* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
Download 1 Chromosome Mapping in Eukaryotes
Polymorphism (biology) wikipedia , lookup
Genetic drift wikipedia , lookup
Heritability of IQ wikipedia , lookup
Neocentromere wikipedia , lookup
Skewed X-inactivation wikipedia , lookup
Gene desert wikipedia , lookup
Pathogenomics wikipedia , lookup
Genetic engineering wikipedia , lookup
Y chromosome wikipedia , lookup
Population genetics wikipedia , lookup
Public health genomics wikipedia , lookup
Nutriepigenomics wikipedia , lookup
Essential gene wikipedia , lookup
Cre-Lox recombination wikipedia , lookup
Polycomb Group Proteins and Cancer wikipedia , lookup
Dominance (genetics) wikipedia , lookup
Artificial gene synthesis wikipedia , lookup
X-inactivation wikipedia , lookup
History of genetic engineering wikipedia , lookup
Gene expression programming wikipedia , lookup
Genome evolution wikipedia , lookup
Site-specific recombinase technology wikipedia , lookup
Designer baby wikipedia , lookup
Minimal genome wikipedia , lookup
Ridge (biology) wikipedia , lookup
Genomic imprinting wikipedia , lookup
Biology and consumer behaviour wikipedia , lookup
Epigenetics of human development wikipedia , lookup
Gene expression profiling wikipedia , lookup
Microevolution wikipedia , lookup
William S. Klug Michael R. Cummings Charlotte A. Spencer Concepts of Genetics Chromosomes consist of linear sequences of genes. STUDY UNIT 4 Chromosome mapping: • Chapter 5 Chromosome Mapping in Eukaryotes • Genes located on the same chromosome are said to be linked and demonstrate linkage in genetic crosses. Copyright © 2006 Pearson Prentice Hall, Inc. •Genes are not cytologically linked •Genes are cytologically linked, complete linkage. • • •Gametes form at equal frequencies •Gametes form at equal frequencies Figure 5-1a Copyright © 2006 Pearson Prentice Hall, Inc. •Genes are cytologically and genetically linked (<50% recombination) Figure 5-1b Copyright © 2006 Pearson Prentice Hall, Inc. 5.1 Genes Linked on the Same Chromosome Segregate Together • 5.1.1 The Linkage Ratio If complete linkage exists between 2 genes because of close proximity, the F2 phenotypic ratio of a dihybrid cross will be unique, designated the . •Four genetically different gametes are formed Genes located on the same chromosome will show evidence of linkage to one another. A can be established for each chromosome (corresponds to n number of chr) •Frequency of crossing over between genes generally proportional to distance between them Figure 5-1c Eg fig 5-2 (Klug et al, 2006) Copyright © 2006 Pearson Prentice Hall, Inc. 1 Drosophila melanogaster: Recessive mutants: brown eye (bw) < bw+ wildtype red eye heavy wing vein (hv) < hv+ wildtype thin wing vein The two genes are cytologically linked. Flies with: Brown eyes, normal thin veins X wildtype red eyes, heavy veins P1: bw bw hv+ X bw + hv hv+ bw + hv F1: bw hv+ bw + hv Figure 5-2-01 5.2 Copyright © 2006 Pearson Prentice Hall, Inc. Crossing Over Serves as the Basis of Determining the Distance between Genes during Chromosome Mapping 5.2.1 Morgan and Crossing Over • Usually crossing over occurs between genes on the same chr • Result: percentage of offspring arise from recombinant gametes • • 1911, Thomas Morgan and Alfred Sturtevant Figure 5-2ab Copyright © 2006 Pearson Prentice Hall, Inc. 5.2.2 Sturtevant and Mapping • • Sturtevant first to realize that Morgan’s results could be used to map the sequence of linked genes Compiled data on recombination between the genes representing yellow, white and miniature (1) yellow, white (2) white, miniature (3) yellow, miniature 0.5% 34.5% 35.4% (1) + (2) approx. = (3) – Recombination frequencies between