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1 Consider two loci and 1 generation of random mating: Random - - PDF document

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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

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Population Genetics 2: Linkage disequilibrium

Genotype frequencies in a population A gene B gene fAA = p2 fAa = 2pq faa = q2 fBB = x2 fBb = 2xy fbb = y2 p + q = 1 x + y = 1

Consider two loci and 1 generation of random mating:

A gene: AA, Aa, and aa B gene: BB, Bb, and bb Random association of alleles at a single locus: HWE What about random association of alleles at different loci after random mating?

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Random association in gametes Alleles at A locus A(p) a(q) B (x) AB (px) aB (qx) Alleles at B locus b (y) Ab (py) ab (qy) remember: p + q =1 and x + y = 1

Consider two loci and 1 generation of random mating:

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

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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: 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:

Example:

AaBB aaBB aaBb AABb AAbb Aabb AABB aabb AaBa

We only see 1/3 after 1 generation of random mating!

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
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LD in individuals:

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

We need a new symbolism

LD in individuals:

Let’s take an AB/ab individual as an example: What types of games can the AB/ab make?

(1) AB (2) ab Parental or non-recombinant gametes (3) Ab (4) aB Non-parental or recombinant gametes

By the way, lets assume physical linage.

A B a b Physical linkage: Notation = AB/ab

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LD in individuals:

A B a b Physical linkage: Notation = AB/ab

When genes are on the same chromosome:

fAB = fab ≥ fAb = faB

f (non-recombinant) ≥ f (recombinant) Recombination fraction (r) is the proportion of recombinant gametes produced by an individual. When r = 0: fAb + faB = 100% [fAb + faB = 0%] When r = 0.5: fAb + faB = 50% [fAb + faB = 50%]

From: iGenetics

  • P. J. Russell (2002)

page 350

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LD in individuals:

When genes are on different chromosomes:

fAB = fab = fAb = faB

f (non-recombinant) = f (recombinant)

A a Un linked genes: 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: fAB = 0.38 (2) ab: fab = 0.38 (3) Ab: fAb = 0.12 (4) aB: faB = 0.12 r = 0.12 + 0.12 = 0.24

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LD in populations:

Random association in gametes Alleles at A locus A(p) a(q) B (x) AB (px) aB (qx) Alleles at B locus b (y) Ab (py) ab (qy) remember: p + q =1 and x + y = 1

faB =qx fAb = py fab = qy fAB =px

fAB + fab + fAb + faB = 1

{

{ {

4 4 8 4 4 7 6 4 4 4 8 4 4 4 7 6 3 2 1

ts recombinan ts recombinan from random at B and A together putting

  • f

prob recomb

  • f

prob ts recombinan

  • non

generation last in gametes AB

  • f

frequency AB ion recombinat no

  • f

y probabilit ' AB

) ( ) 1 ( px r f r f + − =

LD in populations:

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{

{ {

4 4 8 4 4 7 6 4 4 4 8 4 4 4 7 6 3 2 1

ts recombinan ts recombinan from random at B and A together putting

  • f

prob recomb

  • f

prob ts recombinan

  • non

generation last in gametes AB

  • f

frequency AB ion recombinat no

  • f

y probabilit ' AB

) ( ) 1 ( px r f r f + − =

) )( 1 (

AB ' AB

px f r px f − − = −

  • Obs. – exp.

faB =qx - D fAb = py - D fab = qy + D fAB = px + D ) )( 1 (

AB

px f r D − − =

= the linkage disequilibrium parameter

Non-recombinants Recombinants

Remember:

Individual AB/ab produces the following: (1) AB: fAB = 0.38 (2) ab: fab = 0.38 (3) Ab: fAb = 0.12 (4) aB: faB = 0.12 r = 0.12 + 0.12 = 0.24

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Forces that increase D in populations:

1. Migration 2. Natural selection: 3. Genetic Drift

LD in populations:

smaller) is (whichever

  • r

max

py qx D − − =

  • Comparing D among populations is difficult.
  • Standardize D as a fraction of the theoretical maximum for the popn

max

D D

4 3 4 2 1 4 3 4 2 1

t recombinan aB Ab t recombinan non aa AB

f f f f D × − × =

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LD in a population:

fs = y = 0.6920 fS = x = 0.3080 Ss blood group fN = q = 0.4575 fM = p = 0.5425 MN blood group Ns = 773/2000 = 0.3865 NS = 142/2000 = 0.0710 Ms = 611/2000 = 0.3055 MS = 474/2000 = 0.2370 Gamete frequencies

Example: blood group polymorphism in a sample of 1000 British people

D = (0.2370)(0.3865) – (0.3055)(0.0710) = 0.07

  • r

max

py qx D − − =

4 3 4 2 1 4 3 4 2 1

t recombinan aB Ab t recombinan non aa AB

f f f f D × − × = Dmax: qx = 0.14 or py = 0.37; so Dmax = 0.14 D is (0.07/0.14)*100 = 50% of the theoretical maximum

Homework:

SS = 483 Ss = 418 ss = 99 MM = 298 MN = 489 NN = 213 Ss locus MN locus

Genotype counts in the population

Use chi-square test to: 1. determine if each locus is in HWE 2. determine if gamete frequencies are in equilibrium

Ns = 773 NS = 142 Ms = 611 MS = 474 Gametes

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Recombination reduces LD:

Rate of decay of LD under various recombination rates

0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 1 9 17 25 33 41 49 57 65 73 81 89 97

generations Standardized disequilibrium D/Dmax r= 0.001 r= 0.01 r= 0.1 r= 0.5

Recombination reduces LD:

“Hitchhiking” of a mutator gene with and without recombination

Adapted from Sniegowski et al. (2000) BioEssays 22:1057-1066.

No recombination Recombination Mutator allele that increase the mutation rate Beneficial allele subject to strong positive selection

r = 0 r = 0.5

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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:

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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:
  • Admixture of populations
  • Natural selection acting on one or more of the loci
  • Inbreeding in plants that regularly undergo self-fertilization
  • 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.