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Y-STR DNA
Pedigree
(7 generations)
New Technologies
Mitochondrial DNA
Nuclear DNA
• 3 billion base pairs
• 2 copies/cell
• Inherited from both parents
• Unique to individual
XX
Mitochondrial DNA
• ~17,000 base pairs
• >1000 copies/cell
• Maternally inherited
• Not unique to individual
New Technologies
Mitochondrial DNA - Maternal Lineage Testing
• Inherited from the biological mother
• Can be used to identify a maternal lineage
or genetic reconstructions
• Can be used to test hair, bones, teeth
because it is less prone to degradation
• Can not distinguish between individuals
related in the female lineage (brothers, sisters, aunts, uncles)
New Technologies
Mitochondrial DNA
Pedigree
(7 generations)
New Technologies
mtDNA Sequence Comparison
Standard:
GCATATTGCGCCTAGCATATTGCGCCTACCTA
Tested:
GCACATTACGTCTAGGATATTGCGCTTACCTA
References
Basics of Parentage Testing
DNA from each person is unique, except for
identical twins
Each person gets their genomic DNA in equal
amounts from their biological parents
People who are biologically related share more
DNA than those unrelated
The genetic odds in favor of paternity are termed
“Paternity Index”, PI
Paternity Index, PI
• Likelihood Ratio
• Ratio between X and Y, PI = X / Y
• X = chance the alleged father contributed the
paternal allele to the child (0, 0.5, or 1)
• Y = chance a random, unrelated individual
contributed the paternal allele to the child; the
frequency of the paternal allele in the ethnic
group of interest
Example of PI Calculation
•
•
•
•
•
STR analysis at D7S820
Mother = 8,9; Child = 9,10; AF = 10,11
Alleged Father is Caucasian
PI = 0.5 / 0.2906 = 1.72
This means that the Alleged Father is 1.72
times more likely to be the biological father
than a random, unrelated man of the
Caucasian population.
Combined Paternity Index
• Also referred to as the PI or CPI
• The product of the individual paternity indices
for all the genetic loci examined.
• Each locus must be independent, and unlinked
to be multiplied together to produce the
combined paternity index.
Example of a Combined PI
• 13 CODI loci used to examine paternity
• Alleged Father not excluded at any locus
• PI at each locus as follows: CSF1PO = 3.53,
D3S1358 = 2.71; D5S818 = 2.45; D7S820 =
1.72; D8S1179 = 1.95; D13S317 = 3.58;
D16S539 = 2.33; D18S51 = 13.61; D21S11 =
2.20; FGA = 4.53; THO1 = 1.01; TPOX =
0.92; vWA = 2.57
Combined Paternity Index
• Multiplication of all 13 individual paternity
indices
• Combined PI = 212,390
• The alleged father is 212,390 times more
likely to be the biological father than a
random, unrelated man on the same ethnic
group
Probability of Paternity (W)
• Also known as the Likelihood of Paternity,
the Plausibility of Paternity, the Relative
Chance of Paternity, or the Probability Ratio
• Calculated using Bayesian Logic
• Incorporates a Prior Probability (Pr)
Bayesian Logic
• Based on Bayes Theorem (1763)
• Foundation for the probabilistic approach used
for the evaluation of paternity tests.
• Relates the probability of an AF with certain
genetic markers being a member of a
particular group (Biological Fathers) to the
probability that a known member of the group
would have the same genetic markers.
Bayes Theorem
W = Pr . X / (Pr . X + (1 – Pr) . Y)
W = Likelihood of an event occurring
Pr = Prior Probability
X = frequency with which the BF would have
the same phenotype as the AF for a specific M
– C combination
• Y = frequency with which a non-father, ie a
randomly selected man, would have the same
phenotype as the AF
•
•
•
•
Essen-Moller Equation
• W = 1 / (1+ Y/X)
• The mathematical expression of the Relative
Chance of Paternity (RCP) of the AF.
• Bayes theorem applied to parentage testing.
• Hummel modification, W = X/(X + Y),
expressed as a percentage
• Uses a neutral Prior Probability (Pr) of 0.5
Prior Probability
• The probability, after genetic testing, that the
AF is the BF.
• W is determined by integrating Prior
Probability and Paternity Index by means of
Bayes theorem.
• The Hummel modification of the EssenMoller Equation, where a Prior Probability of
0.5 is assumed, is most often used.
