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Eukaryotic Chromosome Mapping Using Genetic Recombination to Estimate Distances Between Genes.

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Presentation on theme: "Eukaryotic Chromosome Mapping Using Genetic Recombination to Estimate Distances Between Genes."— Presentation transcript:

1 Eukaryotic Chromosome Mapping Using Genetic Recombination to Estimate Distances Between Genes

2 Linked Genes Mendel’s experiments Linked Genes Gene location Genes on separate chromosomes Genes on the same chromosome Gamete types Equal numbers of all possible allele combinations More parental combinations than recombinant combinations

3 Independent Assortment vs. Gene Linkage Example from Drosophila Example from Drosophila Red eyes, x Pink eyes Red eyes, x Pink eyes Beige body Ebony body Beige body Ebony body RRBBrrbb RRBBrrbb F1: Red eyes, Beige body F1: Red eyes, Beige bodyRrBb

4 Independent Assortment vs. Gene Linkage Testcross: cross to individual of known genotype F1:Red eyes X Pink eyes F1:Red eyes X Pink eyes Beige body Ebony body Beige body Ebony body RrBbrrbb RrBbrrbb

5 Independent Assortment vs. Gene Linkage F2 phenotype Number of Offspring Expected for Unlinked Genes Red eyes Beige body Pink eyes Ebony body Red eyes Ebony body Pink eyes Beige body

6 Independent Assortment vs. Gene Linkage F1:Red eyes X Pink eyes F1:Red eyes X Pink eyes Beige body Ebony body Beige body Ebony body RrBb rrbb RrBb rrbb RrBbRedBeigeRrbbRedEbonyrrBbPinkBeigerrbbPinkEbony rb RBRb rB rb

7 Independent Assortment vs. Gene Linkage Coupling or Cis Configuration If genes are linked: Red eyes, x Pink eyes Beige body Ebony body Red eyes, x Pink eyes Beige body Ebony body R B r b F1: Red eyes, Beige body R B r b

8 F1: Red eyes, Beige body F1: Red eyes, Beige body Independent Assortment vs. Gene Linkage R B r b R B r b r B X R b Four types of gametes are produced Parental Recombinant

9 Independent Assortment vs. Gene Linkage F1:Red eyes X Pink eyes F1:Red eyes X Pink eyes Beige body Ebony body Beige body Ebony body R B r b R B r b r b r b r b r b R B R B r b r b R b R b r B r B r b R B r b R b r B r b

10 Independent Assortment vs. Gene Linkage F2 phenotype Number of Offspring Chromosome arrangement Red eyes Beige body 398RB//rbParental Pink eyes Ebony body 382rb//rbParental Red eyes Ebony body 108Rb//rbRecombinant Pink eyes Beige body 112rB//rbRecombinant

11 Genetic Map Units 1% recombination = 1 map unit = 1 centimorgan x 100 = 22% x 100 = 22% These genes are located 22 map units apart on the same chromosome. These genes are located 22 map units apart on the same chromosome.

12 Limits of Genetic Mapping Genes that are 50 map units apart will appear to assort independently. 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. The calculated distance between any TWO genes on the same chromosome should be less than 50 map units.

13 Predicting Gamete Frequencies for Linked Genes Red eyes, x Pink eyes Ebony body Beige body Red eyes, x Pink eyes Ebony body Beige body R b r B F1: Red eyes, Beige body R b r B Repulsion or Trans Configuration

14 Predicting Gamete Frequencies for Linked Genes The genes are 22 map units apart, therefore we expect 22% recombinant gametes and 78% parental gametes. The genes are 22 map units apart, therefore we expect 22% recombinant gametes and 78% parental gametes. R b recombinants 0.78 parentals 0.11 R B 0.11 r b 0.39 r B F1: Red eyes Beige body F1: Red eyes Beige body R b r B r B

15 Using 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).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.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.Eight classes of offspring are analyzed to determine recombination frequencies.

