Linkage disequilibrium LD mapping

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Linkage disequilibrium (LD) mapping

• Also looks at common inheritance but in
  populations of unrelated individuals No
  pedigrees required
• Fine mapping with dense markers at least every 60
  kb
• Beyond this distance loci are generally in
  linkage equilibrium
• Also called Association Mapping
• Typically, used to obtain a finer-grained map
  after coarse mapping by linkage analysis

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Quantifying association with contingency tables
Observed o
Expected e
R1
R2
C1
C2
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Quantifying association with contingency tables
Observed o
Expected e
50
50
10
90
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Linkage and LD analysis in tandem
Figure 12.2 from Primrose and Twyman
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LD mapping elucidates our evolutionary origins

• In Northern European populations, LD extends for
  60kb
• In a Nigerian African population, LD extends for
  5kb, a much shorter distance
• What do we conclude from these findings?

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Resources for genetic mapping

• CEPH- (Centre dEtude du Polymorphisme Humaine)
  Large database of pedigrees/nuclear families
• dbSNP- database of SNPs found in the human genome
• Successfully mapped Type I diabetes, cystic
  fibrosis, Breast cancer (BRCAI and II), Crohns
  disease, etc.

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Haplotype mapping

• A haplotype is a pattern of SNPs in a contiguous
  stretch of DNA
• Due to linkage disequilibrium, SNPs are typically
  inherited in discrete haplotype blocks spanning
  10-100kb
• Greatly simplifies LD analysis, because rather
  than screen all SNPs in a region, we just need to
  screen a few and the rest can be inferred
• A complete human haplotype map is still underway

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Example haplotype map
Figure 12.4 from Primrose and Twyman
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Map functions

• A genetic map function M gives a relation rM(d)
  connecting recombination fractions r and genetic
  map distances d.
• The simplest map function (used by TH Morgan) is
  rd.
• However, this function only applies if the case
  if the chance of multiple crossovers is
  negligible.
• For d gt 0.1 (distances gt 10cM), multiple
  crossovers do occur leading to r lt d.
• This happens because even numbers of crossovers
  cancel each other out to produce parental types,
  not recombinants.

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Haldanes Map Function

• Haldanes function gives a correction using
  Poisson statistics.
• Denote the distribution of crossover points in
  the interval d by (p0, p1, p2, p3,) where pk is
  the probability that exactly k crossovers occur
  within the interval. Assume this is Poisson.
• The recombination fraction is the probability of
  an odd number of crossovers
• r p1 p3 p5
• While the map length is
• d p1 2p2 3p3

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The Poisson function
d 1d 4d 10
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Haldanes Map Function

• The recombination fraction is the probability of
  an odd number of crossovers
• r p1 p3 p5
• If p follows Poisson inter-arrival times
• pk e-ddk / k! with E(k) d
• Solving for r we obtain
• r e-dd e-dd3/3! ½ (1 e-2d)
• Rearranging as a function of d d ½ ln(1-2r)

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Genetic phase

• Haplotype alleles received by an individual
  from one parent
• Phase For a doubly heterozygous individual A/a
  B/b, whether the A allele was received in the
  same haplotype as the B or b allele.

PHASE KNOWN
PHASE UNKNOWN
A B
a b
A B
a b
A B
A B
a b
a b
A B
a b
A B
a b
A B
a b
A B
a b
or
Could be
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An exampleMorgans Fly Experiments

• One gene affects eye color(pr, purple, and pr,
  red)The other affects wing length(vg,
  vestigial, and vg, normal).
• Morgan crossed pr/pr vg/vg flies with pr/pr
  vg/vg and then testcrossed the doubly
  heterozygous F1 femalespr/pr vg/vg ?
  pr/pr vg/vg ?.
• Because one parent (tester) contributes gametes
  carrying only recessive alleles, the phenotypes
  of the offspring reveal the gametic contribution
  of the other, doubly heterozygous parent.