• Values range from 0.0 to 1.0
Example of Probability of Paternity
Calculation
Combined Paternity Index = 50
Assume a Prior Probability of 0.5
W = 50(0.5)/50(0.5) + 1- 0.5
= 25.0 / 25.5 = 98.04%
When Prior Probability is 0.5 the Probability of
Paternity is PI / 1 + PI
Examples of Other Prior
Probability Values
• If combined PI is 50 and the Prior Probability
is 0.9 then
• W = 50 (.9)/ 50 (.9) + 1 – 0.9
• = 45 / 45.1 = 99.78 %
• If the combined PI is 50 and the Prior
Probability is 0.1 then
• W = 50 (0.1) / 50 (0.1) + 1- 0.1
• = 5 / 5.9 = 84.75%
Effect of the Prior Probability
• As the Combined PI gets bigger, the value of
the Prior Probability has little effect on the
Probability of Paternity
• If Combined PI = 10,000
• W when Pr is 0.1 = 99.991%
• W when Pr is 0.9 = 99.999%
Exclusion of Parentage
• The AF is excluded as the biological father
when he does not have the obligate paternal
alleles
• With DNA testing, exclusions in two (2)
systems are required before an exclusion is
declared.
• With STR analysis, most laboratories require
exclusions at three (3) loci.
Example of an Exclusion
•
•
•
•
•
•
•
Locus
M
C
AF
D7S820 8,9
9,10 11,12
D18S51 12,14 12,17 12,18
THO1
8,9.3 9.3
6,9.3
D8S1179 9,13 12,13 13,14
Combined Paternity Index = 0.00
Probability of Paternity = 0.00%
PI
0.00
0.00
1.64
0.00
Mutations
Observed when child contains alleles not present
in either biological parent.
Often result of recombinational event in meiosis
in the production of sperm or eggs.
With STR analysis, often leads to the presence
of an allele one (1) repeat unit larger of
smaller than the parent.
Example of a Mutation
•
•
•
•
•
•
•
•
•
Locus
M
C
AF
D7S820 8,9
9,10
11,12
D18S51 12,14 12,17
17,18
THO1
8,9.3 9.3
6,9.3
D8S1179 9,13 12,13
10,12
D21S11 28,30 28,31
31
Combined Paternity Index = 1.08
Probability of Paternity = 51.92%
Additional testing required.
PI
0.0025
5.45
1.64
3.44
14.01
Paternity Index for Mutation
The paternity index at a locus which shows an
exclusion which may be due to a mutation is
calculated as follows.
PI = mutation rate at locus/power of exclusion
Example:
At D7S820 the mutation rate is 0.0015
The power of exclusion is 0.6
PI = 0.0015 / 0.6 = 0.0025
Power of Exclusion (A)
• The ability of a genetic test to exclude a
falsely accused man of paternity.
• It is dependent upon the actual phenotypes of
M and C, and the ethnic group of the M and
AF.
• RMNE (Random Man Not Excluded) is the
frequency with which men selected at random
are not excluded as the BF
• RMNE = 1 - A
When Additional Testing Needed
• When only one or two direct exclusions are
observed with more than 10 loci tested.
• When exclusion is based on indirect
exclusions
• When the combined PI is less than 100
• When close biological relatives are AFs
Direct Exclusion
• When both the child and AF are heterozygous
and the AF does not have the obligate paternal
allele.
• Example
• Locus
M
C
AF
PI
• D7S820 8,9 9,10 11,12
0.00
Indirect Exclusion
• When both the child and AF are homozygous
and the AF does not share the obligate
paternal allele.
• Example
• Locus
M
C
AF
PI
• D7S820 8,9
9
11
0.00
• Here it is possible that an allele is not detected
in C and AF.
Mutation Rates
• Rates vary at different loci.
• Rates vary for maternal and paternal.
• Maternal rates determined from calculation of
number of times the child does not share an allele
with the mother at a locus.
• Paternal rates determined from calculation of the
number of times child does not share an allele with
the BF at a locus.
• Large numbers generated from parentage testing
laboratories compiled by the AABB.
