Answer:
C. 1/256
Explanation:
In the tetrahybrid cross, we have different four groups. Let name them Part (A, B, C & D).
Let Part A be Aa and Aa... If both traits crosses; we have
A a
A AA Aa
a Aa aa
Hence, there is a total of 4 combinations
Let Part B be Bb and Bb... If both traits crosses; we have
B b
B BB Bb
b Bb bb
Hence, there is a total of 4 combinations
Let C be Cc and Cc ... If both traits crossses; we have
C c
C CC Cc
c Cc cc
Hence, there is a total of 4 combinations
Let D be Dd and Dd... If both traits crosses; we have
D d
D DD Dd
d Dd dd
Hence, there is a total of 4 combinations
A genotype has the 4 groups together. Therefore the possible number of ways of forming the genotype = 4 x 4 x 4 x 4= 256
Now, Probability that the F2 offspring would be expected to be genotypically AABBCCdd = [tex]\frac{number of genotype AABBCCdd}{Total number of genotype in the offsprings}[/tex]
= [tex]\frac{1}{256}[/tex]
