BMS_GNA_2042_Lecture 5 Notes Linkage Mapping (1).pdf

Linkage, Recombination & Genetic Mapping Dr. Callum Vidor Textbook References: Klug et al., 12th Ed, Chapter 7 pg 176 to 192, 196-197  Acknowledgement of country Bundjil and Waan...

Linkage, Recombination & Genetic Mapping Dr. Callum Vidor Textbook References: Klug et al., 12th Ed, Chapter 7 pg 176 to 192, 196-197  Acknowledgement of country Bundjil and Waang The Kulin Nation Note: this map of language groups is representative only and not recognised as official. https://www.timeout.com/melbourne/things-to-do/an-introduction-to-boon-wurrung-language-from-aunty-fay-stewart-muir https://taungurung.com.au/creation-stories/ A genetic map of Drosophila Klug et al., 12th Ed., Pearson Learning outcomes from this Lecture Describe the terms linkage and recombination Define a genetic map and their potential uses Map genes based on a phenotype using dihybrid and trihybrid test crosses Take into account the effect of double recombinants and interference in chromosome mapping Outline rarer forms of chromosomal recombination i.e. mitotic and intergenic Linkage, Recombination & Genetic Mapping Linkage and Recombination Dr. Callum Vidor So far we have covered independent assortment of genes Mendelian traits found to be due to segregation of alleles and the independent assortment of chromosomes during meiosis If a dihybrid cross (AaBb x AaBb): 9 A-B- : 3 A-bb : 3 aaB- : 1 aabb phenotypic ratio If a test cross (AaBb x aabb) 1 AaBb : 1 Aabb : 1 aaBb : 1 aabb phenotypic ratio These genes assort independently because they are on different chromosomes What if they are on the same chromosome? Morris et al, 3rd Ed, Macmillain Alleles from linked genes assort together This displays complete linkage: If two genes are completely linked* the alleles from the parent will assort together 100% of the time If a dihybrid cross (AaBb x AaBb): 3 A-B- : 1 aabb phenotypic ratio If a test cross (AaBb x aabb) 1 AaBb : 1 aabb phenotypic ratio * Complete linkage is almost never seen due to recombination Klug et al., 12th Ed., Pearson Recombination during meiosis reassort alleles on homologous chromosomes Chiasmata provide physical evidence for crossover events Recombination during meiosis reassort alleles on homologous chromosomes Linked alleles can be separated by recombination during meiosis producing: Parental genotypes (P) Recombinant genotypes If a dihybrid cross (AaBb x AaBb): ? A-B- : ? A-bb : ? aaB- : ? aabb phenotypic ratio If a test cross (AaBb x aabb) ? AaBb : ? Aabb : ? aaBb : ? aabb phenotypic ratio Ratios depend on the ‘degree’ of linkage i.e. how closely are genes linked? Klug et al., 12th Ed., Pearson The discovery of linkage displays how testcrosses highlight linked genes Linkage was discovered by one of Thomas Morgan’s students, Alfred H. Sturtevant. Here is an example of what they saw: eye colour: pr+ - wildtype, dominant to pr - purple wing length: vg+ - wildtype, dominant to vg - vestigial P1 pr+ pr+; vg+ vg+ x P2 pr pr ; vg vg wild type purple, vestigial F1 pr+ pr; vg+ vg wild type F1 were testcrossed: pr+ pr; vg+ vg x pr pr ; vg vg What ratio would we expect under independent assortment? The discovery of linkage displays how testcrosses highlight linked genes P1 pr+ pr+; vg+ vg+ x P2 pr pr ; vg vg wild type purple, vestigial F1 were testcrossed: pr+ pr; vg+ vg x pr pr ; vg vg pr+ vg+ pr+ vg pr vg+ pr vg pr vg pr+ pr vg+ vg pr+ pr vg vg pr pr vg+ vg pr pr vg vg wildtype vestigial purple Purple, vestigial Phenotype? Proportions? The discovery of linkage displays how testcrosses highlight linked genes P1 pr+ pr+; vg+ vg+ x P2 pr pr ; vg vg wild type purple, vestigial F1 were testcrossed: pr+ pr; vg+ vg x pr pr ; vg vg pr+ vg+ pr+ vg pr vg+ pr vg pr vg pr+ pr vg+ vg pr+ pr vg vg pr pr vg+ vg pr pr vg vg wildtype vestigial purple Purple, vestigial Phenotype? Proportions? 1339 154 151 1195 Total 2839 The discovery of linkage displays how testcrosses highlight linked genes P1 pr+ pr+; vg+ vg+ x P2 pr pr ; vg vg wild type purple, vestigial F1 were testcrossed: pr+ pr; vg+ vg x pr pr ; vg vg pr+ vg+ pr+ vg pr vg+ pr vg pr vg pr+ pr vg+ vg pr+ pr vg vg pr pr vg+ vg pr pr vg vg wildtype vestigial purple Purple, vestigial Phenotype? Proportions? 