Question

) Consider a pea plant- it has a gene controlling seed color- yellow Vs green; another gene controlling seed texture-smooth Vs wrinkled. A true breeding smooth green plant is crossed to a true breeding wrinkled yellow. All the F1 progeny were smooth yellow. The F1 generations were self-fertilized. Assume the two genes (seed color and seed texture) are VERY TIGHTLY LINKED. In the F2 what is the expected ratio of the phenotypes- Smooth & yellow: smooth & green: wrinkled & yellow: wrinkled & green 9:3:3:1 1:0:1:0 B

Answer 1

Yellow smooth - 9

Yellow wrinkle - 3

Green smooth- 3

Green Wrinkle - 1

Step-by-step explanation:

Let the green color of the seed be depicted by "G" and the yellow color of the seed be depicted by "g"

Let the smooth the seed be depicted by "R" and wrinkled seed be depicted by "r"

F1 cross -

true breeding smooth green plant ( RRGG) and true breeding wrinkled yellow (rrgg)

F1 gamete will be RG, RG, rg, rg

F1 offspring will be RrGg , Thus all F1 offspring will be heterozygous smooth and yellow.

Thus, R is dominant over r and g is dominant over G

F2 Generation –

RrGg x RrGg

Gametes will be RG, Rg, rG, rg

RG Rg rG rg

RG RRGG RRGg RrGG RrGg

Rg RRGg RRgg RrGg Rrgg

rG RrGG RrGg rrGG rrGg

rg RrGg Rrgg rrGg rrgg

R is dominant over r and g is dominant over G

Genotypes are –

RRGG - 1 (Smooth Green)

RRGg-2 (Smooth yellow)

RrGG-2 (Smooth Green)

RrGg-4 (Smooth yellow)

RRgg- 1 (smooth yellow)

Rrgg – 2 (Smooth yellow)

rrGG – 1 (wrinkled Green)

rrGg – 2 (Wrinkled yellow)

rrgg – 1 (wrinkled yellow)

Yellow smooth - 9

Yellow wrinkle - 3

Green smooth- 3

Green Wrinkle - 1