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Which of the following could be considered a true-breeding strain for the seedshape

phenotype? (See Figure Q19-38.)
(a) all of the round-seeded plants produced in the F2 generation
(b) all of the wrinkle-seeded plants produced in the F2 generation
(c) all of the round-seeded plants produced in the F1 generation
(d) half of the round-seeded plants produced in the F2 generation

User Shabi
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Final answer:

The true-breeding strain for the wrinkled seed trait in pea plants would be all of the wrinkle-seeded plants produced in the F2 generation, as they would be homozygous recessive (rr) and consistently produce wrinkled seeds when self-fertilized.

Step-by-step explanation:

When discussing a true-breeding strain for a given trait such as seed shape in pea plants, we are referring to plants that, when self-fertilized, produce offspring that are uniform and consistent for that trait. In this context, true-breeding would mean that the strain consistently produces seeds with the same shape as the parent generation. Given that round seed shape (R) is dominant to wrinkled seed shape (r), true-breeding strains would be homozygous for the trait they express (RR for round seeds and rr for wrinkled seeds).

Looking at Mendel's F2 generation outcomes for pea plant crosses, three-fourths (75%) of the F2 progeny from a cross of F1 heterozygotes (Rr) are expected to have round seeds, while one-fourth (25%) would have wrinkled seeds. Thus, the true-breeding strain for the wrinkled seed trait could only be amongst those plants that are homozygous recessive (rr), which would be all of the wrinkle-seeded plants produced in the F2 generation.

User Paulvs
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