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LINKAGE AND GENETIC MAPPING IN EUKARYOTES. In eukaryotic species, each linear chromosome contains a long piece of DNA A typical chromosome contains many.

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Presentation on theme: "LINKAGE AND GENETIC MAPPING IN EUKARYOTES. In eukaryotic species, each linear chromosome contains a long piece of DNA A typical chromosome contains many."— Presentation transcript:

1 LINKAGE AND GENETIC MAPPING IN EUKARYOTES

2 In eukaryotic species, each linear chromosome contains a long piece of DNA A typical chromosome contains many hundred or even a few thousand different genes The term linkage has two related meanings 1. Two or more genes can be located on the same chromosome 2. Genes that are close together tend to be transmitted as a unit 5-3 LINKAGE AND CROSSING OVER

3 Chromosomes are called linkage groups They contain a group of genes that are linked together The number of linkage groups is the number of types of chromosomes of the species For example, in humans 22 autosomal linkage groups An X chromosome linkage group A Y chromosome linkage group Genes that are far apart on the same chromosome may independently assort from each other This is due to crossing-over 5-4

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5 Crossing Over May Produce Recombinant Phenotypes In diploid eukaryotic species, linkage can be altered during meiosis as a result of crossing over Crossing over Occurs during prophase I of meiosis at the bivalent stage Non-sister chromatids of homologous chromosomes exchange DNA segments 5-5 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display

6 Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Diploid cell after chromosome replication Meiosis Possible haploid cells (a) Without crossing over, linked alleles segregate together. b a b a b a B A b a B A Diploid cell after chromosome replication Meiosis Possible haploid cells (b) Crossing over can reassort linked alleles. b A B A b a B a B A B A B A b a

7 5-7 Figure 5.1 These haploid cells contain a combination of alleles NOT found in the original chromosomes These are termed parental or non- recombinant cells This new combination of alleles is a result of genetic recombination These are termed nonparental or recombinant cells

8 Bateson and Punnett Discovered Two Traits That Did Not Assort Independently In 1905, William Bateson and Reginald Punnett conducted a cross in sweet pea involving two different traits Flower color and pollen shape This is a dihybrid cross that is expected to yield a 9:3:3:1 phenotypic ratio in the F 2 generation However, Bateson and Punnett obtained surprising results 5-8 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display

9 Figure 5.2 5-9 A much greater proportion of the two types found in the parental generation

10 Morgan Provided Evidence for the Linkage of Several X-linked Genes The first direct evidence of linkage came from studies of Thomas Hunt Morgan Morgan investigated several traits that followed an X-linked pattern of inheritance Figure 5.3 illustrates an experiment involving three traits Body color Eye color Wing length 5-11 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display

11 yy ww mm y + y w + w m + m F 1 generation x y w m Y x y + w + m + Y F 1 generation contains wild-type females and yellow-bodied, white-eyed, miniature-winged males.

12 5-13 Morgan observed a much higher proportion of the combinations of traits found in the parental generation P Males P Females Morgan’s explanation: All three genes are located on the X chromosome Therefore, they tend to be transmitted together as a unit

13 1. Why did the F 2 generation have a significant number of nonparental combinations? 2. Why was there a quantitative difference between the various nonparental combinations? 5-14 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display Morgan Provided Evidence for the Linkage of Several X-linked Genes

14 5-15 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display Gray body, red eyes1,159 Yellow body, white eyes1,017 Gray body, white eyes 17 Yellow body, red eyes 12 Total2,205 Let’s reorganize Morgan’s data by considering the pairs of genes separately Red eyes, normal wings 770 White eyes, miniature wings 716 Red eyes, miniature wings 401 White eyes, normal wings 318 Total2,205 It was fairly common to get this nonparental combination But this nonparental combination was rare

15 Morgan made three important hypotheses to explain his results 1. The genes for body color, eye color and wing length are all located on the X- chromosome They tend to be inherited together 2. Due to crossing over, the homologous X chromosomes (in the female) can exchange pieces of chromosomes This created new combination of alleles 3. The likelihood of crossing over depends on the distance between the two genes Crossing over is more likely to occur between two genes that are far apart from each other 5-17

16 Figure 5-5 Copyright © 2006 Pearson Prentice Hall, Inc.

17 5-18 Figure 5.4 These parental phenotypes are the most common offspring because the genes are far apart These recombinant offspring are not uncommon

18 5-19 Figure 5.4 because the genes are very close together These recombinant offspring are fairly uncommon These recombinant offspring are very unlikely 1 out of 2,205

19 This method is frequently used to determine if the outcome of a dihybrid cross is consistent with linkage or independent assortment 5-20 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display Chi Square Analysis

20 Parental chromosomes Nonparental chromosomes CWx cwx cWx Cwx Crossing over (b) Crossing over between normal and abnormal chromosome 9 Normal chromosome 9 Abnormal chromosome 9 Knob (a) Normal and abnormal chromosome 9 Interchanged piece from chromosome 8 Creighton and McClintock Experiment They demonstrated physical evidence of cross-overs. C = Colored c = colorless Wx = Starchy endosperm wx = waxy endosperm

