Download Population Genetics 2: Linkage disequilibrium Consider two loci and

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
2
2
fAA = p
fAa = 2pq
2
faa = q
fBB = x
fBb = 2xy
2
fbb = y
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?
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Consider two loci and 1 generation of random mating:
Random association in gametes
Alleles at B locus
Alleles at A locus
A(p)
a(q)
B
(x)
AB
(px)
aB
(qx)
b
(y)
Ab
(py)
ab
(qy)
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
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Consider two loci (on different chromosomes) 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
AABb
AAbb
Aabb
AABB
aabb
AaBb
We only see 1/3 after 1
generation of random mating!
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
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LD in individuals (BIOL 2030 stuff):
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:
Let’s take an AB/ab individual as an
example:
Physical linkage:
A B
What types of gametes can the AB/ab make?
a b
By the way, lets assume physical linkage.
(1)  AB
(2)  ab
(3) Ab
(4) aB
Notation = AB/ab
Parental or non-recombinant gametes
Non-parental or recombinant gametes
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Review meiosis in a single meiocyte:
Gametes
Notation:
Parental (non-recombinant): A B
Parental
configuration
Recombinant: a B
Recombinant: A b
Parental (non-recombinant): a b
1 crossing over event: 50% parental and 50% recombinant !!!
LD in individuals:
When genes are on the same chromosome:
Physical linkage:
A B
fAB = fab ≥ fAb = faB
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: fAB + fab = 100%
[fAb + faB = 0%]
When r = 0.5: fAB + fab = 50% [fAb + faB = 50%]
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From:
iGenetics
P. J. Russell (2002)
page 350
LD in individuals:
When genes are on different chromosomes:
Un linked genes:
A
fAB = fab = fAb = faB
f (non-recombinant) = f (recombinant)
a
B
b
r = 0.5
•  when genes are on different chromosomes
•  when genes are on same chromosome and recombination is high
enough for independent assortment
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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
What do we expect for individuals if allelic association is random?
What do we expect in a population if allelic association is random?
LD in populations:
Random association in gametes
Alleles at B locus
Alleles at A locus
A(p)
a(q)
B
(x)
AB
(px)
aB
(qx)
b
(y)
Ab
(py)
ab
(qy)
remember: p + q =1 and x + y = 1
fAB =px
fab = qy
fAb = py
faB =qx
fAB + fab + fAb + faB = 1
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LD in populations:
The frequency of an AB gamete in a population has two sources:
A B
AB
•  Some individuals have this configuration: non-recombinant [parental]
•  Some individuals produce this as a recombinant configuration.
LD in populations:
f AB
'
- recombinants
recombinants
non

 

= (1 − r )
f AB
+ (
r)
px



frequency of AB
probability of
no recombination gametes in last
generation
prob of prob of putting
recomb together A and B
at random from
recombinants
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'
f AB =
- recombinants
recombinants
non

 

(1 − r )
f AB
+ (
r)
px



frequency of AB
probability of
no recombination gametes in last
generation
Obs. – exp.
prob of prob of putting
recomb together A and B
at random from
recombinants
'
f AB − px = (1 − r )( f AB − px )
D = (1 − r )( f AB − px) = the linkage disequilibrium parameter
fAB = px + D
In excess due to LD
fab = qy + D
fAb = py - D
faB =qx - D
Deficient due to LD
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
What do we expect for an individuals if association is random?
What do we expect in a population if association is random?
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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
Dmax
D = f AB × f ab − f Ab × f aB






excess due to LD
deficient due to LD
Dmax = qx or py (whichever is smaller)
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LD in a population:
Example: blood group polymorphism in a sample of 1000 British people
MN blood group
Gamete frequencies
expected
fM = p = 0.5425
MS = 474/2000 = 0.2370 (+)
px = 0.1671
fn = q = 0.4575
Ms = 611/2000 = 0.3055 (-)
py = 0.3751
nS = 142/2000 = 0.0710 (-)
qx = 0.1409
ns = 773/2000 = 0.3865 (+)
qy = 0.3166
Ss blood group
D = ef
×
ef
× ef
MS
ns − ef
Ms
nS




non recombinant
recombinant
Dmax = qx or py
€
fS = x = 0.3080
fs = y = 0.6920
D = (0.2370)(0.3865) – (0.3055)(0.0710) = 0.07
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:
Genotype counts in the population
Gametes
MN locus
Ss locus
MS = 474
MM = 298
MN = 489
NN = 213
SS = 483
Ss = 418
ss = 99
NS = 142
Ms = 611
Ns = 773
Use chi-square test to:
1. 
determine if each locus is in HWE
2. 
determine if gamete frequencies are in equilibrium
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Recombination reduces LD:
Rate of decay of LD under various recombination rates
Standardized disequilibrium
D/Dmax
1
0.9
r= 0.001
0.8
0.7
0.6
r= 0.01
0.5
0.4
0.3
0.2
r= 0.1
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.
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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:
r = 0.01 (1 centiMorgan) is about 1million bp in humans
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female-limited Batesian mimicry
5-locus linkage group
(supergene):
§  Tail length
§  Hind-wing pattern
§  Forewing pattern
§  Epaulet color
§  Body color
Selection:
Bitter tasting
Only certain complex color
morphs provide gains in
fitness
Tasty mimics
(Papilio memnon
females)
LD:
Few maladaptive patterns
produced per generation due
to linkage
male
Class I
Class III
Class II
The MHC locus on human chromosome 6
LD over 10s-100s million of bp due
to selection for combinations of loci
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Sexual reproduction reduces LD
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 said to be un-linked; such loci are very far apart on the same chromosome, or
in different chromosomes. When r < 0.5 the genes are said to be linked. When r =0 the loci are in
permanent disequilibrium.
•
Disequilibrium can arise from sources other than linkage:
o
o
o
o
•
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.
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