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The test cross format
P pr/pr vg/vg pr/pr vg/vg
F1 pr/pr vg/vg
Tester pr/pr vg/vg pr/pr vg/vg
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Reverse phase experiment
P pr/pr vg/vg pr/pr vg/vg
F1 pr/pr vg/vg
Tester pr/pr vg/vg pr/pr vg/vg
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Another example showing the importance of phase
information
1
2
No Disease
HC/Y
HC/hc?
Colorblind
Colorblind Hemophilia
1
2
HC/Y
HC/hc
1
2
3
4
5
6
HC
hc/Y
Hc/Y
HC/Y
hc/Y
hc/Y
HC/Y
What is the genetic distance between these genes?
Could this computation be done without the
grandparents?
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SNPs and Pharmacogenomics

• Refers to the complete list of genes that
  determine the overall efficacy and toxicity of a
  drug
• Tries to account for all genes that influence
• Drug metabolism
• Drug transport/export
• Receptors
• Signaling pathways, etc.
• Your genotype would allow a physician to
  determine the optimal dose and medication for
  optimal therapy
• Pharmas are spending a lot of money to discover
  clinically relevant SNPs

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Population Genetics 101Measuring Genetic
Variation

• Hardy-Weinberg equilibrium (HWE)
• Genotype frequencies depend only on gene
  frequencies
• pA frequency of allele A
• pB frequency of allele B
• P(A/A) pA2 P(A/B) pB2 P(A/B) 2pApB
• pA pB 1
• pA2 2pApB pB2 1

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Population Genetics 101Measuring Genetic
Variation

• Observed vs. expected heterozygosity
• Ho Observed fraction of heterozygous
  individuals
• He Expected fraction based on allele
  frequencies
• The frequency f(X) of allele X is the fraction of
  times it occurs over all loci (2 per individual)
• He 1 the probability of homozygosity
• 1 f2(X) f2(Y) for all alleles
  (X,Y,)

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Example 10 Unique Genotypes(in bp lengths of
microsatellite)
Ho 0.30 He 0.69
H 1 high diversity H 0 asexual
mitotic reproduction Ho ltlt He indicates
selective pressure or non-random mating
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Components of the genetic model

• Components of the genetic model include
  inheritance pattern (dominant vs. recessive,
  sex-linked vs. autosomal), trait allele frequency
  (a common or rare disease?), and the frequency of
  new mutation at the trait locus.
• Another important component of the genetic model
  is the penetrance of the trait allele. Knowing
  the penetrance of the disease allele is crucial
  because it specifies the probability that an
  unaffected individual is unaffected because he's
  a non-gene carrier or because he's a
  non-penetrant gene carrier. The frequency of
  phenocopies is an important component, too.
• Rough estimates of the disease allele frequency
  and penetrance can often be obtained from the
  literature or from computer databases, such as
  Online Mendelian Inheritance in Man
  (http//www3.ncbi.nlm.nih.gov/Omim/). Estimates
  of the rate of phenocopies and new mutation are
  frequently guesses, included as a nuisance
  parameter in some cases to allow for the fact
  that these can exist.
• Linkage analysis is relatively robust to modest
  misspecification of the disease allele frequency
  and penetrance, but misspecification of whether
  the disease is dominant or recessive can lead to
  incorrect conclusions of linkage or non-linkage.

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Steps to linkage analysis

• In pedigrees in which the genetic model is known,
  linkage analysis can be broken down into five
  steps
• State the components of the genetic model.
• Assign underlying disease genotypes given
  information in the genetic model.
• Determine putative linkage phase.
• Score the meiotic events as recombinant or
  non-recombinant.
• Calculate and interpret LOD scores.
• Let's take a look at each of these steps in
  detail.

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State the components of the model

• In this example, the disease allele will be
  assumed to be rare and to function in an
  autosomal dominant fashion with complete
  penetrance, and the disease locus will be assumed
  to have two alleles
• N (for normal or wild-type)
• A (for affected or disease)

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Assign underlying disease genotypes

• The assumption of complete penetrance of the
  disease allele allows all unaffected individuals
  in the pedigree to be assigned a disease genotype
  of NN. Since the disease allele is assumed rare,
  the disease genotype for affected individuals can
  be assigned as AN.

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Determine putative phase

• Individual II-1 has inherited the disease trait
  together with marker allele 2 from his affected
  father. Thus, the A allele at the disease locus
  and the 2 allele at the marker locus were
  inherited in the gamete transmitted to II-1.
  Once the putative linkage phase (the disease
  allele "segregates" with marker allele 2) has
  been established, this phase can be tested in
  subsequent generations.

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Score the meiotic events asrecombinant (R) or
non-recombinant (NR)

• There are four possible gametes from the
  affected parent II-1 N1, N2, A1, and A2. Based
  on the putative linkage phase assigned in step 3,
  gametes A2 and N1 are non-recombinant.

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Calculate LOD scores

• In this example, the highest LOD score is -0.09
  at q 0.40. At no value of q is the lod score
  positive, let alone gt3.0, so this pedigree has no
  evidence in favor of linkage between the disease
  and marker loci.