likelihood ratios with x examples. - familias · likelihood ratios with x examples. workshop...

Likelihood ratios with X examples.

Workshop program: Familial testing, X-files, FamLinkX

XXI Jornadas del GHEP-ISFG, 06-09-2016, Bayahíbe, República Dominicana

Nádia Pinto IPATIMUP/i3S, Porto, Portugal [email protected]

mailto:[email protected]

Human Karyotype

Female Male

2 copies 1 copy



M F

X Chromosomal markers

M F

Autosomal markers

C

Chromosomal Transmission

C



M F

1 1/2

D

M F

Autosomal markers

1/2 1/2

D




X & Y Chromosomal markers

M F

X & Y Chromosomal markers

1 1/2

M F

Autosomal markers

1/2 1/2

S S




2. FORMALIZING GENEALOGIES UNDER THE PERSPECTIVE OF X-CHROMOSOMAL MARKERS

Autosomal

X X X X X

X X

X X X

X

X

Y Y

Y

Y

Heterosomal

Grandparents

Parents

Grandchildren




2. FORMALIZING GENEALOGIES UNDER THE PERSPECTIVE OF X-CHROMOSOMAL MARKERS




Autosomal

X X X X X

X X

X X X

X

X

Y Y

Y

Y

Heterosomal

Grandparents

Parents

Grandchildren

M F


1

D

M F

Autosomal markers

1/4≤P≤1/2 0 ≤P≤1/4

C

1/4≤P≤1/2




M F


1

D

M F

Autosomal markers

1/4≤P≤1/2 0 ≤P≤1/4

C

0≤P≤1/4




F

1

D

F

1

S

.

.

. X

.

.

. X

Paternal X & Y Chromosomal Transmission



Verbally:

Posterior odds = Prior odds × LR

Mathematically (the proof follows from Bayes theorem):

2

1

2

1

02

01

|

|

|

|

HGP

HGP

HP

HP

GHP

GHP

O

O

genetic configuration observed :OG

individuals linked by genetic relationship R1 :1H

:2H individuals linked by genetic relationship R2

Statistical Analyses



Assuming the hypotheses as a priori equally likely…

2

1

2

1

02

01

|

|

|

|

HGP

HGP

HP

HP

GHP

GHP

O

O




= 1




Assuming the hypotheses as a priori equally likely…

2

1

02

01

|

|

|

|

HGP

HGP

GHP

GHP

O

O




Numerically: Posterior odds = LR




2

1

|

|

HGP

HGPLR

O

O

Genetic configuration observed :OG



R1 and R2 are mutually exclusive and exhaustive

n

k

kfinal LRLR1

n: number of analyzed unassociated loci/set of loci

likelihood ratio associated with locus/set of loci k :kLR

Assuming independent alleles/haplotypes in different markers:




Alleles IBD IBS

No Yes

No No

Yes Yes

No Yes

Two alleles are called identical-by-descent (or IBD) if they have descended from a single ancestral allele.

Example:

M

C

1 2 1 3

F

1

Note that IBD is always measured with respect to some “founding” group of ancestors that are assumed to be unrelated.

2

1 F

1

1

1

C

1 F

1

2

1

F

F

C

C

C

Identity-by-descent (IBD) ≠ Identity-by-state (IBS)

IBD

IBS IBD

IBS , unless mutation

Two individuals are said to be related if the allele of one can be IBD to that of the other.

AUTOSOMAL REVIEW



Patterns of IBD for two individuals

Δ4 I1:

I2:

Δ5 I1:

I2:

Δ6 I1:

I2:

Δ2 I1:

I2:

Δ3 I1:

I2:

Δ1 I1:

I2:

Δ7 I1:

I2:

Δ9 I1:

I2:

Δ8 I1:

I2:

Jacquard’s coefficients (1970)

AUTOSOMAL REVIEW



Inbred individual C (Example):

A and B First Cousins

B

C

A




B

C

A


Alleles:




... ...

