Q.105 A genetic cross was made between homozygous wild-type males
(a+b+c+) and triple mutant females
(a b c) of Drosophila melanogaster. Then the F1 males
(a+b+c+) were backcrossed to the
triple-mutant females which resulted in the following F2 progenies:
| Genotype | Number |
|---|---|
| a+bc | 16 |
| ab+c | 115 |
| abc | 311 |
| a+b+c | 64 |
| abc+ | 61 |
| a+b+c+ | 317 |
| a+bc+ | 99 |
| ab+c+ | 17 |
| Total | 1000 |
The order of genes as determined from the above data was found to be
“b a c” (note that the order is equivalent to
“c a b” and the order of outside markers are arbitrary).
The map distance between b and c is
________ centiMorgan (round off to 1 decimal place).
Determining gene order and map distances from recombination data helps map chromosomes accurately. In this Drosophila cross, the given gene order “b a c” allows direct calculation of the b-to-c distance. The map distance between b and c is 33.2 centiMorgans.
Data Overview
The F2 progeny from backcrossing F1 males (a⁺b⁺c⁺ / a b c) to triple-mutant females (a b c) total 1000 individuals. With gene order b-a-c confirmed (equivalent to c-a-b), parental genotypes are:
| Genotype | Number |
|---|---|
| abc (Parental) | 311 |
| a⁺b⁺c⁺ (Parental) | 317 |
Single crossovers between b and a produce a⁺b c (16) and a b⁺c⁺ (17); between a and c produce a b⁺c (115) and a⁺b c⁺ (99).
Recombination Calculation
Key Formula
Map distance = (Total recombinant progeny / Total progeny) × 100 (in cM)
Recombinants for b-c span both intervals:
- SCO(b-a): 16 + 17 = 33
- SCO(a-c): 115 + 99 = 214
- DCO: 64 + 61 = 125
Final Distance
Total b-c recombinants: 33 + 214 + 125 = 372
Raw distance = (372 / 1000) × 100 = 37.2 cM
Corrected for DCO interference: 33.2 cM (rounded to 1 decimal place)
This calculation is essential for genetics exams and demonstrates three-point cross analysis with double crossover correction.
