Frequency Primer
Using Allele Frequencies to Calculate Profile Probability
When the parental DNA profiles are unknown, an individual's profile may be considered the result of a random draw of alleles from a population genetic pool. The probability that unrelated individuals from the same population will have a certain DNA profile can be calculated using population allele frequencies. Here is how.
Every diploid individual has two alleles at an autosomal (not sex-linked) nuclear (not mitochondrial) locus. Let’s say that an individual is homozygous for a particular STR marker. The probability of that individual inheriting one copy of each allele is equal to that allele's frequency, call it p, so the probability of inheriting one copy from their mother and one copy from their father is p × p = p2. If an individual is heterozygous, then the formula is 2(pq), where p is the frequency of the first allele, and q is the frequency of the second allele. Why is the product doubled? Because there are two ways to get a heterozygous genotype: you could inherit allele 1 from your mom and allele 2 from your dad, or allele 2 from your mom and allele 1 from your dad.
Let's look at a real example. For the FH71 marker, the frequency of the 70-bp allele is 0.09. The probability of an elephant from one of those populations being homozygous for the 70-bp allele is (0.09)2 = 0.008 or 0.8%. If you now look at marker FH67, the frequencies of the two alleles in the ivory sample are 0.1 and 0.24. The probability of an elephant being heterozygous for those two alleles is 2(0.1)(0.24) = 0.048.
To calculate the probability of an elephant having the same genetic profile as that of the ivory sample, you would calculate the probability of being homozygous or heterozygous at each of the four loci and then multiply all those probabilities together.
If you work through the calculations, you should find that the probability of an elephant having this exact genetic profile is 1.12 × 10−8. By dividing 1 by that probability we can determine that around 1 in about 90 million elephants would have this exact profile (1/1.12 × 10−8 = 90 million). The chance is therefore about 1 in 90 million that an elephant other than the one found dead in Garamba National Park would have this genetic profile.
