Presentation on theme: "Eukaryotic Chromosome Mapping"— Presentation transcript:
1Eukaryotic Chromosome Mapping Using Genetic Recombination to Estimate Distances Between Genes
2Linked Genes Mendel’s experiments Linked Genes Gene location Genes on separate chromosomesGenes on the same chromosomeGamete typesEqual numbers of all possible allele combinationsMore parental combinations than recombinant combinations
3Independent Assortment vs. Gene Linkage Example from DrosophilaRed eyes, x Pink eyesBeige body Ebony bodyRRBB rrbbF1: Red eyes, Beige bodyRrBb
4Independent Assortment vs. Gene Linkage Testcross: cross to individual of known genotypeF1:Red eyes X Pink eyesBeige body Ebony bodyRrBb rrbb
5Independent Assortment vs. Gene Linkage F2 phenotypeNumber of OffspringExpected for Unlinked GenesRed eyesBeige body398250Pink eyesEbony body382108112
6Independent Assortment vs. Gene Linkage F1:Red eyes X Pink eyesBeige body Ebony bodyRrBb rrbbRBRbrBrbRrBbRedBeigeRrbbEbonyrrBbPinkrrbbrb
7Independent Assortment vs. Gene Linkage If genes are linked:Red eyes, x Pink eyes Beige body Ebony bodyF1: Red eyes, Beige bodyR Br bR Br bCouplingor CisConfiguration
8Independent Assortment vs. Gene Linkage F1: Red eyes, Beige bodyR Br bXR bFour types of gametes are produced Parental RecombinantR Br br B
9Independent Assortment vs. Gene Linkage F1:Red eyes X Pink eyesBeige body Ebony bodyR B r br b r bR Br bR br BR Br bR br Br br br br br b
10Independent Assortment vs. Gene Linkage F2 phenotypeNumber of OffspringChromosome arrangementRed eyesBeige body398RB//rbParentalPink eyesEbony body382rb//rb108Rb//rbRecombinant112rB//rb
11Genetic Map Units 1% recombination = 1 map unit = 1 centimorgan x 100 = 22%1000These genes are located 22 map units apart on the same chromosome.
12Limits of Genetic Mapping Genes that are 50 map units apart will appear to assort independently.The calculated distance between any TWO genes on the same chromosome should be less than 50 map units.
13Predicting Gamete Frequencies for Linked Genes Red eyes, x Pink eyes Ebony body Beige bodyF1: Red eyes, Beige bodyR br BR br BRepulsionor TransConfiguration
14Predicting Gamete Frequencies for Linked Genes F1: Red eyes Beige bodyR br BThe genes are 22 map units apart, therefore we expect 22% recombinant gametes and 78% parental gametes.R BR b0.110.39r b0.11r B0.390.22 recombinants0.78 parentals
15Using a Three-point Testcross to Determine Genetic Distance A cross between two parental strains is used to produce a tri-hybrid (heterozygous for three genes).The tri-hybrid is crossed to an organism that is homozygous recessive for all three genes.Eight classes of offspring are analyzed to determine recombination frequencies.
16Problem 1, Page 2-1In corn, a strain homozygous for the recessive alleles a (green), d (dwarf) and rg (normal leaves) was crossed to a strain homozygous for the dominant alleles of each of these genes, namely A (red), D (tall) and Rg (ragged leaves). Offspring of this cross were then crossed to plants that were green, dwarf and had normal leaves. The following phenotypic classes were observed.
18Problem 1, Page 2-1 A Red a Green D Tall d Dwarf Rg Ragged leaves rg Normal leaves
19Problem 1, Page 2-1 With Arbitrary Gene Order A D Rga d rgXa d rgA D RgF1a d rgA D RgTestcrossXF2
20Problem 1, Page 2-1 With Arbitrary Gene Order a d rgA D RgXF2Parentals: A D Rga d rgRecombinants: A d rga D RgA D rga d RgA d Rga D rga d rgUsed as a geneticbackground to seethe contribution fromthe tri-hybrid.
21Problem 1, Page 2-1 Determine which classes are parentals The two parental classes will represent the largest number of offspring in the F2 generation.Information on the parents may be given in the problem description itself.Parentals: red, tall, raggedgreen, dwarf, normal
22Problem 1, Page 2-1 Determine which classes are double recombinants Double recombinants have two crossovers: one between the first and middle gene and one between the middle and third geneThese will be the two smallest classes.Double Recombinants: red, tall, normalgreen, dwarf, ragged
23Problem 1, Page 2-1 A rg D A Rg D a rg d a Rg d X Determine the gene orderThe middle gene is the one that changes places in the double recombinants when compared to the parental combinations.A rg Da Rg dRed, tall, normalGreen, dwarf, raggedA Rg Da rg dX
24Problem 1, Page 2-1 This shows why other gene orders are incorrect. A d RgA D RgRed, dwarf, raggedXXa d rga D rgGreen, tall, normalD a RgD A RgGreen, tall, raggedXXd a rgd A rgRed, dwarf, normal
25Contribution of F1 parent Problem 1, Page 2-1Assign genotypes to all classesUse correct gene orderContribution of F1 parentred, tall, raggedgreen, dwarf, normalred, tall, normalgreen, dwarf, raggedred, dwarf, normalgreen, tall, raggedred, dwarf, raggedgreen, tall, normalPA Rg D26527524169070120140Pa rg dDCA rg DDCa Rg dA RgA rg dA Rga Rg DRg DA Rg dRg Da rg D
26A rg d A Rg D a rg d a Rg D A rg D A Rg D a rg d a Rg d Problem 1, Page 2-1Recombination between A and RgSingle CrossoversA rg da Rg DRed, dwarf, normalGreen, tall, raggedA Rg Da rg dX9070Double CrossoversA rg Da Rg dRed, tall, normalGreen, dwarf, raggedA Rg Da rg dX2416Total = 200Recombination = (200/1000) x 100 = 20 %
27A Rg d A Rg D a rg d a rg D A rg D A Rg D a rg d a Rg d Problem 1, Page 2-1Recombination between Rg and DSingle CrossoversA Rg da rg DRed, dwarf, raggedGreen, tall, normalA Rg Da rg dX120140Double CrossoversA rg Da Rg dRed, tall, normalGreen, dwarf, raggedA Rg Da rg dX2416Total = 300Recombination = (300/1000) x 100 = 30 %
28Problem 1, Page 2-1 Two maps are possible: A Rg D 20 map units map unitsorD Rg A30 map units map units
29InterferenceInterference: crossover in one region inhibits crossover in an adjacent regionInterference = 1 – (coefficient of coincidence)Coefficient of coincidence =Observed double crossoversExpected double crossovers
30Calculating Interference Coefficient of coincidence =Observed double crossovers =Expected double crossovers= 40 = 0.6670.2 x 0.3 xInterference =1–(coefficient of coincidence) == 0.333