1) You have an electropherogram with a clean, three-locus profile:
STR locus #1 is 10, 14
STR locus #2 is 30 (only)
STR locus #3 is 3, 6
What is the random match probability of this profile? Fill in the blank:
"Approximately one in every
people, drawn randomly from the human
population, would be expected to have this three-locus profile.
Answers
Following the product probability rule and assuming independence among loci, the random match probability of this three-locus profile is 0.000853. See below the procedure for question number 2.
What does the product probability rule state?The product probability rule allows us to calcuate the probability of occurrence of two or more events at the same time. It is about a joint probability of two or more events that might happen simultaneously, not excluding each other.
This rule is based on the dependence or independence of the events. Two events, A and B, are independent of each other if one of them does not affect the occurrence of the other one.
The rule establishes that the occurrence probability of two independent events -A and B- together is the multiplication product of the probability of occurring A by the probability of occurring B.
P(A∩B) = P(A) x P(B)
In the exposed example,
STR locus #1 has three alleles → 10, 11, 14STR locus #2 has four alleles → 27, 28, 28.2, 30STR locus #3 has three alleles → 3, 6, 7The frequency of each allele is detailed in the table.
According to the electropherogram, the profiles are
STR locus #1 → 10, 14 → Heter0zyg0usSTR locus #2 → 30 → H0m0zyg0usSTR locus #3 → 3, 6 → Heter0zyg0usKnowing the allele frequencies, we can get the genotypic frequencies,
Locus Allele frequency Genotypic frequency
STR locus #1
10 0.22 2xpxq = 2 x 0.22 x 0.18 = 0.079
14 0.18
----------------------------------------------------------------------------------------------------
STR locus #2
30 0.22 p² = 0.22² = 0.048
----------------------------------------------------------------------------------------------------
STR locus #3
3 0.45 2xpxq = 2 x 0.45 x 0.25 = 0.225
3 0.25
----------------------------------------------------------------------------------------------------
So, the probability, P, of getting these genotypes are,
P ₍₁₀₋₁₄₎ = 0.079P₍₃₀₋₃₀₎ = 0.048P₍₃₋₆₎ = 0.225The random match probability of this profile is the probability of getting the three genotypes together.
Following the product probability rule and assuming that loci are independent from each other, the profile probability, PP, is
PP = P₍₁₎ x P₍₂₎ x P₍₃₎ x .... P₍ₙ₎
PP = (P₍₁₀₋₁₄₎ ) (P₍₃₀₋₃₀₎) (P₍₃₋₆₎)
PP = 0.079 x 0.048 x 0.225
PP = 0.000853
The random match probability of this three-locus profile is 0.000853.
To fill in the blank, you need to consider the population size -N- and the probability of getting this three-locus profile.
Since I do not have the total number of individuals in the population, I will say that my population size, N = 500 individuals and I will make the calcs according to this number. You should follow the same procedure but use your N.
We know that the probability of getting this three-locus profile is 0.000853. This is the frequency of individuals carrying this genotype in the population.
To get the number of individuals with this genotype in the population, we need to multiply this frequency by the total number of individuals, N.
Nº of individuals with the three-locus profile = frequency x N
Nº of individuals with the three-locus profile = 0.000853 x 500
Nº of individuals with the three-locus profile = 0.4265 individuals.
Now we know that in a population of 500 individuals, there are 0.4265 individuals with the three-locus profile. Now we need to know the population size in which we can find 1 individual with this genotype.
0.4265 individuals with the three-locus profile-------- 500 individuals
1 individual -------------------------- X = (1 x 500) / 0.4265 = 1,172.33 individuals.
According to these calculations, approximately one in every 1,172.33 individuals, drawn randomly from the population, would be expected to have this three-locus profile.
Note: Remember to use the population size provided to you in this problem. The procedure is the same, but the population size -N- will be different.
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