One gene in a population of snails determines whether they are poisonous. being poisonous (P) is dominant over not being poisonous (p) when you count a sample of 200 snails, 8 are not poisonous. what are the allele frequencies of the parent population

F (pp) = 0.04

F (Pp) = 0.32

F (PP) = 0.64

Explanation:

Studying population genetics, we used the Hardy-Weinberg equilibrium model. In the case of snails, the H-W principle predicts that the total number of cases in the population should follow the following equation: PP + 2.Pp + pp = 200.

If a population has 200 snails and 8 are affected by a recessive condition, we have 4% affected, or 0.04.

This means that 0.04 corresponds to p², and p corresponds to 0.2.

Therefore, P is equal to 0.8 because p + P = 1.

Knowing the allele frequencies, we go to the frequency of the genotypes.

F (pp) = p² = 0.04 = 8 snails

F (Pp) = 2.Pp = 0.32 = 64 snails

F (PP) = P² = 0.64 = 128 snails

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i believe your answer would be c.. : )