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In feathered serpents, two genes control the overall serpentine look: feathering (feathers, F, are dominant to no feathers, f) and color (blue body, B, is dominant to white body, b). You testcross a dihybrid blue, feathered serpents and obtain 12 blue, feathered babies; 2 blue, featherless babies; 14 white, featherless babies, and 1 white, feathered baby. Which babies are recombinants?

a. blue and featherless
b. all blue babies
c. all white babies
d. blue and feathered

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namordu

Answer:

a. blue and featherless

Explanation:

  • Feathers (F) are dominant to no feathers (f)
  • Blue body (B) is dominant to white body (b)

If we testcross a dihybrid, we are crossing a homozygous recessive individual with a heterozygous individual.

Recombination is a rare event in meiosis, so the more abundant offspring are always the parentals, and the less abundant offspring are the recombinant ones.

In this case, bf/bf and BF/bf are the parentals and Bf/bf and bF/bf are the recombinants.

The option a. Blue and featherless (Bf/bf) offspring are the recombinant babies.

The babies that are recombinants should be option a. blue and featherless

Recombinants:

Feathers (F) should be considered dominant to no feathers (f) and Blue body (B) should be considered dominant to white body (b). Recombination should be rare event in meiosis. In the given situation, bf/bf and BF/bf are the parentals and Bf/bf and bF/bf are the recombinants.

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