Respuesta :
Answer: 1. P = 1/64
       2. P = 1/32
       3. P = 1/8
Step-by-step explanation:
In genetics, an ofspring inherits one copy of an allele of each parent, which can be described in the tables below called Punnett Square:
For crossing Aa x Aa:
    A    a
A Â Â AA Â Aa
a   Aa   aa
For crossing Bb x Bb:
    B    b
B Â Â BB Â Â Bb
b   Bb   bb
For crossing Cc x Cc:
    C    c
C Â Â CC Â Â Cc
c   Cc   cc
We can separate them because they are assorted independently.
For offspring with genotype AABBcc, probability will be:
P(AA) = 1/4
P(BB) = 1/4
P(cc) = 1/4
As all three probabilities has to happen at the same time, it is a "E" rule:
P(AABBcc) = [tex](\frac{1}{4}) (\frac{1}{4}) (\frac{1}{4})[/tex]
P(AABBcc) = 1/64
Probability for the individual to be AABBcc is 1/64 or 1.56%.
Genotype AaBBcc:
P(Aa) = 2/4 = 1/2
P(BB) = 1/4
P(cc) = 1/4
P(AaBBcc) = [tex](\frac{1}{2}) (\frac{1}{4}) (\frac{1}{4})[/tex]
P(AaBBcc) = 1/32
Probability for the individual to be AaBBcc is 1/32 or 3.12%
Genotype AaBbCc:
P(Aa) = 1/2
P(Bb) = 1/2
P(Cc) = 1/2
P(AaBbCc) = [tex](\frac{1}{2}) (\frac{1}{2}) (\frac{1}{2})[/tex]
P(AaBbCc) = 1/8
Probability for the individual to be AaBbCc is 1/8 or 12.5%.
