The following data was taken from people in Britain. Among the 1000 people sampled, the following genotype frequencies were observed SS = 99, Ss = 418 and ss = 483. Calculate the frequency of S and s in this population and carry out a X2 test. Is there any reason to reject the hypothesis of Hardy-Weinberg proportions.

The genotype proportions p2, 2pq, and q2 are called the Hardy–Weinberg proportions. Note that the sum of all genotype frequencies of this case is the binomial expansion of the square of the sum of p and q, and such a sum, as it represents the total of all possibilities, must be equal to 1. Therefore, (p + q)2 = p2 + 2pq + q2 = 1. A solution of this equation is q = 1 - p.