- In a gene stacking experiment, a transgenic plant generated using bar as a selection marker was crossed with another transgenic line containing hpt as the selection marker. Analysis of the F1 progeny of this cross showed that 50% of the progeny was resistant to both selection agents (Basta and hygromycin) while the remaining 50% of the progeny was resistant to only hygromycin. Which one of the following statements is a possible explanation of the above results?
(1) Both the transgenic lines used as parents for the cross are single-copy events that are homozygous for the transgenes.
(2) The Basta-resistant transgenic plant is a single- copy, homozygous event and the hygromycin- resistant plant is a double-copy event that is homozygous for two unlinked copies of the transgene.
(3) The bar containing transgenic plant is a hemizygous single-copy event and the hpt containing transgenic plant is a homozygous, single-copy event
(4) Both the transgenic plants used as parents for the cross are hemizygous, single-copyThe correct explanation is (3) The bar-containing transgenic plant is a hemizygous single-copy event and the hpt-containing transgenic plant is a homozygous, single-copy event.
Stepwise reasoning
Let:
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bar = Basta-resistance gene.
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hpt = hygromycin-resistance gene.
Observed F₁ phenotypes:
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50% resistant to both Basta and hygromycin.
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50% resistant to only hygromycin (i.e., no Basta resistance).
No progeny are sensitive to hygromycin.
This means:
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All F₁ plants carry hpt (hygromycin resistance fixed).
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bar segregates present vs absent in 1:1 ratio.
So:
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Parent with hpt must be homozygous (H/H).
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Parent with bar must be heterozygous (B/b) for a single functional locus.
Cross: B/b (bar parent) × H/H (hpt parent) with loci unlinked and each parent homozygous or heterozygous only at its own locus:
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All progeny receive H from each parent → all F₁ are H/H (hygromycin resistant).
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bar locus: gametes from bar parent are ½ B and ½ b; other parent is b/b (no bar).
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F₁: ½ B/b (Basta + hygromycin resistant), ½ b/b (hygromycin only).
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Thus the segregation exactly matches the observation.
Option-wise analysis
(1) Both parents homozygous single-copy
If both were homozygous for their markers:
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bar parent: B/B
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hpt parent: H/H
F₁: B/B H/H (all plants double-resistant).
No 50:50 split is possible. So (1) is wrong.(2) bar plant homozygous; hpt plant homozygous for two unlinked copies
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bar parent: B/B
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hpt parent: H₁/H₁ H₂/H₂ (two unlinked hpt copies).
Any cross gives all F₁ with at least one B and both H copies → 100% double-resistant, not 50%.
So (2) is wrong.
(3) bar hemizygous, hpt homozygous – Correct
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bar parent: B/b (single-copy heterozygous).
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hpt parent: H/H.
F₁: ½ B/b H/H (double-resistant) and ½ b/b H/H (hygromycin only).
Matches observed 50% +/+ and 50% hpt only.
So (3) is the correct explanation.
(4) Both parents hemizygous, single-copy
bar parent: B/b, hpt parent: H/h.
Assuming each contributes only its own marker:-
bar+: ½ of progeny.
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hpt+: ½ of progeny.
But some F₁ would lack one or both markers, and you would not get “all hygromycin resistant” progeny; you’d expect four phenotypic classes (double+, bar only, hpt only, sensitive). So (4) cannot produce the observed 50% double-resistant, 50% hygromycin-only pattern.
SEO‑oriented introduction (for article use)
In gene stacking experiments, analyzing segregation patterns of selectable markers is crucial for inferring zygosity of parental lines. A 1:1 ratio of F₁ plants that are double-resistant (Basta + hygromycin) versus hygromycin-only resistant indicates that hygromycin resistance is fixed (homozygous hpt) while Basta resistance (bar) is contributed by a single heterozygous locus. This pattern is best explained when the bar plant is hemizygous single-copy and the hpt plant is homozygous single-copy, corresponding to option (3).
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