Experts please answer fast....
Dear Student,
Given,
p= 0.6= frequency of the dominant allele
q= 0.4= frequency of recessive allele
We know that,
According to Hardy Weinberg principle,
p2 + 2pq + q2 = 1
where,
p2 = frequency of homozygous dominant allele
2pq = frequency of heterozygous
q2 = frequency of homozygous recessive
Thus,
Frequency of heterozygous = 2pq = 2 x 0.6 x 0.4
= 0.48
Hence, option (4) is correct.
Given,
p= 0.6= frequency of the dominant allele
q= 0.4= frequency of recessive allele
We know that,
p + q = 1
According to Hardy Weinberg principle,
p2 + 2pq + q2 = 1
where,
p2 = frequency of homozygous dominant allele
2pq = frequency of heterozygous
q2 = frequency of homozygous recessive
Thus,
Frequency of heterozygous = 2pq = 2 x 0.6 x 0.4
= 0.48
Hence, option (4) is correct.
Regards