B

C

A


Alleles:

... ...




... ...

1/2 1/2

1/2 1/2

1/2 1/2 B

C

A


Alleles:

... ...




... ...

1/2 1/2

1/2 1/2

1/2 1/2 B

C

A


Alleles:

42

16

P(C is inbred) =

... ...




... ...

1/2 1/2

1/2 1/2

1/2 1/2 B

C

A


Alleles:

42

16

P(C is inbred) =

= %25,62

14

... ...

P(C being autozygous) = 6.25%



Δ4 I1:

I2:

Δ5 I1:

I2:

Δ6 I1:

I2:

Δ2 I1:

I2:

Δ3 I1:

I2:

Δ1 I1:

I2:

k2 I1:

I2:

k0 I1:

I2:

k1 I1:

I2:

Jacquard’s coefficients (1970)


AUTOSOMAL REVIEW

An individual is called inbred if he / she can carry

two identical-by-descent alleles at a locus.

An individual is called (autosomally) inbred if her

/ his parents are related.



k2 I1:

I2:

k0 I1:

I2:

k1 I1:

I2:

M F

MT1 MT2

Identical Twins

k2=1; k1=0; k0=0;

M F

C

Parent - Child

k2=0; k1=1; k0=0;

M F

C

Unrelated

k2=0; k1=0; k0=1;


AUTOSOMAL REVIEW



M F

C1 C2

M F

C1 C2

M F

C1 C2

C1:

C2:

C1:

C2:

C1:

C2:

IBD patterns: Example

k2=1/4 k1=1/2 k0=1/4

AUTOSOMAL REVIEW



Joint Genotypic Probabilities

AUTOSOMAL REVIEW


Sample 1: AiAk

Sample 2: AiAk

H0 k2 k1 k0 P(G=AiAk,AiAk|H0)

Identical twins 1 0 0

Parent-child 0 1 0

Unrelated 0 0 1

Full-siblings 1/4 1/2 1/4

Joint Genotypic Probabilities: Example

AUTOSOMAL REVIEW



JACQUARD’S COEFFICIENTS (1970)

DESAFIO TEÓRICO DE PARENTESCO – GHEP 2016

P3

P1

P4

P2

M1

F1 F2

M2

M4

M3

Δ9 = 12/64

Δ4 I1:

I2:

Δ5 I1:

I2:

Δ6 I1:

I2:

Δ2 I1:

I2:

Δ3 I1:

I2:

Δ1 I1:

I2:

Δ7 I1:

I2:

Δ9 I1:

I2:

Δ8 I1:

I2:

Δ8 = 30/64 Δ7 = 15/64

Δ6 = 1/64 Δ5 = 2/64 Δ4 = 1/64

Δ3 = 2/64 Δ2 = 0 Δ1 = 1/64

JACQUARD’S COEFFICIENTS (1970)


Δ9 = 28/64

Δ4 I1:

I2:

Δ5 I1:

I2:

Δ6 I1:

I2:

Δ2 I1:

I2:

Δ3 I1:

I2:

Δ1 I1:

I2:

Δ7 I1:

I2:

Δ9 I1:

I2:

Δ8 I1:

I2:

