Use the Hardy Weinburg equation of allelic frequency.
P2+2pq+q2=1
174/1378 = 0.126 = q2
Therefore q2 = 0.126; q = 0.355
If q = 0.355 and we understand that p + q = 1, then
P = 1 – 0.355 = 0.645
To find homozygous dominant population (p2);
(0.645)2 x 1378 = 0.416 x 1378 = 573
Heterozygous popualiton (2pq);
(2 x 0.645 x 0.355) x 1378 = 0.458 x 1378= 631
Recessive indviduals (q2) = 174