Question

Assume that plant height and flower size in Four O'Clock plants display antogonistic phenotypes. Four O'Clock plants can be tall or short and the flowers can be large or small. Assume each trait is governed by alleles of a single gene and that each follows Mendel’s first and second laws. You start with two pure breeding strains, one that is tall with large flowers and a second that is short with small flowers. You cross these two strains to each other to generate F1 plants. You self-cross an F1 plant and grow 342 of the resulting seeds into plants. You find that 195 plants are short with large flowers, 62 plants are short with small flowers, 59 plants are tall with large flowers, and 26 plants are tall with small flowers.

For the cross above- what is the phenotype of the F1 individuals?

A. Short with small flowers
B. Tall with small flowers
C. Medium height with medium size flowers
D. Short with large flowers
E.Tall with large flowers
F. Can't tell from the information provided

Please explain I'm really confused

Answer 1

Lets say,

Tall character is denoted by allele TT(pure) and short character is denoted by recessive allele tt (pure).

Large flower character is denoted by allele LL (pure) and small flower character is denoted by ll (pure).

Now a pure tall plant with large flower will have a genotype TTLL and a pure short plant with small flower will have a genotype ttll. Following will be the gamates formed

	tl	tl
TL	TtLl	TtLl
TL	TtLl	TtLl

So in F1 generation only one genotype [TtLl]plant will be produced, which will have a phenotype of medium height and medium size flower.(as the plant is antagonistic).

So for self crossing we will take the F1 plant with phenotype medium height and medium sized flower. Ans will be C.