Δ8 = 32/64 Δ7 = 0

Δ6 = 0 Δ5 = 0 Δ4 = 2/64

Δ3 = 2/64 Δ2 = 0 Δ1 = 0

H2 H1

H2 H1 H1 H2

H0 H1 Δ1= 0

Δ2= 0

Δ3= 2/64

Δ4= 2/64

Δ5= 0

Δ6= 0

Δ7= 0

Δ8= 32/64

Δ9= 28/64

JOINT GENOTYPIC PROBABILITIES - WEIR ET AL., NATURE REVIEWS GENETICS 2006, 7:775


Δ1= 1/64

Δ2= 0

Δ3= 2/64

Δ4= 1/64

Δ5= 2/64

Δ6= 1/64

Δ7= 15/64

Δ8= 30/64

Δ9= 12/64

Relationships k2 k1 k0

Identical twins 1 0 0

Parent-child 0 1 0

Full-siblings 1/4 1/2 1/4

Half-siblings Avuncular

Grandparent-Grandchild 0 1/2 1/2

First Cousins Great-grandparent / Great-grandchild

0 1/4 3/4

Unrelated 0 0 1

IBD partitions: Examples



Autosomal kinship classes: Example

A

B

H1: Avuncular H2: Half-siblings

A

B

H3: Grandparent / Grandchild

B A

≡ ≡

k0 = k1 = 1/2; k2=0




A

B


A

B


B A ≡ ≡

k0 = k1 = 1/2; k2=0

G k2 k1 k0 P(G|H)

AiAi, AiAi

1/4 1/2 1/4 AiAi, AkAk

AiAk,AiAk

… …




A

B


A

B


B A ≡ ≡

k0 = k1 = 1/2; k2=0

For any genotypic configuration G and alternative hypotheses:

The hypotheses are equally likely and, therefore, theorectically indistinguishable.



Autosomal kinship classes: (Sub-)Example

Inheritance claiming

W

M ?

$





W

M ?

$

M W

vs M

W





W

M ?

$

M W

vs M

W

Theorectically indistinguishable by unlinked autosomal markers (despite the number of analysed markers)



F2

F1

X X

X X

1/2

X X

X

X X

F1

F2

X X

X

X X

F1

F2

X

1/2 1/2

IBD X-patterns among two females



F2

X

F1

X

X

X X

1 1

F1 F2

X X X

X X X X

1/2

1/2 1 1

1 1/2

F1

X X X

F2

X X X

X X

1

1/2 1/2

F1 F2

X X X

X X X X

1/2

1/2 1 1

4.2 HIPÓTESIS - INDIQUE LAS DOS HIPÓTESIS EMPLEADAS


Total de respostas: 54 labs

23 labs

H0: H1 y H2 son hermanas de padre y madre H1: H1 y H2 son hermanas de madre

H0: Daughter 1 and Daughter 2 are half sisters H1: Daughter 1 and Daughter 2 are full sisters

Vs.

X X X

5 Labs H0: Hija1 e Hija2 comparten el mismo padre biológico H1: Hija1 e Hija2 tienen diferente padre biológico H0: Probabilidad de que la hija 1 y la hija 2 sean hijas de diferente padre H1: Probabilidad de que la hija 1 y la hija 2 sean hijas del mismo padre

Vs.

H0: Filha 1 e Filha 2 são filhas da mesma mãe e do mesmo pai (Full siblings) H1: Filha 1 e Filha 2 são filhas da mesma mãe mas não do mesmo pai (Half siblings)



Total de laboratórios que indicam existir parentesco entre mãe/pai: 26 labs

H1 H2

H2 H1

3 labs: Ho primos / H1 primos 2 labs: Ho/H1 primos

H0: La hija 1 y la hija 2 son hermanas de padre y madre, y estos son primos H1: La hija 1 y la hija 2 son hermanas de la misma madre y distinto padre

8 labs: Ho primos en primer grado 9 labs: Ho/H1 primos en primer grado

H0: Daughter 1 and Daughter 2 are full sisters (mother and first-degree cousin are the parents) H1: Daughter 1 and Daughter 2 are half sisters (mother and first-degree cousin are the parents of Daughter 1)

1 lab: Ho primos en primer grado via materna 1 lab: Ho/H1 primos en primer grado via materna

Una mujer ha tenido una hija (Hija 1) de un primer matrimonio con un primo en primer grado (hijo de su tío materno, hermano de su madre).


H0 - Hija 1 e Hija 2 são irmãs inteiras e a avó materna é irmã inteira do avô paterno. H1 – Hija 1 e Hija 2 são meias-irmãs por via materna e a avó materna é irmã inteira do avô paterno de Hija 1 e o pai de Hija 2 é não aparentado.

