Twin Zygosity Studies with DNA•VIEW's Kinship Module
Twins V and C have the same DNA types in tested autosomal loci. Are they identical?
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 pseudomom  [race=c  16
 17
 16
 13

2 B pseudodad  type=kinship]  16
 17
 23
 14

The trick is the introduction of constructed "parents" A and B, whose homozygous genotypes at every locus
are chosen to guarantee that all of their children would look like V.
Running the example
 Download twin.gen or find it in the
\dnaview\examples\
directory
 In DNA•VIEW, Import/Export Genotyper Import
to import the data and create case number 2
 If you already have a case #2, then first either delete it, or modify the
twin.gen
file.
 When Genotyper Import volunteers to create the case, say y.
 In the Casework menu, select Automatic kinship
 select case 2
 edit race list type desired race letter, Enter
 immigration/kinship
 Type in scenario in either of these ways:
(concise way)  (simple way)


C : Mom + Dad V : A/Mom + B/Dad

C : Mom + Dad ; identical twins
V : A + B
/ C,V : Mom + Dad ; dizygotic twins

and then Esc
 Calculate and report LR's, one race
 Show summary of calculation
Result
Locus  Likelihood ratio  frequencies 
D3S1358  4 / (1+2p+pp)  =2.67  p=0.224 
VWA  4 / (1+2p+pp)  =2.55  p=0.252 
FGA  4 / (1+p+q+2pq)  =2.91  p=0.18 q=0.142 
D8S1179  4 / (1+p+q+2pq)  =2.39  p=0.349 q=0.19 
D21S11  4 / (1+p+q+2pq)  =3.51  p=0.0574 q=0.0748 
D18S51  4 / (1+p+q+2pq)  =3.12  p=0.145 q=0.107 
D5S818  4 / (1+2p+pp)  =2.06  p=0.394 
D13S317  4 / (1+p+q+2pq)  =2.88  p=0.314 q=0.0449 
D7S820  4 / (1+p+q+2pq)  =2.87  p=0.176 q=0.161 
D16S539  4 / (1+p+q+2pq)  =2.16  p=0.299 q=0.344 
TH01  4 / (1+p+q+2pq)  =2.24  p=0.254 q=0.352 
TPOX  4 / (1+2p+pp)  =1.6  p=0.584 
CSF1PO  4 / (1+p+q+2pq)  =2.4  p=0.269 q=0.259 
cumulative LR   164000   



Why it works
The typed in scenarios parse as
Scenario H_{1}  Scenario H_{0} 
C : Mom + Dad V : A + B 
C : Mom + Dad V : Mom + Dad 
The scenarios we wish to compare are:
 H_{1}: C and V are genetically one person
 H_{0}: C and V are genetically ordinary full siblings
Recall that the necessary likelihood ratio is the ratio
P(E  H_{1}) / P(E  H_{0}),
where E is the event
E: Genotypes as stated are observed in four people C, V, A, B.
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  H_{1}) = 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  H_{1}) = 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.
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.
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