Answer:
The probability that the first offspring will be of AabbccDd genotype is 0.03125.
Step-by-step explanation:
The parents genotype are: AaBbccDd and AaBbCcDD.
A a
A AA Aa
a Aa aa
P (Aa) = 2/4 = 1/2
B b
B BB Bb
b Bb bb
P (bb) = 1/4
c c
C Cc Cc
c cc cc
P (cc) = 2/4 = 1/2
D d
D DD Dd
D DD Dd
P (Dd) = 2/4 = 1/2
Then the probability that the first offspring will be of AabbccDd genotype is:
P (AabbccDd) = (1/2) × (1/4) × (1/2) × (1/2) = 1/32
Thus, the probability that the first offspring will be of AabbccDd genotype is 0.03125.