In: Biology
Supposing the population size to be 100, according to the given information, the number of recessive individuals are 49 out of 100 (making it 49%). According to Hardy Weinberg law the allelic frequencies are constant if there are no evolutionary forces acting upon that population. The allelic frequencies are denoted by p (dominant allele) and q (recessive allele) and p+q=1. The genotypic frequencies are denoted by p2 (homozygous dominant), 2pq (heterozygous dominant) and q2 (homozygous recessive) and p2+2pq+q2 =1.
We have the percentage of homozygous recessive individuals i.e., 49% which means the frequency is 49/100 = 0.49
q2 = 0.49 hence, q= 0.7
Now since p+q=1, p=1-q
p=1-0.7
= 0.3
Since p represents the frequency of dominant allele, the answer would be 0.3.
To cross check our answer let's calculates 2pq
2pq= 2×0.7×0.3 = 0.42
Let's check if p2+2pq+q2 =1
p2 = 0.3×0.3 = 0.09
2pq = 2×0.3×0.7 = 0.42
q2 = 0.49
0.09+0.42+0.49=1
Hence, the answer 0.3 is right for frequency of dominant allele i.e., p.