1339 154 151 1195 Total 2839 Higher proportion of parental types produced by F1 than expected Parental and recombinant chromosome arrangements It was proposed that the two genes were on the same chromosome Parental genotypes were the arrangement of the alleles in the parents Recombinant genotypes were due to crossing over between the parental chromosomes Parents pr+ vg+ x pr vg pr+ vg+ pr vg wild type purple, vestigial F1 pr+ vg+ Physical evidence shown later by Barbara McClintock and pr vg collaborators in Maize. wild type No crossover occurs Crossover occurs between genes between genes pr+ vg+ pr vg pr+ vg pr vg+ Parental gametes Recombinant gametes Parental allele combinations can be in one of two phases In our previous example the dominant and recessive phenotypes were together on the parental chromosome. This isn’t always the case, you can have parental chromosomes in either: Coupling phase Repulsion phase A B A b a b a B What affects the amount of recombination between two genes on the same chromosome? What can this level of recombination tell us? Linkage, Recombination & Genetic Mapping Recombination Frequency and Mapping Dr. Callum Vidor  Distance between genes impacts recombination frequency Recombination between homologous chromosomes is a random event Genes close on same chromosome: Genes distant on same chromosome: Less chance of recombination switching alleles Greater chance of recombination switching alleles Smaller proportion of recombinant gametes Greater proportion of recombinant gametes ∴ Recombination frequency ≈ physical distance between genes Adapted from Khan Academy (www.khanacademy.org) Calculating recombination frequency The proportion of recombinant gametes depends on how often crossovers occur between the genes Recombination frequency (RF) = Number of recombinants / Total progeny x 100 If crossovers never occur between two gene loci, RF=0% The further two genes are apart RF increases The maximum RF is 50% ∴ Recombination frequency can help define linkage: Genes that have RF < 50% are linked Genes that have RF = 50% are not linked Klug et al., 12th Ed., Pearson  Recombination frequency and map distance Close genes = small chance of crossover between them = few recombinant offspring Far genes =high chance of crossover between them = many recombinant offspring Genetic mapping principle: The RF (percentage of recombinant offspring) is correlated with the distance between two genes Map distance is measured in map units (mu) aka centiMorgans (cM) 1 mu/cm = distance that will produce 1% RF Adapted from Khan Academy (www.khanacademy.org) Calculating recombination frequency / map distance RF = (76 + 75) / 1230 x 100 = 12.3% ∴ The purple gene and vestigial gene are 12.3 cM apart. Mapping genes with two point (dihybrid) crosses Sturtevant used the recombination frequencies between genes to produce the first genetic map in 1911. He proposed that map distances are additive, and a two dimensional map could be generated by combining data from different crosses that shared loci. e.g. RF between genes A and B is 21.2 RF between genes A and C is 5.8 RF between genes B and C is 15.4 A C B 5.8 mu 15.4 mu 21.2 mu Sturtevant’s Data and Map Alleles Concerned Number Recombinant / Total Number Percent Recombinant Offspring y and w/w-e 214/21,736 1.0 Carried out tests crosses of dihybrid F1, y and v 1,464/4,551 32.2 two traits at a time y and r 115/324 35.5 y and m 260/693 37.5 w/w-e and v 471/1,584 29.7 Constructed by beginning with y and w/w-e and r 2,062/6,116 33.7 mapping left to right w/w-e and m 406/898 45.2 v and r 17/573 3.0 See any issues between the data and v and m 109/405 26.9 the map? y and m RF = 37.5% Map distance = 57.6 mu w and m RF = 45.2% Brooker, McGraw Hill Map distance = 56.6 mu Double crossovers make mapping longer distances inaccurate Double crossovers between loci revert them to the parental arrangement Cancels out a recombinant offspring The further two genes are apart, the