Question

A plant species under study has both tall and short varieties. The plant makes a growth hormone (made by enzyme a) in the leaves that is transported ( by transport proteins b) To the stems and acts to make them elongate. A true breeding tall plant is crossed with a second plant homozygous for the recessive nonfunctioning allele of the transport protein but homozygous wild-type genotype for enzyme A. What fraction of the F2 offspring will be expected to be dwarf?

Answer 1

A. The question says a true breeding tall plant , so if its tall plant it has both the genes dominant so lets take it as AABB and the second statement says that crossed with homozygous wild type genotype for enzyme means it is AA and says the homozygous recessive for a allele lets take it as bb so the genotype of the second plant will be AAbb now a cross to be made

AABB x AAbb (parental )

AB Ab (Gametes)

AABb F1 Hybrid (TAll plants)

now on self pollination we get AABb x AABb gametes we get will be AB and Ab from both

so on fertilization we get three genotypes AABB (1) and AABb (2) and AAbb (1), first two would be tall and the third will be dwarf . so in terms of fraction 1/4 th of the total will be dwarf. i.e 25% will be Dwarf.