- Given below is a schematic representation of the T- DNA region of a transgene construct and the Southern blot analysis (using Pst l digested genomic DNA) of five independent transgenic lines (labelled as A to E) developed using the construct. The probe used for hybridization is shown as a black box below the construct (UT: Untransformed Plant)
Based on the above data, which one of the following options gives the correct list of single insertion transgenic events?
(1) B and C only (2) A, B and E only
(3) A and B only (4) E onlyThe correct answer is (4) E only – line E is the only single‑insertion event.
Step 1: Interpreting the construct and probe
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T‑DNA: LB – promoter – gene – polyA – RB.
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Two PstI sites flank the gene region; digestion with PstI cuts once on each side of the probed gene.
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The probe (black box) lies within the gene, so every integrated T‑DNA copy yields one PstI fragment that hybridizes.
Therefore:
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A single-copy insertion → one band on the Southern blot.
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Multiple insertions (different loci or configurations) → two or more bands.
Step 2: Reading the Southern blot (lanes UT, A–E)
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UT (untransformed): no hybridizing band (negative control).
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Lane A: two distinct bands → at least 2 T‑DNA copies.
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Lane B: two bands → multiple insertions.
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Lane C: three bands → multiple insertions.
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Lane D: two bands → multiple insertions.
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Lane E: one single band → one T‑DNA copy.
Thus only E represents a single-insertion transgenic event.
Evaluation of options
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B and C only – incorrect
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B and C each show more than one hybridizing band, so they are multi-copy events.
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A, B and E only – incorrect
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A and B are multi-copy; only E is single-copy.
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A and B only – incorrect
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Again, both are multi-copy.
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E only – correct
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Matches the observation that only E has a single band and therefore a single T‑DNA insertion.
SEO-oriented introduction (for article use)
Southern blotting with a gene-specific probe is a standard way to determine transgene copy number in independently transformed plant lines. In this PstI digest, each integrated T-DNA copy yields one hybridizing fragment, so lines A–D, which show multiple bands, clearly harbor multiple insertions. Only line E displays a single band, identifying it as the single-copy insertion event, and therefore option “E only” is correct.
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