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The statistical weight of mixed samples with allelic drop out First serious attempt by Gill et al. 2006, Forensic Science International 160:90 An important.

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Presentation on theme: "The statistical weight of mixed samples with allelic drop out First serious attempt by Gill et al. 2006, Forensic Science International 160:90 An important."— Presentation transcript:

1 The statistical weight of mixed samples with allelic drop out First serious attempt by Gill et al. 2006, Forensic Science International 160:90 An important general paper about mixtures: Curran et al. 1999, J. Forensic Science 44:987

2 Mixed sample with drop out

3 Standard Mixture Analysis Assume there are 2 people and 3 alleles: A 1, A 2, A 3 There must be a total of 4 alleles allowing for the following possible combinations: (A 1,A 1,A 2,A 3 ) and (A 1,A 2,A 2,A 3 ) and (A 1,A 2,A 3,A 3 ). Let the frequency of the 3 alleles be p 1 p 2 p 3

4 Details of (A 1,A 1,A 2,A 3 ) Possible pairs of sampled genotypes are: [A 1 /A 1 and A 2 /A 3 ] or [A 2 /A 3 and A 1 /A 1 ] [A 1 /A 2 and A 1 /A 3 ] or [A 1 /A 3 and A 1 /A 2 ] These pairs are chosen with frequencies 2[p p 2 p 3 ] 2[2 p 1 p 2 2 p 1 p 3 ] The sum of these is 12p 1 2 p 2 p 3 Repeating this for the other two orderings and adding them all up gives 12p 1 p 2 p 3 (p 1 +p 2 +p 3 )

5 General formula let c=number of distinct alleles x= number of people in the mixture u i = number of copies of allele i the frequency of any particular combination

6 Mixtures with drop out Let Q be the dropped out allele The frequency of Q is 1-sum(distinct alleles) Suppose evidence is A 1,A 2,Q Possible orderings are (A 1,A 1,A 2,Q) and (A 1,A 2,A 2,Q) but not (A 1,A 2,Q,Q) since we have assumed only one allele dropped out frequency is 12p 1 p 2 p Q (p 1 +p 2 )

7 Two people mixtures Number of alleles out evidencefrequency 1A 1,A 2, A 3,Q24p 1 p 2 p 3 p Q 1A 1,A 2,Q12p 1 p 2 p Q (p 1 +p 2 ) 1A 1,Q4p13 pQ4p13 pQ 2A 1,A 2,Q12p 1 p 2 p Q 2 2A 1,Q6p12 pQ26p12 pQ2 3 4p 1 p Q 3

8 Likelihood Ratios Compare the probability of two hypotheses, the prosecution and the defense Each hypothesis must compute the probability of the observed genetic evidence Let L = Prob[evidence|prosecution] / Prob[evidence|defense]

9 Example Three person mixture Evidence: 9 Suspect: 11, 14 Two alleles dropped out Let D be the probability that one allele will drop out. In this sample the State assumes at least two alleles dropped out, and four alleles did not: This probability is: (1-D) 4 D 2

10 Example: state hypothesis (1-D) 4 D 2 {prob[two people with only the 9 allele]} (1-D) 4 D 2 p 9 4

11 Example: defense hypothesis There are several possibilities No drop out: (1-D) 6 p 9 6 One allele dropped out, five did not: (1-D) 5 D prob[three people with only the 9 allele and one allele dropped out] = (1-D) 5 D 6p 9 5 p Q Two alleles dropped out, four did not: (1- D) 4 D 2 prob[three people with only the 9 allele and two alleles dropped out] =(1-D) 4 D 2 15p 9 4 p Q 2

12 Example Results 13 loci with a total of 5 alleles dropped out and a minimum of three people in the mixture, 1 known, 2 unknown The lab CPI for Caucasians was 1 in 42 million DL


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