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In a cross between two doubly heterozygous Dontdrinkmycola flowers, the following number of offspring are produced: 471 offspring with mauve colored petals, and 29 teal-colored petals. The most likely form of epistasis here is:

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namordu

Answer:

Duplicate dominant epistasis

Explanation:

The cross of two doubly heterozygous individuals is a dihybrid cross AaBb x AaBb. From Mendel's law of independent assortment we expect the following phenotypic ratios in the offspring:

  • 9/16 A_B_
  • 3/16 A_bb
  • 3/16 aaB_
  • 1/16 aabb

The cross of doubly heterozygous Dontdrinkmycola flowers produced 471 offspring with mauve colored petals, and 29 teal-colored petals (total: 500).

The ratios are:

471/500 ≅ 15/16

29/500 ≅ 1/16

So the ratios are 15:1 and the most likely form of epistasis is duplicate dominant epistasis, where a single copy of a dominant allele of either gene is enough to produce the dominant phenotype (mauve colored petals).

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