Mutation Rates at CODIS Loci
•
•
•
•
•
•
•
•
Locus
D3S1358
D5S818
D7S820
D8S1179
D13S317
D16S539
D18S51
Maternal
< 0.0002
0.0004
0.0003
0.0008
0.0006
0.0030
0.0010
Paternal
0.0011
0.0015
0.0015
0.0027
0.0015
0.0090
0.0026
Mutation Rates at CODIS Loci
•
•
•
•
•
•
•
Locus
D21S11
FGA
CSF1PO
THO1
TPOX
vWA
Maternal
0.0018
0.0001
0.0003
0.0001
0.0001
0.0004
Paternal
0.0024
0.0029
0.0013
0.0002
0.0002
0.0034
Number of STR Repeat Unit
Changes in Mutations
# Repeats
1 repeat
2 repeats
3 repeats
4 repeats
Other
Male
91.9
4.9
1.2
1.6
0.4
Female
91.9
5.8
0.7
0.7
0.9
Total
91.9
5.1
1.1
1.4
0.5
Accreditation of Parentage Testing
Laboratories
• Accreditation offered by
• American Association of Blood Banks
(AABB)
• American Society for Histocompatibility and
Immunogenetics (ASHI)
• College of American Pathologists (CAP)
PT Laboratories - 2000
44 laboratories accredited by the AABB
About 300,00 cases tested (typical case is trio of M, C,
and AF)
One big lab accounts for about 1/3 of testing
70% of testing done using PCR DNA analysis
30% of testing done using RFLP DNA analysis
Proficiency offered through CAP with PI survey
AABB Accreditation
• Accreditation documents sent to AABB
• AABB Standards for Parentage Testing
Laboratories, 4th Edition
• Standards are in Quality Systems Essentials
Format, which is ISO compatible
• Assessment performed using Assessment Tool
• Accreditation given to laboratories which
follow AABB Standards
Paternity Index Calculation
•
•
•
•
Mother
Child
AF
PI = X /Y
BD
AB
AC
Paternity Index
• X is the probability that 1) M is BD; 2) AF is
AC; and 3) C is AB
• X is the frequency of this set of three
phenotypes among true trios
Paternity Index
• Y is the probability that 1) M is BD; 2) AF is
AC; and 3) C, fathered by the alternative
father, is AB
• Y is the frequency of this set of three
phenotypes among false trios
Paternity Index
• p(BD) = probability of the BD phenotype
• p(AC) = probability of the AC phenotype
• Calculated using the Hardy-Weinberg
Equation (2pq)
• The conditional probability of the C
phenotype, given the phenotypes of its
parents, is computed from Mendel’s First
Law, the Principle of Segregation.
Paternity Index
• The formula for the numerator X is:
• X = p(BD) . p(AC) . [(0 . 0) + (0.5 . 0.5)]
•
= p(BD) . P(AC) . 0.25
• The third term is the probability that a C of a
BD M and and AC father will be phenotype
AB. When the C is heterozygous there are two
components, in that an AB C can inherit A
from the M and B from the father and vice
versa.
Paternity Index
• The formula for the denominator Y is:
• Y = p(BD) . p(AC) . [(0 . b) + (0.5 . a)]
•
= p(BD) . p(AC) . 0.5 . a
• The third term is the probability the C of a BD
mother will be AB.
• It is the probability she will contribute an A
allele (0) and the alternative father will
contribute a B allele (b), plus vice versa.
Paternity Index
• When mating is random, the probability that
the alternative father will contribute a specific
allele to the C is equal to the allele frequency
in his ethnic group.
• This is true whether or not the population is in
H-W equilibrium at the locus.
Paternity Index
•
PI = [p(BD) . p(AC) . 0.25] / [p(BD) . p(AC) . (0.5 . a)]
• = 1 / 2a
• The formula does not contain phenotype
frequencies, thus is does not assume H-W
equilibrium.
• This is true for any system in which the
genotypes can be determined unambiguously
from the phenotype, like DNA.
