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. PhenotypeNumber of F2 Flies EarsDorsalEyes 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.

1. 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.
PhenotypeNumber of F2 Flies
EarsDorsalEyes
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.

2. 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