Q = recessive allele frequency = 0.3, and thus in H-W equilibrium there are ONLY two alleles, q (recessive) and
p (dominant). Therefore all of the p and q present for this gene in a population must account for 100% of this gene's alleles. And 100% = 1.00.
So p, the dominant allele frequency, must be equal to 1 - q --> p = 1 - q
p = 1 - 0.3 = 0.7.
Since heterozygotes are a combination of the p and q, we must again look at the frequencies of each genotype: p + q = 1, then (p+q)^2 = 1^2
So multiplying out (p+q)(p+q) = 1, we get: p^2+2pq+q^2 = 1 (all genotypes), where p^2 = frequency of homozygous dominant individuals, 2pq = frequency of heterozygous individuals, and q^2 = frequency of homozygous recessive individuals.
Therefore if the population is in H-W equilibrium, then the expected frequency of heterozygous individuals = 2pq = 2(0.7)(0.3)
2pq = 2(0.21) = 0.42, or 42% of the population.
Hope that helps you to understand how to solve population genetics problems!
<span>Organisms
that expend energy to regulate some or all of their internal conditions are
known as regulators. Regulators are organisms that can control their internal
environment regardless of external environmental change. Regulators achieve this
by homeostatic mechanisms and it requires a lot of energy. Examples of
regulators are mammals and birds</span>
