F2 Proportion of All Recessive Phenotypes in 5-Gene Cross

The cross between AABBCCDD EE and aabbccddee produces F1 heterozygotes (AaBbCcDdEe), and selfing yields an F2 generation where the proportion showing all recessive phenotypes (aabbccddee) is 1/1024.​

Genetic Cross Breakdown

Parental genotypes represent five independently assorting genes: AABBCCDD EE (all dominant) crossed with aabbccddee (all recessive). F1 offspring are uniformly heterozygous AaBbCcDdEe, displaying all dominant phenotypes due to Mendel’s dominance principle.​

Each F1 parent produces 2^5 = 32 gamete types through independent assortment. The F2 Punnett square conceptually expands to 32 x 32 = 1024 combinations, with only one matching aabbccddee.

Correct Answer Explanation

For all recessive phenotypes, each gene requires aa genotype: AA x aa → Aa (F1), then Aa x Aa → 1/4 aa (F2). With five independent genes, multiply probabilities: (1/4)^5 = 1/1024.​

This matches option (D).

Option Analysis

• (A) 1/64: Equals (1/4)^3 or (1/2)^6, for 3 genes or 6 coin flips; incorrect for 5 genes.​

• (B) 1/256: Equals (1/4)^4, for 4 genes; undercounts loci here.​

• (C) 1/512: Equals (1/2)^9; mismatched, as recessive needs homozygous aa (1/4), not 1/2.​

• (D) 1/1024: Correct, as (1/4)^5 precisely for five recessive traits.​