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Elongated and Five are two genes in a housefly that are 24 centimorgans (map units) apart. The E allele of Elongated produces elongated wings, and the e allele produces small wings (E is dominant). The F allele of Five produces five stripes on the back of the fly, and the f allele produces four stripes (F is dominant). In a cross of EfeF to efef, what percentage of the offspring should have elongated wings and five stripes?

Respuesta :

namordu

Answer:

12%

Explanation:

The genes E/e and F/f are 24 cM apart.

  • E_: elongated wings
  • ee: small wings
  • F_: five stripes
  • ff: four stripes

A distance of 24 cM means that the frequency of recombination between the two genes is 24%.

The cross Ef/eF x ef/ef would produce the following offspring:

  • Ef/ef: Parental; elongated wings and four stripes
  • eF/ef: Parental; small wings and five stripes
  • EF/ef: Recombinant; elongated wings and five stripes
  • ef/ef: Recombinant; small wings and four stripes

The total percentage of recombinant gametes is 24%, and there are two possible recombinant gametes, so each of them appears in 12% of the cases.

The individuals with elongated wings and five stripes have the recombinant genotype EF/ef, therefore the percentage of the offspring with this phenotype is 12%.

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