Q.57 In a plant species, flower colour purple is dominant over white. One such purple-flowered plant
upon selfing produced 35 viable plants, of which 9 were white-flowered and the rest were purple-
flowered. What fraction of these purple-flowered progeny is expected to be pure purple-flowered
line?
(A) 1/2 (B) 1/3 (C) 1/4 (D) 2/3
Correct Answer: (B) 1/3
A purple-flowered plant producing 9 white : 26 purple (9:26 ≈ 1:3 ratio) upon selfing must be heterozygous (Pp). Among the 26 purple progeny, 1/3 represent homozygous dominant (PP) pure-breeding lines, while 2/3 are heterozygous (Pp).
Detailed Solution
Step 1: Determine parental genotype
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Total viable plants = 35
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White-flowered (pp) = 9
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Purple-flowered = 26
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Ratio = 9:26 ≈ 1:3 (matches Pp × Pp selfing)
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Parental genotype: Pp (heterozygous purple)
Step 2: Punnett square for Pp selfing
| P | p
---|-------|-------
P | PP | Pp
---|-------|-------
p | Pp | pp
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Phenotypic ratio: 3 purple : 1 white ✓
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Genotypic ratio among purple: 1 PP : 2 Pp
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Fraction pure purple (PP) = 1/3 of purple progeny
Option Analysis
(A) 1/2: Incorrect. Assumes 1:1 PP:Pp ratio (testcross), not 1:2 from selfing.
(B) 1/3: Correct. PP:Pp = 1:2 among purple; pure line fraction = 1/3.
(C) 1/4: Incorrect. Total homozygous dominant fraction (not purple-specific).
(D) 2/3: Incorrect. Heterozygous fraction among purple progeny.
In classical Mendelian genetics, a heterozygous purple-flowered plant (Pp) upon selfing produces 3:1 purple:white ratio. Among purple progeny, exactly 1/3 breed true as pure purple-flowered lines (PP), critical for stable cultivar development.
Genetic Analysis
Observed data fits χ² test:
Expected (3:1): Purple=26.25, White=8.75
χ² = Σ(O-E)²/E = 0.047 (p>0.8, excellent fit)
Breeding Implications
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Pure line selection: Self purple progeny 3x to isolate PP (100% purple).
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F1 hybrid vigor: Pp × PP maximizes heterozygosity.
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Marker-assisted: Linked markers confirm PP vs Pp pre-flowering.
Exam Strategy
9:26 ratio → Pp parent → 1/3 pure purple among purple. Standard NEET pattern.
