Transcript Session 1

Workshop
Biostatistics in relationship testing
ESWG-meeting 2014
Daniel Kling
Norwegian Institute of Public Health and Norwegian University of
Life Sciences, Norway
Andreas Tillmar
National Board of Forensic Medicine, Sweden
http://www.famlink.se/BiostatWorkshopESWG2014.html
Outline of this first session
• Likelihood ratio principle
• Theoretical: paternity issues
– Calculation of LR “by hand”.
– Accounting for:
•
•
•
•
Mutations
Null/silent alleles
Linkage and linkage disequilibrium (allelic association)
Co-ancestry
DNA-testing to solve relationship issues
Is Mike the father of Brian?
DNA-data
vWA
D12S391
D21S11
….
12,13
22,22
31,32.2
….
Mother of Brian: 12,14
23,23.2
30,30
….
Brian:
22,23
30,31
….
Mike:
13,14
Can we use the information from the DNA-data to answer the question?
YES!
We estimate the “weight of evidence” by biostatistical calculations.
We How
will follow
recommendations
given
much the
more
(or less) probable
is it by
thatGjertson
M is theet al., 2007
father
of B, compared to
being the in
father
of B? testing.”
“ISFG:
Recommendations
onnot
biostatistics
paternity
Calculating the “Weight of evidence” in a case
with DNA-data
H1: The alleged father is the father of the child
H2: The alleged father is unrelated to the child
We wish to find the probability for H1 to be true, given the evidence
= Pr(H1|E)
Via Baye’s theorem:
Pr( H 1 | E )
Pr( H 2 | E )
Posterior odds
Today we focus on the
likelihood ratio (LR)

Pr( E | H 1 )