H0 - Hija1 e Hija2 son hermanas biológicas por linea paterna, siendo ambas hermanas por linea materna y padre hijo del tío materno de madre. H1 – Hija2 NO es hermana por linea paterna de hija1, siendo ambas hermanas por linea materna, hija1 hija de P y P hijo del tío materno de madre.

H2 H1 H1 H2


Labs. 575389 y 575456

F1:

F2:

F1:

F2:

F1:

F2:

F1:

F2:

F1:

F2:

ΛFF1 F1:

F2:

F1:

F2:

F1:

F2:

F1:

F2:

ΛFF2 ΛFF

3

ΛFF4 ΛFF

5 ΛFF6

ΛFF7 ΛFF

8 ΛFF9

Patterns of IBD for X Chromosome markers and two females



F1:

F2:

F1:

F2:

F1:

F2:

F1:

F2:

F1:

F2:

ΛFF1 F1:

F2:

F1:

F2:

F1:

F2:

F1:

F2:

ΛFF2 ΛFF

3

ΛFF4 ΛFF

5 ΛFF6

ΛFF7 ΛFF

8 ΛFF9

A female is called X inbred if she can carry two

identical-by-descent X alleles at a locus.




... ...

1/2 1/2

1/2 1/2

1/2 1/2 B A


Alleles:

42

16

P(C is inbred) =

= %25,62

14

... ...

P(C is autozygous) = 6.25%

Inbreeding from Autosomal and X-Chromosomal perspective



Autosomal Transmission

C

X ...

1 1

1/2 1

1/2 1 B A


Y Alleles:

X

Inbreeding from Autosomal and X-Chromosomal perspective


X-Chromosomal Transmission


Y ...

Y X

X

X

X X X

X

1/2 1

1/2 1 B A

Y Alleles:

X

Y

Y X

X

X

X X X

X

1/2

X

1/2

P(inbreeding) = 0

C C

X X


F1:

F2:

F1:

F2:

F1:

F2:

xFF2 xFF

1 xFF0

M F

Identical Twins

k2=1; k1=0; k0=0;

AUTOSOMAL REVIEW



F1:

F2:

F1:

F2:

F1:

F2:

xFF2 xFF

1 xFF0

M

X X X

F

X X X X

Female Identical Twins

x2=1; x1=0; x0=0;

X CHROMOSOMAL

M F

Identical Twins

k2=1; k1=0; k0=0;

AUTOSOMAL REVIEW



F1:

F2:

F1:

F2:

F1:

F2:

xFF2 xFF

1 xFF0

AUTOSOMAL REVIEW

Parent - Child

k2=0; k1=1; k0=0;

P

C


X Chromosomal markers – Alleged father unavailable for testing

AF

F1

F2

X X

X

X

1/2

1 P(F1 & F2 sharing one IBD X-allele) = 1

AF

F2

X

X

1 F1

X

1

Mendelian incompatibilities can be found

Ex. Ex.

X X

X

X Chromosomal markers in kinship testing

Formalizing relatedness through X Chromosomal perspective

Note that in autosomal transmission among non-inbred individuals the sharing of IBD alleles is only mandatory in the cases parent/child and identical twins.




AF

F1

F2 P(F1 & F2 sharing one IBD X-allele) = 1

AF

F2 F1

X

Ex. Ex.

X X

X Chromosomal markers in kinship testing

Formalizing relatedness through X Chromosomal perspective

X

MUTATION? ABSENCE OF RELATEDNESS?


