y-str dna pedigree (7 generations) new technologies

Y-STR DNA

Pedigree(7 generations)

New Technologies

Mitochondrial DNA

XX

Nuclear DNA• 3 billion base pairs• 2 copies/cell• Inherited from both parents• Unique to individual

Mitochondrial DNA• ~17,000 base pairs• >1000 copies/cell• Maternally inherited• Not unique to individual

New Technologies

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

Standard: GCATATTGCGCCTAGCATATTGCGCCTACCTA

Tested: GCACATTACGTCTAGGATATTGCGCTTACCTA

mtDNA Sequence Comparison

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 Essen-Moller 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 PI

• D7S820 8,9 9,10 11,12 0.00• D18S51 12,14 12,17 12,18 0.00• THO1 8,9.3 9.3 6,9.3 1.64• D8S1179 9,13 12,13 13,14 0.00• Combined Paternity Index = 0.00• Probability of Paternity = 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 PI

• D7S820 8,9 9,10 11,12 0.0025• D18S51 12,14 12,17 17,18 5.45• THO1 8,9.3 9.3 6,9.3 1.64• D8S1179 9,13 12,13 10,12 3.44• D21S11 28,30 28,31 31 14.01• Combined Paternity Index = 1.08• Probability of Paternity = 51.92%• Additional testing required.

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 exclusionExample:At D7S820 the mutation rate is 0.0015The 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 Maternal Paternal• D3S1358 < 0.0002 0.0011• D5S818 0.0004 0.0015• D7S820 0.0003 0.0015• D8S1179 0.0008 0.0027• D13S317 0.0006 0.0015• D16S539 0.0030 0.0090• D18S51 0.0010 0.0026

Mutation Rates at CODIS Loci

• Locus Maternal Paternal• D21S11 0.0018 0.0024• FGA 0.0001 0.0029• CSF1PO 0.0003 0.0013• THO1 0.0001 0.0002• TPOX 0.0001 0.0002• vWA 0.0004 0.0034

Number of STR Repeat Unit Changes in Mutations

# Repeats Male Female Total

1 repeat 91.9 91.9 91.9

2 repeats 4.9 5.8 5.1

3 repeats 1.2 0.7 1.1

4 repeats 1.6 0.7 1.4

Other 0.4 0.9 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 BD• Child AB• AF AC• PI = X /Y

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 A 1.00 a 1/a a(2-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 A 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)

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 RMNE

• AB AC 0.5b 2ab 1/4a (a+b)(2-a-b)

• AB AB 0.5(a+b) 2ab (a+b)/4ab (a+b)(2-a-b)

• AB 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.

ReconstructionThe 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

AX = all A heterozygotes (X = 1-a = 0.958)

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

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

Reconstruction

• 1 6 7 8 9

• P(Parents & Relative Conditional

• AF Children) Probability p (A) p (A)

• AA 2.9044E-06 8.06% 1.00 0.0806

• AX 3.3124E-05 91.94% 0.50 0.4597

• 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

Reconstruction

• 1 6 7 8 9

• P(Parents & Relative Conditional

• AF Children) Probability p (A) p (A)

• AA 2.9044E-06 8.06% 1.00 0.0806

• AX 3.3124E-05 91.94% 0.50 0.4597

• 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

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 F R• Parent & Child 0.25 0.50• Siblings 0.25 0.50• Half Siblings 0.125 0.25• Grandparent & GC0.125 0.25• Uncle & Niece/Neph 0.125 0.25• First Cousins 0.0625 0.125• Second Cousins 0.015625 0.032125• Third Cousins 0.003906 0.00781• Identical Twins 0.5 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 0.25 BC 0.25 BD 0.25

At each locus, 2, 1, or 0 alleles shared

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 Hardy-Weinberg 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 2k1 k1 k0

• Parent-Child ¼ 0 1 ½ 0

• Full Siblings ¼ ¼ ½ ¼ ¼• Half Sibling 1/8 0 ½ ¼ ½• GP-GC 1/8 0 ½ ¼ ½• Uncle-Niece 1/8 0 ½ ¼ ½• First Cousins 1/16 0 ¼ 1/8 ¾• Second Cousins 1/64 0 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 (k1 + k0a) / a

AA BB 0 * k2 + 0 * k1 + b2 + k0 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

Note that the likelihood ratio, X/Y, involves only k coefficients and the frequencies of alleles that are present in

both s1 and s2

Kinship Analysis Example

• What is the likelihood the two individuals are full siblings?

Locus 1 Locus 2 Locus 3• Al Sib 1 A,B C,D F,G• Al Sib 2 A,B C,E H,I• Frequency of A allele = 0.05• Frequency of B allele = 0.05• Frequency of C allele = 0.05

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 half-sibling 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.