21 5-30 Figure 5.6

22 Interpreting the Data 5-32 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display Parent AParent B C wx (nonrecombinant) c Wx (nonrecombinant) C Wx (recombinant) c wx (recombinant) c Wx c wx By combining these gametes into a Punnett square, the following types of offspring can be produced

23 5-34 The colored, waxy phenotype (Cc wxwx) can occur only if Recombination did not occur in parent A AND Parent A passed the knobbed, translocated chromosome to its offspring This was the case, as shown in the data table below Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display

24 5-35 The colorless, waxy phenotype (cc wxwx) can occur only if Recombination did occur in parent A AND Parent A passed a chromosome 9 that had a translocation but was knobless This was the case, as shown in the data table below Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display

25 The Data 5-31 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display

26 5-36 These observations were consistent with the idea that a cross over occurred between the C and wx genes As stated by Creighton and McClintock: “Pairing chromosomes, heteromorphic in two regions, have been shown to exchange parts at the same time they exchange genes assigned to these regions.”

27 Genetic mapping is also known as gene mapping or chromosome mapping Its purpose is to determine the linear order of linked genes along the same chromosome Figure 5.8 illustrates a simplified genetic linkage map of Drosophila melanogaster 5-42 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display GENETIC MAPPING IN PLANTS AND ANIMALS

28 5-43 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display Figure 5.8 Each gene has its own unique locus at a particular site within a chromosome

29 Physical Maps Use nucleotide sequences to map genes

30 5-45 Experimentally, the percentage of recombinant offspring is correlated with the distance between the two genes If the genes are far apart  many recombinant offspring If the genes are close  very few recombinant offspring Map distance = Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display Number of recombinant offspring Total number of offspring X 100 The units of distance are called map units (mu) They are also referred to as centiMorgans (cM) One map unit is equivalent to 1% recombination frequency

31 5-47 Figure 5.9 Chromosomes are the product of a crossover during meiosis in the heterozygous parent Recombinant offspring are fewer in number than nonrecombinant offspring

32 5-48 The data at the bottom of Figure 5.9 can be used to estimate the distance between the two genes Map distance = Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display Number of recombinant offspring Total number of offspring X 100 76 + 75 542 + 537 + 76 + 75 X 100= = 12.3 map units

33 The first genetic map was constructed in 1911 by Alfred Sturtevant He was an undergraduate who spent time in the laboratory of Thomas Hunt Morgan Sturtevant wrote: “In conversation with Morgan … I suddenly realized that the variations in the length of linkage, already attributed by Morgan to differences in the spatial orientation of the genes, offered the possibility of determining sequences [of different genes] in the linear dimension of the chromosome. I went home and spent most of the night (to the neglect of my undergraduate homework) in producing the first chromosome map, which included the sex-linked genes, y, w, v, m, and r, in the order and approximately the relative spacing that they still appear on the standard maps.” 5-49 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display Alfred Sturtevant’s Experiment

34 5-52 Figure 5.10

35 The Data 5-53 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display Alleles Concerned Number Recombinant/ Total Number Percent Recombinant Offspring y and w/w-e214/21,736 1.0 y and v1,464/4,55132.2 y and r115/32435.5 y and m260/69337.5 w/w-e and v471/1,58429.7 w/w-e and r2,062/6,11633.7 w/w-e and m406/89845.2 v and r17/573 3.0 v and m109/40526.9

36 Interpreting the Data 5-54 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display In some dihybrid crosses, the percentage of nonparental (recombinant) offspring was rather low For example, there’s only 1% recombinant offspring in the crosses involving the y and w or w-e alleles This suggests that these two genes are very close together Other dihybrid crosses showed a higher percentage of nonparental offspring For example, crosses between the v and m alleles produced 26.9% recombinant offspring This suggests that these two genes are farther apart

37 5-55 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display Sturtevant assumed that the map distances would be more accurate among genes that are closely linked. Therefore, his map is based on the following distances y – w (1.0), w – v (29.7), v – r (3.0) and v – m (26.9) Sturtevant also considered map distances amongst gene pairs to deduce the order of genes Percentage of crossovers between w and r was 33.7 Percentage of crossovers between w and v was 29.7 Percentage of crossovers between v and r was 3.0 Therefore, the gene order is w – v – r Where v is closer to r than it is to w

38 5-56 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display Sturtevant collectively considered all these data and proposed the following genetic map Sturtevant began at the y gene and mapped the genes from left to right

39 5-57 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display A close look at Sturtevant’s data reveals two points that do not agree very well with his genetic map The y and m dihybrid cross yielded 37.5% recombinants But the map distance is 57.6 The w and m dihybrid cross yielded 45.2% recombinants But the map distance is 56.6 So what’s up? As the percentage of recombinant offspring approaches a value of 50 % This value becomes a progressively more inaccurate measure of map distance Refer to Figure 5.11

40 5-58 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display When the distance between two genes is large The likelihood of multiple crossovers increases This causes the observed number of recombinant offspring to underestimate the distance between the two genes Figure 5.11

41 Figure 5-12a Copyright © 2006 Pearson Prentice Hall, Inc.

42 5-59 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display Trihybrid Crosses

43 Data from trihybrid crosses can also yield information about map distance and gene order The following experiment outlines a common strategy for using trihybrid crosses to map genes In this example, we will consider fruit flies that differ in body color, eye color and wing shape 5-59 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display Trihybrid Crosses b = black body color b + = gray body color pr = purple eye color pr + = red eye color vg = vestigial wings vg + = normal wings

44 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display 5-60 Step 1: Cross two true-breeding strains that differ with regard to three alleles. Female is mutant for all three traits Male is homozygous wildtype for all three traits The goal in this step is to obtain aF1 individuals that are heterozygous for all three genes Order of genes not important here.

