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

Answers

Answer 1

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:

Available data:

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 bristlesDominant trait: normal wings and normal bristles

Let us say that:

B is the dominant allele for normal wingsb is the recessive allele for blistery wingsS is the dominant allele for normal bristless 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    


Related Questions

The DNA sequence CAT would be transcribed into which mRNA codon?

Answers

it would transcribe into GUA

Answer:

GUA

Explanation:

Hello There!

These are the base pairs for transcription

IMPORTANT

when transcribing; Adenine transcribes to Uracil not Thymine

so it would be

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Thymine to Adenine

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so we want to find the mRNA codon of the dna sequence CAT

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Explain the difference between herbivores, carnivores and omnivores​

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Answer:

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Answer:

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Answer:

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Explanation:

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Answers

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Which gas is most necessary for plants to do photosynthesis?

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Answers

QUESTION :::::

Which gas is most necessary for plants to do photosynthesis?

A. carbon dioxide

B. nitrogen

C. oxygen

ANSWER ::::

carbon dioxide

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Explanation:

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Answer:

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Answer:

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Plants respire all the time, whether it is dark or light, because their cells need energy to stay alive. But they can only photosynthesize when they have light.

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Answer:

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Explanation:

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1. a) father (Rr), mother (rr)

b) from the attached file(1)

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Rr - 50%rr - 50%

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RR - 25%Rr - 50%rr - 25%

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Answers

Answer:

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where as

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Answer:

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Answers

Answer:

Both realized niche and fundamental niche are similar in size.

Explanation:

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Answer:

looks great!

Explanation:

Answer:

it looks great sorry

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have a great day

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Answer:

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Answer:

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Answer:

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Answer:

Hey mate.......

Explanation:

This is ur answer.....

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A que altura debe moverse un costal de cemento de 50kg para asegurar que su energía potencial sea de 10,500 j?

Answers

Answer:

21 m

Explanation:

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La energía potencial se da como;

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