1. Bio 2970 Lab 5: Linkage Mapping Sarah VanVickle-Chavez

2. Gender

3. 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

4. Hypothesis A cross between the F 1 flies should give us: 3 red eye : 1 white eye F2F2 PhenotypesRed eyeWhite eye Numbers3470 82% 782 18% So far so good

5. An interesting observation F2F2 PhenotypesRed- eyed males Red- eyed females White- eyed males White- eyed females Numbers101124597820 24%58%18%0% © 2007 Paul Billiet ODWSODWS

6. 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

7. 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 1321298688

8. 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.

9. Mutants – Linkage Mapping b pr rw + + + Phenotype b pr rw + + + + pr rw b + + b + rw + pr + b pr + + + rw

10. Mutants

11. 1. Record offspring with each of 8 phenotypes: E.g., cross using three genes A, B, and C PhenotypeNumber +3223 A632 B1051 C226 A;B229 A;C1019 B;C617 A;B;C3091 TOTAL:10088

12. 2. Rearrange into pairs, identifying double recombinants, single recombinant, and nonrecombinant (parental) PhenotypeNumber +3223 (NR) A632 (SR) B1051 (SR) C226 (DR) A;B229 (DR) A;C1019 (SR) B;C617 (SR) A;B;C3091 (NR) TOTAL:10088

13. 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 +3223 (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).

14. 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 +3223 (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.

15. 5. Calculate the distance between each pair of genes. B-C = C-A =

16. 6. Calculate the expected double crossover frequency, the obtained double crossover frequency, the coefficient of coincidence, and the interference.