Paternity Index
• M C AF
X
Y
PI
RMNE
• BD AB AC
0.25
0.5a
1/2a
a(2-a)
• BC AB AC
0.25 0.5a
1/2a
a(2-a)
• BC AB AB
0.25 0.5a
1/2a
a(2-a)
• BC AB
A
0.50
0.5a
1/a
a(2-a)
•
B
AB AC
0.50
a
1/2a
a(2-a)
•
B
AB AB
0.50
a
1/2a
a(2-a)
•
B
AB
1.00
a
1/a
a(2-a)
A
Paternity Index
• M C AF X
Y
PI
RMNE
•
AB AB AC
0.25
0.5(a+b) 1/[2(a+b)] (a+b)(2-a-b)
•
AB AB AB
0.50
0.5(a+b)
1/(a+b)
(a+b)(2-a-b)
•
AB AB
0.50 0.5(a+b) 1/(a+b)
(a+b)(2-a-b)
•
AB
A
AC
0.25
0.5a
1/2a
a(2-a)
•
AB
A
AB
0.25
0.5a
1/2a
a(2-a)
•
AB
A
A
0.50
0.5a
1/a
a(2-a)
•
A
A
AB
0.50
a
1/2a
a(2-a)
•
A
A
A
1.00
a
1/a
a(2-a)
A
Motherless Paternity Index
Calculations
• The assumption in calculating X is that the AF
is the BF
• The assumption in calculating Y is that the AF
and C are unrelated
• Let the phenotypes of the AF = AC and the C
= AB
Motherless Paternity Index
Calculations
• X is the probability that 1) the AF is phenotype AC;
and 2) the C is phenotype AB
• X = p(AC) . (0.5 . b + 0 . a) = p(AC) . 0.5 . b
• p(AC) is the probability of the AF phenotype; 0.5 is
the probability that an AC man will contribute an A
allele; b is the probability that an untested woman
will contribute a B allele; 0 is the probability than an
AC man will contribute a B allele; a is the
probability the woman will contribute an A allele.
Motherless Paternity Index
Calculations
• Y is the probability that 1) a man chosen at
random is phenotype AC; and 2) the child, the
offspring of two untested parents, neither of
whom are related to the AF, is phenotype AB
(the probability of getting an A allele from
one parent and a B allele from the other,
assuming independence).
• Y = p(AC) . 2ab
Motherless Paternity Index
Calculaation
• PI = [p(AC) . 0.5 . b] / [p(AC) . 2ab]
•
= 1 / 4a
• This formula does not assume H-W
equilibrium because it involves only allele
frequencies, not genotype or phenotype
frequencies.
Motherless Paternity Index
• C AF
X
Y
PI
• AB AC
0.5b
2ab
1/4a
• AB AB
• AB
RMNE
(a+b)(2-a-b)
0.5(a+b) 2ab (a+b)/4ab (a+b)(2-a-b)
A
b
2ab
1/2a
(a+b)(2-a-b)
•
A
AC
0.5a
a2
1/2a
a(2-a)
•
A
A
a
a2
1/a
a(2-a)
Power of Exclusion
• 1 – RMNE
• Combined Probability of Exclusion (CPE)
refers to the Average Power of Exclusion
• PEAVG = h2 . [1 – 2hH2]
• H = frequency of Homozygosity
• h = frequency of Heterozygosity
Reconstructions
• If the parents of the AF are available for
testing, a grandparental paternity index can be
calculated.
• If the AF has no known offspring, his
genotype generally can not be determined.
• When both parents of the AF are available, the
AF’s genotype can be reduced to 4 or less.
Reconstruction
The No-Father Case
• M = BC and C = AB
• Paternal allele = A
• AF deceased, but some of his relatives are
available for testing
• The goal is to determine as precisely as
possible the AF genotype then calculate the
probability of transmitting the A allele
• p(A) = probability of transmitting A allele
Paternity Index
No-Father Case
• p(A) = probability AF will transmit A allele to
his offspring
• X = 0.5 p(A)
• Y = 0.5 a
• PI = 0.5 p(A) / 0.5 a = p(A) / a
Reconstruction
• AF M C1 C2
•
BC AB AC
• What is the PI of the deceased AF based on
this data?
