	Contents examples genotype table Running the example Result Why it works What if the real parents are typed?
	Links Symbolic Kinship Program – discussion publication reference Operation of the program Detailed example Other information about DNA•VIEW Forensic mathematics home page

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.

A text file ("allele table" format) with the data (partly) shown in the table just below.
Excel version of the same file

Sample Info	D3S1358 1	D3S1358 2	VWA 1	VWA 2	FGA 1	FGA 2	D8S1179 1	D8S1179 2
2 C twin	16	16	17	17	21	23	13	14
2 V ident twin?	16	16	17	17	21	23	13	14
2 A pseudo-mom	16	16	17	17	21	21	13	13
2 B pseudo-dad	16	16	17	17	23	23	14	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 as follows:
  
  C : Mom + Dad V : A/Mom + B/Dad
  
  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 H₁	Scenario H₀
`C : Mom + Dad V : A + B`	`C : Mom + Dad V : Mom + Dad`

The scenarios we wish to compare are:

H₁: C and V are genetically one person
H₀: C and V are genetically ordinary full siblings

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.

Go to the top of this page