1. Linkage Analysis I -- Parametric
I-Ping Tu

3. Book reference
Genetic Linkage Web Resource:

4. 1 Introduction Quality Trait: e.g. tall/short, green/yellow,
affected/unaffected
Assume Genetic Model 
parametric linkage analysis
lod score method
large pedigrees
No genetic model assumption
Nonparametric linkage analysis
Affected relative pairs

5. Parametric vs. Non-parametric linkage analysis
Assume genetic model known
Non-parametric
No assumptions about the genetic model
The parametric model is more powerful when the genetic model is correctly specified.
Problem size limitations
Parametric – large pedigrees, small number of markers
Non-parametric – small pedigrees, many markers

6. Phenotype Binary Affected, unaffected, and unknown Quantitative
affected or unaffected
Left handed or right handed
Affected, unaffected, and unknown
Unknown – possibly part of the syndrome
Quantitative
Insulin resistance
Blood Pressure

7. Definitions Locus Marker Allele Position on a chromosome Marker locus
Disease locus
Marker
A measurable unit on a chromosome
Dinucleotide repeat (CA)n
Single nucleotide polymorphism(SNP)
Allele
The measurement at a marker locus
2 alleles per locus (one per chromosome)
Marker alleles
1 and 4
Allelesat the disease locus A and a

10. The recombination fraction Θ
Θ = Probability of recombination between two loci.
Θ = 0.5 if ”large” distance.
Θ < if ”short” distanc
An odd number of crossovers = recombination
An even number = no recombination

11. Haldane’s Mapping function

12. Recombination fraction – An example
No! Recombination fractions are not additive for large distances.

14. Penetrance( Gentic Model)
Probability of being affected
Penetrance parameters: f = (f0 f1 f2)
Definition: fk = Probability of being affected if you have k disease alleles k=0, 1, 2.
fk = P(affected conditional on k disease alleles) k=0, 1, 2.
fk = P(affected | k disease alleles) k=0, 1, 2.
Notation: A = Disease allele
a = Normal allele
Disease genotypes: aa, Aa, or AA

15. Penetrance continued Recessive Dominant Full p. Reduced p.
f0 = P(aff| aa)
f1 = P(aff | Aa) 1
0.8
f2 = P(aff| AA)
0.7
Dominant with
phenocopies and
reduced penetrance
Additive penetrances
f0 = 0.01
f0 = 0
f1 = 0.8
f1 = 0.4
Age dependent
penetrances
f2 = 0.8

16. Population prevalence
Kp = Proportion of affected individuals in a population = P(aff) aa Aa AA
= Affected
Disease allele frequency p = 0.05
Assume that the population is in HWE
P(aa) = (1-p)2 = =
P(Aa) = 2p(1-p) =0.095
P(AA) = p2 =
Definition of conditional probability
Kp = P(aff) = ?

17. Population prevalence contd.aa Aa AA
Kp = Area of the red square / Total area (aa + Aa + AA) =
= P(aff ∩ aa) + P(aff ∩ Aa) + P(aff ∩ AA) =
= P(aff | aa)P(aa) + P(aff | Aa)P(Aa) + P(aff | AA)P(AA) =
= f0*(1-p)2 +f1*2p(1-p) + f2*p2 =
= 0.03* * * =
The Law of Total Probability

19. Estimation of the genetic model
Segregation analysis
It is possible to estimate
mode of inheritance
number of loci contributing to a segregating phenotype.
penetrance parameters
Relative frequency (p) of the disease allele in the population
Problems?
Large population based samples required
Ascertainment bias
In parametric linkage analysis we assume that the genetic model is known.

