Using the hardy-weinburg equation; 2p2 + 2pq + q2 = 1
2p2+2pq = 91% = 0.91
q2 = 9% = 0.09
Therefore q = 0.3; and since p + q = 1; then p = 1 – 0.3 = 0.7
P = 0.7
Heterozygous indivisuals in the population (i.e 2pq) therefore is derived as following;
2pq = 2 * 0.7 * 0.3 = 0.42 = 42%
The answer is 42%