Q.11 In a genetic cross between plants bearing violet flowers and green seeds (VVGG), and white flower
and yellow seeds (vvgg), the following phenotypic distribution was obtained in the F2 progeny
(assume both parents to be pure breeding for both the traits, and self–cross at F1 generation):
i) 2340 plants with violet flowers and green seeds
ii) 47 plants with violet flowers and yellow seeds
iii) 43 plants with white flowers and green seeds
iv) 770 plants with white flowers and yellow seeds
Which one of the following interpretations explains the above phenotypic distribution?
(A) Same genes control both flower and seed colors
(B) Genes for flower and seed colors are genetically interacting
(C) Genes for flower and seed colors are present on the same chromosome
(D) Flower color in this plant species is a polygenic trait
Genes for flower and seed colors are linked on the same chromosome, leading to more parental phenotypes than recombinants in the F2 generation. This phenotypic distribution deviates from Mendel’s 9:3:3:1 dihybrid ratio due to genetic linkage.
Phenotypic Data Analysis
Total F2 progeny equals 3200 plants. Parental phenotypes (violet-green: 2340; white-yellow: 770) total 3110, comprising 97.19% of progeny. Recombinant phenotypes (violet-yellow: 47; white-green: 43) total 90, or 2.81%.
Expected vs. Observed Ratios
Independent assortment yields a 9:3:3:1 ratio (approximately 1800:600:600:200 for 3200 progeny). Observed ratios show excess parentals and deficient recombinants, indicating linkage rather than free assortment.
Option Explanations
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(A) Same genes control both traits: Incorrect, as distinct phenotypes (four combinations) appear, ruling out pleiotropy where one gene affects multiple traits.
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(B) Genetic interaction: Incorrect, as no modified ratios like 9:6:1 or 12:3:1 occur; interaction alters dominant recessives differently.
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(C) Genes on same chromosome: Correct. Low recombination (2.81%) confirms linkage; parental types dominate due to rare crossovers.
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(D) Polygenic flower color: Incorrect, as discrete classes and linkage pattern fit monogenic traits, not continuous polygenic variation.
Recombination Calculation
Recombination frequency = (90/3200) × 100 = 2.81%, corresponding to 2.81 cM map distance. This low value confirms tight linkage on the same chromosome.
In genetic linkage studies, the genetic linkage violet flowers green seeds F2 cross reveals how linked genes deviate from Mendel’s ratios. Pure breeding parents VVGG (violet flowers, green seeds) crossed with vvgg (white flowers, yellow seeds) produce F1 VvGg heterozygotes. F2 selfing yields 2340 violet-green, 47 violet-yellow, 43 white-green, and 770 white-yellow plants.
Linkage Evidence in Data
Parental combinations dominate (3110/3200 = 97.19%), while recombinants are rare (90/3200 = 2.81%). This confirms genes for flower and seed colors same chromosome, with recombination frequency of 2.81 cM.
CSIR NET Exam Relevance
For CSIR NET Life Sciences, recognize linkage when F2 shows >90% parentals. Calculate RF = (recombinants/total) × 100. Option C is correct; others fail as explained above.
Keywords: genetic linkage, violet flowers green seeds, F2 phenotypic ratio, flower seed colors same chromosome, CSIR NET genetics, recombination frequency.
