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Y-STR DNA Pedigree (7 generations)

New Technologies. 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. X. X. Mitochondrial DNA ~17,000 base pairs > 1000 copies/cell Maternally inherited

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Y-STR DNA Pedigree (7 generations)

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  1. New Technologies Y-STR DNA Pedigree (7 generations)

  2. New Technologies Mitochondrial DNA • Nuclear DNA • 3 billion base pairs • 2 copies/cell • Inherited from both parents • Unique to individual X X • Mitochondrial DNA • ~17,000 base pairs • >1000 copies/cell • Maternally inherited • Not unique to individual

  3. 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)

  4. New Technologies Mitochondrial DNA Pedigree (7 generations)

  5. New Technologies mtDNA Sequence Comparison Standard:GCATATTGCGCCTAGCATATTGCGCCTACCTA Tested:GCACATTACGTCTAGGATATTGCGCTTACCTA

  6. References

  7. 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 Basics of Parentage Testing

  8. 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 Paternity Index, PI

  9. 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. Example of PI Calculation

  10. 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. Combined Paternity Index

  11. 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 Example of a Combined PI

  12. 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 Combined Paternity Index

  13. 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) Probability of Paternity (W)

  14. 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. Bayesian Logic

  15. 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 Bayes Theorem

  16. 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 Essen-Moller Equation

  17. 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 Prior Probability

  18. 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 Example of Probability of Paternity Calculation

  19. 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% Examples of Other Prior Probability Values

  20. 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% Effect of the Prior Probability

  21. 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. Exclusion of Parentage

  22. 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% Example of an Exclusion

  23. 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. Mutations

  24. 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. Example of a Mutation

  25. 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 Paternity Index for Mutation

  26. 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 Power of Exclusion (A)

  27. 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 When Additional Testing Needed

  28. 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 Direct Exclusion

  29. 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. Indirect Exclusion

  30. 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

  31. 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

  32. 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 Mutation Rates at CODIS Loci

  33. # 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 Number of STR Repeat Unit Changes in Mutations

  34. Accreditation offered by American Association of Blood Banks (AABB) American Society for Histocompatibility and Immunogenetics (ASHI) College of American Pathologists (CAP) Accreditation of Parentage Testing Laboratories

  35. 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 PT Laboratories - 2000

  36. 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 AABB Accreditation

  37. Mother BD Child AB AF AC PI = X /Y Paternity Index Calculation

  38. 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

  39. 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

  40. 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

  41. 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

  42. 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

  43. 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

  44. 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

  45. 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

  46. 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) Paternity Index

  47. 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

  48. 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

  49. 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 Calculations

  50. 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 Calculaation

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