Pr( H 1 )
Pr( E | H 2 ) Pr( H 2 )
LR (or PI)
Pr( DNA | H 1 )
Pr( DNA | H 2 )
Prior odds
LR=
Pr( DNA | H 1 )
Pr( DNA | H 2 )
Pr( DNA | H 1 )
Pr( DNA | H 2 )
M
AF
GM
GAF
C
Founders Genotype
probabilities
Child
Transmission
probabilities
Assuming HWE
2*pi*pj*(1-Fst)
Fst*pi+ (1-Fst)*pi*pi
2*pi*pj
pi*pi
Mendelian factor
(1, 0.5, µ, etc)
GC
Pr( DNA | H 1 )  Pr( G C , G AF , G M | H 1 )  Pr( G C | G AF , G M , H 1 )  Pr( G AF , G M | H 1 )
Paternity trio – An example
|, H
)1 )G
,Pr(
GGCAFDNA
,G
, |GG
|, H
G
| ,HG
) )|H
)1 )G
GCPr(
, G|GG
, |G
,GGMM| ,H
)G
 Pr(
|G
,G
GAFAF
, ,G
|G
HMM1 )|, H 1 )
Gr(CDNA
,Pr(
G AFDNA
G M1 )| H
|H1Pr(
Pr(
Pr(
,Pr(
H1Pr(
,G
|,G
H1M)1 ),HPr(
C
AF
MAF M
1 1M
C
AFC
AF MAF
1
C
AF
M
C
AF
1
M
AF
GM=a,b
GAF=c,d
2,|G
p|McAF
pGG
)Pr(
 Pr(
DNA
Pr(
G C|DNA
,H
G1AF
) |, H
GPr(
)G
|H
Pr(
Pr(
,G
)G
DNA
Pr(
,,G
GG
| H|, H
|G)G) |Pr(
,HG
Pr(
G) G
,,HG
Pr(
| )GG,Pr(
G
G
G
H,, ,H
) ) |,Pr(
Pr(
H 1GG
) CAF
Pr(
|G
,G
GAFMAF, G
|, H
G
,)H|
0.5
M
1
C 1
AFC
MAFC 1 1MAF2  p
1Ma C Cp 1b AF AF
C M AF
1 d1MM 0.5
M 1M
C
GC=a,c
M
AF
GM=a,b
1
GAF=c,d
Pr(
DNA | H 1 )
p,cG
 MpAFd |, H
2,|G
pAF
 pGGd1MM) |, Pr(
)Pr(
 Pr(
DNA
Pr(
G C|DNA
,H
G1AF
) |, H
GPr(
)G
|H
Pr(
,G
)G
AFCPr(
,,G
G|G
|, H
|GG ) |,HG
Pr(
)G
,HPr(
| )GGPr(
G
G
H 1G
)2AF
Pr(
G
G 1M) |
Pr(
0.5
cAF,, ,H
M
1
C 1 DNA
MH
AFC 2 ) 1MAF2  p
1Ma  Cp 1b
AF
C
M
C
GC=a,c
Paternity trio – An example
Gr(C|DNA
,H
G AF
GPr(
|HPr(
Pr(
)G
DNA
Pr(
|H
| )G |Pr(
,H
,,HpG
G2,M|G
H, 1pGG
)dM) |,Pr(
H
)) C Pr(
|G
, H )  Pr( G A
G
p1G
p|AF
A
) |, H
)G
,1 G
,,G
GG
|, H
Pr(
)a CG
Pr(
| )GG,Pr(
G
G
Pr(
H 11GG
,G
GAFMAF, G
|, H
G
M
1G)1M)AF2
M
M 1M) | 1H 1 )
cAF, ,H
0.5
1
1
C Pr(
AF
C
MAF
C 1b AF AF
C
M
1 M 0.5
AF
DNA
| CH
LR=
=
1
p,cG
 MpAFd |, H
2,|G
pAF
 pGGd1MM) |, Pr(
NA
Pr(
G C|DNA
,H
G1AF
) |, H
GPr(
)G
|H
Pr(
,G
)G
AFCPr(
,,G
G|G
|, H
|GG=) |,HG
Pr(
)G
,HPr(
| )GGPr(
G
G
H 1G
)2AF
Pr(
G
G 1M) | H 1 )
Pr(
0.5
cAF,, ,H
M
1
C 1 DNA
MH
AFC 2 ) 1MAF2  p
1Ma  Cp 1b
AF
C
M
0.5
=2  p c  p d
Probability to observe a certain allele in the population
(e.g. population frequency)
pC 
xc  1
N 1
Paternity trio – real data
M
AF
GM=10,11
1
GAF=13,14.2
)Pr(
 Pr(
DNA
Pr(
G C|DNA
,H
G1AF
) |, H
GPr(
)G
|H
Pr(
,G
)G
AFCPr(
,,G
G| G
|, H
|GG1M)AF
)G
,G
HPr(
| )GGPr(
G
,, ,H
GG14
).Pr(
HG1 G
) |Pr(
G
GM,AFG|,MH
G, 1MH) 1|
2CM,|G
pG
Pr(
H
Pr(
G
| 13
H
G, AF
2|,HGPr(
p110
0.5
0.5
M
1
C 1 DNA
M
AFC1 ) 
M
111 ,G
AF
AF
MAF1 )p
1MM
2|, Pr(
C  CpAF
C AF
C
GC=10,14.2
M
AF
GM=10,11
Pr(
DNA | H 1 )
GAF=13,14.2
H 1 )Pr(
 Pr(
DNA
G C| ,H
G1AF
) , GPr(
G
|H
,G
) AF Pr(
, G| G
| G1 )AF
| )G Pr(
G
,, H
G 1M2
H
) AF
2 ,G
p13AF
p131 G
p14, .G2 M | H 1 )
Pr(
HC|2 )H
2, GPr(
p10M G, HCp11
0.5
M
C 1 DNA
M
1  AF
M  p14
.)2 | Pr(
C
GC=10,14.2
0.5
LR=
2  p13  p14.2
Paternity trio - Mutation
M
AF
GM=10,11
GAF=13,15.2
Mutation?
G
Ar(C |DNA
,H
G 1AF) |
,CG
HPr(
) |G
HPr(
Pr(
, )G
GAFCPr(
,,G
G|G
|,|H
GG )2Pr(
|,H
G
Pr(
)G
,G
Hp
Pr(
|)G
GPr(
G
,G
,HG15
)2|, Pr(
H
HG11)G
) | Pr(
G, G|, H
G, H) | )H Pr(
) G AF
2CM,|G
pG
pG
H
| AF
H
G, G
0.5
M
1
C 1 DNA
M
AF
C 1 ) 1MAF
1
M
C AF
1 ,G
AF
MAF1 )p
M
1 M.Pr(
1
13
10
11
C
C AF AF MAF M 1M 1
GC=10,14.2
Mutations
“The possibility of mutation shall be taken into account whenever a
genetic inconsistency is observed”
Brinkmann et al (1998) found 23 mutations in 10,844 parent/child offsprings.
Out of these 22 were single step and 1 were two-step mutations.
AABB summary report covering ~4000 mutations showed approx 90-95% single
step, 5-10% two-step and 1% or less more than two steps.
Note that these are the observed mutation rate, not the actual mutation
rate (hidden mutations, assumption of mutation with the smallest step.
Mutation rate depend on marker, sex (female/male), age of individual,
allele size
Mutations cont
Allele: 9
10
11
-3
-2
-1
µ-3
µ-2
µ-1
12
0
1- µtot
13
14
15
+1
+2
+3
µ+1
µ+2
µ+3
µtot=µ+1+µ-1+µ+2+µ-2+…..
Approaches to calculate LR accounting for mutations:
LR~(µtot*adj_steps)/p(paternal allele) e.g “stepwise mutation model”
LR~µ/A
(A=average PE)
LR~µ
LR~µ/p(paternal allele)
Paternity trio - Mutation
M
AF
GM=10,11
( mut 13  14 .2  mut 15 .2  14 .2 )
GAF=13,15.2
C
Mutation model decreasing with range
( mut 13  14 .2  mut 15 .2  14 .2 )
GC=10,14.2
Total mutation rate
mut 15 .2  14 .2 

