Genetic identity and Kinship

Genetic Identity

G1 and G2 are identical by descent (i.b.d) if they are physical copies of the same ancestor, or one of the other.

G1 and G2 are identical by state (i.b.s) if they represent the same allele.

The kinship between two relatives fij is the probability that random gene from autosomal loci in I and j are i.b.d.

The interbreeding coefficient is the probability that his or her two genes from autosomal loci are i.b.d.

•Every mutation creates a new allele•Identity in state = identity by descent (IBD)

A1A1

A2A2

A1A2A1A1

A1A2 A1A2

Mutation occurred once

A1A1

A2A2

A1A2A1A1

A1A2 A1A2

A1A1

A2A2

A1A2 A1A1

A1A2A1A2

A2 A2 IBD A2 A2 IBD

A2 A2 alike in state (AIS)

not identical by descent

The same mutation arises independently

Identity by descent

A - B C - D | | P - Q | X

Let fAC be the coancestry of A with C etc., i.e. the probability of 2 gametes taken at random, 1 from A and one from B, being IBD.

Probability of taking two gametes, 1 from P and one from Q, as IBD, FX

FX fPQ 1

4fAD

1

4fAC

1

4fBC

1

4fBD

Identity by descent

Example, imagine a full-sib matingA - B / \P - Q | X

Indv. X has 2 alleles, what is the probability of IBD?

FX fPQ 1

4fAD

1

4fAC

1

4fBC

1

4fBD

1

42 fAB fAA fBB 1

40 1

2 1

2

1

4

Identity by descent

Example, imagine a half-sib matingA - B - C | | P - Q | X

FX fPQ 1

4fAD

1

4fAC

1

4fBC

1

4fBD

1

42 fAB fAC fBC fBB 1

40 0 0

1

2

1

8

Kinship and Interbreeding

ii=0.5(1+fi) fi=kl, where k and l are the parents of i. If fi>0 then i is said to be inbred. The question is how to compute kinship,

given a pedigree. Let us look for example at brothers and

sisters:

Pedigree

Let us compute the kinship coefficients for all member of this pedigree. We assume that 1 and 2 are not inbred and unrelated, we start counting from the oldest generation.

We will develop an algorithm to compute ij for all members of this pedigree.

1 2

3 4

5 6

Kinship coefficient algorithm

1 2

3 4

5 6

123456

11/20

201/2

3

4

5

6

Kinship coefficient algorithm II

1 2

3 4

5 6

123456

11/20

201/2

3

4

5

6

Kinship coefficient algorithm III

1 2

3 4

5 6

123456

11/20

201/2

31/2

41/2

5

6

If i originates from k and l

ii= ½+ kl

Kinship coefficient algorithm III

1 2

3 4

5 6

123456

1½0¼¼

20½¼¼

3¼¼½¼

4¼¼¼½

5

6

If i originates from k and lij= ji

= ½(jk + jl)

Kinship coefficient algorithm IV

1 2

3 4

5 6

123456

1½0¼¼

20½¼¼

3¼¼½¼

4¼¼¼½

5

6

One can now reapply the algorithm on the next generation

Kinship coefficient algorithm V

1 2

3 4

5 6

123456

1½0¼¼¼¼

20½¼¼¼¼

3¼¼½¼3/83/8

4¼¼¼½3/83/8

5¼¼3/83/85/83/8

6¼¼3/83/83/85/8

The final result is:

Identity coefficients:

We can now summarize the kinship coefficient of some basic family relations:

Relation

Parent-Offspring¼

Half Sibling1/8

Full Sibling¼

First Cousins1/16

Double First Cousins1/8

Second Cousins1/64

Uncle-Nephew1/8

Detailed Identity States – I

I

J

Allele 1 Allele 2

Detailed Identity States – I I

Detailed Identity States – I I I

Summary

Many of these relations are redundant. If I is not inbred 1,2,3 and 4 will be zero.

One can define kinship in a condensed mater, if we can interchange the maternal and paternal genes.

Condensed Identity States

Condensed Identity States

S3=S*2S*

2

S5=S*4S*

5

S7=S*9S*

12

S8=S*10S*

11

S*13S*

14

Condensed Identity States II

1234 are 0, when i is not inbred.

1256 are 0, when j is not inbred.

1357 and 8 are 0, when i and j are unrelated.

ji=1+1/2(357)+1/4 8

Kinship and identity coefficients

Relation

Parent-Offspring010¼

Half Sibling0½½1/8

Full Sibling1/4½¼¼

First Cousins0¼¾1/16

Double First Cousins1/166/169/161/8

Second Cousins01/1615/161/64

Uncle-Nephew0½½1/8

Genotype prediction.

