A three point cross was performed using fruit flies The thre

A three point cross was performed using fruit flies. The three loci are physically linked on the same autosome. The d allele, in the homozygous state, results in the recessive ‘double dorsal’ phenotype. Similarly, the r allele results in the recessive ‘round ears’ phenotype and the e allele in the recessive ‘endless eye’ phenotype. In contrast, the D, R, and E alleles encode for the dominant wild type phenotypes at each of these loci.

True breeding (homozygous) flies were crossed to produce F1 offspring. Due to poor book-keeping, the genotypes of the parent flies were not recorded – but one parent fly had one mutant phenotype, while the other parent had the other two mutant phenotypes.

  Phenotype Number of F2 Flies

Ears Dorsal       Eyes

1) Round ears    Double Dorsal    Endless Eyes 230

2) Wild-type        Wild-type Wild-type               260

3) Round ears    Wild-type Wild-type 4,390

4) Wild-type       Double Dorsal       Endless Eyes 4,100

5) Round ears       Double Dorsal       Wild-type 15

6) Wild-type        Wild-type        Endless Eyes 15

7) Round ears    Wild-type        Endless Eyes 510

8) Wild-type        Double Dorsal    Wild-type               480

10,000 total

a) What is the phenotype of the F1 flies?

b) What is the genotype of the F1 flies? Your answer should indicate the phase of linkage.

c) What are the genotypes and phenotypes of the original parent flies (previously not recorded)? Your answer should indicate the phase of linkage.

d) Calculate the map distance in cM between each of the three loci and determine the gene order. Show all work for full credit. (6 points)

Solution

a. The phenotype of the F1 flies is wild-type heterozygous at all locus.

b. The genotype of F1 flies

r D E/ R d e

c. The genotypes of parent flies: r D E and R d e

The phenotypes of parent flies: Round ears, Wild-type, Wild-type and Wild-type, Double Dorsal, Endless Eyes

d.

Gene order is R------D------E

The distance between R and D

510+480+15+15 / 10,000 * 100 = 10.2cM

The distance between D and E

230+260+15+15 / 10,000 * 100 = 5.2cM