Question

A suspect in a rape case has been typed for three STR loci. His alleles match those of the evidentiary sample (semen taken from the rape victim) for each locus. He is a heterozygote for the first two loci and a homozygote for the third. The allele frequencies for locus 1 in the general population are .05 and .10. For locus 2 they are .07 and .02. for locus 3, the allele frequency in the general population is .08. what is the probability that a random individual in the general population would match the evidentiary sample?

Answer 1

solution:

Locus 1:
Frequency of Allele 1 = p₁ = 0.5
Frequency of Allele 2 = q₁ = 0.1

Locus 2:
Frequency of Allele 1 = p₂ = 0.7
Frequency of Allele 2 = q₂ = 0.2

Locus 1:
Frequency of Allele 1 = p₃ = 0.8

Therefore,
Genotypic frequency of Locus 1 with Allele 1 and Allele 2 = 2p₁q₁ = 2 * 0.5 * 0.1 = 0.1
Genotypic frequency of Locus 2 with Allele 1 and Allele 2 = 2p₂q₂ = 2 * 0.7 * 0.2 = 0.28
Genotypic frequency of Locus 3 with Allele 1 = (p₁)² = 0.8 * 0.8 = 0.64

Therefore, frequency of co-occurence of these three Genotypes = 0.1 * 0.28 * 0.64 = 0.01792

Therefore the probability that a random individual from the population would match the evidentiary sample is 0.01792 or 1.792%

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