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Dec 11, 2023 •457 likes •607 views

Population Genetics 2: Linkage disequilibrium Consider two loci and 1 generation of random mating: A gene : AA, Aa, and aa B gene : BB, Bb, and bb Genotype frequencies in a population A gene B gene f AA = p 2 f BB = x 2 f Aa = 2pq f Bb = 2 xy f

  1. Population Genetics 2: Linkage disequilibrium Consider two loci and 1 generation of random mating: A gene : AA, Aa, and aa B gene : BB, Bb, and bb Genotype frequencies in a population A gene B gene f AA = p 2 f BB = x 2 f Aa = 2pq f Bb = 2 xy f aa = q 2 f bb = y 2 p + q = 1 x + y = 1 Random association of alleles at a single locus: HWE What about random association of alleles at different loci after random mating? 1

  2. Consider two loci and 1 generation of random mating: Random association in gametes Alleles at A locus A( p ) a( q ) Alleles at B locus B AB aB ( x ) ( p x) ( qx ) b Ab ab ( y ) ( py ) ( q y) remember: p + q =1 and x + y = 1 Random association of alleles at a different locus: LINKAGE EQUILIBRIUM GAMETIC PHASE EQUILIBRIUM Consider two loci and 1 generation of random mating: Surprisingly common result: Gene A: HWE Gene B: HWE Gene A + Gene B: disequilibrium 2

  3. Consider two loci and 1 generation of random mating: Example: Population 1: 100% AABB Population 2: 100% aabb Mix populations equally: 50% AABB + 50% aabb 1 generation of random mating (only three matings possible) : AABB x AABB = AABB aabb x aabb = aabb AABB x aabb = AaBb Nine genotypes are possible: AABB AaBB AABb We only see 1/3 after 1 aaBB AAbb aabb generation of random mating! AaBa aaBb Aabb They did not reach equilibrium after one generation of random mating. With continued random mating the “missing” genotypes would appear, but not immediately at their equilibrium frequencies! Consider two loci and 1 generation of random mating: • attainment of linkage equilibrium is gradual • about 50% of disequilibrium “breaks down” per generation • linkage disequilibrium (LD) persists in populations for many generations • LD = gametic phase disequilibrium 3

  4. LD in individuals: We need a new symbolism Case 1: AB gamete + ab gamete = AaBb Case 2: Ab gamete + aB gamete = AaBb New symbolism: AB/ab indicates the union of AB gamete + ab gamete LD in individuals: Physical linkage: Let’s take an AB/ab individual as an example: A B What types of games can the AB/ab make? a b By the way, lets assume physical linage. Notation = AB/ab (1) AB Parental or non-recombinant gametes (2) ab (3) Ab Non-parental or recombinant gametes (4) aB 4

  5. LD in individuals: When genes are on the same chromosome: Physical linkage: A B f AB = f ab ≥ f Ab = f aB a b f (non-recombinant) ≥ f (recombinant) Notation = AB/ab Recombination fraction (r) is the proportion of recombinant gametes produced by an individual. When r = 0: f Ab + f aB = 100% [ f Ab + f aB = 0%] When r = 0.5: f Ab + f aB = 50% [ f Ab + f aB = 50%] From: i Genetics P. J. Russell (2002) page 350 5

  6. LD in individuals: Un linked genes: When genes are on different chromosomes: A f AB = f ab = f Ab = f aB a f (non-recombinant) = f (recombinant) B b r = 0.5 • when genes are on different chromosomes • when genes are on same chrom and recombination is high enough for independent assortment LD in individuals: Individual AB/ab produces the following: (1) AB: f AB = 0.38 (2) ab: f ab = 0.38 (3) Ab: f Ab = 0.12 (4) aB: f aB = 0.12 r = 0.12 + 0.12 = 0.24 6

  7. LD in populations: Random association in gametes f AB = p x Alleles at A locus f ab = q y A( p ) a( q ) Alleles at B locus f Ab = py B AB aB f aB = qx ( x ) ( p x) ( qx ) b Ab ab ( y ) ( py ) ( q y) f AB + f ab + f Ab + f aB = 1 remember: p + q =1 and x + y = 1 LD in populations: 6 4 4 4 7 4 4 4 8 6 4 7 4 4 8 4 non - recombinan ts recombinan ts = − + ' f ( 1 r ) f ( r ) px 1 2 3 { { { AB AB frequency of AB prob of prob of putting probabilit y of recomb together A and B gametes in last no recombinat ion at random from generation recombinan ts 7