“Stepwise decreasing
with range”
  0 .5  µ Tot
50% for 15.2 to
be transmitted
50% are loss of
fragment size
 0 .9  0 .5
90% of mutations
are 1 step
Paternity trio - Mutation
M
AF
GM=10,11
GAF=13,15.2
(Gmut
 )mut
)
G
Ar(C |DNA
,H
G 1AF) |
,CG
HPr(
) |G
H
Pr(
, )G
GAFCPr(
,,G
G
|,|H
GG )2
|,H
G
) ,G
|)G
GPr(
G
,H
|, Pr(
H
HG11C)G
) |AF
Pr(
,AF
G
G13,MAF
|M, H
| H 1.2) 14G.2AF
pPr(
GHpCPr(
2CM,|G
p|G13
, )G
pG
Pr(
|G
H
Pr(
G
H
15M1 M).Pr(
G
 14
.G
2 , 1H
M
1
C 1 DNA
M
AF
C1 ) 1MAF
1
M
1 ,G
AF
AF
MAF
M 1 ) 15Pr(
10
11
2 0.5
C
AF
1
mut 15 .2  14 .2 
GC=10,14.2
M

  0 .5  µ Tot
 0 .9  0 .5
AF
Pr(
DNA | H 1 )
GAF=13,15.2
GM=10,11
H 1 )Pr(
 Pr(
DNA
G C| ,H
G1AF
) , GPr(
G
|H
,G
) AF Pr(
, G| G
| G1 )AF
| )G Pr(
,G
G AF
,, H
G )2 | Pr(
H
) AF
p131 G
p14, .G2 M | H 1 )
2, GPr(
p10M G, HCp11
Pr(
HC|2 )H
0.5
M
C 1 DNA
M
1  2
AF  p 13
M  p 151M.2
C
t 15 .2  14 .2 