X X

X X




F1:

F2:

F1:

F2:

F1:

F2:

xFF2 xFF

1 xFF0

x2=0; x1=1; x0=0;

X CHROMOSOMAL AUTOSOMAL REVIEW

Parent - Child

k2=0; k1=1; k0=0;

P

C

Mother - Daughter

D

M

X X

X X

AF

F2

X

F1

X

X

X X

Paternal Half-Sisters

AF

X X

X

X X

F1

F2

Paternal Grandmother-Granddaughter



F1:

F2:

F1:

F2:

F1:

F2:

xFF2 xFF

1 xFF0

AUTOSOMAL REVIEW

Unrelated

k2=0; k1=0; k0=1;

M F

B

M F

A



F1:

F2:

F1:

F2:

F1:

F2:

xFF2 xFF

1 xFF0

x2=0; x1=0; x0=1;

X CHROMOSOMAL AUTOSOMAL REVIEW

Unrelated

k2=0; k1=0; k0=1;

Unrelated

M F

B

M F

A

F1

X

F2

X X X

B

A X

X

X X

Y

X X

X X


C1:

C2:

C1:

C2:

C1:

C2:

X IBD patterns: Example – Full-sisters

x2 = 1/2 x1 = 1/2 x0 = 0

M F

X X

A

X X

B

M F

X X

A

X X

B

M F

X X

A

X X

B



C1:

C2:

C1:

C2:

C1:

C2:

X IBD patterns: Example – Paternal Aunt - Niece

x2 = 0 x1 = 1/2 x0 = 1/2

M F

X X

N

A

X X

M F

X X

N

A

X X

M F

X X

N

A

X X



Joint X Chromosomal Genotypic Probabilities (Female-Female pair)

Genotypes Joint Genotypic Probabilities for X-markers and two females

General Non-inbred females

AiAi, AiAi Λ1

FFpi + (Λ2FF+ Λ3

FF+ Λ5FF+ Λ7

FF)pi2+(Λ4

FF+

Λ6FF+ Λ8

FF)pi3+ Λ9

FFpi4

x2FFpi

2 + x1FFpi

3+ x0FFpi

4

AiAi, AjAj Λ2FFpipj + Λ4

FFpipj2+ Λ6

FFpi2pj + Λ9

FFpi2pj

2 x0FFpi

2pj2

AiAi, AiAj Λ3FFpipj +(2 Λ4

FF + Λ8FF)pi

2pj+ 2Λ9FFpi

3pj x1FFpi

2pj+ 2x0FFpi

3pj

AiAj, AiAi Λ5FFpipj +(2 Λ6

FF + Λ8FF)pi

2pj+ 2Λ9FFpi

3pj x1FFpi

2pj+ 2x0FFpi

3pj

AiAi, AjAl 2Λ4FFpipjpl + 2Λ9

FFpi2pjpl 2x0

FFpi2pjpl

AjAl, AiAi 2Λ6FFpipjpl + 2Λ9

FFpi2pjpl 2x0

FFpi2pjpl

AiAj, AiAj 2Λ7FFpipj + Λ8

FFpipj(pi + pj) + 4Λ9FFpi

2pj2

2x2FFpipj + x1

FFpipj(pi + pj) +

4x0FFpi

2pj2

AiAj, AiAk Λ8FFpipjpk + 4Λ9

FFpi2pjpk x1

FFpipjpk + 4x0FFpi

2pjpk

AiAj, AkAl 4Λ9FFpipjpkpl 4x0

FFpipjpkpl



Patterns of IBD for X Chromosome markers and a pair female/male

ΛFM1 F:

M:

F:

M:

ΛFM2

ΛFM3 F:

M:

ΛFM4 F:

M:



Patterns of IBD for X Chromosome markers and a pair female/male

ΛFM1 F:

M:

F:

M:

ΛFM2

xFM1 F:

M:

xFM0 F:

M:

Assuming non-inbred female:



Joint X Chromosomal Genotypic Probabilities (Female-Male pair)

