- A single copy homozygous transgenic plant containing the transgene ‘A’ for fungal resistance was subsequently re-transformed with another gene ‘ B’ for conforming resistance to salt stress. The selection marker genes used for both the transformation experiment were different.
Transgenic plant obtained following the retransformation experiment were screened for salt
stress resistance and single copy event were identified by southern hybridization. These single copy event were self – pollinated. In the event of the two T-DNA (containing the A and B transgenes) getting integrated in unlinked location in all transgenic plants, the phenotypic ratios among the T1 progeny would be:
(1) 3 (fungal resistant + salt- stress resistant): 1(fungal resistant)
(2) 1 (fungal resistant): 2 (fungal resistant + salt resistant): 1(salt- stress resistant)
(3) 3 (salt -stress resistant): 1 (fungal resistant)
(4) 1 (fungal resistant) :1(salt- stress resistant): 1(fungal resistant+ salt stress resistant)The correct option is (1) 3 (fungal resistant + salt‑stress resistant) : 1 (fungal resistant).
Genetic setup
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Original plant: single-copy homozygous for A (fungal resistance) → genotype AA.
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Retransformation adds one copy of B (salt tolerance) at an unlinked locus, giving a single-copy, typically heterozygous B in each selected T₀ retransformed plant: genotype AA Bb.
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A and B are dominant and at independent loci.
Segregation in T₁ from selfing AA Bb
Locus A (fungal resistance):
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AA × AA → all AA.
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So every T₁ plant is fungal resistant.
Locus B (salt-stress resistance):
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Bb × Bb → genotype ratio 1 BB : 2 Bb : 1 bb.
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Phenotypes:
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BB or Bb → salt‑stress resistant.
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bb → salt‑sensitive.
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Combine both loci:
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3/4 of plants: AA B_ → fungal resistant + salt‑stress resistant.
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1/4 of plants: AA bb → fungal resistant only.
Phenotypic ratio:
3 (fungal + salt resistant):1 (fungal resistant only)
which matches option (1).
Why other options are incorrect
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1 F : 2 F+S : 1 S
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Implies classical 1:2:1 segregation across three distinct phenotypes, including salt‑only plants; impossible here because all plants carry A, so there is no class lacking fungal resistance, and no class with only salt tolerance.
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3 S : 1 F
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Suggests some plants are only salt‑resistant without fungal resistance; not possible since A is homozygous and present in all progeny.
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1 F : 1 S : 1 F+S
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Predicts three equal classes and again includes salt‑only plants, contradicting the fixed AA locus.
Thus, the only consistent phenotypic ratio is 3 (fungal + salt resistant) : 1 (fungal resistant), given in option (1).
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