• Since both children have inherited an A allele
from the AF, this increases the probability the
AF is AA
Reconstruction
•
•
•
•
•
Allele frequencies used for calculations:
A = 0.042
B = 0.089
C = 0.037
D = 0.122
Reconstruction
1
2
3
4
5
6
AF Wife Genotype Frequencies p (AB) & p (AC) p (Parents &
AF
Wife
Children
Children)
AA BC 0.0018
0.0066
1/4
2.9044E-06
AX BC 0.0805
0.0066
1 / 16
3.3124E-05
AA genotype = 0.04202 = 0.0018
AX genotype = 2 . 0.0420 . 0.9580 = 0.805
Reconstruction
1
2
3
4
5
6
AF Wife Genotype Frequencies p (AB) & p (AC) p (Parents &
AF
Wife
Children
Children)
AA BC 0.0018
AX BC 0.0805
0.0066
0.0066
1/4
1 / 16
BC genotype = 2 . 0.0890 . 0.0370 = 0.0066
Column 5 calculated by Mendelian Law
Column 6 product of columns 3 x 4 x 5
2.9044E-06
3.3124E-05
Reconstruction
•
•
•
1
6
P(Parents &
AF
Children)
•
•
AA
AX
•
Column 7 = relative probability of AF genotype in column 1; divide the
number in column 6 by the sum of column 6 (ie 2.9044E06 / 3.60284E-05
= 8.06 %)
•
Column 8 = conditional probability AF will transmit A allele given the
genotype in column 1
2.9044E-06
3.3124E-05
7
Relative
Probability
8
Conditional
p (A)
9
p (A)
8.06%
91.94%
1.00
0.50
0.0806
0.4597
Reconstruction
•
•
•
1
6
P(Parents &
AF
Children)
•
•
AA
AX
•
Column 9 = probability that the AF will transmit the A allele given the
genotype in column 1; multiply column 7 x 8
•
p(A) = 0.5403 = (0.0806 + 0.4597)
•
PI = p(A) / a = 0.5403 / 0.0420 = 12.86
2.9044E-06
3.3124E-05
7
Relative
Probability
8
Conditional
p (A)
9
p (A)
8.06%
91.94%
1.00
0.50
0.0806
0.4597
Avuncular Index (AI)
• AI is the likelihood ratio for the hypothesis
that the tested man’s brother is the true BF, as
opposed to a random man.
• Typical scenario is an AF is excluded and the
residual PI is >1,000 (multiplication of all PI s
from non-excluding loci).
• The tested AF is not the BF, but it is likely
that the BF may be a relative of the AF
Avuncular Index
• PI = X /Y
• On average, the AF and his brother share 50%
of their alleles, the remaining 50% of the
untested brother’s alleles are selected at
random from the population.
• Thus the X value for the untested brother is
the average of the X value for the tested AF
and the X value for an untested man.
Avuncular Index
• Thus X = X + Y / 2 = 0.5X + 0.5 Y
• The denominator is Y.
• AI = 0.5X + 0.5Y / Y
•
= PI +1 / 2
Calculation of Likelihood Ratio for
any Relative of Tested Man
• = R * X + (1- R) * Y / Y = R * PI + 1 – R
• R = Coefficient of Relationship; it tells the
proportion of the relative’s alleles expected to
be identical by decent with those of the tested
man. The remainder (1 – R) are random.
Coefficient of Kinship (F)
• From each of two individuals, randomly select
an allele from a given locus.
• F is the probability that the two alleles are
identical by descent.
Coefficient of Relationship (R)
• For two individuals, R is the proportion of
alleles at a given locus that are identical by
descent.
• For two individuals who are not inbred,
R = 2F
Coefficients of Kinship and
Relationship
•
•
•
•
•
•
•
•
•
•
Relationship
Parent & Child
Siblings
Half Siblings
Grandparent & GC
Uncle & Niece/Neph
First Cousins
Second Cousins
Third Cousins
Identical Twins
F
0.25
0.25
0.125
0.125
0.125
0.0625
0.015625
0.003906
0.5
R
0.50
0.50
0.25
0.25
0.25
0.125
0.032125
0.00781
1.00
Kinship Analysis
• Determine whether two individuals are blood
relatives when no other family members are
available for testing.
• Requires a randomly mating population, and
testing of several independent, highly
polymorphic loci.
• Absence of silent alleles (null alleles) allows
phenotypes to denote genotypes.
Kinship Analysis
•
If two parents are genotype AB and CD, they can have
offspring with the four equally frequency genotypes:
•
•
Parent 1
Parent 2
Possible
Genotype Genotype
Offspring
•
•
Frequency
Genotype
AB
CD
AC
0.25
AD
BC
BD
At each locus, 2, 1, or 0 alleles shared
0.25
0.25
0.25
Kinship Analysis
• Consider two non-inbred individuals,
s1
and s2.
• To evaluate the hypothesis that S1 and S2 are
related, calculate the likelihood ratio:
• LR = X/Y
• X = probability of s2 if s1 and s2 related
• Y = probability of s2 if s1 and s2 unrelated
Kinship Analysis
• Calculation of Y is simply the genotype
frequency of s2 calculated from the HardyWeinberg equation: p2 + 2pq +q2 = 1
Kinship Analysis
• Calculation of X
• If s1 and s2 are related, then the observed
genotype of s2 could arise in one of three
ways.
