borderline significantp = 0.04
In the case where heterozygote rs12252:AG (37/80 or 46% abundance) is expected to behave similarly to homozygote rs12252:AA (19%), so that homozygote rs12252:GG (35%) appears as the risk genotype, then a chi-square statistic would report a non-significant p = 0.36, when q (frequency of rs12252:G) = 0.54 and q 2 = 0.29 according to Hardy-Weinberg equilibrium, or borderline significant p = 0.04, when q (frequency of rs12252:G) = 0.47, with q 2 = 0.22.