Forensic Mathematics home page
STR mutation model in DNAVIEW
Multiple mutations, covert mutations, and false exclusions in paternity casework
STR mutations data from the AABB
STR loci, multiplexes, and DNAVIEW
Mutational processes of simple-sequence repeat loci in human populations
(Di Rienzo et al, PNAS 1994)
The "two exclusion" rule Treatment of mutations STR Approach
If a battery of a dozen STR systems is used for paternity testing, occasionally two inconsistencies can be expected even when the man is the father. On the other hand, we can also expect occasionally to see only two inconsistencies when the man is not the father.
Therefore calculating a paternity index for mutation is vital.
If two inconsistent loci are deemed (nearly) enough to rule "exclusion" for the whole case, in effect two inconsistent loci together amount (nearly) to a combined PI=0. One inconsistent locus therefore deserves a very small PI, 0<PI<<1.
For RFLP's, I use the simple formula PI=µ, where µ is the observed rate of mutations/meiosis for the locus, when a locus shows a result inconsistent with paternity.
The rule that PI=µ is based on this simple model of mutation, assuming for the sake of illustration an obligatory paternal gene of Q:
X = P(man without gene Q will contribute Q)
= P(contributed gene will mutate) P(mutated gene will be a Q)
= µ P(Q).
Y = P(random sperm is Q) = P(Q).
X/Y = µ.
That is, the model assumes that the chance a gene will mutate to Q is proportional to the prevalence of Q genes in general, without regard to such possible complications as different probabilities for a small vs. a large or positive vs. negative mutation size change.
The AABB recommends a slightly different formula, for my facile analysis above is wrong. The evaluation of Y is slipshod. Better is
Y = P(paternal gene is Q and man has no Q)
= P(paternal gene is Q) · P(man has no Q)
= P(Q) · A
X/Y = µ/A,
where A is the probability of exclusion; the probability that a random man would have a pattern inconsistent with paternity at this locus. The extra complication does not change the result very much (since normally A is nearly 1) and the relative improvement is anyway dwarfed by the inherent inaccuracy of the method for estimating X.
The AABB however uses not the case-specific power of exclusion, A, but the mean power of exclusion, . I'm not sure why, but in either case the formula is correct on average (which mine is not), and in neither case is any of the three formulas anywhere near accurate.
Late breaking news: I have been told that as of December 2004
the official AABB recommendation is the model discussed on this page.
Later: But when I looked they seem equivocal, accepting of either my method or of "Fimmers" which would be the old RFLP way.
The general rule is that mutations change allele length by a small amount. Thus when the man mismatches the child allele by a small amount there is a lively chance of a mutation but the formula underestimates it, tending to help true fathers get off the hook. Conversely, when the alleles mismatch by a large amount, the same computation unfairly inflates the possibility of a mutation, perhaps unfairly overstating the evidence against a non-father.
Nonetheless, as regards RFLP, I don't have a better formula to suggest. Vigilance.
Therefore as a rule of thumb I suggest assuming that
Suppose the mother is PP, the child is PQ, and the man is Q'R, where Q' is s=1 or 2 steps smaller (or larger) than Q. As usual, let µ be the overall mutation rate for the locus, measured in mutations/meiosis.
Then X = P(genotypes | true trio) is proportional to
P(Q'R man transmits Q) = P(Q' is transmitted) x P(mutation) x P(mutation increases length) x P(s steps)
= (1/2) µ (1/2) (1/10)s-1, and
Y = P(genotypes | false trio) is proportional to
P(Q), the frequency of the Q allele among random sperm.
|Note Apr 2010 This model is almost the same as the 1994 model of Di Rienzo et al (link above).|
Therefore I propose the approximations that the paternity index X/Y
= µ / (4 P(Q)) if the mutation is 1 step,
X/Y = µ /(40 P(Q)) if the mutation is 2 steps, etc.
|PP||PQ||Q'R||µ / (4 PQ) if the mutation is 1 step, µ /(40 PQ) if the mutation is 2 steps, etc.|
|17||17 18||19||µ / (2 P18)|
|12 14||12 14||15 16||µ / (4(P12 + P14)||ambiguous paternal allele|
|12 14||12 14||13 15||3 µ / 4(P12 + P14)||3 possible paternal mutations|
|15 16||15 17||15 16||µ(1+1/3.5) / (4 P17)||paternal or (less likely) maternal mutation|
|none||20.2 23||21.2 23.2||µ / (8 P20.2)||23.2 > 23 mutation too rare to consider (see Lower bound)|
|17||17 18||19||µ / (23.5 P18)||maternity case|
|12 15||13 16||16||1/(P16 + P13).||Since same (maternal) mutation possibilities are involved in both
numerator and denominator of the LR formula, the µ's cancel
and it is the same formula as if mother=child (i.e. no mutation).
Thanks to Li Haixia of Sun Yat-Sen University for a correction. 29July2009
|12 14||13 15||15||1/(P15+P13/2)||Again the µ's all cancel because both numerator and denominator require maternal mutation. But there are 3 possible mutations for the denominator and only 2 for the numerator. So this case is not equivalent to any "no-mutation" pattern.|