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?
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.
Certainly there is no reason to expect any relationship between the distribution of doublet band sizes and of non-doublet band sizes.
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 cleavage-mutation). Sometimes we
will distinguish different doublets with subscripts
MPS meaning a doublet with
individual bands P and
Solution to Trio Problem
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 MRS.
The tested man certainly has M, but what kind of M? His genotype can be any of RMST, SMRT, or TMRS. 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 TMRS is t/(r+s+t).
The fact that MRS 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 MRS. Probably the data will not be available to estimate that frequency. However, obviously P(MRS)<P(M), so an adequate (conservative) estimate for Y is P(M). In summary,
The child and man are related only if
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!