greater chance of not only one crossover, but a second Klug et al., 12th Ed., Pearson Testing if two loci are likely linked based on RF If genes are far apart, RF no longer accurately measures distance When we get a high RF (e.g. >25%) we need to test if our genes are likely to be linked or not Perform a c2 goodness of fit test based on the genetic hypothesis that the genes are unlinked If we support the Null, variation is due to chance and supports Klug et al., 12th Ed., Pearson genes being unlinked If we reject the Null, variation is likely due to another reason, supports the genes being linked Full genetic maps Because far gene recombination frequencies are inaccurate, maps were built linking intermediate genes Can this process be carried out in humans? Yes, but it’s difficult because Small number of progeny Can’t perform desired crosses Restricted to pedigree analysis Klug et al., 12th Ed., Pearson Thing are always more complicated in reality In this unit we will assume that any crossovers occur only between two non-sister chromatids In reality double crossovers can occur between three or four strands of the tetrad Griffiths Fig 4.18 8th edition Linkage, Recombination & Genetic Mapping Mapping using trihybrid test crosses Dr. Callum Vidor Mapping genes with three point (trihybrid) test crosses Previously we were mapping two genes at a time Mapping three genes at a time has advantages: Faster More accurate We will explore how this is done using an example in Drosophilia: Body colour: wild type (b+-) or black (bb) Wing length: wild type (vg+-) or vestigial (vgvg) Eye colour: wild type (pr+-) or purple (prpr) Parent 1 X Parent 2 b+ vg+ pr+ b vg pr b+ vg+ pr+ b vg pr Test-cross F1: b+ vg+ pr+ X b vg pr Must use females from F1 as b vg pr b vg pr no recombination occurs in (female) (male) Drosophila males Example: Trihybrid F1 test cross progeny Body Wings Eyes # wild type wild type wild type 411 black vestigial purple 412 wild type vestigial purple 30 black wild type wild type 28 wild type vestigial wild type 61 black wild type purple 60 wild type wild type purple 2 black vestigial wild type 1 Total 1005 Note: Here we have obvious linkage. If it was not obvious, we would need to do a c2 test for linkage.  Possible gametes of the F1 (female) b+ vg+ pr+ b vg pr b+ vg pr b vg+ pr+ b+ vg pr+ b vg+ pr b+ vg+ pr b vg pr+  Full genotypes of progeny b+ b vg+ vg pr+ pr All gametes from testcross male are the same, b vg pr b b vg vg pr pr b+ b vg vg pr pr b b vg+ vg pr+ pr b+ b vg vg pr+ pr b b vg+ vg pr pr b+ b vg+ vg pr pr b b vg vg pr+ pr Body Wings Eyes Testcross progeny phenotypes are determined wild type wild type wild type 411 by female gamete only b+ vg+ pr+ black vestigial purple 412 b vg pr wild type vestigial purple 30 b+ vg pr black wild type wild type 28 b vg+ pr+ wild type vestigial wild type 61 b+ vg pr+ black wild type purple 60 b vg+ pr wild type wild type purple 2 b+ vg+ pr black vestigial wild type 1 b vg pr+ Total 1005 Body Wings Eyes wild type wild type wild type 411 First consider 2 genes at a time (ignoring the 3rd gene), b+ vg+ pr+ and determine the map distance between them. black vestigial purple 412 b vg pr 1. black body and purple eye wild type vestigial purple 30 b+ pr+ : 411 + 61 parental b+ vg pr b pr : 412 + 60 parental black wild type wild type 28 b+ pr : 30 + 2 recombinant b vg+ pr+ b pr+ : 28 + 1 recombinant wild type vestigial wild type 61 Total recombinants b+ vg pr+ = 61 / 1,005 = 6.1% black wild type purple 60 = 6.1 mu b vg+ pr wild type wild type purple 2 b+ vg+ pr black vestigial wild type 1 b vg pr+ Total 1005 Body Wings Eyes Follow with the other two combinations of traits wild type wild type wild type 411 b+ vg+ pr+ 2. black body and vestigial wing black vestigial purple 412 b vg pr b+ vg+ : 411 + 2 parental wild type vestigial purple 30 b vg : 412 + 1 parental b+ vg pr b+ vg : 30 + 61 recombinant b vg+ : 28 + 60 recombinant black wild type wild type 28 b vg+ pr+ Total recomb. = 179 / 1,005 = 17.8% = 17.8 mu wild type vestigial wild type 61 b+ vg pr+ 3. purple eye and vestigial wing black wild type