45 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display 5-61 Step 2: Perform a testcross by mating F 1 female heterozygotes to male flies that are homozygous recessive for all three alleles During gametogenesis in the heterozygous female F 1 flies, crossovers may produce new combinations of the 3 alleles

46 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display 5-62 Step 3: Collect data for the F 2 generation Phenotype Number of Observed Offspring Gray body, red eyes, normal wings + + + 411 parental Gray body, red eyes, vestigial wings + + vg 61 pr/vg Gray body, purple eyes, normal wings + pr + 2 b/pr and pr/vg Gray body, purple eyes, vestigial wings + pr vg 30 b/pr Black body, red eyes, normal wings b + + 28 b/pr Black body, red eyes, vestigial wings B + vg 1 b/pr and pr/vg Black body, purple eyes, normal wings B pr + 60 pr/vg Black body, purple eyes, vestigial wings B pr vg 412 parental

47 5-63 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display The three genes exist as two alleles each Therefore, there are 2 3 = 8 possible combinations of F 2 offspring If the genes assorted independently, all eight combinations would occur in equal proportions It is obvious that they are far from equal In the offspring of crosses involving linked genes, Parental phenotypes occur most frequently Double crossover phenotypes occur least frequently Single crossover phenotypes occur with “intermediate” frequency

48 5-64 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display The combination of traits in the double crossover tells us which gene is in the middle A double crossover separates the gene in the middle from the other two genes at either end In the double crossover categories, the recessive purple eye color is separated from the other two recessive alleles Thus, the gene for eye color lies between the genes for body color and wing shape

49 Which are the double cross-overs? The ones with the least amount.

50 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display 5-65 Step 4: Calculate the map distance between pairs of genes Number of recombs between pr and vg: 61 + 60+ 2 + 1 = 124 Number of recombs between b and pr: 30 + 28 + 2 + 1 = 61 Number of recombs between b and vg, all but double cross-overs: 61 + 60 + 30 + 28 = 178

51 Map Distance pr/vg = 124/1005 x 100 = 12.3 b/pr = 61/1005 x 100 = 6 b and vg = 179/1005 x 100 = 17.8 _____6____________12.3____________ b pr vg The distance between b and vg was found to be 17.8. The actual distance is 18.3 mu.

52 Interference The slightly smaller lower value was a small underestimate because we did not consider the double crossovers in the calculation between b and vg. The lower than expected value is due to a common genetic phenomenon, termed positive interference. The first crossover decreases the probability that a second crossover will occur nearby.

53 Much of our earliest understanding of genetic recombination came from the genetic analyses of fungi Fungi may be unicellular or multicellular organisms They are typically haploid (1n) They reproduce asexually and, in many cases, sexually The sac fungi (ascomycetes) have been particularly useful to geneticists because of their unique style of sexual reproduction 5-78 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display GENETIC MAPPING IN HAPLOID EUKARYOTES

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55 5-79 Meiosis produces four haploid cells, termed spores Figure 5.12 These are enclosed in a sac termed an ascus

56 The cells of a tetrad or octad are contained within a sac In other words, the products of a single meiotic division are contained within one sac 5-80 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display

57 The arrangement of spores within an ascus varies from species to species Unordered tetrads or octads Ascus provides enough space for the spores to randomly mix together Ordered tetrads or octads Ascus is very tight, thereby preventing spores from randomly moving around 5-81 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display Types of Tetrads or Octads

58 5-82 Figure 5.13 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display Yeast Unicellular alga Mold Ascus provides space for spores to randomly mix together Tight ascus prevents mixing of spores

59 Ordered tetrads or octads have the following key feature The position and order of spores within the ascus is determined by the divisions of meiosis and mitosis In crosses of tan and black Neurospora cultures, the spores appear tan or black in a certain order. All black spores or all tan spores indicate no hybridization. 5-83 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display Ordered Tetrad Analysis

60 5-84 Figure 5.13 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display Pairs of daughter cells are located next to each other All eight cells are arranged in a linear, ordered fashion

61 020

62 Non-crossovers

63 Cross-overs

64 Non cross-overs Cross-overs Non cross-overs

65 To calculate this distance, the experimenter must count the number of cross-over asci, as well as the total number of asci In cross-over asci, only half of the spores are actually the product of a crossover Therefore 5-89 (1/2) (Number of SDS asci) Total number of asci X 100 Map distance = Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display

66 Genetics of Corn http://media.hhmi.org/biointeractive/films/PoppedSecret.html


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