Genotypes Joint Genotypic Probabilities for X-markers and a pair Female / Male

General Pair Female / Male Non-inbred Female

AiAi, Ai Λ1FMpi + (Λ2

FM+ Λ3FM)pi

2 + Λ4FMpi

3 x1FM pi

2+ x0FM pi

3

AiAi, Aj Λ2FMpipj + Λ4

FMpi2pj x0

FM pi2pj

AiAj, Ai Λ3FMpipj + 2Λ4

FM pi2pj x1

FM pipj + 2x0FM pi

2pj

AiAj, Ak 2Λ4FMpipjpk 2x0

FM pipjpk



Examples (considering a non-inbred female):

paternal uncle - niece

x0FM = 1/2

x1FM = 1/2

unrelated

x0FM = 1

x1FM = 0

Genotypes

Joint Genotypic Probabilities for X-markers and a

female (non-inbred) / male pair

General paternal uncle - niece unrelated

AiAi, Ai x1FM pi

2+ x0FM pi

3 1/2pi2+ x0

FM 1/2pi3 pi

3

AiAi, Aj x0FM pi

2pj 1/2pi2pj pi

2pj

AiAj, Ai x1

FM pipj + 2x0FM

pi2pj

1/2pipj + pi2pj 2pi

2pj

AiAj, Ak 2x0FM pipjpk pipjpk 2pipjpk

Joint X Chromosomal Genotypic Probabilities (Female-Male pair)



xMM1 M1:

M2:

xMM0 M1:

M2:


Genotypes Joint Genotypic Probabilities for X-chromosome

markers and two males

Ai, Ai x1MMpi + x0

MMpi2

Ai, Aj x0MMpipj

Joint X Chromosomal Genotypic Probabilities (Male-Male pair)



Examples:

maternal uncle - nephew

x1MM = 1/4

x0MM = 3/4

unrelated

x0MM = 1

x1MM = 0

Genotypes

Joint Genotypic Probabilities for X-markers and two

males

General Maternal

Uncle – Nephew Unrelated

Ai, Ai x1MMpi + x0

MMpi2 1/4pi + 3/4pi

2 pi2

Ai, Aj x0MMpipj 1/4pipj

pipj

Joint X Chromosomal Genotypic Probabilities (Male-Male pair)



Other examples for the use of X-Chromosomal markers



“Are two females related as paternal grandmother granddaughter?”

1. Paternal Grandmaternity

“Have two females the same father?”

2. Paternal Sisterhood

Examples for the use of X-Chromosomal markers





Examples for the use of X-Chromosomal markers


P(F1 & F2 sharing at least one IBD X-allele) = 1


F2 1 F1 1

X X

F1

F2

X X

X

X

1/2

1

X

F1 F2

X X X

X X X X

1/2 1/2

1 1 X

X X

Power of exclusion: Probability of finding, in randomly chosen individuals, inconsistencies with Mendelian rules of transmission.

Autosomes X Chromosome Markers

Duos

=

<

Trios

=

<

A

B

A

B

A

B B

A

B A B

A

C

A B B A

C

B

A

C

A B

C



Example (Paternal Grandmaternity - Motherless case):



The profiles correspond to a pair of genetically unrelated females :2H

The profiles correspond to a pair of females genetically related as paternal grandmother-granddaughter.

:1H

B

A

vs. A & B unrelated

Hypotheses:

1. The putative paternal grandmother has no monozygotic (‘identical’) twin(s).

2. The putative father has no monozygotic (‘identical’) twin(s).

3. The putative father and the real father are genetically unrelated with the mother

of the child.

4. The putative father is either the true father or genetically unrelated with him.

5. The tested individuals are assumed to belong to the population sampled for the

estimation of gene frequencies (reference and/or description of the population

sample used, including sampling criteria).

6. No gametic association (linkage disequilibrium) exists between the analyzed set

of markers.




Assumptions:

Conclusions

Assuming the conditions described in Annex XXX, the results obtained show that the genetic profile configuration is YYYYY times more likely assuming the hypothesis that So-and-so is the biological paternal grandmother of Child than under the hypothesis of the individuals being genetically unrelated (results per set of markers presented in Table ZZZZZ).




