1. INBREEDING AND RELATEDNESS
Inbreeding results when two related individuals are mated.
Two individuals are related if they have at least one ancestor in common (not very remote ancestor)

2. HOW TO QUANTIFY INBREEDING?
Proportion of heterozygous
Inbreeding coefficient as a probability

3. Proportion of heterozygous
Proportion of heterozygous under inbreeding vs
Proportion of heterozygous under random mating (RM)
HIP=Freq. of heterozygous genotypes in a
inbred population (IP)
HRM=Freq. of heterozygous genotypes in a
RM population

4. Proportion of heterozygous
Effect of inbreeding can be defined as the proportion reduction in heterozygosity relative to RM.
Mathematically is
F measures the fractional reduction in heterozygosity of an inbred population relative to a RM population with the same allele freq.

5. Proportion of heterozygous
HRM=2pq proportion of Aa under RM
HIP=HRM-FHRM=HRM(1-F)=
=2pq(1-F) proportion of Aa under
inbreeding (IP)
Freq. of homozygous genotypes in an inbred population can be expressed in terms of F

6. Proportion of heterozygous
Suppose that the proportion AA genotype is denoted as PAA
Because ‘p’ is the allele freq. of A, by
F=1-(HIP/HRM) then
p=PAA + HIP/2 but HIP=2pq(1-F) so
PAA = p-2pq(1-F)/2= p-pq(1-F)=p-pq+pqF
= p-p(1-p)+pqF= p2+pqF
PAA = p2+pqF= p2+p(1-p)F= p2+pF-p2F
=p2(1-F)+pF
Paa =q2(1-F)+qF

7. Summary Proportion of heterozygous
PAA = p2(1-F)+pF = p2+pqF
PAa = 2pq-2pqF
Paa = q2(1-F)+qF = q2+pqF

8. Summary Proportion of heterozygous
F= F=1
PAA = p2(1-F)+pF p p
PAa = 2pq pq
Paa = q2(1-F)+qF q q

9. Inbreeding coefficient as a probability
IBD=two alleles are from the same ancestor allele by DNA replication in one of the common ancestors = IDENTICAL BY DESCENT
IBS=two alleles might not be replicates of a single ancestor allele in which case the alleles are not IBD. They are IDENTICAL BY STATE=IBS
F=probability that two alleles of a gene in an inbred organism are IBD.

10. Inbreeding coefficient as a probability
F=measures the probability of IBD relative to some ancestral population – assuming the ancestral population is not inbred, that is F≠0
A1A1 IBD HOMOZYGOUS
A2A2 IBD HOMOZYGOUS
A1 A2 A1
A1 A1 A1
A2 A1 A1
A2 A2 A1
A1A1 IBS HOMOZYGOUS
A1A2 non-IBD HETEROZYGOUS
Alleles in the ancestral
population all assumed
not to be IBD
Genotypes in the present
population

11. the two alleles are non-IBD
ORIGIN OF AN ALLELE
IDENTICAL BY DESCENT (IBD)
ALIKE IN STATE (IBS) ---- not IBD A1 A1 A1 A2 A1 A2 A1
A1A1
INBRED
because the two
alleles are IBD
A1A1
non-IBD because
the two alleles are non-IBD

12. COEFFICIENT OF PARENTAGE COEFFICIENT OF COANCESTRY
It measures the IBD within one individual (fXX) or between two individuals (fXY)
COP between X and Y is the probability that at a single locus a random allele from X is IBD with a random allele from Y
fXY=0 no relationship;
fXY=1 at a given locus indicates that the 2 individuals are homozygous for copies of the same allele found in an ancestry;
fXY=1 across all loci indicates that the 2 individuals are fully inbred and genetically identical

13. COEFFICIENT OF INBREEDING (F)
F probability that a single locus, the two alleles in the same individual are identical by descent
F measures IBD within an individual
Inbreeding only occurs when the 2 parents have alleles that are IBD
F is the proportion by which heterozygosity is reduced upon inbreeding relative to a HW equilibrium

14. Relationship among the four parent of two individual
A x B=X C x D=Y Alleles in parent X
and Y become IBD through four events
Alleles in A and C are IBD
Alleles in A and D are IBD
Alleles in B and C are IBD
Alleles in B and D are IBD
fxy=1/4[fAC + fAD + fBC + fBD]

15. fxy(FS) Coancestry between full sibs ≥ 1/4
X and Y are full-sibs if both have the same parent A x B
A x B=X A x B=Y
fxy(FS)=1/4[fAA + fAB + fBA + fBB]
=1/4[1/2(1+FA)+ fAB + fBA + 1/2(1+FB)]

16. fxy(FS) Coancestry between full sibs ≥ 1/4
fxy(FS)= 1/4[1/2(1+FA)+ fAB + fBA + 1/2(1+FB)] If FA=FB=0 (parents non-inbreds) and no related fAB=0 then fxy(FS)= 1/4[1/ /2]=1/4 If FA=FB=1 (parents full inbreds) and no related fAB=0 then fxy(FS)= 1/4[ ]=1/2 and >1/2 if parents are related

17. fxy(HS) Coancestry between half sibs ≥ 1/8
X and Y are half-sibs if both have one parent
A x B=X A x D=Y
fxy(HS)=1/4[fAA + fAD + fBA + fBD]
=1/4[1/2(1+FA)+ fAD + fBA + fBD]
If A, B and D are unrelated then
fxy(HS)=1/4[1/2(1+FA)]=1/8(1+FA)
If FA=0 (A is not inbred) then fxy(HS)=1/8
If FA=1 (A is an inbred) then fxy(HS)=1/4

18. fxy(HS) Coancestry between half sibs ≥ 1/8
fxy(FS)= 1/4[1/2(1+FA)+ fAB + fBA + 1/2(1+FB)] If FA=FB=0 (parents non-inbreds) and no related fAB=0 then fxy(FS)= 1/4[1/2+1/2(0) + 2(0) + ½(0)]=1/4 If FA=FB=1 (parents full inbreds) and no related fAB=0 then fxy(FS)= 1/4[1+1]=1/2 and >1/2 if parents are related