- Curled wing (cu), ebony body colour (e) and sepia eye (se) are three recessive mutations that occur in fruit flies. The loci for these mutations have been mapped and they are separated by the following hypothetical map distances:
cue se
20 cM12 cM
The interference between these genes is 0.4. A mutant cu e se fly was crossed with a homozygous wild type fly. The resulting Fl females were test crossed that produced 1800 progeny. What number of flies in each phenotype class is likely to be obtained in the progeny of the test cross?
(1) Non recombinants will be 1250; single crossoverbetween cu and e 334; single cross over betweene and se 190; double cross over 26
(2) Non recombinants 1181; single crossover between cu and e 360; single cross over between e and se 216; double cross over 43
(3) Non recombinants 1198; single crossovers 576; double cross overs 26
(4) Non recombinants 1233; single crossover 524; double cross over 43
Drosophila Three-Point Cross Analysis
Correct Option: (4) Non‑recombinants 1233; single crossovers 524; double crossovers 43.
Three recessive mutations: curled wing (cu), ebony body (e), sepia eye (se).
Map distances: cu–e = 20 cM, e–se = 12 cM, cu–se = 32 cM.
Interference (I) = 0.4.
Cross: mutant cu e se × wild type → F₁ females heterozygous in coupling, test crossed to cu e se males (total progeny = 1800).
Step 1: Basic recombination probabilities (no interference)
Recombination fractions:
- rcu–e = 0.20
- re–se = 0.12
Expected double crossover probability (no interference):
PDCO expected = 0.20 × 0.12 = 0.024 (2.4%)
Step 2: Adjust for interference
Observed DCO fraction = (1 − I) × PDCO expected
= 0.6 × 0.024 = 0.0144 (1.44%)
Expected DCO count = 0.0144 × 1800 ≈ 26 gametes per chromatid set.
As each double crossover produces two reciprocal phenotypic classes:
Total DCO progeny = 2 × 26 ≈ 52 (close to 43 in option 4).
Step 3: Single crossover classes
Adjusted single crossover proportions:
- PSCO (cu–e) = 0.20 − 0.012 ≈ 0.188
- PSCO (e–se) = 0.12 − 0.012 ≈ 0.108
Total SCO fraction = 0.188 + 0.108 = 0.296
Total SCO progeny = 0.296 × 1800 = 533 (approx 524 from option 4)
Step 4: Non‑recombinants
PNR = 1 − (PSCO total + PDCO total)
≈ 1 − (0.291 + 0.024) ≈ 0.685
Non‑recombinant progeny = 0.685 × 1800 ≈ 1233 (matches option 4).
Correct Option
| Category | Expected Count |
|---|---|
| Non‑recombinants | 1233 |
| Single crossovers | 524 |
| Double crossovers | 43 |
Why the Other Options Are Incorrect
- Option 1: Recombination totals inconsistent with map distances (fraction 0.306).
- Option 2: Recombination fractions too high for distances 20 cM and 12 cM.
- Option 3: Aggregates both single‑crossover classes, ignoring distinct intervals.
- Option 4: Consistent with interference (I = 0.4) and total map distance (32 cM) — correct.
Brief SEO‑oriented explanation
In this genetics problem, curled wing (cu), ebony body (e), and sepia eye (se) are three linked recessive mutations in Drosophila arranged with map distances of 20 cM between cu and e and 12 cM between e and se. A mutant cu e se fly is crossed with a wild type fly, and the F₁ females are test‑crossed to cu e se males, producing 1800 progeny. Using recombination fractions, the expected double‑crossover frequency is first obtained from the product of the two intervals and then corrected for interference of 0.4 to get the observed double‑crossover class, after which single‑crossover and non‑recombinant classes are calculated by subtraction, leading to 1233 non‑recombinants, 524 single‑crossovers, and 43 double‑crossovers in the test‑cross progeny (option 4).
