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 Aa Aa bBBb ABabAbaB Independent Assortment

V. Linkage A. Overview Aa Aa bBBb ABabAbaB Independent Assortment Aa Bb Linkage ABab

V. Linkage A. Overview Linkage Aa Bb ABab Aa Bb ABab In Prophase I of Meiosis – Crossing-over A a b B AbaB

X AABBaabb AB ab V. Linkage A.Overview B.Complete Linkage Test Cross

- 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. X AABBaabb AB ab ABab Gametes AB ab F1 V. Linkage A.Overview B.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. X ABab Gametes AB ab F1 x F1 ab V. Linkage A.Overview B.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 AaBb aabb 1:1 ratio A:a 1:1 ratio B:b 1:1 ratio AB:ab NOT 1:1:1:1 V. Linkage A.Overview B.Complete Linkage Phenotypes AB ab aB ? Ab ?

C. Incomplete Linkage B ab A b ab a

- 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 ab b ab a AB ab A

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 ab b ab a AB ab A aB Ab

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 ab b ab a AB ab A aB Ab gamete genotype phenotype abaabbab AaBbAB abaaBbaB abAabbAb LOTS of these FEW of these

V. Linkage A.Overview B.Complete Linkage C.Incomplete Linkage 1. Determining if the genes are linked, or are assorting independently:

Offspring Number AB43 Ab12 aB 8 ab37 AaBb x aabb V. Linkage A.Overview B.Complete Linkage C.Incomplete Linkage 1. Determining if the genes are linked, or are assorting independently: - test cross

Offspring Number AB43 Ab12 aB 8 ab37 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 V. Linkage A.Overview B.Complete Linkage C.Incomplete Linkage 1. Determining if the genes are linked, or are assorting independently: - test cross - determine expectations under the hypothesis of independent assortment

Offspring Number AB43 Ab12 aB 8 ab37 AaBb x aabb BbRow Total A4312 a837 Col. Total V. Linkage A.Overview B.Complete Linkage C.Incomplete Linkage 1. Determining if the genes are linked, or are assorting independently: - test cross - determine expectations under the hypothesis of independent assortment Easy with a 2 x 2 contingency table

Offspring Number AB43 Ab12 aB 8 ab37 AaBb x aabb BbRow Total A a83745 Col. Total V. Linkage A.Overview B.Complete Linkage C.Incomplete Linkage 1. Determining if the genes are linked, or are assorting independently: - test cross - determine expectations under the hypothesis of independent assortment Easy with a 2 x 2 contingency table Compute Row, Columns, and Grand Totals

Offspring Number AB43 Ab12 aB 8 ab37 AaBb x aabb BExp.bRow Total A a83745 Col. Total V. Linkage A.Overview B.Complete Linkage C.Incomplete Linkage 1. Determining if the genes are linked, or are assorting independently: - test cross - determine expectations under the hypothesis of independent assortment Easy with a 2 x 2 contingency table Compute Row, Column, and Grand Totals E = (RT x CT)/GT

BExp.b Row Total A a Col. Total Offspring Number AB43 Ab12 aB 8 ab37 AaBb x aabb V. Linkage A.Overview B.Complete Linkage C.Incomplete Linkage 1. Determining if the genes are linked, or are assorting independently: - test cross - determine expectations under the hypothesis of independent assortment Easy with a 2 x 2 contingency table Compute Row, Column, and Grand Totals E = (RT x CT)/GT

Phenotype ObsExp(o-e)(o-e) 2 /e AB Ab aB ab X 2 =36.38 BExp.b Row Total A a Col. Total V. Linkage A.Overview B.Complete Linkage C.Incomplete Linkage 1. Determining if the genes are linked, or are assorting independently: - Chi-Square Test of Independence

Offspring Number AB43 Ab12 aB 8 ab37 AaBb x aabb V. Linkage A.Overview B.Complete Linkage C.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?

Offspring Number AB43 Ab12 aB 8 ab37 AaBb x aabb AB ab V. Linkage A.Overview B.Complete Linkage C.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’

AB ab aB Ab V. Linkage A.Overview B.Complete Linkage C.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’

Offspring Number AB43 Ab12 aB 8 ab37 AaBb x aabb AB ab 20 map units V. Linkage A.Overview B.Complete Linkage C.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

V. Linkage A.Overview B.Complete Linkage C.Incomplete Linkage D.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.