Final answer:
In the context of the Hardy-Weinberg equation, 36% of squirrels being gray (recessive trait carriers) corresponds to the homozygous recessive allele frequency, denoted as q², which equals 0.36.
Step-by-step explanation:
In a squirrel population where 36% of the squirrels display a recessive trait, this percentage represents the homozygous recessive allele frequency, which is denoted as q² in the Hardy-Weinberg equation.
The Hardy-Weinberg equation is p² + 2pq + q² = 1, where p is the frequency of the dominant allele, and q is the frequency of the recessive allele, with the sum p + q = 1.
Since we know that 36% of the squirrels are homozygous for the recessive gray trait, that gives us q² = 0.36. To find the frequency of the recessive allele q, we take the square root of q², which is √0.36, resulting in q = 0.6.
Consequently, we can calculate the frequency of the dominant allele p by subtracting q from 1, which would be p = 1 - 0.6 = 0.4.