- In a breeding experiment; two homozygous parental lines (P1 and P2) were crossed to produce F1 hybrids. Due to an experimental error, seeds of these hybrids got mixed up with the seeds of two other germplasm lines (P3 and P4) and hybrid seeds derived from them.
A marker-based fingerprinting exercise was performed using six randomly selected seeds (F1-F6) from the mixed material and the four parental lines. Results of this analysis are shown below:Based on the above data, which one of the following options represents the correct set of parents and their F1 progeny?
(1) P1 X P2 = F3 (2) P3 X P4 = F2
(3) P1 X P2 = F1 (4) P3 X P4 = F6The correct option is (4) P3 × P4 = F6.
How to read the banding pattern
Each vertical lane (P1, P2, P3, P4, F1–F6) shows DNA fragments at several marker loci.
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Each parent is homozygous, so each locus gives one band per parent.
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A true F₁ from two different parents must show, at every polymorphic locus, both parental bands (additive pattern).
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A lane that shows only one parental band across loci is either a parental line or a selfed derivative, not an F₁.
Step-by-step comparison
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Identify unique bands of each parent (P1–P4).
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Compare each F lane:
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F1, F2, F3, F4, F5:
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Each of these either matches a single parent at several loci or lacks bands from one of the supposed parents, so they cannot be clean P1×P2 or P3×P4 F₁s.
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F6:
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At each marker position, F6 has both a P3-type band and a P4-type band.
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It never shows P1- or P2-specific bands.
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This is the classic F₁ additive pattern for cross P3 × P4.
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Thus, the only lane that looks like a proper heterozygous combination of two homozygous parents is F6, derived from P3 and P4.
Why the options are wrong or right
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P1 × P2 = F3 – F3 does not consistently carry both P1 and P2 bands at each locus; some markers appear only from one parent.
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P3 × P4 = F2 – F2 does not show the complete additive pattern of P3 and P4; some loci lack one parental band.
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P1 × P2 = F1 – F1 also lacks the full set of both parents’ bands and is not a clear heterozygote.
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P3 × P4 = F6 – Only F6 consistently displays both P3 and P4 bands at all polymorphic loci, as expected for the true F₁ hybrid. Hence this option is correct.
SEO‑oriented introduction (for article use)
Marker-based DNA fingerprinting is an effective way to identify true F₁ hybrids in a mixed seed lot. By comparing banding patterns of unknown seeds (F1–F6) with those of homozygous parents (P1–P4), a genuine hybrid must display the combined bands of both parents at every polymorphic marker. In this problem, only lane F6 carries both P3 and P4 bands across all loci, confirming that F6 is the P3 × P4 F₁ hybrid, which corresponds to option (4).
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