacterial segment may also aid purification by enabling the fusion protein to be recovered by affinity chromatography technique.
7. E.coli glutathione-S-transferase protein can be purified by adsorption onto _________
A. Charcoal
B. Sucrose beads
A. Agarose beads
D. Glucose beads
Answer: C
Clarification: Fusion involving the E.coli glutathione-S-transferase protein can be purified by adsorption onto agarose beads.
8. What is a disadvantage of fusion systems?
A. Property alteration
B. Reduced amount
A. Cost inefficiency
D. Host range
Answer: A
Clarification: The disadvantage with fusion systems is that the presence of E.coli segment may alter the properties of the recombinant protein.
9. Cyanogen bromide can be used to cleave __________
A. Arginine
B. Methionine
A. Aspartame
D. Glycine
Answer: B
Clarification: If a methionine is present at the junction, the fusion protein can be cleaved with cyanogen bromide which cuts polypeptide.
10. For cleaving arginine residues, which of the following can be used?
A. Methionine
B. IPTG
A. Thrombin
D. XA
Answer: C
Clarification: Methods for cleaving the bacterial segment are needed so that the properties of the fusion protein are not altered. Thrombin cleaves adjacent to arginine residues.
11. For cleaving off the bacterial segment, it is necessary that recognition sequence occurs within the protein.
A. True
B. False
Answer: B
Clarification: The important consideration is that recognition sequences for the cleavage agent must not occur within the recombinant protein.
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