DNA profiling in paternity testing

The inheritance of alleles follows the basic pattern of inheritance discussed in Section 6.2.6. A child can only receive one of the father’s alleles and one of the mother’s. By looking at the DNA profiles of mother, father and child there should be a clear relationship. The technique can be used to establish paternity where the father might be denying it or where he wants to claim it. More unusually it can prove maternity. In cases where babies are mixed up in hospitals it can be returned to the correct parent. A wide variety of issues revolve around familial relationships and where these might be in question, for whatever circumstances, DNA profiling is a powerful tool allowing their resolution.

Table 6.6 gives summarised profile data for a hypothetical paternity case where there are two alleged fathers. Comparing the daughter's genotypes for each locus to the mother and the alleged fathers allows a decision to be made. Consider just locus D3S1358, the child has genotype 16, 17. Allele 16 must be from the mother as neither alleged father has this allele. Allele 17 of the child must be from the father and only alleged father 2 has allele 17. This effectively excludes alleged father 1 as being the father. Applying the same analysis for all the other loci confirms this. For each allele that is not maternal in origin, alleged father 2 can be the source and so the evidence favours alleged father 2 as being the father.

Table 6.6. Paternity case.

Locus / Mother
genotype / Childs genotype / Alleged
Father 1 / Alleged
Father 2
D3S1358 / 15, 16 / 16, 17 / 14,15 / 13,17
vWA / 18, 19 / 18, 18 / 14, 16 / 13, 18
D16S539 / 10, 11 / 9, 10 / 9, 10 / 9, 12
D2S1338 / 19, 23 / 23, 24 / 20,23 / 19, 24
D8S1179 / 13, 13 / 13, 14 / 12, 13 / 14, 18
D21S11 / 28, 29 / 28, 30 / 28,31 / 30, 33
D18S51 / 13, 16 / 13, 14 / 12, 15 / 14,17
D19S433 / 12, 15.2 / 15.2, 16 / 15, 16 / 16, 16
THO1 / 9.3, 9.3 / 7, 9.3 / 7, 9.3 / 7, 10
FGA / 19, 22 / 22, 22 / 24,26 / 22, 22
Amelogenin / XX / XX / X,Y / X,Y

As discussed in Section 6.4.3 however this is not necessarily proof of paternity, it depends on how common the alleles donated by the father are in the population. A quantity called the Paternity Index (PI) can be calculated for each locus and then these are combined. The paternity index is a ratio of the likelihood that the alleged father could have contributed the non-maternal allele to the likelihood that the allele was from another, random, male.

Taking locus D3S1358 as an example, allele 17 of the child must be from the father. Alleged father 2’s genotype is 13, 17 and hence the likelihood that he contributed allele 17 is 0.5 (he is equally likely to donate allele 13). The chance that a random male in the population could have been the source of allele 17 of the child depends on the frequency of the allele in the population. From allele frequency tables this value is 0.22. The paternity index is therefore 0.5/ 0.22 = 2.27. This means that it is 2.27 times more likely that the allele was from the alleged father than from another male.

D19S433 in the alleged father is homozygous for allele 16, the likelihood that he passed this onto the child if he is the father is 1.0; he has no other allele to pass on. The chance that allele 16 was from a random male is the allele frequency again in this case 0.053. The paternity index for this locus is therefore 1.0/0.053 = 18.9. These values are calculated for each locus in the profile as shown in Table 6.7.

Table 6.7. Paternity indices (PI) for alleged father 2 in Table 6.5.

Locus / Childs genotype, the paternal allele is underlined / Alleged
Father 2 / Frequency of paternal allele (f) / PI = 0.5/f / PI = 1.0/f
D3S1358 / 16, 17 / 13,17 / 0.223 / 2.24
vWA / 18, 18 / 13, 18 / 0.258 / 1.94
D16S539 / 9, 10 / 9, 12 / 0.108 / 4.63
D2S1338 / 23, 24 / 19, 24 / 0.100 / 5.00
D8S1179 / 13, 14 / 14, 18 / 0.188 / 2.66
D21S11 / 28, 30 / 30, 33 / 0.263 / 1.90
D18S51 / 13, 14 / 14,17 / 0.168 / 2.98
D19S433* / 15.2, 16 / 16, 16 / 0.053 / 18.9*
THO1 / 7, 9.3 / 7, 10 / 0.160 / 3.13
FGA* / 22, 22 / 22, 22 / 0.165 / 6.06*

* these loci are homozygous in the alleged father and the PI is calculated using PI = 1.0/f.

From the individual PIs a combined PI can be calculated by multiplying all the values together. In this example the combined PI = 2.24 x 1.94 x 4.63 x 5.00 x 2.66 x 1.90 x 2.98 x 18.9 x 3.13 x 6.06 = 543164. It is about 500,000 times more likely that alleged father 2 is the biological father than another male. This value is rather unwieldy and is often converted to the probability of paternity (PrP) using the formula, PrP = combined PI/(1 + combined PI). This relationship is based on a Bayesian approach . The value obtained above corresponds to PrP = 543164/ (1 + 543164) = 0.99998. This would strongly support the hypothesis of the actual paternity of alleged father 2.

. This analysis sounds very convoluted and becomes more complicated than perhaps the initial matching of maternal and paternal alleles would seem to warrant. In paternity, clearly other evidence will be relevant, perhaps alleged fathers 1 or 2 were the only possible fathers. Evidence of associations and sexual relationships would be pertinent. The use of statistics attempts to introduce objectivity into the interpretation of the DNA profiles

The web site http://paternity.forensic.gov.uk/index.php?content_id=30 at the Forensic Science Service, UK gives a good background to Paternity Testing.