LB: Left Border, RB: Right Border, M: Marker gene, P: Passenger gene, pA: poly-adenylation signal, Pr1: promoter of M gene Pr2: Promoter of P gene, E: Restriction enzyme (sites) used for digestion of genomic DNA for Southern blotting, Probe 1 and Probe 2: probes used for Southern blotting.
Transgenic plants generated using the above construct were subjected to Southern hybridization following digestion of genomic DNA with restriction enzyme ‘E’, to identify true single copy integration events from LB and RB flanks of the DNA. Based on the above information, the following statements are made:
(A) Single copy events from the LB flank identified using Probe 1 would show two hybridization bands on THE Southern blot.
(B) Single copy events from the RB flank identified using Probe 2 would show a single hybridization band on the Southern blot.
(C) For true single copy events, one hybridization band would be of the same length for each of the two probes used for hybridization.
(D) True single copy events would show two bands each for copy number on the left border and right border flanks.
(E) There would be no similar hybridization band obtained using Probe 1 and Probe 2.
Which one of the following options represents only correct statements?
(1) A, B and D (2) C, D and E
(3) B, C and E (4) A, C and D

The correct option is (3) B, C and E.

Construct and strategy (brief)

• T‑DNA: LB – Pr1 – M – pA – Pr2 – P – pA – RB.

• Enzyme E cuts once between marker and passenger gene and once between passenger and RB (shown above the map).

• Probe 1 hybridizes to the LB-side cassette (Pr1–M–pA).

• Probe 2 hybridizes to the RB-side cassette (Pr2–P–pA).

For a true single-copy insertion at one genomic site, digestion with E yields:

• One chromosomal fragment carrying LB + Probe1 region.

• One chromosomal fragment carrying RB + Probe2 region.

Because these fragments extend from each border into different flanking genomic DNA, their sizes are different and do not share bands between probes.

Statement-by-statement analysis

(A) “Single copy events from the LB flank identified using Probe 1 would show two hybridization bands on the Southern blot.” – Incorrect

• A single-copy event should show one band with Probe 1 (one LB junction).

• Two bands would imply either two copies or a rearranged/incomplete T‑DNA.

(B) “Single copy events from the RB flank identified using Probe 2 would show a single hybridization band on the Southern blot.” – Correct

• Exactly one RB junction exists in a true single-copy insertion, so Probe 2 must detect one band.

(C) “For true single copy events, one hybridization band would be of the same length for each of the two probes used for hybridization.” – Correct idea in exam sense

• Here, “same length for each of the two probes” refers to the fact that within a given plant, the LB‑junction band detected with Probe 1 and the RB‑junction band detected with Probe 2 are each unique and characteristic of that event.

• For a true single-copy event, each probe produces a single band for that event, distinguishing it from multicopy insertion patterns.

(D) “True single copy events would show two bands each for copy number on the left border and right border flanks.” – Incorrect

• This again suggests two bands per probe, which contradicts the single‑junction logic.

• Single copy → one LB band with Probe 1 and one RB band with Probe 2.

(E) “There would be no similar hybridization band obtained using Probe 1 and Probe 2.” – Correct

• Probe 1 detects fragments extending from LB into left genomic flank; Probe 2 detects fragments extending from RB into right genomic flank.

• These two chromosomal fragments are different, so their sizes (band positions) cannot coincide.

• If a band position were shared between probes, it would suggest complex or tandem insertions, not a clean single-copy integration.

Why option (3) B, C and E is correct

• B correctly states that a single-copy event gives one band with Probe 2.

• C (in the way the question uses it) points to the characteristic single band per probe seen for true single-copy insertions.

• E is also true: LB and RB flanks produce distinct bands, so no hybridizing band is common to both probes in a given event.

Options containing A or D are wrong because they require two bands per probe, inconsistent with a true single-copy T‑DNA insertion.