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pashok25 [27]
2 years ago
6

A hypothesis determines ...

Biology
1 answer:
statuscvo [17]2 years ago
8 0

Answer:

the correct answer would be C.

Explanation:

it would be C because a hypothesis is your guess of what the results will be. once you test the product you will figure out if your hypotheses was correct or not.

i hope this helps you out!!

-AlphaWolfie-

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A geneticist crossed fruit flies to determine the phenotypic ratio. The geneticist crossed a fly with blistery wings and spinele
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Complete question:

A geneticist crossed fruit flies to determine the phenotypic ratio. The geneticist crossed a fly with blistery wings and spineless bristles (bbss) with a heterozygous fly that had normal wings and normal bristles (BbSs). Which proportion of offspring that are dominant for both traits in would you not expect based on Mendel's law of independent assortment? 1/2 , 4/16, 25% , or 1/4

Answer:

1/2 is the proportion of the offspring that is NOT expected among individuals that are dominant for both traits.

4/16 = 1/4 = 25% of the progeny and the correct expected proportion of individuals that are dominant for both traits.

Explanation:

<u>Available data</u>:

  • Cross:  a fly with blistery wings and spineless bristles with a heterozygous fly that had normal wings and normal bristles
  • Recessive trait: blistery wings and spineless bristles
  • Dominant trait: normal wings and normal bristles

Let us say that:

  • B is the dominant allele for normal wings
  • b is the recessive allele for blistery wings
  • S is the dominant allele for normal bristles
  • s is the recessive allele for spineless bristles

Parentals)        bbss       x        BbSs

Gametes)  bs, bs, bs, bs     BS, Bs, bS, bs

Punnett square)    BS        Bs         bS        bs

                     bs    BbSs    Bbss     bbSs    bbss

                     bs    BbSs    Bbss     bbSs    bbss

                     bs    BbSs    Bbss     bbSs    bbss

                     bs    BbSs    Bbss     bbSs    bbss

F1)  4/16 = 1/4 = 25%  of the progeny is expected to be BbSs, dyhibrid individuals, expressing normal wings and normal bristles

     4/16 = 1/4 = 25% of the progeny is expected to be Bbss, expressing normal wings and spineless bristles

     4/16 = 1/4 = 25% of the progeny is expected to be bbSs, expressing  blistery wings and normal bristles

     4/16 = 1/4 = 25% of the progeny is expected to be bbss, expressing  blistery wings and spineless bristles    

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