Bio 2970 Lab 5: Linkage Mapping Sarah VanVickle-Chavez Gender The case of the white-eyed mutant CharacterTraits Eye colourRed eye (wild type) White eye (mutant) P Phenotypes Wild type (red-eyed) female x White-eyed male F 1 Phenotypes All red-eyed Red eye is dominant to white eye Hypothesis A cross between the F 1 flies should give us: 3 red eye : 1 white eye F2F2 PhenotypesRed eyeWhite eye Numbers % % So far so good An interesting observation F2F2 PhenotypesRed- eyed males Red- eyed females White- eyed males White- eyed females Numbers %58%18%0% 2007 Paul Billiet ODWSODWS A reciprocal cross Morgan tried the cross the other way round white-eyed female x red-eyed male Result All red-eyed females and all white-eyed males This confirmed what Morgan suspected The gene for eye colour is linked to the X chromosome A test cross PhenotypesF1 Red-eyed female x White-eyed male Expected result 50% red-eyed offspring: 50% white-eyed offspring Regardless of the sex Observed Results Red-eyed Males Red-eyed Females White-eyed Males White-eyed Females Determining if a mutant is dominant or recessive, and if it is X-linked or autosomal To determine if a mutant is dominant or recessive, and if it is X-linked or autosomal, you perform a pair of reciprocal crosses (where the gender of the parents is reversed). If the gene is autosomal identical results in both crosses. If the gene is X-linked results of the two crosses are different. The expected results for reciprocal crosses is summarized in the table below. Mutants Linkage Mapping b pr rw Phenotype b pr rw pr rw b + + b + rw + pr + b pr rw Mutants 1. Record offspring with each of 8 phenotypes: E.g., cross using three genes A, B, and C PhenotypeNumber A632 B1051 C226 A;B229 A;C1019 B;C617 A;B;C3091 TOTAL:10088 2. Rearrange into pairs, identifying double recombinants, single recombinant, and nonrecombinant (parental) PhenotypeNumber (NR) A632 (SR) B1051 (SR) C226 (DR) A;B229 (DR) A;C1019 (SR) B;C617 (SR) A;B;C3091 (NR) TOTAL:10088 3. Compare the NR class with the DR class to determine the one gene that is switched in the double crossover class. This is the gene that is in the middle. PhenotypeNumber (NR) A632 (SR) B1051 (SR) C226 (DR) A;B229 (DR) A;C1019 (SR) B;C617 (SR) A;B;C3091 (NR) TOTAL:10088 In this example, it is (C), so the order is (A-C-B). You could also state the gene order as the reverse (B-C-A). 4. Each of the remaining classes is a single crossover class between a gene on the end and the gene in the middle: either (B-C singles) or (C-A singles). PhenotypeNumber (NR) A632 (SR) = C-A B1051 (SR) = B-C C226 (DR) A;B229 (DR) A;C1019 (SR) = B-C B;C617 (SR) =C-A A;B;C3091 (NR) TOTAL:10088 Compare each singles class with the nonrecombinant class to see which gene is switched to determine which one is the B-C singles class, and which is the C-A singles class. 5. Calculate the distance between each pair of genes. B-C = C-A = 6. Calculate the expected double crossover frequency, the obtained double crossover frequency, the coefficient of coincidence, and the interference.