biology : class12 Vector Dephosphorylation : DNA Isolation, Restriction Enzymes, Vector, Ligation, Host Cell : Recombinant DNA Technology MCQ Questions and Answers
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Vector Dephosphorylation
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MCQ Questions
1. A plasmid vector is linearized with a restriction enzyme and then treated with calf intestinal alkaline phosphatase before ligation. What is the primary molecular consequence of this treatment?
A.
The 3′-hydroxyl groups of the vector are removed
B.
The 5′-phosphate groups of the vector are removed
C.
The phosphodiester bonds at the restriction sites are restored
D.
The cohesive ends of the vector are converted into blunt ends
Correct Answer:
The 5′-phosphate groups of the vector are removed
Explanation:
Alkaline phosphatase removes 5′-terminal phosphate groups from the linearized vector. The resulting vector ends retain 3′-hydroxyl groups but cannot efficiently self-ligate because the required 5′ phosphate is absent.
2. Why is dephosphorylation of a restriction-digested vector especially useful when the vector and insert have compatible cohesive ends?
A.
It prevents the vector from annealing to the insert
B.
It prevents the vector from recircularizing without an insert
C.
It converts the insert into a phosphorylated molecule
D.
It destroys the recognition sequence of the restriction enzyme
Correct Answer:
It prevents the vector from recircularizing without an insert
Explanation:
Compatible cohesive ends can allow a linearized vector to reanneal and self-ligate. Removing the vector's 5′ phosphates greatly reduces formation of a covalently closed vector lacking an insert.
3. A dephosphorylated vector is ligated with an insert whose 5′ ends have also been dephosphorylated. Which outcome is most directly expected?
A.
Efficient formation of covalently sealed recombinant molecules
B.
Efficient vector self-ligation because both molecules lack phosphate
C.
Reduced ligation efficiency because phosphates are unavailable at ligation junctions
D.
Conversion of all cohesive ends into blunt ends
Correct Answer:
Reduced ligation efficiency because phosphates are unavailable at ligation junctions
Explanation:
DNA ligase requires a 5′ phosphate and a 3′ hydroxyl to form a phosphodiester bond. Dephosphorylating both vector and insert removes the phosphate needed for efficient covalent joining.
4. Which experimental control most directly demonstrates that vector dephosphorylation has reduced background caused by vector self-ligation?
A.
Transform cells with untreated, uncut vector
B.
Transform cells with the dephosphorylated, linearized vector without insert
C.
Transform cells with insert DNA alone
D.
Incubate the insert with ligase without vector
Correct Answer:
Transform cells with the dephosphorylated, linearized vector without insert
Explanation:
A no-insert control containing the treated vector measures colonies arising from residual uncut plasmid or re-ligated vector. A low colony count supports effective dephosphorylation and digestion.
5. After dephosphorylation of a linearized plasmid vector, which group of chemical termini is predominantly present at each vector end?
A.
A 5′ phosphate and a 3′ phosphate
B.
A 5′ hydroxyl and a 3′ hydroxyl
C.
A 5′ hydroxyl and a 3′ phosphate
D.
A 5′ phosphate and a 3′ hydroxyl
Correct Answer:
A 5′ hydroxyl and a 3′ hydroxyl
Explanation:
Phosphatase removes the terminal 5′ phosphate, leaving a 5′ hydroxyl. The native 3′ hydroxyl is not removed, so the dephosphorylated vector has 5′-OH and 3′-OH termini.
6. Why can a phosphorylated insert still be used with a dephosphorylated vector even though the initially ligated recombinant molecule may contain nicks?
A.
The insert permanently replaces the vector's missing phosphates
B.
Host-cell repair systems can seal remaining nicks after transformation
C.
Restriction enzymes automatically seal all nicks during transformation
D.
Dephosphorylation activates DNA ligase inside the vector
Correct Answer:
Host-cell repair systems can seal remaining nicks after transformation
Explanation:
Phosphorylated insert ends permit ligase to create the essential phosphodiester bonds at recombinant junctions, although some nicks may remain. After transformation, host-cell DNA repair and replication systems can repair these nicks.
7. A researcher dephosphorylates a vector but does not remove the phosphatase before adding DNA ligase. What is the main concern?
A.
The phosphatase may remove phosphates from DNA ends during the ligation reaction
B.
The phosphatase will convert the vector into RNA
C.
The phosphatase will add phosphate groups to the insert
D.
The phosphatase will cut the vector at its restriction sites
Correct Answer:
The phosphatase may remove phosphates from DNA ends during the ligation reaction
Explanation:
Residual phosphatase can dephosphorylate phosphorylated insert ends or other DNA termini, reducing the availability of 5′ phosphates required for ligation. It is therefore commonly removed or inactivated before ligation.
8. Which observation most strongly suggests incomplete dephosphorylation of a linearized plasmid vector?
A.
A high number of colonies in the vector-only negative control
B.
No colonies in a transformation using intact plasmid
C.
Failure of the restriction enzyme to cut the insert
D.
Formation of RNA from the vector during ligation
Correct Answer:
A high number of colonies in the vector-only negative control
Explanation:
If the dephosphorylated vector still produces many colonies without insert, residual phosphorylated vector ends may have permitted self-ligation, or undigested plasmid may be present. This result prompts evaluation of digestion and phosphatase treatment.
9. Which statement best distinguishes vector dephosphorylation from restriction digestion during recombinant DNA cloning?
A.
Dephosphorylation creates the DNA ends, whereas restriction digestion removes their phosphates
B.
Restriction digestion generates vector ends, whereas dephosphorylation modifies their terminal phosphate status
C.
Both procedures cut the vector, but only digestion changes its 5′ termini
D.
Dephosphorylation inserts foreign DNA, whereas digestion joins vector ends
Correct Answer:
Restriction digestion generates vector ends, whereas dephosphorylation modifies their terminal phosphate status
Explanation:
Restriction enzymes cleave the vector to create ends, often with cohesive termini. Phosphatase treatment does not cut DNA; it removes 5′ phosphates from those pre-existing vector ends to suppress self-ligation.
10. A cloning reaction uses a dephosphorylated vector and a phosphorylated insert with compatible cohesive ends. Which event is specifically favored by vector dephosphorylation?
A.
Formation of vector-insert junctions while suppressing vector-vector circularization
B.
Replication of the insert before it enters the host cell
C.
Removal of the insert's restriction sites
D.
Conversion of every recombinant molecule into a completely nick-free circle before transformation
Correct Answer:
Formation of vector-insert junctions while suppressing vector-vector circularization
Explanation:
The phosphorylated insert supplies 5′ phosphates needed for ligation at recombinant junctions, whereas the dephosphorylated vector is much less able to self-ligate. The treatment does not guarantee that every product is fully nick-free before transformation.
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