linked genes are additive Figure 5-3 Copyright © 2006 Pearson Prentice Hall, Inc. 2 5.2.3 Single Crossovers Frequency of exchange gives an estimate of relative distance between loci • A limited number of crossovers occur in each tetrad during meiosis • Occur randomly along the length of the tetrad In honour of Morgan’s work, map units referred to as centimorgans (cM) • • The farther apart two linked loci are from each other, the more likely a random crossover event will occur between them. Figure 5-4 Copyright © 2006 Pearson Prentice Hall, Inc. • • The other 2 nonsister chromatids enter the gamete unchanged (parental/noncrossover gametes) •If a single crossover occurs 100% of the time between 2 loci, recombination will only be observed in 50% of the potential gametes formed. Figure 5-5a & b Copyright © 2006 Pearson Prentice Hall, Inc. Figure 5-6 Copyright © 2006 Pearson Prentice Hall, Inc. Consider only single crossovers: • Means crossing over occurs in 40% of the tetrads When two linked genes are more than 50 map units apart, a crossover can be expected to occur between them in 100% of the tetrads. General rule for single crossovers between two loci: • • Observe 20% recombinant gametes Expect equal proportions of the 4 gametes • Therefore, the theoretical limit of recombination due to crossing over is 50% If there is 50% or less recombination between 2 loci, the 2 loci are said to be genetically linked. 3 How do we use this information for mapping purposes? TWO-POINT MAPPING Linkage analysis of two genes Linkage notation: Genetic map distances measured in map units or centimorgans (cM) where 1% recombination = 1 mu / 1 cM Map distances between many linked genes are combined to construct a genetic map. Consider individual AaBb. Ratio of parental gametes : recombination gametes May produce the following gametes: depends on how frequently crossing over occurred. For mapping? AB ab Want to determine the freq of recombination gametes as an indication of distance between A and B. Impossible to analyse the genotype of a gamete, therefore, Ab aB Linkage best analysed in testcross: Reason? • heterozygote x homozygote recessive • ∴ each of the 4 phenotypes in progeny represents the alleles received from one parent (AaBb) • Cross-over in aabb parent not detected 4 Example • can distinguish recombinant from parental progeny 1. Perform testcross • phenotypic freq of progeny reflects the frequency of C1 c1 Wx wx gametes formed by heterozygote parent, x c1c1 wx wx coloured, starchy colourless, waxy • i.e. recombination frequency of heterozygote Progeny: coloured, starchy (C1 c1 Wx wx) coloured, waxy (C1 c1 wx wx) 310 colourless, starchy (c1 c1 Wx wx) 311 colourless, waxy 858 (c1 c1 wx wx) 781 Total 2. Determine if loci are indeed linked 2260 3. Identify recombinants in progeny - measure deviation from independent assortment parental phenotypes expect ratio of = 2 classes recombinant phenotypes = 2 less abundant classes in progeny coloured, starchy (C1 c1 Wx wx) if independent assortment had occurred in coloured, waxy (C1 c1 wx wx) 310 gametes of heterozygous parent. colourless, starchy (c1 c1 Wx wx) 311 colourless, waxy - use chi – squared analysis (c1 c1 wx wx) Total ___ 2260 - if ratio is not 1 : 1 : 1 : 1 ⇒ indicates linkage 4. Identify coupling or repulsion conformation 5. Calculate recombination frequency (or % ) where 1% recombination = 1 cM Coupling (cis): Rf = Repulsion (trans): 1 dominant and 1 recessive allele number recombinants = 310 + 311 total number progeny 2260 linked in parental chromosomes = 0.275 C1 W1 c1 w1 = 27.5 % 6. ∴ Coupling phase ⇒ 27.5 cM Compile genetic map C1 27.5 Wx 5 1. In maize, the recessive allele of a locus