Q.99 In a new species of moth, the genes for body colour (black, B, is dominant over grey, b), wing size
(normal wing, W, is dominant over vestigial, w) and eye colour (red, R, is dominant over white, r)
are linked. In this species, only one cross-over event has been observed between any two
homologous chromosomes during meiosis. In a cross between BB; ww and bb; WW, 5 % of the
progeny were black with normal wings. In a separate cross between RR; WW and rr; ww, 15 % of
the progeny were red-eyed with vestigial wings. In a third cross between BB; rr and bb; RR, 10 %
of the progeny were black-coloured with red eyes. Which among the following is the correct order
of these three genetic loci?
(A) Body colour – Eye colour – Wing size (B) Eye colour – wing size – Body colour
(C) Wing size – Body colour – Eye colour (D) Eye colour – Body colour – Wing size
In moth genetic loci order three point cross analysis, recombination frequencies between linked genes reveal sequence on chromosome. This GATE genetics problem uses three pairwise crosses to map B (body), W (wing), R (eye) loci via double crossover minimization.
Cross Data and Map Distances
Since one crossover max per meiosis, observed recombinant % = map distance (cM).
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Cross 1: BB ww × bb WW → 5% black normal (B_W_) = B–W distance = 5 cM
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Cross 2: RR WW × rr ww → 15% red vestigial (R_ww) = R–W distance = 15 cM
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Cross 3: BB rr × bb RR → 10% black red (B_R_) = B–R distance = 10 cM
Three-Point Test Logic
Correct order makes largest distance = sum of smaller ones (double CO in flanking intervals):
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15 cM (R-W) ≈ 10 cM (B-R) + 5 cM (B-W) → Order: R – B – W (eye – body – wing)
Verification: R to B (10 cM) + B to W (5 cM) = 15 cM total R-W.
Option Analysis
(A) Body–Eye–Wing (B-R-W): B-R=10, R-W=? (not 5+10=15 mismatch) → Wrong
(B) Eye–Wing–Body (R-W-B): R-W=15, W-B=? (not matching data) → Wrong
(C) Wing–Body–Eye (W-B-R): W-B=5, B-R=10 → W-R=15 ✓ but doesn’t fit Cross 2 recombinant type
(D) Eye–Body–Wing (R-B-W): R-B=10 + B-W=5 = R-W=15 ✓ Matches all → Correct
| Proposed Order | Predicted Distances | Matches Data? |
|---|---|---|
| R-B-W | R-B=10, B-W=5, R-W=15 | Yes (additive) |
| B-R-W | B-R=10, R-W=5? No | Fails |
| R-W-B | R-W=15, W-B=5? No | Fails |
| W-B-R | W-B=5, B-R=10=15? | Partial fail |
Punnett Square Insight (Cross 1 Example)
BB ww (B W / B w, but pure ww so B w / B w) × bb WW (b W / b W, pure WW)
Parental: B w ; b W
Recombinant B W (black normal): 5% = CO between B and W.
GATE Genetics Strategy
Additivity principle from Morgan’s fly work: test orders where largest RF = sum of intervals. Practice: identify parental ditypes, score SCO/DCO ratios. Similar to Neurospora 3-point crosses.
