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GENETICS ESSENTIALS Concepts and Connections SECOND EDITION GENETICS ESSENTIALS Concepts and Connections SECOND EDITION Benjamin A. Pierce © 2013 W. H.

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Presentation on theme: "GENETICS ESSENTIALS Concepts and Connections SECOND EDITION GENETICS ESSENTIALS Concepts and Connections SECOND EDITION Benjamin A. Pierce © 2013 W. H."— Presentation transcript:

1 GENETICS ESSENTIALS Concepts and Connections SECOND EDITION GENETICS ESSENTIALS Concepts and Connections SECOND EDITION Benjamin A. Pierce © 2013 W. H. Freeman and Company CHAPTER 5 Linkage, Recombination, and Eukaryotic Gene Mapping

2 CHAPTER 5 OUTLINE 5.1 Linked Genes Do Not Assort Independently, 114 5.2 Linked Genes Segregate Together and Crossing Over Produces Recombination Between Them, 116 5.3 A Three-Point Testcross Can Be Used To Map Three Linked Genes, 128 5.4 Genes Can Be Located with Genomewide Association Studies, 137

3 THE CASE OF MALE BALDNESS X-linked or not? Study of linkage of a SNP and the association with male pattern baldness shows that they do not assort independently, they are transmitted together

4 5.1 LINKED GENES DO NOT ASSORT INDEPENDENTLY Linked genes: genes located close together on the same chromosome -Belong to the same linkage group

5 RECOMBINATION AND ALLELES

6 FLOWER COLOR AND POLLEN SHAPE

7 F1 GENERATION GIVES PURPLE COLOR AND LONG POLLEN Appears that the purple is dominant to red and long is dominant to short pollen What with the F2, it is expected to get 9:3:3:1 BUT…

8 THERE IS NO 9:3:3:1 RATIO; WHAT HAPPENED

9 5.2 Linked Genes Segregate Together and Crossing Over Produces Recombination Between Them Fig. 5.3 Notation for crosses with linkage Complete linkage leads to nonrecombinant Gametes and nonrecombinant progeny Crossing over with linked genes leads to recombinant gametes and recombinant progeny Figs. 5.4 & 5.5

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11 LABELING ALLELES OF TWO GENES ON A SAME CHROMOSOME Cross AA BB x aa bb can be represented by A B a b A B x a b The F1 will be: A B a b Also it can be represented A B a b

12 THE TWO TEST CROSS CASES LINKED V.S. NOT LINKED Linked genes give only nonrecombinant gametes Unlinked genes give both nonrecombinant and recombinant

13 CONCEPT CHECK 1 For single crossovers, the frequency of recombinant gametes is half the frequency of crossing over because: a. a test cross between a homozygote and heterozygote produces ½ heterozygous and ½ homozygous progeny. b. the frequency of recombination is always 50%. c. each crossover takes place between only two of the four chromatids of a homologous pair. d. crossovers take place in about 50% of meiosis.

14 WHAT HAPPENS IN RECOMBINATION AND THE GAMETE FORMATION

15 5.2 LINKED GENES SEGREGATE TOGETHER AND CROSSING OVER PRODUCES RECOMBINATION BETWEEN THEM Calculating Recombination Frequency Recombination frequency = (No. recombinant progeny/Total No. of progeny)  100% Fig. 5.6 Coupling and Repulsion Configuration of Linked Genes Coupling (cis configuration): wild-type alleles are found on one chromosome; mutant alleles are found on the other chromosome.

16 CALCULATING FREQUENCIES

17 Rec. Frequency = x100% Rec. Frequency = x100% 8 + 7 55+53+8+7 # of recombinant progeny Total # of progeny Rec. Frequency = x100% 15 123 Rec. Frequency = 12.2 % OR 0.122

18 5.2 LINKED GENES SEGREGATE TOGETHER AND CROSSING OVER PRODUCES RECOMBINATION BETWEEN THEM Coupling and Repulsion Configuration of Linked Genes Repulsion (trans configuration): wild-type allele and mutant allele are found on the same chromosome. Fig. 5.7 Testing for Independent Assortment Figs. 5.8 & 5.9

19 COUPLING AND REPULSION p+ b+ p b p+ b p b+

20 COUPLING AND REPULSION

21 Gene Mapping with Recombination Frequencies: Genetic maps are determined by recombinant frequency. Map unit and centiMorgans Constructing a Genetic Map with the Use of Two-Point Testcrosses Fig. 5.10 5.2 Linked Genes Segregate Together and Crossing Over Produces Recombination Between Them

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23 RECOMBINATION FREQUENCIES AND DISTANCES OF GENES Based on recombination frequencies genetic maps are generated Actual distances are on physical maps Recombination maps are approximations and are measured in map units or cM (centiMorgans) One map unit equals to 1% recombination frequency

24 Genetic distances measured with recombination rates are approximately additive: if the distance from gene A to gene B is 5 m.u., the distance from gene B to gene C is 10 m.u., and the distance from gene A to gene C is 15 m.u.

