Question

. Calculate the genotypic ratio and phenotypic ratio of a short pea plant having full pod with a tall pea plant having constructed pod. Represent all its 16 possibility with the help of a figure

Answer 1

Answer:

Let,

The Short pea plant = t (Recessive)

The tall pea plant = T (Dominant)

The full pod = F (Dominant)

The constricted pod = f (Recessive)

Hence, The parents plants will be:-

FFtt and ffTT

Gametes for FFtt :- Ft, Ft Ft and Ft.

Gametes for FFtt :- Ft, Ft Ft and Ft.Gametes for ffTT :- fT, fT, fT and fT.

Crossing them to produce F1 generation,

$\boxed{\begin{tabular}{c|c|c}\bf{-} & \bf{Ft} & \bf{Ft}\\\cline{1-3}\bf{fT} & FfTt & FfTt\\\cline{1-3}\bf{fT} & FfTt & FfTt\end{tabular}}$

All genotypically same and phenotypically same too.

Genotypically, all of them are heterozygous and Phenotypically all are Full pod and tall.

Now, F1 generation - selfing,

Parents:- FfTt and FfTt

Parents:- FfTt and FfTtGametes:- FT, Ft, fT and ft.

$\boxed{\begin{tabular}{c|c|c|c|c}\bf{-} & \bf{FT} & \bf{Ft} & \bf{fT} & \bf{ft}\\\cline{1-5}\bf{FT} & FFTT & FFTt & FfTT & FfTt\\\cline{1-5}\bf{Ft} & FFTt & FFtt & FfTt & Fftt \\\cline{1-5} \bf{fT} & FfTT & FfTt & ffTT & ffTt \\\cline{1-5} \bf{ft} & FfTt & Fftt & ffTt & fftt\end{tabular}}$

Phenotypic ratio:-

Full pod and Tall = 9

Full pod amd short = 3

Constrictive pod and tall = 3

Constrictive pod and short = 1

Hence, 9:3:3:1.

Genotypic ratio:-

FFTT = 1

FFTt = 2

FFtt = 1

FfTT = 2

FfTt = 4

Fftt = 2

ffTT = 1

ffTt = 2

fftt = 1

Hence, 1:2:1:2:4:2:1:2:1

$\rule{200}2$