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Genetics question: multiple genes at different loci?
A variety of opium poppy having lacerate leaves was crossed with a variety with normal leaves. All the F1 had lacerate leaves. Two F1 plants were interbred to produce the F2. Of the F2, 249 had lacerate leaves and 16 had normal leaves. Give the genotypes for all the plants in the P, F1, and F2 generations. Explain how lacerate leaves are determined in the opium poppy.
3 Answers
- AbuLv 61 decade agoFavorite Answer
My hypothesis for the ratio of the progeny is 15 : 1 (two gene pairs). This is the example of duplicate gene action; meaning one dominant gene is enough to produce lacerate leaf.
P: AABB (lacerate) x aabb (normal)
F1: AaBb (lacerate)
F2: 9 A-B- (lacerate) : 3 A-bb (lacerate) : 3 aaB- (lacerate) : 1 aabb (normal) or 15 lacerate : 1 normal
- Anonymous5 years ago
Here yellow coluor is recessive and brown is dominant. so lets assume that a yellow female has a genotype yy. In first case, the genotype of male would be Yy (brown colour), so after mating yy X Yy the genotype of offsprings would be Yy (brown), Yy (brown), yy (yellow), yy (yellow). so 50% offsprings are yellow and 50% are brown. In second casr, the genotype of brown male would be YY (homozygous dominant), so after mating yy X YY the genotype of offsprings would be Yy (brown), Yy (brown), Yy (brown), Yy (brown). so all are brown
