## Table of contentsHaplotype DNA evidenceAnalysis of Y-haplotype information in a kinship case Forensic mathematics home page Comments are welcome (see home page for email) |

- Several markers are linked, i.e. physically chained and inherited together, so they must be considered as a unit. No recombination. The product rule doesn't apply at all.
- The genetic rules are simpler – the trait is either known to be passed or known not to be passed (depending on the sexes involved) to each offspring; there are no choices or 50% probabilities of transmission as with nuclear DNA.

- Y-haplotype in indentification and paternity I discuss here basic principles in using Y-haplotype information for identity or paternity.
- Identity Suppose suspect and crime stain have the same Y-chromosome haplotype. That result is normal and expect (i.e. 100%) if the suspect is the donor; it is the probably of seeing the haplotype among random men if the suspect is a random man.
- Paternity – ordinary case Typically father and son share a Y-haplotype just as if the son were a crime scene. Therefore in the typical case the equation above also gives the paternity index:
- Paternity – mutation Of course that's not 100% true; there are mutations. Available data supports that the mutation rates and behavior for STR loci on the Y-chromosome are typical for the genome; so around
- child-centric approach The child has
- father-centric approach In a symmetrical way we could begin with the alleged father, and obtain instead the formula
- Which approach is right? How to estimate
c and/oru ?
Deep questions. What is right depends on such things as what you think
the population database represents – grandfather's generation?
the child's? If the population were in drift and mutation equilibrium,
then I suppose all methods would give the same answer.
- Pragmatic estimate of the Y-haplotype evidence
- Approach to "frequencies" Frequencies of an unobserved trait is impossible to know. Fortunately frequency isn't the question. Probability is.

The strength of the evidence is therefore simply expressed as matching odds (or equivalently as a likelihood ratio) of

matching odds = 1 / P(haplotype).

Suppose a man M has Y-haplotype which we call _{M}_{C}_{M}

Obviously, mutation cannot be ignored in this case. Since _{C}_{M}**/2**.

There are several possible approaches. We use the notation PI for the
paternity index, and

PI = X/Y, where

X = Prob(observed haplotypes | F father C) and

Y = Prob(observed haplotypes | F unrelated to C).

To evaluate Y, we can write

Y = _{M}_{C}

X is a little more problematic.

Hence

X = **• μ/2** and

LR = X/Y = X/

LR =

Note that all formulas are equivalent if

Hence LR = 3•0.009/2(2/171) = 1.15. |

The meaning of this neutral result is that the chance to see so rare a haplotype by mutation is about the same as the chance to see it at random in an unrelated individual.

My recent paper on rare haplotypes offers several approaches. Bottom line: simple counting (but add 1) is very conservative. A pretty accurate method that is not complicated is also given. June 2009

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