(mendelian genetics) consider three gene pairs aa, bb, and cc. the uppercase letter is the dominant allele and each gene pair assorts independently of each other. what is the probability of obtaining an offspring displaying the a b c phenotype (that is having the dominant phenotypes for each of the three genes) from a cross of aa bb cc x aa bb cc parents?

9/16

Explanation:

The cross was Aa Bb CC X Aa Bb cc

Because genes assort independently, we can analyze each of them separately in the cross, calculate the probability of getting offpsring with a dominant allele, and then multiply each of those probabilities to obtain the overall probability of having A_B_C_ progeny.

A/a

Aa x Aa ---> 3/4 A_, 1/4 aa

B/b

Bb x Bb ---> 3/4 B_, 1/4 bb

C/c

CC x cc ---> 1 C_

3/4 × 3/4 × 1 = 9/16 A_B_C_ offspring.

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