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Sagot :
If the parental genotypes are homzygous, the F1 birds would be heterozygous while all the F2 offspring would have the wild-type phenotype.
Monohybrid crossing
Based on the assumption that the wild-type genotype is AA and the slow feathering genotype is aa.
A cross between AA and aa:
AA x aa
F1 Aa Aa Aa Aa
One F1 male (Aa) is then crossed with a wild-type female (AA):
Aa x AA
AA AA Aa Aa
Genotypic proportion: 2 AA:2 Aa
Phenotypic proportion = 100% wild-type
More on monohybrid crossing can be found here: https://brainly.com/question/1185199
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