 Solution

Each diploid individual contributes two marker alleles. Count $A$ copies as twice the $AA$ count plus the $Aa$ count, and count $a$ copies analogously. The observed chromosome-level contingency table, followed by its null expected table, isThere are 200 chromosomes in each group; pooled totals are 140 $A$ and 260 $a$. The usual allelic Pearson chi-squared test of independence givesWith the ordinary independent-chromosome null model, use one degree of freedom; its p-value is about $1.04\times 10^{-25}$, providing very strong evidence of genetic association.

The two alleles from the same person are not automatically two independent observations. The usual one-degree-of-freedom calibration is justified, for example, for unrelated individuals with Hardy-Weinberg equilibrium under the null. If that assumption is not suitable, retain individuals as the sampling units and test their allele dosages using an appropriate variance or permutation of case-control labels, or use the genotype-level test from part (iii). No Hardy-Weinberg equilibrium assumption was needed for that genotype-level test.