Short tails are recessive to long tails in a hypothetical mouse population that is in hardy-weinberg equilibrium. if 96% of the population has long tails, the frequency of the short-tail allele is roughly

the value of the dominant allele is roughly 0.02.

Explanation:

The Hardy-Weinberg equation should be used to test whether a population is evolving or not. Since there are two alleles for a given locus, we will call the dominant allele (p) A, and the recessive allele (q) will be called a. Thus, p will be the allele frequency of A and q will be the frequency of a, so we will have p + q = 1 since the sum of these two alleles will equal 100%.

So if 96% of the population of the question is recessive, we can find the value of q, which will be the result of the square root of 0.96 which is approximately 0.98.

From this value, we use the formula (p + 0.98 = 1) to find the value of the dominant allele, which is 0.02.

This means that 0.004% of the population has the AA genotype and 0.016% has the Aa genotype, while 96% have the aa genotype.

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