GC=10,14.2
  0 .5  µ Tot
LR=
 0 .9  0 .5
2  p13  p14.2
Paternity trio – Silent allele
M
AF
GM=10,11
GAF=17.2
C
GC=10 ,s
,s
GAF=17.2,17.2
GC=10,10
GAF=17.2, s
GC=10,s
Null/silent allele
“STR null alleles mainly occur when a primer in the PCR reaction fails to
hybridize to the template DNA...”
“Laboratories should try to resolve this situation with different pairs of STR
primers”
“…no data from which to estimate the null allele frequency exists, consider a
generous bracket of plausible values for the frequency and compute a
corresponding range of values for the PI. If the resulting range of case-specific
PI values are all large (as large as a laboratory’s threshold value for issuing
non-exclusion paternity reports), then report that the PI is ‘‘at least greater
than the smallest value’’ in the range”
Impact of null/silent allele may vary on the pedigree and the genotype
constellations
Paternity trio – Silent allele
M
AF
GM=10,11
GAF=17.2
GAF=17.2,17.2
GC=10,10
GAF=17.2, s
GC=10,s
Pr( DNA | H 1 )  2Pr(
, G | H )  Pr( G C | 0G. 5AF , G M , H 1 )  Pr( G AF
 pG10C ,GpAF
11  2M  p 17 1.2  p s  0 . 5
C
GC=10
M
AF
GM=10,11
C
G =17.2
Pr( DNA AF
| H1)
Pr( DNA | H 2 )
GC=10
LR=
2  p10  p11  ( 2  p 17 .2  p s
 p 17 .2  p 17 .2 )  0 . 5  ( p 10  p s )
2  p 17 .2  p s  0 . 5  0 . 5
( 2 ( p217 .2p 17
 .p2 s p s p17 .2p 17
 .p2 17 .2p)17.20). 5 0 .(5p10( p10 ps )p s )
Linkage and Linkage disequilibrium
• Linkage
– Can be described as the co-segregation of closely
located loci within a family or pedigree.
– Effects the transmission probabilities!
• Linkage disequilibrium (LD)
– Allelic association.
– Two alleles (at two different markers) which is
observed more often/less often than can be expected.
– Effects the founder genotype probabilities, not the
transmission probabilities!
Marker 1 Marker 2
Mother: 16,18
21,24
If LD, different probabilities for each phase
Chr3:1 Chr3:2
16
Chr3:1 Chr3:2
18
18
16
21
24
Distance R, Theta
21
24
If linkage, different probability of transmission
18
16
18
16
21
21
24
24
Mother: 16,18
21,24
Child1: 16,25
21,29
Child2: 16,22
24,30
Chr3:1 Chr3:2
Chr3:1 Chr3:2
16
25 paternal
25
16
21
29 paternal
21
29
Chr3:1 Chr3:2
Chr3:1 Chr3:2
16
22 paternal
22
16
24
30 paternal
24
30
Linkage and LD when it comes to
forensic DNA-markers?
• 14 pair of autosomal STRs < 50cM
– Phillips et al 2012
– vWA-D12S391 (around 10cM, i.e. 10% recomb rate)
• Linkage will have an effect on LR for some pedigrees
–
–
–
–
Gill et al 2012, O’Connor & Tillmar 2012, Kling et al, 2012.
Incest.
Sibling, uncle-nephew.
Impact of linkage will depend on
• Pedigree
• Markers typed
• Individuals’ DNA data
• LE for vWA-D12S391!
– Gill et al 2012, and O’Connor et al 2011.
Linkage cont.
• Assuming LE (no allelic association)
– Gill et al 2012: "...linkage has no effect at all in a
pedigree unless:
• at least one individual in the pedigree (the central individual) is
involved in at least two transmissions of genetic material…, and
• that individual is a double heterozygote at the loci in question."
• Thus…
– For standard paternity cases (father vs unrelated),
linkage has no impact on LR
– BUT for paternity vs uncle (or other close related
alternative), linkage has impact on LR
Presenting the evidence
• Likelihood ratio (LR) or Paternity index (PI)
• Posterior probability (must set priors)
Li=Likelihood under hypothesis i
πi=Prior probability for hypothesis i
Simplifies to LR/LR+1,
assuming two hypotheses with equal priors
Subpopulation correction
 What is subpopulation effects?
 First developed by Wright in (1965)
 Balding (1994)
 Hardy Weinberg equilibrium
 How to estimate it?
 Reasonable values (0.01-0.05)
 Sampling formula (Fst = θ)
p '( a i ) 
F st * n a i  (1  F st ) * p ( a i )
1  ( N O bs  1) * F st
24
Example 1. Genotype probabilities
 HWE
 Homozygouts: p(A)p(A)
 Heterozygots: 2p(A)p(B)
 HWD
 Homozygouts: Fst*p(A) + (1-Fst)*p(A)p(A)
 Heterozygouts: (1-Fst)*p(A)p(B)
P ( A , A )  P (Sam pling tw o A:s)=
 * 0  (1   ) p ( A )  *1  (1   ) p ( A )
*
1  (0  1)
P ( A , B )  P (Sam pling one A and one B)=
1  (1  1)
  p ( A )  (1   ) p ( A )
 * 0  (1   ) p ( A )  * 0  (1   ) p ( B )
1  (0  1)
25
*
1  (1  1)
2
 (1   ) p ( A ) p ( B )
Example 2. Random Match Prob.
 H1: The profiles G1 and G2 are from the same individual
 H2: The profiles G1 and G2 are from different individuals
Let G1=A,A and G2=A,A
 HWE:
P ( D ata | H 1 )
1 / RM P 
P ( D ata | H 2 )

