Given the parents AABBCc × AabbCc, assume simple dominance for each trait and independent assortment. What proportion of the progeny will be expected to phenotypically resemble the AABBCc parent? Given the parents AABBCc × AabbCc, assume simple dominance for each trait and independent assortment. What proportion of the progeny will be expected to phenotypically resemble the AABBCc parent? 1 1/8 3/8 1/4 3/4

3/4

Explanation:

If we assume simple dominance and independent assortment for each trait, we can use Mendel's Law of Segregation to predict the phenotypic proportions in the offspring of the parental cross AABBCc x AabbCc.

Gene A

AA x Aa

F1 genotypes: 1/2 AA, 1/2 AaF1 phenotypes: all AGene B

BB x bb

F1 genotypes: 1 BbF1 phenotypes: all BGene C

Cc x Cc

F1 genotypes:  1/4 CC, 2/4 Cc, 1/4 ccF1 phenotypes: 3/4 C, 1/4 cc

We want to know the proportion of progeny with all dominant phenotype (A_B_C_). Since the genes are independent, we can multiply the probabilities of each gene to obtain the overall probability of having a ABC progeny:

1 A_ x 1 B_ x 3/4 C_ = 3/4 A_B_C_

The answer would be a. aluminum can

The correct answer is c