Haplotype DNA evidence

1. 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.

   May2018, note added 5-10 years after initially writing this web page: This page discusses only how to handle the complication of relatives and possible mutation. It doesn't deal with the difficult questions of evaluating what I loosely call "Pr(haplotype)" below, or we'd have many further reasons to list:

2. 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.
3. Autosomal STR alleles matching probabilities or frequencies can reasonably be estimated as sample frequencies; for Y haplotypes that is far from true.
4. When using population data to estimate the significance of an autosomal STR allele match, it's close to adequate to consider only the allele in question. For Y by contrast, it's important to look at the entire database.
5. Putting confidence intervals on a matching probability is mathematically ignorant. However, while it's a fairly harmless error in the autosomal domain, it's a crippling mistake when dealing with Y haplotype matching probabilities.
6. Obviously all men are related. Forgetting that when dealing with Y misses the fundamental point that nearly all Y matching is from identity by descent. That's not so with a single STR locus as in the autosomal situation.
7. "Theta" — the chance of two alleles or haplotypes being IBD — in autosomal practice is a minor adjustment to the matching chance. With Y haplotypes it's the main thing; it's a good approximation to the matching chance and anything else is a minor adjustment.
8. Explicit models — a careful mathematical approach laying out premises and deriving results from them — can be overlooked in autosomal practice without tragedy. In Y haplotype practice they are vital, and their persistent absence from all early papers many recent ones results in nonsensical recommendations and practice.
9. Geographical clustering is a vital concern with Y haplotypes; in autosomal work not so much.

Y-haplotype in identification 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 expected (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.

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

matching odds = 1 / Pr(haplotype).

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:

PI = 1 / Pr(haplotype).

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 μ=1/400 per locus per generation for single step mutations, but with a lot of variation depending on the locus.

Suppose a man M has Y-haplotype which we call Y_M and a boy C has the type Y_C which differs from Y_M by a single step at just one locus.

Obviously, mutation cannot be ignored in this case. Since μ is the probability of any mutation, but nearly all (90-95%) STR mutations are one-step and expansion and contraction are about equally common, to a reasonable approximation the probability to mutate in either direction between Y_C and Y_M is μ/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 of C) and
Y = Prob(observed haplotypes | F unrelated to C).

To evaluate Y, we can write
Y = mc where
m=Prob(Y_M) and
c=Prob(Y_C).

X is a little more problematic.

child-centric approach

The child has Y_C, inherited from his father. A mutation between Y_C and Y_M may have occurred, with probability μ/2. Therefore, given that a child is type c the probability is approximately μ/2 that his father is type Y_M.

Hence
X = c•μ/2 and
LR = X/Y = X/cu = 3μ/2u.

It remains to estimate u.

father-centric approach

In a symmetrical way we could begin with the alleged father, and obtain instead the formula
LR = 3μ/2c.

Which approach is right? How to estimate c and/or u?

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

Note that all formulas are equivalent if c = u. Therefore to be conservative let's take the uncle-centric view and take c=2/171.

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.

Approach to "frequencies"

Frequencies of an unobserved trait is impossible to know. Fortunately frequency isn't the question. Probability is.

My papers on rare haplotypes offer 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

Mitochondrial ID of a sister

Feb2014
Body B and reference sister S have mitotypes mT_B and mT_S which differ only by one base, perhaps a mutation. What is the LR supporting the hypothesis that B is the lost sister of S?

Formulate mitochondria LR for identification

LR = X/Y where
X = Pr(sisters have types mT_B and mT_S),
Y = Pr(randomly chosen people have types mT_B and mT_S).

The difficulty in evaluating X is that we don't know which sister mitochondria represents a mutation. We can take a mother-centric approach, considering that their common mother had one type or the other and assuming that one of the sister mitochondria is a mutation. Then there are two possibilities for the mother type, so

X = Pr(mother=mT_B and S is a product of mutation) +Pr(mother=mT_S and B is a product of mutation)
   = Pr(mT_B)μ + Pr(mT_S)μ
   = μ[Pr(mT_B) + Pr(mT_S)].

Probability of type mT_B

As part of the answer we will need mitochondrial random matching probabilities. As an example evaluate

b=Pr(random person = mT_B | mT_B observed in body B).

Let's say mT_B occurs x times in our reference database of N mitotypes. Per AlleleProbability.htm, b < (x+1)/(N+1) (where the occurrences of +1 represent conditioning on the casework observation of mT_B).

We can improve on that by accepting the logic of the "κ method" according to which b = λ(x+1)/(N+1). Here λ is a factor less than one and independent of x. (Specifically λ = 1 – κ, i.e. λ is the proportion of the N database mitotypes observations that are not singletons.)

Solving the LR expression

What Y is

Evaluating Y is easy. Under the assumption that B and S are just unconnected random observations,

Y = Pr( ... )
ugh! Can NOT condition on the B & S types!