According to the Hardy-Weinberg equilibrium,in a population in which 1% does NOT have freckles (recessive),what percentage of the next population is likely to NOT have freckles?

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The Hardy-Weinberg equilibrium predicts that the frequency of alleles in a population will remain constant from generation to generation, assuming there is no selection, mutation, migration, or genetic drift, and that mating is random. In this case, we know that 1% of the population does not have freckles, which means that the frequency of the recessive allele (for not having freckles) is q = 0.1. If we assume that the population is large enough that genetic drift can be ignored, we can use the Hardy-Weinberg equation to calculate the expected frequencies of genotypes:

p^2 + 2pq + q^2 = 1

where p and q are the frequencies of the dominant and recessive alleles, respectively, and p^2, 2pq, and q^2 are the frequencies of the three possible genotypes.

Since q = 0.1, we can calculate p as:

p = 1 - q = 0.9

And we can calculate the expected frequencies of the genotypes as:

p^2 (freckles/freckles) = 0.9^2 = 0.81
2pq (freckles/no freckles) = 2(0.9)(0.1) = 0.18
q^2 (no freckles/no freckles) = 0.1^2 = 0.01

Therefore, we can expect that 1% of the next population will not have freckles, since q^2 = 0.01 or 1%.

According to the Hardy-Weinberg Equilibrium,in a Population in Which 1