25 LIMITATIONS OF GENE MAPPING If the genes are far apart and the frequency is 50% we cannot say if the genes are on different chromosomes or just to far on same chromosome Two genes that are relatively far will give underestimate of recombination events due to possibility of multiple crosses between the genes.

26 EXAMPLE: TWO POINT CROSS TO DETERMINE MAP OF 4 GENES

27 5.3 A THREE-POINT TESTCROSS CAN BE USED TO MAP THREE LINKED GENES Constructing a Genetic Map with the Three-Point Testcross Fig. 5.12 Determining the gene order Determining the location of crossovers Fig. 5.13

28 DOUBLE CROSSOVER :THREE GENES

29 A THREE-POINT TESTCROSS CAN BE USED TO MAP THREE LINKED GENES

30 Calculating the recombination frequencies Interference and coefficient of coincidence Effect of multiple crossovers 5.3 A Three-Point Testcross Can Be Used to Map Three Linked Genes

31 CALCULATING THE RECOMBINATION FREQUENCIES Based on the numbers in the fruit fly testcross for three loci calculate the distances between the loci. Recombinant progeny with a chromosome that underwent crossing over between the eye-color locus (st) and the bristle locus (ss) include the single crossovers ( st + / ss e and st / ss + e + ) and the two double crossovers ( st + / ss / e + and st / ss + / e ); Total of 755 progeny; so the recombination frequency between ss and st is:

32 CALCULATING THE RECOMBINATION FREQUENCIES Based on the numbers in the fruit fly testcross for three loci calculate the distances between the loci. The map distance between the bristle locus (ss) and the body locus (e) is determined in the same manner. The recombinant progeny that possess a crossover between ss and e are the single crossovers st + ss + / e and st ss / e + and the double crossovers st + / ss / e + and st / ss + / e. The recombination frequency is:

33 CALCULATING THE RECOMBINATION FREQUENCIES Based on the numbers in the fruit fly testcross for three loci calculate the distances between the loci. Finally, calculate the map distance between the outer two loci, st and e. This map distance can be obtained by summing the map distances between st and ss and between ss and e (14.6 m.u. + 12.2 m.u. = 26.8 m.u.). We can now use the map distances to draw a map of the three genes on the chromosome:

34 INTERFERENCE AND COEFFICIENT OF COINCIDENCE Calculate the proportion of double-recombinant gametes by using the multiplication rule of probability Applying this rule, the proportion (probability) of gametes with double crossovers between st and e is equal to the probability of recombination between st and ss multiplied by the probability of recombination between ss and e, or 0.146 × 0.122 = 0.0178. Multiplying this probability by the total number of progeny gives us the expected number of double-crossover progeny from the cross: 0.0178 × 755 = 13.4. Only 8 double crossovers—considerably fewer than the 13 expected—were observed in the progeny of the cross

35 INTERFERENCE AND COEFFICIENT OF COINCIDENCE Crossovers are frequently not independent events: the occurrence of one crossover tends to inhibit additional crossovers in the same region of the chromosome, and so double crossovers are less frequent than expected. The degree to which one crossover interferes with additional crossovers in the same region is termed the interference. To calculate the interference, we first determine the coefficient of coincidence, which is the ratio of observed double crossovers to expected double crossovers:

36 IN OUR CASE indicates that we are actually observing only 60% of the double crossovers that we expected on the basis of the single-crossover frequencies. Calculation of interference Interference = 1 − coefficient of coincidence So the interference for our three-point cross is: interference = 1 − 0.6 = 0.4 40% of recombinations are not observed due to intefrerence

37 5.4 GENES CAN BE LOCATED WITH GENOMEWIDE ASSOCIATION STUDIES Linkage analysis (genetic markers; anonymous markers) Genome wide association studies Haplotype Linkage disequilibrium

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