Q.57 In a genetic cross between the genotypes WWXX and wwxx, the following phenotypic
distributions were observed among the F2 progeny: WX, 562; wx, 158; Wx, 38; and wX, 42.
Likewise, a cross between XXYY and xxyy yielded the following results: XY, 675; xy, 175; Xy,
72; and xY, 78. Similarly, a cross between WWYY and wwyy yielded: WY, 292; wy, 88; Wy, 12;
and wY, 8. In all the genotypes, capital letters denote the dominant allele. Assume that the F1
progeny were self-fertilized in all three crosses. Also, double cross-over does not occur in this
species. Which of the following is correct?
(A) Relative position: W-X-Y Distances: W-X = 5 map units, X-Y = 17 map units
(B) Relative position: X-Y-W Distances: X-Y = 15 map units, Y-W = 11 map units
(C) Relative position: Y-W-X Distances: Y-W = 5 map units, W-X = 11 map units
(D) Relative position: X-W-Y Distances: X-W = 5 map units, W-Y = 10 map units
Solving Genetic Linkage Mapping: WWXX x wwxx Cross Analysis
Genetic linkage mapping determines gene order and distances from F2 phenotypic ratios in three-point testcrosses. The correct answer is option (C), with gene order Y-W-X and distances Y-W = 5 map units, W-X = 11 map units.
Phenotypic Data Overview
Total progeny counts reveal parental and recombinant classes for each dihybrid cross:
- WWXX × wwxx: WX (562, parental), wx (158, parental), Wx (38, single crossover), wX (42, single crossover), total 800
- XXYY × xxyy: XY (675), xy (175), Xy (72), xY (78), total 1000
- WWYY × wwyy: WY (292), wy (88), Wy (12), wY (8), total 400
Recombination frequencies (RF) calculate as (single crossovers / total) × 100, representing map distances since no double crossovers occur.
Recombination Frequency Calculations
- WW-Xx cross: RF = (38 + 42)/800 × 100 = 10 map units (W-X distance)
- Xx-Yy cross: RF = (72 + 78)/1000 × 100 = 15 map units (X-Y distance)
- Ww-Yy cross: RF = (12 + 8)/400 × 100 = 5 map units (W-Y distance)
These pairwise distances indicate W-Y closest (5 mu), followed by W-X (10 mu) and X-Y farthest (15 mu), consistent with additive distances on a chromosome.
Gene Order Determination
Order Y-W-X fits data: Y-W (5 mu) + W-X (10 mu) ≈ X-Y (15 mu). Parental classes dominate each cross (WX/wx, XY/xy, WY/wy), confirming linkage. Smallest RF (W-Y) identifies closest genes; Y farthest from X positions it at one end.
Option Analysis
| Option | Gene Order | Distances (mu) | Validity |
|---|---|---|---|
| (A) | W-X-Y | W-X=5, X-Y=17 | Incorrect; W-X=10, not 5; sum=22 ≠15 |
| (B) | X-Y-W | X-Y=15, Y-W=11 | Incorrect; Y-W=5, not 11; order mismatches closest pair |
| (C) | Y-W-X | Y-W=5, W-X=11 | Correct; matches all RFs (5+10=15) |
| (D) | X-W-Y | X-W=5, W-Y=10 | Incorrect; X-W=10, not 5; sum=15 but order wrong |
Option (C) uniquely satisfies additive distance rule without contradictions.
