Heredity, Gene Regulation, and Development I. Mendel's Contributions

1. Heredity, Gene Regulation, and Development I. Mendel's Contributions II. Meiosis and the Chromosomal Theory III. Allelic, Genic, and Environmental Interactions IV. Sex Determination and Sex Linkage V. Linkage A. Overview Independent Assortment A A b B a a B b AB ab Ab aB

2. Independent Assortment V. Linkage A. Overview A a A A b B b B a a Linkage AB ab Ab aB AB ab B b

3. V. Linkage A. Overview Linkage A A a a In Prophase I of Meiosis – Crossing-over A a b B AB AB ab ab B B b b Ab aB

4. V. Linkage • Overview • Complete Linkage • Test Cross AABB aabb AB ab X ab AB

5. V. Linkage • Overview • Complete Linkage - if genes are immediate neighbors, they are almost never separated by crossing over and are ‘always’ inherited together. The pattern mimics that of a single gene. AABB aabb AB ab X ab AB ab Gametes AB ab AB F1

6. V. Linkage • Overview • Complete Linkage - if genes are immediate neighbors, they are almost never separated by crossing over and are ‘always’ inherited together. The pattern mimics that of a single gene. F1 x F1 X ab ab Gametes AB ab AB ab ab

7. V. Linkage • Overview • Complete Linkage - if genes are immediate neighbors, they are almost never separated by crossing over and are ‘always’ inherited together. The pattern mimics that of a single gene. 1:1 ratio A:a 1:1 ratio B:b 1:1 ratio AB:ab NOT 1:1:1:1 F1 x F1 X ab ab ab Gametes Phenotypes AB ab ab AB AaBb AB ab aabb aB ? Ab ?

8. C. Incomplete Linkage b b A a a a B b

9. C. Incomplete Linkage - So, since crossing-over is rare (in a particular region), most of the time it WON’T occur and the homologous chromosomes will be passed to gametes with these genes in their original combination…these gametes are the ‘parental types’ and they should be the most common types of gametes produced. b b A a a a B b a b A B

10. C. Incomplete Linkage - But during Prophase I, homologous chromosomes can exchange pieces of DNA. - This “Crossing over” creates new combinations of genes… These are the ‘recombinant types’ b b A a a a b B A a B b a b A B

11. C. Incomplete Linkage As the other parent only contributed recessive alleles, the phenotype of the offspring is determined by the gamete received from the heterozygote… b b A a LOTS of these a a b B a A b B a b FEW of these A B

12. V. Linkage • Overview • Complete Linkage • Incomplete Linkage  • 1. Determining if the genes are linked, • or are assorting independently:

13. V. Linkage • Overview • Complete Linkage • Incomplete Linkage  • 1. Determining if the genes are linked, • or are assorting independently: • - test cross AaBb x aabb

14. V. Linkage • Overview • Complete Linkage • Incomplete Linkage  • 1. Determining if the genes are linked, • or are assorting independently: • - test cross • - determine expectations under the • hypothesis of independent assortment AaBb x aabb The frequency of ‘AB’ should = f(A) x f(B) x N = 55/100 x 51/100 x 100 = 28 The frequency of ‘Ab’ should = f(A) x f(B) x N = 55/100 x 49/100 x 100 = 27 The frequency of ‘aB’ should = f(a) x f(B) x N = 45/100 x 51/100 x 100 = 23 The frequency of ‘ab’ should = f(a) x f(b) x N = 45/100 x 49/100 x 100 = 22

15. V. Linkage • Overview • Complete Linkage • Incomplete Linkage  • 1. Determining if the genes are linked, • or are assorting independently: • - test cross • - determine expectations under the • hypothesis of independent assortment AaBb x aabb Easy with a 2 x 2 contingency table

16. V. Linkage • Overview • Complete Linkage • Incomplete Linkage  • 1. Determining if the genes are linked, • or are assorting independently: • - test cross • - determine expectations under the • hypothesis of independent assortment AaBb x aabb Easy with a 2 x 2 contingency table Compute Row, Columns, and Grand Totals

17. V. Linkage • Overview • Complete Linkage • Incomplete Linkage  • 1. Determining if the genes are linked, • or are assorting independently: • - test cross • - determine expectations under the • hypothesis of independent assortment AaBb x aabb Easy with a 2 x 2 contingency table Compute Row, Column, and Grand Totals E = (RT x CT)/GT

18. V. Linkage • Overview • Complete Linkage • Incomplete Linkage  • 1. Determining if the genes are linked, • or are assorting independently: • - test cross • - determine expectations under the • hypothesis of independent assortment AaBb x aabb Easy with a 2 x 2 contingency table Compute Row, Column, and Grand Totals E = (RT x CT)/GT

19. V. Linkage • Overview • Complete Linkage • Incomplete Linkage  • 1. Determining if the genes are • linked, or are assorting independently: • - Chi-Square Test of Independence

20. V. Linkage • Overview • Complete Linkage • Incomplete Linkage  • 1. Determining if the genes are • linked, or are assorting independently: • 2. Detemining the arrangement of • alleles in the F1 individual; which alleles are • paired on each homolog? AaBb x aabb

21. V. Linkage • Overview • Complete Linkage • Incomplete Linkage  • 1. Determining if the genes are • linked, or are assorting independently: • 2. Detemining the arrangement of • alleles in the F1 individual; which alleles are • paired on each homolog? • - most abundant types are ‘parental types’ AaBb x aabb A a B b

22. V. Linkage • Overview • Complete Linkage • Incomplete Linkage  • 1. Determining if the genes are • linked, or are assorting independently: • 2. Detemining the arrangement of • alleles in the F1 individual; which alleles are • paired on each homolog? • - most abundant types are ‘parental types’ • - least abundant are products of crossing-over: • ‘recombinant types’ A a a B b B A b

23. V. Linkage • Overview • Complete Linkage • Incomplete Linkage  • 1. Determining if the genes are • linked, or are assorting independently: • 2. Detemining the arrangement of • alleles in the F1 individual; which alleles are • paired on each homolog? • 3. Determining the distance between loci: • Add the recombinant types and divide by total • offspring; this is the percentage of recombinant • types. Multiply by 100 (to clear the decimal) and • this is the index of distance, in ‘map units’ or • centiMorgans. • 20/100 = 0.20 x100 = 20.0 centiMorgans AaBb x aabb A a B b 20 map units

24. V. Linkage • Overview • Complete Linkage • Incomplete Linkage  • Summary • - by studying the combined patterns of heredity among linked genes, linkage maps can be created that show the relative positions of genes on chromosomes.