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2 years ago

Which of the following will cause a Frameshift mutation?

Biology

2 answers:

lozanna [386]2 years ago

7 0

Answer: i think it an insertion or a deletion

Explanation: Insertion and deletion result in a frameshift mutation. Frameshift mutations caused by "indel" mutations, meaning insertion or deletion mutations. These mutations shift the way the DNA is read. A frameshift mutation can throw off all of the nucleotide that follow in a DNA sequence, making them particularly likely to lead to significant consequences.

Hope it helps if not sorry

Lyrx [107]2 years ago

4 0

D. an insertion or a deletion

this is your answer

hope it is helpful to you

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

kondor19780726 [428]

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

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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