genotype AaBbCcDdEeFfGg and her dad has the genotype aaBbccDdeeFfGg. What is the likelihood that her genotype is aabbccDDEeffgg

The question is incomplete. Here is the complete question.

Stacy's mom has the genotype AaBbCcDdEeFfGg and her dad has the genotype aaBbccDdeeFfGg. What is the likelihood that her genotype is aabbccDDEeffgg?

(a) None of the answers are correct

(b) ~0.05%

(c) ~0.005%

(d) ~1%

(e) ~0.001%

(f) ~0.5%

(g) ~0.01%

(a) None of the answers are correct

Explanation: In Genetics, genes are assorted independently. So, to determine multiple genes crosses, we can separate each pair of genes.

First we find the likelihood of each genotype:

For gene A: Aa x aa

        A      a

a      Aa     aa

a      Aa     aa

Likelihood of genotype aa is 1/2 or 0.5.

For gene B: Bb x Bb

          B       b

B       BB     Bb

b       Bb     bb

Likelihood of genotype bb is 1/4 or 0.25.

Gene C: Cc x cc and gene E are similar to gene A, so, probability is 0.5 each.

For gene D: Dd x Dd, gene F: Ff x Ff and gene G: Gg x Gg, they are similar to gene B, so, probability is 0.25 each.

With all the probabilities, we multiply each result:

P (aabbccDDEeffgg) = 0.5*0.25*0.5*0.25*0.5*0.25*0.25

P(aabbccDDEeffgg) ≈ 0.0005

Note that, following "probabilities rules", we multiply the results because all the genes have to happen at the same time, i.e., it is an "AND" situation.

So, likelihood or probability of Stacy having genotype aabbccDDEeffgg is 0.0005%.