causes brownmidrib (bm) and the recessive allele of another locus brevis (bv) (dwarf). A double heterozygote was crossed with a brevis plant with a brown midrib and 200 seeds were germinated. The following results are obtained: + + 93 bm + 7 + bv 5 bm bv 95 2. In Drosophila, ebony (e) is recessive and the dominant allele (e+) results in tan body colour. The recessive mutation (d) causes dumpy (shortened) wings while the dominant allele (d+) causes long, normal wings. A tan female with normal wings is crossed with an ebony male with dumpy wings. The resulting progeny are as follows: 44 tan, long 41 ebony, dumpy 15 tan, dumpy 17 ebony, long a) Give the genotypes of the parents. (a) Show the crossing and progeny symbolically. b) Give the gamete ratios of the parents. (b) What is the % recombination between the 2 loci? c) Set up a chromosome linkage map. 5.3 Determining the Gene Sequence during Mapping Relies on the Analysis of Multiple Crossovers 5.3.1 Multiple Exchanges • Study single crossovers between 2 linked genes = distance Figure 5-7ab Copyright © 2006 Pearson Prentice Hall, Inc. In a single tetrad – possible to have 2, 3 or more crossovers Double exchanges of genetic material are as a result of double crossovers (DCOs) • P of a single exchange between A and B and B and C is directly related to the physical distance between them 5.3.2 Three-Point Mapping in Drosophila Double crossover – 2 separate and individual events must occur simultaneously Criteria Product law Example: SCO between A and B 20% ~ p = 0.20 SCO between B and C 30% ~ p = 0.30 DCO between A and B and B and C is predicted to be (0.20)(0.30) = 0.06 (6%) 1. Organism being mapped must be heterozygous at all loci 2. test cross – progeny reflect heterozygote parents gametes hemizygous male must carry recessive alleles 3. Large number of progeny must be examined If 3 linked genes relatively close to one another, expected frequency of DCO gametes is very low 6 5.3 Determining the Gene Sequence during Mapping Relies on the Analysis of Multiple Crossovers yellow body colour (y) white eye colour (w) echinus eye shape (ec) 5.3.3 Determining the Gene Sequence The gene order must be determined before proceeding with the problem 2 methods… Figure 5-8 Copyright © 2006 Pearson Prentice Hall, Inc. Method I Based on the fact that there are only 3 possible orders 3 loci: y, w and ec Method II According to fig 5-8, NCO: y w ec DCO: y w+ ec & & y+ w+ ec+ y+ w ec+ – – Identify the single allele that has been switched so that it no longer is associated with its original neighboring alleles. – This allele is in the middle. Figure 5-9 Copyright © 2006 Pearson Prentice Hall, Inc. THREE-POINT MAPPING 8 types of gametes possible in progeny: Linkage analysis of 3 genes in a single testcross Crossovers can be detected in progeny of an individual that is heterozygous for the 3 linked genes. • parental class P = most abundant NCO • crossover in one interval = single recombination class 1 SCO I 7 Note: • crossover in other If genes are not linked, all 8 gametes will be • interval = single recombination equally frequent class 2 ∴ would expect 1:1:1:1:1:1:1:1 phenotypic ratio (independent assortment) • double crossover = • double recomb. class 1. DCO 2. Perform testcross C1c1 Shsh Wxwx x coloured, full, starchy Determine if loci are indeed linked Ratio ≠ 1:1:1:1:1:1:1:1 ⇒ indicates linkage c1c1 shsh wxwx colourless, shrunken, waxy 3. Phenotypes Gametes heterozygote coloured, full, starchy C1 Sh Wx coloured, shrunken, starchy C1 sh Wx coloured, full, waxy C1 Sh wx coloured, shrunken, waxy C1 sh wx colourless, full, starchy c1 Sh Wx colourless, shrunken, starchy c1 sh Wx colourless, full, waxy c1 