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Solving this problem with the Symbolic Kinship Program will require a little cleverness.
To illustrate the method, here is a worked example. You can download the files and try it out.
Sample Info | Comment | D3S1358 | VWA | FGA | D8S1179 | ||||||||||||
2 C twin | Zygosity | 16 16 | 17 17 | 21 23 | 13 14 | ||||||||||||
2 V ident twin? | test example | 16 16
17 17
| 21 23
| 13 14
| 2 A pseudo-mom | [race=c | 16
| 17
| 16
| 13
| 2 B pseudo-dad | type=kinship] | 16
| 17
| 23
| 14
| |
\dnaview\examples\
directory
twin.gen
file.
(concise way) | (simple way) |
---|---|
C : Mom + Dad |
|
Scenario H1 | Scenario H0 |
---|---|
C : Mom + Dad |
C : Mom + Dad |
The scenarios we wish to compare are:
Since A and B are ancestral under both hypotheses, the chance to see people such
as A and B is the same under both hypotheses and that chance, whatever it is, cancels
in evaluation of the likelihood ratio. Therefore the likelihood ratio is the same if
we instead define the E as the event
E: Genotypes as stated are observed in C and in V.
Then the numerator becomes
P(E | H1) = P( C ) P( V | A, B as parents).
However (and here's the trick), P( V | A, B as parents) = 1. Therefore the
numerator reduces simply to
P(E | H1) = P( C ).
The denominator is in effect P(two siblings like C), and the ratio of these
two is what we need; nameely, the likelihood ratio favoring monozygosity over
dizygosity.
What if the real parents are typed?
If typing are available for one or both of the real parents, or for that matter
of a sibling or other relative, just add the types for that person to the
table.
Don't change A and B leave them just as defined above. Remember, they were never the real parents to begin with. They're just a coding construct, a part of a programming trick.