Answer:
A) 9:3:3:1
Explanation:
Lets consider a true breeding white squash (WWYY) is crossed with a true breeding green squash (wwyy) giving rise to a dihybrid offspring WwYy.
First, let determine the F1 offspring.
If WWYY self crsossed we have: WY, WY as the traits required for the F1 cross,
On the other hand wwyy = wy, wy
Therefore, If a true breeding white squash WWYY crossed with true breeding green squash wwyy, we have the following as the F1 offspring.
WY WY
wy WwYy WwYy
wy WwYy WwYy
Now for the F2 offspring. we will have a dihybrid cross between WwYy * WwYy
WY Wy wY wy
WY WWYY WWYy WwYY WwYy
Wy WWYy WWyy WwYy Wwyy
wY WwYY WwYy wwYY wwYy
wy WwYy Wwyy wwYy wwyy
∴ From the above cross, the phenotypic ratio = 9:3:3:1
9 (WWYY WWYy WwYY WwYy WWYy WwYy WwYY WwYy WwYy)
3 (WWyy Wwyy Wwyy)
3 (wwYY wwYy wwYy)
1 (wwyy)
