35. A phenotypically normal fruit fly was crossed to another fly whose phenotype was not recorded. Of the progeny, 3/8 were wild type, 3/8 had ebony body color, 1/8 had vestigeal wings and 1/8 had ebony body color and vestigeal wings. Ebony body color and vestigeal wings are recessive characters and their genes are located on two different autosomes. Based on this information which one of the following is the likely genotype of the parents?
(1) ee vg vg and e⁺e⁺ vg⁺vg
(2) ee vg⁺vg and e⁺e vg⁺vg
(3) e⁺e vg vg and e⁺e⁺ vg⁺vg
(4) e⁺e vg⁺vg and ee vg⁺vg

Step-by-step solution

Step 1: Interpret the phenotypes

Genes on different autosomes (independent assortment) [web:99][web:102]:

• Ebony body: recessive e
• Wild body: e⁺–
• Vestigial wings: recessive vg
• Wild wings: vg⁺–

Observed progeny proportions:

• 3/8 wild
• 3/8 ebony
• 1/8 vestigial
• 1/8 ebony + vestigial

Total ebony = 3/8 + 1/8 = 1/2; Total vestigial = 1/8 + 1/8 = 1/4.

Step 2: Infer genotypes at each locus

Ebony locus (P(ebony) = 1/2):

Parent 1: e⁺e (phenotypically normal)

Parent 2: ee (ebony)

Vestigial locus (P(vestigial) = 1/4):

Both parents: vg⁺vg

Step 3: Full parental genotypes

One parent (phenotypically normal): e⁺e vg⁺vg

Other parent (ebony, normal wings): ee vg⁺vg

This matches option (4) [web:103].

Verification of ratios

• Wild: (1/2 non-ebony) × (3/4 non-vestigial) = 3/8
• Ebony only: (1/2 ebony) × (3/4 non-vestigial) = 3/8
• Vestigial only: (1/2 non-ebony) × (1/4 vestigial) = 1/8
• Ebony + vestigial: (1/2 ebony) × (1/4 vestigial) = 1/8

Exactly matches observed phenotypes.

Why other options fail

• Option 1: ee vg vg × e⁺e⁺ vg⁺vg → 0 ebony
• Option 2: Both e⁺e → no 1/2 ebony
• Option 3: vg vg parent → 1/2 vestigial, not 1/4