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spayn [35]
2 years ago
6

Elongated and Five are two genes in a housefly that are 24 centimorgans (map units) apart. The E allele of Elongated produces el

ongated 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?
Biology
1 answer:
sammy [17]2 years ago
5 0

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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Ket [755]

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1 year ago
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UkoKoshka [18]

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        The positive results are that this is the reason why the plants grow smaller, so, with less organic material, the plants won't develop in the same way as it would happen with enough organic material. The negative results are that this is not the reason why these variations exist, so we need to test another factor to check why this variation is happening.

8 0
1 year ago
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vova2212 [387]

Answer:

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Explanation:

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8 0
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blsea [12.9K]

Answer:

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2 years ago
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Tom [10]

Answer:

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