  8. 6 4 non 4 4 - recombinan 7 4 4 ts 4 8 6 4 7 recombinan 4 4 8 ts 4 = − + ' f ( 1 r ) f ( r ) px 1 2 3 { { { AB AB prob of prob of putting probabilit y of frequency of AB gametes in last recomb together A and B no recombinat ion generation at random from recombinan ts ' − = − − f px ( 1 r )( f px ) Obs. – exp. AB AB = − − D ( 1 r )( f px ) = the linkage disequilibrium parameter AB f AB = p x + D Non-recombinants f ab = q y + D f Ab = py - D Recombinants f aB = qx - D Remember: Individual AB/ab produces the following: (1) AB: f AB = 0.38 (2) ab: f ab = 0.38 (3) Ab: f Ab = 0.12 (4) aB: f aB = 0.12 r = 0.12 + 0.12 = 0.24 8

  9. Forces that increase D in populations: 1. Migration 2. Natural selection: 3. Genetic Drift LD in populations: • Comparing D among populations is difficult. • Standardize D as a fraction of the theoretical maximum for the popn D D max = × − × D f f f f 1 4 2 4 3 1 4 2 4 3 AB aa Ab aB non recombinan t recombinan t = − − D qx or py (whichever is smaller) max 9

  10. LD in a population: Example: blood group polymorphism in a sample of 1000 British people MN blood group Gamete frequencies = × − × D f f f f 1 4 2 4 3 1 4 2 4 3 AB aa Ab aB f M = p = 0.5425 MS = 474/2000 = 0.2370 non recombinan t recombinan t f N = q = 0.4575 Ms = 611/2000 = 0.3055 NS = 142/2000 = 0.0710 = − − D qx or py max Ns = 773/2000 = 0.3865 Ss blood group f S = x = 0.3080 f s = y = 0.6920 D = (0.2370)(0.3865) – (0.3055)(0.0710) = 0.07 D max : qx = 0.14 or py = 0.37; so D max = 0.14 D is (0.07/0.14)*100 = 50% of the theoretical maximum Homework: Genotype counts in the population Gametes MN locus Ss locus MS = 474 MM = 298 SS = 483 Ms = 611 MN = 489 Ss = 418 NN = 213 ss = 99 NS = 142 Ns = 773 Use chi-square test to: 1. determine if each locus is in HWE 2. determine if gamete frequencies are in equilibrium 10

  11. Recombination reduces LD: Rate of decay of LD under various recombination rates 1 0.9 Standardized disequilibrium r= 0.001 0.8 0.7 0.6 D/Dmax r= 0.01 0.5 0.4 0.3 0.2 0.1 r= 0.1 r= 0.5 0 1 9 17 25 33 41 49 57 65 73 81 89 97 generations Recombination reduces LD: “Hitchhiking” of a mutator gene with and without recombination No recombination r = 0 Recombination r = 0.5 Mutator allele that increase the mutation rate Beneficial allele subject to strong positive selection Adapted from Sniegowski et al. (2000) BioEssays 22:1057-1066. 11

  12. Mapping disease genes: 1. Family studies • Uses family pedigrees • Co-segregation of disease and a marker on the pedigree 2. Allelic association studies (LD mapping) • Uses population data • Relies on strong LD only among closely linked loci • Sample affected and unaffected individuals • Very large samples are required! • Look for markers with more LD in affected individuals Both approached have powers and pitfalls Mapping disease genes: 12

  13. Linage disequilibrium: Keynotes • Attainment of equilibrium at different loci is gradual; > 1 generation of random mating. • Physical linkage slows the rate to equilibrium even more! • “r” determines the rate to equilibrium, the lower the fraction, the longer to equilibrium. • When r = 0.5 the loci are in equilibrium. When r =0 the loci are in complete disequilibrium. • Disequilibrium can arise from sources other than linkage: o Admixture of populations o Natural selection acting on one or more of the loci o Inbreeding in plants that regularly undergo self-fertilization o Genes located in a chromosomal inversion ( SUPERGENE ) • The term LINKAGE DISEQUILIBRIUM is used to describe any source of disequilibrium, regardless of whether the two genes are physically linked or not. 13

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