Utilize the Hardy Weinburg equation pertaining to allelic frequency.
P2+2pq+q2=1
Calculating gives us 174/1378 = 0.126 = q2
Consequently, q2 equals 0.126, which means q = 0.355.
<pGiven that q = 0.355 and noting that p + q = 1, therefore
P = 1 – 0.355 = 0.645.
To determine the population of homozygous dominant (p2);
(0.645)2 x 1378 = 0.416 x 1378 = 573.
For the heterozygous population (2pq);
(2 x 0.645 x 0.355) x 1378 = 0.458 x 1378 = 631.
The number of recessive individuals (q2) equals 174.