# Deconvoluting Mixtures Using Proportional Allele Sharing What does it mean and how do you do it?

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Deconvoluting Mixtures Using Proportional Allele Sharing What does it mean and how do you do it?

What is a mixture? If you start with two single source profiles and combine them, you have a mixture Two sources combined and looks like a two person mixture Male Ref Female Ref Mixture

Same references mixed differently Two sources combined that looks like a single source profile Her birthday is at Thanksgiving Dad Mom Ellie

Start with the mixture, and then go backwards to the single source profiles. What is deconvolution?

Deconvolution Some loci are easy to deconvolute Major/minor contributors

Deconvolution Some loci are harder to deconvolute But if you can assume a contributor… It can become easier Mixture - Victim= Foreign

How do you deal with shared alleles? Common to find shared alleles in any mixture of two people. How is the shared allele distributed between the two contributors? Lots of papers published and various software programs deal with this issue.

AB BC Sharing Donor 1 is AB (10,12) Donor 2 is BC (12,13) Mixture is ABC (10,12,13)

AB BC Sharing But how do you work backwards from this To this?

AB BC Sharing Validation studies give PHR expectations – These expectations show up in protocols for interpretation – Used in setting stochastic thresholds – Used in determining number of contributors At times you may have a major contributor that helps At times you can assume a contributor

AB BC Sharing – Test 1 Assume 50% PHR rule for AB (10,12) Can you then assume 250 rfu from the 12 goes with the 10? So for 10,12 type: 5001500790 250 500 ÷= 0.5 PHR 250 500 + + 1500 + 790 = 0.27 P (proportion) 250500 250500 1500790

AB BC Sharing – Test 1 So for 12,13 type: 5001500790 1250 790 1250 ÷= 0.63 PHR 1250 + 790500 + 1500 + 790 = 0.73 P 790 1250 5001500

AB BC Sharing – Test 2 Or Assume 1000 rfu of the 12 goes with the 10 So for 10,12 type: Note the PHR is the same, but P is double 5001500790 1000 500 ÷ 1000 = 0.5 PHR 1000 + 500 + 1500 + 790 = 0.54 P 500 1500790 1000

AB BC Sharing – Test 2 12,13 type is now: Note that PHR is the same as Test 1 Proportion is about 1/3 less than Test 1 5001500790 500 ÷ 790 = 0.5 PHR 500 + 7905001500790 ++ = 0.46 P 500 1500790

AB BC Sharing Summary Test 1 AB PHR = 0.5 AB portion = 0.27 BC PHR = 0.63 BC portion = 0.73 1:4 mixture Test 2 AB PHR = 0.5 AB portion = 0.54 BC PHR = 0.63 BC portion = 0.46 1:1 mixture

Time to Vote 1.Test 1 is correct 2.Test 2 is correct 3.The truth is somewhere in between 4.You cannot make a determination at all 5.I just want lunch

AA AB Sharing Donor 1 is AA (8,8) Donor 2 is AB (8,12) Mixture is AB (8,12)

AA AB Sharing How do you work backward from this To this

AA AB Sharing – Test 1 Assume 50% PHR rule for AB (8,12) Can you then assume 250 rfu from the 8 goes with the 12? So for 8,12 type: 250 1300500 250 ÷ 500 = 0.5 PHR 250 + 5001300 + 500 = 0.42 P 250 500 1300

AA AB Sharing – Test 1 So for 8,8 type: 1050 1300500 No PHR 1050 1300 + 500 = 0.58 P 1050 1300 500

AA AB Sharing – Test 2 Or Assume 1000 rfu from the 8 goes with the 12? So for 8,12 type: Note PHR is the same, but P doubles 1000 1300500 = 0.5 PHR÷ 1000 500 1000 500 + 1300 + 500 = 0.83 P 500 1300 1000

AA AB Sharing – Test 2 So for 8,8 type: Note that P is smaller by a factor of ~3 ½ 300 1300500 No PHR 300 5001300 + = 0.17 P 1300500 300

AA AB Sharing Summary Test 1 AB PHR = 0.5 AB portion = 0.42 AA portion = 0.58 ≈1:1 mixture Test 2 AB PHR = 0.5 AB portion = 0.83 AA portion = 0.17 ≈1:6 mixture

Time to Vote 1.Test 1 is correct 2.Test 2 is correct 3.The truth is out there 4.Who cares, a 1:1 mixture looks like a 1:6 mixture anyway 5.I said I wanted lunch

How do you deal with shared alleles? Don’t rely on a major or assumed contributor – That inserts the analyst into the amp tube – The enzyme certainly doesn’t care whose DNA it amps Calculate the PHR and P without bias – Use the same set of rules every time – Calculate every possible combination, then see what fits

Should we do this? Section 3.5 – SWGDAM 3.5. Interpretation of DNA Typing Results for Mixed Samples An individual’s contribution to a mixed biological sample is generally proportional to their quantitative representation within the DNA typing results. Accordingly, depending on the relative contribution of the various contributors to a mixture, the DNA typing results may potentially be further refined.

