1. 1 Lecture 5

2. 2 Linked genes, recombination, and chromosomal mapping Mendel's Law of Independent Assortment is a consequence of the fact that chromosomes segregate independently in meiosis These results are readily explained by the two alternative ways the chromosomes can line up on the metaphase plate during meiosis I: Take two individuals One heterozygous and one homozygous AaBbxaabb

3. 3 MeiosisI These results are readily explained by the two alternative ways the chromosomes can line up on the metaphase plate during meiosis I: A a B b OR A a b B ABabAbaB Because the A and B genes assort independently, AaBb dihybrids constructed from different parental genotypes will behave the same. AABBxaabbAAbbxaaBB

4. 4 AaBbxaabb AABBxaabb AAbbxaaBB AaBbxaabb A a B b A a b B AaBb OR

5. 5 Why 50:50 Why not 25:25:25:25 A dihybrid cross involving the genes A and C produced the following results: ·A= Talla= short ·C= Cream c = white Cross I:Cross II:

6. 6 In these crosses, independent assortment is not occurring. For example in the first cross, the alleles Tall and Cream behave as if they are linked to one another. Similarly in the second cross the alleles Tall and white appear as if they are linked to one another. These results are readily explained if the genes A and C lie next to one another on a chromosome:

7. 7 Cross I: A= Talla= short C= Cream c = white A-C a-c A-C a-c

8. 8 Cross II: A= Talla= short C= Cream c = white A-c a-C A-c a-C a-c

9. 9 Coupling/repulsion Results like these led Morgan to suggest that the A and C genes are located on the same pair of homologous chromosomes. Therefore when the A and C alleles are introduced from one parent they are physically located on the same chromosome and they do not assort independently. We say that they are linked. In the above cross we say that the A and C genes are linked. Therefore when we write the genotype of a dihybrid for two linked genes, there are two possible conformations: AaCc

10. 10 Purple vestigial Morgan performed the following experiments in Drosophila to determine if the genes pr and vg were linked. pr+ = normal red eyes pr = purple eyes vg+ = normal wings vg = vestigial wings P F1 If they are on different chromosomes they should assort independently If they are next to one another on the same chromosomes they should not assort independently

11. 11 When Morgan performed this cross, he obtained the following result:

12. 12 The chromosomes that have gone through this crossover are known as crossover products or recombinants. The original chromosomes and those that have not undergone a crossover are known as parental. Evidence for the model that chromosomes physically exchange during meiosis is found in meiotic structures known as chiasmata. During meiosisI when homologs pair, non-sister chromatids appear to cross with each other. The resulting cross-shaped structure is known as a chiasmata.

13. 13 Crossing-over through the microscope Duplicated homologous chromosomes Synapsis Crossing over between Non-sister chromatids AnaphaseI Segregation of homologous chromosomes Haploid gametes

14. 14 Answer: To explain this we need to define the terms parental and recombinant: Parents: Gametes: F1: Meiosis produces the following gametes: How does one determine whether two genes reside on different chromosomes or reside on the same chromosome as linked genes? Recombinant gametes are those with different allelic combinations than those gametes of the previous generation.

15. 15 If genes A and B are on different chromosomes: Test cross progeny F1 diploid

16. 16 Genes A and B are linked on the same chromosome Test cross progeny F1 diploid

17. 17 Recombination frequency If a crossover occurred between linked genes each time homologs paired, the recombinant frequency would be 50% This is because crossing-over involves only two of the four chromatids on the metaphase pair (each of the paired homologs consists of two sister chromatids). For example, the frequency of recombinant gametes between linked genes A and B is 50% if crossing-over occurred each time the homologs paired. AB AB ab ab AB Ab aB ab parental recombinant

18. 18 However there are many instances in which the homologs pair and crossing over does not occur between genes A and B. It occurs somewhere else Consequently the overall frequency of recombinants is significantly reduced from 50% AB AB ab ab AB AB ab ab parental

19. 19 Distance The larger the distance between two genes residing on the same chromosome, the higher the probability there is that a crossover event will occur between them. That is for any chromosome, there is a fixed probability per a given distance on the chromosome that a crossover event will event. Sturtevant realized that this property could be used to map genes with respect to one another. For each pair of genes on a chromosome a recombination frequency can be determined. By determining the recombination frequency between many pairs of genes on a chromosome, the relative distance between genes and the relative order of the genes on the chromosome can be determined.

20. 20 For example Sturtevant identified three recessive mutations that reside on the X chromosome of Drosophila By calculating recombination frequencies between each pair of genes we can begin to establish where these three genes reside on the X chromosome with respect to one another

21. 21 To determine the distance between the w gene and the sn gene Pw sn/w snxw+ sn+/Y F1w sn/w+ sn+xw sn/Y F2

22. 22 Recombination frequencies Recombination frequency equals the number of recombinants over total number of progeny # recombinant progeny = # total progeny 1 map unit (m.u.) = 1% recombination frequency Therefore _________________separate the w and sn genes. This is a relative distance- depends upon recombination between two genes. Not an absolute distance like bp In the above cross, we could have determined recombination frequency by counting only males (or only females)

23. 23 The next issue is where does cv map with respect to w and sn: