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