purple 60 b vg+ pr pr+ vg+ : 411 + 28 parental pr vg : 412 + 30 parental wild type wild type purple 2 pr+ vg : 61 + 1 recombinant b+ vg+ pr pr vg+ : 60 + 2 recombinant black vestigial wild type 1 Total recomb. = 124 / 1,005 = 12.3% = 12.3 mu b vg pr+ Total 1005 Example: Map from initial recombination frequencies We can now draw the following map of our three loci: b pr vg 6.1 12.3 The order of b and vg can be switched 17.8 Note that the distance between the outer loci is less than the sum of the internal regions. Double crossovers in trihybrid cross mapping Crossovers can occur either side of the middle locus b+ pr+ vg+ b+ pr vg+ 2 b pr+ vg 1 b pr vg Each of these progeny represent 2 crossovers between the outer loci Neither was counted as they appeared as a parental arrangement Double crossovers appear as parental for the outer loci, but recombinant for the middle loci Taking into account double crossovers Each double recombinant represents two crossovers, so the true RF for b and vg : = (original recombinants + 2 x double recombinants) / total = [(179 + 2 x 3) / 1005] x 100 =18.4% Therefore, b and vg are 18.4 mu apart (the sum of the inner regions) Interference changes the proportion of double crossovers Does the occurrence of one crossover have an effect of another nearby i.e. are there more or fewer double crossovers than we would expect? We can calculate the expected number of double crossovers from our map distances Chances of DCO = chance of crossover between b and pr X chance of crossover between pr and vg = 0.061 x 0.123 = 0.0075 Among 1,005 offspring: No. Expt. DCO = 1005 x 0.0075 = 7.5 Coefficient of coincidence and interference The coefficient of coincidence (C) = Observed / Expected For our example, C = 3/7.5 = 0.4 or 40% i.e. only 40% of expected double crossovers occurred As the number was less than expected it suggests one crossover interferes with one nearby This can also be represented as the degree of interference (I) I=1-C Complete interference Positive interference No interference Negative interference C=0, I=1 C > 0, I < 1 C=1, I=0 C > 1, I < 0 Linkage, Recombination & Genetic Mapping Uses of mapping & other forms of recombination Dr. Callum Vidor Why generate genetic maps and consider linkage? Traditional uses: Construct a view of the organism genome Identify and clone genes using their map position Determine if mutations affect different genes Continuing uses: Used to identify disease causing rare disease causing alleles Molecular markers mostly used in humans Linkage must be considered in genetic counselling/ risk calculations Assist in genome sequence assembly Example: mapping disease loci Markers across the chromosome used in hand with pedigree information to identify regions likely to encode pathogenic variants Lee et al., Nature Genetics volume 44, pages 193–199 (2012) Example: considering linkage in risk calculation When determining the risk of someone inheriting multiple conditions, you may need to consider linkage e.g. A female has two dominant autosomal disorders caused by mutations in two different genes – they are heterozygous for both. What is the chance their children will inherit: - both mutations? - one or the other mutation? - neither mutation? Mitotic recombination During mitosis homologous chromosomes don’t form bivalents Therefore expect no crossing over to take place Occasionally, it does Implications for development of cancers Can be used as a tool to track cell lineages Matic and Radman, Encyclopedia of Biological Chemistry, 2004 Mitotic recombination example: twin spots in Drosophila Mitotic recombination example: twin spots in Drosophila Mitotic recombination example: twin spots in Drosophila Intragenic recombination Is the unit of recombination smaller than a gene? i.e. can recombination occur within a gene – intragenically Might expect it to occur very infrequently, if at all X-linked recessive mutations Analysed by Green et al. 1949 – used two lozenge alleles with different phenotypes. lzBS Allele 1 lz9 Allele 2 intragenic recombination Double mutant males 20/16,000 progeny Wildtype males