What is the probability that i has a given genotype, given the genotype of j ?

For example, If my uncle has a genetic disease, what is the probability that I will also have it?

What are the probabilities of brothers from inbred parents to be homozygous for a disease causing gene?

……

Genotype prediction.

4 )1(

4 0

4 2)1(

24 0

)/Pr(

)/,Pr()/|Pr(

)/|Pr(*

)/,|/Pr()/|/Pr(9

1

rf

r

rppf

ppr

lki

lkiSlkiS

lkiS

lkiSnmjlkinmj

i

r

lki

lkrr

r

r

rr

If I is heterozygous, with an inbreeding coefficient fi

Genotype prediction II

4 )1(

4 )1(

4 )1(

4 )1(

)/Pr(

)/,Pr()/|Pr(

2

2

2

2

rpff

p

rpff

rpfpf

p

rpfpf

p

kki

kkiSkkiS

kii

kr

kii

r

kiki

kr

kiki

kr

rr

If I is homozygous, with an inbreeding coefficient fi

Genotype prediction III

l/l k/kj

ij

ijS

ijS

lkiSnmj r

or either is 5S

with gene one shares 3,8S

oft independen is 9,6,4,2

7,1

)/,|/Pr(

When j is independent of i, it only follows the H,W equilibrium.When j is equivalent to i, the probability is one if m/n=k/l and zero otherwise.When j shares one allele with I, m/n and k/l must overlap with one allele and the other one has H.W distribution.

Example

What is the blood type of non-inbred siblings?

BABA pppp

BAiSBAjBAiSBAj

BAiSBAjBAiBAj

24

1)

2

1

2

1(

2

11*

4

1

)/,|/Pr(4

1)/,|/Pr(

2

1

)/,|/Pr(4

1)/|/Pr(

98

7

Example I

What is the blood type of non-inbred siblings?

oAA pppp

OOiSOAjOOiSOAj

OOiSOAj

OOiAAj

OOiOAjOOiAj

A2

4

1

2

12

4

10*

2

10*

4

1

.......................

)/,|/Pr(4

1)/,|/Pr(

2

1

)/,|/Pr(4

1

)/|/Pr(

)/|/Pr()/|Pr(

2

98

7

Risk Ratios and Genetic Model Discrimination.

Let us assume that each person in the population is assigned a factor of X=1 if he/she is affected by a condition and X=0 otherwise.

The Prevalence of the condition is K=E(X). Given two non-inbred relatives i and j and given

that i is affected, what is the probability that J is affected?

KR=P(Xj=1|Xi=1( P(Xj=1,Xi=1) = P(Xj=1|Xi=1(P(Xi=1) = KRK =

E(XiXj)

Risk Ratios and Genetic Model Discrimination.

P(Xj=1|Xi=1) = E(XiXj)/K = (cov(Xi,Xj)+K2)/K = cov(Xi,Xj)/K+K

This result simply represents the fact that the extra risk for j results from the covariance of X between i and j.

The risk ratio can thus be defined as: R= cov(Xi,Xj)/K2

Let us compute this covariance, and following it the risk ratio.

Covariance

In a more general way, let us assume that i and j are non-inbred relatives. The covariance between their genes is defined only by condensed identity states 7,8 and 9

Covariance

Let us assume that a given property is defined by a single gene with multiple alleles.

For the sake of simplicity let us normalize E(x)=0, and divide:

lk l

kkl ppXE )(

0

0 ;

kkkl

kkkkllkkl

p

p

Covariance

27

2

27

287

28

227

9

82

7

2

24

1

2

12

22

)()(

)()()(

),|,()(

dijaij

lk l

kklijkk

kijij

kk

kijlk l

kklkk

kij

nmlk l

kmnnmm n

kllkij

mlk l

kkmmkm

kllkijlk l

kkllkij

lk l

km

kln

mnji

ppp

pppp

pppp

ppppp

pplknmpXXE

Risk Ratio

RRelative TypeRisk Ratio

MIdentical Twina2/K2+ d

2/K2

SSiblinga2/2K2+ d

2/4K2

1First Degreea2/2K2

2Second Degreea2/4K2

3Third Degreea2/8K2

Summary

ConceptsKinship coefficient.

Inbreeding coefficientIdentity states

Condensed identity statesIdentity coefficientsGenotype prediction