Question

There are 5 alleles at the BXR008 locus. In a large sample from the population from which the suspects come, the frequencies of the alleles (starting from lowest to highest number of repeats) are;0.1,0.2,0.5,0.15,0.05; construct a table to calculate the matching probability and answer the following;

Number of repeats	Frequency
2	0.1
3	0.2
5	0.5
7	0.15
10	0.05

What is probability of the heterozygous genotype consisting of the least common allele for the BXR008 locus and the second most common.

Answer 1

Answer –

The alleles of locus BXR008 is five different types depending upon the number of repeats. Let’s represent these alleles be A, B, C, D and E. According the question A allele has 2 number of repeats, B allele has 3 number of repeats, C allele has 5 number of repeats, D allele has 7 number of repeats and E allele has 10 number of repeats in their BXR008 locus respectively and their frequencies in the population ae as follows –

Allele	Number of repeats	Frequency in the population
A	2	0.1
B (second most common)	3	0.2
C	5	0.5
D	7	0.15
E (least common)	10	0.05

Table with matching probability –

	A (0.1)	B (0.2)	C (0.5)	D (0.15)	E (0.05)
A (0.1)	AA (0.01)	AB (0.02)	AC (0.05)	AD (0.015)	AE (0.005)
B (0.2)	BA (0.02)	BB (0.04)	BC (0.1)	BD (0.03)	BE (0.01)
C (0.5)	CA (0.05)	CB (0.1)	CC (0.25)	CD (0.075)	CE (0.025)
D (0.15)	DA (0.015)	DB (0.03)	DC (0.075)	DD (0.0225)	DE (0.0075)
E (0.05)	EA (0.005)	EB (0.01)	EC (0.025)	ED (0.0075)	EE (0.0025)

Now, from the frequency table given in the question, the least common allele for the BXR008 locus is allele with 10 number of repeats i.e. allele E with frequency of 0.05 while second most common allele for the BXR008 locus is allele with 3 number of repeats i.e. allele B with frequency of 0.2.

Thus, the probability of heterozygous genotype consisting of the least common allele for the BXR008 locus and the second most common = probability of heterozygous with genotype BE

Therefore, the probability of heterozygous genotype consisting of the least common allele for the BXR008 locus and the second most common = 0.01 + 0.01

= 0.02

Hence, 0.02 is the probability of heterozygous genotype consisting of the least common allele for the BXR008 locus and the second most common.