After an abortion female foetal material is collected for analysis. The woman has a mental disease and there is the possibility of incest. Some suspicions were raised that the mother was abused by her own father and brother. For the investigation genetic material was collected from the woman and female foetus, and three hypotheses of kinship were considered: (a) the father is unrelated with the mother, (b) the father is also the father of the woman, and (c) the father is also the brother of the woman.

Examples for the use of X-Chromosomal markers: Incest case



How X Chromosomal markers can be useful?







Undistinguishable through unlinked autosomal markers.

=




=

Undistinguishable through unlinked autosomal markers.

M:

D:

M:

D:

M:

D:

1/4 1/4 1/2




≠ ≠




≠ ≠

Probability of the foetal being X-autozygous (homozygous for IBD X-alleles) for one locus:

M:

D:




≠ ≠

0

Probability of the foetal being X-autozygous (homozygous for IBD X-alleles) for one locus:

1/4 1/2

M:

D:




≠ ≠

Probability of the foetal sharing two pairs of IBD X-alleles with the mother for one locus:

M:

D:




≠ ≠

0

Probability of the foetal sharing two pairs of IBD X-alleles with the mother for one locus:

1/4 1/2

M:

D:




≠ ≠

Probability of the foetal sharing exactly one pair of IBD X-alleles with the mother for one locus:

M:

D:




≠ ≠

1

Probability of the foetal sharing exactly one pair of IBD X-alleles with the mother for one locus:

1/2 0

M:

D:



Examples for the use of X-Chromosomal markers: DVI cases

Case I:

Missing cousins

Relative available for comparision

One part of a body is recovered.




Case I:

Missing cousins

Relative available for comparision

One part of a body is recovered. Autosomal markers are useless.

P(A&B sharing a pair of IBD X-alleles at one locus) = 3/4

P(A&C sharing a pair of IBD X-alleles at one locus) = 0




Case II:

F

M F

M




Case II:

F

M F

M

0

P(F&M sharing one pair of IBD X-alleles):

1/2




Case III:

F

M F

M




Case III:

F

M F

M

3/4

P(F&M sharing one pair of IBD X-alleles):

1/4

Muchas Gracias!





Referencias: Jacquard, A. The Genetic Structure of Populations, Biomathematics. vol. 5. Springer-

Verlag, Berlin-Heidelberg-New York; 1974 (pp. 102–107). Weir, B.S., Anderson, A.D., Hepler, A.B. Genetic relatedness analysis: modern data and new

challenges. Nat. Rev. Genet. 2006;7:771–780. Egeland, T., Sheehan, N. On identification problems requiring linked autosomal markers. Forensic

Sci. Int. Genet. 2008;2:219–225 Pinto N, Gusmão L, Amorim A, Likelihood ratios in kinship analysis: Contrasting kinship classes,

not genealogies, Forensic Science International: Genetics (2010) 4(3): 218-219 Pinto N, Gusmão L, Amorim A, X-chromosome markers in kinship testing: A generalisation of the

IBD approach identifying situations where their contribution is crucial, Forensic Science International: Genetics (2011) 5(1):27-32.

Pinto N, Gusmão L, Egeland T, Amorim A, Paternity exclusion power: comparative behaviour of autosomal and X-chromosomal markers in standard and deficient cases with inbreeding, Forensic Science International: Genetics (2013) 7: 290-295

Referencias adicionais: Pinto N, Silva PV, Amorim A, General Derivation of the Sets of Pedigrees with the Same Kinship

Coefficients, Human Heredity (2010) 70(3): 194-204 Pinto N, Silva PV, Amorim A, A general method to assess the utility of the X chromosomal

markers in kinship testing, Forensic Science International: Genetics (2012) 6:198-207

likelihood ratios with x examples. - familias · likelihood ratios with x examples. workshop...

Documents