Should we do this? Section 3.5.2 – SWGDAM 3.5.2. The laboratory should define and document what, if any, assumptions are used in a particular mixture deconvolution. 3.5.2.1. If no assumptions are made as to the number of contributors, at a minimum, the laboratory should assign to a major contributor an allele (e.g., homozygous) or pair of alleles (e.g., heterozygous) of greater amplitude at a given locus that do not meet peak height ratio expectations with any other allelic peak(s). 3.5.2.2. If assumptions are made as to the number of contributors, additional information such as the number of alleles at a given locus and the relative peak heights can be used to distinguish major and minor contributors.

Should we do this? Section 3.5.3 – SWGDAM So don’t forget about proportion between contributors (mixture ratio) (PHR is proportion within a contributor) We’ll mention 3.5.3.1 in a minute 3.5.3. A laboratory may define other quantitative characteristics of mixtures (e.g., mixture ratios) to aid in further refining the contributors.

All combinations for 2 people Grouped by number of alleles Sub-grouped by homozygotes/heterzygotes and sharing/no-sharing

All combinations for 2 people Grouped as “family” or “category” of like types Highlighted top row is generic form Other possible combinations in white

All combinations for 2 people Some don’t have much to calculate

All combinations for 2 people Some have no sharing so easy to calculate

All combinations for 2 people We just saw two categories that take a bit more effort to calculate

Proportional Allele Sharing Method No crazy math Easy to follow the logic of the model Supported by numerous in-house studies It just works All models are wrong, but some are wrong more often than others

Rule 1 – AB BC Sharing Whenever possible, shared alleles are shared proportionately 5001500790

Rule 1 – AB BC Sharing First, consider the alleles that are unshared to get the proportion of the two donors Essentially, you calculate the proportion for two homozygotes 5001500790

Rule 1 – AB BC Sharing Proportion = 5001500790500 + 790 = 0.39 for 10 (,12) 790500 + 790 = 0.61 for (12,) 13 500 790

Rule 1 – AB BC Sharing So based 39%/61% ratio: For 10, 12 donor For 12, 13 donor 5001500790.61.39 1500 .39 = 585 rfu 500 ÷ 585 = 0.85 PHR  ÷ 790.61 = 915 rfu 1500 915 = 0.85 PHR.391500 500585.611500 915790

Rule 1 – AB BC Sharing PHR = 0.85 for both contributors PHR is always the same for proportional sharing model The enzyme doesn’t arbitrarily give one person a good PHR and the other a bad PHR 5001500790

AB BC Sharing Summary Test 1 AB PHR = 0.5 AB portion = 0.27 BC PHR = 0.63 BC portion = 0.73 1:4 mixture Test 2 AB PHR = 0.5 AB portion = 0.54 BC PHR = 0.63 BC portion = 0.46 1:1 mixture Proportional AB PHR = 0.85 AB portion = 0.39 BC PHR = 0.85 BC portion = 0.61 1:2.5 mixture

Rule 2 – AA AB Sharing Whenever possible, PHRs are assumed to be 1.0

Rule 2 – AA AB Sharing Which way? 50% “down” 50% “up”

Rule 2 – AA AB Sharing PHR = 1.0 is the middle ground

Rule 2 – AA AB Sharing PHR = 1.0 is the middle ground Replicate amps Calculate the PHR Not very close to 1.0 1093/1320 = 0.82 797/1013 = 0.79 823/1031 = 0.80 602/894 = 0.68 Ave PHR = 0.805Ave PHR = 0.74

Rule 2 – AA AB Sharing PHR is (typically) smallest/tallest Do it again with first/second as the smallest and tallest switched Pretty close to 1.0 with only 2 replicates (Some folks do HMW/LMW) 1093/1320 = 0.82 1013/797 = 1.27 1031/823 = 1.25 602/894 = 0.68 Ave = 1.04Ave = 0.965

Rule 2 – AA AB Sharing So two reasons for Rule 2 – PHR = 1.0 is the middle ground – PHR = 1.0 fits replicate amps May not fit quite as well at large loci and/or big steps between alleles – eg: 11,23 at D18

Rule 2 – AA AB Sharing For AB (heterozygote): For AA (homozygote): 1300500 800 500 = 1.0 PHR÷ + = 0.56 P 500 = 0.44 P 5001300 + 500 No PHR 800 1300 + 500 1300 800 500 PHR Defined as 1.0!