16 Problem 1, Page 2-1 In 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. In 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.

17 Problem 1, Page 2-1 Offspring Resulting from Three-Point Testcross red, tall, ragged green, dwarf, normal red, tall, normal green, dwarf, ragged red, dwarf, normal green, tall, ragged red, dwarf, ragged green, tall, normal

18 Problem 1, Page 2-1 ARed aGreen DTall dDwarf Rg Ragged leaves rg Normal leaves

19 Problem 1, Page 2-1 With Arbitrary Gene Order A D Rg a d rg X A D Rg TestcrossX F2 a d rg A D Rg F1

20 Problem 1, Page 2-1 With Arbitrary Gene Order a d rg A D Rg X F2 Parentals: A D Rg a d rg a d rg Recombinants: A d rg a D Rg a D Rg A D rg A D rg a d Rg a d Rg A d Rg A d Rg a D rg a D rg a d rg Used as a genetic background to see the contribution from the tri-hybrid.

21 Problem 1, Page 2-1 Determine which classes are parentalsDetermine 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, ragged green, dwarf, normal green, dwarf, normal

22 Problem 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 gene These will be the two smallest classes. Double Recombinants: red, tall, normal green, dwarf, ragged green, dwarf, ragged

23 Problem 1, Page 2-1 Determine the gene orderDetermine the gene order The middle gene is the one that changes places in the double recombinants when compared to the parental combinations. A Rg D a rg d XX A rg D a Rg d Red, tall, normal Green, dwarf, ragged

24 Problem 1, Page 2-1 This shows why other gene orders are incorrect. A D Rg a d rg X A d Rg a D rg Green, tall, normal Red, dwarf, ragged X D A Rg d a rg X D a Rg d A rg Red, dwarf, normal Green, tall, ragged X

25 Problem 1, Page 2-1 AssignAssign genotypes to all classes Use correct gene order P P DC DC A  Rg Rg  D A Rg D a rg d A rg D a Rg d A rg d a Rg D A Rg d a rg D red, tall, ragged green, dwarf, normal red, tall, normal green, dwarf, ragged red, dwarf, normal green, tall, ragged red, dwarf, ragged green, tall, normal Contribution of F1 parent

26 Problem 1, Page 2-1 Recombination between A and Rg A Rg D a rg d X A rg d a Rg D Red, dwarf, normal Green, tall, ragged Single Crossovers Double Crossovers A Rg D a rg d XX A rg D a Rg d Red, tall, normal Green, dwarf, ragged Total = 200 Recombination = (200/1000) x 100 = 20 % Recombination = (200/1000) x 100 = 20 %

27 Problem 1, Page 2-1 Recombination between Rg and D A Rg D a rg d X A Rg d a rg D Red, dwarf, ragged Green, tall, normal Single Crossovers Double Crossovers A Rg D a rg d XX A rg D a Rg d Red, tall, normal Green, dwarf, ragged Total = 300 Recombination = (300/1000) x 100 = 30 % Recombination = (300/1000) x 100 = 30 %

28 Problem 1, Page 2-1 A Rg D A Rg D 20 map units 30 map units 20 map units 30 map units D Rg A D Rg A 30 map units 20 map units 30 map units 20 map units or Two maps are possible:

29 Interference Interference: crossover in one region inhibits crossover in an adjacent regionInterference: crossover in one region inhibits crossover in an adjacent region Interference = 1 – (coefficient of coincidence)Interference = 1 – (coefficient of coincidence) Coefficient of coincidence =Coefficient of coincidence = Observed double crossovers Observed double crossovers Expected double crossovers Expected double crossovers

30 Calculating Interference Coefficient of coincidence =Coefficient of coincidence = Observed double crossovers = Observed double crossovers = Expected double crossovers Expected double crossovers = 40 = = 40 = x 0.3 x x 0.3 x Interference =Interference = 1–(coefficient of coincidence) = 1–(coefficient of coincidence) = = = 0.333


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