• 1. Two alleles of s1 and s2 are identical (k2)
• 2. One allele of s1 and s2 are identical (2k1)
• 3. No alleles of s1 and s2 are identical (k0)
Kinship Analysis
• The k coefficients for two related people, S
and T is as follows:
• Let A and B be the parents of S.
• Let C and D be the parents of T.
• k2 is determined by the coefficients of kinship
(F) between parents S and T.
Kinship Analysis
•
•
•
•
•
Calculation of the k Coefficients
k2 = (FAC* FBD) + (FAD* FBC)
2k1 = 4FST - 2k2
k1 = 2FST - k2
k0 = 1 – k2 – 2k1
Kinship Analysis
• F and k Coefficients
•
Relationship
F
k2
•
Parent-Child
•
Full Siblings
¼
¼
•
Half Sibling
1/8
0
•
GP-GC
1/8
0
•
Uncle-Niece
1/8
0
•
•
First Cousins
Second Cousins
1/16
0
0
1/64
2k1
k1
k0
0
1
½
0
¼
½
½
½
½
¼
¼
¼
¼
¼
1/8
¼
½
½
½
¾
1/16
1/32
15/16
Kinship Anlysis
s1
s2
X
Y
X/Y
AA AA
1* k2 + a * k1 + a * k1 + a2 * k0
a2
(k2 + 2k1a + k0a2) / a2
AA AB
0 * k2 + b * k1 + b * k1 + 2ab * k0 2ab
AA BB
0 * k 2 + 0 * k1 + b2 + k 0
b2
k0
AA BC
0 * k2 + 0 * k1 + 2bc + k0
2bc
k0
AB AB
1* k2 + b * k1 + a * k1 + 2ab * k0 2ab (k2 + k1a + k1b+k02ab) / a2
AB AA
0 * k2 + a * k1 + a2 * k0
a2
(k1 + k0a) / a
AB AC
0 * k2 + c * k1 +2ac * k0
2ac
(k1 +2 k0a) /2 a
AB CD
0 * k2 + 0 * k1 +2cd * k0
2cd
k0
(k1 + k0a) / a
Kinship Analysis Example
• What is the likelihood the two individuals are
full siblings?
Locus 1
•
•
•
•
•
Locus 2
Al Sib 1 A,B
C,D
Al Sib 2 A,B
C,E
Frequency of A allele = 0.05
Frequency of B allele = 0.05
Frequency of C allele = 0.05
Locus 3
F,G
H,I
Kinship Analysis Example
• Locus 1:
• X/Y = (k2 + k1a + k1b + k02ab) / 2ab
•
= (0.25) + (0.25)(0.05) + (0.25)(0.05) + (0.25)(0.005) / 0.005
• = 0.27625 / 0.005
• = 55.25
Kinship Analysis Example
•
•
•
•
•
Locus 2:
X / Y = k1 + 2k0a / 2a
= 0.25 + (2)(0.25)(0.05) / 0.10
= 0.275 / 0.10
= 2.75
Kinship Analysis Example
• Locus 3
• X / Y = k0
• = 0.25
Kinship Analysis Example
• The combined Sib Index, SI, (likelihood of the
two alleged sibs being full siblings) is the
multiplication of each of the individual
indices.
• SI = 55.25 * 2.75 * 0.25 = 37.98
• This is a strong indication that the two are
related as full siblings.
Kinship Analysis Example
• What is the likelihood that the two individuals
are related as half siblings?
• Use the same formulas, but use the halfsibling values for the k coefficients.
Kinship Analysis Example
• Locus 1:
• X/Y = (k2 + k1a + k1b + k02ab) / 2ab
•
= (0) + (0.25)(0.05) + (0.25)(0.05) + (0.50)(0.005) / 0.005
• = 0.0275 / 0.005
• = 5.5
Kinship Analysis Example
•
•
•
•
•
Locus 2:
X / Y = k1 + 2k0a / 2a
= 0.25 + (2)(0.50)(0.05) / 0.10
= 0.3 / 0.10
= 3.0
Kinship Analysis Example
• Locus 3
• X / Y = k0
• = 0.50
Kinship Analysis Example
• The combined Sib Index, SI, (likelihood of the two
alleged sibs being half siblings) is the multiplication
of each of the individual indices.
• SI = 5.5 * 3.0 * 0.5 = 8.25
• This is weak evidence that the two are related as half
siblings.
• The two are 4.6 times more likely to be full siblings
(37.98 / 8.25) than half siblings.