Does the Genotypes for the F2 generation have to be 9:3:3:1The F2 two generation crossed AaWwxAaWw AA:normal antenna aa:Aristopedia antenna WW: normal wings Ww:vestigial wings My results for the F2...

Does the Genotypes for the F2 generation have to be 9:3:3:1

The F2 two generation crossed AaWwxAaWw

AA:normal antenna aa:Aristopedia antenna WW: normal wings Ww:vestigial wings

My results for the F2 were a 9:4:3:1 instead of the 9:3:3:1

What could have caused this to occur

Asked on by t521863

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trophyhunter1 | College Teacher | (Level 1) Educator Emeritus

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When you perform a dihybrid cross, the results of the Punnett square are always in the ratio of 9:3:3:1. The parents are hybrid for both their wings and their antennae. Therefore, with a genotype of AaWw for each parent, their gametes will be of four different types: AW, Aw, aW, aw, due to the Law of Independent Assortment. The problem will have 16 boxes, each represents a potential fertilization. Nine boxes will have the two dominant traits--normal antenna and normal wings. Three boxes will have one dominant and one recessive trait--Normal antennae and vestigial wings. Three boxes will have the other dominant and recessive traits--normal wings and arostopedia antenna and only one box will have both recessive traits--aristopedia antenna and vestigial wings. It is impossible to have a 9:4:3:1 result. If you add that up, you have an extra box in your Punnett square which means you counted something twice. ``  

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