AA AB Sharing Summary Test 1 AB PHR = 0.5 AB portion = 0.42 AA portion = 0.58 ≈1:1 mixture Test 2 AB PHR = 0.5 AB portion = 0.83 AA portion = 0.17 ≈1:6 mixture PHR = 1.0 AB PHR = 0.5 AB portion = 0.56 AA portion = 0.44 ≈1:1 mixture Similar ratio, but major/minor (sort of) has flipped

An advantage of this approach The proportion of contributors calculated at a specific locus is not dependent upon something calculated at some other locus This allows for consideration of degradation – (When we look at degraded samples using this approach, they kind of “self-correct” meaning with known mixtures, the true types are still usually the best fit.)

An advantage of this approach Section 3.5.3.1 – SWGDAM But you can’t predict from one locus how degradation will affect the next This approach helps, as each locus is independent 3.5.3. A laboratory may define other quantitative characteristics of mixtures (e.g., mixture ratios) to aid in further refining the contributors. 3.5.3.1. Differential degradation of the contributors to a mixture may impact the mixture ratio across the entire profile.

Rule 3 – Minimum Peak Height Minimum peak heights (mph) are always maintained and supersede Rules 1 and 2. We won’t discuss this much now Analogous to your peak calling threshold (75 rfu or 100 rfu, etc) Or based on the mixture ratio (proportions)

Rule 3 – Minimum Peak Height Comes into play for certain combinations Think of looking for an AB and BB contributor – 12,12 homozygote? How many RFU? – We just saw this example – The other homozygote option MPH gives a starting point

Three Person Mixtures These simple rules work for three person mixtures also Most (well, lots anyway) 3 person mixtures break down into simple patterns that we just discussed for 2 person mixtures – Rule 1 – Rule 2

Three Person Mixtures PHR and P for Donor 1 is straight forward – 6,7 Donors 2 and 3 is AA AB pattern (Rule 2) – 9,9.3 – 9.3,9.3 Donor 1 Donors 2 and 3

Three Person Mixtures PHR and P for Donor 1 is straight forward – 24,26 Donors 2 and 3 is AB BC pattern (Rule 1) – 21,22 – 22,25 Donor 1 = 24,26 Donors 2 and 3 are 21,22 and 22,25

Three Person Mixtures You just have to realize some calculated PHR and P results have two contributors added together – Victim ≈ 15%, Consensual ≈ 35%, Foreign ≈ 50% – AB AB CD locus (V and C are both AB) P for AB = 48% (combined known V and C) P for CD = 52% for F

Three Person Mixtures This is where it gets a bit tricky for three person mixtures

Three Person Mixtures 4 types where we cannot calculate PHR and P B CA A B A CB C A BA CB D B A A B

Three Person Mixtures Two alleles shared by two people (twice) – “Circular” sharing – “Double” sharing In these cases, upper and lower boundaries can be calculated based on PHR to determine if viable – “Not Excluded” result – Increase PHR stringency to “test” fit – If type with defined PHR and P dropout but not the “Not Excluded” option…

A computer can help Calculate the PHR and P for every possible combination using Rules 1, 2, and 3 – 3 Contributors in a 4 allele pattern: 6 “families” of types 52 total combinations – 3 Contributors in a 5 allele pattern Only 2 “families” But one contains 30 combinations

A computer can help Filter the possibilities shown to the analyst: – Don’t show combinations with low PHR (eg: <50%) – Don’t show combinations with proportions of 5% when you know your minor is at least 20% (4-fold difference) – Don’t show combinations that do not include a known donor (V on own panties) Starts to become fairly manageable

A computer can help Assumes good data – Can’t do much with a 3 person mixture that only had 100pg of DNA in the first place – Deconvolution works best when you are in a range that your validation says PHR’s are robust Even if you can’t deconvolute the mixture, you may be able to limit the possible types present to a manageable number – (Statistics…)

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