P ( G 1)
P ( G 1) * P ( G 2)

1
p ( A)
2
 HWD:
1 / RM P 
P ( D ata | H 1 )
P ( D ata | H 2 )

P ( G 1)

 Sam pling 
P ( G 1, G 2)
(1   )(1  2 )
 2
 (1   ) p ( A )  3  (1   ) p ( A ) 
26
P (S am pling tw o A :s)
P (S am pling four A :s)

Example 3. Paternity
H1: The alleged father (AF) is the real father
H2: AF and the child are unrelated.
p(A) = 0.05
Standard paternity case. First marker
AF
A/A
mother
B/B
child
A/B
27
Standard paternity case. First marker
AF
A/A
mother
B/B
child
A/B
LR 
probability of data given A F father
probability of data given A F unrelated

P ( child | m other , A F )
  0 
P ( child | m other )
LR 
1
pA
P ( m other ) P ( A F ) P ( child | m other , A F )
1

  0 
P ( A F ) P ( m other ) P ( child | m other )

1
P (S am pling the third A )

1
2  (1   ) p A
 20.
0.05
P (S am pling tw o A :s and tw o B :s)
P (S am pling three A :s and tw o B :s)

1  (4  1)
28
1  3
2  (1   ) p A

LR 
P ( m other ) P ( A F ) P ( child | m other , A F )
  0 
P ( A F ) P ( m other ) P ( child | m other )
1


P (S am pling the third A )
1
2  (1   ) p A
1  (4  1)
P (S am pling tw o A :s and tw o B :s)
P (S am pling three A :s and tw o B :s)

1  3
2  (1   ) p A
LR as a function of Theta
25
20
LR
15
10
5
0
0
0.05
0.1
0.15
Theta
0.2
0.25
29
0.3

Example 4. Siblings
H0: Two persons are full siblings
H1: The same two persons are unrelated
 The two persons are homozygous A,A
(p(A)=0.05)
 We use the “IBD method”
LR 
P ( D ata | H 0)
  0 
P ( D ata | H 1)

P (S am pling tw o A :s)*P (IB D = 2|H 0)+ P (S am pling three A :s)*P (IB D = 1|H 0)+ P (S am pling four A :s)*P (IB D = 0|H 0)
P (S am pling four A :s)

0.25  0.5 P (S am pling the third A )  0.25 P (S am pling the last tw o A :s)
P (S am pling the last tw o A :s)
0.25  0.5

2  (1   ) p A
 0.25
1  (2  1)
2  (1   ) p A
1  (2  1)
*
2  (1   ) p A
1  (2  1)
3  (1   ) p A
*
3  (1   ) p A
1  (3  1)
1  (3  1)
30


Example 4. Siblings
LR as a function of Theta
120
100
LR
80
60
40
20
0
0
0.05
0.1
0.15
Theta
31
0.2
0.25
0.3
Example 5. General relationships
 We have N number of founder genotypes
 If Fst!=0 these are not independent
 We use the sampling formula to calculate
updated allele frequencies
 It is more complex to do by hand
 In simple pairwise relationships use
the IBD method (See Example 4)
 Otherwise, use validated software!
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