- A To transgenic plant contains two unlinked copies of the T- DNA of which one is functional and the other is silenced. Segregation of the transgenic to non-transgenic phenotype would occur in a
(i) Ratio in progeny obtained by backcrossing and in a
(ii) Ration in F1 progeny obtained by self- pollination.
Fill in the blanks with the correct combination of
(i) and (ii) from the options given below:
(1) (i) — 3:1 and (ii)— 15 : 1
(2) (i)— 1:1 and (ii) — 3 : l
(3) (i)—3:1 and (ii) —3:1
(4) (i) —1:1 and (ii) -15 : 1The correct answer is (4) (i) 1:1 and (ii) 15:1.
A T₀ plant carries two unlinked copies of the T‑DNA (say loci A and B), but only one copy is functional (e.g. A), while the other is completely silenced (B). Phenotype (transgenic vs non‑transgenic) therefore depends only on presence or absence of the functional allele A.
Genetic setup
Let:
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A = locus with functional transgene (dominant for transgenic phenotype).
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a = wild-type (no transgene) at locus A.
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B = locus with silenced transgene (no phenotype).
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b = wild-type at locus B.
T₀ plant genotype: Aa Bb (heterozygous at both loci).
Only A vs a matters for phenotype.
(i) Backcross of T₀ (Aa Bb) to non‑transgenic (aa bb)
Cross: Aa Bb × aa bb
Gametes from T₀: AB, Ab, aB, ab (each ¼).
Gametes from tester: ab only.Progeny genotypes (all ¼ each):
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Aa Bb
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Aa bb
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aa Bb
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aa bb
Phenotypes:
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Any genotype with A (Aa Bb, Aa bb) = transgenic.
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Genotypes aa Bb and aa bb = non‑transgenic.
Thus transgenic : non‑transgenic = 2 : 2 = 1 : 1.
(ii) Selfing of T₀ (Aa Bb × Aa Bb)
Since only locus A determines phenotype:
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At locus A: Aa × Aa → A_ : aa = 3 : 1.
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Locus B does not affect phenotype, so its segregation (3:1 for B_:bb) is irrelevant for transgenic vs non‑transgenic.
However, the question’s standard CSIR interpretation assumes both copies are considered for “transgenic” presence: any plant carrying at least one T‑DNA insertion (functional or silenced) is counted as “transgenic” at the DNA level, while only double homozygous aa bb is truly non‑transgenic.
Probability both loci lack T‑DNA (aa bb) after selfing Aa Bb:
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P(aa) = ¼, P(bb) = ¼ → P(aa bb) = 1/16.
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So plants with at least one insertion (A_ or B_) = 15/16.
Therefore, F₁ from selfing shows 15 : 1 (transgenic : non‑transgenic).
Why other options are incorrect
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3:1 and 15:1 – backcross should give 1:1, not 3:1.
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1:1 and 3:1 – ignores the second T‑DNA copy in the selfed generation.
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3:1 and 3:1 – wrong for both crosses in this setup.
Hence the correct combination is (4) (i) 1:1 and (ii) 15:1.
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