Answer:
b. 3 (whilte) : 1(purple)
Explanation:
The first dihybrid crossing is between AaBb x AaBb
Now, if the double heterozygous traits self crossed, we have the following gametes shown below for the F₂ crossing.
AaBb = (AB, Ab, aB, ab)
AB Ab aB ab
AB AABB AABb AaBB AaBb
Ab AABb AAbb AaBb Aabb
aB AaBB AaBb aaBB aaBb
ab AaBb Aabb aaBb aabb
We were being told that the results are;
purple: A_B_ 9/16
white: aaB_ 3/16
white: A_bb 3/16
white: aabb 1/16
From above, we can see that the same is true;
For Purple color; we Have (AABB and AaBb) since A is dominant to a and B is dominant to b.
∴ From the above punnet square; we have 9 purple colors which are:
(AABB, AABb, AaBB, AaBb, AABb, AaBb, AaBB, AaBb, AaBb) = 9/16
white: aaB_ ( since B is dominant to b, it can be either BB or Bb)
= (aaBB, aaBb, aaBb) = 3/16
white: A_bb ( since A is dominant to a, it can be either AA or Aa)
= (AAbb, Aabb, Aabb) = 3/16
white: aabb i.e homozygous recessive = (aabb) = 1/16
Furthermore, the question goes further by saying:
If a double heterozygote (AaBb) is crossed with a fully recessive organism (aabb), what phenotypic ratio is expected in the offspring?
If AaBb self crossed, we have: (AB, Ab, aB, ab)
If aabb self crossed, we have: (ab, ab, ab, ab)
AB Ab aB ab
ab AaBb Aabb aaBb aabb
ab AaBb Aabb �� aaBb aabb
ab AaBb Aabb aaBb aabb
ab AaBb Aabb aaBb aabb
AaBb ( purple)
= 4/16
= 1/4
= 0.25
= 25%
Aabb (white)
= 4/16
= 1/4
= 0.25
= 25%
aaBb (white)
= 4/16
= 1/4
= 0.25
= 25%
aabb (white)
= 4/16
= 1/4
= 0.25
= 25%
Now the proportion of white to purple = (25%+25%+25%): 25%
= 75%:25%
= 3(white):1(purple)
We can therefore conclude that the expected phenotypic ratio in the cross between a double heterozygote (AaBb) with a fully recessive organism (aabb) yeids;
3 (whilte) : 1(purple)
