Aparticular population of chickens has a frequency for the dominant allele as 0.70 and a frequency for the recessive allele as 0.30. which expression is the correct way to calculate the frequency of individuals that are heterozygous?
a) (0.70) x (0.30)
b) (0.70) - (0.30)
c)2 x (0.70) x (0.30)
d)2 x (0.70 - 0.30)

The Hardy-Weinberg equation is as follows:

p² + 2pq + q² = 1

Where p is the frequency of the homozygous dominant genotype, q is the frequency of the homozygous recessive genotype and 2pq is the frequency of the heterozygous genotype. Using the 2pq term here,

Frequency of heterozygous = 2 * 0.7 * 0.3

Therefore, option C is correct.

The zone your looking for is the intertidal zone.