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 InfoCommentD3S1358VWAFGAD8S1179
2 C twin Zygosity16 1617 1721 2313 14
2 V ident twin? test example16 16 17 17 21 23 13 14
2 A pseudo-mom[race=c 16 17 16 13
2 B pseudo-dadtype=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

  1. Download twin.gen or find it in the \dnaview\examples\ directory
  2. In DNA•VIEW, Import/Export Genotyper Import to import the data and create case number 2
  3. In the Casework menu, select Automatic kinship


LocusLikelihood ratiofrequencies
D3S13584 / (1+2p+pp)=2.67p=0.224
VWA4 / (1+2p+pp)=2.55p=0.252
FGA4 / (1+p+q+2pq)=2.91p=0.18 q=0.142
D8S11794 / (1+p+q+2pq)=2.39p=0.349 q=0.19
D21S114 / (1+p+q+2pq)=3.51p=0.0574 q=0.0748
D18S514 / (1+p+q+2pq)=3.12p=0.145 q=0.107
D5S8184 / (1+2p+pp)=2.06p=0.394
D13S3174 / (1+p+q+2pq)=2.88p=0.314 q=0.0449
D7S8204 / (1+p+q+2pq)=2.87p=0.176 q=0.161
D16S5394 / (1+p+q+2pq)=2.16p=0.299 q=0.344
TH014 / (1+p+q+2pq)=2.24p=0.254 q=0.352
TPOX4 / (1+2p+pp)=1.6p=0.584
CSF1PO4 / (1+p+q+2pq)=2.4p=0.269 q=0.259
cumulative LR164000

Why it works

The typed in scenarios parse as
Scenario H1Scenario H0
C : Mom + Dad
V : A + B
C : Mom + Dad
V : Mom + Dad

The scenarios we wish to compare are:

Recall that the necessary likelihood ratio is the ratio P(E | H1) / P(E | H0), 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 | 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.

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