Q. 89 The sex of a Drosophila melanogaster, which has 4 copies of X-chromosomes and 4 sets of autosomes
will be
(A) female.
(B) male.
(C) metafemale.
(D) metamale.
In Drosophila melanogaster, sex is determined by the X chromosome to autosome (X:A) ratio, not the presence of Y. A fly with 4 X-chromosomes and 4 sets of autosomes has an X:A ratio of 1.0, resulting in a metafemale phenotype. The correct answer is (C) metafemale.
X:A Ratio Basics
Normal females have 2X:2A (ratio 1.0), producing balanced female traits through activation of the Sex-lethal (Sxl) gene. Males show 1X:2A (ratio 0.5), keeping Sxl off for male development. Ratios above 1.0 (e.g., 3X:2A = 1.5) yield metafemales, while below 0.5 (e.g., 1X:3A ≈ 0.33) produce metamales.
Option Breakdown
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(A) Female: Matches 2X:2A (ratio 1.0) for standard females, but 4X:4A exceeds this balance, causing exaggeration.
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(B) Male: Occurs at 1X:2A (ratio 0.5), like XY; irrelevant here with four X chromosomes.
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(C) Metafemale: Defines 4X:4A (ratio 1.0 in tetraploid context, but hyperfeminized due to excess X signals); viable but weak, sterile, with reduced viability.
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(D) Metamale: Fits low ratios like 1X:3A (≈0.33); opposite of this high-X case.
Key Implications
This genic balance system, discovered by Calvin Bridges, highlights X-linked female determinants outweighing autosomal male factors at high ratios. Metafemales (superfemales) often show slender bristles and delayed development, underscoring dosage sensitivity in sex differentiation.
