31. A DNA segment was cloned into the active site region of lacZ gene and the recombinant plasmid introduced into lac T strain of E. coli and plated on a medium containing X-gal. The colonies showed blue color.
Which one of the following statements is correct?
(1) The nature of the cloned DNA segment need not be special as cloning of any DNA in lacZ will result in disruption of its reading frame and production of blue colour on X-gal plates.
(2) The cloned DNA segment could be a Group I intron whose removal from the precursor lac Z transcript in E. coli results in production of mature lac Z mRNA which can then produce active Lac Z protein.
(3) The cloned sequence is likely to be lacY sequence which is naturally a part of lac operon in E. coli.
(4) The cloned sequence is likely to be an anti-terminator sequence which allows full length transcription of lacZ.
The correct statement is:
(2) The cloned DNA segment could be a Group I intron whose removal from the precursor lacZ transcript in E. coli results in production of mature lacZ mRNA which can then produce active LacZ protein.
Explanation:
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Blue-white screening principle:
In the commonly used blue-white screening system, the lacZ gene encodes the enzyme β-galactosidase, which cleaves the substrate X-gal to produce a blue pigment. -
Why blue colonies here?
The fact that colonies are blue despite cloning into the lacZ gene means the reading frame and function of lacZ are not disrupted. This can happen if the inserted DNA is a self-splicing intron, such as a Group I intron, which is removed from the precursor lacZ mRNA in E. coli. The mature mRNA is then translated into functional β-galactosidase, producing blue colonies9. -
Why other options are incorrect:
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(1) Cloning any DNA into lacZ disrupts the reading frame and typically produces white colonies, not blue. So this is false.
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(3) lacY is a different gene in the lac operon; cloning lacY into lacZ does not explain blue colonies.
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(4) An anti-terminator sequence allowing full-length transcription does not explain blue colonies if the lacZ reading frame is disrupted.
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Summary Table
| Option | Explanation | Correctness |
|---|---|---|
| (1) | Any DNA insertion disrupts lacZ → white colonies | Incorrect |
| (2) | Group I intron self-splices → lacZ functional → blue colonies | Correct |
| (3) | lacY gene cloned into lacZ irrelevant to blue color | Incorrect |
| (4) | Anti-terminator sequence unrelated to lacZ function | Incorrect |
Conclusion
The blue colonies indicate that the cloned DNA does not disrupt lacZ function, consistent with the presence of a Group I intron that self-splices out of the transcript, restoring functional β-galactosidase production.
9 Comments
Sapna yadav
June 5, 2025👍
Saloni
June 5, 2025✅
Saloni
June 5, 2025Clearly explained ✅
Komal Sharma
June 6, 2025Done
Komal Sharma
June 6, 2025Done it’s better to understand
Prami Masih
June 8, 2025👍👍
Neelam Sharma
September 9, 2025The blue colonies indicate that the cloned DNA does not disrupt lacZ function, consistent with the presence of a Group I intron that self-splices out of the
transcript, restoring functional β galactosidase production.
Kyunki yah enzyme X gal ko degrade krke bule colur produce krega and intron splicing se intron remove ho jaynge mature mRNA of lac z functional hoga or transcript hojayega blue colour hoga
Neelam Sharma
September 9, 2025The blue colonies indicate that the cloned DNA does not disrupt lacZ function, consistent with the presence of a Group I intron that self-splices out of the transcript, restoring functional β-galactosidase production.
Heena Mahlawat
November 5, 2025Option 2 where cloned dna is intron that’s spliced out and maintaining function of b -gal