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A plant that produces wrinkled, green seeds was mated to a plant that produces phenotypically round, yellow seeds. In the F1 generation, half of the plants produced round, yellow seeds and the other half produced round, green seeds. The round, yellow F1 seeds were planted and those plants were allowed to self-fertilize. In the F2 generation what fraction of the seeds will be genotypically identical to the plants of the parental generation?A. 3/16
B. 1/16
C. 1/4
D. 9/16

Respuesta :

Answer:

A. 3/16

Explanation:

The plants with wrinkled, green seeds (rryy) and  round, yellow seeds  were crossed. Since the progeny did not show a "wrinkled" trait, the parent plant with "round" seeds was homozygous for seed shape and heterozygous for the seed color (RRYy). The genotype of the double recessive parent plant would be "rryy".

A cross between RRYy x rryy gives progeny in following ratio: 1/2 round, yellow seeds( RrYy ): 1/2 round, green seeds (Rryy)

The F2 progeny was obtained by self crossing RrYy plants. In F2 progeny, 3/16 plants (2/16 RRYy and 1/16 rryy) were genotypically identical to the parent plant.  

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