p2 2pq q2 1 Homozygous with the AluI SINE insertion AluIA

(p² + 2pq + q²) = 1

Homozygous with the AluI SINE insertion (AluI⁺/AluI⁺):    23
Heterozygous for the AluI SINE insertion (AluI⁺/AluI^–):    49
Homozygous with no AluI SINE insertion (AluI^–/AluI^–):     30
Total sample:                                                                    102

For these data, determine the following:

The sample size is 102 and each person has two copies of the TPA gene (either with or without an AluI insertion) so there are 204 copies of TPA in this sample.

Question: How many copies of the AluI⁺ allele, how many copies of the AluI^– allele, the allele frequency (let\'s call this p) of AluI⁺, and the allele frequency (let\'s call this q) of AluI^–.

Solution

copies of Alu+ allele = (2 x 23 + 49) = 95

copies of Alu- allele = (2 x 30 + 49) = 109

frequency of Alu+ or (p) = no. of Alu+ alleles / total number of alleles = (2 x 23 + 49) / 204 = 0.465

sililarly, Alu- or q = (2 x 30 + 49) / 204 = 0.534