20. 2. Parametric two-point linkage analysis
Let q be the recombination freq between the diseased gene and the observed marker.
H0: q = 0.5 VS HA: q < 0.5

21. Estimation of the recombination fraction θ
Example: N = 4 trios with affected mother and daughter
Assume : that all the 12 individuals have been genotyped for a specific DNA marker
that all the mothers are heterozygous at the marker locus
that mothers and fathers have disease genotypes (Aa) and (aa), respectively
that each daughter has inherited a disease allele from her mother
that parental marker genotypes are not identical
that the phase is known for all the mothers (unrealistic)
Data : Trio 1-3: No recombination between marker and disease locus
Trio 4: Recombination between marker and disease locus
Estimate : θ* = 1/4

22. Estimation of θ continued
Assume that all meioses can be scored unequivocally as recombinant or non-recombinant with regard to a marker locus and a disease locus
n = Number of meioses
r = Number of recombinant meioses
Estimate : θ* = r/n
Estimates above 0.5 are not relevant from a biological point of view
Definition: θ * = min(0.5, r/n)

23. The binomial distribution
The number of recombinants r among n independent meioses follows a binomial distribution.
The probability of r recombinants out of n is a function of the recombination fraction θ. Let us denote this function L(θ).
Note that L(θ) is the probability (likelihood) of the observed data if the recombination fraction is θ.
The maximum likelihood estimate (MLE) of θ is the value θ* for which L(θ) reaches its maximum.
MLE: θ*= r/n

26. Lod score history Score proposed by Haldane & Smith 1947
Newton E. Morton analysed the distribution of the lod score statistic under various assumptions
Lod scores below -2 are generally accepted as significant evidence against linkage.
Common in replicating studies.

29. More complicated situations
Phase Unknown
Marker or Disease gene homozygosity
Reduced penetrane
Varying penetrance
age, sex, phenotype, diagnostic uncertinty
Phenocopies
Missing marker data
Extended pedigrees
Pedigree loops
Multilocus genotypes

38. Recessive mode of inheritance
Prerequisites
Autosomal recessive inheritance
100% penetrance f0=f1=0, f2=1
No phenocopies
Nuclear family typed for one informative marker
All four meioses are informative

43. More complicated situations
Reduced penetrane
Varying penetrance
age, sex, phenotype, diagnostic uncertinty
Phenocopies
Missing marker data
Extended pedigrees
Pedigree loops
Multilocus genotypes

44. Lod score assignment

46. The pedigree likelihood contd.
g = (G1, G2, G3, G4) in the recessive example.
P(y|g) depends on the penetrance parameters f = (f0, f1, f2)
P(g|θ) depends on disease and marker allele frequencies
Ex: G1 in the recessive example: (1A|2a , 3A|4a)
P(g|θ) = 2pq*2p1p2 for the father
2pq*2p3p4 for the mother
θ2/4 for the affected daughter3
θ2/4 for the affecteddaughter4

47. P(g|q) P(y|g): genetic model P(g|q)=PP(gi) PP(gj|gFjgMj)
i means founder
j means non-founder
Genotypes g includes those of marker and disease genes
Missing data, multilocus markers…

49. More on missing marker data
Good estimates of the allele frequencies necessary
Assuming a uniform allele frequency distribution is usually no good idea
Bias
See e.g. Ott (1999)
Allele frequencies for markers available on Web-sites.
Genotype say 50 unrelated controls from the same population
Possible to use also alleles from individuals in the study without introducing bias.

50. Heterogeneity Allelic heterogeneity Genetic heterogeneity
Ex: Different mutations in BRCA1 will lead to the same phenotype
Genetic heterogeneity
Only a proportion of the families in a study can be explained by one disease locus.
Test for heterogeneity
Smith (1963) - The admixture test
Implemented in HOMOG (a program in the
LINKAGE package)
Estimates the proportion of linked families

53. Age-dependent penetrance contd.
Assume that a 45 year old woman comes to the clinic. What is the odds that she is a disease gene carrier?
Odds to be a diseasegene carrier indifferent age bands:
<30
1:2
30-39
1:3
40-49
1:8
50-59
1:12
60-69
1:27
70-79
1:36
Penetrance if
aa:
0.0012
Aa:
0.0235 :
150* i.e. about 1:8

54. General pedigrees The Elston-Stewart algorithm (1971)
Start at the bottom of the pedigree and solve the problem for each nuclear family.
The likelihood for each branch is ’peeled’ on the individual linking the sub-tree to the part of the pedigree

56. Two-point vs. Multipoint Linkage
Two-point linkage analysis
Analyze marker-disease co-segregation one locus at a time
One two-point lod score for each marker
IBS-sharing of a marker allele might lead to false positive lod scores if possible look at haplotypes.
Multipoint (often sliding n-point)
Regard the marker positions as fixed
Vary the location (x) of the disease locus across each sub-map of n adjacent markers.
Compare each multilocus likelihood to a likelihood corresponding to ’x off the map’ ( θ = 0.5).