Sh wx colourless, shrunken, waxy c1 sh wx Total: Number 4 2538 113 601 626 116 2708 2 6708 Identify coupling or repulsion phases of the 3 genes Consider parental class (NCO) = most coloured, shrunken, starchy C1 sh Wx 2538 colourless, full, waxy c1 Sh wx 2708 ∴ noncrossover chrs are and c1 Sh wx ∴ C1 and Wx are in coupling phase and sh is in repulsion to them. 4. Determine gene order : Parental compare P/NCO class with DCO C1 sh Wx (order not yet known) C1 sh Wx C1 c1 Sh wx c1 sh Wx - this indicates which gene is in the middle. Sh wx Parental / No Crossing Over class coloured, shrunken, starchy C1 sh Wx 2538 colourless, full, waxy 2708 c1 Sh wx Double Crossover class (least) : DCO C1 c1 coloured, full, starchy C1 Sh Wx 4 colourless, shrunken, waxy c1 sh wx 2 sh Sh Wx wx C1 c1 Sh sh Wx wx Sh in the middle produces the correct double crossover phenotypes. 8 5. Identify single recombination classes as being SCOI or 6. Calculate recombination frequency (or %) ( compare to NCO ) % recomb in region I = coloured, shrunken, starchy coloured, full, waxy coloured, shrunken, waxy C1 sh Wx C1 Sh wx 2538 113 c1 sh Wx c1 Sh wx + % DCO (113 + 116) + (4+2) 6708 = 0.035 = 3.5 % colourless, full, starchy colourless, shrunken, starchy colourless, full, waxy % SCO I = ⇒ 3.5 cM ∴ distance between C1 and Sh = 3.5 cM 116 2708 % recomb in region 2 = % SCO II Single crossover class I: = + % DCO (601 + 626) + (4+2) 6708 - crossover between C1 and Sh = 0.184 = 18.4% ⇒ 18.4 cM - crossover between Sh and Wx ∴ distance between Sh and Wx = 18.4 cM 5.4 6. Compile genetic map C1 3.5 Sh Interference Affects the Recovery of Multiple Exchanges INTERFERENCE 18.4 Wx Until now we have assumed that a crossover at one position is independent of another in the same bivalent. Not necessarily true. Interference: one chiasma interferes with the formation of a second chiasma in the same vicinity. ∴ From example: C1 3.5 Sh 18.4 Wx Interference measured as coefficient of coincidence (C) observed dco was C = expected dco observed DCO expected DCO ∴C (4+2) / 6708 = 0.00089 = 0.035 x 0.184 = 0.00644 0.00089 / 0.00644 = 0.139 Interference = 1 – C Expected dco (DCOexp)? Observed freq crossover in area I = 1 – 0.139 x Observed freq crossover in area II = 0.861 If Interference > 0 Interference will be = 0 if C = 1 (when observed dco = expected dco) Interference = 1 9 5.5 As the Distance between Two Genes Increases, Mapping Experiments Become More Inaccurate • When two genes are close together, the accuracy of mapping is high. As the distance between them increases, the accuracy of mapping decreases. • In most cases experimentally derived distance is an underestimate. • Figure 5-12 & 5-14 Copyright © 2006 Pearson Prentice Hall, Inc. 5.12 Gene Mapping Is Now Possible Using Molecular Analysis of DNA DNA markers • Followed in pedigrees • Associated with specific chr • Human genes analyzed in relationship to markers Gene Mapping Using Annotated Computer Databases • Genome projects – physical maps in bp 3. A mielie, which was homozygous for the recessive genes a ("green"), d ("dwarf") and rg (normal leaves) was crossed with a mielie homozygous for the dominant genes A (red), D (tall) en Rg ("ragged" leaves). The offspring were then test crossed and the following results obtained: Pt No. Pt No. Rg A D 265 rg A d 90 rg a d 275 Rg a D 70 rg A D 24 Rg A d 120 Rg a d 16 rg a D 140 4. Given a chromosome map: R M 8 T 20 Accept the given linkage setup. Accept CC = 0.5 Calculate the size of each phenotypic class of the offspring if a trihybrid testcross is performed. (a) Which classes represent the single crossover classes? (b) Set up a chromosome map. (c) Calculate the coefficient of coincidence. 10