Paternity calculation with 3banded pattern
Sometimes the child and alleged father in a paternity case each
appear to have three alleles at a particular locus. For example,
suppose the genetic evidence E is
Apparently the man and the child share a genetic trait – quite a
peculiar one in fact. Therefore the types at this locus are evidence
favoring paternity. But how much evidence? How do we quantify it?
Considerations
It is impossible to make any sort of computation without knowing (or
at least assuming) something about the underlying biology.
What is it?
Fortunately it can now pretty well be assumed that at least the vast
majority of instances of a 3banded pattern are due to two bands
coming from one chromosome (i.e. not trisomy). In the RFLP case, we
can assume that a mutation has introduced an extra cleavage site
within the tandem repeat region.
"... can now pretty well be assumed ..."
February 2014 — I was wrong about that.
I was half right — it's not necessarily (or even often?) trisomy.
But it didn't occur to me that there's a third possibility until I read the brilliant article
"The nature of triallelic TPOX genotypes in African populations", Lane AB
Forensic Sci Int Genet. 2008 Mar;2(2):1347. doi: 10.1016/j.fsigen.2007.10.051. Epub 2007 Nov 26.
It reads like a statistical detective story, gradually revealing the explanation for the observed data.
In summary, I can now think of at least three explanations for triallelism:
 Tandem duplication. The person has the usual two chromosomes one of which has an extra allele.
That's the model this web page base was based on.
 Trisomy (as Downs)
 Extra allele on a completely different chromosome (Lane's explanation)
and each would lead to a different calculation

In any case, we will assume that two of the bands are transmitted as a unit. Let's call these two
linked bands a doublet.
Frequencies of traits
I have heard reports that, in the population of chromosomes that show
the extra band, certain band sizes are very common and typical. This
would be expected if the cleavage mutation occurred fairly recently
and only once or a few times, and if there has therefore not been
much opportunity for subsequent lengthaltering mutations to create a
great variety of band sizes among bands that are part of a doublet.
Certainly there is no reason to expect any relationship between
the distribution of doublet band sizes and of nondoublet band sizes.
Bad strategy
Therefore I don't agree with the approach that sometimes seems
tempting, to calculate a paternity index based on treating one or the
other of the doublet bands as a normal band. The frequency that will
be obtained for that one band will surely come from a database of
normal chromosomes only, and therefore conceivable it will be
unfairly small. It certainly will be irrelevant.
Preferred strategy
A better strategy, I think, is to concentrate on the obvious genetic
peculiarity – the extra cleavage site (or whatever causes the
doublet). That is a sufficiently rare trait that by regarding it, and
even ignoring the specific band sizes of the doublet (except if they
happen to exclude the man), we probably can obtain sufficient and
useful information from the locus while making justifiable and
conservative assumptions.
In other words, think of the all instances of the locus that have
a doublet as belonging to a classes of alleles that we shall call
M (for cleavagemutation). Sometimes we
will distinguish different doublets with subscripts –
M_{PS} meaning a doublet with
individual bands P and
S.
For the typical case mentioned above,
with evidence E, we note that the mother passed a
Q to the child, and the biological father
therefore passed a chromosome containing
RS. That is, we take the view that he
passed the trait M, and more specifically
that he passed M_{RS}.
The tested man certainly has M, but what
kind of M? His genotype can be any of
RM_{ST},
SM_{RT}, or
TM_{RS}. If we had no data about
the relative frequencies of the different varieties of
M, then we would assume that they are all
equally common. In that case it would be right to say that the
relative frequencies of the three possible genotypes are in
proportion r:s:t (where r,
s, and t represent the frequencies of the corresponding
alleles) – hence that
the chance that the man is
TM_{RS} is
t/(r+s+t).
The fact that M_{RS} has been
observed at least once – in the child – means that (1) is "conservative". It is an underestimate for the
chance that the man has the requisite paternal variety of
M. The probability for the man to transmit
this requisite allele is therefore at least
P(man transmits RS) >
t/(2·(r+s+t)),
and we can regard (2) as the numerator,
X, of the PI (paternity index) in this case.
The denominator, Y, is the frequency in the population of
chromosomes of the trait M_{RS}.
Probably the data will not be available to estimate that frequency.
However, obviously
P(M_{RS})<P(M),
so an adequate (conservative) estimate for Y is
P(M). In summary,
.
Rough estimate
There is no reason to expect any of T,
R, or S to be
more or less common than any of the others, so on average
t/(r+s+t) = 1/3. If, as seems
likely, we can be confident that fewer than one person in 100
exhibits M, then
P(M)<1/200. Therefore the
recommended approach will probably give PI>33 – no doubt
a vast underestimate, but useful.
The motherless problem
Suppose the genetic evidence E is
Solution to Motherless Problem
It is possible that the child received M
from its mother and only a single band from the man. However, we
ignore this unlikely possibility.
The child and man are related only if
 the child is QM_{RS}
 the man is TM_{RS}
 the mother transmitted Q
 the man transmitted M_{RS}
Evaluating these possibilities and multiplying together,
,
and for the genotype frequency of the child,
.
Therefore we can say that
.
In the typical case that
q=r=s=t, this amounts to
.
A sharper estimate for P(M)
than 1/200 will be well worthwhile here!
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