58. Software Jurg Otts website at Rockefeller University
For parametric linkage analysis
LINKAGE
FASTLINK
VITESSE

59. Linkage Analysis II --Nonparametric

60. IBS or IBD The affected sibs have one allele in
1 4 42
The affected sibs have one allele in
common (4), but the 4-alleles come
from different parents.
Definition: Two alleles are said to be identical by state
(IBS) if they are of the same kind.
If two alleles have the same ancestral origin they are said to be identical by descent (IBD)
IBS-count: 1
IBS is a weaker concept than IBD
IBD-count: 0

62. Notation Let us first assume that x is the disease locus
x A fixedlocus on the genome
N = N(x) = The number of alleles shared IBD by an affected sib pair at locus x
Let us first assume that x is the disease locus

63. ASP linkage analysis Collect affected sib pairs
How many depends on the genetic effect
Power calculations
Genotype all 4 members of each pedigree
Estimate the conditional IBD probabilities
Compare with the IBD probabilities under the null hypothesis of no linkage:

65. P(N = k) k=0, 1, 2 ? Possible parental disease locus genotypes
The corresponding genotype probabilities under the assumption of HWE and
independence between the parents are:
This matrix is symmetric so it is sufficient to consider6 different mating types

66. P(N = k) k=0, 1, 2 Mating type P(Ci) C1 aa,aa q4 C2 Aa,aa 4pq3 C3
Before we go on, remember the genetic model: Recessive disease with f = (0, 0, 1)
Why? Because both affected sibs must have2 disease alleles and these pairs of alleles must be of different parental origin. ThusP((2 aff sibs| IBD=0)|Ci) = 0 for i = 1-5.
Finally we calculate the denominator P(2 aff sibs).

68. IBD probabilities for a few genetic models
Table 2.1 page 30 in the compendium
λs= Sibling relative risk = 0.25/z0 (strength of the genetic component)

71. The Maximum Lod Score (MLS)
Assumptions:
n affected sib pairs
Null hypothesis
a marker at a specific test locus x has been genotyped
perfect marker information (N = N(x) known)
H0: ~
= (0.25, 0.5, 0.25)
Alternative
H1: ~
= (z0, z1, z2) !=(0.25, 0.5, 0.25)
(a fixed alternative) 2
1 4
1 4
Pedigree number i: Ni = 2
The support for the alternative
hypothesis is
Ex: LR = 4 at the disease locus if z2=1 (recessive disease with full penetranceand no phenocopies)

72. MLS continued
Note: Both the observed IBD-count (j) and the IBD-probabilities Ψdepend on x.
n affected sib pairs
# 0 IBD = n0= no(x)
# 1 IBD = n1= n1(x)
# 2 IBD = n2= n2(x)
Combined evidence in favor of H1:
Base10

73. MLS continued The maximum lod score = is known as the MLS-score
Constrained maximization over Holman’s triangle leads to increased power.
The derivation is more complicated under incomplete marker
The MMLS-score is defined as the maximum of the MLS-scores over x.

74. NPL Score Example: Half Sib Pair
Xij,t : indicator function for i-th pair shares j copy of IBD allele
X1,t = SiXi1,t , l= recombination rate, t : trait locus
P(Xi1,t |affected half sib)=(1+ae-2l|t-t| )/2
Log-Likelihood = Xtlog(1+a)+(N-Xt)log(1-a)
Score Statistic for testing H0: a =0 is X1,t
For t unknown, we use maxtYt ,, Yt =X1,t
Remark: Yt is a Markov Chain

75. NPL = Non Parametric Linkage
The NPL Score
NPL = Non Parametric Linkage
Before we define the score let us repeat the definitions of expectation and variance :

76. The NPL score continued
Note: E(Zi) = 0 underH0
E(Zi) > 0 under H1

77. The NPL score at a locus x
Properties: E( Z(x) ) = 0 under H0
V( Z(x) ) = 1 under H0
Large NPL scores lead to rejection of H0
E( Z(x) ) > 0 under H1
E( Z(x) ) increases with the sample size under H1