14. Two mutants of Drosophila melanogaster. one showing light red eye colour and the other showing dark brown eye colour, were crossed. In F₁ all flies showed normal red eye colour. This indicated that the two mutations are
(1) allelic.        (2) non allelic.
(3) linked.      (4) unlinked.

Genetic reasoning

• Each parent is homozygous recessive for its own eye‑color mutation:

  • light‑red mutant: $aa$

  • dark‑brown mutant: $bb$ (symbols just for illustration).

• F₁ genotype from crossing them: a+a b+ba+ab+b – heterozygous at both loci.

• Because each F₁ carries at least one wild‑type allele of each gene, pigment synthesis is restored and all offspring show normal red eyes.

This is the classic outcome of a complementation test: two recessive mutants with the same general phenotype produce wild‑type progeny when the mutations are in different genes (non‑allelic).

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Option‑wise explanation

1. Allelic

   • If both mutations were different alleles of the same gene (allelic), the F₁ (heteroallelic, e.g., a¹a²) would still have no wild‑type copy and would remain mutant in eye colour.

   • Since the F₁ are wild type, the mutations cannot be allelic.

2. Non allelic – correct

   • Non‑allelic means the mutations lie in different genes that both affect eye pigmentation.

   • In the F₁, each gene gets a wild‑type allele from the other parent, so they complement and restore normal red eye colour.

3. Linked

   • Linkage describes physical proximity on the same chromosome and affects segregation ratios, not whether F₁ are mutant or wild type in this complementation context.

   • The result (all F₁ wild type) is about gene function and allelism, not about map distance.

4. Unlinked

   • Unlinked genes assort independently; again this refers to chromosomal location, not to the complementation outcome.

   • The key information here is that the two mutations complement (→ non‑allelic), not whether the loci are on the same or different chromosomes.

Therefore, the two eye‑colour mutations are non‑allelic.